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+# 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* 
+
+# <span id="page-11-0"></span>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.

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.
+
+<span id="page-15-0"></span>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

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                                                                                   | <b>PRACTICE</b>                                                  |    |
+| START EER - dieb, tird - all Countries excellents                                                            |                                                                  |    |
+| LETT CHEZ-Qui Internations<br>ESATI SEC - SHE SHE   FUNTOJ SPANIN AV 1411 (EN 44)                            | <b>SAIT</b>                                                      |    |
+| TIT   LAFE   Chapter &                                                                                       | <b>ISLANDATION</b>                                               |    |
+|                                                                                                              |                                                                  |    |
+| INNE YA'SA' - PONTES attentiar, rely cass increases                                                          |                                                                  |    |
+| CH 05 States of Consciousness<br>atade to us + box, taux - everywhere not see tow to                         |                                                                  |    |
+| <b>Guid - Extra Credit</b><br>ESAPTI SERA », publicada (" Pediministrato", excitore ta                       | GUIZ                                                             |    |
+| company.   CA-52, Sn la aniversidad: Vecebularia<br>high- aged : - Principal Mellerson 400 - Edit Pales, Add | 1,59                                                             |    |
+|                                                                                                              | START ALL A BAR GUIS ALCOHOL: 201<br>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* 

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+# <span id="page-21-0"></span>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
+
+<span id="page-22-0"></span>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).

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
+
+# <span id="page-24-0"></span>**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

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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+# <span id="page-25-0"></span>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
+
+# <span id="page-26-0"></span>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.

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.
+
+<span id="page-27-0"></span>
+
+# **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      |
+|            |        |        |

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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+# <span id="page-28-0"></span>**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<sup>−</sup>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<sup>−</sup>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<sup>−</sup>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:<br>−19 C. Hence 4,600 electrons will<br>Each electron has<br>−1.602 × 10<br>have −1.602 × 10−19 C/electron × 4,600 electrons =<br>−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<br>C.                                                                                                                                  |
+|                      |                                                                                                                                                              |
+| Example 1.2          | The total char ge entering a terminal is gi<br>ven by q<br>= 5<br>t sin 4<br>πt mC.<br>Calculate the current at t<br>= 0.5 s.                                |
+|                      | Solution:                                                                                                                                                    |
+|                      | dq___<br>__d<br>i<br>=<br>dt =<br>dt (5t sin 4πt) mC/s = (5 sin 4πt<br>+ 20πt cos 4πt) mA                                                                    |
+|                      | At t<br>= 0.5,                                                                                                                                               |
+|                      | i<br>= 5 sin 2π<br>+ 10π cos 2π<br>= 0 + 10π<br>= 31.42 mA                                                                                                   |
+| Practice Problem 1.2 | = (10 − 10e−2t<br>If in Example 1.2, q<br>) mC, find the current at t<br>= 1.0 s.                                                                            |
+|                      | Answer: 2.707 mA.                                                                                                                                            |
+|                      |                                                                                                                                                              |
+| Example 1.3          | Determine the total charge entering a terminal between t = 1 s and<br>2 −<br>t = 2 s if the current passing the terminal is i = (3t<br>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
+$$
+i = \begin{cases} 4 \text{ A}, & 0 < t < 1 \\ 4t^2 \text{ A}, & t > 1 \end{cases}
+$$
+
+<span id="page-31-0"></span>Calculate the charge entering the element from *t* = 0 to *t* = 2 s.
+
+**Answer:** 13.333 C.

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
+
+<span id="page-32-0"></span>© 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.

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 <sup>p</sup> = +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<sup>−</sup><sup>3</sup> ) = 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.
+
+<span id="page-36-0"></span>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

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*<sup>1</sup> = 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*<sup>2</sup> = 12(5) = 60 W Absorbed power *p*<sup>3</sup> = 8(6) = 48 W Absorbed power
+
+For Example 1.7.
+
+<span id="page-38-0"></span>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*<sup>1</sup> + *p*<sup>2</sup> + *p*<sup>3</sup> + *p*<sup>4</sup> = −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*<sup>1</sup> = −45 W, *p*<sup>2</sup> = 18 W, *p*<sup>3</sup> = 12 W, *p*<sup>4</sup> = 15 W.

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.
+
+<sup>2</sup> The dagger sign preceding a section heading indicates the section that may be skipped, explained briefly, or assigned as homework.
+
+<sup>3</sup> 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<sup>−</sup>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.
+
+<span id="page-41-0"></span>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 +<sup>400</sup> <sup>=</sup> 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.

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*<sup>8</sup>Ω.
+
+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*<sup>8</sup>Ω 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*<sup>8</sup>Ω 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*<sup>8</sup>Ω using nodal analysis.
+
+4. *Attempt a problem solution*. We first write down all of the equations we will need in order to find *i*<sup>8</sup>Ω.
+
+$$
+i_{8\Omega} = i_2,
+$$
+  $i_2 = \frac{v_1}{8},$   $i_{8\Omega} = \frac{v_1}{8}$   
+ $\frac{v_1 - 5}{2} + \frac{v_1 - 0}{8} + \frac{v_1 + 3}{4} = 0$ 
+
+**Figure 1.19** Illustrative example.
+
+**Figure 1.20** Problem definition.
+
+<span id="page-44-0"></span>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*<sup>1</sup> + *i*<sup>2</sup> + *i*<sup>3</sup> = −**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

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.
+
+<span id="page-45-0"></span>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) 10<sup>6</sup> (b) 103 (c) 10<sup>−</sup><sup>3</sup> (d) 10<sup>−</sup><sup>6</sup>
+
+**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.*

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/015_Review Questions.md

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+# <span id="page-46-0"></span>Problems
+
+# Section 1.3 Charge and Current
+
+- **1.1** How much charge is represented by these number of electrons?
+  - (a) 6.482 × 10<sup>17</sup>
+  - (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*<sup>−</sup>2*<sup>t</sup>* 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*<sup>1</sup> = 60 W absorbed, *p*<sup>3</sup> = −145 W absorbed, and *p*<sup>4</sup> = 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.
+
+# <span id="page-48-0"></span>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?

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/016_Problems.md

@@ -0,0 +1,27 @@
+# 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**

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/017_Chapter 2 - Basic Laws.md

@@ -0,0 +1,25 @@
+# <span id="page-51-0"></span>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.
+
+# <span id="page-52-0"></span>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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/018_2.1 Introduction.md

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+# **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.

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 <sup>−</sup>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.
+
+<span id="page-57-0"></span>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 <sup>2</sup> *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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/020_2.3 Nodes, Branches, and Loops.md

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+# **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.
+
+<span id="page-59-0"></span>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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/021_2.4 Kirchhoff's Laws.md

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+# **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 <sup>k</sup>* (*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*<sup>5</sup> 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.
+
+<sup>F</sup>or the circuit in Fig. 2.21(a), find voltages *v*1 and *v*<sup>2</sup> 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 *<sup>v</sup>* Example 2.6 *<sup>o</sup>* 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),
+
+$$
+-30 + v_1 + v_2 = 0
+$$
+
+<span id="page-65-0"></span>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*<sup>2</sup> + *v*<sup>3</sup> = 0 ⇒ *v*<sup>3</sup> = *v*<sup>2</sup> **(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.
+
+<span id="page-66-0"></span>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.

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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+**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*<sup>1</sup> = *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*<sup>1</sup> = *R*<sup>2</sup> = ⋯=*RN* = *R*, then
+
+$$
+R_{\text{eq}} = \frac{R}{N} \tag{2.39}
+$$

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

@@ -0,0 +1,308 @@
+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*<sup>1</sup> = 1/*R*1, *G*<sup>2</sup> = 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*<sup>2</sup> = 0; that is, *R*2 is a short circuit, as sho wn in Fig. 2.33(a). From Eq. (2.43), *R*<sup>2</sup> = 0 implies that *i*<sup>1</sup> = 0, *i*<sup>2</sup> = *i*. This means that the entire current *i* bypasses *R*1 and flows through the short circuit *R*<sup>2</sup> = 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*<sup>2</sup> = 0 in Eq. (2.37).]
+- 2. The entire current flows through the short circuit.
+
+As another e xtreme case, suppose *R*<sup>2</sup> = ∞, 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*<sup>2</sup> → ∞,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*<sup>2</sup> and the power dissipated in the 12-Ω and 40-Ω resistors.
+
+**Answer:** *v*<sup>1</sup> = 10 V, *i*1 = 833.3 mA, *p*1 = 8.333 W, *v*<sup>2</sup> = 20 V, *i*2 =500 mA, *p*<sup>2</sup> = 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.
+
+$$
+i_1 = \frac{18,000}{9,000 + 18,000} (30 \text{ mA}) = 20 \text{ mA}
+$$
+$$
+i_2 = \frac{9,000}{9,000 + 18,000} (30 \text{ mA}) = 10 \text{ mA}
+$$
+
+<span id="page-73-0"></span>Notice that the voltage across the 9-kΩ and 18-kΩ resistors is the same, and *vo* = 9,000*i*<sup>1</sup> = 18,000*i*<sup>2</sup> = 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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/024_2.7 Wye-Delta Transformations.md

@@ -0,0 +1,272 @@
+# **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.
+
+<span id="page-79-0"></span>For the bridge network in Fig. 2.54, find *Rab* and *i*.
+
+**Answer:** 60 Ω, 4 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.
+
+<span id="page-85-0"></span>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.

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*<sup>1</sup> = *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*<sup>1</sup> = *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
+
+<span id="page-86-0"></span>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**
+
+<span id="page-87-0"></span>**2.8** In the circuit in Fig. 2.65, *V* is: (a) 30 V (b) 14 V (c) 10 V (d) 6 V

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.

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*<sup>3</sup>
+  - (b) voltage across *R*<sup>3</sup>
+  - (c) voltage across *R*<sup>1</sup>
+  - (d) power dissipated in *R*<sup>2</sup>
+  - (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*<sup>1</sup> = *R*<sup>2</sup> = *R*<sup>3</sup> = *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*<sup>1</sup> = 80 Ω, cost = \$0.60 (standard size) *R*<sup>2</sup> = 90 Ω, cost = \$0.90 (standard size) *R*<sup>3</sup> = 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%
+
+\_\_\_\_\_ *<sup>I</sup>*− *I*<sup>ʹ</sup> *<sup>I</sup>*
+
+**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*<sup>1</sup> = 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*<sup>1</sup> = 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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/029_Comprehensive Problems.md

@@ -0,0 +1,40 @@
+# <span id="page-99-0"></span>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*<sup>2</sup> needed to power the devices using a 36-V battery.
+
+**Figure 2.141**
+
+For Prob. 2.82.
+
+# **chapter**
+
+3

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/030_Chapter 3 - Methods of Analysis.md

@@ -0,0 +1,82 @@
+# <span id="page-101-0"></span>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
+
+# <span id="page-102-0"></span>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, . . . , *v<sup>n</sup>*− 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**

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/031_3.1 Introduction.md

@@ -0,0 +1,358 @@
+
+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*<sup>2</sup> = 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
+
+<sup>1</sup>= 5
+
+2 Ω 6 Ω 10 A
+
+(a)
+
+2 Ω 6 Ω 10 A
+
+2 i 5
+
+i i <sup>4</sup>= 10 <sup>2</sup>
+
+5 A
+
+4 Ω
+
+<sup>3</sup> <sup>i</sup>
+
+*<sup>v</sup>*<sup>2</sup> *<sup>v</sup>*<sup>1</sup>
+
+<sup>1</sup>= 5 i
+
+5 A
+
+4 Ω
+
+1
+
+i
+
+i
+
+**Answer:** *v*<sup>1</sup> = 30 V, *v*<sup>2</sup> = −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
+
+Δ<sup>3</sup> =
+
+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
+
+<span id="page-108-0"></span>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*<sup>1</sup> = 4.8 V, *v*<sup>2</sup> = 2.4 V, and *v*<sup>3</sup> = −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*<sup>1</sup> = 32 V, *v*<sup>2</sup> = −25.6 V, *v*<sup>3</sup> = 62.4 V.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/032_3.2 Nodal Analysis.md

@@ -0,0 +1,433 @@
+# **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*<sup>1</sup> − 2 + *v*<sup>2</sup> = 0 ⇒ *v*<sup>2</sup> = *v*<sup>1</sup> + 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*<sup>2</sup> = *v*<sup>1</sup> + 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*<sup>x</sup> + – + – 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*<sup>3</sup> − *v*<sup>2</sup> 6 + 10 = \_\_\_\_\_\_ *v*<sup>1</sup> − *v*<sup>4</sup> 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 *v<sup>x</sup>* = *v*<sup>1</sup> − *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*<sup>3</sup> = *v*<sup>3</sup> − *v*2 and *v<sup>x</sup>* = *v*<sup>1</sup> �� *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*<sup>2</sup> = *v*<sup>1</sup> − 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 <sup>−</sup>16 ] [ *v*1 *v*3 *v*4 ] = [ 0 80 <sup>40</sup>]
+
+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*<sup>2</sup> = *v*<sup>1</sup> − 20 = 6.667 V. We have not used Eq. (3.4.5); it can be used to cross-check results.
+
+<span id="page-113-0"></span>Find *v*1, *v*2, and *v*3 in the circuit of Fig. 3.14 using nodal analysis. Practice Problem 3.4
+
+**Answer:** *v*<sup>1</sup> = 7.608 V, *v*<sup>2</sup> = −17.39 V, *v*<sup>3</sup> = 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*<sup>3</sup> *i*<sup>1</sup> + (*R*<sup>2</sup> + *R*3) *i*<sup>2</sup> = −*V*<sup>2</sup> **(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*<sup>1</sup> = 2*i*<sup>2</sup> − 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 ] <sup>=</sup> [ <sup>1</sup> 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*<sup>1</sup> = 4.6 A, *i*<sup>2</sup> = 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*<sup>2</sup> + 4 (*i*<sup>2</sup> − *i*3) + 10 (*i*<sup>2</sup> − *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*<sup>1</sup> − *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*<sup>1</sup> − *i*<sup>2</sup> = 1.5 A.
+
+# <span id="page-118-0"></span>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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/033_3.4 Mesh Analysis.md

@@ -0,0 +1,105 @@
+# **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*<sup>2</sup> = −5 A and write a mesh equation for the other mesh in the usual way; that is,
+
+−10 + 4*i*<sup>1</sup> + 6(*i*<sup>1</sup> − *i*2) = 0 ⇒ *i*<sup>1</sup> = −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*<sup>1</sup> + 10*i*<sup>2</sup> + 4*i*<sup>2</sup> = 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.
+
+<sup>F</sup>or 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*<sup>1</sup> + 4*i*<sup>3</sup> + 8(*i*<sup>3</sup> − *i*4) + 6*i*<sup>2</sup> = 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*:
+
+$$
+i_2 = i_3 + 3I_o
+$$
+
+<span id="page-120-0"></span>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:** 
+$$
+i_1 = 12.379 \text{ A}, i_2 = 378.9 \text{ mA}, i_3 = 3.284 \text{ A}.
+$$

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

@@ -0,0 +1,209 @@
+# **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<br>v1                      |  |
+|--------|------|-------|-------|------------------------------|--|
+| −0.2   | 0.2  | 0     | 0     | 5<br>v2                      |  |
+| −1     | 0    | 1.25  | −0.25 | =<br>−3<br>v3                |  |
+| [<br>0 | 0    | −0.25 | 1.25  | ]<br>[<br>]<br>[<br>2]<br>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<br>] |   | i4      | =<br>[ | 0<br>] |  |
+|   | 0  | −1 | 0  | −3 | 4       | [ | ]<br>i5 |        | −6     |  |
+
+From this, we can use *MATLAB* to obtain mesh currents *i*1, *i*2, *i*3, *i*4, and *i*5.
+
+# <span id="page-124-0"></span>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<br>] | i4<br>[ | =<br>[<br>] | 20<br>] |  |
+|   | 0   | 0   | −20 | 0   | 80     | i5      |             | −20     |  |

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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+# **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.
+
+<span id="page-125-0"></span>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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/036_3.8 Circuit Analysis with PSpice.md

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+# **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<br>= 120 V, V2<br>= 81.29 V, V3<br>= 89.032 V. |              |  |                           |  |         |  |
+
+Practice Problem 3.10 For the circuit in Fig. 3.33, use *PSpice* to find the node voltages.
+
+**Answer:** *V*<sup>1</sup> = −10 V, *V*<sup>2</sup> = 14.286 V, *V*<sup>3</sup> = 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.
+
+# <span id="page-127-0"></span>**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*<sup>1</sup> = *i*<sup>2</sup> = 1.333 A and *i*<sup>3</sup> = 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*<sup>1</sup> = −428.6 mA, *i*<sup>2</sup> = 2.286 A, *i*<sup>3</sup> = 2 A.
+
+# **3.9** †

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/037_3.9 Applications - DC Transistor Circuits.md

@@ -0,0 +1,179 @@
+# 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 *v<sup>o</sup>* = *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*<sup>1</sup> + 200k(*I*<sup>1</sup> − I2) = 0 or 3*I*<sup>1</sup> − 2*I*<sup>2</sup> = 2 × 10<sup>−</sup><sup>5</sup> **(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
+
+<span id="page-132-0"></span>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*<sup>1</sup> and *I*2. Adding Eq. (3.13.1) to (3.13.2) we get;
+
+*I*<sup>1</sup> = 1.3 × 10<sup>−</sup><sup>5</sup> A and *I*<sup>2</sup> = (−0.7 + 2.6)10<sup>−</sup><sup>5</sup> ∕2 = 9.5 *µ*A Since *I*<sup>3</sup> = −150*I*<sup>2</sup> = −1.425 mA, we can now solve for *vo* using loop 3: −*v<sup>o</sup>* + 1 k*I*<sup>3</sup> + 16 = 0 or *v<sup>o</sup>* = −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*<sup>1</sup> = 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* + (*v<sup>o</sup>* − 16)∕1k = 0 or *v<sup>o</sup>* = 16 − 150 × 103 × 9.5 × 10<sup>−</sup><sup>6</sup> = **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 *v<sup>o</sup>* 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.
+
+- <span id="page-133-0"></span>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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/038_3.10 Summary.md

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+# 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
+
+- <span id="page-134-0"></span>**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

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/039_Review Questions.md

@@ -0,0 +1,23 @@
+# **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.*

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/040_Problems.md

@@ -0,0 +1,383 @@
+# 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*.
+
+<span id="page-146-0"></span>**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 *v<sup>o</sup>* = 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.

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

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/042_Chapter 4 - Circuit Theorems.md

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+# <span id="page-147-0"></span>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
+
+# <span id="page-148-0"></span>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.

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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+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 *v<sup>s</sup>* = 10 V gives *i* = 2 A. According to the linearity principle, *v<sup>s</sup>* = 1 V will give *i* = 0.2 A. By the same token, *i* = 1 mA must be due to *v<sup>s</sup>* = 5 mV.
+
+For example, when current i1 flows through resistor R, the power is <sup>p</sup><sup>1</sup> = Ri<sup>1</sup> 2 , and when current i2 flows through R, the power is <sup>p</sup><sup>2</sup> = Ri<sup>2</sup> 2 . If current i<sup>1</sup> + <sup>i</sup>2 flows through R, the power absorbed is <sup>p</sup><sup>3</sup> = <sup>R</sup> (i<sup>1</sup> + <sup>i</sup>2) <sup>2</sup> = Ri<sup>1</sup> <sup>2</sup> + Ri<sup>2</sup> <sup>2</sup> + 2Ri1i<sup>2</sup> ≠ <sup>p</sup><sup>1</sup> + <sup>p</sup>2. Thus, the power relation is nonlinear.
+
+# **Figure 4.1**
+
+A linear circuit with input *vs* and output *i*.
+
+<sup>F</sup>or the circuit in Fig. 4.2, find *Io* when *v<sup>s</sup>* <sup>=</sup> 12 V and *v<sup>s</sup>* <sup>=</sup> 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 *v<sup>x</sup>* = 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 *v<sup>s</sup>* = 12 V,
+
+When *v<sup>s</sup>* = 24 V,
+
+$$
+I_o = i_2 = \frac{24}{76}
+$$
+ A
+
+\_\_\_ 12 76 A
+
+*Io* = *i*<sup>2</sup> =
+
+showing that when the source value is doubled, *Io* doubles.
+
+<span id="page-150-0"></span>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*<sup>1</sup> = (3 + 5)*Io* = 8 V and *I*<sup>1</sup> = *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*<sup>2</sup> = 4*i*<sup>3</sup> = 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*′ *<sup>o</sup>* and *i o* ″ are due to the 4-A current source and 20-V voltage source respectively. To obtain *i*′ *<sup>o</sup>*, 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*′ *<sup>o</sup>*. 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*′ *<sup>o</sup>*, (b) obtain *i* ″ *o*.
+
+For loop 3,
+
+−5*i*1 − 1*i*2 + 10*i*3 + 5*i*′ *<sup>o</sup>* = 0 **(4.4.4)** But at node 0, *i*3 = *i*1 − *i*′ *<sup>o</sup>* = 4 − *i*′ *<sup>o</sup>* **(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*″ *<sup>o</sup>*, 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*″ *<sup>o</sup>*. 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*<sup>1</sup> = 3 and
+
+$$
+i_3 = \frac{v_1}{3} = 1 \text{ A}
+$$
+
+Thus,
+
+$$
+i = i_1 + i_2 + i_3 = 2 - 1 + 1 = 2 \text{ A}
+$$
+
+<span id="page-155-0"></span>Find *I* in the circuit of Fig. 4.14 using the superposition principle. Practice Problem 4.5
+
+2 Ω
+
+2 A
+
+– 6 V <sup>+</sup>
+
+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 <sup>v</sup><sup>s</sup> in series with a resistor R by a current source is in parallel with a resistor <sup>R</sup>, 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, *v<sup>x</sup>* = 3 − *i* = 7.5 V.
+
+For Practice Prob. 4.7.
+
+# <span id="page-159-0"></span>**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 <sup>R</sup>Th = <sup>v</sup>oc ∕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).

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/044_4.3 Superposition.md

@@ -0,0 +1,425 @@
+■ **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 *v<sup>o</sup>* = 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*<sup>1</sup> = 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 *v<sup>x</sup>* = *i*1 − *i*<sup>2</sup>
+
+But −4*i*<sup>2</sup> = *v<sup>x</sup>* = *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*<sup>3</sup> = 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*<sup>2</sup> = 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 <sup>x</sup> <sup>+</sup> – 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 *v<sup>o</sup>* = 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
+
+$$
+i_x = i_2 - i_1
+$$
+
+<span id="page-165-0"></span>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 <sup>R</sup>N, where IN is the short-circuit current through the terminals and <sup>R</sup>N 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 *v<sup>o</sup>* = 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 <sup>5</sup><sup>Ω</sup> <sup>=</sup> 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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/045_4.5 Thevenin's Theorem.md

@@ -0,0 +1,48 @@
+# <span id="page-169-0"></span>**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*<sup>1</sup> 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*<sup>0</sup> = *A*1*vs*<sup>1</sup> + *A*2*vs*<sup>2</sup> + *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*<sup>0</sup> = 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.

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

@@ -0,0 +1,15 @@
+## <span id="page-170-0"></span>**148** Chapter 4 Circuit Theorems
+
+*i* = *D*<sup>0</sup> = −*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*<sup>0</sup> = 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.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/047_4.8 Maximum Power Transfer.md

@@ -0,0 +1,87 @@
+# **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 (<sup>R</sup><sup>L</sup> = <sup>R</sup>Th).
+
+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*<sup>2</sup> Th [ (*R*Th + *RL*) 2 <sup>−</sup>2*RL*(*R*Th + *RL*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*<sup>R</sup>*Th + *RL*) 4 ] = *V*<sup>2</sup> Th [ (*R*Th + *RL* − 2*RL*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*<sup>R</sup>*Th + *RL*) 3 ] <sup>=</sup><sup>0</sup>
+
+**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*<sup>2</sup> *p*∕*dR*<sup>2</sup> *<sup>L</sup>*< 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 <sup>R</sup><sup>L</sup> = <sup>R</sup>Th.
+
+**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.
+
+<span id="page-172-0"></span>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*<sup>1</sup> = −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.