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- "title": "Fundamentals of Electric Circuits",
- "author": "Charles K. Alexander and Matthew N.O. Sadiku",
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- "title": "Preface",
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- "title": "Acknowledgments",
- "page": 16
- },
- {
- "level": 1,
- "title": "About the Authors",
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- "level": 3,
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- "level": 3,
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- "title": "PROBLEMS",
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- "level": 1,
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- "title": "2.4-1 The Convolution Integral",
- "page": 190
- },
- {
- "level": 3,
- "title": "2.4-2 Graphical Understanding of Convolution Operation",
- "page": 198
- },
- {
- "level": 3,
- "title": "2.4-3 Interconnected Systems",
- "page": 210
- },
- {
- "level": 3,
- "title": "2.4-4 A Very Special Function for LTIC Systems: The Everlasting Exponential e^{st}",
- "page": 213
- },
- {
- "level": 3,
- "title": "2.4-5 Total Response",
- "page": 215
- },
- {
- "level": 2,
- "title": "2.5 SYSTEM STABILITY",
- "page": 216
- },
- {
- "level": 3,
- "title": "2.5-1 External (BIBO) Stability",
- "page": 216
- },
- {
- "level": 3,
- "title": "2.5-2 Internal (Asymptotic) Stability",
- "page": 218
- },
- {
- "level": 3,
- "title": "2.5-3 Relationship Between BIBO and Asymptotic Stability",
- "page": 219
- },
- {
- "level": 2,
- "title": "2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR",
- "page": 223
- },
- {
- "level": 3,
- "title": "2.6-1 Dependence of System Behavior on Characteristic Modes",
- "page": 223
- },
- {
- "level": 3,
- "title": "2.6-2 Response Time of a System: The System Time Constant",
- "page": 225
- },
- {
- "level": 3,
- "title": "2.6-3 Time Constant and Rise Time of a System",
- "page": 226
- },
- {
- "level": 3,
- "title": "2.6-4 Time Constant and Filtering",
- "page": 227
- },
- {
- "level": 3,
- "title": "2.6-5 Time Constant and Pulse Dispersion (Spreading)",
- "page": 229
- },
- {
- "level": 3,
- "title": "2.6-6 Time Constant and Rate of Information Transmission",
- "page": 229
- },
- {
- "level": 3,
- "title": "2.6-7 The Resonance Phenomenon",
- "page": 230
- },
- {
- "level": 2,
- "title": "2.7 MATLAB: M-FILES",
- "page": 232
- },
- {
- "level": 3,
- "title": "2.7-1 Script M-Files",
- "page": 233
- },
- {
- "level": 3,
- "title": "2.7-2 Function M-Files",
- "page": 234
- },
- {
- "level": 3,
- "title": "2.7-3 For-Loops",
- "page": 235
- },
- {
- "level": 3,
- "title": "2.7-4 Graphical Understanding of Convolution",
- "page": 237
- },
- {
- "level": 2,
- "title": "2.8 APPENDIX: DETERMINING THE IMPULSE RESPONSE",
- "page": 240
- },
- {
- "level": 2,
- "title": "2.9 SUMMARY",
- "page": 241
- },
- {
- "level": 2,
- "title": "REFERENCES",
- "page": 243
- },
- {
- "level": 2,
- "title": "PROBLEMS",
- "page": 243
- },
- {
- "level": 1,
- "title": "3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS",
- "page": 257
- },
- {
- "level": 2,
- "title": "3.1 INTRODUCTION",
- "page": 257
- },
- {
- "level": 3,
- "title": "3.1-1 Size of a Discrete-Time Signal",
- "page": 258
- },
- {
- "level": 2,
- "title": "3.2 USEFUL SIGNAL OPERATIONS",
- "page": 260
- },
- {
- "level": 2,
- "title": "3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS",
- "page": 265
- },
- {
- "level": 3,
- "title": "3.3-1 Discrete-Time Impulse Function δ[n]",
- "page": 265
- },
- {
- "level": 3,
- "title": "3.3-2 Discrete-Time Unit Step Function u[n]",
- "page": 266
- },
- {
- "level": 3,
- "title": "3.3-3 Discrete-Time Exponential γ^n",
- "page": 267
- },
- {
- "level": 3,
- "title": "3.3-4 Discrete-Time Sinusoid cos(Omega n+θ)",
- "page": 271
- },
- {
- "level": 3,
- "title": "3.3-5 Discrete-Time Complex Exponential e^{jOmega n}",
- "page": 272
- },
- {
- "level": 2,
- "title": "3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS",
- "page": 273
- },
- {
- "level": 3,
- "title": "3.4-1 Classification of Discrete-Time Systems",
- "page": 282
- },
- {
- "level": 2,
- "title": "3.5 DISCRETE-TIME SYSTEM EQUATIONS",
- "page": 285
- },
- {
- "level": 3,
- "title": "3.5-1 Recursive (Iterative) Solution of Difference Equation",
- "page": 286
- },
- {
- "level": 2,
- "title": "3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS: THE ZERO-INPUT RESPONSE",
- "page": 290
- },
- {
- "level": 2,
- "title": "3.7 THE UNIT IMPULSE RESPONSE h[n]",
- "page": 297
- },
- {
- "level": 3,
- "title": "3.7-1 The Closed-Form Solution of h[n]",
- "page": 298
- },
- {
- "level": 2,
- "title": "3.8 SYSTEM RESPONSE TO EXTERNAL INPUT: THE ZERO-STATE RESPONSE",
- "page": 300
- },
- {
- "level": 3,
- "title": "3.8-1 Graphical Procedure for the Convolution Sum",
- "page": 308
- },
- {
- "level": 3,
- "title": "3.8-2 Interconnected Systems",
- "page": 314
- },
- {
- "level": 3,
- "title": "3.8-3 Total Response",
- "page": 317
- },
- {
- "level": 2,
- "title": "3.9 SYSTEM STABILITY",
- "page": 318
- },
- {
- "level": 3,
- "title": "3.9-1 External (BIBO) Stability",
- "page": 318
- },
- {
- "level": 3,
- "title": "3.9-2 Internal (Asymptotic) Stability",
- "page": 319
- },
- {
- "level": 3,
- "title": "3.9-3 Relationship Between BIBO and Asymptotic Stability",
- "page": 321
- },
- {
- "level": 2,
- "title": "3.10 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR",
- "page": 325
- },
- {
- "level": 2,
- "title": "3.11 MATLAB: DISCRETE-TIME SIGNALS AND SYSTEMS",
- "page": 326
- },
- {
- "level": 3,
- "title": "3.11-1 Discrete-Time Functions and Stem Plots",
- "page": 326
- },
- {
- "level": 3,
- "title": "3.11-2 System Responses Through Filtering",
- "page": 328
- },
- {
- "level": 3,
- "title": "3.11-3 A Custom Filter Function",
- "page": 330
- },
- {
- "level": 3,
- "title": "3.11-4 Discrete-Time Convolution",
- "page": 331
- },
- {
- "level": 2,
- "title": "3.12 APPENDIX: IMPULSE RESPONSE FOR A SPECIAL CASE",
- "page": 333
- },
- {
- "level": 2,
- "title": "3.13 SUMMARY",
- "page": 333
- },
- {
- "level": 2,
- "title": "PROBLEMS",
- "page": 334
- },
- {
- "level": 1,
- "title": "4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM",
- "page": 350
- },
- {
- "level": 2,
- "title": "4.1 THE LAPLACE TRANSFORM",
- "page": 350
- },
- {
- "level": 3,
- "title": "4.1-1 Finding the Inverse Transform",
- "page": 358
- },
- {
- "level": 2,
- "title": "4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM",
- "page": 369
- },
- {
- "level": 3,
- "title": "4.2-1 Time Shifting",
- "page": 369
- },
- {
- "level": 3,
- "title": "4.2-2 Frequency Shifting",
- "page": 373
- },
- {
- "level": 3,
- "title": "4.2-3 The Time-Differentiation Property",
- "page": 374
- },
- {
- "level": 3,
- "title": "4.2-4 The Time-Integration Property",
- "page": 376
- },
- {
- "level": 3,
- "title": "4.2-5 The Scaling Property",
- "page": 377
- },
- {
- "level": 3,
- "title": "4.2-6 Time Convolution and Frequency Convolution",
- "page": 377
- },
- {
- "level": 2,
- "title": "4.3 SOLUTION OF DIFFERENTIAL AND INTEGRO-DIFFERENTIAL EQUATIONS",
- "page": 380
- },
- {
- "level": 3,
- "title": "4.3-1 Comments on Initial Conditions at 0^− and at 0^+",
- "page": 383
- },
- {
- "level": 3,
- "title": "4.3-2 Zero-State Response",
- "page": 386
- },
- {
- "level": 3,
- "title": "4.3-3 Stability",
- "page": 391
- },
- {
- "level": 3,
- "title": "4.3-4 Inverse Systems",
- "page": 393
- },
- {
- "level": 2,
- "title": "4.4 ANALYSIS OF ELECTRICAL NETWORKS: THE TRANSFORMED NETWORK",
- "page": 393
- },
- {
- "level": 3,
- "title": "4.4-1 Analysis of Active Circuits",
- "page": 402
- },
- {
- "level": 2,
- "title": "4.5 BLOCK DIAGRAMS",
- "page": 406
- },
- {
- "level": 2,
- "title": "4.6 SYSTEM REALIZATION",
- "page": 408
- },
- {
- "level": 3,
- "title": "4.6-1 Direct Form I Realization",
- "page": 409
- },
- {
- "level": 3,
- "title": "4.6-2 Direct Form II Realization",
- "page": 410
- },
- {
- "level": 3,
- "title": "4.6-3 Cascade and Parallel Realizations",
- "page": 413
- },
- {
- "level": 3,
- "title": "4.6-4 Transposed Realization",
- "page": 416
- },
- {
- "level": 3,
- "title": "4.6-5 Using Operational Amplifiers for System Realization",
- "page": 419
- },
- {
- "level": 2,
- "title": "4.7 APPLICATION TO FEEDBACK AND CONTROLS",
- "page": 424
- },
- {
- "level": 3,
- "title": "4.7-1 Analysis of a Simple Control System",
- "page": 426
- },
- {
- "level": 2,
- "title": "4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM",
- "page": 432
- },
- {
- "level": 3,
- "title": "4.8-1 Steady-State Response to Causal Sinusoidal Inputs",
- "page": 438
- },
- {
- "level": 2,
- "title": "4.9 BODE PLOTS",
- "page": 439
- },
- {
- "level": 3,
- "title": "4.9-1 Constant Ka_1a_2/b_1b_3",
- "page": 442
- },
- {
- "level": 3,
- "title": "4.9-2 Pole (or Zero) at the Origin",
- "page": 442
- },
- {
- "level": 3,
- "title": "4.9-3 First-Order Pole (or Zero)",
- "page": 444
- },
- {
- "level": 3,
- "title": "4.9-4 Second-Order Pole (or Zero)",
- "page": 446
- },
- {
- "level": 3,
- "title": "4.9-5 The Transfer Function from the Frequency Response",
- "page": 455
- },
- {
- "level": 2,
- "title": "4.10 FILTER DESIGN BY PLACEMENT OF POLES AND ZEROS OF H(s)",
- "page": 456
- },
- {
- "level": 3,
- "title": "4.10-1 Dependence of Frequency Response on Poles and Zeros of H(s)",
- "page": 456
- },
- {
- "level": 3,
- "title": "4.10-2 Lowpass Filters",
- "page": 459
- },
- {
- "level": 3,
- "title": "4.10-3 Bandpass Filters",
- "page": 461
- },
- {
- "level": 3,
- "title": "4.10-4 Notch (Bandstop) Filters",
- "page": 461
- },
- {
- "level": 3,
- "title": "4.10-5 Practical Filters and Their Specifications",
- "page": 464
- },
- {
- "level": 2,
- "title": "4.11 THE BILATERAL LAPLACE TRANSFORM",
- "page": 465
- },
- {
- "level": 3,
- "title": "4.11-1 Properties of the Bilateral Laplace Transform",
- "page": 471
- },
- {
- "level": 3,
- "title": "4.11-2 Using the Bilateral Transform for Linear System Analysis",
- "page": 472
- },
- {
- "level": 2,
- "title": "4.12 MATLAB: CONTINUOUS-TIME FILTERS",
- "page": 475
- },
- {
- "level": 3,
- "title": "4.12-1 Frequency Response and Polynomial Evaluation",
- "page": 476
- },
- {
- "level": 3,
- "title": "4.12-2 Butterworth Filters and the Find Command",
- "page": 479
- },
- {
- "level": 3,
- "title": "4.12-3 Using Cascaded Second-Order Sections for Butterworth Filter Realization",
- "page": 481
- },
- {
- "level": 3,
- "title": "4.12-4 Chebyshev Filters",
- "page": 483
- },
- {
- "level": 2,
- "title": "4.13 SUMMARY",
- "page": 486
- },
- {
- "level": 2,
- "title": "REFERENCES",
- "page": 488
- },
- {
- "level": 2,
- "title": "PROBLEMS",
- "page": 488
- },
- {
- "level": 1,
- "title": "5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM",
- "page": 508
- },
- {
- "level": 2,
- "title": "5.1 THE z-TRANSFORM",
- "page": 508
- },
- {
- "level": 3,
- "title": "5.1-1 Inverse Transform by Partial Fraction Expansion and Tables",
- "page": 515
- },
- {
- "level": 3,
- "title": "5.1-2 Inverse z-Transform by Power Series Expansion",
- "page": 519
- },
- {
- "level": 2,
- "title": "5.2 SOME PROPERTIES OF THE z-TRANSFORM",
- "page": 521
- },
- {
- "level": 3,
- "title": "5.2-1 Time-Shifting Properties",
- "page": 521
- },
- {
- "level": 3,
- "title": "5.2-2 z-Domain Scaling Property (Multiplication by γ^n)",
- "page": 525
- },
- {
- "level": 3,
- "title": "5.2-3 z-Domain Differentiation Property (Multiplication by n)",
- "page": 526
- },
- {
- "level": 3,
- "title": "5.2-4 Time-Reversal Property",
- "page": 526
- },
- {
- "level": 3,
- "title": "5.2-5 Convolution Property",
- "page": 527
- },
- {
- "level": 2,
- "title": "5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS",
- "page": 530
- },
- {
- "level": 3,
- "title": "5.3-1 Zero-State Response of LTID Systems: The Transfer Function",
- "page": 534
- },
- {
- "level": 3,
- "title": "5.3-2 Stability",
- "page": 538
- },
- {
- "level": 3,
- "title": "5.3-3 Inverse Systems",
- "page": 539
- },
- {
- "level": 2,
- "title": "5.4 SYSTEM REALIZATION",
- "page": 539
- },
- {
- "level": 2,
- "title": "5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS",
- "page": 546
- },
- {
- "level": 3,
- "title": "5.5-1 The Periodic Nature of Frequency Response",
- "page": 552
- },
- {
- "level": 3,
- "title": "5.5-2 Aliasing and Sampling Rate",
- "page": 556
- },
- {
- "level": 2,
- "title": "5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS",
- "page": 558
- },
- {
- "level": 2,
- "title": "5.7 DIGITAL PROCESSING OF ANALOG SIGNALS",
- "page": 567
- },
- {
- "level": 2,
- "title": "5.8 THE BILATERAL z-TRANSFORM",
- "page": 574
- },
- {
- "level": 3,
- "title": "5.8-1 Properties of the Bilateral z-Transform",
- "page": 579
- },
- {
- "level": 3,
- "title": "5.8-2 Using the Bilateral z-Transform for Analysis of LTID Systems",
- "page": 580
- },
- {
- "level": 2,
- "title": "5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS",
- "page": 583
- },
- {
- "level": 2,
- "title": "5.10 MATLAB: DISCRETE-TIME IIR FILTERS",
- "page": 585
- },
- {
- "level": 3,
- "title": "5.10-1 Frequency Response and Pole-Zero Plots",
- "page": 586
- },
- {
- "level": 3,
- "title": "5.10-2 Transformation Basics",
- "page": 587
- },
- {
- "level": 3,
- "title": "5.10-3 Transformation by First-Order Backward Difference",
- "page": 588
- },
- {
- "level": 3,
- "title": "5.10-4 Bilinear Transformation",
- "page": 589
- },
- {
- "level": 3,
- "title": "5.10-5 Bilinear Transformation with Prewarping",
- "page": 590
- },
- {
- "level": 3,
- "title": "5.10-6 Example: Butterworth Filter Transformation",
- "page": 591
- },
- {
- "level": 3,
- "title": "5.10-7 Problems Finding Polynomial Roots",
- "page": 592
- },
- {
- "level": 3,
- "title": "5.10-8 Using Cascaded Second-Order Sections to Improve Design",
- "page": 592
- },
- {
- "level": 2,
- "title": "5.11 SUMMARY",
- "page": 594
- },
- {
- "level": 2,
- "title": "REFERENCES",
- "page": 595
- },
- {
- "level": 2,
- "title": "PROBLEMS",
- "page": 595
- },
- {
- "level": 1,
- "title": "6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES",
- "page": 613
- },
- {
- "level": 2,
- "title": "6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES",
- "page": 613
- },
- {
- "level": 3,
- "title": "6.1-1 The Fourier Spectrum",
- "page": 618
- },
- {
- "level": 3,
- "title": "6.1-2 The Effect of Symmetry",
- "page": 627
- },
- {
- "level": 3,
- "title": "6.1-3 Determining the Fundamental Frequency and Period",
- "page": 629
- },
- {
- "level": 2,
- "title": "6.2 EXISTENCE AND CONVERGENCE OF THE FOURIER SERIES",
- "page": 632
- },
- {
- "level": 3,
- "title": "6.2-1 Convergence of a Series",
- "page": 633
- },
- {
- "level": 3,
- "title": "6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping",
- "page": 635
- },
- {
- "level": 2,
- "title": "6.3 EXPONENTIAL FOURIER SERIES",
- "page": 641
- },
- {
- "level": 3,
- "title": "6.3-1 Exponential Fourier Spectra",
- "page": 644
- },
- {
- "level": 3,
- "title": "6.3-2 Parseval’s Theorem",
- "page": 652
- },
- {
- "level": 3,
- "title": "6.3-3 Properties of the Fourier Series",
- "page": 655
- },
- {
- "level": 2,
- "title": "6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS",
- "page": 657
- },
- {
- "level": 2,
- "title": "6.5 GENERALIZED FOURIER SERIES:SIGNALS AS VECTORS",
- "page": 661
- },
- {
- "level": 3,
- "title": "6.5-1 Component of a Vector",
- "page": 662
- },
- {
- "level": 3,
- "title": "6.5-2 Signal Comparison and Component of a Signal",
- "page": 663
- },
- {
- "level": 3,
- "title": "6.5-3 Extension to Complex Signals",
- "page": 665
- },
- {
- "level": 3,
- "title": "6.5-4 Signal Representation by an Orthogonal Signal Set",
- "page": 667
- },
- {
- "level": 2,
- "title": "6.6 NUMERICAL COMPUTATION OF D_n",
- "page": 679
- },
- {
- "level": 2,
- "title": "6.7 MATLAB: FOURIER SERIES APPLICATIONS",
- "page": 681
- },
- {
- "level": 3,
- "title": "6.7-1 Periodic Functions and the Gibbs Phenomenon",
- "page": 681
- },
- {
- "level": 3,
- "title": "6.7-2 Optimization and Phase Spectra",
- "page": 684
- },
- {
- "level": 2,
- "title": "6.8 SUMMARY",
- "page": 687
- },
- {
- "level": 2,
- "title": "REFERENCES",
- "page": 688
- },
- {
- "level": 2,
- "title": "PROBLEMS",
- "page": 689
- },
- {
- "level": 1,
- "title": "7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM",
- "page": 700
- },
- {
- "level": 2,
- "title": "7.1 APERIODIC SIGNAL REPRESENTATION BY THE FOURIER INTEGRAL",
- "page": 700
- },
- {
- "level": 3,
- "title": "7.1-1 Physical Appreciation of the Fourier Transform",
- "page": 707
- },
- {
- "level": 2,
- "title": "7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS",
- "page": 709
- },
- {
- "level": 3,
- "title": "7.2-1 Connection Between the Fourier and Laplace Transforms",
- "page": 720
- },
- {
- "level": 2,
- "title": "7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM",
- "page": 721
- },
- {
- "level": 2,
- "title": "7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS",
- "page": 741
- },
- {
- "level": 3,
- "title": "7.4-1 Signal Distortion During Transmission",
- "page": 743
- },
- {
- "level": 3,
- "title": "7.4-2 Bandpass Systems and Group Delay",
- "page": 746
- },
- {
- "level": 2,
- "title": "7.5 IDEAL AND PRACTICAL FILTERS",
- "page": 750
- },
- {
- "level": 2,
- "title": "7.6 SIGNAL ENERGY",
- "page": 753
- },
- {
- "level": 2,
- "title": "7.7 APPLICATION TO COMMUNICATIONS: AMPLITUDE MODULATION",
- "page": 756
- },
- {
- "level": 3,
- "title": "7.7-1 Double-Sideband, Suppressed-Carrier (DSB-SC) Modulation",
- "page": 757
- },
- {
- "level": 3,
- "title": "7.7-2 Amplitude Modulation (AM)",
- "page": 762
- },
- {
- "level": 3,
- "title": "7.7-3 Single-Sideband Modulation (SSB)",
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- "level": 2,
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- "page": 779
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- "page": 780
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- {
- "level": 2,
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- "level": 2,
- "title": "REFERENCES",
- "page": 783
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- "level": 3,
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- "level": 2,
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- "level": 2,
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- "level": 2,
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- "level": 3,
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- "level": 3,
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- "page": 953
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- "level": 3,
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- "level": 3,
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- "level": 3,
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- {
- "level": 2,
- "title": "REFERENCES",
- "page": 990
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- {
- "level": 2,
- "title": "PROBLEMS",
- "page": 990
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- "level": 1,
- "title": "INDEX",
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- "web_url": "/book?id=linear-systems-and-signals-3rd-edition-bp-lathi"
- },
- {
- "doc_id": "technical_analysis_of_the_financial_market_-_john_j_murphy",
- "file_name": "Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy.pdf",
- "title": "Technical Analysis of the Financial Markets",
- "author": "John J. Murphy",
- "category": "trading",
- "category_slug": "trading",
- "publication_year": "2020",
- "total_pages": 494,
- "tags": [
- "_john_j_murphy",
- "technical_analysis_of_the_financial_market_"
- ],
- "toc": [
- {
- "level": 1,
- "title": "Title Page",
- "page": 3
- },
- {
- "level": 1,
- "title": "Copyright",
- "page": 4
- },
- {
- "level": 1,
- "title": "Dedication",
- "page": 6
- },
- {
- "level": 1,
- "title": "Contents",
- "page": 7
- },
- {
- "level": 1,
- "title": "About the Author",
- "page": 19
- },
- {
- "level": 1,
- "title": "About the Contributors",
- "page": 20
- },
- {
- "level": 1,
- "title": "Introduction",
- "page": 21
- },
- {
- "level": 1,
- "title": "Acknowledgments",
- "page": 23
- },
- {
- "level": 1,
- "title": "1 Philosophy of Technical Analysis",
- "page": 24
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 24
- },
- {
- "level": 2,
- "title": "Philosophy or Rationale",
- "page": 24
- },
- {
- "level": 2,
- "title": "Technical versus Fundamental Forecasting",
- "page": 27
- },
- {
- "level": 2,
- "title": "Analysis versus Timing",
- "page": 28
- },
- {
- "level": 2,
- "title": "Flexibility and Adaptability of Technical Analysis",
- "page": 29
- },
- {
- "level": 2,
- "title": "Technical Analysis Applied to Different Trading Mediums",
- "page": 30
- },
- {
- "level": 2,
- "title": "Technical Analysis Applied to Different Time Dimensions",
- "page": 30
- },
- {
- "level": 2,
- "title": "Economic Forecasting",
- "page": 31
- },
- {
- "level": 2,
- "title": "Technician or Chartist?",
- "page": 31
- },
- {
- "level": 2,
- "title": "A Brief Comparison of Technical Analysis in Stocks and Futures",
- "page": 32
- },
- {
- "level": 2,
- "title": "Less Reliance on Market Averages and Indicators",
- "page": 35
- },
- {
- "level": 2,
- "title": "Some Criticisms of the Technical Approach",
- "page": 35
- },
- {
- "level": 2,
- "title": "Random Walk Theory",
- "page": 38
- },
- {
- "level": 2,
- "title": "Universal Principles",
- "page": 40
- },
- {
- "level": 1,
- "title": "2 Dow Theory",
- "page": 41
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 41
- },
- {
- "level": 2,
- "title": "Basic Tenets",
- "page": 42
- },
- {
- "level": 2,
- "title": "The Use of Closing Prices and the Presence of Lines",
- "page": 47
- },
- {
- "level": 2,
- "title": "Some Criticisms of Dow Theory",
- "page": 48
- },
- {
- "level": 2,
- "title": "Stocks as Economic Indicators",
- "page": 49
- },
- {
- "level": 2,
- "title": "Dow Theory Applied to Futures Trading",
- "page": 49
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 50
- },
- {
- "level": 1,
- "title": "3 Chart Construction",
- "page": 51
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 51
- },
- {
- "level": 2,
- "title": "Types of Charts Available",
- "page": 51
- },
- {
- "level": 2,
- "title": "Candlesticks",
- "page": 53
- },
- {
- "level": 2,
- "title": "Arithmetic versus Logarithmic Scale",
- "page": 54
- },
- {
- "level": 2,
- "title": "Construction of the Daily Bar Chart",
- "page": 55
- },
- {
- "level": 2,
- "title": "Volume",
- "page": 56
- },
- {
- "level": 2,
- "title": "Futures Open Interest",
- "page": 57
- },
- {
- "level": 2,
- "title": "Weekly and Monthly Bar Charts",
- "page": 59
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 61
- },
- {
- "level": 1,
- "title": "4 Basic Concepts of Trend",
- "page": 62
- },
- {
- "level": 2,
- "title": "Definition of Trend",
- "page": 62
- },
- {
- "level": 2,
- "title": "Trend Has Three Directions",
- "page": 64
- },
- {
- "level": 2,
- "title": "Trend Has Three Classifications",
- "page": 65
- },
- {
- "level": 2,
- "title": "Support and Resistance",
- "page": 67
- },
- {
- "level": 2,
- "title": "Trendlines",
- "page": 75
- },
- {
- "level": 2,
- "title": "The Fan Principle",
- "page": 83
- },
- {
- "level": 2,
- "title": "The Importance of the Number Three",
- "page": 85
- },
- {
- "level": 2,
- "title": "The Relative Steepness of the Trendline",
- "page": 85
- },
- {
- "level": 2,
- "title": "The Channel Line",
- "page": 89
- },
- {
- "level": 2,
- "title": "Percentage Retracements",
- "page": 94
- },
- {
- "level": 2,
- "title": "Speed Resistance Lines",
- "page": 96
- },
- {
- "level": 2,
- "title": "Gann and Fibonacci Fan Lines",
- "page": 98
- },
- {
- "level": 2,
- "title": "Internal Trendlines",
- "page": 99
- },
- {
- "level": 2,
- "title": "Reversal Days",
- "page": 99
- },
- {
- "level": 2,
- "title": "Price Gaps",
- "page": 102
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 105
- },
- {
- "level": 1,
- "title": "5 Major Reversal Patterns",
- "page": 106
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 106
- },
- {
- "level": 2,
- "title": "Price Patterns",
- "page": 106
- },
- {
- "level": 2,
- "title": "Two Types of Patterns: Reversal and Continuation",
- "page": 107
- },
- {
- "level": 2,
- "title": "The Head and Shoulders Reversal Pattern",
- "page": 109
- },
- {
- "level": 2,
- "title": "The Importance of Volume",
- "page": 113
- },
- {
- "level": 2,
- "title": "Finding a Price Objective",
- "page": 113
- },
- {
- "level": 2,
- "title": "The Inverse Head and Shoulders",
- "page": 115
- },
- {
- "level": 2,
- "title": "Complex Head and Shoulders Patterns",
- "page": 117
- },
- {
- "level": 2,
- "title": "Triple Tops and Bottoms",
- "page": 119
- },
- {
- "level": 2,
- "title": "Double Tops and Bottoms",
- "page": 121
- },
- {
- "level": 2,
- "title": "Variations from the Ideal Pattern",
- "page": 124
- },
- {
- "level": 2,
- "title": "Saucers and Spikes",
- "page": 128
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 129
- },
- {
- "level": 1,
- "title": "6 Continuation Patterns",
- "page": 130
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 130
- },
- {
- "level": 2,
- "title": "Triangles",
- "page": 131
- },
- {
- "level": 2,
- "title": "The Symmetrical Triangle",
- "page": 132
- },
- {
- "level": 2,
- "title": "The Ascending Triangle",
- "page": 135
- },
- {
- "level": 2,
- "title": "The Descending Triangle",
- "page": 137
- },
- {
- "level": 2,
- "title": "The Broadening Formation",
- "page": 139
- },
- {
- "level": 2,
- "title": "Flags and Pennants",
- "page": 140
- },
- {
- "level": 2,
- "title": "The Wedge Formation",
- "page": 144
- },
- {
- "level": 2,
- "title": "The Rectangle Formation",
- "page": 145
- },
- {
- "level": 2,
- "title": "The Measured Move",
- "page": 149
- },
- {
- "level": 2,
- "title": "The Continuation Head and Shoulders Pattern",
- "page": 150
- },
- {
- "level": 2,
- "title": "Confirmation and Divergence",
- "page": 152
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 153
- },
- {
- "level": 1,
- "title": "7 Volume and Open Interest",
- "page": 154
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 154
- },
- {
- "level": 2,
- "title": "Volume and Open Interest as Secondary Indicators",
- "page": 154
- },
- {
- "level": 2,
- "title": "Interpretation of Volume for All Markets",
- "page": 158
- },
- {
- "level": 2,
- "title": "Interpretation of Open Interest in Futures",
- "page": 164
- },
- {
- "level": 2,
- "title": "Summary of Volume and Open Interest Rules",
- "page": 169
- },
- {
- "level": 2,
- "title": "Blowoffs and Selling Climaxes",
- "page": 169
- },
- {
- "level": 2,
- "title": "Commitments of Traders Report",
- "page": 169
- },
- {
- "level": 2,
- "title": "Watch the Commercials",
- "page": 170
- },
- {
- "level": 2,
- "title": "Net Trader Positions",
- "page": 171
- },
- {
- "level": 2,
- "title": "Open Interest In Options",
- "page": 171
- },
- {
- "level": 2,
- "title": "Put/Call Ratios",
- "page": 172
- },
- {
- "level": 2,
- "title": "Combine Option Sentiment With Technicals",
- "page": 173
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 173
- },
- {
- "level": 1,
- "title": "8 Long Term Charts",
- "page": 175
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 175
- },
- {
- "level": 2,
- "title": "The Importance of Longer Range Perspective",
- "page": 175
- },
- {
- "level": 2,
- "title": "Construction of Continuation Charts for Futures",
- "page": 176
- },
- {
- "level": 2,
- "title": "The Perpetual Contract™",
- "page": 177
- },
- {
- "level": 2,
- "title": "Long Term Trends Dispute Randomness",
- "page": 178
- },
- {
- "level": 2,
- "title": "Patterns on Charts: Weekly and Monthly Reversals",
- "page": 178
- },
- {
- "level": 2,
- "title": "Long Term to Short Term Charts",
- "page": 178
- },
- {
- "level": 2,
- "title": "Why Should Long Range Charts Be Adjusted for Inflation?",
- "page": 179
- },
- {
- "level": 2,
- "title": "Long Term Charts Not Intended for Trading Purposes",
- "page": 180
- },
- {
- "level": 2,
- "title": "Examples of Long Term Charts",
- "page": 181
- },
- {
- "level": 1,
- "title": "9 Moving Averages",
- "page": 187
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 187
- },
- {
- "level": 2,
- "title": "The Moving Average: A Smoothing Device with a Time Lag",
- "page": 188
- },
- {
- "level": 2,
- "title": "Moving Average Envelopes",
- "page": 197
- },
- {
- "level": 2,
- "title": "Bollinger Bands",
- "page": 198
- },
- {
- "level": 2,
- "title": "Using Bollinger Bands as Targets",
- "page": 200
- },
- {
- "level": 2,
- "title": "Band Width Measures Volatility",
- "page": 200
- },
- {
- "level": 2,
- "title": "Moving Averages Tied to Cycles",
- "page": 201
- },
- {
- "level": 2,
- "title": "Fibonacci Numbers Used as Moving Averages",
- "page": 202
- },
- {
- "level": 2,
- "title": "Moving Averages Applied to Long Term Charts",
- "page": 202
- },
- {
- "level": 2,
- "title": "The Weekly Rule",
- "page": 204
- },
- {
- "level": 2,
- "title": "To Optimize or Not",
- "page": 209
- },
- {
- "level": 2,
- "title": "Summary",
- "page": 209
- },
- {
- "level": 2,
- "title": "The Adaptive Moving Average",
- "page": 210
- },
- {
- "level": 2,
- "title": "Alternatives to the Moving Average",
- "page": 210
- },
- {
- "level": 1,
- "title": "10 Oscillators and Contrary Opinion",
- "page": 212
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 212
- },
- {
- "level": 2,
- "title": "Oscillator Usage in Conjunction with Trend",
- "page": 213
- },
- {
- "level": 2,
- "title": "Measuring Momentum",
- "page": 214
- },
- {
- "level": 2,
- "title": "Measuring Rate of Change (ROC)",
- "page": 219
- },
- {
- "level": 2,
- "title": "Constructing an Oscillator Using Two Moving Averages",
- "page": 219
- },
- {
- "level": 2,
- "title": "Commodity Channel Index",
- "page": 222
- },
- {
- "level": 2,
- "title": "The Relative Strength Index (RSI)",
- "page": 224
- },
- {
- "level": 2,
- "title": "Using the 70 and 30 Lines to Generate Signals",
- "page": 230
- },
- {
- "level": 2,
- "title": "Stochastics (K%D)",
- "page": 230
- },
- {
- "level": 2,
- "title": "Larry Williams %R",
- "page": 233
- },
- {
- "level": 2,
- "title": "The Importance of Trend",
- "page": 234
- },
- {
- "level": 2,
- "title": "When Oscillators are Most Useful",
- "page": 235
- },
- {
- "level": 2,
- "title": "Moving Average Convergence/Divergence (MACD)",
- "page": 236
- },
- {
- "level": 2,
- "title": "MACD Histogram",
- "page": 238
- },
- {
- "level": 2,
- "title": "Combine Weeklies and Dailies",
- "page": 239
- },
- {
- "level": 2,
- "title": "The Principle of Contrary Opinion in Futures",
- "page": 240
- },
- {
- "level": 2,
- "title": "Investor Sentiment Readings",
- "page": 243
- },
- {
- "level": 2,
- "title": "Investors Intelligence Numbers",
- "page": 244
- },
- {
- "level": 1,
- "title": "11 Point and Figure Charting",
- "page": 246
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 246
- },
- {
- "level": 2,
- "title": "The Point and Figure Versus the Bar Chart",
- "page": 246
- },
- {
- "level": 2,
- "title": "Construction of the Intraday Point and Figure Chart",
- "page": 250
- },
- {
- "level": 2,
- "title": "The Horizontal Count",
- "page": 253
- },
- {
- "level": 2,
- "title": "Price Patterns",
- "page": 254
- },
- {
- "level": 2,
- "title": "3 Box Reversal Point and Figure Charting",
- "page": 256
- },
- {
- "level": 2,
- "title": "Construction of the 3 Point Reversal Chart",
- "page": 257
- },
- {
- "level": 2,
- "title": "The Drawing of Trendlines",
- "page": 261
- },
- {
- "level": 2,
- "title": "Measuring Techniques",
- "page": 264
- },
- {
- "level": 2,
- "title": "Trading Tactics",
- "page": 265
- },
- {
- "level": 2,
- "title": "Advantages of Point and Figure Charts",
- "page": 266
- },
- {
- "level": 2,
- "title": "P&F Technical Indicators",
- "page": 270
- },
- {
- "level": 2,
- "title": "Computerized P&F Charting",
- "page": 270
- },
- {
- "level": 2,
- "title": "P&F Moving Averages",
- "page": 272
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 273
- },
- {
- "level": 1,
- "title": "12 Japanese Candlesticks",
- "page": 274
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 274
- },
- {
- "level": 2,
- "title": "Candlestick Charting",
- "page": 274
- },
- {
- "level": 2,
- "title": "Basic Candlesticks",
- "page": 276
- },
- {
- "level": 2,
- "title": "Candle Pattern Analysis",
- "page": 278
- },
- {
- "level": 2,
- "title": "Filtered Candle Patterns",
- "page": 283
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 284
- },
- {
- "level": 2,
- "title": "Candle Patterns",
- "page": 285
- },
- {
- "level": 1,
- "title": "13 Elliott Wave Theory",
- "page": 294
- },
- {
- "level": 2,
- "title": "Historical Background",
- "page": 294
- },
- {
- "level": 2,
- "title": "Basic Tenets of the Elliott Wave Principle",
- "page": 294
- },
- {
- "level": 2,
- "title": "Connection Between Elliott Wave and Dow Theory",
- "page": 298
- },
- {
- "level": 2,
- "title": "Corrective Waves",
- "page": 299
- },
- {
- "level": 2,
- "title": "The Rule of Alternation",
- "page": 307
- },
- {
- "level": 2,
- "title": "Channeling",
- "page": 308
- },
- {
- "level": 2,
- "title": "Wave 4 as a Support Area",
- "page": 309
- },
- {
- "level": 2,
- "title": "Fibonacci Numbers as the Basis of the Wave Principle",
- "page": 309
- },
- {
- "level": 2,
- "title": "Fibonacci Ratios and Retracements",
- "page": 310
- },
- {
- "level": 2,
- "title": "Fibonacci Time Targets",
- "page": 313
- },
- {
- "level": 2,
- "title": "Combining All Three Aspects of Wave Theory",
- "page": 313
- },
- {
- "level": 2,
- "title": "Elliott Wave Applied to Stocks Versus Commodities",
- "page": 315
- },
- {
- "level": 2,
- "title": "Summary and Conclusions",
- "page": 315
- },
- {
- "level": 2,
- "title": "Reference Material",
- "page": 317
- },
- {
- "level": 1,
- "title": "14 Time Cycles",
- "page": 318
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 318
- },
- {
- "level": 2,
- "title": "Cycles",
- "page": 319
- },
- {
- "level": 2,
- "title": "How Cyclic Concepts Help Explain Charting Techniques",
- "page": 328
- },
- {
- "level": 2,
- "title": "Dominant Cycles",
- "page": 331
- },
- {
- "level": 2,
- "title": "Combining Cycle Lengths",
- "page": 333
- },
- {
- "level": 2,
- "title": "The Importance of Trend",
- "page": 334
- },
- {
- "level": 2,
- "title": "Left and Right Translation",
- "page": 335
- },
- {
- "level": 2,
- "title": "How to Isolate Cycles",
- "page": 335
- },
- {
- "level": 2,
- "title": "Seasonal Cycles",
- "page": 340
- },
- {
- "level": 2,
- "title": "Stock Market Cycles",
- "page": 344
- },
- {
- "level": 2,
- "title": "The January Barometer",
- "page": 344
- },
- {
- "level": 2,
- "title": "The Presidential Cycle",
- "page": 345
- },
- {
- "level": 2,
- "title": "Combining Cycles with Other Technical Tools",
- "page": 345
- },
- {
- "level": 2,
- "title": "Maximum Entropy Spectral Analysis",
- "page": 345
- },
- {
- "level": 2,
- "title": "Cycle Reading and Software",
- "page": 346
- },
- {
- "level": 1,
- "title": "15 Computers and Trading Systems",
- "page": 347
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 347
- },
- {
- "level": 2,
- "title": "Some Computer Needs",
- "page": 348
- },
- {
- "level": 2,
- "title": "Grouping Tools and Indicators",
- "page": 349
- },
- {
- "level": 2,
- "title": "Using the Tools and Indicators",
- "page": 349
- },
- {
- "level": 2,
- "title": "Welles Wilder’s Parabolic and Directional Movement Systems",
- "page": 350
- },
- {
- "level": 2,
- "title": "Pros and Cons of System Trading",
- "page": 356
- },
- {
- "level": 2,
- "title": "Need Expert Help?",
- "page": 357
- },
- {
- "level": 2,
- "title": "Test Systems or Create Your Own",
- "page": 358
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 358
- },
- {
- "level": 1,
- "title": "16 Money Management and Trading Tactics",
- "page": 360
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 360
- },
- {
- "level": 2,
- "title": "The Three Elements of Successful Trading",
- "page": 360
- },
- {
- "level": 2,
- "title": "Money Management",
- "page": 361
- },
- {
- "level": 2,
- "title": "Reward to Risk Ratios",
- "page": 363
- },
- {
- "level": 2,
- "title": "Trading Multiple Positions: Trending versus Trading Units",
- "page": 364
- },
- {
- "level": 2,
- "title": "What to Do After Periods of Success and Adversity",
- "page": 364
- },
- {
- "level": 2,
- "title": "Trading Tactics",
- "page": 365
- },
- {
- "level": 2,
- "title": "Combining Technical Factors and Money Management",
- "page": 367
- },
- {
- "level": 2,
- "title": "Types of Trading Orders",
- "page": 368
- },
- {
- "level": 2,
- "title": "From Daily Charts to Intraday Price Charts",
- "page": 369
- },
- {
- "level": 2,
- "title": "The Use of Intraday Pivot Points",
- "page": 371
- },
- {
- "level": 2,
- "title": "Summary of Money Management and Trading Guidelines",
- "page": 372
- },
- {
- "level": 2,
- "title": "Application to Stocks",
- "page": 373
- },
- {
- "level": 2,
- "title": "Asset Allocation",
- "page": 373
- },
- {
- "level": 2,
- "title": "Managed Accounts and Mutual Funds",
- "page": 373
- },
- {
- "level": 2,
- "title": "Market Profile",
- "page": 374
- },
- {
- "level": 1,
- "title": "17 The Link Between Stocks and Futures: Intermarket Analysis",
- "page": 375
- },
- {
- "level": 2,
- "title": "Intermarket Analysis",
- "page": 376
- },
- {
- "level": 2,
- "title": "Program Trading: The Ultimate Link",
- "page": 376
- },
- {
- "level": 2,
- "title": "The Link Between Bonds and Stocks",
- "page": 378
- },
- {
- "level": 2,
- "title": "The Link Between Bonds and Commodities",
- "page": 379
- },
- {
- "level": 2,
- "title": "The Link Between Commodities and the Dollar",
- "page": 379
- },
- {
- "level": 2,
- "title": "Stock Sectors and Industry Groups",
- "page": 381
- },
- {
- "level": 2,
- "title": "The Dollar and Large Caps",
- "page": 382
- },
- {
- "level": 2,
- "title": "Intermarket Analysis and Mutual Funds",
- "page": 382
- },
- {
- "level": 2,
- "title": "Relative Strength Analysis",
- "page": 383
- },
- {
- "level": 2,
- "title": "Relative Strength and Sectors",
- "page": 385
- },
- {
- "level": 2,
- "title": "Relative Strength and Individual Stocks",
- "page": 386
- },
- {
- "level": 2,
- "title": "Top-Down Market Approach",
- "page": 386
- },
- {
- "level": 2,
- "title": "Deflation Scenario",
- "page": 387
- },
- {
- "level": 2,
- "title": "Intermarket Correlation",
- "page": 388
- },
- {
- "level": 2,
- "title": "Intermarket Neural Network Software",
- "page": 389
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 389
- },
- {
- "level": 1,
- "title": "18 Stock Market Indicators",
- "page": 391
- },
- {
- "level": 2,
- "title": "Measuring Market Breadth",
- "page": 391
- },
- {
- "level": 2,
- "title": "Sample Data",
- "page": 391
- },
- {
- "level": 2,
- "title": "Comparing Market Averages",
- "page": 392
- },
- {
- "level": 2,
- "title": "The Advance-Decline Line",
- "page": 393
- },
- {
- "level": 2,
- "title": "AD Divergence",
- "page": 394
- },
- {
- "level": 2,
- "title": "Daily Versus Weekly AD Lines",
- "page": 394
- },
- {
- "level": 2,
- "title": "Variations in AD Line",
- "page": 395
- },
- {
- "level": 2,
- "title": "McClellan Oscillator",
- "page": 395
- },
- {
- "level": 2,
- "title": "McClellan Summation Index",
- "page": 396
- },
- {
- "level": 2,
- "title": "New Highs Versus New Lows",
- "page": 397
- },
- {
- "level": 2,
- "title": "New High-New Low Index",
- "page": 398
- },
- {
- "level": 2,
- "title": "Upside Versus Downside Volume",
- "page": 399
- },
- {
- "level": 2,
- "title": "The Arms Index",
- "page": 400
- },
- {
- "level": 2,
- "title": "TRIN Versus TICK",
- "page": 401
- },
- {
- "level": 2,
- "title": "Smoothing the Arms Index",
- "page": 401
- },
- {
- "level": 2,
- "title": "Open Arms",
- "page": 402
- },
- {
- "level": 2,
- "title": "Equivolume Charting",
- "page": 402
- },
- {
- "level": 2,
- "title": "Candlepower",
- "page": 404
- },
- {
- "level": 2,
- "title": "Comparing Market Averages",
- "page": 405
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 407
- },
- {
- "level": 1,
- "title": "19 Pulling It All Together—A Checklist",
- "page": 408
- },
- {
- "level": 2,
- "title": "Technical Checklist",
- "page": 408
- },
- {
- "level": 2,
- "title": "How to Coordinate Technical and Fundamental Analysis",
- "page": 410
- },
- {
- "level": 2,
- "title": "Chartered Market Technician (CMT)",
- "page": 410
- },
- {
- "level": 2,
- "title": "Market Technicians Association (MTA)",
- "page": 411
- },
- {
- "level": 2,
- "title": "The Global Reach of Technical Analysis",
- "page": 411
- },
- {
- "level": 2,
- "title": "Technical Analysis by Any Name",
- "page": 412
- },
- {
- "level": 2,
- "title": "Federal Reserve Finally Approves",
- "page": 412
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 413
- },
- {
- "level": 1,
- "title": "A Advanced Technical Indicators",
- "page": 416
- },
- {
- "level": 2,
- "title": "Demand Index (DI)",
- "page": 416
- },
- {
- "level": 2,
- "title": "Herrick Payoff Index (HPI)",
- "page": 418
- },
- {
- "level": 2,
- "title": "Starc Bands and Keltner Channels",
- "page": 420
- },
- {
- "level": 2,
- "title": "Formula for Demand Index",
- "page": 423
- },
- {
- "level": 1,
- "title": "B Market Profile",
- "page": 425
- },
- {
- "level": 2,
- "title": "Introduction",
- "page": 425
- },
- {
- "level": 2,
- "title": "Market Profile Graphic",
- "page": 427
- },
- {
- "level": 2,
- "title": "Market Structure",
- "page": 428
- },
- {
- "level": 2,
- "title": "Market Profile Organizing Principles",
- "page": 429
- },
- {
- "level": 2,
- "title": "Range Development and Profile Patterns",
- "page": 433
- },
- {
- "level": 2,
- "title": "Tracking Longer Term Market Activity",
- "page": 434
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 438
- },
- {
- "level": 1,
- "title": "C The Essentials of Building a Trading System",
- "page": 440
- },
- {
- "level": 2,
- "title": "5-Step Plan",
- "page": 441
- },
- {
- "level": 2,
- "title": "Step 1: Start with a Concept (an Idea)",
- "page": 441
- },
- {
- "level": 2,
- "title": "Step 2: Turn Your Idea into a Set of Objective Rules",
- "page": 443
- },
- {
- "level": 2,
- "title": "Step 3: Visually Check It Out on the Charts",
- "page": 443
- },
- {
- "level": 2,
- "title": "Step 4: Formally Test It with a Computer",
- "page": 443
- },
- {
- "level": 2,
- "title": "Step 5: Evaluate Results",
- "page": 446
- },
- {
- "level": 2,
- "title": "Money Management",
- "page": 446
- },
- {
- "level": 2,
- "title": "Conclusion",
- "page": 447
- },
- {
- "level": 1,
- "title": "D Continuous Futures Contracts",
- "page": 450
- },
- {
- "level": 2,
- "title": "Nearest Contract",
- "page": 450
- },
- {
- "level": 2,
- "title": "Next Contract",
- "page": 451
- },
- {
- "level": 2,
- "title": "Gann Contract",
- "page": 451
- },
- {
- "level": 2,
- "title": "Continuous Contracts",
- "page": 452
- },
- {
- "level": 2,
- "title": "Constant Forward Continuous Contracts",
- "page": 452
- },
- {
- "level": 1,
- "title": "Glossary",
- "page": 454
- },
- {
- "level": 1,
- "title": "Selected Bibliography",
- "page": 462
- },
- {
- "level": 1,
- "title": "Selected Resources",
- "page": 466
- },
- {
- "level": 1,
- "title": "Index",
- "page": 469
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- "level": 1,
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- "level": 1,
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- "level": 2,
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- "level": 3,
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- "level": 3,
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- "level": 3,
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- "page": 107
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- {
- "level": 3,
- "title": "Trading Rules",
- "page": 109
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- {
- "level": 3,
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- "level": 2,
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- "level": 3,
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- "level": 3,
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- "page": 117
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- "page": 128
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- "level": 2,
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- },
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- "level": 3,
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- "page": 128
- },
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- "level": 3,
- "title": "Trading Psychology",
- "page": 129
- },
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- "level": 3,
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- {
- "level": 2,
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- "level": 3,
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- "page": 134
- },
- {
- "level": 3,
- "title": "Trading Rules",
- "page": 136
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- "level": 3,
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- "page": 137
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- {
- "level": 2,
- "title": "32. Time",
- "page": 137
- },
- {
- "level": 3,
- "title": "Cycles",
- "page": 138
- },
- {
- "level": 3,
- "title": "Indicator Seasons",
- "page": 138
- },
- {
- "level": 3,
- "title": "Market Time",
- "page": 140
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- "page": 141
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- {
- "level": 2,
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- },
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- "level": 3,
- "title": "Investing",
- "page": 142
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- {
- "level": 3,
- "title": "Swing Trading",
- "page": 144
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- {
- "level": 3,
- "title": "Day-Trading",
- "page": 147
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- {
- "level": 1,
- "title": "PART 6 General Market Indicators",
- "page": 149
- },
- {
- "level": 2,
- "title": "34. The New High–New Low Index",
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- {
- "level": 3,
- "title": "How to Construct NH-NL",
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- "level": 3,
- "title": "Crowd Psychology",
- "page": 150
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- {
- "level": 3,
- "title": "Trading Rules for NH-NL",
- "page": 151
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- "level": 3,
- "title": "NH-NL in Multiple Timeframes and Look-Back Periods",
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- {
- "level": 2,
- "title": "35. Stocks above 50-Day MA",
- "page": 155
- },
- {
- "level": 2,
- "title": "36. Other Stock Market Indicators",
- "page": 156
- },
- {
- "level": 3,
- "title": "Advance/Decline",
- "page": 156
- },
- {
- "level": 2,
- "title": "37. Consensus and Commitment Indicators",
- "page": 158
- },
- {
- "level": 3,
- "title": "Tracking Advisory Opinion",
- "page": 159
- },
- {
- "level": 3,
- "title": "Signals from the Press",
- "page": 160
- },
- {
- "level": 3,
- "title": "Signals from Advertisers",
- "page": 160
- },
- {
- "level": 3,
- "title": "Commitments of Futures Traders",
- "page": 161
- },
- {
- "level": 3,
- "title": "Legal Insider Trading",
- "page": 163
- },
- {
- "level": 3,
- "title": "Short Interest",
- "page": 163
- },
- {
- "level": 1,
- "title": "PART 7 Trading Systems",
- "page": 165
- },
- {
- "level": 2,
- "title": "38. System Testing, Paper Trading, and the Three Key Demands for Every Trade",
- "page": 167
- },
- {
- "level": 3,
- "title": "Paper Trading",
- "page": 168
- },
- {
- "level": 3,
- "title": "Three Key Demands for Every Trade",
- "page": 169
- },
- {
- "level": 2,
- "title": "39. Triple Screen Trading System",
- "page": 170
- },
- {
- "level": 3,
- "title": "Trend-Following Indicators and Oscillators",
- "page": 171
- },
- {
- "level": 3,
- "title": "Choosing Timeframes—the Factor of Five",
- "page": 171
- },
- {
- "level": 3,
- "title": "First Screen—Market Tide",
- "page": 172
- },
- {
- "level": 3,
- "title": "Second Screen—Market Wave",
- "page": 174
- },
- {
- "level": 3,
- "title": "Third Screen—Entry Technique",
- "page": 174
- },
- {
- "level": 3,
- "title": "Triple Screen Summary",
- "page": 176
- },
- {
- "level": 3,
- "title": "Triple Screen in Day-Trading",
- "page": 177
- },
- {
- "level": 3,
- "title": "Stops and Profit Targets",
- "page": 178
- },
- {
- "level": 2,
- "title": "40. The Impulse System",
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- },
- {
- "level": 3,
- "title": "Entries",
- "page": 180
- },
- {
- "level": 3,
- "title": "Exits",
- "page": 182
- },
- {
- "level": 2,
- "title": "41. Channel Trading Systems",
- "page": 182
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- {
- "level": 3,
- "title": "Two Ways to Construct a Channel",
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- {
- "level": 3,
- "title": "Symmetrical Channels",
- "page": 183
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- {
- "level": 3,
- "title": "Mass Psychology",
- "page": 184
- },
- {
- "level": 3,
- "title": "Trading Rules",
- "page": 186
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- {
- "level": 3,
- "title": "Standard Deviation Channels (Bollinger Bands)",
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- {
- "level": 1,
- "title": "PART 8 Trading Vehicles",
- "page": 189
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- "level": 2,
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- },
- {
- "level": 2,
- "title": "44. Options",
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- },
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- "level": 3,
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- },
- {
- "level": 3,
- "title": "Writing Options",
- "page": 197
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- "level": 3,
- "title": "Writer’s Choice",
- "page": 198
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- {
- "level": 3,
- "title": "Limiting Risk",
- "page": 199
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- {
- "level": 3,
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- {
- "level": 2,
- "title": "45. CFDs",
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- {
- "level": 2,
- "title": "46. Futures",
- "page": 203
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- {
- "level": 3,
- "title": "Futures and Cash Trades",
- "page": 204
- },
- {
- "level": 3,
- "title": "Hedging",
- "page": 205
- },
- {
- "level": 3,
- "title": "Supply, Demand, and Seasonality",
- "page": 206
- },
- {
- "level": 3,
- "title": "Floors and Ceilings",
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- },
- {
- "level": 3,
- "title": "Contango, Inversion, and Spreads",
- "page": 207
- },
- {
- "level": 3,
- "title": "Commitments of Traders",
- "page": 208
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- {
- "level": 3,
- "title": "Margins and Risk Control",
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- {
- "level": 2,
- "title": "47. Forex",
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- {
- "level": 1,
- "title": "PART 9 Risk Management",
- "page": 213
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- {
- "level": 2,
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- {
- "level": 3,
- "title": "Why Johnny Can’t Sell",
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- {
- "level": 3,
- "title": "Probability and Innumeracy",
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- {
- "level": 3,
- "title": "A Positive Expectation",
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- {
- "level": 3,
- "title": "Businessman’s Risk or Loss",
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- {
- "level": 2,
- "title": "49. The Two Main Rules of Risk Control",
- "page": 218
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- {
- "level": 3,
- "title": "The Two Worst Mistakes",
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- },
- {
- "level": 2,
- "title": "50. The Two Percent Rule",
- "page": 219
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- {
- "level": 3,
- "title": "The Iron Triangle of Risk Control",
- "page": 220
- },
- {
- "level": 3,
- "title": "The 2% Rule in the Futures Markets",
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- },
- {
- "level": 2,
- "title": "51. The Six Percent Rule",
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- {
- "level": 3,
- "title": "The Concept of Available Risk",
- "page": 224
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- {
- "level": 2,
- "title": "52. A Comeback from a Drawdown",
- "page": 226
- },
- {
- "level": 3,
- "title": "A Trading Manager",
- "page": 228
- },
- {
- "level": 1,
- "title": "PART 10 Practical Details",
- "page": 231
- },
- {
- "level": 2,
- "title": "53. How to Set Profit Targets: “Enough” Is the Power Word",
- "page": 231
- },
- {
- "level": 2,
- "title": "54. How to Set Stops: Say No to Wishful Thinking",
- "page": 235
- },
- {
- "level": 3,
- "title": "Place Stops outside the Zone of “Market Noise”",
- "page": 236
- },
- {
- "level": 3,
- "title": "Don’t Place Your Stops at Obvious Levels",
- "page": 236
- },
- {
- "level": 3,
- "title": "Don’t Let a Winning Trade Turn into a Loss",
- "page": 239
- },
- {
- "level": 3,
- "title": "Move Your Stop Only in the Direction of Your Trade",
- "page": 240
- },
- {
- "level": 3,
- "title": "Catastrophic Stops: A Professional’s Life Jacket",
- "page": 240
- },
- {
- "level": 3,
- "title": "Stops and Overnight Gaps: Only for the Pros",
- "page": 241
- },
- {
- "level": 2,
- "title": "55. Is This an A-Trade?",
- "page": 241
- },
- {
- "level": 2,
- "title": "56. Scanning for Possible Trades",
- "page": 246
- },
- {
- "level": 1,
- "title": "PART 11 Good Record-Keeping",
- "page": 249
- },
- {
- "level": 2,
- "title": "57. Your Daily Homework",
- "page": 250
- },
- {
- "level": 3,
- "title": "Are You Ready to Trade Today?",
- "page": 252
- },
- {
- "level": 2,
- "title": "58. Creating and Scoring Trade Plans",
- "page": 254
- },
- {
- "level": 3,
- "title": "Scoring Your Trade Plans (a Trade Apgar)",
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- },
- {
- "level": 3,
- "title": "Using a Tradebill",
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- },
- {
- "level": 2,
- "title": "59. Trade Journal",
- "page": 259
- },
- {
- "level": 3,
- "title": "Three Benefits",
- "page": 261
- },
- {
- "level": 1,
- "title": "CONCLUSION A Journey without an End: How to Continue Learning",
- "page": 265
- },
- {
- "level": 1,
- "title": "Sources",
- "page": 269
- },
- {
- "level": 1,
- "title": "Acknowledgments",
- "page": 273
- },
- {
- "level": 1,
- "title": "About the Author",
- "page": 275
- },
- {
- "level": 1,
- "title": "INDEX",
- "page": 277
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- {
- "level": 1,
- "title": "EULA",
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diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/001_Cover.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/001_Cover.md
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+++ /dev/null
@@ -1,393 +0,0 @@
-# Fundamentals of Electric Circuits
-
-# Charles K. Alexander
-
-Department of Electrical and Computer Engineering *Cleveland State University*
-
-# Matthew N. O. Sadiku
-
-Department of Electrical and Computer Engineering
-
-*Prairie View A&M University*
-
-## FUNDAMENTALS OF ELECTRIC CIRCUITS, SIXTH EDITION
-
-Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2017 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2013, 2009, and 2007. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.
-
-Some ancillaries, including electronic and print components, may not be available to customers outside the United States.
-
-This book is printed on acid-free paper.
-
-1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5
-
-ISBN 978-0-07-802822-9 MHID 0-07-802822-1
-
-Senior Vice President, Products & Markets: *Kurt L. Strand* Vice President, General Manager, Products & Markets: *Marty Lange* Vice President, Content Design & Delivery: *Kimberly Meriwether David* Managing Director: *Thomas Timp* Global Brand Manager: *Raghu Srinivasan* Director, Product Development: *Rose Koos* Product Developer: *Vincent Bradshaw* Marketing Manager: *Nick McFadden* Digital Product Analyst: *Patrick Diller* Associate Director of Digital Content: *Amy Bumbaco, Ph.D.* Director, Content Design & Delivery: *Linda Avenarius* Program Manager: *Faye M. Herrig* Content Project Managers: *Melissa M. Leick, Tammy Juran, Sandra Schnee* Buyer: *Sandy Ludovissy* Design: *Studio Montage, Inc.* Content Licensing Specialist: *Lorraine Buczek* Cover Image: *Courtesy NASA/JPL - Caltech* Compositor: *MPS Limited* Printer: *R. R. Donnelley*
-
-All credits appearing on page or at the end of the book are considered to be an extension of the copyright page.
-
-## **Library of Congress Cataloging-in-Publication Data**
-
-Alexander, Charles K., author.
-
- Fundamentals of electric circuits / Charles K. Alexander, Department of Electrical and Computer Engineering, Cleveland State University, Matthew N. O. Sadiku, Department of Electrical Engineering, Prairie View A&M University. — Sixth edition.
-
- pages cm Includes index. ISBN 978-0-07-802822-9 (alk. paper) — ISBN 0-07-802822-1 (alk. paper) 1. Electric circuits. I. Sadiku, Matthew N. O., author. II. Title.
-
- TK454.A452 2017 621.3815—dc23 2015035301
-
-The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites.
-
-Dedicated to our wives, Kikelomo and Hannah, whose understanding and support have truly made this book possible.
-
-Matthew and Chuck
-
-*This page intentionally left blank*
-
-# Contents
-
-*[Preface xi](#page-11-0) [Acknowledgments xv](#page-15-0) [About the Authors xxi](#page-21-0)*
-
-# **PART 1** [DC Circuits 2](#page-24-0) **Chapter 1** [Basic Concepts 3](#page-25-0) [1.1 Introduction 4](#page-26-0) [1.2 Systems of Units 5](#page-27-0) [1.3 Charge and Current 6](#page-28-0) [1.4 Voltage 9](#page-31-0) [1.5 Power and Energy 10](#page-32-0) [1.6 Circuit Elements 14](#page-36-0) [1.7 Applications 16](#page-38-0) 1.7.1 TV Picture Tube 1.7.2 Electricity Bills [1.8 Problem Solving 19](#page-41-0) [1.9 Summary 22](#page-44-0) [Review Questions 23](#page-45-0) [Problems 24](#page-46-0) [Comprehensive Problems 26](#page-48-0)
-
-# **Chapter 2** [Basic Laws 29](#page-51-0)
-
-- [2.1 Introduction 30](#page-52-0)
-- [2.2 Ohm's Law 30](#page-52-0)
-- [2.3 Nodes, Branches, and Loops 35](#page-57-0)
-- [2.4 Kirchhoff's Laws 37](#page-59-0)
-- [2.5 Series Resistors and Voltage Division 43](#page-65-0)
-- [2.6 Parallel Resistors and Current Division 44](#page-66-0)
-- [2.7 Wye-Delta Transformations 51](#page-73-0) Delta to Wye Conversion Wye to Delta Conversion
-- [2.8 Applications 57](#page-79-0) 2.8.1 Lighting Systems [2.8.2 Design of DC Meters](#page-85-0) 2.9 Summary 63 [Review Questions 64](#page-86-0)
-
-# [Problems 65](#page-87-0)
-
-[Comprehensive Problems 77](#page-99-0)
-
-# **Chapter 3** [Methods of Analysis 79](#page-101-0)
-
-- [3.1 Introduction 80](#page-102-0)
-- [3.2 Nodal Analysis 80](#page-102-0)
-- [3.3 Nodal Analysis with Voltage Sources 86](#page-108-0)
-- [3.4 Mesh Analysis 91](#page-113-0)
-- [3.5 Mesh Analysis with Current Sources 96](#page-118-0)
-- [3.6 Nodal and Mesh Analyses](#page-120-0) by Inspection 98
-- [3.7 Nodal Versus Mesh Analysis 102](#page-124-0)
-- [3.8 Circuit Analysis with](#page-125-0) PSpice 103
-- [3.9 Applications: DC Transistor Circuits 105](#page-127-0) [3.10 Summary 110](#page-132-0) [Review Questions 111](#page-133-0)
- - [Problems 112](#page-134-0) [Comprehensive Problem 124](#page-146-0)
-
-# **Chapter 4** [Circuit Theorems 125](#page-147-0)
-
-- [4.1 Introduction 126](#page-148-0)
-- [4.2 Linearity Property 126](#page-148-0)
-- [4.3 Superposition 128](#page-150-0)
-- [4.4 Source Transformation 133](#page-155-0)
-- [4.5 Thevenin's Theorem 137](#page-159-0)
-- [4.6 Norton's Theorem 143](#page-165-0)
-- [4.7 Derivations of Thevenin's](#page-169-0) and Norton's Theorems 147
-- [4.8 Maximum Power Transfer 148](#page-170-0)
-- [4.9 Verifying Circuit Theorems](#page-172-0) with PSpice 150
-- [4.10 Applications 153](#page-175-0) 4.10.1 Source Modeling [4.10.2 Resistance Measurement](#page-180-0)
-- 4.11 Summary 158 [Review Questions 159](#page-181-0) [Problems 160](#page-182-0) [Comprehensive Problems 171](#page-193-0)
-
-# **Chapter 5** [Operational Amplifiers 173](#page-195-0)
-
-- [5.1 Introduction 174](#page-196-0)
-- [5.2 Operational Amplifiers 174](#page-196-0)
-
-[5.3 Ideal Op Amp 178](#page-200-0) [5.4 Inverting Amplifier 179](#page-201-0) [5.5 Noninverting Amplifier 181](#page-203-0) [5.6 Summing Amplifier 183](#page-205-0) [5.7 Difference Amplifier 185](#page-207-0) [5.8 Cascaded Op Amp Circuits 189](#page-211-0) [5.9 Op Amp Circuit Analysis with](#page-214-0) PSpice 192 [5.10 Applications 194](#page-216-0) [5.10.1 Digital-to-Analog Converter](#page-219-0) 5.10.2 Instrumentation Amplifiers 5.11 Summary 197 [Review Questions 199](#page-221-0) [Problems 200](#page-222-0) [Comprehensive Problems 211](#page-233-0)
-
-# **Chapter 6** [Capacitors and Inductors 213](#page-235-0)
-
-- [6.1 Introduction 214](#page-236-0)
-- [6.2 Capacitors 214](#page-236-0)
-- [6.3 Series and Parallel Capacitors 220](#page-242-0)
-- [6.4 Inductors 224](#page-246-0)
-- [6.5 Series and Parallel Inductors 228](#page-250-0)
-- [6.6 Applications 231](#page-253-0) 6.6.1 Integrator 6.6.2 Differentiator 6.6.3 Analog Computer
-- [6.7 Summary 238](#page-260-0) [Review Questions 239](#page-261-0) [Problems 240](#page-262-0) [Comprehensive Problems 249](#page-271-0)
-
-# **Chapter 7** [First-Order Circuits 251](#page-273-0)
-
-- [7.1 Introduction 252](#page-274-0)
-- [7.2 The Source-Free](#page-275-0) RC Circuit 253
-- [7.3 The Source-Free](#page-279-0) RL Circuit 257
-- [7.4 Singularity Functions 263](#page-285-0)
-- [7.5 Step Response of an](#page-293-0) RC Circuit 271
-- [7.6 Step Response of an](#page-300-0) RL Circuit 278
-- [7.7 First-Order Op Amp Circuits 282](#page-304-0)
-- [7.8 Transient Analysis with](#page-309-0) PSpice 287
-- [7.9 Applications 291](#page-313-0)
- - 7.9.1 Delay Circuits
- - 7.9.2 Photoflash Unit
- - 7.9.3 Relay Circuits
- - [7.9.4 Automobile Ignition Circuit](#page-319-0)
-- 7.10 Summary 297 [Review Questions 298](#page-320-0) [Problems 299](#page-321-0) [Comprehensive Problems 309](#page-331-0)
-
-# **Chapter 8** [Second-Order Circuits 311](#page-333-0)
-
-- [8.1 Introduction 312](#page-334-0)
-- [8.2 Finding Initial and Final Values 313](#page-335-0)
-- [8.3 The Source-Free Series](#page-339-0) RLC Circuit 317
-- [8.4 The Source-Free Parallel](#page-346-0) RLC Circuit 324
-- [8.5 Step Response of a Series](#page-351-0) RLC Circuit 329
-- [8.6 Step Response of a Parallel](#page-356-0) RLC Circuit 334
-- [8.7 General Second-Order Circuits 337](#page-359-0)
-- [8.8 Second-Order Op Amp Circuits 342](#page-364-0)
-- 8.9 PSpice Analysis of RLC [Circuits 344](#page-366-0)
-- [8.10 Duality 348](#page-370-0)
-- [8.11 Applications 351](#page-373-0) [8.11.1 Automobile Ignition System](#page-376-0) 8.11.2 Smoothing Circuits
-- 8.12 Summary 354 [Review Questions 355](#page-377-0) [Problems 356](#page-378-0) [Comprehensive Problems 365](#page-387-0)
-
-```
-PART 2 AC Circuits 366
-```
-
-# **Chapter 9** [Sinusoids and Phasors 367](#page-389-0)
-
-- [9.1 Introduction 368](#page-390-0)
-- [9.2 Sinusoids 369](#page-391-0)
-- [9.3 Phasors 374](#page-396-0)
-- [9.4 Phasor Relationships for](#page-405-0) Circuit Elements 383
-- [9.5 Impedance and Admittance 385](#page-407-0)
-- [9.6 Kirchhoff's Laws in the Frequency](#page-409-0) Domain 387
-- [9.7 Impedance Combinations 388](#page-410-0)
-- [9.8 Applications 394](#page-416-0) 9.8.1 Phase-Shifters 9.8.2 AC Bridges
-- [9.9 Summary 400](#page-422-0) [Review Questions 401](#page-423-0) [Problems 401](#page-423-0) [Comprehensive Problems 409](#page-431-0)
-
-# **Chapter 10** [Sinusoidal Steady-State](#page-433-0) Analysis 411
-
-- [10.1 Introduction 412](#page-434-0)
-- [10.2 Nodal Analysis 412](#page-434-0)
-- [10.3 Mesh Analysis 415](#page-437-0)
-
-- [10.4 Superposition Theorem 419](#page-441-0)
-- [10.5 Source Transformation 422](#page-444-0) [10.6 Thevenin and Norton](#page-446-0) Equivalent Circuits 424
-- [10.7 Op Amp AC Circuits 429](#page-451-0)
-- [10.8 AC Analysis Using](#page-453-0) PSpice 431
-- [10.9 Applications 435](#page-457-0) [10.9.1 Capacitance Multiplier](#page-461-0) 10.9.2 Oscillators
-- 10.10 Summary 439 [Review Questions 439](#page-461-0) [Problems 441](#page-463-0)
-
-# **Chapter 11** [AC Power Analysis 455](#page-477-0)
-
-- [11.1 Introduction 456](#page-478-0)
-- [11.2 Instantaneous and Average Power 456](#page-478-0)
-- [11.3 Maximum Average Power Transfer 462](#page-484-0)
-- [11.4 Effective or RMS Value 465](#page-487-0) [11.5 Apparent Power and](#page-490-0)
- - Power Factor 468
-- [11.6 Complex Power 471](#page-493-0)
-- [11.7 Conservation of AC Power 475](#page-497-0)
-- [11.8 Power Factor Correction 479](#page-501-0) [11.9 Applications 481](#page-503-0) 11.9.1 Power Measurement [11.9.2 Electricity Consumption Cost](#page-508-0)
-- 11.10 Summary 486 [Review Questions 488](#page-510-0) [Problems 488](#page-510-0) [Comprehensive Problems 498](#page-520-0)
-
-# **Chapter 12** [Three-Phase Circuits 501](#page-523-0)
-
-- [12.1 Introduction 502](#page-524-0)
-- [12.2 Balanced Three-Phase Voltages 503](#page-525-0)
-- [12.3 Balanced Wye-Wye Connection 507](#page-529-0)
-- [12.4 Balanced Wye-Delta Connection 510](#page-532-0)
-- [12.5 Balanced Delta-Delta](#page-534-0) Connection 512
-- [12.6 Balanced Delta-Wye Connection 514](#page-536-0)
-- [12.7 Power in a Balanced System 517](#page-539-0)
-- [12.8 Unbalanced Three-Phase](#page-545-0) Systems 523
-- 12.9 PSpice [for Three-Phase Circuits 527](#page-549-0)
-- [12.10 Applications 532](#page-554-0) [12.10.1 Three-Phase Power Measurement](#page-22-0) 12.10.2 Residential Wiring
-
-[12.11 Summary 541](#page-563-0) [Review Questions 541](#page-563-0) [Problems 542](#page-564-0) [Comprehensive Problems 551](#page-573-0)
-
-# **Chapter 13** [Magnetically Coupled](#page-575-0) Circuits 553
-
-- [13.1 Introduction 554](#page-576-0)
-- [13.2 Mutual Inductance 555](#page-577-0)
-- [13.3 Energy in a Coupled Circuit 562](#page-584-0)
-- [13.4 Linear Transformers 565](#page-587-0)
-- [13.5 Ideal Transformers 571](#page-593-0)
-- [13.6 Ideal Autotransformers 579](#page-601-0)
-- [13.7 Three-Phase Transformers 582](#page-604-0)
-- 13.8 PSpice [Analysis of Magnetically](#page-606-0) Coupled Circuits 584
-- [13.9 Applications 589](#page-611-0)
- - [13.9.1 Transformer as an Isolation Device](#page-617-0) 13.9.2 Transformer as a Matching Device
-
- - 13.9.3 Power Distribution
-- 13.10 Summary 595 [Review Questions 596](#page-618-0) [Problems 597](#page-619-0) [Comprehensive Problems 609](#page-631-0)
-
-# **Chapter 14** [Frequency Response 611](#page-633-0)
-
-- [14.1 Introduction 612](#page-634-0)
-- [14.2 Transfer Function 612](#page-634-0)
-- [14.3 The Decibel Scale 615](#page-637-0)
-- [14.4 Bode Plots 617](#page-639-0)
-- [14.5 Series Resonance 627](#page-649-0)
-- [14.6 Parallel Resonance 632](#page-654-0)
-- [14.7 Passive Filters 635](#page-657-0)
- - 14.7.1 Low-Pass Filter
- - 14.7.2 High-Pass Filter
- - 14.7.3 Band-Pass Filter
- - 14.7.4 Band-Stop Filter
-- [14.8 Active Filters 640](#page-662-0) 14.8.1 First-Order Low-Pass Filter
- - 14.8.2 First-Order High-Pass Filter
- - 14.8.3 Band-Pass Filter
- - [14.8.4 Band-Reject \(or Notch\) Filter](#page-668-0)
-- 14.9 Scaling 646
- - 14.9.1 Magnitude Scaling
- - 14.9.2 Frequency Scaling
- - [14.9.3 Magnitude and Frequency Scaling](#page-22-0)
-
-- [14.10 Frequency Response Using](#page-672-0) PSpice 650
-- [14.11 Computation Using](#page-675-0) MATLAB 653
-- [14.12 Applications 655](#page-677-0)
- - 14.12.1 Radio Receiver
- - [14.12.2 Touch-Tone Telephone](#page-683-0)
- - 14.12.3 Crossover Network
-- 14.13 Summary 661 [Review Questions 662](#page-684-0) [Problems 663](#page-685-0) [Comprehensive Problems 671](#page-693-0)
-
-# **PART 3** [Advanced Circuit](#page-694-0) Analysis 672
-
-**Chapter 15** [Introduction to the Laplace](#page-695-0) Transform 673
-
-- [15.1 Introduction 674](#page-696-0)
-- [15.2 Definition of the Laplace](#page-697-0) Transform 675
-- [15.3 Properties of the Laplace](#page-699-0) Transform 677
-- [15.4 The Inverse Laplace Transform 688](#page-710-0) 15.4.1 Simple Poles 15.4.2 Repeated Poles 15.4.3 Complex Poles
-- [15.5 The Convolution Integral 695](#page-717-0)
-- [15.6 Application to Integrodifferential](#page-725-0) Equations 703
-- [15.7 Summary 706](#page-728-0) [Review Questions 706](#page-728-0) [Problems 707](#page-729-0)
-
-# **Chapter 16** [Applications of the Laplace](#page-735-0) Transform 713
-
-- [16.1 Introduction 714](#page-736-0)
-- [16.2 Circuit Element Models 715](#page-737-0)
-- [16.3 Circuit Analysis 720](#page-742-0)
-- [16.4 Transfer Functions 724](#page-746-0)
-- [16.5 State Variables 728](#page-750-0)
-- [16.6 Applications 735](#page-757-0) 16.6.1 Network Stability 16.6.2 Network Synthesis
-- [16.7 Summary 743](#page-765-0) [Review Questions 744](#page-766-0) [Problems 745](#page-767-0) [Comprehensive Problems 756](#page-778-0)
-
-# **Chapter 17** [The Fourier Series 757](#page-779-0)
-
-- [17.1 Introduction 758](#page-780-0)
-- [17.2 Trigonometric Fourier Series 759](#page-781-0)
-- [17.3 Symmetry Considerations 766](#page-788-0)
- - 17.3.1 Even Symmetry
- - 17.3.2 Odd Symmetry
- - 17.3.3 Half-Wave Symmetry
-- [17.4 Circuit Applications 776](#page-798-0)
-- [17.5 Average Power and RMS Values 780](#page-802-0)
-- [17.6 Exponential Fourier Series 783](#page-805-0)
-- [17.7 Fourier Analysis with](#page-811-0) PSpice 789 [17.7.1 Discrete Fourier Transform](#page-817-0)
-- 17.7.2 Fast Fourier Transform 17.8 Applications 795 17.8.1 Spectrum Analyzers 17.8.2 Filters
-- [17.9 Summary 798](#page-820-0) [Review Questions 800](#page-822-0) [Problems 800](#page-822-0) [Comprehensive Problems 809](#page-831-0)
-
-# **Chapter 18** [Fourier Transform 811](#page-833-0)
-
-- [18.1 Introduction 812](#page-834-0)
-- [18.2 Definition of the Fourier Transform 812](#page-834-0)
-- [18.3 Properties of the Fourier](#page-840-0) Transform 818
-- [18.4 Circuit Applications 831](#page-853-0)
-- [18.5 Parseval's Theorem 834](#page-856-0)
-- [18.6 Comparing the Fourier and](#page-859-0) Laplace Transforms 837
-- [18.7 Applications 838](#page-860-0) [18.7.1 Amplitude Modulation](#page-863-0) 18.7.2 Sampling
-- 18.8 Summary 841 [Review Questions 842](#page-864-0) [Problems 843](#page-865-0) [Comprehensive Problems 849](#page-871-0)
-
-# **Chapter 19** [Two-Port Networks 851](#page-873-0)
-
-- [19.1 Introduction 852](#page-874-0)
-- [19.2 Impedance Parameters 853](#page-875-0)
-- [19.3 Admittance Parameters 857](#page-879-0)
-- [19.4 Hybrid Parameters 860](#page-882-0)
-- [19.5 Transmission Parameters 865](#page-887-0)
-- [19.6 Relationships Between](#page-892-0) Parameters 870
-
-## Contents **ix**
-
-*[Index I-1](#page-981-0)*
-
-- [19.7 Interconnection of Networks 873](#page-895-0)
-- [19.8 Computing Two-Port Parameters](#page-901-0) Using PSpice 879
-- [19.9 Applications 882](#page-904-0) 19.9.1 Transistor Circuits [19.9.2 Ladder Network Synthesis](#page-913-0)
-- 19.10 Summary 891 [Review Questions 892](#page-914-0) [Problems 892](#page-914-0) [Comprehensive Problem 903](#page-925-0)
-- **Appendix A** [Simultaneous Equations and Matrix](#page-926-0) Inversion A **Appendix B** [Complex Numbers A-9](#page-935-0) **Appendix C** [Mathematical Formulas A-16](#page-942-0) **Appendix D** [Answers to Odd-Numbered](#page-947-0) Problems A-21 *[Selected Bibliography B-1](#page-979-0)*
-
-*This page intentionally left blank*
-
-# Preface
-
-In keeping with our focus on space for covers for our book, we have chosen the NASA Voyager spacecraft for the sixth edition. The reason for this is that like any spacecraft there are many circuits that play criti cal roles in their functionality. The beginning of the Voyager 1 and 2 odyssey began on August 20, 1977, for Voyager 2 and on September 5, 1977, for Voyager 1. Both were launched from NASA's Kennedy Space Center in Florida. The Voyager 1 was launched on a faster orbit so it eventually became the first man-made object to leave our solar system. There is some debate over whether it has actually left the solar system, but it certainly will at some point in time. Voyager 2 and two Pioneer spacecraft will also leave the solar system at some point in time.
-
-Voyager 1 is still functioning and sending back data, a truly significant achievement for NASA engineers. The design processes that make the Voyager operate so reliably are based on the fundamentals discussed in this textbook. Finally, space is vast so that Voyager 1 will fly past other solar systems; the odds of actually coming into contact with something are so remote that it may virtually fly through the universe forever! For more about Voyager 1, go to NASA's website: www.nasa.gov/.
-
-# Features
-
-# New to This Edition
-
-We have added learning objectives to each chapter to reflect what we believe are the most important items to learn from each chapter. These should help you focus more carefully on what you should be learning.
-
-There are more than 580 revised end-of-chapter problems, new endof-chapter problems, and revised practice problems. We continue to try and make our problems as practical as possible.
-
-In addition, we have improved Connect for this edition by increasing the number of problems available substantially. Now, professors may select from more than a thousand problems as they build thier online homework assignments.
-
-We have also built SmartBook for this edition. With SmartBook, stu dents get the same text as the print version, along with personalized tips on what to study next, thanks to SmartBook's adaptive technology.
-
-# Retained from Previous Editions
-
-A course in circuit analysis is perhaps the first exposure students have to electrical engineering. This is also a place where we can enhance some of the skills that they will later need as they learn how to design. An important part of this book is our 121 *design a problem* problems. These problems were developed to enhance skills that are an impor tant part of the design process. We know it is not possible to fully develop a student's design skills in a fundamental course like circuits. To fully develop design skills a student needs a design experience
-
-normally reserved for their senior year. This does not mean that some of those skills cannot be developed and exercised in a circuits course. The text already included open-ended questions that help students use creativity, which is an important part of learning how to design. We already have some questions that are open-ended but we desired to add much more into our text in this important area and have devel oped an approach to do just that. When we develop problems for the student to solve our goal is that in solving the problem the student learns more about the theory and the problem solving process. Why not have the students design problems like we do? That is exactly what we do in each chapter. Within the normal problem set, we have a set of problems where we ask the student to design a problem to help other students better understand an important concept. This has two very important results. The first will be a better understanding of the basic theory and the second will be the enhancement of some of the student's basic design skills. We are making effective use of the principle of learning by teaching. Essentially we all learn better when we teach a subject. Designing effective problems is a key part of the teaching process. Students should also be encouraged to develop problems, when appropriate, which have nice numbers and do not necessarily overemphasize complicated mathematical manipulations.
-
-A very important advantage to our textbook, we have a total of 2,481 Examples, Practice Problems, Review Questions, and End-of-Chapter Problems! Answers are provided for all practice problems and the odd numbered end-of-chapter problems.
-
-The main objective of the sixth edition of this book remains the same as the previous editions—to present circuit analysis in a manner that is clearer, more interesting, and easier to understand than other cir cuit textbooks, and to assist the student in beginning to see the "fun" in engineering. This objective is achieved in the following ways:
-
-# • **Chapter Openers and Summaries**
-
-Each chapter opens with a discussion about how to enhance skills which contribute to successful problem solving as well as success ful careers or a career-oriented talk on a subdiscipline of electrical engineering. This is followed by an introduction that links the chap ter with the previous chapters and states the chapter objectives. The chapter ends with a summary of key points and formulas.
-
-• **Problem-Solving Methodology**
-
-Chapter 1 introduces a six-step method for solving circuit problems which is used consistently throughout the book and media supple ments to promote best-practice problem-solving procedures.
-
-• **Student-Friendly Writing Style**
-
-All principles are presented in a lucid, logical, step-by-step man ner. As much as possible, we avoid wordiness and giving too much detail that could hide concepts and impede overall understanding of the material.
-
-# • **Boxed Formulas and Key Terms**
-
-Important formulas are boxed as a means of helping students sort out what is essential from what is not. Also, to ensure that students clearly understand the key elements of the subject matter, key terms are defined and highlighted.
-
-# • **Margin Notes**
-
-Marginal notes are used as a pedagogical aid. They serve multiple uses such as hints, cross-references, more exposition, warnings, reminders not to make some particular common mistakes, and prob lem-solving insights.
-
-# • **Worked Examples**
-
-Thoroughly worked examples are liberally given at the end of ev ery section. The examples are regarded as a part of the text and are clearly explained without asking the reader to fill in missing steps. Thoroughly worked examples give students a good understanding of the solution process and the confidence to solve problems them selves. Some of the problems are solved in two or three different ways to facilitate a substantial comprehension of the subject mate rial as well as a comparison of different approaches.
-
-# • **Practice Problems**
-
-To give students practice opportunity, each illustrative example is immediately followed by a practice problem with the answer. The student can follow the example step-by-step to aid in the solution of the practice problem without flipping pages or looking at the end of the book for answers. The practice problem is also intended to test a student's understanding of the preceding example. It will reinforce their grasp of the material before the student can move on to the next section. Complete solutions to the practice problems are avail able to students on the website.
-
-# • **Application Sections**
-
-The last section in each chapter is devoted to practical application aspects of the concepts covered in the chapter. The material covered in the chapter is applied to at least one or two practical problems or devices. This helps students see how the concepts are applied to real-life situations.
-
-# • **Review Questions**
-
-Ten review questions in the form of multiple-choice objective items are provided at the end of each chapter with answers. The review questions are intended to cover the little "tricks" that the examples and end-of-chapter problems may not cover. They serve as a self test device and help students determine how well they have mas tered the chapter.
-
-# • **Computer Tools**
-
-In recognition of the requirements by ABET ® on integrating computer tools, the use of *PSpice, Multisim, MATLAB, KCIDE for Circuits*, and developing design skills are encouraged in a studentfriendly manner. *PSpice* is covered early on in the text so that stu dents can become familiar and use it throughout the text. Tutorials on all of these are available on Connect. *MATLAB* is also introduced early in the book.
-
-# • **Design a Problem Problems**
-
-Finally, *design a problem* problems are meant to help the student de velop skills that will be needed in the design process.
-
-# • **Historical Tidbits**
-
-Historical sketches throughout the text provide profiles of important pioneers and events relevant to the study of electrical engineering.
-
-# • **Early Op Amp Discussion**
-
- The operational amplifier (op amp) as a basic element is introduced early in the text.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/002_Copyright.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/002_Copyright.md
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-# • **Fourier and Laplace Transforms Coverage**
-
- To ease the transition between the circuit course and signals and systems courses, Fourier and Laplace transforms are covered lu cidly and thoroughly. The chapters are developed in a manner that the interested instructor can go from solutions of first-order circuits to Chapter 15. This then allows a very natural progression from Laplace to Fourier to AC.
-
-# • **Four-Color Art Program**
-
- An interior design and four-color art program bring circuit drawings to life and enhance key pedagogical elements throughout the text.
-
-• **Extended Examples**
-
- Examples worked in detail according to the six-step problem solv ing method provide a road map for students to solve problems in a consistent fashion. At least one example in each chapter is devel oped in this manner.
-
-• **EC 2000 Chapter Openers**
-
- Based on ABET's skill-based CRITERION 3, these chapter openers are devoted to discussions as to how students can acquire the skills that will lead to a significantly enhanced career as an engineer. Be cause these skills are so very important to the student while still in college as well after graduation, we use the heading, *"Enhancing your Skills and your Career.* "
-
-• **Homework Problems**
-
- There are 580 new or revised end-of-chapter problems and changed practice problems which will provide students with plenty of practice as well as reinforce key concepts.
-
-• **Homework Problem Icons**
-
- Icons are used to highlight problems that relate to engineering de sign as well as problems that can be solved using *PSpice, Multisim, KCIDE,* or *MATLAB* .
-
-# Organization
-
-This book was written for a two-semester or three-quarter course in linear circuit analysis. The book may also be used for a one-semester course by a proper selection of chapters and sections by the instructor. It is broadly divided into three parts.
-
-- Part 1, consisting of Chapters 1 to 8, is devoted to dc circuits. It covers the fundamental laws and theorems, circuits techniques, and passive and active elements.
-- Part 2, which contains Chapter 9 to 14, deals with ac circuits. It introduces phasors, sinusoidal steady-state analysis, ac power, rms values, three-phase systems, and frequency response.
-- Part 3, consisting of Chapters 15 to 19, are devoted to advanced techniques for network analysis. It provides students with a solid introduction to the Laplace transform, Fourier series, Fourier trans form, and two-port network analysis.
-
-The material in the three parts is more than sufficient for a two-semester course, so the instructor must select which chapters or sections to cover. Sections marked with the dagger sign (†) may be skipped, explained briefly, or assigned as homework. They can be omitted without loss of continuity. Each chapter has plenty of problems grouped according to the sections of the related material and diverse enough that the instructor can choose some as examples and assign some as homework. As stated ear lier, we are using three icons with this edition. We are using to de note problems that either require *PSpice* in the solution process, where the circuit complexity is such that *PSpice* or *Multisim* would make the solution process easier, and where *PSpice* or *Multisim* makes a good check to see if the problem has been solved correctly. We are using to denote problems where *MATLAB* is required in the solution process, where *MATLAB* makes sense because of the problem makeup and its complexity, and where *MATLAB* makes a good check to see if the problem has been solved correctly. Finally, we use to identify problems that help the student develop skills that are needed for engineering design. More difficult problems are marked with an asterisk (\*).
-
-Comprehensive problems follow the end-of-chapter problems. They are mostly applications problems that require skills learned from that particular chapter.
-
-# Prerequisites
-
-As with most introductory circuit courses, the main prerequisites, for a course using this textbook, are physics and calculus. Although familiar ity with complex numbers is helpful in the later part of the book, it is not required. A very important asset of this text is that ALL the mathemati cal equations and fundamentals of physics needed by the student, are included in the text.
-
-# Acknowledgments
-
-We would like to express our appreciation for the loving support we have received from our wives (Hannah and Kikelomo), daughters (Christina, Tamara, Jennifer, Motunrayo, Ann, and Joyce), son (Baixi), and our extended family members. We sincerely appreciate the invaluable help given us by Richard Rarick in helping us make the sixth edition a significantly more relevant book. He has checked all the new and revised problems and offered advice on making them more accurate and clear.
-
-At McGraw-Hill, we would like to thank the following editorial and production staff: Raghu Srinivasan, global brand manager; Vincent Bradshaw, product developer; Nick McFadden, marketing manager; and Melissa Leick, content project manager.
-
-The sixth edition has benefited greatly from the many outstanding individuals who have offered suggestions for improvements in both the text as well as the various problems. In particular, we thank Nicholas Reeder, Professor of Electronics Engineering Technology, Sinclair Community College, Dayton, Ohio, and Douglas De Boer, Professor of Engineering, Dordt College, Sioux Center, Iowa, for their detailed and careful corrections and suggestions for clarification which have
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/003_Preface.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/003_Preface.md
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-## **xvi** Preface
-
-contributed to making this a better edition. In addition, the follow ing have made important contributions to this edition (in alphabetical order):
-
-# Zekeriya Aliyazicioglu, *California State Polytechnic University— Pomona*
-
-Rajan Chandra, *California State Polytechnic University—Pomona* Mohammad Haider, *University of Alabama—Birmingham* John Heathcote, *Reedley College* Peter LoPresti, *University of Tulsa* Robert Norwood, *John Brown University* Aaron Ohta, *University of Hawaii—Manoa* Salomon Oldak, *California State Polytechnic University—Pomona* Hesham Shaalan, *U.S. Merchant Marine Academy* Surendra Singh, *University of Tulsa*
-
-Finally, we sincerely appreciate the feedback received from instructors and students who used the previous editions. We want this to continue, so please keep sending us e-mails or direct them to the publisher. We can be reached at c.alexander@ieee.org for Charles Alexander and sadiku@ieee .org for Matthew Sadiku.
-
-C. K. Alexander and M. N. O. Sadiku
-
-# Supplements
-
-# Instructor and Student Resources
-
-Available on Connect are a number of additional instructor and student resources to accompany the text. These include complete solutions for all practice and end-of-chapter problems, solutions in *PSpice* and *Multisim* problems, lecture PowerPoints®, and text image files. In addition, instructors can use COSMOS, a complete online solutions manual organization system to create custom homework, quizzes, and tests using end-of-chapter problems from the text.
-
-# Knowledge Capturing Integrated Design Environment for Circuits (KCIDE for Circuits)
-
-This software, developed at Cleveland State University and funded by NASA, is designed to help the student work through a circuits problem in an organized manner using the six-step problem-solving methodology in the text. *KCIDE for Circuits* allows students to work a circuit problem in *PSpice* and *MATLAB*, track the evolution of their solution, and save a record of their process for future reference. In addition, the software automatically generates a Word document and/or a PowerPoint presentation. The software package can be downloaded for free.
-
-It is hoped that the book and supplemental materials supply the in structor with all the pedagogical tools necessary to effectively present the material.
-
-# McGraw-Hill Create®
-
-Craft your teaching resources to match the way you teach! With McGraw-Hill Create, http://create.mheducation.com, you can easily rearrange chapters, combine material from other content sources, and quickly upload content you have written like your course syllabus or teaching notes. Find the content you need in Create by searching through thousands of lead ing McGraw-Hill textbooks. Arrange your book to fit your teaching style. Create even allows you to personalize your book's appearance by select ing the cover and adding your name, school, and course information. Or der a Create book and you'll receive a complimentary print review copy in three to five business days or a complimentary electronic review copy (eComp) via e-mail in minutes. Go to http://create.mheducation.com to day and register to experience how McGraw-Hill Create empowers you to teach *your* students *your* way.
-
-**Required=Results**
-
-# **McGraw-Hill Connect® Learn Without Limits**
-
-Connect is a teaching and learning platform that is proven to deliver better results for students and instructors.
-
-Connect empowers students by continually adapting to deliver precisely what they need, when they need it and how they need it, so your class time is more engaging and effective.
-
-88% of instructors who use **Connect** require it; instructor satisfaction **increases** by 38% when **Connect** is required.
-
-# Analytics
-
-# **Connect Insight®**
-
-Connect Insight is Connect's new one-of-a-kind visual analytics dashboard—now available for both instructors and students—that provides at-a-glance information regarding student
-
-performance, which is immediately actionable. By presenting assignment, assessment, and topical performance results together with a time metric that is easily visible for aggregate or individual results, Connect Insight gives the user the ability to take a just-intime approach to teaching and learning, which was never before available. Connect Insight presents data that empowers students and helps instructors improve class performance in a way that is efficient and effective.
-
-Mobile
-
-Connect's new, intuitive mobile interface gives students and instructors flexible and convenient, anytime–anywhere access to all components of the Connect platform.
-
-Students can view their results for any **Connect** course.
-
-| | 10.00 | |
-|--------------------------------------------------------------------------------------------------------------|------------------------------------------------------------------|----|
-| LATE Acounting week 1 quiz | PRACTICE | |
-| START EER - dieb, tird - all Countries excellents | | |
-| LETT CHEZ-Qui Internations
ESATI SEC - SHE SHE FUNTOJ SPANIN AV 1411 (EN 44) | SAIT | |
-| TIT LAFE Chapter & | ISLANDATION | |
-| | | |
-| INNE YA'SA' - PONTES attentiar, rely cass increases | | |
-| CH 05 States of Consciousness
atade to us + box, taux - everywhere not see tow to | | |
-| Guid - Extra Credit
ESAPTI SERA », publicada (" Pediministrato", excitore ta | GUIZ | |
-| company. CA-52, Sn la aniversidad: Vecebularia
high- aged : - Principal Mellerson 400 - Edit Pales, Add | 1,59 | |
-| | START ALL A BAR GUIS ALCOHOL: 201
Ch Oh. En casa Variabalaria | 18 |
-
-# Adaptive
-
-More students earn **A's** and **B's** when they use McGraw-Hill Education **Adaptive** products.
-
-# **SmartBook®**
-
-Proven to help students improve grades and study more efficiently, SmartBook contains the same content within the print book, but actively tailors that content to the needs of the individual. SmartBook's adaptive technology provides precise, personalized instruction on what the student should do next, guiding the student to master and remember key concepts, targeting gaps in knowledge and offering customized feedback, driving the student toward comprehension and retention of the subject matter. Available on smartphones and tablets, SmartBook puts learning at the student's fingertips—anywhere, anytime.
-
-Over **4 billion questions** have been answered making McGraw-Hill Education products more intelligent, reliable & precise.
-
-THE FIRST AND ONLY **ADAPTIVE READING EXPERIENCE** DESIGNED TO TRANSFORM THE WAY STUDENTS READ
-
-*This page intentionally left blank*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/004_About the Authors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/004_About the Authors.md
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-# About the Authors
-
-**Charles K. Alexander** is professor of electrical and computer engineering in the Washkewicz College of Engineering at Cleveland State University, Cleveland, Ohio. He is also the director of the Center for Research in Electronics and Aerospace Technology (CREATE). From 2002 until 2006 he was dean of the Fenn College of Engineering. He has held the position of dean of engineering at Cleveland State University, California State University, Northridge, and Temple University (acting dean for six years). He has held the position of department chair at Temple University and Tennessee Technological University as well as the position of Stocker Visiting Professor (an endowed chair) at Ohio University. He has held faculty status at all of the aforementioned universities.
-
-Dr. Alexander has secured funding for two centers of research at Ohio University and Cleveland State University. He has been the director of three additional research centers at Temple and Tennessee Tech and has obtained research funding of approximately \$100 million (in today's dollars). He has served as a consultant to 23 private and governmental organizations including the Air Force and the Navy.
-
-He received the honorary Dr. Eng. from Ohio Northern University (2009), his PhD (1971) and M.S.E.E. (1967) from Ohio University, and the B.S.E.E. (1965) from Ohio Northern University.
-
-Dr. Alexander has authored many publications, including a work book and a videotape lecture series, and is coauthor of *Fundamentals of Electric Circuits*, currently in its fifth edition, *Engineering Skills for Career Success, Problem Solving Made* ALMOST *Easy*, the fifth edition of the *Standard Handbook of Electronic Engineering,* and *Applied Circuit Analysis,* all with McGraw-Hill. He has delivered more than 500 paper, professional, and technical presentations.
-
-Dr. Alexander is a Life Fellow of the IEEE and served as its president and CEO in 1997. In addition he has held several volunteer posi tions within the IEEE during his more than 45 years of service. This includes serving from 1991 to 1999 on the IEEE board of directors.
-
-He has received several local, regional, national, and international awards for teaching and research, including an honorary Doctor of Engineering degree, Fellow of the IEEE, the IEEE-USA Jim Watson Student Professional Awareness Achievement Award, the IEEE Undergraduate Teaching Award, the Distinguished Professor Award, the Distinguished Engineering Education Achievement Award, the Distinguished Engi neering Education Leadership Award, the IEEE Centennial Medal, and the IEEE/RAB Innovation Award.
-
-Charles K. Alexander
-
-Matthew N. O. Sadiku
-
-**Matthew N. O. Sadiku** received his PhD from Tennessee Technological University, Cookeville. From 1984 to 1988, he was an assistant professor at Florida Atlantic University, where he did graduate work in computer science. From 1988 to 2000, he was at Temple University, Philadelphia, Pennsylvania, where he became a full professor. From 2000 to 2002, he was with Lucent/Avaya, Holmdel, New Jersey, as a system engineer and with Boeing Satellite Systems as a senior scientist. He is currently a professor at Prairie View A&M University.
-
-Dr. Sadiku is the author of more than 240 professional papers and over 60 books, including *Elements of Electromagnetics* (Oxford Uni versity Press, 6th ed., 2015), *Numerical Techniques in Electromagnetics with MATLAB* (CRC, 3rd ed., 2009), and *Metropolitan Area Net works* (CRC Press, 1995). Some of his books have been translated into French, Korean, Chinese (and Chinese Long Form in Taiwan), Italian, Portuguese, and Spanish. He was the recipient of the 2000 McGraw-Hill/ Jacob Millman Award for outstanding contributions in the field of electrical engineering. He was also the recipient of Regents Professor award for 2012 to 2013 by the Texas A&M University System.
-
-His current research interests are in the areas of numerical modeling of electromagnetic systems and computer communication networks. He is a registered professional engineer and a fellow of the Institute of Electrical and Electronics Engineers (IEEE) "for contributions to computa tional electromagnetics and engineering education." He was the IEEE Region 2 Student Activities Committee Chairman. He was an associ ate editor for *IEEE Transactions on Education* and is a member of the Association for Computing Machinery (ACM).
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/005_PART 1 - DC Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/005_PART 1 - DC Circuits.md
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-# Fundamentals of Electric Circuits
-
-# **PART ONE**
-
-# DC Ci rcuits
-
-# OUTLINE
-
-- 1 Basic Concepts
-- 2 Basic Laws
-- 3 Methods of Analysis
-- 4 Circuit Theorems
-- 5 Operational Amplifiers
-- 6 Capacitors and Inductors
-- 7 First-Order Circuits
-- 8 Second-Order Circuits
-
-# **chapter**
-
-1
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/006_Chapter 1 - Basic Concepts.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/006_Chapter 1 - Basic Concepts.md
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-# Basic Concepts
-
-*Some books are to be tasted, others to be swallowed, and some few to be chewed and digested.*
-
-—Francis Bacon
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.a), "an ability to apply knowledge of mathematics, science, and engineering."**
-
-As students, you are required to study mathematics, science, and engineering with the purpose of being able to apply that kno wledge to the solution of engineering problems. The skill here is the ability to apply the fundamentals of these areas in the solution of a problem. So how do you develop and enhance this skill?
-
-The best approach is to w ork as man y problems as possible in all of your courses. However, if you are really going to be successful with this, you must spend time analyzing where and when and why you have difficulty in easily arriving at successful solutions. You may be surprised to learn that most of your problem-solving problems are with mathematics rather than your understanding of theory . You may also learn that you start working the problem too soon. Taking time to think about the problem and ho w you should solv e it will al ways save you time and frustration in the end.
-
-What I have found that works best for me is to apply our sixstep problem-solving technique. Then I carefully identify the areas where I ha ve dif ficulty solving the problem. Many times, my actual deficiencies are in my understanding and ability to use correctly certain mathematical principles. I then return to my fundamental math texts and carefully review the appropriate sections, and in some cases, work some example problems in that text. This brings me to another important thing you should always do: Keep nearby all your basic mathematics, science, and engineering textbooks.
-
-This process of continually looking up material you thought you had acquired in earlier courses may seem v ery tedious at first; however, as your skills de velop and your kno wledge increases, this process will become easier and easier. On a personal note, it is this very process that led me from being a much less than a verage student to someone who could earn a Ph.D. and become a successful researcher.
-
-Photo by Charles Alexander
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Understand the different units with which engineers work.
-- 2. Understand the relationship between charge and current and how to use both in a variety of applications.
-- 3. Understand voltage and how it can be used in a variety of applications.
-- 4. Develop an understanding of power and energy and their relationship with current and voltage.
-- 5. Begin to understand the volt-amp characteristics of a variety of circuit elements.
-- 6. Begin to understand an organized approach to problem solving and how it can be used to assist in your efforts to solve circuit problems.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/007_1.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/007_1.1 Introduction.md
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-# **1.1** Introduction
-
-Electric circuit theory and electromagnetic theory are the tw o funda mental theories upon which all branches of electrical engineering are built. Many branches of electrical engineering, such as po wer, electric machines, control, electronics, communications, and instrumentation, are based on electric circuit theory . Therefore, the basic electric circuit theory course is the most important course for an electrical engineering student, and al ways an e xcellent starting point for a be ginning student in electrical engineering education. Circuit theory is also v aluable to students specializing in other branches of the ph ysical sciences because circuits are a good model for the study of energy systems in general, and because of the applied mathematics, physics, and topology involved.
-
-In electrical engineering, we are often interested in communicating or transferring energy from one point to another . To do this requires an interconnection of electrical devices. Such interconnection is referred to as an *electric circuit*, and each component of the circuit is kno wn as an *element.*
-
-An electric circuit is an interconnection of electrical elements.
-
-A simple electric circuit is sho wn in Fig. 1.1. It consists of three basic elements: a battery, a lamp, and connecting wires. Such a simple circuit can e xist by itself; it has se veral applications, such as a flashlight, a search light, and so forth.
-
-A complicated real circuit is displayed in Fig. 1.2, representing the schematic diagram for a radio receiver. Although it seems complicated, this circuit can be analyzed using the techniques we co ver in this book. Our goal in this text is to learn various analytical techniques and computer software applications for describing the behavior of a circuit like this.
-
-Electric circuits are used in numerous electrical systems to accomplish different tasks. Our objecti ve in this book is not the study of various uses and applications of circuits. Rather, our major concern is the analysis of the circuits. By the analysis of a circuit, we mean a study of the behavior of the
-
-**Figure 1.1** A simple electric circuit.
-
-
-
-# **Figure 1.2**
-
-Electric circuit of a radio transmitter.
-
-circuit: How does it respond to a gi ven input? How do the interconnected elements and devices in the circuit interact?
-
-We commence our study by defining some basic concepts. These concepts include char ge, current, v oltage, circuit elements, po wer, and energy. Before defining these concepts, we must first establish a system of units that we will use throughout the text.
-
-# **1.2** Systems of Units
-
-As electrical engineers, we must deal with measurable quantities. Our measurements, ho wever, must be communicated in a standard language that virtually all professionals can understand, irrespecti ve of the country in which the measurement is conducted. Such an international measurement language is the International System of Units (SI), adopted by the General Conference on Weights and Measures in 1960. In this system, there are seven base units from which the units of all other ph ysical quantities can be de rived. Table 1.1 shows six base units and one derived unit (the coulomb) that are related to this text. SI units are commonly used in electrical engineering.
-
-One great advantage of the SI unit is that it uses prefixes based on the power of 10 to relate larger and smaller units to the basic unit. Table 1.2 shows the SI prefixes and their symbols. For example, the following are expressions of the same distance in meters (m):
-
-| 600,000,000 mm | 600,000 m | 600 km |
-|----------------|-----------|--------|
-| | | |
-
-## **TABLE 1.1**
-
-Six basic SI units and one derived unit relevant to this text.
-
-| Quantity | Basic unit | Symbol |
-|---------------------------|------------|--------|
-| Length | meter | m |
-| Mass | kilogram | kg |
-| Time | second | s |
-| Electric current | ampere | A |
-| Thermodynamic temperature | kelvin | K |
-| Luminous intensity | candela | cd |
-| Charge | coulomb | C |
-
-## **TABLE 1.2**
-
-# The SI prefixes.
-
-| Multiplier | Prefix | Symbol |
-|------------|--------|--------|
-| 1018 | exa | E |
-| 1015 | peta | P |
-| 1012 | tera | T |
-| 109 | giga | G |
-| 106 | mega | M |
-| 103 | kilo | k |
-| 102 | hecto | h |
-| 10 | deka | da |
-| 10−1 | deci | d |
-| 10−2 | centi | c |
-| 10−3 | milli | m |
-| 10−6 | micro | μ |
-| 10−9 | nano | n |
-| 10−12 | pico | p |
-| 10−15 | femto | f |
-| 10−18 | atto | a |
-| | | |
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/008_1.2 Systems of Units.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/008_1.2 Systems of Units.md
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-# **1.3** Charge and Current
-
-The concept of electric charge is the underlying principle for explaining all electrical phenomena. Also, the most basic quantity in an electric circuit is the *electric charge.* We all experience the effect of electric charge when we try to remove our wool sweater and have it stick to our body or walk across a carpet and receive a shock.
-
-Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C).
-
-We know from elementary physics that all matter is made of fundamental building blocks known as atoms and that each atom consists of electrons, protons, and neutrons. We also know that the char ge *e* on an electron is negative and equal in magnitude to 1.602 × 10−19 C, while a proton carries a positi ve charge of the same magnitude as the electron. The presence of equal numbers of protons and electrons leaves an atom neutrally charged.
-
-The following points should be noted about electric charge:
-
-- 1. The coulomb is a large unit for charges. In 1 C of charge, there are 1∕(1.602 × 10−19) = 6.24 × 1018 electrons. Thus realistic or laboratory values of charges are on the order of pC, nC, or *μ*C.1
-- 2. According to e xperimental observ ations, the only char ges that occur in nature are inte gral multiples of the electronic char ge *e* = −1.602 × 10−19 C.
-- 3. The *law of conservation of charge* states that charge can neither be created nor destroyed, only transferred. Thus, the algebraic sum of the electric charges in a system does not change.
-
-We now consider the flow of electric char ges. A unique feature of electric charge or electricity is the f act that it is mobile; that is, it can be transferred from one place to another , where it can be con verted to another form of energy.
-
-When a conducting wire (consisting of se veral atoms) is connected to a battery (a source of electromotive force), the charges are compelled to move; positive charges move in one direction while ne gative charges move in the opposite direction. This motion of char ges creates elec tric current. It is conventional to take the current flow as the movement of positive charges. That is, opposite to the flow of ne gative charges, as Fig. 1.3 illustrates. This con vention w as introduced by Benjamin Franklin (1706–1790), the American scientist and in ventor. Although we now know that current in metallic conductors is due to ne gatively charged electrons, we will follo w the uni versally accepted con vention that current is the net flow of positive charges. Thus,
-
-Electric current is the time rate of change of charge, measured in amperes (A).
-
-Mathematically, the relationship between current *i*, charge *q*, and time *t* is
-
-$$
-i \triangleq \frac{dq}{dt} \tag{1.1}
-$$
-
-I – + – –– –
-
-# **Figure 1.3**
-
-Electric current due to flow of electronic charge in a conductor.
-
-A convention is a standard way of describing something so that others in the profession can understand what we mean. We will be using IEEE conventions throughout this book.
-
-1 However, a large power supply capacitor can store up to 0.5 C of charge.
-
-# Historical
-
-**Andre-Marie Ampere** (1775–1836), a French mathematician and physicist, laid the foundation of electrodynamics. He defined the electric current and developed a way to measure it in the 1820s.
-
-Born in Lyons, France, Ampere at age 12 mastered Latin in a few weeks, as he was intensely interested in mathematics and many of the best mathematical works were in Latin. He was a brilliant scientist and a prolific writer. He formulated the laws of electromagnetics. He in vented the electromagnet and the ammeter. The unit of electric current, the ampere, was named after him.
-
-© Apic/Getty Images
-
-where current is measured in amperes (A), and
-
-# 1 ampere = 1 coulomb/second
-
-The charge transferred between time *t*0 and *t* is obtained by inte grating both sides of Eq. (1.1). We obtain
-
-$$
-Q \triangleq \int_{t_0}^t i \, dt \tag{1.2}
-$$
-
-The way we define current as *i* in Eq. (1.1) suggests that current need not be a constant-valued function. As many of the examples and problems in this chapter and subsequent chapters suggest, there can be se veral types of current; that is, charge can vary with time in several ways.
-
-There are different ways of looking at direct current and alternating current. The best definition is that there are two ways that current can flow: It can always flow in the same direction, where it does not reverse direction, in which case we have *direct current* (dc). These currents can be constant or time varying. If the current flows in both directions, then we have *alternating current* (ac).
-
-A direct current (dc) flows only in one direction and can be constant or time varying.
-
-By convention, we will use the symbol *I* to represent a constant current. If the current v aries with respect to time (either dc or ac) we will use the symbol *i*. A common use of this w ould be the output of a rectifier (dc) such as *i*(*t*) = ∣5 sin(377t)∣ amps or a sinusoidal current (ac) such as *i*(*t*) = 160 sin(377t) amps.
-
-An alternating current (ac) is a current that changes direction with respect to time.
-
-An example of alternating current (ac) is the current you use in your house to run the air conditioner , refrigerator , w ashing machine, and other electric appliances. Figure 1.4 depicts tw o common e xamples of
-
-# **Figure 1.4**
-
-Two common types of current: (a) direct current (dc), (b) alternating current (ac).
-
-**Figure 1.5** Conventional current flow: (a) positive current flow, (b) negative current flow.
-
-dc (coming from a battery) and ac (coming from your home outlets). We will consider other types later in the book.
-
-Once we define current as the movement of charge, we expect current to have an associated direction of flow. As mentioned earlier, the direction of current flow is conventionally taken as the direction of positive charge movement. Based on this convention, a current of 5 A may be represented positively or negatively as shown in Fig. 1.5. In other w ords, a negative current of −5 A flowing in one direction as shown in Fig. 1.5(b) is the same as a current of +5 A flowing in the opposite direction.
-
-| Example 1.1 | How much charge is represented by 4,600 electrons? |
-|----------------------|--------------------------------------------------------------------------------------------------------------------------------------------------------------|
-| | Solution:
−19 C. Hence 4,600 electrons will
Each electron has
−1.602 × 10
have −1.602 × 10−19 C/electron × 4,600 electrons =
−7.369 × 10−16 C |
-| Practice Problem 1.1 | Calculate the amount of charge represented by 6.667 billion protons. |
-| | Answer: 1.0681 × 10−9
C. |
-| | |
-| Example 1.2 | The total char ge entering a terminal is gi
ven by q
= 5
t sin 4
πt mC.
Calculate the current at t
= 0.5 s. |
-| | Solution: |
-| | dq___
__d
i
=
dt =
dt (5t sin 4πt) mC/s = (5 sin 4πt
+ 20πt cos 4πt) mA |
-| | At t
= 0.5, |
-| | i
= 5 sin 2π
+ 10π cos 2π
= 0 + 10π
= 31.42 mA |
-| Practice Problem 1.2 | = (10 − 10e−2t
If in Example 1.2, q
) mC, find the current at t
= 1.0 s. |
-| | Answer: 2.707 mA. |
-| | |
-| Example 1.3 | Determine the total charge entering a terminal between t = 1 s and
2 −
t = 2 s if the current passing the terminal is i = (3t
t) A. |
-
-# **Solution:**
-
-$$
-Q = \int_{t=1}^{2} i \, dt = \int_{1}^{2} (3t^2 - t) \, dt
-$$
-$$
-= \left(t^3 - \frac{t^2}{2}\right)\Big|_{1}^{2} = (8 - 2) - \left(1 - \frac{1}{2}\right) = 5.5 \text{ C}
-$$
-
-ough an element is
-
-\n
-$$
-i = \begin{cases} 4 \text{ A}, & 0 < t < 1 \\ 4t^2 \text{ A}, & t > 1 \end{cases}
-$$
-
-Calculate the charge entering the element from *t* = 0 to *t* = 2 s.
-
-**Answer:** 13.333 C.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/009_1.4 Voltage.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/009_1.4 Voltage.md
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-# **1.4** Voltage
-
-As explained briefly in the previous section, to mo ve the electron in a conductor in a particular direction requires some w ork or energy transfer. This work is performed by an e xternal electromotive force (emf), typically represented by the battery in Fig. 1.3. This emf is also kno wn as *voltage* or *potential difference*. The voltage *vab* between two points *a* and *b* in an electric circuit is the energy (or work) needed to move a unit charge from *b* to *a*; mathematically,
-
-$$
-v_{ab} \triangleq \frac{dw}{dq} \tag{1.3}
-$$
-
-where *w* is energy in joules (J) and *q* is charge in coulombs (C). The voltage *vab* or simply *v* is measured in volts (V), named in honor of the Italian physicist Alessandro Antonio Volta (1745–1827), who invented the first voltaic battery. From Eq. (1.3), it is evident that
-
-1 volt = 1 joule/coulomb = 1 newton-meter/coulomb
-
-Thus,
-
-Voltage (or potential difference) is the energy required to move a unit charge from a reference point (−) to another point (+), measured in volts (V).
-
-Figure 1.6 sho ws the v oltage across an element (represented by a rectangular block) connected to points *a* and *b*. The plus ( +) and minus (−) signs are used to define reference direction or voltage polarity. The *vab* can be interpreted in two ways: (1) Point *a* is at a potential of *vab* volts higher than point *b*, or (2) the potential at point *a* with respect to point *b* is *vab*. It follows logically that in general
-
-$$
-v_{ab} = -v_{ba} \tag{1.4}
-$$
-
-For example, in Fig. 1.7, we ha ve two representations of the same v oltage. In Fig. 1.7(a), point *a* is +9 V above point *b*; in Fig. 1.7(b), point *b* is −9 V above point *a*. We may say that in Fig. 1.7(a), there is a 9-V *voltage drop* from *a* to *b* or equivalently a 9-V *voltage rise* from *b* to *a*. In other words, a voltage drop from *a* to *b* is equivalent to a voltage rise from *b* to *a*.
-
-Current and voltage are the tw o basic variables in electric circuits. The common term *signal* is used for an electric quantity such as a current or a voltage (or even electromagnetic wave) when it is used for conveying
-
-Polarity of voltage *vab*.
-
-Two equivalent representations of the same voltage *vab*: (a) Point *a* is 9 V above point *b*; (b) point *b* is −9 V above point *a*.
-
-# Practice Problem 1.3
-
-# Historical
-
-© UniversalImagesGroup/ Getty Images
-
-**Alessandro Antonio Volta** (1745–1827), an Italian physicist, in vented the electric battery—which provided the first continuous flow of electricity—and the capacitor.
-
-Born into a noble family in Como, Italy, Volta was performing electrical experiments at age 18. His invention of the battery in 1796 revolutionized the use of electricity. The publication of his work in 1800 marked the beginning of electric circuit theory. Volta received many honors during his lifetime. The unit of voltage or potential difference, the volt, was named in his honor.
-
-Keep in mind that electric current is always through an element and that electric voltage is always across the element or between two points.
-
-information. Engineers prefer to call such v ariables signals rather than mathematical functions of time because of their importance in commu nications and other disciplines. Lik e electric current, a constant v oltage is called a *dc voltage* and is represented by *V*, whereas a sinusoidally time-varying voltage is called an *ac voltage* and is represented by *v*. A dc voltage is commonly produced by a battery; ac v oltage is produced by an electric generator.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/010_1.5 Power and Energy.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/010_1.5 Power and Energy.md
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-# **1.5** Power and Energy
-
-Although current and v oltage are the tw o basic v ariables in an electric circuit, the y are not suf ficient by themselves. F or practical purposes, we need to kno w how much *power* an electric de vice can handle. We all know from e xperience that a 100-w att bulb gives more light than a 60-watt bulb. We also know that when we pay our bills to the electric utility companies, we are paying for the electric *energy* consumed over a certain period of time. Thus, power and energy calculations are important in circuit analysis.
-
-To relate power and ener gy to v oltage and current, we recall from physics that:
-
-Power is the time rate of expending or absorbing energy, measured in watts (W).
-
-We write this relationship as
-
-$$
-p \triangleq \frac{dw}{dt} \tag{1.5}
-$$
-
-where *p* is power in watts (W), *w* is energy in joules (J), and *t* is time in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that
-
-$$
-p = \frac{dw}{dt} = \frac{dw}{dq} \cdot \frac{dq}{dt} = vi \tag{1.6}
-$$
-
-or
-
-$$
-p = vi \tag{1.7}
-$$
-
-The power *p* in Eq. (1.7) is a time-varying quantity and is called the *instantaneous power*. Thus, the power absorbed or supplied by an element is the product of the voltage across the element and the current through it. If the power has a + sign, power is being delivered to or absorbed by the element. If, on the other hand, the power has a − sign, power is being supplied by the element. But how do we know when the power has a negative or a positive sign?
-
-Current direction and voltage polarity play a major role in determining the sign of po wer. It is therefore important that we pay attention to the relationship between current *i* and voltage *v* in Fig. 1.8(a). The voltage polarity and current direction must conform with those sho wn in Fig. 1.8(a) in order for the power to have a positive sign. This is known as the *passive sign convention.* By the passive sign convention, current enters through the positive polarity of the voltage. In this case, *p* = +*vi* or *vi* > 0 implies that the element is absorbing po wer. However, if *p* = −*vi* or *vi* < 0, as in Fig. 1.8(b), the element is releasing or supplying power.
-
-Passive sign convention is satisfied when the current enters through the positive terminal of an element and p = +vi. If the current enters through the negative terminal, p = −vi.
-
-Unless otherwise stated, we will follow the passive sign convention throughout this text. For example, the element in both circuits of Fig. 1.9 has an absorbing power of +12 W because a positive current enters the positive terminal in both cases. In Fig. 1.10, ho wever, the element is supplying power of +12 W because a positive current enters the negative terminal. Of course, an absorbing power of −12 W is equivalent to a supplying power of +12 W. In general,
-
-# +Power absorbed = −Power supplied
-
-In fact, the *law of conservation of energy* must be obeyed in any electric circuit. For this reason, the algebraic sum of po wer in a circuit, at any instant of time, must be zero:
-
-$$
-\sum p = 0 \tag{1.8}
-$$
-
-This again confirms the fact that the total po wer supplied to the circuit must balance the total power absorbed.
-
-From Eq. (1.6), the energy absorbed or supplied by an element from time *t*0 to time *t* is
-
-$$
-w = \int_{t_0}^{t} p \, dt = \int_{t_0}^{t} v i \, dt \tag{1.9}
-$$
-
-# **Figure 1.8**
-
-Reference polarities for power using the passive sign convention: (a) absorbing power, (b) supplying power.
-
-When the voltage and current directions conform to Fig. 1.8(b), we have the active sign convention and p = +vi.
-
-# **Figure 1.9**
-
-Two cases of an element with an absorbing power of 12 W: (a) *p* = 4 × 3 = 12 W, (b) *p* = 4 × 3 = 12 W.
-
-# **Figure 1.10** Two cases of an element with a supplying power of 12 W: (a) *p* = −4 × 3 = −12 W, (b) *p* = −4 × 3 = −12 W.
-
-Energy is the capacity to do work, measured in joules (J).
-
-The electric po wer utility companies measure ener gy in w att-hours (Wh), where
-
-$$
-1 \text{ Wh} = 3,600 \text{ J}
-$$
-
-Example 1.4 An energy source forces a constant current of 2 A for 10 s to flow through a light bulb. If 2.3 kJ is gi ven off in the form of light and heat ener gy, calculate the voltage drop across the bulb.
-
-# **Solution:**
-
-The total charge is
-
-$$
-\Delta q = i \Delta t = 2 \times 10 = 20 \text{ C}
-$$
-
-The voltage drop is
-
-$$
-v = \frac{\Delta w}{\Delta q} = \frac{2.3 \times 10^3}{20} = 115 \text{ V}
-$$
-
-To move charge *q* from point *b* to point *a* requires 25 J. Find the volt age drop *vab* (the voltage at *a* positive with respect to *b*) if: (a) *q* = 5 C, (b) *q* = −10 C. Practice Problem 1.4
-
-**Answer:** (a) 5 V, (b) −2.5 V.
-
-Example 1.5 Find the power delivered to an element at *t* = 3 ms if the current entering its positive terminal is
-
-*i* = 5 cos 60*π t* A
-
-and the voltage is: (a) *v* = 3*i*, (b) *v* = 3 *di*∕*dt*.
-
-# **Solution:**
-
-(a) The voltage is *v* = 3*i* = 15 cos 60*π t*; hence, the power is
-
-$$
-p = vi = 75 \cos^2 60 \pi t \,\mathrm{W}
-$$
-
-$$
-At t = 3 ms,
-$$
-
-*p* = 75 cos2 (60*π* × 3 × 10−3 ) = 75 cos2 0.18*π* = 53.48 W
-
-(b) We find the voltage and the power as
-
-$$
-v = 3\frac{di}{dt} = 3(-60\pi)5 \sin 60\pi t = -900\pi \sin 60\pi t \text{ V}
-$$
-
-$$
-p = vi = -4500\pi \sin 60\pi t \cos 60\pi t \text{ W}
-$$
-
-At *t* = 3 ms,
-
-*p* = −4500*π* sin 0.18*π* cos 0.18*π* W
-
-= −14137.167 sin 32.4° cos 32.4° = −6.396 kW
-
-# Historical
-
-**1884 Exhibition** In the United States, nothing promoted the future of electricity like the 1884 International Electrical Exhibition. Just imagine a world without electricity, a world illuminated by candles and gaslights, a world where the most common transportation was by walking and riding on horseback or by horse-drawn carriage. Into this world an exhibi tion was created that highlighted Thomas Edison and reflected his highly developed ability to promote his inventions and products. His exhibit featured spectacular lighting displays powered by an impressive 100-kW "Jumbo" generator.
-
-Edward Weston's dynamos and lamps were featured in the United States Electric Lighting Company's display. Weston's well known col lection of scientific instruments was also shown.
-
-Other prominent exhibitors included Frank Sprague, Elihu Thompson, and the Brush Electric Company of Cleveland. The American Institute of Electrical Engineers (AIEE) held its first technical meeting on October 7–8 at the Franklin Institute during the exhibit. AIEE merged with the Institute of Radio Engineers (IRE) in 1964 to form the Institute of Electrical and Electronics Engineers (IEEE).
-
-Practice Problem 1.5
-
-Source: IEEE History Center
-
-Find the power delivered to the element in Example 1.5 at *t* = 5 ms if the current remains the same but the voltage is: (a) *v* = 2*i* V,
-
-(b)
-$$
-v = \left(10 + 5 \int_0^t i \, dt \right) V.
-$$
-
-**Answer:** (a) 17.27 W, (b) 29.7 W.
-
-Example 1.6 How much energy does a 100-W electric bulb consume in two hours?
-
-# **Solution:**
-
-*w* = *pt* = 100 (W) × 2 (h) × 60 (min/h) × 60 (s/min) = 720,000 J = 720 kJ
-
-This is the same as
-
-$$
-w = pt = 100
-$$
- W $\times$ 2 h = 200 Wh
-
-Practice Problem 1.6
-
-A home electric heater dra ws 10 A when connected to a 115 V outlet. How much energy is consumed by the heater over a period of 6 hours?
-
-**Answer:** 6.9 k watt-hours
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/011_1.6 Circuit Elements.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/011_1.6 Circuit Elements.md
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-# **1.6** Circuit Elements
-
-As we discussed in Section 1.1, an element is the basic building block of a circuit. An electric circuit is simply an interconnection of the elements. Circuit analysis is the process of determining voltages across (or the currents through) the elements of the circuit.
-
-There are tw o types of elements found in electric circuits: *passive* elements and *active* elements. An active element is capable of generating energy while a passive element is not. Examples of passive elements are resistors, capacitors, and inductors. Typical active elements include generators, batteries, and operational amplifiers. Our aim in this section is to g ain familiarity with some important acti ve elements.
-
-The most important acti ve elements are v oltage or current sources that generally deliver power to the circuit connected to them. There are two kinds of sources: independent and dependent sources.
-
-An ideal independent source is an active element that provides a speci fied voltage or current that is completely independent of other circuit elements.
-
-**Figure 1.11**
-
-Symbols for independent voltage sources: (a) used for constant or time-varying voltage, (b) used for constant voltage (dc).
-
-In other words, an ideal independent voltage source delivers to the circuit whatever current is necessary to maintain its terminal v oltage. Physical sources such as batteries and generators may be regarded as approximations to ideal v oltage sources. Figure 1.11 sho ws the symbols for inde pendent voltage sources. Notice that both symbols in Fig. 1.11(a) and (b) can be used to represent a dc v oltage source, b ut only the symbol in Fig. 1.11(a) can be used for a time-v arying voltage source. Similarly, an ideal independent current source is an active element that provides a specified current completely independent of the voltage across the source. That is, the current source deli vers to the circuit whate ver
-
-voltage is necessary to maintain the designated current. The symbol for an independent current source is displayed in Fig. 1.12, where the arrow indicates the direction of current *i*.
-
-An ideal dependent (or controlled) source is an active element in which the source quantity is controlled by another voltage or current.
-
-Dependent sources are usually designated by diamond-shaped sym bols, as shown in Fig. 1.13. Since the control of the dependent source is achieved by a voltage or current of some other element in the circuit, and the source can be voltage or current, it follows that there are four possible types of dependent sources, namely:
-
-- 1. A voltage-controlled voltage source (VCVS).
-- 2. A current-controlled voltage source (CCVS).
-- 3. A voltage-controlled current source (VCCS).
-- 4. A current-controlled current source (CCCS).
-
-Dependent sources are useful in modeling elements such as transistors, operational amplifiers, and integrated circuits. An example of a currentcontrolled voltage source is sho wn on the right-hand side of Fig. 1.14, where the v oltage 10*i* of the v oltage source depends on the current *i* through element *C*. Students might be surprised that the value of the dependent voltage source is 10 *i* V (and not 10 *i* A) because it is a v oltage source. The key idea to keep in mind is that a voltage source comes with polarities (+ −) in its symbol, while a current source comes with an arrow, irrespective of what it depends on.
-
-It should be noted that an ideal v oltage source (dependent or in dependent) will produce any current required to ensure that the termi nal voltage is as stated, whereas an ideal current source will produce the necessary voltage to ensure the stated current flow. Thus, an ideal source could in theory supply an infinite amount of energy. It should also be noted that not only do sources supply po wer to a circuit, the y can absorb power from a circuit too. For a voltage source, we know the voltage but not the current supplied or drawn by it. By the same token, we know the current supplied by a current source b ut not the v oltage across it.
-
-i **Figure 1.12**
-
-Symbols for: (a) dependent voltage source, (b) dependent current source.
-
-**Figure 1.14** The source on the right-hand side is a current-controlled voltage source.
-
-Calculate the power supplied or absorbed by each element in Fig. 1.15. Example 1.7
-
-# **Solution:**
-
-We apply the sign convention for power shown in Figs. 1.8 and 1.9. For *p*1, the 5-A current is out of the positive terminal (or into the negative terminal); hence,
-
-*p*1 = 20(−5) = −100 W Supplied power
-
-For *p*2 and *p*3, the current flows into the positive terminal of the element in each case.
-
-> *p*2 = 12(5) = 60 W Absorbed power *p*3 = 8(6) = 48 W Absorbed power
-
-For Example 1.7.
-
-For *p*4, we should note that the voltage is 8 V (positive at the top), the same as the voltage for *p*3 since both the passive element and the dependent source are connected to the same terminals. (Remember that v oltage is always measured across an element in a circuit.) Since the current flows out of the positive terminal,
-
-$$
-p_4 = 8(-0.2I) = 8(-0.2 \times 5) = -8
-$$
- W Supplement power
-
-We should observ e that the 20-V independent v oltage source and 0.2*I* dependent current source are supplying power to the rest of the network, while the two passive elements are absorbing power. Also,
-
-*p*1 + *p*2 + *p*3 + *p*4 = −100 + 60 + 48 − 8 = 0
-
-In agreement with Eq. (1.8), the total po wer supplied equals the total power absorbed.
-
-# Practice Problem 1.7
-
-**Figure 1.16** For Practice Prob. 1.7.
-
-Compute the power absorbed or supplied by each component of the circuit in Fig. 1.16.
-
-**Answer:** *p*1 = −45 W, *p*2 = 18 W, *p*3 = 12 W, *p*4 = 15 W.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/012_1.7 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/012_1.7 Applications.md
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-# **1.7** Applications2
-
-In this section, we will consider tw o practical applications of the con cepts developed in this chapter. The first one deals with the TV picture tube and the other with how electric utilities determine your electric bill.
-
-# **1.7.1** TV Picture Tube
-
-One important application of the motion of electrons is found in both the transmission and reception of TV signals. At the transmission end, a TV camera reduces a scene from an optical image to an electrical signal. Scanning is accomplished with a thin beam of electrons in an iconoscope camera tube.
-
-At the receiving end, the image is reconstructed by using a cathoderay tube (CR T) located in the TV recei ver.3 The CRT is depicted in Fig. 1.17. Unlike the iconoscope tube, which produces an electron beam of constant intensity, the CRT beam varies in intensity according to the incoming signal. The electron gun, maintained at a high potential, fires the electron beam. The beam passes through two sets of plates for vertical and horizontal deflections so that the spot on the screen where the beam strikes can move right and left and up and down. When the electron beam strikes the fluorescent screen, it gives off light at that spot. Thus, the beam can be made to "paint" a picture on the TV screen.
-
-2 The dagger sign preceding a section heading indicates the section that may be skipped, explained briefly, or assigned as homework.
-
-3 Modern TV tubes use a different technology.
-
-Cathode-ray tube.
-
-# Historical
-
-# Karl Ferdinand Braun and Vladimir K. Zworykin
-
-**Karl Ferdinand Braun** (1850–1918), of the University of Strasbourg, invented the Braun cathode-ray tube in 1879. This then became the basis for the picture tube used for so many years for televisions. It is still the most economical device today, although the price of flat-screen systems is rapidly becoming competitive. Before the Braun tube could be used in television, it took the inventiveness of **Vladimir K. Zworykin** (1889–1982) to develop the iconoscope so that the modern television would become a reality. The iconoscope developed into the orthicon and the image orthicon, which allowed images to be captured and converted into signals that could be sent to the television receiver. Thus, the television camera was born.
-
-The electron beam in a TV picture tube carries 10 Example 1.8 15 electrons per second. As a design engineer, determine the voltage *Vo* needed to accelerate the electron beam to achieve 4 W.
-
-# **Solution:**
-
-The charge on an electron is
-
-*e* = −1.6 × 10−19 C
-
-If the number of electrons is *n*, then *q* = *ne* and
-
-$$
-i = \frac{dq}{dt} = e \frac{dn}{dt} = (-1.6 \times 10^{-19})(10^{15}) = -1.6 \times 10^{-4} \text{ A}
-$$
-
-The negative sign indicates that the current flows in a direction opposite to electron flow as shown in Fig. 1.18, which is a simplified diagram of the CRT for the case when the vertical deflection plates carry no charge. The beam power is
-
-$$
-p = V_o i
-$$
- or $V_o = \frac{p}{i} = \frac{4}{1.6 \times 10^{-4}} = 25,000 \text{ V}$
-
-Thus, the required voltage is 25 kV.
-
-# Practice Problem 1.8
-
-If an electron beam in a TV picture tube carries 10 13 electrons/second and is passing through plates maintained at a potential difference of 30 kV, calculate the power in the beam.
-
-**Answer:** 48 mW.
-
-# **1.7.2** Electricity Bills
-
-The second application deals with ho w an electric utility compan y charges their customers. The cost of electricity depends upon the amount of ener gy consumed in kilo watt-hours (kWh). (Other f actors that affect the cost include demand and po wer factors; we will ignore these for now.) However, even if a consumer uses no energy at all, there is a minimum service char ge the customer must pay because it costs money to stay connected to the po wer line. As ener gy consumption increases, the cost per kWh drops. It is interesting to note the a verage monthly consumption of household appliances for a family of five, shown in Table 1.3.
-
-## **TABLE 1.3**
-
-Typical average monthly consumption of household appliances.
-
-| Appliance | kWh consumed | Appliance | kWh consumed |
-|---------------|--------------|-------------------|--------------|
-| Water heater | 500 | Washing machine | 120 |
-| Freezer | 100 | Stove | 100 |
-| Lighting | 100 | Dryer | 80 |
-| Dishwasher | 35 | Microwave oven | 25 |
-| Electric iron | 15 | Personal computer | 12 |
-| TV | 10 | Radio | 8 |
-| Toaster | 4 | Clock | 2 |
-
-A simplified diagram of the cathode-ray tube; for Example 1.8.
-
-A homeowner consumes 700 kWh in January. Determine the electricity Example 1.9 bill for the month using the following residential rate schedule:
-
-Base monthly charge of \$12.00.
-
-First 100 kWh per month at 16 cents/kWh.
-
-Next 200 kWh per month at 10 cents/kWh.
-
-Over 300 kWh per month at 6 cents/kWh.
-
-# **Solution:**
-
-We calculate the electricity bill as follows.
-
-Base monthly charge = \$12.00 First 100 kWh @ \$0.16/k Wh = \$16.00 Next 200 kWh @ \$0.10/k Wh = \$20.00 Remaining 400 kWh @ \$0.06/k Wh = \$24.00 Total charge = \$72.00 Average cost = \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \$72 100 + 200 +400 = 10.2 cents/kWh
-
-Referring to the residential rate schedule in Example 1.9, calculate the average cost per kWh if only 350 kWh are consumed in July when the family is on vacation most of the time.
-
-Practice Problem 1.9
-
-**Answer:** 14.571 cents/kWh.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/013_1.8 Problem Solving.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/013_1.8 Problem Solving.md
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-# **1.8** Problem Solving
-
-Although the problems to be solved during one's career will vary in complexity and magnitude, the basic principles to be follo wed remain the same. The process outlined here is the one developed by the authors over many years of problem solving with students, for the solution of engi neering problems in industry, and for problem solving in research.
-
-We will list the steps simply and then elaborate on them.
-
-- 1. Carefully **define** the problem.
-- 2. **Present** everything you know about the problem.
-- 3. Establish a set of **alternative** solutions and determine the one that promises the greatest likelihood of success.
-- 4. **Attempt** a problem solution.
-- 5. **Evaluate** the solution and check for accuracy.
-- 6. Has the problem been solved **satisfactorily**? If so, present the solution; if not, then return to step 3 and continue through the process again.
-
-1. *Carefully define the problem* . This may be the most important part of the process, because it becomes the foundation for all the rest of the steps. In general, the presentation of engineering problems is
-
-somewhat incomplete. You must do all you can to make sure you understand the problem as thoroughly as the presenter of the problem understands it. Time spent at this point clearly identifying the problem will save you considerable time and frustration later. As a student, you can clarify a problem statement in a textbook by asking your professor. A problem presented to you in industry may require that you consult several individuals. At this step, it is important to develop questions that need to be addressed before continuing the solution process. If you have such questions, you need to consult with the appropriate individuals or resources to obtain the answers to those questions. With those answers, you can now refine the problem, and use that refinement as the problem statement for the rest of the solution process.
-
-2. *Present everything you know about the problem*. You are now ready to write down everything you know about the problem and its possible solutions. This important step will save you time and frustration later.
-
-3. *Establish a set of alternative solutions and determine the one that promises the greatest likelihood of success* . Almost every problem will have a number of possible paths that can lead to a solution. It is highly desirable to identify as many of those paths as possible. At this point, you also need to determine what tools are available to you, such as *PSpice* and *MATLAB* and other software packages that can greatly reduce effort and increase accuracy. Again, we want to stress that time spent carefully defining the problem and investigating alternative ap proaches to its solution will pay big dividends later. Evaluating the al ternatives and determining which promises the greatest likelihood of success may be difficult but will be well worth the effort. Document this process well since you will want to come back to it if the first approach does not work.
-
-4. *Attempt a problem solution*. Now is the time to actually begin solving the problem. The process you follow must be well documented in order to present a detailed solution if successful, and to evaluate the process if you are not successful. This detailed evaluation may lead to corrections that can then lead to a successful solution. It can also lead to new alternatives to try. Many times, it is wise to fully set up a solution before putting numbers into equations. This will help in checking your results.
-
-5. *Evaluate the solution and check for accuracy*. You now thoroughly evaluate what you have accomplished. Decide if you have an acceptable solution, one that you want to present to your team, boss, or professor.
-
-6. *Has the problem been solved satisfactorily? If so, present the solu tion; if not, then return to step 3 and continue through the process again.* Now you need to present your solution or try another alternative. At this point, presenting your solution may bring closure to the process. Often, however, presentation of a solution leads to further refinement of the problem definition, and the process continues. Following this process will eventually lead to a satisfactory conclusion.
-
-Now let us look at this process for a student taking an electrical and computer engineering foundations course. (The basic process also ap plies to almost e very engineering course.) K eep in mind that although the steps have been simplified to apply to academic types of problems, the process as stated always needs to be followed. We consider a simple example.
-
-# **Solution:**
-
-1. *Carefully define the problem*. This is only a simple example, but we can already see that we do not know the polarity on the 3-V source. We have the following options. We can ask the professor what the polarity should be. If we cannot ask, then we need to make a decision on what to do next. If we have time to work the problem both ways, we can solve for the current when the 3-V source is plus on top and then plus on the bottom. If we do not have the time to work it both ways, assume a polarity and then carefully document your decision. Let us assume that the professor tells us that the source is plus on the bottom as shown in Fig. 1.20.
-
-2. *Present everything you know about the problem*. Presenting all that we know about the problem involves labeling the circuit clearly so that we define what we seek.
-
-Given the circuit shown in Fig. 1.20, solve for *i*8Ω.
-
-We now check with the professor , if reasonable, to see if the prob lem is properly defined.
-
-3. *Establish a set of alternative solutions and determine the one that promises the greatest likelihood of success* . There are essentially three techniques that can be used to solve this problem. Later in the text you will see that you can use circuit analysis (using Kirchhoff's laws and Ohm's law), nodal analysis, and mesh analysis.
-
-To solve for *i*8Ω using circuit analysis will eventually lead to a solution, but it will likely take more work than either nodal or mesh analysis. To solve for *i*8Ω using mesh analysis will require writing two simultaneous equations to find the two loop currents indicated in Fig. 1.21. Using nodal analysis requires solving for only one unknown. This is the easiest approach.
-
-**Figure 1.21** Using nodal analysis.
-
-Therefore, we will solve for *i*8Ω using nodal analysis.
-
-4. *Attempt a problem solution*. We first write down all of the equations we will need in order to find *i*8Ω.
-
-$$
-i_{8\Omega} = i_2,
-$$
- $i_2 = \frac{v_1}{8},$ $i_{8\Omega} = \frac{v_1}{8}$
- $\frac{v_1 - 5}{2} + \frac{v_1 - 0}{8} + \frac{v_1 + 3}{4} = 0$
-
-**Figure 1.19** Illustrative example.
-
-**Figure 1.20** Problem definition.
-
-Now we can solve for *v*1.
-
-$$
-8\left[\frac{v_1 - 5}{2} + \frac{v_1 - 0}{8} + \frac{v_1 + 3}{4}\right] = 0
-$$
-
-leads to (4 *v* 1 − 20) + (*v* 1) + (2 *v* 1 + 6) = 0
-
-$$
-7v_1 = +14
-$$
-, $v_1 = +2$ V, $i_{8\Omega} = \frac{v_1}{8} = \frac{2}{8} = 0.25$ A
-
-5. *Evaluate the solution and check for accuracy* . We can now use Kirchhoff's voltage law (KVL) to check the results.
-
-$$
-i_1 = \frac{v_1 - 5}{2} = \frac{2 - 5}{2} = -\frac{3}{2} = -1.5 \text{ A}
-$$
-
-$$
-i_2 = i_{8\Omega} = 0.25 \text{ A}
-$$
-
-$$
-i_3 = \frac{v_1 + 3}{4} = \frac{2 + 3}{4} = \frac{5}{4} = 1.25 \text{ A}
-$$
-
-*i*1 + *i*2 + *i*3 = −**1.5** + **0.25** + **1.25** = **0** (Checks.)
-
-Applying KVL to loop 1,
-
-$$
--5 + v_{2\Omega} + v_{8\Omega} = -5 + (-i_1 \times 2) + (i_2 \times 8)
-$$
-
-= -5 + [ -(-1.5)2] + (0.25 \times 8)
-= -5 + 3 + 2 = 0 (Checks.)
-
-Applying KVL to loop 2,
-
-$$
--v_{8\Omega} + v_{4\Omega} - 3 = -(i_2 \times 8) + (i_3 \times 4) - 3
-$$
-
-= -(0.25 \times 8) + (1.25 \times 4) - 3
-= -2 + 5 - 3 = 0 (Checks.)
-
-So we now have a v ery high de gree of confidence in the accuracy of our answer.
-
-6. *Has the problem been solved satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again.* This problem has been solved satisfactorily.
-
-The current through the 8-Ω resistor is 0.25 A flowing down through the 8-Ω resistor.
-
-Try applying this process to some of the more difficult problems at the end of the chapter. Practice Problem 1.10
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/014_1.9 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/014_1.9 Summary.md
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-# **1.9** Summary
-
-- 1. An electric circuit consists of electrical elements connected together.
-- 2. The International System of Units (SI) is the international mea surement language, which enables engineers to communicate their results. From the se ven principal units, the units of other ph ysical quantities can be derived.
-
-3. Current is the rate of char ge flow past a gi ven point in a gi ven direction.
-
-$$
-i = \frac{dq}{dt}
-$$
-
-4. Voltage is the energy required to move 1 C of char ge from a reference point (−) to another point (+).
-
-> *vab* = \_\_\_ *dw dq*
-
-5. Power is the energy supplied or absorbed per unit time. It is also the product of voltage and current.
-
-$$
-p = \frac{dw}{dt} = vi
-$$
-
-- 6. According to the passi ve sign con vention, power assumes a posi tive sign when the current enters the positive polarity of the voltage across an element.
-- 7. An ideal v oltage source produces a specific potential difference across its terminals re gardless of what is connected to it. An ideal current source produces a specific current through its terminals regardless of what is connected to it.
-- 8. Voltage and current sources can be dependent or independent. A dependent source is one whose value depends on some other circuit variable.
-- 9. Two areas of application of the concepts covered in this chapter are the TV picture tube and electricity billing procedure.
-
-# Review Questions
-
-**1.1** One millivolt is one millionth of a volt.
-
-```
-(a) True (b) False
-```
-
-**1.2** The prefix *micro* stands for:
-
-(a) 106 (b) 103 (c) 10−3 (d) 10−6
-
-**1.3** The voltage 2,000,000 V can be expressed in powers of 10 as:
-
-(a) 2 mV (b) 2 kV (c) 2 MV (d) 2 GV
-
-**1.4** A charge of 2 C flowing past a given point each second is a current of 2 A.
-
-(a) True (b) False
-
-**1.5** The unit of current is:
-
-(a) coulomb (b) ampere (c) volt (d) joule
-
-**1.6** Voltage is measured in:
-
-(a) watts (b) amperes
-
-(a) True (b) False
-
-- (c) volts (d) joules per second
-- **1.7** A 4-A current charging a dielectric material will accumulate a charge of 24 C after 6 s.
-
-**1.8** The voltage across a 1.1-kW toaster that produces a current of 10 A is:
-
-(a) 11 kV (b) 1100 V (c) 110 V (d) 11 V
-
-**1.9** Which of these is not an electrical quantity?
-
-| (a) charge | (b) time | (c) voltage |
-|-------------|-----------|-------------|
-| (d) current | (e) power | |
-
-- **1.10** The dependent source in Fig. 1.22 is:
- - (a) voltage-controlled current source
- - (b) voltage-controlled voltage source
- - (c) current-controlled voltage source
- - (d) current-controlled current source
-
-**Figure 1.22** For Review Question 1.10.
-
-*Answers: 1.1b, 1.2d, 1.3c, 1.4a, 1.5b, 1.6c, 1.7a, 1.8c, 1.9b, 1.10d.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/015_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/015_Review Questions.md
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-# Problems
-
-# Section 1.3 Charge and Current
-
-- **1.1** How much charge is represented by these number of electrons?
- - (a) 6.482 × 1017
- - (b) 1.24 × 1018
- - (c) 2.46 × 1019
- - (d) 1.628 × 1020
-- **1.2** Determine the current flowing through an element if the charge flow is given by
- - (a) *q*(*t*) = (3) mC
-
-(b)
-$$
-q(t) = (4t^2 + 20t - 4)
-$$
- C
-
-(c)
-$$
-q(t) = (15e^{-3t} - 2e^{-18t}) nC
-$$
-
-(d) $q(t) = 5t^2(3t^3 + 4) pC$
-
-(d)
-$$
-q(t) = 5t^2(3t^3 + 4) \text{ pC}
-$$
-
-(e) $q(t) = 2e^{-3t} \sin(20\pi t) \mu\text{C}$
-
-**1.3** Find the charge *q*(*t*) flowing through a device if the current is:
-
-(a)
-$$
-i(t) = 3
-$$
- A, $q(0) = 1$ C
-\n(b) $i(t) = (2t + 5)$ mA, $q(0) = 0$
-\n(c) $i(t) = 20 \cos(10t + \pi/6) \mu$ A, $q(0) = 2 \mu$ C
-\n(d) $i(t) = 10e^{-30t} \sin 40t$ A, $q(0) = 0$
-
-- **1.4** A total charge of 300 C flows past a given cross section of a conductor in 30 seconds. What is the value of the current?
-- **1.5** Determine the total charge transferred over the time interval of 0 ≤ *t* ≤ 10 s when *i*(*t*) = \_\_1 2 *t* A.
-- **1.6** The charge entering a certain element is shown in Fig. 1.23. Find the current at:
-
-(a)
-$$
-t = 1
-$$
- ms (b) $t = 6$ ms (c) $t = 10$ ms
-
-**1.7** The charge flowing in a wire is plotted in Fig. 1.24. Sketch the corresponding current.
-
-# **Figure 1.24**
-
-For Prob. 1.7.
-
-**1.8** The current flowing past a point in a device is shown in Fig. 1.25. Calculate the total charge through the point.
-
-# **Figure 1.25**
-
-For Prob. 1.8.
-
-**1.9** The current through an element is shown in Fig. 1.26. Determine the total charge that passed through the element at:
-
-For Prob. 1.9.
-
-# Sections 1.4 and 1.5 Voltage, Power, and Energy
-
-- **1.10** A lightning bolt with 10 kA strikes an object for 15 *μ*s. How much charge is deposited on the object?
-- **1.11** A rechargeable flashlight battery is capable of delivering 90 mA for about 12 h. How much charge can it release at that rate? If its terminal voltage is 1.5 V, how much energy can the battery deliver?
-- **1.12** If the current flowing through an element is given by
-
-$$
-i(t) = \begin{cases} 3tA, & 0 \le t < 6 \text{ s} \\ 18A, & 6 \le t < 10 \text{ s} \\ -12A, & 10 \le t < 15 \text{ s} \\ 0, & t \ge 15 \text{ s} \end{cases}
-$$
-
-Plot the charge stored in the element over 0 < *t* < 20 s.
-
-**1.13** The charge entering the positive terminal of an element is
-
-$$
-q = 5\sin 4\pi t \,\mathrm{mC}
-$$
-
- while the voltage across the element (plus to minus) is
-
-$$
-v = 3\cos 4\pi t \,\mathrm{V}
-$$
-
-- (a) Find the power delivered to the element at *t* = 0.3 s.
-- (b) Calculate the energy delivered to the element between 0 and 0.6 s.
-- **1.14** The voltage *v*(*t*) across a device and the current *i*(*t*) through it are
-
-$$
-v(t) = 20 \sin(4t)
-$$
- V and $i(t) = 10(1 + e^{-2t})$ mA
-
-Calculate:
-
-- (a) the total charge in the device at *t* = 1 *s*, *q*(0) = 0.
-- (b) the power consumed by the device at *t* = 1 s.
-- **1.15** The current entering the positive terminal of a device is *i*(*t*) = 6*e*−2*t* mA and the voltage across the device is *v*(*t*) = 10*di*∕*dt*V.
- - (a) Find the charge delivered to the device between *t* = 0 and *t* = 2 s.
- - (b) Calculate the power absorbed.
- - (c) Determine the energy absorbed in 3 s.
-
-# Section 1.6 Circuit Elements
-
-- **1.16** Figure 1.27 shows the current through and the voltage across an element.
- - (a) Sketch the power delivered to the element for *t* > 0.
- - (b) Fnd the total energy absorbed by the element for the period of 0 < *t* < 4s.
-
-**Figure 1.27** For Prob. 1.16.
-
-**1.17** Figure 1.28 shows a circuit with four elements, *p*1 = 60 W absorbed, *p*3 = −145 W absorbed, and *p*4 = 75 W absorbed. How many watts does element 2 absorb?
-
-# **Figure 1.28**
-
-For Prob. 1.17.
-
-**1.18** Find the power absorbed by each of the elements in Fig. 1.29.
-
-# **Figure 1.29**
-
-For Prob. 1.18.
-
-**1.19** Find *I* and the power absorbed by each element in the network of Fig. 1.30.
-
-# **Figure 1.30**
-
-For Prob. 1.19.
-
-**1.20** Find *Vo* and the power absorbed by each element in the circuit of Fig. 1.31.
-
-For Prob. 1.20.
-
-# Section 1.7 Applications
-
-- **1.21** A 60-W incandescent bulb operates at 120 V. How many electrons and coulombs flow through the bulb in one day?
-- **1.22** A lightning bolt strikes an airplane with 40 kA for 1.7 ms. How many coulombs of charge are deposited on the plane?
-- **1.23** A 1.8-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents/kWh, what is the cost of its operation for 30 days?
-- **1.24** A utility company charges 8.2 cents/kWh. If a consumer operates a 60-W light bulb continuously for one day, how much is the consumer charged?
-- **1.25** A 1.2-kW toaster takes roughly 4 minutes to heat four slices of bread. Find the cost of operating the toaster twice per day for 2 weeks (14 days). Assume energy costs 9 cents/kWh.
-- **1.26** A cell phone battery is rated at 3.85 V and can store 10.78 watt-hours of energy.
- - (a) How much average current can it deliver over a period of 3 hours if it is fully discharged at the end of that time?
- - (b) How much average power is delivered in part (a)?
- - (c) What is the ampere-hour rating of the battery?
-- **1.27** A constant current of 3 A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + *t*∕2V, where *t* is in hours,
- - (a) how much charge is transported as a result of the charging?
-
-- (b) how much energy is expended?
-- (c) how much does the charging cost? Assume electricity costs 9 cents/kWh.
-- **1.28** A 150-W incandescent outdoor lamp is connected to a 120-V source and is left burning continuously for an average of 12 hours per day. Determine:
- - (a) the current through the lamp when it is lit.
- - (b) the cost of operating the light for one non-leap year if electricity costs 9.5 cents per kWh.
-- **1.29** An electric stove with four burners and an oven is used in preparing a meal as follows.
-
-| Burner 1: 20 minutes | Burner 2: 40 minutes |
-|----------------------|----------------------|
-| Burner 3: 15 minutes | Burner 4: 45 minutes |
-| Oven: 30 minutes | |
-
-If each b urner is rated at 1.2 kW and the o ven at 1.8 kW, and electricity costs 12 cents per kWh, calculate the cost of electricity used in preparing the meal.
-
-**1.30** Reliant Energy (the electric company in Houston, Texas) charges customers as follows:
-
-> Monthly charge \$6 First 250 kWh @ \$0.02/kWh All additional kWh @ \$0.07/kWh
-
-If a customer uses 2,436 kWh in one month, ho w much will Reliant Energy charge?
-
-**1.31** In a household, a business is run for an average of 6 h/day. The total power consumed by the computer and its printer is 230 W. In addition, a 75-W light runs during the same 6 h. If their utility charges 11.75 cents per kWh, how much do the owners pay every 30 days?
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/016_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/016_Problems.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/016_Problems.md
+++ /dev/null
@@ -1,27 +0,0 @@
-# Comprehensive Problems
-
-- **1.32** A telephone wire has a current of 20*μ*A flowing through it. How long does it take for a charge of 15 C to pass through the wire?
-- **1.33** A lightning bolt carried a current of 2 kA and lasted for 3 ms. How many coulombs of charge were contained in the lightning bolt?
-- **1.34** Figure 1.32 shows the power consumption of a certain household in 1 day. Calculate:
- - (a) the total energy consumed in kWh,
- - (b) the average power over the total 24 hour period.
-
-For Prob. 1.34.
-
-**1.35** The graph in Fig. 1.33 represents the power drawn by an industrial plant between 8:00 and 8:30 a.m. Cal culate the total energy in MWh consumed by the plant.
-
-For Prob. 1.35.
-
-**1.36** A battery can be rated in ampere-hours (Ah) or watt hours (Wh). The ampere hours can be obtained from the watt hours by dividing watt hours by a nominal
-
-voltage of 12 V. If an automobile battery is rated at 20 Ah:
-
-- (a) What is the maximum current that can be supplied for 15 minutes?
-- (b) How many days will it last if it is discharged at a rate of 2 mA?
-- **1.37** A total of 2 MJ are delivered to an automobile battery (assume 12 V) giving it an additional charge. How much is that additional charge? Express your answer in ampere-hours.
-- **1.38** How much energy does a 10-hp motor deliver in 30 minutes? Assume that 1 horsepower = 746 W.
-- **1.39** A 600-W TV receiver is turned on for 4 h with nobody watching it. If electricity costs 10 cents/kWh, how much money is wasted?
-
-*This page intentionally left blank*
-
-# **chapter**
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/017_Chapter 2 - Basic Laws.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/017_Chapter 2 - Basic Laws.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/017_Chapter 2 - Basic Laws.md
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@@ -1,25 +0,0 @@
-# Basic Laws 2
-
-*There are too many people praying for mountains of difficulty to be removed, when what they really need is the courage to climb them!* —Unknown
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.b), "an ability to design and conduct experiments, as well as to analyze and interpret data."**
-
-Engineers must be able to design and conduct e xperiments, as well as analyze and interpret data. Most students ha ve spent man y hours per forming experiments in high school and in college. During this time, you have been asked to analyze the data and to interpret the data. Therefore, you should already be skilled in these tw o activities. My recommendation is that, in the process of performing e xperiments in the future, you spend more time in analyzing and interpreting the data in the conte xt of the experiment. What does this mean?
-
-If you are looking at a plot of voltage versus resistance or current versus resistance or power versus resistance, what do you actually see? Does the curve make sense? Does it agree with what the theory tells you? Does it differ from e xpectation, and, if so, wh y? Clearly, practice with analyzing and interpreting data will enhance this skill.
-
-Since most, if not all, the e xperiments you are required to do as a student involve little or no practice in designing the experiment, how can you develop and enhance this skill?
-
-Actually, developing this skill under this constraint is not as difficult as it seems. What you need to do is to take the experiment and analyze it. Just break it down into its simplest parts, reconstruct it trying to under stand why each element is there, and finally, determine what the author of the experiment is trying to teach you. Even though it may not always seem so, every experiment you do w as designed by someone who w as sincerely motivated to teach you something.
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Know and understand the voltage current relationship of resistors (Ohm's law).
-- 2. Understand the basic structure of electrical circuits, essentially nodes, loops, and branches.
-- 3. Understand Kirchhoff's voltage and current laws and their importance in analyzing electrical circuits.
-- 4. Understand series resistances and voltage division, and parallel resistances and current division.
-- 5. Know how to convert delta-connected circuits to wye-connected circuits and how to convert wye-connected circuits to deltaconnected circuits.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/018_2.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/018_2.1 Introduction.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/018_2.1 Introduction.md
+++ /dev/null
@@ -1,5 +0,0 @@
-# **2.1** Introduction
-
-Chapter 1 introduced basic concepts such as current, voltage, and power in an electric circuit. To actually determine the v alues of these v ariables in a gi ven circuit requires that we understand some fundamen tal laws that govern electric circuits. These laws, known as Ohm's law and Kirchhoff's laws, form the foundation upon which electric circuit analysis is built.
-
-In this chapter , in addition to these la ws, we shall discuss some techniques commonly applied in circuit design and analysis. These techniques include combining resistors in series or parallel, voltage division, current division, and delta-to-wye and wye-to-delta transformations. The application of these laws and techniques will be restricted to resistive circuits in this chapter. We will finally apply the laws and techniques to real-life problems of electrical lighting and the design of dc meters.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/019_2.2 Ohm's Law.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/019_2.2 Ohm's Law.md
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-# **2.2** Ohm's Law
-
-Materials in general ha ve a characteristic beha vior of resisting the flow of electric charge. This physical property, or ability to resist current, is known as *resistance* and is represented by the symbol *R*. The resistance of any material with a uniform cross-sectional area *A* depends on *A* and its length ℓ, as sho wn in Fig. 2.1(a). We can represent resistance (as measured in the laboratory), in mathematical form,
-
-$$
-R = \rho \frac{\ell}{A} \tag{2.1}
-$$
-
-where *ρ* is known as the *resistivity* of the material in ohm-meters. Good conductors, such as copper and aluminum, ha ve low resistivities, while insulators, such as mica and paper, have high resistivities. Table 2.1 presents the values of *ρ* for some common materials and shows which materials are used for conductors, insulators, and semiconductors.
-
-The circuit element used to model the current-resisting beha vior of a material is the *resistor*. For the purpose of constructing circuits, resistors are
-
-**Figure 2.1** (a) Resistor, (b) Circuit symbol for resistance.
-
-| TABLE 2.1 | | |
-|-----------|--|--|
-| | | |
-
-| Material | Resistivity (Ω∙m) | Usage |
-|-----------|-------------------|---------------|
-| Silver | 1.64 × 10−8 | Conductor |
-| Copper | 1.72 × 10−8 | Conductor |
-| Aluminum | 2.8 × 10−8 | Conductor |
-| Gold | 2.45 × 10−8 | Conductor |
-| Carbon | 4 × 10−5 | Semiconductor |
-| Germanium | 47 × 10−2 | Semiconductor |
-| Silicon | 6.4 × 102 | Semiconductor |
-| Paper | 1010 | Insulator |
-| Mica | 5 × 1011 | Insulator |
-| Glass | 1012 | Insulator |
-| Teflon | 3 × 1012 | Insulator |
-
-Resistivities of common materials.
-
-usually made from metallic alloys and carbon compounds. The circuit symbol for the resistor is shown in Fig. 2.1(b), where *R* stands for the resistance of the resistor. The resistor is the simplest passive element.
-
-Georg Simon Ohm (1787–1854), a German ph ysicist, is credited with finding the relationship between current and voltage for a resistor. This relationship is known as *Ohm's law*.
-
-Ohm's law states that the voltage v across a resistor is directly proportional to the current i flowing through the resistor.
-
-That is,
-
-$$
-v \propto i \tag{2.2}
-$$
-
-Ohm defined the constant of proportionality for a resistor to be the resistance, *R*. (The resistance is a material property which can change if the internal or external conditions of the element are altered, e.g., if there are changes in the temperature.) Thus, Eq. (2.2) becomes
-
-$$
-v = iR \tag{2.3}
-$$
-
-# Historical
-
-**Georg Simon Ohm** (1787–1854), a German physicist, in 1826 experimentally determined the most basic law relating voltage and cur rent for a resistor. Ohm's work was initially denied by critics.
-
-Born of humble beginnings in Erlangen, Bavaria, Ohm threw himself into electrical research. His efforts resulted in his famous law. He was awarded the Copley Medal in 1841 by the Royal Society of London. In 1849, he was given the Professor of Physics chair by the University of Munich. To honor him, the unit of resistance was named the ohm.
-
-(b) **Figure 2.2**
-
-(a) Short circuit (*R* =0), (b) Open circuit (*R* =∞).
-
-**Figure 2.3** Fixed resistors: (a) wirewound type, (b) carbon film type. © McGraw-Hill Education/Mark Dierker, photographer
-
-**Figure 2.4** Circuit symbol for: (a) a variable resistor in general, (b) a potentiometer.
-
-which is the mathematical form of Ohm's law. *R* in Eq. (2.3) is mea sured in the unit of ohms, designated Ω. Thus,
-
-The resistance R of an element denotes its ability to resist the flow of electric current; it is measured in ohms (Ω).
-
-We may deduce from Eq. (2.3) that
-
-$$
-R = \frac{v}{i} \tag{2.4}
-$$
-
-so that
-
-$$
-1 \Omega = 1 \text{ V/A}
-$$
-
-To apply Ohm' s la w as stated in Eq. (2.3), we must pay careful attention to the current direction and v oltage polarity. The direction of current *i* and the polarity of voltage *v* must conform with the passive sign convention, as shown in Fig. 2.1(b). This implies that current flows from a higher potential to a lower potential in order for *v* = *i R*. If current flows from a lower potential to a higher potential, *v* = −*i R*.
-
-Since the value of *R* can range from zero to infinity, it is important that we consider the tw o extreme possible values of *R*. An element with *R* = 0 is called a *short circuit*, as shown in Fig. 2.2(a). For a short circuit,
-
-$$
-v = iR = 0 \tag{2.5}
-$$
-
-showing that the v oltage is zero b ut the current could be an ything. In practice, a short circuit is usually a connecting wire assumed to be a perfect conductor. Thus,
-
-A short circuit is a circuit element with resistance approaching zero.
-
-Similarly, an element with *R* =∞ is known as an *open circuit*, as shown in Fig. 2.2(b). For an open circuit,
-
-ircuit,
-\n
-$$
-i = \lim_{R \to \infty} \frac{v}{R} = 0
-$$
-\n(2.6)
-
-indicating that the current is zero though the v oltage could be anything. Thus,
-
-An open circuit is a circuit element with resistance approaching infinity.
-
-A resistor is either fixed or v ariable. Most resistors are of the fixed type, meaning their resistance remains constant. The two common types of fixed resistors (wirewound and composition) are shown in Fig. 2.3. The composition resistors are used when large resistance is needed. The circuit symbol in Fig. 2.1(b) is for a fixed resistor. Variable resistors have adjustable resistance. The symbol for a variable resistor is shown in Fig. 2.4(a). A common variable resistor is known as a *potentiometer* or *pot* for short, with the symbol shown in Fig. 2.4(b). The pot is a three-terminal element with a sliding contact or wiper . By sliding the wiper , the resistances be tween the wiper terminal and the fixed terminals v ary. Like fixed resistors, variable resistors can be of either wire wound or composition type, as shown in Fig. 2.5. Although resistors like those in Figs. 2.3 and 2.5 are used in circuit designs, today most circuit components including resistors are either surface mounted or integrated, as typically shown in Fig. 2.6.
-
-**Figure 2.5** Variable resistors: (a) composition type, (b) slider pot. © McGraw-Hill Education/Mark Dierker, photographer
-
-It should be pointed out that not all resistors obe y Ohm's law. A resistor that obeys Ohm's law is known as a *linear* resistor. It has a constant resistance and thus its current-voltage characteristic is as illustrated in Fig. 2.7(a): Its *i*-*v* graph is a straight line passing through the ori gin. A *nonlinear* resistor does not obe y Ohm's law. Its resistance varies with current and its *i*-*v* characteristic is typically sho wn in Fig. 2.7(b). Examples of devices with nonlinear resistance are the light bulb and the diode. Although all practical resistors may e xhibit nonlinear beha vior under certain conditions, we will assume in this book that all elements actually designated as resistors are linear.
-
-A useful quantity in circuit analysis is the reciprocal of resistance *R*, known as *conductance* and denoted by *G*:
-
-$$
-G = \frac{1}{R} = \frac{i}{v} \tag{2.7}
-$$
-
-The conductance is a measure of how well an element will conduct electric current. The unit of conductance is the *mho* (ohm spelled backward) or reciprocal ohm, with symbol ℧, the inverted omega. Although engineers often use the mho, in this book we prefer to use the siemens (S), the SI unit of conductance:
-
-$$
-1 S = 1 \, \mathbf{U} = 1 \, \mathbf{A} / \mathbf{V} \tag{2.8}
-$$
-
-Thus,
-
-Conductance is the ability of an element to conduct electric current; it is measured in mhos (℧) or siemens (S).
-
-The same resistance can be e xpressed in ohms or siemens. F or example, 10 Ω is the same as 0.1 S. From Eq. (2.7), we may write
-
-$$
-i = Gv \tag{2.9}
-$$
-
-The power dissipated by a resistor can be e xpressed in terms of *R*. Using Eqs. (1.7) and (2.3),
-
-$$
-p = vi = i^{2}R = \frac{v^{2}}{R}
-$$
- (2.10)
-
-**Figure 2.6** Resistors in an integrated circuit board.
-
-**Figure 2.7** The *i*-*v* characteristic of: (a) a linear resistor, (b) a nonlinear resistor.
-
-The power dissipated by a resistor may also be e xpressed in terms of *G* as
-
-$$
-p = vi = v^2 G = \frac{i^2}{G}
-$$
- (2.11)
-
-We should note two things from Eqs. (2.10) and (2.11):
-
-- 1. The power dissipated in a resistor is a nonlinear function of either current or voltage.
-- 2. Since *R* and *G* are positive quantities, the power dissipated in a resistor is al ways positive. Thus, a resistor al ways absorbs power from the circuit. This confirms the idea that a resistor is a passive element, incapable of generating energy.
-
-# **Solution:**
-
-The voltage across the resistor is the same as the source voltage (30 V) because the resistor and the voltage source are connected to the same pair of terminals. Hence, the current is
-
-$$
-i = \frac{v}{R} = \frac{30}{5 \times 10^3} = 6 \text{ mA}
-$$
-
-The conductance is
-
-$$
-G = \frac{1}{R} = \frac{1}{5 \times 10^3} = 0.2 \text{ mS}
-$$
-
-We can calculate the power in various ways using either Eqs. (1.7), (2.10), or (2.11).
-
-*p* = *vi* = 30 (6 × 10 −3) = 180 mW
-
-or
-
-$$
-p = i^2 R = (6 \times 10^{-3})^2 5 \times 10^3 = 180
-$$
- mW
-
-or
-
-$$
-p = v^2 G = (30)^2 0.2 \times 10^{-3} = 180
-$$
- mW
-
-**Figure 2.8** For Example 2.2.
-
-For the circuit shown in Fig. 2.9, calculate the voltage *v*, the conductance *G*, and the power *p*.
-
-**Answer:** 30 V, 100 *µ*S, 90 mW.
-
-For Practice Prob. 2.2
-
-A voltage source of 20 sin *πt* V is connected across a 5-kΩ resistor. Find Example 2.3 the current through the resistor and the power dissipated.
-
-# **Solution:**
-
-$$
-i = \frac{v}{R} = \frac{20 \sin \pi t}{5 \times 10^3} = 4 \sin \pi t \text{ mA}
-$$
-
-Hence,
-
-$$
-p = vi = 80 \sin^2 \pi t \text{ mW}
-$$
-
-A resistor absorbs an instantaneous power of 30 cos Practice Problem 2.3 2 *t* mW when con nected to a voltage source *v* = 15 cos *t* V. Find *i* and *R*.
-
-**Answer:** 2 cos *t* mA, 7.5 kΩ.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/020_2.3 Nodes, Branches, and Loops.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/020_2.3 Nodes, Branches, and Loops.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/020_2.3 Nodes, Branches, and Loops.md
+++ /dev/null
@@ -1,51 +0,0 @@
-# **2.3** Nodes, Branches, and Loops
-
-Since the elements of an electric circuit can be interconnected in se veral ways, we need to understand some basic concepts of network topology. To differentiate between a circuit and a network, we may regard a network as an interconnection of elements or devices, whereas a circuit is a network providing one or more closed paths. The convention, when addressing network topology, is to use the w ord network rather than circuit. We do this even though the word network and circuit mean the same thing when used in this context. In network topology, we study the properties relating to the placement of elements in the network and the geometric configuration of the network. Such elements include branches, nodes, and loops.
-
-A branch represents a single element such as a voltage source or a resistor.
-
-In other words, a branch represents an y two-terminal element. The circuit in Fig. 2.10 has five branches, namely, the 10-V voltage source, the 2-A current source, and the three resistors.
-
-A node is the point of connection between two or more branches.
-
-A node is usually indicated by a dot in a circuit. If a short circuit (a connecting wire) connects two nodes, the two nodes constitute a single node. The circuit in Fig. 2.10 has three nodes *a*, *b*, and *c*. Notice that
-
-Practice Problem 2.2
-
-**Figure 2.11** The three-node circuit of Fig. 2.10 is redrawn.
-
-the three points that form node *b* are connected by perfectly conducting wires and therefore constitute a single point. The same is true of the four points forming node *c*. We demonstrate that the circuit in Fig. 2.10 has only three nodes by redrawing the circuit in Fig. 2.11. The two circuits in Figs. 2.10 and 2.11 are identical. However, for the sake of clarity, nodes *b* and *c* are spread out with perfect conductors as in Fig. 2.10.
-
-A loop is any closed path in a circuit.
-
-A loop is a closed path formed by starting at a node, passing through a set of nodes, and returning to the starting node without passing through any node more than once. A loop is said to be *independent* if it contains at least one branch which is not a part of an y other independent loop. Independent loops or paths result in independent sets of equations.
-
-It is possible to form an independent set of loops where one of the loops does not contain such a branch. In Fig. 2.11, *abca* with the 2 Ω resistor is independent. A second loop with the 3 Ω resistor and the current source is independent. The third loop could be the one with the 2Ω resistor in parallel with the 3Ω resistor. This does form an independent set of loops.
-
-A network with *b* branches, *n* nodes, and *l* independent loops will satisfy the fundamental theorem of network topology:
-
-$$
-b = l + n - 1 \tag{2.12}
-$$
-
-As the next two definitions show, circuit topology is of great value to the study of voltages and currents in an electric circuit.
-
-Two or more elements are in series if they exclusively share a single node and consequently carry the same current.
-
-Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them.
-
-Elements are in series when the y are chain-connected or connected se quentially, end to end. F or example, two elements are in series if they share one common node and no other element is connected to that common node. Elements in parallel are connected to the same pair of terminals. Elements may be connected in a w ay that they are neither in series nor in parallel. In the circuit shown in Fig. 2.10, the voltage source and the 5- Ω resistor are in series because the same current will flow through them. The 2-Ω resistor, the 3-Ω resistor, and the current source are in parallel because they are connected to the same two nodes *b* and *c* and consequently have the same voltage across them. The 5-Ω and 2-Ω
-
-Example 2.4
-
-Determine the number of branches and nodes in the circuit shown in Fig. 2.12. Identify which elements are in series and which are in parallel.
-
-resistors are neither in series nor in parallel with each other.
-
-# **Solution:**
-
-Since there are four elements in the circuit, the circuit has four branches: 10 V, 5 Ω, 6 Ω, and 2 A. The circuit has three nodes as identified in Fig. 2.13. The 5-Ω resistor is in series with the 10-V voltage source because the same current would flow in both. The 6-Ω resistor is in parallel with the 2-A current source because both are connected to the same nodes 2 and 3.
-
-How many branches and nodes does the circuit in Fig. 2.14 have? Identify Practice Problem 2.4 the elements that are in series and in parallel.
-
-**Answer:** Five branches and three nodes are identified in Fig. 2.15. The 1-Ω and 2-Ω resistors are in parallel. The 4-Ω resistor and 10-V source are also in parallel.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/021_2.4 Kirchhoff's Laws.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/021_2.4 Kirchhoff's Laws.md
deleted file mode 100644
index 6087c0a8233dcbe5b1584994ddf3e64be0720c72..0000000000000000000000000000000000000000
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@@ -1,378 +0,0 @@
-# **2.4** Kirchhoff's Laws
-
-Ohm's law by itself is not sufficient to analyze circuits. However, when it is coupled with Kirchhoff's two laws, we have a sufficient, powerful set of tools for analyzing a large variety of electric circuits. Kirchhoff's laws were first introduced in 1847 by the German physicist Gustav Robert Kirchhoff (1824–1887). These laws are formally kno wn as Kirchhoff's current law (KCL) and Kirchhoff's voltage law (KVL).
-
-Kirchhoff's first law is based on the la w of conservation of charge, which requires that the algebraic sum of charges within a system cannot change.
-
-Kirchhoff's current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero.
-
-Mathematically, KCL implies that
-
-$$
-\sum_{n=1}^{N} i_n = 0
-$$
- (2.13)
-
-where *N* is the number of branches connected to the node and *in* is the *n*th current entering (or lea ving) the node. By this la w, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa.
-
-# Historical
-
-**Gustav Robert Kirchhoff** (1824–1887), a German physicist, stated two basic laws in 1847 concerning the relationship between the cur rents and voltages in an electrical network. Kirchhoff's laws, along with Ohm's law, form the basis of circuit theory.
-
-Born the son of a lawyer in Konigsberg, East Prussia, Kirchhoff entered the University of Konigsberg at age 18 and later became a lecturer in Berlin. His collaborative work in spectroscopy with German chemist Robert Bunsen led to the discovery of cesium in 1860 and rubidium in 1861. Kirchhoff was also credited with the Kirchhoff law of radiation. Thus, Kirchhoff is famous among engineers, chemists, and physicists.
-
-To prove KCL, assume a set of currents *i k* (*t*) , *k* = 1, 2,…, flow into a node. The algebraic sum of currents at the node is
-
-$$
-i_T(t) = i_1(t) + i_2(t) + i_3(t) + \cdots
-$$
- (2.14)
-
-Integrating both sides of Eq. (2.14) gives
-
-$$
-q_T(t) = q_1(t) + q_2(t) + q_3(t) + \cdots
-$$
- (2.15)
-
-where *qk* (*t*) = ∫ *ik* (*t*) *d t* and *qT* (*t*) = ∫ *iT* (*t*) *d t* .But the law of conservation of electric charge requires that the algebraic sum of electric charges at the node must not change; that is, the node stores no net charge. Thus, *qT* (*t*) = 0 → *iT* (*t*) = 0, confirming the validity of KCL.
-
-Consider the node in Fig. 2.16. Applying KCL gives
-
-$$
-i_1 + (-i_2) + i_3 + i_4 + (-i_5) = 0 \tag{2.16}
-$$
-
-since currents *i*1, *i*3, and *i*4 are entering the node, while currents *i*2 and *i*5 are leaving it. By rearranging the terms, we get
-
-$$
-i_1 + i_3 + i_4 = i_2 + i_5 \tag{2.17}
-$$
-
-Equation (2.17) is an alternative form of KCL:
-
-The sum of the currents entering a node is equal to the sum of the currents leaving the node.
-
-Note that KCL also applies to a closed boundary . This may be re garded as a generalized case, because a node may be regarded as a closed surface shrunk to a point. In tw o dimensions, a closed boundary is the same as a closed path. As typically illustrated in the circuit of Fig. 2.17, the total current entering the closed surface is equal to the total current leaving the surface.
-
-A simple application of KCL is combining current sources in parallel. The combined current is the algebraic sum of the current supplied by the indi vidual sources. F or example, the current sources sho wn in
-
-**Figure 2.16** Currents at a node illustrating KCL.
-
-**Figure 2.17** Applying KCL to a closed boundary.
-
-Two sources (or circuits in general) are said to be equivalent if they have the same i-v relationship at a pair of terminals.
-
-Fig. 2.18(a) can be combined as in Fig. 2.18(b). The combined or equivalent current source can be found by applying KCL to node *a*.
-
-$$
-I_T + I_2 = I_1 + I_3
-$$
-
-or
-
-$$
-I_T = I_1 - I_2 + I_3 \tag{2.18}
-$$
-
-A circuit cannot contain two different currents, *I*1 and *I*2, in series, unless *I*1 =*I*2; otherwise KCL will be violated.
-
-Kirchhoff's second law is based on the principle of conservation of energy:
-
-Kirchhoff's voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero.
-
-Expressed mathematically, KVL states that
-
-$$
-\sum_{m=1}^{M} v_m = 0
-$$
- (2.19)
-
-where *M* is the number of voltages in the loop (or the number of branches in the loop) and *vm* is the *m*th voltage.
-
-To illustrate KVL, consider the circuit in Fig. 2.19. The sign on each voltage is the polarity of the terminal encountered first as we travel around the loop. We can start with an y branch and go around the loop either clockwise or counterclockwise. Suppose we start with the v oltage source and go clockwise around the loop as sho wn; then v oltages would be −*v*1, +*v*2, +*v*3, +*v*4, and −*v*5, in that order. For example, as we reach branch 3, the positive terminal is met first; hence, we have +*v*3. For branch 4, we reach the ne gative terminal first; hence, −*v*4. Thus, KVL yields
-
-$$
--v_1 + v_2 + v_3 - v_4 + v_5 = 0 \tag{2.20}
-$$
-
-Rearranging terms gives
-
-$$
-v_2 + v_3 + v_5 = v_1 + v_4 \tag{2.21}
-$$
-
-which may be interpreted as
-
-| Sum of voltage drops = Sum of voltage rises | (2.22) |
-|---------------------------------------------|--------|
-|---------------------------------------------|--------|
-
-This is an alternative form of KVL. Notice that if we had traveled counterclockwise, the result w ould have been +*v*1, −*v*5, +*v*4, −*v*3, and −*v*2, which is the same as before e xcept that the signs are re versed. Hence, Eqs. (2.20) and (2.21) remain the same.
-
-When voltage sources are connected in series, KVL can be applied to obtain the total v oltage. The combined v oltage is the algebraic sum of the v oltages of the indi vidual sources. F or example, for the v oltage sources shown in Fig. 2.20(a), the combined or equivalent voltage source in Fig. 2.20(b) is obtained by applying KVL.
-
-$$
--V_{ab} + V_1 + V_2 - V_3 = 0
-$$
-
-# **Figure 2.18** Current sources in parallel: (a) original circuit, (b) equivalent circuit.
-
-KVL can be applied in two ways: by taking either a clockwise or a counterclockwise trip around the loop. Either way, the algebraic sum of voltages around the loop is zero.
-
-**Figure 2.19** A single-loop circuit illustrating KVL.
-
-$$
-V_{ab} = V_1 + V_2 - V_3 \tag{2.23}
-$$
-
-To avoid violating KVL, a circuit cannot contain tw o different voltages *V*1 and *V*2 in parallel unless *V*1 =*V*2.
-
-# **Figure 2.20**
-
-Voltage sources in series: (a) original circuit, (b) equivalent circuit.
-
-For the circuit in Fig. 2.21(a), find voltages *v*1 and *v*2 Example 2.5 .
-
-# **Solution:**
-
-To find *v*1 and *v*2 we apply Ohm's law and Kirchhoff's voltage law. Assume that current *i* flows through the loop as shown in Fig. 2.21(b). From Ohm's law,
-
-$$
-v_1 = 2i, \qquad v_2 = -3i \tag{2.5.1}
-$$
-
-Applying KVL around the loop gives
-
-$$
--20 + v_1 - v_2 = 0 \tag{2.5.2}
-$$
-
-Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain
-
-$$
--20 + 2i + 3i = 0 \qquad \text{or} \qquad 5i = 20 \qquad \Rightarrow \qquad i = 4 \text{ A}
-$$
-
-Substituting *i* in Eq. (2.5.1) finally gives
-
-$$
-v_1 = 8 \text{ V}, \qquad v_2 = -12 \text{ V}
-$$
-
-**Answer:** 16 V, −8 V.
-
-Determine *v* Example 2.6 *o* and *i* in the circuit shown in Fig. 2.23(a).
-
-# **Figure 2.23**
-
-For Example 2.6.
-
-# **Solution:**
-
-We apply KVL around the loop as shown in Fig. 2.23(b). The result is
-
-$$
--12 + 4i + 2v_o - 4 + 6i = 0 \tag{2.6.1}
-$$
-
-Applying Ohm's law to the 6-Ω resistor gives
-
-$$
-v_o = -6i \tag{2.6.2}
-$$
-
-Substituting Eq. (2.6.2) into Eq. (2.6.1) yields
-
-$$
--16 + 10i - 12i = 0 \qquad \Rightarrow \qquad i = -8 \text{ A}
-$$
-
-and *vo* = 48 V.
-
-**Answer:** 20 V, −10 V.
-
-For Example 2.7.
-
-Find *vo* and *io* in the circuit of Fig. 2.26.
-
-Practice Problem 2.7
-
-Example 2.8 Find currents and voltages in the circuit shown in Fig. 2.27(a).
-
-For Example 2.8.
-
-# **Solution:**
-
-We apply Ohm's law and Kirchhoff's laws. By Ohm's law,
-
-$$
-v_1 = 8i_1, \qquad v_2 = 3i_2, \qquad v_3 = 6i_3 \tag{2.8.1}
-$$
-
-Since the voltage and current of each resistor are related by Ohm's law as shown, we are really looking for three things: (*v*1, *v*2, *v*3) or (*i*1, *i*2, *i*3). At node *a*, KCL gives
-
-$$
-i_1 - i_2 - i_3 = 0 \tag{2.8.2}
-$$
-
-Applying KVL to loop 1 as in Fig. 2.27(b),
-
-$$
--30 + v_1 + v_2 = 0
-$$
-
-We express this in terms of *i*1 and *i*2 as in Eq. (2.8.1) to obtain
-
-$$
--30 + 8i_1 + 3i_2 = 0
-$$
-
-or
-
-$$
-i_1 = \frac{(30 - 3i_2)}{8} \tag{2.8.3}
-$$
-
-Applying KVL to loop 2,
-
-−*v*2 + *v*3 = 0 ⇒ *v*3 = *v*2 **(2.8.4)**
-
-as expected since the two resistors are in parallel. We express *v*1 and *v*2 in terms of *i*1 and *i*2 as in Eq. (2.8.1). Equation (2.8.4) becomes
-
-$$
-6i_3 = 3i_2
-$$
- $\Rightarrow$ $i_3 = \frac{i_2}{2}$ (2.8.5)
-
-Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives
-
-$$
-\frac{30 - 3i_2}{8} - i_2 - \frac{i_2}{2} = 0
-$$
-
-or *i*2 = 2 A. From the v alue of *i*2, we now use Eqs. (2.8.1) to (2.8.5) to obtain
-
-$$
-i_1 = 3
-$$
- A, $i_3 = 1$ A, $v_1 = 24$ V, $v_2 = 6$ V, $v_3 = 6$ V
-
-Find the currents and voltages in the circuit shown in Fig. 2.28. Practice Problem 2.8
-
-**Answer:**
-$$
-v_1 = 6
-$$
- V, $v_2 = 4$ V, $v_3 = 10$ V, $i_1 = 3$ A, $i_2 = 500$ mA, $i_3 = 2.5$ A.
-
-# **2.5** Series Resistors and Voltage Division
-
-The need to combine resistors in series or in parallel occurs so frequently that it warrants special attention. The process of combining the resistors is facilitated by combining two of them at a time. With this in mind, consider the single-loop circuit of Fig. 2.29. The two resistors are in series, since the same current *i* flows in both of them. Applying Ohm's law to each of the resistors, we obtain
-
-$$
-v_1 = iR_1, \qquad v_2 = iR_2 \tag{2.24}
-$$
-
-If we apply KVL to the loop (mo ving in the clockwise direction), we have
-
-$$
--v + v_1 + v_2 = 0 \tag{2.25}
-$$
-
-Combining Eqs. (2.24) and (2.25), we get
-
-$$
-v = v_1 + v_2 = i(R_1 + R_2)
-$$
- (2.26)
-
-*v* + – R1 *v*1 R2 *v*2 i + – + – a b
-
-# **Figure 2.29** A single-loop circuit with two resistors in series.
-
-or
-
-$$
-i = \frac{v}{R_1 + R_2} \tag{2.27}
-$$
-
-**Figure 2.28** For Practice Prob. 2.8.
-
-Notice that Eq. (2.26) can be written as
-
-$$
-v = iR_{\text{eq}} \tag{2.28}
-$$
-
-implying that the two resistors can be replaced by an equivalent resistor *R*eq; that is,
-
-$$
-R_{\text{eq}} = R_1 + R_2 \tag{2.29}
-$$
-
-Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30. The two circuits in Figs. 2.29 and 2.30 are equivalent because they exhibit the same voltage-current relationships at the terminals *a*-*b*. An equivalent circuit such as the one in Fig. 2.30 is useful in simplifying the analysis of a circuit. In general,
-
-The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances.
-
-For *N* resistors in series then,
-
-$$
-R_{\text{eq}} = R_1 + R_2 + \dots + R_N = \sum_{n=1}^{N} R_n \tag{2.30}
-$$
-
-To determine the voltage across each resistor in Fig. 2.29, we substitute Eq. (2.26) into Eq. (2.24) and obtain
-
-$$
-v_1 = \frac{R_1}{R_1 + R_2} v, \qquad v_2 = \frac{R_2}{R_1 + R_2} v
-$$
- (2.31)
-
-Notice that the source v oltage *v* is divided among the resistors in direct proportion to their resistances; the larger the resistance, the larger the voltage drop. This is called the *principle of voltage division*, and the circuit in Fig. 2.29 is called a *voltage divider*. In general, if a voltage divider has *N* resistors (*R*1, *R*2, … , *RN*) in series with the source voltage *v*, the *n*th resistor (*Rn*) will have a voltage drop of
-
-... ,
-$$
-R_N
-$$
-) in series with the source voltage *v*, the *n*th
-a voltage drop of
-$$
-v_n = \frac{R_n}{R_1 + R_2 + \dots + R_N} v
-$$
-(2.32)
-
-# **2.6** Parallel Resistors and Current Division
-
-Consider the circuit in Fig. 2.31, where tw o resistors are connected in parallel and therefore ha ve the same v oltage across them. From Ohm's law,
-
-$$
-v = i_1 R_1 = i_2 R_2
-$$
-
-$$
-i_1 = \frac{v}{R_1}
-$$
-, $i_2 = \frac{v}{R_2}$ (2.33)
-
-Applying KCL at node *a* gives the total current *i* as
-
-$$
-i = i_1 + i_2 \tag{2.34}
-$$
-
-Substituting Eq. (2.33) into Eq. (2.34), we get
-
-or
-
-$$
-i = \frac{v}{R_1} + \frac{v}{R_2} = v\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{v}{R_{\text{eq}}}
-$$
- (2.35)
-
-**Figure 2.31** Two resistors in parallel.
-
-Equivalent circuit of the Fig. 2.29 circuit.
-
- Resistors in series behave as a single resistor whose resistance is equal to the sum of the resistances of the
-
-individual resistors.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/022_2.5 Series Resistors and Voltage Division.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/022_2.5 Series Resistors and Voltage Division.md
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@@ -1,38 +0,0 @@
-**Figure 2.30**
-
-where *R*eq is the equivalent resistance of the resistors in parallel:
-
-$$
-\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \tag{2.36}
-$$
-
-or
-
-$$
-\frac{1}{R_{\text{eq}}} = \frac{R_1 + R_2}{R_1 R_2}
-$$
-
-or
-
-$$
-R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \tag{2.37}
-$$
-
-Thus,
-
-The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum.
-
-It must be emphasized that this applies only to tw o resistors in parallel. From Eq. (2.37), if *R*1 = *R*2 then *R*eq = *R*1/*R*2.
-
-We can extend the result in Eq. (2.36) to the general case of a circuit with *N* resistors in parallel. The equivalent resistance is
-
-$$
-\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_N}
-$$
- (2.38)
-
-Note that *R*eq is always smaller than the resistance of the smallest resistor in the parallel combination. If *R*1 = *R*2 = ⋯=*RN* = *R*, then
-
-$$
-R_{\text{eq}} = \frac{R}{N} \tag{2.39}
-$$
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/023_2.6 Parallel Resistors and Current Division.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/023_2.6 Parallel Resistors and Current Division.md
deleted file mode 100644
index 37b537be68764021d0d32c079ee3f01ecf891662..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/023_2.6 Parallel Resistors and Current Division.md
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@@ -1,308 +0,0 @@
-For e xample, if four 100- Ω resistors are connected in parallel, their equivalent resistance is 25 Ω.
-
-It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equi valent conductance for *N* resistors in parallel is
-
-$$
-G_{\text{eq}} = G_1 + G_2 + G_3 + \dots + G_N \quad (2.40)
-$$
-
-where *G*eq = 1/*R*eq, *G*1 = 1/*R*1, *G*2 = 1/*R*2, *G*3 = 1/ *R*3, … , *GN* = 1/ *RN*. Equation (2.40) states:
-
-The equivalent conductance of resistors connected in parallel is the sum of their individual conductances.
-
-This means that we may replace the circuit in Fig. 2.31 with that in Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The equivalent conductance of parallel resistors is obtained the same w ay as the equivalent resistance of series resistors. In the same manner , the equivalent conductance of resistors in series is obtained just the same Conductances in parallel behave as a single conductance whose value is equal to the sum of the individual conductances.
-
-**Figure 2.32** Equivalent circuit to Fig. 2.31.
-
-way as the resistance of resistors in parallel. Thus, the equivalent conductance *G*eq of *N* resistors in series (such as shown in Fig. 2.29) is
-
-$$
-\frac{1}{G_{\text{eq}}} = \frac{1}{G_1} + \frac{1}{G_2} + \frac{1}{G_3} + \dots + \frac{1}{G_N}
-$$
- (2.41)
-
-Given the total current *i* entering node *a* in Fig. 2.31, ho w do we obtain current *i*1 and *i*2? We know that the equi valent resistor has the same voltage, or
-
-$$
-v = iR_{\text{eq}} = \frac{iR_1R_2}{R_1 + R_2} \tag{2.42}
-$$
-
-Combining Eqs. (2.33) and (2.42) results in
-
-$$
-i_1 = \frac{R_2 i}{R_1 + R_2}, \qquad i_2 = \frac{R_1 i}{R_1 + R_2}
-$$
- (2.43)
-
-which shows that the total current *i* is shared by the resistors in in verse proportion to their resistances. This is known as the *principle of current division*, and the circuit in Fig. 2.31 is known as a *current divider*. Notice that the lar ger current flows through the smaller re sistance.
-
-As an extreme case, suppose one of the resistors in Fig. 2.31 is zero, say *R*2 = 0; that is, *R*2 is a short circuit, as sho wn in Fig. 2.33(a). From Eq. (2.43), *R*2 = 0 implies that *i*1 = 0, *i*2 = *i*. This means that the entire current *i* bypasses *R*1 and flows through the short circuit *R*2 = 0, the path of least resistance. Thus when a circuit is short circuited, as sho wn in Fig. 2.33(a), two things should be kept in mind:
-
-- 1. The equivalent resistance *R*eq = 0. [See what happens when *R*2 = 0 in Eq. (2.37).]
-- 2. The entire current flows through the short circuit.
-
-As another e xtreme case, suppose *R*2 = ∞, that is, *R*2 is an open circuit, as shown in Fig. 2.33(b). The current still flows through the path of least resistance, *R*1. By taking the limit of Eq. (2.37) as *R*2 → ∞,we obtain *R*eq = *R*1 in this case.
-
-If we divide both the numerator and denominator by *R*1*R*2, Eq. (2.43) becomes
-
-$$
-i_1 = \frac{G_1}{G_1 + G_2} i \tag{2.44a}
-$$
-
-$$
-i_2 = \frac{G_2}{G_1 + G_2} i \tag{2.44b}
-$$
-
-Thus, in general, if a current divider has *N* conductors (*G*1, *G*2, … , *GN*) in parallel with the source current *i*, the *n*th conductor (*Gn*) will have current
-
-e current *i*, the *n*th conductor (
-$$
-G_n
-$$
-) will have current
-\n
-$$
-i_n = \frac{G_n}{G_1 + G_2 + \dots + G_N} i
-$$
-\n(2.45)
-
-**Figure 2.33** (a) A shorted circuit, (b) an open circuit.
-
-In general, it is often convenient and possible to combine resistors in series and parallel and reduce a resisti ve network to a single *equivalent resistance R*eq. Such an equi valent resistance is the resistance between the designated terminals of the netw ork and must e xhibit the same *i*-*v* characteristics as the original network at the terminals.
-
-Find *R*eq for the circuit shown in Fig. 2.34. Example 2.9
-
-# **Solution:**
-
-To get *R*eq, we combine resistors in series and in parallel. The 6- Ω and 3-Ω resistors are in parallel, so their equivalent resistance is
-
-$$
-6 \Omega \parallel 3\Omega = \frac{6 \times 3}{6 + 3} = 2 \Omega
-$$
-
-(The symbol ∥ is used to indicate a parallel combination.) Also, the 1-Ω and 5-Ω resistors are in series; hence their equivalent resistance is
-
-1 Ω + 5 Ω = 6 Ω
-
-Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the tw o 2-Ω resistors are in series, so the equivalent resistance is
-
-$$
-2\Omega + 2\Omega = 4\Omega
-$$
-
-This 4-Ω resistor is now in parallel with the 6-Ω resistor in Fig. 2.35(a); their equivalent resistance is
-
-$$
-4 \Omega || 6 \Omega = \frac{4 \times 6}{4 + 6} = 2.4 \Omega
-$$
-
-The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.35(b), the three resistors are in series. Hence, the equivalent resistance for the circuit is
-
-$$
-R_{\text{eq}} = 4 \ \Omega + 2.4 \ \Omega + 8 \ \Omega = 14.4 \ \Omega
-$$
-
-By combining the resistors in Fig. 2.36, find *R*eq. Practice Problem 2.9
-
-**Answer:** 11 Ω.
-
-For Practice Prob. 2.9.
-
-Calculate the equivalent resistance *Rab* Example 2.10 in the circuit in Fig. 2.37.
-
-# For Example 2.10.
-
-# **Solution:**
-
-The 3-Ω and 6-Ω resistors are in parallel because they are connected to the same two nodes *c* and *b*. Their combined resistance is
-
-$$
-3 \Omega \parallel 6 \Omega = \frac{3 \times 6}{3 + 6} = 2 \Omega
-$$
- (2.10.1)
-
-Similarly, the 12-Ω and 4-Ω resistors are in parallel since the y are connected to the same two nodes *d* and *b*. Hence
-
-$$
-12 \Omega || 4 \Omega = \frac{12 \times 4}{12 + 4} = 3 \Omega
-$$
- (2.10.2)
-
-Also the 1-Ω and 5-Ω resistors are in series; hence, their equivalent re sistance is
-
-$$
-1 \Omega + 5 \Omega = 6 \Omega \tag{2.10.3}
-$$
-
-With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig. 2.38(a), 3-Ω in parallel with 6-Ω gives 2-Ω, as calculated in Eq. (2.10.1). This 2-Ω equivalent resistance is now in series with the 1-Ω resistance to give a combined resistance of 1 Ω +2Ω=3Ω. Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In Fig. 2.38(b), we combine the 2-Ω and 3-Ω resistors in parallel to get
-
-$$
-2 \Omega || 3 \Omega = \frac{2 \times 3}{2 + 3} = 1.2 \Omega
-$$
-
-This 1.2-Ω resistor is in series with the 10-Ω resistor, so that
-
-$$
-R_{ab} = 10 + 1.2 = 11.2 \ \Omega
-$$
-
-Find *Rab* for the circuit in Fig. 2.39.
-
-**Answer:** 19 Ω.
-
-For Practice Prob. 2.10.
-
-Find the equivalent conductance *G*eq for the circuit in Fig. 2.40(a).
-
-# **Solution:**
-
-The 8-S and 12-S resistors are in parallel, so their conductance is
-
-$$
-8 S + 12 S = 20 S
-$$
-
-This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b) so that the combined conductance is
-
-$$
-\frac{20 \times 5}{20 + 5} = 4
-$$
-
-This is in parallel with the 6-S resistor. Hence,
-
-$$
-G_{\text{eq}} = 6 + 4 = 10 \text{ S}
-$$
-
-We should note that the circuit in Fig. 2.40(a) is the same as that in Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in siemens, those in Fig. 2.40(c) are expressed in ohms. To show that the circuits are the same, we find *R*eq for the circuit in Fig. 2.40(c).
-
-$$
-R_{\text{eq}} = \frac{1}{6} \left\| \left( \frac{1}{5} + \frac{1}{8} \right) \frac{1}{12} \right\| = \frac{1}{6} \left\| \left( \frac{1}{5} + \frac{1}{20} \right) \right\| = \frac{1}{6} \left\| \frac{1}{4} \right\|
-$$
-$$
-= \frac{\frac{1}{6} \times \frac{1}{4}}{\frac{1}{6} + \frac{1}{4}} = \frac{1}{10} \Omega
-$$
-$$
-G_{\text{eq}} = \frac{1}{R_{\text{eq}}} = 10 \text{ S}
-$$
-
-This is the same as we obtained previously.
-
-Calculate *G*eq in the circuit of Fig. 2.41. Practice Problem 2.11
-
-# **Answer:** 8 S.
-
-For Practice Prob. 2.11.
-
-Find *io* and *vo* in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3-Ω resistor.
-
-# **Solution:**
-
-The 6-Ω and 3-Ω resistors are in parallel, so their combined resistance is
-
-$$
-6 \Omega \parallel 3 \Omega = \frac{6 \times 3}{6 + 3} = 2 \Omega
-$$
-
-Thus, our circuit reduces to that shown in Fig. 2.42(b). Notice that *vo* is not affected by the combination of the resistors because the resistors are
-
-# Example 2.11
-
-# **Figure 2.40**
-
-For Example 2.11: (a) original circuit, (b) its equivalent circuit, (c) same circuit as in (a) but resistors are expressed in ohms.
-
-Example 2.12
-
-For Example 2.12: (a) original circuit, (b) its equivalent circuit.
-
-in parallel and therefore have the same voltage *vo*. From Fig. 2.42(b), we can obtain *vo* in two ways. One way is to apply Ohm's law to get
-
-$$
-i = \frac{12}{4+2} = 2 \text{ A}
-$$
-
-and hence, *vo* = 2*i* = 2 × 2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4-Ω and 2-Ω resistors. Hence,
-
-$$
-v_o = \frac{2}{2+4} (12 \text{ V}) = 4 \text{ V}
-$$
-
-Similarly, *io* can be obtained in two ways. One approach is to apply Ohm's law to the 3-Ω resistor in Fig. 2.42(a) now that we know *vo*; thus,
-
-$$
-v_o = 3i_o = 4 \qquad \Rightarrow \qquad i_o = \frac{4}{3} \text{ A}
-$$
-
-Another approach is to apply current division to the circuit in Fig. 2.42(a) now that we know *i*, by writing
-
-$$
-i_o = \frac{6}{6+3} i = \frac{2}{3} (2 \text{ A}) = \frac{4}{3} \text{ A}
-$$
-
-The power dissipated in the 3-Ω resistor is
-
-$$
-p_o = v_o i_o = 4 \left(\frac{4}{3}\right) = 5.333 \text{ W}
-$$
-
-Find *v*1 and *v*2 in the circuit sho wn in Fig. 2.43. Also calculate *i*1 and *i*2 and the power dissipated in the 12-Ω and 40-Ω resistors.
-
-**Answer:** *v*1 = 10 V, *i*1 = 833.3 mA, *p*1 = 8.333 W, *v*2 = 20 V, *i*2 =500 mA, *p*2 = 10 W.
-
-Practice Problem 2.12
-
-# For Practice Prob. 2.12.
-
-**Figure 2.43**
-
-# Example 2.13
-
-For the circuit sho wn in Fig. 2.44(a), determine: (a) the v oltage *vo*, (b) the power supplied by the current source, (c) the power absorbed by each resistor.
-
-# **Solution:**
-
-(a) The 6-k Ω and 12-k Ω resistors are in series so that their combined value is 6 + 12 = 18 kΩ. Thus the circuit in Fig. 2.44(a) reduces to that shown in Fig. 2.44(b). We now apply the current division technique to find *i*1 and *i*2.
-
-$$
-i_1 = \frac{18,000}{9,000 + 18,000} (30 \text{ mA}) = 20 \text{ mA}
-$$
-$$
-i_2 = \frac{9,000}{9,000 + 18,000} (30 \text{ mA}) = 10 \text{ mA}
-$$
-
-Notice that the voltage across the 9-kΩ and 18-kΩ resistors is the same, and *vo* = 9,000*i*1 = 18,000*i*2 = 180 V, as expected. (b) Power supplied by the source is
-
-$$
-p_o = v_o i_o = 180(30) \text{ mW} = 5.4 \text{ W}
-$$
-
-(c) Power absorbed by the 12-kΩ resistor is
-
-$$
-p = iv = i_2(i_2 R) = i_2^2 R = (10 \times 10^{-3})^2 (12,000) = 1.2 W
-$$
-
-Power absorbed by the 6-kΩ resistor is
-
-$$
-p = i_2^2 R = (10 \times 10^{-3})^2 (6,000) = 0.6 W
-$$
-
-Power absorbed by the 9-kΩ resistor is
-
-$$
-p = \frac{v_o^2}{R} = \frac{(180)^2}{9,000} = 3.6 \text{ W}
-$$
-
-or
-
-$$
-p = v_o i_1 = 180(20) \text{ mW} = 3.6 \text{ W}
-$$
-
-Notice that the power supplied (5.4 W) equals the power absorbed (1.2 + 0.6 + 3.6 = 5.4 W). This is one way of checking results.
-
-For the circuit shown in Fig. 2.45, find: (a) *v*1 and *v*2, (b) the power dis- Practice Problem 2.13 sipated in the 3-k Ω and 20-kΩ resistors, and (c) the power supplied by the current source.
-
-**Answer:** (a) 135 V, 180 V, (b) 2.025 W, 540 mW, (c) 5.4 W.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/024_2.7 Wye-Delta Transformations.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/024_2.7 Wye-Delta Transformations.md
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-# **2.7** Wye-Delta Transformations
-
-Situations often arise in circuit analysis when the resistors are neither in parallel nor in series. For example, consider the bridge circuit in Fig. 2.46. How do we combine resistors *R*1 through *R*6 when the resistors are neither in series nor in parallel? Man y circuits of the type sho wn in Fig. 2.46 can be simplified by using three-terminal equivalent networks. These are the wye (Y) or tee (T) netw ork shown in Fig. 2.47 and the delta (Δ) or pi (Π) network shown in Fig. 2.48. These networks occur by themselves or as part of a lar ger network. They are used in three-phase networks, electrical filters, and matching networks. Our main interest
-
-**Figure 2.46** The bridge network.
-
-Two forms of the same network: (a) Y, (b) T.
-
-here is in how to identify them when they occur as part of a network and how to apply wye-delta transformation in the analysis of that network.
-
-# Delta to Wye Conversion
-
-Suppose it is more convenient to work with a wye network in a place where the circuit contains a delta configuration. We superimpose a wye network on the existing delta network and find the equivalent resistances in the wye network. To obtain the equivalent resistances in the wye network, we compare the tw o networks and mak e sure that the resistance between each pair of nodes in the Δ (or Π) network is the same as the resistance between the same pair of nodes in the Y (or T) network. For terminals 1 and 2 in Figs. 2.47 and 2.48, for example,
-
-$$
-R_{12}(Y) = R_1 + R_3 \tag{2.46}
-$$
-
-$$
-R_{12}(\Delta) = R_b || (R_a + R_c)
-$$
-
-# **Figure 2.48**
-
-Two forms of the same network: (a) Δ, (b) Π.
-
-Setting
-$$
-R_{12}(Y) = R_{12}(\Delta)
-$$
- gives
-
-$$
-R_{12} = R_1 + R_3 = \frac{R_b (R_a + R_c)}{R_a + R_b + R_c}
-$$
- (2.47a)
-
-Similarly,
-
-$$
-R_{13} = R_1 + R_2 = \frac{R_c (R_a + R_b)}{R_a + R_b + R_c}
-$$
- (2.47b)
-
-$$
-R_{34} = R_2 + R_3 = \frac{R_a (R_b + R_c)}{R_a + R_b + R_c}
-$$
- (2.47c)
-
-Subtracting Eq. (2.47c) from Eq. (2.47a), we get
-
-$$
-R_1 - R_2 = \frac{R_c (R_b - R_a)}{R_a + R_b + R_c}
-$$
- (2.48)
-
-Adding Eqs. (2.47b) and (2.48) gives
-
-$$
-R_1 = \frac{R_b R_c}{R_a + R_b + R_c} \tag{2.49}
-$$
-
-and subtracting Eq. (2.48) from Eq. (2.47b) yields
-
-$$
-R_2 = \frac{R_c R_a}{R_a + R_b + R_c} \tag{2.50}
-$$
-
-Subtracting Eq. (2.49) from Eq. (2.47a), we obtain
-
-$$
-R_3 = \frac{R_a R_b}{R_a + R_b + R_c}
-$$
- (2.51)
-
-We do not need to memorize Eqs. (2.49) to (2.51). To transform a ∆ network to Y, we create an extra node *n* as shown in Fig. 2.49 and follow this conversion rule:
-
-Each resistor in the Y network is the product of the resistors in the two adjacent Δ branches, divided by the sum of the three Δ resistors.
-
-One can follow this rule and obtain Eqs. (2.49) to (2.51) from Fig. 2.49.
-
-# Wye to Delta Conversion
-
-To obtain the conversion formulas for transforming a wye network to an equivalent delta network, we note from Eqs. (2.49) to (2.51) that
-
-the conversion formulas for transforming a wye network to an
-it delta network, we note from Eqs. (2.49) to (2.51) that
-$$
-R_1 R_2 + R_2 R_3 + R_3 R_1 = \frac{R_a R_b R_c (R_a + R_b + R_c)}{(R_a + R_b + R_c)^2}
-$$
-$$
-= \frac{R_a R_b R_c}{R_a + R_b + R_c}
-$$
-(2.52)
-
-Dividing Eq. (2.52) by each of Eqs. (2.49) to (2.51) leads to the follo wing equations:
-
-$$
-R_a = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1}
-$$
- (2.53)
-
-$$
-R_1
-$$
-\n
-$$
-R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2}
-$$
-\n(2.54)
-
-$$
-R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3}
-$$
-\n(2.55)
-
-From Eqs. (2.53) to (2.55) and Fig. 2.49, the con version rule for Y to Δ is as follows:
-
-Each resistor in the Δ network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor.
-
-The Y and Δ networks are said to be *balanced* when
-
-$$
-R_1 = R_2 = R_3 = R_Y, \qquad R_a = R_b = R_c = R_\Delta \tag{2.56}
-$$
-
-Under these conditions, conversion formulas become
-
-$$
-R_{\rm Y} = \frac{R_{\rm \Delta}}{3} \qquad \text{or} \qquad R_{\rm \Delta} = 3R_{\rm Y} \tag{2.57}
-$$
-
-One may w onder wh y *R*Y is less than *R*Δ. Well, we notice that the Y-connection is like a "series" connection while the Δ-connection is like a "parallel" connection.
-
-Note that in making the transformation, we do not take anything out of the circuit or put in anything new. We are merely substituting different but mathematically equivalent three-terminal network patterns to create a circuit in which resistors are either in series or in parallel, allo wing us to calculate *R*eq if necessary.
-
-# **Figure 2.49** Superposition of Y and Δ networks as an aid in transforming one to the other.
-
-Example 2.14 Convert the Δ network in Fig. 2.50(a) to an equivalent Y network.
-
-**Figure 2.50** For Example 2.14: (a) original Δ network, (b) Y equivalent network.
-
-# **Solution:**
-
-Using Eqs. (2.49) to (2.51), we obtain
-
-$$
-R_1 = \frac{R_b R_c}{R_a + R_b + R_c} = \frac{10 \times 25}{15 + 10 + 25} = \frac{250}{50} = 5 \text{ }\Omega
-$$
-\n
-$$
-R_2 = \frac{R_c R_a}{R_a + R_b + R_c} = \frac{25 \times 15}{50} = 7.5 \text{ }\Omega
-$$
-\n
-$$
-R_3 = \frac{R_a R_b}{R_a + R_b + R_c} = \frac{15 \times 10}{50} = 3 \text{ }\Omega
-$$
-
-The equivalent Y network is shown in Fig. 2.50(b).
-
-Practice Problem 2.14 Transform the wye network in Fig. 2.51 to a delta network.
-
-**Answer:** *Ra* = 140 Ω, *Rb* = 70 Ω, *Rc* = 35 Ω.
-
-**Figure 2.51** For Practice Prob. 2.14.
-
-Example 2.15
-
-Obtain the equivalent resistance *Rab* for the circuit in Fig. 2.52 and use it to find current *i*.
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly defined. Please note, this part normally will deservedly take much more time.
-- 2. **Present.** Clearly, when we remove the voltage source, we end up with a purely resistive circuit. Since it is composed of deltas and wyes, we have a more comple x process of combining the elements together .
-
-We can use wye-delta transformations as one approach to find a solution. It is useful to locate the wyes (there are tw o of them, one at *n* and the other at *c*) and the deltas (there are three: *can, abn, cnb*).
-
-3. **Alternative.** There are different approaches that can be used to solve this problem. Since the focus of Sec. 2.7 is the wye-delta transfor mation, this should be the technique to use. Another approach would be to solve for the equi valent resistance by injecting one amp into the circuit and finding the voltage between *a* and *b*; we will learn about this approach in Chap. 4.
-
-The approach we can apply here as a check w ould be to use a wye-delta transformation as the first solution to the problem. Later we can check the solution by starting with a delta-wye transformation.
-
-4. **Attempt.** In this circuit, there are two Y networks and three Δ networks. Transforming just one of these will simplify the circuit. If we convert the Y network comprising the 5-Ω, 10-Ω, and 20-Ω resistors, we may select
-
-$$
-R_1 = 10 \Omega, \qquad R_2 = 20 \Omega, \qquad R_3 = 5 \Omega
-$$
-
-Thus from Eqs. (2.53) to (2.55) we have
-
-$$
-R_1 = 16 \text{ cm}, \quad R_2 = 26 \text{ cm}, \quad R_3 = 5 \text{ cm}
-$$
-\n
-$$
-R_a = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1} = \frac{10 \times 20 + 20 \times 5 + 5 \times 10}{10}
-$$
-\n
-$$
-= \frac{350}{10} = 35 \text{ }\Omega
-$$
-\n
-$$
-R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2} = \frac{350}{20} = 17.5 \text{ }\Omega
-$$
-\n
-$$
-R_c = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} = \frac{350}{5} = 70 \text{ }\Omega
-$$
-
-With the Y converted to Δ, the equivalent circuit (with the voltage source removed for now) is shown in Fig. 2.53(a). Combining the three pairs of resistors in parallel, we obtain
-
-**Figure 2.53** Equivalent circuits to Fig. 2.52, with the voltage source removed.
-
-so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find
-
-$$
-R_{ab} = (7.292 + 10.5) \| 21 = \frac{17.792 \times 21}{17.792 + 21} = 9.632 \Omega
-$$
-
-Then
-
-$$
-i = \frac{v_s}{R_{ab}} = \frac{120}{9.632} = 12.458 \text{ A}
-$$
-
-We observe that we have successfully solved the problem. Now we must evaluate the solution.
-
-5. **Evaluate.** Now we must determine if the answer is correct and then evaluate the final solution.
-
-It is relatively easy to check the answer; we do this by solving the problem starting with a delta-wye transformation. Let us trans form the delta, *can*, into a wye.
-
-Let *Rc* = 10 Ω, *Ra* = 5 Ω, and *Rn* = 12.5 Ω. This will lead to (let *d* represent the middle of the wye):
-
-$$
-R_{ad} = \frac{R_c R_n}{R_a + R_c + R_n} = \frac{10 \times 12.5}{5 + 10 + 12.5} = 4.545 \ \Omega
-$$
-$$
-R_{cd} = \frac{R_a R_n}{27.5} = \frac{5 \times 12.5}{27.5} = 2.273 \ \Omega
-$$
-$$
-R_{nd} = \frac{R_a R_c}{27.5} = \frac{5 \times 10}{27.5} = 1.8182 \ \Omega
-$$
-
-This now leads to the circuit sho wn in Figure 2.53(c). Looking at the resistance between *d* and *b*, we have two series combination in parallel, giving us
-
-parallel, giving us
-\n
-$$
-R_{db} = \frac{(2.273 + 15)(1.8182 + 20)}{2.273 + 15 + 1.8182 + 20} = \frac{376.9}{39.09} = 9.642 \ \Omega
-$$
-
-This is in series with the 4.545-Ω resistor, both of which are in parallel with the 30-Ω resistor. This then gives us the equivalent resistance of the circuit.
-
-tance of the circuit.
-\n
-$$
-R_{ab} = \frac{(9.642 + 4.545)30}{9.642 + 4.545 + 30} = \frac{425.6}{44.19} = 9.631 \ \Omega
-$$
-
-This now leads to
-
-$$
-i = \frac{v_s}{R_{ab}} = \frac{120}{9.631} = 12.46 \text{ A}
-$$
-
-We note that using tw o variations on the wye-delta transformation leads to the same results. This represents a very good check.
-
-6. **Satisfactory?** Since we have found the desired answer by deter mining the equi valent resistance of the circuit first and the answer checks, then we clearly ha ve a satisfactory solution. This represents what can be presented to the indi vidual assigning the problem.
-
-For the bridge network in Fig. 2.54, find *Rab* and *i*.
-
-**Answer:** 60 Ω, 4 A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/025_2.8 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/025_2.8 Applications.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/025_2.8 Applications.md
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-# **2.8** Applications
-
-Resistors are often used to model de vices that con vert electrical ener gy into heat or other forms of energy. Such devices include conducting wire, light bulbs, electric heaters, stoves, ovens, and loudspeakers. In this section, we will consider two real-life problems that apply the concepts developed in this chapter: electrical lighting systems and design of dc meters.
-
-# 2.8.1 Lighting Systems
-
-Lighting systems, such as in a house or on a Christmas tree, often consist of *N* lamps connected either in parallel or in series, as sho wn in Fig. 2.55. Each lamp is modeled as a resistor . Assuming that all the lamps are identical and *Vo* is the power-line voltage, the voltage across each lamp is *Vo* for the parallel connection and *Vo* /*N* for the series connection. The series connection is easy to manufa cture but is seldom used in practice, for at least two reasons. First, it is less reliable; when a lamp fa ils, all the lamps go out. Second, it is harder to maintain; when a lamp is bad, one must test all the lamps one by one to detect the faulty one.
-
-**Figure 2.54** For Practice Prob. 2.15.
-
-So far, we have assumed that connecting wires are perfect conductors (i.e., conductors of zero resistance). In real physical systems, however, the resistance of the connecting wire may be appreciably large, and the modeling of the system must include that resistance.
-
-# Historical
-
-**Thomas Alva Edison** (1847–1931) was perhaps the greatest American inventor. He patented 1093 inventions, including such history-making inventions as the incandescent electric bulb, the phonograph, and the first commercial motion pictures.
-
-Born in Milan, Ohio, the youngest of seven children, Edison received only three months of formal education because he hated school. He was home-schooled by his mother and quickly began to read on his own. In 1868, Edison read one of Faraday's books and found his calling. He moved to Menlo Park, New Jersey, in 1876, where he man aged a well-staffed research laboratory. Most of his inventions came out of this laboratory. His laboratory served as a model for modern research organ iza tions. Because of his diverse interests and the over whelming number of his inventions and patents, Edison began to estab lish manufacturing companies for making the devices he invented. He designed the first electric power station to supply electric light. Formal electrical engineering education began in the mid-1880s with Edison as a role model and leader.
-
-Library of Congress
-
-(a) Parallel connection of light bulbs, (b) series connection of light bulbs.
-
-Example 2.16
-
-Three light bulbs are connected to a 9-V battery as shown in Fig. 2.56(a). Calculate: (a) the total current supplied by the battery , (b) the current through each bulb, (c) the resistance of each bulb.
-
-# **Solution:**
-
-(a) The total po wer supplied by the battery is equal to the total po wer absorbed by the bulbs; that is,
-
-$$
-p = 15 + 10 + 20 = 45
-$$
- W
-
-Since *p* = *V I,* then the total current supplied by the battery is
-
-$$
-I = \frac{p}{V} = \frac{45}{9} = 5 \text{ A}
-$$
-
-(b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since *R*1 (20-W bulb) is in parallel with the battery as well as the series com bination of *R*2 and *R*3,
-
-$$
-V_1 = V_2 + V_3 = 9 \text{ V}
-$$
-
-The current through *R*1 is
-
-$$
-I_1 = \frac{p_1}{V_1} = \frac{20}{9} = 2.222 \text{ A}
-$$
-
-By KCL, the current through the series combination of *R*2 and *R*3 is
-
-$$
-I_2 = I - I_1 = 5 - 2.222 = 2.778 \text{ A}
-$$
-
-(c) Since *p* = *I* 2 *R*,
-
-$$
-R_1 = \frac{p_1}{l_1^2} = \frac{20}{2.222^2} = 4.05 \ \Omega
-$$
-
-\n
-$$
-R_2 = \frac{p_2}{l_2^2} = \frac{15}{2.777^2} = 1.945 \ \Omega
-$$
-
-\n
-$$
-R_3 = \frac{p_3}{l_3^2} = \frac{10}{2.777^2} = 1.297 \ \Omega
-$$
-
-Refer to Fig. 2.55 and assume there are six light b ulbs that can be con nected in parallel and six dif ferent light bulbs that can be connected in series. In either case, each light bulb is to operate at 40 W. If the voltage at the plug is 115 V for the parallel and series connections, calculate the current through and the voltage across each bulb for both cases.
-
-**Answer:** 115 V and 347.8 mA (parallel), 19.167 V and 2.087 A (series).
-
-# 2.8.2 Design of DC Meters
-
-By their nature, resistors are used to control the flow of current. We take advantage of this property in se veral applications, such as in a poten tiometer (Fig. 2.57). The word *potentiometer*, derived from the w ords *potential* and *meter*, implies that potential can be metered out. The potentiometer (or pot for short) is a three-terminal de vice that operates on the principle of v oltage division. It is essentially an adjustable v oltage divider. As a voltage regulator, it is used as a volume or level control on radios, TVs, and other devices. In Fig. 2.57,
-
-$$
-V_{\text{out}} = V_{bc} = \frac{R_{bc}}{R_{ac}} V_{\text{in}}
-$$
- (2.58)
-
-where *Rac* = *Rab* + *Rbc*. Thus, *V*out decreases or increases as the sliding contact of the pot moves toward *c* or *a*, respectively.
-
-Another application where resistors are used to control current flow is in the analog dc meters—the ammeter, voltmeter, and ohmmeter, which measure current, voltage, and resistance, respectively. Each of these meters employs the d'Arsonval meter movement, shown in Fig. 2.58. The movement consists essentially of a movable iron-core coil mounted on a pivot between the poles of a permanent magnet. When current flows through the coil, it creates a torque which causes the pointer to deflect. The amount of current through the coil determines the deflection of the pointer, which is registered on a scale attached to the meter movement. For example, if the meter movement is rated 1 mA, 50 Ω, it w ould take 1 mA to cause a full-scale deflection of the meter movement. By introducing additional circuitry to the d'Arsonval meter mo vement, an am meter, voltmeter, or ohmmeter can be constructed.
-
-Consider Fig. 2.59, where an analog v oltmeter and ammeter are con nected to an element. The voltmeter measures the voltage across a *load* and
-
-Practice Problem 2.16
-
-**Figure 2.57** The potentiometer controlling potential levels.
-
- An instrument capable of measuring voltage, current, and resistance is called a multimeter or a volt-ohm meter (VOM).
-
- A load is a component that is receiving energy (an energy sink), as opposed to a generator supplying energy (an energy source). More about loading will be discussed in Section 4.9.1.
-
-A d'Arsonval meter movement.
-
-Connection of a voltmeter and an ammeter to an element.
-
-is therefore connected in parallel with the element. As shown in Fig. 2.60(a), the voltmeter consists of a d'Arson val movement in series with a resistor whose resistance *Rm* is deliberately made very large (theoretically, infinite), to minimize the current drawn from the circuit. To extend the range of voltage that the meter can measure, series multiplier resistors are often connected with the voltmeters, as shown in Fig. 2.60(b). The multiple-range voltmeter in Fig. 2.60(b) can measure v oltage from 0 to 1 V, 0 to 10 V, or 0 to 100 V, depending on whether the switch is connected to *R*1, *R*2, or *R*3, respectively.
-
-Let us calculate the multiplier resistor *Rn* for the single-range voltmeter in Fig. 2.60(a), or *Rn* = *R*1, *R*2, or *R*3 for the multiple-range voltmeter in Fig. 2.60(b). We need to determine the value of *Rn* to be connected in series with the internal resistance *Rm* of the voltmeter. In any design, we consider the worst-case condition. In this case, the worst case occurs when the fullscale current *I*fs =*Im* flows through the meter. This should also correspond
-
-**Figure 2.60** Voltmeters: (a) single-range type, (b) multiple-range type.
-
-to the maximum v oltage reading or the full-scale v oltage *V*fs. Since the multiplier resistance *Rn* is in series with the internal resistance *Rm*,
-
-$$
-V_{\text{fs}} = I_{\text{fs}} (R_n + R_m) \tag{2.59}
-$$
-
-From this, we obtain
-
-$$
-R_n = \frac{V_{\text{fs}}}{I_{\text{fs}}} - R_m \tag{2.60}
-$$
-
-Similarly, the ammeter measures the current through the load and is connected in series with it. As shown in Fig. 2.61(a), the ammeter consists of a d'Arsonval movement in parallel with a resistor whose resistance *Rm* is deliberately made very small (theoretically, zero) to minimize the v oltage drop across it. To allow multiple ranges, shunt resistors are often connected in parallel with *Rm* as sho wn in Fig. 2.61(b). The shunt resistors allow the meter to measure in the range 0–10 mA, 0–100 mA, or 0–1 A, depending on whether the switch is connected to *R*1, *R*2, or *R*3, respectively.
-
-Now our objective is to obtain the multiplier shunt *Rn* for the singlerange ammeter in Fig. 2.61(a), or *Rn* =*R*1, *R*2, or *R*3 for the multiple-range ammeter in Fig. 2.61(b). We notice that *Rm* and *Rn* are in parallel and that at full-scale reading *I* = *I*fs = *Im* + *In*, where *In* is the current through the shunt resistor *Rn*. Applying the current division principle yields
-
-$$
-I_m = \frac{R_n}{R_n + R_m} I_{\text{fs}}
-$$
-
-$$
-\sum_{i=1}^{n} x_i
-$$
-
-$$
-R_n = \frac{I_m}{I_{\text{fs}} - I_m} R_m \tag{2.61}
-$$
-
-The resistance *Rx* of a linear resistor can be measured in tw o ways. An indirect way is to measure the current *I* that flows through it by connecting an ammeter in series with it and the v oltage *V* across it by con necting a voltmeter in parallel with it, as shown in Fig. 2.62(a). Then
-
-$$
-R_x = \frac{V}{I} \tag{2.62}
-$$
-
-The direct method of measuring resistance is to use an ohmmeter . An ohmmeter consists basically of a d'Arson val movement, a v ariable resistor or potentiometer, and a battery, as shown in Fig. 2.62(b). Applying KVL to the circuit in Fig. 2.62(b) gives
-
-$$
-E = (R + R_m + R_x)I_m
-$$
-
-or
-
-$$
-R_x = \frac{E}{I_m} - (R + R_m)
-$$
- (2.63)
-
-The resistor *R* is selected such that the meter gives a full-scale deflection; that is, *Im* = *I*fs when *Rx* = 0. This implies that
-
-$$
-E = (R + R_m) I_{\text{fs}} \tag{2.64}
-$$
-
-Substituting Eq. (2.64) into Eq. (2.63) leads to
-
-$$
-R_x = \left(\frac{I_{\text{fs}}}{I_m} - 1\right)(R + R_m) \tag{2.65}
-$$
-
-As mentioned, the types of meters we ha ve discussed are kno wn as *analog* meters and are based on the d'Arsonval meter movement. Another type of meter , called a *digital meter*, is based on acti ve circuit elements
-
-# **Figure 2.61**
-
-Ammeters: (a) single-range type, (b) multiple-range type.
-
-# **Figure 2.62** Two ways of measuring resistance: (a) using an ammeter and a voltmeter, (b) using an ohmmeter.
-
-# Historical
-
-Library of Congress
-
-**Samuel F. B. Morse** (1791–1872), an American painter, invented the telegraph, the first practical, commercialized application of electricity.
-
-Morse was born in Charlestown, Massachusetts, and studied at Yale and the Royal Academy of Arts in London to become an artist. In the 1830s, he became intrigued with developing a telegraph. He had a working model by 1836 and applied for a patent in 1838. The U.S. Senate appro priated funds for Morse to construct a telegraph line between Baltimore and Washington, D.C. On May 24, 1844, he sent the famous first message: "What hath God wrought!" Morse also developed a code of dots and dashes for letters and numbers, for sending messages on the telegraph. The development of the telegraph led to the invention of the telephone.
-
-such as op amps. For example, a digital multimeter displays measurements of dc or ac voltage, current, and resistance as discrete numbers, instead of using a pointer deflection on a continuous scale as in an analog multimeter. Digital meters are what you would most likely use in a modern lab. However, the design of digital meters is beyond the scope of this book.
-
-Example 2.17 Following the v oltmeter setup of Fig. 2.60, design a v oltmeter for the following multiple ranges:
-
-> (a) 0–1 V (b) 0–5 V (c) 0–50 V (d) 0–100 V Assume that the internal resistance *Rm* = 2 kΩ and the full-scale current *I*fs =100*μ*A.
-
-# **Solution:**
-
-We apply Eq. (2.60) and assume that *R*1, *R*2, *R*3, and *R*4 correspond with ranges 0–1 V, 0–5 V, 0–50 V, and 0–100 V, respectively. (a) For range 0–1 V,
-
-$$
-R_1 = \frac{1}{100 \times 10^{-6}} - 2000 = 10,000 - 2000 = 8 \text{ k}\Omega
-$$
-
-(b) For range 0–5 V,
-
-$$
-R_2 = \frac{5}{100 \times 10^{-6}} - 2000 = 50,000 - 2000 = 48 \text{ k}\Omega
-$$
-
-(c) For range 0–50 V,
-
-$$
-R_3 = \frac{50}{100 \times 10^{-6}} - 2000 = 500,000 - 2000 = 498 \text{ k}\Omega
-$$
-
-(d) For range 0–100 V,
-
-$$
-R_4 = \frac{100 \text{ V}}{100 \times 10^{-6}} - 2000 = 1,000,000 - 2000 = 998 \text{ k}\Omega
-$$
-
-Note that the ratio of the total resistance (*Rn* + *Rm*) to the full-scale voltage *V*fs is constant and equal to 1/*I*fs for the four ranges. This ratio (given in ohms per v olt, or Ω/ V) is known as the *sensitivity* of the v oltmeter. The larger the sensitivity, the better the voltmeter.
-
-Following the ammeter setup of Fig. 2.61, design an ammeter for the following multiple ranges: (a) 0–1 A (b) 0–100 mA (c) 0–10 mA Take the full-scale meter current as *Im* = 1 mA and the internal resistance of the ammeter as *Rm* = 50 Ω.
-
-**Answer:** Shunt resistors: 50 mΩ, 505 mΩ, 5.556 Ω.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/026_2.9 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/026_2.9 Summary.md
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-# **2.9** Summary
-
-1. A resistor is a passi ve element in which the v oltage *v* across it is directly proportional to the current *i* through it. That is, a resistor is a device that obeys Ohm's law,
-
-*v* = *iR*
-
-where *R* is the resistance of the resistor.
-
-- 2. A short circuit is a resistor (a perfectly , conducting wire) with zero resistance (*R* = 0). An open circuit is a resistor with infinite resistance (*R* = *∞*).
-- 3. The conductance *G* of a resistor is the reciprocal of its resistance:
-
-$$
-G = \frac{1}{R}
-$$
-
-4. A branch is a single tw o-terminal element in an electric circuit. A node is the point of connection between tw o or more branches. A loop is a closed path in a circuit. The number of branches *b*, the number of nodes *n*, and the number of independent loops *l* in a network are related as
-
-$$
-b = l + n - 1
-$$
-
-- 5. Kirchhoff's current law (KCL) states that the currents at an y node algebraically sum to zero. In other w ords, the sum of the currents entering a node equals the sum of currents leaving the node.
-- 6. Kirchhoff's v oltage la w (KVL) states that the v oltages around a closed path algebraically sum to zero. In other w ords, the sum of voltage rises equals the sum of voltage drops.
-- 7. Two elements are in series when the y are connected sequentially , end to end. When elements are in series, the same current flows through them (*i*1 = *i*2). They are in parallel if the y are connected to the same two nodes. Elements in parallel always have the same voltage across them (*v*1 = *v*2).
-- 8. When two resistors *R*1 (=1/*G*1) and *R*2 (=1/*G*2) are in series, their equivalent resistance *R*eq and equivalent conductance *G*eq are
-
-$$
-R_{\text{eq}} = R_1 + R_2,
-$$
- $G_{\text{eq}} = \frac{G_1 G_2}{G_1 + G_2}$
-
-9. When two resistors *R*1 (=1/*G*1) and *R*2 (=1/*G*2) are in parallel, their equivalent resistance *R*eq and equivalent conductance *G*eq are
-
-$$
-R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}, \qquad G_{\text{eq}} = G_1 + G_2
-$$
-
-Practice Problem 2.17
-
-10. The voltage division principle for two resistors in series is
-
-$$
-v_1 = \frac{R_1}{R_1 + R_2} v
-$$
-, $v_2 = \frac{R_2}{R_1 + R_2} v$
-
-11. The current division principle for two resistors in parallel is
-
-$$
-i_1 = \frac{R_2}{R_1 + R_2} i
-$$
-, $i_2 = \frac{R_1}{R_1 + R_2} i$
-
-12. The formulas for a delta-to-wye transformation are
-
-$$
-R_1 = \frac{R_b R_c}{R_a + R_b + R_c}, \qquad R_2 = \frac{R_c R_a}{R_a + R_b + R_c}
-$$
-$$
-R_3 = \frac{R_a R_b}{R_a + R_b + R_c}
-$$
-
-13. The formulas for a wye-to-delta transformation are
-
-The formulas for a wye-to-delta transformation are
-\n
-$$
-R_a = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1}, \qquad R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2}
-$$
-\n
-$$
-R_c = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3}
-$$
-
-14. The basic laws covered in this chapter can be applied to the problems of electrical lighting and design of dc meters.
-
-# Review Questions
-
-| 2.1 | | The reciprocal of resistance is: | |
-|-----|--|----------------------------------|--|
-| | | | |
-
-| (a) voltage | (b) current |
-|-----------------|--------------|
-| (c) conductance | (d) coulombs |
-
-**2.2** An electric heater draws 10 A from a 120-V line. The resistance of the heater is:
-
-| (a) 1200 Ω | (b) 120 Ω |
-|------------|-----------|
-| (c) 12 Ω | (d) 1.2 Ω |
-
-**2.3** The voltage drop across a 1.5-kW toaster that draws 12 A of current is:
-
-| (a) 18 kV | (b) 125 V |
-|-----------|-------------|
-| (c) 120 V | (d) 10.42 V |
-
-**2.4** The maximum current that a 2W, 80 kΩ resistor can safely conduct is:
-
-| (a) 160 kA | (b) 40 kA |
-|------------|-----------|
-| (c) 5 mA | (d) 25 μA |
-
-**2.5** A network has 12 branches and 8 independent loops. How many nodes are there in the network?
-
-(a) 19 (b) 17 (c) 5 (d) 4
-
-**2.6** The current *I* in the circuit of Fig. 2.63 is:
-
-| (a) −0.8 A | (b) −0.2 A |
-|------------|------------|
-| (c) 0.2 A | (d) 0.8 A |
-
-# **Figure 2.63**
-
-For Review Question 2.6.
-
-**2.7** The current *Io* of Fig. 2.64 is:
-
-(a)
-$$
--4
-$$
- A (b) $-2$ A (c) $4$ A (d) $16$ A
-
-Problems **65**
-
-**2.8** In the circuit in Fig. 2.65, *V* is: (a) 30 V (b) 14 V (c) 10 V (d) 6 V
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/027_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/027_Review Questions.md
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-# **Figure 2.65**
-
-For Review Question 2.8.
-
-**2.9** Which of the circuits in Fig. 2.66 will give you *Vab* =7V?
-
-# **Figure 2.66**
-
-# For Review Question 2.9.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/028_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/028_Problems.md
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-# Problems
-
-# Section 2.2 Ohm's Law
-
-**2.1** Design a problem, complete with a solution, to help students to better understand Ohm's law. Use at least two resistors and one voltage source. Hint, you could use both resistors at once or one at a time, it is up to you. Be creative.
-
-- **2.10** In the circuit of Fig. 2.67, a decrease in *R*3 leads to a decrease of, select all that apply:
- - (a) current through *R*3
- - (b) voltage across *R*3
- - (c) voltage across *R*1
- - (d) power dissipated in *R*2
- - (e) none of the above
-
-# **Figure 2.67** For Review Question 2.10.
-
-*Answers: 2.1c, 2.2c, 2.3b, 2.4c, 2.5c, 2.6b, 2.7a, 2.8d, 2.9d, 2.10b, d.*
-
-- **2.2** Find the hot resistance of a light bulb rated 60 W, 120 V.
-- **2.3** A bar of silicon is 4 cm long with a circular cross sec tion. If the resistance of the bar is 240 Ω at room tem perature, what is the cross-sectional radius of the bar?
-
-- **2.4** (a) Calculate current *i* in Fig. 2.68 when the switch is in position 1.
- - (b) Find the current when the switch is in position 2.
-
-# **Figure 2.68**
-
-For Prob. 2.4.
-
-# Section 2.3 Nodes, Branches, and Loops
-
-**2.5** For the network graph in Fig. 2.69, find the number of nodes, branches, and loops.
-
-**Figure 2.69** For Prob. 2.5.
-
-> **2.6** In the network graph shown in Fig. 2.70, determine the number of branches and nodes.
-
-**2.7** Determine the number of branches and nodes in the circuit of Fig. 2.71.
-
-**Figure 2.71** For Prob. 2.7.
-
-# Section 2.4 Kirchhoff's Laws
-
-**2.8** Design a problem, complete with a solution, to help
-
-other students better understand Kirchhoff's Current Law. Design the problem by specifying values of *ia*, *ib*, and *ic*, shown in Fig. 2.72, and asking them to solve for values of *i*1, *i*2, and *i*3. Be careful to specify realistic currents.
-
-# **Figure 2.72**
-
-For Prob. 2.8.
-
-**2.9** Find *i*1, *i*2, and *i*3 in Fig. 2.73.
-
-**Figure 2.73** For Prob. 2.9.
-
-**2.10** Determine *i*1 and *i*2 in the circuit of Fig. 2.74.
-
-**Figure 2.74** For Prob. 2.10.
-
-**2.11** In the circuit of Fig. 2.75, calculate *V*1 and *V*2.
-
-For Prob. 2.11.
-
-**2.12** In the circuit in Fig. 2.76, obtain *v*1, *v*2, and *v*3.
-
-For Prob. 2.12.
-
-**2.13** For the circuit in Fig. 2.77, use KCL to find the branch currents *I*1 to *I*4.
-
-**Figure 2.77** For Prob. 2.13.
-
-**2.14** Given the circuit in Fig. 2.78, use KVL to find the branch voltages *V*1 to *V*4.
-
-**Figure 2.78** For Prob. 2.14.
-
-**2.15** Calculate *v* and *ix* in the circuit of Fig. 2.79.
-
-For Prob. 2.15.
-
-**2.16** Determine *Vo* in the circuit in Fig. 2.80.
-
-**2.17** Obtain *v*1 through *v*3 in the circuit of Fig. 2.81.
-
-**2.18** Find *I* and *V* in the circuit of Fig. 2.82.
-
-**Figure 2.82** For Prob. 2.18.
-
-For Prob. 2.19.
-
-**2.19** From the circuit in Fig. 2.83, find *I*, the power dissipated by the resistor, and the power supplied by each source.
-
-**2.20** Determine *io* in the circuit of Fig. 2.84.
-
-- For Prob. 2.20.
-- **2.21** Find *Vx* in the circuit of Fig. 2.85.
-
-**2.22** Find *Vo* in the circuit in Fig. 2.86 and the power absorbed by the dependent source.
-
-- **2.23** In the circuit shown in Fig. 2.87, determine *Vx* and
- - the power absorbed by the 60-Ω resistor.
-
-**2.24** For the circuit in Fig. 2.88, find *Vo* / *Vs* in terms of *α*, *R*1, *R*2, *R*3, and *R*4. If *R*1 = *R*2 = *R*3 = *R*4, what value of *α* will produce *|Vo* / *Vs|* =10?
-
-**Figure 2.88** For Prob. 2.24.
-
-**2.25** For the network in Fig. 2.89, find the current, voltage, and power associated with the 20-kΩ resistor.
-
-# Sections 2.5 and 2.6 Series and Parallel Resistors
-
-**2.26** For the circuit in Fig. 2.90, *io* = 3 A. Calculate *ix* and the total power absorbed by the entire circuit.
-
-**Figure 2.90** For Prob. 2.26.
-
-**2.27** Calculate *Io* in the circuit of Fig. 2.91.
-
-**Figure 2.91** For Prob. 2.27.
-
-## Problems **69**
-
-**2.28** Design a problem, using Fig. 2.92, to help other students better understand series and parallel circuits.
-
-**Figure 2.92** For Prob. 2.28.
-
-**2.29** All resistors (R) in Fig. 2.93 are 10 Ω each. Find *R*eq.
-
-**Figure 2.93** For Prob. 2.29.
-
-**2.30** Find *R*eq for the circuit in Fig. 2.94.
-
-**2.31** For the circuit in Fig. 2.95, determine *i*1 to *i*5.
-
-For Prob. 2.31.
-
-**2.32** Find *i*1 through *i*4 in the circuit in Fig. 2.96.
-
-# **Figure 2.96**
-
-For Prob. 2.32.
-
-**2.33** Obtain *v* and *i* in the circuit of Fig. 2.97.
-
-# **Figure 2.97**
-
-For Prob. 2.33.
-
-**2.34** Using series/parallel resistance combination, find the equivalent resistance seen by the source in the circuit of Fig. 2.98. Find the overall absorbed power by the resistor network.
-
-**Figure 2.98** For Prob. 2.34.
-
-**2.35** Calculate *Vo* and *Io* in the circuit of Fig. 2.99.
-
-**2.36** Find *i* and *Vo* in the circuit of Fig. 2.100.
-
-**2.37** Given the circuit in Fig. 2.101 and that the resistance, *R*eq, looking into the circuit from the left is equal to 100 Ω, determine the value of *R*1.
-
-**2.39** Evaluate *R*eq looking into each set of terminals for each of the circuits shown in Fig. 2.103.
-
-**2.40** For the ladder network in Fig. 2.104, find *I* and *R*eq.
-
-**Figure 2.104**
-
-For Prob. 2.40.
-
-# **Figure 2.105**
-
-For Prob. 2.41.
-
-**2.42** Reduce each of the circuits in Fig. 2.106 to a single resistor at terminals *a*-*b*.
-
-For Prob. 2.37.
-
-**2.38** Find *R*eq and *io* in the circuit of Fig. 2.102.
-
-# **Figure 2.106**
-
-For Prob. 2.42.
-
-**2.43** Calculate the equivalent resistance *Rab* at terminals *a-b* for each of the circuits in Fig. 2.107.
-
-**2.45** Find the equivalent resistance at terminals *a*-*b* of each circuit in Fig. 2.109.
-
-**Figure 2.109** For Prob. 2.45.
-
-**2.44** For the circuits in Fig. 2.108, obtain the equivalent resistance at terminals *a*-*b*.
-
-**Figure 2.108** For Prob. 2.44.
-
-**2.46** Find *I* in the circuit of Fig. 2.110.
-
-For Prob. 2.46.
-
-**2.47** Find the equivalent resistance *Rab* in the circuit of Fig. 2.111.
-
-# Section 2.7 Wye-Delta Transformations
-
-**2.48** Convert the circuits in Fig. 2.112 from Y to Δ.
-
-**2.50** Design a problem to help other students better understand wye-delta transformations using Fig. 2.114.
-
-# **Figure 2.114**
-
-For Prob. 2.50.
-
-**2.51** Obtain the equivalent resistance at the terminals *a*-*b* for each of the circuits in Fig. 2.115.
-
-**Figure 2.115** For Prob. 2.51.
-
-**2.52** For the circuit shown in Fig. 2.116, find the equivalent resistance. All resistors are 3Ω. **\***
-
-\* An asterisk indicates a challenging problem.
-
-## Problems **73**
-
-**2.53** Obtain the equivalent resistance *R* **2.56** Determine *V* in the circuit of Fig. 2.120. *ab* in each of the circuits of Fig. 2.117. In (b), all resistors have a value of 30 Ω. **\***
-
-**2.54** Consider the circuit in Fig. 2.118. Find the equivalent resistance at terminals: (a) *a*-*b*, (b) *c*-*d*.
-
-**2.55** Calculate *Io* in the circuit of Fig. 2.119.
-
-For Prob. 2.55.
-
-For Prob. 2.56.
-
-**2.57** Find *R*eq **\*** and *I* in the circuit of Fig. 2.121.
-
-# **Figure 2.121**
-
-For Prob. 2.57.
-
-# Section 2.8 Applications
-
-**2.58** The 150 W light bulb in Fig. 2.122 is rated at 110 volts. Calculate the value of *V*s to make the light bulb operate at its rated conditions.
-
-**2.59** An enterprising young man travels to Europe carrying three light bulbs he had purchased in North America. The light bulbs he has are a 100-W light bulb, a 60-W light bulb, and a 40-W light bulb. Each light bulb is rated at 110 V. He wishes to connect these to a 220-V system that is found in Europe. For reasons we are not sure of, he connects the 40-W
-
-light bulb in series with a parallel combination of the 60-W light bulb and the 100-W light bulb as shown in Fig. 2.123. How much power is actually being delivered to each light bulb? What does he see when he first turns on the light bulbs?
-
-Is there a better way to connect these light bulbs in order to have them work more effectively?
-
-- **2.60** If the three bulbs of Prob. 2.59 are connected in parallel to the 120-V source, calculate the current through each bulb.
-- **2.61** As a design engineer, you are asked to design a
-- lighting system consisting of a 70-W power supply and two light bulbs as shown in Fig. 2.124. You must select the two bulbs from the following three available bulbs.
-
-*R*1 = 80 Ω, cost = \$0.60 (standard size) *R*2 = 90 Ω, cost = \$0.90 (standard size) *R*3 = 100 Ω, cost = \$0.75 (nonstandard size)
-
-The system should be designed for minimum cost such that *I* lies within the range *I* = 1.2 A ± 5 percent.
-
-For Prob. 2.61.
-
-**2.62** A three-wire system supplies two loads *A* and *B* as shown in Fig. 2.125. Load *A* consists of a motor drawing a current of 8 A, while load *B* is a PC drawing 2 A. Assuming 10 h/day of use for 365 days and 6 cents/kWh, calculate the annual energy cost of the system.
-
-**Figure 2.125**
-
-- **2.63** If an ammeter with an internal resistance of 100 Ω and a current capacity of 2 mA is to measure 5 A, determine the value of the resistance needed. Calculate the power dissipated in the shunt resistor.
-- **2.64** The potentiometer (adjustable resistor) *Rx* in Fig. 2.126 is to be designed to adjust current *ix* from 10 mA to 1 A. Calculate the values of *R* and *Rx* to achieve this.
-
-**Figure 2.126** For Prob. 2.64.
-
-- **2.65** Design a circuit that uses a d'Arsonval meter (with an internal resistance of 2 kΩ that requires a current of 5 mA to cause the meter to deflect full scale) to build a voltmeter to read values of voltages up to 100 volts.
-- **2.66** A 20-kΩ/V voltmeter reads 10 V full scale.
- - (a) What series resistance is required to make the meter read 50 V full scale?
- - (b) What power will the series resistor dissipate when the meter reads full scale?
-- **2.67** (a) Obtain the voltage *Vo* in the circuit of Fig. 2.127(a).
- - (b) Determine the voltage *V*ʹ *o* measured when a voltmeter with 6-kΩ internal resistance is connected as shown in Fig. 2.127(b).
-
-(c) The finite resistance of the meter introduces an error into the measurement. Calculate the percent error as
-
-$$
-\left|\frac{V_o - V'_o}{V_o}\right| \times 100\,\%
-$$
-
-(d) Find the percent error if the internal resistance were 36 kΩ.
-
-# **Figure 2.127**
-
-For Prob. 2.67.
-
-- **2.68** (a) Find the current *I* in the circuit of Fig. 2.128(a). (b) An ammeter with an internal resistance of 1 Ω is inserted in the network to measure *I*ʹas shown in Fig. 2.128(b). What is *I*ʹ?
- - (c) Calculate the percent error introduced by the meter as
-
- *|*×100%
-
-\_\_\_\_\_ *I*− *I*ʹ *I*
-
-**2.69** A voltmeter is used to measure *Vo* in the circuit in Fig. 2.129. The voltmeter model consists of an ideal voltmeter in parallel with a 250-kΩ resistor. Let *Vs* = 95 V, *Rs* = 25 kΩ, and *R*1 = 40 kΩ. Calculate *Vo* with and without the voltmeter when
-
-(a)
-$$
-R_2 = 5 \text{ k}\Omega
-$$
- (b) $R_2 = 25 \text{ k}\Omega$
-
-# **Figure 2.129** For Prob. 2.69.
-
-- **2.70** (a) Consider the Wheatstone bridge shown in Fig. 2.130. Calculate *va*, *vb*, and *vab*.
- - (b) Rework part (a) if the ground is placed at *a* instead of *o*.
-
-**2.71** Figure 2.131 represents a model of a solar photovoltaic panel. Given that *Vs* = 95 V, *R*1 = 25 Ω, and *iL* =2 A, find *RL*.
-
-**Figure 2.131** For Prob. 2.71.
-
-**2.72** Find *Vo* in the two-way power divider circuit in Fig. 2.132.
-
-For Prob. 2.72.
-
-**2.73** An ammeter model consists of an ideal ammeter in series with a 20-Ω resistor. It is connected with a current source and an unknown resistor *Rx* as shown in Fig. 2.133. The ammeter reading is noted. When a potentiometer *R* is added and adjusted until the ammeter reading drops to one half its previous reading, then *R* = 65 Ω. What is the value of *Rx* ?
-
-**2.74** The circuit in Fig. 2.134 is to control the speed of a motor such that the motor draws currents 5 A, 3 A, and 1 A when the switch is at high, medium, and low positions, respectively. The motor can be modeled as a load resistance of 20 mΩ. Determine the series dropping resistances *R*1, *R*2, and *R*3.
-
-**2.75** Find *Rab* in the four-way power divider circuit in Fig. 2.135. Assume each *R* = 4 Ω.
-
-**Figure 2.133** For Prob. 2.73.
-
-**Figure 2.135** For Prob. 2.75.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/029_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/029_Comprehensive Problems.md
deleted file mode 100644
index a36d80784351ac51bccf804ff80861f25533ff69..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/029_Comprehensive Problems.md
+++ /dev/null
@@ -1,40 +0,0 @@
-# Comprehensive Problems
-
-**2.76** Repeat Prob. 2.75 for the eight-way divider shown in Fig. 2.136.
-
-# **Figure 2.136** For Prob. 2.76.
-
-**2.77** Suppose your circuit laboratory has the following standard commercially available resistors in large quantities:
-
-1.8 Ω 20 Ω 300 Ω 24 kΩ 56 kΩ
-
-Using series and parallel combinations and a minimum number of available resistors, how would you obtain the following resistances for an electronic circuit design?
-
-| (a) 5 Ω | (b) 311.8 Ω |
-|-----------|--------------|
-| (c) 40 kΩ | (d) 52.32 kΩ |
-
-**2.78** In the circuit in Fig. 2.137, the wiper divides the potentiometer resistance between *αR* and (1 − *α*)*R*, 0 ≤ α ≤1.Find *vo* / *vs*.
-
-**2.79** An electric pencil sharpener rated 240 mW, 6 V is connected to a 9-V battery as shown in Fig. 2.138. Calculate the value of the series-dropping resistor *Rx* needed to power the sharpener.
-
-**2.80** A loudspeaker is connected to an amplifier as shown in Fig. 2.139. If a 10-Ω loudspeaker draws the maximum power of 12 W from the amplifier, determine the maximum power a 4-Ω loudspeaker will draw.
-
-For Prob. 2.80.
-
-**2.81** For a specific application, the circuit shown in Fig. 2.140 was designed so that *IL* = 83.33 mA and that *Rin* = 5 kΩ. What are the values of *R*1 and *R*2?
-
-- **2.82** The pin diagram of a resistance array is shown in Fig. 2.141. Find the equivalent resistance between the following:
- - (a) 1 and 2
- - (b) 1 and 3
- - (c) 1 and 4
-
-**2.83** Two delicate devices are rated as shown in Fig. 2.142. Find the values of the resistors *R*1 and *R*2 needed to power the devices using a 36-V battery.
-
-**Figure 2.141**
-
-For Prob. 2.82.
-
-# **chapter**
-
-3
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/030_Chapter 3 - Methods of Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/030_Chapter 3 - Methods of Analysis.md
deleted file mode 100644
index 11091d82604d634bcc0a38217f7062e2e0c3e0bb..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/030_Chapter 3 - Methods of Analysis.md
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@@ -1,82 +0,0 @@
-# Methods of Analysis
-
-*No great work is ever done in a hurry. To develop a great scientific discovery, to print a great picture, to write an immortal poem, to become a minister, or a famous general—to do anything great requires time, patience, and perseverance. These things are done by degrees, "little by little."*
-
-—W. J. Wilmont Buxton
-
-# Enhancing Your Career
-
-# **Career in Electronics**
-
-One area of application for electric circuit analysis is electronics. The term *electronics* was originally used to distinguish circuits of very low current levels. This distinction no longer holds, as power semiconductor devices operate at high le vels of current. Today, electronics is regarded as the science of the motion of char ges in a g as, vacuum, or semicon ductor. Modern electronics in volves transistors and transistor circuits. The earlier electronic circuits were assembled from components. Man y electronic circuits are now produced as integrated circuits, fabricated in a semiconductor substrate or chip.
-
-Electronic circuits find applications in many areas, such as automation, broadcasting, computers, and instrumentation. The range of devices that use electronic circuits is enormous and is limited only by our imagination. Radio, television, computers, and stereo systems are but a few.
-
-An electrical engineer usually performs di verse functions and is likely to use, design, or construct systems that incorporate some form of electronic circuits. Therefore, an understanding of the operation and analysis of electronics is essential to the electrical engineer . Electronics has become a specialty distinct from other disciplines within electrical engi neering. Because the field of electronics is ever advancing, an electronics engineer must update his/her knowledge from time to time. The best way to do this is by being a member of a professional organization such as the Institute of Electrical and Electronics Engineers (IEEE). With a membership of over 300,000, the IEEE is the largest professional organization in the world. Members benefit immensely from the numerous magazines, journals, transactions, and conference/symposium proceedings published yearly by IEEE. You should consider becoming an IEEE member.
-
-Troubleshooting an electronic circuit board. © Brand X Pictures/PunchStock RF
-
-# Learning Objectives
-
-By using the information and exercises in this chapter you will be able to:
-
-- 1. Understand Kirchhoff's current law.
-- 2. Understand Kirchhoff's voltage law.
-- 3. Develop an understanding of how to use Kirchhoff's current law to write nodal equations and then to solve for unknown node voltages.
-- 4. Develop an understanding of how to use Kirchhoff's voltage law to write mesh equations and then to solve for unknown loop currents.
-- 5. Explain how to use *PSpice* to solve for unknown node voltages and currents.
-
-# **3.1** Introduction
-
-Having understood the fundamental laws of circuit theory (Ohm's law and Kirchhoff's laws), we are now prepared to apply these laws to develop two powerful techniques for circuit analysis: nodal analysis, which is based on a systematic application of Kirchhof f's current law (KCL), and mesh analysis, which is based on a systematic application of Kirchhoff's voltage law (KVL). The two techniques are so important that this chapter should be re garded as the most important in the book. Students are therefore encouraged to pay careful attention.
-
-With the two techniques to be developed in this chapter, we can analyze any linear circuit by obtaining a set of simultaneous equations that are then solved to obtain the required values of current or voltage. One method of solving simultaneous equations involves Cramer's rule, which allows us to calculate circuit variables as a quotient of determinants. The examples in the chapter will illustrate this method; Appendix A also briefly summarizes the essentials the reader needs to kno w for applying Cramer' s rule. Another method of solving simultaneous equations is to use *MATLAB*, a computer software discussed in Appendix E.
-
-Also in this chapter, we introduce the use of *PSpice for Windows*, a circuit simulation computer software program that we will use throughout the text. Finally, we apply the techniques learned in this chapter to analyze transistor circuits.
-
-# **3.2** Nodal Analysis
-
-Nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit v ariables. Choosing node v oltages instead of element voltages as circuit variables is convenient and reduces the number of equations one must solve simultaneously.
-
-To simplify matters, we shall assume in this section that circuits do not contain voltage sources. Circuits that contain voltage sources will be analyzed in the next section.
-
-In *nodal analysis*, we are interested in finding the node voltages. Given a circuit with n nodes without voltage sources, the nodal analysis of the circuit involves taking the following three steps.
-
-Nodal analysis is also known as the node-voltage method.
-
-# Steps to Determine Node Voltages:
-
-- 1. Select a node as the reference node. Assign voltages *v*1, *v*2, . . . , *vn*− 1 to the remaining *n* − 1 nodes. The voltages are referenced with respect to the reference node.
-- 2. Apply KCL to each of the *n* − 1 nonreference nodes. Use Ohm's law to express the branch currents in terms of node voltages.
-- 3. Solve the resulting simultaneous equations to obtain the un known node voltages.
-
-We shall now explain and apply these three steps.
-
-The first step in nodal analysis is selecting a node as the *reference* or *datum node*. The reference node is commonly called the *ground* since it is assumed to have zero potential. A reference node is indicated b y any of the three symbols in Fig. 3.1. The type of ground in Fig. 3.1(c) is called a *chassis ground* and is used in de vices where the case, enclosure, or chassis acts as a reference point for all circuits. When the potential of the earth is used as reference, we use the *earth ground* in Fig. 3.1(a) or (b). We shall always use the symbol in Fig. 3.1(b).
-
-Once we ha ve selected a reference node, we assign v oltage designations to nonreference nodes. Consider , for e xample, the circuit in Fig. 3.2(a). Node 0 is the reference node ( *v* = 0), while nodes 1 and 2 are assigned voltages *v*1 and *v*2, respectively. Keep in mind that the node voltages are defined with respect to the reference node. As illustrated in Fig. 3.2(a), each node voltage is the voltage rise from the reference node to the corresponding nonreference node or simply the v oltage of that node with respect to the reference node.
-
-As the second step, we apply KCL to each nonreference node in the circuit. To avoid putting too much information on the same circuit, the circuit in Fig. 3.2(a) is redra wn in Fig. 3.2(b), where we no w add *i*1, *i*2, and *i*3 as the currents through resistors *R*1, *R*2, and *R*3, respectively. At node 1, applying KCL gives
-
-$$
-I_1 = I_2 + i_1 + i_2 \tag{3.1}
-$$
-
-At node 2,
-
-$$
-I_2 + i_2 = i_3 \tag{3.2}
-$$
-
-We now apply Ohm's law to express the unknown currents *i*1, *i*2, and *i*3 in terms of node voltages. The key idea to bear in mind is that, since resistance is a passive element, by the passive sign convention, current must always flow from a higher potential to a lower potential.
-
-Current flows from a higher potential to a lower potential in a resistor.
-
-We can express this principle as
-
-$$
-i = \frac{v_{\text{higher}} - v_{\text{lower}}}{R}
-$$
- (3.3)
-
-# **Figure 3.1**
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/031_3.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/031_3.1 Introduction.md
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@@ -1,358 +0,0 @@
-
-Common symbols for indicating a reference node, (a) common ground, (b) ground, (c) chassis ground.
-
-The number of nonreference nodes is equal to the number of independent equations that we will derive.
-
-**Figure 3.2** Typical circuit for nodal analysis.
-
-Note that this principle is in agreement with the w ay we defined resistance in Chapter 2 (see Fig. 2.1). With this in mind, we obtain from Fig. 3.2(b),
-
-$$
-i_1 = \frac{v_1 - 0}{R_1} \quad \text{or} \quad i_1 = G_1 v_1
-$$
-
-\n
-$$
-i_2 = \frac{v_1 - v_2}{R_2} \quad \text{or} \quad i_2 = G_2 (v_1 - v_2)
-$$
-
-\n
-$$
-i_3 = \frac{v_2 - 0}{R_3} \quad \text{or} \quad i_3 = G_3 v_2
-$$
-
-\n(3.4)
-
-Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in
-
-$$
-I_1 = I_2 + \frac{v_1}{R_1} + \frac{v_1 - v_2}{R_2}
-$$
- (3.5)
-
-$$
-I_2 + \frac{v_1 - v_2}{R_2} = \frac{v_2}{R_3}
-$$
- (3.6)
-
-In terms of the conductances, Eqs. (3.5) and (3.6) become
-
-$$
-I_1 = I_2 + G_1 v_1 + G_2 (v_1 - v_2)
-$$
-\n(3.7)
-
-$$
-I_2 + G_2(v_1 - v_2) = G_3v_2 \tag{3.8}
-$$
-
-The third step in nodal analysis is to solve for the node voltages. If we apply KCL to *n* − 1 nonreference nodes, we obtain *n* − 1 simulta neous equations such as Eqs. (3.5) and (3.6) or (3.7) and (3.8). F or the circuit of Fig. 3.2, we solve Eqs. (3.5) and (3.6) or (3.7) and (3.8) to obtain the node voltages *v*1 and *v*2 using any standard method, such as the substitution method, the elimination method, Cramer' s rule, or matrix inversion. To use either of the last two methods, one must cast the simultaneous equations in matrix form. For example, Eqs. (3.7) and (3.8) can be cast in matrix form as
-
-$$
-\begin{bmatrix} G_1 + G_2 & -G_2 \ -G_2 & G_2 + G_3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} I_1 - I_2 \ I_2 \end{bmatrix}
-$$
- (3.9)
-
-which can be solved to get *v*1 and *v*2. Equation 3.9 will be generalized in Section 3.6. The simultaneous equations may also be solv ed using calculators or with softw are packages such as *MATLAB, Mathcad, Maple,* and *Quattro Pro*.
-
-Example 3.1 Calculate the node voltages in the circuit shown in Fig. 3.3(a).
-
-# **Solution:**
-
-Consider Fig. 3.3(b), where the circuit in Fig. 3.3(a) has been prepared for nodal analysis. Notice how the currents are selected for the application of KCL. Except for the branches with current sources, the labeling of the currents is arbitrary but consistent. (By consistent, we mean that if, for example, we assume that *i*2 enters the 4-Ω resistor from the left-hand side, *i*2 must leave the resistor from the right-hand side.) The reference node is selected, and the node voltages *v*1 and *v*2 are now to be determined.
-
-At node 1, applying KCL and Ohm's law gives
-
-$$
-i_1 = i_2 + i_3
-$$
- $\Rightarrow$ $5 = \frac{v_1 - v_2}{4} + \frac{v_1 - 0}{2}$
-
-Multiplying each term in the last equation by 4, we obtain
-
-$$
-20 = v_1 - v_2 + 2v_1
-$$
-
-Appendix A discusses how to use Cramer's rule.
-
-$$
-3v_1 - v_2 = 20 \tag{3.1.1}
-$$
-
-At node 2, we do the same thing and get
-
-$$
-i_2 + i_4 = i_1 + i_5
-$$
- $\Rightarrow$ $\frac{v_1 - v_2}{4} + 10 = 5 + \frac{v_2 - 0}{6}$
-
-Multiplying each term by 12 results in
-
-$$
-3v_1 - 3v_2 + 120 = 60 + 2v_2
-$$
-
-or
-
-$$
--3v_1 + 5v_2 = 60 \tag{3.1.2}
-$$
-
-Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve the equations using any method and obtain the values of *v*1 and *v*2.
-
-■ **METHOD 1** Using the elimination technique, we add Eqs. (3.1.1) and (3.1.2).
-
-$$
-4v_2 = 80 \qquad \Rightarrow \qquad v_2 = 20 \text{ V}
-$$
-
-Substituting *v*2 = 20 in Eq. (3.1.1) gives
-
-$$
-3v_1 - 20 = 20
-$$
- $\Rightarrow$ $v_1 = \frac{40}{3} = 13.333$ V
-
-■ **METHOD 2** To use Cramer's rule, we need to put Eqs. (3.1.1) and (3.1.2) in matrix form as
-
-$$
-\begin{bmatrix} 3 & -1 \\ -3 & 5 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 20 \\ 60 \end{bmatrix}
-$$
- (3.1.3)
-
-The determinant of the matrix is
-
-$$
-\Delta = \begin{bmatrix} 3 & -1 \\ -3 & 5 \end{bmatrix} = 15 - 3 = 12
-$$
-
-We now obtain *v*1 and *v*2 as
-
-$$
-v_1 = \frac{\Delta_1}{\Delta} = \frac{\begin{vmatrix} 20 & -1 \\ 60 & 5 \end{vmatrix}}{\Delta} = \frac{100 + 60}{12} = 13.333 \text{ V}
-$$
-$$
-v_2 = \frac{\Delta_2}{\Delta} = \frac{\begin{vmatrix} 3 & 20 \\ -3 & 60 \end{vmatrix}}{\Delta} = \frac{180 + 60}{12} = 20 \text{ V}
-$$
-
-giving us the same result as did the elimination method.
-
-If we need the currents, we can easily calculate them from the values of the nodal voltages.
-
-$$
-i_1 = 5 \text{ A},
-$$
- $i_2 = \frac{v_1 - v_2}{4} = -1.6668 \text{ A},$ $i_3 = \frac{v_1}{2} = 6.666 \text{ A}$
- $i_4 = 10 \text{ A},$ $i_5 = \frac{v_2}{6} = 3.333 \text{ A}$
-
-The fact that *i*2 is negative shows that the current flows in the direction opposite to the one assumed.
-
-Obtain the node voltages in the circuit of Fig. 3.4. Practice Problem 3.1
-
-2
-
-1= 5
-
-2 Ω 6 Ω 10 A
-
-(a)
-
-2 Ω 6 Ω 10 A
-
-2 i 5
-
-i i 4= 10 2
-
-5 A
-
-4 Ω
-
-3 i
-
-*v*2 *v*1
-
-1= 5 i
-
-5 A
-
-4 Ω
-
-1
-
-i
-
-i
-
-**Answer:** *v*1 = 30 V, *v*2 = −2.5 V.
-
-14 A 7 A 5 Ω 4 Ω 5 Ω 1 2
-
-Example 3.2 Determine the voltages at the nodes in Fig. 3.5(a).
-
-# **Solution:**
-
-**Figure 3.4** For Practice Prob. 3.1.
-
-The circuit in this example has three nonreference nodes, unlike the previous example which has two nonreference nodes. We assign voltages to the three nodes as shown in Fig. 3.5(b) and label the currents.
-
-At node 1,
-
-$$
-3 = i_1 + i_x \qquad \Rightarrow \qquad 3 = \frac{v_1 - v_3}{4} + \frac{v_1 - v_2}{2}
-$$
-
-Multiplying by 4 and rearranging terms, we get
-
-$$
-3v_1 - 2v_2 - v_3 = 12 \tag{3.2.1}
-$$
-
-At node 2,
-
-$$
-i_x = i_2 + i_3
-$$
- $\Rightarrow$ $\frac{v_1 - v_2}{2} = \frac{v_2 - v_3}{8} + \frac{v_2 - 0}{4}$
-
-Multiplying by 8 and rearranging terms, we get
-
-$$
--4v_1 + 7v_2 - v_3 = 0 \tag{3.2.2}
-$$
-
-At node 3,
-
-$$
-i_1 + i_2 = 2i_x
-$$
- $\Rightarrow$ $\frac{v_1 - v_3}{4} + \frac{v_2 - v_3}{8} = \frac{2(v_1 - v_2)}{2}$
-
-Multiplying by 8, rearranging terms, and dividing by 3, we get
-
-$$
-2v_1 - 3v_2 + v_3 = 0 \tag{3.2.3}
-$$
-
-We have three simultaneous equations to solve to get the node voltages *v*1, *v*2, and *v*3. We shall solve the equations in three ways.
-
-■ **METHOD 1** Using the elimination technique, we add Eqs. (3.2.1) and (3.2.3).
-
-$$
-5v_1 - 5v_2 = 12
-$$
-
-or
-
-$$
-v_1 - v_2 = \frac{12}{5} = 2.4
-$$
- (3.2.4)
-
-Adding Eqs. (3.2.2) and (3.2.3) gives
-
-$$
--2v_1 + 4v_2 = 0 \Rightarrow v_1 = 2v_2 \tag{3.2.5}
-$$
-
-Substituting Eq. (3.2.5) into Eq. (3.2.4) yields
-
-$$
-2v_2 - v_2 = 2.4
-$$
- $\Rightarrow$ $v_2 = 2.4$ , $v_1 = 2v_2 = 4.8$ V
-
-From Eq. (3.2.3), we get
-
-$$
-v_3 = 3v_2 - 2v_1 = 3v_2 - 4v_2 = -v_2 = -2.4
-$$
- V
-
-Thus,
-
-$$
-v_1 = 4.8 \text{ V}, \qquad v_2 = 2.4 \text{ V}, \qquad v_3 = -2.4 \text{ V}
-$$
-
-■ **METHOD 2** To use Cramer's rule, we put Eqs. (3.2.1) to (3.2.3) in matrix form.
-
-$$
-\begin{bmatrix} 3 & -2 & -1 \ -4 & 7 & -1 \ 2 & -3 & 1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 12 \ 0 \ 0 \end{bmatrix}
-$$
- (3.2.6)
-
-From this, we obtain
-
-$$
-v_1 = \frac{\Delta_1}{\Delta}
-$$
-, $v_2 = \frac{\Delta_2}{\Delta}$ , $v_3 = \frac{\Delta_3}{\Delta}$
-
-where Δ, Δ1, Δ2, and Δ3 are the determinants to be calculated as follows. As explained in Appendix A, to calculate the determinant of a 3 by 3 matrix, we repeat the first two rows and cross multiply.
-
-Similarly, we obtain
-
-Δ3 =
-
-ǀ
-
-− − −
-
- 2 3 −4
-
-−3 −2 7
-
- 0 12 0
-
-ǀ
-
-+ + +
-
-$$
-\Delta_1 = \frac{\begin{vmatrix} 12 & -2 & -1 \\ 0 & 3 & -1 \\ -1 & 12 & 2 \end{vmatrix}}{\begin{vmatrix} 2 & -2 & -1 \\ -1 & 2 & -1 \\ 0 & 7 & -1 \end{vmatrix} + \begin{vmatrix} 1 & -84 + 0 + 0 - 0 - 36 - 0 = 48 \\ + \end{vmatrix}}
-$$
-$$
-\Delta_2 = \frac{\begin{vmatrix} 3 & 12 & -1 \\ 3 & 12 & -1 \\ -3 & 3 & -2 \end{vmatrix}}{\begin{vmatrix} 3 & -2 & -12 \\ -3 & 0 & -1 \end{vmatrix} + \begin{vmatrix} 3 & -2 & -12 \\ + \end{vmatrix}}
-$$
-
-= 0 + 144 + 0 − 168 − 0 − 0 = −24
-
-Thus, we find
-
-$$
-v_1 = \frac{\Delta_1}{\Delta} = \frac{48}{10} = 4.8 \text{ V}, \qquad v_2 = \frac{\Delta_2}{\Delta} = \frac{24}{10} = 2.4 \text{ V}
-$$
-
- $v_3 = \frac{\Delta_3}{\Delta} = \frac{-24}{10} = -2.4 \text{ V}$
-
-as we obtained with Method 1.
-
-■ **METHOD 3** We now use *MATLAB* to solve the matrix. Equation (3.2.6) can be written as
-
-$$
-AV = B \qquad \Rightarrow \qquad V = A^{-1}B
-$$
-
-where **A** is the 3 by 3 square matrix, **B** is the column vector, and **V** is a column vector comprised of *v*1, *v*2, and *v*3 that we want to determine. We use *MATLAB* to determine **V** as follows:
-
->>A =
-$$
-\begin{bmatrix} 3 & -2 & -1 \\ 4 & 7 & -1 \\ 2 & -3 & 1 \end{bmatrix}
-$$
-;
->>B = $\begin{bmatrix} 12 & 0 & 0 \\ 4 & 8 \end{bmatrix}$ ;
->>V = inv(A) \* B
-= 4.8000
-= 2.4000
-= 2.4000
-
-Thus, *v*1 = 4.8 V, *v*2 = 2.4 V, and *v*3 = −2.4 V, as obtained previously.
-
-Practice Problem 3.2 Find the voltages at the three nonreference nodes in the circuit of Fig. 3.6.
-
-4 A 2 Ω 3 Ω 4 Ω 6 Ω i x 4i x 1 3 2
-
-**Figure 3.6** For Practice Prob. 3.2.
-
-**Answer:** *v*1 = 32 V, *v*2 = −25.6 V, *v*3 = 62.4 V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/032_3.2 Nodal Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/032_3.2 Nodal Analysis.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/032_3.2 Nodal Analysis.md
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-# **3.3** Nodal Analysis with Voltage Sources
-
-We now consider ho w voltage sources af fect nodal analysis. W e use the circuit in Fig. 3.7 for illustration. Consider the following two possibilities.
-
-■ **CASE 1** If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the non reference node equal to the voltage of the voltage source. In Fig. 3.7, for example,
-
-$$
-v_1 = 10 \text{ V} \tag{3.10}
-$$
-
-Thus, our analysis is somewhat simplified by this knowledge of the voltage at this node.
-
-■ **CASE 2** If the voltage source (dependent or independent) is con nected between two nonreference nodes, the two nonreference nodes
-
-form a *generalized node* or *supernode*; we apply both KCL and KVL to determine the node voltages.
-
-A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.
-
-In Fig. 3.7, nodes 2 and 3 form a supernode. (W e could have more than two nodes forming a single supernode. F or example, see the circuit in Fig. 3.14.) We analyze a circuit with supernodes using the same three steps mentioned in the pre vious section e xcept that the supernodes are treated dif ferently. Why? Because an essential component of nodal analysis is applying KCL, which requires kno wing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be satisfied at a supernode like any other node. Hence, at the super node in Fig. 3.7,
-
-$$
-i_1 + i_4 = i_2 + i_3 \tag{3.11a}
-$$
-
-or
-
-$$
-\frac{v_1 - v_2}{2} + \frac{v_1 - v_3}{4} = \frac{v_2 - 0}{8} + \frac{v_3 - 0}{6}
-$$
- (3.11b)
-
-To apply Kirchhoff's voltage law to the supernode in Fig. 3.7, we redraw the circuit as shown in Fig. 3.8. Going around the loop in the clockwise direction gives
-
-$$
--v_2 + 5 + v_3 = 0 \qquad \Rightarrow \qquad v_2 - v_3 = 5 \tag{3.12}
-$$
-
-- From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages. Note the following properties of a supernode:
- - 1. The v oltage source inside the supernode pro vides a constraint equation needed to solve for the node voltages.
- - 2. A supernode has no voltage of its own.
- - 3. A supernode requires the application of both KCL and KVL.
-
-**Figure 3.8** Applying KVL to a supernode.
-
- A supernode may be regarded as a closed surface enclosing the voltage source and its two nodes.
-
-For Example 3.3.
-
-For the circuit shown in Fig. 3.9, find the node voltages.
-
-# **Solution:**
-
-The supernode contains the 2-V source, nodes 1 and 2, and the 10- Ω resistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives
-
-$$
-2 = i_1 + i_2 + 7
-$$
-
-Expressing *i*1 and *i*2 in terms of the node voltages
-
-$$
-2 = \frac{v_1 - 0}{2} + \frac{v_2 - 0}{4} + 7 \qquad \Rightarrow \qquad 8 = 2v_1 + v_2 + 28
-$$
-
-or
-
-$$
-v_2 = -20 - 2v_1 \tag{3.3.1}
-$$
-
-To get the relationship between *v*1 and *v*2, we apply KVL to the circuit in Fig. 3.10(b). Going around the loop, we obtain
-
-−*v*1 − 2 + *v*2 = 0 ⇒ *v*2 = *v*1 + 2 **(3.3.2)**
-
-From Eqs. (3.3.1) and (3.3.2), we write
-
-$$
-v_2 = v_1 + 2 = -20 - 2v_1
-$$
-
-or
-
-$$
-3v_1 = -22 \qquad \Rightarrow \qquad v_1 = -7.333 \text{ V}
-$$
-
-and *v*2 = *v*1 + 2 = −5.333 V. Note that the 10-Ω resistor does not make any difference because it is connected across the supernode.
-
-Applying: (a) KCL to the supernode, (b) KVL to the loop.
-
-Find the node voltages in the circuit of Fig. 3.12. Example 3.4
-
-# **Figure 3.12** For Example 3.4. 20 V 2 Ω 4 Ω 6 Ω 3 Ω 1 Ω *v*x 3*v*x + – + – 10 A 1 4 2 3 + –
-
-# **Solution:**
-
-Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to the two supernodes as in Fig. 3.13(a). At supernode 1-2,
-
-$$
-i_3 + 10 = i_1 + i_2
-$$
-
-Expressing this in terms of the node voltages,
-
-\_\_\_\_\_\_ *v*3 − *v*2 6 + 10 = \_\_\_\_\_\_ *v*1 − *v*4 3 + \_\_ *v*1 2
-
-or
-
-$$
-5v_1 + v_2 - v_3 - 2v_4 = 60 \tag{3.4.1}
-$$
-
-At supernode 3-4,
-
-$$
-i_1 = i_3 + i_4 + i_5
-$$
- $\Rightarrow$ $\frac{v_1 - v_4}{3} = \frac{v_3 - v_2}{6} + \frac{v_4}{1} + \frac{v_3}{4}$
-
-or
-
-$$
-4v_1 + 2v_2 - 5v_3 - 16v_4 = 0 \tag{3.4.2}
-$$
-
-**Figure 3.13** Applying: (a) KCL to the two supernodes, (b) KVL to the loops.
-
-We now apply KVL to the branches involving the voltage sources as shown in Fig. 3.13(b). For loop 1,
-
-$$
--v_1 + 20 + v_2 = 0 \Rightarrow v_1 - v_2 = 20 \tag{3.4.3}
-$$
-
-For loop 2,
-
-$$
--v_3 + 3v_x + v_4 = 0
-$$
-
-But *vx* = *v*1 − *v*4 so that
-
-$$
-3v_1 - v_3 - 2v_4 = 0 \tag{3.4.4}
-$$
-
-For loop 3,
-
-$$
-v_x - 3v_x + 6i_3 - 20 = 0
-$$
-
-But 6*i*3 = *v*3 − *v*2 and *vx* = *v*1 − *v*4. Hence,
-
-$$
--2v_1 - v_2 + v_3 + 2v_4 = 20 \tag{3.4.5}
-$$
-
-We need four node v oltages, *v*1, *v*2, *v*3, and *v*4, and it requires only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fifth equation is redundant, it can be used to check results. We can solve Eqs. (3.4.1) to (3.4.4) directly using *MATLAB*. We can eliminate one node voltage so that we solv e three simultaneous equations instead of four . From Eq. (3.4.3), *v*2 = *v*1 − 20. Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives
-
-$$
-6v_1 - v_3 - 2v_4 = 80 \tag{3.4.6}
-$$
-
-and
-
-$$
-6v_1 - 5v_3 - 16v_4 = 40 \tag{3.4.7}
-$$
-
-Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as
-
-3 6 6 −1 −1 −5 −2 −2 −16 ] [ *v*1 *v*3 *v*4 ] = [ 0 80 40]
-
-Using Cramer's rule gives
-
-$$
-\Delta = \begin{vmatrix} 3 & -1 & -2 \\ 6 & -1 & -2 \\ 6 & -5 & -16 \end{vmatrix} = -18, \qquad \Delta_1 = \begin{vmatrix} 0 & -1 & -2 \\ 80 & -1 & -2 \\ 40 & -5 & -16 \end{vmatrix} = -480,
-$$
-
-$$
-\Delta_3 = \begin{vmatrix} 3 & 0 & -2 \\ 6 & 80 & -2 \\ 6 & 40 & -16 \end{vmatrix} = -3120, \qquad \Delta_4 = \begin{vmatrix} 3 & -1 & 0 \\ 6 & -1 & 80 \\ 6 & -5 & 40 \end{vmatrix} = 840
-$$
-
-Thus, we arrive at the node voltages as
-
-[
-
-$$
-v_1 = \frac{\Delta_1}{\Delta} = \frac{-480}{-18} = 26.67 \text{ V}, \qquad v_3 = \frac{\Delta_3}{\Delta} = \frac{-3120}{-18} = 173.33 \text{ V},
-$$
-
- $v_4 = \frac{\Delta_4}{\Delta} = \frac{840}{-18} = -46.67 \text{ V}$
-
-and *v*2 = *v*1 − 20 = 6.667 V. We have not used Eq. (3.4.5); it can be used to cross-check results.
-
-Find *v*1, *v*2, and *v*3 in the circuit of Fig. 3.14 using nodal analysis. Practice Problem 3.4
-
-**Answer:** *v*1 = 7.608 V, *v*2 = −17.39 V, *v*3 = 1.6305 V.
-
-# **3.4** Mesh Analysis
-
-Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it.
-
-Nodal analysis applies KCL to find unknown voltages in a gi ven circuit, while mesh analysis applies KVL to find unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is *planar*. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is *nonplanar*. A circuit may have crossing branches and still be planar if it can be redra wn such that it has no crossing branches. F or example, the circuit in Fig. 3.15(a) has tw o crossing branches, b ut it can be redrawn as in Fig. 3.15(b). Hence, the circuit in Fig. 3.15(a) is planar. However, the circuit in Fig. 3.16 is nonplanar , because there is no w ay to redra w it and a void the branches crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be considered in this text.
-
-**Figure 3.14** For Practice Prob. 3.4.
-
- Mesh analysis is also known as loop analysis or the mesh-current method.
-
-# **Figure 3.15**
-
-(a) A planar circuit with crossing branches, (b) the same circuit redrawn with no crossing branches.
-
-To understand mesh analysis, we should first explain more about what we mean by a mesh.
-
-A mesh is a loop that does not contain any other loops within it.
-
-**Figure 3.17** A circuit with two meshes.
-
-In Fig. 3.17, for example, paths *abefa* and *bcdeb* are meshes, but path *abcdefa* is not a mesh. The current through a mesh is known as *mesh current*. In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit.
-
-In this section, we will apply mesh analysis to planar circuits that do not contain current sources. In the next section, we will consider circuits with current sources. In the mesh analysis of a circuit with *n* meshes, we take the following three steps.
-
-# Steps to Determine Mesh Currents:
-
-- 1. Assign mesh currents *i*1, *i*2, . . . , *in* to the *n* meshes.
-- 2. Apply KVL to each of the *n* meshes. Use Ohm's law to express the voltages in terms of the mesh currents.
-- 3. Solve the resulting *n* simultaneous equations to get the mesh currents.
-
-To illustrate the steps, consider the circuit in Fig. 3.17. The first step requires that mesh currents *i*1 and *i*2 are assigned to meshes 1 and 2. Although a mesh current may be assigned to each mesh in an arbi trary direction, it is conventional to assume that each mesh current flows clockwise.
-
-As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain
-
-$$
--V_1 + R_1 i_1 + R_3 (i_1 - i_2) = 0
-$$
-
-or
-
-$$
-(R_1 + R_3)i_1 - R_3i_2 = V_1 \tag{3.13}
-$$
-
-−*R*3 *i*1 + (*R*2 + *R*3) *i*2 = −*V*2 **(3.14)**
-
-For mesh 2, applying KVL gives
-
-$$
-R_2 i_2 + V_2 + R_3 (i_2 - i_1) = 0
-$$
-
-or
-
-The shortcut way will not apply if one mesh current is assumed clockwise and the other assumed counterclockwise, although this is permissible. Note in Eq. (3.13) that the coefficient of *i*1 is the sum of the resistances in the first mesh, while the coefficient of *i*2 is the negative of the resistance common to meshes 1 and 2. Now observe that the same is true in Eq. (3.14). This can serve as a shortcut way of writing the mesh equa tions. We will exploit this idea in Section 3.6.
-
-Although path abcdefa is a loop and not a mesh, KVL still holds. This is the reason for loosely using the terms loop analysis and mesh analysis to mean the same thing.
-
-The direction of the mesh current is arbitrary—(clockwise or counterclockwise)—and does not affect the validity of the solution.
-
-The third step is to solve for the mesh currents. Putting Eqs. (3.13) and (3.14) in matrix form yields
-
-$$
-\begin{bmatrix} R_1 + R_3 & -R_3 \ -R_3 & R_2 + R_3 \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \end{bmatrix} = \begin{bmatrix} V_1 \\ -V_2 \end{bmatrix}
-$$
- (3.15)
-
-which can be solved to obtain the mesh currents *i*1 and *i*2. We are at liberty to use any technique for solving the simultaneous equations. According to Eq. (2.12), if a circuit has *n* nodes, *b* branches, and *l* independent loops or meshes, then *l* = *b* − *n* + 1. Hence, *l* independent simultaneous equations are required to solve the circuit using mesh analysis.
-
-Notice that the branch currents are different from the mesh currents unless the mesh is isolated. To distinguish between the two types of currents, we use *i* for a mesh current and *I* for a branch current. The current elements *I*1, *I*2, and *I*3 are algebraic sums of the mesh currents. It is e vident from Fig. 3.17 that
-
-$$
-I_1 = i_1,
-$$
- $I_2 = i_2,$ $I_3 = i_1 - i_2$ (3.16)
-
-For the circuit in Fig. 3.18, find the branch currents *I*1, *I*2, and *I*3 using mesh analysis.
-
-# **Solution:**
-
-We first obtain the mesh currents using KVL. For mesh 1,
-
-$$
--15 + 5i_1 + 10(i_1 - i_2) + 10 = 0
-$$
-
-or
-
-$$
-3i_1 - 2i_2 = 1 \t\t(3.5.1)
-$$
-
-For mesh 2,
-
-$$
-6i_2 + 4i_2 + 10(i_2 - i_1) - 10 = 0
-$$
-
-or
-
-$$
-i_1 = 2i_2 - 1 \tag{3.5.2}
-$$
-
-■ **METHOD 1** Using the substitution method, we substitute Eq. (3.5.2) into Eq. (3.5.1), and write
-
-$$
-6i_2 - 3 - 2i_2 = 1 \qquad \Rightarrow \qquad i_2 = 1 \text{ A}
-$$
-
-From Eq. (3.5.2), *i*1 = 2*i*2 − 1 = 2 − 1 = 1 A. Thus,
-
-$$
-I_1 = i_1 = 1
-$$
- A, $I_2 = i_2 = 1$ A, $I_3 = i_1 - i_2 = 0$
-
-■ **METHOD 2** To use Cramer's rule, we cast Eqs. (3.5.1) and (3.5.2) in matrix form as
-
-> [ 3 −1 −2 2 ] [*i*1 *i*2 ] = [ 1 1]
-
-For Example 3.5.
-
-# We obtain the determinants
-
-$$
-\Delta = \begin{vmatrix} 3 & -2 \\ -1 & 2 \end{vmatrix} = 6 - 2 = 4
-$$
-
-$$
-\Delta_1 = \begin{vmatrix} 1 & -2 \\ 1 & 2 \end{vmatrix} = 2 + 2 = 4, \qquad \Delta_2 = \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} = 3 + 1 = 4
-$$
-
-Thus,
-
-$$
-i_1 = \frac{\Delta_1}{\Delta} = 1 \text{ A}, \qquad i_2 = \frac{\Delta_2}{\Delta} = 1 \text{ A}
-$$
-
-as before.
-
-Practice Problem 3.5
-
-Calculate the mesh currents *i*1 and *i*2 of the circuit of Fig. 3.19.
-
-**Answer:** *i*1 = 4.6 A, *i*2 = 200 mA.
-
-**Figure 3.19** For Practice Prob. 3.5.
-
-Example 3.6 Use mesh analysis to find the current *Io* in the circuit of Fig. 3.20.
-
-# **Solution:**
-
-We apply KVL to the three meshes in turn. For mesh 1,
-
-$$
--24 + 10(i_1 - i_2) + 12(i_1 - i_3) = 0
-$$
-
-$$
-11i_1 - 5i_2 - 6i_3 = 12 \tag{3.6.1}
-$$
-
-For mesh 2,
-
-24*i*2 + 4 (*i*2 − *i*3) + 10 (*i*2 − *i*1) = 0
-
-$$
--5i_1 + 19i_2 - 2i_3 = 0 \tag{3.6.2}
-$$
-
-For mesh 3,
-
-**Figure 3.20** For Example 3.6.
-
-$$
-4I_o + 12(i_3 - i_1) + 4(i_3 - i_2) = 0
-$$
-
-But at node A, *Io* = *i*1 − *i*2, so that
-
-$$
-4(i1 - i2) + 12(i3 - i1) + 4(i3 - i2) = 0
-$$
-
-or
-
-$$
--i_1 - i_2 + 2i_3 = 0 \tag{3.6.3}
-$$
-
-In matrix form, Eqs. (3.6.1) to (3.6.3) become
-
-$$
-\begin{bmatrix} 11 & -5 & -6 \ -5 & 19 & -2 \ -1 & -1 & 2 \ \end{bmatrix} \begin{bmatrix} i_1 \ i_2 \ i_3 \end{bmatrix} = \begin{bmatrix} 12 \ 0 \ 0 \end{bmatrix}
-$$
-
-We obtain the determinants as
-
-−2
-
-+ +
-
-$$
-\Delta_3 = \begin{pmatrix}\n11 & -5 & 12 \\
--5 & 19 & 0 \\
--11 & -5 & 19\n\end{pmatrix} + 60 + 228 = 288
-$$
-
-We calculate the mesh currents using Cramer's rule as
-
-−5
-
-− −
-
-0
-
-$$
-i_1 = \frac{\Delta_1}{\Delta} = \frac{432}{192} = 2.25 \text{ A}, \qquad i_2 = \frac{\Delta_2}{\Delta} = \frac{144}{192} = 0.75 \text{ A},
-$$
-
- $i_3 = \frac{\Delta_3}{\Delta} = \frac{288}{192} = 1.5 \text{ A}$
-
-Thus, *Io* = *i*1 − *i*2 = 1.5 A.
-
-# Practice Problem 3.6
-
-**Figure 3.21** For Practice Prob. 3.6.
-
-**Figure 3.22** A circuit with a current source.
-
-Using mesh analysis, find *Io* in the circuit of Fig. 3.21.
-
-**Answer:** −4 A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/033_3.4 Mesh Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/033_3.4 Mesh Analysis.md
deleted file mode 100644
index 0eb0ef97781485e52ce64d89614fdc0a16fa662d..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/033_3.4 Mesh Analysis.md
+++ /dev/null
@@ -1,105 +0,0 @@
-# **3.5** Mesh Analysis with Current Sources
-
-Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Consider the following two possible cases.
-
-■ **CASE 1** When a current source exists only in one mesh: Consider the circuit in Fig. 3.22, for example. We set *i*2 = −5 A and write a mesh equation for the other mesh in the usual way; that is,
-
-−10 + 4*i*1 + 6(*i*1 − *i*2) = 0 ⇒ *i*1 = −2 A **(3.17)**
-
-■ **CASE 2** When a current source exists between two meshes: Consider the circuit in Fig. 3.23(a), for example. We create a *supermesh* by excluding the current source and any elements connected in series with it, as shown in Fig. 3.23(b). Thus,
-
-A supermesh results when two meshes have a (dependent or independent) current source in common.
-
-(a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current source.
-
-> As shown in Fig. 3.23(b), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) Why treat the supermesh differently? Because mesh analysis applies KVL which requires that we know the voltage across each branch—and we do not know the voltage across a current source in advance. However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in Fig. 3.23(b) gives
-
-−20 + 6*i*1 + 10*i*2 + 4*i*2 = 0
-
-$$
-6i_1 + 14i_2 = 20 \tag{3.18}
-$$
-
-We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in Fig. 3.23(a) gives
-
-$$
-i_2 = i_1 + 6 \tag{3.19}
-$$
-
-Solving Eqs. (3.18) and (3.19), we get
-
-$$
-i_1 = -3.2 \text{ A}, \qquad i_2 = 2.8 \text{ A}
-$$
- (3.20)
-
-Note the following properties of a supermesh:
-
-- 1. The current source in the supermesh pro vides the constraint equation necessary to solve for the mesh currents.
-- 2. A supermesh has no current of its own.
-- 3. A supermesh requires the application of both KVL and KCL.
-
-For the circuit in Fig. 3.24, find *i*1 to *i*4 using mesh analysis. Example 3.7
-
-**Figure 3.24**
-
-For Example 3.7.
-
-# **Solution:**
-
-Note that meshes 1 and 2 form a supermesh because they have an independent current source in common. Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown. Applying KVL to the larger supermesh,
-
-2*i*1 + 4*i*3 + 8(*i*3 − *i*4) + 6*i*2 = 0
-
-or
-
-$$
-i_1 + 3i_2 + 6i_3 - 4i_4 = 0 \tag{3.7.1}
-$$
-
-For the independent current source, we apply KCL to node *P*:
-
-$$
-i_2 = i_1 + 5 \tag{3.7.2}
-$$
-
-For the dependent current source, we apply KCL to node *Q*:
-
-$$
-i_2 = i_3 + 3I_o
-$$
-
-But *Io* = −*i*4, hence,
-
-or
-
-$$
-i_2 = i_3 - 3i_4 \tag{3.7.3}
-$$
-
-Applying KVL in mesh 4,
-
-$$
-2i_4 + 8(i_4 - i_3) + 10 = 0
-$$
-
-$$
-5i_4 - 4i_3 = -5 \tag{3.7.4}
-$$
-
-From Eqs. (3.7.1) to (3.7.4),
-
-$$
-i_1 = -7.5 \text{ A}, \qquad i_2 = -2.5 \text{ A}, \qquad i_3 = 3.93 \text{ A}, \qquad i_4 = 2.143 \text{ A}
-$$
-
-**Figure 3.26** (a) The circuit in Fig. 3.2, (b) the circuit in Fig. 3.17.
-
-(b)
-
-Use mesh analysis to determine *i*1, *i*2, and *i*3 in Fig. 3.25.
-
-**Answer:**
-$$
-i_1 = 12.379 \text{ A}, i_2 = 378.9 \text{ mA}, i_3 = 3.284 \text{ A}.
-$$
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/034_3.6 Nodal and Mesh Analyses by Inspection.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/034_3.6 Nodal and Mesh Analyses by Inspection.md
deleted file mode 100644
index bb78f69c2eaa6d95d7185f01c38153d834e9a6dc..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/034_3.6 Nodal and Mesh Analyses by Inspection.md
+++ /dev/null
@@ -1,209 +0,0 @@
-# **3.6** Nodal and Mesh Analyses by Inspection
-
-This section presents a generalized procedure for nodal or mesh analysis. It is a shortcut approach based on mere inspection of a circuit.
-
-When all sources in a circuit are independent current sources, we do not need to apply KCL to each node to obtain the node-v oltage equations as we did in Section 3.2. We can obtain the equations by mere inspection of the circuit. As an e xample, let us ree xamine the circuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.2 as
-
-$$
-\begin{bmatrix} G_1 + G_2 & -G_2 \ -G_2 & G_2 + G_3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} I_1 - I_2 \ I_2 \end{bmatrix}
-$$
- (3.21)
-
-Observe that each of the diagonal terms is the sum of the conductances connected directly to node 1 or 2, while the off-diagonal terms are the negatives of the conductances connected between the nodes. Also, each term on the right-hand side of Eq. (3.21) is the algebraic sum of the currents entering the node.
-
-In general, if a circuit with independent current sources has *N* nonreference nodes, the node-v oltage equations can be written in terms of the conductances as
-
-$$
-\begin{bmatrix} G_{11} & G_{12} & \cdots & G_{1N} \\ G_{21} & G_{22} & \cdots & G_{2N} \\ \vdots & \vdots & \vdots & \vdots \\ G_{N1} & G_{N2} & \cdots & G_{NN} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{bmatrix} = \begin{bmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{bmatrix}
-$$
- (3.22)
-
-or simply
-
-$$
-Gv = i \tag{3.23}
-$$
-
-where
-
-*Gkk* = Sum of the conductances connected to node *k*
-
- *Gk j* = *Gjk* = Negative of the sum of the conductances directly connecting nodes *k* and *j*, *k* ≠ *j*
-
- *vk*= Unknown voltage at node *k*
-
- *ik* = Sum of all independent current sources directly connected to node *k*, with currents entering the node treated as positive
-
-**G** is called the *conductance matrix*; **v** is the output v ector; and **i** is the input vector. Equation (3.22) can be solved to obtain the unknown node voltages. Keep in mind that this is v alid for circuits with only indepen dent current sources and linear resistors.
-
-Similarly, we can obtain mesh-current equations by inspection when a linear resistive circuit has only independent voltage sources. Consider the circuit in Fig. 3.17, shown again in Fig. 3.26(b) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.4 as
-
-$$
-\begin{bmatrix} R_1 + R_3 & -R_3 \ -R_3 & R_2 + R_3 \end{bmatrix} \begin{bmatrix} i_1 \ i_2 \end{bmatrix} = \begin{bmatrix} v_1 \ -v_2 \end{bmatrix}
-$$
- (3.24)
-
-We notice that each of the diagonal terms is the sum of the resistances in the related mesh, while each of the off-diagonal terms is the negative of the resistance common to meshes 1 and 2. Each term on the right-hand side of Eq. (3.24) is the algebraic sum taken clockwise of all independent voltage sources in the related mesh.
-
-In general, if the circuit has *N* meshes, the mesh-current equations can be expressed in terms of the resistances as
-
-$$
-\begin{bmatrix} R_{11} & R_{12} & \cdots & R_{1N} \\ R_{21} & R_{22} & \cdots & R_{2N} \\ \vdots & \vdots & \vdots & \vdots \\ R_{N1} & R_{N2} & \cdots & R_{NN} \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{bmatrix}
-$$
- (3.25)
-
-or simply
-
-$$
-\mathbf{Ri} = \mathbf{v} \tag{3.26}
-$$
-
-where
-
-*Rkk* = Sum of the resistances in mesh *k*
-
-- *Rkj* = *Rjk* = Negative of the sum of the resistances in common with meshes *k* and *j*, *k* ≠ *j*
-- *ik* = Unknown mesh current for mesh *k* in the clockwise direction
-- *vk* = Sum taken clockwise of all independent voltage sources in mesh *k*, with voltage rise treated as positive
-
-**R** is called the *resistance matrix* ; **i** is the output v ector; and **v** is the input v ector. We can solv e Eq. (3.25) to obtain the unkno wn mesh currents.
-
-Example 3.8 Write the node-v oltage matrix equations for the circuit in Fig. 3.27 by inspection.
-
-**Figure 3.27** For Example 3.8.
-
-# **Solution:**
-
-The circuit in Fig. 3.27 has four nonreference nodes, so we need four node equations. This implies that the size of the conductance matrix **G**, is 4 by 4. The diagonal terms of **G**, in siemens, are
-
-$$
-G_{11} = \frac{1}{5} + \frac{1}{10} = 0.3, \qquad G_{22} = \frac{1}{5} + \frac{1}{8} + \frac{1}{1} = 1.325
-$$
-
-$$
-G_{33} = \frac{1}{8} + \frac{1}{8} + \frac{1}{4} = 0.5, \qquad G_{44} = \frac{1}{8} + \frac{1}{2} + \frac{1}{1} = 1.625
-$$
-
-The off-diagonal terms are
-
-$$
-G_{12} = -\frac{1}{5} = -0.2, \qquad G_{13} = G_{14} = 0
-$$
-
-\n
-$$
-G_{21} = -0.2, \qquad G_{23} = -\frac{1}{8} = -0.125, \qquad G_{24} = -\frac{1}{1} = -1
-$$
-
-\n
-$$
-G_{31} = 0, \qquad G_{32} = -0.125, \qquad G_{34} = -\frac{1}{8} = -0.125
-$$
-
-\n
-$$
-G_{41} = 0, \qquad G_{42} = -1, \qquad G_{43} = -0.125
-$$
-
-The input current vector **i** has the following terms, in amperes:
-
-$$
-i_1 = 3
-$$
-, $i_2 = -1 - 2 = -3$ , $i_3 = 0$ , $i_4 = 2 + 4 = 6$
-
-Thus, the node-voltage equations are
-
-$$
-\begin{bmatrix} 0.3 & -0.2 & 0 & 0 \ -0.2 & 1.325 & -0.125 & -1 \ 0 & -0.125 & 0.5 & -0.125 \ 0 & -1 & -0.125 & 1.625 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \ v_4 \end{bmatrix} = \begin{bmatrix} 3 \ -3 \ 0 \ 6 \end{bmatrix}
-$$
-
-which can be solved using *MATLAB* to obtain the node voltages *v*1, *v*2, *v*3, and *v*4.
-
-By inspection, obtain the node-voltage equations for the circuit in Fig. 3.28.
-
-**Answer:**
-
-| 1.25 | −0.2 | −1 | 0 | 0
v1 | |
-|--------|------|-------|-------|------------------------------|--|
-| −0.2 | 0.2 | 0 | 0 | 5
v2 | |
-| −1 | 0 | 1.25 | −0.25 | =
−3
v3 | |
-| [
0 | 0 | −0.25 | 1.25 | ]
[
]
[
2]
v4 | |
-
-By inspection, write the mesh-current equations for the circuit in Fig. 3.29. Example 3.9
-
-# **Figure 3.29**
-
-For Example 3.9.
-
-# **Solution:**
-
-We have five meshes, so the resistance matrix is 5 by 5. The diagonal terms, in ohms, are:
-
-$$
-R_{11} = 5 + 2 + 2 = 9, \qquad R_{22} = 2 + 4 + 1 + 1 + 2 = 10,
-$$
-
-$$
-R_{33} = 2 + 3 + 4 = 9, \qquad R_{44} = 1 + 3 + 4 = 8, \qquad R_{55} = 1 + 3 = 4
-$$
-
-The off-diagonal terms are:
-
-$$
-R_{12} = -2, \t R_{13} = -2, \t R_{14} = 0 = R_{15},
-$$
-
-\n
-$$
-R_{21} = -2, \t R_{23} = -4, \t R_{24} = -1, \t R_{25} = -1,
-$$
-
-\n
-$$
-R_{31} = -2, \t R_{32} = -4, \t R_{34} = 0 = R_{35},
-$$
-
-\n
-$$
-R_{41} = 0, \t R_{42} = -1, \t R_{43} = 0, \t R_{45} = -3,
-$$
-
-\n
-$$
-R_{51} = 0, \t R_{52} = -1, \t R_{53} = 0, \t R_{54} = -3
-$$
-
-The input voltage vector **v** has the following terms in volts:
-
-$$
-v_1 = 4
-$$
-, $v_2 = 10 - 4 = 6$ ,
- $v_3 = -12 + 6 = -6$ , $v_4 = 0$ , $v_5 = -6$
-
-Thus, the mesh-current equations are:
-
-| | 9 | −2 | −2 | 0 | 0 | | i1 | | 4 | |
-|---|----|----|----|----|---------|---|---------|--------|--------|--|
-| | −2 | 10 | −4 | −1 | −1 | | i2 | | 6 | |
-| | −2 | −4 | 9 | 0 | 0 | | i3 | | −6 | |
-| [ | 0 | −1 | 0 | 8 | −3
] | | i4 | =
[ | 0
] | |
-| | 0 | −1 | 0 | −3 | 4 | [ | ]
i5 | | −6 | |
-
-From this, we can use *MATLAB* to obtain mesh currents *i*1, *i*2, *i*3, *i*4, and *i*5.
-
-# Practice Problem 3.9
-
-By inspection, obtain the mesh-current equations for the circuit in Fig. 3.30.
-
-**Figure 3.30** For Practice Prob. 3.9.
-
-# **Answer:**
-
-| | 150 | −40 | 0 | −80 | 0 | i1 | | 30 | |
-|---|-----|-----|-----|-----|--------|---------|-------------|---------|--|
-| | −40 | 65 | −30 | −15 | 0 | i2 | | 0 | |
-| | −0 | −30 | 50 | 0 | −20 | i3 | | −12 | |
-| [ | 80 | −15 | 0 | 95 | 0
] | i4
[ | =
[
] | 20
] | |
-| | 0 | 0 | −20 | 0 | 80 | i5 | | −20 | |
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/035_3.7 Nodal Versus Mesh Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/035_3.7 Nodal Versus Mesh Analysis.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/035_3.7 Nodal Versus Mesh Analysis.md
+++ /dev/null
@@ -1,9 +0,0 @@
-# **3.7** Nodal Versus Mesh Analysis
-
-Both nodal and mesh analyses provide a systematic way of analyzing a complex network. Someone may ask: Gi ven a network to be analyzed, how do we know which method is better or more efficient? The choice of the better method is dictated by two factors.
-
-The first factor is the nature of the particular network. Networks that contain many series-connected elements, voltage sources, or supermeshes are more suitable for mesh analysis, whereas networks with parallelconnected elements, current sources, or supernodes are more suitable for nodal analysis. Also, a circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. The key is to select the method that results in the smaller number of equations.
-
-The second factor is the information required. If node voltages are required, it may be expedient to apply nodal analysis. If branch or mesh currents are required, it may be better to use mesh analysis.
-
-It is helpful to be familiar with both methods of analysis, for at least two reasons. First, one method can be used to check the results from the other method, if possible. Second, since each method has its limitations, only one method may be suitable for a particular problem. For example, mesh analysis is the only method to use in analyzing transistor circuits, as we shall see in Section 3.9. But mesh analysis cannot easily be used to solve an op amp circuit, as we shall see in Chapter 5, because there is no direct way to obtain the voltage across the op amp itself. For nonplanar networks, nodal analysis is the only option, because mesh analysis only applies to planar networks. Also, nodal analysis is more amenable to solution by computer, as it is easy to program. This allows one to analyze complicated circuits that defy hand calculation. A computer software package based on nodal analysis is introduced next.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/036_3.8 Circuit Analysis with PSpice.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/036_3.8 Circuit Analysis with PSpice.md
deleted file mode 100644
index e1e4689a6597eabf4ee2f4af4b33c8318135ace5..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/036_3.8 Circuit Analysis with PSpice.md
+++ /dev/null
@@ -1,42 +0,0 @@
-# **3.8** Circuit Analysis with PSpice
-
-*PSpice* is a computer software circuit analysis program that we will gradually learn to use throughout the course of this text. This section illustrates how to use *PSpice for Windows* to analyze the dc circuits we have studied so far.
-
-The reader is expected to review the tutorial before proceeding in this section. It should be noted that *PSpice* is only helpful in determining branch voltages and currents when the numerical values of all the circuit components are known.
-
-A tutorial on using PSpice for Windows can be found in Connect.
-
-Use *PSpice* to find the node voltages in the circuit of Fig. 3.31. Example 3.10
-
-# **Solution:**
-
-The first step is to draw the given circuit using Schematics. If one follows the instructions given in the PSpice tutorial found in Connect, the schematic in Fig. 3.32 is produced. Because this is a dc analysis, we use voltage source VDC and current source IDC. The pseudocomponent VIEWPOINTS are added to display the required node voltages. Once the circuit is drawn and saved as *exam310.sch*, we run *PSpice* by selecting **Analysis/Simulate**. The circuit is simulated and the results are displayed
-
-**Figure 3.32** For Example 3.10; the schematic of the circuit in Fig. 3.31.
-
-on VIEWPOINTS and also saved in output file *exam310.out*. The output file includes the following:
-
-| | NODE VOLTAGE | | NODE VOLTAGE NODE VOLTAGE | | | |
-|-------------------------------------------------------------------|--------------|--|---------------------------|--|---------|--|
-| (1) | 120.0000 (2) | | 81.2900 (3) | | 89.0320 | |
-| indicating that V1
= 120 V, V2
= 81.29 V, V3
= 89.032 V. | | | | | | |
-
-Practice Problem 3.10 For the circuit in Fig. 3.33, use *PSpice* to find the node voltages.
-
-**Answer:** *V*1 = −10 V, *V*2 = 14.286 V, *V*3 = 50 V.
-
-Example 3.11 In the circuit of Fig. 3.34, determine the currents *i*1, *i*2, and *i*3.
-
-For Example 3.11.
-
-# **Solution:**
-
-The schematic is shown in Fig. 3.35. (The schematic in Fig. 3.35 includes the output results, implying that it is the schematic displayed on the screen *after* the simulation.) Notice that the voltage-controlled voltage source E1 in Fig. 3.35 is connected so that its input is the voltage across the 4- Ω resistor; its gain is set equal to 3. In order to dis play the required currents, we insert pseudocomponent IPROBES in the appropriate branches. The schematic is saved as *exam311.sch* and simulated by selecting **Analysis/Simulate**. The results are displayed on IPROBES as shown in Fig. 3.35 and saved in output file *exam311.out*. From the output file or the IPROBES, we obtain *i*1 = *i*2 = 1.333 A and *i*3 = 2.667 A.
-
-**Figure 3.35** The schematic of the circuit in Fig. 3.34.
-
-Use *PSpice* to determine currents *i*1, *i*2, and *i*3 in the circuit of Fig. 3.36.
-
-**Answer:** *i*1 = −428.6 mA, *i*2 = 2.286 A, *i*3 = 2 A.
-
-# **3.9** †
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/037_3.9 Applications - DC Transistor Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/037_3.9 Applications - DC Transistor Circuits.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/037_3.9 Applications - DC Transistor Circuits.md
+++ /dev/null
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-# Applications: DC Transistor Circuits
-
-Most of us deal with electronic products on a routine basis and ha ve some e xperience with personal computers. A basic component for the integrated circuits found in these electronics and computers is the active, three-terminal device known as the *transistor*. Understanding the transistor is essential before an engineer can start an electronic circuit design.
-
-Figure 3.37 depicts various kinds of transistors commercially available. There are tw o basic types of transistors: *bipolar junction tr ansistors* (BJTs) and *field-effect transistors* (FETs). Here, we consider only the BJTs, which were the first of the two and are still used today . Our objective is to present enough detail about the BJT to enable us to apply the techniques developed in this chapter to analyze dc transistor circuits.
-
-Practice Problem 3.11
-
-Courtesy of Lucent Technologies/Bell Labs
-
-# Historical
-
-**William Schockley** (1910–1989), **John Bardeen** (1908–1991), and **Walter Brattain** (1902–1987) co-invented the transistor.
-
-Nothing has had a greater impact on the transition from the "Industrial Age" to the "Age of the Engineer" than the transistor. I am sure that Dr. Shockley, Dr. Bardeen, and Dr. Brattain had no idea they would have this incredible effect on our history. While working at Bell Laboratories, they successfully demonstrated the point-contact transistor, invented by Bardeen and Brattain in 1947, and the junction transistor, which Shockley conceived in 1948 and successfully produced in 1951.
-
-It is interesting to note that the idea of the field-effect transistor, the most commonly used one today, was first conceived in 1925–1928 by J. E. Lilienfeld, a German immigrant to the United States. This is evident from his patents of what appears to be a field-effect transistor. Unfortunately, the technology to realize this device had to wait until 1954 when Shockley's field-effect transistor became a reality. Just think what today would be like if we had this transistor 30 years earlier!
-
-For their contributions to the creation of the transistor, Dr. Shockley, Dr. Bardeen, and Dr. Brattain received, in 1956, the Nobel Prize in physics. It should be noted that Dr. Bardeen is the only individual to win two Nobel prizes in physics; the second came later for work in superconductivity at the University of Illinois.
-
-**Figure 3.38** Two types of BJTs and their circuit symbols: (a) *npn*, (b) *pnp*.
-
-**Figure 3.37** Various types of transistors. *(© McGraw-Hill Education/Mark Dierker, photographer)*
-
-There are tw o types of BJTs: *npn* and *pnp*, with their circuit sym bols as shown in Fig. 3.38. Each type has three terminals, designated as emitter (E), base (B), and collector (C). For the *npn* transistor, the currents and voltages of the transistor are specified as in Fig. 3.39. Applying KCL to Fig. 3.39(a) gives
-
-$$
-I_E = I_B + I_C \tag{3.27}
-$$
-
-where *IE*, *IC*, and *IB* are emitter, collector, and base currents, respectively. Similarly, applying KVL to Fig. 3.39(b) gives
-
-$$
-V_{CE} + V_{EB} + V_{BC} = 0 \tag{3.28}
-$$
-
-where *VCE*, *VEB*, and *VBC* are collector -emitter, emitter-base, and basecollector voltages. The BJT can operate in one of three modes: acti ve, cutoff, and saturation. When transistors operate in the active mode, typically *VBE* ≃ 0.7 V,
-
-$$
-I_C = \alpha I_E \tag{3.29}
-$$
-
-where *α* is called the *common-base curr ent gain*. In Eq. (3.29), *α* denotes the fraction of electrons injected by the emitter that are col lected by the collector. Also,
-
-$$
-I_C = \beta I_B \tag{3.30}
-$$
-
-where *β* is known as the *common-emitter current gain*. The *α* and *β* are characteristic properties of a given transistor and assume constant values for that transistor. Typically, *α* takes values in the range of 0.98 to 0.999, while *β* takes values in the range of 50 to 1000. From Eqs. (3.27) to (3.30), it is evident that
-
-$$
-I_E = (1 + \beta)I_B \tag{3.31}
-$$
-
-and
-
-$$
-\beta = \frac{\alpha}{1 - \alpha} \tag{3.32}
-$$
-
-These equations show that, in the active mode, the BJT can be modeled as a dependent current-controlled current source. Thus, in circuit analysis, the dc equi valent model in Fig. 3.40(b) may be used to replace the *npn* transistor in Fig. 3.40(a). Since *β* in Eq. (3.32) is large, a small base current controls lar ge currents in the output circuit. Consequently , the bipolar transistor can serve as an amplifier, producing both current gain and voltage gain. Such amplifiers can be used to furnish a considerable amount of power to transducers such as loudspeakers or control motors.
-
-(a) An *npn* transistor, (b) its dc equivalent model.
-
-It should be observ ed in the follo wing e xamples that one cannot directly analyze transistor circuits using nodal analysis because of the potential difference between the terminals of the transistor . Only when the transistor is replaced by its equivalent model can we apply nodal analysis.
-
-# **Figure 3.39**
-
-The terminal variables of an *npn* transistor: (a) currents, (b) voltages.
-
-In fact, transistor circuits provide motivation to study dependent sources.
-
-Example 3.12 Find *IB*, *IC*, and *vo* in the transistor circuit of Fig. 3.41. Assume that the transistor operates in the active mode and that *β* = 50.
-
-**Figure 3.41** For Example 3.12.
-
-# **Solution:**
-
-For the input loop, KVL gives
-
-$$
--4 + I_B (20 \times 10^3) + V_{BE} = 0
-$$
-
-Since *VBE* = 0.7 V in the active mode,
-
-$$
-I_B = \frac{4 - 0.7}{200 \times 10^3} = 16.5 \,\mu\text{A}
-$$
-
-But
-
-$$
-I_C = \beta I_B = 50 \times 16.5 \,\mu A = 0.825 \text{ mA}
-$$
-
-For the output loop, KVL gives
-
-$$
--v_o - 100I_C + 6 = 0
-$$
-
-or
-
-+
-
-$$
-v_o = 6 - 100I_C = 6 - 0.0825 = 5.917
-$$
- V
-
-Note that *vo* = *VCE* in this case.
-
-**Answer:** 4.691 V, 888.5 mV.
-
-Practice Problem 3.12
-
-For Practice Prob. 3.12.
-
-For the BJT circuit in Fig. 3.43, *β* = 150 and *VBE* = 0.7 V. Find *vo*.
-
-# **Solution:**
-
-- 1. **Define.** The circuit is clearly defined and the problem is clearly stated. There appear to be no additional questions that need to be asked.
-- 2. **Present.** We are to determine the output voltage of the circuit shown in Fig. 3.43. The circuit contains an ideal transistor with *β* = 150 and *VBE* = 0.7 V.
-- 3. **Alternative.** We can use mesh analysis to solv e for *vo*. We can re place the transistor with its equivalent circuit and use nodal analysis. We can try both approaches and use them to check each other . As a third check, we can use the equi valent circuit and solv e it using *PSpice*.
-
-4. **Attempt.**
-
-■ **METHOD 1** Working with Fig. 3.44(a), we start with the first loop.
-
-−2 + 100k*I*1 + 200k(*I*1 − I2) = 0 or 3*I*1 − 2*I*2 = 2 × 10−5 **(3.13.1)**
-
-**Figure 3.44** Solution of the problem in Example 3.13: (a) Method 1, (b) Method 2, (c) Method 3.
-
-Example 3.13
-
-1 kΩ
-
-Now for loop 2.
-
-$$
-200k(I_2 - I_1) + V_{BE} = 0 \qquad \text{or} \qquad -2I_1 + 2I_2 = -0.7 \times 10^{-5}
-$$
-\n(3.13.2)
-
-Since we have two equations and two unknowns, we can solve for *I*1 and *I*2. Adding Eq. (3.13.1) to (3.13.2) we get;
-
-*I*1 = 1.3 × 10−5 A and *I*2 = (−0.7 + 2.6)10−5 ∕2 = 9.5 *µ*A Since *I*3 = −150*I*2 = −1.425 mA, we can now solve for *vo* using loop 3: −*vo* + 1 k*I*3 + 16 = 0 or *vo* = −1.425 + 16 = **14.575 V**
-
-■ **METHOD 2** Replacing the transistor with its equivalent circuit produces the circuit shown in Fig. 3.44(b). We can now use nodal analysis to solve for *vo*.
-
-At node number 1: *V*1 = 0.7 V
-
-$$
-(0.7 - 2)/100k + 0.7/200k + I_B = 0 \qquad \text{or} \qquad I_B = 9.5 \,\mu\text{A}
-$$
-
-At node number 2 we have:
-
-150*IB* + (*vo* − 16)∕1k = 0 or *vo* = 16 − 150 × 103 × 9.5 × 10−6 = **14.575 V**
-
-- 5. **Evaluate.** The answers check, b ut to further check we can use *PSpice* (Method 3), which gi ves us the solution sho wn in Fig. 3.44(c).
-- 6. **Satisfactory?** Clearly, we have obtained the desired answer with a very high confidence level. We can now present our work as a solution to the problem.
-
-Practice Problem 3.13
-
-The transistor circuit in Fig. 3.45 has *β* = 80 and *VBE* = 0.7 V. Find *vo* and *Io*.
-
-**Figure 3.45** For Practice Prob. 3.13. **Answer:** 12 V, 600 *µ* A.
-
-- 1. Nodal analysis is the application of Kirchhoff's current law at the nonreference nodes. (It is applicable to both planar and nonplanar circuits.) We express the result in terms of the node voltages. Solv ing the simultaneous equations yields the node voltages.
-- 2. A supernode consists of two nonreference nodes connected by a (dependent or independent) voltage source.
-- 3. Mesh analysis is the application of Kirchhoff's voltage law around meshes in a planar circuit. We express the result in terms of mesh currents. Solving the simultaneous equations yields the mesh currents.
-
-- 4. A supermesh consists of two meshes that have a (dependent or independent) current source in common.
-- 5. Nodal analysis is normally used when a circuit has fewer node equations than mesh equations. Mesh analysis is normally used when a circuit has fewer mesh equations than node equations.
-- 6. Circuit analysis can be carried out using *PSpice*.
-- 7. DC transistor circuits can be analyzed using the techniques covered in this chapter.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/038_3.10 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/038_3.10 Summary.md
deleted file mode 100644
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@@ -1,54 +0,0 @@
-# Review Questions
-
-**3.1** At node 1 in the circuit of Fig. 3.46, applying KCL gives:
-
-(a)
-$$
-2 + \frac{12 - v_1}{3} = \frac{v_1}{6} + \frac{v_1 - v_2}{4}
-$$
-
-\n(b) $2 + \frac{v_1 - 12}{3} = \frac{v_1}{6} + \frac{v_2 - v_1}{4}$
-\n(c) $2 + \frac{12 - v_1}{3} = \frac{0 - v_1}{6} + \frac{v_1 - v_2}{4}$
-\n(d) $2 + \frac{v_1 - 12}{3} = \frac{0 - v_1}{6} + \frac{v_2 - v_1}{4}$
-
-# **Figure 3.46**
-
-- For Review Questions 3.1 and 3.2.
-- **3.2** In the circuit of Fig. 3.46, applying KCL at node 2 gives:
-
-**3.3** For the circuit in Fig. 3.47, *v*1 and *v*2 are related as:
-
-(a)
-$$
-v_1 = 6i + 8 + v_2
-$$
-
-\n(b) $v_1 = 6i - 8 + v_2$
-\n(c) $v_1 = -6i + 8 + v_2$
-\n(d) $v_1 = -6i - 8 + v_2$
-
-# **Figure 3.47**
-
-For Review Questions 3.3 and 3.4.
-
-**3.4** In the circuit of Fig. 3.47, the voltage *v*2 is:
-
-| (a) −8 V | (b) −1.6 V |
-|-----------|------------|
-| (c) 1.6 V | (d) 8 V |
-
-**3.5** The current *i* in the circuit of Fig. 3.48 is:
-
-| (a) −2.667 A | (b) −0.667 A |
-|--------------|--------------|
-| (c) 0.667 A | (d) 2.667 A |
-
-# **Figure 3.48**
-
-For Review Questions 3.5 and 3.6.
-
-- **3.6** The loop equation for the circuit in Fig. 3.48 is:
- - (a) −10 + 4*i* + 6 + 2*i* = 0 (b) 10 + 4*i* + 6 + 2*i* = 0 (c) 10 + 4*i* − 6 + 2*i* = 0 (d) −10 + 4*i* − 6 + 2*i* = 0
-
-- **3.7** In the circuit of Fig. 3.49, current *i*1 is:
- - (a) 4 A (b) 3 A (c) 2 A (d) 1 A
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/039_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/039_Review Questions.md
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+++ /dev/null
@@ -1,23 +0,0 @@
-# **Figure 3.49**
-
-For Review Questions 3.7 and 3.8.
-
-**3.8** The voltage *v* across the current source in the circuit of Fig. 3.49 is:
-
-| (a) 20 V | (b) 15 V | (c) 10 V | (d) 5 V |
-|----------|----------|----------|---------|
-|----------|----------|----------|---------|
-
-**3.9** The *PSpice* part name for a current-controlled voltage source is:
-
-(a) EX (b) FX (c) HX (d) GX
-
-- **3.10** Which of the following statements are not true of the pseudocomponent IPROBE:
- - (a) It must be connected in series.
- - (b) It plots the branch current.
- - (c) It displays the current through the branch in which it is connected.
- - (d) It can be used to display voltage by connecting it in parallel.
- - (e) It is used only for dc analysis.
- - (f) It does not correspond to a particular circuit element.
-
-*Answers: 3.1a, 3.2c, 3.3a, 3.4c, 3.5c, 3.6a, 3.7d, 3.8b, 3.9c, 3.10b,d.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/040_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/040_Problems.md
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@@ -1,383 +0,0 @@
-# Problems
-
-# Sections 3.2 and 3.3 Nodal Analysis
-
-**3.1** Using Fig. 3.50, design a problem to help other students better understand nodal analysis.
-
-# **Figure 3.50**
-
-For Prob. 3.1 and Prob. 3.39.
-
-**3.2** For the circuit in Fig. 3.51, obtain *v*1 and *v*2.
-
-**Figure 3.51** For Prob. 3.2.
-
-**3.3** Find the currents *I*1 through *I*4 and the voltage *vo* in the circuit of Fig. 3.52.
-
-**Figure 3.52** For Prob. 3.3.
-
-**3.4** Given the circuit in Fig. 3.53, calculate the currents *i*1 through *i*4.
-
-For Prob. 3.4.
-
-**3.5** Obtain *vo* in the circuit of Fig. 3.54.
-
-For Prob. 3.5.
-
-**3.6** Solve for *V*1 in the circuit of Fig. 3.55 using nodal analysis.
-
-**3.7** Apply nodal analysis to solve for *Vx* in the circuit of Fig. 3.56.
-
-For Prob. 3.7.
-
-**3.8** Using nodal analysis, find *vo* in the circuit of Fig. 3.57.
-
-**Figure 3.57** For Prob. 3.8 and Prob. 3.37.
-
-**3.9** Determine *Ib* in the circuit in Fig. 3.58 using nodal analysis.
-
-**Figure 3.58** For Prob. 3.9.
-
-**3.10** Find *Io* in the circuit of Fig. 3.59.
-
-**Figure 3.59** For Prob. 3.10.
-
-**3.11** Find *Vo* and the power dissipated in all the resistors in the circuit of Fig. 3.60.
-
-**Figure 3.60** For Prob. 3.11.
-
-**3.12** Using nodal analysis, determine *Vo* in the circuit in Fig. 3.61.
-
-**Figure 3.61** For Prob. 3.12.
-
-**3.13** Calculate *v*1 and *v*2 in the circuit of Fig. 3.62 using nodal analysis.
-
-**Figure 3.62**
-
-For Prob. 3.13.
-
-**3.14** Using nodal analysis, find *vo* in the circuit of Fig. 3.63.
-
-# **Figure 3.63**
-
-For Prob. 3.14.
-
-**3.15** Apply nodal analysis to find *io* and the power dissipated in each resistor in the circuit of Fig. 3.64.
-
-For Prob. 3.15.
-
-**3.16** Determine voltages *v*1 through *v*3 in the circuit of Fig. 3.65 using nodal analysis.
-
-For Prob. 3.16.
-
-**3.17** Using nodal analysis, find current *io* in the circuit of Fig. 3.66.
-
-**Figure 3.66** For Prob. 3.17.
-
-> **3.18** Determine the node voltages in the circuit in Fig. 3.67 using nodal analysis.
-
-For Prob. 3.18.
-
-**3.19** Use nodal analysis to find *v*1, *v*2, and *v*3 in the circuit of Fig. 3.68.
-
-**Figure 3.68** For Prob. 3.19.
-
-**3.20** For the circuit in Fig. 3.69, find *v*1, *v*2, and *v*3 using nodal analysis.
-
-# **Figure 3.69**
-
-For Prob. 3.20.
-
-**3.21** For the circuit in Fig. 3.70, find *v*1 and *v*2 using nodal analysis.
-
-# **Figure 3.70**
-
-For Prob. 3.21.
-
-**3.22** Determine *v*1 and *v*2 in the circuit of Fig. 3.71.
-
-# **Figure 3.71** For Prob. 3.22.
-
-**3.23** Use nodal analysis to find *Vo* in the circuit of Fig. 3.72.
-
-For Prob. 3.23.
-
-**3.24** Use nodal analysis and *MATLAB* to find *Vo* in the circuit of Fig. 3.73.
-
-# **Figure 3.73** For Prob. 3.24.
-
-- **3.25** Use nodal analysis along with *MATLAB* to determine
- - the node voltages in Fig. 3.74.
-
-**3.26** Calculate the node voltages *v*1, *v*2, and *v*3 in the circuit of Fig. 3.75.
-
-**3.27** Use nodal analysis to determine voltages *v*1, *v*2, and *v*3 in the circuit of Fig. 3.76. \*
-
-# **Figure 3.76**
-
-For Prob. 3.27.
-
-**3.28** Use *MATLAB* to find the voltages at nodes *a*, *b*, *c*, and *d* in the circuit of Fig. 3.77. \*
-
-# **Figure 3.77**
-
-For Prob. 3.28.
-
-**3.29** Use *MATLAB* to solve for the node voltages in the circuit of Fig. 3.78.
-
-# **Figure 3.78**
-
-For Prob. 3.29.
-
-\* An asterisk indicates a challenging problem.
-
-**3.30** Using nodal analysis, find *vo* and *io* in the circuit of Fig. 3.79.
-
-**3.31** Find the node voltages for the circuit in Fig. 3.80.
-
-**Figure 3.80** For Prob. 3.31.
-
-**3.32** Obtain the node voltages *v*1, *v*2, and *v*3 in the circuit of Fig. 3.81.
-
-**Figure 3.81** For Prob. 3.32.
-
-# Sections 3.4 and 3.5 Mesh Analysis
-
-**3.33** Which of the circuits in Fig. 3.82 is planar? For the planar circuit, redraw the circuits with no crossing branches.
-
-**3.34** Determine which of the circuits in Fig. 3.83 is planar and redraw it with no crossing branches.
-
-# **Figure 3.83**
-
-For Prob. 3.34.
-
-- **3.35** Rework Prob. 3.5 using mesh analysis.
-- **3.36** Use mesh analysis to obtain *ia*, *ib*, and *ic* in the circuit in Fig. 3.84.
-
-**Figure 3.84** For Prob. 3.36.
-
-**3.37** Solve Prob. 3.8 using mesh analysis.
-
-**3.38** Apply mesh analysis to the circuit in Fig. 3.85 and obtain *Io*.
-
-**Figure 3.85** For Prob. 3.38.
-
-**3.39** Using Fig. 3.50 from Prob. 3.1, design a problem to help other students better understand mesh analysis.
-
-**3.40** For the bridge network in Fig. 3.86, find *io* using mesh analysis.
-
-**3.41** Apply mesh analysis to find *i* in Fig. 3.87.
-
-**Figure 3.87** For Prob. 3.41.
-
-# **Figure 3.88**
-
-For Prob. 3.42.
-
-**3.43** Use mesh analysis to find *vab* and *io* in the circuit of Fig. 3.89.
-
-**Figure 3.89** For Prob. 3.43.
-
-**3.44** Use mesh analysis to obtain *io* in the circuit of Fig. 3.90.
-
-**Figure 3.90** For Prob. 3.44.
-
-**3.45** Find current *i* in the circuit of Fig. 3.91.
-
-For Prob. 3.45.
-
-**3.46** Calculate the mesh currents *i*1 and *i*2 in Fig. 3.92.
-
-For Prob. 3.46.
-
-**3.47** Rework Prob. 3.19 using mesh analysis.
-
-**3.48** Determine the current through the 10-kΩ resistor in the circuit of Fig. 3.93 using mesh analysis.
-
-**3.49** Find *vo* and *io* in the circuit of Fig. 3.94.
-
-**Figure 3.94** For Prob. 3.49.
-
-**3.50** Use mesh analysis to find the current *io* in the circuit of Fig. 3.95.
-
-For Prob. 3.50.
-
-**3.51** Apply mesh analysis to find *vo* in the circuit of Fig. 3.96.
-
-**Figure 3.96** For Prob. 3.51.
-
-**3.52** Use mesh analysis to find *i*1, *i*2, and *i*3 in the circuit of Fig. 3.97.
-
-**3.53** Find the mesh currents in the circuit of Fig. 3.98 using *MATLAB*.
-
-**Figure 3.98** For Prob. 3.53.
-
-**3.54** Find the mesh currents *i*1, *i*2, and *i*3 in the circuit in Fig. 3.99.
-
-**Figure 3.99**
-
-For Prob. 3.54.
-
-# **Figure 3.100** For Prob. 3.55.
-
-**3.56** Determine *v*1 and *v*2 in the circuit of Fig. 3.101.
-
-# **Figure 3.101**
-
-For Prob. 3.56.
-
-**3.57** In the circuit of Fig. 3.102, find the values of *R*, *V*1, and *V*2 given that *io* = 15 mA.
-
-**3.58** Find *i*1, *i*2, and *i*3 in the circuit of Fig. 3.103.
-
-# **Figure 3.103**
-
-For Prob. 3.58.
-
-**3.59** Rework Prob. 3.30 using mesh analysis.
-
-**3.60** Calculate the power dissipated in each resistor in the circuit of Fig. 3.104.
-
-# **Figure 3.104**
-
-For Prob. 3.60.
-
-**3.61** Calculate the current gain *io*∕*is* in the circuit of Fig. 3.105.
-
-**Figure 3.105**
-
-For Prob. 3.61.
-
-**3.62** Find the mesh currents *i*1, *i*2, and *i*3 in the network of Fig. 3.106.
-
-For Prob. 3.62.
-
-**3.63** Find *vx* and *ix* in the circuit shown in Fig. 3.107.
-
-**3.66** Write a set of mesh equations for the circuit in Fig. 3.110. Use *MATLAB* to determine the mesh currents.
-
-# **Figure 3.110** For Prob. 3.66.
-
-# Section 3.6 Nodal and Mesh Analyses by Inspection
-
-**3.67** Obtain the node-voltage equations for the circuit in Fig. 3.111 by inspection. Then solve for *Vo*.
-
-# **Figure 3.111**
-
-For Prob. 3.67.
-
-**3.68** Using Fig. 3.112, design a problem, to solve for *Vo*, to help other students better understand nodal analysis. Try your best to come up with values to make the calculations easier.
-
-**Figure 3.108** For Prob. 3.64.
-
-**3.65** Use *MATLAB* to solve for the mesh currents in the circuit of Fig. 3.109. 6 V
-
-**Figure 3.109** For Prob. 3.65.
-
-**3.69** For the circuit shown in Fig. 3.113, write the nodevoltage equations by inspection.
-
-For Prob. 3.69.
-
-**3.70** Write the node-voltage equations by inspection and then determine values of *V*1 and *V*2 in the circuit of Fig. 3.114.
-
-For Prob. 3.70.
-
-**3.71** Write the mesh-current equations for the circuit in Fig. 3.115. Next, determine the values of *i*1, *i*2, and *i*3.
-
-For Prob. 3.71.
-
-**3.72** By inspection, write the mesh-current equations for the circuit in Fig. 3.116.
-
-# **Figure 3.116**
-
-For Prob. 3.72.
-
-**3.73** Write the mesh-current equations for the circuit in Fig. 3.117.
-
-# **Figure 3.117**
-
-For Prob. 3.73.
-
-**3.74** By inspection, obtain the mesh-current equations for the circuit in Fig. 3.118.
-
-# **Figure 3.118** For Prob. 3.74.
-
-Section 3.8 Circuit Analysis with PSpice or MultiSim
-
-- **3.75** Use *PSpice* or *MultiSim* to solve Prob. 3.58.
-- **3.76** Use *PSpice* or *MultiSim* to solve Prob. 3.27.
-
-**3.77** Solve for *V*1 and *V*2 in the circuit of Fig. 3.119 using *PSpice* or *MultiSim*.
-
-# **Figure 3.119**
-
-For Prob. 3.77.
-
-- **3.78** Solve Prob. 3.20 using *PSpice* or *MultiSim*.
-- **3.79** Rework Prob. 3.28 using *PSpice* or *MultiSim*.
-- **3.80** Find the nodal voltages *v*1 through *v*4 in the circuit of Fig. 3.120 using *PSpice* or *MultiSim*.
-
-# **Figure 3.120**
-
-For Prob. 3.80.
-
-- **3.81** Use *PSpice* or *MultiSim* to solve the problem in Example 3.4.
-- **3.82** If the Schematics Netlist for a network is as follows, draw the network.
-
-| R_R1 | 1 | 2 | 2K | |
-|---------|---|---|--------------|--------------|
-| R_R2 | 2 | 0 | 4K | |
-| R_R3 | 3 | 0 | 8K | |
-| R R4 | 3 | 4 | 6K | |
-| R_R5 | 1 | 3 | 3K | |
-| V_VS | 4 | 0 | DC | 100 |
-| I IS. | 0 | 1 | DC | 4 |
-| $F_F1$ | 1 | 3 | VF F1 | $\mathsf{Z}$ |
-| $VF_F1$ | 5 | 0 | 0V | |
-| E E1 | 3 | 2 | $\mathbf{1}$ | 3 |
-| | | | | |
-
-**3.83** The following program is the Schematics Netlist of a particular circuit. Draw the circuit and determine the voltage at node 2.
-
-| $\frac{1}{2}$ | | | | |
-|---------------|--------------|---|-----|--|
-| R R1 | 1 | 2 | 20 | |
-| R R2 | $\mathsf{Z}$ | Ø | 50 | |
-| R R3 | 2 | 3 | 70 | |
-| R R4 | 3. | 0 | 30 | |
-| v vs | 1 | a | 20V | |
-| I IS | 2 | a | DC. | |
-
-# Section 3.9 Applications
-
-**3.84** Calculate *vo* and *Io* in the circuit of Fig. 3.121.
-
-- **3.85** An audio amplifier with a resistance of 9 Ω supplies power to a speaker. What should be the resistance of the speaker for maximum power to be delivered?
-- **3.86** For the simplified transistor circuit of Fig. 3.122, calculate the voltage *vo*.
-
-**Figure 3.122** For Prob. 3.86.
-
-**3.87** For the circuit in Fig. 3.123, find the gain *vo*∕*vs*.
-
-**3.88** Determine the gain *vo*∕*vs* of the transistor amplifier circuit in Fig. 3.124. \*
-
-**Figure 3.124**
-
-- For Prob. 3.88.
- - **3.89** For the transistor circuit shown in Fig. 3.125, find *IB* and *VCE*. Let *β* = 100, and *VBE* = 0.7 V.
-
-**Figure 3.125**
-
-For Prob. 3.89.
-
-**3.90** Calculate *vs* for the transistor in Fig. 3.126 given that *vo* = 6 V, *β* = 90, *VBE* = 0.7 V.
-
-**Figure 3.126** For Prob. 3.90.
-
-\***3.93** Rework Example 3.11 with hand calculation.
-
-**3.91** For the transistor circuit of Fig. 3.127, find *IB*, *VCE*, and *vo*. Take *β* = 150, *VBE* = 0.7 V.
-
-**3.92** Using Fig. 3.128, design a problem to help other students better understand transistors. Make sure you use reasonable numbers!
-
-**Figure 3.128** For Prob. 3.92.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/041_Comprehensive Problem.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/041_Comprehensive Problem.md
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-# **chapter**
-
-4
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/042_Chapter 4 - Circuit Theorems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/042_Chapter 4 - Circuit Theorems.md
deleted file mode 100644
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-# Circuit Theorems
-
-*Your success as an engineer will be directly proportional to your ability to communicate!*
-
-—Charles K. Alexander
-
-# Enhancing Your Skills and Your Career
-
-# **Enhancing Your Communication Skills**
-
-Taking a course in circuit analysis is one step in preparing yourself for a career in electrical engineering. Enhancing your communication skills while in school should also be part of that preparation, as a large part of your time will be spent communicating.
-
-People in industry have complained again and again that graduating engineers are ill-prepared in written and oral communication. An engineer who communicates effectively becomes a valuable asset.
-
-You can probably speak or write easily and quickly. But how *effectively* do you communicate? The art of effective communication is of the utmost importance to your success as an engineer.
-
-For engineers in industry , communication is k ey to promotability . Consider the result of a survey of U.S. corporations that asked what factors influence managerial promotion. The survey includes a listing of 22 personal qualities and their importance in adv ancement. You may be surprised to note that "technical skill based on experience" placed fourth from the bottom. Attributes such as self-confidence, ambition, flexibility, maturity, ability to mak e sound decisions, getting things done with and through people, and capacity for hard work all ranked higher. At the top of the list w as "ability to communicate. " The higher your professional career progresses, the more you will need to communicate. Therefore, you should regard effective communication as an important tool in your engineering tool chest.
-
-Learning to communicate ef fectively is a lifelong task you should always work toward. The best time to begin is while still in school. Continually look for opportunities to de velop and strengthen your reading, writing, listening, and speaking skills. You can do this through classroom presentations, team projects, acti ve participation in student or ganizations, and enrollment in communication courses. The risks are less now than later in the workplace.
-
-*Ability to communicate effectively is regarded by many as the most important step to an executive promotion.* © IT Stock/PunchStock RF
-
-# Learning Objectives
-
-By using the information and exercises in this chapter you will be able to:
-
-- 1. Develop and enhance your skills in using nodal analysis and mesh analysis to analyze basic circuits.
-- 2. Understand how linearity works with basic circuits.
-- 3. Explain the principle of superposition and how it can be used to help analyze circuits.
-- 4. Understand the value of source transformation and how it can be used to simplify circuits.
-- 5. Recognize Thevenin's and Norton's theorems and know how they can lead to greatly simplified circuits.
-- 6. Explain the maximum power transfer concept.
-
-# **4.1** Introduction
-
-A major adv antage of analyzing circuits using Kirchhof f's laws as we did in Chapter 3 is that we can analyze a circuit without tampering with its original configuration. A major disadvantage of this approach is that, for a large, complex circuit, tedious computation is involved.
-
-The growth in areas of application of electric circuits has led to an evolution from simple to complex circuits. To handle the complexity, engineers over the years have developed some theorems to simplify circuit analysis. Such theorems include Thevenin's and Norton's theorems. Since these theorems are applicable to *linear* circuits, we first discuss the concept of circuit linearity. In addition to circuit theorems, we discuss the concepts of superposition, source transformation, and maximum power transfer in this chapter. The concepts we de velop are applied in the last section to source modeling and resistance measurement.
-
-# **4.2** Linearity Property
-
-Linearity is the property of an element describing a linear relationship between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter. The property is a combination of both the homogeneity (scaling) property and the additivity property.
-
-The homogeneity property requires that if the input (also called the *excitation*) is multiplied by a constant, then the output (also called the *response*) is multiplied by the same constant. For a resistor, for example, Ohm's law relates the input *i* to the output *v*,
-
-$$
-v = iR \tag{4.1}
-$$
-
-If the current is increased by a constant *k*, then the voltage increases correspondingly by *k*; that is,
-
-$$
-kiR = kv \tag{4.2}
-$$
-
-The additivity property requires that the response to a sum of inputs is the sum of the responses to each input applied separately . Using the voltage-current relationship of a resistor, if
-
-$$
-v_1 = i_1 R \tag{4.3a}
-$$
-
-and
-
-$$
-v_2 = i_2 R \tag{4.3b}
-$$
-
-then applying (*i*1 + *i*2) gives
-
-$$
-v = (i_1 + i_2)R = i_1R + i_2R = v_1 + v_2
-$$
-\n(4.4)
-
-We say that a resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and the additivity properties.
-
-In general, a circuit is linear if it is both additive and homogeneous. A linear circuit consists of only linear elements, linear dependent sources, and independent sources.
-
-A linear circuit is one whose output is linearly related (or directly proportional) to its input.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/043_4.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/043_4.1 Introduction.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/043_4.1 Introduction.md
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@@ -1,482 +0,0 @@
-Throughout this book we consider only linear circuits. Note that since *p* = *i* 2 *R* = *v* 2 ∕*R* (making it a quadratic function rather than a linear one), the relationship between power and voltage (or current) is nonlinear . Therefore, the theorems covered in this chapter are not applicable to power.
-
-To illustrate the linearity principle, consider the linear circuit shown in Fig. 4.1. The linear circuit has no independent sources inside it. It is excited by a v oltage source *vs*, which serv es as the input. The circuit is terminated by a load *R*. We may tak e the current *i* through *R* as the output. Suppose *vs* = 10 V gives *i* = 2 A. According to the linearity principle, *vs* = 1 V will give *i* = 0.2 A. By the same token, *i* = 1 mA must be due to *vs* = 5 mV.
-
-For example, when current i1 flows through resistor R, the power is p1 = Ri1 2 , and when current i2 flows through R, the power is p2 = Ri2 2 . If current i1 + i2 flows through R, the power absorbed is p3 = R (i1 + i2) 2 = Ri1 2 + Ri2 2 + 2Ri1i2 ≠ p1 + p2. Thus, the power relation is nonlinear.
-
-# **Figure 4.1**
-
-A linear circuit with input *vs* and output *i*.
-
-For the circuit in Fig. 4.2, find *Io* when *vs* = 12 V and *vs* = 24 V. Example 4.1
-
-# **Solution:**
-
-Applying KVL to the two loops, we obtain
-
-$$
-12i1 - 4i2 + vs = 0
-$$
-\n(4.1.1)
-\n
-$$
--4i1 + 16i2 - 3vx - vs = 0
-$$
-\n(4.1.2)
-
-But *vx* = 2*i*1. Equation (4.1.2) becomes
-
-$$
--10i_1 + 16i_2 - v_s = 0 \tag{4.1.3}
-$$
-
-Adding Eqs. (4.1.1) and (4.1.3) yields
-
-$$
-2i_1 + 12i_2 = 0 \qquad \Rightarrow \qquad i_1 = -6i_2
-$$
-
-Substituting this in Eq. (4.1.1), we get
-
-$$
--76i_2 + v_s = 0 \qquad \Rightarrow \qquad i_2 = \frac{v_s}{76}
-$$
-
-When *vs* = 12 V,
-
-When *vs* = 24 V,
-
-$$
-I_o = i_2 = \frac{24}{76}
-$$
- A
-
-\_\_\_ 12 76 A
-
-*Io* = *i*2 =
-
-showing that when the source value is doubled, *Io* doubles.
-
-For Practice Prob. 4.1.
-
-Example 4.2
-
-Assume *Io* = 1 A and use linearity to find the actual value of *Io* in the
-
-# **Figure 4.4**
-
-circuit of Fig. 4.4.
-
-**Answer:** 40 V, 60 V.
-
-# **Solution:**
-
-**Answer:** 16 V.
-
-If *Io* = 1 A, then *V*1 = (3 + 5)*Io* = 8 V and *I*1 = *V*1∕4 = 2 A. Applying KCL at node 1 gives
-
-$$
-I_2 = I_1 + I_o = 3 \text{ A}
-$$
-
-$$
-V_2 = V_1 + 2I_2 = 8 + 6 = 14 \text{ V}, \qquad I_3 = \frac{V_2}{7} = 2 \text{ A}
-$$
-
-Applying KCL at node 2 gives
-
-$$
-I_4 = I_3 + I_2 = 5 \text{ A}
-$$
-
-Therefore, *Is* = 5 A. This shows that assuming *Io* = 1 gives *Is* = 5 A, the actual source current of 15 A will give *Io* = 3 A as the actual value.
-
-Assume that *Vo* = 1 V and use linearity to calculate the actual value of *Vo* in the circuit of Fig. 4.5.
-
-Practice Problem 4.2
-
-For Practice Prob. 4.2.
-
-# **4.3** Superposition
-
-If a circuit has tw o or more independent sources, one w ay to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis as in Chapter 3. Another way is to determine the contribution of each independent source to the v ariable and then add them up. The latter approach is known as the *superposition principle*.
-
-The idea of superposition rests on the linearity property.
-
-The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.
-
-The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contrib ution of each independent source separately. However, to apply the superposition principle, we must keep two things in mind:
-
-- 1. We consider one independent source at a time while all other independent sources are *turned off*. This implies that we replace e very voltage source by 0 V (or a short circuit), and e very current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.
-- 2. Dependent sources are left intact because the y are controlled by circuit variables.
-
-With these in mind, we apply the superposition principle in three steps:
-
-# Steps to Apply Superposition Principle:
-
-- 1. Turn off all independent sources e xcept one source. Find the output (voltage or current) due to that active source using the techniques covered in Chapters 2 and 3.
-- 2. Repeat step 1 for each of the other independent sources.
-- 3. Find the total contribution by adding algebraically all the contributions due to the independent sources.
-
-Analyzing a circuit using superposition has one major disadvantage: It may very likely involve more work. If the circuit has three independent sources, we may ha ve to analyze three simpler circuits each pro viding the contribution due to the respective individual source. However, superposition does help reduce a comple x circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits.
-
-Keep in mind that superposition is based on linearity . For this reason, it is not applicable to the ef fect on po wer due to each source, be cause the po wer absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition.
-
- Superposition is not limited to circuit analysis but is applicable in many fields where cause and effect bear a linear relationship to one another.
-
-Other terms such as killed, made inactive, deadened, or set equal to zero are often used to convey the same idea.
-
-**Figure 4.6** For Example 4.3.
-
-8 Ω
-
-i 2 i 3
-
-4 Ω
-
-(b)
-
-Example 4.3 Use the superposition theorem to find *v* in the circuit of Fig. 4.6.
-
-# **Solution:**
-
-Since there are two sources, let
-
-$$
-v = v_1 + v_2
-$$
-
-where *v*1 and *v*2 are the contributions due to the 6-V v oltage source and the 3-A current source, respecti vely. To obtain *v*1, we set the current source to zero, as sho wn in Fig. 4.7(a). Applying KVL to the loop in Fig. 4.7(a) gives
-
-$$
-12i_1 - 6 = 0 \qquad \Rightarrow \qquad i_1 = 0.5 \text{ A}
-$$
-
-Thus,
-
-$$
-v_1 = 4i_1 = 2 \text{ V}
-$$
-
-We may also use voltage division to get *v*1 by writing
-
-$$
-v_1 = \frac{4}{4+8}(6) = 2 \text{ V}
-$$
-
-To get *v*2, we set the voltage source to zero, as in Fig. 4.7(b). Using current division,
-
-$$
-i_3 = \frac{8}{4+8}(3) = 2 \text{ A}
-$$
-
-Hence,
-
-3 A
-
-*v*2
-
-+ –
-
-*v*2 = 4*i*3 = 8 V
-
-And we find
-
-**Answer:** 16 V.
-
-$$
-v = v_1 + v_2 = 2 + 8 = 10 \text{ V}
-$$
-
-**Figure 4.7** For Example 4.3: (a) calculating *v*1, (b) calculating *v*2.
-
-**Figure 4.8** For Practice Prob. 4.3.
-
-Find *io* in the circuit of Fig. 4.9 using superposition.
-
-# **Solution:**
-
-The circuit in Fig. 4.9 in volves a dependent source, which must be left intact. We let
-
-$$
-i_o = i'_o + i''_o \tag{4.4.1}
-$$
-
-where *i*′ *o* and *i o* ″ are due to the 4-A current source and 20-V voltage source respectively. To obtain *i*′ *o*, we turn off the 20-V source so that we have the circuit in Fig. 4.10(a). We apply mesh analysis in order to obtain *i*′ *o*. For loop 1,
-
-$$
-i_1 = 4 \text{ A} \tag{4.4.2}
-$$
-
-For loop 2,
-
-$$
-l_1 = 4 \,\mathrm{A}
-$$
-
-$$
--3i_1 + 6i_2 - 1i_3 - 5i'_o = 0 \tag{4.4.3}
-$$
-
-# **Figure 4.10**
-
-For Example 4.4: Applying superposition to (a) obtain *i*′ *o*, (b) obtain *i* ″ *o*.
-
-For loop 3,
-
-−5*i*1 − 1*i*2 + 10*i*3 + 5*i*′ *o* = 0 **(4.4.4)** But at node 0, *i*3 = *i*1 − *i*′ *o* = 4 − *i*′ *o* **(4.4.5)**
-
-Substituting Eqs. (4.4.2) and (4.4.5) into Eqs. (4.4.3) and (4.4.4) gi ves two simultaneous equations
-
-$$
-3i_2 - 2i'_o = 8 \tag{4.4.6}
-$$
-
-$$
-i_2 + 5i'_o = 20 \tag{4.4.7}
-$$
-
-which can be solved to get
-
-$$
-i'_{o} = \frac{52}{17} A
-$$
- (4.4.8)
-
-To obtain *i*″ *o*, we turn of f the 4-A current source so that the circuit becomes that shown in Fig. 4.10(b). For loop 4, KVL gives
-
-$$
-6i_4 - i_5 - 5i''_o = 0 \tag{4.4.9}
-$$
-
-and for loop 5,
-
-$$
--i_4 + 10i_5 - 20 + 5i''_o = 0 \tag{4.4.10}
-$$
-
-But *i*5 = −*i*″ *o*. Substituting this in Eqs. (4.4.9) and (4.4.10) gives
-
-$$
-6i_4 - 4i''_o = 0 \tag{4.4.11}
-$$
-
-$$
-i_4 + 5i''_o = -20 \tag{4.4.12}
-$$
-
-which we solve to get
-
-$$
-i_o'' = -\frac{60}{17} \,\mathrm{A} \tag{4.4.13}
-$$
-
-Now substituting Eqs. (4.4.8) and (4.4.13) into Eq. (4.4.1) gives
-
-$$
-i_o = -\frac{8}{17} = -0.4706 \text{ A}
-$$
-
-Example 4.5
-
-For the circuit in Fig. 4.12, use the superposition principle to find *i*.
-
-# **Solution:**
-
-In this case, we have three sources. Let
-
-$$
-i = i_1 + i_2 + i_3
-$$
-
-where *i*1, *i*2, and *i*3 are due to the 12-V , 24-V, and 3-A sources respec tively. To get *i*1, consider the circuit in Fig. 4.13(a). Combining 4 Ω (on the right-hand side) in series with 8 Ω gives 12 Ω. The 12 Ω in parallel with 4 Ω gives 12 × 4∕16 = 3 Ω. Thus,
-
-$$
-i_1 = \frac{12}{6} = 2 \text{ A}
-$$
-
-To get *i*2, consider the circuit in Fig. 4.13(b). Applying mesh analysis gives
-
-$$
-16i_a - 4i_b + 24 = 0 \qquad \Rightarrow \qquad 4i_a - i_b = -6 \tag{4.5.1}
-$$
-
-$$
-\mathcal{T}_b - 4i_a = 0 \qquad \Rightarrow \qquad i_a = \frac{7}{4} i_b \tag{4.5.2}
-$$
-
-Substituting Eq. (4.5.2) into Eq. (4.5.1) gives
-
-$$
-i_2 = i_b = -1
-$$
-
-To get *i*3, consider the circuit in Fig. 4.13(c). Using nodal analysis gives
-
-$$
-3 = \frac{v_2}{8} + \frac{v_2 - v_1}{4} \qquad \Rightarrow \qquad 24 = 3v_2 - 2v_1 \tag{4.5.3}
-$$
-
-$$
-\frac{v_2 - v_1}{4} = \frac{v_1}{4} + \frac{v_1}{3} \qquad \Rightarrow \qquad v_2 = \frac{10}{3}v_1 \tag{4.5.4}
-$$
-
-Substituting Eq. (4.5.4) into Eq. (4.5.3) leads to *v*1 = 3 and
-
-$$
-i_3 = \frac{v_1}{3} = 1 \text{ A}
-$$
-
-Thus,
-
-$$
-i = i_1 + i_2 + i_3 = 2 - 1 + 1 = 2 \text{ A}
-$$
-
-Find *I* in the circuit of Fig. 4.14 using the superposition principle. Practice Problem 4.5
-
-2 Ω
-
-2 A
-
-– 6 V +
-
-8 Ω
-
-I
-
-–
-
-8 V
-
-+
-
-**Figure 4.14** For Practice Prob. 4.5.
-
-# **4.4** Source Transformation
-
-6 Ω
-
-We have noticed that series-parallel combination and wye-delta transformation help simplify circuits. *Source transformation* is another tool for simplifying circuits. Basic to these tools is the concept of *equivalence*. We recall that an equivalent circuit is one whose *v*-*i* characteristics are identical with the original circuit.
-
-In Section 3.6, we sa w that node-v oltage (or mesh-current) equa tions can be obtained by mere inspection of a circuit when the sources are all independent current (or all independent v oltage) sources. It is therefore expedient in circuit analysis to be able to substitute a v oltage source in series with a resistor for a current source in parallel with a
-
-resistor, or vice versa, as shown in Fig. 4.15. Either substitution is known as a *source transformation*.
-
-**Figure 4.15** Transformation of independent sources.
-
-A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa.
-
-The tw o circuits in Fig. 4.15 are equi valent—provided the y ha ve the same voltage-current relation at terminals *a*-*b*. It is easy to sho w that they are indeed equi valent. If the sources are turned off, the equivalent resistance at terminals *a*-*b* in both circuits is *R*. Also, when terminals *a*-*b* are short-circuited, the short-circuit current flowing from *a* to *b* is *isc* = *vs*∕*R* in the circuit on the left-hand side and *isc* = *is* for the circuit on the right-hand side. Thus, *vs*∕*R* = *is* in order for the two circuits to be equivalent. Hence, source transformation requires that
-
-$$
-v_s = i_s R \qquad \text{or} \qquad i_s = \frac{v_s}{R} \tag{4.5}
-$$
-
-Source transformation also applies to dependent sources, pro vided we carefully handle the dependent v ariable. As shown in Fig. 4.16, a dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa where we make sure that Eq. (4.5) is satisfied.
-
-**Figure 4.16** Transformation of dependent sources.
-
-Like the wye-delta transformation we studied in Chapter 2, a source transformation does not af fect the remaining part of the circuit. When applicable, source transformation is a po werful tool that allo ws circuit manipulations to ease circuit analysis. However, we should keep the following points in mind when dealing with source transformation.
-
-- 1. Note from Fig. 4.15 (or Fig. 4.16) that the arrow of the current source is directed toward the positive terminal of the voltage source.
-- 2. Note from Eq. (4.5) that source transformation is not possible when *R* = 0, which is the case with an ideal voltage source. However, for a practical, nonideal voltage source, *R* ≠ 0. Similarly, an ideal current source with *R* = ∞ cannot be replaced by a finite voltage source. More will be said on ideal and nonideal sources in Section 4.10.1.
-
-Use source transformation to find *vo* in the circuit of Fig. 4.17.
-
-# **Solution:**
-
-We first transform the current and voltage sources to obtain the cir cuit in Fig. 4.18(a). Combining the 4-Ω and 2-Ω resistors in series and transforming the 12-V v oltage source gi ves us Fig. 4.18(b). We now combine the 3-Ω and 6-Ω resistors in parallel to get 2-Ω. We also combine the 2-A and 4-A current sources to get a 2-A source. Thus, by repeatedly applying source transformations, we obtain the circuit in Fig. 4.18(c).
-
-Example 4.6
-
-We use current division in Fig. 4.18(c) to get
-
-$$
-i = \frac{2}{2+8}(2) = 0.4 \text{ A}
-$$
-
-and
-
-$$
-v_o = 8i = 8(0.4) = 3.2
-$$
- V
-
-Alternatively, since the 8-Ω and 2-Ω resistors in Fig. 4.18(c) are in parallel, they have the same voltage *vo* across them. Hence,
-
-$$
-v_o = (8 \parallel 2)(2 \text{ A}) = \frac{8 \times 2}{10}(2) = 3.2 \text{ V}
-$$
-
-Find *io* in the circuit of Fig. 4.19 using source transformation. Practice Problem 4.6
-
-**Answer:** 1.78 A.
-
-# Example 4.7
-
-**Figure 4.20** For Example 4.7. Find *vx* in Fig. 4.20 using source transformation.
-
-# **Solution:**
-
-The circuit in Fig. 4.20 in volves a v oltage-controlled dependent cur rent source. We transform this dependent current source as well as the 6-V independent v oltage source as sho wn in Fig. 4.21(a). The 18-V voltage source is not transformed because it is not connected in series with any resistor. The two 2-Ω resistors in parallel combine to gi ve a 1-Ω resistor, which is in parallel with the 3-A current source. The current source is transformed to a v oltage source as shown in Fig. 4.21(b). Notice that the terminals for *vx* are intact. Applying KVL around the loop in Fig. 4.21(b) gives
-
-**Figure 4.21** For Example 4.7: Applying source transformation to the circuit in Fig. 4.20.
-
-Applying KVL to the loop containing only the 3-V v oltage source, the 1-Ω resistor, and *vx* yields
-
-$$
--3 + 1i + v_x = 0 \qquad \Rightarrow \qquad v_x = 3 - i \tag{4.7.2}
-$$
-
-Substituting this into Eq. (4.7.1), we obtain
-
-$$
-15 + 5i + 3 - i = 0 \qquad \Rightarrow \qquad i = -4.5 \text{ A}
-$$
-
-Alternatively, we may apply KVL to the loop containing *vx*, the 4-Ω resistor, the v oltage-controlled dependent v oltage source, and the 18-V voltage source in Fig. 4.21(b). We obtain
-
-$$
--v_x + 4i + v_x + 18 = 0 \qquad \Rightarrow \qquad i = -4.5 \text{ A}
-$$
-
-Thus, *vx* = 3 − *i* = 7.5 V.
-
-For Practice Prob. 4.7.
-
-# **4.5** Thevenin's Theorem
-
-It often occurs in practice that a particular element in a circuit is variable (usually called the *load*) while other elements are fixed. As a typical example, a household outlet terminal may be connected to dif ferent appliances constituting a v ariable load. Each time the v ariable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenin's theorem pro vides a technique by which the fixed part of the circuit is replaced by an equivalent circuit.
-
-According to Thevenin's theorem, the linear circuit in Fig. 4.23(a) can be replaced by that in Fig. 4.23(b). (The load in Fig. 4.23 may be a single resistor or another circuit.) The circuit to the left of the terminals *a*-*b* in Fig. 4.23(b) is kno wn as the *Thevenin equivalent cir cuit*; it w as developed in 1883 by M. Leon Thevenin (1857–1926), a French telegraph engineer.
-
-Thevenin's theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh, where VTh is the open-circuit voltage at the terminals and RTh is the input or equivalent resistance at the terminals when the independent sources are turned off.
-
-The proof of the theorem will be gi ven later, in Section 4.7. Our major concern right now is how to find the Thevenin equivalent voltage *V*Th and resistance *R*Th. To do so, suppose the tw o circuits in Fig. 4.23 are equivalent. Two circuits are said to be *equivalent* if the y have the same voltage-current relation at their terminals. Let us find out what will make the tw o circuits in Fig. 4.23 equi valent. If the terminals *a*-*b* are made open-circuited (by remo ving the load), no current flows, so that the open-circuit voltage across the terminals *a*-*b* in Fig. 4.23(a) must be equal to the voltage source *V*Th in Fig. 4.23(b), since the two circuits are equivalent. Thus, *V*Th is the open-circuit v oltage across the terminals as shown in Fig. 4.24(a); that is,
-
-Finding *V*Th and *R*Th.
-
-Again, with the load disconnected and terminals *a*-*b* opencir cuited, we turn of f all independent sources. The input resistance (or equi valent resistance) of the dead circuit at the terminals *a*-*b* in Fig. 4.23(a) must be equal to *R*Th in Fig. 4.23(b) because the two circuits are equivalent. Thus, *R*Th is the input resistance at the terminals when the independent sources are turned off, as shown in Fig. 4.24(b); that is,
-
-$$
-R_{\rm Th} = R_{\rm in} \tag{4.7}
-$$
-
-# **Figure 4.23** Replacing a linear two-terminal circuit
-
-by its Thevenin equivalent: (a) original circuit, (b) the Thevenin equivalent circuit.
-
-# **Figure 4.25**
-
-Finding *R*Th when circuit has dependent sources.
-
-Later we will see that an alternative way of finding RTh is RTh = voc ∕isc.
-
-# **Figure 4.26**
-
-A circuit with a load: (a) original circuit, (b) Thevenin equivalent.
-
-To apply this idea in finding the Thevenin resistance *R*Th, we need to consider two cases.
-
-■ **CASE 1** If the network has no dependent sources, we turn of f all independent sources. *R*Th is the input resistance of the netw ork looking between terminals *a* and *b*, as shown in Fig. 4.24(b).
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/044_4.3 Superposition.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/044_4.3 Superposition.md
deleted file mode 100644
index c0f55b0c002dcf89f3a2380e7bb4d4e8d0583c04..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/044_4.3 Superposition.md
+++ /dev/null
@@ -1,425 +0,0 @@
-■ **CASE 2** If the network has dependent sources, we turn of f all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage source *vo* at terminals *a* and *b* and determine the resulting current *io*. Then *R*Th = *vo*∕*io*, as shown in Fig. 4.25(a). Alternatively, we may insert a current source *io* at terminals *a*-*b* as shown in Fig. 4.25(b) and find the terminal voltage *vo*. Again *R*Th = *vo*∕*io*. Either of the two approaches will give the same result. In either approach we may assume an y value of *vo* and *io*. For example, we may use *vo* = 1 V or *io* = 1 A, or even use unspecified values of *vo* or *io*.
-
-It often occurs that *R*Th tak es a ne gative v alue. In this case, the negative resistance (*v* = −*iR*) implies that the circuit is supplying power. This is possible in a circuit with dependent sources; Example 4.10 will illustrate this.
-
-Thevenin's theorem is v ery important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a single indepen dent voltage source and a single resistor. This replacement technique is a powerful tool in circuit design.
-
-As mentioned earlier, a linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same w ay externally as the original circuit. Con sider a linear circuit terminated by a load *RL*, as shown in Fig. 4.26(a). The current *IL* through the load and the voltage *VL* across the load are easily determined once the Thevenin equivalent of the circuit at the load' s terminals is obtained, as sho wn in Fig. 4.26(b). From Fig. 4.26(b), we obtain
-
-$$
-I_L = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_L} \tag{4.8a}
-$$
-
-$$
-V_{L} = R_{L}I_{L} = \frac{R_{L}}{R_{\text{Th}} + R_{L}}V_{\text{Th}}
-$$
-(4.8b)
-
-Note from Fig. 4.26(b) that the Thevenin equivalent is a simple v oltage divider, yielding *VL* by mere inspection.
-
-Find the Thevenin equivalent circuit of the circuit shown in Fig. 4.27, to the left of the terminals *a*-*b*. Then find the current through *RL* = 6, 16, and 36 Ω.
-
-# **Solution:**
-
-We find *R*Th by turning of f the 32-V v oltage source (replacing it with a short circuit) and the 2-A current source (replacing it with an
-
-For Example 4.8.
-
-open circuit). The circuit becomes what is sho wn in Fig. 4.28(a). Thus,
-
-For Example 4.8: (a) finding *R*Th, (b) finding *V*Th.
-
-To find *V*Th, consider the circuit in Fig. 4.28(b). Applying mesh analysis to the two loops, we obtain
-
-$$
--32 + 4i_1 + 12(i_1 - i_2) = 0, \qquad i_2 = -2 \text{ A}
-$$
-
-Solving for *i*1, we get *i*1 = 0.5 A. Thus,
-
-$$
-V_{\text{Th}} = 12(i_1 - i_2) = 12(0.5 + 2.0) = 30 \text{ V}
-$$
-
-Alternatively, it is even easier to use nodal analysis. We ignore the 1-Ω resistor since no current flows through it. At the top node, KCL gives
-
-$$
-\frac{32 - V_{\text{Th}}}{4} + 2 = \frac{V_{\text{Th}}}{12}
-$$
-
-or
-
-$$
-96 - 3V_{\text{Th}} + 24 = V_{\text{Th}} \qquad \Rightarrow \qquad V_{\text{Th}} = 30 \text{ V}
-$$
-
-as obtained before. We could also use source transformation to find *V*Th.
-
-The Thevenin equivalent circuit is sho wn in Fig. 4.29. The current through *RL* is
-
-$$
-I_L = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_L} = \frac{30}{4 + R_L}
-$$
-
-When *RL* = 6,
-
-$$
-I_L = \frac{30}{10} = 3 \text{ A}
-$$
-
-When *RL* = 16,
-
-$$
-I_L = \frac{30}{20} = 1.5 \text{ A}
-$$
-
-When *RL* = 36,
-
-$$
-I_L = \frac{30}{40} = 0.75 \text{ A}
-$$
-
-**Figure 4.29** The Thevenin equivalent circuit for Example 4.8.
-
-# **Figure 4.30**
-
-For Practice Prob. 4.8.
-
-Example 4.9 Find the Thevenin equivalent of the circuit in Fig. 4.31 at terminals *a*-*b*.
-
-# **Solution:**
-
-This circuit contains a dependent source, unlik e the circuit in the pre vious e xample. To find *R*Th, we set the independent source equal to zero b ut lea ve the dependent source alone. Because of the presence of the dependent source, ho wever, we e xcite the netw ork with a v oltage source *vo* connected to the terminals as indicated in Fig. 4.32(a). We may set *vo* = 1 V to ease calculation, since the circuit is linear . Our goal is to find the current *io* through the terminals, and then obtain *R*Th = 1∕*io*. (Alternatively, we may insert a 1-A current source, find the corresponding voltage *vo*, and obtain *R*Th = *vo*∕1.)
-
-**Figure 4.31** For Example 4.9.
-
-**Figure 4.32** Finding *R*Th and *V*Th for Example 4.9.
-
-Applying mesh analysis to loop 1 in the circuit of Fig. 4.32(a) results in
-
-−2*vx* + 2(*i*1 − *i*2) = 0 or *vx* = *i*1 − *i*2
-
-But −4*i*2 = *vx* = *i*1 − *i*2; hence,
-
-$$
-i_1 = -3i_2 \tag{4.9.1}
-$$
-
-For loops 2 and 3, applying KVL produces
-
-4*i*2 + 2(*i*2 − *i*1) + 6(*i*2 − *i*3) = 0 **(4.9.2)**
-
-$$
-6(i_3 - i_2) + 2i_3 + 1 = 0 \tag{4.9.3}
-$$
-
-Solving these equations gives
-
-$$
-i_3 = -\frac{1}{6} \,\mathrm{A}
-$$
-
-But *io* = −*i*3 = 1∕6 A. Hence,
-
-$$
-R_{\text{Th}} = \frac{1 \text{ V}}{i_o} = 6 \text{ }\Omega
-$$
-
-To get *V*Th, we find *voc* in the circuit of Fig. 4.32(b). Applying mesh analysis, we get
-
-$$
-i_1 = 5 \tag{4.9.4}
-$$
-
-$$
--2v_x + 2(i_3 - i_2) = 0 \Rightarrow v_x = i_3 - i_2
-$$
- (4.9.5)
-$$
-4(i_2 - i_1) + 2(i_2 - i_3) + 6i_2 = 0
-$$
-
-or
-
-$$
-12i_2 - 4i_1 - 2i_3 = 0 \tag{4.9.6}
-$$
-
-But 4(*i*1 − *i*2) = *vx*. Solving these equations leads to *i*2 = 10∕3. Hence,
-
-$$
-V_{\rm Th} = v_{oc} = 6i_2 = 20 \, \text{V}
-$$
-
-The Thevenin equivalent is as shown in Fig. 4.33.
-
-# **Figure 4.33**
-
-The Thevenin equivalent of the circuit in Fig. 4.31.
-
-Find the Thevenin equivalent circuit of the circuit in Fig. 4.34 to the left of the terminals.
-
-**Answer:** *V*Th = 5.333 V, *R*Th = 444.4 mΩ.
-
-6 V 5 Ω 3 Ω 4 Ω 1.5I x + – I x
-
-a
-
-b
-
-Practice Problem 4.9
-
-Determine the Thevenin equi valent of the circuit in Fig. 4.35(a) at Example 4.10 terminals *a*-*b*.
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly defined; we are to determine the Thevenin equivalent of the circuit shown in Fig. 4.35(a).
-- 2. **Present.** The circuit contains a 2-Ω resistor in parallel with a 4-Ω resistor. These are, in turn, in parallel with a dependent current source. It is important to note that there are no independent sources.
-- 3. **Alternative.** The first thing to consider is that, since we have no independent sources in this circuit, we must excite the circuit externally. In addition, when you have no independent sources the value for *V*Th will be equal to zero so you will only have to find *R*Th.
-
-For Example 4.10.
-
-The simplest approach is to excite the circuit with either a 1-V voltage source or a 1-A current source. Because we will end up with an equivalent resistance (either positive or negative), I prefer to use the current source and nodal analysis which will yield a voltage at the output terminals equal to the resistance (with 1 A flowing in, *vo* is equal to 1 times the equivalent resistance).
-
- As an alternative, the circuit could also be excited by a 1-V voltage source and mesh analysis could be used to find the equivalent resistance.
-
-4. **Attempt.** We start by writing the nodal equation at *a* in Fig. 4.35(b) assuming *io* = 1 A.
-
-$$
-2i_x + (v_o - 0)/4 + (v_o - 0)/2 + (-1) = 0 \tag{4.10.1}
-$$
-
-Given that we have two unknowns and only one equation, we will need a constraint equation.
-
-$$
-i_x = (0 - v_o)/2 = -v_o/2 \tag{4.10.2}
-$$
-
-Substituting Eq. (4.10.2) into Eq. (4.10.1) yields
-
-$$
-2(-v_o/2) + (v_o - 0)/4 + (v_o - 0)/2 + (-1) = 0
-$$
-
-= $(-1 + \frac{1}{4} + \frac{1}{2})v_o - 1$ or $v_o = -4$ V
-
-Since *vo* = 1 × *R*Th, then *R*Th = *vo* ∕1 = −4 Ω.
-
-The negative value of the resistance tells us that, according to the passive sign convention, the circuit in Fig. 4.35(a) is supplying power. Of course, the resistors in Fig. 4.35(a) cannot supply power (they absorb power); it is the dependent source that supplies the power. This is an example of how a dependent source and resistors could be used to simulate negative resistance.
-
-5. **Evaluate.** First of all, we note that the answer has a negative value. We know this is not possible in a passive circuit, but in this circuit we do have an active device (the dependent current source). Thus, the equivalent circuit is essentially an active circuit that can supply power.
-
- Now we must evaluate the solution. The best way to do this is to perform a check, using a different approach, and see if we obtain the same solution. Let us try connecting a 9-Ω resistor in series with a 10-V voltage source across the output terminals of the original circuit and then the Thevenin equivalent. To make the circuit easier to solve, we can take and change the parallel current source and 4-Ω resistor to a series voltage source and 4-Ω resistor by using source transformation. This, with the new load, gives us the circuit shown in Fig. 4.35(c).
-
-We can now write two mesh equations.
-
-$$
-8i_x + 4i_1 + 2(i_1 - i_2) = 0
-$$
-
-$$
-2(i_2 - i_1) + 9i_2 + 10 = 0
-$$
-
-Note, we only have two equations but have three unknowns, so we need a constraint equation. We can use
-
-$$
-i_x = i_2 - i_1
-$$
-
-This leads to a new equation for loop 1. Simplifying leads to
-
-$$
-(4 + 2 - 8)i1 + (-2 + 8)i2 = 0
-$$
-
-or
-
-$$
--2i_1 + 6i_2 = 0 \t or \t i_1 = 3i_2
-$$
-
-$$
--2i_1 + 11i_2 = -10
-$$
-
-Substituting the first equation into the second gives
-
-$$
--6i_2 + 11i_2 = -10
-$$
- or $i_2 = -10/5 = -2$ A
-
-Using the Thevenin equivalent is quite easy since we have only one loop, as shown in Fig. 4.35(d).
-
-$$
--4i + 9i + 10 = 0
-$$
- or $i = -10/5 = -2$ A
-
-5. **Satisfactory?** Clearly we have found the value of the equivalent circuit as required by the problem statement. Checking does validate that solution (we compared the answer we obtained by using the equivalent circuit with one obtained by using the load with the original circuit). We can present all this as a solution to the problem.
-
-Obtain the Thevenin equivalent of the circuit in Fig. 4.36.
-
-Answer:
-$$
-V_{\text{Th}} = 0 \text{ V}, R_{\text{Th}} = -7.5 \Omega.
-$$
-
-b
-
-Practice Problem 4.10
-
-# **4.6** Norton's Theorem
-
-In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an American engineer at Bell Telephone Laboratories, proposed a similar theorem.
-
-Norton's theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN, where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off.
-
-Thus, the circuit in Fig. 4.37(a) can be replaced by the one in Fig. 4.37(b).
-
-The proof of Norton' s theorem will be gi ven in the ne xt section. For now, we are mainly concerned with ho w to get *RN* and *IN*. We find *RN* in the same w ay we find *R*Th. In f act, from what we kno w about source transformation, the Thevenin and Norton resistances are equal; that is,
-
-$$
-R_N = R_{\text{Th}} \tag{4.9}
-$$
-
-To find the Norton current *IN*, we determine the short-circuit current flowing from terminal *a* to *b* in both circuits in Fig. 4.37. It is e vident
-
-**Figure 4.36** For Practice Prob. 4.10.
-
-–
-
-# **Figure 4.37** (a) Original circuit, (b) Norton equivalent circuit.
-
-## **144** Chapter 4 Circuit Theorems
-
-Finding Norton current *IN*.
-
-The Thevenin and Norton equivalent circuits are related by a source transformation.
-
-that the short-circuit current in Fig. 4.37(b) is *IN*. This must be the same short-circuit current from terminal *a* to *b* in Fig. 4.37(a), since the tw o circuits are equivalent. Thus,
-
-$$
-I_N = i_{sc} \tag{4.10}
-$$
-
-shown in Fig. 4.38. Dependent and independent sources are treated the same way as in Thevenin's theorem.
-
-Observe the close relationship between Norton' s and Thevenin's theorems: *RN* = *R*Th as in Eq. (4.9), and
-
-$$
-I_N = \frac{V_{\text{Th}}}{R_{\text{Th}}} \tag{4.11}
-$$
-
-This is essentially source transformation. F or this reason, source trans formation is often called Thevenin-Norton transformation.
-
-Since *V*Th, *IN*, and *R*Th are related according to Eq. (4.11), to deter mine the Thevenin or Norton equivalent circuit requires that we find:
-
-- The open-circuit voltage *voc* across terminals *a* and *b*.
-- The short-circuit current *isc* at terminals *a* and *b*.
-- The equivalent or input resistance *R*in at terminals *a* and *b* when all independent sources are turned off.
-
-We can calculate an y two of the three using the method that tak es the least effort and use them to get the third using Ohm's law. Example 4.11 will illustrate this. Also, since
-
-$$
-V_{\rm Th} = v_{oc} \tag{4.12a}
-$$
-
-$$
-I_N = i_{sc} \tag{4.12b}
-$$
-
-$$
-R_{\text{Th}} = \frac{v_{oc}}{i_{sc}} = R_N \tag{4.12c}
-$$
-
-the open-circuit and short-circuit tests are sufficient to find any Thevenin or Norton equivalent, of a circuit which contains at least one independent source.
-
-Example 4.11 Find the Norton equi valent circuit of the circuit in Fig. 4.39 at terminals *a*-*b*.
-
-# **Solution:**
-
-We find *RN* in the same way we find *R*Th in the Thevenin equivalent circuit. Set the independent sources equal to zero. This leads to the circuit in Fig. 4.40(a), from which we find *RN*. Thus,
-
-$$
-R_N = 5 \parallel (8 + 4 + 8) = 5 \parallel 20 = \frac{20 \times 5}{25} = 4 \text{ }\Omega
-$$
-
-To find *IN*, we short-circuit terminals *a* and *b*, as shown in Fig. 4.40(b). We ignore the 5-Ω resistor because it has been short-circuited. Applying mesh analysis, we obtain
-
-$$
-i_1 = 2 \text{ A}, \qquad 20i_2 - 4i_1 - 12 = 0
-$$
-
-From these equations, we obtain
-
-$$
-i_2=1 \; \mathrm{A}=i_{sc}=I_N
-$$
-
-**Figure 4.39** For Example 4.11.
-
-**Figure 4.40** For Example 4.11; finding: (a) *RN*, (b) *IN* = *isc*, (c) *V*Th = *voc*.
-
-Alternatively, we may determine *IN* from *V*Th ∕*R*Th. We obtain *V*Th as the open-circuit voltage across terminals *a* and *b* in Fig. 4.40(c). Using mesh analysis, we obtain
-
-$$
-i_3 = 2 \text{ A}
-$$
-
- $25i_4 - 4i_3 - 12 = 0 \Rightarrow i_4 = 0.8 \text{ A}$
-
-and
-
-$$
-v_{oc} = V_{\text{Th}} = 5i_4 = 4 \text{ V}
-$$
-
-Hence,
-
-$$
-I_N = \frac{V_{\text{Th}}}{R_{\text{Th}}} = \frac{4}{4} = 1 \text{ A}
-$$
-
-as obtained pre viously. This also serv es to confirm Eq. (4.12c) that *R*Th = *voc*∕*isc* = 4 ∕1 = 4 Ω. Thus, the Norton equi valent circuit is as shown in Fig. 4.41.
-
-Find the Norton equi valent circuit for the circuit in Fig. 4.42, at Practice Problem 4.11 terminals *a*-*b*.
-
-**Answer:** *RN* = 90 Ω, *IN* = 4.5 A.
-
-For Practice Prob. 4.11.
-
-**Figure 4.43** For Example 4.12.
-
-Example 4.12 Using Norton's theorem, find *RN* and *IN* of the circuit in Fig. 4.43 at terminals *a*-*b*.
-
-# **Solution:**
-
-To find *RN*, we set the independent voltage source equal to zero and connect a voltage source of *vo* = 1 V (or any unspecified voltage *vo*) to the terminals. We obtain the circuit in Fig. 4.44(a). We ignore the 4-Ω resistor because it is short-circuited. Also due to the short circuit, the 5- Ω resistor, the voltage source, and the dependent current source are all in parallel. Hence, *ix* = 0. At node *a*, *io* = \_\_\_\_ 1 V 5Ω = 0.2 A, and
-
-$$
-R_N = \frac{v_o}{i_o} = \frac{1}{0.2} = 5 \ \Omega
-$$
-
-To find *IN*, we short-circuit terminals *a* and *b* and find the current *isc*, as indicated in Fig. 4.44(b). Note from this figure that the 4 Ω resistor, the 10-V voltage source, the 5-Ω resistor, and the dependent current source are all in parallel. Hence,
-
-$$
-i_x = \frac{10}{4} = 2.5 \text{ A}
-$$
-
-At node *a*, KCL gives
-
-$$
-i_{sc} = \frac{10}{5} + 2i_x = 2 + 2(2.5) = 7 \text{ A}
-$$
-
-Thus,
-
-$$
-I_N = 7 \,\mathrm{A}
-$$
-
-**Figure 4.44** For Example 4.12: (a) finding *RN*, (b) finding *IN*.
-
-For Practice Prob. 4.12.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/045_4.5 Thevenin's Theorem.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/045_4.5 Thevenin's Theorem.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/045_4.5 Thevenin's Theorem.md
+++ /dev/null
@@ -1,48 +0,0 @@
-# **4.7** Derivations of Thevenin's and Norton's Theorems
-
-In this section, we will pro ve Thevenin's and Norton's theorems using the superposition principle.
-
-Consider the linear circuit in Fig. 4.46(a). It is assumed that the circuit contains resistors and dependent and independent sources. We have access to the circuit via terminals *a* and *b*, through which current from an external source is applied. Our objective is to ensure that the voltagecurrent relation at terminals *a* and *b* is identical to that of the Thevenin equivalent in Fig. 4.46(b). F or the sak e of simplicity , suppose the lin ear circuit in Fig. 4.46(a) contains tw o independent voltage sources *vs*1 and *vs*2 and two independent current sources *is*1 and *is*2. We may obtain any circuit variable, such as the terminal v oltage *v*, by applying super position. That is, we consider the contrib ution due to each independent source including the e xternal source *i*. By superposition, the terminal voltage *v* is
-
-$$
-v = A_0 i + A_1 v_{s1} + A_2 v_{s2} + A_3 i_{s1} + A_4 i_{s2}
-$$
- (4.13)
-
-where *A*0, *A*1, *A*2, *A*3, and *A*4 are constants. Each term on the right-hand side of Eq. (4.13) is the contrib ution of the related independent source; that is, *A*0*i* is the contrib ution to *v* due to the e xternal current source *i*, *A*1*vs*1 is the contribution due to the voltage source *vs*1, and so on. We may collect terms for the internal independent sources together as *B*0, so that Eq. (4.13) becomes
-
-$$
-v = A_0 i + B_0 \tag{4.14}
-$$
-
-where *B*0 = *A*1*vs*1 + *A*2*vs*2 + *A*3*is*1 + *A*4*is*2. We now want to evaluate the values of constants *A*0 and *B*0. When the terminals *a* and *b* are opencircuited, *i* = 0 and *v* = *B*0. Thus, *B*0 is the open-circuit voltage *voc*, which is the same as *V*Th, so
-
-$$
-B_0 = V_{\text{Th}} \tag{4.15}
-$$
-
-When all the internal sources are turned off, *B*0 = 0. The circuit can then be replaced by an equivalent resistance *R*eq, which is the same as *R*Th, and Eq. (4.14) becomes
-
-$$
-v = A_0 i = R_{\text{Th}} i \qquad \Rightarrow \qquad A_0 = R_{\text{Th}} \tag{4.16}
-$$
-
-Substituting the values of *A*0 and *B*0 in Eq. (4.14) gives
-
-$$
-v = R_{\text{Th}} i + V_{\text{Th}} \tag{4.17}
-$$
-
-which expresses the voltage-current relation at terminals *a* and *b* of the circuit in Fig. 4.46(b). Thus, the two circuits in Fig. 4.46(a) and 4.46(b) are equivalent.
-
-When the same linear circuit is dri ven by a v oltage source *v* as shown in Fig. 4.47(a), the current flowing into the circuit can be obtained by superposition as
-
-$$
-i = C_0 v + D_0 \tag{4.18}
-$$
-
-where *C*0*v* is the contrib ution to *i* due to the e xternal voltage source *v* and *D*0 contains the contrib utions to *i* due to all internal independent sources. When the terminals *a*-*b* are short-circuited, *v* = 0 so that
-
-Derivation of Thevenin equivalent: (a) a current-driven circuit, (b) its Thevenin equivalent.
-
-**Figure 4.47** Derivation of Norton equivalent: (a) a voltage-driven circuit, (b) its Norton equivalent.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/046_4.7 Derivations of Thevenin's and Norton's Theorems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/046_4.7 Derivations of Thevenin's and Norton's Theorems.md
deleted file mode 100644
index 11fe5c2bd6d8531fad10427c7c14ba7df021e524..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/046_4.7 Derivations of Thevenin's and Norton's Theorems.md
+++ /dev/null
@@ -1,15 +0,0 @@
-## **148** Chapter 4 Circuit Theorems
-
-*i* = *D*0 = −*isc*, where *isc* is the short-circuit current flowing out of terminal *a*, which is the same as the Norton current *IN*, i.e.,
-
-$$
-D_0 = -I_N \tag{4.19}
-$$
-
-When all the internal independent sources are turned of f, *D*0 = 0 and the circuit can be replaced by an equivalent resistance *R*eq (or an equivalent conductance *G*eq = 1∕*R*eq), which is the same as *R*Th or *RN*. Thus, Eq. (4.19) becomes
-
-$$
-i = \frac{v}{R_{\text{Th}}} - I_N \tag{4.20}
-$$
-
-This expresses the v oltage-current relation at terminals *a*-*b* of the cir cuit in Fig. 4.47(b), confirming that the two circuits in Fig. 4.47(a) and 4.47(b) are equivalent.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/047_4.8 Maximum Power Transfer.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/047_4.8 Maximum Power Transfer.md
deleted file mode 100644
index 0a0486a4825ac6c85b8b7d92ddad47135e4c70c8..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/047_4.8 Maximum Power Transfer.md
+++ /dev/null
@@ -1,87 +0,0 @@
-# **4.8** Maximum Power Transfer
-
-In many practical situations, a circuit is designed to provide power to a load. There are applications in areas such as communications where it is desirable to maximize the power delivered to a load. We now address the problem of deli vering the maximum po wer to a load when gi ven a system with known internal losses. It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load.
-
-The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load. We assume that we can adjust the load resistance *RL*. If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig. 4.48, the power delivered to the load is
-
-$$
-p = i^2 R_L = \left(\frac{V_{\text{Th}}}{R_{\text{Th}} + R_L}\right)^2 R_L
-$$
- (4.21)
-
-For a given circuit, *V*Th and *R*Th are fixed. By varying the load resistance *RL*, the power delivered to the load varies as sketched in Fig. 4.49. We notice from Fig. 4.49 that the power is small for small or large values of *RL* but maximum for some value of *RL* between 0 and ∞. We now want to show that this maximum power occurs when *RL* is equal to *R*Th. This is known as the *maximum power theorem.*
-
-Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load (RL = RTh).
-
-To prove the maximum power transfer theorem, we differentiate *p* in Eq. (4.21) with respect to *RL* and set the result equal to zero. We obtain
-
- *dp \_\_\_\_ dRL* = *V*2 Th [ (*R*Th + *RL*) 2 −2*RL*(*R*Th + *RL*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*R*Th + *RL*) 4 ] = *V*2 Th [ (*R*Th + *RL* − 2*RL*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*R*Th + *RL*) 3 ] =0
-
-**Figure 4.48**
-
-The circuit used for maximum power transfer.
-
-**Figure 4.49** Power delivered to the load as a function of *RL*.
-
-This implies that
-
-$$
-0 = (R_{\text{Th}} + R_L - 2R_L) = (R_{\text{Th}} - R_L)
-$$
-\n(4.22)
-
-which yields
-
-$$
-R_L = R_{\text{Th}} \tag{4.23}
-$$
-
-showing that the maximum power transfer takes place when the load resistance *RL* equals the Thevenin resistance *R*Th. We can readily confirm that Eq. (4.23) gives the maximum power by showing that *d*2 *p*∕*dR*2 *L*< 0.
-
-The maximum po wer transferred is obtained by substituting Eq. (4.23) into Eq. (4.21), for
-
-> *p*max = *V*2 \_\_\_\_Th 4*R*Th **(4.24)**
-
-Equation (4.24) applies only when *RL* = *R*Th. When *RL* ≠ *R*Th, we compute the power delivered to the load using Eq. (4.21).
-
-Find the v alue of *RL* for maximum po wer transfer in the circuit of Example 4.13 Fig. 4.50. Find the maximum power.
-
-The source and load are said to be
-
-matched when RL = RTh.
-
-**Solution:**
-
-We need to find the Thevenin resistance *R*Th and the Thevenin voltage *V*Th across the terminals *a*-*b*. To get *R*Th, we use the circuit in Fig. 4.51(a) and obtain
-
-$$
-R_{\text{Th}} = 2 + 3 + 6 \parallel 12 = 5 + \frac{6 \times 12}{18} = 9 \ \Omega
-$$
-
-For Example 4.13: (a) finding *R*Th, (b) finding *V*Th.
-
-To get *V*Th, we consider the circuit in Fig. 4.51(b). Applying mesh analysis gives
-
-$$
--12 + 18i_1 - 12i_2 = 0, \qquad i_2 = -2 \text{ A}
-$$
-
-Solving for *i*1, we get *i*1 = −2∕3. Applying KVL around the outer loop to get *V*Th across terminals *a*-*b*, we obtain
-
-−12 + 6*i*1 + 3*i*2 + 2(0) + *V*Th = 0 ⇒ *V*Th = 22 V
-
-For maximum power transfer,
-
-$$
-R_L = R_{\text{Th}} = 9 \ \Omega
-$$
-
-and the maximum power is
-
-$$
-p_{\text{max}} = \frac{V_{\text{Th}}^2}{4R_L} = \frac{22^2}{4 \times 9} = 13.44 \text{ W}
-$$
-
-Determine the value of *RL* that will draw the maximum power from the rest of the circuit in Fig. 4.52. Calculate the maximum power.
-
-**Answer:** 126.67 Ω, 96.71 mW.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/048_4.9 Verifying Circuit Theorems with PSpice.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/048_4.9 Verifying Circuit Theorems with PSpice.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/048_4.9 Verifying Circuit Theorems with PSpice.md
+++ /dev/null
@@ -1,97 +0,0 @@
-# **4.9** Verifying Circuit Theorems with PSpice
-
-In this section, we learn how to use *PSpice* to verify the theorems covered in this chapter. Specifically, we will consider using DC Sweep analysis to find the Thevenin or Norton equi valent at an y pair of nodes in a circuit and the maximum po wer transfer to a load. The reader is advised to read Section D.3 of Appendix D in preparation for this section.
-
-To find the Thevenin equivalent of a circuit at a pair of open terminals using *PSpice*, we use the schematic editor to dra w the circuit and insert an independent probing current source, say , Ip, at the terminals. The probing current source must have a part name ISRC. We then perform a DC Sweep on Ip, as discussed in Section D.3. Typically, we may let the current through Ip vary from 0 to 1 A in 0.1-A increments. After saving and simulating the circuit, we use Probe to display a plot of the voltage across Ip versus the current through Ip. The zero intercept of the plot gives us the Thevenin equivalent voltage, while the slope of the plot is equal to the Thevenin resistance.
-
-To find the Norton equivalent involves similar steps except that we insert a probing independent v oltage source (with a part name VSRC), say, Vp, at the terminals. We perform a DC Sweep on Vp and let Vp vary from 0 to 1 V in 0.1-V increments. A plot of the current through Vp versus the voltage across Vp is obtained using the Probe menu after simulation. The zero intercept is equal to the Norton current, while the slope of the plot is equal to the Norton conductance.
-
-To find the maximum power transfer to a load using *PSpice* in volves performing a DC parametric Sweep on the component v alue of *RL* in Fig. 4.48 and plotting the po wer delivered to the load as a function of *RL*. According to Fig. 4.49, the maximum po wer occurs
-
-**Figure 4.52** For Practice Prob. 4.13.
-
-when *RL* = *R*Th. This is best illustrated with an example, and Example 4.15 provides one.
-
-We use VSRC and ISRC as part names for the independent voltage and current sources, respectively.
-
-Consider the circuit in Fig. 4.31 (see Example 4.9). Use *PSpice* to find the Thevenin and Norton equivalent circuits.
-
-# **Solution:**
-
-(a) To find the Thevenin resistance *R*Th and Thevenin voltage *V*Th at the terminals *a*-*b* in the circuit in Fig. 4.31, we first use Schematics to draw the circuit as shown in Fig. 4.53(a). Notice that a probing current source I2 is inserted at the terminals. Under **Analysis**∕**Setput**, we select DC Sweep. In the DC Sweep dialog box, we select Linear for the *Sweep Type* and Current Source for the *Sweep Var*. *Type*. We enter I2 under the *Name* box, 0 as *Start Value*, 1 as *End Value*, and 0.1 as *Increment*. After simulation, we add trace V(I2:–) from the *PSpice* A∕D window and obtain the plot shown in Fig. 4.53(b). From the plot, we obtain *V*Th = Zero intercept = 20 V, *R*Th = Slope = \_\_\_\_\_\_\_ 26 − 20 1 = 6 Ω These agree with what we got analytically in Example 4.9.
-
-(b) To find the Norton equivalent, we modify the schematic in Fig. 4.53(a) by replaying the probing current source with a probing v oltage source V1. The result is the schematic in Fig. 4.54(a). Again, in the DC Sweep dialog box, we select Linear for the *Sweep Type* and Voltage Source for the *Sweep Var*. *Type*. We enter V1 under *Name* box, 0 as *Start Value*, 1 as *End Value*, and 0.1 as *Increment*. Under the *PSpice* A∕D Window, we add trace I (V1) and obtain the plot in Fig. 4.54(b). From the plot, we obtain
-
-$$
-I_N
-$$
- = Zero intercept = 3.335 A
- $G_N$ = Slope = $\frac{3.335 \text{ } \bigcirc \text{ } 3.165}{1} = 0.17 \text{ S}$
-
-Example 4.14
-
-**Figure 4.54** For Example 4.14: (a) schematic and (b) plot for finding *GN* and *IN*.
-
-| Practice Problem 4.14 | Rework Practice Prob. 4.9 using PSpice. |
-|-----------------------|-------------------------------------------------|
-| | Answer:
VTh
= 5.333 V, RTh
= 444.4 mΩ. |
-| | |
-
-**Figure 4.55**
-
-For Example 4.15.
-
-# **Figure 4.56**
-
-Schematic for the circuit in Fig. 4.55.
-
-Example 4.15 Refer to the circuit in Fig. 4.55. Use *PSpice* to find the maximum power transfer to *RL*.
-
-# **Solution:**
-
-We need to perform a DC Sweep on *RL* to determine when the po wer across it is maximum. We first draw the circuit using Schematics as shown in Fig. 4.56. Once the circuit is dra wn, we tak e the follo wing three steps to further prepare the circuit for a DC Sweep.
-
-The first step involves defining the value of *RL* as a parameter, since we want to vary it. To do this:
-
-- 1. **DCLICKL** the value 1k of R2 (representing *RL*) to open up the *Set Attribute Value* dialog box.
-- 2. Replace 1k with {RL} and click **OK** to accept the change.
-
-# Note that the curly brackets are necessary.
-
-The second step is to define parameter. To achieve this:
-
-- 1. Select **Draw**∕**Get New Part∕Libraries** ⋯∕**special.slb**.
-- 2. Type PARAM in the *PartName* box and click **OK**.
-- 3. **DRAG** the box to any position near the circuit.
-- 4. **CLICKL** to end placement mode.
-- 5. **DCLICKL** to open up the *PartName: PARAM* dialog box.
-- 6. **CLICKL** on *NAME1* = and enter RL (with no curly brackets) in the *Value* box, and **CLICKL Save Attr** to accept change.
-- 7. **CLICKL** on *VALUE1*= and enter 2k in the *Value* box, and **CLICKL Save Attr** to accept change.
-- 8. Click **OK**.
-
-The value 2k in item 7 is necessary for a bias point calculation; it cannot be left blank.
-
-The third step is to set up the DC Sweep to sweep the parameter. To do this:
-
-- 1. Select **Analysis**∕**Setput** to bring up the DC Sweep dialog box.
-- 2. For the *Sweep Type*, select Linear (or Octave for a wide range of *RL*).
-- 3. For the *Sweep Var*. *Type*, select Global Parameter.
-- 4. Under the *Name* box, enter RL.
-- 5. In the *Start Value* box, enter 100.
-- 6. In the *End Value* box, enter 5k.
-- 7. In the *Increment* box, enter 100.
-- 8. Click **OK** and **Close** to accept the parameters.
-
-After taking these steps and sa ving the circuit, we are ready to simulate. Select **Analysis/Simulate**. If there are no errors, we select **Add Trace** in the *PSpice* A/D windo w and type −V(R2:2)∗ I(R2) in the *Trace Command* box. [The ne gative sign is needed since I(R2) is negative.] This gives the plot of the power delivered to *RL* as *RL* varies from 100 Ω to 5 k Ω. We can also obtain the power absorbed by *RL* by typing V(R2:2)∗ V(R2:2)/RL in the *Trace Command* box. Either way, we obtain the plot in Fig. 4.57. It is evident from the plot that the maximum power is 250 *μ*W. Notice that the maximum occurs when *RL* = 1 kΩ, as expected analytically.
-
-150 uW 200 uW 100 uW 50 uW 0 2.0 K 4.0 K 6.0 K –V(R2:2)\*I(R2) RL
-
-# **Figure 4.57**
-
-250 uW
-
-For Example 4.15: the plot of power across *RL*.
-
-Find the maximum po wer transferred to *RL* if the 1-k Ω resistor in Practice Problem 4.15 Fig. 4.55 is replaced by a 2-kΩ resistor.
-
-**Answer:** 125 *μ*W.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/049_4.10 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/049_4.10 Applications.md
deleted file mode 100644
index 05799d580b2625103fe3c201f2740a96e864c80b..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/049_4.10 Applications.md
+++ /dev/null
@@ -1,201 +0,0 @@
-# **4.10** Applications
-
-In this section we will discuss tw o important practical applications of the concepts co vered in this chapter: source modeling and resistance measurement.
-
-# **4.10.1** Source Modeling
-
-Source modeling provides an example of the usefulness of the Thevenin or the Norton equivalent. An active source such as a battery is often characterized by its Thevenin or Norton equivalent circuit. An ideal voltage source provides a constant v oltage irrespective of the current dra wn by the load, while an ideal current source supplies a constant current regardless of the load voltage. As Fig. 4.58 shows, practical voltage and current sources are not ideal, due to their *internal resistances* or *source resistances Rs* and *Rp*. They become ideal as *Rs* → 0 and *Rp* → ∞. To show that this is the case, consider the effect of the load on voltage sources,
-
-## **154** Chapter 4 Circuit Theorems
-
-as shown in Fig. 4.59(a). By the voltage division principle, the load voltage is
-
-$$
-v_L = \frac{R_L}{R_s + R_L} v_s \tag{4.25}
-$$
-
-As *RL* increases, the load v oltage approaches a source v oltage *vs*, as illustrated in Fig. 4.59(b). From Eq. (4.25), we should note that:
-
-1. The load v oltage will be constant if the internal resistance *Rs* of the source is zero or , at least, *Rs* ≪ *RL*. In other w ords, the smaller *Rs* is compared with *RL*, the closer the v oltage source is to being ideal.
-
-(a) Practical voltage source connected to a load *RL*, (b) load voltage decreases as *RL* decreases.
-
-2. When the load is disconnected (i.e., the source is open-circuited so that *RL* → ∞), *voc* = *vs*. Thus, *vs* may be re garded as the *unloaded source* voltage. The connection of the load causes the terminal voltage to drop in magnitude; this is known as the *loading effect*.
-
-The same argument can be made for a practical current source when connected to a load as sho wn in Fig. 4.60(a). By the current di vision principle,
-
-$$
-i_L = \frac{R_p}{R_p + R_L} i_s \tag{4.26}
-$$
-
-Figure 4.60(b) shows the variation in the load current as the load resis tance increases. Again, we notice a drop in current due to the load (loading effect), and load current is constant (ideal current source) when the internal resistance is very large (i.e., *Rp* → ∞ or, at least, *Rp* ≫ *RL*).
-
-Sometimes, we need to know the unloaded source voltage *vs* and the internal resistance *Rs* of a v oltage source. To find *vs* and *Rs*, we follo w the procedure illustrated in Fig. 4.61. First, we measure the open-circuit voltage *voc* as in Fig. 4.61(a) and set
-
-$$
-v_s = v_{oc} \tag{4.27}
-$$
-
-Then, we connect a v ariable load *RL* across the terminals as in Fig. 4.61(b). We adjust the resistance *RL* until we measure a load voltage of exactly one-half of the open-circuit voltage, *vL* = *voc*∕2, be cause now *RL* = *R*Th = *Rs*. At that point, we disconnect *RL* and measure it. We set
-
-$$
-R_s = R_L \tag{4.28}
-$$
-
-For example, a car battery may have *vs* = 12 V and *Rs* = 0.05 Ω.
-
-# **Figure 4.60**
-
-(a) Practical current source connected to a load *RL*, (b) load current decreases as *RL* increases.
-
-(a) Measuring *voc*, (b) measuring *vL*.
-
-The terminal v oltage of a v oltage source is 12 V when connected to a Example 4.16 2-W load. When the load is disconnected, the terminal v oltage rises to 12.4 V. (a) Calculate the source v oltage *vs* and internal resistance *Rs*. (b) Determine the voltage when an 8-Ω load is connected to the source.
-
-# **Solution:**
-
-(a) We replace the source by its Thevenin equivalent. The terminal voltage when the load is disconnected is the open-circuit voltage,
-
-$$
-v_s = v_{oc} = 12.4 \text{ V}
-$$
-
-When the load is connected, as shown in Fig. 4.62(a), *vL* = 12 V and *pL* = 2 W. Hence,
-
-$$
-p_L = \frac{v_L^2}{R_L}
-$$
- $\Rightarrow$ $R_L = \frac{v_L^2}{p_L} = \frac{12^2}{2} = 72 \Omega$
-
-The load current is
-
-$$
-i_L = \frac{v_L}{R_L} = \frac{12}{72} = \frac{1}{6}
-$$
- A
-
-The voltage across *Rs* is the difference between the source voltage *vs* and the load voltage *vL*, or
-
-$$
-12.4 - 12 = 0.4 = R_s i_L, \qquad R_s = \frac{0.4}{I_L} = 2.4 \ \Omega
-$$
-
-(b) Now that we have the Thevenin equivalent of the source, we connect the 8-Ω load across the Thevenin equivalent as shown in Fig. 4.62(b). Using voltage division, we obtain
-
-$$
-v = \frac{8}{8 + 2.4} (12.4) = 9.538 \text{ V}
-$$
-
-**Figure 4.62** For Example 4.16.
-
-The measured open-circuit voltage across a certain amplifier is 16 V. The voltage drops to 8 V when a 8-Ω loudspeaker is connected to the amplifier. Calculate the voltage when a 24-Ω loudspeaker is used instead.
-
-Practice Problem 4.16
-
-**Answer:** 12 V.
-
-# **4.10.2** Resistance Measurement
-
-Although the ohmmeter method pro vides the simplest w ay to mea sure resistance, more accurate measurement may be obtained using the Wheatstone bridge. While ohmmeters are designed to measure resistance in low, mid, or high range, a Wheatstone bridge is used to mea sure resistance in the mid range, say, between 1 Ω and 1 MΩ. Very low values of resistances are measured with a *milliohmmeter,* while very high values are measured with a *Megger tester*.
-
-The Wheatstone bridge (or resistance bridge) circuit is used in a number of applications. Here we will use it to measure an unkno wn resistance. The unkno wn resistance *Rx* is connected to the bridge as shown in Fig. 4.63. The variable resistance is adjusted until no current flows through the galvanometer, which is essentially a d'Arsonval movement operating as a sensiti ve current-indicating device like an ammeter in the microamp range. Under this condition *v*1 = *v*2, and the bridge is said to be *balanced*. Since no current flows through the galvanometer, *R*1 and *R*2 behave as though the y were in series; so do *R*3 and *Rx*. The fact that no current flows through the galvanometer also implies that *v*1 = *v*2. Applying the voltage division principle,
-
-$$
-v_1 = \frac{R_2}{R_1 + R_2} v = v_2 = \frac{R_x}{R_3 + R_x} v
-$$
-\n(4.29)
-
-Hence, no current flows through the galvanometer when
-
-$$
-\frac{R_2}{R_1 + R_2} = \frac{R_x}{R_3 + R_x} \qquad \Rightarrow \qquad R_2 R_3 = R_1 R_x
-$$
-
-$$
-R_x = \frac{R_3}{R_1} R_2 \tag{4.30}
-$$
-
-If *R*1 = *R*3, and *R*2 is adjusted until no current flows through the galvanometer, then *Rx* = *R*2.
-
-How do we find the current through the galvanometer when the Wheatstone bridge is *unbalanced*? We find the Thevenin equivalent (*V*Th and *R*Th) with respect to the g alvanometer terminals. If *Rm* is the resis tance of the g alvanometer, the current through it under the unbalanced condition is
-
-$$
-I = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_m} \tag{4.31}
-$$
-
-Example 4.18 will illustrate this.
-
-In Fig. 4.63, *R*1 = 500 Ω and *R*3 = 200 Ω. The bridge is balanced when *R*2 is adjusted to be 125 Ω. Determine the unknown resistance *Rx*.
-
-# **Solution:**
-
-or
-
-Using Eq. (4.30) gives
-
-$$
-R_x = \frac{R_3}{R_1} R_2 = \frac{200}{500} 125 = 50 \ \Omega
-$$
-
-Historical note: The bridge was invented by Charles Wheatstone (1802–1875), a British professor who also invented the telegraph, as Samuel Morse did independently in the
-
-United States.
-
-**Figure 4.63** The Wheatstone bridge; *Rx* is the resistance to be measured.
-
-A Wheatstone bridge has *R*1 = *R*3 = 2 kΩ. *R*2 is adjusted until no current flows through the galvanometer. At that point, *R*2 = 6.3 kΩ. What is the value of the unknown resistance?
-
-# **Answer:** 6.3 kΩ.
-
-The circuit in Fig. 4.64 represents an unbalanced bridge. If the galvanometer has a resistance of 40 Ω, find the current through the galvanometer.
-
-Unbalanced bridge of Example 4.18.
-
-# **Solution:**
-
-We first need to replace the circuit by its Thevenin equivalent at ter minals *a* and *b*. The Thevenin resistance is found using the circuit in Fig. 4.65(*a*). Notice that the 3-kΩ and 1-KΩ resistors are in parallel; so are the 400 and 600- Ω resistors. The two parallel combinations form a series combination with respect to terminals *a* and *b*. Hence,
-
-$$
-R_{\text{Th}} = 3000 \parallel 1000 + 400 \parallel 600
-$$
-
-=
-$$
-\frac{3000 \times 1000}{3000 + 1000} + \frac{400 \times 600}{400 + 600} = 750 + 240 = 990 \Omega
-$$
-
-To find the Thevenin voltage, we consider the circuit in Fig. 4.65(b). Using the voltage division principle gives
-
-sing the voltage division principle gives
-$$
-v_1 = \frac{1000}{1000 + 3000} (220) = 55 \text{ V}, \qquad v_2 = \frac{600}{600 + 400} (220) = 132 \text{ V}
-$$
-
-Applying KVL around loop *ab* gives
-
-$$
--v_1 + V_{\text{Th}} + v_2 = 0
-$$
- or $V_{\text{Th}} = v_1 - v_2 = 55 - 132 = -77$ V
-
-Having determined the Thevenin equivalent, we find the current through the galvanometer using Fig. 4.65(c).
-
-$$
-I_G = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_m} = \frac{-77}{990 + 40} = -74.76 \text{ mA}
-$$
-
-The negative sign indicates that the current flows in the direction oppo site to the one assumed, that is, from terminal *b* to terminal *a*.
-
-Practice Problem 4.17
-
-Example 4.18
-
-For Example 4.18: (a) Finding *R*Th, (b) finding *V*Th, (c) determining the current through the galvanometer.
-
-Practice Problem 4.18 14 Ω 60 Ω 16 V 40 Ω 20 Ω 30 Ω G
-
-**Figure 4.66** For Practice Prob. 4.18.
-
-Obtain the current through the g alvanometer, ha ving a resistance of 14 Ω, in the Wheatstone bridge shown in Fig. 4.66.
-
-**Answer:** 64 mA.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/050_4.11 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/050_4.11 Summary.md
deleted file mode 100644
index 9a16178f0a48e6c9fe267e4392b36b0e06b4e112..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/050_4.11 Summary.md
+++ /dev/null
@@ -1,75 +0,0 @@
-# **4.11** Summary
-
-- 1. A linear netw ork consists of linear elements, linear dependent sources, and linear independent sources.
-- 2. Network theorems are used to reduce a complex circuit to a simpler one, thereby making circuit analysis much simpler.
-- 3. The superposition principle states that for a circuit ha ving multiple independent sources, the voltage across (or current through) an element is equal to the algebraic sum of all the individual voltages (or currents) due to each independent source acting one at a time.
-- 4. Source transformation is a procedure for transforming a v oltage source in series with a resistor to a current source in parallel with a resistor, or vice versa.
-- 5. Thevenin's and Norton's theorems allow us to isolate a portion of a network while the remaining portion of the netw ork is replaced by an equivalent network. The Thevenin equivalent consists of a v oltage source *V*Th in series with a resistor *R*Th, while the Norton equivalent consists of a current source *IN* in parallel with a resistor *RN*. The two theorems are related by source transformation.
-
-$$
-R_N = R_{\text{Th}}, \qquad I_N = \frac{V_{\text{Th}}}{R_{\text{Th}}}
-$$
-
-- 6. For a gi ven Thevenin equivalent circuit, maximum po wer transfer occurs when *RL* = *R*Th; that is, when the load resistance is equal to the Thevenin resistance.
-- 7. The maximum po wer transfer theorem states that the maximum power is delivered by a source to the load *RL* when *RL* is equal to *R*Th, the Thevenin resistance at the terminals of the load.
-- 8. *PSpice* can be used to v erify the circuit theorems co vered in this chapter.
-- 9. Source modeling and resistance measurement using the Wheatstone bridge provide applications for Thevenin's theorem.
-
-# Review Questions
-
-**4.1** The current through a branch in a linear network is 2 A when the input source voltage is 10 V. If the voltage is reduced to 1 V and the polarity is reversed, the current through the branch is:
-
-> (a) − 2 A (b) − 0.2 A (c) 0.2 A (d) 2 A (e) 20 A
-
-**4.2** For superposition, it is not required that only one independent source be considered at a time; any number of independent sources may be considered simultaneously.
-
-(a) True (b) False
-
-**4.3** The superposition principle applies to power calculation.
-
-(a) True (b) False
-
-- **4.4** Refer to Fig. 4.67. The Thevenin resistance at terminals *a* and *b* is:
- - (a) 25 Ω (b) 20 Ω
-
-# **Figure 4.67**
-
-For Review Questions 4.4 to 4.6.
-
-**4.5** The Thevenin voltage across terminals *a* and *b* of the circuit in Fig. 4.67 is:
-
-| (a) 50 V | (b) 40 V |
-|----------|----------|
-| (c) 20 V | (d) 10 V |
-
-**4.6** The Norton current at terminals *a* and *b* of the circuit in Fig. 4.67 is:
-
-| (a) 10 A | (b) 2.5 A |
-|----------|-----------|
-| (c) 2 A | (d) 0 A |
-
-**4.7** The Norton resistance *RN* is exactly equal to the Thevenin resistance *R*Th.
-
-(a) True (b) False
-
-**4.8** Which pair of circuits in Fig. 4.68 are equivalent?
-
-| (a) a and b | (b) b and d |
-|-------------|-------------|
-| (c) a and c | (d) c and d |
-
-# **Figure 4.68**
-
-For Review Question 4.8.
-
-**4.9** A load is connected to a network. At the terminals to which the load is connected, *R*Th = 10 Ω and *V*Th = 40 V. The maximum possible power supplied to the load is:
-
-| (a) 160 W | (b) 80 W |
-|-----------|----------|
-| (c) 40 W | (d) 1 W |
-
-**4.10** The source is supplying the maximum power to the load when the load resistance equals the source resistance.
-
-(a) True (b) False
-
-*Answers: 4.1b, 4.2a, 4.3b, 4.4d, 4.5b, 4.6a, 4.7a, 4.8c, 4.9c, 4.10a.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/051_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/051_Review Questions.md
deleted file mode 100644
index 782cb9c901384a859a5c627a5bd41a09bc3c0a0b..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/051_Review Questions.md
+++ /dev/null
@@ -1,411 +0,0 @@
-# Problems
-
-# Section 4.2 Linearity Property
-
-**4.1** Calculate the current *io* in the circuit of Fig. 4.69. What value of input voltage is necessary to make *io* equal to 5 amps?
-
-# **Figure 4.69**
-
-For Prob. 4.1.
-
-- **4.3** (a) In the circuit of Fig. 4.71, calculate *vo* and *io* when *vs* = 1 V.
- - (b) Find *vo* and *io* when *vs* = 10 V.
- - (c) What are *vo* and *io* when each of the 1-Ω resistors is replaced by a 10-Ω resistor and *vs* = 10 V?
-
-For Prob. 4.3.
-
-**4.4** Use linearity to determine *io* in the circuit of Fig. 4.72.
-
-**4.5** For the circuit in Fig. 4.73, assume *vo* = 1 V, and use linearity to find the actual value of *vo*.
-
-# **Figure 4.73**
-
-For Prob. 4.5.
-
-**4.6** For the linear circuit shown in Fig. 4.74, use linearity to complete the following table.
-
-# **Figure 4.74**
-
-For Prob. 4.6.
-
-**4.7** Use linearity and the assumption that *Vo* = 1 V to find the actual value of *Vo* in Fig. 4.75.
-
-# Section 4.3 Superposition
-
-**4.8** Using superposition, find *Vo* in the circuit of Fig. 4.76. Check with *PSpice or MultiSim*.
-
-## Problems **161**
-
-**4.9** Given that *I* = 6 amps when *Vs* = 160 volts and *Is* = −10 amps and *I* = 5 amp when *Vs* = 200 volts and *Is* = 0, use superposition and linearity to determine the value of *I* when *Vs* = 120 volts and *Is* = 5 amps.
-
-**4.10** Using Fig. 4.78, design a problem to help other students better understand superposition. Note, the letter *k* is a gain you can specify to make the problem easier to solve but must not be zero.
-
-**Figure 4.78** For Prob. 4.10.
-
-**4.11** Use the superposition principle to find *io* and *vo* in the circuit of Fig. 4.79.
-
-For Prob. 4.11.
-
-**4.12** Determine *vo* in the circuit of Fig. 4.80 using the superposition principle.
-
-**4.13** Use superposition to find *vo* in the circuit of Fig. 4.81.
-
-**4.14** Apply the superposition principle to find *vo* in the circuit of Fig. 4.82.
-
-# **Figure 4.82**
-
-For Prob. 4.14.
-
-**4.15** For the circuit in Fig. 4.83, use superposition to find *i*. Calculate the power delivered to the 3-Ω resistor.
-
-# **Figure 4.83** For Probs. 4.15 and 4.56.
-
-**4.16** Given the circuit in Fig. 4.84, use superposition to obtain *io*.
-
-For Prob. 4.16.
-
-**Figure 4.85** For Prob. 4.17.
-
-**4.18** Use superposition to find *Vo* in the circuit of Fig. 4.86.
-
-**Figure 4.86**
-
-- For Prob. 4.18.
-- **4.19** Use superposition to solve for *vx* in the circuit of Fig. 4.87.
-
-**Figure 4.87**
-
-For Prob. 4.19.
-
-# Section 4.4 Source Transformation
-
-**4.20** Use source transformation to reduce the circuit between terminals *a* and *b* shown in Fig. 4.88 to a single voltage source in series with a single resistor.
-
-**Figure 4.88** For Prob. 4.20.
-
-**4.21** Using Fig. 4.89, design a problem to help other students better understand source transformation.
-
-**Figure 4.89** For Prob. 4.21.
-
-**4.22** For the circuit in Fig. 4.90, use source transformation to find *i*.
-
-# **Figure 4.90**
-
-For Prob. 4.22.
-
-**4.23** Referring to Fig. 4.91, use source transformation to determine the current and power absorbed by the 8-Ω resistor.
-
-**Figure 4.91** For Prob. 4.23.
-
-**4.24** Use source transformation to find the voltage *Vx* in the circuit of Fig. 4.92.
-
-## Problems **163**
-
-**4.25** Obtain *vo* in the circuit of Fig. 4.93 using source transformation. Check your result using *PSpice or MultiSim*.
-
-# **Figure 4.93**
-
-For Prob. 4.25.
-
-**4.26** Use source transformation to find *io* in the circuit of Fig. 4.94.
-
-# **Figure 4.94**
-
-For Prob. 4.26.
-
-**4.27** Apply source transformation to find *vx* in the circuit of Fig. 4.95.
-
-**4.28** Use source transformation to find *Io* in Fig. 4.96.
-
-For Prob. 4.28.
-
-**4.29** Use source transformation to find *vo* in the circuit of Fig. 4.97.
-
-# **Figure 4.97**
-
-For Prob. 4.29.
-
-**4.30** Use source transformation on the circuit shown in Fig 4.98 to find *ix*.
-
-**4.31** Determine *vx* in the circuit of Fig. 4.99 using source transformation.
-
-**Figure 4.99** For Prob. 4.31.
-
-**4.32** Use source transformation to find *ix* in the circuit of Fig. 4.100.
-
-# Sections 4.5 and 4.6 Thevenin's and Norton's Theorems
-
-- **4.33** Determine the Thevenin equivalent circuit, shown in Fig. 4.101, as seen by the 7-ohm resistor.
- - Then calculate the current flowing through the 7-ohm resistor.
-
-For Prob. 4.33.
-
-**4.34** Using Fig. 4.102, design a problem that will help other students better understand Thevenin equivalent circuits.
-
-# **Figure 4.104**
-
-For Prob. 4.37.
-
-**4.38** Apply Thevenin's theorem to find *Vo* in the circuit of Fig. 4.105.
-
-**Figure 4.105** For Prob. 4.38.
-
-**4.39** Obtain the Thevenin equivalent at terminals *a*-*b* of the circuit shown in Fig. 4.106.
-
-# **Figure 4.106**
-
-For Prob. 4.39.
-
-**4.40** Find the Thevenin equivalent at terminals *a*-*b* of the circuit in Fig. 4.107.
-
-For Prob. 4.40.
-
-**Figure 4.102**
-
-For Probs. 4.34 and 4.49.
-
-- **4.35** Use Thevenin's theorem to find *vo* in Prob. 4.12.
-- **4.36** Solve for the current *i* in the circuit of Fig. 4.103 using Thevenin's theorem. (*Hint:* Find the Thevenin equivalent seen by the 12-Ω resistor.)
-
-**4.41** Find the Thevenin and Norton equivalents at terminals *a*-*b* of the circuit shown in Fig. 4.108.
-
-For Prob. 4.41.
-
-**4.42** For the circuit in Fig. 4.109, find the Thevenin equivalent between terminals *a* and *b*. **\***
-
-# **Figure 4.109**
-
-For Prob. 4.42.
-
-**4.43** Find the Thevenin equivalent looking into terminals *a*-*b* of the circuit in Fig. 4.110 and solve for *ix*.
-
-# **Figure 4.110**
-
-For Prob. 4.43.
-
-**4.44** For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals:
-
-**Figure 4.111**
-
-For Prob. 4.44.
-
-\* An asterisk indicates a challenging problem.
-
-**4.45** Find the Thevenin equivalent of the circuit in Fig. 4.112 as seen by looking into terminals *a* and *b*.
-
-# **Figure 4.112**
-
-For Prob. 4.45.
-
-**4.46** Using Fig. 4.113, design a problem to help other students better understand Norton equivalent circuits.
-
-**Figure 4.113** For Prob. 4.46.
-
-**4.47** Obtain the Thevenin and Norton equivalent circuits of the circuit in Fig. 4.114 with respect to terminals *a* and *b*.
-
-# **Figure 4.114**
-
-For Prob. 4.47.
-
-**4.48** Determine the Norton equivalent at terminals *a*-*b* for the circuit in Fig. 4.115.
-
-**Figure 4.115** For Prob. 4.48.
-
-**4.49** Find the Norton equivalent looking into terminals *a*-*b* of the circuit in Fig. 4.102. Let *V* = 40 V, *I* = 3 A, *R*1 = 10 Ω, *R*2 = 40 Ω, and *R*3 = 20 Ω.
-
-**4.50** Obtain the Norton equivalent of the circuit in Fig. 4.116 to the left of terminals *a*-*b*. Use the result to find current *i*.
-
-# **Figure 4.116**
-
-For Prob. 4.50.
-
-**4.51** Given the circuit in Fig. 4.117, obtain the Norton equivalent as viewed from terminals:
-
-**Figure 4.117**
-
-For Prob. 4.51.
-
-**4.52** For the transistor model in Fig. 4.118, obtain the Thevenin equivalent at terminals *a*-*b*.
-
-**4.53** Find the Norton equivalent at terminals *a*-*b* of the circuit in Fig. 4.119.
-
-**4.54** Find the Thevenin equivalent between terminals *a*-*b* of the circuit in Fig. 4.120.
-
-For Prob. 4.54.
-
-**4.55** Obtain the Norton equivalent at terminals *a*-*b* of the circuit in Fig. 4.121. **\***
-
-For Prob. 4.55.
-
-**4.56** Use Norton's theorem to find *Vo* in the circuit of Fig. 4.122.
-
-**Figure 4.122** For Prob. 4.56.
-
-**4.57** Obtain the Thevenin and Norton equivalent circuits at terminals *a*-*b* for the circuit in Fig. 4.123.
-
-**Figure 4.123** For Probs. 4.57 and 4.79.
-
-**4.58** The network in Fig. 4.124 models a bipolar transistor common-emitter amplifier connected to a load. Find the Thevenin resistance seen by the load.
-
-**Figure 4.124**
-
-For Prob. 4.58.
-
-**4.59** Determine the Thevenin and Norton equivalents at terminals *a*-*b* of the circuit in Fig. 4.125.
-
-# **Figure 4.125**
-
-For Probs. 4.59 and 4.80.
-
-**4.60** For the circuit in Fig. 4.126, find the Thevenin and Norton equivalent circuits at terminals *a*-*b*. **\***
-
-# **Figure 4.126**
-
-For Probs. 4.60 and 4.81.
-
-**4.61** Obtain the Thevenin and Norton equivalent circuits at terminals *a*-*b* of the circuit in Fig. 4.127. **\***
-
-For Prob. 4.61.
-
-**4.62** Find the Thevenin equivalent of the circuit in Fig. 4.128. **\***
-
-# **Figure 4.128** For Prob. 4.62.
-
-**4.63** Find the Norton equivalent for the circuit in Fig. 4.129.
-
-# **Figure 4.129**
-
-For Prob. 4.63.
-
-**4.64** Obtain the Thevenin equivalent seen at terminals *a*-*b* of the circuit in Fig. 4.130.
-
-# **Figure 4.130**
-
-For Prob. 4.64.
-
-For Prob. 4.65.
-
-**4.65** For the circuit shown in Fig. 4.131, determine the relationship between *Vo* and *Io*.
-
-# Section 4.8 Maximum Power Transfer
-
-**4.66** Find the maximum power that can be delivered to the resistor *R* in the circuit of Fig. 4.132.
-
-# **Figure 4.132**
-
-For Prob. 4.66.
-
-- **4.67** The variable resistor *R* in Fig. 4.133 is adjusted until it absorbs the maximum power from the circuit.
- - (a) Calculate the value of *R* for maximum power. (b) Determine the maximum power absorbed by *R*.
-
-# **Figure 4.133**
-
-For Prob. 4.67.
-
-**4.68** Consider the 30-Ω resistor in Fig. 4.134. First compute the Thevenin equivalent circuit as seen by the 30-Ω resistor. Compute the value of *R* that results in Thevenin equivalent resistance equal to the 30-Ω resistance and then calculate power delivered to the 30-Ω resistor. Now let *R* = 0 Ω, 110 Ω, and ∞, calculate the power delivered to the 30-Ω resistor in each case. What can you say about the value of *R* that will result in the maximum power that can be delivered to the 30-Ω resistor? **\***
-
-**Figure 4.134**
-
-# For Prob. 4.68.
-
-**4.69** Find the maximum power transferred to resistor *R* in the circuit of Fig. 4.135.
-
-**Figure 4.135** For Prob. 4.69.
-
-**4.70** Determine the maximum power delivered to the variable resistor *R* shown in the circuit of Fig. 4.136.
-
-# **Figure 4.136**
-
-**4.71** For the circuit in Fig. 4.137, what resistor connected across terminals *a*-*b* will absorb maximum power from the circuit? What is that power?
-
-**Figure 4.137** For Prob. 4.71.
-
-- **4.72** (a) For the circuit in Fig. 4.138, obtain the Thevenin equivalent at terminals *a*-*b*.
- - (b) Calculate the current in *RL* = 13 Ω.
- - (c) Find *RL* for maximum power deliverable to *RL*.
- - (d) Determine that maximum power.
-
-**Figure 4.138** For Prob. 4.72.
-
-**4.73** Determine the maximum power that can be delivered to the variable resistor *R* in the circuit of Fig. 4.139.
-
-# **Figure 4.139**
-
-- For Prob. 4.73.
- - **4.74** For the bridge circuit shown in Fig. 4.140, find the load *RL* for maximum power transfer and the maximum power absorbed by the load.
-
-**Figure 4.140** For Prob. 4.74.
-
-- **4.75** For the circuit in Fig. 4.141, determine the value of *R* such that the maximum power delivered to the load is 12 mW. **\***
-
-For Prob. 4.75.
-
-# Section 4.9 Verifying Circuit Theorems with PSpice
-
-- **4.76** Solve Prob. 4.34 using *PSpice or MultiSim*. Let *V* = 40 V, *I* = 3 A, *R*1 = 10 Ω, *R*2 = 40 Ω, and *R*3 = 20 Ω.
-- **4.77** Use *PSpice or MultiSim* to solve Prob. 4.44.
-- **4.78** Use *PSpice or MultiSim* to solve Prob. 4.52.
-- **4.79** Obtain the Thevenin equivalent of the circuit in Fig. 4.123 using *PSpice or MultiSim*.
-
-- **4.80** Use *PSpice or MultiSim* to find the Thevenin equivalent circuit at terminals *a*-*b* of the circuit in Fig. 4.125.
-- **4.81** For the circuit in Fig. 4.126, use *PSpice or MultiSim* to find the Thevenin equivalent at terminals *a*-*b*.
-
-# Section 4.10 Applications
-
-- **4.82** An automobile battery has an open circuit voltage of 14.7 V which drops to 12 V when connected to two 65-W headlights. What is the resistance of the headlights and the value of the internal resistance of the battery?
-- **4.83** The following results were obtained from measurements taken between the two terminals of a resistive network.
-
-| Terminal Voltage | 72 V | 0 V |
-|------------------|------|-----|
-| Terminal Current | 0 A | 9 A |
-
-Find the Thevenin equivalent of the network.
-
-- **4.84** When connected to a 4- Ω resistor, a battery has a terminal voltage of 10.8 V but produces 12 V on an open circuit. Determine the Thevenin equivalent circuit for the battery.
-- **4.85** The Thevenin equivalent at terminals *a*-*b* of the linear network shown in Fig. 4.142 is to be determined by measurement. When a 10-kΩ resistor is connected to terminals *a-b*, the voltage *Vab* is measured as 20 V. When a 30-kΩ resistor is connected to the terminals, *Vab* is measured as 40 V. Determine: (a) the Thevenin equivalent at terminals *a*-*b*, (b) *Vab* when a 20-kΩ resistor is connected to terminals *a*-*b*.
-
-# **Figure 4.142**
-
-For Prob. 4.85.
-
-**4.86** A black box with a circuit in it is connected to a variable resistor. An ideal ammeter (with zero resistance) and an ideal voltmeter (with infinite resistance) are used to measure current and voltage as shown in Fig. 4.143. The results are shown in the table on the next page.
-
-- (a) Find *i* when *R* = 12 Ω.
-- (b) Determine the maximum power from the box.
-
-| R(Ω) | V(V) | i(A) |
-|------|------|------|
-| 2 | 6 | 3 |
-| 8 | 16 | 2 |
-| 14 | 21 | 1.5 |
-
-**4.87** A transducer is modeled with a current source *Is* and a parallel resistance *Rs*. The current at the terminals of the source is measured to be 9.975 mA when an ammeter with an internal resistance of 20 Ω is used.
-
-- (a) If adding a 2-kΩ resistor across the source terminals causes the ammeter reading to fall to 9.876 mA, calculate *Is* and *Rs*.
-- (b) What will the ammeter reading be if the resistance between the source terminals is changed to 4 kΩ?
-- **4.88** Consider the circuit in Fig. 4.144. An ammeter with internal resistance *Ri* is inserted between *A* and *B* to measure *Io*. Determine the reading of the ammeter if: (a) *Ri* = 500 Ω, (b) *Ri* = 0 Ω. (*Hint:* Find the Thevenin equivalent circuit at terminals *a*-*b*.)
-
-**Figure 4.144**
-
-For Prob. 4.88.
-
-**4.89** Consider the circuit in Fig. 4.145. (a) Replace the resistor *RL* by a zero resistance ammeter and determine the ammeter reading. (b) To verify the reciprocity theorem, interchange the ammeter and the 12-V source and determine the ammeter reading again.
-
-**Figure 4.145** For Prob. 4.89.
-
-**4.90** The Wheatstone bridge circuit shown in Fig. 4.146 is used to measure the resistance of a strain gauge. The adjustable resistor has a linear taper with a maximum value of 100 Ω. If the resistance of the strain gauge is found to be 42.6 Ω, what fraction of the full slider travel is the slider when the bridge is balanced?
-
-For Prob. 4.90.
-
-- can measure *Rx* in the range of 0–25 Ω.
-- (b) Repeat for the range of 0–250 Ω.
-
-# **Figure 4.147**
-
-For Prob. 4.91.
-
-**4.92** Consider the bridge circuit of Fig. 4.148. Is the bridge balanced? If the 10-kΩ resistor is replaced by an 18-kΩ resistor, what resistor connected between terminals *a*-*b* absorbs the maximum power? What is this power? **\***
-
-For Prob. 4.92.
-
-**\***
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/052_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/052_Comprehensive Problems.md
deleted file mode 100644
index 7ac6979824781c2553ccad9f0d6ca5275ffd00ad..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/052_Comprehensive Problems.md
+++ /dev/null
@@ -1,46 +0,0 @@
-# Comprehensive Problems
-
-**4.93** The circuit in Fig. 4.149 models a common-emitter transistor amplifier. Find *ix* using source transformation.
-
-# **Figure 4.149**
-
-For Prob. 4.93.
-
-**4.94** An attenuator is an interface circuit that reduces the voltage level without changing the output resistance.
-
-> (a) By specifying *Rs* and *Rp* of the interface circuit in Fig. 4.150, design an attenuator that will meet the following requirements:
-
-$$
-\frac{V_o}{V_g} = 0.125, \qquad R_{\text{eq}} = R_{\text{Th}} = R_g = 100 \ \Omega
-$$
-
-(b) Using the interface designed in part (a), calculate the current through a load of *RL* = 50 Ω when *Vg* = 12 V.
-
-For Prob. 4.94.
-
-**4.95** A dc voltmeter with a sensitivity of 10 kΩ∕V is used to find the Thevenin equivalent of a linear network. Readings on two scales are as follows: **\***
-
-(a) 0–10 V scale: 8 V (b) 0–50 V scale: 10 V
-
-Obtain the Thevenin voltage and the Thevenin resistance of the network.
-
-**4.96** A resistance array is connected to a load resistor *R* and a 9-V battery as shown in Fig. 4.151.
-
-- (a) Find the value of *R* such that *Vo* = 1.8 V.
-- (b) Calculate the value of *R* that will draw the maximum current. What is the maximum current?
-
-# **Figure 4.151** For Prob. 4.96.
-
-**4.97** A common-emitter amplifier circuit is shown in Fig. 4.152. Obtain the Thevenin equivalent to the left of points *B* and *E*.
-
-# **Figure 4.152**
-
-For Prob. 4.97.
-
-**4.98** For Practice Prob. 4.18, determine the current through the 40-Ω resistor and the power dissipated by the resistor. **\***
-
-*This page intentionally left blank*
-
-# **chapter**
-
-5
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/053_Chapter 5 - Operational Amplifiers.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/053_Chapter 5 - Operational Amplifiers.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/053_Chapter 5 - Operational Amplifiers.md
+++ /dev/null
@@ -1,28 +0,0 @@
-# Operational Amplifiers
-
-*He who will not reason is a bigot; he who cannot is a fool; and he who dares not is a slave.*
-
-—Lord Byron
-
-# Enhancing Your Career
-
-# **Career in Electronic Instrumentation**
-
-Engineering involves applying physical principles to design devices for the benefit of humanity. But ph ysical principles cannot be understood without measurement. In f act, ph ysicists often say that ph ysics is the science that measures reality. Just as measurements are a tool for understanding the physical world, instruments are tools for measurement. The operational amplifier introduced in this chapter is a building block of modern electronic instrumentation. Therefore, mastery of operational amplifier fundamentals is paramount to any practical application of electronic circuits.
-
-Electronic instruments are used in all fields of science and engineering. They have proliferated in science and technology to the e xtent that it would be ridiculous to have a scientific or technical education without exposure to electronic instruments. For example, physicists, physiologists, chemists, and biologists must learn to use electronic instruments. For electrical engineering students in particular, the skill in operating digital and analog electronic instruments is crucial. Such instruments include ammeters, v oltmeters, ohmmeters, oscilloscopes, spectrum analyzers, and signal generators.
-
-Beyond de veloping the skill for operating the instruments, some electrical engineers specialize in designing and constructing electronic instruments. These engineers deri ve pleasure in b uilding their o wn instruments. Most of them invent and patent their inventions. Specialists in electronic instruments find employment in medical schools, hospitals, research laboratories, aircraft industries, and thousands of other indus tries where electronic instruments are routinely used.
-
-Electronic Instrumentation used in medical research. © Royalty-Free/Corbis
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Comprehend how real operational amplifiers (op amps) function.
-- 2. Understand that ideal op amps function nearly identically to real ones and that they can be used to model them effectively in a variety of circuit applications.
-- 3. Realize how the basic inverting op amp is the workhorse of the op amp family.
-- 4. Use the inverting op amp to create summers.
-- 5. Use the op amp to create a difference amplifier.
-- 6. Explain how to cascade a variety of op amp circuits.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/054_5.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/054_5.1 Introduction.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/054_5.1 Introduction.md
+++ /dev/null
@@ -1,9 +0,0 @@
-# **5.1** Introduction
-
-Having learned the basic la ws and theorems for circuit analysis, we are now ready to study an acti ve circuit element of paramount importance: the *operational amplifier,* or *op amp* for short. The op amp is a versatile circuit building block.
-
-The op amp is an electronic unit that behaves like a voltage-controlled voltage source.
-
-It can also be used in making a v oltage- or current -controlled current source. An op amp can sum signals, amplify a signal, inte grate it, or differentiate it. The ability of the op amp to perform these mathematical operations is the reason it is called an *operational amplifier*. It is also the reason for the widespread use of op amps in analog design. Op amps are popular in practical circuit designs because the y are versatile, inexpensive, easy to use, and fun to work with.
-
-We begin by discussing the ideal op amp and later consider the nonideal op amp. Using nodal analysis as a tool, we consider ideal op amp circuits such as the in verter, voltage follower, summer, and dif ference amplifier. We will also analyze op amp circuits with *PSpice*. Finally, we learn how an op amp is used in digital-to-analog converters and instrumentation amplifiers.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/055_5.2 Operational Amplifiers.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/055_5.2 Operational Amplifiers.md
deleted file mode 100644
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@@ -1,160 +0,0 @@
-# **5.2** Operational Amplifiers
-
-An operational amplifier is designed so that it performs some mathematical operations when e xternal components, such as resistors and capaci tors, are connected to its terminals. Thus,
-
-An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration.
-
-The op amp is an electronic device consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. A full discussion of what is inside the op amp is beyond the scope of this book. It will suffice
-
-The term operational amplifier was introduced in 1947 by John Ragazzini and his colleagues, in their work on analog computers for the National Defense Research Council after World War II. The first op amps used vacuum tubes rather than transistors.
-
-An op amp may also be regarded as a voltage amplifier with very high gain.
-
-to treat the op amp as a circuit b uilding block and simply study what takes place at its terminals.
-
-Op amps are commercially a vailable in integrated circuit packages in several forms. Figure 5.1 sho ws a typical op amp package. A typical one is the eight-pin dual in-line package (or DIP), shown in Fig. 5.2(a). Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to us. The five important terminals are:
-
-- 1. The inverting input, pin 2.
-- 2. The noninverting input, pin 3.
-- 3. The output, pin 6.
-- 4. The positive power supply *V*+, pin 7.
-- 5. The negative power supply *V*−, pin 4.
-
-The circuit symbol for the op amp is the triangle in Fig. 5.2(b); as shown, the op amp has tw o inputs and one output. The inputs are mark ed with minus (−) and plus (+) to specify *inverting* and *noninverting* inputs, respectively. An input applied to the noninverting terminal will appear with the same polarity at the output, while an input applied to the in verting terminal will appear inverted at the output.
-
-As an active element, the op amp must be powered by a voltage supply as typically shown in Fig. 5.3. Although the power supplies are often ignored in op amp circuit diagrams for the sake of simplicity, the power supply currents must not be overlooked. By KCL,
-
-$$
-i_o = i_1 + i_2 + i_+ + i_- \tag{5.1}
-$$
-
-The equivalent circuit model of an op amp is shown in Fig. 5.4. The output section consists of a v oltage-controlled source in series with the
-
-**Figure 5.2** A typical op amp: (a) pin configuration, (b) circuit symbol.
-
-Powering the op amp.
-
-**Figure 5.4** The equivalent circuit of the nonideal op amp.
-
-# **Figure 5.1**
-
-A typical operational amplifier. © McGraw-Hill Education/Mark Dierker, photographer
-
-The pin diagram in Fig. 5.2(a) corresponds to the 741 generalpurpose op amp made by Fairchild Semiconductor.
-
-output resistance *Ro*. It is evident from Fig. 5.4 that the input resis tance *Ri* is the Thevenin equivalent resistance seen at the input terminals, while the output resistance *Ro* is the Thevenin equivalent resistance seen at the output. The differential input voltage *vd* is given by
-
-$$
-v_d = v_2 - v_1 \tag{5.2}
-$$
-
-where *v*1 is the voltage between the inverting terminal and ground and *v*2 is the voltage between the noninverting terminal and ground. The op amp senses the difference between the two inputs, multiplies it by the gain *A*, and causes the resulting voltage to appear at the output. Thus, the output *vo* is given by
-
-$$
-v_o = Av_d = A(v_2 - v_1)
-$$
- (5.3)
-
-Sometimes, voltage gain is expressed in decibels (dB), as discussed in Chapter 14.
-
-*A* dB = 20 log10 A
-
-*A* is called the *open-loop voltage gain* because it is the g ain of the op amp without any external feedback from output to input. Table 5.1 shows typical values of voltage gain *A*, input resistance *Ri*, output resistance *Ro*, and supply voltage *VCC*.
-
-The concept of feedback is crucial to our understanding of op amp circuits. A negative feedback is achieved when the output is fed back to the inverting terminal of the op amp. As Example 5.1 shows, when there is a feedback path from output to input, the ratio of the output voltage to the input voltage is called the *closed-loop gain*. As a result of the ne gative feedback, it can be shown that the closed-loop gain is almost insensitive to the open-loop gain *A* of the op amp. For this reason, op amps are used in circuits with feedback paths.
-
-A practical limitation of the op amp is that the magnitude of its output voltage cannot e xceed |*VCC*|. In other w ords, the output v oltage is dependent on and is limited by the po wer supply voltage. Figure 5.5 illustrates that the op amp can operate in three modes, depending on the differential input voltage *vd*:
-
-- 1. Positive saturation, *vo* = *VCC*.
-- 2. Linear region, −*VCC* ≤ *vo* = *Avd* ≤ *VCC*.
-- 3. Negative saturation, *vo* = −*VCC*.
-
-If we attempt to increase *vd* beyond the linear range, the op amp becomes saturated and yields *vo* = *VCC* or *vo* = −*VCC*. Throughout this book, we will assume that our op amps operate in the linear mode. This means that the output voltage is restricted by
-
-$$
--V_{CC} \le v_o \le V_{CC} \tag{5.4}
-$$
-
-# **TABLE 5.1**
-
-# Typical ranges for op amp parameters.
-
-| Parameter | Typical range | Ideal values |
-|----------------------------------------------|--------------------------|--------------|
-| Open-loop gain, A | 105
to 108 | ` |
-| | 105 | `Ω |
-| Output resistance, Ro
Supply voltage, VCC | 10 to 100 Ω
5 to 24 V | 0Ω |
-| | Input resistance, Ri | to 1013 Ω |
-
-Op amp output voltage *vo* as a function of the differential input voltage *vd*.
-
-, input resis tance
-
-Although we shall al ways operate the op amp in the linear re gion, the possibility of saturation must be borne in mind when one designs with op amps, to a void designing op amp circuits that will not w ork in the laboratory.
-
-of 2 MΩ, and output resistance of 50 Ω. The op amp is used in the circuit of Fig. 5.6(a). Find the closed-loop gain *vo*∕*vs*. Determine current *i* when
-
-Throughout this book, we assume that an op amp operates in the linear range. Keep in mind the voltage constraint on the op amp in this mode.
-
-Example 5.1
-
-**Figure 5.6**
-
-For Example 5.1: (a) original circuit, (b) the equivalent circuit.
-
-A 741 op amp has an open-loop voltage gain of 2 × 105
-
-# **Solution:**
-
-Using the op amp model in Fig. 5.4, we obtain the equivalent circuit of Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit in Fig. 5.6(b) by using nodal analysis. At node 1, KCL gives
-
-$$
-\frac{v_s - v_1}{10 \times 10^3} = \frac{v_1}{2000 \times 10^3} + \frac{v_1 - v_o}{20 \times 10^3}
-$$
-
-Multiplying through by 2000 × 103 , we obtain
-
-$$
-200v_s = 301v_1 - 100v_o
-$$
-
-or
-
-$$
-2v_s \simeq 3v_1 - v_o \quad \Rightarrow \quad v_1 = \frac{2v_s + v_o}{3} \tag{5.1.1}
-$$
-
-At node *O*,
-
-$$
-\frac{v_1 - v_o}{20 \times 10^3} = \frac{v_o - Av_d}{50}
-$$
-
-But *vd* = −*v*1 and *A* = 200,000. Then
-
-$$
-v_1 - v_o = 400(v_o + 200,000v_1)
-$$
-\n(5.1.2)
-
-Substituting *v*1 from Eq. (5.1.1) into Eq. (5.1.2) gives
-
-$$
-0 \simeq 26,667,067v_o + 53,333,333v_s \quad \Rightarrow \quad \frac{v_o}{v_s} = -1.9999699
-$$
-
-This is closed -loop gain, because the 20 -kΩ feedback resistor closes the loop between the output and input terminals. When *vs* = 2 V, *vo* = −3.9999398 V. From Eq. (5.1.1), we obtain *v*1 = 20.066667 *μ*V. Thus,
-
-$$
-i = \frac{v_1 - v_o}{20 \times 10^3} = 0.19999 \text{ mA}
-$$
-
-It is evident that working with a nonideal op amp is tedious, as we are dealing with very large numbers.
-
-# Practice Problem 5.1
-
-**Figure 5.7** For Practice Prob. 5.1.
-
-If the same 741 op amp in Example 5.1 is used in the circuit of Fig. 5.7, calculate the closed-loop gain *v*o∕*v*s. Find *io* when *vs* = 1 V.
-
-**Answer:** 9.00041, 657 *μ*A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/056_5.3 Ideal Op Amp.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/056_5.3 Ideal Op Amp.md
deleted file mode 100644
index 289c6e32210486392bee804a13586f4c94cd686b..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/056_5.3 Ideal Op Amp.md
+++ /dev/null
@@ -1,88 +0,0 @@
-# **5.3** Ideal Op Amp
-
-To facilitate the understanding of op amp circuits, we will assume ideal op amps. An op amp is ideal if it has the following characteristics:
-
-- 1. Infinite open-loop gain, *A* ≃ ∞.
-- 2. Infinite input resistance, *Ri* ≃ ∞.
-- 3. Zero output resistance, *Ro* ≃ 0.
-
-An ideal op amp is an amplifier with infinite open-loop gain, infinite input resistance, and zero output resistance.
-
-Although assuming an ideal op amp provides only an approximate analysis, most modern amplifiers have such large gains and input im pedances that the approximate analysis is a good one. Unless stated otherwise, we will assume from now on that every op amp is ideal.
-
-For circuit analysis, the ideal op amp is illustrated in Fig. 5.8, which is derived from the nonideal model in Fig. 5.4. Two important properties of the ideal op amp are:
-
-1. The currents into both input terminals are zero:
-
-$$
-i_1 = 0, \quad i_2 = 0
-$$
- (5.5)
-
-This is due to infinite input resistance. An infinite resistance between the input terminals implies that an open circuit e xists there and current cannot enter the op amp. But the output current is not necessarily zero according to Eq. (5.1).
-
-2. The voltage across the input terminals is equal to zero; i.e.,
-
-$$
-v_d = v_2 - v_1 = 0 \tag{5.6}
-$$
-
-or
-
-$$
-v_1 = v_2 \tag{5.7}
-$$
-
-The two characteristics can be exploited by noting that for voltage calculations the input port behaves as a short circuit, while for current calculations the input port behaves as an open circuit.
-
-Thus, an ideal op amp has zero current into its tw o input terminals and the v oltage between the tw o input terminals is equal to zero. Equations (5.5) and (5.7) are e xtremely important and should be regarded as the key handles to analyzing op amp circuits.
-
-Rework Practice Prob. 5.1 using the ideal op amp model. Example 5.2
-
-# **Solution:**
-
-We may replace the op amp in Fig. 5.7 by its equivalent model in Fig. 5.9 as we did in Example 5.1. But we do not really need to do this. We just need to keep Eqs. (5.5) and (5.7) in mind as we analyze the circuit in Fig. 5.7. Thus, the Fig. 5.7 circuit is presented as in Fig. 5.9. Notice that
-
-$$
-v_2 = v_s \tag{5.2.1}
-$$
-
-Since *i*1 = 0, the 40 - and 5 -kΩ resistors are in series; the same current flows through them. *v*1 is the voltage across the 5 -kΩ resistor. Hence, using the voltage division principle,
-
-$$
-v_1 = \frac{5}{5+40} v_o = \frac{v_o}{9}
-$$
- (5.2.2)
-
-According to Eq. (5.7),
-
-$$
-v_2 = v_1 \tag{5.2.3}
-$$
-
-Substituting Eqs. (5.2.1) and (5.2.2) into Eq. (5.2.3) yields the closed loop gain,
-
-$$
-v_s = \frac{v_o}{9} \quad \Rightarrow \quad \frac{v_o}{v_s} = 9 \tag{5.2.4}
-$$
-
-which is very close to the value of 9.00041 obtained with the nonideal model in Practice Prob. 5.1. This shows that negligibly small error re sults from assuming ideal op amp characteristics.
-
-At node *O*,
-
-$$
-i_o = \frac{v_o}{40 + 5} + \frac{v_o}{20} \text{ mA}
-$$
- (5.2.5)
-
-From Eq. (5.2.4), when *vs* = 1 V, *vo* = 9 V. Substituting for *vo* = 9 V in Eq. (5.2.5) produces
-
-$$
-i_o = 0.2 + 0.45 = 0.65 \text{ mA}
-$$
-
-This, again, is close to the value of 0.657 mA obtained in Practice Prob. 5.1 with the nonideal model.
-
-Repeat Example 5.1 using the ideal op amp model. Practice Problem 5.2
-
-**Answer:** −2, 200 *μ*A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/057_5.4 Inverting Amplifier.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/057_5.4 Inverting Amplifier.md
deleted file mode 100644
index 6801d9a5c02772e11b332d8eb2c6c2f7300c9d6a..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/057_5.4 Inverting Amplifier.md
+++ /dev/null
@@ -1,136 +0,0 @@
-# **5.4** Inverting Amplifier
-
-In this and the following sections, we consider some useful op amp cir cuits that often serv e as modules for designing more comple x circuits. The first of such op amp circuits is the inverting amplifier shown in Fig. 5.10. In this circuit, the nonin verting input is grounded, *vi* is con nected to the in verting input through *R*1, and the feedback resistor *Rf* is connected between the inverting input and output. Our goal is to obtain
-
-+ – *v*2 0 V + +
-
-– 1
-
-i 2
-
-**Figure 5.10** The inverting amplifier.
-
-*v*i
-
-R1
-
-i 1
-
-*v*1 0 A
-
-For Example 5.2.
-
-Rf
-
-–
-
-*v*o +
-
-–
-
-the relationship between the input v oltage *v*i and the output v oltage *vo*. Applying KCL at node 1,
-
-$$
-i_1 = i_2 \Rightarrow \frac{v_i - v_1}{R_1} = \frac{v_1 - v_o}{R_f}
-$$
- (5.8)
-
-But *v*1 = *v*2 = 0 for an ideal op amp, since the nonin verting terminal is grounded. Hence,
-
-or
-
-$$
-\frac{v_i}{R_1} = -\frac{v_o}{R_f}
-$$
-$$
-v_o = -\frac{R_f}{R_1}v_i
-$$
-(5.9)
-
-The voltage gain is *Av* = *vo*∕*vi* = −*Rf*∕*R*1. The designation of the circuit in Fig. 5.10 as an *inverter* arises from the negative sign. Thus,
-
-An inverting amplifier reverses the polarity of the input signal while amplifying it.
-
-Notice that the gain is the feedback resistance di vided by the in put resistance which means that the gain depends only on the external elements connected to the op amp. In view of Eq. (5.9), an equivalent circuit for the inverting amplifier is shown in Fig. 5.11. The inverting amplifier is used, for example, in a current-to-voltage converter.
-
-Example 5.3 Refer to the op amp in Fig. 5.12. If *vi* = 0.5 V, calculate: (a) the output voltage *vo*, and (b) the current in the 10-kΩ resistor.
-
-# **Solution:**
-
-(a) Using Eq. (5.9),
-
-$$
-\frac{v_o}{v_i} = -\frac{R_f}{R_1} = -\frac{25}{10} = -2.5
-$$
-
-$$
-v_o = -2.5v_i = -2.5(0.5) = -1.25
-$$
- V
-
-(b) The current through the 10-kΩ resistor is
-
-$$
-i = \frac{v_i - 0}{R_1} = \frac{0.5 - 0}{10 \times 10^3} = 50 \,\mu\text{A}
-$$
-
-Practice Problem 5.3 Find the output of the op amp circuit shown in Fig. 5.13. Calculate the current through the feedback resistor.
-
-**Answer:** −3.15 V, 11.25 *μ*A.
-
-A key feature of the inverting amplifier is that both the input signal and the feedback are applied at the inverting terminal of the op amp.
-
-Note there are two types of gains: The one here is the closed-loop voltage gain Av, while the op amp itself has an
-
-open-loop voltage gain A.
-
-# **Figure 5.11**
-
-An equivalent circuit for the inverter in Fig. 5.10.
-
-**Figure 5.13** For Practice Prob. 5.3. Determine *vo* in the op amp circuit shown in Fig. 5.14.
-
-# **Solution:**
-
-Applying KCL at node *a*,
-
-$$
-\frac{v_a - v_o}{40 \text{ k}\Omega} = \frac{6 - v_a}{20 \text{ k}\Omega}
-$$
-
-$$
-v_a - v_o = 12 - 2v_a \implies v_o = 3v_a - 12
-$$
-
-But *va* = *vb* = 2 V for an ideal op amp, because of the zero voltage drop across the input terminals of the op amp. Hence,
-
-$$
-v_o = 6 - 12 = -6
-$$
- V
-
-Notice that if *vb* = 0 = *va*, then *vo* = −12, as expected from Eq. (5.9).
-
-For Example 5.4.
-
-Two kinds of current -to-voltage converters (also known as *transresis-* Practice Problem 5.4 *tance amplifiers*) are shown in Fig. 5.15.
-
-(a) Show that for the converter in Fig. 5.15(a),
-
-$$
-\frac{v_o}{i_s} = -R
-$$
-
-(b) Show that for the converter in Fig. 5.15(b),
-
-$$
-\frac{v_o}{i_s} = -R_1 \left( 1 + \frac{R_3}{R_1} + \frac{R_3}{R_2} \right)
-$$
-
-**Answer:** Proof.
-
-Another important application of the op amp is the nonin verting amplifier shown in Fig. 5.16. In this case, the input v oltage *vi* is applied directly at the nonin verting input terminal, and resistor *R*1 is connected
-
-**Figure 5.16** The noninverting amplifier.
-
-Example 5.4
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/058_5.5 Noninverting Amplifier.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/058_5.5 Noninverting Amplifier.md
deleted file mode 100644
index b06be8c55136b2e02c8f511048ea44e494d6d8c0..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/058_5.5 Noninverting Amplifier.md
+++ /dev/null
@@ -1,84 +0,0 @@
-
-between the ground and the in verting terminal. We are interested in the output voltage and the voltage gain. Application of KCL at the inverting terminal gives
-
-$$
-i_1 = i_2 \Rightarrow \frac{0 - v_1}{R_1} = \frac{v_1 - v_o}{R_f}
-$$
- (5.10)
-
-But *v*1 = *v*2 = *vi*. Equation (5.10) becomes
-
-$$
-\frac{-v_i}{R_1} = \frac{v_i - v_o}{R_f}
-$$
-
-or
-
-$$
-v_o = \left(1 + \frac{R_f}{R_1}\right) v_i
-$$
- (5.11)
-
-The voltage gain is *Av* = *vo*∕*vi* = 1 + *Rf*∕*R*1, which does not have a negative sign. Thus, the output has the same polarity as the input.
-
-A noninverting amplifier is an op amp circuit designed to provide a positive voltage gain.
-
-Again we notice that the gain depends only on the external resistors.
-
-Notice that if feedback resistor *Rf* = 0 (short circuit) or *R*1 = ∞ (open circuit) or both, the gain becomes 1. Under these conditions (*Rf* = 0 and *R*1 = ∞), the circuit in Fig. 5.16 becomes that shown in Fig. 5.17, which is called a *voltage follower* (or *unity gain amplifier*) because the output follows the input. Thus, for a voltage follower
-
-$$
-v_o = v_i \tag{5.12}
-$$
-
-Such a circuit has a ve ry high input impedance and is therefore use ful as an intermediate-stage (or buffer) amplifier to isolate one circuit from another, as portrayed in Fig. 5.18. The voltage follower minimizes interaction between the tw o stages and eliminates interstage loading.
-
-Example 5.3 For the op amp circuit in Fig. 5.19, calculate the output voltage *vo*.
-
-# **Solution:**
-
-We may solve this in two ways: using superposition and using nodal analysis.
-
-■ **METHOD 1** Using superposition, we let
-
-**Figure 5.17** The voltage follower.
-
-**Figure 5.18** A voltage follower used to isolate two cascaded stages of a circuit.
-
-where *vo*1 is due to the 6-V voltage source, and *vo*2 is due to the 4-V input. To get *vo*1, we set the 4-V source equal to zero. Under this condition, the circuit becomes an inverter. Hence Eq. (5.9) gives
-
-$$
-v_{o1} = -\frac{10}{4}(6) = -15
-$$
- V
-
-To get *vo*2, we set the 6 -V source equal to zero. The circuit becomes a noninverting amplifier so that Eq. (5.11) applies.
-
-$$
-v_{o2} = \left(1 + \frac{10}{4}\right)4 = 14 \text{ V}
-$$
-
-Thus,
-
-$$
-v_o = v_{o1} + v_{o2} = -15 + 14 = -1
-$$
- V
-
-■ **METHOD 2** Applying KCL at node *a*,
-
-$$
-\frac{6-v_a}{4} = \frac{v_a - v_o}{10}
-$$
-
-But *va*= *vb*=4, and so
-
-$$
-\frac{6-4}{4} = \frac{4-v_o}{10} \Rightarrow 5 = 4-v_o
-$$
-
-or *vo*= −1 V, as before.
-
-Calculate *vo* in the circuit of Fig. 5.20. Practice Problem 5.5
-
-**Answer:** 21 V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/059_5.6 Summing Amplifier.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/059_5.6 Summing Amplifier.md
deleted file mode 100644
index ab98e25a924b73d0187636f8329361e2d162c67f..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/059_5.6 Summing Amplifier.md
+++ /dev/null
@@ -1,94 +0,0 @@
-# **5.6** Summing Amplifier
-
-Besides amplification, the op amp can perform addition and subtraction. The addition is performed by the summing amplifier covered in this section; the subtraction is performed by the difference amplifier covered in the next section.
-
-A summing amplifier is an op amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs.
-
-The summing amplifier, shown in Fig. 5.21, is a v ariation of the inverting amplifier. It takes advantage of the fact that the inverting configuration can handle many inputs at the same time. We keep in mind
-
-**Figure 5.19** For Example 5.5.
-
-+
-
-4 kΩ
-
-6 V *v*o
-
-–
-
-a b
-
-4 V
-
-– +
-
-+
-
-# **Figure 5.20**
-
-4 kΩ
-
-For Practice Prob. 5.5.
-
-**Figure 5.21** The summing amplifier.
-
-+
-
-–
-
-10 kΩ
-
-+ – that the current entering each op amp input is zero. Applying KCL at node *a* gives
-
-$$
-i = i_1 + i_2 + i_3 \tag{5.13}
-$$
-
-But
-
-$$
-i_1 = \frac{v_1 - v_a}{R_1}, \quad i_2 = \frac{v_2 - v_a}{R_2}
-$$
-
-\n
-$$
-i_3 = \frac{v_3 - v_a}{R_3}, \quad i = \frac{v_a - v_o}{R_f}
-$$
- (5.14)
-
-We note that *va* = 0 and substitute Eq. (5.14) into Eq. (5.13). We get
-
-$$
-v_o = -\left(\frac{R_f}{R_1}v_1 + \frac{R_f}{R_2}v_2 + \frac{R_f}{R_3}v_3\right)
-$$
- (5.15)
-
-indicating that the output v oltage is a weighted sum of the inputs. F or this reason, the circuit in Fig. 5.21 is called a *summer*. Needless to say, the summer can have more than three inputs.
-
-# Example 5.6
-
-Calculate *vo* and *io* in the op amp circuit in Fig. 5.22.
-
-**Figure 5.22** For Example 5.6.
-
-# **Solution:**
-
-This is a summer with two inputs. Using Eq. (5.15) gives
-
-$$
-v_o = -\left[\frac{10}{5}(2) + \frac{10}{2.5}(1)\right] = -(4 + 4) = -8
-$$
- V
-
-The current *io* is the sum of the currents through the 10- and 2-kΩ resistors. Both of these resistors have voltage *vo* = −8 V across them, since *va* = *vb* = 0. Hence,
-
-$$
-i_o = \frac{v_o - 0}{10} + \frac{v_o - 0}{2}
-$$
- mA = -0.8 - 4 = -4.8 mA
-
-Find *vo* and *io* in the op amp circuit shown in Fig. 5.23.
-
-**Answer:** −3.8 V, −1.425 mA.
-
-or
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/060_5.7 Difference Amplifier.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/060_5.7 Difference Amplifier.md
deleted file mode 100644
index 84b035b058931bf5276a645b7b377b2bfcdfd847..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/060_5.7 Difference Amplifier.md
+++ /dev/null
@@ -1,175 +0,0 @@
-# **5.7** Difference Amplifier
-
-Difference (or dif ferential) amplifiers are used in various applications where there is a need to amplify the dif ference between tw o input sig nals. They are first cousins of the *instrumentation amplifier,* the most useful and popular amplifier, which we will discuss in Section 5.10.
-
-A difference amplifier is a device that amplifies the difference between two inputs but rejects any signals common to the two inputs.
-
-Consider the op amp circuit sho wn in Fig. 5.24. K eep in mind that zero currents enter the op amp terminals. Applying KCL to node *a*,
-
-> \_\_\_\_\_\_ *v*1 − *va R*1 = \_\_\_\_\_\_ *va* − *vo R*2
-
-$$
-v_o = \left(\frac{R_2}{R_1} + 1\right) v_a - \frac{R_2}{R_1} v_1 \tag{5.16}
-$$
-
-Difference amplifier.
-
-The difference amplifier is also known as the subtractor, for reasons to be shown later.
-
-# Practice Problem 5.6
-
-Applying KCL to node *b*,
-
-$$
-\frac{v_2 - v_b}{R_3} = \frac{v_b - 0}{R_4}
-$$
-
-or
-
-$$
-v_b = \frac{R_4}{R_3 + R_4} v_2 \tag{5.17}
-$$
-
-But *va* = *vb*. Substituting Eq. (5.17) into Eq. (5.16) yields
-
-$$
-v_o = \left(\frac{R_2}{R_1} + 1\right) \frac{R_4}{R_3 + R_4} v_2 - \frac{R_2}{R_1} v_1
-$$
-
-or
-
-$$
-v_o = \frac{R_2(1 + R_1/R_2)}{R_1(1 + R_3/R_4)} v_2 - \frac{R_2}{R_1} v_1
-$$
- (5.18)
-
-Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that *vo* = 0 when *v*1 = *v*2. This property exists when
-
-$$
-\frac{R_1}{R_2} = \frac{R_3}{R_4}
-$$
-\n(5.19)
-
-Thus, when the op amp circuit is a dif ference amplifier, Eq. (5.18) becomes
-
-$$
-v_o = \frac{R_2}{R_1}(v_2 - v_1)
-$$
-\n(5.20)
-
-If *R*2 = *R*1 and *R*3 = *R*4, the difference amplifier becomes a *subtractor,* with the output
-
-$$
-v_o = v_2 - v_1 \tag{5.21}
-$$
-
-Example 5.7 Design an op amp circuit with inputs *v*1 and *v*2 such that *vo* = −5*v*1 + 3*v*2.
-
-# **Solution:**
-
-The circuit requires that
-
-$$
-v_o = 3v_2 - 5v_1 \tag{5.7.1}
-$$
-
-This circuit can be realized in two ways.
-
-**Design 1** If we desire to use only one op amp, we can use the op amp circuit of Fig. 5.24. Comparing Eq. (5.7.1) with Eq. (5.18), we see
-
-$$
-\frac{R_2}{R_1} = 5 \quad \Rightarrow \quad R_2 = 5R_1 \tag{5.7.2}
-$$
-
-Also,
-
-$$
-5\frac{(1+R_1/R_2)}{(1+R_3/R_4)} = 3 \quad \Rightarrow \quad \frac{\frac{6}{5}}{1+R_3/R_4} = \frac{3}{5}
-$$
-
-or
-
-$$
-2 = 1 + \frac{R_3}{R_4} \Rightarrow R_3 = R_4 \tag{5.7.3}
-$$
-
-If we choose *R*1 = 10 k Ω and *R*3 = 20 k Ω, then *R*2 = 50 k Ω and *R*4 = 20 kΩ.
-
-**Design 2** If we desire to use more than one op amp, we may cascade an inverting amplifier and a two-input inverting summer, as shown in Fig. 5.25. For the summer,
-
-$$
-v_o = -v_a - 5v_1 \tag{5.7.4}
-$$
-
-and for the inverter,
-
-$$
-v_a = -3v_2 \tag{5.7.5}
-$$
-
-Combining Eqs. (5.7.4) and (5.7.5) gives
-
-*vo* = 3*v*2 − 5*v*1
-
-which is the desired result. In Fig. 5.25, we may select *R*1 = 10 kΩ and *R*3 = 20 kΩ or *R*1 = *R*3 = 10 kΩ.
-
-Design a difference amplifier with gain 7.5. Practice Problem 5.7
-
-**Answer:** Typical: *R*1 = *R*3 = 20 kΩ, *R*2 = *R*4 = 150 kΩ.
-
-An *instrumentation amplifier* shown in Fig. 5.26 is an amplifier of lowlevel signals used in process control or measurement applications and commercially available in single-package units. Show that
-
-$$
-v_o = \frac{R_2}{R_1} \left( 1 + \frac{2R_3}{R_4} \right) (v_2 - v_1)
-$$
-
-# **Solution:**
-
-We recognize that the amplifier *A*3 in Fig. 5.26 is a difference amplifier. Thus, from Eq. (5.20),
-
-$$
-v_o = \frac{R_2}{R_1}(v_{o2} - v_{o1})
-$$
-\n(5.8.1)
-
-Since the op amps *A*1 and *A*2 draw no current, current *i* flows through the three resistors as though they were in series. Hence,
-
-$$
-v_{o1} - v_{o2} = i(R_3 + R_4 + R_3) = i(2R_3 + R_4)
-$$
- (5.8.2)
-
-**Figure 5.25** For Example 5.7.
-
-Example 5.8
-
-**Figure 5.26** Instrumentation amplifier; for Example 5.8.
-
-But
-
-$$
-i = \frac{v_a - v_b}{R_4}
-$$
-
-and *va* = *v*1, *vb* = *v*2. Therefore,
-
-$$
-i = \frac{v_1 - v_2}{R_4}
-$$
- (5.8.3)
-
-Inserting Eqs. (5.8.2) and (5.8.3) into Eq. (5.8.1) gives
-
-$$
-v_o = \frac{R_2}{R_1} \left( 1 + \frac{2R_3}{R_4} \right) (v_2 - v_1)
-$$
-
-as required. We will discuss the instrumentation amplifier in detail in Section 5.10.
-
-Obtain *io* Practice Problem 5.8 in the instrumentation amplifier circuit of Fig. 5.27.
-
-# **Figure 5.27**
-
-Instrumentation amplifier; for Practice Prob. 5.8.
-
-**Answer:** 800 *μ*A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/061_5.8 Cascaded Op Amp Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/061_5.8 Cascaded Op Amp Circuits.md
deleted file mode 100644
index 9f154d7288bfeacde6ec98d39947670c5e1eca0c..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/061_5.8 Cascaded Op Amp Circuits.md
+++ /dev/null
@@ -1,109 +0,0 @@
-# **5.8** Cascaded Op Amp Circuits
-
-As we know, op amp circuits are modules or building blocks for designing complex circuits. It is often necessary in practical applications to connect op amp circuits in cascade (i.e., head to tail) to achie ve a large overall gain. In general, tw o circuits are cascaded when the y are con nected in tandem, one behind another in a single file.
-
-A cascade connection is a head-to-tail arrangement of two or more op amp circuits such that the output of one is the input of the next.
-
-When op amp circuits are cascaded, each circuit in the string is called a *stage;* the original input signal is increased by the g ain of the individual stage. Op amp circuits have the advantage that they can be cascaded without changing their input -output relationships. This is due to the fact that each (ideal) op amp circuit has infinite input resistance and zero output resistance. Figure 5.28 displays a block diagram represen tation of three op amp circuits in cascade. Since the output of one stage is the input to the next stage, the overall gain of the cascade connection is the product of the gains of the individual op amp circuits, or
-
-$$
-A = A_1 A_2 A_3 \tag{5.22}
-$$
-
-Although the cascade connection does not afect the op amp input-output relationships, care must be e xercised in the design of an actual op amp circuit to ensure that the load due to the ne xt stage in the cascade does not saturate the op amp.
-
-**Figure 5.28**
-
-A three-stage cascaded connection.
-
-Find *vo* and *io* in the circuit in Fig. 5.29. Example 5.9
-
-# **Solution:**
-
-This circuit consists of two noninverting amplifiers cascaded. At the output of the first op amp,
-
-$$
-v_a = \left(1 + \frac{12}{3}\right)(20) = 100 \text{ mV}
-$$
-
-At the output of the second op amp,
-
-$$
-v_o = \left(1 + \frac{10}{4}\right) v_a = (1 + 2.5)100 = 350
-$$
- mV
-
-The required current *io* is the current through the 10-kΩ resistor.
-
-$$
-i_o = \frac{v_o - v_b}{10}
-$$
- mA
-
-# **Figure 5.29** For Example 5.9.
-
-But
-$$
-v_b = v_a = 100
-$$
- mV. Hence,
-
-mV. Hence,
-$$
-i_o = \frac{(350 - 100) \times 10^{-3}}{10 \times 10^3} = 25 \,\mu\text{A}
-$$
-
-```
-For Practice Prob. 5.9.
-```
-
-| Example 5.10 |
-|--------------|
-|--------------|
-
-If *v*1 = 1 V and *v*2 = 2 V, find *vo* in the op amp circuit of Fig. 5.31.
-
-For Example 5.10.
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly defined.
-- 2. **Present.** With an input of *v*1 of 1 V and of *v*2 of 2 V, determine the output voltage of the circuit shown in Figure 5.31. The op amp circuit is actually composed of three circuits. The first circuit acts as an amplifier of gain −3(−6 kΩ∕2 kΩ) for *v*1 and the second functions as an amplifier of gain −2(−8 kΩ∕4 kΩ) for *v*2. The last circuit serves as a summer of two different gains for the output of the other two circuits.
-- 3. **Alternative.** There are different ways of working with this circuit. Because it involves ideal op amps, then a purely mathematical
-
-approach will work quite easily. A second approach would be to use *PSpice* as a confirmation of the math.
-
-4. **Attempt.** Let the output of the first op amp circuit be designated as *v*11 and the output of the second op amp circuit be designated as *v*22. Then we get
-
-$$
-v_{11} = -3v_1 = -3 \times 1 = -3 \text{ V},
-$$
-
-$$
-v_{22} = -2v_2 = -2 \times 2 = -4 \text{ V}
-$$
-
-In the third circuit we have
-
-$$
-v_o = -(10 \text{ k}\Omega/5 \text{ k}\Omega)v_{11} + [-(10 \text{ k}\Omega/15 \text{ k}\Omega)v_{22}]
-$$
-
-= -2(-3) - (2/3)(-4)
-= 6 + 2.667 = **8.667 V**
-
- 5. **Evaluate.** To properly evaluate our solution, we need to identify a reasonable check. Here we can easily use *PSpice* to provide that check.
-
- Now we can simulate this in *PSpice*. The results are shown in Fig. 5.32.
-
- We obtain the same results using two entirely different techniques (the first is to treat the op amp circuits as just gains and a summer and the second is to use circuit analysis with *PSpice*). This is a very good method of assuring that we have the correct answer.
-
- 6. **Satisfactory?** We are satisfied we have obtained the asked for results. We can now present our work as a solution to the problem.
-
-Practice Problem 5.10
-
-If *v*1 = 5 V and *v*2 = 5 V, find *v*o in the op amp circuit of Fig. 5.33.
-
-For Practice Prob. 5.10.
-
-**Answer:** 35 V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/062_5.9 Op Amp Circuit Analysis with PSpice.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/062_5.9 Op Amp Circuit Analysis with PSpice.md
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-# **5.9** Op Amp Circuit Analysis with PSpice
-
-*PSpice for Windows* does not have a model for an ideal op amp, although one may create one as a subcircuit using the *Create Subcircuit* line in the *Tools* menu. Rather than creating an ideal op amp, we will use one of the four nonideal, commercially a vailable op amps supplied in the *PSpice library eval.slb*. The op amp models have the part names LF411, LM111, LM324, and uA741, as shown in Fig. 5.34. Each of them can be obtained from **Draw/Get New Part/libraries . . . /eval.lib** or by simply selecting **Draw/Get New Part** and typing the part name in the *PartName* dialog box, as usual. Note that each of them requires dc supplies, without which the op amp will not work. The dc supplies should be connected as shown in Fig. 5.3.
-
-(a) JFET–input op amp subcircuit
-
-(b) Op amp subcircuit
-
-(c) Five– connection op amp subcircuit
-
-(d) Five–connection op amp subcircuit
-
-**Figure 5.34**
-
-Nonideal op amp model available in *PSpice*.
-
-Use *PSpice* to solve the op amp circuit for Example 5.1.
-
-# **Solution:**
-
-Using Schematics, we draw the circuit in Fig. 5.6(a) as shown in Fig.�5.35. Notice that the positive terminal of the voltage source *v*s is connected to the inverting terminal (pin 2) via the 10-kΩ resistor, while the noninverting terminal (pin 3) is grounded as required in Fig. 5.6(a). Also, notice how the op amp is powered; the positive power supply terminal V+ (pin 7) is connected to a 15 -V dc voltage source, while the negative power supply terminal V− (pin 4) is connected to −15 V. Pins 1 and 5 are left floating because they are used for offset null adjustment, which does not concern us in this chapter. Besides adding the dc power supplies to the original circuit in Fig. 5.6(a), we have also added pseudocomponents VIEWPOINT and IPROBE to respectively measure the output voltage *v*o at pin 6 and the required current *i* through the 20-kΩ resistor.
-
-Schematic for Example 5.11.
-
-After sa ving the schematic, we simulate the circuit by selecting **Analysis/Simulate** and ha ve the results displayed on VIEWPOINT and IPROBE. From the results, the closed-loop gain is
-
-$$
-\frac{v_o}{v_s} = \frac{-3.9983}{2} = -1.99915
-$$
-
-and *i* = 0.1999 mA, in agreement with the results obtained analytically in Example 5.1.
-
-**Answer:** 9.0027, 650.2 *μ*A.
-
-Rework Practice Prob. 5.1 using *PSpice*. Practice Problem 5.11
-
-## **5.10** † Applications
-
-The op amp is a fundamental b uilding block in modern electronic instrumentation. It is used e xtensively in man y de vices, along with resistors and other passi ve elements. Its numerous practical applica tions include instrumentation amplifiers, digital-to-analog converters, analog computers, level shifters, filters, calibration circuits, inverters, summers, integrators, differentiators, subtractors, log arithmic amplifiers, comparators, gyrators, oscillators, rectifiers, regulators, voltageto- current con verters, current -to-voltage converters, and clippers. Some of these we have already considered. We will consider two more applications here: the digital-to-analog converter and the instrumentation amplifier.
-
-# **5.10.1** Digital-to-Analog Converter
-
-The digital-to-analog converter (DAC) transforms digital signals into analog form. A typical e xample of a four -bit DAC is illustrated in Fig. 5.36(a). The four-bit DAC can be realized in many ways. A simple realization is the *binary weighted ladder*, shown in Fig. 5.36(b). The bits are weights according to the magnitude of their place v alue, by descending value of *Rf*∕*Rn* so that each lesser bit has half the weight of the ne xt higher. This is ob viously an in verting summing amplifier. The output is related to the inputs as sho wn in Eq. (5.15). Thus,
-
-$$
--V_o = \frac{R_f}{R_1}V_1 + \frac{R_f}{R_2}V_2 + \frac{R_f}{R_3}V_3 + \frac{R_f}{R_4}V_4
-$$
- (5.23)
-
-Input *V*1 is called the *most significant bit* (MSB), while input *V*4 is the *least significant bit* (LSB). Each of the four binary inputs *V*1, . . . , *V*4 can assume only two voltage levels: 0 or 1 V. By using the proper input and feedback resistor values, the DAC provides a single output that is pro portional to the inputs.
-
-# Example 5.12
-
-In the op amp circuit of Fig. 5.36(b), let *Rf* = 10 k Ω, *R*1 = 10 k Ω, *R*2 = 20 kΩ, *R*3 = 40 kΩ, and *R*4 = 80 kΩ. Obtain the analog output for binary inputs [0000], [0001], [0010], . . . , [1111].
-
-# **Solution:**
-
-Substituting the given values of the input and feedback resistors in Eq. (5.23) gives
-
-$$
--V_o = \frac{R_f}{R_1}V_1 + \frac{R_f}{R_2}V_2 + \frac{R_f}{R_3}V_3 + \frac{R_f}{R_4}V_4
-$$
-$$
-= V_1 + 0.5V_2 + 0.25V_3 + 0.125V_3
-$$
-
-Using this equation, a digital input [ *V*1*V*2*V*3*V*4] = [0000] produces an analog output of −*Vo* = 0 V; [*V*1*V*2*V*3*V*4] = [0001] gives −*Vo* = 0.125 V.
-
-(b) binary weighted ladder type.
-
- In practice, the voltage levels may be typically 0 and ± 5 V.
-
-Similarly,
-
-$$
-[V_1 V_2 V_3 V_4] = [0010] \Rightarrow -V_o = 0.25 \text{ V}
-$$
-
-\n
-$$
-[V_1 V_2 V_3 V_4] = [0011] \Rightarrow -V_o = 0.25 + 0.125 = 0.375 \text{ V}
-$$
-
-\n
-$$
-[V_1 V_2 V_3 V_4] = [0100] \Rightarrow -V_o = 0.5 \text{ V}
-$$
-
-\n:
-\n
-$$
-[V_1 V_2 V_3 V_4] = [1111] \Rightarrow -V_o = 1 + 0.5 + 0.25 + 0.125
-$$
-
-\n
-$$
-= 1.875 \text{ V}
-$$
-
-Table 5.2 summarizes the result of the digital -to-analog conversion. Note that we have assumed that each bit has a value of 0.125 V. Thus, in this system, we cannot represent a voltage between 1.000 and 1.125, for example. This lack of resolution is a major limitation of digital-to-analog conversions. For greater accuracy, a word representation with a greater number of bits is required. Even then a digital representation of an ana log voltage is never exact. In spite of this inexact representation, digital representation has been used to accomplish remarkable things such as audio CDs and digital photography.
-
-# **TABLE 5.2**
-
-Input and output values of the four-bit DAC.
-
-| Binary input | | Output |
-|--------------|---------------|--------|
-| [V1V2V3V4] | Decimal value | −Vo |
-| 0000 | 0 | 0 |
-| 0001 | 1 | 0.125 |
-| 0010 | 2 | 0.25 |
-| 0011 | 3 | 0.375 |
-| 0100 | 4 | 0.5 |
-| 0101 | 5 | 0.625 |
-| 0110 | 6 | 0.75 |
-| 0111 | 7 | 0.875 |
-| 1000 | 8 | 1.0 |
-| 1001 | 9 | 1.125 |
-| 1010 | 10 | 1.25 |
-| 1011 | 11 | 1.375 |
-| 1100 | 12 | 1.5 |
-| 1101 | 13 | 1.625 |
-| 1110 | 14 | 1.75 |
-| 1111 | 15 | 1.875 |
-| | | |
-
-A three-bit DAC is shown in Fig. 5.37.
-
-- (a) Determine |*Vo*| for [*V*1*V*2*V*3] = [010].
-- (b) Find |*Vo*| if [*V*1*V*2*V*3] = [110].
-- (c) If |*Vo*| = 1.25 V is desired, what should be [*V*1*V*2*V*3]?
-- (d) To get |*Vo*| = 1.75 V, what should be [*V*1*V*2*V*3]?
-
-**Answer:** 0.5 V, 1.5 V, [101], [111].
-
-Three-bit DAC; for Practice Prob. 5.12.
-
-# **5.10.2** Instrumentation Amplifiers
-
-One of the most useful and v ersatile op amp circuits for precision mea surement and process control is the *instrumentation amplifier* (IA), so called because of its widespread use in measurement systems. Typical applications of IAs include isolation amplifiers, thermocouple amplifiers, and data acquisition systems.
-
-The instrumentation amplifier is an extension of the difference amplifier in that it amplifies the difference between its input signals. As shown in Fig. 5.26 (see Example 5.8), an instrumentation amplifier typically consists of three op amps and seven resistors. For convenience, the amplifier is shown again in Fig. 5.38(a), where the resistors are made equal except for the external gain-setting resistor *RG*, connected between the gain set terminals. Figure 5.38(b) shows its schematic symbol. Example 5.8 showed that
-
-$$
-v_o = A_v(v_2 - v_1)
-$$
- (5.24)
-
-where the voltage gain is
-
-$$
-A_v = 1 + \frac{2R}{R_G} \tag{5.25}
-$$
-
-As shown in Fig. 5.39, the instrumentation amplifier amplifies small differential signal v oltages superimposed on lar ger common -mode
-
-Instrumentation amplifier Amplified dierential signal,
-
-no common-mode signal
-
-# **Figure 5.39**
-
-The IA rejects common voltages but amplifies small signal voltages.
-
-voltages. Since the common-mode voltages are equal, they cancel each other.
-
-The IA has three major characteristics:
-
-- 1. The voltage gain is adjusted by *one* external resistor *RG*.
-- 2. The input impedance of both inputs is v ery high and does not vary as the gain is adjusted.
-- 3. The output *vo* depends on the dif ference between the inputs *v*1 and *v*2, not on the voltage common to them (common-mode voltage).
-
-Due to the widespread use of IAs, manuf acturers have developed these amplifiers on single-package units. A typical e xample is the LH0036, developed by National Semiconductor . The gain can be v aried from 1 to 1,000 by an e xternal resistor whose v alue may vary from 100 to 10 kΩ.
-
-In Fig. 5.38, let *R* = 10 kΩ, *v*1 = 2.011 V, and *v*2 = 2.017 V. If *RG* is Example 5.13 adjusted to 500 Ω, determine: (a) the voltage gain, (b) the output volt age *vo*.
-
-# **Solution:**
-
-(a) The voltage gain is
-
-$$
-A_v = 1 + \frac{2R}{R_G} = 1 + \frac{2 \times 10,000}{500} = 41
-$$
-
-(b) The output voltage is
-
-$$
-v_o = A_v(v_2 - v_1) = 41(2.017 - 2.011) = 41(6)
-$$
- mV = 246 mV
-
-Determine the value of the external gain-setting resistor *RG* required for the IA in Fig. 5.38 to produce a gain of 200 when *R* = 25 kΩ.
-
-Practice Problem 5.13
-
-**Answer:** 251.3 Ω.
-
-**5.11** Summary
-
-- 1. The op amp is a high -gain amplifier that has high input resistance and low output resistance.
-- 2. Table 5.3 summarizes the op amp circuits considered in this chapter. The expression for the gain of each amplifier circuit holds whether the inputs are dc, ac, or time-varying in general.
-
-# **TABLE 5.3**
-
-Summary of basic op amp circuits.
-
-- 3. An ideal op amp has an infinite input resistance, a zero output resistance, and an infinite gain.
-- 4. For an ideal op amp, the current into each of its two input terminals is zero, and the voltage across its input terminals is negligibly small.
-- 5. In an inverting amplifier, the output voltage is a negative multiple of the input.
-- 6. In a noninverting amplifier, the output is a positive multiple of the input.
-- 7. In a voltage follower, the output follows the input.
-- 8. In a summing amplifier, the output is the weighted sum of the inputs.
-- 9. In a difference amplifier, the output is proportional to the difference of the two inputs.
-- 10. Op amp circuits may be cascaded without changing their inputoutput relationships.
-- 11. *PSpice* can be used to analyze an op amp circuit.
-- 12. Typical applications of the op amp considered in this chapter include the digital-to-analog converter and the instrumentation amplifier.
-
-# Review Questions
-
-- **5.1** The two input terminals of an op amp are labeled as:
- - (a) high and low.
- - (b) positive and negative.
- - (c) inverting and noninverting.
- - (d) differential and nondifferential.
-- **5.2** For an ideal op amp, which of the following statements are not true?
- - (a) The differential voltage across the input terminals is zero.
- - (b) The current into the input terminals is zero.
- - (c) The current from the output terminal is zero.
- - (d) The input resistance is zero.
- - (e) The output resistance is zero.
-- **5.3** For the circuit in Fig. 5.40, voltage *vo* is:
-
-| (a) −6 V | (b) −5 V |
-|------------|------------|
-| (c) −1.2 V | (d) −0.2 V |
-
-# **Figure 5.40**
-
-For Review Questions 5.3 and 5.4.
-
-**5.4** For the circuit in Fig. 5.40, current *ix* is:
-
-| (a) 600 μA | (b) 500 μA |
-|------------|-------------|
-| (c) 200 μA | (d) 1∕12 μA |
-
-**5.5** If *vs* = 0 in the circuit of Fig. 5.41, current *io* is:
-
-| (a) −10 μA | (b) −2.5 μA |
-|--------------|--------------|
-| (c) 10∕12 μA | (d) 10∕14 μA |
-
-**5.6** If *vs* = 8 mV in the circuit of Fig. 5.41, the output
-
-**Figure 5.41** For Review Questions 5.5, 5.6, and 5.7.
-
-voltage is:
-
-| (a) −44 mV | (b) −8 mV |
-|------------|-----------|
-| (c) 4 mV | (d) 7 mV |
-
-**5.7** Refer to Fig. 5.41. If *vs* = 8 mV, voltage *va* is:
-
-| (a) −8 mV | (b) 0 mV |
-|-------------|----------|
-| (c) 10∕3 mV | (d) 8 mV |
-
-**5.8** The power absorbed by the 4-kΩ resistor in Fig. 5.42 is:
-
-| (a) 9 mW | (b) 4 mW |
-|----------|----------|
-| (c) 2 mW | (d) 1 mW |
-
-**Figure 5.42**
-
-For Review Questions 5.8.
-
-- **5.9** Which of these amplifiers is used in a digital-toanalog converter?
- - (a) noninverter
- - (b) voltage follower
- - (c) summer
- - (d) difference amplifier
-- **5.10** Difference amplifiers are used in (please check all that apply):
- - (a) instrumentation amplifiers
- - (b) voltage followers
- - (c) voltage regulators
- - (d) buffers
- - (e) summing amplifiers
- - (f ) subtracting amplifiers
-
-*Answers: 5.1c, 5.2c,d, 5.3b, 5.4b, 5.5a, 5.6c, 5.7d, 5.8b, 5.9c, 5.10a, f.*
-
-# Problems
-
-# Section 5.2 Operational Amplifiers
-
-- **5.1** The equivalent model of a certain op amp is shown in Fig. 5.43. Determine:
- - (a) the input resistance
- - (b) the output resistance
- - (c) the voltage gain in dB
-
-# **Figure 5.43**
-
-For Prob. 5.1.
-
-- **5.2** The open-loop gain of an op amp is 50,000. Calculate the output voltage when there are inputs of +10 *μ*V on the inverting terminal and +20 *μ*V on the noninverting terminal.
-- **5.3** Determine the voltage input to the inverting terminal of an op amp when −40 *μ*V is applied to the noninverting terminal and the output through an openloop gain of 150,000 is 15 V.
-- **5.4** The output voltage of an op amp is −4 V when the noninverting input is 1 mV. If the open-loop gain of the op amp is 2 × 106 , what is the inverting input?
-- **5.5** For the op amp circuit of Fig. 5.44, the op amp has an open-loop gain of 100,000, an input resistance of 10 kΩ, and an output resistance of 100 Ω. Find the voltage gain *vo*∕*vi* using the nonideal model of the op amp.
-
-**5.6** Using the same parameters for the 741 op amp in Example 5.1, find *vo* in the op amp circuit of Fig. 5.45.
-
-# **Figure 5.45**
-
-For Prob. 5.6.
-
-**5.7** The op amp in Fig. 5.46 has *Ri* = 100 kΩ, *Ro* = 100 Ω, *A* = 100,000. Find the differential voltage *vd* and the output voltage *vo*.
-
-**Figure 5.46** For Prob. 5.7.
-
-# Section 5.3 Ideal Op Amp
-
-**5.8** Obtain *vo* for each of the op amp circuits in Fig. 5.47.
-
-For Prob. 5.8.
-
-**5.9** Determine *vo* for each of the op amp circuits in Fig. 5.48.
-
-**5.10** Find the gain *vo*∕*vs* of the circuit in Fig. 5.49.
-
-For Prob. 5.10.
-
-**5.11** Using Fig. 5.50, design a problem to help other students better understand how ideal op amps work.
-
-# **Figure 5.51**
-
-For Prob. 5.12.
-
-# **Figure 5.52**
-
-For Prob. 5.13.
-
-**5.14** Determine the output voltage *vo* in the circuit of Fig. 5.53.
-
-# **Figure 5.53**
-
-For Prob. 5.14.
-
-# Section 5.4 Inverting Amplifier
-
-- **5.15** (a) Determine the ratio *vo*∕*is* in the op amp circuit of Fig. 5.54.
-- (b) Evaluate the ratio for *R*1 = 20 kΩ, *R*2 = 25 kΩ, *R*3 = 40 kΩ.
-
-**5.16** Using Fig. 5.55, design a problem to help students better understand inverting op amps.
-
-**Figure 5.55**
-
-- **5.17** Calculate the gain *vo*∕*vi* when the switch in Fig. 5.56 is in:
- - (a) position 1 (b) position 2 (c) position 3.
-
-For Prob. 5.17.
-
-**5.18** For the circuit shown in Figure 5.57, solve for the Thevenin equivalent circuit looking into terminals A and B. \*
-
-For Prob. 5.18.
-
-\* An asterisk indicates a challenging problem.
-
-**5.19** Determine *io* in the circuit of Fig. 5.58.
-
-# **Figure 5.58**
-
-**5.20** In the circuit of Fig. 5.59, calculate *vo* of *vs* = 2 V.
-
-**Figure 5.59**
-
-For Prob. 5.20.
-
-**5.21** Calculate *vo* in the op amp circuit of Fig. 5.60.
-
-# **Figure 5.60**
-
-For Prob. 5.21.
-
-- **5.22** Design an inverting amplifier with a gain of −15.
-- **5.23** For the op amp circuit in Fig. 5.61, find the voltage gain *vo*∕*vs*.
-
-**Figure 5.61** For Prob. 5.23.
-
-**5.24** In the circuit shown in Fig. 5.62, find *k* in the voltage transfer function *vo* = *kvs*.
-
-# **Figure 5.62**
-
-For Prob. 5.24.
-
-# Section 5.5 Noninverting Amplifier
-
-**Figure 5.63** For Prob. 5.25.
-
-**5.26** Using Fig. 5.64, design a problem to help other students better understand noninverting op amps.
-
-**Figure 5.64** For Prob. 5.26.
-
-**5.27** Find *vo* in the op amp circuit of Fig. 5.65.
-
-**Figure 5.65** For Prob. 5.27.
-
-**5.28** Find *io* in the op amp circuit of Fig. 5.66.
-
-# **Figure 5.66**
-
-For Prob. 5.28.
-
-**5.29** Determine the voltage gain *vo*∕*vi* of the op amp circuit in Fig. 5.67.
-
-# **Figure 5.67** For Prob. 5.29.
-
-**5.30** In the circuit shown in Fig. 5.68, find *ix* and the power absorbed by the 20-kΩ resistor.
-
-# **Figure 5.68**
-
-For Prob. 5.30.
-
-**5.31** For the circuit in Fig. 5.69, find *ix*.
-
-**Figure 5.69** For Prob. 5.31.
-
-**5.32** Calculate *ix* and *vo* in the circuit of Fig. 5.70. Find the power dissipated by the 60-kΩ resistor.
-
-**5.33** Refer to the op amp circuit in Fig. 5.71. Calculate *ix* and the power absorbed by the 3-kΩ resistor.
-
-**Figure 5.71**
-
-For Prob. 5.33.
-
-**5.34** Given the op amp circuit shown in Fig. 5.72, express *vo* in terms of *v*1 and *v*2.
-
-**5.35** Design a noninverting amplifier with a gain of 7.5.
-
-**5.36** For the circuit shown in Fig. 5.73, find the Thevenin equivalent at terminals *a*-*b*. (*Hint:* To find *R*Th, apply a current source *io* and calculate *vo*.)
-
-# **Figure 5.73**
-
-For Prob. 5.36.
-
-# Section 5.6 Summing Amplifier
-
-**5.37** Determine the output of the summing amplifier in Fig. 5.74.
-
-# **Figure 5.74**
-
-For Prob. 5.37.
-
-**5.38** Using Fig. 5.75, design a problem to help other students better understand summing amplifiers.
-
-# **Figure 5.75**
-
-For Prob. 5.38.
-
-**5.39** For the op amp circuit in Fig. 5.76, determine the value of *v*2 in order to make *vo* = −16.5 V.
-
-**5.40** Referring to the circuit shown in Fig. 5.77, determine *V*o in terms of *V*1 and *V*2.
-
-**Figure 5.77** For Prob. 5.40.
-
-**5.41** An *averaging amplifier* is a summer that provides an output equal to the average of the inputs. By using proper input and feedback resistor values, one can get
-
-$$
--v_{\text{out}} = \frac{1}{4}(v_1 + v_2 + v_3 + v_4)
-$$
-
- Using a feedback resistor of 10 kΩ, design an averaging amplifier with four inputs.
-
-- **5.42** The feedback resistor of a three-input averaging summing amplifier is 50 kΩ. What are the values of *R*1, *R*2, and *R*3?
-- **5.43** The feedback resistor of a five-input averaging summing amplifier is 40 kΩ. What are the values of *R*1, *R*2, *R*3, *R*4, and *R*5?
-- **5.44** Show that the output voltage *vo* of the circuit in Fig. 5.78 is
-
-$$
-v_o = \frac{(R_3 + R_4)}{R_3(R_1 + R_2)}(R_2v_1 + R_1v_2)
-$$
-
-**Figure 5.78** For Prob. 5.44.
-
-**5.45** Design an op amp circuit to perform the following operation:
-
-$$
-v_o = 3.5v_1 - 2.5v_2
-$$
-
-All resistances must be ≤ 100 kΩ.
-
-**5.46** Using only two op amps, design a circuit to solve
-
-$$
--v_{\text{out}} = \frac{v_3 - v_1}{5} + \frac{v_1 - v_2}{2}
-$$
-
-# Section 5.7 Difference Amplifier
-
-**5.47** The circuit in Fig. 5.79 is for a difference amplifier. Find *vo* given that *v*1 = 1 V and *v*2 = 2 V.
-
-**5.48** The circuit in Fig. 5.80 is a differential amplifier driven by a bridge. Find *vo*.
-
-**5.49** Design a difference amplifier to have a gain of 4 and a common-mode input resistance of 20 kΩ at each input.
-
-**5.50** Design a circuit to amplify the difference between two inputs by 2.5.
-
-(a) Use only one op amp.
-
-(b) Use two op amps.
-
-**5.52** Design an op amp circuit such that \*
-
-$$
-v_o = 4v_1 + 6v_2 - 3v_3 - 5v_4
-$$
-
-Let all the resistors be in the range of 20 to 200 kΩ.
-
-**5.53** The ordinary difference amplifier for fixed-gain operation is shown in Fig. 5.81(a). It is simple and reliable unless gain is made variable. One way of providing gain adjustment without losing simplicity and accuracy is to use the circuit in Fig. 5.81(b). Another way is to use the circuit in Fig. 5.81(c). Show that: \*
-
-(a) for the circuit in Fig. 5.81(a),
-
-$$
-\frac{v_o}{v_i} = \frac{R_2}{R_1}
-$$
-
-(b) for the circuit in Fig. 5.81(b),
-
-$$
-\frac{v_o}{v_i} = \frac{R_2}{R_1} \frac{1}{1 + \frac{R_1}{2R_G}}
-$$
-
-(c) for the circuit in Fig. 5.81(c),
-
-$$
-\frac{v_o}{v_i} = \frac{R_2}{R_1} \left( 1 + \frac{R_2}{2R_G} \right)
-$$
-
-# **Figure 5.81**
-
-For Prob. 5.53.
-
-# Section 5.8 Cascaded Op Amp Circuits
-
-**5.54** Determine the voltage transfer ratio *vo*∕*vs* in the op amp circuit of Fig. 5.82, where R = 10 kΩ.
-
-**Figure 5.82**
-
-For Prob. 5.54.
-
-- **5.55** In a certain electronic device, a three-stage amplifier is desired, whose overall voltage gain is 42 dB. The individual voltage gains of the first two stages are to be equal, while the gain of the third is to be onefourth of each of the first two. Calculate the voltage gain of each.
-
-**5.56** Using Fig. 5.83, design a problem to help other students better understand cascaded op amps.
-
-**5.57** Find *vo* in the op amp circuit of Fig. 5.84.
-
-# **Figure 5.84**
-
-For Prob. 5.57.
-
-**5.58** Calculate *io* in the op amp circuit of Fig. 5.85.
-
-**Figure 5.85**
-
-- For Prob. 5.58.
- - **5.59** In the op amp circuit of Fig. 5.86, determine the voltage gain *vo*∕*vs*. Take *R* = 10 kΩ.
-
-**Figure 5.86** For Prob. 5.59.
-
-**5.60** Calculate *vo*∕*vi* in the op amp circuit of Fig. 5.87.
-
-For Prob. 5.60.
-
-**5.61** Determine *vo* in the circuit of Fig. 5.88.
-
-# **Figure 5.88**
-
-For Prob. 5.61.
-
-**5.62** Obtain the closed-loop voltage gain *vo*∕*vi* of the circuit in Fig. 5.89.
-
-**Figure 5.89** For Prob. 5.62.
-
-**5.63** Determine the gain *vo*∕*vi* of the circuit in Fig. 5.90.
-
-**Figure 5.90** For Prob. 5.63.
-
-**5.64** For the op amp circuit shown in Fig. 5.91, find *vo*∕*vs*.
-
-**Figure 5.91** For Prob. 5.64.
-
-**5.65** Find *vo* in the op amp circuit of Fig. 5.92.
-
-For Prob. 5.65.
-
-**5.66** For the circuit in Fig. 5.93, find *vo*.
-
-**Figure 5.93** For Prob. 5.66.
-
-**5.67** Obtain the output *vo* in the circuit of Fig. 5.94.
-
-**Figure 5.96** For Prob. 5.70.
-
-**5.68** Find *vo* in the circuit of Fig. 5.95, assuming that *Rf* = ∞ (open circuit).
-
-**Figure 5.95** For Probs. 5.68 and 5.69.
-
-- **5.69** Repeat the previous problem if *Rf* = 10 kΩ.
-- **5.70** Determine *vo* in the op amp circuit of Fig. 5.96.
-
-30 kΩ
-
-40 kΩ
-
-**5.71** Determine *vo* in the op amp circuit of Fig. 5.97.
-
-- **5.72** Find the load voltage *vL* in the circuit of Fig. 5.98.
-
-**5.73** Determine the load voltage *vL* in the circuit of Fig. 5.99.
-
-**5.74** Find *io* in the op amp circuit of Fig. 5.100.
-
-# **Figure 5.100** For Prob. 5.74.
-
-# Section 5.9 Op Amp Circuit Analysis with PSpice
-
-- **5.75** Rework Example 5.11 using the nonideal op amp LM324 instead of uA741.
-- **5.76** Solve Prob. 5.19 using *PSpice* or *MultiSim* and op amp uA741.
-- **5.77** Solve Prob. 5.48 using *PSpice* or *MultiSim* and op amp LM324.
-- **5.78** Use *PSpice* or *MultiSim* to obtain *vo* in the circuit of Fig. 5.101.
-
-**Figure 5.101** For Prob. 5.78.
-
-**5.79** Determine *vo* in the op amp circuit of Fig. 5.102, using *PSpice* or *MultiSim*.
-
-**Figure 5.102** For Prob. 5.79.
-
-- **5.80** Use *PSpice* or *MultiSim* to solve Prob. 5.70.
-- **5.81** Use *PSpice* or *MultiSim* to verify the results in Example 5.9. Assume nonideal op amps LM324.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/063_5.10 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/063_5.10 Applications.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/063_5.10 Applications.md
+++ /dev/null
@@ -1,31 +0,0 @@
-# Section 5.10 Applications
-
-**5.82** A four-bit DAC covers a voltage range of 0 to 10 V.
-
-Calculate the resolution of the DAC in volts per discrete binary step.
-
-**5.83** Design a six-bit digital-to-analog converter.
-
-- (a) If |*Vo*| = 1.1875 V is desired, what should [*V*1*V*2*V*3*V*4*V*5*V*6] be?
-- (b) Calculate |*Vo*| if [*V*1*V*2*V*3*V*4*V*5*V*6] = [011011].
-- (c) What is the maximum value |*Vo*| can assume?
-- **5.84** A four-bit *R-2R ladder* DAC is presented in Fig. 5.103. \*
- - (a) Show that the output voltage is given by
-
-$$
--V_o = R_f \left( \frac{V_1}{2R} + \frac{V_2}{4R} + \frac{V_3}{8R} + \frac{V_4}{16R} \right)
-$$
-
-(b) If *Rf* = 12 kΩ and *R* = 10 kΩ, find |*Vo*| for [*V*1*V*2*V*3*V*4] = [1011] and [*V*1*V*2*V*3*V*4] = [0101].
-
-- For Prob. 5.84.
- - **5.85** In the op amp circuit of Fig. 5.104, find the value of *R* so that the power absorbed by the 10-kΩ resistor is 10 mW. Determine the power gain.
-
-**Figure 5.104** For Prob. 5.85.
-
-- **5.86** Design a voltage controlled ideal current source (within the operating limits of the op amp) where the output current is equal to 200 *vs*(*t*) *μ*A.
-- **5.87** Figure 5.105 displays a two-op-amp instrumentation amplifier. Derive an expression for *vo* in terms of *v*1 and *v*2. How can this amplifier be used as a subtractor?
-
-**Figure 5.105** For Prob. 5.87.
-
-**5.88** Figure 5.106 shows an instrumentation amplifier driven by a bridge. Obtain the gain *vo*∕*vi* of the amplifier. \*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/064_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/064_Problems.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/064_Problems.md
+++ /dev/null
@@ -1,25 +0,0 @@
-# Comprehensive Problems
-
-**5.89** Design a circuit that provides a relationship between output voltage *vo* and input voltage *vs* such that *vo* = 12*vs* − 10. Two op amps, a 6-V battery, and several resistors are available.
-
-**5.90** The op amp circuit in Fig. 5.107 is a *current amplifier*. Find the current gain *io*∕*is* of the amplifier.
-
-# **Figure 5.107**
-
-## For Prob. 5.90.
-
-**5.91** A noninverting current amplifier is portrayed in Fig. 5.108. Calculate the gain *io*∕*is*. Take *R*1 = 8 kΩ and *R*2 = 1 kΩ.
-
-**5.92** Refer to the *bridge amplifier* shown in Fig. 5.109. Determine the voltage gain *vo*∕*vi*.
-
-# **Figure 5.109**
-
-For Prob. 5.92.
-
-**5.93** A voltage-to-current converter is shown in Fig. 5.110, which means that *iL*= *Avi* if *R*1*R*2 = *R*3*R*4. Find the constant term *A*. \*
-
-*This page intentionally left blank*
-
-# **chapter**
-
-6
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/065_Chapter 6 - Capacitors and Inductors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/065_Chapter 6 - Capacitors and Inductors.md
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+++ /dev/null
@@ -1,34 +0,0 @@
-# Capacitors and Inductors
-
-*But in science the credit goes to the man who convinces the world, not to the man to whom the idea first occurs.*
-
-—Francis Darwin
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.c), "an ability to design a system, component, or process to meet desired needs."**
-
-The "ability to design a system, component, or process to meet desired needs" is why engineers are hired. That is why this is the most important *technical* skill that an engineer has. Interestingly, your success as an engineer is directly proportional to your ability to communicate but your being able to design is wh y you will be hired in the first place.
-
-Design takes place when you ha ve what is termed an open-ended problem that eventually is defined by the solution. Within the context of this course or te xtbook, we can only e xplore some of the elements of design. Pursuing all of the steps of our problem-solving technique teaches you se veral of the most important elements of the design process.
-
-Probably the most important part of design is clearly defining what the system, component, process, or , in our case, problem is. Rarely is an engineer given a perfectly clear assignment. Therefore, as a student, you can develop and enhance this skill by asking yourself, your colleagues, or your professors questions designed to clarify the problem statement.
-
-Exploring alternati ve solutions is another important part of the design process. Again, as a student, you can practice this part of the de sign process on almost every problem you work.
-
-Evaluating your solutions is critical to any engineering assignment. Again, this is a skill that you as a student can practice on every problem you work.
-
-Photo by Charles Alexander
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Fully understand the volt-amp characteristics of capacitors and inductors and their use in basic circuits.
-- 2. Explain how capacitors behave when combined in parallel and in series.
-- 3. Understand how inductors behave when combined in parallel and in series.
-- 4. Know how to create integrators using capacitors and op amps.
-- 5. Learn how to create differentiators and their limitations.
-- 6. Learn how to create analog computers and to understand how they can be used to solve linear differential equations.
-
-In contrast to a resistor, which spends or dissipates energy irreversibly, an inductor or capacitor stores or releases energy (i.e., has a memory).
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/066_6.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/066_6.1 Introduction.md
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+++ /dev/null
@@ -1,9 +0,0 @@
-# **6.1** Introduction
-
-So far we have limited our study to resistive circuits. In this chapter, we shall introduce two new and important passive linear circuit elements: the capacitor and the inductor. Unlike resistors, which dissipate ener gy, capacitors and inductors do not dissipate but store energy, which can be retrieved at a later time. F or this reason, capacitors and inductors are called *storage* elements.
-
-The application of resistive circuits is quite limited. With the introduction of capacitors and inductors in this chapter , we will be able to analyze more important and practical circuits. Be assured that the circuit analysis techniques covered in Chapters 3 and 4 are equally applicable to circuits with capacitors and inductors.
-
-We begin by introducing capacitors and describing how to combine them in series or in parallel. Later, we do the same for inductors. As typical applications, we explore how capacitors are combined with op amps to form integrators, differentiators, and analog computers.
-
-**Figure 6.1** A typical capacitor.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/067_6.2 Capacitors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/067_6.2 Capacitors.md
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@@ -1,290 +0,0 @@
-# **6.2** Capacitors
-
-A capacitor is a passi ve element designed to store ener gy in its elec tric field. Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics, communications, computers, and power systems. For example, they are used in the tuning circuits of radio recei vers and as dynamic memory elements in computer systems.
-
-A capacitor is typically constructed as depicted in Fig. 6.1.
-
-A capacitor consists of two conducting plates separated by an insulator (or dielectric).
-
-In many practical applications, the plates may be aluminum foil while the dielectric may be air, ceramic, paper, or mica.
-
-# Historical
-
-**Michael Faraday** (1791–1867), an Engl ish c hemist a nd phy sicist, was probably the greatest experimentalist who ever lived.
-
-Born near London, Faraday realized his boyhood dream by work ing with the great chemist Sir Humphry Davy at the Royal Institu tion, where he worked for 54 years. He made several contributions in all areas of physical science and coined such words as electrolysis, anode, and cathode. His discovery of electromagnetic induction in 1831 was a major breakthrough in engineering because it provided a way of generating electricity. The electric motor and generator operate on this principle. The unit of capacitance, the farad, was named in his honor.
-
-© Stock Montage/Getty Images
-
-When a voltage source *v* is connected to the capacitor, as in Fig. 6.2, the source deposits a positive charge *q* on one plate and a negative charge −*q* on the other . The capacitor is said to store the electric char ge. The amount of charge stored, represented by *q*, is directly proportional to the applied voltage *v* so that
-
-$$
-q = Cv \tag{6.1}
-$$
-
-where *C*, the constant of proportionality , is kno wn as the *capacitance* of the capacitor. The unit of capacitance is the farad (F), in honor of the English physicist Michael Faraday (1791–1867). From Eq. (6.1), we may derive the following definition.
-
-Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F).
-
-Note from Eq. (6.1) that 1 farad = 1 coulomb/volt.
-
-Although the capacitance *C* of a capacitor is the ratio of the charge *q* per plate to the applied voltage *v*, it does not depend on *q* or *v*. It depends on the physical dimensions of the capacitor. For example, for the parallelplate capacitor shown in Fig. 6.1, the capacitance is given by
-
-$$
-C = \frac{\epsilon A}{d} \tag{6.2}
-$$
-
-where *A* is the surf ace area of each plate, *d* is the distance between the plates, and *ϵ* is the permittivity of the dielectric material between the plates. Although Eq. (6.2) applies to only parallel-plate capacitors, we may infer from it that, in general, three f actors determine the v alue of the capacitance:
-
-- 1. The surface area of the plates—the lar ger the area, the greater the capacitance.
-- 2. The spacing between the plates—the smaller the spacing, the greater the capacitance.
-
-**Figure 6.2** A capacitor with applied voltage *v*.
-
-Alternatively, capacitance is the amount of charge stored per plate for a unit voltage difference in a capacitor.
-
- Capacitor voltage rating and capacitance are typically inversely rated due to the relationships in Eqs. (6.1) and (6.2). Arcing occurs if d is small and V is high.
-
-**Figure 6.3**
-
-Circuit symbols for capacitors: (a) fixed capacitor, (b) variable capacitor.
-
-3. The permitti vity of the material—the higher the permitti vity, the greater the capacitance.
-
-Capacitors are commercially a vailable in different values and types. Typically, capacitors have values in the picofarad (pF) to microfarad (*μ*F) range. They are described by the dielectric material the y are made of and by whether they are of fixed or variable type. Figure 6.3 shows the circuit symbols for fixed and variable capacitors. Note that according to the passive sign convention, if *v* > 0 and *i* > 0 or if *v* < 0 and *i* < 0, the capacitor is being charged, and if *v* **·** *i* < 0, the capacitor is discharging.
-
-# **Figure 6.4**
-
-**Figure 6.5** Variable capacitors: (a) trimmer capacitor, (b) filmtrim capacitor. Courtesy of Johanson.
-
-According to Eq. (6.4), for a capacitor to carry current, its voltage must vary with time. Hence, for constant voltage, i = 0.
-
-Figure 6.4 shows common types of fixed-value capacitors. Polyester capacitors are light in weight, stable, and their change with tempera ture is predictable. Instead of polyester, other dielectric materials such as mica and polystyrene may be used. Film capacitors are rolled and housed in metal or plastic films. Electrolytic capacitors produce very high capacitance. Figure 6.5 shows the most common types of variable capacitors. The capacitance of a trimmer (or padder) capacitor is often placed in parallel with another capacitor so that the equi valent capacitance can be varied slightly. The capacitance of the variable air capacitor (meshed plates) is varied by turning the shaft. Variable capacitors are used in radio receivers allowing one to tune to various stations. In addition, capacitors are used to block dc, pass ac, shift phase, store energy, start motors, and suppress noise.
-
-To obtain the current-voltage relationship of the capacitor , we take the derivative of both sides of Eq. (6.1). Since
-
-$$
-i = \frac{dq}{dt} \tag{6.3}
-$$
-
-differentiating both sides of Eq. (6.1) gives
-
-$$
-i = C \frac{dv}{dt}
-$$
- (6.4)
-
-This is the current-v oltage relationship for a capacitor , assuming the passive sign convention. The relationship is illustrated in Fig. 6.6 for a capacitor whose capacitance is independent of v oltage. Capacitors that satisfy Eq. (6.4) are said to be *linear*. For a *nonlinear capacitor*, the plot of the current-voltage relationship is not a straight line. Although some capacitors are nonlinear, most are linear. We will assume linear capaci tors in this book.
-
-The v oltage-current relation of the capacitor can be obtained by integrating both sides of Eq. (6.4). We get
-
-$$
-v(t) = \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau
-$$
- (6.5)
-
-or
-
-$$
-v(t) = \frac{1}{C} \int_{t_0}^t i(\tau) \, d\tau + v(t_0)
-$$
-\n(6.6)
-
-where *v*(*t*0) = *q*(*t*0)∕*C* is the voltage across the capacitor at time *t*0. Equation (6.6) shows that capacitor voltage depends on the past history of the capacitor current. Hence, the capacitor has memory—a property that is often exploited.
-
-The instantaneous power delivered to the capacitor is
-
-$$
-p = vi = Cv \frac{dv}{dt}
-$$
- (6.7)
-
-The energy stored in the capacitor is therefore
-
-$$
-w = \int_{-\infty}^{t} p(\tau) d\tau = C \int_{-\infty}^{t} v \frac{dv}{d\tau} d\tau = C \int_{v(-\infty)}^{v(t)} v dv = \frac{1}{2} C v^{2} \Big|_{v(-\infty)}^{v(t)} \quad (6.8)
-$$
-
-We note that *v*(−∞) = 0, because the capacitor was uncharged at *t* = −∞. Thus,
-
-$$
-w = \frac{1}{2} C v^2 \tag{6.9}
-$$
-
-Using Eq. (6.1), we may rewrite Eq. (6.9) as
-
-$$
-w = \frac{q^2}{2C} \tag{6.10}
-$$
-
-Equation (6.9) or (6.10) represents the energy stored in the electric field that exists between the plates of the capacitor. This energy can be retrieved, since an ideal capacitor cannot dissipate energy. In fact, the word *capacitor* is derived from this element's capacity to store energy in an electric field.
-
-We should note the following important properties of a capacitor:
-
-1. Note from Eq. (6.4) that when the v oltage across a capacitor is not changing with time (i.e., dc voltage), the current through the capacitor is zero. Thus,
-
-A capacitor is an open circuit to dc.
-
-However, if a battery (dc voltage) is connected across a capacitor, the capacitor charges.
-
- An alternative way of looking at this is using Eq. (6.9), which indicates that energy is proportional to voltage squared. Since injecting or extracting energy can only be done over some finite time, voltage cannot change instantaneously across a capacitor.
-
-**Figure 6.6** Current-voltage relationship of a capacitor.
-
-Voltage across a capacitor: (a) allowed, (b) not allowable; an abrupt change is not possible.
-
-**Figure 6.8** Circuit model of a nonideal capacitor.
-
-Example 6.1
-
-2. The voltage on the capacitor must be continuous.
-
-The voltage on a capacitor cannot change abruptly.
-
-The capacitor resists an abrupt change in the v oltage across it. According to Eq. (6.4), a discontinuous change in v oltage requires an infinite current, which is physically impossible. For example, the voltage across a capacitor may tak e the form shown in Fig. 6.7(a), whereas it is not physically possible for the capacitor voltage to take the form shown in Fig. 6.7(b) because of the abrupt changes. Con versely, the current through a capacitor can change instantaneously.
-
-- 3. The ideal capacitor does not dissipate ener gy. It tak es power from the circuit when storing ener gy in its field and returns previously stored energy when delivering power to the circuit.
-- 4. A real, nonideal capacitor has a parallel-model leakage resistance, as shown in Fig. 6.8. The leakage resistance may be as high as 100 MΩ and can be neglected for most practical applications. For this reason, we will assume ideal capacitors in this book.
-
-(a) Calculate the charge stored on a 3-pF capacitor with 20 V across it. (b) Find the energy stored in the capacitor.
-
-# **Solution:**
-
-(a) Since *q* = *Cv*,
-
-$$
-q = 3 \times 10^{-12} \times 20 = 60 \text{ pC}
-$$
-
-(b) The energy stored is
-
-$$
-w = \frac{1}{2} Cv^2 = \frac{1}{2} \times 3 \times 10^{-12} \times 400 = 600 \text{ pJ}
-$$
-
-What is the voltage across a 4.5- *μ*F capacitor if the charge on one plate is 0.12 mC? How much energy is stored? Practice Problem 6.1
-
-**Answer:** 26.67 V, 1.6 mJ.
-
-Example 6.2 The voltage across a 5-*μ*F capacitor is
-
-*v*(*t*) = 10 cos 6000*t* V
-
-Calculate the current through it.
-
-# **Solution:**
-
-By definition, the current is
-
-$$
-i(t) = C \frac{dv}{dt} = 5 \times 10^{-6} \frac{d}{dt} (10 \cos 6000t)
-$$
-
-= -5 × 10-6 × 6000 × 10 sin 6000t = -0.3 sin 6000t A
-
-If a 10-*μ*F capacitor is connected to a voltage source with
-
-*v*(*t*) = 20 cos(200*t*) V
-
-determine the current through the capacitor.
-
-**Answer:** −40 sin (200*t*) mA.
-
-Determine the voltage across a 2-*μ*F capacitor if the current through it is Example 6.3
-
-$$
-i(t) = 6e^{-3000t} \text{ mA}
-$$
-
-Assume that the initial capacitor voltage is zero.
-
-# **Solution:**
-
-Since
-$$
-v = \frac{1}{C} \int_0^t i \, d\tau + v(0)
-$$
- and $v(0) = 0$ ,
-\n
-$$
-v = \frac{1}{2 \times 10^{-6}} \int_0^t 6e^{-3000\tau} \, d\tau \cdot 10^{-3}
-$$
-\n
-$$
-= \frac{3 \times 10^3}{-3000} e^{-3000\tau} \Big|_0^t = (1 - e^{-3000t}) \text{ V}
-$$
-
-The current through a 100-*μ*F capacitor is *i*(*t*) = 50 sin 120*πt* mA. Calculate the voltage across it at *t* = 1 ms and *t* = 5 ms. Take *v*(0) = 0.
-
-**Answer:** 93.14 mV, 1.736 V.
-
-Determine the current through a 200- *μ*F capacitor whose voltage is Example 6.4 shown in Fig. 6.9.
-
-# **Solution:**
-
-The voltage waveform can be described mathematically as
-
-$$
-v(t) = \begin{cases} 50t \text{ V} & 0 < t < 1 \\ 100 - 50t \text{ V} & 1 < t < 3 \\ -200 + 50t \text{ V} & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases}
-$$
-
-Since *i* = *C dv*∕*dt* and *C* = 200 *μ*F, we take the derivative of *v* to obtain
-
-$$
-i(t) = 200 \times 10^{-6} \times \begin{cases} 50 & 0 < t < 1 \\ -50 & 1 < t < 3 \\ 50 & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases}
-$$
-$$
-= \begin{cases} 10 \text{ mA} & 0 < t < 1 \\ -10 \text{ mA} & 1 < t < 3 \\ 10 \text{ mA} & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases}
-$$
-
-Thus, the current waveform is as shown in Fig. 6.10.
-
-Practice Problem 6.3
-
-**Figure 6.9** For Example 6.4.
-
-Practice Problem 6.2
-
-For Practice Prob. 6.4.
-
-An initially uncharged 1-mF capacitor has the current shown in Fig. 6.11 across it. Calculate the voltage across it at *t* = 2 ms and *t* = 5 ms.
-
-**Answer:** 100 mV, 400 mV.
-
-Example 6.5 Obtain the ener gy stored in each capacitor in Fig. 6.12(a) under dc conditions.
-
-# **Solution:**
-
-Under dc conditions, we replace each capacitor with an open circuit, as shown in Fig. 6.12(b). The current through the series combination of the 2-kΩ and 4-kΩ resistors is obtained by current division as
-
-$$
-i = \frac{3}{3 + 2 + 4}(6 \text{ mA}) = 2 \text{ mA}
-$$
-
-Hence, the voltages *v*1 and *v*2 across the capacitors are
-
-$$
-v_1 = 2000i = 4 \text{ V} \qquad v_2 = 4000i = 8 \text{ V}
-$$
-
-and the energies stored in them are
-
-$$
-w_1 = \frac{1}{2}C_1v_1^2 = \frac{1}{2}(2 \times 10^{-3})(4)^2 = 16 \text{ mJ}
-$$
-
-$$
-w_2 = \frac{1}{2}C_2v_2^2 = \frac{1}{2}(4 \times 10^{-3})(8)^2 = 128 \text{ mJ}
-$$
-
-**Figure 6.12**
-
-For Example 6.5.
-
-# Practice Problem 6.5
-
-**Figure 6.13** For Practice Prob. 6.5.
-
-Under dc conditions, find the energy stored in the capacitors in Fig. 6.13.
-
-**Answer:** 20.25 mJ, 3.375 mJ.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/068_6.3 Series and Parallel Capacitors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/068_6.3 Series and Parallel Capacitors.md
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-# **6.3** Series and Parallel Capacitors
-
-We know from resisti ve circuits that the series-parallel combination is a powerful tool for reducing circuits. This technique can be e xtended to series-parallel connections of capacitors, which are sometimes encoun tered. We desire to replace these capacitors by a single equivalent capacitor *C*eq.
-
-In order to obtain the equi valent capacitor *C*eq of *N* capacitors in parallel, consider the circuit in Fig. 6.14(a). The equi valent circuit is in Fig. 6.14(b). Note that the capacitors ha ve the same voltage *v* across them. Applying KCL to Fig. 6.14(a),
-
-$$
-i = i_1 + i_2 + i_3 + \dots + i_N \tag{6.11}
-$$
-
-But *ik* = *Ck dv*∕*dt*. Hence,
-
-$$
-i = C_1 \frac{dv}{dt} + C_2 \frac{dv}{dt} + C_3 \frac{dv}{dt} + \dots + C_N \frac{dv}{dt}
-$$
-
-=
-$$
-\left(\sum_{k=1}^{N} C_k\right) \frac{dv}{dt} = C_{eq} \frac{dv}{dt}
-$$
- (6.12)
-
-where
-
-$$
-C_{\text{eq}} = C_1 + C_2 + C_3 + \dots + C_N \tag{6.13}
-$$
-
-The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.
-
-We observe that capacitors in parallel combine in the same manner as resistors in series.
-
-We no w obtain *C*eq of *N* capacitors connected in series by comparing the circuit in Fig. 6.15(a) with the equi valent circuit in Fig. 6.15(b). Note that the same current *i* flows (and consequently the same charge) through the capacitors. Applying KVL to the loop in Fig. 6.15(a),
-
-$$
-v = v_1 + v_2 + v_3 + \dots + v_N \tag{6.14}
-$$
-
-But *vk*=\_\_\_1 *Ck* ∫ *t*0 *t i*(*τ*) *dτ*+*vk*(*t*0). Therefore,
-
-$$
-v = \frac{1}{C_1} \int_{t_0}^t i(\tau) d\tau + v_1(t_0) + \frac{1}{C_2} \int_{t_0}^t i(\tau) d\tau + v_2(t_0)
-$$
-
-+ $\cdots + \frac{1}{C_N} \int_{t_0}^t i(\tau) d\tau + v_N(t_0)$
-= $\left(\frac{1}{\tau_0} + \frac{1}{\tau_0} + \cdots + \frac{1}{\tau_N}\right) \int_0^t i(\tau) d\tau + v_1(t_0) + v_2$
-
-$$
-= \left(\frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_N}\right) \int_{t_0}^t i(\tau) \, d\tau + v_1(t_0) + v_2(t_0) + \dots + v_N(t_0)
-$$
-
-$$
-=\frac{1}{C_{\text{eq}}}\int_{t_0}^t i(\tau)\,d\tau + v(t_0)
-$$
-
-where
-
-$$
-\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_N}
-$$
- (6.16)
-
-# **Figure 6.14**
-
-(a) Parallel-connected *N* capacitors, (b) equivalent circuit for the parallel capacitors.
-
-(a)
-
-## **(6.15) Figure 6.15**
-
-(a) Series-connected *N* capacitors, (b) equivalent circuit for the series capacitor.
-
-The initial voltage *v*(*t*0) across *C*eq is required by KVL to be the sum of the capacitor voltages at *t*0. Or according to Eq. (6.15),
-
-$$
-v(t_0) = v_1(t_0) + v_2(t_0) + \dots + v_N(t_0)
-$$
-
-Thus, according to Eq. (6.16),
-
-The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.
-
-Note that capacitors in series combine in the same manner as resistors in parallel. For *N* = 2 (i.e., two capacitors in series), Eq. (6.16) becomes
-
-> \_\_\_1 *C*eq = \_\_\_1 *C*1 + \_\_\_1 *C*2
-
-or
-
-$$
-C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} \tag{6.17}
-$$
-
-Example 6.6 Find the equi valent capacitance seen between terminals *a* and *b* of the circuit in Fig. 6.16.
-
-# **Solution:**
-
-The 20- *μ*F and 5- *μ*F capacitors are in series; their equivalent capaci tance is
-
-$$
-\frac{20 \times 5}{20 + 5} = 4 \ \mu\text{F}
-$$
-
-This 4-*μ*F capacitor is in parallel with the 6- *μ*F and 20- *μ*F capacitors; their combined capacitance is
-
-$$
-4 + 6 + 20 = 30 \,\mu\text{F}
-$$
-
-This 30-*μ*F capacitor is in series with the 60- *μ*F capacitor. Hence, the equivalent capacitance for the entire circuit is
-
-$$
-C_{\text{eq}} = \frac{30 \times 60}{30 + 60} = 20 \,\mu\text{F}
-$$
-
-Find the equivalent capacitance seen at the terminals of the circuit in Practice Problem 6.6 Fig. 6.17.
-
-# **Answer:** 40 *μ*F.
-
-For Practice Prob. 6.6.
-
-For the circuit in Fig. 6.18, find the voltage across each capacitor. Example 6.7
-
-# **Solution:**
-
-We first find the equivalent capacitance *C*eq, shown in Fig. 6.19. The two parallel capacitors in Fig. 6.18 can be combined to get 40 + 20 = 60 mF. This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors. Thus,
-
-$$
-C_{\text{eq}} = \frac{1}{\frac{1}{60} + \frac{1}{30} + \frac{1}{20}} \text{ mF} = 10 \text{ mF}
-$$
-
-The total charge is
-
-$$
-q = C_{\text{eq}}v = 10 \times 10^{-3} \times 30 = 0.3 \text{ C}
-$$
-
-This is the charge on the 20-mF and 30-mF capacitors, because they are in series with the 30-V source. (A crude way to see this is to imagine that charge acts like current, since *i* = *dq*∕*dt*.) Therefore,
-
-$$
-v_1 = \frac{q}{C_1} = \frac{0.3}{20 \times 10^{-3}} = 15 \text{ V}
-$$
- $v_2 = \frac{q}{C_2} = \frac{0.3}{30 \times 10^{-3}} = 10 \text{ V}$
-
-Having determined *v*1 and *v*2, we now use KVL to determine *v*3 by
-
-$$
-v_3 = 30 - v_1 - v_2 = 5
-$$
- V
-
-Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage *v*3 and their combined capacitance is 40 + 20 = 60 mF. This combined capacitance is in series with the 20-mF and 30-mF capacitors and consequently has the same charge on it. Hence,
-
-$$
-v_3 = \frac{q}{60 \text{ mF}} = \frac{0.3}{60 \times 10^{-3}} = 5 \text{ V}
-$$
-
-Find the voltage across each of the capacitors in Fig. 6.20. Practice Problem 6.7
-
-**Answer:**
-$$
-v_1 = 75 \text{ V}, v_2 = 75 \text{ V}, v_3 = 25 \text{ V}, v_4 = 50 \text{ V}.
-$$
-
-**Figure 6.18** For Example 6.7.
-
-**Figure 6.19** Equivalent circuit for Fig. 6.18.
-
-**Figure 6.20** For Practice Prob. 6.7.
-
-**Figure 6.21** Typical form of an inductor.
-
-In view of Eq. (6.18), for an inductor to have voltage across its terminals, its current must vary with time. Hence, v = 0 for constant current through the inductor.
-
-# **Figure 6.22**
-
-Various types of inductors: (a) solenoidal wound inductor, (b) toroidal inductor, (c) axial lead inductor.
-
-© McGraw-Hill Education/Mark Dierker, photographer
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/069_6.4 Inductors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/069_6.4 Inductors.md
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-# **6.4** Inductors
-
-An inductor is a passive element designed to store energy in its magnetic field. Inductors find numerous applications in electronic and power systems. They are used in power supplies, transformers, radios, TVs, radars, and electric motors.
-
-Any conductor of electric current has inductive properties and may be regarded as an inductor. But in order to enhance the inducti ve effect, a practical inductor is usually formed into a c ylindrical coil with man y turns of conducting wire, as shown in Fig. 6.21.
-
-An inductor consists of a coil of conducting wire.
-
-If current is allowed to pass through an inductor, it is found that the voltage across the inductor is directly proportional to the time rate of change of the current. Using the passive sign convention,
-
-$$
-v = L \frac{di}{dt}
-$$
- (6.18)
-
-where *L* is the constant of proportionality called the *inductance* of the inductor. The unit of inductance is the henry (H), named in honor of the American inventor Joseph Henry (1797–1878). It is clear from Eq. (6.18) that 1 henry equals 1 volt-second per ampere.
-
-Inductance is the property whereby an inductor exhibits opposition to the change of current flowing through it, measured in henrys (H).
-
-The inductance of an inductor depends on its ph ysical dimension and construction. F ormulas for calculating the inductance of inductors of different shapes are deri ved from electromagnetic theory and can be found in standard electrical engineering handbooks. For example, for the inductor, (solenoid) shown in Fig. 6.21,
-
-$$
-L = \frac{N^2 \mu A}{\ell} \tag{6.19}
-$$
-
-where *N* is the number of turns, ℓ is the length, A is the cross-sectional area, and *μ* is the permeability of the core. We can see from Eq. (6.19) that inductance can be increased by increasing the number of turns of coil, using material with higher permeability as the core, increasing the cross-sectional area, or reducing the length of the coil.
-
-Like capacitors, commercially a vailable inductors come in dif ferent values and types. Typical practical inductors ha ve inductance values ranging from a few microhenrys (*μ*H), as in communication systems, to tens of henrys (H) as in po wer systems. Inductors may be fixed or variable. The core may be made of iron, steel, plastic, or air . The terms *coil* and *choke* are also used for inductors. Common inductors are sho wn in Fig. 6.22. The circuit symbols for inductors are shown in Fig. 6.23, following the passive sign convention.
-
-Equation (6.18) is the v oltage-current relationship for an inductor . Figure 6.24 sho ws this relationship graphically for an inductor whose
-
-# Historical
-
-**Joseph Henry** (1797–1878), an American physicist, discovered inductance and constructed an electric motor.
-
-Born in Albany, New York, Henry graduated from Albany Academy and taught philosophy at Princeton University from 1832 to 1846. He was the first secretary of the Smithsonian Institution. He conducted several experiments on electromagnetism and developed powerful electromagnets that could lift objects weighing thousands of pounds. Interestingly, Joseph Henry discovered electromagnetic induction before Faraday but failed to publish his findings. The unit of inductance, the henry, was named after him.
-
-inductance is independent of current. Such an inductor is kno wn as a *linear inductor*. For a *nonlinear inductor,* the plot of Eq. (6.18) will not be a straight line because its inductance v aries with current. We will assume linear inductors in this textbook unless stated otherwise.
-
-The current-voltage relationship is obtained from Eq. (6.18) as
-
-$$
-di = \frac{1}{L}v \, dt
-$$
-
-Integrating gives
-
-$$
-i = \frac{1}{L} \int_{-\infty}^{t} v(\tau) d\tau
-$$
- (6.20)
-
-$$
-(6.20)
-$$
-
-or
-
-$$
-i = \frac{1}{L} \int_{t_0}^t v(\tau) d\tau + i(t_0)
-$$
- (6.21)
-
-where *i*(*t*0) is the total current for −∞ < *t* < *t*0 and *i*(−∞) = 0. The idea of making *i*(−∞) = 0 is practical and reasonable, because there must be a time in the past when there was no current in the inductor.
-
-The inductor is designed to store ener gy in its magnetic field. The energy stored can be obtained from Eq. (6.18). The power delivered to the inductor is
-
-$$
-p = vi = \left(L\frac{di}{dt}\right)i\tag{6.22}
-$$
-
-The energy stored is
-
-$$
-w = \int_{-\infty}^{t} p(\tau) d\tau = L \int_{-\infty}^{t} \frac{di}{d\tau} i d\tau
-$$
-
-= $L \int_{-\infty}^{t} i di = \frac{1}{2} L i^{2}(t) - \frac{1}{2} L i^{2}(-\infty)$ (6.23)
-
-# **Figure 6.23**
-
-Circuit symbols for inductors: (a) air-core, (b) iron-core, (c) variable iron-core.
-
-**Figure 6.24** Voltage-current relationship of an inductor.
-
-Since *i*(−∞) = 0,
-
-$$
-w = \frac{1}{2}Li^2 \tag{6.24}
-$$
-
-We should note the following important properties of an inductor.
-
-1. Note from Eq. (6.18) that the v oltage across an inductor is zero when the current is constant. Thus,
-
-An inductor acts like a short circuit to dc.
-
-2. An important property of the inductor is its opposition to the change in current flowing through it.
-
-The current through an inductor cannot change instantaneously.
-
-According to Eq. (6.18), a discontinuous change in the current through an inductor requires an infinite voltage, which is not physically possible. Thus, an inductor opposes an abrupt change in the current through it. For example, the current through an inductor may take the form sho wn in Fig. 6.25(a), whereas the inductor current cannot take the form shown in Fig. 6.25(b) in real-life situations due to the discontinuities. Ho wever, the voltage across an inductor can change abruptly.
-
-- 3. Like the ideal capacitor , the ideal inductor does not dissipate energy. The energy stored in it can be retrie ved at a later time. The inductor tak es power from the circuit when storing ener gy and delivers po wer to the circuit when returning pre viously stored energy.
-- 4. A practical, nonideal inductor has a significant resistive component, as shown in Fig. 6.26. This is due to the fact that the inductor is made of a conducting material such as copper , which has some resistance. This resistance is called the *winding resistance Rw*, and it appears in series with the inductance of the inductor . The presence of *Rw* makes it both an ener gy storage device and an energy dissipation device. Since *Rw* is usually v ery small, it is ignored in most cases. The nonideal inductor also has a *winding capacitance Cw* due to the capacitive coupling between the conducting coils. *Cw* is very small and can be ignored in most cases, except at high frequencies. We will assume ideal inductors in this book.
-
-Example 6.8 The current through a 0.1-H inductor is *i*(*t*) = 10*te*−5*t* A. Find the voltage across the inductor and the energy stored in it.
-
-# **Solution:**
-
-Since *v* = *L di*∕*dt* and *L* = 0.1 H,
-
-$$
-v = 0.1 \frac{d}{dt} (10te^{-5t}) = e^{-5t} + t(-5)e^{-5t} = e^{-5t}(1 - 5t)
-$$
- V
-
-**Figure 6.25**
-
-Current through an inductor: (a) allowed, (b) not allowable; an abrupt change is not possible.
-
-Since an inductor is often made of a highly conducting wire, it has a very small resistance.
-
-**Figure 6.26** Circuit model for a practical inductor.
-
-The energy stored is
-
-$$
-w = \frac{1}{2}Li^2 = \frac{1}{2}(0.1)100t^2e^{-10t} = 5t^2e^{-10t} \text{ J}
-$$
-
-If the current through a 1-mH inductor is *i*(*t*) = 90 sin (200*t*) mA, find the Practice Problem 6.8 terminal voltage and the energy stored.
-
-**Answer:** 18 cos (200*t*) mV, 4.05 sin2 (200*t*) *μ*J.
-
-Find the current through a 5-H inductor if the voltage across it is Example 6.9
-
-$$
-v(t) = \begin{cases} 30t^2, & t > 0 \\ 0, & t < 0 \end{cases}
-$$
-
-Also, find the energy stored at *t* = 5 s. Assume *i*(*v*) > 0.
-
-# **Solution:**
-
-Since *i* = \_\_1 *L* ∫ *t*0 *t v*(*τ*) *dτ* + *i*(*t*0) and *L* = 5 H, *i* = \_\_1 5 ∫ 0 *t* 30*τ* 2 *dτ* + 0 = 6 × *t* 3 \_\_ 3 = 2*t* 3 A
-
-The power *p* = *vi* = 60*t* 5 , and the energy stored is then
-
-$$
-w = \int p \, dt = \int_0^5 60t^5 \, dt = 60 \frac{t^6}{6} \Big|_0^5 = 156.25 \, \text{kJ}
-$$
-
-Alternatively, we can obtain the energy stored using Eq. (6.24), by writing
-
-$$
-w\Big|_0^5 = \frac{1}{2}Li^2(5) - \frac{1}{2}Li(0) = \frac{1}{2}(5)(2 \times 5^3)^2 - 0 = 156.25 \text{ kJ}
-$$
-
-as obtained before.
-
-The terminal voltage of a 2-H inductor is *v* = 10(1 − *t*) V. Find the Practice Problem 6.9 current flowing through it at *t* = 4 s and the energy stored in it at *t* = 4 s. Assume *i*(0) = 2 A.
-
-**Figure 6.27** For Example 6.10.
-
-Example 6.10 Consider the circuit in Fig. 6.27(a). Under dc conditions, find: (a) *i*, *vC*, and *iL*, (b) the energy stored in the capacitor and inductor.
-
-# **Solution:**
-
-(a) Under dc conditions, we replace the capacitor with an open circuit and the inductor with a short circuit, as in Fig. 6.27(b). It is evident from Fig. 6.27(b) that
-
-$$
-i = i_L = \frac{12}{1+5} = 2
-$$
- A
-
-The voltage *vC* is the same as the voltage across the 5-Ω resistor. Hence,
-
-$$
-v_C = 5i = 10 \text{ V}
-$$
-
-(b) The energy in the capacitor is
-
-$$
-w_C = \frac{1}{2}Cv_C^2 = \frac{1}{2}(1)(10^2) = 50 \text{ J}
-$$
-
-and that in the inductor is
-
-$$
-w_L = \frac{1}{2}Li_L^2 = \frac{1}{2}(2)(2^2) = 4 \text{ J}
-$$
-
-For Practice Prob. 6.10.
-
-**Figure 6.29** (a) A series connection of *N* inductors, (b) equivalent circuit for the series inductors.
-
-Practice Problem 6.10 Determine *vC*, *iL*, and the energy stored in the capacitor and inductor in the circuit of Fig. 6.28 under dc conditions.
-
-**Answer:** 15 V, 7.5 A, 450 J, 168.75 J.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/070_6.5 Series and Parallel Inductors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/070_6.5 Series and Parallel Inductors.md
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@@ -1,170 +0,0 @@
-# **6.5** Series and Parallel Inductors
-
-Now that the inductor has been added to our list of passive elements, it is necessary to extend the powerful tool of series-parallel combination. We need to know how to find the equivalent inductance of a series-connected or parallel-connected set of inductors found in practical circuits.
-
-Consider a series connection of *N* inductors, as shown in Fig. 6.29(a), with the equivalent circuit shown in Fig. 6.29(b). The inductors have the same current through them. Applying KVL to the loop,
-
-$$
-v = v_1 + v_2 + v_3 + \dots + v_N \tag{6.25}
-$$
-
-Substituting *vk*=*Lk di*∕*dt* results in
-
-$$
-v = L_1 \frac{di}{dt} + L_2 \frac{di}{dt} + L_3 \frac{di}{dt} + \dots + L_N \frac{di}{dt}
-$$
-
-= $(L_1 + L_2 + L_3 + \dots + L_N) \frac{di}{dt}$
-= $\left(\sum_{k=1}^{N} L_k\right) \frac{di}{dt} = L_{eq} \frac{di}{dt}$ (6.26)
-
-where
-
-Thus,
-
-The equivalent inductance of series-connected inductors is the sum of the individual inductances.
-
-Inductors in series are combined in exactly the same way as resistors in series.
-
-We now consider a parallel connection of *N* inductors, as shown in Fig. 6.30(a), with the equi valent circuit in Fig. 6.30(b). The inductors have the same voltage across them. Using KCL,
-
-$$
-i = i_1 + i_2 + i_3 + \dots + i_N \tag{6.28}
-$$
-
-But *ik* = \_\_1 *Lk* ∫ *t*0 *t v dt* + *ik*(*t*0); hence, *i* = \_\_1 *L*1 ∫ *t*0 *t v dt* + *i*1(*t*0) + \_\_1 *L*2 ∫ *t*0 *t v dt* + *i*2(*t*0) + ⋯ + \_\_\_1 *LN* ∫ *t*0 *t v dt* + *iN* (*t*0) = ( \_\_1 *L*1 + \_\_1 *L*2 + ⋯ + \_\_\_1 *LN*)∫ *t*0 *t v dt* + *i*1(*t*0) + *i*2(*t*0) + ⋯ + *iN* (*t*0) *N t N t*
-
-$$
-= \left(\sum_{k=1}^{N} \frac{1}{L_k}\right) \int_{t_0}^{t} v \, dt + \sum_{k=1}^{N} i_k(t_0) = \frac{1}{L_{\text{eq}}} \int_{t_0}^{t} v \, dt + i(t_0) \tag{6.29}
-$$
-
-where
-
-$$
-\frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + \dots + \frac{1}{L_N}
-$$
- (6.30)
-
-The initial current *i*(*t*0) through *L*eq at *t*=*t*0 is expected by KCL to be the sum of the inductor currents at *t*0. Thus, according to Eq. (6.29),
-
-$$
-i(t_0) = i_1(t_0) + i_2(t_0) + \dots + i_N(t_0)
-$$
-
-According to Eq. (6.30),
-
-The equivalent inductance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances.
-
-Note that the inductors in parallel are combined in the same way as resistors in parallel.
-
-For two inductors in parallel (*N* = 2), Eq. (6.30) becomes
-
-$$
-\frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} \qquad \text{or} \qquad L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2} \tag{6.31}
-$$
-
-As long as all the elements are of the same type, the ∆-Y transformations for resistors discussed in Section 2.7 can be e xtended to capacitors and inductors.
-
-# **Figure 6.30**
-
-(a) A parallel connection of *N* inductors, (b) equivalent circuit for the parallel inductors.
-
-# **TABLE 6.1**
-
-Important characteristics of the basic elements.†
-
-| Relation | Resistor (R) | Capacitor (C) | Inductor (L) |
-|---------------------------------|----------------------------------------|-------------------------------------------------|----------------------------------------------------|
-| v-i: | v = iR | t
__1
v =
C ∫
i(τ) dτ + v(t0)
t0 | v = L __di
dt |
-| i-v: | i = v∕R | i = C ___ dv
dt | t
__1
i =
∫
v(τ) dτ + i(t0)
L
t0 |
-| p or w: | v2
__
2
p = i
R =
R | __1
Cv2
w =
2 | __1
Li2
w =
2 |
-| Series: | Req
= R1
+ R2 | _______ C1C2
Ceq
=
C1
+ C2 | Leq
= L1
+ L2 |
-| Parallel: | _______ R1R2
Req
=
R1
+ R2 | Ceq
= C1
+ C2 | _______ L1L2
Leq
=
L1
+ L2 |
-| At dc: | Same | Open circuit | Short circuit |
-| Circuit variable
that cannot | | | |
-| change abruptly: | Not applicable | v | i |
-
-† *Passive sign convention is assumed.*
-
-It is appropriate at this point to summarize the most important characteristics of the three basic circuit elements we have studied. The summary is given in Table 6.1.
-
-Example 6.11 Find the equivalent inductance of the circuit shown in Fig. 6.31.
-
-# **Solution:**
-
-The 10-H, 12-H, and 20-H inductors are in series; thus, combining them gives a 42-H inductance. This 42-H inductor is in parallel with the 7-H inductor so that they are combined, to give
-
-$$
-\frac{7 \times 42}{7 + 42} = 6
-$$
-
-This 6-H inductor is in series with the 4-H and 8-H inductors. Hence,
-
-$$
-L_{\text{eq}} = 4 + 6 + 8 = 18 \text{ H}
-$$
-
-Practice Problem 6.11 Calculate the equivalent inductance for the inductive ladder network in Fig. 6.32.
-
-For the circuit in Fig. 6.33, *i*(*t*) = 4(2 − *e* Example 6.12 −10*t* ) mA. If *i*2(0) = −1 mA, find: (a) *i*1(0); (b) *v*(*t*), *v*1(*t*), and *v*2(*t*); (c) *i*1(*t*) and *i*2(*t*).
-
-# **Solution:**
-
-(a) From *i*(*t*) = 4(2 − *e*−10*t* ) mA, *i*(0) = 4(2 − 1) = 4 mA. Since *i* = *i*1 + *i*2,
-
-$$
-i_1(0) = i(0) - i_2(0) = 4 - (-1) = 5 \text{ mA}
-$$
-
-(b) The equivalent inductance is
-
-$$
-L_{\text{eq}} = 2 + 4 || 12 = 2 + 3 = 5 \text{ H}
-$$
-
-Thus,
-
-$$
-v(t) = L_{eq} \frac{di}{dt} = 5(4)(-1)(-10)e^{-10t} \text{ mV} = 200e^{-10t} \text{ mV}
-$$
-
-and
-
-$$
-v_1(t) = 2\frac{di}{dt} = 2(-4)(-10)e^{-10t}
-$$
- mV = 80 $e^{-10t}$ mV
-
-Since *v* = *v*1 + *v*2,
-
-$$
-v_2(t) = v(t) - v_1(t) = 120e^{-10t} \text{ mV}
-$$
-
-(c) The current *i*1 is obtained as
-
-$$
-i_1(t) = \frac{1}{4} \int_0^t v_2 dt + i_1(0) = \frac{120}{4} \int_0^t e^{-10t} dt + 5 \text{ mA}
-$$
-
-= $-3e^{-10t} \Big|_0^t + 5 \text{ mA} = -3e^{-10t} + 3 + 5 = 8 - 3e^{-10t} \text{ mA}$
-
-Similarly,
-
-$$
-i_2(t) = \frac{1}{12} \int_0^t v_2 dt + i_2(0) = \frac{120}{12} \int_0^t e^{-10t} dt - 1 \text{ mA}
-$$
-$$
-= -e^{-10t} \Big|_0^t - 1 \text{ mA} = -e^{-10t} + 1 - 1 = -e^{-10t} \text{ mA}
-$$
-
-Note that *i*1(*t*) + *i*2(*t*) = *i*(*t*).
-
-In the circuit of Fig. 6.34, *i*1(*t*) = 3*e* Practice Problem 6.12 −2*t* A. If *i*(0) = 7 A, find: (a) *i*2(0); (b) *i*2(*t*) and *i*(*t*); (c) *v*1(*t*), *v*2(*t*), and *v*(*t*).
-
-**Answer:** (a) 4 A, (b)
-$$
-(-2 + 6e^{-2t})
-$$
- A, $(-2 + 9e^{-2t})$ A, (c) $-36e^{-2t}$ V, $-144e^{-2t}$ V, $-180e^{-2t}$ V.
-
-**Figure 6.34** For Practice Prob. 6.12.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/071_6.6 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/071_6.6 Applications.md
deleted file mode 100644
index 6e0cb70c2b8f5252b2ee81e3392ef19c1a7ae530..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/071_6.6 Applications.md
+++ /dev/null
@@ -1,270 +0,0 @@
-# **6.6** Applications
-
-Circuit elements such as resistors and capacitors are commercially available in either discrete form or integrated-circuit (IC) form. Unlike capacitors and resistors, inductors with appreciable inductance are dif ficult to produce on IC substrates. Therefore, inductors (coils) usually
-
-**Figure 6.33** For Example 6.12.
-
-come in discrete form and tend to be more b ulky and e xpensive. For this reason, inductors are not as versatile as capacitors and resistors, and they are more limited in applications. However, there are several applications in which inductors have no practical substitute. They are routinely used in relays, delays, sensing devices, pick-up heads, telephone circuits, radio and TV receivers, power supplies, electric motors, microphones, and loudspeakers, to mention a few.
-
-Capacitors and inductors possess the following three special properties that make them very useful in electric circuits:
-
-- 1. The capacity to store ener gy makes them useful as temporary v oltage or current sources. Thus, they can be used for generating a large amount of current or voltage for a short period of time.
-- 2. Capacitors oppose an y abrupt change in v oltage, while inductors oppose any abrupt change in current. This property makes inductors useful for spark or arc suppression and for converting pulsating dc voltage into relatively smooth dc voltage.
-- 3. Capacitors and inductors are frequenc y sensiti ve. This property makes them useful for frequency discrimination.
-
-The first two properties are put to use in dc circuits, while the third one is taken advantage of in ac circuits. We will see how useful these properties are in later chapters. F or now, consider three applications involving capacitors and op amps: integrator, differentiator, and analog computer.
-
-# **6.6.1** Integrator
-
-Important op amp circuits that use energy-storage elements include integrators and dif ferentiators. These op amp circuits often in volve resistors and capacitors; inductors (coils) tend to be more b ulky and expensive.
-
-The op amp integrator is used in numerous applications, especially in analog computers, to be discussed in Section 6.6.3.
-
-An integrator is an op amp circuit whose output is proportional to the integral of the input signal.
-
-If the feedback resistor *Rf* in the f amiliar in verting amplifier of Fig. 6.35(a) is replaced by a capacitor , we obtain an ideal inte grator, as shown in Fig. 6.35(b). It is interesting that we can obtain a mathematical representation of integration this way. At node *a* in Fig. 6.35(b),
-
-$$
-i_R = i_C \tag{6.32}
-$$
-
-But
-
-$$
-i_R = \frac{v_i}{R}, \qquad i_C = -C \frac{dv_o}{dt}
-$$
-
-Substituting these in Eq. (6.32), we obtain
-
-$$
-\frac{v_i}{R} = -C \frac{dv_o}{dt}
-$$
- (6.33a)
-
-$$
-dv_o = -\frac{1}{RC}v_i dt
-$$
- (6.33b)
-
-Replacing the feedback resistor in the inverting amplifier in (a) produces an integrator in (b).
-
-Integrating both sides gives
-
-$$
-v_o(t) - v_o(0) = -\frac{1}{RC} \int_0^t v_i(\tau) \, d\tau \tag{6.34}
-$$
-
-To ensure that *vo*(0) = 0, it is always necessary to discharge the integrator's capacitor prior to the application of a signal. Assuming *vo*(0) = 0,
-
-$$
-v_o = -\frac{1}{RC} \int_0^t v_i(\tau) \, d\tau \tag{6.35}
-$$
-
-which shows that the circuit in Fig. 6.35(b) pro vides an output v oltage proportional to the integral of the input. In practice, the op amp integrator requires a feedback resistor to reduce dc gain and prevent saturation. Care must be tak en that the op amp operates within the linear range so that it does not saturate.
-
-If *v*1 = 10 cos 2 *t* mV and *v*2 = 0.5*t* mV, find *vo* in the op amp circuit in Example 6.13 Fig. 6.36. Assume that the voltage across the capacitor is initially zero.
-
-# **Solution:**
-
-This is a summing integrator, and
-
-$$
-v_o = -\frac{1}{R_1C} \int v_1 dt - \frac{1}{R_2C} \int v_2 dt
-$$
-
-=
-$$
--\frac{1}{3 \times 10^6 \times 2 \times 10^{-6}} \int_0^t 10 \cos(2\tau) d\tau
-$$
-
-$$
--\frac{1}{100 \times 10^3 \times 2 \times 10^{-6}} \int_0^t 0.5\tau d\tau
-$$
-
-=
-$$
--\frac{1}{6} \frac{10}{2} \sin 2t - \frac{1}{0.2} \frac{0.5t^2}{2} = -0.833 \sin 2t - 1.25t
-$$
-
-The integrator in Fig. 6.35(b) has *R* = 100 kΩ, *C* = 20 *μ*F. Determine the Practice Problem 6.13 output voltage when a dc voltage of 2.5 mV is applied at *t* = 0. Assume that the op amp is initially nulled.
-
-**Answer:** −1.25*t* m V.
-
-# **6.6.2** Differentiator
-
-A differentiator is an op amp circuit whose output is proportional to the rate of change of the input signal.
-
-In Fig. 6.35(a), if the input resistor is replaced by a capacitor , the resulting circuit is a dif ferentiator, shown in Fig. 6.37. Applying KCL at node *a*,
-
-$$
-i_R = i_C \tag{6.36}
-$$
-
-2 mV But
-
-$$
-i_R = -\frac{v_o}{R}, \qquad i_C = C \frac{dv_i}{dt}
-$$
-
-Substituting these in Eq. (6.36) yields
-
-$$
-v_o = -RC \frac{dv_i}{dt} \tag{6.37}
-$$
-
-showing that the output is the deri vative of the input. Differentiator circuits are electronically unstable because an y electrical noise within the circuit is exaggerated by the differentiator. For this reason, the differentiator circuit in Fig. 6.37 is not as useful and popular as the integrator. It is seldom used in practice.
-
-Example 6.14 Sketch the output voltage for the circuit in Fig. 6.38(a), given the input voltage in Fig. 6.38(b). Take *vo* = 0 at *t* = 0.
-
-# **Solution:**
-
-This is a differentiator with
-
-$$
-RC = 5 \times 10^3 \times 0.2 \times 10^{-6} = 10^{-3} \text{ s}
-$$
-
-For 0 < *t* < 4 ms, we can express the input voltage in Fig. 6.38(b) as
-
-*vi* = { 2000*t* 8 − 2000*t* 0 < *t* < 2 ms 2 < *t* < 4 ms
-
-This is repeated for 4 < *t* < 8 ms. Using Eq. (6.37), the output is ob tained as
-
-$$
-v_o = -RC \frac{dv_i}{dt} = \begin{cases} -2\text{ V} & 0 < t < 2\text{ ms} \\ 2\text{ V} & 2 < t < 4\text{ ms} \end{cases}
-$$
-
-Thus, the output is as sketched in Fig. 6.39.
-
-*v*o (V)
-
-8642 2 0 –2 t (ms)
-
-**Figure 6.39** Output of the circuit in Fig. 6.38(a).
-
-Practice Problem 6.14 The differentiator in Fig. 6.37 has *R* = 100 kΩ and *C* = 0.1 *μ*F. Given that *vi* = 5*t* V, determine the output *vo*.
-
-**Answer:** −50 mV.
-
-**Figure 6.37** An op amp differentiator.
-
-For Example 6.14.
-
-# **6.6.3** Analog Computer
-
-Op amps were initially de veloped for electronic analog computers. Analog computers can be programmed to solv e mathematical models of mechanical or electrical systems. These models are usually expressed in terms of differential equations.
-
-To solv e simple dif ferential equations using the analog computer requires cascading three types of op amp circuits: inte grator circuits, summing amplifiers, and inverting/noninverting amplifiers for negative/ positive scaling. The best way to illustrate how an analog computer solves a differential equation is with an example.
-
-Suppose we desire the solution *x*(*t*) of the equation
-
-$$
-a\frac{d^2x}{dt^2} + b\frac{dx}{dt} + cx = f(t), \qquad t > 0
-$$
-\n(6.38)
-
-where *a*, *b*, and *c* are constants, and *f*(*t*) is an arbitrary forcing func tion. The solution is obtained by first solving the highest-order derivative term. Solving for *d*2 *x*∕*dt*2 yields
-
-$$
-\frac{d^2x}{dt^2} = \frac{f(t)}{a} - \frac{b}{a}\frac{dx}{dt} - \frac{c}{a}x\tag{6.39}
-$$
-
-To obtain *dx*∕*dt*, the *d*2 *x*∕*dt*2 term is inte grated and inverted. Finally, to obtain *x*, the *dx*∕*dt* term is integrated and inverted. The forcing function is injected at the proper point. Thus, the analog computer for solving Eq. (6.38) is implemented by connecting the necessary summers, inverters, and inte grators. A plotter or oscilloscope may be used to vie w the output *x*, or *dx*∕*dt*, or *d*2 *x*∕*dt*2 , depending on where it is connected in the system.
-
-Although the above example is on a second-order differential equation, any differential equation can be simulated by an analog computer comprising integrators, inverters, and inverting summers. But care must be exercised in selecting the v alues of the resistors and capacitors, to ensure that the op amps do not saturate during the solution time interval.
-
-The analog computers with vacuum tubes were built in the 1950s and 1960s. Recently their use has declined. They ha ve been superseded by modern digital computers. Ho wever, we still study analog computers for two reasons. First, the a vailability of integrated op amps has made it pos sible to build analog computers easily and cheaply. Second, understanding analog computers helps with the appreciation of the digital computers.
-
-Design an analog computer circuit to solve the differential equation: Example 6.15
-
-$$
-\frac{d^2v_o}{dt^2} + 2\frac{dv_o}{dt} + v_o = 10\sin 4t, \qquad t > 0
-$$
-
-subject to *vo*(0) = −4, *vo* ′ (0) = 1, where the prime refers to the time derivative.
-
-# **Solution:**
-
-1. **Define.** We have a clearly defined problem and expected solution. I might remind the student that many times the problem is not so well defined and this portion of the problem-solving process could
-
-require much more effort. If this is so, then you should always keep in mind that time spent here will result in much less effort later and most likely save you a lot of frustration in the process.
-
-- 2. **Present.** Clearly, using the devices developed in Section 6.6.3 will allow us to create the desired analog computer circuit. We will need the integrator circuits (possibly combined with a summing capability) and one or more inverter circuits.
-- 3. **Alternative.** The approach for solving this problem is straightforward. We will need to pick the correct values of resistances and capacitors to allow us to realize the equation we are representing. The final output of the circuit will give the desired result.
-- 4. **Attempt.** There are an infinite number of possibilities for picking the resistors and capacitors, many of which will result in correct solutions. Extreme values of resistors and capacitors will result in incorrect outputs. For example, low values of resistors will overload the electronics. Picking values of resistors that are too large will cause the op amps to stop functioning as ideal devices. The limits can be determined from the characteristics of the real op amp.
-
-We first solve for the second derivative as
-
-$$
-\frac{d^2v_o}{dt^2} = 10\sin 4t - 2\frac{dv_o}{dt} - v_o
-$$
- (6.15.1)
-
-Solving this requires some mathematical operations, includ ing summing, scaling, and integration. Integrating both sides of Eq. (6.15.1) gives
-
-$$
-\frac{dv_o}{dt} = -\int_0^t \left(-10\sin(4\tau) + 2\frac{dv_o(\tau)}{d\tau} + v_o(\tau)\right) d\tau + v'_o(0) \quad (6.15.2)
-$$
-
-where *v o*′ (0) = 1. We implement Eq. (6.15.2) using the summing integrator shown in Fig. 6.40(a). The values of the resistors and capacitors have been chosen so that *RC* = 1 for the term
-
-$$
--\frac{1}{RC}\int_0^t v_o(\tau)\,d\tau
-$$
-
-Other terms in the summing integrator of Eq. (6.15.2) are implemented accordingly. The initial condition *dv o*(0) ∕*dt* = 1 is imple mented by connecting a 1-V battery with a switch across the capacitor as shown in Fig. 6.40(a).
-
-The next step is to obtain *v o* by integrating *dv o* ∕*dt* and inverting the result,
-
-$$
-v_o = -\int_0^t \frac{dv_o(\tau)}{d\tau} d\tau + v(0)
-$$
- (6.15.3)
-
-This is implemented with the circuit in Fig. 6.40(b) with the battery giving the initial condition of −4 V. We now combine the two circuits in Fig. 6.40(a) and (b) to obtain the complete circuit shown in Fig. 6.40(c). When the input signal 10 sin 4 *t* is applied, we open the switches at *t* = 0 to obtain the output waveform *v o*, which may be viewed on an oscilloscope.
-
-For Example 6.15.
-
-5. **Evaluate.** The answer looks correct, but is it? If an actual solution for *vo* is desired, then a good check would be to first find the solution by realizing the circuit in *PSpice*. This result could then be compared with a solution using the differential solution capability of *MATLAB.*
-
-Since all we need to do is check the circuit and confirm that it represents the equation, we have an easier technique to use. We just go through the circuit and see if it generates the desired equation.
-
-However, we still have choices to make. We could go through the circuit from left to right but that would involve differentiating the result to obtain the original equation. An easier approach would be to go from right to left. This is the approach we will use to check the answer.
-
- Starting with the output, *vo*, we see that the right-hand op amp is nothing more than an inverter with a gain of one. This means that the output of the middle circuit is −*vo*. The following represents the action of the middle circuit.
-
-$$
--v_o = -\left(\int_0^t \frac{dv_o}{d\tau} d\tau + v_o(0)\right) = -\left(v_o \Big|_0^t + v_o(0)\right)
-$$
-
-= -\left(v\_o(t) - v\_o(0) + v\_o(0)\right)
-
-where *vo*(0) = −4 V is the initial voltage across the capacitor.
-
-We check the circuit on the left the same way.
-
-$$
-\frac{dv_o}{dt} = -\left(\int_0^t -\frac{d^2v_o}{dt^2}d\tau - v'_o(0)\right) = -\left(-\frac{dv_o}{dt} + v'_o(0) - v'_o(0)\right)
-$$
-
-Now, all we need to verify is that the input to the first op amp is −*d*2 *vo*∕*dt*2 .
-
-Looking at the input we see that it is equal to
-
-$$
--10\sin(4t) + v_o + \frac{1/10^{-6}}{0.5\,\text{M}\Omega} \frac{dv_o}{dt} = -10\sin(4t) + v_o + 2\frac{dv_o}{dt}
-$$
-
-which does produce −*d*2 *vo*∕*dt*2 from the original equation.
-
-6. **Satisfactory?** The solution we have obtained is satisfactory. We can now present this work as a solution to the problem.
-
-Practice Problem 6.15 Design an analog computer circuit to solve the differential equation:
-
-$$
-\frac{d^2v_o}{dt^2} + 3\frac{dv_o}{dt} + 2v_o = 4\cos 10t, \qquad t > 0
-$$
-
-subject to *vo*(0) = 2, *vo* ′ (0) = 0.
-
-**Answer:** See Fig. 6.41, where *RC* = 1 s.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/072_6.7 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/072_6.7 Summary.md
deleted file mode 100644
index f71d3c7a475ecd46de08571771f3791fe884a59f..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/072_6.7 Summary.md
+++ /dev/null
@@ -1,38 +0,0 @@
-# **6.7** Summary
-
-1. The current through a capacitor is directly proportional to the time rate of change of the voltage across it.
-
-$$
-i = C \frac{dv}{dt}
-$$
-
-The current through a capacitor is zero unless the v oltage is changing. Thus, a capacitor acts like an open circuit to a dc source.
-
-2. The voltage across a capacitor is directly proportional to the time integral of the current through it.
-
-$$
-v = \frac{1}{C} \int_{-\infty}^{t} i \, d\tau = \frac{1}{C} \int_{t_0}^{t} i \, d\tau + v(t_0)
-$$
-
-The voltage across a capacitor cannot change instantly.
-
-- 3. Capacitors in series and in parallel are combined in the same w ay as conductances.
-- 4. The voltage across an inductor is directly proportional to the time rate of change of the current through it.
-
-$$
-v = L\frac{di}{dt}
-$$
-
-The voltage across the inductor is zero unless the current is chang ing. Thus, an inductor acts like a short circuit to a dc source.
-
-5. The current through an inductor is directly proportional to the time integral of the voltage across it.
-
-$$
-i = \frac{1}{L} \int_{-\infty}^{t} v \, d\tau = \frac{1}{L} \int_{t_0}^{t} v \, d\tau + i(t_0)
-$$
-
-The current through an inductor cannot change instantly.
-
-- 6. Inductors in series and in parallel are combined in the same w ay resistors in series and in parallel are combined.
-- 7. At any given time *t*, the energy stored in a capacitor is \_1 2 *Cv*2 , while the energy stored in an inductor is \_1 2 *Li*2 .
-- 8. Three application circuits, the inte grator, the differentiator, and the analog computer , can be realized using resistors, capacitors, and op amps.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/073_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/073_Review Questions.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/073_Review Questions.md
+++ /dev/null
@@ -1,77 +0,0 @@
-# Review Questions
-
-**6.1** What charge is on a 5-F capacitor when it is connected across a 120-V source?
-
-| (a) 600 C | (b) 300 C |
-|-----------|-----------|
-| (c) 24 C | (d) 12 C |
-
-**6.2** Capacitance is measured in:
-
-| (a) coulombs | (b) joules |
-|--------------|------------|
-| (c) henrys | (d) farads |
-
-**6.3** When the total charge in a capacitor is doubled, the energy stored:
-
-| (a) remains the same | (b) is halved |
-|----------------------|-------------------|
-| (c) is doubled | (d) is quadrupled |
-
-**6.4** Can the voltage waveform in Fig. 6.42 be associated with a real capacitor?
-
-(a) Yes (b) No
-
-# **Figure 6.42**
-
-For Review Question 6.4.
-
-**6.5** The total capacitance of two 40-mF series-connected capacitors in parallel with a 4-mF capacitor is:
-
-| (a) 3.8 mF | (b) 5 mF | (c) 24 mF |
-|------------|-----------|-----------|
-| (d) 44 mF | (e) 84 mF | |
-
-**6.6** In Fig. 6.43, if *i* = cos 4*t* and *v* = sin 4*t*, the element is:
-
-(a) a resistor (b) a capacitor (c) an inductor
-
-# **Figure 6.43**
-
-For Review Question 6.6.
-
-**6.7** A 5-H inductor changes its current by 3 A in 0.2 s. The voltage produced at the terminals of the inductor is:
-
-| (a) 75 V | (b) 8.888 V |
-|----------|-------------|
-| (c) 3 V | (d) 1.2 V |
-
-**6.8** If the current through a 10-mH inductor increases from zero to 2 A, how much energy is stored in the inductor?
-
-| (a) 40 mJ | (b) 20 mJ |
-|-----------|-----------|
-| (c) 10 mJ | (d) 5 mJ |
-
-**6.9** Inductors in parallel can be combined just like resistors in parallel.
-
-$$
-(a) True \t\t (b) False
-$$
-
-**6.10** For the circuit in Fig. 6.44, the voltage divider formula is:
-
-(a)
-$$
-v_1 = \frac{L_1 + L_2}{L_1} v_s
-$$
-
-\n(b) $v_1 = \frac{L_1 + L_2}{L_2} v_s$
-\n(c) $v_1 = \frac{L_2}{L_1 + L_2} v_s$
-\n(d) $v_1 = \frac{L_1}{L_1 + L_2} v_s$
-\n $v_s$
-
-# **Figure 6.44**
-
-For Review Question 6.10.
-
-*Answers: 6.1a, 6.2d, 6.3d, 6.4b, 6.5c, 6.6b, 6.7a, 6.8b, 6.9a, 6.10d.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/074_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/074_Problems.md
deleted file mode 100644
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@@ -1,342 +0,0 @@
-# Problems
-
-# Section 6.2 Capacitors
-
-- **6.1** If the voltage across a 7.5-F capacitor is 2*te*−3*t* V, find the current and the power.
-- **6.2** A 50-*μ*F capacitor has energy *w*(*t*) = 10 cos2 377*t* J. Determine the current through the capacitor.
-- **6.3** Design a problem to help other students better understand how capacitors work.
-- **6.4** A voltage across a capacitor is equal to [2 – 2 cos(4*t*)] V and the current flowing through it is equal to 2 sin (4*t*) *μ*A. Determine the value of the capacitance. Calculate the power being stored by the capacitor.
-- **6.5** The voltage across a 4-*μ*F capacitor is shown in Fig. 6.45. Find the current waveform.
-
-For Prob. 6.5.
-
-**6.6** The voltage waveform in Fig. 6.46 is applied across a 55-*μ*F capacitor. Draw the current waveform through it.
-
-# **Figure 6.46**
-
-For Prob. 6.6.
-
-- **6.7** At *t* = 0, the voltage across a 25-mF capacitor is 10 V. Calculate the voltage across the capacitor for *t* > 0 when current 5*t* mA flows through it.
-- **6.8** A 4-mF capacitor has the terminal voltage
-
-$$
-v = \begin{cases} 50 \text{ V}, & t \le 0 \\ Ae^{-100t} + Be^{-600t} \text{ V}, & t \ge 0 \end{cases}
-$$
-
-If the capacitor has an initial current of 2 A, find:
-
-- (a) the constants *A* and *B*,
-- (b) the energy stored in the capacitor at *t* = 0,
-- (c) the capacitor current for *t* > 0.
-
-Problems **241**
-
-- **6.9** The current through a 0.5-F capacitor is 6(1 − *e* −*t* ) A. Determine the voltage and power at *t* = 2 s. Assume *v*(0) = 0.
-- **6.10** The voltage across a 5-mF capacitor is shown in Fig. 6.47. Determine the current through the capacitor.
-
-# **Figure 6.47**
-
-For Prob. 6.10.
-
-**6.11** A 4-mF capacitor has the current waveform shown in Fig. 6.48. Assuming that *v*(0) = 10 V, sketch the voltage waveform *v*(*t*).
-
-# **Figure 6.48**
-
-For Prob. 6.11.
-
-- **6.12** A voltage of 45*e*−2000*t* V appears across a parallel combination of a 100-mF capacitor and a 12-Ω resistor. Calculate the power absorbed by the parallel combination.
-- **6.13** Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc conditions.
-
-**Figure 6.49**
-
-For Prob. 6.13.
-
-# Section 6.3 Series and Parallel Capacitors
-
-**6.14** Series-connected 20- and 60-pF capacitors are placed in parallel with series-connected 30- and 70-pF capacitors. Determine the equivalent capacitance.
-
-**6.15** Two capacitors (25 and 75 *μ*F) are connected to a 100-V source. Find the energy stored in each capacitor if they are connected in:
-
-(a) parallel (b) series
-
-**6.16** The equivalent capacitance at terminals *a*-*b* in the circuit of Fig. 6.50 is 20 *μ*F. Calculate the value of *C*.
-
-# **Figure 6.50**
-
-For Prob. 6.16.
-
-**6.17** Determine the equivalent capacitance for each of the circuits of Fig. 6.51.
-
-# **Figure 6.51**
-
-For Prob. 6.17.
-
-**6.18** Find *C*eq in the circuit of Fig. 6.52 if all capacitors are 4 *μ*F.
-
-**6.19** Find the equivalent capacitance between terminals *a* and *b* in the circuit of Fig. 6.53. All capacitances are in *μ*F.
-
-**Figure 6.53** For Prob. 6.19.
-
-> **6.20** Find the equivalent capacitance at terminals *a*-*b* of the circuit in Fig. 6.54.
-
-**6.21** Determine the equivalent capacitance at terminals *a*-*b* of the circuit in Fig. 6.55.
-
-**6.22** Obtain the equivalent capacitance of the circuit in Fig. 6.56.
-
-# **Figure 6.56**
-
-For Prob. 6.22.
-
-# **Figure 6.57**
-
-For Prob. 6.23.
-
-**6.24** In the circuit shown in Fig. 6.58 assume that the capacitors were initially uncharged and that the current source has been connected to the circuit long enough for all the capacitors to reach steady state (no current flowing through the capacitors). Determine the voltage across each capacitor and the energy
-
-# **Figure 6.58** For Prob. 6.24.
-
-**6.25** (a) Show that the voltage-division rule for two capacitors in series as in Fig. 6.59(a) is
-
-$$
-v_1 = \frac{C_2}{C_1 + C_2} v_s
-$$
-, $v_2 = \frac{C_1}{C_1 + C_2} v_s$
-
-assuming that the initial conditions are zero.
-
-**Figure 6.59** For Prob. 6.25.
-
-(b) For two capacitors in parallel as in Fig. 6.59(b), show that the current-division rule is
-
-$$
-i_1 = \frac{C_1}{C_1 + C_2} i_s
-$$
-, $i_2 = \frac{C_2}{C_1 + C_2} i_s$
-
-assuming that the initial conditions are zero.
-
-- **6.26** Three capacitors, *C*1 = 5 *μ*F, *C*2 = 10 *μ*F, and *C*3 = 20 *μ*F, are connected in parallel across a 200-V source. Determine:
- - (a) the total capacitance,
- - (b) the charge on each capacitor,
- - (c) the total energy stored in the parallel combination.
-- **6.27** Given that four 10-*μ*F capacitors can be connected
-- in series and in parallel, find the minimum and maximum values that can be obtained by such series/parallel combinations.
-- **6.28** Obtain the equivalent capacitance of the network shown in Fig. 6.60. **\***
-
-**Figure 6.60** For Prob. 6.28.
-
-**6.29** Determine *C*eq for each circuit in Fig. 6.61.
-
-For Prob. 6.29.
-
-**6.30** Assuming that the capacitors are initially uncharged, find *vo*(*t*) in the circuit of Fig. 6.62.
-
-**6.31** If *v*(0) = 0, find *v*(*t*), *i*1(*t*), and *i*2(*t*) in the circuit of Fig. 6.63.
-
-**Figure 6.63** For Prob. 6.31.
-
-**6.32** In the circuit in Fig. 6.64, let *is* = 4.5*e*−2*t* mA and the voltage across each capacitor is equal to zero at *t* = 0. Determine *v*1 and *v*2 and the energy stored in each capacitor for all *t* > 0.
-
-For Prob. 6.32.
-
-\* An asterisk indicates a challenging problem.
-
-**6.33** Obtain the Thevenin equivalent at the terminals, *a*-*b*, of the circuit shown in Fig. 6.65. Please note that Thevenin equivalent circuits do not generally exist for circuits involving capacitors and resistors. This is a special case where the Thevenin equivalent circuit does exist.
-
-**Figure 6.65**
-
-For Prob. 6.33.
-
-# Section 6.4 Inductors
-
-- **6.34** The current through a 25-mH inductor is 10*e*−*t*∕2 A. Find the voltage and the power at *t* = 3 s.
-- **6.35** An inductor has a linear change in current from 100 mA to 200 mA in 2 ms and induces a voltage of 160 mV. Calculate the value of the inductor.
-- **6.36** Design a problem to help other students better understand how inductors work.
-- **6.37** The current through a 12-mH inductor is 4 sin 100*t* A. Find the voltage, and the energy stored at *t* = \_\_\_*π* 200s.
-- **6.38** The current through a 40-mH inductor is
-
-$$
-i(t) = \begin{cases} 0, & t < 0 \\ te^{-2t} \, \mathbf{A}, & t > 0 \end{cases}
-$$
-
-Find the voltage *v*(*t*).
-
-**6.39** The voltage across a 50-mH inductor is given by
-
-$$
-v(t) = [5e^{-2t} + 2t + 4] \text{ V} \qquad \text{for } t > 0.
-$$
-
- Determine the current *i*(*t*) through the inductor. Assume that *i*(0) = 0 A.
-
-**6.40** The current through a 5-mH inductor is shown in Fig. 6.66. Determine the voltage across the inductor at *t* = 1, 3, and 5 ms.
-
-**Figure 6.66** For Prob. 6.40.
-
-- **6.41** The voltage across a 2-H inductor is 20(1 − *e* −2*t* ) V. If the initial current through the inductor is 0.3 A, find the current and the energy stored in the inductor at *t* = 1 s.
-- **6.42** If the voltage waveform in Fig. 6.67 is applied across the terminals of a 5-H inductor, calculate the current through the inductor. Assume *i*(0) = −1 A.
-
-# **Figure 6.67**
-
-For Prob. 6.42.
-
-- **6.43** The current in a 150-mH inductor increases from 0 to 60 mA (steady state). How much energy is stored in the inductor?
-- **6.44** A 100-mH inductor is connected in parallel with a 2-kΩ resistor. The current through the inductor is *i*(*t*) = 35*e*−400*t* mA. (a) Find the voltage *vL* across the inductor. (b) Find the voltage *vR* across the resistor. (c) Does *vR*(*t*) + *vL*(*t*) = 0? (d) Calculate the energy stored in the inductor at *t* = 0. **\***
-- **6.45** If the voltage waveform in Fig. 6.68 is applied to a 25-mH inductor, find the inductor current *i*(*t*) for 0 < *t* < 2 seconds. Assume *i*(0) = 0.
-
-# **Figure 6.68**
-
-For Prob. 6.45.
-
-**6.46** Find *vC*, *iL*, and the energy stored in the capacitor and inductor in the circuit of Fig. 6.69 under dc conditions.
-
-For Prob. 6.46.
-
-# **Figure 6.70**
-
-For Prob. 6.47.
-
-**6.48** Under steady-state dc conditions, find *i* and *v* in the circuit in Fig. 6.71.
-
-For Prob. 6.48.
-
-# Section 6.5 Series and Parallel Inductors
-
-**6.49** Find the equivalent inductance of the circuit in Fig. 6.72. Assume all inductors are 40 mH.
-
-# **Figure 6.72**
-
-For Prob. 6.49.
-
-**6.50** An energy-storage network consists of seriesconnected 16- and 14-mH inductors in parallel with series-connected 24- and 36-mH inductors. Calculate the equivalent inductance.
-
-**6.51** Determine *L*eq at terminals *a*-*b* of the circuit in Fig. 6.73.
-
-**Figure 6.73** For Prob. 6.51.
-
-**6.53** Find *L*eq at the terminals of the circuit in Fig. 6.75.
-
-**6.54** Find the equivalent inductance looking into the terminals of the circuit in Fig. 6.76.
-
-**6.55** Find *L*eq in each of the circuits in Fig. 6.77.
-
-**6.56** Find *L*eq in the circuit of Fig. 6.78.
-
-For Prob. 6.56.
-
-**6.57** Determine *L*eq that may be used to represent the inductive network of Fig. 6.79 at the terminals. **\***
-
-For Prob. 6.57.
-
-**6.58** The current waveform in Fig. 6.80 flows through a 3-H inductor. Sketch the voltage across the inductor over the interval 0 < *t* < 6 s.
-
-**Figure 6.80** For Prob. 6.58.
-
-**6.59** (a) For two inductors in series as in Fig. 6.81(a), show that the voltage division principle is
-
-$$
-v_1 = \frac{L_1}{L_1 + L_2} v_s
-$$
-, $v_2 = \frac{L_2}{L_1 + L_2} v_s$
-
-assuming that the initial conditions are zero.
-
-(b) For two inductors in parallel as in Fig. 6.81(b), show that the current-division principle is
-
-$$
-i_1 = \frac{L_2}{L_1 + L_2} i_s
-$$
-, $i_2 = \frac{L_1}{L_1 + L_2} i_s$
-
-assuming that the initial conditions are zero.
-
-For Prob. 6.59.
-
-**6.60** In the circuit of Fig. 6.82, *io*(0) = 2 A. Determine *io*(*t*) and *vo*(*t*) for *t* > 0.
-
-For Prob. 6.60.
-
-**6.61** Consider the circuit in Fig. 6.83. Find: (a) *L*eq, *i*1(*t*), and *i*2(*t*) if *is* = 3*e*−*t* mA, (b) *vo*(*t*), (c) energy stored in the 20-mH inductor at *t* = 1 s.
-
-**6.62** Consider the circuit in Fig. 6.84. Given that *v*(*t*) = 12*e*−3*t* mV for *t* > 0 and *i*1(0) = −30 mA, find: (a) *i*2(0), (b) *i*1(*t*) and *i*2(*t*).
-
-For Prob. 6.62.
-
-**6.63** In the circuit of Fig. 6.85, sketch *vo*.
-
-**6.64** The switch in Fig. 6.86 has been in position *A* for a long time. At *t* = 0, the switch moves from position *A* to *B*. The switch is a make-before-break type so that there is no interruption in the inductor current. Find:
-
-(a) *i*(*t*) for *t* > 0,
-
-(b) *v just after* the switch has been moved to position *B*, (c) *v*(*t*) long after the switch is in position *B*.
-
-**6.65** The inductors in Fig. 6.87 are initially charged and are connected to the black box at *t* = 0. If *i*1(0) = 4 A, *i*2(0) = −2 A, and *v*(*t*) = 50*e*−200*t* mV, *t* ≥ 0, find:
-
-(a) the energy initially stored in each inductor,
-
-- (b) the total energy delivered to the black box from *t* = 0 to *t* = ∞,
-- (c) *i*1(*t*) and *i*2(*t*), *t* ≥ 0, (d) *i*(*t*), *t* ≥ 0.
-
-**6.66** The current *i*(*t*) through a 20-mH inductor is equal, in magnitude, to the voltage across it for all values of time. If *i*(0) = 2 A, find *i*(*t*).
-
-# Section 6.6 Applications
-
-**6.67** An op amp integrator has *R* = 50 kΩ and *C* = 0.04 *μ*F. If the input voltage is *vi* = 10 sin 50*t* mV, obtain the output voltage. Assume that at *t* equal to zero, the output is equal to zero.
-
-- **6.68** A 6-V dc voltage is applied to an integrator with *R* = 50 kΩ, *C* = 100 *μ*F at *t* = 0. How long will it take for the op amp to saturate if the saturation voltages are +12 V and −12 V? Assume that the initial capacitor voltage was zero.
-- **6.69** An op amp integrator with *R* = 4 MΩ and *C* = 1 *μ*F has the input waveform shown in Fig. 6.88. Plot the output waveform.
-
-**6.70** Using a single op amp, a capacitor, and resistors of 100 kΩ or less, design a circuit to implement
-
-$$
-v_o = -2 \int_0^t v_i(\tau) \, d\tau
-$$
-
-Assume *vo*=0 at *t*=0.
-
-**6.71** Show how you would use a single op amp to generate
-
-$$
-v_o = -\int_0^t (v_1 + 4v_2 + 10v_3) \, dt
-$$
-
-If the inte grating capacitor is *C* = 5 *μ*F, obtain the other component values.
-
-**6.72** At *t* = 1.5 ms, calculate *vo* due to the cascaded integrators in Fig. 6.89. Assume that the integrators are reset to 0 V at *t* = 0.
-
-For Prob. 6.72.
-
-**6.73** Show that the circuit in Fig. 6.90 is a noninverting integrator.
-
-# **Figure 6.90** For Prob. 6.73.
-
-**6.74** The triangular waveform in Fig. 6.91(a) is applied to the input of the op amp differentiator in Fig. 6.91(b). Plot the output.
-
-For Prob. 6.74.
-
-- **6.75** An op amp differentiator has *R* = 250 kΩ and *C* = 10 *μ*F. The input voltage is a ramp *r*(*t*) = 7*t* mV. Find the output voltage.
-- **6.76** A voltage waveform has the following characteristics: a positive slope of 20 V/s for 5 ms followed by a negative slope of 10 V/s for 10 ms. If the waveform is applied to a differentiator with *R* = 50 kΩ, *C* = 10 *μ*F, sketch the output voltage waveform.
-
-**6.77** The output *vo* of the op amp circuit in Fig. 6.92(a) is shown in Fig. 6.92(b). Let *Ri* = *Rf* = 1 MΩ and *C* = 1 *μ*F. Determine the input voltage waveform and sketch it. **\***
-
-**6.79** Design an analog computer circuit to solve for *v*(*t*), given the following equation and a value for *f*(*t*) and that *v*(0) = 0 V.
-
-$$
-\left(\frac{dv(t)}{dt}\right) + 3vt = f t dt
-$$
-
-**6.80** Figure 6.93 presents an analog computer designed to solve a differential equation. Assuming *f*(*t*) is known, set up the equation for *f*(*t*).
-
-**Figure 6.93** For Prob. 6.80.
-
-**6.81** Design an analog computer to simulate the following equation to solve for *v*(*t*) (assume the initial conditions are zero):
-
-$$
-(d^{3}v(t)/dt^{3}) + 3(dv(t)/dt) = 4f(t)
-$$
-
-**6.82** Design an op amp circuit such that
-
-$$
-v_o = 10v_s + 2 \int v_s dt
-$$
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/075_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/075_Comprehensive Problems.md
deleted file mode 100644
index 156504d7e72e821fa8219049205916c9e51fc377..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/075_Comprehensive Problems.md
+++ /dev/null
@@ -1,23 +0,0 @@
- where *vs* and *vo* are the input v oltage and output voltage, respectively.
-
-**6.83** Your laboratory has available a large number of 5-*μ*F capacitors rated at 150 V. To design a capacitor bank of 10 *μ*F rated at 600 V, how many 5-*μ*F capacitors are needed and how would you connect them?
-
-**6.84** An 8-mH inductor is used in a fusion power experiment. If the current through the inductor is *i*(*t*) = 10 cos2 (*πt*) mA, for all *t* > 0 s, find the power being delivered to the inductor and the energy stored in it at *t* = 0.5 s.
-
-–4
-
-**6.78** Design an analog computer to simulate *d*2 \_\_\_\_*vo dvo*
-
-$$
-\frac{d v_o}{dt^2} + 2\frac{d v_o}{dt} + v_o = 10 \sin 2t
-$$
-
-(b)
-
-where *v*0(0)=−6 V and *v*0 ′ (0)=0. **6.85** A square-wave generator produces the voltage waveform shown in Fig. 6.94(a). What kind of a circuit component is needed to convert the voltage waveform to the triangular current waveform shown in Fig. 6.94(b)? Calculate the value of the component, assuming that it is initially uncharged.
-
-**6.86** An electric motor can be modeled as a series combination of a 12-Ω resistor and 200-mH inductor. If a current *i*(*t*) = 2*te*−10*t* A flows through the series combination, find the voltage across the combination.
-
-# **chapter**
-
-7
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/076_Chapter 7 - First-Order Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/076_Chapter 7 - First-Order Circuits.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/076_Chapter 7 - First-Order Circuits.md
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-# First-Order Circuits
-
-*We live in deeds, not years; in thoughts, not breaths; in feelings, not in figures on a dial. We should count time in heart-throbs. He most lives who thinks most, feels the noblest, acts the best.*
-
-—P. J. Bailey
-
-# Enhancing Your Career
-
-# **Careers in Computer Engineering**
-
-Electrical engineering education has gone through drastic changes in recent decades. Most departments have come to be known as Department of Electrical and Computer Engineering, emphasizing the rapid changes due to computers. Computers occup y a prominent place in modern so ciety and education. They have become commonplace and are helping to change the f ace of research, de velopment, production, business, and entertainment. The scientist, engineer , doctor, attorney, teacher, airline pilot, businessperson—almost anyone benefits from a computer's abilities to store large amounts of information and to process that information in very short periods of time. The internet, a computer communication network, is essential in b usiness, education, and library science. Com puter usage continues to grow by leaps and bounds.
-
-An education in computer engineering should provide breadth in software, hardware design, and basic modeling techniques. It should include courses in data structures, digital systems, computer architecture, micro processors, interfacing, software engineering, and operating systems.
-
-Electrical engineers who specialize in computer engineering find jobs in computer industries and in numerous fields where computers are being used. Companies that produce software are growing rapidly in number and size and providing employment for those who are skilled in programming. An excellent way to advance one's knowledge of computers is to join the IEEE Computer Society, which sponsors diverse magazines, journals, and conferences.
-
-Computer design of very large scale integrated (VLSI) circuits. Courtesy Brian Fast, Cleveland State University
-
-# Learning Objectives
-
-By using the information and exercises in this chapter you will be able to:
-
-- 1. Understand solutions to unforced, first-order linear differential equations.
-- 2. Comprehend singularity equations and know their importance in solving linear differential equations.
-- 3. Understand the effect of unit step sources on first-order linear differential equations.
-- 4. Explain how dependent sources and op amps influence simple first-order linear differential equations.
-- 5. Use *PSpice* to solve simple transient circuits with an inductor or a capacitor.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/077_7.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/077_7.1 Introduction.md
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-# **7.1** Introduction
-
-Now that we ha ve considered the three passi ve elements (resistors, ca pacitors, and inductors) and one active element (the op amp)individually, we are prepared to consider circuits that contain various combinations of two or three of the passi ve elements. In this chapter , we shall e xamine two types of simple circuits: a circuit comprising a resistor and capacitor and a circuit comprising a resistor and an inductor . These are called *RC* and *RL* circuits, respectively. As simple as these circuits are, the y find continual applications in electronics, communications, and control sys tems, as we shall see.
-
-We carry out the analysis of *RC* and *RL* circuits by applying Kirchhoff's laws, as we did for resisti ve circuits. The only dif ference is that applying Kirchhoff's laws to purely resistive circuits results in algebraic equations, while applying the la ws to *RC* and *RL* circuits produces differential equations, which are more difficult to solve than algebraic equations. The differential equations resulting from analyzing *RC* and *RL* circuits are of the first order. Hence, the circuits are collectively known as *first-order* circuits.
-
-A first-order circuit is characterized by a first-order differential equation.
-
-In addition to there being tw o types of first-order circuits *(RC* and *RL)*, there are tw o ways to e xcite the circuits. The first way is by ini tial conditions of the storage elements in the circuits. In these so-called *source-free circuits*, we assume that ener gy is initially stored in the ca pacitive or inductive element. The energy causes current to flow in the circuit and is gradually dissipated in the resistors. Although source-free circuits are by definition free of independent sources, they may ha ve dependent sources. The second way of exciting first-order circuits is by independent sources. In this chapter, the independent sources we will consider are dc sources. (In later chapters, we shall consider sinusoidal and exponential sources.) The two types of first-order circuits and the two ways of exciting them add up to the four possible situations we will study in this chapter.
-
-Finally, we consider four typical applications of *RC* and *RL* circuits: delay and relay circuits, a photoflash unit, and an automobile ignition circuit.
-
-# **7.2** The Source-Free RC Circuit
-
-A source-free *RC* circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the resistors.
-
-Consider a series combination of a resistor and an initially char ged capacitor, as shown in Fig. 7.1. (The resistor and capacitor may be the equivalent resistance and equivalent capacitance of combinations of re sistors and capacitors.) Our objective is to determine the circuit response, which, for pedagogic reasons, we assume to be the voltage *v*(*t*) across the capacitor. Since the capacitor is initially charged, we can assume that at time *t* = 0, the initial voltage is
-
-$$
-v(0) = V_0 \tag{7.1}
-$$
-
-with the corresponding value of the energy stored as
-
-$$
-w(0) = \frac{1}{2}CV_0^2
-$$
- (7.2)
-
-Applying KCL at the top node of the circuit in Fig. 7.1 yields
-
-$$
-i_C + i_R = 0 \tag{7.3}
-$$
-
-By definition, *iC* = *C dv*∕*dt* and *iR* = *v*∕*R*. Thus,
-
-$$
-C\frac{dv}{dt} + \frac{v}{R} = 0
-$$
-\n(7.4a)
-
-or
-
-$$
-\frac{dv}{dt} + \frac{v}{RC} = 0\tag{7.4b}
-$$
-
-This is a *first-order differential equation,* since only the first derivative of *v* is involved. To solve it, we rearrange the terms as
-
-$$
-\frac{dv}{v} = -\frac{1}{RC}dt\tag{7.5}
-$$
-
-Integrating both sides, we get
-
-$$
-\ln v = -\frac{t}{RC} + \ln A
-$$
-
-where ln *A* is the integration constant. Thus,
-
-$$
-\ln\frac{v}{A} = -\frac{t}{RC}
-$$
- (7.6)
-
-Taking powers of *e* produces
-
-$$
-v(t) = Ae^{-t/RC}
-$$
-
-But from the initial conditions, *v*(0) = *A* = *V*0. Hence,
-
-$$
-v(t) = V_0 e^{-t/RC}
-$$
-\n
-$$
-(7.7)
-$$
-
-This shows that the voltage response of the *RC* circuit is an exponential decay of the initial voltage. Since the response is due to the initial energy stored and the physical characteristics of the circuit and not due to some e xternal voltage or current source, it is called the *natural response* of the circuit.
-
-**Figure 7.1** A source-free *RC* circuit.
-
-A circuit response is the manner in which the circuit reacts to an excitation.
-
-The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation.
-
-The natural response depends on the nature of the circuit alone, with no external sources. In fact, the circuit has a response only because of the energy initially stored in the capacitor.
-
-| (t)∕V0 = e−t∕τ
Values of v
t
τ
2τ
3τ
4τ
5τ | TABLE 7.1 | |
-|-----------------------------------------------------------------|-----------|---------|
-| | | |
-| | | v(t)∕V0 |
-| | | 0.36788 |
-| | | 0.13534 |
-| | | 0.04979 |
-| | | 0.01832 |
-| | | 0.00674 |
-
-Graphical determination of the time constant *τ* from the response curve.
-
-The natural response is illustrated graphically in Fig. 7.2. Note that at *t* = 0, we have the correct initial condition as in Eq. (7.1). As *t* increases, the voltage decreases toward zero. The rapidity with which the v oltage decreases is e xpressed in terms of the *time constant*, denoted by *τ*, the lowercase Greek letter tau.
-
-The time constant of a circuit is the time required for the response to decay to a factor of 1∕e or 36.8 percent of its initial value.1
-
-This implies that at *t* = *τ*, Eq. (7.7) becomes
-
-$$
-V_0 e^{-\tau/RC} = V_0 e^{-1} = 0.368 V_0
-$$
-
-or
-
-$$
-\tau = RC \tag{7.8}
-$$
-
-In terms of the time constant, Eq. (7.7) can be written as
-
-$$
-v(t) = V_0 e^{-t/\tau} \tag{7.9}
-$$
-
-With a calculator it is easy to sho w that the v alue of *v*(*t*)∕*V*0 is as shown in Table 7.1. It is evident from Table 7.1 that the voltage *v*(*t*) is less than 1 percent of *V*0 after 5*τ* (five time constants). Thus, it is customary to assume that the capacitor is fully dischar ged (or charged) after five time constants. In other words, it takes 5*τ* for the circuit to reach its final state or steady state when no changes take place with time. Notice that for every time interval of *τ*, the voltage is reduced by 36.8 percent of its pre vious value, *v*(*t* + *τ*) = *v*(*t*)∕*e* = 0.368*v*(*t*), regardless of the value of *t*.
-
-Observe from Eq. (7.8) that the smaller the time constant, the more rapidly the v oltage decreases, that is, the f aster the response. This is illustrated in Fig. 7.4. A circuit with a small time constant gi ves a f ast response in that it reaches the steady state (or final state) quickly due to quick dissipation of ener gy stored, whereas a circuit with a lar ge time constant gives a slo w response because it tak es longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants.
-
-With the voltage *v*(*t*) in Eq. (7.9), we can find the current *iR*(*t*),
-
-$$
-i_R(t) = \frac{v(t)}{R} = \frac{V_0}{R} e^{-t/\tau}
-$$
-\n(7.10)
-
-1 The time constant may be viewed from another perspective. Evaluating the derivative of *v*(*t*) in Eq. (7.7) at *t* = 0, we obtain
-
-$$
-\frac{d}{dt}\left(\frac{v}{V_0}\right)\Big|_{t=0} = -\frac{1}{\tau}e^{-t/\tau}\Big|_{t=0} = -\frac{1}{\tau}
-$$
-
-Thus, the time constant is the initial rate of decay, or the time taken for *v*∕*V*0 to decay from unity to zero, assuming a constant rate of decay. This initial slope interpretation of the time constant is often used in the laboratory to find *τ* graphically from the response curve displayed on an oscilloscope. To find *τ* from the response curve, draw the tangent to the curve at *t* = 0, as shown in Fig. 7.3. The tangent intercepts with the time axis at *t* = *τ*.
-
-Plot of *v*∕*V*0 = *e* −*t*∕*τ* for various values of the time constant.
-
-The power dissipated in the resistor is
-
-$$
-p(t) = vi_R = \frac{V_0^2}{R}e^{-2t/\tau}
-$$
-\n(7.11)
-
-The energy absorbed by the resistor up to time *t* is
-
-$$
-w_R(t) = \int_0^t p(\lambda) d\lambda = \int_0^t \frac{V_0^2}{R} e^{-2\lambda/\tau} d\lambda
-$$
-
-= $-\frac{\tau V_0^2}{2R} e^{-2\lambda/\tau} \Big|_0^t = \frac{1}{2} C V_0^2 (1 - e^{-2t/\tau}), \qquad \tau = RC$ (7.12)
-
-Notice that as *t* → ∞, *wR* (∞) → \_1 2*CV* 0 2 , which is the same as *wC* (0), the energy initially stored in the capacitor. The energy that was initially stored in the capacitor is eventually dissipated in the resistor.
-
-In summary:
-
-The Key to Working with a Source-Free RC Circuit Is Finding:
-
-- 1. The initial voltage *v*(0) = *V*0 across the capacitor.
-- 2. The time constant *τ*.
-
-With these tw o items, we obtain the response as the capacitor v oltage *vC* (*t*) = *v*(*t*) = *v*(0) *e*−*t*∕*τ* . Once the capacitor voltage is first obtained, other variables (capacitor current *iC*, resistor voltage *vR*, and resistor current *iR*) can be determined. In finding the time constant *τ* = *RC*, *R* is often the Thevenin equivalent resistance at the terminals of the capacitor; that is, we take out the capacitor *C* and find *R* = *R*Th at its terminals.
-
-In Fig. 7.5, let *vC* (0) = 15 V. Find *vC*, *vx*, and *ix* for *t* > 0.
-
-# **Solution:**
-
-We first need to make the circuit in Fig. 7.5 conform with the standard *RC* circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin The time constant is the same regardless of what the output is defined to be.
-
-When a circuit contains a single capacitor and several resistors and dependent sources, the Thevenin equivalent can be found at the terminals of the capacitor to form a simple RC circuit. Also, one can use Thevenin's theorem when several capacitors can be combined to form a single equivalent capacitor.
-
-# Example 7.1
-
-**Figure 7.5**
-
-For Example 7.1.
-
-**Figure 7.6** Equivalent circuit for the circuit in Fig. 7.5.
-
-The 8-Ω and 12- Ω resistors in series can be combined to give a 20-Ω resistor. This 20-Ω resistor in parallel with the 5-Ω resistor can be combined so that the equivalent resistance is
-
-$$
-R_{\text{eq}} = \frac{20 \times 5}{20 + 5} = 4 \ \Omega
-$$
-
-Hence, the equivalent circuit is as shown in Fig. 7.6, which is analogous to Fig. 7.1. The time constant is
-
-Thus,
-
-$$
-\tau = R_{\text{eq}}C = 4(0.1) = 0.4 \text{ s}
-$$
-
-$$
-^{as,}
-$$
-
-$$
-v = v(0)e^{-t/\tau} = 15e^{-t/0.4} \text{ V}, \qquad v_C = v = 15e^{-2.5t} \text{ V}
-$$
-
-From Fig. 7.5, we can use voltage division to get *vx*; so
-
-$$
-v_x = \frac{12}{12 + 8}v = 0.6(15e^{-2.5t}) = 9e^{-2.5t}
-$$
- V
-
-Finally,
-
-$$
-i_x = \frac{v_x}{12} = 0.75e^{-2.5t} \text{ A}
-$$
-
-# Practice Problem 7.1
-
-Refer to the circuit in Fig. 7.7. Let *vC* (0) = 60 V. Determine *vC*, *vx*, and *io* for *t* ≥ 0.
-
-For Practice Prob. 7.1.
-
-**Answer:** 60*e*−0.25*t* V, 20*e*−0.25*t*V, −5*e*−0.25*t*A.
-
-**Figure 7.8** For Example 7.2.
-
-Example 7.2 The switch in the circuit in Fig. 7.8 has been closed for a long time, and it is opened at *t* = 0. Find *v*(*t*) for *t* ≥ 0. Calculate the initial energy stored in the capacitor.
-
-# **Solution:**
-
-For *t* < 0, the switch is closed; the capacitor is an open circuit to dc, as represented in Fig. 7.9(a). Using voltage division
-
-$$
-v_C(t) = \frac{9}{9+3}(20) = 15 \text{ V}, \qquad t < 0
-$$
-
-Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at *t* = 0− is the same at *t* = 0, or
-
-$$
-v_C(0) = V_0 = 15 \text{ V}
-$$
-
-For *t* > 0, the switch is opened, and we have the *RC* circuit shown in Fig. 7.9(b). [Notice that the *RC* circuit in Fig. 7.9(b) is source free; the independent source in Fig. 7.8 is needed to provide *V*0 or the initial energy in the capacitor.] The 1-Ω and 9-Ω resistors in series give
-
-$$
-R_{\text{eq}} = 1 + 9 = 10 \Omega
-$$
-
-The time constant is
-
-$$
-\tau = R_{\text{eq}}C = 10 \times 20 \times 10^{-3} = 0.2 \text{ s}
-$$
-
-Thus, the voltage across the capacitor for *t* ≥ 0 is
-
-$$
-v(t) = v_C(0)e^{-t/\tau} = 15e^{-t/0.2} \text{ V}
-$$
-
-or
-
-$$
-v(t) = 15e^{-5t} \,\mathrm{V}
-$$
-
-The initial energy stored in the capacitor is
-
-$$
-w_C(0) = \frac{1}{2}Cv_C^2(0) = \frac{1}{2} \times 20 \times 10^{-3} \times 15^2 = 2.25 \text{ J}
-$$
-
-If the switch in Fig. 7.10 opens at *t* = 0, find *v*(*t*) for *t* ≥ 0 and *wC* (0). Practice Problem 7.2
-
-**Answer:** 8 *e*−2*t* V, 5.333 J.
-
-9 Ω 1 Ω *v*C(0) 3 Ω + 20 V (a) 9 Ω 1 Ω (b) + ‒ Vo = 15 V 20 mF + ‒
-
-**Figure 7.9** For Example 7.2: (a) *t* < 0, (b) *t* > 0.
-
-**Figure 7.10** For Practice Prob. 7.2.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/078_7.2 The Source-Free RC Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/078_7.2 The Source-Free RC Circuit.md
deleted file mode 100644
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@@ -1,350 +0,0 @@
-# **7.3** The Source-Free RL Circuit
-
-Consider the series connection of a resistor and an inductor, as shown in Fig. 7.11. Our goal is to determine the circuit response, which we will assume to be the current *i*(*t*) through the inductor. We select the induc tor current as the response in order to take advantage of the idea that the inductor current cannot change instantaneously. At *t* = 0, we assume that the inductor has an initial current *I*0, or
-
-$$
-i(0) = I_0 \tag{7.13}
-$$
-
-with the corresponding energy stored in the inductor as
-
-$$
-w(0) = \frac{1}{2} L I_0^2 \tag{7.14}
-$$
-
-Applying KVL around the loop in Fig. 7.11,
-
-$$
-v_L + v_R = 0 \tag{7.15}
-$$
-
-But *vL* = *L di*∕*dt* and *vR* = *iR*. Thus,
-
-$$
-L\frac{di}{dt} + Ri = 0
-$$
-
-**Figure 7.11** A source-free *RL* circuit.
-
-$$
-\frac{di}{dt} + \frac{R}{L}i = 0\tag{7.16}
-$$
-
-Rearranging terms and integrating gives
-
-$$
-\int_{I_0}^{i(t)} \frac{di}{i} = -\int_0^t \frac{R}{L} dt
-$$
-
-\n
-$$
-\ln i \Big|_{I_0}^{i(t)} = -\frac{Rt}{L} \Big|_0^t \implies \ln i(t) - \ln I_0 = -\frac{Rt}{L} + 0
-$$
-
-or
-
-$$
-\ln \frac{i(t)}{I_0} = -\frac{Rt}{L} \tag{7.17}
-$$
-
-Taking the powers of *e*, we have
-
-$$
-i(t) = I_0 e^{-Rt/L}
-$$
-\n(7.18)
-
-This shows that the natural response of the *RL* circuit is an e xponential decay of the initial current. The current response is shown in Fig. 7.12. It is evident from Eq. (7.18) that the time constant for the *RL* circuit is
-
-$$
-\tau = \frac{L}{R} \tag{7.19}
-$$
-
-with *τ* ag ain ha ving the unit of seconds. Thus, Eq. (7.18) may be written as
-
-$$
-i(t) = I_0 e^{-t/\tau}
-$$
-\n(7.20)
-
-With the current in Eq. (7.20), we can find the voltage across the resistor as
-
-$$
-v_R(t) = iR = I_0 Re^{-t/\tau}
-$$
- (7.21)
-
-The power dissipated in the resistor is
-
-$$
-p = v_R i = I_0^2 Re^{-2t/\tau}
-$$
- (7.22)
-
-The energy absorbed by the resistor is
-
-$$
-w_R(t) = \int_0^t p(\lambda) d\lambda = \int_0^t I_0^2 Re^{-2\lambda/\tau} d\lambda = -\frac{\tau}{2} I_0^2 Re^{-2\lambda/\tau} \Big|_0^t, \qquad \tau = \frac{L}{R}
-$$
-
-or
-
-$$
-w_R(t) = \frac{1}{2} L \, I_0^2 (1 - e^{-2t/\tau}) \tag{7.23}
-$$
-
-Figure 7.12 shows an initial slope interpretation may be given to *τ*.
-
-Note that as *t* → ∞,*wR*(∞) → \_1 2 *L I*2 0, which is the same as *wL*(0), the initial energy stored in the inductor as in Eq. (7.14). Again, the energy initially stored in the inductor is eventually dissipated in the resistor.
-
-The smaller the time constant *τ* of a circuit, the faster the rate of decay of the response. The larger the time constant, the slower the rate of decay of the response. At any rate, the response decays to less than 1 percent of its initial value (i.e., reaches steady
-
-state) after 5*τ*.
-
-In summary:
-
-# The Key to Working with a Source-Free RL Circuit Is to Find:
-
-- 1. The initial current *i*(0) = *I*0 through the inductor.
-- 2. The time constant *τ* of the circuit.
-
-With the tw o items, we obtain the response as the inductor current *iL*(*t*) = *i*(*t*) = *i*(0)*e*−*t*∕*τ* . Once we determine the inductor current *iL*, other variables (inductor voltage *vL*, resistor voltage *vR*, and resistor current *iR*) can be obtained. Note that in general, *R* in Eq. (7.19) is the Thevenin resistance at the terminals of the inductor.
-
-Assuming that *i*(0) = 10 A, calculate *i*(*t*) and *ix*(*t*) in the circuit of Example 7.3 Fig. 7.13.
-
-# **Solution:**
-
-There are two ways we can solve this problem. One way is to obtain the equivalent resistance at the inductor terminals and then use Eq. (7.20). The other way is to start from scratch by using Kirchhoff's voltage law. Whichever approach is taken, it is always better to first obtain the inductor current.
-
-■ **METHOD 1** The equivalent resistance is the same as the Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with *vo* = 1 V at the inductor terminals *a*-*b*, as in Fig. 7.14(a). (We could also insert a 1-A current source at the terminals.) Applying KVL to the two loops results in
-
-$$
-2(i_1 - i_2) + 1 = 0 \qquad \Rightarrow \qquad i_1 - i_2 = -\frac{1}{2} \tag{7.3.1}
-$$
-
-$$
-6i_2 - 2i_1 - 3i_1 = 0 \qquad \Rightarrow \qquad i_2 = \frac{5}{6}i_1 \tag{7.3.2}
-$$
-
-Substituting Eq. (7.3.2) into Eq. (7.3.1) gives
-
-$$
-i_1 = -3
-$$
- A, $i_o = -i_1 = 3$ A
-
-**Figure 7.14** Solving the circuit in Fig. 7.13. When a circuit has a single inductor and several resistors and dependent sources, the Thevenin equivalent can be found at the terminals of the inductor to form a simple RL circuit. Also, one can use Thevenin's theorem when several inductors can be combined to form a single equivalent inductor.
-
-Hence,
-
-The time constant is
-
-$$
-R_{\text{eq}} = R_{\text{Th}} = \frac{v_o}{i_o} = \frac{1}{3} \Omega
-$$
-
-*τ* = \_\_\_*L R*eq = \_1 2 \_ \_1 3 =\_\_3 2 s
-
-Thus, the current through the inductor is
-
-$$
-i(t) = i(0)e^{-t/\tau} = 10e^{-(2/3)t} \text{ A}, \qquad t > 0
-$$
-
-■ **METHOD 2** We may directly apply KVL to the circuit as in Fig. 7.14(b). For loop 1,
-
-$$
-\frac{1}{2}\frac{di_1}{dt} + 2(i_1 - i_2) = 0
-$$
-
-or
-
-$$
-\frac{di_1}{dt} + 4i_1 - 4i_2 = 0 \tag{7.3.3}
-$$
-
-For loop 2,
-
-$$
-6i_2 - 2i_1 - 3i_1 = 0 \qquad \Rightarrow \qquad i_2 = \frac{5}{6}i_1 \tag{7.3.4}
-$$
-
-Substituting Eq. (7.3.4) into Eq. (7.3.3) gives
-
-$$
-\frac{di_1}{dt} + \frac{2}{3}i_1 = 0
-$$
-
-Rearranging terms,
-
-$$
-\frac{di_1}{i_1} = -\frac{2}{3}dt
-$$
-
-Since *i*1 = *i*, we may replace *i*1 with *i* and integrate:
-
-$$
-\ln i \Big|_{i(0)}^{i(t)} = -\frac{2}{3}t \Big|_0^t
-$$
-
-or
-
-$$
-\ln \frac{i(t)}{i(0)} = -\frac{2}{3}t
-$$
-
-Taking the powers of *e*, we finally obtain
-
-$$
-i(t) = i(0)e^{-(2/3)t} = 10e^{-(2/3)t} \text{ A}, \qquad t > 0
-$$
-
-which is the same as by Method 1. The voltage across the inductor is
-
-$$
-v = L \frac{di}{dt} = 0.5(10) \left( -\frac{2}{3} \right) e^{-(2/3)t} = -\frac{10}{3} e^{-(2/3)t}
-$$
- V
-
-Since the inductor and the 2-Ω resistor are in parallel,
-
-$$
-i_x(t) = \frac{v}{2} = -1.6667e^{-(2/3)t} \text{A}, \qquad t > 0
-$$
-
-Find *i* and
-$$
-v_x
-$$
- in the circuit of Fig. 7.15. Let $i(0) = 7$ A.
-
-**Answer:** 7 *e*−2*t* A, −7 *e*−2*t* V, *t* > 0.
-
-Practice Problem 7.3
-
-The switch in the circuit of Fig. 7.16 has been closed for a long time. At Example 7.4 *t* = 0, the switch is opened. Calculate *i*(*t*) for *t* > 0.
-
-# **Solution:**
-
-When *t* < 0, the switch is closed, and the inductor acts as a short circuit to dc. The 16-Ω resistor is short-circuited; the resulting circuit is shown in Fig. 7.17(a). To get *i*1 in Fig. 7.17(a), we combine the 4- Ω and 12-Ω resistors in parallel to get
-
-$$
-\frac{4 \times 12}{4 + 12} = 3 \ \Omega
-$$
-
-Hence,
-
-$$
-i_1 = \frac{40}{2+3} = 8 \text{ A}
-$$
-
-We obtain *i*(*t*) from *i*1 in Fig. 7.17(a) using current division, by writing
-
-$$
-i(t) = \frac{12}{12 + 4} i_1 = 6 \text{ A}, \qquad t < 0
-$$
-
-Since the current through an inductor cannot change instantaneously,
-
-$$
-i(0) = i(0^-) = 6\,\mathrm{A}
-$$
-
- When *t* > 0, the switch is open and the voltage source is disconnected. We now have the source-free *RL* circuit in Fig. 7.17(b). Combining the resistors, we have
-
-$$
-R_{\text{eq}} = (12 + 4) || 16 = 8 \Omega
-$$
-
-The time constant is
-
-$$
-\tau = \frac{L}{R_{\text{eq}}} = \frac{2}{8} = \frac{1}{4} \,\text{s}
-$$
-
-Thus,
-
-$$
-i(t) = i(0)e^{-t/\tau} = 6e^{-4t} A
-$$
-
-**Figure 7.16** For Example 7.4.
-
-Solving the circuit of Fig. 7.16: (a) for *t* < 0, (b) for *t* > 0.
-
-For Practice Prob. 7.4.
-
-**Figure 7.19** For Example 7.5.
-
-**Figure 7.20**
-
-The circuit in Fig. 7.19 for: (a) *t* < 0, (b) *t* > 0.
-
-Example 7.5 In the circuit shown in Fig. 7.19, find *io*, *vo*, and *i* for all time, assuming that the switch was open for a long time.
-
-# **Solution:**
-
-It is better to first find the inductor current *i* and then obtain other quantities from it.
-
-For *t* < 0, the switch is open. Since the inductor acts like a short circuit to dc, the 6-Ω resistor is short-circuited, so that we ha ve the circuit shown in Fig. 7.20(a). Hence, *io* = 0, and
-
-$$
-i(t) = \frac{10}{2+3} = 2 \text{ A}, \qquad t < 0
-$$
-$$
-v_o(t) = 3i(t) = 6 \text{ V}, \qquad t < 0
-$$
-
-Thus, *i*(0) = 2.
-
- For *t* > 0, the switch is closed, so that the v oltage source is shortcircuited. We now have a source-free *RL* circuit as shown in Fig. 7.20(b). At the inductor terminals,
-
-$$
-R_{\text{Th}} = 3 \parallel 6 = 2 \text{ }\Omega
-$$
-
-so that the time constant is
-
-$$
-\tau = \frac{L}{R_{\text{Th}}} = 1 \text{ s}
-$$
-
-Hence,
-
-$$
-i(t) = i(0)e^{-t/\tau} = 2e^{-t}A, \qquad t > 0
-$$
-
-Because the inductor is in parallel with the 6- and 3-Ω resistors,
-
-$$
-v_o(t) = -v_L = -L\frac{di}{dt} = -2(-2e^{-t}) = 4e^{-t} \text{ V}, \qquad t > 0
-$$
-
-and
-
-$$
-i_o(t) = \frac{v_L}{6} = -\frac{2}{3}e^{-t}A, \qquad t > 0
-$$
-
-Thus, for all time,
-
-$$
-i_o(t) = \begin{cases} 0 \text{ A}, & t < 0 \\ -\frac{2}{3}e^{-t} \text{ A}, & t > 0 \end{cases}, \quad v_o(t) = \begin{cases} 6 \text{ V}, & t < 0 \\ 4e^{-t} \text{ V}, & t > 0 \end{cases}
-$$
-$$
-i(t) = \begin{cases} 2 \text{ A}, & t < 0 \\ 2e^{-t} \text{ A}, & t \ge 0 \end{cases}
-$$
-
-o We notice that the inductor current is continuous at *t* = 0, while the (t) current through the 6-Ω resistor drops from 0 to −2∕3 at *t* = 0, and the voltage across the 3-Ω resistor drops from 6 to 4 at *t* = 0. We also notice that the time constant is the same regardless of what the output is defined to be. Figure 7.21 plots *i* and *io*.
-
-Determine *i*, *io*, and *vo* for all *t* in the circuit shown in Fig. 7.22. Assume that the switch was closed for a long time. It should be noted that opening a switch in series with an ideal current source creates an infinite voltage at the current source terminals. Clearly this is impossible. For the purposes of problem solving, we can place a shunt resistor in parallel with the source (which now makes it a voltage source in series with a resistor). In more practical circuits, devices that act like current sources are, for the most part, electronic circuits. These circuits will allow the source to act like an ideal current source over its operating range but voltage-limit it when the load resistor becomes too large (as in an open circuit).
-
-# **Answer:**
-
-$$
-i = \begin{cases} 16 \text{ A}, & t < 0 \\ 16e^{-2t} \text{ A}, & t \ge 0 \end{cases}, \quad i_o = \begin{cases} 8 \text{ A}, & t < 0 \\ -5.333e^{-2t} \text{ A}, & t > 0 \end{cases},
-$$
-$$
-v_o = \begin{cases} 32 \text{ V}, & t < 0 \\ 10.667e^{-2t} \text{ V}, & t > 0 \end{cases}
-$$
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/079_7.4 Singularity Functions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/079_7.4 Singularity Functions.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/079_7.4 Singularity Functions.md
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@@ -1,380 +0,0 @@
-# **7.4** Singularity Functions
-
-Before going on with the second half of this chapter, we need to digress and consider some mathematical concepts that will aid our understand ing of transient analysis. A basic understanding of singularity functions will help us make sense of the response of first-order circuits to a sudden application of an independent dc voltage or current source.
-
-Singularity functions (also called *switching functions*) are very useful in circuit analysis. They serve as good approximations to the switching signals that arise in circuits with switching operations. They are helpful in the neat, compact description of some circuit phenomena, especially the step response of *RC* or *RL* circuits to be discussed in the next sections. By definition,
-
-Singularity functions are functions that either are discontinuous or have discontinuous derivatives.
-
-**Figure 7.21** A plot of *i* and *io*.
-
-**Figure 7.22** For Practice Prob. 7.5.
-
-The three most widely used singularity functions in circuit analysis are the *unit step*, the *unit impulse*, and the *unit ramp* functions.
-
-The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.
-
-In mathematical terms,
-
-$$
-u(t) = \begin{cases} 0, & t < 0 \\ 1, & t > 0 \end{cases}
-$$
- (7.24)
-
-The unit step function is undefined at *t* = 0, where it changes abrupt ly from 0 to 1. It is dimensionless, lik e other mathematical functions such as sine and cosine. Figure 7.23 depicts the unit step function. If the abrupt change occurs at *t* = *t*0 (where *t*0 > 0) instead of *t* = 0, the unit step function becomes
-
-$$
-u(t - t_0) = \begin{cases} 0, & t < t_0 \\ 1, & t > t_0 \end{cases}
-$$
- (7.25)
-
-which is the same as saying that *u*(*t*) is delayed by *t*0 seconds, as shown in Fig. 7.24(a). To get Eq. (7.25) from Eq. (7.24), we simply replace every *t* by *t* − *t*0. If the change is at *t* = −*t*0, the unit step function becomes
-
-$$
-u(t+t_0) = \begin{cases} 0, & t < -t_0 \\ 1, & t > -t_0 \end{cases} \tag{7.26}
-$$
-
-meaning that *u*(*t*) is advanced by *t*0 seconds, as shown in Fig. 7.24(b).
-
-We use the step function to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control systems and digital computers. For example, the voltage
-
-$$
-v(t) = \begin{cases} 0, & t < t_0 \\ V_0, & t > t_0 \end{cases} \tag{7.27}
-$$
-
-may be expressed in terms of the unit step function as
-
-$$
-v(t) = V_0 u(t - t_0)
-$$
-\n(7.28)
-
-(b)
-
-t = 0
-
-b
-
-a
-
-If we let *t*0 = 0, then *v*(*t*) is simply the step v oltage *V*0 *u*(*t*). A voltage source of *V*0 *u*(*t*) is shown in Fig. 7.25(a); its equivalent circuit is shown in Fig. 7.25(b). It is e vident in Fig. 7.25(b) that terminals *a-b* are shortcircuited (*v* = 0) for *t* < 0 and that *v* = *V*0 appears at the terminals for
-
-V0
-
-= +
-
-**Figure 7.25** (a) Voltage source of *V*0 *u*(*t*), (b) its equivalent circuit.
-
-‒ +
-
-b
-
-a
-
-(a)
-
-‒
-
-V0u(t)
-
-Alternatively, we may derive Eqs. (7.25) and (7.26) from Eq. (7.24) by writing u [f (t)] = 1, f (t) > 0, where f(t) may be t − t0 or t + t0.
-
-**Figure 7.24**
-
-The unit step function.
-
-(a) The unit step function delayed by *t*0, (b) the unit step advanced by *t*0.
-
-*t* > 0. Similarly, a current source of *I*0 *u*(*t*) is shown in Fig. 7.26(a), while its equivalent circuit is in Fig. 7.26(b). Notice that for *t* < 0, there is an open circuit (*i* = 0), and that *i* = *I*0 flows for *t* > 0.
-
-**Figure 7.26**
-
-(a) Current source of *I*0*u*(*t*), (b) its equivalent circuit.
-
-The derivative of the unit step function *u*(*t*) is the *unit impulse function δ*(*t*), which we write as
-
-$$
-\delta(t) = \frac{d}{dt}u(t) = \begin{cases} 0, & t < 0 \\ \text{Undefined}, & t = 0 \\ 0, & t > 0 \end{cases} \tag{7.29}
-$$
-
-The unit impulse function—also known as the *delta* function—is shown in Fig. 7.27.
-
-**Figure 7.27** 0 t *δ*(t) (∞)
-
-The unit impulse function.
-
-The unit impulse function *δ*(t) is zero everywhere except at t = 0, where it is undefined.
-
-Impulsive currents and v oltages occur in electric circuits as a result of switching operations or impulsi ve sources. Although the unit im pulse function is not ph ysically realizable ( just like ideal sources, ideal resistors, etc.), it is a very useful mathematical tool.
-
-The unit impulse may be regarded as an applied or resulting shock. It may be visualized as a very short duration pulse of unit area. This may be expressed mathematically as
-
-$$
-\int_{0^{-}}^{0^{+}} \delta(t) dt = 1 \tag{7.30}
-$$
-
-where *t* = 0− denotes the time just before *t* = 0 and *t* = 0 + is the time just after *t* = 0. For this reason, it is customary to write 1 (denoting unit area) beside the arrow that is used to symbolize the unit impulse function, as in Fig. 7.27. The unit area is known as the *strength* of the impulse function. When an impulse function has a strength other than unity , the area of the impulse is equal to its strength. F or example, an impulse func tion 10*δ*(*t*) has an area of 10. Figure 7.28 sho ws the impulse functions 5*δ*(*t* + 2),10*δ*(*t*), and −4*δ*(*t* − 3).
-
-To illustrate how the impulse function affects other functions, let us evaluate the integral
-
-$$
-\int_{a}^{b} f(t)\delta(t - t_0)dt
-$$
-\n(7.31)
-
-Three impulse functions.
-
-where *a* < *t*0 < *b*. Since δ(*t* − *t*0) = 0 except at *t* = *t*0, the integrand is zero except at *t*0. Thus,
-
-$$
-\int_{a}^{b} f(t)\delta(t - t_0) dt = \int_{a}^{b} f(t_0)\delta(t - t_0) dt
-$$
-$$
-= f(t_0) \int_{a}^{b} \delta(t - t_0) dt = f(t_0)
-$$
-
-or
-
-$$
-\int_{a}^{b} f(t)\delta(t - t_0) dt = f(t_0)
-$$
-\n(7.32)
-
-This shows that when a function is integrated with the impulse function, we obtain the value of the function at the point where the impulse occurs. This is a highly useful property of the impulse function kno wn as the *sampling* or *sifting* property. The special case of Eq. (7.31) is for *t*0 = 0. Then Eq. (7.32) becomes
-
-$$
-\int_{0^{-}}^{0^{+}} f(t) \delta(t) dt = f(0)
-$$
-\n(7.33)
-
-Integrating the unit step function *u*(*t*) results in the *unit ramp function r*(*t*); we write
-
-$$
-r(t) = \int_{-\infty}^{t} u(\lambda)d\lambda = tu(t)
-$$
- (7.34)
-
-$$
-\overline{\text{or}}
-$$
-
-$$
-r(t) = \begin{cases} 0, & t \le 0 \\ t, & t \ge 0 \end{cases} \tag{7.35}
-$$
-
-The unit ramp function is zero for negative values of t and has a unit slope for positive values of t.
-
-Figure 7.29 shows the unit ramp function. In general, a ramp is a func tion that changes at a constant rate.
-
-The unit ramp function may be delayed or adv anced as sho wn in Fig. 7.30. For the delayed unit ramp function,
-
-$$
-r(t - t_0) = \begin{cases} 0, & t \le t_0 \\ t - t_0, & t \ge t_0 \end{cases}
-$$
- (7.36)
-
-and for the advanced unit ramp function,
-
-$$
-r(t + t_0) = \begin{cases} 0, & t \le -t_0 \\ t + t_0, & t \ge -t_0 \end{cases}
-$$
-(7.37)
-
-We should keep in mind that the three singularity functions (impulse, step, and ramp) are related by differentiation as
-
-$$
-\delta(t) = \frac{du(t)}{dt}, \qquad u(t) = \frac{dr(t)}{dt}
-$$
- (7.38)
-
-The unit ramp function: (a) delayed by *t*0, (b) advanced by *t*0.
-
-1
-
-r(t ‒ t0)
-
-or by integration as
-
-$$
-u(t) = \int_{-\infty}^{t} \delta(\lambda) d\lambda, \quad r(t) = \int_{-\infty}^{t} u(\lambda) d\lambda
-$$
- (7.39)
-
-Although there are man y more singularity functions, we are only inter ested in these three (the impulse function, the unit step function, and the ramp function) at this point.
-
-Express the voltage pulse in Fig. 7.31 in terms of the unit step. Calculate Example 7.6 its derivative and sketch it.
-
-# **Solution:**
-
-The type of pulse in Fig. 7.31 is called the *gate function*. It may be regarded as a step function that switches on at one value of *t* and switches off at another value of *t*. The gate function shown in Fig. 7.31 switches on at *t* = 2 s and switches off at *t* = 5 s. It consists of the sum of two unit step functions as shown in Fig. 7.32(a). From the figure, it is evident that
-
-$$
-v(t) = 10u(t - 2) - 10u(t - 5) = 10[u(t - 2) - u(t - 5)]
-$$
-
-Taking the derivative of this gives
-
-$$
-\frac{dv}{dt} = 10[\delta(t-2) - \delta(t-5)]
-$$
-
-which is shown in Fig. 7.32(b). We can obtain Fig. 7.32(b) directly from Fig. 7.31 by simply observing that there is a sudden increase by 10 V at *t* = 2 s leading to 10 *δ*(*t* − 2). At *t* = 5 s, there is a sudden decrease by 10 V leading to −10V *δ*(*t* − 5).
-
-Gate functions are used along with switches to pass or block another signal.
-
-For Example 7.6.
-
-# Practice Problem 7.6
-
-Express the current pulse in Fig. 7.33 in terms of the unit step. Find its integral and sketch it.
-
-**Answer:** 10[*u* (*t*) − 2*u* (*t* − 2) + *u* (*t* − 4)] A, 10[*r* (*t*) − 2*r* (*t* − 2) + *r*(*t* − 4)] amp-sec. See Fig. 7.34.
-
-Example 7.7 Express the *sawtooth* function shown in Fig. 7.35 in terms of singularity functions.
-
-# **Solution:**
-
-There are three ways of solving this problem. The first method is by mere observation of the given function, while the other methods involve some graphical manipulations of the function.
-
-■ **METHOD 1** By looking at the sketch of *v*(*t*) in Fig. 7.35, it is not hard to notice that the given function *v*(*t*) is a combination of singularity functions. So we let
-
-$$
-v(t) = v_1(t) + v_2(t) + \cdots
-$$
- (7.7.1)
-
-The function *v*1(*t*) is the ramp function of slope 5, shown in Fig. 7.36(a); that is,
-
-$$
-v_1(t) = 5r(t) \tag{7.7.2}
-$$
-
-Partial decomposition of *v*(*t*) in Fig. 7.35.
-
-Since *v*1(*t*) goes to infinity, we need another function at *t* = 2s in order to get *v*(*t*). We let this function be *v*2, which is a ramp function of slope −5, as shown in Fig. 7.36(b); that is,
-
-$$
-v_2(t) = -5r(t-2)
-$$
-\n(7.7.3)
-
-Adding *v*1 and *v*2 gives us the signal in Fig. 7.36(c). Obviously, this is not the same as *v*(*t*) in Fig. 7.35. But the difference is simply a constant 10 units for *t* > 2 s. By adding a third signal *v*3, where
-
-$$
-v_3 = -10u(t - 2) \tag{7.7.4}
-$$
-
-we get *v*(*t*), as shown in Fig. 7.37. Substituting Eqs. (7.7.2) through (7.7.4) into Eq. (7.7.1) gives
-
-Complete decomposition of *v*(*t*) in Fig. 7.35.
-
-■ **METHOD 2** A close observation of Fig. 7.35 reveals that *v*(*t*) is a multiplication of two functions: a ramp function and a gate function. Thus,
-
-$$
-v(t) = 5t[u(t) - u(t-2)]
-$$
-
-= 5tu(t) - 5tu(t - 2)
-= 5r(t) - 5(t - 2 + 2)u(t - 2)
-= 5r(t) - 5(t - 2)u(t - 2) - 10u(t - 2)
-= 5r(t) - 5r(t - 2) - 10u(t - 2)
-
-the same as before.
-
-■ **METHOD 3** This method is similar to Method 2. We observe from Fig. 7.35 that *v*(*t*) is a multiplication of a ramp function and a unit step function, as shown in Fig. 7.38. Thus,
-
-$$
-v(t) = 5r(t)u(-t+2)
-$$
-
-If we replace *u*(−*t*) by 1 −*u*(*t*), then we can replace *u*(−*t* + 2) by 1 −*u*(*t* − 2). Hence,
-
-$$
-v(t) = 5r(t)[1 - u(t - 2)]
-$$
-
-which can be simplified as in Method 2 to get the same result.
-
-Decomposition of *v*(*t*) in Fig. 7.35.
-
-For Practice Prob. 7.7.
-
-Example 7.8 Given the signal
-
-| | 3, | t < 0 |
-|--------|-----------------|--------------|
-| g(t) = | −2, | 0 <
t < 1 |
-| | {
2t
− 4, | t
> 1 |
-
-express *g*(*t*) in terms of step and ramp functions.
-
-# **Solution:**
-
-The signal *g*(*t*) may be regarded as the sum of three functions specified within the three intervals *t* < 0, 0 < *t* < 1, and *t* > 1.
-
- For *t* < 0, *g*(*t*) may be regarded as 3 multiplied by *u*(−*t*), where *u*(−*t*) = 1 for *t* < 0 and 0 for *t* > 0. Within the time interval 0 < *t* < 1, the function may be considered as −2 multiplied by a gated function [*u*(*t*) − *u*(*t* − 1)]. For *t* > 1, the function may be regarded as 2*t* − 4 multiplied by the unit step function *u*(*t* − 1). Thus,
-
-$$
-g(t) = 3u(-t) - 2[u(t) - u(t-1)] + (2t - 4)u(t-1)
-$$
-
-= 3u(-t) - 2u(t) + (2t - 4 + 2)u(t-1)
-= 3u(-t) - 2u(t) + 2(t-1)u(t-1)
-= 3u(-t) - 2u(t) + 2r(t-1)
-
-One may avoid the trouble of using *u*(−*t*) by replacing it with 1 − *u*(*t*). Then
-
-$$
-g(t) = 3[1 - u(t)] - 2u(t) + 2r(t - 1) = 3 - 5u(t) + 2r(t - 1)
-$$
-
-Alternatively, we may plot *g*(*t*) and apply Method 1 from Example 7.7.
-
-If
-$$
-h(t) = \begin{cases} 0, & t < 0 \\ -4, & 0 < t < 2 \\ 3t - 8, & 2 < t < 6 \\ 0, & t > 6 \end{cases}
-$$
-Practice Problem 7.8
-
-express *h*(*t*) in terms of the singularity functions.
-
-**Answer:**
-$$
--4u(t) + 2u(t-2) + 3r(t-2) - 10u(t-6) - 3r(t-6)
-$$
-.
-
-Evaluate the following integrals involving the impulse function: Example 7.9
-
-$$
-\int_0^{10} (t^2 + 4t - 2)\delta(t - 2) dt
-$$
-
-$$
-\int_{-\infty}^{\infty} [\delta(t - 1)e^{-t} \cos t + \delta(t + 1)e^{-t} \sin t] dt
-$$
-
-# **Solution:**
-
-For the first integral, we apply the sifting property in Eq. (7.32).
-
-$$
-\int_0^{10} (t^2 + 4t - 2) \delta(t - 2) dt = (t^2 + 4t - 2)|_{t=2} = 4 + 8 - 2 = 10
-$$
-
-Similarly, for the second integral,
-
-$$
-\int_{-\infty}^{\infty} \left[ \delta(t-1)e^{-t} \cos t + \delta(t+1)e^{-t} \sin t \right] dt
-$$
-
-= $e^{-t} \cos t|_{t=1} + e^{-t} \sin t|_{t=-1}$
-= $e^{-t} \cos 1 + e^{t} \sin (-1) = 0.1988 - 2.2873 = -2.0885$
-
-Evaluate the following integrals:
-
-$$
-\int_{-\infty}^{\infty} (t^3 + 5t^2 + 10)\delta(t+3) \, dt, \qquad \int_{0}^{10} \delta(t-\pi) \cos 3t \, dt
-$$
-
-**Answer:** 28, −1.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/080_7.5 Step Response of an RC Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/080_7.5 Step Response of an RC Circuit.md
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index c65e76bcc95485bd593983344c6df65ba4514015..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/080_7.5 Step Response of an RC Circuit.md
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@@ -1,385 +0,0 @@
-# **7.5** Step Response of an RC Circuit
-
-When the dc source of an *RC* circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a *step response*.
-
-The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.
-
-Practice Problem 7.9
-
-The step response is the response of the circuit due to a sudden applica tion of a dc voltage or current source.
-
-Consider the *RC* circuit in Fig. 7.40(a) which can be replaced by the circuit in Fig. 7.40(b), where *Vs* is a constant dc v oltage source. Again, we select the capacitor voltage as the circuit response to be determined. We assume an initial v oltage *V*0 on the capacitor , although this is not necessary for the step response. Since the v oltage of a capacitor cannot change instantaneously,
-
-$$
-v(0^{-}) = v(0^{+}) = V_0 \tag{7.40}
-$$
-
-where *v*(0−) is the voltage across the capacitor just before switching and *v*( 0 +) is its voltage immediately after switching. Applying KCL, we have
-
-$$
-C\frac{dv}{dt} + \frac{v - V_s u(t)}{R} = 0
-$$
-
-$$
-\frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC}u(t)
-$$
-\n(7.41)
-
-where *v* is the voltage across the capacitor. For *t* > 0, Eq. (7.41) becomes
-
-$$
-\frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC}
-$$
- (7.42)
-
-Rearranging terms gives
-
-$$
-\frac{dv}{dt} = -\frac{v - V_s}{RC}
-$$
-
-\_\_\_\_\_\_ *dv v*− *Vs* = − \_\_\_ *dt RC* **(7.43)**
-
-Integrating both sides and introducing the initial conditions,
-
-$$
-\ln(v - V_s) \Big|_{V_0}^{v(t)} = -\frac{t}{RC} \Big|_{0}^{t}
-$$
-$$
-\ln(v(t) - V_s) - \ln(V_0 - V_s) = -\frac{t}{RC} + 0
-$$
-
-or
-
-or
-
-or
-
-$$
-\ln \frac{v - V_s}{V_0 - V_s} = -\frac{t}{RC}
-$$
-\n(7.44)
-
-Taking the exponential of both sides
-
-$$
-\frac{v - V_s}{V_0 - V_s} = e^{-t/\tau}, \qquad \tau = RC
-$$
-$$
-v - V_s = (V_0 - V_s)e^{-t/\tau}
-$$
-
-or
-
-$$
-v(t) = V_s + (V_0 - V_s)e^{-t/\tau}, \qquad t > 0 \tag{7.45}
-$$
-
-Thus,
-
-$$
-v(t) = \begin{cases} V_0, & t < 0 \\ V_s + (V_0 - V_s)e^{-t/\tau}, & t > 0 \end{cases}
-$$
- (7.46)
-
-This is known as the *complete response* (or total response) of the *RC* circuit to a sudden application of a dc voltage source, assuming the capacitor is initially charged. The reason for the term "complete" will become evident a little later . Assuming that *Vs* > *V*0, a plot of *v*(*t*) is sho wn in Fig. 7.41.
-
-If we assume that the capacitor is uncharged initially, we set *V*0 = 0 in Eq. (7.46) so that
-
-$$
-v(t) = \begin{cases} 0, & t < 0 \\ V_s(1 - e^{-t/\tau}), & t > 0 \end{cases}
-$$
-(7.47)
-
-which can be written alternatively as
-
-$$
-v(t) = V_s(1 - e^{-t/\tau})u(t)
-$$
-\n(7.48)
-
-This is the complete step response of the *RC* circuit when the capacitor is initially uncharged. The current through the capacitor is obtained from Eq. (7.47) using *i*(*t*) = *C dv ∕ dt*. We get
-
-$$
-i(t) = C\frac{dv}{dt} = \frac{C}{\tau}V_s e^{-t/\tau}, \qquad \tau = RC, \qquad t > 0
-$$
-$$
-i(t) = \frac{V_s}{R} e^{-t/\tau} u(t) \tag{7.49}
-$$
-
-Figure 7.42 shows the plots of capacitor v oltage *v*(*t*) and capacitor current *i*(*t*).
-
-*R*
-
-Rather than going through the derivations above, there is a systematic approach—or rather, a shortcut method—for finding the step response of an *RC* or *RL* circuit. Let us reexamine Eq. (7.45), which is more general than Eq. (7.48). It is evident that *v*(*t*) has two components. Classically there are two ways of decomposing this into tw o components. The first is to break it into a "natural response and a forced response'' and the second is to break it into a "transient response and a steady-state response.'' Starting with the natural response and forced response, we write the total or complete response as
-
-> Complete response = natural response + forced response stored energy independent source
-
-> > *v* = *vn* + *vf* **(7.50)**
-
-or
-
-or
-
-where
-
-and
-
-$$
-v_f = V_s(1 - e^{-t/\tau})
-$$
-
-*vn* = *Vo e*−*t*∕*τ*
-
-We are familiar with the natural response *vn* of the circuit, as discussed in Section 7.2. *vf* is known as the *forced* response because it is produced by the circuit when an external "force'' (a voltage source in this case) is applied. It represents what the circuit is forced to do by the input excitation. The natural response e ventually dies out along with the transient component of the forced response, leaving only the steady-state component of the forced response.
-
-Step response of an *RC* circuit with initially uncharged capacitor: (a) voltage response, (b) current response.
-
-Response of an *RC* circuit with initially charged capacitor.
-
-## **274** Chapter 7 First-Order Circuits
-
-Another way of looking at the complete response is to break into two components—one temporary and the other permanent, that is
-
-Complete response = transient response + steady-state response temporary part permanent part
-
-or
-
-$$
-v = v_t + v_{ss} \tag{7.51}
-$$
-
-where
-
-$$
-v_t = (V_o - V_s)e^{-t/\tau}
-$$
- (7.52a)
-
-and
-
-$$
-v_{ss} = V_s \tag{7.52b}
-$$
-
-The *transient response vt* is temporary; it is the portion of the complete response that decays to zero as time approaches infinity. Thus,
-
-The transient response is the circuit's temporary response that will die out with time.
-
-The *steady-state response vss* is the portion of the complete response that remains after the transient reponse has died out. Thus,
-
-The steady-state response is the behavior of the circuit a long time after an external excitation is applied.
-
-The first decomposition of the complete response is in terms of the source of the responses, while the second decomposition is in terms of the permanency of the responses. Under certain conditions, the natural response and transient response are the same. The same can be said about the forced response and steady-state response.
-
-Whichever way we look at it, the complete response in Eq. (7.45) may be written as
-
-$$
-v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}
-$$
- (7.53)
-
-where *v*(0) is the initial voltage at *t* = 0 + and *v*(∞) is the final or steadystate value. Thus, to find the step response of an *RC* circuit requires three things:
-
-- 1. The initial capacitor voltage *v*(0).
-- 2. The final capacitor voltage *v*(∞).
-- 3. The time constant *τ*.
-
-Once we know x (0), x (∞), and *τ*, almost all the circuit problems in this chapter can be solved using the formula
-
-*x*(*t*) = *x*(∞) + [*x*(0) − *x*(∞)] *e*−*t*∕*τ*
-
-We obtain item 1 from the given circuit for *t* < 0 and items 2 and 3 from the circuit for *t* > 0. Once these items are determined, we obtain the
-
-This is the same as saying that the complete response is the sum of the transient response and the steady-state response.
-
-response using Eq. (7.53). This technique equally applies to *RL* circuits, as we shall see in the next section.
-
-Note that if the switch changes position at time *t* = *t*0 instead of at *t* = 0, there is a time delay in the response so that Eq. (7.53) becomes
-
-$$
-v(t) = v(\infty) + [v(t_0) - v(\infty)]e^{-(t - t_0)/\tau}
-$$
- (7.54)
-
-where *v*(*t*0) is the initial value at *t* = *t*0 + . Keep in mind that Eq. (7.53) or (7.54) applies only to step responses, that is, when the input e xcitation is constant.
-
-The switch in Fig. 7.43 has been in position *A* for a long time. At *t* = 0, the switch moves to *B*. Determine *v*(*t*) for *t* > 0 and calculate its value at *t* = 1 and 4 s.
-
-# **Solution:**
-
-For *t* < 0, the switch is at position *A*. The capacitor acts like an open circuit to dc, but *v* is the same as the voltage across the 5-k Ω resistor. Hence, the voltage across the capacitor just before *t* = 0 is obtained by voltage division as
-
-$$
-v(0^{-}) = \frac{5}{5+3}(24) = 15 \text{ V}
-$$
-
-Using the fact that the capacitor voltage cannot change instantaneously,
-
-$$
-v(0) = v(0^-) = v(0^+) = 15
-$$
- V
-
-For *t* > 0, the switch is in position *B*. The Thevenin resistance connected to the capacitor is *R*Th = 4 kΩ, and the time constant is
-
-$$
-\tau = R_{\text{Th}} C = 4 \times 10^3 \times 0.5 \times 10^{-3} = 2 \text{ s}
-$$
-
-Since the capacitor acts like an open circuit to dc at steady state, *v*(∞) = 30 V. Thus,
-
-$$
-v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}
-$$
-
-= 30 + (15 - 30)e^{-t/2} = (30 - 15e^{-0.5t}) V
-
-At *t* = 1,
-
-$$
-v(1) = 30 - 15e^{-0.5} = 20.9 \text{ V}
-$$
-
-At *t* = 4,
-
-$$
-v(4) = 30 - 15e^{-2} = 27.97
-$$
- V
-
-Example 7.10
-
-# Practice Problem 7.10
-
-Find *v*(*t*) for *t* > 0 in the circuit of Fig. 7.44. Assume the switch has been open for a long time and is closed at *t* = 0. Calculate *v*(*t*) at *t* = 0.5.
-
-**Answer:** (9.375 + 5.625*e*−2*t* ) V for all *t* > 0, 11.444 V.
-
-Example 7.11 In Fig. 7.45, the switch has been closed for a long time and is opened at *t* = 0. Find *i* and *v* for all time.
-
-# **Solution:**
-
-The resistor current *i* can be discontinuous at *t* = 0, while the capacitor voltage *v* cannot. Hence, it is always better to find *v* and then obtain *i* from *v*.
-
-By definition of the unit step function,
-
-$$
-30u(t) = \begin{cases} 0, & t < 0\\ 30, & t > 0 \end{cases}
-$$
-
-For *t* < 0, the switch is closed and 30 *u*(*t*) = 0, so that the 30*u*(*t*) voltage source is replaced by a short circuit and should be regarded as contributing nothing to *v*. Since the switch has been closed for a long time, the capacitor voltage has reached steady state and the capacitor acts like an open circuit. Hence, the circuit becomes that shown in Fig. 7.46(a) for *t* < 0. From this circuit we obtain
-
-$$
-v = 10 \text{ V}, \qquad i = -\frac{v}{10} = -1 \text{ A}
-$$
-
-Since the capacitor voltage cannot change instantaneously,
-
-$$
-v(0) = v(0^-) = 10 \text{ V}
-$$
-
- For *t* > 0, the switch is opened and the 10-V voltage source is disconnected from the circuit. The 30 *u*(*t*) voltage source is now operative, so the circuit becomes that shown in Fig. 7.46(b). After a long time, the circuit reaches steady state and the capacitor acts like an open circuit again. We obtain *v*(∞) by using voltage division, writing
-
-$$
-v(\infty) = \frac{20}{20 + 10}(30) = 20 \text{ V}
-$$
-
-**Figure 7.46** Solution of Example 7.11: (a) for *t* < 0, (b) for *t* > 0.
-
-The Thevenin resistance at the capacitor terminals is
-
-$$
-R_{\text{Th}} = 10||20 = \frac{10 \times 20}{30} = \frac{20}{3}\Omega
-$$
-
-and the time constant is
-
-$$
-\tau = R_{\text{Th}} C = \frac{20}{3} \cdot \frac{1}{4} = \frac{5}{3} \text{ s}
-$$
-
-Thus,
-
-$$
-v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}
-$$
-
-= 20 + (10 - 20)e^{-(3/5)t} = (20 - 10e^{-0.6t}) V
-
-To obtain *i*, we notice from Fig. 7.46(b) that *i* is the sum of the currents through the 20-Ω resistor and the capacitor; that is,
-
-$$
-i = \frac{v}{20} + C\frac{dv}{dt}
-$$
-
-= 1 - 0.5e-0.6t + 0.25(-0.6)(-10)e-0.6t = (1 + e-0.6t) A
-
-Notice from Fig. 7.46(b) that *v* + 10*i* = 30 is satisfied, as expected. Hence,
-
-$$
-v = \begin{cases} 10 \text{ V}, & t < 0 \\ (20 - 10e^{-0.6t}) \text{ V}, & t \ge 0 \end{cases}
-$$
-$$
-i = \begin{cases} -1 \text{ A}, & t < 0 \\ (1 + e^{-0.6t}) \text{ A}, & t > 0 \end{cases}
-$$
-
-Notice that the capacitor voltage is continuous while the resistor current is not.
-
-The switch in Fig. 7.47 is closed at *t* = 0. Find *i*(*t*) and *v*(*t*) for all time. Note that *u*(−*t*) = 1 for *t* < 0 and 0 for *t* > 0. Also, *u*(−*t*) = 1 − *u*(*t*). Practice Problem 7.11
-
-
-
-# **7.6** Step Response of an RL Circuit
-
-Consider the *RL* circuit in Fig. 7.48(a), which may be replaced by the circuit in Fig. 7.48(b). Again, our goal is to find the inductor current *i* as the circuit response. Rather than apply Kirchhoff's laws, we will use the simple technique in Eqs. (7.50) through (7.53). Let the response be the sum of the transient response and the steady-state response,
-
-$$
-i = i_t + i_{ss} \tag{7.55}
-$$
-
-We know that the transient response is al ways a decaying e xponential, that is,
-
-$$
-i_t = Ae^{-t/\tau}, \qquad \tau = \frac{L}{R} \tag{7.56}
-$$
-
-where *A* is a constant to be determined.
-
-The steady-state response is the value of the current a long time after the switch in Fig. 7.48(a) is closed. We know that the transient response essentially dies out after five time constants. At that time, the inductor becomes a short circuit, and the v oltage across it is zero. The entire source v oltage *Vs* appears across *R*. Thus, the steady-state response is
-
-$$
-i_{ss} = \frac{V_s}{R}
-$$
- (7.57)
-
-Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) gives
-
-$$
-i = Ae^{-t/\tau} + \frac{V_s}{R}
-$$
-\n
-$$
-(7.58)
-$$
-
-We now determine the constant *A* from the initial v alue of *i*. Let *I*0 be the initial current through the inductor, which may come from a source other than *Vs*. Since the current through the inductor cannot change instantaneously,
-
-$$
-i(0^+) = i(0^-) = I_0 \tag{7.59}
-$$
-
-Thus, at *t* = 0, Eq. (7.58) becomes
-
-$$
-I_0 = A + \frac{V_s}{R}
-$$
-
-From this, we obtain *A* as
-
-$$
-A = I_0 - \frac{V_s}{R}
-$$
-
-Substituting for *A* in Eq. (7.58), we get
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/081_7.6 Step Response of an RL Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/081_7.6 Step Response of an RL Circuit.md
deleted file mode 100644
index f1209809aa1a149a763367e11f1a4a0e8d291c0d..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/081_7.6 Step Response of an RL Circuit.md
+++ /dev/null
@@ -1,209 +0,0 @@
-$$
-i(t) = \frac{V_s}{R} + \left(I_0 - \frac{V_s}{R}\right)e^{-t/\tau}
-$$
-\n(7.60)
-
-This is the complete response of the *RL* circuit. It is illustrated in Fig. 7.49. The response in Eq. (7.60) may be written as
-
-$$
-i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}
-$$
- (7.61)
-
-**Figure 7.49** Total response of the *RL* circuit with initial inductor current *I*0.
-
-where *i*(0) and *i*(∞) are the initial and final values of *i*, respectively. Thus, to find the step response of an *RL* circuit requires three things:
-
-- 1. The initial inductor current *i*(0) at *t* = 0. 2. The final inductor current *i*(∞).
-- 3. The time constant *τ*.
-
-We obtain item 1 from the given circuit for *t* < 0 and items 2 and 3 from the circuit for *t* > 0. Once these items are determined, we obtain the response using Eq. (7.61). Keep in mind that this technique applies only for step responses.
-
-Again, if the switching tak es place at time *t* = *t*0 instead of *t* = 0, Eq. (7.61) becomes
-
-$$
-i(t) = i(\infty) + [i(t_0) - i(\infty)]e^{-(t-t_0)/\tau}
-$$
-\n(7.62)
-
-If *I*0 = 0, then
-
-$$
-i(t) = \begin{cases} 0, & t < 0 \\ \frac{V_s}{R}(1 - e^{-t/\tau}), & t > 0 \end{cases}
-$$
-(7.63a)
-
-or
-
-$$
-i(t) = \frac{V_s}{R} (1 - e^{-t/\tau}) u(t)
-$$
-\n(7.63b)
-
-This is the step response of the *RL* circuit with no initial inductorcurrent. The v oltage across the inductor is obtained from Eq. (7.63) using *v* = *L di*∕*dt*. We get
-
-$$
-v(t) = L\frac{di}{dt} = V_s \frac{L}{\tau R} e^{-t/\tau}, \qquad \tau = \frac{L}{R}, \qquad t > 0
-$$
-
-or
-
-$$
-v(t) = V_s e^{-t/\tau} u(t) \tag{7.64}
-$$
-
-Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64).
-
-Step responses of an *RL* circuit with no initial inductor current: (a) current response, (b) voltage response.
-
-For Example 7.12.
-
-Find *i*(*t*) in the circuit of Fig. 7.51 for *t* > 0. Assume that the switch has been closed for a long time.
-
-# **Solution:**
-
-When *t* < 0, the 3-Ω resistor is short-circuited, and the inductor acts like a short circuit. The current through the inductor at *t* = 0− (i.e., just before *t* = 0) is
-
-$$
-i(0^{-}) = \frac{10}{2} = 5 \text{ A}
-$$
-
-Because the inductor current cannot change instantaneously,
-
-$$
-i(0) = i(0^+) = i(0^-) = 5 \text{ A}
-$$
-
-When *t* > 0, the switch is open. The 2- and 3- Ω resistors are in series, so that
-
-$$
-i(\infty) = \frac{10}{2+3} = 2 \text{ A}
-$$
-
-The Thevenin resistance across the inductor terminals is
-
-$$
-R_{\rm Th} = 2 + 3 = 5 \ \Omega
-$$
-
-For the time constant,
-
-*τ* = \_\_\_*L R*Th = \_1 3 \_\_ 5 =\_\_\_1 15 s
-
-Thus,
-
-$$
-i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}
-$$
-
-= 2 + (5 - 2)e-15t = 2 + 3e-15t A, t > 0
-
-*Check:* In Fig. 7.51, for *t* > 0, KVL must be satisfied; that is,
-
-$$
-10 = 5i + L\frac{di}{dt}
-$$
-
-$$
-5i + L\frac{di}{dt} = [10 + 15e^{-15t}] + \left[\frac{1}{3}(3)(-15)e^{-15t}\right] = 10
-$$
-
-This confirms the result.
-
-At *t* = 0, switch 1 in Fig. 7.53 is closed, and switch 2 is closed 4 s later. Example 7.13 Find *i*(*t*) for *t* > 0. Calculate *i* for *t* = 2 s and *t* = 5 s.
-
-**Figure 7.53** For Example 7.13.
-
-# **Solution:**
-
-We need to consider the three time intervals *t* ≤ 0, 0 ≤ *t* ≤ 4, and *t* ≥ 4 separately. For *t* < 0, switches *S*1 and *S*2 are open so that *i* = 0. Since the inductor current cannot change instantly,
-
-$$
-i(0^-) = i(0) = i(0^+) = 0
-$$
-
-For 0 ≤ *t* ≤ 4, *S*1 is closed so that the 4- and 6- Ω resistors are in series. (Remember, at this time, *S*2 is still open.) Hence, assuming for now that *S*1 is closed forever,
-
-$$
-i(\infty) = \frac{40}{4+6} = 4 \text{ A}, \qquad R_{\text{Th}} = 4+6 = 10 \text{ }\Omega
-$$
-$$
-\tau = \frac{L}{R_{\text{Th}}} = \frac{5}{10} = \frac{1}{2} \text{ s}
-$$
-
-Thus,
-
-$$
-i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}
-$$
-
-= 4 + (0 - 4)e-2t = 4(1 - e-2t) A, 0 \le t \le 4
-
- For *t* ≥ 4, *S*2 is closed; the 10-V voltage source is connected, and the circuit changes. This sudden change does not affect the inductor current because the current cannot change abruptly. Thus, the initial current is
-
-$$
-i(4) = i(4^-) = 4(1 - e^{-8}) \simeq 4 \text{ A}
-$$
-
-To find *i*(∞), let *v* be the voltage at node *P* in Fig. 7.53. Using KCL,
-
-$$
-\frac{40 - v}{4} + \frac{10 - v}{2} = \frac{v}{6} \implies v = \frac{180}{11} \text{ V}
-$$
-$$
-i(\infty) = \frac{v}{6} = \frac{30}{11} = 2.727 \text{ A}
-$$
-
-The Thevenin resistance at the inductor terminals is
-
-$$
-R_{\text{Th}} = 4||2 + 6 = \frac{4 \times 2}{6} + 6 = \frac{22}{3} \,\Omega
-$$
-
-and
-
-$$
-\tau = \frac{L}{R_{\text{Th}}} = \frac{5}{\frac{22}{3}} = \frac{15}{22} \text{ s}
-$$
-
-Hence,
-
-$$
-i(t) = i(\infty) + [i(4) - i(\infty)]e^{-(t-4)/\tau}, \qquad t \ge 4
-$$
-
-We need (*t* − 4) in the exponential because of the time delay. Thus,
-
-$$
-i(t) = 2.727 + (4 - 2.727)e^{-(t-4)/\tau}, \qquad \tau = \frac{15}{22}
-$$
-$$
-= 2.727 + 1.273e^{-1.4667(t-4)}, \qquad t \ge 4
-$$
-
-Putting all this together,
-
-$$
-i(t) = \begin{cases} 0, & t \le 0 \\ 4(1 - e^{-2t}), & 0 \le t \le 4 \\ 2.727 + 1.273e^{-1.4667(t-4)}, & t \ge 4 \end{cases}
-$$
-
-At *t* = 2,
-
-At *t* = 5,
-
-$$
-i(2) = 4(1 - e^{-4}) = 3.93 \text{ A}
-$$
-
-$$
-i(5) = 2.727 + 1.273e^{-1.4667} = 3.02
-$$
- A
-
-Switch *S*1 in Fig. 7.54 is closed at *t* = 0, and switch *S*2 is closed at *t* = 2s.
-
-**Answer:** *i*(*t*) = { 0, 4(1 − *e*−9*t* ), 7.2 − 3.2 *e*−5(*t*−2) , *t* < 0 0 < *t* < 2 *t* > 2 *i*(1) = 4 A, *i*(3) = 7.178 A.
-
-Calculate *i*(*t*) for all *t*. Find *i*(1) and *i*(3).
-
-# **Figure 7.54** For Practice Prob. 7.13.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/082_7.7 First-Order Op Amp Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/082_7.7 First-Order Op Amp Circuits.md
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-# **7.7** † First-Order Op Amp Circuits
-
-An op amp circuit containing a storage element will exhibit first-order behavior. Differentiators and integrators treated in Section 6.6 are examples of first-order op amp circuits. Again, for practical reasons, inductors are hardly ever used in op amp circuits; therefore, the op amp circuits we consider here are of the *RC* type.
-
-As usual, we analyze op amp circuits using nodal analysis. Some times, the Thevenin equivalent circuit is used to reduce the op amp circuit to one that we can easily handle. The following three examples illustrate the concepts. The first one deals with a source-free op amp circuit, while the other two involve step responses. The three examples have been carefully selected to co ver all possible *RC* types of op amp circuits, depending on the location of the capacitor with respect to the op amp; that is, the capacitor can be located in the input, the output, or the feedback loop.
-
-For the op amp circuit in Fig. 7.55(a), find *vo* for *t* > 0, gi ven that Example 7.14 *v*(0) = 3 V. Let *Rf* = 80 kΩ, *R*1 = 20 kΩ, and *C* = 5 *µ*F.
-
-For Example 7.14.
-
-# **Solution:**
-
-This problem can be solved in two ways:
-
-■ **METHOD 1** Consider the circuit in Fig. 7.55(a). Let us derive the appropriate differential equation using nodal analysis. If *v*1 is the voltage at node 1, at that node, KCL gives
-
-$$
-\frac{0 - v_1}{R_1} = C \frac{dv}{dt}
-$$
- (7.14.1)
-
-Because nodes 2 and 3 must be at the same potential, the potential at node 2 is zero. Thus, *v*1 − 0 = *v* or *v*1 = *v* and Eq. (7.14.1) becomes
-
-$$
-\frac{dv}{dt} + \frac{v}{CR_1} = 0
-$$
- (7.14.2)
-
-This is similar to Eq. (7.4b) so that the solution is obtained the same way as in Section 7.2, i.e.,
-
-$$
-v(t) = V_0 e^{-t/\tau}, \qquad \tau = R_1 C \tag{7.14.3}
-$$
-
-where *V*0 is the initial voltage across the capacitor. But *v*(0) = 3 = *V*0 and *τ* = 20 × 103 × 5 × 10*−*6 = 0*.*1*.* Hence,
-
-$$
-v(t) = 3e^{-10t}
-$$
- (7.14.4)
-
-Applying KCL at node 2 gives
-
-$$
-C\frac{dv}{dt} = \frac{0 - v_o}{R_f}
-$$
-
-or
-
-$$
-v_o = -R_f C \frac{dv}{dt} \tag{7.14.5}
-$$
-
-Now we can find *v*0 as
-
-$$
-v_o = -80 \times 10^3 \times 5 \times 10^{-6} (-30e^{-10t}) = 12e^{-10t} \text{ V}, \qquad t > 0
-$$
-
-■ **METHOD 2** Let us apply the shortcut method from Eq. (7.53). We need to find *vo*( 0+)*, vo*(∞)*,* and *τ.* Since *v*( 0+) = *v*(0*−*) = 3 V*,* we apply KCL at node 2 in the circuit of Fig. 7.55(b) to obtain
-
-$$
-\frac{3}{20,000} + \frac{0 - v_o(0^+)}{80,000} = 0
-$$
-
-or *vo*( 0+) = 12 V*.* Since the circuit is source free, *v*(∞) = 0 V*.* To find *τ,* we need the equivalent resistance *Req* across the capacitor terminals. If we remove the capacitor and replace it by a 1-A current source, we have the circuit shown in Fig. 7.55(c). Applying KVL to the input loop yields
-
-$$
-20,000(1) - v = 0 \qquad \Rightarrow \qquad v = 20 \text{ kV}
-$$
-
-Then
-
-$$
-R_{\text{eq}} = \frac{v}{1} = 20 \text{ k}\Omega
-$$
-
-and *τ* = *R*eq*C* = 0*.*1*.* Thus,
-
-$$
-v_o(t) = v_o(\infty) + [v_o(0) - v_o(\infty)]e^{-t/\tau}
-$$
-
-= 0 + (12 - 0)e^{-10t} = 12e^{-10t} V, \qquad t > 0
-
-as before.
-
-For the op amp circuit in Fig. 7.56, find *vo* for *t >* 0 if *v*(0) = 4 V*.* Assume that *Rf* = 50 kΩ*, R*1 = 10 kΩ*,* and *C* = 10 *µ*F*.* Practice Problem 7.14
-
-> **Answer:** −4 *e* −2*t* V*, t >* 0*.*
-
-**Figure 7.56** For Practice Prob. 7.14.
-
-Example 7.15 Determine *v*(*t*) and *vo*(*t*) in the circuit of Fig. 7.57.
-
-# **Solution:**
-
-This problem can be solved in two ways, just like the previous example. However, we will apply only the second method. Since what we are looking for is the step response, we can apply Eq. (7.53) and write
-
-$$
-v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}, \qquad t > 0 \tag{7.15.1}
-$$
-
-*v*o +
-
-where we need only find the time constant *τ,* the initial value *v*(0)*,* and the final value *v*(∞). Notice that this applies strictly to the capacitor voltage due a step input. Since no current enters the input terminals of the op amp, the elements on the feedback loop of the op amp constitute an *RC* circuit, with
-
-$$
-\tau = RC = 50 \times 10^3 \times 10^{-6} = 0.05 \tag{7.15.2}
-$$
-
-For *t* < 0*,* the switch is open and there is no voltage across the capacitor. Hence, *v*(0) = 0*.* For *t* > 0*,* we obtain the voltage at node 1 by voltage division as
-
-$$
-v_1 = \frac{20}{20 + 10}3 = 2 \text{ V}
-$$
- (7.15.3)
-
-Since there is no storage element in the input loop, *v*1 remains constant for all *t*. At steady state, the capacitor acts like an open circuit so that the op amp circuit is a noninverting amplifier. Thus,
-
-$$
-v_o(\infty) = \left(1 + \frac{50}{20}\right)v_1 = 3.5 \times 2 = 7 \text{ V}
-$$
- (7.15.4)
-
-But
-
-$$
-v_1 - v_o = v \tag{7.15.5}
-$$
-
-so that
-
-$$
-v(\infty) = 2 - 7 = -5
-$$
- V
-
-Substituting *τ, v*(0)*,* and *v*(∞) into Eq. (7.15.1) gives
-
-$$
-v(t) = -5 + [0 - (-5)]e^{-20t} = 5(e^{-20t} - 1) \text{ V}, \qquad t > 0 \tag{7.15.6}
-$$
-
-From Eqs. (7.15.3), (7.15.5), and (7.15.6), we obtain
-
-$$
-v_o(t) = v_1(t) - v(t) = 7 - 5e^{-20t} \text{ V}, \qquad t > 0 \tag{7.15.7}
-$$
-
-Find *v*(*t*) and *vo*(*t*) in the op amp circuit of Fig. 7.58.
-
-**Answer:** (Note, the v oltage across the capacitor and the output v oltage must be both equal to zero, for *t* < 0, since the input w as zero for all *t* < 0.) 40(1 − *e*−10*t* ) *u*(*t*) mV, 40( *e*−10*t* − 1) *u*(*t*) mV.
-
-Practice Problem 7.15
-
-*v*o +
-
-1 *μ*F
-
-100 kΩ
-
-‒
-
-**Figure 7.58** For Practice Prob. 7.15.
-
-**Figure 7.57** For Example 7.15.
-
-Example 7.16 Find the step response *vo*(*t*) for *t* > 0 in the op amp circuit of Fig. 7.59. Let *vi* = 2*u*(*t*) V, *R*1 = 20 kΩ, *Rf* = 50 kΩ, *R*2 = *R*3 = 10 kΩ, *C* = 2 *µ*F.
-
-# **Solution:**
-
-Notice that the capacitor in Example 7.14 is located in the input loop, while the capacitor in Example 7.15 is located in the feedback loop. In this example, the capacitor is located in the output of the op amp. Again, we can solve this problem directly using nodal analysis. However, using the Thevenin equivalent circuit may simplify the problem.
-
-We temporarily remove the capacitor and find the Thevenin equivalent at its terminals. To obtain *V*Th, consider the circuit in Fig. 7.60(a). Since the circuit is an inverting amplifier,
-
-$$
-V_{ab} = -\frac{R_f}{R_1} v_i
-$$
-
-By voltage division,
-
-# **Figure 7.60**
-
-Obtaining *V*Th and *R*Th across the capacitor in Fig. 7.59.
-
-To obtain *R*Th, consider the circuit in Fig. 7.60(b), where *Ro* is the output resistance of the op amp. Since we are assuming an ideal op amp, *Ro* = 0, and
-
-$$
-R_{\text{Th}} = R_2 \parallel R_3 = \frac{R_2 R_3}{R_2 + R_3}
-$$
-
-Substituting the given numerical values,
-
-$$
-V_{\text{Th}} = -\frac{R_3}{R_2 + R_3} \frac{R_f}{R_1} v_i = -\frac{10}{20} \frac{50}{20} 2u(t) = -2.5u(t)
-$$
-$$
-R_{\text{Th}} = \frac{R_2 R_3}{R_2 + R_3} = 5k\Omega
-$$
-
-$$
-v_o(t) = -2.5(1 - e^{-t/\tau})u(t)
-$$
-
-where *τ* = *R*Th*C* = 5 × 103 × 2 × 10−6 = 0.01. Thus, the step response for *t* > 0 is
-
-$$
-v_o(t) = 2.5(e^{-100t} - 1)u(t)
-$$
- V
-
-**Figure 7.61** Thevenin equivalent circuit of the circuit in Fig. 7.59.
-
-**Answer:** 27(1 − *e*−50*t* )*u*(*t*) V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/083_7.8 Transient Analysis with PSpice.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/083_7.8 Transient Analysis with PSpice.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/083_7.8 Transient Analysis with PSpice.md
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@@ -1,82 +0,0 @@
-# **7.8** Transient Analysis with PSpice
-
-As we discussed in Section 7.5, the transient response is the temporary response of the circuit that soon disappears. *PSpice* can be used to obtain the transient response of a circuit with storage elements. Section D.4 in Appendix D provides a review of transient analysis using *PSpice for Windows*. It is recommended that you read Section D.4 before continuing with this section.
-
-If necessary, dc *PSpice* analysis is first carried out to determine the initial conditions. Then the initial conditions are used in the transient *PSpice* analysis to obtain the transient responses. It is rec om mended but not necessary that during this dc analysis, all capac itors should be opencircuited while all inductors should be short-circuited.
-
-Practice Problem 7.16
-
-**Figure 7.62** For Practice Prob. 7.16.
-
-PSpice uses "transient" to mean "function of time." Therefore, the transient response in PSpice may not actually die out as expected.
-
-Use *PSpice* to find the response *i*(*t*) for *t* > 0 in the circuit of Fig. 7.63. Example 7.17
-
-# **Solution:**
-
-Solving this problem by hand gives *i*(0) = 0, *i*(∞) = 2A, *R*Th = 6, *τ* = 3∕6 = 0.5 s, so that
-
-$$
-i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} = 2(1 - e^{-2t}), \qquad t > 0
-$$
-
-To use *PSpice,* we first draw the schematic as shown in Fig. 7.64. We recall from Appendix D that the part name for a closed switch is Sw\_tclose. We do not need to specify the initial condition of the induc tor because *PSpice* will determine that from the circuit. By selecting **Analysis/Setup/Transient**, we set *Print Step* to 25 ms and *Final Step* to 5*τ* = 2.5 s. After saving the circuit, we simulate by selecting **Analysis/ Simulate**. In the *PSpice* A/D window, we select **Trace/Add** and display –I(L1) as the current through the inductor. Figure 7.65 shows the plot of *i*(*t*), which agrees with that obtained by hand calculation.
-
-**Figure 7.64** The schematic of the circuit in Fig. 7.63.
-
-**Figure 7.63** For Example 7.17.
-
-**Figure 7.65** For Example 7.17; the response of the circuit in Fig. 7.63.
-
-**Answer:** *v*(*t*) = 8(1 − *e*−*t*
-
-that in Fig. 7.65.
-
-Note that the negative sign on I(L1) is needed because the current enters through the upper terminal of the inductor, which happens to be the negative terminal after one counterclock wise rotation. A way to avoid the negative sign is to ensure that current enters pin 1 of the inductor. To obtain this desired direction of positive current flow, the initially horizontal inductor symbol should be rotated counterclockwise 270° and placed in the desired location.
-
-) V, *t* > 0. The response is similar in shape to
-
-For the circuit in Fig. 7.66, use *Pspice* to find *v*(*t*) for *t* > 0.
-
-Practice Problem 7.17
-
-**Figure 7.66** For Practice Prob. 7.17.
-
-12 Ω 30 V 6 Ω 6 Ω 3 Ω 0.1 F 4 A + ‒ t=0 t=0 (a) *v*(t) + ‒ 6 Ω 6 Ω 12 Ω 0.1 F + *v*(t) ‒ 30 V (b) 10 Ω 0.1 F + *v*(t) ‒ 10 V (c) + ‒ + ‒ Example 7.18 In the circuit of Fig. 7.67(a), determine the response *v*(*t*).
-
-**Figure 7.67** For Example 7.18. Original circuit (a), circuit *for t* > 0 (b), and reduced circuit for *t* > 0 (c).
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly stated and the circuit is clearly labeled.
-- 2. **Present.** Given the circuit shown in Fig. 7.67(a), determine the response *v* ( *t*).
-- 3. **Alternative.** We can solve this circuit using circuit analysis techniques, nodal analysis, mesh analysis, or *PSpice*. Let us solve the problem using circuit analysis techniques (this time Thevenin equivalent circuits) and then check the answer using two methods of *PSpice* .
-- 4. **Attempt.** For time < 0, the switch on the left is open and the switch on the right is closed. Assume that the switch on the right has been closed long enough for the circuit to reach steady state; then the capacitor acts like an open circuit and the current from the 4-A source flows through the parallel combination of the 6- Ω and 3-Ω resistors (6 ∥ 3 = 18 ∕ 9 = 2), producing a voltage equal to 2 × 4 = 8 V = − *v*(0).
-
- At *t* = 0, the switch on the left closes and the switch on the right opens, producing the circuit shown in Fig. 7.67(b).
-
- The easiest way to complete the solution is to find the Thevenin equivalent circuit as seen by the capacitor. The opencircuit voltage (with the capacitor removed) is equal to the voltage drop across the 6-Ω resistor on the left, or 10 V (the voltage drops uniformly across the 12- Ω resistor, 20 V, and across the 6- Ω resistor, 10 V). This is *V*Th . The resistance looking in where the capacitor was is equal to 12 ∥ 6 + 6 = 72 ∕18 + 6 = 10 Ω, which is *R*eq. This produces the Thevenin equivalent circuit shown in Fig. 7.67(c). Matching up the boundary conditions ( *v*(0) = −8 V and *v* ( ∞ ) = 10 V) and *τ* = *RC* = 1, we get
-
-$$
-v(t) = 10 - 18e^{-t} V
-$$
-
-## 5. **Evaluate.** There are two ways of solving the problem using *PSpice* .
-
-■ **METHOD 1** One way is to first do the dc *PSpice* analysis to determine the initial capacitor voltage. The schematic of the revelant circuit is in Fig. 7.68(a). Two pseudocomponent VIEWPOINTs are inserted to measure the voltages at nodes 1 and 2. When the circuit is simulated, we obtain the displayed values in Fig. 7.68(a) as *V* 1 = 0 V and *V*2 = 8 V. Thus, the initial capacitor voltage is *v*(0) = *V* 1 − *V*2 = −8 V. The *PSpice* transient analysis uses this value along with the schematic in Fig. 7.68(b). Once the circuit in Fig. 7.68(b) is drawn, we insert the capacitor initial voltage as IC = −8. We select **Analysis/ Setup/Transient** and set *Print Step* to 0.1 s and *Final Step* to 4 *τ* = 4 s. After saving the circuit, we select **Analysis/Simulate** to simulate the circuit. In the *PSpice* A/D window, we select **Trace/Add** and display V(R2:2) – V(R3:2) or V(C1:1) − V(C1:2) as the capacitor voltage *v* ( *t*). The plot of *v* ( *t*) is shown in Fig. 7.69. This agrees with the result ob tained by hand calculation, *v* ( *t* ) = 10 − 18 *e* − *t* V.
-
-**Figure 7.69** Response *v*(*t*) for the circuit in Fig. 7.67.
-
-**Figure 7.68**
-
-(a) Schematic for dc analysis to get *v*(0), (b) schematic for transient analysis used in getting the response *v*(*t*).
-
-■ **METHOD 2** We can simulate the circuit in Fig. 7.67 directly, since *PSpice* can handle the open and closed switches and determine the initial conditions automatically. Using this approach, the schemat ic is drawn as shown in Fig. 7.70. After drawing the circuit, we select **Analysis/Setup/Transient** and set *Print Step* to 0.1 s and *Final Step* to 4*τ* = 4 s. We save the circuit, then select **Analysis/Simulate** to simulate the circuit. In the *PSpice* A/D window, we select **Trace/Add** and display V(R2:2) − V(R3:2) as the capacitor voltage *v*(*t*). The plot of *v*(*t*) is the same as that shown in Fig. 7.69.
-
-6. **Satisfactory?** Clearly, we have found the value of the output response *v*(*t*), as required by the problem statement. Checking does validate that solution. We can present all this as a complete solution to the problem.
-
-The switch in Fig. 7.71 was open for a long time but closed at *t* = 0. If *i*(0) = 10A, find *i*(*t*) for *t* > 0 by hand and also by *PSpice*.
-
-# **Answer:** *i*(*t*) = 6 + 4 *e*−5*t* A. The plot of *i*(*t*) obtained by *PSpice* analysis is shown in Fig. 7.72.
-
-# Practice Problem 7.18
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/084_7.9 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/084_7.9 Applications.md
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-# **7.9** †Applications
-
-The various devices in which *RC* and *RL* circuits find applications include filtering in dc power supplies, smoothing circuits in digital communications, differentiators, integrators, delay circuits, and relay circuits. Some of these applications take advantage of the short or long time constants of the *RC* or *RL* circuits. We will consider four simple applications here. The first two are *RC* circuits, the last two are *RL* circuits.
-
-# **7.9.1** Delay Circuits
-
-An *RC* circuit can be used to pro vide various time delays. Figure 7.73 shows such a circuit. It basically consists of an *RC* circuit with the ca pacitor connected in parallel with a neon lamp. The voltage source can provide enough voltage to fire the lamp. When the switch is closed, the capacitor voltage increases gradually toward 110 V at a rate determined by the circuit's time constant, (*R*1 + *R*2)*C*. The lamp will act as an open
-
-An *RC* delay circuit.
-
-circuit and not emit light until the voltage across it exceeds a particular level, say 70 V. When the voltage level is reached, the lamp fires (goes on), and the capacitor dischar ges through it. Due to the lo w resistance of the lamp when on, the capacitor voltage drops fast and the lamp turns off. The lamp acts again as an open circuit and the capacitor recharges. By adjusting *R*2, we can introduce either short or long time delays into the circuit and mak e the lamp fire, recharge, and fire repeatedly every time constant *τ* = (*R*1 + *R*2)*C*, because it takes a time period *τ* to get the capacitor voltage high enough to fire or low enough to turn off.
-
-The warning blinkers commonly found on road construction sites are one example of the usefulness of such an *RC* delay circuit.
-
-Example 7.19 Consider the circuit in Fig. 7.73, and assume that *R*1 = 1.5 MΩ, 0 < *R*2 < 2.5 MΩ. (a) Calculate the extreme limits of the time constant of the circuit. (b) How long does it take for the lamp to glow for the first time after the switch is closed? Let *R*2 assume its largest value.
-
-# **Solution:**
-
-(a) The smallest value for *R*2 is 0 Ω, and the corresponding time constant for the circuit is
-
-$$
-\tau = (R_1 + R_2)C = (1.5 \times 10^6 + 0) \times 0.1 \times 10^{-6} = 0.15 \text{ s}
-$$
-
-The largest value for *R*2 is 2.5 MΩ, and the corresponding time constant for the circuit is
-
-$$
-\tau = (R_1 + R_2)C = (1.5 + 2.5) \times 10^6 \times 0.1 \times 10^{-6} = 0.4 \text{ s}
-$$
-
-Thus, by proper circuit design, the time constant can be adjusted to introduce a proper time delay in the circuit.
-
-(b) Assuming that the capacitor is initially uncharged, *vC*(0) = 0, while *vC*(∞) = 110. But
-
-$$
-v_C(t) = v_C(\infty) + [v_C(0) - v_C(\infty)]e^{-t/\tau} = 110[1 - e^{-t/\tau}]
-$$
-
-where *τ* = 0.4 s, as calculated in part (a). The lamp glows when *vC* = 70 V. If *vC*(*t*) = 70 V at *t* = *t*0, then
-
-$$
-70 = 110[1 - e^{-t_0/\tau}] \qquad \Rightarrow \qquad \frac{7}{11} = 1 - e^{-t_0/\tau}
-$$
-
-or
-
-$$
-e^{-t_0/\tau} = \frac{4}{11} \qquad \Rightarrow \qquad e^{t_0/\tau} = \frac{11}{4}
-$$
-
-Taking the natural logarithm of both sides gives
-
-$$
-t_0 = \tau \ln \frac{11}{4} = 0.4 \ln 2.75 = 0.4046 \text{ s}
-$$
-
-A more general formula for finding *t*0 is
-
-$$
-t_0 = \tau \ln \frac{-v(\infty)}{v(t_0) - v(\infty)}
-$$
-
-The lamp will fire repeatedly every *t*0 seconds if and only if *v*(*t*0) < *v*(∞).
-
-The *RC* circuit in Fig. 7.74 is designed to operate an alarm which acti - Practice Problem 7.19 vates when the current through it exceeds 90 *μ*A. If 0 ≤ *R* ≤ 6 kΩ, find the range of the time delay that the variable resistor can create.
-
-**Answer:** Between 34.47 and 89.26 ms.
-
-# **7.9.2** Photoflash Unit
-
-An electronic flash unit pro vides a common e xample of an *RC* circuit. This application exploits the ability of the capacitor to oppose any abrupt change in voltage. Figure 7.75 shows a simplified circuit. It consists essentially of a high-voltage dc supply, a current-limiting large resistor *R*1, and a capacitor *C* in parallel with the flashlamp of low resistance *R*2. When the switch is in position 1, the capacitor charges slowly due to the large time constant ( *τ*1 = *R*1*C*). As shown in Fig. 7.76(a), the capacitor voltage rises gradually from zero to *Vs*, while its current decreases gradually from *I*1 = *Vs*∕*R*1 to zero. The charging time is approximately five times the time constant,
-
-$$
-t_{\text{charge}} = 5R_1C \tag{7.65}
-$$
-
-With the switch in position 2, the capacitor v oltage is dischar ged. The low resistance *R*2 of the photolamp permits a high discharge current with peak *I*2 = *Vs*∕*R*2 in a short duration, as depicted in Fig. 7.76(b). Discharging takes place in approximately five times the time constant,
-
-$$
-t_{\text{discharge}} = 5R_2C \tag{7.66}
-$$
-
-(a) Capacitor voltage showing slow charge and fast discharge, (b) capacitor current showing low charging current *I*1 = *Vs*∕*R*1 and high discharge current *I*2 = *Vs*∕*R*2.
-
-Thus, the simple *RC* circuit of Fig. 7.75 provides a short-duration, highcurrent pulse. Such a circuit also finds applications in electric spot welding and the radar transmitter tube.
-
-# **Figure 7.75**
-
-Circuit for a flash unit providing slow charge in position 1 and fast discharge in position 2.
-
-Example 7.20 An electronic flashgun has a current-limiting 6-kΩ resistor and 2000-*μ*F electrolytic capacitor char ged to 240 V. If the lamp resistance is 12 Ω, find: (a) the peak charging current, (b) the time required for the capaci tor to fully charge, (c) the peak discharging current, (d) the total energy stored in the capacitor, and the average power dissipated by the lamp.
-
-# **Solution:**
-
-(a) The peak charging current is
-
-$$
-I_1 = \frac{V_s}{R_1} = \frac{240}{6 \times 10^3} = 40 \text{ mA}
-$$
-
-(b) From Eq. (7.65),
-
-$$
-t_{\text{charge}} = 5R_1C = 5 \times 6 \times 10^3 \times 2000 \times 10^{-6} = 60 \text{ s} = 1 \text{ minute}
-$$
-
-(c) The peak discharging current is
-
-$$
-I_2 = \frac{V_s}{R_2} = \frac{240}{12} = 20 \text{ A}
-$$
-
-(d) The energy stored is
-
-$$
-W = \frac{1}{2}CV_s^2 = \frac{1}{2} \times 2000 \times 10^{-6} \times 240^2 = 57.6 \text{ J}
-$$
-
- The energy stored in the capacitor is dissipated across the lamp during the discharging period. From Eq. (7.66),
-
-$$
-t_{\text{discharge}} = 5R_2C = 5 \times 12 \times 2000 \times 10^{-6} = 0.12 \text{ s}
-$$
-
-Thus, the average power dissipated is
-
-$$
-p = \frac{W}{t_{\text{discharge}}} = \frac{57.6}{0.12} = 480 \text{ watts}
-$$
-
-The flash unit of a camera has a 2-mF capacitor charged to 40 V. Practice Problem 7.20
-
-- (a) How much charge is on the capacitor?
-- (b) What is the energy stored in the capacitor?
-- (c) If the flash fires in 0.8 ms, what is the average current through the flashtube?
-- (d) How much power is delivered to the flashtube? After a picture has been taken, the capacitor needs to be recharged by a power unit that supplies a maximum of 5 mA. How much time does it take to charge the capacitor?
-
-**Answer:** (a) 80 mC, (b) 1.6 J, (c) 100 A, (d) 2 kW, 16 s.
-
-# **7.9.3** Relay Circuits
-
-A magnetically controlled switch is called a *relay*. A relay is essentially an electromagnetic de vice used to open or close a switch that controls another circuit. Figure 7.77(a) sho ws a typical relay circuit. The coil
-
-circuit is an *RL* circuit like that in Fig. 7.77(b), where *R* and *L* are the resistance and inductance of the coil. When switch *S*1 in Fig. 7.77(a) is closed, the coil circuit is energized. The coil current gradually increases and produces a magnetic field. Eventually the magnetic field is sufficiently strong to pull the mo vable contact in the other circuit and close switch *S*2. At this point, the relay is said to be *pulled in*. The time interval *td* between the closure of switches *S*1 and *S*2 is called the *relay delay time*.
-
-Relays were used in the earliest digital circuits and are still used for switching high-power circuits.
-
-The coil of a certain relay is operated by a 12-V battery. If the coil has a Example 7.21 resistance of 150 Ω and an inductance of 30 mH and the current needed to pull in is 50 mA, calculate the relay delay time.
-
-# **Solution:**
-
-The current through the coil is given by
-
-$$
-i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}
-$$
-
-where
-
-$$
-i(0) = 0,
-$$
- $i(\infty) = \frac{12}{150} = 80 \text{ mA}$
-
-$$
-\tau = \frac{L}{R} = \frac{30 \times 10^{-3}}{150} = 0.2 \text{ ms}
-$$
-
-Thus,
-
-$$
-i(t) = 80[1 - e^{-t/\tau}] \text{ mA}
-$$
-
-If *i*(*td*) = 50 mA, then
-
-50 = 80[1 − *e*−*td*∕*τ* ] ⇒ \_\_5 8 = 1 − *e*−*td*∕*τ*
-
-or
-
-$$
-e^{-t_d/\tau} = \frac{3}{8} \qquad \Rightarrow \qquad e^{t_d/\tau} = \frac{8}{3}
-$$
-
-By taking the natural logarithm of both sides, we get
-
-$$
-t_d = \tau \ln \frac{8}{3} = 0.2 \ln \frac{8}{3} \text{ ms} = 0.1962 \text{ ms}
-$$
-
-Alternatively, we may find *td* using
-
-$$
-t_d = \tau \ln \frac{i(0) - i(\infty)}{i(t_d) - i(\infty)}
-$$
-
-# Practice Problem 7.21
-
-A relay has a resistance of 200 Ω and an inductance of 500 mH. The relay contacts close when the current through the coil reaches 175 *μ*A. What time elapses between the application of 110 V to the coil and contact closure?
-
-**Answer:** 957.5 *μ*s.
-
-# **7.9.4** Automobile Ignition Circuit
-
-The ability of inductors to oppose rapid change in current mak es them useful for arc or spark generation. An automobile ignition system tak es advantage of this feature.
-
-The g asoline engine of an automobile requires that the fuel-air mixture in each c ylinder be ignited at proper times. This is achie ved by means of a spark plug (Fig. 7.78), which essentially consists of a pair of electrodes separated by an air g ap. By creating a lar ge voltage (thousands of v olts) between the electrodes, a spark is formed across the air g ap, thereby igniting the fuel. But ho w can such a lar ge voltage be obtained from the car battery, which supplies only 12 V? This is achieved by means of an inductor (the spark coil) *L*. Since the v oltage across the inductor is *v* = *L di*∕*dt*, we can make *di*∕*dt* large by creating a large change in current in a very short time. When the ignition switch in Fig. 7.78 is closed, the current through the inductor increases gradu ally and reaches the final value of *i* = *Vs*∕*R*, where *Vs* = 12 V. Again, the time taken for the inductor to char ge is five times the *time constant* of the circuit (*τ* = *L*∕*R*),
-
-$$
-t_{\text{charge}} = 5 \frac{L}{R} \tag{7.67}
-$$
-
-Since at steady state, *i* is constant, *di*∕*dt* = 0, and the inductor v oltage *v* = 0. When the switch suddenly opens, a lar ge voltage is de veloped across the inductor (due to the rapidly collapsing field) causing a spark or arc in the air g ap. The spark continues until the ener gy stored in the inductor is dissipated in the spark dischar ge. In laboratories, when one is working with inductive circuits, this same ef fect causes a v ery nasty shock, and one must exercise caution.
-
-A solenoid with resistance 4 Ω and inductance 6 mH is used in an auto- Example 7.22 mobile ignition circuit similar to that in Fig. 7.78. If the battery supplies 12 V, determine: the final current through the solenoid when the switch is closed, the energy stored in the coil, and the voltage across the air gap, assuming that the switch takes 1 *μ*s to open.
-
-# **Solution:**
-
-The final current through the coil is
-
-*I* = \_\_ *Vs R* =\_\_\_ 12 4 = 3 A
-
-The energy stored in the coil is
-
-$$
-W = \frac{1}{2}L\ l^2 = \frac{1}{2} \times 6 \times 10^{-3} \times 3^2 = 27 \text{ mJ}
-$$
-
-The voltage across the gap is
-
-$$
-V = L\frac{\Delta I}{\Delta t} = 6 \times 10^{-3} \times \frac{3}{1 \times 10^{-6}} = 18 \text{ kV}
-$$
-
-The spark coil of an automobile ignition system has a 20-mH induc - Practice Problem 7.22 tance and a 5-Ω resistance. With a supply voltage of 12 V, calculate: the time needed for the coil to fully charge, the energy stored in the coil, and the voltage developed at the spark gap if the switch opens in 2 *μ*s.
-
-**Answer:** 20 ms, 57.6 mJ, and 24 kV.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/085_7.10 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/085_7.10 Summary.md
deleted file mode 100644
index 290d45f56a0d331c10105151b2c3d4c51f1ccec4..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/085_7.10 Summary.md
+++ /dev/null
@@ -1,53 +0,0 @@
-# **7.10** Summary
-
-- 1. The analysis in this chapter is applicable to an y circuit that can be reduced to an equi valent circuit comprising a resistor and a single energy-storage element (inductor or capacitor). Such a circuit is first-order because its behavior is described by a first-order differential equation. When analyzing *RC* and *RL* circuits, one must always keep in mind that the capacitor is an open circuit to steady-state dc conditions while the inductor is a short circuit to steady-state dc conditions.
-- 2. The natural response is obtained when no independent source is present. It has the general form
-
-$$
-x(t) = x(0)e^{-t/\tau}
-$$
-
- where *x* represents current through (or v oltage across) a resistor, a capacitor, or an inductor, and *x*(0) is the initial value of *x*. Be cause most practical resistors, capacitors, and inductors always have losses, the natural response is a transient response, i.e. it dies out with time.
-
-3. The time constant *τ* is the time required for a response to decay to 1∕e of its initial value. For *RC* circuits, *τ* = *RC* and for *RL* circuits, *τ* = *L*∕*R*.
-
-4. The singularity functions include the unit step, the unit ramp func tion, and the unit impulse functions. The unit step function *u*(*t*) is
-
-$$
-u(t) = \begin{cases} 0, & t < 0 \\ 1, & t > 0 \end{cases}
-$$
-
-The unit impulse function is
-
-$$
-\delta(t) = \begin{cases}\n0, & t < 0 \\
-\text{Undefined}, & t = 0 \\
-0, & t > 0\n\end{cases}
-$$
-
-The unit ramp function is
-
-$$
-r(t) = \begin{cases} 0, & t \le 0 \\ t, & t \ge 0 \end{cases}
-$$
-
-- 5. The steady-state response is the beha vior of the circuit after an in dependent source has been applied for a long time. The transient response is the component of the complete response that dies out with time.
-- 6. The total or complete response consists of the steady-state response and the transient response.
-- 7. The step response is the response of the circuit to a sudden application of a dc current or v oltage. Finding the step response of a firstorder circuit requires the initial v alue *x*( 0 +), the final value *x*(∞), and the time constant *τ*. With these three items, we obtain the step response as
-
-$$
-x(t) = x(\infty) + [x(0^+) - x(\infty)]e^{-t/\tau}
-$$
-
-A more general form of this equation is
-
-$$
-x(t) = x(\infty) + [x(t_0^+) - x(\infty)]e^{-(t-t_0)/\tau}
-$$
-
-Or we may write it as
-
-Instantaneous value = Final + [Initial − Final] *e*−(*t*−*t*0)∕*τ*
-
-- 8. *PSpice* is very useful for obtaining the transient response of a circuit.
- - 9. Four practical applications of *RC* and *RL* circuits are: a delay circuit, a photoflash unit, a relay circuit, and an automobile ignition circuit.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/086_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/086_Review Questions.md
deleted file mode 100644
index 22342583558fe805a71ae21d6d4c69eb0c3fb2b8..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/086_Review Questions.md
+++ /dev/null
@@ -1,59 +0,0 @@
-# Review Questions
-
-**7.1** An *RC* circuit has *R* = 2 Ω and *C* = 4 F. The time constant is:
-
-> (a) 0.5 s (b) 2 s (c) 4 s (d) 8 s 15 s
-
-**7.2** The time constant for an *RL* circuit with *R* = 2 Ω and *L* = 4 H is:
-
-> (a) 0.5 s (b) 2 s (c) 4 s (d) 8 s 15 s
-
-**7.3** A capacitor in an *RC* circuit with *R* = 2 Ω and *C* = 4 F is being charged. The time required for the capacitor voltage to reach 63.2 percent of its steadystate value is:
-
-(a) 2 s (b) 4 s (c) 8 s (d) 16 s none of the above
-
-**7.4** An *RL* circuit has *R* = 2 Ω and *L* = 4 H. The time needed for the inductor current to reach 40 percent of its steady-state value is:
-
-> (a) 0.5 s (b) 1 s (c) 2 s (d) 4 s none of the above
-
-Problems **299**
-
-- **7.5** In the circuit of Fig. 7.79, the capacitor voltage just before *t* = 0 is:
- - (a) 10 V (b) 7 V (c) 6 V (d) 4 V 0 V
-
-# **Figure 7.79**
-
-For Review Questions 7.5 and 7.6.
-
-**7.6** In the circuit in Fig. 7.79, *v*(∞) is:
-
-(a)
-$$
-10 \text{ V}
-$$
- (b) $7 \text{ V}$ (c) $6 \text{ V}$
-(d) $4 \text{ V}$ 0 V
-
-**7.7** For the circuit in Fig. 7.80, the inductor current just before *t* = 0 is:
-
-> (a) 8 A (b) 6 A (c) 4 A (d) 2 A 0 A
-
-# **Figure 7.80**
-
-For Review Questions 7.7 and 7.8.
-
-**7.8** In the circuit of Fig. 7.80, *i*(∞) is:
-
-(a) 10 A (b) 6 A (c) 4 A (d) 2 A 0 A
-
-**7.9** If *vs* changes from 2 V to 4 V at *t* = 0, we may express *vs* as:
-
-> (a) *δ*(*t*) V (b) 2*u*(*t*) V (c) 2*u*(−*t*) + 4*u*(*t*) V (d) 2 + 2*u*(*t*) V 4*u*(*t*) − 2 V
-
-**7.10** The pulse in Fig. 7.116(a) can be expressed in terms of singularity functions as:
-
-| (a) 2u(t) + 2u(t − 1) V | (b) 2u(t) − 2u(t − 1) V |
-|-------------------------|-------------------------|
-| (c) 2u(t) − 4u(t − 1) V | (d) 2u(t) + 4u(t − 1) V |
-
-*Answers: 7.1d, 7.2b, 7.3c, 7.4b, 7.5d, 7.6a, 7.7c, 7.8e , 7.9c,d, 7.10b.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/087_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/087_Problems.md
deleted file mode 100644
index 7d3dd5dd90b43bc151d9af52c4647a4fbaf087de..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/087_Problems.md
+++ /dev/null
@@ -1,506 +0,0 @@
-# Problems
-
-# Section 7.2 The Source-Free RC Circuit
-
-**7.1** In the circuit shown in Fig. 7.81
-
-$$
-v(t) = 56e^{-200t} \,\text{V}, \quad t > 0
-$$
-
-$$
-i(t) = 8e^{-200t} \text{ mA}, \quad t > 0
-$$
-
-- (a) Find the values of *R* and *C*.
-- (b) Calculate the time constant *τ*.
-- (c) Determine the time required for the voltage to decay half its initial value at *t* = 0.
-
-**7.2** Find the time constant for the *RC* circuit in Fig. 7.82.
-
-# **Figure 7.82**
-
-For Prob. 7.2.
-
-For Prob. 7.3.
-
-**7.3** Determine the time constant for the circuit in Fig. 7.83.
-
-**7.4** The switch in Fig. 7.84 has been in position *A* for a long time. Assume the switch moves instantaneously from *A* to *B* at *t* = 0. Find *v* for *t* > 0.
-
-For Prob. 7.4.
-
-**7.5** Using Fig. 7.85, design a problem to help other students better understand source-free *RC* circuits.
-
-**Figure 7.85**
-
-For Prob. 7.5.
-
-**7.6** The switch in Fig. 7.86 has been closed for a long time, and it opens at *t* = 0. Find *v*(*t*) for *t* ≥ 0.
-
-For Prob. 7.6.
-
-**7.7** Assuming that the switch in Fig. 7.87 has been in position *A* for a long time and is moved to position *B* at *t* = 0, Then at *t* = 1 second, the switch moves from *B* to *C*. Find *vC*(*t*) for *t* ≥ 0.
-
-**Figure 7.87** For Prob. 7.7.
-
-**7.8** For the circuit in Fig. 7.88, if
-
-$$
-v = 10e^{-4t}
-$$
- V and $i = 0.2e^{-4t}$ A, $t > 0$
-
-- (a) Find *R* and *C*.
-- (b) Determine the time constant.
-- (c) Calculate the initial energy in the capacitor.
-- (d) Obtain the time it takes to dissipate 50 percent of the initial energy.
-
-# **Figure 7.88**
-
-For Prob. 7.8.
-
-**7.9** The switch in Fig. 7.89 opens at *t* = 0. Find *vo* for *t* > 0.
-
-# **Figure 7.89**
-
-For Prob. 7.9.
-
-**7.10** For the circuit in Fig. 7.90, find *vo*(*t*) for *t* > 0. Determine the time necessary for the capacitor voltage to decay to one-third of its value at *t* = 0.
-
-# **Figure 7.90**
-
-For Prob. 7.10.
-
-# Section 7.3 The Source-Free RL Circuit
-
-**7.11** For the circuit in Fig. 7.91, find *i*o for *t* > 0.
-
-**Figure 7.91** For Prob. 7.11.
-
-# **Figure 7.92**
-
-For Prob. 7.12.
-
-**7.13** In the circuit of Fig. 7.93,
-
-$$
-v(t) = 80e^{-10^{3}t} \text{V}, \quad t > 0
-$$
-$$
-i(t) = 5e^{-10^{3}t} \text{mA}, \quad t > 0
-$$
-
-- (a) Find *R*, *L*, and *τ*.
-- (b) Calculate the energy dissipated in the resistance for 0 < *t* < 0.5 ms.
-
-# **Figure 7.93**
-
-For Prob. 7.13.
-
-# **Figure 7.94**
-
-- For Prob. 7.14.
-- **7.15** Find the time constant for each of the circuits in Fig. 7.95.
-
-For Prob. 7.15.
-
-**7.16** Determine the time constant for each of the circuits in Fig. 7.96.
-
-For Prob. 7.16.
-
-**7.17** Consider the circuit of Fig. 7.97. Find *vo*(*t*) if *i*(0) = 15 A and *v*(*t*) = 0.
-
-**Figure 7.97** For Prob. 7.17.
-
-**7.18** For the circuit in Fig. 7.98, determine *vo*(*t*) when *i*(0) = 5 A and *v*(*t*) = 0.
-
-# **Figure 7.98**
-
-For Prob. 7.18.
-
-**7.19** In the circuit of Fig. 7.99, find *i*(*t*) for *t* > 0 if *i*(0) = 5 A.
-
-**Figure 7.99** For Prob. 7.19.
-
-**7.20** For the circuit in Fig. 7.100,
-
-and
-
-$$
-i = 30e^{-50t}A, \qquad t > 0
-$$
-
-*v* = 90 *e*−50*t*
-
-V
-
-- (a) Find *L* and *R*.
-- (b) Determine the time constant.
-- (c) Calculate the initial energy in the inductor.
-- (d) What fraction of the initial energy is dissipated in 10 ms?
-
-# **Figure 7.100**
-
-For Prob. 7.20.
-
-**7.21** In the circuit of Fig. 7.101, find the value of *R* for which the steady-state energy stored in the inductor will be 2 J.
-
-# **Figure 7.101**
-
-For Prob. 7.21.
-
-**7.22** Find *i*(*t*) and *v*(*t*) for *t* > 0 in the circuit of Fig. 7.102 if *i*(0) = 10 A.
-
-For Prob. 7.22.
-
-**7.23** Consider the circuit in Fig. 7.103. Given that *vo*(0) = 10 V, find *vo* and *vx* for *t* > 0.
-
-**Figure 7.103** For Prob. 7.23.
-
-# Section 7.4 Singularity Functions
-
-**7.24** Express the following signals in terms of singularity functions.
-
-**7.25** Design a problem to help other students better understand singularity functions.
-
-**7.26** Express the signals in Fig. 7.104 in terms of singularity functions.
-
-**Figure 7.104** For Prob. 7.26.
-
-**7.27** Express *v*(*t*) in Fig. 7.105 in terms of step functions.
-
-Problems **303**
-
-‒1 0 3 21 15 10 5 ‒10 ‒5 t *v*(t)
-
-# **Figure 7.105**
-
-For Prob. 7.27.
-
-**7.28** Sketch the waveform represented by
-
-$$
-i(t) = [r(t) - r(t-1) - u(t-2) - r(t-2)
-$$
-
-+
-$$
-r(t-3) + u(t)(t-4)]
-$$
- A
-
-**7.29** Sketch the following functions:
-
-(a)
-$$
-x(t) = 10e^{-t}u(t - 1)
-$$
-,
-\n(b) $y(t) = 10e^{-(t-1)}u(t)$ ,
-\n(c) $z(t) = \cos 4t\delta(t - 1)$
-
-**7.30** Evaluate the following integrals involving the impulse functions:
-
-(a)
-$$
-\int_{-\infty}^{\infty} 4t^2 \delta(t-1) dt
-$$
-
-(b)
-$$
-\int_{-\infty}^{\infty} 4t^2 \cos 2\pi t \delta(t-0.5) dt
-$$
-
-**7.31** Evaluate the following integrals:
-
-(a)
-$$
-\int_{-\infty}^{\infty} e^{-4t^2} \delta(t - 2) dt
-$$
-
-(b)
-$$
-\int_{-\infty}^{\infty} [5\delta(t) + e^{-t} \delta(t) + \cos 2\pi t \delta(t)] dt
-$$
-
-**7.32** Evaluate the following integrals:
-
-(a)
-$$
-\int_{1}^{t} u(\lambda) d\lambda
-$$
-
-\n(b)
-$$
-\int_{0}^{4} r(t-1) dt
-$$
-
-\n(c)
-$$
-\int_{1}^{5} (t-6)^{2} \delta(t-2) dt
-$$
-
-- **7.33** The voltage across a 10-mH inductor is 45*δ*(*t* − 2)mV. Find the inductor current, assuming that the inductor is initially uncharged.
-- **7.34** Evaluate the following derivatives:
-
-(a)
-$$
-\frac{d}{dt}[u(t-1)u(t+1)]
-$$
-
-\n(b)
-$$
-\frac{d}{dt}[r(t-6)u(t-2)]
-$$
-
-\n(c)
-$$
-\frac{d}{dt}[\sin 4tu(t-3)]
-$$
-
-**7.35** Find the solution to the following differential equations:
-
-(a)
-$$
-\frac{dv}{dt} + 2v = 0
-$$
-, $v(0) = -1$ V
-(b) $2\frac{di}{dt} - 3i = 0$ , $i(0) = 2$
-
-**7.36** Solve for *v* in the following differential equations, subject to the stated initial condition.
-
-(a)
-$$
-dv/dt + v = u(t)
-$$
-, $v(0) = 0$
-(b) $2 dv/dt - v = 3u(t)$ , $v(0) = -6$
-
-**7.37** A circuit is described by
-
-$$
-4\frac{dv}{dt} + v = 10
-$$
-
-- (a) What is the time constant of the circuit?
-- (b) What is *v*(∞), the final value of *v*?
-- (c) If *v*(0) = 2, find *v*(*t*) for *t* ≥ 0.
-- **7.38** A circuit is described by
-
-$$
-\frac{di}{dt} + 3i = 2u(t)
-$$
-
-Find *i*(*t*) for *t* > 0 given that *i*(0) = 0.
-
-# Section 7.5 Step Response of an RC Circuit
-
-**7.39** Calculate the capacitor voltage for *t* < 0 and *t* > 0 for each of the circuits in Fig. 7.106.
-
-**Figure 7.106** For Prob. 7.39.
-
-**7.40** Find the capacitor voltage for *t* < 0 and *t* > 0 for each of the circuits in Fig. 7.107.
-
-For Prob. 7.40.
-
-**7.41** Using Fig. 7.108, design a problem to help other students better understand the step response of an *RC* circuit.
-
-**Figure 7.108**
-
-- For Prob. 7.41.
- - **7.42** (a) If the switch in Fig. 7.109 has been open for a long time and is closed at *t* = 0, find *vo*(*t*).
- - (b) Suppose that the switch has been closed for a long time and is opened at *t* = 0. Find *vo*(*t*).
-
-For Prob. 7.42.
-
-**7.43** Consider the circuit in Fig. 7.110. Find *i*(*t*) for *t* < 0 and *t* > 0.
-
-For Prob. 7.43.
-
-**7.44** The switch in Fig. 7.111 has been in position *a* for a long time. At *t* = 0, it moves to position *b*. Calculate *i*(*t*) for all *t* > 0.
-
-**Figure 7.111** For Prob. 7.44.
-
-**7.45** Find *vo* in the circuit of Fig. 7.112 when *vs* = 30*u*(*t*) V. Assume that *vo*(0) = 5 V.
-
-**Figure 7.112** For Prob. 7.45.
-
-**7.46** For the circuit in Fig. 7.113, *is*(*t*) = 5*u*(*t*). Find *v*(*t*).
-
-**Figure 7.113** For Prob. 7.46.
-
-**7.47** Determine *v*(*t*) for *t* > 0 in the circuit of Fig. 7.114 if *v*(0) = 0.
-
-For Prob. 7.47.
-
-Problems **305**
-
-**7.48** Find *v*(*t*) and *i*(*t*) in the circuit of Fig. 7.115.
-
-# **Figure 7.115**
-
-For Prob. 7.48.
-
-**7.49** If the waveform in Fig. 7.116(a) is applied to the circuit of Fig. 7.116(b), find *v*(*t*). Assume *v*(0) = 0.
-
-# **Figure 7.116**
-
-For Prob. 7.49 and Review Question 7.10.
-
-\***7.50** In the circuit of Fig. 7.117, find *ix* for *t* > 0. Let *R*1 = *R*2 = 1 kΩ, *R*3 = 2 kΩ, and *C* = 0.25 mF.
-
-# **Figure 7.117** For Prob. 7.50.
-
-# Section 7.6 Step Response of an RL Circuit
-
-- **7.51** Rather than applying the shortcut technique used in Section 7.6, use KVL to obtain Eq. (7.60).
-- **7.52** Using Fig. 7.118, design a problem to help other students better understand the step response of an *RL* circuit.
-
-**Figure 7.118** For Prob. 7.52.
-
-**7.53** Determine the inductor current *i*(*t*) for both *t* < 0 and *t* > 0 for each of the circuits in Fig. 7.119.
-
-**Figure 7.119** For Prob. 7.53.
-
-**7.54** Obtain the inductor current for both *t* < 0 and *t* > 0 in each of the circuits in Fig. 7.120.
-
-For Prob. 7.54.
-
-\* An asterisk indicates a challenging problem.
-
-**7.55** Find *v*(*t*) for *t* < 0 and *t* > 0 in the circuit of Fig. 7.121.
-
-# **Figure 7.121**
-
-For Prob. 7.55.
-
-**7.56** For the network shown in Fig. 7.122, find *v*(*t*) for *t* > 0.
-
-**Figure 7.122**
-
-For Prob. 7.56.
-
-\***7.57** Find *i*1(*t*) and *i*2(*t*) for *t* > 0 in the circuit of Fig. 7.123.
-
-**Figure 7.123**
-
-For Prob. 7.57.
-
-- **7.58** Rework Prob. 7.17 if *i*(0) = 10 A and *v*(*t*) = 20*u*(*t*) V.
-- **7.59** Determine the step response *vo*(*t*) to *is* = 6*u*(*t*) A in the circuit of Fig. 7.124.
-
-**Figure 7.124** For Prob. 7.59.
-
-**7.60** Find *v*(*t*) for *t* > 0 in the circuit of Fig. 7.125 if the initial current in the inductor is zero.
-
-# **Figure 7.125**
-
-For Prob. 7.60.
-
-**7.61** In the circuit in Fig. 7.126, *is* changes from 5 A to 10 A at *t* = 0; that is, *is* = 5*u*(−*t*) + 10*u*(*t*). Find *v* and *i*.
-
-# **Figure 7.126**
-
-For Prob. 7.61.
-
-**7.62** For the circuit in Fig. 7.127, calculate *i*(*t*) if *i*(0) = 0.
-
-**Figure 7.127**
-
-For Prob. 7.62.
-
-**7.63** Obtain *v*(*t*) and *i*(*t*) in the circuit of Fig. 7.128.
-
-# **Figure 7.128**
-
-For Prob. 7.63.
-
-**7.64** Determine the value of *iL*(*t*) and the total energy dissipated by the circuit from *t* = 0 sec to *t* = ∞ sec. The value of *iin*(*t*) is equal to [6 – 6*u*(*t*)] A.
-
-**7.65** If the input pulse in Fig. 7.130(a) is applied to the circuit in Fig. 7.130(b), determine the response *i*(*t*).
-
-# Section 7.7 First-order Op Amp Circuits
-
-**7.66** Using Fig. 7.131, design a problem to help other students better understand first-order op amp circuits.
-
-For Prob. 7.66.
-
-**7.67** If *v*(0) = 10 V, find *vo*(*t*) for *t* > 0 in the op amp circuit in Fig. 7.132. Let *R* = 100 kΩ and *C* = 20 *µ*F.
-
-# **Figure 7.133**
-
-For Prob. 7.68.
-
-**7.69** For the op amp circuit in Fig. 7.134, find *vo*(*t*) for *t* > 0.
-
-# **Figure 7.134**
-
-For Prob. 7.69.
-
-**7.70** Determine *vo* for *t* > 0 when *vs* = 20 mV in the op amp circuit of Fig. 7.135.
-
-# **Figure 7.135** For Prob. 7.70.
-
-- **7.71** For the op amp circuit in Fig. 7.136, suppose *vs* = 10*u*(*t*) V. Find *v*(*t*) for *t* > 0.
-
-**7.72** Find *io* in the op amp circuit in Fig. 7.137. Assume that *v*(0) = −2 V, *R* = 10 kΩ, and *C* = 10 *µ*F.
-
-# **Figure 7.137**
-
-For Prob. 7.72.
-
-**7.73** For the op amp circuit of Fig. 7.138, let *R*1 = 10 kΩ, *Rf* = 30 kΩ, *C* = 20 *μ*F, and *v*(0) = 1 V. Find *v*0.
-
-# **Figure 7.138**
-
-For Prob. 7.73.
-
-**7.74** Determine *vo*(*t*) for *t* > 0 in the circuit of Fig. 7.139. Let *is* = 10*u*(*t*) *μ*A and assume that the capacitor is initially uncharged.
-
-# **Figure 7.139**
-
-For Prob. 7.74.
-
-**7.75** In the circuit of Fig. 7.140, find *vo* and *io*, given that *vs* = 10[1 − *e*−*t* ]*u*(*t*) V.
-
-**Figure 7.140** For Prob. 7.75.
-
-# Section 7.8 Transient Analysis with PSpice
-
-- **7.76** Repeat Prob. 7.49 using *PSpice or MultiSim*.
-- **7.77** The switch in Fig. 7.141 opens at *t* = 0. Use *PSpice or MultiSim* to determine *v*(*t*) for *t* > 0.
-
-# **Figure 7.141**
-
-For Prob. 7.77.
-
-**7.78** The switch in Fig. 7.142 moves from position *a* to *b* at *t* = 0. Use *PSpice or MultiSim* to find *i*(*t*) for *t* > 0.
-
-# **Figure 7.142** For Prob. 7.78.
-
-**7.79** In the circuit of Fig. 7.143, determine *io*(*t*).
-
-# **Figure 7.143**
-
-- For Prob. 7.79.
-- **7.80** In the circuit of Fig. 7.144, find the value of *io* for all values of 0 < *t*.
-
-For Prob. 7.80.
-
-**7.81** Repeat Prob. 7.65 using *PSpice or MultiSim*.
-
-# Section 7.9 Applications
-
-- **7.82** In designing a signal-switching circuit, it was found that a 100-*µ*F capacitor was needed for a time constant of 3 ms. What value resistor is necessary for the circuit?
-- **7.83** An *RC* circuit consists of a series connection of a 120-V source, a switch, a 34-MΩ resistor, and a 15-*µ*F capacitor. The circuit is used in estimating the speed of a horse running a 4-km racetrack. The switch closes when the horse begins and opens when the horse crosses the finish line. Assuming that the capacitor charges to 85.6 V, calculate the speed of the horse.
-- **7.84** A capacitor with a value of 10 mF has a leakage resistance of 2 MΩ. How long does it take the voltage across the capacitor to decay to 40% of the initial voltage to which the capacitor is charged? Assume that the capacitor is charged and then set aside by itself.
-- **7.85** A simple relaxation oscillator circuit is shown in Fig. 7.145. The neon lamp fires when its voltage reaches 75 V and turns off when its voltage drops to 30 V. Its resistance is 120 Ω when on and infinitely high when off.
- - (a) For how long is the lamp on each time the capacitor discharges?
- - (b) What is the time interval between light flashes?
-
-**Figure 7.145** For Prob. 7.85.
-
-**7.86** Figure 7.146 shows a circuit for setting the length of time voltage is applied to the electrodes of a welding machine. The time is taken as how long it takes the capacitor to charge from 0 to 8 V. What is the time range covered by the variable resistor?
-
-# **Figure 7.146** For Prob. 7.86.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/088_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/088_Comprehensive Problems.md
deleted file mode 100644
index dd82f4ab0bb358ec032932e68c4d31755ef46eda..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/088_Comprehensive Problems.md
+++ /dev/null
@@ -1,72 +0,0 @@
-# Comprehensive Problems
-
-- **7.88** The circuit in Fig. 7.148(a) can be designed as an approximate differentiator or an integrator, depending on whether the output is taken across the resistor or the capacitor, and also on the time constant *τ* = *RC* of the circuit and the width *T* of the input pulse in Fig. 7.148(b). The circuit is a differentiator if *τ*≪ *T*, say *τ*< 0.1*T*, or an integrator if *τ*≫ *T*, say *τ*> 10*T*.
-- (a) What is the minimum pulse width that will allow a differentiator output to appear across the capacitor?
-- (b) If the output is to be an integrated form of the input, what is the maximum value the pulse width can assume?
-
-For Prob. 7.88.
-
-**7.89** An *RL* circuit may be used as a differentiator if the output is taken across the inductor and *τ*≪ *T* (say *τ*< 0.1*T*), where *T* is the width of the input pulse. If *R* is fixed at 200 kΩ, determine the maximum value of *L* required to differentiate a pulse with *T* = 10 *µ*s.
-
-**7.90** An attenuator probe employed with oscilloscopes was designed to reduce the magnitude of the input voltage *vi* by a factor of 10. As shown in Fig. 7.149, the oscilloscope has internal resistance *Rs* and capacitance *Cs*, while the probe has an internal resistance *Rp*. If *Rp* is fixed at 6 MΩ, find *Rs* and *Cs* for the circuit to have a time constant of 15 *µ*s.
-
-**Figure 7.149** For Prob. 7.90.
-
-**7.91** The circuit in Fig. 7.150 is used by a biology student to study "frog kick." She noticed that the frog kicked a little when the switch was closed but kicked violently for 5 s when the switch was opened. Model the frog as a resistor and calculate its resistance. Assume that it takes 10 mA for the frog to kick violently.
-
-# **Figure 7.150** For Prob. 7.91.
-
-**7.92** To move a spot of a cathode-ray tube across the screen requires a linear increase in the voltage across the deflection plates, as shown in Fig. 7.151. Given that the capacitance of the plates is 4 nF, sketch the current flowing through the plates.
-
-**Figure 7.151** For Prob. 7.92.
-
-# **chapter**
-
-8
-
-# Second-Order Circuits
-
-*Everyone who can earn a masters degree in engineering must earn a masters degree in engineering in order to maximize the success of their career! If you want to do research, state-of-the-art engineering, teach in a university, or start your own business, you really need to earn a doctoral degree!*
-
-—Charles K. Alexander
-
-# Enhancing Your Career
-
-To increase your engineering career opportunities after graduation, develop a strong fundamental understanding in a broad set of engineer ing areas. When possible, this might best be accomplished by working toward a graduate degree immediately upon receiving your undergraduate degree.
-
-Each de gree in engineering represents certain skills the student acquires. At the Bachelor de gree level, you learn the language of engi neering and the fundamentals of engineering and design. At the Master's level, you acquire the ability to do advanced engineering projects and to communicate your work effectively both orally and in writing. The Ph.D. represents a thorough understanding of the fundamentals of electrical engineering and a mastery of the skills necessary both for w orking at the frontiers of an engineering area and for communicating one' s effort to others.
-
-If you have no idea what career you should pursue after graduation, a graduate de gree program will enhance your ability to e xplore career options. Since your undergraduate degree will only provide you with the fundamentals of engineering, a Master' s degree in engineering supple mented by business courses benefits more engineering students than does getting a Master's of Business Administration (MBA). The best time to get your MB A is after you ha ve been a practicing engineer for some years and decide your career path would be enhanced by strengthening your business skills.
-
-Engineers should constantly educate themselv es, formally and informally, taking advantage of all means of education. Perhaps there is no better way to enhance your career than to join a professional society such as IEEE and be an active member.
-
-Enhancing your career involves understanding your goals, adapting to changes, anticipating opportunities, and planning your own niche.
-
-© 2005 Institute of Electrical and Electronics Engineers (IEEE), from *IEEE Potentials* cover, April/May 2005
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Develop a better understanding of the solution of generalsecond order differential equations.
-- 2. Learn how to determine initial and final values.
-- 3. Understand the response in source-free series *RLC* circuits.
-- 4. Understand the response in source-free parallel *RLC* circuits.
-- 5. Understand the step response of series *RLC* circuits.
-- 6. Understand the step response of parallel *RLC* circuits.
-- 7. Understand general second-order circuits.
-- 8. Understand general second-order circuits with op amps.
-
-# **8.1** Introduction
-
-In the previous chapter we considered circuits with a single storage element (a capacitor or an inductor). Such circuits are first-order because the differential equations describing them are first-order. In this chap ter we will consider circuits containing two storage elements. These are known as *second-order* circuits because their responses are described by differential equations that contain second derivatives.
-
-Typical e xamples of second-order circuits are *RLC* circuits, in which the three kinds of passive elements are present. Examples of such circuits are shown in Fig. 8.1(a) and (b). Other examples are *RL* and *RC* circuits, as shown in Fig. 8.1(c) and (d). It is apparent from Fig. 8.1 that a second-order circuit may have two storage elements of different type or the same type (provided elements of the same type cannot be represented by an equi valent single element). An op amp circuit with tw o storage elements may also be a second-order circuit. As with first-order circuits, a second-order circuit may contain se veral resistors and dependent and independent sources.
-
-A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.
-
-Our analysis of second-order circuits will be similar to that used for first-order. We will first consider circuits that are excited by the initial conditions of the storage elements. Although these circuits may contain dependent sources, they are free of independent sources. These sourcefree circuits will give natural responses as expected. Later we will consider circuits that are e xcited by independent sources. These circuits will give both the transient response and the steady-state response. We consider only dc independent sources in this chapter. The case of sinusoidal and exponential sources is deferred to later chapters.
-
-We begin by learning ho w to obtain the initial conditions for the circuit variables and their deri vatives, as this is crucial to analyzing second-order circuits. Then we consider series and parallel *RLC* circuits such as shown in Fig. 8.1 for the two cases of excitation: by initial
-
-
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/089_Chapter 8 - Second-Order Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/089_Chapter 8 - Second-Order Circuits.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/089_Chapter 8 - Second-Order Circuits.md
+++ /dev/null
@@ -1,54 +0,0 @@
-# **Figure 8.1**
-
-Typical examples of second-order circuits: (a) series *RLC* circuit, (b) parallel *RLC* circuit, (c) *RL* circuit, (d) *RC* circuit.
-
-conditions of the ener gy storage elements and by step inputs. Later we e xamine other types of second-order circuits, including op amp circuits. We will consider *PSpice* analysis of second-order circuits. Finally, we will consider the automobile ignition system and smooth ing circuits as typical applications of the circuits treated in this chapter. Other applications such as resonant circuits and filters will be covered in Chapter 14.
-
-# **8.2** Finding Initial and Final Values
-
-Perhaps the major problem students f ace in handling second-order circuits is finding the initial and final conditions on circuit variables. Students are usually comfortable getting the initial and final values of *v* and *i* but often have difficulty finding the initial values of their derivatives: *dv*∕*dt* and *di*∕*dt*. For this reason, this section is e xplicitly devoted to the subtleties of getting *v*(0), *i*(0), *dv*(0)∕*dt*, *di*(0)∕*dt*, *i*(∞), and *v*(∞). Unless otherwise stated in this chapter , *v* denotes capacitor v oltage, while *i* is the inductor current.
-
-There are two key points to keep in mind in determining the initial conditions.
-
-First—as always in circuit analysis—we must carefully handle the polarity of voltage *v*(*t*) across the capacitor and the direction of the current *i*(*t*) through the inductor. Keep in mind that *v* and *i* are defined strictly according to the passive sign convention (see Figs. 6.3 and 6.23). One should carefully observe how these are defined and apply them accordingly.
-
-Second, keep in mind that the capacitor voltage is always continuous so that
-
-$$
-v(0^+) = v(0^-) \tag{8.1a}
-$$
-
-and the inductor current is always continuous so that
-
-$$
-i(0^+) = i(0^-) \tag{8.1b}
-$$
-
-where *t* = 0− denotes the time just before a switching event and *t* = 0+ is the time just after the switching event, assuming that the switching event takes place at *t* = 0.
-
-Thus, in finding initial conditions, we first focus on those variables that cannot change abruptly, capacitor voltage and inductor current, by applying Eq. (8.1). The following examples illustrate these ideas.
-
-The switch in Fig. 8.2 has been closed for a long time. It is open at *t* = 0. Example 8.1 Find: (a) *i*(0+), *v*(0+), (b) *di*(0+)∕*dt*, *dv*(0+)∕*dt*, (c) *i*(∞), *v*(∞).
-
-# **Solution:**
-
-(a) If the switch is closed a long time before *t* = 0, it means that the circuit has reached dc steady state at *t* = 0. At dc steady state, the inductor acts like a short circuit, while the capacitor acts like an open circuit, so we have the circuit in Fig. 8.3(a) at *t* = 0−. Thus,
-
-$$
-i(0^{-}) = \frac{12}{4+2} = 2 \text{ A}, \qquad v(0^{-}) = 2i(0^{-}) = 4 \text{ V}
-$$
-
-As the inductor current and the capacitor voltage cannot change abruptly,
-
-$$
-i(0^+) = i(0^-) = 2
-$$
- A, $v(0^+) = v(0^-) = 4$ V
-
-(b) At *t* = 0+, the switch is open; the equivalent circuit is as shown in Fig. 8.3(b). The same current flows through both the inductor and capacitor. Hence,
-
-$$
-i_C(0^+) = i(0^+) = 2 \text{ A}
-$$
-
-Since *C dv*∕*dt* = *iC*, *dv*∕*dt* = *iC*∕*C*, and
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/090_8.2 Finding Initial and Final Values.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/090_8.2 Finding Initial and Final Values.md
deleted file mode 100644
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+++ /dev/null
@@ -1,172 +0,0 @@
-
-$$
-\frac{dv(0^{+})}{dt} = \frac{i_C(0^{+})}{C} = \frac{2}{0.1} = 20 \text{ V/s}
-$$
-
-Similarly, since *L di*∕*dt* = *vL*, *di*∕*dt* = *vL*∕*L*. We now obtain *vL* by applying KVL to the loop in Fig. 8.3(b). The result is
-
-$$
--12 + 4i(0^+) + v_L(0^+) + v(0^+) = 0
-$$
-
-or
-
-$$
-v_L(0^+) = 12 - 8 - 4 = 0
-$$
-
-Thus,
-
-$$
-\frac{di(0^{+})}{dt} = \frac{v_L(0^{+})}{L} = \frac{0}{0.25} = 0 \text{ A/s}
-$$
-
-(c) For *t* > 0, the circuit undergoes transience. But as *t* → ∞, the circuit reaches steady state again. The inductor acts like a short circuit and the capacitor like an open circuit, so that the circuit in Fig. 8.3(b) becomes that shown in Fig. 8.3(c), from which we have
-
-$$
-i(\infty) = 0 \text{ A}, \qquad v(\infty) = 12 \text{ V}
-$$
-
-Practice Problem 8.1 The switch in Fig. 8.4 was open for a long time but closed at *t* = 0. Determine: (a) *i*(0+), *v*(0+), (b) *di*(0+)∕*dt*, *dv*(0+)∕*dt*, (c) *i*(∞), *v*(∞).
-
-**Answer:** (a) 3.5 A, 7 V, (b) 87.5 A/s, 0 V/s, (c) 21 A, 42 V.
-
-In the circuit of Fig. 8.5, calculate: (a) *iL*(0+), *vC* (0+), *vR*(0+), (b) *diL*(0+)∕*dt*, *dvC*(0+)∕*dt*, *dvR*(0+)∕*dt*, (c) *iL*(∞), *vC*(∞), *vR*(∞). Example 8.2
-
-# **Solution:**
-
-(a) For *t* < 0, 3*u*(*t*) = 0. At *t* = 0−, since the circuit has reached steady state, the inductor can be replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig. 8.6(a). From this figure we obtain
-
-$$
-i_L(0^-) = 0
-$$
-, $v_R(0^-) = 0$ , $v_C(0^-) = -20$ V (8.2.1)
-
-Although the derivatives of these quantities at *t* = 0− are not required, it is evident that they are all zero, since the circuit has reached steady state and nothing changes.
-
-# **Figure 8.6**
-
-The circuit in Fig. 8.5 for: (a) *t* = 0−, (b) *t* = 0+.
-
-For *t* > 0, 3*u*(*t*) = 3, so that the circuit is no w equivalent to that in Fig. 8.6(b). Since the inductor current and capacitor v oltage cannot change abruptly,
-
-$$
-i_L(0^+) = i_L(0^-) = 0,
-$$
- $v_C(0^+) = v_C(0^-) = -20$ V (8.2.2)
-
-Although the voltage across the 4-Ω resistor is not required, we will use it to apply KVL and KCL; let it be called *vo*. Applying KCL at node *a* in Fig. 8.6(b) gives
-
-$$
-3 = \frac{v_R(0^+)}{2} + \frac{v_o(0^+)}{4}
-$$
- (8.2.3)
-
-Applying KVL to the middle mesh in Fig. 8.6(b) yields
-
-$$
--v_R(0^+) + v_o(0^+) + v_C(0^+) + 20 = 0 \tag{8.2.4}
-$$
-
-Since *vC* (0+) = −20 V from Eq. (8.2.2), Eq. (8.2.4) implies that
-
-$$
-v_R(0^+) = v_o(0^+) \tag{8.2.5}
-$$
-
-From Eqs. (8.2.3) and (8.2.5), we obtain
-
-$$
-v_R(0^+) = v_o(0^+) = 4 \text{ V}
-$$
- (8.2.6)
-
-(b) Since *L diL*∕*dt* = *vL*,
-
-$$
-\frac{di_L(0^+)}{dt} = \frac{v_L(0^+)}{L}
-$$
-
-But applying KVL to the right mesh in Fig. 8.6(b) gives
-
-$$
-v_L(0^+) = v_C(0^+) + 20 = 0
-$$
-
-Hence,
-
-$$
-\frac{di_L(0^+)}{dt} = 0\tag{8.2.7}
-$$
-
-Similarly, since *C dvC*∕*dt* = *iC*, then *dvC*∕*dt* = *iC*∕*C*. We apply KCL at node *b* in Fig. 8.6(b) to get *iC*:
-
-$$
-\frac{v_o(0^+)}{4} = i_C(0^+) + i_L(0^+) \tag{8.2.8}
-$$
-
-Since *vo*(0+) = 4 and *iL*(0+) = 0, *iC*(0+) = 4∕4 = 1 A. Then
-
-$$
-\frac{dv_C(0^+)}{dt} = \frac{i_C(0^+)}{C} = \frac{1}{0.5} = 2 \text{ V/s}
-$$
- (8.2.9)
-
-To get *dvR*(0+)∕*dt*, we apply KCL to node *a* and obtain
-
-$$
-3 = \frac{v_R}{2} + \frac{v_o}{4}
-$$
-
-Taking the derivative of each term and setting *t* = 0+ gives
-
-$$
-0 = 2\frac{dv_R(0^+)}{dt} + \frac{dv_o(0^+)}{dt}
-$$
- (8.2.10)
-
-We also apply KVL to the middle mesh in Fig. 8.6(b) and obtain
-
-$$
--v_R + v_C + 20 + v_o = 0
-$$
-
-Again, taking the derivative of each term and setting *t* = 0+ yields
-
-$$
--\frac{dv_R(0^+)}{dt} + \frac{dv_C(0^+)}{dt} + \frac{dv_o(0^+)}{dt} = 0
-$$
-
-Substituting for *dvC* (0+)∕*dt* = 2 gives
-
-$$
-\frac{dv_R(0^+)}{dt} = 2 + \frac{dv_o(0^+)}{dt}
-$$
- (8.2.11)
-
-From Eqs. (8.2.10) and (8.2.11), we get
-
-$$
-\frac{dv_R(0^+)}{dt} = \frac{2}{3} \text{ V/s}
-$$
-
-We can find *diR*(0+)∕*dt* although it is not required. Since *vR* = 2*iR*,
-
-$$
-\frac{di_R(0^+)}{dt} = \frac{1}{2}\frac{dv_R(0^+)}{dt} = \frac{1}{2}\frac{2}{3} = \frac{1}{3} \text{ A/s}
-$$
-
-(c) As *t* → ∞, the circuit reaches steady state. We have the equivalent circuit in Fig. 8.6(a) except that the 3-A current source is now operative. By current division principle,
-
-$$
-i_L(\infty) = \frac{2}{2+4} \cdot 3 \text{ A} = 1 \text{ A}
-$$
-\n
-$$
-v_R(\infty) = \frac{4}{2+4} \cdot 3 \text{ A} \times 2 = 4 \text{ V}, \qquad v_C(\infty) = -20 \text{ V}
-$$
-\n(8.2.12)
-
-For the circuit in Fig. 8.7, find: (a) *iL*(0+), *vC*(0+), *vR*(0+), Practice Problem 8.2 (b) *diL*(0+)∕*dt*, *dvC*(0+)∕*dt*, *dvR*(0+)∕*dt*, (c) *iL*(∞), *vC*(∞), *vR*(∞).
-
-**Answer:** (a) −6 A, 0, 0, (b) 0, 20 V/s, 0, (c) −2 A, 20 V, 20 V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/091_8.3 The Source-Free Series RLC Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/091_8.3 The Source-Free Series RLC Circuit.md
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-# **8.3** The Source-Free Series RLC Circuit
-
-An understanding of the natural response of the series *RLC* circuit is a necessary background for future studies in filter design and communications networks.
-
-Consider the series *RLC* circuit sho wn in Fig. 8.8. The circuit is being excited by the energy initially stored in the capacitor and inductor. The energy is represented by the initial capacitor v oltage *V*0 and initial inductor current *I*0. Thus, at *t* = 0,
-
-$$
-v(0) = \frac{1}{C} \int_{-\infty}^{0} i \, dt = V_0 \tag{8.2a}
-$$
-
-$$
-i(0) = I_0 \tag{8.2b}
-$$
-
-Applying KVL around the loop in Fig. 8.8,
-
-$$
-Ri + L\frac{di}{dt} + \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau = 0
-$$
- (8.3)
-
-A source-free series *RLC* circuit.
-
-To eliminate the integral, we differentiate with respect to *t* and rearrange terms. We get
-
-$$
-\frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0
-$$
-\n(8.4)
-
-This is a *second-order differential equation* and is the reason for calling the *RLC* circuits in this chapter second-order circuits. Our goal is to solve Eq. (8.4). To solv e such a second-order dif ferential equation requires that we have two initial conditions, such as the initial v alue of *i* and its first derivative or initial values of some *i* and *v*. The initial value of *i* is given in Eq. (8.2b). We get the initial v alue of the deri vative of *i* from Eqs. (8.2a) and (8.3); that is,
-
-$$
-Ri(0) + L\frac{di(0)}{dt} + V_0 = 0
-$$
-
-or
-
-$$
-\frac{di(0)}{dt} = -\frac{1}{L} (RI_0 + V_0)
-$$
-\n(8.5)
-
-With the two initial conditions in Eqs. (8.2b) and (8.5), we can no w solve Eq. (8.4). Our e xperience in the preceding chapter on first-order circuits suggests that the solution is of exponential form. So we let
-
-$$
-i = Ae^{st} \tag{8.6}
-$$
-
-where *A* and *s* are constants to be determined. Substituting Eq. (8.6) into Eq. (8.4) and carrying out the necessary differentiations, we obtain
-
-$$
-As2est + \frac{AR}{L}sest + \frac{A}{LC}est = 0
-$$
-
-or
-
-$$
-Ae^{st}(s^2 + \frac{R}{L}s + \frac{1}{LC}) = 0
-$$
- (8.7)
-
-Since *i* = *Aest* is the assumed solution we are trying to find, only the expression in parentheses can be zero:
-
-$$
-s^2 + \frac{R}{L}s + \frac{1}{LC} = 0
-$$
- (8.8)
-
-This quadratic equation is kno wn as the *characteristic equation* of the differential Eq. (8.4), since the roots of the equation dictate the character of *i*. The two roots of Eq. (8.8) are
-
-$$
-s_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 - \frac{1}{LC}}
-$$
- (8.9a)
-
-$$
-s_2 = -\frac{R}{2L} - \sqrt{\left(\frac{R}{2L}\right)^2 - \frac{1}{LC}}
-$$
- (8.9b)
-
-A more compact way of expressing the roots is
-
-$$
-s_1 = -\alpha + \sqrt{\alpha^2 - \omega_0^2}
-$$
-, $s_2 = -\alpha - \sqrt{\alpha^2 - \omega_0^2}$ (8.10)
-
-See Appendix C.1 for the formula to find the roots of a quadratic equation. where
-
-$$
-\alpha = \frac{R}{2L}, \qquad \omega_0 = \frac{1}{\sqrt{LC}} \tag{8.11}
-$$
-
-The roots *s*1 and *s*2 are called *natural frequencies*, measured in nepers per second (Np/s), because they are associated with the natural response of the circuit; *ω*0 is kno wn as the *resonant frequency* or strictly as the *undamped natural frequency*, expressed in radians per second (rad/s); and *α* is the *neper frequency* expressed in nepers per second. In terms of *α* and *ω*0, Eq. (8.8) can be written as
-
-$$
-x^2 + 2\alpha s + \omega_0^2 = 0
-$$
- (8.8a)
-
-The variables *s* and *ω*0 are important quantities we will be discussing throughout the rest of the text.
-
-*s*
-
-The two values of *s* in Eq. (8.10) indicate that there are two possible solutions for *i*, each of which is of the form of the assumed solution in Eq. (8.6); that is,
-
-$$
-i_1 = A_1 e^{s_1 t}, \qquad i_2 = A_2 e^{s_2 t} \tag{8.12}
-$$
-
-Since Eq. (8.4) is a linear equation, an y linear combination of the tw o distinct solutions *i*1 and *i*2 is also a solution of Eq. (8.4). A complete or total solution of Eq. (8.4) would therefore require a linear combination of *i*1 and *i*2. Thus, the natural response of the series *RLC* circuit is
-
-$$
-i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}
-$$
- (8.13)
-
-where the constants *A*1 and *A*2 are determined from the initial values *i*(0) and *di*(0)∕*dt* in Eqs. (8.2b) and (8.5).
-
-From Eq. (8.10), we can infer that there are three types of solutions:
-
-1. If *α* > *ω*0, we have the *overdamped* case.
-
-- 2. If *α* = *ω*0, we have the *critically damped* case.
-- 3. If *α* < *ω*0, we have the *underdamped* case.
-
-We will consider each of these cases separately.
-
-# **Overdamped Case (***α* > *ω***0)**
-
-From Eqs. (8.9) and (8.10), *α* > *ω*0 implies *C* > 4*L*∕*R*2 . When this happens, both roots *s*1 and *s*2 are negative and real. The response is
-
-$$
-i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}
-$$
- (8.14)
-
-which decays and approaches zero as *t* increases. Figure 8.9(a) illustrates a typical overdamped response.
-
-# **Critically Damped Case (***α* **=** *ω***0)**
-
-When *α* = *ω*0, *C* = 4*L*∕*R*2 and
-
-$$
-s_1 = s_2 = -\alpha = -\frac{R}{2L}
-$$
- (8.15)
-
-The response is overdamped when the roots of the circuit's characteristic equation are unequal and real, critically damped when the roots are equal and real, and underdamped when the roots are complex.
-
-The neper (Np) is a dimensionless unit named after John Napier (1550–1617), a Scottish mathematician.
-
-The ratio *α*/*ω*0 is known as the damping
-
-ratio ζ.
-
-For this case, Eq. (8.13) yields
-
-$$
-i(t) = A_1 e^{-\alpha t} + A_2 e^{-\alpha t} = A_3 e^{-\alpha t}
-$$
-
-where *A*3 = *A*1 + *A*2 . This cannot be the solution, because the two initial conditions cannot be satisfied with the single constant *A*3. What then could be wrong? Our assumption of an e xponential solution is incor rect for the special case of critical damping. Let us go back to Eq. (8.4). When *α* = *ω*0 = *R*∕2*L*, Eq. (8.4) becomes
-
-$$
-\frac{d^2i}{dt^2} + 2\alpha \frac{di}{dt} + \alpha^2 i = 0
-$$
-
-$$
-\frac{d}{dt}\left(\frac{di}{dt} + \alpha i\right) + \alpha \left(\frac{di}{dt} + \alpha i\right) = 0
-$$
-\n(8.16)
-
-If we let
-
-or
-
-$$
-f = \frac{di}{dt} + \alpha i \tag{8.17}
-$$
-
-then Eq. (8.16) becomes
-
-$$
-\frac{df}{dt} + \alpha f = 0
-$$
-
-which is a first-order differential equation with solution *f* = *A*1*e*−*αt* , where *A*1 is a constant. Equation (8.17) then becomes
-
-$$
-\frac{di}{dt} + \alpha i = A_1 e^{-\alpha t}
-$$
-
-$$
-e^{\alpha t} \frac{di}{dt} + e^{\alpha t} \alpha i = A_1 \tag{8.18}
-$$
-
-This can be written as
-
-$$
-\frac{d}{dt}(e^{at}i) = A_1 \tag{8.19}
-$$
-
-Integrating both sides yields
-
-$$
-e^{\alpha t}i = A_1t + A_2
-$$
-
-or
-
-or
-
-$$
-i = (A_1 t + A_2)e^{-\alpha t}
-$$
- (8.20)
-
-where *A*2 is another constant. Hence, the natural response of the critically damped circuit is a sum of two terms: a negative exponential and a negative exponential multiplied by a linear term, or
-
-$$
-i(t) = (A_2 + A_1 t)e^{-\alpha t}
-$$
- (8.21)
-
-A typical critically damped response is sho wn in Fig. 8.9(b). In f act, Fig. 8.9(b) is a sk etch of *i*(*t*) = *te*−*αt* , which reaches a maximum v alue of *e* −1 ∕*α* at *t* = 1∕*α*, one time constant, and then decays all the way to zero.
-
-(a) Overdamped response, (b) critically damped response, (c) underdamped response.
-
-(c)
-
-# **Underdamped Case (***α* < *ω***0)**
-
-For *α* < *ω*0, *C* < 4*L*∕*R*2 . The roots may be written as
-
-$$
-s_1 = -\alpha + \sqrt{-(\omega_0^2 - \alpha^2)} = -\alpha + j\omega_d
-$$
- (8.22a)
-
-$$
-s_2 = -\alpha - \sqrt{-(\omega_0^2 - \alpha^2)} = -\alpha - j\omega_d
-$$
- (8.22b)
-
-where *j* = √ \_\_\_ −1 and *ωd* = √ \_\_\_\_\_\_\_ *ω* 0 2− *α*2 ,which is called the *damped frequency*. Both *ω*0 and *ωd* are natural frequencies because they help determine the natural response; while *ω*0 is often called the *undamped natur al fr equency, ωd* is called the *damped natural frequency*. The natural response is
-
-$$
-i(t) = A_1 e^{-(\alpha - j\omega_d)t} + A_2 e^{-(\alpha + j\omega_d)t}
-$$
-
-= $e^{-\alpha t} (A_1 e^{-j\omega_d t} + A_2 e^{-j\omega_d t})$ (8.23)
-
-Using Euler's identities,
-
-$$
-e^{j\theta} = \cos\theta + j\sin\theta, \qquad e^{-j\theta} = \cos\theta - j\sin\theta \qquad (8.24)
-$$
-
-we get
-
-$$
-i(t) = e^{-\alpha t} [A_1(\cos \omega_d t + j \sin \omega_d t) + A_2(\cos \omega_d t - j \sin \omega_d t)]
-$$
-
-= $e^{-\alpha t} [(A_1 + A_2) \cos \omega_d t + j(A_1 - A_2) \sin \omega_d t]$ (8.25)
-
-Replacing constants ( *A*1 + *A*2) and *j*(*A*1 − *A*2) with constants *B*1 and *B*2, we write
-
-$$
-i(t) = e^{-\alpha t} (B_1 \cos \omega_d t + B_2 \sin \omega_d t)
-$$
- (8.26)
-
-With the presence of sine and cosine functions, it is clear that the natural response for this case is e xponentially damped and oscillatory in nature. The response has a time constant of 1 ∕*α* and a period of *T* = 2*π*∕*ωd*. Figure 8.9(c) depicts a typical underdamped response. Part (a) and (b) of Fig. 8.9 assume for each case that *i*(0) = 0.
-
-Once the inductor current *i*(*t*) is found for the *RLC* series circuit as shown above, other circuit quantities such as indi vidual element voltages can easily be found. F or example, the resistor voltage is *vR* = *Ri*, and the inductor voltage is *vL* = *L di*∕*dt*. The inductor current *i*(*t*) is selected as the key variable to be determined first in order to take advantage of Eq. (8.1b).
-
-We conclude this section by noting the follo wing interesting, peculiar properties of an *RLC* network:
-
-- 1. The behavior of such a network is captured by the idea of *damping,* which is the gradual loss of the initial stored ener gy, as evidenced by the continuous decrease in the amplitude of the response. The damping effect is due to the presence of resistance *R*. The neper frequency *α* determines the rate at which the response is damped. If *R* = 0, then *α* = 0, and we have an *LC* circuit with 1∕√ \_\_\_ *LC* as the undamped natural frequency. Since *α* < *ω*0 in this case, the response is not only undamped b ut also oscillatory. The circuit is said to be *loss-less,* because the dissipating or damping element (*R*) is absent. By adjusting the value of *R*, the response may be made undamped, overdamped, critically damped, or underdamped.
-- 2. Oscillatory response is possible due to the presence of the tw o types of storage elements. Ha ving both *L* and *C* allows the flow
-
-R = 0 produces a perfectly sinusoidal response. This response cannot be practically accomplished with L and C because of the inherent losses in them. See Figs 6.8 and 6.26. An electronic device called an oscillator can produce a perfectly sinusoidal response.
-
-Examples 8.5 and 8.7 demonstrate the effect of varying R.
-
-The response of a second-order circuit with two storage elements of the same type, as in Fig. 8.1(c) and (d), cannot be oscillatory.
-
-of energy back and forth between the tw o. The damped oscillation exhibited by the underdamped response is kno wn as *ringing*. It stems from the ability of the storage elements *L* and *C* to transfer energy back and forth between them.
-
-3. Observe from Fig. 8.9 that the w aveforms of the responses dif fer. In general, it is dif ficult to tell from the waveforms the dif ference between the overdamped and critically damped responses. The critically damped case is the borderline between the underdamped and overdamped cases and it decays the fastest. With the same initial conditions, the overdamped case has the longest settling time, because it tak es the longest time to dissipate the initial stored energy. If we desire the response that approaches the final value most rapidly without oscillation or ringing, the critically damped circuit is the right choice.
-
-circuit that is as close as possible to a critically damped circuit.
-
-What this means in most practical circuits is that we seek an overdamped
-
-Example 8.3
-
-In Fig. 8.8, *R* = 40 Ω, *L* = 4 H, and *C* = 1∕4 F. Calculate the charac teristic roots of the circuit. Is the natural response o verdamped, underdamped, or critically damped?
-
-# **Solution:**
-
-We first calculate
-
-$$
-\alpha = \frac{R}{2L} = \frac{40}{2(4)} = 5
-$$
-, $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{4 \times \frac{1}{4}}} = 1$
-
-The roots are
-
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -5 \pm \sqrt{25 - 1}
-$$
-
-or
-
-$$
-s_1 = -0.101, \qquad s_2 = -9.899
-$$
-
-Since *α* > *ω*0, we conclude that the response is overdamped. This is also evident from the fact that the roots are real and negative.
-
-| Practice Problem 8.3 | If R = 10 Ω, L = 5 H, and C = 2 mF in Fig. 8.8, find α, ω0, s1,
and s2. What |
-|----------------------|---------------------------------------------------------------------------------|
-| | type of natural response will the circuit have? |
-
-**Answer:** 1, 10, −1 ± *j*9.95, underdamped.
-
-| Example 8.4 | | Find i(t) in the circuit of Fig. 8.10. Assume that the circuit has reached |
-|-------------|-------------------------|----------------------------------------------------------------------------|
-| | steady state at t = 0−. | |
-
-# **Solution:**
-
-For *t* < 0, the switch is closed. The capacitor acts like an open circuit while the inductor acts like a shunted circuit. The equivalent circuit is shown in Fig. 8.11(a). Thus, at *t* = 0,
-
-$$
-i(0) = \frac{10}{4+6} = 1 \text{ A}, \qquad v(0) = 6i(0) = 6 \text{ V}
-$$
-
-The circuit in Fig. 8.10: (a) for *t* < 0, (b) for *t* > 0.
-
-where *i*(0) is the initial current through the inductor and *v*(0) is the initial voltage across the capacitor.
-
-For *t* > 0, the switch is opened and the v oltage source is discon nected. The equivalent circuit is shown in Fig. 8.11(b), which is a sourcefree series *RLC* circuit. Notice that the 3-Ω and 6-Ω resistors, which are in series in Fig. 8.10 when the switch is opened, have been combined to give *R* = 9 Ω in Fig. 8.11(b). The roots are calculated as follows:
-
-$$
-\alpha = \frac{R}{2L} = \frac{9}{2\left(\frac{1}{2}\right)} = 9, \qquad \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\frac{1}{2} \times \frac{1}{50}}} = 10
-$$
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -9 \pm \sqrt{81 - 100}
-$$
-
-or
-
-*s*1,2 = −9 ± *j*4.359
-
-Hence, the response is underdamped (*α* < *ω*); that is,
-
-$$
-i(t) = e^{-9t} (A_1 \cos 4.359t + A_2 \sin 4.359t)
-$$
- (8.4.1)
-
-We now obtain *A*1 and *A*2 using the initial conditions. At *t* = 0,
-
-$$
-i(0) = 1 = A_1 \tag{8.4.2}
-$$
-
-From Eq. (8.5),
-
-$$
-\left. \frac{di}{dt} \right|_{t=0} = -\frac{1}{L} [Ri(0) + v(0)] = -2[9(1) - 6] = -6 \text{ A/s}
-$$
- (8.4.3)
-
-Note that *v*(0) = *V*0 = −6 V is used, because the polarity of *v* in Fig. 8.11(b) is opposite that in Fig. 8.8. Taking the derivative of *i*(*t*) in Eq. (8.4.1),
-
-$$
-\frac{di}{dt} = -9e^{-9t}(A_1 \cos 4.359t + A_2 \sin 4.359t) \n+ e^{-9t}(4.359)(-A_1 \sin 4.359t + A_2 \cos 4.359t)
-$$
-
-Imposing the condition in Eq. (8.4.3) at *t* = 0 gives
-
-$$
--6 = -9(A_1 + 0) + 4.359(-0 + A_2)
-$$
-
-But *A*1 = 1 from Eq. (8.4.2). Then
-
-$$
--6 = -9 + 4.359A_2
-$$
- $\Rightarrow$ $A_2 = 0.6882$
-
-Substituting the values of *A*1 and *A*2 in Eq. (8.4.1) yields the com plete solution as
-
-$$
-i(t) = e^{-9t}(\cos 4.359t + 0.6882 \sin 4.359t) A
-$$
-
-# Practice Problem 8.4
-
-**Figure 8.12** For Practice Prob. 8.4.
-
-**Figure 8.13** A source-free parallel *RLC* circuit.
-
-The circuit in Fig. 8.12 has reached steady state at *t* = 0−. If the makebefore-break switch moves to position *b* at *t* = 0, calculate *i*(*t*) for *t* > 0.
-
-**Answer:**
-$$
-e^{-2.5t} (10 \cos 1.6583t - 15.076 \sin 1.6583t)
-$$
- A.
-
-# **8.4** The Source-Free Parallel RLC Circuit
-
-Parallel *RLC* circuits find many practical applications, notably in communications networks and filter designs.
-
-Consider the parallel *RLC* circuit shown in Fig. 8.13. Assume initial inductor current *I*0 and initial capacitor voltage *V*0,
-
-$$
-i(0) = I_0 = \frac{1}{L} \int_{-\infty}^{0} v(t) dt
-$$
- (8.27a)
-
-$$
-v(0) = V_0 \tag{8.27b}
-$$
-
-Because the three elements are in parallel, they have the same voltage *v* across them. According to passive sign convention, the current is entering each element; that is, the current through each element is leaving the top node. Thus, applying KCL at the top node gives
-
-$$
-\frac{v}{R} + \frac{1}{L} \int_{-\infty}^{t} v(\tau) \, d\tau + C \frac{dv}{dt} = 0 \tag{8.28}
-$$
-
-Taking the derivative with respect to *t* and dividing by *C* results in
-
-$$
-\frac{d^2v}{dt^2} + \frac{1}{RC}\frac{dv}{dt} + \frac{1}{LC}v = 0
-$$
- (8.29)
-
-We obtain the characteristic equation by replacing the first derivative by *s* and the second derivative by *s* 2 . By following the same reasoning used in establishing Eqs. (8.4) through (8.8), the characteristic equation is obtained as
-
-$$
-s^2 + \frac{1}{RC} s + \frac{1}{LC} = 0
-$$
- (8.30)
-
-The roots of the characteristic equation are
-
-$$
-s_{1,2} = -\frac{1}{2RC} \pm \sqrt{\left(\frac{1}{2RC}\right)^2 - \frac{1}{LC}}
-$$
-
-or
-
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}
-$$
- (8.31)
-
-where
-
-$$
-\alpha = \frac{1}{2RC}, \qquad \omega_0 = \frac{1}{\sqrt{LC}} \tag{8.32}
-$$
-
-The names of these terms remain the same as in the preceding section, as they play the same role in the solution. Again, there are three possible solutions, depending on whether *α* > *ω*0, *α* = *ω*0, or *α* < *ω*0. Let us consider these cases separately.
-
-# **Overdamped Case (***α* **>** *ω***0)**
-
-From Eq. (8.32), *α* > *ω*0 when *L* > 4*R*2 *C*. The roots of the characteristic equation are real and negative. The response is
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/092_8.4 The Source-Free Parallel RLC Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/092_8.4 The Source-Free Parallel RLC Circuit.md
deleted file mode 100644
index 27ebcbd2af96ca304d54840a7cbd1c3bd002527b..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/092_8.4 The Source-Free Parallel RLC Circuit.md
+++ /dev/null
@@ -1,343 +0,0 @@
-
-$$
-v(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}
-$$
- (8.33)
-
-# **Critically Damped Case (***α* **=** *ω***0)**
-
-For *α* = *ω*0, *L* = 4*R*2 *C*. The roots are real and equal so that the response is
-
-$$
-v(t) = (A_1 + A_2 t)e^{-\alpha t}
-$$
- (8.34)
-
-# **Underdamped Case (***α* **<** *ω***0)**
-
-When *α* < *ω*0, *L* < 4*R*2 *C*. In this case the roots are complex and may be expressed as
-
-$$
-s_{1,2} = -\alpha \pm j\omega_d \tag{8.35}
-$$
-
-where
-
-$$
-\omega_d = \sqrt{\omega_0^2 - \alpha^2} \tag{8.36}
-$$
-
-The response is
-
-$$
-v(t) = e^{-at}(A_1 \cos \omega_d t + A_2 \sin \omega_d t)
-$$
- (8.37)
-
-The constants *A*1 and *A*2 in each case can be determined from the initial conditions. We need *v*(0) and *dv*(0)∕*dt*. The first term is known from Eq. (8.27b). We find the second term by combining Eqs. (8.27) and (8.28), as
-
-> *dv*(0) \_\_\_\_\_ *dt* =0
-
-+ *I*0 + *C*
-
- \_\_\_ *V*0 *R*
-
-or
-
-$$
-\frac{dv(0)}{dt} = -\frac{(V_0 + RI_0)}{RC}
-$$
- (8.38)
-
-The voltage waveforms are similar to those sho wn in Fig. 8.9 and will depend on whether the circuit is overdamped, underdamped, or critically damped.
-
-Having found the capacitor voltage *v*(*t*) for the parallel *RLC* circuit as shown above, we can readily obtain other circuit quantities such as indi vidual element currents. F or e xample, the resistor current is *iR* = *v*∕*R* and the capacitor current is *iC* = *C dv*∕*dt*. We have selected the capacitor voltage *v*(*t*) as the key variable to be determined first in order to take advantage of Eq. (8.1a). Notice that we first found the inductor current *i*(*t*) for the *RLC* series circuit, whereas we first found the capacitor voltage *v*(*t*) for the parallel *RLC* circuit.
-
-Example 8.5
-
-In the parallel circuit of Fig. 8.13, find *v*(*t*) for *t* > 0, assuming *v*(0) = 5 V, *i*(0) = 0, *L* = 1 H, and *C* = 10 mF. Consider these cases: *R* = 1.923 Ω, *R* = 5 Ω, and *R* = 6.25 Ω.
-
-# **Solution:**
-
-Solution:
-\n**CASE 1** If
-$$
-R = 1.923 \Omega
-$$
-,
-\n
-$$
-\alpha = \frac{1}{2RC} = \frac{1}{2 \times 1.923 \times 10 \times 10^{-3}} = 26
-$$
-\n
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 10 \times 10^{-3}}} = 10
-$$
-
-Since *α* > *ω*0 in this case, the response is overdamped. The roots of the characteristic equation are
-
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -2, -50
-$$
-
-and the corresponding response is
-
-$$
-v(t) = A_1 e^{-2t} + A_2 e^{-50t}
-$$
- (8.5.1)
-
-We now apply the initial conditions to get *A*1 and *A*2.
-
-$$
-v(0) = 5 = A_1 + A_2 \tag{8.5.2}
-$$
-
-$$
-v(0) = 5 = A_1 + A_2
-$$
-$$
-\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{1.923 \times 10 \times 10^{-3}} = -260
-$$
-
-But differentiating Eq. (8.5.1),
-
-$$
-\frac{dv}{dt} = -2A_1e^{-2t} - 50A_2e^{-50t}
-$$
-
-$$
-At t = 0,
-$$
-
-$$
--260 = -2A_1 - 50A_2 \tag{8.5.3}
-$$
-
-From Eqs. (8.5.2) and (8.5.3), we obtain *A*1 = −0.2083 and *A*2 = 5.208. Substituting *A*1 and *A*2 in Eq. (8.5.1) yields
-
-$$
-v(t) = -0.2083e^{-2t} + 5.208e^{-50t}
-$$
- (8.5.4)
-
-■ **CASE 2** When *R* = 5 Ω,
-
-en
-$$
-R = 5 \Omega
-$$
-,
-\n
-$$
-\alpha = \frac{1}{2RC} = \frac{1}{2 \times 5 \times 10 \times 10^{-3}} = 10
-$$
-
-while *ω*0 = 10 remains the same. Since *α* = *ω*0 = 10, the response is critically damped. Hence, *s*1 = *s*2 = −10, and
-
-$$
-v(t) = (A_1 + A_2 t)e^{-10t}
-$$
-\n(8.5.5)
-
-To get *A*1 and *A*2, we apply the initial conditions
-
-$$
-v(0) = 5 = A_1 \tag{8.5.6}
-$$
-
-$$
-v(0) = 5 = A_1
-$$
-$$
-\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{5 \times 10 \times 10^{-3}} = -100
-$$
-
-But differentiating Eq. (8.5.5),
-
-$$
-\frac{dv}{dt} = (-10A_1 - 10A_2t + A_2)e^{-10t}
-$$
-
-At *t* = 0,
-
-$$
--100 = -10A_1 + A_2 \tag{8.5.7}
-$$
-
-From Eqs. (8.5.6) and (8.5.7), *A*1 = 5 and *A*2 = −50. Thus,
-
-$$
-v(t) = (5 - 50t)e^{-10t} \text{ V}
-$$
- (8.5.8)
-
-■ **CASE 3** When *R* = 6.25 Ω,
-
-$$
-\text{hen } R = 6.25 \, \Omega,
-$$
-\n
-$$
-\alpha = \frac{1}{2RC} = \frac{1}{2 \times 6.25 \times 10 \times 10^{-3}} = 8
-$$
-
-while *ω*0 = 10 remains the same. As *α* < *ω*0 in this case, the response is underdamped. The roots of the characteristic equation are
-
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -8 \pm j6
-$$
-
-Hence,
-
-$$
-v(t) = (A_1 \cos 6t + A_2 \sin 6t)e^{-8t}
-$$
- (8.5.9)
-
-We now obtain *A*1 and *A*2, as
-
-$$
-v(0) = 5 = A_1 \tag{8.5.10}
-$$
-
-$$
-\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{6.25 \times 10 \times 10^{-3}} = -80
-$$
-
-But differentiating Eq. (8.5.9),
-
-$$
-\frac{dv}{dt} = (-8A_1 \cos 6t - 8A_2 \sin 6t - 6A_1 \sin 6t + 6A_2 \cos 6t)e^{-8t}
-$$
-
-At $t = 0$ ,
-$$
--80 = -8A_1 + 6A_2
-$$
- (8.5.11)
-
-From Eqs. (8.5.10) and (8.5.11), *A*1 = 5 and *A*2 = −6.667. Thus,
-
-$$
-v(t) = (5 \cos 6t - 6.667 \sin 6t)e^{-8t}
-$$
- (8.5.12)
-
-Notice that by increasing the value of *R*, the degree of damping decreases and the responses differ. Figure 8.14 plots the three cases.
-
-For Example 8.5: responses for three degrees of damping.
-
-Practice Problem 8.5
-
-In Fig. 8.13, let *R* = 2 Ω, *L* = 0.4 H, *C* = 25 mF, *v*(0) = 0, *i*(0) = 50 mA. Find *v*(*t*) for *t* > 0.
-
-**Answer:** −2*te*−10*t* V.
-
-Example 8.6 Find *v*(*t*) for *t* > 0 in the *RLC* circuit of Fig. 8.15.
-
-# **Solution:**
-
-When *t* < 0, the switch is open; the inductor acts like a short circuit while the capacitor behaves like an open circuit. The initial voltage across the capacitor is the same as the voltage across the 50-Ω resistor; that is,
-
-$$
-v(0) = \frac{50}{30 + 50}(40) = \frac{5}{8} \times 40 = 25 \text{ V}
-$$
- (8.6.1)
-
-The initial current through the inductor is
-
-$$
-i(0) = -\frac{40}{30 + 50} = -0.5 \text{ A}
-$$
-
-The direction of *i* is as indicated in Fig. 8.15 to conform with the direc tion of *I*0 in Fig. 8.13, which is in agreement with the convention t hat current flows into the positive terminal of an inductor (see Fig. 6.23). We need to express this in terms of *dv*∕*dt*, since we are looking for *v*.
-
-need to express this in terms of
-$$
-dv/dt
-$$
-, since we are looking for *v*.
-\n
-$$
-\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{25 - 50 \times 0.5}{50 \times 20 \times 10^{-6}} = 0
-$$
-\n(8.6.2)
-
-When *t* > 0, the switch is closed. The voltage source along with the 30-Ω resistor is separated from the rest of the circuit. The parallel *RLC* circuit acts independently of the voltage source, as illustrated in Fig. 8.16. Next, we determine that the roots of the characteristic equation are
-
-circuit acts independently of the voltage source, as illustrated in
-\nNext, we determine that the roots of the characteristic equation
-\n
-$$
-\alpha = \frac{1}{2RC} = \frac{1}{2 \times 50 \times 20 \times 10^{-6}} = 500
-$$
-\n
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{0.4 \times 20 \times 10^{-6}} = 354
-$$
-\n
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}
-$$
-\n
-$$
-= -500 \pm \sqrt{250,000 - 124,997.6} = -500 \pm 354
-$$
-
-or
-
-$$
-s_1 = -854, \qquad s_2 = -146
-$$
-
-The circuit in Fig. 8.15 when *t* > 0. The parallel *RLC* circuit on the right-hand side acts independently of the circuit on the left-hand side of the junction.
-
-Since *α* > *ω*0, we have the overdamped response
-
-$$
-v(t) = A_1 e^{-854t} + A_2 e^{-146t}
-$$
- (8.6.3)
-
-At *t* = 0, we impose the condition in Eq. (8.6.1),
-
-$$
-v(0) = 25 = A_1 + A_2
-$$
- $\Rightarrow$ $A_2 = 25 - A_1$ (8.6.4)
-
-Taking the derivative of *v*(*t*) in Eq. (8.6.3),
-
-$$
-\frac{dv}{dt} = -854A_1e^{-854t} - 146A_2e^{-146t}
-$$
-
-Imposing the condition in Eq. (8.6.2),
-
-$$
-\frac{dv(0)}{dt} = 0 = -854A_1 - 146A_2
-$$
-
-or
-
-$$
-0 = 854A_1 + 146A_2 \tag{8.6.5}
-$$
-
-Solving Eqs. (8.6.4) and (8.6.5) gives
-
-$$
-A_1 = -5.156, \qquad A_2 = 30.16
-$$
-
-Thus, the complete solution in Eq. (8.6.3) becomes
-
-$$
-v(t) = -5.156e^{-854t} + 30.16e^{-146t}
-$$
- V
-
-Refer to the circuit in Fig. 8.17. Find *v*(*t*) for *t* > 0.
-
-**Answer:** 50(*e*−10*t* − *e*−2.5*t* ) V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/093_8.5 Step Response of a Series RLC Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/093_8.5 Step Response of a Series RLC Circuit.md
deleted file mode 100644
index da8fb471635089f483ba21d7b6a89726b7440776..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/093_8.5 Step Response of a Series RLC Circuit.md
+++ /dev/null
@@ -1,349 +0,0 @@
-# **8.5** Step Response of a Series RLC Circuit
-
-As we learned in the preceding chapter, the step response is obtained by the sudden application of a dc source. Consider the series *RLC* circuit shown in Fig. 8.18. Applying KVL around the loop for *t* > 0,
-
-$$
-L\frac{di}{dt} + Ri + v = V_s \tag{8.39}
-$$
-
-# **Figure 8.17** For Practice Prob. 8.6.
-
-Vs R L C i t = 0 *v* + ‒ + ‒
-
-**Figure 8.18** Step voltage applied to a series *RLC* circuit.
-
-But
-
-$$
-i = C \frac{dv}{dt}
-$$
-
-Substituting for *i* in Eq. (8.39) and rearranging terms,
-
-$$
-\frac{d^2v}{dt^2} + \frac{R}{L}\frac{dv}{dt} + \frac{v}{LC} = \frac{V_s}{LC}
-$$
- (8.40)
-
-which has the same form as Eq. (8.4). More specifically, the coefficients are the same (and that is important in determining the frequency parameters) but the variable is different. (Likewise, see Eq. (8.47).) Hence, the characteristic equation for the series *RLC* circuit is not af fected by the presence of the dc source.
-
-The solution to Eq. (8.40) has two components: the transient re sponse *vt*(*t*) and the steady-state response *vss*(*t*); that is,
-
-$$
-v(t) = v_t(t) + v_{ss}(t)
-$$
- (8.41)
-
-The transient response *vt*(*t*) is the component of the total response that dies out with time. The form of the transient response is the same as the form of the solution obtained in Section 8.3 for the source-free circuit, given by Eqs. (8.14), (8.21), and (8.26). Therefore, the transient response *vt*(*t*) for the overdamped, underdamped, and critically damped cases are:
-
-$$
-v_t(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} \qquad \text{(Overdamped)} \tag{8.42a}
-$$
-
-$$
-v_t(t) = (A_1 + A_2 t)e^{-\alpha t}
-$$
- (Critically damped) (8.42b)
-
-$$
-v_t(t) = (A_1 \cos \omega_d t + A_2 \sin \omega_d t)e^{-\alpha t}
-$$
- (Underdamped) (8.42c)
-
-The steady-state response is the final value of *v*(*t*). In the circuit in Fig. 8.18, the final value of the capacitor voltage is the same as the source voltage *Vs*. Hence,
-
-$$
-v_{ss}(t) = v(\infty) = V_s \tag{8.43}
-$$
-
-Thus, the complete solutions for the o verdamped, underdamped, and critically damped cases are:
-
-| v(t) = Vs + A1es1t
+ A2es2t
(Overdamped) | (8.44a) |
-|------------------------------------------------------------|---------|
-| v(t) = Vs + (A1 + A2t)e−αt
(Critically damped) | (8.44b) |
-| v(t) = Vs + (A1 cos ωdt + A2 sin ωdt)e−αt
(Underdamped) | (8.44c) |
-
-The values of the constants *A*1 and *A*2 are obtained from the initial conditions: *v*(0) and *dv*(0)∕*dt*. Keep in mind that *v* and *i* are, respectively, the voltage across the capacitor and the current through the inductor. Therefore, Eq. (8.44) only applies for finding *v*. But once the capacitor voltage *vC* = *v* is known, we can determine *i* = *C dv*∕*dt*, which is the same current through the capacitor , inductor, and resistor. Hence, the v oltage across the resistor is *vR* = *iR*, while the inductor voltage is *vL* = *L di*∕*dt*.
-
-Alternatively, the complete response for any v ariable *x*(*t*) can be found directly, because it has the general form
-
-$$
-x(t) = x_{ss}(t) + x_t(t)
-$$
- (8.45)
-
-where the *xss* = *x*(∞) is the final value and *xt*(*t*) is the transient response. The final value is found as in Section 8.2. The transient response has the same form as in Eq. (8.42), and the associated constants are determined from Eq. (8.44) based on the values of *x*(0) and *dx*(0)∕*dt*.
-
-# **Solution:**
-
-■ **CASE 1** When *R* = 5 Ω. For *t* < 0, the switch is closed for a long time. The capacitor behaves like an open circuit while the inductor acts like a short circuit. The initial current through the inductor is
-
-$$
-i(0) = \frac{24}{5+1} = 4 \text{ A}
-$$
-
-and the initial voltage across the capacitor is the same as the voltage across the 1-Ω resistor; that is,
-
-$$
-v(0) = 1i(0) = 4 \text{ V}
-$$
-
-For *t* > 0, the switch is opened, so that we ha ve the 1- Ω resistor disconnected. What remains is the series *RLC* circuit with the v oltage source. The characteristic roots are determined as follows:
-
-$$
-\alpha = \frac{R}{2L} = \frac{5}{2 \times 1} = 2.5, \qquad \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 0.25}} = 2
-$$
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -1, -4
-$$
-
-Since *α* > *ω*0, we have the overdamped natural response. The total response is therefore
-
-$$
-v(t) = v_{ss} + (A_1 e^{-t} + A_2 e^{-4t})
-$$
-
-where *vss* is the steady-state response. It is the final value of the capacitor voltage. In Fig. 8.19, *vf* = 24 V. Thus,
-
-$$
-v(t) = 24 + (A_1 e^{-t} + A_2 e^{-4t})
-$$
- (8.7.1)
-
-We now need to find *A*1 and *A*2 using the initial conditions.
-
-$$
-v(0) = 4 = 24 + A_1 + A_2
-$$
-
-or
-
-$$
--20 = A_1 + A_2 \tag{8.7.2}
-$$
-
-The current through the inductor cannot change abruptly and is the same current through the capacitor at *t* = 0+ because the inductor and capacitor are now in series. Hence,
-
-$$
-i(0) = C \frac{dv(0)}{dt} = 4 \qquad \Rightarrow \qquad \frac{dv(0)}{dt} = \frac{4}{C} = \frac{4}{0.25} = 16
-$$
-
-Before we use this condition, we need to take the derivative of *v* in Eq. (8.7.1).
-
-$$
-\frac{dv}{dt} = -A_1 e^{-t} - 4A_2 e^{-4t}
-$$
- (8.7.3)
-
-At *t* = 0,
-
-$$
-\frac{dv(0)}{dt} = 16 = -A_1 - 4A_2 \tag{8.7.4}
-$$
-
-R 1 H
-
-i
-
-24 V
-
-+ ‒
-
-Example 8.7
-
-t = 0
-
-0.25 F 1 Ω
-
-*v* + ‒
-
-From Eqs. (8.7.2) and (8.7.4), *A*1 = −64∕3 and *A*2 = 4∕3. Substituting *A*1 and *A*2 in Eq. (8.7.1), we get
-
-$$
-v(t) = 24 + \frac{4}{3}(-16e^{-t} + e^{-4t}) \text{ V}
-$$
- (8.7.5)
-
-Since the inductor and capacitor are in series for *t* > 0, the inductor current is the same as the capacitor current. Hence,
-
-$$
-i(t) = C \frac{dv}{dt}
-$$
-
-Multiplying Eq. (8.7.3) by *C* = 0.25 and substituting the values of *A*1 and *A*2 gives
-
-$$
-i(t) = \frac{4}{3}(4e^{-t} - e^{-4t}) \text{ A}
-$$
- (8.7.6)
-
-Note that *i*(0) = 4 A, as expected.
-
-■ **CASE 2** When *R* = 4 Ω. Again, the initial current through the inductor is
-
-$$
-i(0) = \frac{24}{4+1} = 4.8 \text{ A}
-$$
-
-and the initial capacitor voltage is
-
-$$
-v(0) = 1i(0) = 4.8 \text{ V}
-$$
-
-For the characteristic roots,
-
-$$
-\alpha = \frac{R}{2L} = \frac{4}{2 \times 1} = 2
-$$
-
-while *ω*0 = 2 remains the same. In this case, *s*1 = *s*2 = −*α* = −2, and we have the critically damped natural response. The total response is therefore
-
-$$
-v(t) = v_{ss} + (A_1 + A_2 t)e^{-2t}
-$$
-
-and, as before *vss* = 24 V,
-
-$$
-v(t) = 24 + (A_1 + A_2 t)e^{-2t}
-$$
- (8.7.7)
-
-To find *A*1 and *A*2, we use the initial conditions. We write
-
-$$
-v(0) = 4.8 = 24 + A_1
-$$
- $\Rightarrow$ $A_1 = -19.2$ (8.7.8)
-
-Since *i*(0) = *C dv*(0)∕*dt* = 4.8 or
-
-$$
-\frac{dv(0)}{dt} = \frac{4.8}{C} = 19.2
-$$
-
-From Eq. (8.7.7),
-
-$$
-\frac{dv}{dt} = (-2A_1 - 2tA_2 + A_2)e^{-2t}
-$$
-\n(8.7.9)
-
-At *t* = 0,
-
-$$
-\frac{dv(0)}{dt} = 19.2 = -2A_1 + A_2 \tag{8.7.10}
-$$
-
-From Eqs. (8.7.8) and (8.7.10), *A*1 = −19.2 and *A*2 = −19.2. Thus, Eq. (8.7.7) becomes
-
-$$
-v(t) = 24 - 19.2(1 + t)e^{-2t}
-$$
- V (8.7.11)
-
-The inductor current is the same as the capacitor current; that is,
-
-$$
-i(t) = C \frac{dv}{dt}
-$$
-
-Multiplying Eq. (8.7.9) by *C* = 0.25 and substituting the values of *A*1 and *A*2 gives
-
-$$
-i(t) = (4.8 + 9.6t)e^{-2t} A
-$$
- (8.7.12)
-
-Note that *i*(0) = 4.8 A, as expected.
-
-■ **CASE 3** When *R* = 1 Ω. The initial inductor current is
-
-$$
-i(0) = \frac{24}{1+1} = 12 \text{ A}
-$$
-
-and the initial voltage across the capacitor is the same as the voltage across the 1-Ω resistor,
-
-$$
-v(0) = 1i(0) = 12 \text{ V}
-$$
-$$
-\alpha = \frac{R}{2L} = \frac{1}{2 \times 1} = 0.5
-$$
-
-Since *α* = 0.5 < *ω*0 = 2, we have the underdamped response \_\_\_\_\_\_\_
-
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -0.5 \pm j1.936
-$$
-
-The total response is therefore
-
-$$
-v(t) = 24 + (A_1 \cos 1.936t + A_2 \sin 1.936t)e^{-0.5t}
-$$
- (8.7.13)
-
-We now determine *A*1 and *A*2. We write
-
-$$
-v(0) = 12 = 24 + A_1 \qquad \Rightarrow \qquad A_1 = -12 \tag{8.7.14}
-$$
-
-Since *i*(0) = *C dv*(0)∕*dt* = 12,
-
-$$
-\frac{dv(0)}{dt} = \frac{12}{C} = 48\tag{8.7.15}
-$$
-
-But
-
-$$
-\frac{dv}{dt} = e^{-0.5t}(-1.936A_1 \sin 1.936t + 1.936A_2 \cos 1.936t)
-$$
-$$
--0.5e^{-0.5t}(A_1 \cos 1.936t + A_2 \sin 1.936t)
-$$
-(8.7.16)
-
-At *t* = 0,
-
-$$
-\frac{dv(0)}{dt} = 48 = (-0 + 1.936A_2) - 0.5(A_1 + 0)
-$$
-
-Substituting *A*1 = −12 gives *A*2 = 21.694, and Eq. (8.7.13) becomes
-
-$$
-v(t) = 24 + (21.694 \sin 1.936t - 12 \cos 1.936t)e^{-0.5t} \text{ V}
-$$
- (8.7.17)
-
-The inductor current is
-
-$$
-i(t) = C \frac{dv}{dt}
-$$
-
-Multiplying Eq. (8.7.16) by *C* = 0.25 and substituting the values of *A*1 and *A*2 gives
-
-$$
-i(t) = (3.1 \sin 1.936t + 12 \cos 1.936t)e^{-0.5t} \text{ A}
-$$
- (8.7.18)
-
-Note that *i*(0) = 12 A, as expected.
-
-Figure 8.20 plots the responses for the three cases. From this figure, we observe that the critically damped response approaches the step input of 24 V the fastest.
-
-For Example 8.7: response for three degrees of damping.
-
-Practice Problem 8.7 Having been in position *a* for a long time, the switch in Fig. 8.21 is moved to position *b* at *t* = 0. Find *v*(*t*) and *vR*(*t*) for *t* > 0.
-
-**Figure 8.21** For Practice Prob. 8.7.
-
-**Answer:** 15 − (1.7321 sin 3.464*t* + 3 cos 3.464*t*)*e*−2*t* V, 3.464*e*−2*t* sin 3.464*t* V.
-
-**Figure 8.22** Parallel *RLC* circuit with an applied current.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/094_8.6 Step Response of a Parallel RLC Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/094_8.6 Step Response of a Parallel RLC Circuit.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/094_8.6 Step Response of a Parallel RLC Circuit.md
+++ /dev/null
@@ -1,215 +0,0 @@
-# **8.6** Step Response of a Parallel RLC Circuit
-
-Consider the parallel *RLC* circuit sho wn in Fig. 8.22. We want to find *i* due to a sudden application of a dc current. Applying KCL at the top node for *t* > 0,
-
-$$
-\frac{v}{R} + i + C\frac{dv}{dt} = I_s \tag{8.46}
-$$
-
-$$
-v = L \frac{di}{dt}
-$$
-
-Substituting for *v* in Eq. (8.46) and dividing by *LC*, we get
-
-$$
-\frac{d^2i}{dt^2} + \frac{1}{RC}\frac{di}{dt} + \frac{i}{LC} = \frac{I_s}{LC}
-$$
- (8.47)
-
-which has the same characteristic equation as Eq. (8.29).
-
-The complete solution to Eq. (8.47) consists of the transient response *it*(*t*) and the steady-state response *iss*; that is,
-
-$$
-i(t) = it(t) + iss(t)
-$$
- (8.48)
-
-The transient response is the same as what we had in Section 8.4. The steady-state response is the final value of *i*. In the circuit in Fig. 8.22, the final value of the current through the inductor is the same as the source current *Is*. Thus,
-
-$$
-i(t) = I_s + A_1 e^{s_1 t} + A_2 e^{s_2 t}
-$$
- (Overdamped)
-\n
-$$
-i(t) = I_s + (A_1 + A_2 t)e^{-\alpha t}
-$$
- (Critically damped)
-\n
-$$
-i(t) = I_s + (A_1 \cos \omega_d t + A_2 \sin \omega_d t)e^{-\alpha t}
-$$
- (Underdamped) (8.49)
-
-The constants *A*1 and *A*2 in each case can be determined from the initial conditions for *i* and *di*∕*dt*. Again, we should keep in mind that Eq. (8.49) only applies for finding the inductor current *i*. But once the inductor current *iL* = *i* is known, we can find *v* = *L di*∕*dt*, which is the same v oltage across inductor, capacitor, and resistor . Hence, the current through the resistor is *iR* = *v*∕*R*, while the capacitor current is *iC* = *C dv*∕*dt*. Alternatively, the complete response for any variable *x*(*t*) may be found directly, using
-
-$$
-x(t) = x_{ss}(t) + x_t(t)
-$$
- (8.50)
-
-where *xss* and *xt* are its final value and transient response, respectively.
-
-# **Solution:**
-
-For *t* < 0, the switch is open, and the circuit is partitioned into two inde pendent subcircuits. The 4-A current flows through the inductor, so that
-
-$$
-i(0) = 4 \text{ A}
-$$
-
-Since 30*u*(− *t*) = 30 when *t* < 0 and 0 when *t* > 0, the voltage source is operative for *t* < 0. The capacitor acts like an open circuit and the voltage across it is the same as the voltage across the 20-Ω resistor connected in parallel with it. By voltage division, the initial capacitor voltage is
-
-$$
-v(0) = \frac{20}{20 + 20}(30) = 15 \text{ V}
-$$
-
-For *t* > 0, the switch is closed, and we ha ve a parallel *RLC* circuit with a current source. The voltage source is zero which means it acts like a short-circuit. The tw o 20-Ω resistors are no w in parallel. They are combined to gi ve *R* = 20 ‖ 20 = 10 Ω. The characteristic roots are determined as follows:
-
-$$
-\alpha = \frac{1}{2RC} = \frac{1}{20 \times 10 \times 8 \times 10^{-3}} = 6.25
-$$
-
-\n
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{20 \times 8 \times 10^{-3}}} = 2.5
-$$
-
-\n
-$$
-s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -6.25 \pm \sqrt{39.0625 - 6.25}
-$$
-
-\n
-$$
-= -6.25 \pm 5.7282
-$$
-
-or
-
-$$
-s_1 = -11.978, \qquad s_2 = -0.5218
-$$
-
-Since *α* > *ω*0, we have the overdamped case. Hence,
-
-$$
-i(t) = I_s + A_1 e^{-11.978t} + A_2 e^{-0.5218t}
-$$
- (8.8.1)
-
-where *Is* = 4 is the final value of *i*(*t*). We now use the initial conditions to determine *A*1 and *A*2. At *t* = 0,
-
-*i*(0) = 4 = 4 + *A*1 + *A*2 ⇒ *A*2 = −*A*1 **(8.8.2)**
-
-Taking the derivative of *i*(*t*) in Eq. (8.8.1),
-
-$$
-\frac{di}{dt} = -11.978A_1e^{-11.978t} - 0.5218A_2e^{-0.5218t}
-$$
-
-so that at *t* = 0,
-
-$$
-\frac{di(0)}{dt} = -11.978A_1 - 0.5218A_2 \tag{8.8.3}
-$$
-
-But
-
-$$
-L\frac{di(0)}{dt} = v(0) = 15
-$$
- $\Rightarrow$ $\frac{di(0)}{dt} = \frac{15}{L} = \frac{15}{20} = 0.75$
-
-Substituting this into Eq. (8.8.3) and incorporating Eq. (8.8.2), we get
-
-$$
-0.75 = (11.978 - 0.5218)A_2 \Rightarrow A_2 = 0.0655
-$$
-
-Thus, *A*1 = −0.0655 and *A*2 = 0.0655. Inserting *A*1 and *A*2 in Eq. (8.8.1) gives the complete solution as
-
-$$
-i(t) = 4 + 0.0655(e^{-0.5218t} - e^{-11.978t})
-$$
- A
-
-From *i*(*t*), we obtain *v*(*t*) = *L di*∕*dt* and
-
-$$
-i_R(t) = \frac{v(t)}{20} = \frac{L}{20} \frac{di}{dt} = 0.785e^{-11.978t} - 0.0342e^{-0.5218t}
-$$
- A
-
-Find *i*(*t*) and *v*(*t*) for *t* > 0 in the circuit of Fig. 8.24.
-
-# **Answer:** 10(1− cos(0.5*t*)) A, 100 sin(0.5*t*) V.
-
-Practice Problem 8.8
-
-‒
-
-# **8.7** General Second-Order Circuits
-
-Now that we have mastered series and parallel *RLC* circuits, we are prepared to apply the ideas to an y second-order circuit having one or more independent sources with constant values. Although the series and parallel *RLC* circuits are the second-order circuits of greatest interest, other second-order circuits including op amps are also useful. Given a secondorder circuit, we determine its step response *x*(*t*) (which may be voltage or current) by taking the following four steps:
-
-- 1. We first determine the initial conditions *x*(0) and *dx*(0)∕*dt* and the final value *x*(∞), as discussed in Section 8.2.
-- 2. We turn off the independent sources and find the form of the transient response *xt* (*t*) by applying KCL and KVL. Once a second-order differential equation is obtained, we determine its characteristic roots. Depending on whether the response is overdamped, critically damped, or underdamped, we obtain *xt*(*t*) with two unknown constants as we did in the previous sections.
-- 3. We obtain the steady-state response as
-
-$$
-x_{ss}(t) = x(\infty) \tag{8.51}
-$$
-
-where *x*(∞) is the final value of *x*, obtained in step 1.
-
-4. The total response is now found as the sum of the transient response and steady-state response
-
-$$
-x(t) = x_t(t) + x_{ss}(t)
-$$
- (8.52)
-
-We finally determine the constants associated with the transient response by imposing the initial conditions *x*(0) and *dx*(0)∕*dt*, determined in step 1.
-
-We can apply this general procedure to find the step response of any second-order circuit, including those with op amps. The following examples illustrate the four steps.
-
-Find the complete response *v* and then *i* for *t* > 0 in the circuit of Fig. 8.25.
-
-# **Solution:**
-
-We first find the initial and final values. At *t* = 0−, the circuit is at steady state. The switch is open; the equivalent circuit is shown in Fig. 8.26(a). It is evident from the figure that
-
-$$
-v(0^-) = 12 \text{ V}, \qquad i(0^-) = 0
-$$
-
-At *t* = 0+, the switch is closed; the equivalent circuit is in Fig. 8.26(b). By the continuity of capacitor voltage and inductor current, we know that
-
-$$
-v(0^+) = v(0^-) = 12 \text{ V}, \qquad i(0^+) = i(0^-) = 0
-$$
- (8.9.1)
-
-# **Figure 8.24** For Practice Prob. 8.8.
-
- A circuit may look complicated at first. But once the sources are turned off in an attempt to find the form of the transient response, it may be reducible to a first-order circuit, when the storage elements can be combined, or to a parallel/series RLC circuit. If it is reducible to a first-order circuit, the solution becomes simply what we had in Chapter 7. If it is reducible to a parallel or series RLC circuit, we apply the techniques of previous sections in this chapter.
-
- Problems in this chapter can also be solved by using Laplace transforms, which are covered in Chapters 15 and 16.
-
-# Example 8.9
-
-Equivalent circuit of the circuit in Fig. 8.25 for: (a) *t* < 0, (b) *t* > 0.
-
-Obtaining the form of the transient response for Example 8.9.
-
-To get *dv*(0+)∕*dt*, we use *C dv*∕*dt* = *iC* or *dv*∕*dt* = *iC*∕*C*. Applying KCL at node *a* in Fig. 8.26(b),
-
-$$
-i(0^{+}) = i_{C}(0^{+}) + \frac{v(0^{+})}{2}
-$$
-$$
-0 = i_{C}(0^{+}) + \frac{12}{2} \implies i_{C}(0^{+}) = -6 \text{ A}
-$$
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/095_8.7 General Second-Order Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/095_8.7 General Second-Order Circuits.md
deleted file mode 100644
index 322725f26cde3945d567757f2deb10eecfc1a7b6..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/095_8.7 General Second-Order Circuits.md
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@@ -1,361 +0,0 @@
-Hence,
-
-$$
-\frac{dv(0^{+})}{dt} = \frac{-6}{0.5} = -12 \text{ V/s}
-$$
- (8.9.2)
-
-The final values are obtained when the inductor is replaced by a short circuit and the capacitor by an open circuit in Fig. 8.26(b), giving
-
-$$
-i(\infty) = \frac{12}{4+2} = 2 \text{ A}, \qquad v(\infty) = 2i(\infty) = 4 \text{ V}
-$$
- (8.9.3)
-
-Next, we obtain the form of the transient response for *t* > 0. By turning off the 12-V voltage source, we have the circuit in Fig. 8.27. Applying KCL at node *a* in Fig. 8.27 gives
-
-$$
-i = \frac{v}{2} + \frac{1}{2} \frac{dv}{dt}
-$$
- (8.9.4)
-
-Applying KVL to the left mesh results in
-
-$$
-4i + 1\frac{di}{dt} + v = 0
-$$
- (8.9.5)
-
-Since we are interested in *v* for the moment, we substitute *i* from Eq. (8.9.4) into Eq. (8.9.5). We obtain
-
-$$
-2v + 2\frac{dv}{dt} + \frac{1}{2}\frac{dv}{dt} + \frac{1}{2}\frac{d^2v}{dt^2} + v = 0
-$$
-
-or
-
-$$
-\frac{d^2v}{dt^2} + 5\frac{dv}{dt} + 6v = 0
-$$
-
-From this, we obtain the characteristic equation as
-
-$$
-s^2 + 5s + 6 = 0
-$$
-
-with roots *s* = −2 and *s* = −3. Thus, the natural response is
-
-$$
-v_n(t) = Ae^{-2t} + Be^{-3t}
-$$
- (8.9.6)
-
-where *A* and *B* are unknown constants to be determined later. The steadystate response is
-
-$$
-v_{ss}(t) = v(\infty) = 4 \tag{8.9.7}
-$$
-
-The complete response is
-
-$$
-v(t) = v_t + v_{ss} = 4 + Ae^{-2t} + Be^{-3t}
-$$
- (8.9.8)
-
-We now determine *A* and *B* using the initial values. From Eq. (8.9.1), *v*(0) = 12. Substituting this into Eq. (8.9.8) at *t* = 0 gives
-
-$$
-12 = 4 + A + B \qquad \Rightarrow \qquad A + B = 8 \tag{8.9.9}
-$$
-
-Taking the derivative of *v* in Eq. (8.9.8),
-
-$$
-\frac{dv}{dt} = -2Ae^{-2t} - 3Be^{-3t}
-$$
- (8.9.10)
-
-Substituting Eq. (8.9.2) into Eq. (8.9.10) at *t* = 0 gives
-
-$$
--12 = -2A - 3B \qquad \Rightarrow \qquad 2A + 3B = 12 \tag{8.9.11}
-$$
-
-From Eqs. (8.9.9) and (8.9.11), we obtain
-
-$$
-A = 12, \qquad B = -4
-$$
-
-so that Eq. (8.9.8) becomes
-
-$$
-v(t) = 4 + 12e^{-2t} - 4e^{-3t} \text{ V}, \qquad t > 0 \tag{8.9.12}
-$$
-
-From *v*, we can obtain other quantities of interest by referring to Fig. 8.26(b). To obtain *i*, for example,
-
-$$
-i = \frac{v}{2} + \frac{1}{2} \frac{dv}{dt} = 2 + 6e^{-2t} - 2e^{-3t} - 12e^{-2t} + 6e^{-3t}
-$$
-
-= 2 - 6e^{-2t} + 4e^{-3t} A, \t t > 0 (8.9.13)
-
-Notice that *i*(0) = 0, in agreement with Eq. (8.9.1).
-
-Determine *v* and *i* for *t* > 0 in the circuit of Fig. 8.28. (See comments Practice Problem 8.9 about current sources in Practice Prob. 7.5.)
-
-**Answer:** 20(1 − *e*−5*t* ) V, 5(1 − *e*−5*t* ) A.
-
-**Figure 8.28** For Practice Prob. 8.9.
-
-Find *vo*(*t*) for *t* > 0 in the circuit of Fig. 8.29.
-
-# **Solution:**
-
-This is an example of a second-order circuit with two inductors. We first obtain the mesh currents *i*1 and *i*2, which happen to be the currents through the inductors. We need to obtain the initial and final values of these currents.
-
-For *t* < 0, 7*u*(*t*) = 0, so that *i*1(0− ) = 0 = *i*2(0− ). For *t* > 0, 7*u*(*t*) = 7, so that the equivalent circuit is as shown in Fig. 8.30(a). Due to the continuity of inductor current,
-
-$$
-i_1(0^+) = i_1(0^-) = 0,
-$$
- $i_2(0^+) = i_2(0^-) = 0$ (8.10.1)
-
-$$
-v_{L_2}(0^+) = v_o(0^+) = 1[(i_1(0^+) - i_2(0^+)] = 0 \tag{8.10.2}
-$$
-
-Applying KVL to the left loop in Fig. 8.30(a) at *t* = 0+,
-
-$$
-7 = 3i_1(0^+) + v_{L_1}(0^+) + v_o(0^+)
-$$
-
-**Figure 8.29** For Example 8.10.
-
-Example 8.10
-
-**Figure 8.30** Equivalent circuit of that in Fig. 8.29 for: (a) *t* > 0, (b) *t* → ∞.
-
-$$
-v_{L_1}(0^+) = 7 \, \text{V}
-$$
-
-Since *L*1 *di*1∕*dt* = *vL*1,
-
-$$
-\frac{di_1(0^+)}{dt} = \frac{v_{L1}}{L_1} = \frac{7}{\frac{1}{2}} = 14 \text{ A/s}
-$$
- (8.10.3)
-
-Similarly, since *L*2 *di*2∕*dt* = *vL*2,
-
-$$
-\frac{di_2(0^+)}{dt} = \frac{v_{L2}}{L_2} = 0
-$$
-\n(8.10.4)
-
-As *t* → ∞, the circuit reaches steady state, and the inductors can be replaced by short circuits, as shown in Fig. 8.30(b). From this figure,
-
-$$
-i_1(\infty) = i_2(\infty) = \frac{7}{3} A
-$$
- (8.10.5)
-
-Next, we obtain the form of the transient responses by removing the voltage source, as shown in Fig. 8.31. Applying KVL to the two meshes yields
-
-$$
-4i_1 - i_2 + \frac{1}{2}\frac{di_1}{dt} = 0
-$$
- (8.10.6)
-
-and
-
-Obtaining the form of the transient response for Example 8.10.
-
-$$
-i_2 + \frac{1}{5} \frac{di_2}{dt} - i_1 = 0 \tag{8.10.7}
-$$
-
-From Eq. (8.10.6),
-
-$$
-i_2 = 4i_1 + \frac{1}{2} \frac{di_1}{dt}
-$$
- (8.10.8)
-
-Substituting Eq. (8.10.8) into Eq. (8.10.7) gives
-
-$$
-4i_1 + \frac{1}{2}\frac{di_1}{dt} + \frac{4}{5}\frac{di_1}{dt} + \frac{1}{10}\frac{d^2i_1}{dt^2} - i_1 = 0
-$$
-$$
-\frac{d^2i_1}{dt^2} + 13\frac{di_1}{dt} + 30i_1 = 0
-$$
-
-From this we obtain the characteristic equation as
-
-$$
-s^2 + 13s + 30 = 0
-$$
-
-which has roots *s* = −3 and *s* = −10. Hence, the form of the transient response is
-
-$$
-i_{1n} = Ae^{-3t} + Be^{-10t}
-$$
- (8.10.9)
-
-**Figure 8.31**
-
-where *A* and *B* are constants. The steady-state response is
-
-$$
-i_{1ss} = i_1(\infty) = \frac{7}{3} \text{ A}
-$$
- (8.10.10)
-
-From Eqs. (8.10.9) and (8.10.10), we obtain the complete response as
-
-$$
-i_1(t) = \frac{7}{3} + Ae^{-3t} + Be^{-10t}
-$$
- (8.10.11)
-
-We finally obtain *A* and *B* from the initial values. From Eqs. (8.10.1) and (8.10.11),
-
-$$
-0 = \frac{7}{3} + A + B \tag{8.10.12}
-$$
-
-Taking the derivative of Eq. (8.10.11), setting *t* = 0 in the derivative, and enforcing Eq. (8.10.3), we obtain
-
-$$
-14 = -3A - 10B \tag{8.10.13}
-$$
-
-From Eqs. (8.10.12) and (8.10.13), *A* = −4∕3 and *B* = −1. Thus,
-
-$$
-i_1(t) = \frac{7}{3} - \frac{4}{3} e^{-3t} - e^{-10t}
-$$
- (8.10.14)
-
-We no w obtain *i*2 from *i*1. Applying KVL to the left loop in Fig. 8.30(a) gives
-
-$$
-7 = 4i_1 - i_2 + \frac{1}{2} \frac{di_1}{dt} \qquad \Rightarrow \qquad i_2 = -7 + 4i_1 + \frac{1}{2} \frac{di_1}{dt}
-$$
-
-Substituting for *i*1 in Eq. (8.10.14) gives
-
-$$
-i_2(t) = -7 + \frac{28}{3} - \frac{16}{3}e^{-3t} - 4e^{-10t} + 2e^{-3t} + 5e^{-10t}
-$$
-
-= $\frac{7}{3} - \frac{10}{3}e^{-3t} + e^{-10t}$ (8.10.15)
-
-From Fig. 8.29,
-
-**Answer:** 14(*e*−*t*
-
-$$
-v_o(t) = 1[i_1(t) - i_2(t)]
-$$
- (8.10.16)
-
-Substituting Eqs. (8.10.14) and (8.10.15) into Eq. (8.10.16) yields
-
-$$
-v_o(t) = 2(e^{-3t} - e^{-10t})
-$$
-\n(8.10.17)
-
-Note that *vo*(0) = 0, as expected from Eq. (8.10.2).
-
-) V, *t* > 0.
-
-− *e*−6*t*
-
-For *t* > 0, obtain *vo*(*t*) in the circuit of Fig. 8.32. ( *Hint:* First find *v*1 Practice Problem 8.10 and *v*2.)
-
-**Figure 8.32** For Practice Prob. 8.10.
-
-The use of op amps in second-order circuits avoids the use of inductors, which are undesirable in some applications.
-
-# **8.8** Second-Order Op Amp Circuits
-
-An op amp circuit with tw o storage elements that cannot be combined into a single equi valent element is second-order. Because inductors are bulky and hea vy, they are rarely used in practical op amp circuits. F or this reason, we will only consider *RC* second-order op amp circuits here. Such circuits find a wide range of applications in devices such as filters and oscillators.
-
-The analysis of a second-order op amp circuit follows the same four steps given and demonstrated in the previous section.
-
-Example 8.11 In the op amp circuit of Fig. 8.33, find *vo*(*t*) for *t* > 0 when *vs* = 10*u*(*t*) mV. Let *R*1 = *R*2 = 10 kΩ, *C*1 = 20 *μ*F, and *C*2 = 100 *μ*F.
-
-**Figure 8.33** For Example 8.11.
-
-# **Solution:**
-
-Although we could follow the same four steps given in the previous section to solve this problem, we will solve it a little differently. Due to the voltage follower configuration, the voltage across *C*1 is *vo*. Applying KCL at node 1,
-
-$$
-\frac{v_s - v_1}{R_1} = C_2 \frac{dv_2}{dt} + \frac{v_1 - v_o}{R_2}
-$$
- (8.11.1)
-
-At node 2, KCL gives
-
-$$
-\frac{v_1 - v_o}{R_2} = C_1 \frac{dv_o}{dt}
-$$
- (8.11.2)
-
-But
-
-$$
-v_2 = v_1 - v_o \tag{8.11.3}
-$$
-
-We now try to eliminate *v*1 and *v*2 in Eqs. (8.11.1) to (8.11.3). Substituting Eqs. (8.11.2) and (8.11.3) into Eq. (8.11.1) yields
-
-$$
-\frac{v_s - v_1}{R_1} = C_2 \frac{dv_1}{dt} - C_2 \frac{dv_o}{dt} + C_1 \frac{dv_o}{dt}
-$$
- (8.11.4)
-
-From Eq. (8.11.2),
-
-$$
-v_1 = v_o + R_2 C_1 \frac{dv_o}{dt}
-$$
- (8.11.5)
-
-Substituting Eq. (8.11.5) into Eq. (8.11.4), we obtain
-
-$$
-\frac{v_s}{R_1} = \frac{v_o}{R_1} + \frac{R_2 C_1}{R_1} \frac{dv_o}{dt} + C_2 \frac{dv_o}{dt} + R_2 C_1 C_2 \frac{d^2 v_o}{dt^2} - C_2 \frac{dv_o}{dt} + C_1 \frac{dv_o}{dt}
-$$
-
-or
-
-$$
-\frac{d^2v_o}{dt^2} + \left(\frac{1}{R_1C_2} + \frac{1}{R_2C_2}\right)\frac{dv_o}{dt} + \frac{v_o}{R_1R_2C_1C_2} = \frac{v_s}{R_1R_2C_1C_2}
-$$
-(8.11.6)
-
-With the given values of *R*1, *R*2, *C*1, and *C*2, Eq. (8.11.6) becomes
-
-$$
-\frac{d^2v_o}{dt^2} + 2\frac{dv_o}{dt} + 5v_o = 5v_s
-$$
- (8.11.7)
-
-To obtain the form of the transient response, set *vs* = 0 in Eq. (8.11.7), which is the same as turning off the source. The characteristic equation is
-
-$$
-s^2 + 2s + 5 = 0
-$$
-
-which has complex roots *s*1,2 = −1 ± *j*2. Hence, the form of the transient response is
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/096_8.8 Second-Order Op Amp Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/096_8.8 Second-Order Op Amp Circuits.md
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+++ /dev/null
@@ -1,105 +0,0 @@
-
-$$
-v_{ot} = e^{-t}(A \cos 2t + B \sin 2t)
-$$
- (8.11.8)
-
-where *A* and *B* are unknown constants to be determined.
-
-As *t* → ∞, the circuit reaches the steady-state condition, and the capacitors can be replaced by open circuits. Since no current flows through *C*1 and *C*2 under steady-state conditions and no current can enter the input terminals of the ideal op amp, current does not flow through *R*1 and *R*2.
-
-Thus,
-
-$$
-v_o(\infty) = v_1(\infty) = v_s
-$$
-
-The steady-state response is then
-
-$$
-v_{\text{oss}} = v_o(\infty) = v_s = 10 \text{ mV}, \qquad t > 0 \tag{8.11.9}
-$$
-
-The complete response is
-
-$$
-v_o(t) = v_{ot} + v_{oss} = 10 + e^{-t}(A \cos 2t + B \sin 2t) \text{ mV} \qquad (8.11.10)
-$$
-
-To determine *A* and *B*, we need the initial conditions. For *t* < 0, *vs* = 0, so that
-
-$$
-v_o(0^-) = v_2(0^-) = 0
-$$
-
-For *t* > 0, the source is operative. However, due to capacitor voltage continuity,
-
-$$
-v_o(0^+) = v_2(0^+) = 0 \tag{8.11.11}
-$$
-
-From Eq. (8.11.3),
-
-$$
-v_1(0^+) = v_2(0^+) + v_o(0^+) = 0
-$$
-
-and, hence, from Eq. (8.11.2),
-
-$$
-\frac{dv_o(0^+)}{dt} = \frac{v_1 - v_o}{R_2 C_1} = 0
-$$
-\n(8.11.12)
-
-We now impose Eq. (8.11.11) on the complete response in Eq. (8.11.10) at *t* = 0, for
-
-$$
-0 = 10 + A \qquad \Rightarrow \qquad A = -10 \tag{8.11.13}
-$$
-
-Taking the derivative of Eq. (8.11.10),
-
-\_\_\_ *dvo dt* = *e*−*t* (−*A* cos 2*t* − *B* sin 2*t* − 2*A* sin 2*t* + 2*B* cos 2*t*)
-
-Setting *t* = 0 and incorporating Eq. (8.11.12), we obtain
-
-$$
-0 = -A + 2B \tag{8.11.14}
-$$
-
-From Eqs. (8.11.13) and (8.11.14), *A* = −10 and *B* = −5. Thus, the step response becomes
-
-$$
-v_o(t) = 10 - e^{-t}(10\cos 2t + 5\sin 2t) \text{ mV}, \qquad t > 0
-$$
-
-In the op amp circuit shown in Fig. 8.34, *vs* = 25*u*(*t*) V, find *vo*(*t*) for *t* > 0. Assume that *R*1 = *R*2 = 10 kΩ, *C*1 = 20 *μ*F, and *C*2 = 100 *μ*F.
-
-**Answer:**
-$$
-(25 - 31.25e^{-t} + 6.25e^{-5t})
-$$
- V, $t > 0$ .
-
-**8.9** PSpice Analysis of RLC Circuits *RLC* circuits can be analyzed with great ease using *PSpice*, just like the
-
-*RC* or *RL* circuits of Chapter 7. The following two examples will illustrate this. The reader may review Section D.4 in Appendix D on *PSpice* for transient analysis.
-
-# Example 8.12
-
-The input voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.35(b). Use *PSpice* to plot *v*(*t*) for 0 < *t* < 4 s.
-
-# **Solution:**
-
-- 1. **Define.** As true with most textbook problems, the problem is clearly defined.
-- 2. **Present.** The input is equal to a single square wave of amplitude 12 V with a period of 2 s. We are asked to plot the output, using *PSpice*.
-- 3. **Alternative.** Since we are required to use *PSpice*, that is the only alternative for a solution. However, we can check it using the technique illustrated in Section 8.5 (a step response for a series *RLC* circuit).
-- 4. **Attempt.** The given circuit is drawn using Schematics as in Fig. 8.36. The pulse is specified using VPWL voltage source, but VPULSE could be used instead. Using the piecewise linear function, we set the attributes of VPWL as T1 = 0, V1 = 0, T2 = 0.001, V2 = 12, and so forth, as shown in Fig. 8.36. Two voltage markers are inserted to plot the input and output voltages. Once the circuit is drawn and the attributes are set, we select **Analysis/Setup/Transient** to open up the *Transient Analysis* dialog box. As a parallel *RLC* circuit, the roots of the characteristic equation are −1 and −9. Thus, we may set *Final Time* as 4 s (four times the magnitude of the lower root). When the schematic
-
-is saved, we select **Analysis/Simulate** and obtain the plots for the input and output voltages under the *PSpice* A/D window as shown in Fig. 8.37.
-
- Now we check using the technique from Section 8.5. We can start by realizing the Thevenin equivalent for the resistor-source combination is *V*Th = 12∕2 (the open circuit voltage divides equally across both resistors) = 6 V. The equivalent resistance is 30 Ω (60 ‖ 60). Thus, we can now solve for the response using *R* = 30 Ω, *L* = 3 H, and *C* = (1∕27) F.
-
-We first need to solve for *α* and *ω*0:
-
-$$
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/097_8.9 PSpice Analysis of RLC Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/097_8.9 PSpice Analysis of RLC Circuits.md
deleted file mode 100644
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@@ -1,89 +0,0 @@
-\alpha = R/(2L) = 30/6 = 5
-$$
- and $\omega_0 = \frac{1}{\sqrt{3\frac{1}{27}}} = 3$
-
- Since 5 is greater than 3, we have the overdamped case \_\_\_\_\_\_
-
-*i*(*t*) = *C*
-
-$$
-s_{1,2} = -5 \pm \sqrt{5^2 - 9} = -1, -9, \qquad v(0) = 0,
-$$
-
- $v(\infty) = 6 \text{ V}, \qquad i(0) = 0$
-
-*dv*(*t*) \_\_\_\_ *dt* ,
-
-where
-
-where
-\n
-$$
-v(t) = A_1 e^{-t} + A_2 e^{-9t} + 6
-$$
-\n
-$$
-v(0) = 0 = A_1 + A_2 + 6
-$$
-\n
-$$
-i(0) = 0 = C(-A_1 - 9A_2)
-$$
-
-which yields *A*1 = −9*A*2. Substituting this into the above, we get 0 = 9*A*2 − *A*2 + 6, or *A*2 = 0.75 an*d A*1 = −6.75.
-
-$$
-v(t) = (-6.75e^{-t} + 0.75e^{-9t} + 6)u(t) V
-$$
- for all $0 < t < 2$ s.
-
- At *t* = 1 s, *v*(1) = −6.75*e*−1 + 0.75*e*−9 + 6 = −2.483 + 0.0001 + 6 = 3.552 V. At *t* = 2 s, *v*(2) = −6.75*e* −2 + 0 + 6 = 5.086 V.
-
- Note that from 2 < *t* < 4 s, *V*Th = 0, which implies that *v*(∞) = 0. Therefore, *v*(*t*) = (*A*3*e*−(*t* − 2) + *A*4*e*−9(*t* − 2))*u*(*t* − 2) V. At *t* = 2 s, *A*3 + *A*4 = 5.086.
-
-$$
-i(t) = (A_3e^{-(t-2)} - 9A_4e^{-9(t-2)})
-$$
-
-\n
-$$
-i(t) = \frac{(-A_3e^{-(t-2)} - 9A_4e^{-9(t-2)})}{27}
-$$
-
-and
-
-$$
-i(2) = \frac{(6.75e^{-2} - 6.75e^{-18})}{27} = 33.83 \text{ mA}
-$$
-
-Therefore, −*A*3 − 9*A*4 = 0.9135.
-
- Combining the two equations, we get −*A*3 − 9(5.086 − *A*3) = 0.9135, which leads to *A*3 = 5.835 and *A*4 = −0.749.
-
-$$
-v(t) = (5.835e^{-(t-2)} - 0.749e^{-9(t-2)}) u(t-2)
-$$
- V
-
- At *t* = 3 s, *v*(3) = (2.147 − 0) = 2.147 V. At *t* = 4 s, *v*(4) = 0.7897 V.
-
-- 5. **Evaluate.** A check between the values calculated above and the plot shown in Figure 8.37 shows good agreement within the obvious level of accuracy.
-- 6. **Satisfactory?** Yes, we have agreement and the results can be presented as a solution to the problem.
-
-Practice Problem 8.12 Find *i*(*t*) using *PSpice* for 0 < *t* < 4 s if the pulse voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.38.
-
-# **Solution:**
-
-When the switch is in position *a*, the 6- Ω resistor is redundant. The schematic for this case is shown in Fig. 8.41(a). To ensure that current
-
-**Figure 8.41** For Example 8.13: (a) for dc analysis, (b) for transient analysis.
-
-*i*(*t*) enters pin 1, the inductor is rotated three times before it is placed in the circuit. The same applies for the capacitor. We insert pseudo -components VIEWPOINT and IPROBE to determine the initial capacitor voltage and initial inductor current. We carry out a dc *PSpice* analysis by select ing **Analysis/Simulate.** As shown in Fig. 8.41(a), we obtain the initial capacitor voltage as 0 V and the initial inductor current *i*(0) as 4 A from the dc analysis. These initial values will be used in the transient analysis.
-
-When the switch is moved to position *b*, the circuit becomes a sourcefree parallel *RLC* circuit with the schematic in Fig. 8.41(b). We set the initial condition IC = 0 for the capacitor and IC = 4 A for the inductor. A current marker is inserted at pin 1 of the inductor. We select **Analysis/ Setup/Transient** to open up the *Transient Analysis* dialog box and set *Final Time* to 3 s. After saving the schematic, we select **Analysis/ Transient**. Figure 8.42 sho ws the plot of *i*(*t*). The plot agrees with *i*(*t*) = 4.8*e*−*t* − 0.8*e*−6*t* A, which is the solution by hand calculation.
-
-Plot of *i*(*t*) for Example 8.13.
-
-Refer to the circuit in Fig. 8.21 (see Practice Prob. 8.7). Use *PSpice* to Practice Problem 8.13 obtain *v*(*t*) for 0 < *t* < 2.
-
-**Answer:** See Fig. 8.43.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/098_8.10 Duality.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/098_8.10 Duality.md
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+++ /dev/null
@@ -1,757 +0,0 @@
-# **8.10** Duality
-
-The concept of duality is a time-sa ving, ef fort-effective measure of solving circuit problems. Consider the similarity between Eqs. (8.4) and (8.29). The two equations are the same, e xcept that we must interchange the following quantities: (1) voltage and current, (2) resistance and conductance, (3) capacitance and inductance. Thus, it sometimes occurs in circuit analysis that two different circuits have the same equations and solutions, except that the roles of certain complementary elements are interchanged. This interchangeability is kno wn as the principle of *duality*.
-
-The duality principle asserts a parallelism between pairs of characterizing equations and theorems of electric circuits.
-
-Dual pairs are sho wn in Table 8.1. Note that po wer does not appear in Table 8.1, because power has no dual. The reason for this is the principle of linearity; since power is not linear, duality does not apply. Also notice from Table 8.1 that the principle of duality extends to circuit elements, configurations, and theorems.
-
-Two circuits that are described by equations of the same form, b ut in which the variables are interchanged, are said to be dual to each other.
-
-Two circuits are said to be duals of one another if they are described by the same characterizing equations with dual quantities interchanged.
-
-The usefulness of the duality principle is self-e vident. Once we know the solution to one circuit, we automatically ha ve the solution for the dual circuit. It is ob vious that the circuits in Figs. 8.8 and 8.13 are dual. Consequently , the result in Eq. (8.32) is the dual of that in Eq. (8.11). We must k eep in mind that the method described here for finding a dual is limited to planar circuits. Finding a dual for a nonplanar circuit is be yond the scope of this te xtbook because nonplanar circuits cannot be described by a system of mesh equations.
-
-To find the dual of a given circuit, we do not need to write do wn the mesh or node equations. We can use a graphical technique. Gi ven a planar circuit, we construct the dual circuit by taking the following three steps:
-
-- 1. Place a node at the center of each mesh of the given circuit. Place the reference node (the ground) of the dual circuit outside the given circuit.
-- 2. Draw lines between the nodes such that each line crosses an ele ment. Replace that element by its dual (see Table 8.1).
-- 3. To determine the polarity of voltage sources and direction of current sources, follow this rule: A voltage source that produces a positi ve (clockwise) mesh current has as its dual a current source whose reference direction is from the ground to the nonreference node.
-
-In case of doubt, one may verify the dual circuit by writing the nodal or mesh equations. The mesh (or nodal) equations of the original circuit are similar to the nodal (or mesh) equations of the dual circuit. The duality principle is illustrated with the following two examples.
-
-# **TABLE 8.1**
-
-# Dual pairs.
-
-| Resistance R | Conductance G |
-|----------------|----------------|
-| Inductance L | Capacitance C |
-| Voltage v | Current i |
-| Voltage source | Current source |
-| Node | Mesh |
-| Series path | Parallel path |
-| Open circuit | Short circuit |
-| KVL | KCL |
-| Thevenin | Norton |
-| | |
-
-Even when the principle of linearity applies, a circuit element or variable may not have a dual. For example, mutual inductance (to be covered in Chapter 13) has no dual.
-
-Construct the dual of the circuit in Fig. 8.44.
-
-# **Solution:**
-
-As shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two meshes and also the ground node 0 for the dual circuit. We draw a line between one node and another crossing an element. We replace the line joining the nodes by the duals of the elements which it crosses. For example, a line between nodes 1 and 2 crosses a 2-H inductor, and we place a 2-F capacitor (an inductor's dual) on the line. A line between nodes 1 and 0 crossing the 6-V voltage source will contain a 6-A current source. By drawing lines crossing all the elements, we construct the dual circuit on the given circuit as in Fig. 8.45(a). The dual circuit is redrawn in Fig. 8.45(b) for clarity.
-
-**Figure 8.44** For Example 8.14.
-
-# **Figure 8.45**
-
-(a) Construction of the dual circuit of Fig. 8.44, (b) dual circuit redrawn.
-
-Draw the dual circuit of the one in Fig. 8.46.
-
-**Answer:** See Fig. 8.47.
-
-Obtain the dual of the circuit in Fig. 8.48. Example 8.15
-
-# **Solution:**
-
-The dual circuit is constructed on the original circuit as in Fig. 8.49(a). We first locate nodes 1 to 3 and the reference node 0. Joining nodes 1 and 2, we cross the 2-F capacitor, which is replaced by a 2-H inductor.
-
-Practice Problem 8.14
-
-For Example 8.15.
-
-Joining nodes 2 and 3, we cross the 20- Ω resistor, which is replaced by a \_\_\_1 20 -Ω resistor. We keep doing this until all the elements are crossed. The result is in Fig. 8.49(a). The dual circuit is redrawn in Fig. 8.49(b).
-
-# **Figure 8.49**
-
-For Example 8.15: (a) construction of the dual circuit of Fig. 8.48, (b) dual circuit redrawn.
-
-To v erify the polarity of the v oltage source and the direction of the current source, we may apply mesh currents *i*1, *i*2, and *i*3 (all in the clockwise direction) in the original circuit in Fig. 8.48. The 10-V voltage source produces positive mesh current *i*1, so that its dual is a 10-A current source directed from 0 to 1. Also, *i*3 = −3 A in Fig. 8.48 has as its dual *v*3 = −3 V in Fig. 8.49(b).
-
-**Answer:** See Fig. 8.51.
-
-For Practice Prob. 8.15.
-
-**Figure 8.51** Dual of the circuit in Fig. 8.50.
-
-# **8.11** Applications
-
-Practical applications of *RLC* circuits are found in control and communications circuits such as ringing circuits, peaking circuits, resonant circuits, smoothing circuits, and filters. Most of these circuits cannot be covered until we treat ac sources. For now, we will limit ourselves to two simple applications: automobile ignition and smoothing circuits.
-
-# **8.11.1** Automobile Ignition System
-
-In Section 7.9.4, we considered the automobile ignition system as a charging system. That was only a part of the system. Here, we consider another part—the voltage generating system. The system is modeled by the circuit shown in Fig. 8.52. The 12-V source is due to the battery and alternator. The 4-Ω resistor represents the resistance of the wiring. The ignition coil is modeled by the 8-mH inductor . The 1-*μ*F capacitor (known as the *condenser* to automechanics) is in parallel with the switch (known as the *breaking points* or *electronic ignition*). In the follo wing example, we determine how the *RLC* circuit in Fig. 8.52 is used in gen erating high voltage.
-
-Automobile ignition circuit.
-
-Assuming that the switch in Fig. 8.52 is closed prior to *t* = 0−, find the inductor voltage *vL* for *t* > 0.
-
-# **Solution:**
-
-If the switch is closed prior to *t* = 0− and the circuit is in steady state, then
-
-$$
-i(0^{-}) = \frac{12}{4} = 3 \text{ A}, \qquad v_C(0^{-}) = 0
-$$
-
-At *t* = 0+, the switch is opened. The continuity conditions require that
-
-$$
-i(0^+) = 3 \text{ A}, \qquad v_C(0^+) = 0
-$$
- (8.16.1)
-
-We obtain *di*(0+)/*dt* from *vL*(0+). Applying KVL to the mesh at *t* = 0+ yields
-
-$$
--12 + 4i(0^{+}) + v_L(0^{+}) + v_C(0^{+}) = 0
-$$
-
-$$
--12 + 4 \times 3 + v_L(0^{+}) + 0 = 0 \implies v_L(0^{+}) = 0
-$$
-
-# Example 8.16
-
-Hence,
-
-$$
-\frac{di(0^{+})}{dt} = \frac{v_L(0^{+})}{L} = 0
-$$
-\n(8.16.2)
-
-As *t* → ∞, the system reaches steady state, so that the capacitor acts like an open circuit. Then
-
-$$
-i(\infty) = 0 \tag{8.16.3}
-$$
-
-If we apply KVL to the mesh for *t* > 0, we obtain
-
-$$
-12 = Ri + L\frac{di}{dt} + \frac{1}{C} \int_0^t i \, dt + v_C(0)
-$$
-
-Taking the derivative of each term yields
-
-$$
-\frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0
-$$
- (8.16.4)
-
-We obtain the form of the transient response by following the procedure in Section 8.3. Substituting *R* = 4 Ω, *L* = 8 mH, and *C* = 1 *μ*F, we get
-
-$$
-\alpha = \frac{R}{2L} = 250
-$$
-, $\omega_0 = \frac{1}{\sqrt{LC}} = 1.118 \times 10^4$
-
-Since *α* < *ω*0, the response is underdamped. The damped natural fre quency is
-
-$$
-\omega_d = \sqrt{\omega_0^2 - \alpha^2} \simeq \omega_0 = 1.118 \times 10^4
-$$
-
-The form of the transient response is
-
-$$
-it(t) = e^{-\alpha} (A \cos \omega_d t + B \sin \omega_d t)
-$$
- (8.16.5)
-
-where *A* and *B* are constants. The steady-state response is
-
-$$
-i_{ss}(t) = i(\infty) = 0 \tag{8.16.6}
-$$
-
-so that the complete response is
-
-$$
-i(t) = it(t) + iss(t) = e^{-250t}(A \cos 11,180t + B \sin 11,180t)
-$$
- (8.16.7)
-
-We now determine *A* and *B*.
-
-$$
-i(0) = 3 = A + 0 \qquad \Rightarrow \qquad A = 3
-$$
-
-Taking the derivative of Eq. (8.16.7),
-
-$$
-\frac{di}{dt} = -250e^{-250t}(A \cos 11,180t + B \sin 11,180t) \n+ e^{-250t}(-11,180A \sin 11,180t + 11,180B \cos 11,180t)
-$$
-
-Setting *t* = 0 and incorporating Eq. (8.16.2),
-
-$$
-0 = -250A + 11{,}180B \qquad \Rightarrow \qquad B = 0.0671
-$$
-
-Thus,
-
-$$
-i(t) = e^{-250t}(3 \cos 11,180t + 0.0671 \sin 11,180t)
-$$
- (8.16.8)
-
-The voltage across the inductor is then
-
-$$
-v_L(t) = L\frac{di}{dt} = -268e^{-250t}\sin 11{,}180t
-$$
- (8.16.9)
-
-This has a maximum value when sine is unity, that is, at 11,180*t*0 = *π*∕2 or *t*0 = 140.5 *μ*s. At *ti*me = *t*0, the inductor voltage reaches its peak, which is
-
-$$
-v_L(t_0) = -268e^{-250t_0} = -259 \text{ V} \tag{8.16.10}
-$$
-
-Although this is far less than the voltage range of 6000 to 10,000 V required to fire the spark plug in a typical automobile, a device known as a *transformer* (to be discussed in Chapter 13) is used to step up the inductor voltage to the required level.
-
-In Fig. 8.52, find the capacitor voltage *vC* for *t* > 0.
-
-**Answer:** 12 − 12*e*−250*t* cos 11,180*t* + 267.7*e*−250*t* sin 11,180*t* V.
-
-# **8.11.2** Smoothing Circuits
-
-In a typical digital communication system, the signal to be transmitted is first sampled. Sampling refers to the procedure of selecting samples of a signal for processing, as opposed to processing the entire signal. Each sample is con verted into a binary number represented by a series of pulses. The pulses are transmitted by a transmission line such as a coaxial cable, twisted pair, or optical fiber. At the receiving end, the signal is applied to a digital-to-analog (D/A) con verter whose output is a "staircase" function, that is, constant at each time interv al. In order to recover the transmitted analog signal, the output is smoothed by letting it pass through a "smoothing" circuit, as illustrated in Fig. 8.53. An *RLC* circuit may be used as the smoothing circuit.
-
-The output of a D/A converter is shown in Fig. 8.54(a). If the *RLC* circuit in Fig. 8.54(b) is used as the smoothing circuit, determine the output
-
-Practice Problem 8.16
-
-# **Figure 8.53**
-
-A series of pulses is applied to the digitalto-analog (D/A) converter, whose output is applied to the smoothing circuit.
-
-# Example 8.17
-
-**Figure 8.54** For Example 8.17: (a) output of a D/A converter, (b) an *RLC* smoothing circuit.
-
-# **Solution:**
-
-voltage *vo*(*t*).
-
-This problem is best solved using *PSpice*. The schematic is shown in Fig. 8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise
-
-**Figure 8.55** For Example 8.17: (a) schematic, (b) input and output voltages.
-
-linear function. The attributes of V1 are set as T1 = 0, V1 = 0, T2 = 0.001, V2 = 4, T3 = 1, V3 = 4, and so on. To be able to plot both input and output voltages, we insert two voltage markers as shown. We select **Analysis/Setup/Transient** to open up the *Transient Analysis* dialog box and set *Final Time* as 6 s. Once the schematic is saved, we select **Analysis/Simulate** to run and obtain the plots shown in Fig. 8.55(b).
-
-Practice Problem 8.17 Rework Example 8.17 if the output of the D/A converter is as shown in Fig. 8.56.
-
-**Answer:** See Fig. 8.57.
-
-# **8.12** Summary
-
-- 1. The determination of the initial v alues *x*(0) and *dx*(0)∕*dt* and final value *x*(∞) is crucial to analyzing second-order circuits.
-- 2. The *RLC* circuit is second-order because it is described by a second-order dif ferential equation. Its characteristic equation is
-
-*s* 2 + 2*αs* + *ω*0 2 = 0, where *α* is the neper frequenc y and *ω*0 is the undamped natural frequency. For a series circuit, *α* = *R*∕2*L*, for a parallel circuit *α* = 1∕2*RC*, and for both cases *ω*0 = 1∕ √ \_\_\_ *LC* .
-
-- 3. If there are no independent sources in the circuit after switching (or sudden change), we regard the circuit as source-free. The complete solution is the natural response.
-- 4. The natural response of an *RLC* circuit is o verdamped, under damped, or critically damped, depending on the roots of the characteristic equation. The response is critically damped when the roots are equal (*s*1 = *s*2 or *α* = *ω*0), overdamped when the roots are real and unequal (*s*1 ≠ *s*2 or *α* > *ω*0), or underdamped when the roots are complex conjugate (*s*1 = *s*2 \* or *α* < *ω*0).
-- 5. If independent sources are present in the circuit after switching, the complete response is the sum of the transient response and the steady-state response.
-- 6. *PSpice* is used to analyze *RLC* circuits in the same w ay as for *RC* or *RL* circuits.
-- 7. Two circuits are dual if the mesh equations that describe one circuit have the same form as the nodal equations that describe the other. The analysis of one circuit gives the analysis of its dual circuit.
-- 8. The automobile ignition circuit and the smoothing circuit are typical applications of the material covered in this chapter.
-
-# Review Questions
-
-**8.1** For the circuit in Fig. 8.58, the capacitor voltage at *t* = 0− (just before the switch is closed) is:
-
-# **Figure 8.58**
-
-For Review Questions 8.1 and 8.2.
-
-**8.2** For the circuit in Fig. 8.58, the initial inductor current (at *t* = 0) is:
-
-(a) 0 A (b) 2 A (c) 6 A (d) 12 A
-
-- **8.3** When a step input is applied to a second-order circuit, the final values of the circuit variables are found by:
- - (a) Replacing capacitors with closed circuits and inductors with open circuits.
- - (b) Replacing capacitors with open circuits and inductors with closed circuits.
- - (c) Doing neither of the above.
-
-**8.4** If the roots of the characteristic equation of an *RLC* circuit are −2 and −3, the response is:
-
-> (a) (*A* cos 2*t* + *B* sin 2*t*)*e*−3*t* (b) (*A* + 2*Bt*)*e*−3*t* (c) *Ae*−2*t* + *Bte*−3*t* (d) *Ae*−2*t* + *Be*−3*t*
-
-where *A* and *B* are constants.
-
-- **8.5** In a series *RLC* circuit, setting *R* = 0 will produce:
- - (a) an overdamped response
- - (b) a critically damped response
- - (c) an underdamped response
- - (d) an undamped response
- - (e) none of the above
-- **8.6** A parallel *RLC* circuit has *L* = 2 H and *C* = 0.25 F. The value of *R* that will produce a unity neper frequency is:
-
-(a) 0.5 Ω (b) 1 Ω (c) 2 Ω (d) 4 Ω
-
-- **8.7** Refer to the series *RLC* circuit in Fig. 8.59. What kind of response will it produce?
- - (a) overdamped
- - (b) underdamped
- - (c) critically damped
- - (d) none of the above
-
-C1
-
-i
-
-
-
-# **Figure 8.59**
-
-For Review Question 8.7.
-
-- **8.8** Consider the parallel *RLC* circuit in Fig. 8.60. What type of response will it produce?
- - (a) overdamped
- - (b) underdamped
- - (c) critically damped
- - (d) none of the above
-
-For Review Question 8.8.
-
-- **8.9** Match the circuits in Fig. 8.61 with the following items:
- - (i) first-order circuit
- - (ii) second-order series circuit
- - (iii) second-order parallel circuit
- - (iv) none of the above
-
-## *v*s R (c) i s C2 C1 L R1 (d) C2 R2 + ‒ R1 R2
-
-# **Figure 8.61**
-
-For Review Question 8.9.
-
-**8.10** In an electric circuit, the dual of resistance is:
-
-| (a) conductance | (b) inductance |
-|-------------------|------------------|
-| (c) capacitance | (d) open circuit |
-| (e) short circuit | |
-
-*Answers: 8.1a, 8.2c, 8.3b, 8.4d, 8.5d, 8.6c, 8.7b, 8.8b, 8.9 (i)-c, (ii)-b, e, (iii)-a, (iv)-d, f, 8.10a.*
-
-# Problems
-
-# Section 8.2 Finding Initial and Final Values
-
-**8.1** For the circuit in Fig. 8.62, find:
-
-(a) *i*(0+) and *v*(0+), (b) *di*(0+)∕*dt* and *dv*(0+)∕*dt*,
-
-(c) *i*(∞) and *v*(∞).
-
-**Figure 8.62** For Prob. 8.1.
-
-**Figure 8.63** For Prob. 8.2.
-
-**8.3** Refer to the circuit shown in Fig. 8.64. Calculate:
-
-(a) *iL*(0+), *vC*(0+), and *vR*(0+), (b) *diL*(0+)∕*dt*, *dvC*(0+)∕*dt*, and *dvR*(0+)∕*dt*, (c) *iL*(∞), *vC*(∞), and *vR*(∞).
-
-**Figure 8.64**
-
-- For Prob. 8.3.
- - **8.4** In the circuit of Fig. 8.65, find: (a) *v*(0+) and *i*(0+), (b) *dv*(0+)∕*dt* and *di*(0+)∕*dt*, (c) *v*(∞) and *i*(∞).
-
-**Figure 8.65**
-
-For Prob. 8.4.
-
-- **8.5** Refer to the circuit in Fig. 8.66. Determine:
- - (a) *i*(0+) and *v*(0+), (b) *di*∕(0+)*dt* and *dv*(0+)∕*dt*,
- - (c) *i*(∞) and *v*(∞).
-
-For Prob. 8.5.
-
-**8.6** In the circuit of Fig. 8.67, find:
-
-(a) *vR*(0+) and *vL*(0+), (b) *dvR*(0+)∕*dt* and *dvL*(0+)∕*dt*, (c) *vR*(∞) and *vL*(∞).
-
-# **Figure 8.67**
-
-For Prob. 8.6.
-
-# Section 8.3 Source-Free Series RLC Circuit
-
-- **8.7** A series *RLC* circuit has *R* = 20 kΩ, *L* = 0.2 mH, and *C* = 5 *μ*F. What type of damping is exhibited by the circuit?
-- **8.8** Design a problem to help other students better understand source-free *RLC* circuits.
-- **8.9** The current in an *RLC* circuit is described by
-
-$$
-\frac{d^2i}{dt^2} + 10\frac{di}{dt} + 25i = 0
-$$
-
-If *i*(0) = 10 A and *di*(0)∕*dt* = 0, find *i*(*t*) for *t* > 0.
-
-**8.10** The differential equation that describes the current in an *RLC* network is
-
-$$
-3\frac{d^2i}{dt^2} + 15\frac{di}{dt} + 12i = 0
-$$
-
-Given that *i*(0) = 0, *di*(0)∕*dt* = 6 mA/s, obtain *i*(*t*).
-
-**8.11** The natural response of an *RLC* circuit is described by the differential equation
-
-$$
-\frac{d^2v}{dt^2} + 2\frac{dv}{dt} + v = 0
-$$
-
- for which the initial conditions are *v*(0) = 10 V and *dv*(0)∕*dt* = 0. Solve for *v*(*t*).
-
-- **8.12** If *R* = 50 Ω, *L* = 1.5 H, what value of *C* will make an *RLC* series circuit:
- - (a) overdamped,
- - (b) critically damped,
- - (c) underdamped?
-- **8.13** For the circuit in Fig. 8.68, calculate the value of *R* needed to have a critically damped response.
-
-**Figure 8.68** For Prob. 8.13.
-
-**8.14** The switch in Fig. 8.69 moves from position *A* to position *B* at *t* = 0 (please note that the switch must connect to point *B* before it breaks the connection at *A*, a make-before-break switch). Let *v*(0) = 0, find *v*(*t*) for *t* > 0.
-
-**8.15** The responses of a series *RLC* circuit are
-
-$$
-v_C(t) = 30 - 10e^{-20t} + 30e^{-10t}
-$$
-V
-$$
-i_L(t) = 40e^{-20t} - 60e^{-10t}
-$$
- mA
-
- where *vC* and *iL* are the capacitor voltage and inductor current, respectively. Determine the values of *R*, *L*, and *C*.
-
-**8.16** Find *i*(*t*) for *t* > 0 in the circuit of Fig. 8.70.
-
-**Figure 8.70**
-
-For Prob. 8.16.
-
-**8.17** In the circuit of Fig. 8.71, the switch instantaneously moves from position *A* to *B* at *t* = 0. Find *v*(*t*) for all *t* ≥ 0.
-
-**8.18** Find the voltage across the capacitor as a function of time for *t* > 0 for the circuit in Fig. 8.72. Assume steady-state conditions exist at *t* = 0−.
-
-**Figure 8.72**
-
-For Prob. 8.18.
-
-**8.19** Obtain *v*(*t*) for *t* > 0 in the circuit of Fig. 8.73.
-
-**Figure 8.73**
-
-For Prob. 8.19.
-
-**8.20** The switch in the circuit of Fig. 8.74 has been closed for a long time but is opened at *t* = 0. Determine *i*(*t*) for *t* > 0.
-
-# **Figure 8.74**
-
-For Prob. 8.20.
-
-\***8.21** Calculate *v*(*t*) for *t* > 0 in the circuit of Fig. 8.75.
-
-**Figure 8.75** For Prob. 8.21.
-
-\* An asterisk indicates a challenging problem.
-
-# Section 8.4 Source-Free Parallel RLC Circuit
-
-**8.22** Assuming *R* = 2 kΩ, design a parallel *RLC* circuit that has the characteristic equation
-
-$$
-s^2 + 100s + 10^6 = 0.
-$$
-
-**8.23** For the network in Fig. 8.76, what value of *C* is needed to make the response underdamped with unity neper frequency (*α* = 1)?
-
-# **Figure 8.76** For Prob. 8.23.
-
-**8.24** The switch in Fig. 8.77 moves from position A to position *B* at *t* = 0 (please note that the switch must connect to point *B* before it breaks the connection at *A*, a make-before-break switch). Determine *i*(*t*) for *t* > 0.
-
-**Figure 8.77** For Prob. 8.24.
-
-**8.25** Using Fig. 8.78, design a problem to help other students better understand source-free *RLC* circuits.
-
-**Figure 8.78** For Prob. 8.25.
-
-# Section 8.5 Step Response of a Series RLC Circuit
-
-**8.26** The step response of an *RLC* circuit is given by
-
-$$
-\frac{d^2i}{dt^2} + 2\frac{di}{dt} + 5i = 10
-$$
-
-Given that *i*(0) = 2 and *di*(0)∕*dt* = 4, solve for *i*(*t*).
-
-**8.27** A branch voltage in an *RLC* circuit is described by
-
-$$
-\frac{d^2v}{dt^2} + 4\frac{dv}{dt} + 8v = 24
-$$
-
- If the initial conditions are *v*(0) = 0 = *dv*(0)∕*dt*, find *v*(*t*).
-
-**8.28** A series *RLC* circuit is described by
-
-$$
-L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{i}{C} = 10
-$$
-
- Find the response when *L* = 0.5 H, *R* = 4 Ω, and *C* = 0.2 F. Let *i*(0) = 1, *di*(0)∕*dt* = 0.
-
-**8.29** Solve the following differential equations subject to the specified initial conditions
-
-(a)
-$$
-d^2v/dt^2 + 4v = 12
-$$
-, $v(0) = 0$ , $dv(0)/dt = 2$
-\n(b) $d^2i/dt^2 + 5 \frac{di}{dt} + 4i = 8$ , $i(0) = -1$ ,
-\n $\frac{di(0)}{dt} = 0$
-\n(c) $\frac{d^2v}{dt^2} + 2 \frac{dv}{dt} + v = 3$ , $v(0) = 5$ ,
-\n $\frac{dv(0)}{dt} = 1$
-\n(d) $\frac{d^2i}{dt^2} + 2 \frac{di}{dt} + 5i = 10$ , $i(0) = 4$ ,
-\n $\frac{di(0)}{dt} = -2$
-
-**8.30** The step responses of a series *RLC* circuit are
-
-$$
-v_C = 40 - 10e^{-2000t} - 10e^{-4000t} \text{ V}, \qquad t > 0
-$$
-
-$$
-i_L(t) = 3e^{-2000t} + 6e^{-4000t} \text{ mA}, \qquad t > 0
-$$
-
-- (a) Find *C*. (b) Determine what type of damping is exhibited by the circuit.
-- **8.31** Consider the circuit in Fig. 8.79. Find *vL*(0+) and *vC*(0+).
-
-For Prob. 8.31.
-
-**Figure 8.80** For Prob. 8.32.
-
-**Figure 8.81**
-
-For Prob. 8.33.
-
-**8.34** Calculate *i*(*t*) for *t* > 0 in the circuit of Fig. 8.82.
-
-For Prob. 8.34.
-
-**8.35** Using Fig. 8.83, design a problem to help other students better understand the step response of series *RLC* circuits.
-
-# **Figure 8.83**
-
-For Prob. 8.35.
-
-**8.36** Obtain *v*(*t*) and *i*(*t*) for *t* > 0 in the circuit of Fig. 8.84.
-
-**Figure 8.84** For Prob. 8.36.
-
-\***8.37** For the network in Fig. 8.85, solve for *i*(*t*) for *t* > 0.
-
-# **Figure 8.85**
-
-For Prob. 8.37.
-
-**8.38** Refer to the circuit in Fig. 8.86. Calculate *i*(*t*) for *t* > 0.
-
-# **Figure 8.86** For Prob. 8.38.
-
-# **Figure 8.87**
-
-For Prob. 8.39.
-
-**8.40** The switch in the circuit of Fig. 8.88 is moved from position *a* to *b* at *t* = 0. Assume that the voltage across the capacitor is equal to zero at *t* = 0 and that the switch is a make before break switch. Determine *i*(*t*) for all *t* > 0.
-
-**\*8.41** For the network in Fig. 8.89, find *i*(*t*) for *t* > 0.
-
-**Figure 8.89** For Prob. 8.41.
-
-**\*8.42** Given the network in Fig. 8.90, find *v*(*t*) for *t* > 0.
-
-- For Prob. 8.42.
- - **8.43** The switch in Fig. 8.91 is opened at *t* = 0 after the circuit has reached steady state. Choose *R* and *C* such that *α* = 8 Np/s and *ωd* = 30 rad/s.
-
-**Figure 8.91**
-
-For Prob. 8.43.
-
-**8.44** A series *RLC* circuit has the following parameters: *R* = 1 kΩ, *L* = 1 H, and *C* = 10 nF. What type of damping does this circuit exhibit?
-
-# Section 8.6 Step Response of a Parallel RLC Circuit
-
-**8.45** In the circuit of Fig. 8.92, find *v*(*t*) and *i*(*t*) for *t* > 0.
-
-For Prob. 8.45.
-
-**8.46** Using Fig. 8.93, design a problem to help other students better understand the step response of a parallel *RLC* circuit.
-
-# **Figure 8.93**
-
-For Prob. 8.46.
-
-**8.47** Find the output voltage *vo*(*t*) in the circuit of Fig. 8.94.
-
-# **Figure 8.94**
-
-For Prob. 8.47.
-
-**8.48** Given the circuit in Fig. 8.95, find *i*(*t*) and *v*(*t*) for *t* > 0.
-
-For Prob. 8.48.
-
-**8.49** Determine *i*(*t*) for *t* > 0 in the circuit of Fig. 8.96.
-
-**Figure 8.97**
-
-For Prob. 8.50.
-
-**8.51** Find *v*(*t*) for *t* > 0 in the circuit of Fig. 8.98.
-
-# **Figure 8.98**
-
-For Prob. 8.51.
-
-**8.52** The step response of a parallel *RLC* circuit is *v* = 10 + 20*e*−300*t* (cos 400*t* − 2 sin 400*t*) V, *t* ≥ 0 when the inductor is 25 mH. Find *R* and *C*.
-
-# Section 8.7 General Second-Order Circuits
-
-**8.53** After being open for a day, the switch in the circuit of Fig. 8.99 is closed at *t* = 0. Find the differential equation describing *i*(*t*), *t* > 0.
-
-# **Figure 8.99**
-
-For Prob. 8.53.
-
-**8.55** For the circuit in Fig. 8.101, find *v*(*t*) for *t* > 0. Assume that *i*(0+) = 2 A.
-
-# **Figure 8.101**
-
-For Prob. 8.55.
-
-**8.56** In the circuit of Fig. 8.102, find *i*(*t*) for *t* > 0.
-
-# **Figure 8.102**
-
-For Prob. 8.56.
-
-**8.57** Given the circuit shown in Fig. 8.103, determine the characteristic equation of the circuit and the values for *i*(*t*) and *v*(*t*) for all *t* > 0.
-
-# **Figure 8.103**
-
-For Prob. 8.57.
-
-- **8.58** In the circuit of Fig. 8.104, the switch has been in position 1 for a long time but moved to position 2 at *t* = 0. Find:
- - (a) *v*(0+), *dv*(0+)∕*dt*, (b) *v*(*t*) for *t* ≥ 0.
-
-**Figure 8.104** For Prob. 8.58.
-
-**8.59** The switch in Fig. 8.105 has been in position 1 for *t* < 0. At *t* = 0, it is moved from position 1 to the top of the capacitor at *t* = 0. Please note that the switch is a make before break switch, it stays in contact with position 1 until it makes contact with the top of the capacitor and then breaks the contact at position 1. Given that the initial voltage across the capacitor is equal to zero, determine *v*(*t*).
-
-# **Figure 8.105**
-
-For Prob. 8.59.
-
-# **Figure 8.106**
-
-For Prob. 8.60.
-
-- **8.61** For the circuit in Prob. 8.5, find *i* and *v* for *t* > 0.
-- **8.62** Find the response *vR*(*t*) for *t* > 0 in the circuit of Fig. 8.107. Let *R* = 8 Ω, *L* = 2 H, and *C* = 125 mF.
-
-# **Figure 8.107**
-
-For Prob. 8.62.
-
-# Section 8.8 Second-Order Op Amp Circuits
-
-**Figure 8.108** For Prob. 8.63.
-
-- **8.64** Using Fig. 8.109, design a problem to help other students better understand second-order op amp circuits.
-
-**8.65** Determine the differential equation for the op amp circuit in Fig. 8.110. If *v*1(0+) = 2 V and *v*2(0+) = 0 V, find *vo* for *t* > 0. Let *R* = 100 kΩ and *C* = 1 *μ*F.
-
-**Figure 8.110** For Prob. 8.65.
-
-**8.66** Obtain the differential equations for *vo*(*t*) in the op amp circuit of Fig. 8.111.
-
-**\*8.67** In the op amp circuit of Fig. 8.112, determine *vo*(*t*) for *t* > 0. Let *v*in = *u*(*t*) V, *R*1 = *R*2 = 10 kΩ, *C*1 = *C*2 = 100 *μ*F.
-
-# **Figure 8.112**
-
-**8.68** For the step function *vs* = *u*(*t*), use *PSpice* or *MultiSim* to find the response *v*(*t*) for 0 < *t* < 6 s in the circuit of Fig. 8.113.
-
-# **Figure 8.113**
-
-For Prob. 8.68.
-
-**8.69** Given the source-free circuit in Fig. 8.114, use *PSpice* or *MultiSim* to get *i*(*t*) for 0 < *t* < 20 s. Take *v*(0) = 30 V and *i*(0) = 2 A.
-
-# **Figure 8.114**
-
-For Prob. 8.69.
-
-**8.70** For the circuit in Fig. 8.115, use *PSpice* or *MultiSim* to obtain *v*(*t*) for 0 < *t* < 4 s. Assume that the capacitor voltage and inductor current at *t* = 0 are both zero.
-
-For Prob. 8.70.
-
-**8.71** Obtain *v*(*t*) for 0 < *t* < 4 s in the circuit of Fig. 8.116 using *PSpice* or *MultiSim*.
-
-# **Figure 8.116**
-
-For Prob. 8.71.
-
-**8.72** The switch in Fig. 8.117 has been in position 1 for a long time. At *t* = 0, it is switched to position 2. Use *PSpice* or *MultiSim* to find *i*(*t*) for 0 < *t* < 0.2 s.
-
-# **Figure 8.117**
-
-For Prob. 8.72.
-
-**8.73** Design a problem, to be solved using *PSpice* or *MultiSim*, to help other students better understand source-free *RLC* circuits.
-
-# Section 8.10 Duality
-
-**8.74** Draw the dual of the circuit shown in Fig. 8.118.
-
-**Figure 8.118** For Prob. 8.74.
-
-**8.75** Obtain the dual of the circuit in Fig. 8.119.
-
-**8.76** Find the dual of the circuit in Fig. 8.120.
-
-# **Figure 8.120**
-
-For Prob. 8.76.
-
-**8.77** Draw the dual of the circuit in Fig. 8.121.
-
-# **Figure 8.121**
-
-For Prob. 8.77.
-
-# Comprehensive Problems
-
-- **8.80** A mechanical system is modeled by a series *RLC* circuit. It is desired to produce an overdamped response with time constants 0.1 and 0.5 ms. If a series 50-kΩ resistor is used, find the values of *L* and *C*.
-- **8.81** An oscillogram can be adequately modeled by a second-order system in the form of a parallel *RLC* circuit. It is desired to give an underdamped voltage across a 200-Ω resistor. If the damped frequency is 4 kHz and the time constant of the envelope is 0.25 s, find the necessary values of *L* and *C*.
-- **8.82** The circuit in Fig. 8.123 is the electrical analog of body functions used in medical schools to study convulsions. The analog is as follows:
- - *C*1 = Volume of fluid in a drug
- - *C*2 = Volume of blood stream in a specified region
- - *R*1 = Resistance in the passage of the drug from the input to the blood stream
- - *R*2 = Resistance of the excretion mechanism, such as kidney, etc.
- - *v*0 = Initial concentration of the drug dosage
- - *v*(*t*) = Percentage of the drug in the blood stream
-
- Find *v*(*t*) for *t* > 0 given that *C*1 = 0.5 *μ*F, *C*2 = 5 *μ*F, *R*1 = 5 MΩ, *R*2 = 2.5 MΩ, and *v*0 = 60*u*(*t*) V.