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engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/099_8.11 Applications.md

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-# Section 8.11 Applications
-
-**8.78** An automobile airbag igniter is modeled by the circuit in Fig. 8.122. Determine the time it takes the voltage across the igniter to reach its first peak after switching from *A* to *B*. Let *R* = 3 Ω, *C* = 1∕30 F, and *L* = 60 mH.
-
-**8.79** A load is modeled as a 100-mH inductor in parallel with a 12-Ω resistor. A capacitor is needed to be connected to the load so that the network is critically damped at 60 Hz. Calculate the size of the capacitor.
-
-8.83 Figure 8.124 shows a typical tunnel-diode oscillator circuit. The diode is modeled as a nonlinear resistor with *iD* = *f* (*vD*), i.e., the diode current is a nonlinear function of the voltage across the diode. Derive the differential equation for the circuit in
-
-terms of *v* and *iD*.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/100_Comprehensive Problems.md

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-# <span id="page-388-0"></span>**PART TWO**

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/101_PART 2 - AC Circuits.md

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-# AC Circuits
-
-# OUTLINE
-
-- 9 Sinusoids and Phasors
-- 10 Sinusoidal Steady-State Analysis
-- 11 AC Power Analysis
-- 12 Three-Phase Circuits
-- 13 Magnetically Coupled Circuits
-- 14 Frequency Response
-
-# **chapter**
-
-9

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/102_Chapter 9 - Sinusoids and Phasors.md

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-# <span id="page-389-0"></span>Sinusoids and Phasors
-
-*He who knows not, and knows not that he knows not, is a fool—shun him. He who knows not, and knows that he knows not, is a c hild—teach him. He who knows, and knows not that he knows, is asleep—wake him up. He who knows, and knows that he knows, is wise—follow him.*
-
-—Persian Proverb
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.d), "an ability to function on multi‑disciplinary teams."**
-
-The "ability to function on multidisciplinary teams" is inherently critical for the working engineer. Engineers rarely, if ever, work by themselves. Engineers will always be part of some team. One of the things I lik e to remind students is that you do not ha ve to like everyone on a team; you just have to be a successful part of that team.
-
-Most frequently, these teams include indi viduals from a v ariety of engineering disciplines, as well as individuals from nonengineering disciplines such as marketing and finance.
-
-Students can easily de velop and enhance this skill by w orking in study groups in every course they take. Clearly, working in study groups in nonengineering courses, as well as engineering courses outside your discipline, will also give you experience with multidisciplinary teams.
-
-Photo by Charles Alexander
-
-# Historical
-
-<span id="page-390-0"></span>George Westinghouse. Photo © Bettmann/Corbis
-
-**Nikola Tesla** (1856–1943) and **George Westinghouse** (1846–1914) helped establish alternating current as the primary mode of electricity transmission and distribution.
-
-Today it is obvious that ac generation is well established as the form of electric power that makes widespread distribution of electric power efficient and economical. However, at the end of the 19th century, which was the better—ac or dc—w as hotly debated and had e xtremely outspoken supporters on both sides. The dc side was led by Thomas Edison, who had earned a lot of respect for his many contributions. Power generation using ac really be gan to b uild after the successful contrib utions of Tesla. The real commercial success in ac came from Geor ge Westinghouse and the outstanding team, including Tesla, he assembled. In addition, tw o other big names were C. F. Scott and B. G. Lamme.
-
-The most significant contribution to the early success of ac w as the patenting of the polyphase ac motor by Tesla in 1888. The induction motor and polyphase generation and distrib ution systems doomed the use of dc as the prime energy source.
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Better understand sinusoids.
-- 2. Understand phasors.
-- 3. Understand the phasor relationships for circuit elements.
-- 4. Know and understand the concepts of impedance and admittance.
-- 5. Understand Kirchhoff's laws in the frequency domain.
-- 6. Comprehend the concept of phase-shift.
-- 7. Understand the concept of AC bridges.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/103_9.1 Introduction.md

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-# **9.1** Introduction
-
-Thus far our analysis has been limited for the most part to dc circuits: those circuits e xcited by constant or time-in variant sources. We ha ve restricted the forcing function to dc sources for the sake of simplicity, for pedagogic reasons, and also for historic reasons. Historically, dc sources were the main means of providing electric power up until the late 1800s. At the end of that century, the battle of direct current versus alternating current began. Both had their advocates among the electrical engineers of the time. Because ac is more efficient and economical to transmit over long distances, ac systems ended up the winner . Thus, it is in k eeping with the historical sequence of events that we considered dc sources first.
-
-We now begin the analysis of circuits in which the source v oltage or current is time-v arying. In this chapter , we are particularly interested in sinusoidally time-varying excitation, or simply, excitation by a *sinusoid*.
-
-## <span id="page-391-0"></span>A sinusoid is a signal that has the form of the sine or cosine function.
-
-A sinusoidal current is usually referred to as *alternating current* (*ac*). Such a current reverses at regular time intervals and has alternately positive and negative values. Circuits driven by sinusoidal current or voltage sources are called *ac circuits*.
-
-We are interested in sinusoids for a number of reasons. First, nature itself is characteristically sinusoidal. We e xperience sinusoidal v ariation in the motion of a pendulum, the vibration of a string, the ripples on the ocean surface, and the natural response of underdamped secondorder systems, to mention but a few. Second, a sinusoidal signal is easy to generate and transmit. It is the form of v oltage generated throughout the world and supplied to homes, factories, laboratories, and so on. It is the dominant form of signal in the communications and electric power industries. Third, through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals. Lastly , a sinu soid is easy to handle mathematically . The derivative and integral of a sinusoid are themselve s sinusoids. For these and other reasons, the sinusoid is an extremely important function in circuit analysis.
-
-A sinusoidal forcing function produces both a transient response and a steady-state response, much like the step function, which we studied in Chapters 7 and 8. The transient response dies out with time so that only the steady-state response remains. When the transient response has become negligibly small compared with the steady-state response, we say that the circuit is operating at sinusoidal steady state. It is this *sinusoidal steady-state response* that is of main interest to us in this chapter.
-
- We begin with a basic discussion of sinusoids and phasors. We then introduce the concepts of impedance and admittance. The basic circuit laws, Kirchhoff's and Ohm's, introduced for dc circuits, will be applied to ac circuits. Finally , we consider applications of ac circuits in phaseshifters and bridges.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/104_9.2 Sinusoids.md

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-# **9.2** Sinusoids
-
-Consider the sinusoidal voltage
-
-$$
-v(t) = V_m \sin \omega t \tag{9.1}
-$$
-
-where
-
-*Vm* = the *amplitude* of the sinusoid
-
-*ω* = the *angular frequency* in radians/s
-
-*ωt* = the *argument* of the sinusoid
-
-The sinusoid is shown in Fig. 9.1(a) as a function of its argument and in Fig. 9.1(b) as a function of time. It is e vident that the sinusoid repeats itself every *T* seconds; thus, *T* is called the *period* of the sinusoid. From the two plots in Fig. 9.1, we observe that *ωT* = 2*π*,
-
-$$
-T = \frac{2\pi}{\omega} \tag{9.2}
-$$
-
-# Historical
-
-© Hulton Archives/Getty Images
-
-**Heinrich Rudorf Hertz** (1857–1894), a German experimental physicist, demonstrated that electromagnetic waves obey the same fundamental laws as light. His work confirmed James Clerk Maxwell's celebrated 1864 theory and prediction that such waves existed.
-
-Hertz w as born into a prosperous f amily in Hamb urg, German y. He attended the Uni versity of Berlin and did his doctorate under the prominent physicist Hermann von Helmholtz. He became a professor at Karlsruhe, where he be gan his quest for electromagnetic w aves. Hertz successfully generated and detected electromagnetic w aves; he was the first to show that light is electromagnetic ener gy. In 1887, Hertz noted for the first time the photoelectric effect of electrons in a molecular structure. Although Hertz only lived to the age of 37, his discovery of electromagnetic waves paved the way for the practical use of such waves in radio, television, and other communication systems. The unit of frequency, the hertz, bears his name.
-
-The fact that *v*(*t*) repeats itself every *T* seconds is shown by replacing *t* by *t* + *T* in Eq. (9.1). We get
-
-$$
-v(t+T) = V_m \sin \omega(t+T) = V_m \sin \omega \left(t + \frac{2\pi}{\omega}\right)
-$$
-  
-=  $V_m \sin (\omega t + 2\pi) = V_m \sin \omega t = v(t)$  (9.3)
-
-Hence,
-
-$$
-v(t+T) = v(t) \tag{9.4}
-$$
-
-that is, *v* has the same value at *t* + *T* as it does at *t* and *v*(*t*) is said to be *periodic*. In general,
-
-A periodic function is one that satisfies <sup>f</sup> (t) = <sup>f</sup> (t + nT ), for all t and for all integers n.
-
-As mentioned, the *period T* of the periodic function is the time of one complete cycle or the number of seconds per c ycle. The reciprocal of this quantity is the number of c ycles per second, kno wn as the *cyclic frequency f* of the sinusoid. Thus,
-
-$$
-f = \frac{1}{T}
-$$
- (9.5)
-
-From Eqs. (9.2) and (9.5), it is clear that
-
-$$
-\omega = 2 \pi f \tag{9.6}
-$$
-
-While *ω* is in radians per second (rad/s), *f* is in hertz (Hz).
-
-Let us now consider a more general expression for the sinusoid,
-
-$$
-v(t) = V_m \sin(\omega t + \phi)
-$$
-\n(9.7)
-
-where (*ωt* + *ϕ*) is the argument and *ϕ* is the *phase*. Both argument and phase can be in radians or degrees.
-
-Let us examine the two sinusoids
-
-$$
-v_1(t) = V_m \sin \omega t \qquad \text{and} \qquad v_2(t) = V_m \sin(\omega t + \phi) \tag{9.8}
-$$
-
-shown in Fig. 9.2. The starting point of *v*2 in Fig. 9.2 occurs first in time. Therefore, we say that *v*<sup>2</sup> *leads v*1 by *ϕ* or that *v*<sup>1</sup> *lags v*2 by *ϕ*. If *ϕ* ≠ 0, we also say that *v*1 and *v*2 are *out of phase*. If *ϕ* = 0, then *v*1 and *v*2 are said to be *in phase;* they reach their minima and maxima at e xactly the same time. We can compare *v*1 and *v*2 in this manner because they operate at the same frequency; they do not need to have the same amplitude.
-
-A sinusoid can be e xpressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. This is achieved by using the following trigonometric identities:
-
-$$
-sin(A \pm B) = sin A cos B \pm cos A sin B
-$$
-  
-\n
-$$
-cos(A \pm B) = cos A cos B \mp sin A sin B
-$$
-\n(9.9)
-
-With these identities, it is easy to show that
-
-$$
-sin(\omega t \pm 180^\circ) = -sin \omega t
-$$
-  
-\n
-$$
-cos(\omega t \pm 180^\circ) = -cos \omega t
-$$
-  
-\n
-$$
-sin(\omega t \pm 90^\circ) = \pm cos \omega t
-$$
-  
-\n
-$$
-cos(\omega t \pm 90^\circ) = \mp sin \omega t
-$$
-\n(9.10)
-
-Using these relationships, we can transform a sinusoid from sine form to cosine form or vice versa.
-
-The unit of f is named after the German physicist Heinrich R. Hertz (1857–1894).
-
-# **Figure 9.3**
-
-A graphical means of relating cosine and sine: (a) cos(*ωt* − 90°) = sin *ωt*, (b) sin(*ωt* + 180°) = −sin *ωt*.
-
-A graphical approach may be used to relate or compare sinusoids as an alternati ve to using the trigonometric identities in Eqs. (9.9) and (9.10). Consider the set of axes shown in Fig. 9.3(a). The horizontal axis represents the magnitude of cosine, while the vertical axis (pointing down) denotes the magnitude of sine. Angles are measured positi vely counterclockwise from the horizontal, as usual in polar coordinates. This graphical technique can be used to relate tw o sinusoids. F or example, we see in Fig. 9.3(a) that subtracting 90° from the ar gument of cos *ωt* gives sin *ωt*, o r cos(*ωt* − 90°) = sin *ωt*. Similarly, adding 1 80° to the argument of sin*ωt* gives −sin *ωt*, or sin(*ωt* + 180°) = −sin *ωt*, as shown in Fig. 9.3(b).
-
-The graphical technique can also be used to add tw o sinusoids of the same frequency when one is in sine form and the other is in cosine form. To add *A* cos *ωt* and *B* sin *ωt*, we note that *A* is the magnitude of cos *ωt* while *B* is the magnitude of sin*ωt*, as shown in Fig. 9.4(a). The magnitude and argument of the resultant sinusoid in cosine form is readily obtained from the triangle. Thus,
-
-$$
-A\cos\omega t + B\sin\omega t = C\cos(\omega t - \theta)
-$$
-\n(9.11)
-
-where
-
-$$
-C = \sqrt{A^2 + B^2}
-$$
-,  $\theta = \tan^{-1} \frac{B}{A}$  (9.12)
-
-For example, we may add 3 cos *ωt* and −4 sin*ωt* as shown in Fig. 9.4(b) and obtain
-
-$$
-3\cos \omega t - 4\sin \omega t = 5\cos(\omega t + 53.1^{\circ})
-$$
- (9.13)
-
-Compared with the trigonometric identities in Eqs. (9.9) and (9.10), the graphical approach eliminates memorization. Ho wever, we must not confuse the sine and cosine ax es with the ax es for comple x numbers to be discussed in the ne xt section. Something else to note in Figs. 9.3 and 9.4 is that although the natural tendenc y is to ha ve the vertical axis point up, the positive direction of the sine function is down in the present case.
-
-(a) Adding *A* cos *ωt* and *B* sin *ωt*, (b) adding 3 cos *ωt* and −4 sin *ωt*.
-
-*v*(*t*) = 12 cos(50*t* + 10°) V.
-
-# **Solution:**
-
-The amplitude is *Vm* = 12 V. The phase is *ϕ* = 10°. The angular frequency is *ω* = 50 rad/s. The period *T* = \_\_\_ <sup>2</sup>*<sup>π</sup> ω* = \_\_\_ <sup>2</sup>*<sup>π</sup>* 50 = 0.1257 s. The frequency is *f* = \_\_1 *T* = 7.958 Hz.
-
-Given the sinusoid 45 cos(5 *πt* + 36°), calculate its amplitude, phase, angular frequency, period, and frequency.
-
-**Answer:** 45, 36°, 15.708 rad/s, 400 ms, 2.5 Hz.
-
-Calculate the phase angle between *v*1 = −10 cos(*ωt* + 50°) and *v*2 = Example 9.2 12 sin(*ωt* − 10°). State which sinusoid is leading.
-
-# **Solution:**
-
-Let us calculate the phase in three ways. The first two methods use trigonometric identities, while the third method uses the graphical approach.
-
-■ **METHOD 1** In order to compare *v*1 and *v*2, we must e xpress them in the same form. If we e xpress them in cosine form with posi tive amplitudes,
-
-*v*1 = −10 cos(*ωt* + 50°) = 10 cos(*ωt* + 50° − 180°) *v*1 = 10 cos(*ωt* − 130°) or *v*1 = 10 cos(*ωt* + 230°) **(9.2.1)** and *v*2 = 12 sin(*ωt* − 10°) = 12 cos(*ωt* − 10° − 90°) *v*2 = 12 cos(*ωt* − 100°) **(9.2.2)**
-
-It can be deduced from Eqs. (9.2.1) and (9.2.2) that the phase difference between *v*1 and *v*2 is 30°. We can write *v*2 as
-
-*v*2 = 12 cos(*ωt* − 130° + 30°) or *v*2 = 12 cos(*ωt* + 260°) **(9.2.3)**
-
-Comparing Eqs. (9.2.1) and (9.2.3) shows clearly that *v*2 leads *v*1 by 30°.
-
-■ **METHOD 2** Alternatively, we may express *v*1 in sine form:
-
-*v*1 = −10 cos(*ωt* + 50°) = 10 sin(*ωt* + 50° − 90°) = 10 sin(*ωt* − 40°) = 10 sin(*ωt* − 10° − 30°)
-
-Practice Problem 9.1
-
-<span id="page-396-0"></span>But *v*2 = 12 sin(*ωt* − 10°). Comparing the tw o shows that *v*1 lags *v*<sup>2</sup> by 30°. This is the same as saying that *v*2 leads *v*1 by 30°.
-
-■ **METHOD 3** We may regard *v*1 as simply −10 cos*ωt* with a phase shift of +50°. Hence, *v*1 is as shown in Fig. 9.5. Similarly, *v*2 is 12 sin*ωt* with a phase shift of −10°, as shown in Fig. 9.5. It is easy to see from Fig. 9.5 that *v*2 leads *v*1 by 30°, that is, 90° − 50° − 10°.
-
-Find the phase angle between
-
-*i*1 = −4 sin(377*t* + 55°) and *i*2 = 5 cos(377*t* − 65°)
-
-Does *i*1 lead or lag *i*2?
-
-**Answer:** 210°, *i*1 leads *i*2.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/105_9.3 Phasors.md

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-# **9.3** Phasors
-
-Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions.
-
-A phasor is a complex number that represents the amplitude and phase of a sinusoid.
-
-Phasors provide a simple means of analyzing linear circuits e xcited by sinusoidal sources; solutions of such circuits would be intractable otherwise. The notion of solving ac circuits using phasors was first introduced by Charles Steinmetz in 1893. Before we completely define phasors and apply them to circuit analysis, we need to be thoroughly f amiliar with complex numbers.
-
-A complex number *z* can be written in rectangular form as
-
-$$
-z = x + jy \tag{9.14a}
-$$
-
-where *j* = √ \_\_\_ −1 ; *x* is the real part of *z*; *y* is the imaginary part of *z*. In this context, the variables *x* and *y* do not represent a location as in tw odimensional vector analysis but rather the real and imaginary parts of *z* in the complex plane. Nevertheless, we note that there are some resemblances between manipulating complex numbers and manipulating twodimensional vectors.
-
-The complex number *z* can also be written in polar or e xponential form as
-
-$$
-z = r/\phi = re^{j\phi} \tag{9.14b}
-$$
-
-Charles Proteus Steinmetz (1865–1923) was a German-Austrian mathematician and electrical engineer.
-
-Appendix B presents a short tutorial on complex numbers.
-
-# Historical
-
-**Charles Proteus Steinmetz** (1865–1923), a German-Austrian mathematician and engineer, introduced the phasor method (covered in this chapter) in ac circuit analysis. He is also noted for his w ork on the theory of hysteresis.
-
-Steinmetz was born in Breslau, Germany, and lost his mother at the age of one. As a youth, he was forced to leave Germany because of his political activities just as he w as about to complete his doctoral dis sertation in mathematics at the Uni versity of Breslau. He migrated to Switzerland and later to the United States, where he w as employed by General Electric in 1893. That same year, he published a paper in which complex numbers were used to analyze ac circuits for the first time. This led to one of his man y textbooks, *Theory and Calculation of ac Phenomena,* published by McGra w-Hill in 1897. In 1901, he became the president of the American Institute of Electrical Engineers, which later became the IEEE.
-
-where *r* is the magnitude of *z*, and *ϕ* is the phase of *z*. We notice that *z* can be represented in three ways:
-
-| z = x + jy | Rectangular form |        |
-|------------|------------------|--------|
-| z = r⧸ϕ    | Polar form       | (9.15) |
-| z = rejϕ   | Exponential form |        |
-
-The relationship between the rectangular form and the polar form is shown in Fig. 9.6, where the *x* axis represents the real part and the *y* axis represents the imaginary part of a comple x number. Given *x* and *y*, we can get *r* and *ϕ* as
-
-$$
-r = \sqrt{x^2 + y^2}
-$$
-,  $\phi = \tan^{-1} \frac{y}{x}$  (9.16a)
-
-On the other hand, if we know *r* and *ϕ*, we can obtain *x* and *y* as
-
-$$
-x = r \cos \phi, \qquad y = r \sin \phi \tag{9.16b}
-$$
-
-Thus, *z* may be written as
-
-$$
-z = x + jy = r/\underline{\phi} = r(\cos\phi + j\sin\phi)
-$$
- (9.17)
-
-Addition and subtraction of complex numbers are better performed in rectangular form; multiplication and division are better done in polar form. Given the complex numbers
-
-$$
-z = x + jy = r/\underline{\phi}, \qquad z_1 = x_1 + jy_1 = r_1/\underline{\phi}_1
-$$
-$$
-z_2 = x_2 + jy_2 = r_2/\underline{\phi}_2
-$$
-
-the following operations are important. **Addition:**
-
-$$
-z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2)
-$$
-\n(9.18a)
-
-# **Subtraction:**
-
-$$
-z_1 - z_2 = (x_1 - x_2) + j(y_1 - y_2)
-$$
-\n(9.18b)
-
-**Multiplication:**
-
-$$
-z_1 z_2 = r_1 r_2 / \phi_1 + \phi_2 \tag{9.18c}
-$$
-
-**Division:**
-
-$$
-\frac{z_1}{z_2} = \frac{r_1}{r_2} / \phi_1 - \phi_2 \tag{9.18d}
-$$
-
-**Reciprocal:**
-
-$$
-\frac{1}{z} = \frac{1}{r} \angle -\phi \tag{9.18e}
-$$
-
-**Square Root:**
-
-$$
-\sqrt{z} = \sqrt{r} \sqrt{\phi/2}
-$$
- (9.18f)
-
-# **Complex Conjugate:**
-
-$$
-z^* = x - jy = r'_\text{p} = re^{-j\phi} \tag{9.18g}
-$$
-
-Note that from Eq. (9.18e),
-
-$$
-\frac{1}{j} = -j \tag{9.18h}
-$$
-
-These are the basic properties of complex numbers we need. Other properties of complex numbers can be found in Appendix B.
-
-The idea of phasor representation is based on Euler' s identity. In general,
-
-$$
-e^{\pm i\phi} = \cos\phi \pm j\sin\phi \qquad (9.19)
-$$
-
-which shows that we may re gard cos *ϕ* and sin *ϕ* as the real and imagi nary parts of *e <sup>j</sup><sup>ϕ</sup>* ; we may write
-
-$$
-\cos \phi = \text{Re}(e^{j\phi}) \tag{9.20a}
-$$
-
-$$
-\sin \phi = \text{Im}(e^{j\phi})\tag{9.20b}
-$$
-
-where Re and Im stand for the *real part of* and the *imaginary part of*. Given a sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), we use Eq. (9.20a) to e xpress *v*(*t*) as
-
-$$
-v(t) = V_m \cos(\omega t + \phi) = \text{Re}(V_m e^{j(\omega t + \phi)})
-$$
- (9.21)
-
-or
-
-$$
-v(t) = \text{Re}(V_m e^{j\phi} e^{j\omega t})
-$$
-\n(9.22)
-
-Thus,
-
-$$
-v(t) = \text{Re}(\mathbf{V}e^{j\omega t})
-$$
- (9.23)
-
-where
-
-$$
-\mathbf{V} = V_m e^{j\phi} = V_m / \phi \tag{9.24}
-$$
-
-**V** is thus the *phasor representation* of the sinusoid *v*(*t*), as we saidearlier. In other words, a phasor is a complex representation of the magnitude and phase of a sinusoid. Either Eq. (9.20a) or Eq. (9.20b) can be used to develop the phasor, but the standard convention is to use Eq. (9.20a).
-
-One way of looking at Eqs. (9.23) and (9.24) is to consider the plot of the *sinor* **V***e jωt* = *Vme j*(*ωt*+*ϕ*) on the comple x plane. As time increases, the sinor rotates on a circle of radius *Vm* at an angular velocity *ω* in the counterclockwise direction, as shown in Fig. 9.7(a). We may regard *v*(*t*) as the projection of the sinor **V***e jωt* on the real axis, as shown in Fig. 9.7(b). The value of the sinor at time *t* = 0 is the phasor **V** of the sinusoid *v*(*t*). The sinor may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term *e jωt* is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency *ω* of the phasor; otherwise we can make serious mistakes.
-
- A phasor may be regarded as a mathematical equivalent of a sinusoid with the time dependence dropped.
-
- If we use sine for the phasor instead of cosine, then <sup>v</sup> (t) = <sup>V</sup> <sup>m</sup> sin(*ω*t + *ϕ*) = Im(Vmej(*ω*t+*ϕ*) ) and the corresponding phasor is the same as that in Eq. (9.24).
-
-Equation (9.23) states that to obtain the sinusoid corresponding to a given phasor **V**, multiply the phasor by the time f actor *ejω<sup>t</sup>* and tak e the real part. As a complex quantity, a phasor may be expressed in rectangular form, polar form, or exponential form. Because a phasor has magnitude and phase ("direction"), it behaves as a vector and is printed in boldface. For example, phasors **V** = *Vm*⧸*ϕ* and **I** = *Im*⧸−*θ* are graphically represented in Fig. 9.8. Such a graphical representation of phasors is known as a *phasor diagram*.
-
-Equations (9.21) through (9.23) re veal that to get the phasor cor responding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we tak e out the time f actor *e<sup>j</sup>ω<sup>t</sup>* , and whate ver is left is the pha sor corresponding to the sinusoid. By suppressing the time f actor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows:
-
-$$
-v(t) = V_m \cos(\omega t + \phi) \qquad \Leftrightarrow \qquad \mathbf{V} = V_m / \underline{\phi}
-$$
- (9.25)  
-\n(Time-domain representation) (Phasor-domain representation)
-
-We use lightface italic letters such as <sup>z</sup> to represent complex numbers but boldface letters such as **V** to represent phasors, because phasors are vectorlike quantities.
-
-A phasor diagram showing **V** = *Vm*⧸*ϕ* and **I** = *Im*⧸−*θ*.
-
-Given a sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), we obtain the corresponding phasor as **V** = *Vm* <sup>⧸</sup>*ϕ*. Equation (9.25) is also demonstrated in Table 9.1, where the sine function is considered in addition to the cosine function. From Eq. (9.25), we see that to get the phasor representation of a sinu soid, we e xpress it in cosine form and tak e the magnitude and phase. Given a phasor, we obtain the time domain representation as the cosine function with the same magnitude as the phasor and the ar gument as *ωt* plus the phase of the phasor. The idea of expressing information in alternate domains is fundamental to all areas of engineering.
-
-## **TABLE 9.1**
-
-Sinusoid-phasor transformation.
-
-| Phasor domain representation |
-|------------------------------|
-| Vm⧸ϕ                         |
-| Vm⧸ϕ − 90°                   |
-| Im⧸θ                         |
-| Im⧸θ − 90°                   |
-|                              |
-
-Note that in Eq. (9.25) the frequenc y (or time) f actor *ejω<sup>t</sup>* is sup pressed, and the frequency is not explicitly shown in the phasor domain representation because *ω* is constant. However, the response depends on *ω*. For this reason, the phasor domain is also known as the *frequency domain.*
-
-From Eqs. (9.23) and (9.24), *v*(*t*) = Re(**V***e jωt* ) = *Vm* cos(*ωt* + *ϕ*), so that
-
-$$
-\frac{dv}{dt} = -\omega V_m \sin(\omega t + \phi) = \omega V_m \cos(\omega t + \phi + 90^\circ)
-$$
-  
-= Re( $\omega V_m e^{j\omega t} e^{j\phi} e^{j90^\circ}$ ) = Re( $j\omega V e^{j\omega t}$ ) (9.26)
-
-This shows that the deri vative *v*(*t*) is transformed to the phasor domain as *jω***V**
-
-$$
-\frac{dv}{dt} \qquad \Leftrightarrow \qquad j\omega V \qquad (9.27)
-$$
-\n(Time domain)
-
-\n(Phasor domain)
-
-Similarly, the inte gral of *v*(*t*) is transformed to the phasor domain as **V**∕*jω*
-
-$$
-\int v \, dt \qquad \Leftrightarrow \qquad \frac{V}{j\omega} \qquad (9.28)
-$$
-\n(Time domain)
-
-\n(Phasor domain)
-
-Equation (9.27) allows the replacement of a derivative with respect to time with multiplication of *jω* in the phasor domain, whereas Eq. (9.28) allows the replacement of an inte gral with respect to time with di vision by *jω* in the phasor domain. Equations (9.27) and (9.28) are useful in finding the steady-state solution, which does not require knowing the initial values of the variable involved. This is one of the important applications of phasors.
-
-Besides time differentiation and integration, another important use of phasors is found in summing sinusoids of the same frequency. This is best illustrated with an example, and Example 9.6 provides one.
-
-The differences between *v*(*t*) and **V** should be emphasized:
-
-- 1. *v*(*t*) is the *instantaneous or time domain* representation, while **V** is the *frequency or phasor domain* representation.
-- 2. *v*(*t*) is time dependent, while **V** is not. (This f act is often for gotten by students.)
-- 3. *v*(*t*) is al ways real with no comple x term, while **V** is generally complex.
-
-Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency.
-
- Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by j*ω*.
-
- Integrating a sinusoid is equivalent to dividing its corresponding phasor by j*ω*.
-
- Adding sinusoids of the same frequency is equivalent to adding their corresponding phasors.
-
-Evaluate these complex numbers: Example 9.3
-
-(a) 
-$$
-(40/50^{\circ} + 20/-30^{\circ})^{1/2}
-$$
-  
-\n(b) 
-$$
-\frac{10/-30^{\circ} + (3-j4)}{(2+j4)(3-j5)^{*}}
-$$
-
-# **Solution:**
-
-(a) Using polar to rectangular transformation,
-
-$$
-40/50^{\circ} = 40(\cos 50^{\circ} + j \sin 50^{\circ}) = 25.71 + j30.64
-$$
-
-$$
-20\angle -30^{\circ} = 20[\cos(-30^{\circ}) + j\sin(-30^{\circ})] = 17.32 - j10
-$$
-
-Adding them up gives
-
-$$
-40/50^{\circ} + 20/-30^{\circ} = 43.03 + j20.64 = 47.72/25.63^{\circ}
-$$
-
-Taking the square root of this,
-
-$$
-(40/50^{\circ} + 20/-30^{\circ})^{1/2} = 6.91/12.81^{\circ}
-$$
-
-(b) Using polar-rectangular transformation, addition, multiplication, and division,
-
-(b) Using polar-rectangular transformation, a  
-division,  
-$$
-\frac{10(-30^\circ + (3 - j4))}{(2 + j4)(3 - j5)^*} = \frac{8.66 - j5 + (3 - j4)}{(2 + j4)(3 + j5)}
-$$
-$$
-= \frac{11.66 - j9}{-14 + j22} = \frac{14.73(-37.66^\circ)}{26.08(122.47^\circ)}
-$$
-$$
-= 0.565(-160.13^\circ)
-$$
-
-| Practice Problem 9.3 | Evaluate the following complex numbers:                                |
-|----------------------|------------------------------------------------------------------------|
-|                      | (a) [(5 + j2)(−1 + j4) − 5⧸ 60°]*                                      |
-|                      | 10 + j5 + 3⧸ 40°<br>______________<br>(b)<br>+ 10⧸ 30° + j5<br>−3 + j4 |
-|                      | Answer: (a) −15.5 −<br>j13.67, (b) 8.293 + j7.2.                       |
-|                      |                                                                        |
-
-Example 9.4 Transform these sinusoids to phasors:
-
-(a) *i* = 6 cos(50*t* − 40°) A (b) *v* = −4 sin(30*t* + 50°) V
-
-# **Solution:**
-
-(a) *i* = 6 cos(50*t* − 40°) has the phasor
-
-$$
-I = 6/–40^{\circ} A
-$$
-
-(b) Since 
-$$
--\sin A = \cos(A + 90^\circ)
-$$
-,  
- $v = -4 \sin(30t + 50^\circ) = 4 \cos(30t + 50^\circ + 90^\circ)$ 
-
-$$
-= 4\cos(30t + 140^\circ) \text{ V}
-$$
-
-The phasor form of *v* is
-
-$$
-V = 4/140^{\circ} V
-$$
-
-Practice Problem 9.4 Express these sinusoids as phasors:
-
-(a) *v* = −14 sin(5*t* − 22°) V (b) *i* = −8 cos(16*t* + 15°) A
-
-**Answer:** (a) **V** = 14⧸ 68° V, (b) **I** = 8⧸−165° A.
-
-Find the sinusoids represented by these phasors: Example 9.5
-
-(a) 
-$$
-\mathbf{I} = -3 + j4 \text{ A}
-$$
-  
-\n(b)  $\mathbf{V} = j8e^{-j20^{\circ}} \text{ V}$ 
-
-# **Solution:**
-
-(a) **I** = −3 + *j*4 = 5⧸ 126.87°. Transforming this to the time domain gives
-
-$$
-i(t) = 5 \cos(\omega t + 126.87^{\circ})
-$$
- A
-
-(b) Because *j* = 1⧸ 90°,
-
-$$
-\mathbf{V} = j8 \underline{/ -20^{\circ}} = (1 \underline{/ 90^{\circ}})(8 \underline{/ -20^{\circ}})
-$$
-$$
-= 8 \underline{/ 90^{\circ} - 20^{\circ}} = 8 \underline{/ 70^{\circ}} \text{ V}
-$$
-
-Converting this to the time domain gives
-
-$$
-v(t) = 8\cos(\omega t + 70^\circ)\text{V}
-$$
-
-Find the sinusoids corresponding to these phasors: Practice Problem 9.5
-
-(a) **V** = −25⧸ 40° V (b) **I** = *j*(12 − *j*5) A **Answer:** (a) *v*(*t*) = 25 cos(*ωt* − 140°) V or 25 cos(*ωt* + 220°) V, (b) *i*(*t*) = 13 cos(*ωt* + 67.38°) A.
-
-Given *i*1(*t*) = 4 cos(*ωt* + 30°) A and *i*2(*t*) = 5 sin(*ω<sup>t</sup>* <sup>−</sup> 20°) A, find Example 9.6 their sum.
-
-# **Solution:**
-
-Here is an important use of phasors—for summing sinusoids of the same frequency. Current *i*1(*t*) is in the standard form. Its phasor is
-
-$$
-\mathbf{I}_1 = 4/30^\circ
-$$
-
-We need to express *i*2(*t*) in cosine form. The rule for converting sine to cosine is to subtract 90°. Hence,
-
-$$
-i_2 = 5\cos(\omega t - 20^\circ - 90^\circ) = 5\cos(\omega t - 110^\circ)
-$$
-
-and its phasor is
-
-**I**2 = 5⧸−110°
-
-If we let *i* = *i*1 + *i*2, then
-
-$$
-\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 = 4/30^\circ + 5/-110^\circ
-$$
-  
-= 3.464 + j2 - 1.71 - j4.698 = 1.754 - j2.698  
-= 3.218/-56.97° A
-
-Transforming this to the time domain, we get
-
-$$
-i(t) = 3.218 \cos(\omega t - 56.97^{\circ}) \text{ A}
-$$
-
-Of course, we can find *i*1 + *i*2 using Eq. (9.9), but that is the hard way.
-
-Practice Problem 9.6 If *v*1 = −10 sin(*ωt* − 30°) V and *v*2 = 20 cos(*ωt* + 45°) V, find *v*<sup>=</sup> *v*1 + *v*2.
-
-**Answer:** *v*(*t*) = 29.77 cos(*ωt* + 49.98°) V.
-
-Example 9.7 Using the phasor approach, determine the current *i*(*t*) in a circuit described by the integrodifferential equation
-
-$$
-4i + 8\int i\,dt - 3\frac{di}{dt} = 50\cos(2t + 75^\circ)
-$$
-
-# **Solution:**
-
-We transform each term in the equation from time domain to phasor domain. Keeping Eqs. (9.27) and (9.28) in mind, we obtain the phasor form of the given equation as
-
-$$
-4\mathbf{I} + \frac{8\mathbf{I}}{j\omega} - 3j\omega\mathbf{I} = 50/75^{\circ}
-$$
-
-But *ω* = 2, so
-
-$$
-I(4 - j4 - j6) = 50 / 75^{\circ}
-$$
-
-$$
-I = \frac{50/75^{\circ}}{4 - j10} = \frac{50/75^{\circ}}{10.77/-68.2^{\circ}} = 4.642/143.2^{\circ}
-$$
- A
-
-Converting this to the time domain,
-
-$$
-i(t) = 4.642 \cos(2t + 143.2^{\circ}) \text{ A}
-$$
-
-Keep in mind that this is only the steady-state solution, and it does not require knowing the initial values.
-
-Practice Problem 9.7 Find the v oltage *v*(*t*) in a circuit described by the inte grodifferential equation
-
-$$
-2\frac{dv}{dt} + 5v + 10 \int v \, dt = 50 \cos(5t - 30^{\circ})
-$$
-
-using the phasor approach.
-
-**Answer:** 
-$$
-v(t) = 5.3 \cos(5t - 88^\circ)
-$$
- V.
-
-# <span id="page-405-0"></span>**9.4** Phasor Relationships for Circuit Elements
-
-Now that we know how to represent a voltage or current in the phasor or frequency domain, one may le gitimately ask ho w we apply this to cir cuits involving the passive elements *R*, *L*, and *C*. What we need to do is to transform the voltage-current relationship from the time domain to the frequency domain for each element. Again, we will assume the passi ve sign convention.
-
-We be gin with the resistor . If the current through a resistor *R* is *i* = *Im* cos(*ωt* + *ϕ*), the voltage across it is given by Ohm's law as
-
-$$
-v = iR = RIm \cos(\omega t + \phi)
-$$
- (9.29)
-
-The phasor form of this voltage is
-
-$$
-\mathbf{V} = R I_m / \underline{\phi} \tag{9.30}
-$$
-
-But the phasor representation of the current is **I** = I*m*<sup>⧸</sup>*ϕ*. Hence,
-
-$$
-V = RI \tag{9.31}
-$$
-
-showing that the v oltage-current relation for the resistor in the phasor domain continues to be Ohm's law, as in the time domain. Figure 9.9 illustrates the voltage-current relations of a resistor. We should note from Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor diagram in Fig. 9.10.
-
-For the inductor *L*, assume the current through it is *i* = *Im* cos(*ωt* + *ϕ*). The voltage across the inductor is
-
-$$
-v = L\frac{di}{dt} = -\omega L I_m \sin(\omega t + \phi)
-$$
-\n(9.32)
-
-Recall from Eq. (9.10) that −sin *A* = cos(*A* + 90°). We can write the voltage as
-
-$$
-v = \omega L I_m \cos(\omega t + \phi + 90^\circ) \tag{9.33}
-$$
-
-which transforms to the phasor
-
-$$
-\mathbf{V} = \omega L I_m e^{j(\phi + 90^\circ)} = \omega L I_m e^{j\phi} e^{j90^\circ} = \omega L I_m / \phi + 90^\circ \tag{9.34}
-$$
-
-But *Im*⧸*ϕ* = **I**, and from Eq. (9.19), *ej*90° = *j*. Thus,
-
-$$
-V = j\omega L I \tag{9.35}
-$$
-
-showing that the v oltage has a magnitude of *ωLIm* and a phase of *ϕ* + 90°. The voltage and current are 90° out of phase. Specifically, the current lags the v oltage by 90°. Figure 9.11 sho ws the v oltage-current relations for the inductor. Figure 9.12 shows the phasor diagram.
-
-For the capacitor *C*, assume the voltage across it is *v* = *Vm* cos(*ωt* + *ϕ*). The current through the capacitor is
-
-$$
-i = C \frac{dv}{dt}
-$$
- (9.36)

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/106_9.4 Phasor Relationships for Circuit Elements.md

@@ -1,117 +0,0 @@
-By following the same steps as we took for the inductor or by applying Eq. (9.27) on Eq. (9.36), we obtain
-
-$$
-\mathbf{I} = j\omega C \mathbf{V} \qquad \Rightarrow \qquad \mathbf{V} = \frac{\mathbf{I}}{j\omega C} \tag{9.37}
-$$
-
-# **Figure 9.9**
-
-Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.
-
-Phasor diagram for the resistor.
-
-# **Figure 9.11**
-
-Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.
-
-**Figure 9.12** Phasor diagram for the inductor; **I** lags **V**.
-
-Although it is equally correct to say that the inductor voltage leads the current by 90°, convention gives the current phase relative to the voltage.
-
-## **384** Chapter 9 Sinusoids and Phasors
-
-showing that the current and voltage are 90° out of phase. To be specific, the current leads the voltage by 90°. Figure 9.13 shows the voltage-current relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2 summarizes the time domain and phasor domain representations of the circuit elements.
-
-# **TABLE 9.2**
-
-Summary of voltage-current relationships.
-
-| Element | Time domain        | Frequency domain  |
-|---------|--------------------|-------------------|
-| R       | v = Ri             | V = RI            |
-| L       | v = L __di<br>dt   | V = jωLI          |
-| C       | i = C ___ dv<br>dt | V = ____ I<br>jωC |
-
-Example 9.8 The voltage *v* = 12cos(60*t* + 45°) is applied to a 0.1-H inductor. Find the steady-state current through the inductor.
-
-# **Solution:**
-
-For the inductor , **V** = *jωL***I**, where *ω* = 60 rad/s and **V** = 12 ⧸45° V. Hence,
-
-$$
-I = \frac{V}{j\omega L} = \frac{12/45^{\circ}}{j60 \times 0.1} = \frac{12/45^{\circ}}{6/90^{\circ}} = 2/45^{\circ}
-$$
- A
-
-Converting this to the time domain,
-
-$$
-i(t) = 2\cos(60t - 45^\circ) \,\mathrm{A}
-$$
-
-Practice Problem 9.8 If voltage *v* = 25 sin(100*t* − 15°) *V* is applied to a 50 *μ*F capacitor, calculate the current through the capacitor.
-
-# **Answer:** 125 sin(100*t* + 75°) mA.
-
-$$
-0.125 \sim
-$$
-
-# <span id="page-407-0"></span>**9.5** Impedance and Admittance
-
-In the preceding section, we obtained the v oltage-current relations for the three passive elements as
-
-$$
-V = RI, \t V = j\omega LI, \t V = \frac{I}{j\omega C}
-$$
-(9.38)
-
-These equations may be written in terms of the ratio of the phasor v oltage to the phasor current as
-
-$$
-\frac{V}{I} = R, \qquad \frac{V}{I} = j\omega L, \qquad \frac{V}{I} = \frac{1}{j\omega C}
-$$
- (9.39)
-
-From these three e xpressions, we obtain Ohm's law in phasor form for any type of element as
-
-$$
-Z = \frac{V}{I} \qquad \text{or} \qquad V = ZI \tag{9.40}
-$$
-
-where **Z** is a frequenc y-dependent quantity known as *impedance*, measured in ohms.
-
-The impedance **Z** of a circuit is the ratio of the phasor voltage **V** to the phasor current **I**, measured in ohms (Ω).
-
-The impedance represents the opposition that the circuit e xhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity.
-
-The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39). Table 9.3 summarizes their impedances. From the table we notice that **Z***L* = *jωL* and **Z***C* = −*j*∕*ωC*. Consider two extreme cases of angular frequenc y. When *ω* = 0 (i.e., for dc sources), **Z***L* = 0 and **Z***C* → ∞, confirming what we already know—that the inductor acts like a short circuit, while the capacitor acts like an open circuit. When *ω* → ∞ (i.e., for high frequencies), **Z***L* → ∞ and **Z***C* = 0, indicating that the inductor is an open circuit to high frequencies, while the capacitor is a short circuit. Figure 9.15 illustrates this.
-
-As a complex quantity, the impedence may be e xpressed in rectangular form as
-
-$$
-Z = R \pm jX \tag{9.41}
-$$
-
-where *R* = Re **Z** is the *resistance* and *X* = Im **Z** is the *reactance*. The reactance, *X,* is just a magnitude, a positi ve value, but when used as a vector, a *j* is associated with inductance and a −*j* is associated with capacitance. Thus, impedance **Z** = *R* + *jX* is said to be *inductive* o r lagging since current lags v oltage, while impedance **Z** = *R* − *jX* i s capacitive or leading because current leads v oltage. The impedance, resistance, and reactance are all measured in ohms. The impedance may also be expressed in polar form as
-
-$$
-\mathbf{Z} = |\mathbf{Z}| \; / \theta \tag{9.42}
-$$
-
-**TABLE 9.3**
-
-Impedances and admittances of passive elements.
-
-| Impedance          | Admittance                                                 |
-|--------------------|------------------------------------------------------------|
-| Z = R              | Y = __1<br>R                                               |
-| Z = jωL            | Y = ____ 1<br>jωL                                          |
-| Z = ____ 1<br>jω C | Y = jωC                                                    |
-|                    | Short circuit at dc<br>Open circuit at<br>high frequencies |
-| (a)                |                                                            |
-|                    | Open circuit at dc                                         |
-| (b)                | Short circuit at<br>high frequencies                       |
-|                    |                                                            |
-
-# **Figure 9.15**

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/107_9.5 Impedance and Admittance.md

@@ -1,126 +0,0 @@
-Equivalent circuits at dc and high frequencies: (a) inductor, (b) capacitor. Comparing Eqs. (9.41) and (9.42), we infer that
-
-$$
-Z = R \pm jX = |Z|/\theta \tag{9.43}
-$$
-
-where
-
-$$
-|Z| = \sqrt{R^2 + X^2}, \qquad \theta = \tan^{-1} \frac{\pm X}{R}
-$$
- (9.44)
-
-and
-
-$$
-R = |\mathbf{Z}| \cos \theta, \qquad X = |\mathbf{Z}| \sin \theta \tag{9.45}
-$$
-
-It is sometimes con venient to w ork with the reciprocal of imped ance, known as *admittance*.
-
-The admittance **Y** is the reciprocal of impedance, measured in siemens (S).
-
-The admittance **Y** of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or
-
-$$
-Y = \frac{1}{Z} = \frac{I}{V}
-$$
- (9.46)
-
-The admittances of resistors, inductors, and capacitors can be obtained from Eq. (9.39). They are also summarized in Table 9.3.
-
-As a complex quantity, we may write **Y** as
-
-$$
-Y = G + jB \tag{9.47}
-$$
-
-where *G* = Re **Y** is called the *conductance* and *B* = Im **Y** is called the *susceptance*. Admittance, conductance, and susceptance are all expressed in the unit of siemens (or mhos). From Eqs. (9.41) and (9.47),
-
-$$
-G + jB = \frac{1}{R + jX} \tag{9.48}
-$$
-
-By rationalization,
-
-$$
-G + jB = \frac{1}{R + jX} \cdot \frac{R - jX}{R - jX} = \frac{R - jX}{R^2 + X^2}
-$$
-(9.49)
-
-Equating the real and imaginary parts gives
-
-$$
-G = \frac{R}{R^2 + X^2}, \qquad B = -\frac{X}{R^2 + X^2}
-$$
-(9.50)
-
-showing that *G* ≠ 1∕*R* as it is in resisti ve circuits. Of course, if *X* = 0, then *G* = 1∕*R*.
-
-<span id="page-409-0"></span>Find *v*(*t*) and *i*(*t*) in the circuit shown in Fig. 9.16. Example 9.9
-
-# **Solution:**
-
-From the voltage source 10 cos 4*t*, *ω* = 4,
-
-$$
-V_s = 10/0^{\circ} V
-$$
-
-The impedance is
-
-$$
-\mathbf{Z} = 5 + \frac{1}{j\omega C} = 5 + \frac{1}{j4 \times 0.1} = 5 - j2.5 \ \Omega
-$$
-
-Hence the current
-
-$$
-\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{10/0^{\circ}}{5 - j2.5} = \frac{10(5 + j2.5)}{5^2 + 2.5^2}
-$$
-(9.9.1)  
-= 1.6 + j0.8 = 1.789/26.57° A
-
-The voltage across the capacitor is
-
-The voltage across the capacitor is  
-\n
-$$
-\mathbf{V} = IZ_C = \frac{I}{j\omega C} = \frac{1.789/26.57^\circ}{j4 \times 0.1} = \frac{1.789/26.57^\circ}{0.4/90^\circ} = 4.47/-63.43^\circ \text{ V}
-$$
-\n(9.9.2)
-
-Converting **I** and **V** in Eqs. (9.9.1) and (9.9.2) to the time domain, we get
-
-$$
-i(t) = 1.789 \cos(4t + 26.57^{\circ}) \text{ A}
-$$
-  
-$$
-v(t) = 4.47 \cos(4t - 63.43^{\circ}) \text{ V}
-$$
-
-Notice that *i*(*t*) leads *v*(*t*) by 90° as expected.
-
-Refer to Fig. 9.17. Determine *v*(*t*) and *i*(*t*). Practice Problem 9.9
-
-**Answer:** 8.944 sin (10*t* + 93.43°) V, 4.472 sin(10*t* + 3.43°) A.
-
-**9.6** Kirchhoff's Laws in the Frequency Domain
-
-We cannot do circuit analysis in the frequenc y domain without Kirch hoff's current and v oltage laws. Therefore, we need to e xpress them in the frequency domain.
-
-For KVL, let *v*1,*v*2, … , *vn* be the voltages around a closed loop. Then
-
-$$
-v_1 + v_2 + \dots + v_n = 0 \tag{9.51}
-$$
-
-In the sinusoidal steady state, each v oltage may be written in cosine form, so that Eq. (9.51) becomes
-
-$$
-V_{m1}\cos(\omega t + \theta_1) + V_{m2}\cos(\omega t + \theta_2)
-$$
-  
-+ ... +  $V_{mn}\cos(\omega t + \theta_n) = 0$  (9.52)

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/108_9.6 Kirchhoff's Laws in the Frequency Domain.md

@@ -1,46 +0,0 @@
-# **Figure 9.17**
-
-For Practice Prob. 9.9.
-
-<span id="page-410-0"></span>This can be written as
-
-$$
-\text{Re}(V_{m1}e^{j\theta_1}e^{j\omega t}) + \text{Re}(V_{m2}e^{j\theta_2}e^{j\omega t}) + \dots + \text{Re}(V_{mn}e^{j\theta_n}e^{j\omega t}) = 0
-$$
-
-or
-
-$$
-\text{Re}[(V_{m1}e^{j\theta_1} + V_{m2}e^{j\theta_2} + \dots + V_{mn}e^{j\theta_n})e^{j\omega t}] = 0 \tag{9.53}
-$$
-
-If we let **V***k* = *Vmkej<sup>θ</sup>k*, then
-
-$$
-Re[(V_1 + V_2 + \dots + V_n)e^{j\omega t}] = 0
-$$
-\n(9.54)
-
-Because *ejω<sup>t</sup>* ≠ 0,
-
-$$
-V_1 + V_2 + \dots + V_n = 0 \tag{9.55}
-$$
-
-indicating that Kirchhoff's voltage law holds for phasors.
-
-By following a similar procedure, we can show that Kirchhoff's current law holds for phasors. If we let *i*1, *i*2, … , *in* be the current leaving or entering a closed surface in a network at time *t*, then
-
-$$
-i_1 + i_2 + \dots + i_n = 0 \tag{9.56}
-$$
-
-If **I**1, **I**2, … , **I***n* are the phasor forms of the sinusoids *i*1,*i*2, … ,*in*, then
-
-$$
-\mathbf{I}_1 + \mathbf{I}_2 + \dots + \mathbf{I}_n = 0 \tag{9.57}
-$$
-
-which is Kirchhoff's current law in the frequency domain.
-
-Once we have shown that both KVL and KCL hold in the frequency domain, it is easy to do man y things, such as impedance combination, nodal and mesh analyses, superposition, and source transformation.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/109_9.7 Impedance Combinations.md

@@ -1,259 +0,0 @@
-# **9.7** Impedance Combinations
-
-Consider the *N* series-connected impedances sho wn in Fig. 9.18. The same current **I** flows through the impedances. Applying KVL around the loop gives
-
-$$
-V = V_1 + V_2 + \dots + V_N = I(Z_1 + Z_2 + \dots + Z_N)
-$$
-(9.58)
-
-*N* impedances in series.
-
-The equivalent impedance at the input terminals is
-
-$$
-\mathbf{Z}_{\text{eq}} = \frac{\mathbf{V}}{\mathbf{I}} = \mathbf{Z}_1 + \mathbf{Z}_2 + \dots + \mathbf{Z}_N
-$$
-
-$$
-Z_{eq} = Z_1 + Z_2 + \dots + Z_N \tag{9.59}
-$$
-
-or
-
-showing that the total or equi valent impedance of series-connected impedances is the sum of the indi vidual impedances. This is similar to the series connection of resistances.
-
-If *N* = 2, as shown in Fig. 9.19, the current through the imped ances is
-
-$$
-\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}_1 + \mathbf{Z}_2} \tag{9.60}
-$$
-
-Because **V**1 = **Z**1**I** and **V**2 = **Z**2**I**, then
-
-$$
-V_1 = \frac{Z_1}{Z_1 + Z_2} V, \qquad V_2 = \frac{Z_2}{Z_1 + Z_2} V
-$$
- (9.61)
-
-which is the *voltage-division* relationship.
-
-In the same manner , we can obtain the equi valent impedance or admittance of the *N* parallel-connected impedances shown in Fig. 9.20. The voltage across each impedance is the same. Applying KCL at the top node,
-
-$$
-\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 + \dots + \mathbf{I}_N = \mathbf{V} \left( \frac{1}{\mathbf{Z}_1} + \frac{1}{\mathbf{Z}_2} + \dots + \frac{1}{\mathbf{Z}_N} \right) \tag{9.62}
-$$
-
-*N* impedances in parallel.
-
-The equivalent impedance is
-
-$$
-\frac{1}{Z_{\text{eq}}} = \frac{I}{V} = \frac{1}{Z_1} + \frac{1}{Z_2} + \dots + \frac{1}{Z_N}
-$$
-(9.63)
-
-and the equivalent admittance is
-
-$$
-\mathbf{Y}_{\text{eq}} = \mathbf{Y}_1 + \mathbf{Y}_2 + \dots + \mathbf{Y}_N \tag{9.64}
-$$
-
-This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances.
-
-When *N* = 2, as sho wn in Fig. 9.21, the equi valent impedance becomes
-
-$$
-Z_{\text{eq}} = \frac{1}{Y_{\text{eq}}} = \frac{1}{Y_1 + Y_2} = \frac{1}{1/Z_1 + 1/Z_2} = \frac{Z_1 Z_2}{Z_1 + Z_2}
-$$
-(9.65)
-
-**Figure 9.21** Current division.
-
-Voltage division.
-
-Also, since
-
-$$
-\mathbf{V} = \mathbf{I}\mathbf{Z}_{\text{eq}} = \mathbf{I}_1\mathbf{Z}_1 = \mathbf{I}_2\mathbf{Z}_2
-$$
-
-the currents in the impedances are
-
-$$
-\mathbf{I}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I}, \qquad \mathbf{I}_2 = \frac{\mathbf{Z}_1}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I}
-$$
- (9.66)
-
-which is the *current-division* principle.
-
-The delta-to-wye and wye-to-delta transformations that we applied to resisti ve circuits are also v alid for impedances. With reference to Fig. 9.22, the conversion formulas are as follows.
-
-**Figure 9.22** Superimposed *Y* and ∆ networks.
-
-*Y*-∆ *Conversion:*
-
-$$
-Z_a = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_1}
-$$
-  
-\n
-$$
-Z_b = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_2}
-$$
-  
-\n
-$$
-Z_c = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_3}
-$$
-  
-\n(9.67)
-
-∆-*Y Conversion:*
-
-$$
-\mathbf{Z}_{1} = \frac{\mathbf{Z}_{b}\mathbf{Z}_{c}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}
-$$
-\n
-$$
-\mathbf{Z}_{2} = \frac{\mathbf{Z}_{c}\mathbf{Z}_{a}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}
-$$
-\n
-$$
-\mathbf{Z}_{3} = \frac{\mathbf{Z}_{a}\mathbf{Z}_{b}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}
-$$
-\n(9.68)
-
-A delta or wye circuit is said to be balanced if it has equal impedances in all three branches.
-
-When a ∆-*Y* circuit is balanced, Eqs. (9.67) and (9.68) become
-
-$$
-\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (9.69)
-$$
-
-where **Z***Y* = **Z**1 = **Z**2 = **Z**3 and **Z**∆ = **Z***a* = **Z***b* = **Z***c*.
-
-As you see in this section, the principles of voltage division, current division, circuit reduction, impedance equi valence, and *Y*-∆ transformation all apply to ac circuits. Chapter 10 will sho w that other c ircuit techniques—such as superposition, nodal analysis, mesh analysis, source transformation, the Thevenin theorem, and the Norton theorem are all applied to ac circuits in a manner similar to their application in dc circuits.
-
-Find the input impedance of the circuit in Fig. 9.23. Assume that the Example 9.10 circuit operates at *ω* = 50 rad/s.
-
-# **Solution:**
-
-Let
-
-- **Z**1 = Impedance of the 2-mF capacitor
-- **Z**2 = Impedance of the 3-Ω resistor in series with the10-mF capacitor
-- **Z**3 = Impedance of the 0.2-H inductor in series with the 8-Ω resistor
-
-Then
-
-$$
-\mathbf{Z}_1 = \frac{1}{j\omega C} = \frac{1}{j50 \times 2 \times 10^{-3}} = -j10 \text{ }\Omega
-$$
-  
-$$
-\mathbf{Z}_2 = 3 + \frac{1}{j\omega C} = 3 + \frac{1}{j50 \times 10 \times 10^{-3}} = (3 - j2) \text{ }\Omega
-$$
-  
-$$
-\mathbf{Z}_3 = 8 + j\omega L = 8 + j50 \times 0.2 = (8 + j10) \text{ }\Omega
-$$
-
-The input impedance is
-
-at impedance is  
-\n
-$$
-\mathbf{Z}_{in} = \mathbf{Z}_1 + \mathbf{Z}_2 \parallel \mathbf{Z}_3 = -j10 + \frac{(3 - j2)(8 + j10)}{11 + j8}
-$$
-\n
-$$
-= -j10 + \frac{(44 + j14)(11 - j8)}{11^2 + 8^2} = -j10 + 3.22 - j1.07 \Omega
-$$
-
-Thus,
-
-$$
-\mathbf{Z}_{\text{in}} = 3.22 - j11.07 \ \Omega
-$$
-
-For Example 9.10.
-
-**Figure 9.25**
-
-**Figure 9.26**
-
-circuit in Fig. 9.25.
-
-For Example 9.11.
-
-The frequency domain equivalent of the
-
-*vs* = 20 cos(4*t* − 15°) ⇒ **V***s* = 20⧸−15° V, *ω* = 4 10 mF <sup>⇒</sup> \_\_\_\_ <sup>1</sup> *<sup>j</sup>ωC* = \_\_\_\_\_\_\_\_\_\_\_\_ 1 *j*4 × 10 × 10<sup>−</sup><sup>3</sup> = −*j*25 Ω 5 H ⇒ *jωL* = *j*4 × 5 = *j*20 Ω
-
- **Z**1 = Impedance of the 60-Ω resistor
-
- **Z**2 = Impedance of the parallel combination of the 10-mF capacitor and the 5-H inductor
-
-Then **Z**1 = 60 Ω and
-
-$$
-\mathbf{Z}_2 = -j25 \parallel j20 = \frac{-j25 \times j20}{-j25 + j20} = j100 \text{ }\Omega
-$$
-
-By the voltage-division principle,
-
-$$
-\mathbf{V}_o = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \frac{j100}{60 + j100} (20/15^\circ)
-$$
-  
-= (0.8575/30.96°)(20/15°) = 17.15/15.96° V
-
-We convert this to the time domain and obtain
-
-*vo*(*t*) = 17.15 cos(4*t* + 15.96°) V
-
-For Practice Prob. 9.11.
-
-# Find current **I** in the circuit of Fig. 9.28. Example 9.12
-
-# **Solution:**
-
-The delta netw ork connected to nodes *a*, *b*, and *c* can be con verted to the *Y* network of Fig. 9.29. We obtain the *Y* impedances as follows using Eq. (9.68):
-
-68):  
-\n
-$$
-\mathbf{Z}_{an} = \frac{j4(2-j4)}{j4+2-j4+8} = \frac{4(4+j2)}{10} = (1.6+j0.8) \ \Omega
-$$
-\n
-$$
-\mathbf{Z}_{bn} = \frac{j4(8)}{10} = j3.2 \ \Omega, \qquad \mathbf{Z}_{cn} = \frac{8(2-j4)}{10} = (1.6-j3.2) \ \Omega
-$$
-
-The total impedance at the source terminals is
-
-$$
-\mathbf{Z} = 12 + \mathbf{Z}_{an} + (\mathbf{Z}_{bn} - j3) \parallel (\mathbf{Z}_{cn} + j6 + 8)
-$$
-  
-= 12 + 1.6 + j0.8 + (j0.2) || (9.6 + j2.8)  
-= 13.6 + j0.8 +  $\frac{j0.2(9.6 + j2.8)}{9.6 + j3}$   
-= 13.6 + j1 = 13.64/4.204° Ω
-
-The desired current is
-
-$$
-I = \frac{V}{Z} = \frac{50/0^{\circ}}{13.64/4.204^{\circ}} = 3.666/-4.204^{\circ} A
-$$
-
-# <span id="page-416-0"></span>Practice Problem 9.12 Find **I** in the circuit of Fig. 9.30.
-
-**Figure 9.30** For Practice Prob. 9.12.
-
-# **Figure 9.31**
-
-Series *RC* shift circuits: (a) leading output, (b) lagging output.
-
-**Answer:** 12.728 ⧸ 63.8° A.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/110_9.8 Applications.md

@@ -1,325 +0,0 @@
-# **9.8** Applications
-
-In Chapters 7 and 8, we sa w certain uses of *RC*, *RL*, and *RLC* circuits in dc applications. These circuits also have ac applications; among them are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac bridge circuits, and transformers. This list of applications is ine xhaustive. We will consider some of them later. It will suffice here to observe two simple ones: *RC* phase-shifting circuits, and ac bridge circuits.
-
-# **9.8.1** Phase-Shifters
-
-A phase-shifting circuit is often emplo yed to correct an undesirable phase shift already present in a circuit or to produce special desired effects. An *RC* circuit is suitable for this purpose because its capacitor causes the circuit current to lead the applied v oltage. Two commonly used *RC* circuits are shown in Fig. 9.31. (*RL* circuits or any reactive circuits could also serve the same purpose.)
-
-In Fig. 9.31(a), the circuit current **I** leads the applied voltage **V***i* by some phase angle *θ*, where 0 < *θ*< 90°, depending on the values of *R* and *C*. If *XC* = −1∕*ωC*, then the total impedance is **Z** = *R* + *jXC*, and the phase shift is given by
-
-$$
-\theta = \tan^{-1} \frac{X_C}{R}
-$$
- (9.70)
-
-This shows that the amount of phase shift depends on the v alues of *R*, *C*, and the operating frequenc y. Since the output v oltage **V***o* across the resistor is in phase with the current, **V***o* leads (positive phase shift) **V***i* as shown in Fig. 9.32(a).
-
-In Fig. 9.31(b), the output is taken across the capacitor. The current **I** leads the input v oltage **V***i* by *θ*, but the output voltage *vo*(*t*) across the capacitor lags (negative phase shift) the input v oltage *vi*(*t*) as illustrated in Fig. 9.32(b).
-
-**Figure 9.32** Phase shift in *RC* circuits: (a) leading output, (b) lagging output.
-
-We should keep in mind that the simple *RC* circuits in Fig. 9.31 also act as voltage dividers. Therefore, as the phase shift *θ* approaches 90°, the output vo ltage **V***o* approaches zero. F or this reason, these simple *RC* circuits are used only when small amounts of phase shift are required. If it is desired to ha ve phase shifts greater than 60°, simple *RC* networks are cascaded, thereby providing a total phase shift equal to the sum of the indi vidual phase shifts. In practice, the phase shifts due to the stages are not equal, because the succeeding stages load down the earlier stages unless op amps are used to sepa rate the stages.
-
-# **Solution:**
-
-If we select circuit components of equal ohmic v alue, say *R* = ∣*XC*∣ = 20 Ω, at a particular frequenc y, according to Eq. (9.70), the phase shift is exactly 45°. By cascading two similar *RC* circuits in Fig. 9.31(a), we obtain the circuit in Fig. 9.33, providing a positive or leading phase shift of 90°, as we shall soon show. Using the series-parallel combination technique, **Z** in Fig. 9.33 is obtained as
-
-$$
-\mathbf{Z} = 20 \| (20 - j20) = \frac{20(20 - j20)}{40 - j20} = 12 - j4 \ \Omega
-$$
- (9.13.1)
-
-Using voltage division,
-
-$$
-\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} - j20} \mathbf{V}_i = \frac{12 - j4}{12 - j24} \mathbf{V}_i = \frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i \quad (9.13.2)
-$$
-
-and
-
-$$
-\mathbf{V}_o = \frac{20}{20 - j20} \mathbf{V}_1 = \frac{\sqrt{2}}{2} \underline{45^\circ} \mathbf{V}_1
-$$
- (9.13.3)
-
-Substituting Eq. (9.13.2) into Eq. (9.13.3) yields
-
-$$
-\mathbf{V}_o = \left(\frac{\sqrt{2}}{2} \angle 45^\circ\right) \left(\frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i\right) = \frac{1}{3} \angle 90^\circ \mathbf{V}_i
-$$
-
-Thus, the output leads the input by 90° b ut its magnitude is only about 33 percent of the input.
-
-Design an *RC* circuit to pro vide a 90° lagging phase shift of the out - Practice Problem 9.13 put voltage relative to the input v oltage. If an ac v oltage of 60 V rms is applied, what is the output voltage?
-
-**Answer:** Figure 9.34 shows a typical design; 20 V rms.
-
-# **Figure 9.33**
-
-An *RC* phase shift circuit with 90° leading phase shift; for Example 9.13.
-
-**Figure 9.35** For Example 9.14.
-
-Example 9.14 For the *RL* circuit shown in Fig. 9.35(a), calculate the amount of phase shift produced at 2 kHz.
-
-# **Solution:**
-
-At 2 kHz, we transform the 10- and 5-mH inductances to the corresponding impedances.
-
-$$
-10 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 10 \times 10^{-3}
-$$
-$$
-= 40\pi = 125.7 \text{ }\Omega
-$$
-$$
-5 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 5 \times 10^{-3}
-$$
-$$
-= 20\pi = 62.83 \text{ }\Omega
-$$
-
-Consider the circuit in Fig. 9.35(b). The impedance **Z** is the parallel combination of *j*125.7 Ω and 100 + *j*62.83 Ω. Hence,
-
-$$
-\mathbf{Z} = j125.7 \parallel (100 + j62.83)
-$$
-  
-\n
-$$
-\mathbf{Z} = j125.7 \parallel (100 + j62.83)
-$$
-  
-\n
-$$
-= \frac{j125.7(100 + j62.83)}{100 + j188.5} = 69.56 / 60.1^{\circ} \ \Omega
-$$
- (9.14.1)
-
-Using voltage division,
-
-$$
-\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} + 150} \mathbf{V}_i = \frac{69.56 / 60.1^{\circ}}{184.7 + j60.3} \mathbf{V}_i
-$$
-  
-= 0.3582 / 42.02°  $\mathbf{V}_i$  (9.14.2)
-
-and
-
-$$
-\mathbf{V}_o = \frac{j62.832}{100 + j62.832} \mathbf{V}_1 = 0.532 \, \text{/} 57.86^{\circ} \, \mathbf{V}_1 \tag{9.14.3}
-$$
-
-Combining Eqs. (9.14.2) and (9.14.3),
-
-**V***o* = (0.532 ⧸ 57.86°)(0.3582 ⧸ 42.02°)**V***i* = 0.1906 ⧸ 100° **V***<sup>i</sup>*
-
-showing that the output is about 19 percent of the input in magnitude but leading the input by 100°. If the circuit is terminated by a load, the load will affect the phase shift.
-
-Practice Problem 9.14 Refer to the *RL* circuit in Fig. 9.36. If 10 V is applied to the input, find the magnitude and the phase shift produced at 5 kHz. Specify whether the phase shift is leading or lagging.
-
-**Answer:** 1.7161 V, 120.39°, lagging.
-
-# **9.8.2** AC Bridges
-
-An ac bridge circuit is used in measuring the inductance *L* of an inductor or the capacitance *C* of a capacitor. It is similar in form to the Wheatstone bridge for measuring an unknown resistance (discussed in Section 4.10) and follows the same principle. To measure *L* and *C*, however, an ac
-
-source is needed as well as an ac meter instead of the galvanometer. The ac meter may be a sensitive ac ammeter or voltmeter.
-
-Consider the general ac bridge circuit displayed in Fig. 9.37. The bridge is *balanced* when no current flows through the meter. This means that **V**1 = **V**2. Applying the voltage division principle,
-
-$$
-\mathbf{V}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \mathbf{V}_2 = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \mathbf{V}_s
-$$
-(9.71)
-
-Thus,
-
-$$
-\frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \qquad \Rightarrow \qquad \mathbf{Z}_2 \mathbf{Z}_3 = \mathbf{Z}_1 \mathbf{Z}_x \tag{9.72}
-$$
-
-or
-
-$$
-Z_x = \frac{Z_3}{Z_1} Z_2 \tag{9.73}
-$$
-
-This is the balanced equation for the ac bridge and is similar to Eq. (4.30) for the resistance bridge except that the *R*'s are replaced by **Z**'s.
-
-Specific ac bridges for measuring *L* and *C* are shown in Fig. 9.38, where *Lx* and *Cx* are the unknown inductance and capacitance to be measured while *Ls* and *Cs* are a standard inductance and capacitance (the values of which are kno wn to great precision). In each case, tw o resistors, *R*1 and *R*2, are varied until the ac meter reads zero. Then the bridge is balanced. From Eq. (9.73), we obtain
-
-$$
-L_x = \frac{R_2}{R_1} L_s \tag{9.74}
-$$
-
-and
-
-$$
-C_x = \frac{R_1}{R_2} C_s \tag{9.75}
-$$
-
-Notice that the balancing of the ac bridges in Fig. 9.38 does not depend on the frequency *f* of the ac source, since *f* does not appear in the relationships in Eqs. (9.74) and (9.75).
-
-Specific ac bridges: (a) for measuring *L*, (b) for measuring *C*.
-
-**Figure 9.37**
-
-A general ac bridge.
-
-Example 9.15 The ac bridge circuit of Fig. 9.37 balances when **Z**1 is a 1-kΩ resistor, **Z**<sup>2</sup> is a 4.2-kΩ resistor, **Z**3 is a parallel combination of a 1.5-MΩ resistor and a 12-pF capacitor, and *f* = 2 kHz. Find: (a) the series com ponents that make up **Z***x*, and (b) the parallel components that make up **Z***x*.
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly stated.
-- 2. **Present.** We are to determine the unknown components subject to the fact that they balance the given quantities. Given that a parallel and series equivalent exists for this circuit, we need to find both.
-- 3. **Alternative.** Although there are alternative techniques that can be used to find the unknown values, a straightforward equality works best. Once we have answers, we can check them by using hand techniques such as nodal analysis or just using *PSpice*.
-- 4. **Attempt.** From Eq. (9.73),
-
-$$
-Z_{x} = \frac{Z_{3}}{Z_{1}} Z_{2}
-$$
- (9.15.1)
-
-where **Z***x* = *Rx* + *jXx*,
-
-$$
-\mathbf{Z}_1 = 1000 \,\Omega, \qquad \mathbf{Z}_2 = 4200 \,\Omega \tag{9.15.2}
-$$
-
-and
-
-$$
-\mathbf{Z}_3 = R_3 \parallel \frac{1}{j\omega C_3} = \frac{\frac{R_3}{j\omega C_3}}{R_3 + 1/j\omega C_3} = \frac{R_3}{1 + j\omega R_3 C_3}
-$$
-
-Since *R*3 = 1.5 MΩ and *C*3 = 12 pF,
-
-Since 
-$$
-R_3 = 1.5 \text{ M}\Omega
-$$
- and  $C_3 = 12 \text{ pF}$ ,  
-\n
-$$
-\mathbf{Z}_3 = \frac{1.5 \times 10^6}{1 + j2\pi \times 2 \times 10^3 \times 1.5 \times 10^6 \times 12 \times 10^{-12}} = \frac{1.5 \times 10^6}{1 + j0.2262}
-$$
-
-or
-
-$$
-Z_3 = 1.427 - j0.3228 M\Omega
-$$
- (9.15.3)
-
-(a) Assuming that **Z***x* is made up of series components, we substitute Eqs. (9.15.2) and (9.15.3) in Eq. (9.15.1) and obtain
-
-$$
-R_x + jX_x = \frac{4200}{1000}(1.427 - j0.3228) \times 10^6
-$$
-  
-= (5.993 - j1.356) MΩ (9.15.4)
-
-Equating the real and imaginary parts yields *Rx* = 5.993 MΩ and a capacitive reactance
-
-$$
-X_x = \frac{1}{\omega C} = 1.356 \times 10^6
-$$
-
-or
-
-$$
-C = \frac{1}{\omega X_x} = \frac{1}{2\pi \times 2 \times 10^3 \times 1.356 \times 10^6} = 58.69 \text{ pF}
-$$
-
-(b) *Zx* remains the same as in Eq. (9.15.4) but *Rx* and *Xx* are in parallel. Assuming an *RC* parallel combination,
-
-$$
-\mathbf{Z}_x = (5.993 - j1.356) \text{ M}\Omega
-$$
-$$
-= R_x \parallel \frac{1}{j\omega C_x} = \frac{R_x}{1 + j\omega R_x C_x}
-$$
-
-By equating the real and imaginary parts, we obtain
-
-By equating the real and imaginary parts, we obtain  
-\n
-$$
-R_x = \frac{\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2}{\text{Real}(\mathbf{Z}_x)} = \frac{5.993^2 + 1.356^2}{5.993} = 6.3 \text{ M}\Omega
-$$
-\n
-$$
-C_x = -\frac{\text{Imag}(\mathbf{Z}_x)}{\omega[\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2]}
-$$
-\n
-$$
-= -\frac{-1.356}{2\pi(2000)(5.917^2 + 1.356^2)} = 2.852 \mu\text{F}
-$$
-
-We have assumed a parallel *RC* combination which works in this case.
-
-5. **Evaluate.** Let us now use *PSpice* to see if we indeed have the correct equalities. Running *PSpice* with the equivalent circuits, an open circuit between the "bridge" portion of the circuit, and a 10-volt input voltage yields the following voltages at the ends of the "bridge" relative to a reference at the bottom of the circuit:
-
-| FREQ        | VM(\$N_0002) | VP(\$N_0002) |
-|-------------|--------------|--------------|
-| 2.000E + 03 | 9.993E + 00  | -8.634E - 03 |
-| 2.000E + 03 | 9.993E + 00  | -8.637E - 03 |
-
-Because the voltages are essentially the same, then no measurable current can flow through the "bridge" portion of the circuit for any element that connects the two points together and we have a balanced bridge, which is to be expected. This indicates we have properly determined the unknowns.
-
-There is a very important problem with what we have done! Do you know what that is? We have what can be called an ideal, "theoretical" answer, but one that really is not very good in the real world. The difference between the magnitudes of the upper impedances and the lower impedances is much too large and would never be accepted in a real bridge circuit. For greatest accuracy, the overall magnitude of the impedances must at least be within the same relative order. To increase the accuracy of the solution of this problem, I would recommend increasing the magnitude of the top impedances to be in the range of 500 kΩ to 1.5 MΩ. One additional real-world comment: The size of these impedances also creates serious problems in making actual measurements, so the appropriate instruments must be used in order to minimize their loading (which would change the actual voltage readings) on the circuit.
-
-6. **Satisfactory?** Because we solved for the unknown terms and then tested to see if they worked, we validated the results. They can now be presented as a solution to the problem.
-
-<span id="page-422-0"></span>Practice Problem 9.15 In the ac bridge circuit of Fig. 9.37, suppose that balance is achie ved when **Z**1 is a 4.8-k Ω resistor , **Z**2 is a 10- Ω resistor in series with a 0.25-*μ*H inductor, **Z**3 is a 12-kΩ resistor, and *f* = 6 MHz. Determine the series components that make up **Z***x*.
-
-**Answer:** A 25-Ω resistor in series with a 0.625-*μ*H inductor.
-
-# **9.9** Summary
-
-1. A sinusoid is a signal in the form of the sine or cosine function. It has the general form
-
-$$
-v(t) = V_m \cos(\omega t + \phi)
-$$
-
-where *Vm* is the amplitude, *ω* = 2*πf* is the angular frequency, (*ωt* + *ϕ*) is the argument, and *ϕ* is the phase.
-
-2. A phasor is a complex quantity that represents both the magnitude and the phase of a sinusoid. Given the sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), its phasor **V** is
-
-$$
-\mathbf{V}=V_m\big/\phi
-$$
-
-- 3. In ac circuits, v oltage and current phasors al ways ha ve a fixed relation to one another at an y moment of time. If *v*(*t*) = *Vm* cos(*ωt* + *ϕv*) represents the v oltage through an element and *i*(*t*) = *Im* cos(*ωt* + *ϕi*) represents the current through the element, then *ϕi* = *ϕv* if the element is a resistor , *ϕi* leads *ϕv* by 90° if the element is a capacitor , and *ϕi* lags *ϕv* by 90° if the element is an inductor.
-- 4. The impedance **Z** of a circuit is the ratio of the phasor voltage across it to the phasor current through it:
-
-$$
-\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = R(\omega) + jX(\omega)
-$$
-
-The admittance **Y** is the reciprocal of impedance:
-
-$$
-\mathbf{Y} = \frac{1}{\mathbf{Z}} = G(\omega) + jB(\omega)
-$$
-
-Impedances are combined in series or in parallel the same w ay as resistances in series or parallel; that is, impedances in series add while admittances in parallel add.
-
-- 5. For a resistor **Z** = *R*, for an inductor **Z** = *jX* = *jωL*, and for a capacitor **Z** = −*jX* = 1∕*jωC*.
-- 6. Basic circuit la ws (Ohm's and Kirchhof f's) apply to ac circuits in the same manner as they do for dc circuits; that is,
-
-$$
-\mathbf{V} = \mathbf{Z}\mathbf{I}
-$$
-$$
-\Sigma \mathbf{I}_k = 0 \quad \text{(KCL)}
-$$
-$$
-\Sigma \mathbf{V}_k = 0 \quad \text{(KVL)}
-$$
-
-- <span id="page-423-0"></span>7. The techniques of voltage/current division, series/parallel combination of impedance/admittance, circuit reduction, and *Y*-∆ transformation all apply to ac circuit analysis.
-- 8. AC circuits are applied in phase-shifters and bridges.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/111_9.9 Summary.md

@@ -1,64 +0,0 @@
-# Review Questions
-
-**9.1** Which of the following is *not* a right way to express the sinusoid *A* cos *ωt*?
-
-> (a) *A* cos 2*π ft* (b) *A* cos(2*πt*∕*T*) (c) *A* cos *ω*(*t* − *T*) (d) *A* sin(*ωt* − 90°)
-
-**9.2** A function that repeats itself after fixed intervals is said to be:
-
-| (a) a phasor | (b) harmonic |
-|--------------|--------------|
-| (c) periodic | (d) reactive |
-
-**9.3** Which of these frequencies has the shorter period?
-
-(a) 1 krad/s (b) 1 kHz
-
-- **9.4** If *v*1 = 30 sin(*ωt* + 10°) and *v*2 = 20 sin(*ωt* + 50°), which of these statements are true?
-  - (a) *v*1 leads *v*2 (b) *v*2 leads *v*<sup>1</sup> (c) *v*2 lags *v*1 (d) *v*1 lags *v*<sup>2</sup>
-  - (e) *v*1 and *v*2 are in phase
-- **9.5** The voltage across an inductor leads the current through it by 90°.
-
-(a) True (b) False
-
-- **9.6** The imaginary part of impedance is called:
-  - (a) resistance (b) admittance (c) susceptance (d) conductance
-  - (e) reactance
-- **9.7** The impedance of a capacitor increases with increasing frequency.
-
-(a) True (b) False
-
-**9.8** At what frequency will the output voltage *vo*(*t*) in Fig. 9.39 be equal to the input voltage *v*(*t*) ?
-
-| (a) 0 rad/s | (b) 1 rad/s           | (c) 4 rad/s |
-|-------------|-----------------------|-------------|
-| (d) ∞ rad/s | (e) none of the above |             |
-
-# **Figure 9.39**
-
-For Review Question 9.8.
-
-**9.9** A series *RC* circuit has ∣*VR*∣ = 12 V and ∣*VC*∣ = 5 V. The magnitude of the supply voltage is:
-
-(a) −7 V (b) 7 V (c) 13 V (d) 17 V
-
-**9.10** A series *RCL* circuit has *R* = 30 Ω, *XC* = 50 Ω, and *XL* = 90 Ω. The impedance of the circuit is:
-
-> (a) 30 + *j*140 Ω (b) 30 + *j*40 Ω (c) 30 − *j*40 Ω (d) −30 − *j*40 Ω (e) −30 + *j*40 Ω
-
-*Answers: 9.1d, 9.2c, 9.3b, 9.4b,d, 9.5a, 9.6e, 9.7b, 9.8d, 9.9c, 9.10b.*
-
-# Problems
-
-# Section 9.2 Sinusoids
-
-- **9.1** Given the sinusoidal voltage *v*(*t*) = 50 cos (30*t* + 10°) V, find: (a) the amplitude *Vm*, (b) the period *T*, (c) the frequency *f*, and (d) *v*(*t*) at *t* = 10 ms.
-- **9.2** A current source in a linear circuit has
-
-*is* = 15 cos (25 *π t* + 25°) A
-
-- (a) What is the amplitude of the current?
-- (b) What is the angular frequency?
-- (c) Find the frequency of the current.
-- (d) Calculate *is* at *t* = 2 ms.
-- **9.3** Express the following functions in cosine form:

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/112_Review Questions.md

@@ -1,513 +0,0 @@
-(a) 10 sin(*ωt* + 30°) (b) −9 sin (8*t*) (c) −20 sin(*ωt* + 45°)
-
-- **9.4** Design a problem to help other students better understand sinusoids.
-- **9.5** Given *v*1 = 45 sin(*ωt* + 30°) V and *v*2 = 50 cos(*ωt* − 30°) V, determine the phase angle between the two sinusoids and which one lags the other.
-- **9.6** For the following pairs of sinusoids, determine which one leads and by how much.
-
-(a) 
-$$
-v(t) = 10 \cos(4t - 60^{\circ})
-$$
- and  
- $i(t) = 4 \sin(4t + 50^{\circ})$ 
-
-- (b) *v*1(*t*) = 4 cos(377*t* + 10°) and *v*2(*t*) = −20 cos 377*t*
-- (c) *x*(*t*) = 13 cos 2*t* + 5 sin 2*t* and *y*(*t*) = 15 cos(2*t* − 11.8°)
-
-# Section 9.3 Phasors
-
-- **9.7** If *f*(*ϕ*) = cos *ϕ* + *j* sin *ϕ*, show that *f*(*ϕ*) = *e j<sup>ϕ</sup>* .
-- **9.8** Calculate these complex numbers and express your results in rectangular form:
-
-(a) 
-$$
-\frac{60/45^{\circ}}{7.5 - j10} + j2
-$$
-  
-\n(b) 
-$$
-\frac{32/20^{\circ}}{(6 - j8)(4 + j2)} + \frac{20}{-10 + j24}
-$$
-  
-\n(c) 
-$$
-20 + (16/-50^{\circ})(5 + j12)
-$$
-
-**9.9** Evaluate the following complex numbers and leave your results in polar form:
-
-(a) 
-$$
-5/30^{\circ}
-$$
-  $\left(6 - j8 + \frac{3/60^{\circ}}{2 + j}\right)$   
-(b)  $\frac{(10/60^{\circ})(35/ -50^{\circ})}{(2 + j6) - (5 + j)}$ 
-
-**9.10** Design a problem to help other students better understand phasors.
-
-**9.11** Find the phasors corresponding to the following signals:
-
-(a) 
-$$
-v(t) = 21 \cos(4t - 15^\circ) \text{V}
-$$
-
-(b) 
-$$
-i(t) = -8 \sin(10t + 70^{\circ})
-$$
- mA
-
-(c) 
-$$
-v(t) = 120 \sin(10t - 50^{\circ})
-$$
- V
-
-(d) 
-$$
-i(t) = -60 \cos(30t + 10^{\circ})
-$$
- mA
-
-**9.12** Let **X** = 4⧸ 40° and **Y** = 20⧸−30°. Evaluate the following quantities and express your results in polar form:
-
-$$
-\text{(a)}\,(X+Y)X^*
-$$
-
-$$
-(b) (X - Y)^*
-$$
-
-(c) (**X** + **Y**)∕**X**
-
-**9.13** Evaluate the following complex numbers:
-
-(a) 
-$$
-\frac{2+j3}{1-j6} + \frac{7-j8}{-5+j11}
-$$
-  
-\n(b) 
-$$
-\frac{(5/10°)(10/-40°)}{(4/-80°)(-6/50°)}
-$$
-  
-\n(c) 
-$$
-\begin{vmatrix} 2+j3 & -j2 \\ -j2 & 8-j5 \end{vmatrix}
-$$
-
-**9.14** Simplify the following expressions:
-
-Simplify the following expressions:  
-\n(a) 
-$$
-\frac{(5 - j6) - (2 + j8)}{(-3 + j4)(5 - j) + (4 - j6)}
-$$
-\n(b) 
-$$
-\frac{(240/75^\circ + 160/–30^\circ)(60 - j80)}{(67 + j84)(20/32^\circ)}
-$$
-\n(c) 
-$$
-\left(\frac{10 + j20}{3 + j4}\right)^2 \sqrt{(10 + j5)(16 - j20)}
-$$
-
-**9.15** Evaluate these determinants:
-
-(a) 
-$$
-\begin{vmatrix} 10 + j6 & 2 - j3 \ -5 & -1 + j \end{vmatrix}
-$$
-  
-\n(b) 
-$$
-\begin{vmatrix} 20 \underline{/-30^{\circ}} & -4 \underline{/-10^{\circ}} \\ 16 \underline{/-0^{\circ}} & 3 \underline{/-40^{\circ}} \end{vmatrix}
-$$
-  
-\n(c) 
-$$
-\begin{vmatrix} 1 - j & -j & 0 \\ j & 1 & -j \\ 1 & j & 1 + j \end{vmatrix}
-$$
-
-**9.16** Transform the following sinusoids to phasors:
-
-(a) −20 cos(4*t* + 135°) (b) 8 sin(20*t* + 30°) (c) 20 cos (2*t*) + 15 sin (2*t*)
-
-- **9.17** Two voltages *v*1 and *v*2 appear in series so that their sum is *v* = *v*1 + *v*2. If *v*1 = 10 cos(50*t* − *π*∕3) V and *v*2 = 12 cos(50*t* + 30°) V, find *v*.
-- **9.18** Obtain the sinusoids corresponding to each of the following phasors:
-
-(a) 
-$$
-V_1 = 60/15^{\circ}
-$$
- V,  $\omega = 1$   
-\n(b)  $V_2 = 6 + j8$  V,  $\omega = 40$   
-\n(c)  $I_1 = 2.8e^{-j\pi/3}$  A,  $\omega = 377$   
-\n(d)  $I_2 = -0.5 - j1.2$  A,  $\omega = 10^3$ 
-
-**9.19** Using phasors, find:
-
-(a) 3 cos(20*t* + 10°) − 5 cos(20*t* − 30°)
-
-- (b) 40 sin 50*t* + 30 cos(50*t* − 45°)
-- (c) 20 sin 400*t* + 10 cos(400*t* + 60°)
-
-$$
--5\sin(400t-20^\circ)
-$$
-
-**9.20** A linear network has a current input 7.5 cos(10*t* + 30°) A and a voltage output 120 cos(10*t* + 75°) V. Determine the associated impedance.
-
-**9.21** Simplify the following:
-
-(a) 
-$$
-f(t) = 5 \cos(2t + 15^\circ) - 4 \sin(2t - 30^\circ)
-$$
-  
-(b)  $g(t) = 8 \sin t + 4 \cos(t + 50^\circ)$ 
-
-- (c) *h*(*t*) = ∫ 0 (10 cos 40*t* + 50 sin 40*t*) *dt*
-- **9.22** An alternating voltage is given by *v*(*t*) = 55 cos(5*t* + 45°) V. Use phasors to find
-
-$$
-10v(t) + 4\frac{dv}{dt} - 2\int_{-\infty}^{t} v(t) dt
-$$
-
-Assume that the v alue of the inte gral is zero at *t* = −∞.
-
-**9.23** Apply phasor analysis to evaluate the following:
-
-(a) 
-$$
-v = [110 \sin(20t + 30^\circ) + 220 \cos(20t - 90^\circ)]
-$$
- V  
-(b)  $i = [30 \cos(5t + 60^\circ) - 20 \sin(5t + 60^\circ)]$  A
-
-**9.24** Find *v*(*t*) in the following integrodifferential equations using the phasor approach:
-
-(a) 
-$$
-v(t) + \int v dt = 10 \cos t
-$$
-  
-\n(b)  $\frac{dv}{dt} + 5v(t) + 4 \int v dt = 20 \sin(4t + 10^{\circ})$ 
-
-**9.25** Using phasors, determine *i*(*t*) in the following equations:
-
-(a) 
-$$
-2\frac{di}{dt} + 3i(t) = 4\cos(2t - 45^{\circ})
-$$
-  
-\n(b)  $10 \int i \, dt + \frac{di}{dt} + 6i(t) = 5\cos(5t + 22^{\circ})$  A
-
-**9.26** The loop equation for a series *RLC* circuit gives
-
-$$
-\frac{di}{dt} + 2i + \int_{-\infty}^{t} i \, dt = \cos 2t \, A
-$$
-
-Assuming that the v alue of the inte gral at *t* = −∞ is zero, find *i*(*t*) using the phasor method.
-
-**9.27** A parallel *RLC* circuit has the node equation
-
-$$
-\frac{dv}{dt} + 50v + 100 \int v \, dt = 110 \cos(377t - 10^{\circ}) \, \text{V}
-$$
-
-Determine *v*(*t*) using the phasor method. You may assume that the value of the integral at *t* = −∞ is zero.
-
-# Section 9.4 Phasor Relationships for Circuit Elements
-
-- **9.28** Determine the current that flows through an 20-Ω resistor connected to a voltage source *v*s = 120 cos (377*t* + 37°) V.
-- **9.29** Given that *vc*(0) = 2 cos(155°) V, what is the instantaneous voltage across a 2-*μ*F capacitor when the current through it is *i* = 4 sin(106 *t* + 25°) A?
-
-- **9.30** A voltage *v*(*t*) = 100 cos(60*t* + 20°) V is applied to a parallel combination of a 40-kΩ resistor and a 50-*μ*F capacitor. Find the steady-state currents through the resistor and the capacitor.
-- **9.31** A series *RLC* circuit has *R* = 80 Ω, *L* = 240 mH, and *C* = 5 mF. If the input voltage is *v*(*t*) = 115 cos 2*t*, find the current flowing through the circuit.
-- **9.32** Using Fig. 9.40, design a problem to help other students better understand phasor relationships for circuit elements.
-
-- **9.33** A series *RL* circuit is connected to a 220-V ac source. If the voltage across the resistor is 170 V, find the voltage across the inductor.
-- **9.34** What value of *ω* will cause the forced response, *vo*, in Fig. 9.41 to be zero?
-
-**Figure 9.41** For Prob. 9.34.
-
-# Section 9.5 Impedance and Admittance
-
-**9.35** Find the steady-state current *i* in the circuit of Fig. 9.42, when *vs*(*t*) = 115 cos 200*t* V.
-
-**Figure 9.42** For Prob. 9.35.
-
-# **9.36** Using Fig. 9.43, design a problem to help other students better understand impedance.
-
-**9.37** Determine the admittance **Y** for the circuit in Fig. 9.44.
-
-# **Figure 9.45**
-
-For Prob. 9.38.
-
-**9.39** For the circuit shown in Fig. 9.46, find *Z*eq and use that to find current **I**. Let *ω* = 10 rad/s.
-
-**Figure 9.46** For Prob. 9.39.
-
-**9.40** In the circuit of Fig. 9.47, find *io* when:
-
-(a) 
-$$
-\omega = 1
-$$
- rad/s (b)  $\omega = 5$  rad/s  
-(c)  $\omega = 10$  rad/s
-
-# **Figure 9.47** For Prob. 9.40.
-
-**9.41** Find *v*(*t*) in the *RLC* circuit of Fig. 9.48.
-
-# **Figure 9.48**
-
-For Prob. 9.41.
-
-**9.42** Calculate *vo*(*t*) in the circuit of Fig. 9.49.
-
-# **Figure 9.49**
-
-For Prob. 9.42.
-
-**9.43** Find current **I***o* in the circuit shown in Fig. 9.50.
-
-**Figure 9.50** For Prob. 9.43.
-
-**9.44** Calculate *i*(*t*) in the circuit of Fig. 9.51.
-
-**Figure 9.51** For prob. 9.44.
-
-**Figure 9.52** For Prob. 9.45.
-
-200 mH 100 mF 2 Ω <sup>2</sup><sup>Ω</sup> <sup>i</sup> o *<sup>v</sup>*<sup>s</sup> <sup>+</sup> ‒
-
-# **Figure 9.53**
-
-For Prob. 9.46.
-
-**9.47** In the circuit of Fig. 9.54, determine the value of *is*(*t*).
-
-# **Figure 9.54**
-
-For Prob. 9.47.
-
-**9.48** Given that *vs*(*t*) = 20 sin(100*t* − 40°) in Fig. 9.55, determine *ix*(*t*).
-
-# **Figure 9.55**
-
-For Prob. 9.48.
-
-**9.49** Find *v <sup>s</sup>* (*t*) in the circuit of Fig. 9.56 if the current *ix* through the 1-Ω resistor is 8 sin 200*t* A.
-
-**Figure 9.56** For Prob. 9.49.
-
-**9.50** Determine *vx* in the circuit of Fig. 9.57. Let *is*(*t*) = 5 cos(100*t* + 40°) A.
-
-# **Figure 9.57**
-
-For Prob. 9.50.
-
-**9.51** If the voltage *vo* across the 2-Ω resistor in the circuit of Fig. 9.58 is 90 cos 2*t* V, obtain *is*.
-
-# **Figure 9.58**
-
-For Prob. 9.51.
-
-**9.52** If **V***o* = 8⧸ 30° V in the circuit of Fig. 9.59, find **I***s*.
-
-# **Figure 9.59**
-
-For Prob. 9.52.
-
-**Figure 9.60**
-
-For Prob. 9.53.
-
-**9.54** In the circuit of Fig. 9.61, find **V***s* if **I***o* = 30⧸ 0° A.
-
-**Figure 9.61** For Prob. 9.54.
-
-# **Figure 9.62**
-
-For Prob. 9.55.
-
-# Section 9.7 Impedance Combinations
-
-**9.56** At *ω* = 377 rad/s, find the input impedance of the circuit shown in Fig. 9.63.
-
-**Figure 9.63** For Prob. 9.56.
-
-**9.57** At *ω* = 1 rad/s, obtain the input admittance in the circuit of Fig. 9.64.
-
-# **Figure 9.64**
-
-For Prob. 9.57.
-
-**9.58** Using Fig. 9.65, design a problem to help other students better understand impedance combinations.
-
-\* An asterisk indicates a challenging problem.
-
-**\* 9.59** For the network in Fig. 9.66, find **Z**in. Let *ω* = 100 rad/s.
-
-**Figure 9.66**
-
-For Prob. 9.59.
-
-**9.60** Obtain **Z**in for the circuit in Fig. 9.67.
-
-**Figure 9.67**
-
-For Prob. 9.60.
-
-**9.61** Find **Z**eq in the circuit of Fig. 9.68.
-
-# **Figure 9.68**
-
-For Prob. 9.61.
-
-**9.62** For the circuit in Fig. 9.69, find the input impedance **Z**in at 10 krad/s.
-
-**Figure 9.69** For Prob. 9.62.
-
-**9.63** For the circuit in Fig. 9.70, find the value of **Z***T*.
-
-**Figure 9.70** For Prob. 9.63.
-
-**9.64** Find **Z***T* and Vo in the circuit in Fig. 9.71. Let the value of the inductance equal *j*20 Ω.
-
-**9.65** Determine **Z***T* and **I** for the circuit in Fig. 9.72.
-
-# **Figure 9.72**
-
-For Prob. 9.65.
-
-**9.66** For the circuit in Fig. 9.73, calculate **Z***T* and **V***ab*.
-
-**Figure 9.73** For Prob. 9.66.
-
-**9.67** At *ω* = 103 rad/s, find the input admittance of each of the circuits in Fig. 9.74.
-
-# **Figure 9.74**
-
-For Prob. 9.67.
-
-**9.68** Determine **Y**eq for the circuit in Fig. 9.75.
-
-# **Figure 9.75**
-
-For Prob. 9.68.
-
-**9.69** Find the equivalent admittance **Y**eq of the circuit in Fig. 9.76.
-
-# **Figure 9.76**
-
-For Prob. 9.69.
-
-**9.70** Find the equivalent impedance of the circuit in Fig. 9.77.
-
-**Figure 9.77** For Prob. 9.70.
-
-**9.71** Obtain the equivalent impedance of the circuit in Fig. 9.78.
-
-# **Figure 9.78**
-
-For Prob. 9.71.
-
-**Figure 9.79**
-
-For Prob. 9.72.
-
-**9.73** Determine the equivalent impedance of the circuit in Fig. 9.80.
-
-# Section 9.8 Applications
-
-- **9.74** Design an *RL* circuit to provide a 90° leading phase shift.
-- **9.75** Design a circuit that will transform a sinusoidal
-- voltage input to a cosinusoidal voltage output.
-- **9.76** For the following pairs of signals, determine if *v*<sup>1</sup> leads or lags *v*2 and by how much.
-
-(a) *v*1 = 10 cos(5*t* − 20°), *v*2 = 8 sin 5*t*
-
-(b) 
-$$
-v_1 = 19 \cos(2t + 90^\circ)
-$$
-,  $v_2 = 6 \sin 2t$ 
-
-(c) 
-$$
-v_1 = -4 \cos 10t
-$$
-,  $v_2 = 15 \sin 10t$ 
-
-- **9.77** Refer to the *RC* circuit in Fig. 9.81.
-  - (a) Calculate the phase shift at 2 MHz.
-  - (b) Find the frequency where the phase shift is 45°.
-
-# **Figure 9.81**
-
-For Prob. 9.77.
-
-- **9.78** A coil with impedance 8 + *j*6 Ω is connected in series with a capacitive reactance *X*. The series combination is connected in parallel with a resistor *R*. Given that the equivalent impedance of the resulting circuit is 5⧸ 0° Ω, find the value of *R* and *X*.
-- **9.79** (a) Calculate the phase shift of the circuit in Fig. 9.82. (b) State whether the phase shift is leading or lagging (output with respect to input).
-  - (c) Determine the magnitude of the output when the input is 120 V.
-
-# **Figure 9.82**
-
-For Prob. 9.79.
-
-- **9.80** Consider the phase-shifting circuit in Fig. 9.83. Let **V***i* = 120 V operating at 60 Hz. Find:
-  - (a) **V***o* when *R* is maximum
-  - (b) **V***o* when *R* is minimum
-  - (c) the value of *R* that will produce a phase shift of 45°
-
-# **Figure 9.83**
-
-For Prob. 9.80.
-
-- **9.81** The ac bridge in Fig. 9.37 is balanced when *R*1 = 400 Ω, *R*2 = 600 Ω, *R*3 = 1.2 kΩ, and *C*2 = 0.3 *μ*F. Find *Rx* and *Cx*. Assume *R*2 and *C*2 are in series.
-- **9.82** A capacitance bridge balances when *R*1 = 100 Ω, *R*2 = 2 kΩ, and *Cs* = 40 *μ*F. What is *Cx*, the capacitance of the capacitor under test?
-- **9.83** An inductive bridge balances when *R*1 = 1.2 kΩ, *R*2 = 500 Ω, and *Ls* = 250 mH. What is the value of *Lx*, the inductance of the inductor under test?
-
-<span id="page-431-0"></span>**9.84** The ac bridge shown in Fig. 9.84 is known as a *Maxwell bridge* and is used for accurate measurement of inductance and resistance of a coil in terms of a standard capacitance *Cs*. Show that when the bridge is balanced,
-
-$$
-L_x = R_2 R_3 C_s \qquad \text{and} \qquad R_x = \frac{R_2}{R_1} R_3
-$$
-
-Find *Lx* and *Rx* for *R*1 = 40 k Ω, *R*2 = 1.6 k Ω, *R*3 = 4 kΩ, and *Cs* = 0.45 *μ*F.
-
-**Figure 9.84** Maxwell bridge; For Prob. 9.84.
-
-# *<sup>f</sup>* = \_\_\_\_\_\_\_\_\_\_\_\_ 1 2*π*<sup>√</sup>
-
-**9.86** The circuit shown in Fig. 9.86 is used in a television receiver. What is the total impedance of this circuit?
-
-Comprehensive Problems
-
-For Prob. 9.86.
-
-**9.87** The network in Fig. 9.87 is part of the schematic describing an industrial electronic sensing device. What is the total impedance of the circuit at 4 kHz?
-
-**Figure 9.87** For Prob. 9.87.
-
-- (a) What is the impedance of the circuit?
-- (b) If the frequency were halved, what would be the impedance of the circuit?
-
-# **Figure 9.88**
-
-For Prob. 9.88.
-
-**9.89** An industrial load is modeled as a series combination of an inductor and a resistance as shown in Fig. 9.89. Calculate the value of a capacitor *C* across the series combination so that the net impedance is resistive at a frequency of 2 kHz.
-
-For Prob. 9.89.
-
-**9.90** An industrial coil is modeled as a series combination of an inductance *L* and resistance *R*, as shown in Fig. 9.90. Since an ac voltmeter measures only the magnitude of a sinusoid, the following

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/113_Comprehensive Problems.md

@@ -1,34 +0,0 @@
-**9.85** The ac bridge circuit of Fig. 9.85 is called a *Wien bridge*. It is used for measuring the frequency of a source. Show that when the bridge is balanced,
-
-> *\_\_\_\_\_\_\_\_ R*2*R*4*C*2*C*4
-
-measurements are taken at 60 Hz when the circuit operates in the steady state:
-
-$$
-|\mathbf{V}_s| = 145 \text{ V}, \qquad |\mathbf{V}_1| = 50 \text{ V}, \qquad |\mathbf{V}_o| = 110 \text{ V}
-$$
-
-Use these measurements to determine the values of *L* and *R*.
-
-# **Figure 9.90**
-
-For Prob. 9.90.
-
-**9.91** Figure 9.91 shows a series combination of an inductance and a resistance. If it is desired to connect a capacitor in parallel with the series combination such that the net impedance is resistive at 10 kHz, what is the required value of *C*?
-
-**Figure 9.91** For Prob. 9.91.
-
-- **9.92** A transmission line has a series impedance of **Z** = 100⧸ 75° Ω and a shunt admittance of **Y** = 450⧸ 48° *μ*S. Find: (a) the characteristic impedance **Z***o* = √ \_\_\_\_\_ **Z**∕**Y** , (b) the propagation constant *γ* = √ \_\_\_ **ZY** .
-- **9.93** A power transmission system is modeled as shown in Fig. 9.92. Given the source voltage and circuit elements
-
-| Vs = 115⧸ 0° V,        | source impedance          |
-|------------------------|---------------------------|
-| Zs = (1 + j0.5) Ω,     | line impedance            |
-| Zt = (0.4 + j0.3) Ω,   | and load impedance        |
-| ZL = (23.2 + j18.9) Ω, | find the load current IL. |
-
-**Figure 9.92** For Prob. 9.93.
-
-# **chapter**
-
-# 10

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/114_Chapter 10 - Sinusoidal Steady-State Analysis.md

@@ -1,68 +0,0 @@
-# <span id="page-433-0"></span>Sinusoidal Steady-State Analysis
-
-*Three men are my friends—he that loves me, he that hates me, he that is indifferent to me. Who loves me, teaches me tenderness; who hates me, teaches me caution; who is indifferent to me, teaches me self-reliance.* —J. E. Dinger
-
-# Enhancing Your Career
-
-# **Career in Software Engineering**
-
-Software engineering is that aspect of engineering that deals with the practical application of scientific knowledge in the design, construction, and v alidation of computer programs and the associated documenta ‑ tion required to de velop, operate, and maintain them. It is a branch of electrical engineering that is becoming increasingly important as more and more disciplines require one form of software package or another to perform routine tasks and as programmable microelectronic systems are used in more and more applications.
-
-The role of a softw are engineer should not be confused with that of a computer scientist; the softw are engineer is a practitioner , not a theoretician. A softw are engineer should ha ve good computer ‑ pro gramming skills and be familiar with programming languages, in par‑ ticular C++, which is becoming increasingly popular. Because hardware and software are closely interlinked, it is essential that a softw are engi‑ neer have a thorough understanding of hardware design. Most important, the software engineer should ha ve some specialized kno wledge of the area in which the software development skill is to be applied.
-
-All in all, the field of software engineering offers a great career to those who enjo y programming and de veloping softw are packages. The higher rewards will go to those having the best preparation, with the most interesting and challenging opportunities going to those with graduate education.
-
-A three‑dimensional printing of the output of an AutoCAD model of a NASA flywheel. © Ansoft Corporation
-
-# <span id="page-434-0"></span>Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Analyze electrical circuits in the frequency domain using nodal analysis.
-- 2. Analyze electrical circuits in the frequency domain using mesh analysis.
-- 3. Apply the superposition principle to frequency domain electrical circuits.
-- 4. Apply source transformation in frequency domain circuits.
-- 5. Understand how Thevenin and Norton equivalent circuits can be used in the frequency domain.
-- 6. Analyze electrical circuits with op amps.
-
-# **10.1** Introduction
-
-In Chapter 9, we learned that the forced or steady ‑state response of cir‑ cuits to sinusoidal inputs can be obtained by using phasors. We also know that Ohm's and Kirchhoff's laws are applicable to ac circuits. In this chapter, we want to see ho w nodal analysis, mesh analysis, Thevenin's theorem, Norton's theorem, superposition, and source transformations are applied in analyzing ac circuits. Since these techniques were already introduced for dc circuits, our major effort here will be to illustrate with examples.
-
-Analyzing ac circuits usually requires three steps.
-
-# Steps to Analyze AC Circuits:
-
-- 1. Transform the circuit to the phasor or frequency domain.
-- 2. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.).
-- 3. Transform the resulting phasor to the time domain.
-
-Step 1 is not necessary if the problem is specified in the frequency domain. In step 2, the analysis is performed in the same manner as dc circuit analysis except that complex numbers are involved. Having read Chapter 9, we are adept at handling step 3.
-
-Toward the end of the chapter , we learn ho w to apply *PSpice* in solving ac circuit problems. We finally apply ac circuit analysis to two practical ac circuits: oscillators and ac transistor circuits.
-
-# **10.2** Nodal Analysis
-
-The basis of nodal analysis is Kirchhof f's current la w. Since KCL is valid for phasors, as demonstrated in Section 9.6, we can analyze ac cir‑ cuits by nodal analysis. The following examples illustrate this.
-
-Frequency domain analysis of an ac circuit via phasors is much easier than analysis of the circuit in the time domain.
-
-Find *ix* in the circuit of Fig. 10.1 using nodal analysis. Example 10.1
-
-For Example 10.1.
-
-# **Solution:**
-
-We first convert the circuit to the frequency domain:
-
-$$
-20 \cos 4t \Rightarrow 20 \underline{/0^{\circ}}, \qquad \Omega = 4 \text{ rad/s}
-$$
-  
-\n
-$$
-1 \text{ H} \Rightarrow j \Omega L = j4
-$$
-  
-\n
-$$

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/115_10.1 Introduction.md

@@ -1,138 +0,0 @@
-0.5 \text{ H} \Rightarrow j \Omega L = j2
-$$
-  
-\n
-$$
-0.1 \text{ F} \Rightarrow \frac{1}{j \Omega C} = -j2.5
-$$
-
-Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2.
-
-# **Figure 10.2**
-
-Frequency domain equivalent of the circuit in Fig. 10.1.
-
-Applying KCL at node 1,
-
-$$
-\frac{20 - V_1}{10} = \frac{V_1}{-j2.5} + \frac{V_1 - V_2}{j4}
-$$
-
-or
-
-(1 + *j*1.5)**V**1 + *j*2.5**V**<sup>2</sup> = 20 **(10.1.1)**
-
-At node 2,
-
-$$
-2\mathbf{I}_x + \frac{\mathbf{V}_1 - \mathbf{V}_2}{j4} = \frac{\mathbf{V}_2}{j2}
-$$
-
-But **I***x* = **V**1∕−*j*2.5. Substituting this gives
-
-$$
-\frac{2V_1}{-j2.5} + \frac{V_1 - V_2}{j4} = \frac{V_2}{j2}
-$$
-
-By simplifying, we get
-
-$$
-11V_1 + 15V_2 = 0 \tag{10.1.2}
-$$
-
-Equations (10.1.1) and (10.1.2) can be put in matrix form as
-
-$$
-\begin{bmatrix} 1+j1.5 & j2.5 \\ 11 & 15 \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} 20 \\ 0 \end{bmatrix}
-$$
-
-We obtain the determinants as
-
-$$
-\Delta = \begin{vmatrix} 1+j1.5 & j2.5 \\ 11 & 15 \end{vmatrix} = 15 - j5
-$$
-
-$$
-\Delta_1 = \begin{vmatrix} 20 & j2.5 \\ 0 & 15 \end{vmatrix} = 300, \qquad \Delta_2 = \begin{vmatrix} 1+j1.5 & 20 \\ 11 & 0 \end{vmatrix} = -220
-$$
-$$
-\mathbf{V}_1 = \frac{\Delta_1}{\Delta} = \frac{300}{15 - j5} = 18.97 \underline{/18.43^\circ} \text{ V}
-$$
-$$
-\mathbf{V}_2 = \frac{\Delta_2}{\Delta} = \frac{-220}{15 - j5} = 13.91 \underline{/198.3^\circ} \text{ V}
-$$
-
-The current **I***x* is given by
-
-$$
-\mathbf{I}_x \text{ is given by}
-$$
-\n
-$$
-\mathbf{I}_x = \frac{\mathbf{V}_1}{-j2.5} = \frac{18.97 \; / 18.43^\circ}{2.5 \; / -90^\circ} = 7.59 \; / 108.4^\circ \; \text{A}
-$$
-
-Transforming this to the time domain,
-
-$$
-i_x = 7.59 \cos(4t + 108.4^\circ)
-$$
- A
-
-Using nodal analysis, find *v*1 and *v*<sup>2</sup> Practice Problem 10.1 in the circuit of Fig. 10.3.
-
-**Figure 10.3** For Practice Prob. 10.1.
-
-**Answer:** *v*1(*t*) = 28.31 cos(2*t* + 60.01°) V, *v*2(*t*) = 82.56 cos(2*t* + 57.12°) V.
-
-Compute **V**1 and **V**<sup>2</sup> Example 10.2 in the circuit of Fig. 10.4.
-
-For Example 10.2.
-
-# **Solution:**
-
-Nodes 1 and 2 form a supernode as shown in Fig. 10.5. Applying KCL at the supernode gives
-
-$$
-3 = \frac{\mathbf{V}_1}{-j3} + \frac{\mathbf{V}_2}{j6} + \frac{\mathbf{V}_2}{12}
-$$
-
-or
-
-$$
-36 = j4V_1 + (1 - j2)V_2 \tag{10.2.1}
-$$
-
-<span id="page-437-0"></span>
-
-# **Figure 10.5**
-
-A supernode in the circuit of Fig. 10.4.
-
-But a voltage source is connected between nodes 1 and 2, so that
-
-$$
-V_1 = V_2 + 10 \frac{\angle 45^{\circ}}{\angle 0.2.2}
-$$
-
-Substituting Eq. (10.2.2) in Eq. (10.2.1) results in
-
-$$
-36 - 40 \underline{1135^\circ} = (1 + i2) \mathbf{V}_2 \implies \mathbf{V}_2 = 31.41 \underline{18^\circ} \text{ V}
-$$
-
-From Eq. (10.2.2),
-
-$$
-V_1 = V_2 + 10 \underline{745^\circ} = 25.78 \underline{770.48^\circ} \text{ V}
-$$
-
-Calculate 
-$$
-V_1
-$$
- and  $V_2$  in the circuit shown in Fig. 10.6. **Practice Problem 10.2**
-
-For Practice Prob. 10.2.
-
-**Answer: V**1 = 96.8 ∕69.66° V, **V**2 = 16.88∕165.72° V.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/116_10.3 Mesh Analysis.md

@@ -1,238 +0,0 @@
-# **10.3** Mesh Analysis
-
-Kirchhoff's voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuits was shown in Section 9.6 and is illustrated in the follo wing examples. Keep in mind that the v ery nature of using mesh analysis is that it is to be applied to planar circuits.
-
-Determine current **I***o* in the circuit of Fig. 10.7 using mesh analysis. Example 10.3
-
-# **Solution:**
-
-Applying KVL to mesh 1, we obtain
-
-$$
-(8+j10-j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - j10\mathbf{I}_3 = 0 \tag{10.3.1}
-$$
-
-**Figure 10.7** For Example 10.3.
-
-For mesh 2,
-
-$$
-(4 - j2 - j2)I2 - (-j2)I1 - (-j2)I3 + 20 / 90o = 0
-$$
- (10.3.2)
-
-For mesh 3, **I**3 = 5. Substituting this in Eqs. (10.3.1) and (10.3.2), we get
-
-$$
-(8+j8)\mathbf{I}_1 + j2\mathbf{I}_2 = j50 \tag{10.3.3}
-$$
-
-$$
-j2\mathbf{I}_1 + (4 - j4)\mathbf{I}_2 = -j20 - j10
-$$
- (10.3.4)
-
-Equations (10.3.3) and (10.3.4) can be put in matrix form as
-
-[ <sup>8</sup> + *j*<sup>8</sup> *j*2 *j*2 <sup>4</sup> − *j*<sup>4</sup> ] [ **<sup>I</sup>**<sup>1</sup> **I**2 ] = [ *<sup>j</sup>*<sup>50</sup> −*j*30]
-
-from which we obtain the determinants
-
-$$
-\Delta = \begin{vmatrix} 8 + j8 & j2 \\ j2 & 4 - j4 \end{vmatrix} = 32(1 + j)(1 - j) + 4 = 68
-$$
-  
-$$
-\Delta_2 = \begin{vmatrix} 8 + j8 & j50 \\ j2 & -j30 \end{vmatrix} = 340 - j240 = 416.17 \underline{/-35.22^{\circ}}
-$$
-  
-$$
-\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{416.17 \underline{/-35.22^{\circ}}}{68} = 6.12 \underline{/-35.22^{\circ}} A
-$$
-
-The desired current is
-
-$$
-I_o = -I_2 = 6.12 \underline{144.78}^{\circ} A
-$$
-
-For Practice Prob. 10.3.
-
-# **Solution:**
-
-As shown in Fig. 10.10, meshes 3 and 4 form a supermesh due to the current source between the meshes. For mesh 1, KVL gives
-
-$$
--10 + (8 - j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - 8\mathbf{I}_3 = 0
-$$
-
-or
-
-$$
-(8 - j2)\mathbf{I}_1 + j2\mathbf{I}_2 - 8\mathbf{I}_3 = 10 \tag{10.4.1}
-$$
-
-For mesh 2,
-
-$$
-I_2 = -3 \tag{10.4.2}
-$$
-
-For the supermesh,
-
-$$
-(8 - j4)\mathbf{I}_3 - 8\mathbf{I}_1 + (6 + j5)\mathbf{I}_4 - j5\mathbf{I}_2 = 0 \tag{10.4.3}
-$$
-
-Due to the current source between meshes 3 and 4, at node A,
-
-$$
-\mathbf{I}_4 = \mathbf{I}_3 + 4 \tag{10.4.4}
-$$
-
-■ **METHOD 1** Instead of solving the above four equations, we re ‑ duce them to two by elimination.
-
-Combining Eqs. (10.4.1) and (10.4.2),
-
-$$
-(8 - j2)\mathbf{I}_1 - 8\mathbf{I}_3 = 10 + j6 \tag{10.4.5}
-$$
-
-Combining Eqs. (10.4.2) to (10.4.4),
-
-$$
--8I1 + (14 + j)I3 = -24 - j35
-$$
- (10.4.6)
-
-**Figure 10.10** Analysis of the circuit in Fig. 10.9.
-
-From Eqs. (10.4.5) and (10.4.6), we obtain the matrix equation
-
-$$
-\begin{bmatrix} 8-j2 & -8 \ -8 & 14+j \end{bmatrix} \begin{bmatrix} I_1 \ I_3 \end{bmatrix} = \begin{bmatrix} 10+j6 \ -24-j35 \end{bmatrix}
-$$
-
-We obtain the following determinants
-
-$$
-\Delta = \begin{vmatrix} 8 - j2 & -8 \\ -8 & 14 + j \end{vmatrix} = 112 + j8 - j28 + 2 - 64 = 50 - j20
-$$
-  
-$$
-\Delta_1 = \begin{vmatrix} 10 + j6 & -8 \\ -24 - j35 & 14 + j \end{vmatrix} = 140 + j10 + j84 - 6 - 192 - j280
-$$
-  
-$$
-= -58 - j186
-$$
-
-Current **I**1 is obtained as
-
-$$
-\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{-58 - j186}{50 - j20} = 3.618 \; \underline{/274.5^\circ} \, \text{A}
-$$
-
-The required voltage **V***o* is
-
-$$
-\mathbf{V}_o = -j2(\mathbf{I}_1 - \mathbf{I}_2) = -j2(3.618 \angle 274.5^\circ + 3)
-$$
-  
-= -7.2134 - j6.568 = 9.756 \angle 222.32^\circ \text{V}
-
-■ **METHOD 2** We can use *MATLAB* to solve Eqs. (10.4.1) to (10.4.4). We first cast the equations as
-
-$$
-\begin{bmatrix} 8-j2 & j2 & -8 & 0 \ 0 & 1 & 0 & 0 \ -8 & -j5 & 8-j4 & 6+j5 \ 0 & 0 & -1 & 1 \ \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \\ I_4 \end{bmatrix} = \begin{bmatrix} 10 \\ -3 \\ 0 \\ 4 \end{bmatrix}
-$$
- (10.4.7a)
-
-or
-
-$$
-AI = B
-$$
-
-By inverting **A**, we can obtain **I** as
-
-$$
-\mathbf{I} = \mathbf{A}^{-1} \mathbf{B} \tag{10.4.7b}
-$$
-
-We now apply *MATLAB* as follows:
-
-```
->> A = [(8-j*2) j*2 -8 0;
- 0 1 0 0;
- -8 -j*5 (8-j*4) (6+j*5);
- 0 0 -1 1];
->> B = [10 -3 0 4]';
->> I = inv(A)*B
-I =
- 0.2828 - 3.6069i
- -3.0000
- -1.8690 - 4.4276i
- 2.1310 - 4.4276i
->> Vo = -2*j*(I(1) - I(2))
-Vo =
- -7.2138 - 6.5655i
-```
-
-as obtained previously.
-
-<span id="page-441-0"></span>Calculate current **I***o* in the circuit of Fig. 10.11.
-
-**Answer:** 6.089∕5.94° A.
-
-# **10.4** Superposition Theorem
-
-Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. The theorem becomes important if the circuit has sources operating at *different* frequencies. In this case, since the impedances depend on frequency, we must have a different frequency domain circuit for each frequency. The total re ‑ sponse must be obtained by adding the individual responses in the *time* domain. It is incorrect to try to add the responses in the phasor or fre‑ quency domain. Why? Because the exponential factor *e<sup>j</sup>ω<sup>t</sup>* is implicit in sinusoidal analysis, and that factor would change for every angular frequency *ω*. It would therefore not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit has sources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain.
-
-Use the superposition theorem to find **I***o* in the circuit in Fig. 10.7.
-
-# **Solution:**
-
-Let
-
-$$
-\mathbf{I}_o = \mathbf{I}_o' + \mathbf{I}_o'' \tag{10.5.1}
-$$
-
-where **I**′ *o* and **I**″ *o* are due to the voltage and current sources, respectively. To find **I**′ *o*, consider the circuit in Fig. 10.12(a). If we let **Z** be the parallel combination of −*j*2 and 8 + *j*10, then
-
-2 and 8 + j10, then  
-\n
-$$
-\mathbf{Z} = \frac{-j2(8+j10)}{-2j+8+j10} = 0.25 - j2.25
-$$
-
-and current **I**′ *o* is
-
-$$
-\mathbf{I}'_o = \frac{j20}{4 - j2 + \mathbf{Z}} = \frac{j20}{4.25 - j4.25}
-$$
-
-or
-
-$$
-\mathbf{I}'_o = -2.353 + j2.353\tag{10.5.2}
-$$
-
-To get **I**″ *o*, consider the circuit in Fig. 10.12(b). For mesh 1,
-
-$$
-(8 + j8)\mathbf{I}_1 - j10\mathbf{I}_3 + j2\mathbf{I}_2 = 0 \tag{10.5.3}
-$$
-
-For mesh 2,
-
-$$
-(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j2\mathbf{I}_3 = 0 \tag{10.5.4}
-$$
-
-For mesh 3,
-
-**I**3 = 5 **(10.5.5)**
-
-**Figure 10.12** Solution of Example 10.5.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/117_10.4 Superposition Theorem.md

@@ -1,143 +0,0 @@
-# Example 10.5
-
-(a)
-
-From Eqs. (10.5.4) and (10.5.5),
-
-$$
-(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j10 = 0
-$$
-
-Expressing **I**1 in terms of **I**2 gives
-
-$$
-\mathbf{I}_1 = (2 + j2)\mathbf{I}_2 - 5\tag{10.5.6}
-$$
-
-Substituting Eqs. (10.5.5) and (10.5.6) into Eq. (10.5.3), we get
-
-$$
-(8+j8)[(2+j2)I2 - 5] - j50 + j2I2 = 0
-$$
-
-or
-
-$$
-I_2 = \frac{90 - j40}{34} = 2.647 - j1.176
-$$
-
-Current **I**″ *o* is obtained as
-
-$$
-\mathbf{I}_{o}'' = -\mathbf{I}_{2} = -2.647 + j1.176
-$$
- (10.5.7)
-
-From Eqs. (10.5.2) and (10.5.7), we write
-
-$$
-I_o = I'_o + I''_o = -5 + j3.529 = 6.12 \underline{144.78^\circ} A
-$$
-
-which agrees with what we got in Example 10.3. It should be noted that applying the superposition theorem is not the best way to solve this prob‑ lem. It seems that we have made the problem twice as hard as the origi ‑ nal one by using superposition. However, in Example 10.6, superposi ‑ tion is clearly the easiest approach.
-
-| Practice Problem 10.5 | Find current | Io in the circuit of Fig. 10.8 using the superposition |
-|-----------------------|--------------|--------------------------------------------------------|
-|                       | theorem.     |                                                        |
-|                       |              |                                                        |
-
-**Answer:** 5.97∕65.45° A.
-
-Because the circuit operates at three different frequencies (*ω* = 0 for the dc voltage source), one way to obtain a solution is to use superposition, which breaks the problem into single‑frequency problems. So we let
-
-$$
-v_o = v_1 + v_2 + v_3 \tag{10.6.1}
-$$
-
-where *v*1 is due to the 5‑V dc voltage source, *v*2 is due to the 10 cos 2*t* V voltage source, and *v*3 is due to the 2 sin 5*t* A current source.
-
-To find *v*1, we set to zero all sources except the 5‑V dc source. We recall that at steady state, a capacitor is an open circuit to dc while an inductor is a short circuit to dc. There is an alternative way of looking at this. Because *ω* = 0, *jωL* = 0, 1∕*jωC* = ∞. Either way, the equivalent circuit is as shown in Fig. 10.14(a). By voltage division,
-
-$$
--v_1 = \frac{1}{1+4} (5) = 1 \text{ V}
-$$
- (10.6.2)
-
-To find *v*2, we set to zero both the 5‑V source and the 2 sin 5*t* current source and transform the circuit to the frequency domain.
-
-10 cos 2t 
-$$
-\Rightarrow
-$$
-  $10\underline{/0}^{\circ}$ ,  $\omega = 2 \text{ rad/s}$   
-2 H  $\Rightarrow$   $j\omega L = j4 \Omega$   
-0.1 F  $\Rightarrow$   $\frac{1}{j\omega C} = -j5 \Omega$ 
-
-The equivalent circuit is now as shown in Fig. 10.14(b). Let
-
-$$
-\mathbf{Z} = -j5 \| 4 = \frac{-j5 \times 4}{4 - j5} = 2.439 - j1.951
-$$
-
-**Figure 10.14**
-
-Solution of Example 10.6: (a) setting all sources to zero except the 5‑V dc source, (b) setting all sources to zero except the ac voltage source, (c) setting all sources to zero except the ac current source.
-
-By voltage division,
-
-oltage division,  
-\n
-$$
-\mathbf{V}_2 = \frac{1}{1 + j4 + \mathbf{Z}} (10/0^\circ) = \frac{10}{3.439 + j2.049} = 2.498 \underline{\text{ } 20.79^\circ}
-$$
-
-In the time domain,
-
-$$
-v_2 = 2.498 \cos(2t - 30.79^\circ) \tag{10.6.3}
-$$
-
-To obtain *v*3, we set the voltage sources to zero and transform what is left to the frequency domain.
-
-$$
-2 \sin 5t \Rightarrow 2(-90^\circ, \omega = 5 \text{ rad/s})
-$$
-  
-$$
-2 \text{ H} \Rightarrow j\omega L = j10 \Omega
-$$
-  
-$$
-0.1 \text{ F} \Rightarrow \frac{1}{j\omega C} = -j2 \Omega
-$$
-
-<span id="page-444-0"></span>The equivalent circuit is in Fig. 10.14(c). Let
-
-$$
-\mathbf{Z}_1 = -j2 \parallel 4 = \frac{-j2 \times 4}{4 - j2} = 0.8 - j1.6 \ \Omega
-$$
-
-By current division,
-
-$$
-\mathbf{I}_1 = \frac{j10}{j10 + 1 + \mathbf{Z}_1} (2 \angle -90^\circ) \text{ A}
-$$
-$$
-\mathbf{V}_3 = \mathbf{I}_1 \times 1 = \frac{j10}{1.8 + j8.4} (-j2) = 2.328 \angle -80^\circ \text{ V}
-$$
-
-In the time domain,
-
-$$
-v_3 = 2.33 \cos(5t - 80^\circ) = 2.33 \sin(5t + 10^\circ) \text{ V} \qquad (10.6.4)
-$$
-
-Substituting Eqs. (10.6.2) to (10.6.4) into Eq. (10.6.1), we have
-
-$$
-v_o(t) = -1 + 2.498 \cos(2t - 30.79^\circ) + 2.33 \sin(5t + 10^\circ) \text{ V}
-$$
-
-Practice Problem 10.6 Calculate *vo* in the circuit of Fig. 10.15 using the superposition theorem.
-
-**Answer:** 11.577 sin(5*t* − 81.12°) + 3.154 cos(10*t* − 86.24°) V.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/118_10.5 Source Transformation.md

@@ -1,83 +0,0 @@
-# **10.5** Source Transformation
-
-As Fig. 10.16 shows, source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the following relationship in mind:
-
-$$
-\mathbf{V}_s = \mathbf{Z}_s \mathbf{I}_s \qquad \Leftrightarrow \qquad \mathbf{I}_s = \frac{\mathbf{V}_s}{\mathbf{Z}_s} \qquad (10.1)
-$$
-
-Calculate **V***x* in the circuit of Fig. 10.17 using the method of source transformation.
-
-For Example 10.7.
-
-# **Solution:**
-
-We transform the voltage source to a current source and obtain the cir ‑ cuit in Fig. 10.18(a), where
-
-$$
-I_s = \frac{20/-90^{\circ}}{5} = 4/-90^{\circ} = -j4 \text{ A}
-$$
-
-The parallel combination of 5‑Ω resistance and (3 + *j*4) impedance gives
-
-$$
-\mathbf{Z}_1 = \frac{5(3+j4)}{8+j4} = 2.5 + j1.25 \ \Omega
-$$
-
-Converting the current source to a voltage source yields the circuit in Fig. 10.18(b), where
-
-$$
-\mathbf{V}_s = \mathbf{I}_s \mathbf{Z}_1 = -j4(2.5 + j1.25) = 5 - j10 \text{ V}
-$$
-
-**Figure 10.18** Solution of the circuit in Fig. 10.17.
-
-By voltage division,
-
-oltage division,  
-$$
-\mathbf{V}_x = \frac{10}{10 + 2.5 + j1.25 + 4 - j13} (5 - j10) = 5.519 \underline{/-28}^\circ \text{ V}
-$$
-
-Example 10.7
-
-<span id="page-446-0"></span>Practice Problem 10.7 Find **I***o* in the circuit of Fig. 10.19 using the concept of source transformation.
-
-**Answer:** 1.9727∕99.46° A.
-
-**10.6** Thevenin and Norton Equivalent Circuits
-
-Thevenin's and Norton's theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency domain version of a Thevenin equivalent circuit is depicted in Fig. 10.20, where a linear circuit is replaced by a voltage source in series with an impedance. The Norton equivalent circuit is illustrated in Fig. 10.21, where a linear cir ‑ cuit is replaced by a current source in parallel with an impedance. Keep in mind that the two equivalent circuits are related as
-
-$$
-\mathbf{V}_{\mathrm{Th}} = \mathbf{Z}_N \mathbf{I}_N, \qquad \mathbf{Z}_{\mathrm{Th}} = \mathbf{Z}_N \tag{10.2}
-$$
-
-just as in source transformation. **V**Th is the open‑circuit voltage while **I***<sup>N</sup>* is the short‑circuit current.
-
-If the circuit has sources operating at dif ferent frequencies (see Example 10.6, for example), the Thevenin or Norton equivalent circuit must be determined at each frequenc y. This leads to entirely dif ferent equivalent circuits, one for each frequenc y, not one equi valent circuit with equivalent sources and equivalent impedances.
-
-Example 10.8 Obtain the Thevenin equivalent at terminals *a*‑*b* of the circuit in Fig. 10.22.
-
-**Figure 10.22** For Example 10.8.
-
-**Figure 10.21** Norton equivalent.
-
-# **Solution:**
-
-We find **Z**Th by setting the voltage source to zero. As shown in Fig. 10.23(a), the 8 ‑Ω resistance is now in parallel with the −*j*6 reac‑ tance, so that their combination gives
-
-$$
-\mathbf{Z}_1 = -j6 \| 8 = \frac{-j6 \times 8}{8 - j6} = 2.88 - j3.84 \ \Omega
-$$
-
-Similarly, the 4 ‑Ω resistance is in parallel with the *j*12 reactance, and their combination gives
-
-$$
-\mathbf{Z}_2 = 4 || j12 = \frac{j12 \times 4}{4 + j12} = 3.6 + j1.2 \ \Omega
-$$
-
-The Thevenin impedance is the series combination of **Z**1 and **Z**2; that is,
-
-$$
-\mathbf{Z}_{\text{Th}} = \mathbf{Z}_1 + \mathbf{Z}_2 = 6.48 - j2.64 \ \Omega

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/119_10.6 Thevenin and Norton Equivalent Circuits.md

@@ -1,935 +0,0 @@
-$$
-
-To find **V**Th, consider the circuit in Fig. 10.23(b). Currents **I**1 and **I**<sup>2</sup> are obtained as
-
-$$
-\mathbf{I}_1 = \frac{120/75^{\circ}}{8 - j6} \text{ A}, \qquad \mathbf{I}_2 = \frac{120/75^{\circ}}{4 + j12} \text{ A}
-$$
-
-Applying KVL around loop *bcdeab* in Fig. 10.23(b) gives
-
-$$
-\mathbf{V}_{\mathrm{Th}} - 4\mathbf{I}_2 + (-j6)\mathbf{I}_1 = 0
-$$
-
-or
-
-$$
-\mathbf{V}_{\text{Th}} = 4\mathbf{I}_2 + j6\mathbf{I}_1 = \frac{480/75^\circ}{4 + j12} + \frac{720/75^\circ + 90^\circ}{8 - j6}
-$$
-$$
-= 37.95/3.43^\circ + 72/201.87^\circ
-$$
-$$
-= -28.936 - j24.55 = 37.95/220.31^\circ \text{ V}
-$$
-
-Practice Problem 10.8 Find the Thevenin equivalent at terminals *a*‑*b* of the circuit in Fig. 10.24.
-
-For Practice Prob. 10.8.
-
-**Answer: Z**Th = 12.4 − *j*3.2 Ω, **V**Th = 47.43∕−51.57° V.
-
-# Example 10.9
-
-Find the Thevenin equivalent of the circuit in Fig. 10.25 as seen from terminals *a*‑*b*.
-
-# **Figure 10.25**
-
-For Example 10.9.
-
-# **Solution:**
-
-To find **V**Th, we apply KCL at node 1 in Fig. 10.26(a).
-
-$$
-15 = Io + 0.5Io \Rightarrow Io = 10 A
-$$
-
-Applying KVL to the loop on the right ‑hand side in Fig. 10.26(a), we obtain
-
-$$
--I_o(2-j4) + 0.5I_o(4+j3) + V_{Th} = 0
-$$
-
-or
-
-$$
-\mathbf{V}_{\text{Th}} = 10(2 - j4) - 5(4 + j3) = -j55
-$$
-
-Thus, the Thevenin voltage is
-
-$$
-V_{\text{Th}} = 55 \angle -90^{\circ} \text{ V}
-$$
-
-**Figure 10.26** Solution of the problem in Fig. 10.25: (a) finding **V**Th, (b) finding **Z**Th.
-
-To obtain **Z**Th, we remove the independent source. Due to the presence of the dependent current source, we connect a 3-A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals *a*-*b* as shown in Fig. 10.26(b). At the node, KCL gives
-
-$$
-3 = I_o + 0.5I_o \qquad \Rightarrow \qquad I_o = 2A
-$$
-
-Applying KVL to the outer loop in Fig. 10.26(b) gives
-
-$$
-\mathbf{V}_s = \mathbf{I}_o(4 + j3 + 2 - j4) = 2(6 - j)
-$$
-
-The Thevenin impedance is
-
-$$
-\mathbf{Z}_{\text{Th}} = \frac{\mathbf{V}_s}{\mathbf{I}_s} = \frac{2(6-j)}{3} = 4 - j0.6667 \ \Omega
-$$
-
-Determine the Thevenin equivalent of the circuit in Fig. 10.27 as seen from the terminals *a*-*b*.
-
-**Answer:** 
-$$
-\mathbb{Z}_{\text{Th}} = 4.473 \underline{\smash{\big)}\, -7.64^{\circ}} \,\Omega, \, \mathbf{V}_{\text{Th}} = 11.763 \underline{\smash{\big)}\, 72.9^{\circ}} \text{ volts.}
-$$
-
-**Figure 10.27** For Practice Prob. 10.9.
-
-Obtain current **I***o* in Fig. 10.28 using Norton's theorem. Example 10.10
-
-For Example 10.10.
-
-# **Solution:**
-
-Our first objective is to find the Norton equivalent at terminals *a*-*b*. **Z***<sup>N</sup>* is found in the same way as **Z**Th. We set the sources to zero as shown in Fig. 10.29(a). As evident from the figure, the (8 − *j*2) and (10 + *j*4) impedances are short-circuited, so that
-
-$$
-\mathbf{Z}_N = 5 \ \Omega
-$$
-
-To get **I***N*, we short-circuit terminals *a*-*b* as in Fig. 10.29(b) and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,
-
-$$
--j40 + (18 + j2)\mathbf{I}_1 - (8 - j2)\mathbf{I}_2 - (10 + j4)\mathbf{I}_3 = 0 \tag{10.10.1}
-$$
-
-Solution of the circuit in Fig. 10.28: (a) finding **Z***N*, (b) finding **V***N*, (c) calculating **I***o*.
-
-For the supermesh,
-
-$$
-(13 - j2)I2 + (10 + j4)I3 - (18 + j2)I1 = 0 \t(10.10.2)
-$$
-
-At node *a*, due to the current source between meshes 2 and 3,
-
-$$
-I_3 = I_2 + 3 \tag{10.10.3}
-$$
-
-Adding Eqs. (10.10.1) and (10.10.2) gives
-
-$$
--j40 + 5\mathbf{I}_2 = 0 \qquad \Rightarrow \qquad \mathbf{I}_2 = j8
-$$
-
-From Eq. (10.10.3),
-
-$$
-\mathbf{I}_3 = \mathbf{I}_2 + 3 = 3 + j8
-$$
-
-The Norton current is
-
-$$
-\mathbf{I}_N = \mathbf{I}_3 = (3 + j8) \text{ A}
-$$
-
-Figure 10.29(c) shows the Norton equivalent circuit along with the im ‑ pedance at terminals *a*‑*b*. By current division,
-
-$$
-\mathbf{I}_o = \frac{5}{5 + 20 + j15} \mathbf{I}_N = \frac{3 + j8}{5 + j3} = 1.465 / 38.48^{\circ} \text{ A}
-$$
-
-# Practice Problem 10.10
-
-Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals *a*‑*b*. Use the equivalent to find **I***o*.
-
-# **Figure 10.30**
-
-For Practice Prob. 10.10 and Prob. 10.35.
-
-**Answer: Z***N* = 3.176 + *j*0.706 Ω, **I***N* = 8.396⧸−32.68° A, **I***o* = 1.9714⧸−2.10° A.
-
-# <span id="page-451-0"></span>**10.7** Op Amp AC Circuits
-
-The three steps stated in Section 10.1 also apply to op amp circuits, as long as the op amp is operating in the linear region. As usual, we will assume ideal op amps. (See Section 5.2.) As discussed in Chapter 5, the key to analyzing op amp circuits is to keep two important properties of an ideal op amp in mind:
-
-- 1. No current enters either of its input terminals.
-- 2. The voltage across its input terminals is zero.
-
-The following examples will illustrate these ideas.
-
-# **Figure 10.31**
-
-For Example 10.11: (a) the original circuit in the time domain, (b) its frequency domain equivalent.
-
-# **Solution:**
-
-We first transform the circuit to the frequency domain, as shown in Fig. 10.31(b), where **V***s* = 3⧸ 0°, *ω* = 1000 rad/s. Applying KCL at node 1, we obtain
-
-$$
-\frac{3/0^{\circ} - V_1}{10} = \frac{V_1}{-j5} + \frac{V_1 - 0}{10} + \frac{V_1 - V_o}{20}
-$$
-
-or
-
-$$
-6 = (5 + j4)\mathbf{V}_1 - \mathbf{V}_o \tag{10.11.1}
-$$
-
-At node 2, KCL gives
-
-$$
-\frac{\mathbf{V}_1 - 0}{10} = \frac{0 - \mathbf{V}_o}{-j10}
-$$
-
-which leads to
-
-$$
-\mathbf{V}_1 = -j\mathbf{V}_o \tag{10.11.2}
-$$
-
-Substituting Eq. (10.11.2) into Eq. (10.11.1) yields
-
-$$
-6 = -j(5 + j4)\mathbf{V}_o - \mathbf{V}_o = (3 - j5)\mathbf{V}_o
-$$
-$$
-\mathbf{V}_o = \frac{6}{3 - j5} = 1.029 \angle 59.04^\circ
-$$
-
-Hence,
-
-$$
-v_o(t) = 1.029 \cos(1000t + 59.04^{\circ}) \text{ V}
-$$
-
-Practice Problem 10.11 Find *vo* and *io* in the op amp circuit of Fig. 10.32. Let *vs*<sup>=</sup> 12 cos 5000*t* V.
-
-**Answer:** 4 sin 5,000*t* V, 400 sin 5,000*t μ*A.
-
-**Figure 10.33** For Example 10.12.
-
-Example 10.12 Compute the closed ‑loop g ain and phase shift for the circuit in Fig. 10.33. Assume that *R*1 = *R*2 = 10 kΩ, *C*1 = 2 *μ*F, *C*2 = 1 *μ*F, and *ω* = 200 rad/s.
-
-# **Solution:**
-
-The feedback and input impedances are calculated as
-
-$$
-\mathbf{Z}_f = R_2 \left\| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2}
-$$
-$$
-\mathbf{Z}_i = R_1 + \frac{1}{j\omega C_1} = \frac{1 + j\omega R_1 C_1}{j\omega C_1}
-$$
-
-Since the circuit in Fig. 10.33 is an inverting amplifier, the closed‑loop gain is given by
-
-$$
-G = \frac{V_o}{V_s} = \frac{Z_f}{Z_i} = \frac{-j\omega C_1 R_2}{(1 + j\omega R_1 C_1)(1 + j\omega R_2 C_2)}
-$$
-
-Substituting the given values of *R*1, *R*2, *C*1, *C*2, and *ω*, we obtain
-
-given values of 
-$$
-R_1
-$$
-,  $R_2$ ,  $C_1$ ,  $C_2$ , and  $\omega$ ,  
-\n
-$$
-G = \frac{-j4}{(1+j4)(1+j2)} = 0.434 / 130.6^{\circ}
-$$
-
-Thus, the closed‑loop gain is 0.434 and the phase shift is 130.6°.
-
-Practice Problem 10.12 Obtain the closed‑loop gain and phase shift for the circuit in Fig. 10.34. Let *R* = 10 kΩ, *C* = 1 *μ*F, and *ω* = 1000 rad/s.
-
-**Answer:** 1.0147, −5.6°.
-
-# <span id="page-453-0"></span>**10.8** AC Analysis Using PSpice
-
-*PSpice* affords a big relief from the tedious task of manipulating com‑ plex numbers in ac circuit analysis. The procedure for using *PSpice* for ac analysis is quite similar to that required for dc analysis. The reader should read Section D.5 in Appendix D for a review of *PSpice* concepts for ac analysis. AC circuit analysis is done in the phasor or frequency domain, and all sources must have the same frequency. Although ac analysis with *PSpice* involves using AC Sweep, our analysis in this chapter requires a single frequency *f* = *ω*∕2*π*. The out‑ put file of *PSpice* contains voltage and current phasors. If necessary, the impedances can be calculated using the voltages and currents in the output file.
-
-Obtain *vo* and *io* in the circuit of Fig. 10.35 using *PSpice*. Example 10.13
-
-# **Solution:**
-
-We first convert the sine function to cosine.
-
-$$
-8 \sin(1000t + 50^{\circ}) = 8 \cos(1000t + 50^{\circ} - 90^{\circ})
-$$
-$$
-= 8 \cos(1000t - 40^{\circ})
-$$
-
-The frequency *f* is obtained from *ω* as
-
-$$
-f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} = 159.155
-$$
- Hz
-
-The schematic for the circuit is shown in Fig. 10.36. Notice that the current‑controlled current source F1 is connected such that its current flows from node 0 to node 3 in conformity with the original circuit in Fig. 10.35. Since we only want the magnitude and phase of *vo* and *io*, we set the attributes of IPRINT and VPRINT1 each to *AC* = *yes*, *MAG* = *yes*, *PHASE* = *yes*. As a single ‑frequency analysis, we select **Analysis/ Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 159.155, and *Final Freq* = 159.155. After saving the schematic, we simulate it by selecting **Analysis/Simulate.** The output file includes the source fre‑ quency in addition to the attributes checked for the pseudocomponents IPRINT and VPRINT1,
-
-| FREQ      |           | IM(V_PRINT3) IP(V_PRINT3) |
-|-----------|-----------|---------------------------|
-| 1.592E+02 | 3.264E–03 | –3.743E+01                |
-| FREQ      | VM(3)     | VP(3)                     |
-| 1.592E+02 | 1.550E+00 | –9.518E+01                |
-
-**Figure 10.36**
-
-The schematic of the circuit in Fig. 10.35.
-
-From this output file, we obtain
-
-**V***o* = 1.55⧸−95.18° V, **I***o* = 3.264⧸−37.43° mA
-
-which are the phasors for
-
-*vo* = 1.55 cos(1000*t* − 95.18°) = 1.55 sin(1000*t* − 5.18°) V
-
-and
-
-*io* = 3.264 cos(1000*t* − 37.43°) mA
-
-For Practice Prob. 10.13.
-
-**Answer:** 3.219 cos(3,000*t* − 154.6°) V, 6.527 cos(3,000*t* − 55.12°) mA.
-
-Example 10.14
-
-Find **V**1 and **V**2 in the circuit of Fig. 10.38.
-
-# **Solution:**
-
-1. **Define.** In its present form, the problem is clearly stated. Again, we must emphasize that time spent here will save lots of time and expense later on! One thing that might have created a problem for you is that, if the reference was missing for this problem, you would then need to ask the individual assigning the problem where
-
-it is to be located. If you could not do that, then you would need to assume where it should be and then clearly state what you did and why you did it.
-
-- 2. **Present.** The given circuit is a frequency domain circuit and the unknown node voltages **V**1 and **V**2 are also frequency domain values. Clearly, we need a process to solve for these unknowns in the frequency domain.
-- 3. **Alternative.** We have two direct alternative solution techniques that we can easily use. We can do a straightforward nodal analysis approach or use *PSpice*. Since this example is in a section dedicated to using *PSpice* to solve problems, we will use *PSpice* to find **V**1 and **V**2. We can then use nodal analysis to check the answer.
-- 4. **Attempt.** The circuit in Fig. 10.35 is in the time domain, whereas the one in Fig. 10.38 is in the frequency domain. Since we are not given a particular frequency and *PSpice* requires one, we select any frequency consistent with the given impedances. For example, if we select *ω* = 1 rad/s, the corresponding frequency is *f* = *ω*∕2*π* = 0.15916 Hz. We obtain the values of the capacitance (*C* = 1∕*ωXC*) and inductances (*L* = *XL*∕*ω*). Making these changes results in the schematic in Fig. 10.39. To ease wiring, we have exchanged the positions of the voltage‑controlled current source
-
-# **Figure 10.39** Schematic for the circuit in the Fig. 10.38.
-
-G1 and the 2 + *j*2 Ω impedance. Notice that the current of G1 flows from node 1 to node 3, while the controlling voltage is across the capacitor C2, as required in Fig. 10.38. The attributes of pseudo components VPRINT1 are set as shown. As a single‑ frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.15916, and *Final Freq* = 0.15916. After saving the schematic, we select **Analysis/Simulate** to simulate the circuit. When this is done, the output file includes
-
-| FREQ      | VM(1)     | VP(1)      |
-|-----------|-----------|------------|
-| 1.592E–01 | 2.708E+00 | –5.673E+01 |
-|           |           |            |
-| FREQ      | VM(3)     | VP(3)      |
-| 1.592E-01 | 4.468E+00 | –1.026E+02 |
-
-from which we obtain,
-
-**V**1 = **2.708**⧸**−56.74° V** and **V**2 = **6.911**⧸**−80.72° V**
-
-5. **Evaluate.** One of the most important lessons to be learned is that when using programs such as *PSpice* you still need to validate the answer. There are many opportunities for making a mistake, including coming across an unknown "bug" in *PSpice* that yields incorrect results.
-
-So, how can we validate this solution? Obviously, we can rework the entire problem with nodal analysis, and perhaps using *MATLAB*, to see if we obtain the same results. There is another way we will use here: Write the nodal equations and substitute the answers obtained in the *PSpice* solution, and see if the nodal equations are satisfied.
-
-The nodal equations for this circuit are given below. Note we have substituted **V**1 = **V***x* into the dependent source.
-
-$$
--3 + \frac{\mathbf{V}_1 - 0}{1} + \frac{\mathbf{V}_1 - 0}{-j1} + \frac{\mathbf{V}_1 - \mathbf{V}_2}{2 + j2} + 0.2\mathbf{V}_1 + \frac{\mathbf{V}_1 - \mathbf{V}_2}{-j2} = 0
-$$
-  
-(1 + j + 0.25 - j0.25 + 0.2 + j0.5) $\mathbf{V}_1$   
--(0.25 - j0.25 + j0.5) $\mathbf{V}_2$  = 3  
-(1.45 + j1.25) $\mathbf{V}_1$  - (0.25 + j0.25) $\mathbf{V}_2$  = 3  
-1.9144/40.76° $\mathbf{V}_1$  - 0.3536/45° $\mathbf{V}_2$  = 3
-
-Now, to check the answer, we substitute the *PSpice* answers into this.
-
-$$
-1.9144 \underline{/40.76^{\circ}} \times 2.708 \underline{/-56.74^{\circ}} - 0.3536 \underline{/45^{\circ}} \times 6.911 \underline{/-80.72^{\circ}}
-$$
-  
-= 5.184 \underline{/-15.98^{\circ}} - 2.444 \underline{/-35.72^{\circ}}  
-= 4.984 - j1.4272 - 1.9842 + j1.4269  
-= 3 - j0.0003 [Answer checks]
-
-6. **Satisfactory?** Although we used only the equation from node 1 to check the answer, this is more than satisfactory to validate the answer from the *PSpice* solution. We can now present our work as a solution to the problem.
-
-<span id="page-457-0"></span>Obtain **V***x* and **I***x* in the circuit depicted in Fig. 10.40. Practice Problem 10.14
-
-For Practice Prob. 10.14.
-
-**Answer:** 39.37⧸ 44.78° V, 10.336⧸158° A.
-
-# **10.9** Applications
-
-The concepts learned in this chapter will be applied in later chapters to calculate electric power and determine frequency response. The con ‑ cepts are also used in analyzing coupled circuits, three‑phase circuits, ac transistor circuits, filters, oscillators, and other ac circuits. In this section, we apply the concepts to develop two practical ac circuits: the capaci ‑ tance multiplier and the sine wave oscillators.
-
-# **10.9.1** Capacitance Multiplier
-
-The op amp circuit in Fig. 10.41 is known as a *capacitance multiplier*, for reasons that will become obvious. Such a circuit is used in integrated‑ circuit technology to produce a multiple of a small physical capacitance *C* when a large capacitance is needed. The circuit in Fig. 10.41 can be used to multiply capacitance values by a factor up to 1,000. For exam‑ ple, a 10‑pF capacitor can be made to behave like a 100‑nF capacitor.
-
-In Fig. 10.41, the first op amp operates as a voltage follower, while the second one is an inverting amplifier. The voltage follower iso‑ lates the capacitance formed by the circuit from the loading imposed by the inverting amplifier. Since no current enters the input terminals of the op amp, the input current **I***i* flows through the feedback capaci‑ tor. Hence, at node 1,
-
-$$
-\mathbf{I}_{i} = \frac{\mathbf{V}_{i} - \mathbf{V}_{o}}{1/j\omega C} = j\omega C(\mathbf{V}_{i} - \mathbf{V}_{o})
-$$
-\n(10.3)
-
-Applying KCL at node 2 gives
-
-$$
-\frac{\mathbf{V}_i - \mathbf{0}}{R_1} = \frac{\mathbf{0} - \mathbf{V}_o}{R_2}
-$$
-
-or
-
-$$
-\mathbf{V}_o = -\frac{R_2}{R_1} \mathbf{V}_i \tag{10.4}
-$$
-
-Substituting Eq. (10.4) into (10.3) gives
-
-$$
-\mathbf{I}_i = j\omega C \bigg( 1 + \frac{R_2}{R_1} \bigg) \mathbf{V}_i
-$$
-
-or
-
-$$
-\frac{\mathbf{I}_i}{\mathbf{V}_i} = j\omega \left( 1 + \frac{R_2}{R_1} \right) C \tag{10.5}
-$$
-
-The input impedance is
-
-$$
-\mathbf{Z}_{i} = \frac{\mathbf{V}_{i}}{\mathbf{I}_{i}} = \frac{1}{j\omega C_{\text{eq}}}
-$$
-(10.6)
-
-where
-
-$$
-C_{\text{eq}} = \left(1 + \frac{R_2}{R_1}\right)C\tag{10.7}
-$$
-
-Thus, by a proper selection of the values of *R*1 and *R*2, the op amp circuit in Fig. 10.41 can be made to produce an effective capacitance between the input terminal and ground, which is a multiple of the physical capaci‑ tance *C*. The size of the effective capacitance is practically limited by the inverted output voltage limitation. Thus, the larger the capacitance multiplication, the smaller is the allowable input voltage to prevent the op amps from reaching saturation.
-
-A similar op amp circuit can be designed to simulate inductance. (See Prob. 10.89.) There is also an op amp circuit configuration to create a resistance multiplier.
-
-Example 10.15 Calculate *C*eq in Fig. 10.41 when *R*1 = 10 kΩ, *R*2 = 1 MΩ, and *C* = 1 nF.
-
-# **Solution:**
-
-From Eq. (10.7) *<sup>C</sup>*eq = (1 + \_\_\_ *R*2 *R*1 )*C* = (1 + 1 × <sup>106</sup> \_\_\_\_\_\_\_\_ 10 × 103 ) 1 nF = 101 nF Determine the equivalent capacitance of the op amp circuit in Fig. 10.41 if *R*1 = 10 kΩ, *R*2 = 10 MΩ, and *C* = 10 nF.
-
-**Answer:** 10 *μ*F.
-
-# **10.9.2** Oscillators
-
-We know that dc is produced by batteries. But how do we produce ac? One way is using *oscillators,* which are circuits that convert dc to ac.
-
-An oscillator is a circuit that produces an ac waveform as output when powered by a dc input.
-
-The only external source an oscillator needs is the dc power supply. Ironically, the dc power supply is usually obtained by con verting the ac supplied by the electric utility company to dc. Having gone through the trouble of conversion, one may wonder why we need to use the oscillator to convert the dc to ac again. The problem is that the ac supplied by the utility company operates at a preset frequenc y of 60 Hz in the United States (50 Hz in some other nations), whereas man y applications such as electronic circuits, communication systems, and micro wave devices require internally generated frequencies that range from 0 to 10 GHz or higher. Oscillators are used for generating these frequencies.
-
-In order for sine w ave oscillators to sustain oscillations, the y must meet the *Barkhausen criteria*:
-
-- 1. The overall gain of the oscillator must be unity or greater. Therefore, losses must be compensated for by an amplifying device.
-- 2. The overall phase shift (from input to output and back to the input) must be zero.
-
-Three common types of sine wave oscillators are phase ‑shift, twin *T*, and Wien ‑bridge oscillators. Here we consider only the Wien ‑bridge oscillator.
-
-The *Wien-bridge oscillator* is widely used for generating sinusoids in the frequency range below 1 MHz. It is an *RC* op amp circuit with only a few components, easily tunable and easy to design. As shown in Fig. 10.42, the oscillator essentially consists of a noninverting amplifier with two feedback paths: The positive feedback path to the noninverting input creates oscillations, while the ne gative feedback path to the in verting input controls the gain. If we define the impedances of the *RC* series and parallel combinations as **Z***s* and **Z***p*, then
-
-$$
-Z_s = R_1 + \frac{1}{j\omega C_1} = R_1 - \frac{j}{\omega C_1}
-$$
-(10.8)  
-$$
-Z_p = R_2 \Big| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2}
-$$
-(10.9)
-
-The feedback ratio is
-
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{\mathbf{Z}_p}{\mathbf{Z}_s + \mathbf{Z}_p}
-$$
-
-Practice Problem 10.15
-
-Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives
-
-Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives  
-\n
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{R_2}{R_2 + \left(R_1 - \frac{j}{\omega C_1}\right)(1 + j\omega R_2 C_2)}
-$$
-\n
-$$
-= \frac{\omega R_2 C_1}{\omega (R_2 C_1 + R_1 C_1 + R_2 C_2) + j(\omega^2 R_1 C_1 R_2 C_2 - 1)}
-$$
-\n(10.11)
-
-To satisfy the second Barkhausen criterion, **V**2 must be in phase with **V***o*, which implies that the ratio in Eq. (10.11) must be purely real. Hence, the imaginary part must be zero. Setting the imaginary part equal to zero gives the oscillation frequency *ωo* as
-
-$$
-\omega_o^2 R_1 C_1 R_2 C_2 - 1 = 0
-$$
-
-or
-
-$$
-\omega_o = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}\tag{10.12}
-$$
-
-In most practical applications, *R*1 = *R*2 = *R* and *C*1 = *C*2 = *C*, so that
-
-$$
-\omega_o = \frac{1}{RC} = 2\pi f_o \tag{10.13}
-$$
-
-or
-
-$$
-f_o = \frac{1}{2\pi RC}
-$$
- (10.14)
-
-Substituting Eq. (10.13) and *R*1 = *R*2 = *R*, *C*1 = *C*2 = *C* into Eq. (10.11) yields
-
-$$
-\frac{V_2}{V_o} = \frac{1}{3}
-$$
- (10.15)
-
-Thus, in order to satisfy the first Barkhausen criterion, the op amp must compensate by providing a gain of 3 or greater so that the overall gain is at least 1 or unity. We recall that for a noninverting amplifier,
-
-$$
-\frac{\mathbf{V}_o}{\mathbf{V}_2} = 1 + \frac{R_f}{R_g} = 3\tag{10.16}
-$$
-
-or
-
-$$
-R_f = 2R_g \tag{10.17}
-$$
-
-Due to the inherent delay caused by the op amp, Wien‑bridge oscil‑ lators are limited to operating in the frequency range of 1 MHz or less.
-
-# Example 10.16
-
-Design a Wien‑bridge circuit to oscillate at 100 kHz.
-
-# **Solution:**
-
-Using Eq. (10.14), we obtain the time constant of the circuit as
-
-14), we obtain the time constant of the circuit as  
-$$
-RC = \frac{1}{2\pi f_o} = \frac{1}{2\pi \times 100 \times 10^3} = 1.59 \times 10^{-6}
-$$
- (10.16.1)
-
-If we select *R* = 10 k Ω, then we can select *C* = 159 pF to satisfy Eq. (10.16.1). Since the gain must be 3, *Rf*∕*Rg* = 2. We could select *Rf* = 20 kΩ while *Rg* = 10 kΩ.
-
-<span id="page-461-0"></span>In the Wien‑bridge oscillator circuit in Fig. 10.42, let *R*1 = *R*2 = 2.5 kΩ, *C*1 = *C*2 = 1 nF. Determine the frequency *fo* of the oscillator. Practice Problem 10.16
-
-**Answer:** 63.66 kHz.
-
-# **10.10** Summary
-
-- 1. We apply nodal and mesh analysis to ac circuits by applying KCL and KVL to the phasor form of the circuits.
-- 2. In solving for the steady ‑state response of a circuit that has inde ‑ pendent sources with different frequencies, each independent source *must* be considered separately. The most natural approach to analyz‑ ing such circuits is to apply the superposition theorem. A separate phasor circuit for each frequency *must* be solved independently, and the corresponding response should be obtained in the time domain. The overall response is the sum of the time domain responses of all the individual phasor circuits.
-- 3. The concept of source transformation is also applicable in the fre ‑ quency domain.
-- 4. The Thevenin equivalent of an ac circuit consists of a voltage source **V**Th in series with the Thevenin impedance **Z**Th.
-- 5. The Norton equivalent of an ac circuit consists of a current source **I***<sup>N</sup>* in parallel with the Norton impedance **Z***N* (=**Z**Th).
-- 6. *PSpice* is a simple and powerful tool for solving ac circuit problems. It relieves us of the tedious task of working with the complex num‑ bers involved in steady‑state analysis.
-- 7. The capacitance multiplier and the ac oscillator provide two typical applications for the concepts presented in this chapter . A capaci‑ tance multiplier is an op amp circuit used in producing a multiple of a physical capacitance. An oscillator is a device that uses a dc input to generate an ac output.
-
-‒
-
-# Review Questions
-
-10 0° V ‒j1 Ω **V**<sup>o</sup> +
-
-**10.1** The voltage **V***o* across the capacitor in Fig. 10.43 is:
-
-‒
-
-For Review Question 10.1.
-
-**10.2** The value of the current **I***o* in the circuit of Fig. 10.44 is:
-
-(a) 
-$$
-4\angle 0^{\circ}
-$$
- A  
-(b)  $2.4\angle -90^{\circ}$  A  
-(c)  $0.6\angle 0^{\circ}$  A  
-(d)  $-1$  A
-
-**Figure 10.44** For Review Question 10.2.
-
-**10.3** Using nodal analysis, the value of **V***o* in the circuit of Fig. 10.45 is:
-
-(a) 
-$$
--24 \text{ V}
-$$
- (b)  $-8 \text{ V}$ 
-
-$$
-(c) 8 V \t\t (d) 24 V
-$$
-
-# **10.6** For the circuit in Fig. 10.48, the Thevenin impedance at terminals *a*‑*b* is:
-
-| (a) 1 Ω          | (b) 0.5 − j0.5 Ω |
-|------------------|------------------|
-| (c) 0.5 + j0.5 Ω | (d) 1 + j2 Ω     |
-| (e) 1 − j2 Ω     |                  |
-
-# **Figure 10.48**
-
-For Review Questions 10.6 and 10.7.
-
-- - (a) 10 cos *t* A (b) 10 sin *t* A (c) 5 cos *t* A
-
-**10.4** In the circuit of Fig. 10.46, current *i*(*t*) is:
-
-(d) 5 sin *t* A (e) 4.472 cos(*t* − 63.43°) A
-
-**Figure 10.46**
-
-**Figure 10.45**
-
-For Review Question 10.3.
-
-For Review Question 10.4.
-
-- **10.5** Refer to the circuit in Fig. 10.47 and observe that the two sources do not have the same frequency. The current *ix*(*t*) can be obtained by:
-  - (a) source transformation
-  - (b) the superposition theorem
-  - (c) *PSpice*
-
-**Figure 10.47** For Review Question 10.5. **10.7** In the circuit of Fig. 10.48, the Thevenin voltage at terminals *a*‑*b* is:
-
-(a) 
-$$
-3.535 \angle -45^{\circ} \text{ V}
-$$
- (b)  $3.535 \angle 45^{\circ} \text{ V}$   
-(c)  $7.071 \angle -45^{\circ} \text{ V}$  (d)  $7.071 \angle 45^{\circ} \text{ V}$ 
-
-**10.8** Refer to the circuit in Fig. 10.49. The Norton equivalent impedance at terminals *a*‑*b* is:
-
-| (a) −j4 Ω | (b) −j2 Ω |
-|-----------|-----------|
-| (c) j2 Ω  | (d) j4 Ω  |
-
-# **Figure 10.49**
-
-For Review Questions 10.8 and 10.9.
-
-**10.9** The Norton current at terminals *a*‑*b* in the circuit of Fig. 10.49 is:
-
-(a) 
-$$
-1/\underline{0^{\circ}}
-$$
- A  
-(b)  $1.5/\underline{-90^{\circ}}$  A  
-(c)  $1.5/90^{\circ}$  A  
-(d)  $3/90^{\circ}$  A
-
-- **10.10** *PSpice* can handle a circuit with two independent sources of different frequencies.
-  - (a) True (b) False
-
-*Answers: 10.1c, 10.2a, 10.3d, 10.4a, 10.5b, 10.6c, 10.7a, 10.8a, 10.9d, 10.10b.*
-
-# <span id="page-463-0"></span>Problems
-
-# Section 10.2 Nodal Analysis
-
-**10.1** Determine *i* in the circuit of Fig. 10.50.
-
-# **Figure 10.50**
-
-For Prob. 10.1.
-
-# **Figure 10.51**
-
-For Prob. 10.2.
-
-**10.3** Determine *vo* in the circuit of Fig. 10.52.
-
-# **Figure 10.52** For Prob. 10.3.
-
-# **Figure 10.53**
-
-For Prob. 10.4.
-
-For Prob. 10.5.
-
-**10.6** Determine **V***x* in Fig. 10.55.
-
-# **Figure 10.55** For Prob. 10.6.
-
-**10.7** Use nodal analysis to find **V** in the circuit of Fig. 10.56.
-
-# **Figure 10.56** For Prob. 10.7.
-
-**10.8** Use nodal analysis to find current *io* in the circuit of Fig. 10.57. Let *is* = 6 cos(200*t* + 15°) A.
-
-# **Figure 10.57**
-
-For Prob. 10.8.
-
-For Prob. 10.9.
-
-‒
-
-**Figure 10.59**
-
-**10.11** Using nodal analysis, find *io*(*t*) in the circuit in Fig. 10.60.
-
-For Prob. 10.11.
-
-**10.12** Using Fig. 10.61, design a problem to help other students better understand nodal analysis.
-
-**Figure 10.61** For Prob. 10.12.
-
-**10.13** Determine **V***x* in the circuit of Fig. 10.62 using any method of your choice.
-
-For Prob. 10.13.
-
-**10.14** Calculate the voltage at nodes 1 and 2 in the circuit of Fig. 10.63 using nodal analysis.
-
-# **Figure 10.63**
-
-For Prob. 10.14.
-
-**10.15** Solve for the current **I** in the circuit of Fig. 10.64 using nodal analysis.
-
-**Figure 10.64**
-
-For Prob. 10.15.
-
-# **Figure 10.65**
-
-For Prob. 10.16.
-
-**10.17** By nodal analysis, obtain current **I***o* in the circuit of Fig. 10.66.
-
-**Figure 10.66** For Prob. 10.17.
-
-**10.19** Obtain **V***o* in Fig. 10.68 using nodal analysis.
-
-**10.20** Refer to Fig. 10.69. If *vs*(*t*) = *Vm* sin *ωt* and *vo*(*t*) = *A* sin(*ωt* + *ϕ*), derive the expressions for *A* and *ϕ*.
-
-**Figure 10.69** For Prob. 10.20.
-
-**10.21** For each of the circuits in Fig. 10.70, find **V***o*∕**V***i* for *ω* = 0, *ω* → ∞, and *ω*<sup>2</sup> = 1∕*LC*.
-
-**Figure 10.70** For Prob. 10.21.
-
-**10.22** For the circuit in Fig. 10.71, determine **V***o*∕**V***s*.
-
-**Figure 10.71** For Prob. 10.22.
-
-**10.23** Using nodal analysis obtain **V** in the circuit of Fig. 10.72.
-
-**Figure 10.72**
-
-For Prob. 10.23.
-
-# Section 10.3 Mesh Analysis
-
-**10.24** Design a problem to help other students better understand mesh analysis.
-
-**10.25** Solve for *io* in Fig. 10.73 using mesh analysis.
-
-For Prob. 10.25.
-
-**10.26** Use mesh analysis to find current *io* in the circuit of Fig. 10.74.
-
-**10.29** Using Fig. 10.77, design a problem to help other students better understand mesh analysis.
-
-R3
-
-For Prob. 10.26.
-
-**10.27** Using mesh analysis, find **I**1 and **I**2 in the circuit of Fig. 10.75.
-
-jXL<sup>1</sup>
-
-**Figure 10.77** For Prob. 10.29.
-
-For Prob. 10.30.
-
-**Figure 10.76** For Prob. 10.28.
-
-**10.32** Determine **V***o* and **I***o* in the circuit of Fig. 10.80 using mesh analysis.
-
-**Figure 10.80** For Prob. 10.32.
-
-**10.33** Compute **I** in Prob. 10.15 using mesh analysis.
-
-**10.34** Use mesh analysis to find **I***o* in Fig. 10.28 (for Example 10.10).
-
-**10.35** Calculate **I***o* in Fig. 10.30 (for Practice Prob. 10.10) using mesh analysis.
-
-**10.36** Compute **V***o* in the circuit of Fig. 10.81 using mesh analysis.
-
-**Figure 10.81** For Prob. 10.36.
-
-**10.37** Use mesh analysis to find currents **I**1, **I**2, and **I**3 in the circuit of Fig. 10.82.
-
-**10.38** Using mesh analysis, obtain **I***o* in the circuit shown in Fig. 10.83.
-
-For Prob. 10.38.
-
-**10.39** Find **I**1, **I**2, **I**3, and **I***x* in the circuit of Fig. 10.84.
-
-**Figure 10.84** For Prob. 10.39.
-
-# Section 10.4 Superposition Theorem
-
-**10.40** Find *io* in the circuit shown in Fig. 10.85 using superposition.
-
-# **Figure 10.85**
-
-For Prob. 10.40.
-
-**10.41** Find *vo* for the circuit in Fig. 10.86, assuming that *is*(*t*) = 2 sin (2*t*) + 3 cos (4*t*) A.
-
-**Figure 10.86** For Prob. 10.41.
-
-**10.42** Using Fig. 10.87, design a problem to help other students better understand the superposition theorem.
-
-**Figure 10.87** For Prob. 10.42.
-
-For Prob. 10.43.
-
-**10.43** Using the superposition principle, find *ix* in the circuit of Fig. 10.88.
-
-For Prob. 10.46.
-
-**10.47** Determine *io* in the circuit of Fig. 10.92, using the superposition principle.
-
-For Prob. 10.47.
-
-**10.44** Use the superposition principle to obtain *vx* in the circuit of Fig. 10.89. Let *vs* = 50 sin 2*t* V and *is* = 12 cos(6*t* + 10°) A.
-
-**10.45** Use superposition to find *i*(*t*) in the circuit of Fig. 10.90.
-
-**Figure 10.90** For Prob. 10.45.
-
-**10.48** Find *io* in the circuit of Fig. 10.93 using superposition.
-
-**Figure 10.93**
-
-For Prob. 10.48.
-
-# Section 10.5 Source Transformation
-
-**10.49** Using source transformation, find *i* in the circuit of Fig. 10.94.
-
-For Prob. 10.49.
-
-**Figure 10.95**
-
-For Prob. 10.50.
-
-- **10.51** Use source transformation to find **I***o* in the circuit of Prob. 10.42.
-- **10.52** Use the method of source transformation to find **I***x* in the circuit of Fig. 10.96.
-
-**Figure 10.96** For Prob. 10.52.
-
-**10.53** Use the concept of source transformation to find **V***<sup>o</sup>* in the circuit of Fig. 10.97.
-
-For Prob. 10.53.
-
-**10.54** Rework Prob. 10.7 using source transformation.
-
-# Section 10.6 Thevenin and Norton Equivalent Circuits
-
-**10.55** Find the Thevenin and Norton equivalent circuits at terminals *a*‑*b* for each of the circuits in Fig. 10.98.
-
-**Figure 10.98** For Prob. 10.55.
-
-**10.56** For each of the circuits in Fig. 10.99, obtain Thevenin and Norton equivalent circuits at terminals *a*‑*b*.
-
-# **Figure 10.99**
-
-For Prob. 10.56.
-
-**10.57** Using Fig. 10.100, design a problem to help other students better understand Thevenin and Norton equivalent circuits.
-
-# **Figure 10.100** For Prob. 10.57.
-
-**10.58** For the circuit depicted in Fig. 10.101, find the Thevenin equivalent circuit at terminals *a*‑*b*.
-
-**Figure 10.101** For Prob. 10.58.
-
-**10.59** Calculate the output impedance of the circuit shown in Fig. 10.102.
-
-For Prob. 10.59.
-
-**10.60** Find the Thevenin equivalent of the circuit in Fig. 10.103 as seen from:
-
-**10.61** Find the Thevenin equivalent at terminals *a*-*b* of the circuit in Fig. 10.104.
-
-**Figure 10.104** For Prob. 10.61.
-
-**10.62** Using Thevenin's theorem, find *vo* in the circuit of Fig. 10.105.
-
-For Prob. 10.62.
-
-**10.63** Obtain the Norton equivalent of the circuit depicted in Fig. 10.106 at terminals *a*-*b*.
-
-For Prob. 10.63.
-
-**10.64** For the circuit shown in Fig. 10.107, find the Norton equivalent circuit at terminals *a*-*b*.
-
-# **Figure 10.107**
-
-For Prob. 10.64.
-
-**10.65** Using Fig. 10.108, design a problem to help other students better understand Norton's theorem.
-
-Problems **449**
-
-For Prob. 10.70.
-
-**10.71** Find *vo* in the op amp circuit of Fig. 10.114.
-
-**Figure 10.114**
-
-For Prob. 10.71.
-
-**10.72** Compute *io*(*t*) in the op amp circuit in Fig. 10.115 if *vs* = 4 cos(104 *t*) V.
-
-# **Figure 10.115**
-
-For Prob. 10.72.
-
-**10.73** If the input impedance is defined as **Z**in = **V***s*∕**I***s*, find the input impedance of the op amp circuit in Fig. 10.116 when *R*1 = 10 kΩ, *R*2 = 20 kΩ, *C*1 = 10 nF, *C*2 = 20 nF, and *ω* = 5000 rad/s.
-
-**Figure 10.116** For Prob. 10.73.
-
-**Figure 10.111** For Prob. 10.68.
-
-**Figure 10.110** For Prob. 10.67.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/120_10.7 Op Amp AC Circuits.md

@@ -1,109 +0,0 @@
-# Section 10.7 Op Amp AC Circuits
-
-**10.69** For the integrator shown in Fig. 10.112, obtain **V***o*∕**V***s*. Find *vo*(*t*) when *vs*(*t*) = **V***<sup>m</sup>* sin *ωt* and *ω* = 1∕*RC*.
-
-**Figure 10.112** For Prob. 10.69.
-
-**10.70** Using Fig. 10.113, design a problem to help other students better understand op amps in AC circuits. **10.74** Evaluate the voltage gain **A***v* = **V***o*∕**V***s* in the op amp circuit of Fig. 10.117. Find **A***<sup>v</sup>* at *ω* = 0, *ω* → ∞, *ω* = 1∕*R*1*C*1, and *ω* = 1∕*R*2*C*2.
-
-10 kΩ
-
-**V**<sup>o</sup> +
-
-20 kΩ
-
-i o
-
-‒ +
-
-+
-
-‒
-
-‒
-
-6 30° V ‒j2 kΩ
-
-‒j4 kΩ
-
-**10.75** In the op amp circuit of Fig. 10.118, find the closed‑ loop gain and phase shift of the output voltage with respect to the input voltage if *C*1 = *C*2 = 1 nF, *R*1 = *R*2 = 100 kΩ, *R*3 = 20 kΩ, *R*4 = 40 kΩ, and *ω* = 2000 rad/s.
-
-*<sup>v</sup>*<sup>s</sup> *<sup>v</sup>*<sup>o</sup>
-
-R2
-
-C2
-
-R1
-
-R4
-
-+
-
-‒
-
-R3
-
-‒ +
-
-C1
-
-+ ‒
-
-**Figure 10.118** For Prob. 10.75.
-
-**10.77** Compute the closed‑loop gain **V***o*∕**V***s* for the op amp circuit of Fig. 10.120.
-
-**10.78** Determine *vo*(*t*) in the op amp circuit in Fig. 10.121 below.
-
-**10.79** For the op amp circuit in Fig. 10.122, obtain **V***o*.
-
-For Prob. 10.79.
-
-**10.80** Obtain *vo*(*t*) for the op amp circuit in Fig. 10.123 if *vs* = 12 cos(1000*t* − 60°) V.
-
-**Figure 10.123** For Prob. 10.80.
-
-**10.81** Use *PSpice or MultiSim* to determine **V***o* in the circuit of Fig. 10.124. Assume *ω* = 1 rad/s.
-
-**Figure 10.124** For Prob. 10.81.
-
-**10.82** Solve Prob. 10.19 using *PSpice or MultiSim*.
-
-**10.83** Use *PSpice or MultiSim* to find *vo*(*t*) in the circuit of Fig. 10.125. Let *is* = 2 cos(103 *t*) A.
-
-# **Figure 10.125**
-
-For Prob. 10.83.
-
-i
-
-**10.84** Obtain **V***o* in the circuit of Fig. 10.126 using *PSpice or MultiSim*.
-
-# **Figure 10.126**
-
-For Prob. 10.84.
-
-**10.85** Using Fig. 10.127, design a problem to help other students better understand performing AC analysis with *PSpice or MultiSim*.
-
-# **Figure 10.127** For Prob. 10.85.
-
-**10.86** Use *PSpice or MultiSim* to find **V**1, **V**2, and **V**3 in the network of Fig. 10.128.
-
-**Figure 10.128** For Prob. 10.86.
-
-C
-
-<sup>+</sup> <sup>+</sup> **<sup>V</sup>**<sup>o</sup> ‒ ‒
-
-<sup>R</sup> <sup>R</sup><sup>1</sup>
-
-<sup>R</sup><sup>2</sup> <sup>R</sup>
-
-C
-
-**10.88** Use *PSpice or MultiSim* to find *vo* and *io* in the circuit of Fig. 10.130 below.
-
-**Figure 10.132** For Prob. 10.90.
-
-**V**i

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/121_10.9 Applications.md

@@ -1,72 +0,0 @@
-# Section 10.9 Applications
-
-**10.89** The op amp circuit in Fig. 10.131 is called an *inductance simulator*. Show that the input impedance is given by
-
-$$
-\mathbf{Z}_{in} = \frac{\mathbf{V}_{in}}{\mathbf{I}_{in}} = j\omega L_{eq}
-$$
-
-where
-
-$$
-L_{\text{eq}} = \frac{R_1 R_3 R_4}{R_2 C}
-$$
-
-**Figure 10.131** For Prob. 10.89.
-
-- **10.91** Consider the oscillator in Fig. 10.133.
-  - (a) Determine the oscillation frequency.
-  - (b) Obtain the minimum value of *R* for which oscillation takes place.
-
-**Figure 10.133** For Prob. 10.91.
-
-**10.87** Determine **V**1, **V**2, and **V**3 in the circuit of Fig. 10.129 using *PSpice or MultiSim*.
-
-- **10.92** The oscillator circuit in Fig. 10.134 uses an ideal op amp.
-  - (a) Calculate the minimum value of *Ro* that will cause oscillation to occur.
-  - (b) Find the frequency of oscillation.
-
-# **Figure 10.134**
-
-**10.93** Figure 10.135 shows a *Colpitts oscillator*. Show that the oscillation frequency is
-
-$$
-f_o = \frac{1}{2\pi\sqrt{LC_T}}
-$$
-
-where *CT* = *C*1*C*2∕(*C*1 + C2). Assume *Ri* ≫ *XC*<sup>2</sup> .
-
-# **Figure 10.135**
-
-A Colpitts oscillator; for Prob. 10.93.
-
-(*Hint:* Set the imaginary part of the impedance in the feedback circuit equal to zero.)
-
-**10.94** Design a Colpitts oscillator that will operate at 50 kHz.
-
-**10.95** Figure 10.136 shows a *Hartley oscillator*. Show that the frequency of oscillation is
-
-# **Figure 10.136** A Hartley oscillator; for Prob. 10.95.
-
-**10.96** Refer to the oscillator in Fig. 10.137.
-
-(a) Show that
-
-$$
-\mathbf{v} \text{ that}
-$$
-\n
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{1}{3 + j(oL/R - R/oL)}
-$$
-
-- (b) Determine the oscillation frequency *fo*.
-- (c) Obtain the relationship between *R*1 and *R*2 in order for oscillation to occur.
-
-**Figure 10.137** For Prob. 10.96.
-
-*This page intentionally left blank* 
-
-# **chapter**
-
-11

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/122_Chapter 11 - AC Power Analysis.md

@@ -1,71 +0,0 @@
-# <span id="page-477-0"></span>AC Power Analysis
-
-*Four things come not back: the spoken word; the sped arrow; time past; the neglected opportunity.*
-
-—Al Halif Omar Ibn
-
-# Enhancing Your Career
-
-# **Career in Power Systems**
-
-The discovery of the principle of an ac generator by Michael Faraday in 1831 was a major breakthrough in engineering; it provided a convenient way of generating the electric po wer that is needed in e very electronic, electrical, or electromechanical device we use now.
-
-Electric power is obtained by converting energy from sources such as fossil fuels (gas, oil, and coal), nuclear fuel (uranium), h ydro energy (water falling through a head), geothermal energy (hot water, steam), wind energy, tidal ener gy, and biomass ener gy (wastes). These various ways of generating electric po wer are studied in detail in the field of power engineering, which has become an indispensable subdiscipline of electrical engineering. An electrical engineer should be f amiliar with the analysis, generation, transmission, distrib ution, and cost of electric power.
-
-The electric po wer industry is a v ery large employer of electrical engineers. The industry includes thousands of electric utility systems ranging from large, interconnected systems serving large regional areas to small power companies serving indi vidual communities or f actories. Due to the comple xity of the po wer industry, there are numerous elec trical engineering jobs in dif ferent areas of the industry: po wer plant (generation), transmission and distribution, maintenance, research, data acquisition and flow control, and management. Since electric po wer is used e verywhere, electric utility companies are e verywhere, of fering exciting training and steady emplo yment for men and w omen in thou sands of communities throughout the world.
-
-A pole-type transformer with a lowvoltage, three-wire distribution system. © Dennis Wise/Getty Images RF
-
-# <span id="page-478-0"></span>Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Fully understand instantaneous and average power.
-- 2. Understand the basics of maximum average power.
-- 3. Understand effective or rms values and how to calculate them and to understand their importance.
-- 4. Understand apparent power (complex power), power, and reactive power and power factor.
-- 5. Understand power factor correction and the importance of its use.
-
-# **11.1** Introduction
-
-Our effort in ac circuit analysis so f ar has been focused mainly on cal culating voltage and current. Our major concern in this chapter is power analysis.
-
-Power analysis is of paramount importance. Po wer is the most important quantity in electric utilities, electronic, and communication systems, because such systems in volve transmission of po wer from one point to another . Also, e very industrial and household electrical device—every fan, motor, lamp, pressing iron, TV, personal computer has a po wer rating that indicates ho w much po wer the equipment re quires; e xceeding the po wer rating can do permanent damage to an appliance. The most common form of electric po wer is 50- or 60-Hz ac power. The choice of ac o ver dc allowed high-voltage power transmission from the power generating plant to the consumer.
-
-We will be gin by defining and deriving *instantaneous power* and *average power*. We will then introduce other power concepts. As practical applications of these concepts, we will discuss how power is measured and reconsider how electric utility companies charge their customers.
-
-# **11.2** Instantaneous and Average Power
-
-As mentioned in Chapter 2, the *instantaneous power p*(*t*) absorbed by an element is the product of the instantaneous v oltage *v*(*t*) across the ele ment and the instantaneous current *i*(*t*) through it. Assuming the passive sign convention,
-
-$$
-p(t) = v(t)i(t)
-$$
- (11.1)
-
-The instantaneous power (in watts) is the power at any instant of time.
-
-It is the rate at which an element absorbs energy.
-
-Consider the general case of instantaneous po wer absorbed by an arbitrary combination of circuit elements under sinusoidal excitation, as
-
-We can also think of the instantaneous power as the power absorbed by the element at a specific instant of time. Instantaneous quantities are denoted by lowercase letters.
-
-shown in Fig. 11.1. Let the v oltage and current at the terminals of the circuit be
-
-$$
-v(t) = V_m \cos(\omega t + \theta_v)
-$$
- (11.2a)
-
-$$
-i(t) = I_m \cos(\omega t + \theta_i)
-$$
- (11.2b)
-
-where *Vm* and *Im* are the amplitudes (or peak values), and *θv* and *θi* are the phase angles of the voltage and current, respectively. The instantaneous power absorbed by the circuit is
-
-$$
-p(t) = v(t)i(t) = V_m I_m \cos(\omega t + \theta_v) \cos(\omega t + \theta_i)
-$$
- (11.3)

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/123_11.1 Introduction.md

@@ -1,295 +0,0 @@
-We apply the trigonometric identity
-
-$$
-\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)] \tag{11.4}
-$$
-
-and express Eq. (11.3) as
-
-$$
-p(t) = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) + \frac{1}{2}V_m I_m \cos(2\omega t + \theta_v + \theta_i)
-$$
- (11.5)
-
-This shows us that the instantaneous power has two parts. The first part is constant or time independent. Its value depends on the phase dif ference between the voltage and the current. The second part is a sinusoidal function whose frequency is 2*ω*, which is twice the angular frequency of the voltage or current.
-
-A sketch of *p*(*t*) in Eq. (11.5) is shown in Fig. 11.2, where *T* = 2*π*∕*ω* is the period of v oltage or current. We observ e that *p*(*t*) is periodic, *p*(*t*) = *p*(*t* + *T*0), and has a period of *T*0 = *T*∕2, since its frequency is twice that of voltage or current. We also observe that *p*(*t*) is positive for some part of each c ycle and ne gative for the rest of the c ycle. When *p*(*t*) is positive, power is absorbed by the circuit. When *p*(*t*) is negative, power is absorbed by the source; that is, power is transferred from the circuit to the source. This is possible because of the storage elements (capacitors and inductors) in the circuit.
-
-**Figure 11.2** The instantaneous power *p*(*t*) entering a circuit.
-
-The instantaneous power changes with time and is therefore difficult to measure. The *average* power is more convenient to measure. In f act, the wattmeter, the instrument for measuring power, responds to average power.
-
-The average power, in watts, is the average of the instantaneous power over one period.
-
-# **Figure 11.1**
-
-Sinusoidal source and passive linear circuit.
-
-Thus, the average power is given by
-
-$$
-P = \frac{1}{T} \int_0^T p(t) \, dt \tag{11.6}
-$$
-
-Although Eq. (11.6) shows the averaging done over *T*, we would get the same result if we performed the integration over the actual period of *p* ( *t*) which is *T*0 = *T*∕2.
-
- Substituting *p* ( *t*) in Eq. (11.5) into Eq. (11.6) gives
-
-$$
-P = \frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) dt
-$$
-  
-+  $\frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(2\omega t + \theta_v + \theta_i) dt$   
-=  $\frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) \frac{1}{T} \int_0^T dt$   
-+  $\frac{1}{2} V_m I_m \frac{1}{T} \int_0^T \cos(2\omega t + \theta_v + \theta_i) dt$  (11.7)
-
-The first integrand is constant, and the average of a constant is the same constant. The second integrand is a sinusoid. We know that the average of a sinusoid over its period is zero because the area under the sinusoid during a positi ve half-cycle is canceled by the area under it during the following negative half-cycle. Thus, the second term in Eq. (11.7) v an ishes and the average power becomes
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)
-$$
- (11.8)
-
-Since cos( *θv*  − *θ i* ) = cos( *θi*  − *θv*), what is important is the diference in the phases of the voltage and current.
-
-Note that *p* ( *t*) is time-varying while *P* does not depend on time. To find the instantaneous power, we must necessarily have *v* ( *t*) and *i* ( *t*) in the time domain. But we can find the average power when voltage and cur rent are expressed in the time domain, as in Eq. (11.8), or when they are expressed in the frequency domain. The phasor forms of *v* ( *t*) and *i* ( *t*) in Eq. (11.2) are **V** = *V<sup>m</sup>* ⧸ *θv* and **I** = *Im* ⧸ *θ <sup>i</sup>*, respectively. *P* is calculated using Eq. (11.8) or using phasors **V** and **I**. To use phasors, we notice that
-
-$$
-\frac{1}{2}\mathbf{V}\mathbf{I}^* = \frac{1}{2}V_m I_m / \theta_v - \theta_i
-$$
-  
-= 
-$$
-\frac{1}{2}V_m I_m [\cos(\theta_v - \theta_i) + j \sin(\theta_v - \theta_i)]
-$$
- (11.9)
-
-We recognize the real part of this e xpression as the a verage power *P* according to Eq. (11.8). Thus,
-
-$$
-P = \frac{1}{2} \text{Re}[\mathbf{V} \mathbf{I}^*] = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)
-$$
- (11.10)
-
-Consider two special cases of Eq. (11.10). When *θ <sup>v</sup>* = *θ <sup>i</sup>*, the voltage and current are in phase. This implies a purely resistive circuit or resis tive load *R*, and
-
-$$
-P = \frac{1}{2} V_m I_m = \frac{1}{2} I_m^2 R = \frac{1}{2} |\mathbf{I}|^2 R
-$$
- (11.11)
-
-where ∣**I**∣ 2 = **I** × **I**\*. Equation (11.11) shows that a purely resistive circuit absorbs power at all times. When *θv* − *θ<sup>i</sup>* = ±90°, we have a purely reactive circuit, and
-
-$$
-P = \frac{1}{2} V_m I_m \cos 90^\circ = 0 \tag{11.12}
-$$
-
-showing that a purely reacti ve circuit absorbs no a verage po wer. In summary,
-
-A resistive load (R) absorbs power at all times, while a reactive load (L or C ) absorbs zero average power.
-
-$$
-v(t) = 120 \cos(377t + 45^{\circ})
-$$
- V and  $i(t) = 10 \cos(377t - 10^{\circ})$  A
-
-find the instantaneous power and the a verage po wer absorbed by the passive linear network of Fig. 11.1.
-
-# **Solution:**
-
-The instantaneous power is given by
-
-$$
-p = vi = 1200 \cos(377t + 45^{\circ}) \cos(377t - 10^{\circ})
-$$
-
-Applying the trigonometric identity
-
-$$
-\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]
-$$
-
-gives
-
-$$
-p = 600[\cos(754t + 35^\circ) + \cos 55^\circ]
-$$
-
-or
-
-$$
-p(t) = 344.2 + 600 \cos(754t + 35^\circ)
-$$
- W
-
-The average power is
-
-$$
-P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}120(10) \cos[45^\circ - (-10^\circ)]
-$$
-  
-= 600 cos 55° = 344.2 W
-
-which is the constant part of *p*(*t*) above.
-
-Calculate the instantaneous power and average power absorbed by the passive linear network of Fig. 11.1 if Practice Problem 11.1
-
-*v*(*t*) = 330 cos(10*t* + 20°) V and *i*(*t*) = 33 sin(10*t* + 60°) A
-
-**Answer:** 3.5 + 5.445 cos(20*t* − 10°) kW, 3.5 kW.
-
-Calculate the average power absorbed by an impedance **Z** = 30 − *j*70 Ω Example 11.2 when a voltage **V** = 120⧸ 0**°** is applied across it.
-
-**Solution:**
-
-The current through the impedance is
-
-ent through the impedance is  
-\n
-$$
-I = \frac{V}{Z} = \frac{120/0^{\circ}}{30 - j70} = \frac{120/0^{\circ}}{76.16/-66.8^{\circ}} = 1.576/66.8^{\circ} A
-$$
-
-Given that Example 11.1
-
-The average power is
-
-$$
-P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}(120)(1.576)\cos(0 - 66.8^\circ) = 37.24 \text{ W}
-$$
-
-# Practice Problem 11.2
-
-A current **I** = 33⧸ 30**°** A flows through an impedance **Z** = 40⧸ −22° Ω. Find the average power delivered to the impedance.
-
-**Answer:** 20.19 kW.
-
-Example 11.3 For the circuit shown in Fig. 11.3, find the average power supplied by the source and the average power absorbed by the resistor.
-
-# **Solution:**
-
-The current **I** is given by
-
-1 is given by  
-\n
-$$
-I = \frac{5/30^{\circ}}{4 - j2} = \frac{5/30^{\circ}}{4.472/-26.57^{\circ}} = 1.118/56.57^{\circ} A
-$$
-
-The average power supplied by the voltage source is
-
-$$
-P = \frac{1}{2}(5)(1.118)\cos(30^\circ - 56.57^\circ) = 2.5 \text{ W}
-$$
-
-The current through the resistor is
-
-$$
-I_R = I = 1.118 / 56.57^{\circ}
-$$
- A
-
-and the voltage across it is
-
-$$
-V_R = 4I_R = 4.472/56.57^\circ
-$$
- V
-
-The average power absorbed by the resistor is
-
-$$
-P = \frac{1}{2}(4.472)(1.118) = 2.5
-$$
- W
-
-which is the same as the average power supplied. Zero average power is absorbed by the capacitor.
-
-Practice Problem 11.3
-
-For Practice Prob. 11.3.
-
-In the circuit of Fig. 11.4, calculate the average power absorbed by the resistor and inductor. Find the average power supplied by the voltage source.
-
-**Answer:** 29.04 kW, 0 W, 29.04 kW.
-
-Determine the average power generated by each source and the average Example 11.4 power absorbed by each passive element in the circuit of Fig. 11.5(a).
-
-For Example 11.4.
-
-# **Solution:**
-
-We apply mesh analysis as shown in Fig. 11.5(b). For mesh 1,
-
-$$
-\mathbf{I}_1 = 4 \text{ A}
-$$
-
-For mesh 2,
-
-$$
-(j10 - j5)
-$$
-**I**<sub>2</sub> -  $j10$ **I**<sub>1</sub> + 60/ $\cancel{30^{\circ}}$  = 0, **I**<sub>1</sub> = 4 A
-
-or
-
-$$
-j5I_2 = -60/30^{\circ} + j40
-$$
-  $\Rightarrow$   $I_2 = -12/-60^{\circ} + 8$   
-= 10.58/79.1° A
-
-For the voltage source, the current flowing from it is **I**<sup>2</sup> = 10.58⧸ 79.1**°** A and the voltage across it is 60⧸ 30**°** V, so that the average power is
-
-$$
-P_5 = \frac{1}{2} (60)(10.58) \cos(30^\circ - 79.1^\circ) = 207.8 \text{ W}
-$$
-
-Following the passive sign convention (see Fig. 1.8), this average power is absorbed by the source, in view of the direction of **I**2 and the polarity of the voltage source. That is, the circuit is delivering average power to the voltage source.
-
-For the current source, the current through it is **I**<sup>1</sup> = 4⧸ 0**°** and the voltage across it is
-
-$$
-\mathbf{V}_1 = 20\mathbf{I}_1 + j10(\mathbf{I}_1 - \mathbf{I}_2) = 80 + j10(4 - 2 - j10.39)
-$$
-  
-= 183.9 + j20 = 184.984/(6.21° V)
-
-The average power supplied by the current source is
-
-$$
-P_1 = \frac{1}{2} (184.984)(4) \cos(6.21^\circ - 0) = -367.8 \text{ W}
-$$
-
-It is negative according to the passive sign convention, meaning that the current source is supplying power to the circuit.
-
-For the resistor, the current through it is **I**<sup>1</sup> = 4⧸ 0**°** and the voltage across it is 20**I**<sup>1</sup> = 80⧸ 0**°**, so that the power absorbed by the resistor is
-
-$$
-P_2 = \frac{1}{2} (80)(4 \pm 160 \text{ W})
-$$
-
-<span id="page-484-0"></span>For the capacitor, the current through it is **I**<sup>2</sup> = 10.58⧸ 79.1**°** and the volt age across it is −*j*5**I**<sup>2</sup> = (5⧸ −90**°**)(10.58⧸ 79.1**°**) = 52.9⧸ 79.1**°**− 90°. The average power absorbed by the capacitor is
-
-$$
-P_4 = \frac{1}{2} (52.9)(10.58)\cos(-90^\circ) = 0
-$$
-
-For the inductor, the current through it is **I**<sup>1</sup> − **I**<sup>2</sup> = 2 − *j*10.39 = 10.58⧸ −79.1**°**. The voltage across it is *j*10(**I**<sup>1</sup> − **I**2) = 105.8⧸ −79.1**°** + 90°. Hence, the average power absorbed by the inductor is
-
-$$
-P_3 = \frac{1}{2} (105.8)(10.58) \text{ as } 90^\circ = 0
-$$
-
-Notice that the inductor and the capacitor absorb zero average power and that the total power supplied by the current source equals the power absorbed by the resistor and the voltage source, or
-
-*P*<sup>1</sup> + *P*2 + *P*<sup>3</sup> + *P*<sup>4</sup> + *P*<sup>5</sup> = −367.8 + 160 + 0 + 0 + 207.8 = 0
-
-indicating that power is conserved.
-
-Calculate the average power absorbed by each of the five elements in the circuit of Fig. 11.6. Practice Problem 11.4
-
-For Practice Prob. 11.4.
-
-**Answer:** 40-V Voltage source: −60 W; *j*20-V Voltage source: −40 W; resistor: 100 W; others: 0 W.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/124_11.3 Maximum Average Power Transfer.md

@@ -1,192 +0,0 @@
-# **11.3** Maximum Average Power Transfer
-
-In Section 4.8 we solv ed the problem of maximizing the po wer delivered by a power-supplying resistive network to a load *RL*. Representing the circuit by its Thevenin equi valent, we pro ved that the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance *RL* = *R*Th. We now extend that result to ac circuits.
-
-Consider the circuit in Fig. 11.7, where an ac circuit is connected to a load **Z***L* and is represented by its Thevenin equivalent. The load is usually represented by an impedance, which may model an electric motor, an antenna, a TV, and so forth. In rectangular form, the Thevenin impedance **Z**Th and the load impedance **Z***L* are
-
-$$
-\mathbf{Z}_{\mathrm{Th}} = R_{\mathrm{Th}} + jX_{\mathrm{Th}} \tag{11.13a}
-$$
-
-$$
-\mathbf{Z}_L = R_L + jX_L \tag{11.13b}
-$$
-
-The current through the load is
-
-ough the load is  
-\n
-$$
-I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + Z_L} = \frac{V_{\text{Th}}}{(R_{\text{Th}} + jX_{\text{Th}}) + (R_L + jX_L)}
-$$
-\n(11.14)
-
-From Eq. (11.11), the average power delivered to the load is
-
-1), the average power delivered to the load is  
-\n
-$$
-P = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{|\mathbf{V}_{\text{Th}}|^2 R_L/2}{(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2}
-$$
-\n(11.15)
-
-Our objective is to adjust the load parameters *RL* and *XL* so that *P* is maximum. To do this we set *∂P*∕*∂RL* and *∂P*∕*∂XL* equal to zero. From Eq. (11.15), we obtain
-
-num. To do this we set 
-$$
-\partial P/\partial R_L
-$$
- and  $\partial P/\partial X_L$  equal to zero. From  
-1.15), we obtain  
-$$
-\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}|^2 R_L (X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}
-$$
-(11.16a)
-
-$$
-\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}| \mathbf{r}_L(X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}
-$$
-(11.16a)  
-$$
-\frac{\partial P}{\partial R_L} = \frac{|\mathbf{V}_{\text{Th}}|^2 [(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2 - 2R_L(R_{\text{Th}} + R_L)]}{2[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}
-$$
-(11.16b)
-
-Setting *∂P*∕*∂XL* to zero gives
-
-$$
-X_L = -X_{\text{Th}} \tag{11.17}
-$$
-
-and setting *∂P*∕*∂RL* to zero results in
-
-gives us the maximum average power as
-
-ero results in  
-\n
-$$
-R_L = \sqrt{R_{\text{Th}}^2 + (X_{\text{Th}} + X_L)^2}
-$$
-\n(11.18)
-
-Combining Eqs. (11.17) and (11.18) leads to the conclusion that for maximum average power transfer, **Z***L* must be selected so that *XL* = −*X*Th and *RL* = *R*Th, i.e.,
-
-$$
-Z_L = R_L + jX_L = R_{Th} - jX_{Th} = Z_{Th}^*
-$$
- (11.19)
-
-For maximum average power transfer, the load impedance **Z**L must be equal to the complex conjugate of the Thevenin impedance **Z**Th.
-
-This result is known as the *maximum average power transfer theorem* for the sinusoidal steady state. Setting *RL* = *R*Th and *XL* = −*X*Th in Eq. (11.15)  When **Z**L = **Z**\*Th, we say that the load is matched to the source.
-
-In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Eq. (11.18) by setting *XL* = 0; that is,
-
-*<sup>P</sup>*max = ∣**V**Th<sup>∣</sup>
-
-2 \_\_\_\_\_ 8*R*Th
-
-$$
-R_L = \sqrt{R_{\text{Th}}^2 + X_{\text{Th}}^2} = |\mathbf{Z}_{\text{Th}}|
-$$
- (11.21)
-
-**(11.20)**
-
-# **Figure 11.7** Finding the maximum average power
-
-transfer: (a) circuit with a load, (b) the Thevenin equivalent.
-
-This means that for maximum a verage power transfer to a purely resis tive load, the load impedance (or resistance) is equal to the magnitude of the Thevenin impedance.
-
-**Figure 11.8** For Example 11.5.
-
-Example 11.5 Determine the load impedance **Z***L* that maximizes the a verage po wer drawn from the circuit of Fig. 11.8. What is the maximum a verage power?
-
-# **Solution:**
-
-First we obtain the Thevenin equivalent at the load terminals. To get **Z**Th, consider the circuit shown in Fig. 11.9(a). We find
-
-$$
-\mathbf{Z}_{\text{Th}} = j5 + 4 || (8 - j6) = j5 + \frac{4(8 - j6)}{4 + 8 - j6} = 2.933 + j4.467 \ \Omega
-$$
-
-# **Figure 11.9**
-
-Finding the Thevenin equivalent of the circuit in Fig. 11.8.
-
-To find **V**Th, consider the circuit in Fig. 11.8(b). By voltage division,
-
-$$
-\mathbf{V}_{\text{Th}} = \frac{8 - j6}{4 + 8 - j6} (10) = 7.454 \underline{\text{/}} - 10.3^{\circ} \text{ V}
-$$
-
-The load impedance draws the maximum power from the circuit when
-
-$$
-\mathbf{Z}_L = \mathbf{Z}_{\text{Th}}^* = 2.933 - j4.467 \ \Omega
-$$
-
-According to Eq. (11.20), the maximum average power is
-
-$$
-P_{\text{max}} = \frac{|\mathbf{V}_{\text{Th}}|^2}{8R_{\text{Th}}} = \frac{(7.454)^2}{8(2.933)} = 2.368 \text{ W}
-$$
-
-For the circuit shown in Fig. 11.10, find the load impedance **Z***L* that absorbs the maximum average power. Calculate that maximum average power.
-
-8 Ω 5 Ω ‒j4 Ω j10 Ω **Z**<sup>L</sup> 12 A **Figure 11.10**
-
-Practice Problem 11.5
-
-For Practice Prob. 11.5.
-
-**Answer:** 3.415 − *j*0.7317 Ω, 51.47 W.
-
-<span id="page-487-0"></span>In the circuit in Fig. 11.11, find the value of *RL* that will absorb the Example 11.6 maximum average power. Calculate that power.
-
-# **Solution:**
-
-We first find the Thevenin equivalent at the terminals of *RL*.
-
-rst find the Thevenin equivalent at the terminals of *R<sub>L</sub>*.  
-**Z**<sub>Th</sub> = (40 − *j*30) 
-$$
-||j20 = \frac{j20(40 - j30)}{j20 + 40 - j30} = 9.412 + j22.35 Ω
-$$
-
-By voltage division,
-
-division,  
-\n
-$$
-\mathbf{V}_{\text{Th}} = \frac{j20}{j20 + 40 - j30} (150/30^{\circ}) = 72.76/134^{\circ} \text{ V}
-$$
-
-The value of *RL* that will absorb the maximum average power is
-
-$$
-V_L \text{ that will absorb the maximum average po}
-$$
-\n
-$$
-R_L = |\mathbf{Z}_{\text{Th}}| = \sqrt{9.412^2 + 22.35^2} = 24.25 \ \Omega
-$$
-
-The current through the load is
-
-t through the load is  
-\n
-$$
-I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + R_L} = \frac{72.76/134^{\circ}}{33.66 + j22.35} = 1.8/100.42^{\circ} \text{ A}
-$$
-
-The maximum average power absorbed by *RL* is
-
-$$
-P_{\text{max}} = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{1}{2} (1.8)^2 (24.25) = 39.29 \text{ W}
-$$
-
-In Fig. 11.12, the resistor *RL* is adjusted until it absorbs the maximum average power. Calculate *RL* and the maximum average power absorbed by it.
-
-**Answer:** 30 Ω, 23.06 W.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/125_11.4 Effective or RMS Value.md

@@ -1,164 +0,0 @@
-# **11.4** Effective or RMS Value
-
-The idea of *effective value* arises from the need to measure the efectiveness of a voltage or current source in delivering power to a resistive load.
-
-The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.
-
-Practice Problem 11.6
-
-For Example 11.6.
-
-Finding the effective current: (a) ac circuit, (b) dc circuit.
-
-In Fig. 11.13, the circuit in (a) is ac while that of (b) is dc. Our objecti ve is to find *I*eff that will transfer the same po wer to resistor *R* as the sinusoid *i*. The average power absorbed by the resistor in the ac circuit is
-
-$$
-P = \frac{1}{T} \int_0^T i^2 R \, dt = \frac{R}{T} \int_0^T i^2 \, dt \tag{11.22}
-$$
-
-while the power absorbed by the resistor in the dc circuit is
-
-$$
-P = I_{\text{eff}}^2 R \tag{11.23}
-$$
-
-Equating the expressions in Eqs. (11.22) and (11.23) and solving for *I*eff, we obtain
-
-$$
-I_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt}
-$$
- (11.24)
-
-The effective value of the v oltage is found in the same w ay as current; that is,
-
-$$
-V_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T v^2 dt}
-$$
- (11.25)
-
-This indicates that the effective value is the (square) *root* of the *mean* (or average) of the *square* of the periodic signal. Thus, the effective value is often known as the *root-mean-square* value, or *rms* value for short; and we write
-
-$$
-I_{\rm eff} = I_{\rm rms}, \qquad V_{\rm eff} = V_{\rm rms} \tag{11.26}
-$$
-
-For any periodic function *x*(*t*) in general, the rms value is given by
-
-$$
-X_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T x^2 dt}
-$$
- (11.27)
-
-The effective value of a periodic signal is its root mean square (rms) value.
-
-Equation 11.27 states that to find the rms value of *x*(*t*), we first find its *square x*<sup>2</sup> and then find the *mean* of that, or
-
-$$
-\frac{1}{T} \int_0^T x^2 dt
-$$
-
-and then the square *root* ( √ \_\_\_\_\_\_ ) of that mean. The rms value of a constant is the constant itself. For the sinusoid *i*(*t*) = *Im* cos *ωt*, the effective or rms value is
-
-$$
-I_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T I_m^2 \cos^2 \omega t \, dt}
-$$
-$$
-= \sqrt{\frac{I_m^2}{T} \int 0 \, T \, \frac{1}{2} (1 + \cos 2\omega t) \, dt} = \frac{I_m}{\sqrt{2}} \tag{11.28}
-$$
-
-Similarly, for *v*(*t*) = *Vm* cos *ωt*,
-
-$$
-V_{\rm rms} = \frac{V_m}{\sqrt{2}}\tag{11.29}
-$$
-
-Keep in mind that Eqs. (11.28) and (11.29) are only valid for sinusoidal signals.
-
-The average power in Eq. (11.8) can be written in terms of the rms values.
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} \cos(\theta_v - \theta_i)
-$$
-  
-=  $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i)$  (11.30)
-
-Similarly, the average power absorbed by a resistor *R* in Eq. (11.11) can be written as
-
-$$
-P = I_{\rm rms}^2 R = \frac{V_{\rm rms}^2}{R}
-$$
- (11.31)
-
-When a sinusoidal voltage or current is specified, it is often in terms of its maximum (or peak) v alue or its rms v alue, since its a verage value is zero. The power industries specify phasor magnitudes in terms of their rms values rather than peak v alues. For instance, the 110 V available at every household is the rms value of the voltage from the power company. It is convenient in power analysis to express voltage and current in their rms values. Also, analog v oltmeters and ammeters are designed to read directly the rms value of voltage and current, respectively.
-
-Determine the rms v alue of the current w aveform in Fig. 11.14. If the Example 11.7 current is passed through a 2- Ω resistor, find the average power absorbed by the resistor.
-
-# **Solution:**
-
-The period of the waveform is *T* = 4. Over a period, we can write the current waveform as
-
-$$
-i(t) = \begin{cases} 5t, & 0 < t < 2 \\ -10, & 2 < t < 4 \end{cases}
-$$
-
-The rms value is
-
-$$
-I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt} = \sqrt{\frac{1}{4} \left[ \int_0^2 (5t)^2 dt + \int_2^4 (-10)^2 dt \right]}
-$$
-$$
-= \sqrt{\frac{1}{4} \left[ 25 \frac{t^3}{3} \right]_0^2 + 100t \Big|_2^4} = \sqrt{\frac{1}{4} \left( \frac{200}{3} + 200 \right)} = 8.165 \text{ A}
-$$
-
-The power absorbed by a 2-Ω resistor is
-
-$$
-P = I_{\text{rms}}^2 R = (8.165)^2 (2) = 133.3 \text{ W}
-$$
-
-Find the rms value of the current waveform of Fig. 11.15. If the current flows through a 9-Ω resistor, calculate the average power absorbed by the resistor.
-
-**Answer:** 9.238 A, 768 W.
-
-**Figure 11.14** For Example 11.7.
-
-<span id="page-490-0"></span>For Example 11.8.
-
-Example 11.8 The waveform shown in Fig. 11.16 is a half-w ave rectified sine wave. Find the rms value and the amount of average power dissipated in a 10-Ω resistor.
-
-# **Solution:**
-
-The period of the voltage waveform is *T* = 2*π*, and
-
-$$
-v(t) = \begin{cases} 10 \sin t, & 0 < t < \pi \\ 0, & \pi < t < 2\pi \end{cases}
-$$
-
-The rms value is obtained as
-
-$$
-V_{\text{rms}}^2 = \frac{1}{T} \int_0^T v^2(t) \, dt = \frac{1}{2\pi} \left[ \int_0^{\pi} (10 \sin t)^2 \, dt + \int_{\pi}^{2\pi} 0^2 \, dt \right]
-$$
-
-But sin<sup>2</sup> *t* = \_\_1 <sup>2</sup> (1 − cos 2*t*). Hence,
-
-$$
-V_{\text{rms}}^2 = \frac{1}{2\pi} \int_0^{\pi} \frac{100}{2} (1 - \cos 2t) dt = \frac{50}{2\pi} \left( t - \frac{\sin 2t}{2} \right) \Big|_0^{\pi}
-$$
-$$
-= \frac{50}{2\pi} \left( \pi - \frac{1}{2} \frac{\sin 2\pi - 0}{2} \right) = 25, \qquad V_{\text{rms}} = 5 \text{ V}
-$$
-The average power absorbed is 
-$$
-\sqrt{\pi} \alpha \int_0^{\pi} \alpha \sqrt{1 - \frac{1}{2} \sin 2\pi} dt
-$$
-
-$$
-P = \frac{V_{\text{rms}}^2}{R} = \frac{5^2}{10} = 2.5 \text{ W}
-$$
-
-# Practice Problem 11.8
-
-Find the rms value of the full-wave rectified sine wave in Fig. 11.17. Calculate the average power dissipated in a 6-Ω resistor.
-
-**Answer:** 70.71 V, 833.3 W.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/126_11.5 Apparent Power and Power Factor.md

@@ -1,152 +0,0 @@
-# For Practice Prob. 11.8. **11.5** Apparent Power and Power Factor
-
-In Section 11.2 we saw that if the voltage and current at the terminals of a circuit are
-
-$$
-v(t) = V_m \cos(\omega t + \theta_v)
-$$
- and  $i(t) = I_m \cos(\omega t + \theta_i)$  (11.32)
-
-or, in phasor form, **V** = *Vm*<sup>⧸</sup>*θv* and **I** = *Im*⧸*θi* , the average power is
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)
-$$
- (11.33)
-
-In Section 11.4, we saw that
-
-$$
-P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) = S \cos(\theta_{\nu} - \theta_{i})
-$$
- (11.34)
-
-We have added a new term to the equation:
-
-$$
-S = V_{\rm rms} I_{\rm rms}
-$$
- (11.35)
-
-The average power is a product of two terms. The product *V*rms*I*rms is known as the *apparent power S* . The factor cos( *θv* − *θi*) is called the *power factor* (pf).
-
-The apparent power (in VA) is the product of the rms values of voltage and current.
-
-The apparent power is so called because it seems apparent that the power should be the v oltage-current product, by analogy with dc resisti ve circuits. It is measured in volt-amperes or VA to distinguish it from the average or real power, which is measured in watts. The power factor is dimensionless, since it is the ratio of the average power to the apparent power,
-
-$$
-pf = \frac{P}{S} = \cos(\theta_v - \theta_i)
-$$
- (11.36)
-
-The angle *θv* − *θi* is called the *power factor angle,* because it is the angle whose cosine is the power factor. The power factor angle is equal to the angle of the load impedance if **V** is the voltage across the load and **I** is the current through it. This is evident from the fact that
-
-$$
-Z = \frac{V}{I} = \frac{V_m/\theta_v}{I_m/\theta_i} = \frac{V_m}{I_m}/\theta_v - \theta_i
-$$
- (11.37)
-
-Alternatively, since
-
-$$
-\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.38a}
-$$
-
-and
-
-$$
-\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.38b}
-$$
-
-the impedance is
-
-$$
-Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \underbrace{\beta_{\nu} - \theta_{i}} \tag{11.39}
-$$
-
-The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance.
-
-From Eq. (11.36), the power factor may be seen as that f actor by which the apparent power must be multiplied to obtain the real or average power. The value of pf ranges between zero and unity . For a purely resisti ve load, the voltage and current are in phase, so that *θv* − *θi* = 0 and pf = 1. This implies that the apparent po wer is equal to the a verage power. For a purely reactive load, *θv* − *θi* = ±90° and pf = 0. In this case the a verage power is zero. In between these tw o extreme cases, pf is said to be *leading* or *lagging*. Leading power factor means that current leads v oltage, which implies a capaciti ve load. Lagging po wer factor means that current lags voltage, implying an inductive load. Power factor affects the From Eq. (11.36), the power factor may also be regarded as the ratio of the real power dissipated in the load to the apparent power of the load.
-
-electric bills consumers pay the electric utility companies, as we will see in Section 11.9.2.
-
-Example 11.9 A series-connected load dra ws a current *i*(*t*) = 4 cos(100 *πt* + 10°) A when the applied v oltage is *v*(*t*) = 120 cos(100*πt* − 20°) V. Find the apparent power and the power factor of the load. Determine the element values that form the series-connected load.
-
-# **Solution:**
-
-The apparent power is
-
-$$
-S = V_{\text{rms}} I_{\text{rms}} = \frac{120}{\sqrt{2}} \frac{4}{\sqrt{2}} = 240 \text{ VA}
-$$
-
-The power factor is
-
-$$
-pf = \cos(\theta_v - \theta_i) = \cos(-20^\circ - 10^\circ) = 0.866 \quad \text{(leading)}
-$$
-
-The pf is leading because the current leads the voltage. The pf may also be obtained from the load impedance.
-
-$$
-\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = \frac{120/-20^{\circ}}{4/10^{\circ}} = 30/-30^{\circ} = 25.98 - j15 \ \Omega
-$$
-\n
-$$
-\text{pf} = \cos(-30^{\circ}) = 0.866 \qquad \text{(leading)}
-$$
-
-The load impedance **Z** can be modeled by a 25.98-Ω resistor in series with a capacitor with
-
-$$
-X_C = -15 = -\frac{1}{\omega C}
-$$
-
-or
-
-$$
-C = \frac{1}{15\omega} = \frac{1}{15 \times 100\pi} = 212.2 \,\mu\text{F}
-$$
-
-| Practice Problem 11.9 | Obtain the power factor and the apparent power of a load whose                                 |
-|-----------------------|------------------------------------------------------------------------------------------------|
-|                       | impedance is Z = 60 + j40 Ω when the applied voltage<br>is v(t) =<br>155.56 cos(377t + 10°) V. |
-|                       |                                                                                                |
-
-**Answer:** 0.8321 lagging, 167.69⧸ 33.69° VA.
-
-# **Solution:**
-
-The total impedance is
-
-$$
-\mathbf{Z} = 6 + 4 \left( (-j2) \right) = 6 + \frac{-j2 \times 4}{4 - j2} = 6.8 - j1.6 = 7 \underline{\text{/} -13.24^{\circ}} \,\Omega
-$$
-
-<span id="page-493-0"></span>The power factor is
-
-$$
-pf = \cos(-13.24) = 0.9734 \text{ (leading)}
-$$
-
-since the impedance is capacitive. The rms value of the current is
-
-$$
-\mathbf{I}_{\rm rms} = \frac{\mathbf{V}_{\rm rms}}{\mathbf{Z}} = \frac{30/0^{\circ}}{7/-13.24^{\circ}} = 4.286/13.24^{\circ} \text{A}
-$$
-
-The average power supplied by the source is
-
-*P* = *V*rms*I*rmspf = (30)(4.286)0.9734 = 125 W
-
-or
-
-$$
-P = I_{\text{rms}}^2 R = (4.286)^2 (6.8) = 125 \text{ W}
-$$
-
-where *R* is the resistive part of **Z**.
-
-Calculate the power factor of the entire circuit of Fig. 11.19 as seen by the source. What is the average power supplied by the source?
-
-**Answer:** 0.936 lagging, 2.008 kW.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/127_11.6 Complex Power.md

@@ -1,227 +0,0 @@
-# For Practice Prob. 11.10. **11.6** Complex Power
-
-Considerable effort has been expended over the years to express power relations as simply as possible. Po wer engineers have coined the term *complex power,* which they use to find the total effect of parallel loads. Complex power is important in po wer analysis because it contains *all* the information pertaining to the power absorbed by a given load.
-
-Consider the ac load in Fig. 11.20. Gi ven the phasor form **V** = *Vm*<sup>⧸</sup>*θv* and **I** = *Im*⧸*θi* of voltage *v*(*t*) and current *i*(*t*), the *complex power*  **S** absorbed by the ac load is the product of the v oltage and the comple x conjugate of the current, or
-
-$$
-S = \frac{1}{2}VI^*
-$$
-\n(11.40)
-
-assuming the passi ve sign con vention (see Fig. 11.20). In terms of the rms values,
-
-$$
-S = V_{\rm rms}I_{\rm rms}^* \tag{11.41}
-$$
-
-where
-
-$$
-\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.42}
-$$
-
-and
-
-$$
-\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.43}
-$$
-
-10 Ω 8 Ω 165 0° V rms <sup>+</sup> <sup>j</sup>4 <sup>Ω</sup> ‒j6 <sup>Ω</sup> ‒ Practice Problem 11.10
-
-# **Figure 11.19**
-
-# **Figure 11.20**
-
-The voltage and current phasors associated with a load.
-
-When working with the rms values of currents or voltages, we may drop the subscript rms if no confusion will be caused by doing so.
-
-Thus, we may write Eq. (11.41) as
-
-$$
-\mathbf{S} = V_{\text{rms}} I_{\text{rms}} \underline{\beta_v - \theta_i}
-$$
-  
-=  $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i) + j V_{\text{rms}} I_{\text{rms}} \sin(\theta_v - \theta_i)$  (11.44)
-
-This equation can also be obtained from Eq. (11.9). We notice from Eq. (11.44) that the magnitude of the complex power is the apparent power; hence, the comple x power is measured in v olt-amperes (VA). Also, we notice that the angle of the complex power is the power factor angle.
-
-The complex power may be expressed in terms of the load impedance **Z**. From Eq. (11.37), the load impedance **Z** may be written as
-
-$$
-Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \frac{\beta_v - \theta_i}{\beta}
-$$
-(11.45)
-
-Thus, **V**rms = **ZI**rms. Substituting this into Eq. (11.41) gives
-
-$$
-S = I_{\rm rms}^2 Z = \frac{V_{\rm rms}^2}{Z^*} = V_{\rm rms}I_{\rm rms}^*
-$$
- (11.46)
-
-Since **Z** = *R* + *jX*, Eq. (11.46) becomes
-
-$$
-S = I_{\text{rms}}^2(R + jX) = P + jQ \tag{11.47}
-$$
-
-where *P* and *Q* are the real and imaginary parts of the complex power; that is,
-
-$$
-P = \text{Re}(\mathbf{S}) = I_{\text{rms}}^2 R \tag{11.48}
-$$
-
-$$
-Q = \text{Im}(\mathbf{S}) = I_{\text{rms}}^2 X \tag{11.49}
-$$
-
-*P* is the a verage or real po wer and it depends on the load' s resistance *R*. *Q* depends on the load' s reactance *X* and is called the *reactive* (or quadrature) power.
-
-Comparing Eq. (11.44) with Eq. (11.47), we notice that
-
-$$
-P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}), \qquad Q = V_{\text{rms}} I_{\text{rms}} \sin(\theta_{\nu} - \theta_{i}) \tag{11.50}
-$$
-
-The real power *P* is the a verage power in watts delivered to a load; it is the only useful po wer. It is the actual po wer dissipated by the load. The reactive power *Q* is a measure of the energy exchange between the source and the reactive part of the load. The unit of *Q* is the *volt-ampere reactive* (VAR) to distinguish it from the real po wer, whose unit is the w att. We know from Chapter 6 that ener gy storage elements neither dissipate nor supply power, but exchange power back and forth with the rest of the network. In the same w ay, the reactive power is being transferred back and forth between the load and the source. It represents a lossless interchange between the load and the source. Notice that:
-
-- 1. *Q* = 0 for resistive loads (unity pf).
-- 2. *Q* < 0 for capacitive loads (leading pf).
-- 3. *Q* > 0 for inductive loads (lagging pf).
-
-Thus,
-
-Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. As a complex quantity, its real part is real power P and its imaginary part is reactive power Q.
-
-Introducing the complex power enables us to obtain the real and reactive powers directly from voltage and current phasors.
-
-Complex Power = 
-$$
-\mathbf{S} = P + jQ = \mathbf{V}_{\text{rms}}(\mathbf{I}_{\text{rms}})^*
-$$
-  
-\n=  $|\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| / \theta_v - \theta_i$   
-\nApparent Power =  $S = |\mathbf{S}| = |\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| = \sqrt{P^2 + Q^2}$   
-\nReal Power =  $P = \text{Re}(\mathbf{S}) = S \cos(\theta_v - \theta_i)$   
-\nReactive Power =  $Q = \text{Im}(\mathbf{S}) = S \sin(\theta_v - \theta_i)$   
-\nPower Factor =  $\frac{P}{S} = \cos(\theta_v - \theta_i)$ 
-
-This sho ws ho w the comple x po wer contains *all* the rele vant po wer information in a given load.
-
-It is a standard practice to represent **S**, *P*, and *Q* in the form of a triangle, known as the *power triangle,* shown in Fig. 11.21(a). This is similar to the impedance triangle sho wing the relationship between **Z**, *R*, and *X*, illustrated in Fig. 11.21(b). The power triangle has four items—the apparent/complex power, real power, reactive power, and the power factor angle. Given two of these items, the other two can easily be obtained from the triangle. As shown in Fig. 11.22, when **S** lies in the first quadrant, we have an inductive load and a lagging pf. When **S** lies in the fourth quadrant, the load is capacitive and the pf is leading. It is also possible for the comple x power to lie in the second or third quadrant. This requires that the load impedance ha ve a negative resistance, which is possible with active circuits.
-
-**S** contains all power information of a load. The real part of **S** is the real power P; its imaginary part is the reactive power Q; its magnitude is the apparent power S; and the cosine of its phase angle is the power factor pf.
-
-The voltage across a load is *v*(*t*) = 60 cos(*ωt* − 10°) V and the cur - Example 11.11 rent through the element in the direction of the v oltage drop is *i*(*t*) = 1.5 cos( *ωt* + 50°) A. Find: (a) the comple x and apparent po wers, (b) the real and reacti ve powers, and (c) the po wer factor and the load impedance.
-
-# **Solution:**
-
-(a) For the rms values of the voltage and current, we write
-
-$$
-\mathbf{V}_{\rm rms} = \frac{60}{\sqrt{2}} \angle 10^{\circ}, \qquad \mathbf{I}_{\rm rms} = \frac{1.5}{\sqrt{2}} \angle 50^{\circ}
-$$
-
-The complex power is
-
-$$
-\mathbf{S} = \mathbf{V}_{\rm rms} \mathbf{I}_{\rm rms}^* = \left(\frac{60}{\sqrt{2}} \angle 10^\circ \right) \left(\frac{1.5}{\sqrt{2}} \angle 50^\circ \right) = 45 \angle 60^\circ \text{ VA}
-$$
-
-The apparent power is
-
-$$
-S = |\mathbf{S}| = 45 \text{ VA}
-$$
-
-(b) We can express the complex power in rectangular form as
-
-$$
-S = 45 \underline{/-60^{\circ}} = 45 [\cos(-60^{\circ}) + j \sin(-60^{\circ})] = 22.5 - j38.97
-$$
-
-Since **S** = *P* + *jQ*, the real power is
-
-$$
-P = 22.5 \, \mathrm{W}
-$$
-
-while the reactive power is
-
-$$
-Q = -38.97
-$$
- **VAR**
-
-(c) The power factor is
-
-$$
-pf = \cos(-60^\circ) = 0.5 \text{ (leading)}
-$$
-
-It is leading, because the reactive power is negative. The load impedance is
-
-$$
-Z = \frac{V}{I} = \frac{60/-10^{\circ}}{1.5/+50^{\circ}} = 40/-60^{\circ} \,\Omega
-$$
-
-which is a capacitive impedance.
-
-For a load, **V**rms = 110⧸ 85° V, **I**rms = 3⧸ 15° A. Determine: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance. Practice Problem 11.11
-
-> **Answer:** (a) 330 <sup>⧸</sup> 70° VA, 44 VA, (b) 112.87 W, 310.1 VAR, (c) 0.342 lagging, (12.541 + *j*34.46) Ω.
-
-Example 11.12 A load **Z** dra ws 12 kV A at a po wer f actor of 0.856 lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance.
-
-# **Solution:**
-
-(a) Given that pf = cos *θ* = 0.856, we obtain the power angle as *θ* = cos−1 0.856 = 31.13°. If the apparent power is *S* = 12,000 VA, then the average or real power is
-
-*P* = *S* cos *θ* = 12,000 × 0.856 = 10.272 kW
-
-<span id="page-497-0"></span>while the reactive power is
-
-$$
-Q = S \sin \theta = 12,000 \times 0.517 = 6.204
-$$
-kVA
-
-(b) Since the pf is lagging, the complex power is
-
-$$
-S = P + jQ = 10.272 + j6.204
-$$
- kVA
-
-From **S** = **V**rms**I**\*rms, we obtain
-
-$$
-\text{Im } \mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* \text{, we obtain}
-$$
-\n
-$$
-\mathbf{I}_{\text{rms}}^* = \frac{\mathbf{S}}{\mathbf{V}_{\text{rms}}} = \frac{10,272 + j6204}{120/0^{\circ}} = 85.6 + j51.7 \text{ A} = 100/31.13^{\circ} \text{ A}
-$$
-
-Thus **I**rms = 100⧸ −31.13° and the peak current is
-
-$$
-I_m = \sqrt{2}I_{\text{rms}} = \sqrt{2}(100) = 141.4 \text{ A}
-$$
-
-(c) The load impedance
-
-update
-
-\n
-$$
-\mathbf{Z} = \frac{\mathbf{V}_{\text{rms}}}{\mathbf{I}_{\text{rms}}} = \frac{120/0^{\circ}}{100/-31.13^{\circ}} = 1.2/31.13^{\circ} \ \Omega
-$$
-
-which is an inductive impedance.
-
-A sinusoidal source supplies 100 kVAR reactive power to load **Z** = 250⧸ −75° Ω. Determine: (a) the power factor, (b) the apparent power delivered to the load, and (c) the rms voltage.
-
-**Answer:** (a) 0.2588 leading, (b) 103.53 kVA, (c) 5.087 kV.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/128_11.7 Conservation of AC Power.md

@@ -1,207 +0,0 @@
-# **11.7** Conservation of AC Power
-
-The principle of conservation of power applies to ac circuits as well as to dc circuits (see Section 1.5).
-
-To see this, consider the circuit in Fig. 11.23(a), where two load impedances **Z**1 and **Z**2 are connected in parallel across an ac source **V**. KCL gives
-
-$$
-\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 \tag{11.52}
-$$
-
-The complex power supplied by the source is (from now on, unless otherwise specified, all values of voltages and currents will be assumed to be rms values)
-
-$$
-S = VI^* = V(I_1^* + I_2^*) = VI_1^* + VI_2^* = S_1 + S_2 \qquad (11.53)
-$$
-
-11.3 and 11.4 that average power is conserved in ac circuits.
-
-In fact, we already saw in Examples
-
-Practice Problem 11.12
-
-An ac voltage source supplied loads connected in: (a) parallel, (b) series.
-
-where **S**1 and **S**2 denote the comple x powers delivered to loads **Z**1 and **Z**2, respectively.
-
-If the loads are connected in series with the voltage source, as shown in Fig. 11.23(b), KVL yields
-
-$$
-\mathbf{V} = \mathbf{V}_1 + \mathbf{V}_2 \tag{11.54}
-$$
-
-The complex power supplied by the source is
-
-$$
-S = VI^* = (V_1 + V_2)I^* = V_1I^* + V_2I^* = S_1 + S_2 \quad (11.55)
-$$
-
-where **S**1 and **S**2 denote the comple x powers delivered to loads **Z**1 and **Z**2, respectively.
-
-We conclude from Eqs. (11.53) and (11.55) that whether the loads are connected in series or in parallel (or in general), the total po wer *supplied* by the source equals the total power *delivered* to the load. Thus, in general, for a source connected to *N* loads,
-
-$$
-S = S_1 + S_2 + \dots + S_N \tag{11.56}
-$$
-
-This means that the total comple x power in a network is the sum of the complex powers of the individual components. (This is also true of real power and reactive power, but not true of apparent power.) This expresses the principle of conservation of ac power:
-
-The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads.
-
-From this we imply that the real (or reactive) power flow from sources in a network equals the real (or reactive) power flow into the other elements in the network.
-
-Example 11.13 Figure 11.24 sho ws a load being fed by a v oltage source through a transmission line. The impedance of the line is represented by the (4 + *j*2) Ω impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load.
-
-# **Solution:**
-
-The total impedance is
-
-$$
-\mathbf{Z} = (4+j2) + (15-j10) = 19 - j8 = 20.62 \underline{\smash{\big)}\,22.83^\circ}
-$$
-  $\Omega$ 
-
-In fact, all forms of ac power are conserved: instantaneous, real, reactive, and complex.
-
-The current through the circuit is
-
-through the circuit is  
-\n
-$$
-\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{220/0^{\circ}}{20.62/-22.83^{\circ}} = 10.67/22.83^{\circ} \text{ A rms}
-$$
-
-(a) For the source, the complex power is
-
-$$
-S_s = V_s I^* = (220/0^\circ)(10.67/-22.83^\circ)
-$$
-  
-= 2347.4/-22.83° = (2163.5 - j910.8) VA
-
-From this, we obtain the real power as 2163.5 W and the reactive power as 910.8 VAR (leading).
-
-(b) For the line, the voltage is
-
-$$
-\mathbf{V}_{\text{line}} = (4 + j2)\mathbf{I} = (4.472 \underline{/ 26.57^{\circ}})(10.67 \underline{/ 22.83^{\circ}})
-$$
-$$
-= 47.72 \underline{/ 49.4^{\circ}} \text{ V rms}
-$$
-
-The complex power absorbed by the line is
-
-$$
-S_{line} = V_{line}I^* = (47.72/49.4^{\circ})(10.67/-22.83^{\circ})
-$$
-  
-= 509.2/26.57° = 455.4 + j227.7 VA
-
-or
-
-$$
-S_{\text{line}} = |I|^2 Z_{\text{line}} = (10.67)^2 (4 + j2) = 455.4 + j227.7 VA
-$$
-
-That is, the real power is 455.4 W and the reactive power is 227.76 VAR (lagging).
-
-(c) For the load, the voltage is
-
-$$
-\mathbf{V}_L = (15 - j10)\mathbf{I} = (18.03 \text{/} - 33.7^{\circ})(10.67 \text{/} 22.83^{\circ})
-$$
-  
-= 192.38 \text{/} - 10.87° V rms
-
-The complex power absorbed by the load is
-
-$$
-\mathbf{S}_L = \mathbf{V}_L \mathbf{I}^* = (192.38 \text{/} - 10.87^\circ)(10.67 \text{/} - 22.83^\circ)
-$$
-  
-= 2053 \text{/} - 33.7^\circ = (1708 - j1139) VA
-
-The real power is 1708 W and the reactive power is 1139 VAR (leading). Note that **S***s* = **S**line + **S***L*, as expected. We have used the rms values of voltages and currents.
-
-In the circuit in Fig. 11.25, the 60- Ω resistor absorbs an average power of 240 W. Find **V** and the complex power of each branch of the circuit. What is the overall complex power of the circuit? (Assume the current through the 60-Ω resistor has no phase shift.)
-
-**Answer:** 240.7 <sup>⧸</sup> 21.45° V (rms); the 20- Ω resistor: 656 VA; the (30 − *j*10) Ω impedance: 480 − *j*160 VA; the (60 + *j*20) Ω impedance: 240 + *j*80 VA; overall: 1376 − *j*80 VA.
-
-Practice Problem 11.13
-
-For Practice Prob. 11.13.
-
-For Example 11.14.
-
-# **Solution:**
-
-The current through **Z**1 is
-
-$$
-\mathbf{I}_1 = \frac{\mathbf{V}}{\mathbf{Z}_1} = \frac{120/10^{\circ}}{60/-30^{\circ}} = 2/40^{\circ} \text{ A rms}
-$$
-
-while the current through **Z**2 is
-
-$$
-I_2 = \frac{V}{Z_2} = \frac{120/10^{\circ}}{40/45^{\circ}} = 3/-35^{\circ}
-$$
- A rms
-
-The complex powers absorbed by the impedances are
-
-$$
-\mathbf{S}_1 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_1^*} = \frac{(120)^2}{60/30^\circ} = 240/-30^\circ = 207.85 - j120 \text{ VA}
-$$
-\n
-$$
-\mathbf{S}_2 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_2^*} = \frac{(120)^2}{40/-45^\circ} = 360/45^\circ = 254.6 + j254.6 \text{ VA}
-$$
-
-The total complex power is
-
-$$
-S_t = S_1 + S_2 = 462.4 + j134.6 VA
-$$
-
-(a) The total apparent power is
-
-arent power is  
-\n
-$$
-|\mathbf{S}_t| = \sqrt{462.4^2 + 134.6^2} = 481.6 \text{ VA}.
-$$
-
-(b) The total real power is
-
-$$
-P_t = \text{Re}(S_t) = 462.4 \text{ W or } P_t = P_1 + P_2.
-$$
-
-(c) The total reactive power is
-
-$$
-Q_t = \text{Im}(S_t) = 134.6 \text{ VAR or } Q_t = Q_1 + Q_2.
-$$
-
-(d) The pf = *Pt*∕∣**S***t*∣ = 462.4∕481.6 = 0.96 (lagging).
-
-We may cross check the result by finding the complex power **S***s* supplied by the source.
-
-$$
-\mathbf{I}_t = \mathbf{I}_1 + \mathbf{I}_2 = (1.532 + j1.286) + (2.457 - j1.721)
-$$
-  
-= 4 - j0.435 = 4.024 $\underline{/$ -6.21° A rms  
-$$
-\mathbf{S}_s = \mathbf{V}\mathbf{I}_t^* = (120/10°)(4.024/6.21°)
-$$
-  
-= 482.88/16.21° = 463 + j135 VA
-
-which is the same as before.
-
-Two loads connected in parallel are respectively 3 kW at a pf of 0.75 leading and 6 kW at a pf of 0.95 lagging. Calculate the pf of the com bined two loads. Find the complex power supplied by the source. Practice Problem 11.14
-
-**Answer:** 0.9972 (leading), 9 − *j*0.6742 kVA.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/129_11.8 Power Factor Correction.md

@@ -1,109 +0,0 @@
-# <span id="page-501-0"></span>**11.8** Power Factor Correction
-
-Most domestic loads (such as w ashing machines, air conditioners, and refrigerators) and industrial loads (such as induction motors) are induc tive and operate at a lo w lagging po wer factor. Although the inducti ve nature of the load cannot be changed, we can increase its power factor.
-
-The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction.
-
-Since most loads are inducti ve, as shown in Fig. 11.27(a), a load' s power factor is improved or corrected by deliberately installing a capacitor in parallel with the load, as shown in Fig. 11.27(b). The effect of adding the capacitor can be illustrated using either the power triangle or the phasor diagram of the currents in volved. Figure 11.28 sho ws the latter, where it is assumed that the circuit in Fig. 11.27(a) has a power factor of cos *θ*1, while the one in Fig. 11.27(b) has a power factor of cos *θ*2. It is evident from Fig. 11.28 that adding the capacitor has caused the phase angle between the supplied v oltage and current to reduce from *θ*1 to *θ*2, thereby increasing the power factor. We also notice from the magnitudes of the vectors in Fig. 11.28 that with the same supplied v oltage, the circuit in Fig. 11.27(a) dra ws larger current *IL* than the current *I* drawn by the circuit in Fig. 11.27(b). Power companies charge more for larger currents, because they result in increased power losses (by a squared factor, since *P* = *IL* 2 *R*). Therefore, it is beneficial to both the power company and the consumer that every effort is made to minimize current level or keep the power factor as close to unity as possible. By choosing a suitable size for the capacitor, the current can be made to be completely in phase with the voltage, implying unity power factor.
-
-Alternatively, power factor correction may be viewed as the addition of a reactive element (usually a capacitor) in parallel with the load in order to make the power factor closer to unity.
-
- An inductive load is modeled as a series combination of an inductor and a resistor.
-
-**Figure 11.27** Power factor correction: (a) original inductive load, (b) inductive load with improved power factor.
-
-We can look at the power factor correction from another perspective. Consider the power triangle in Fig. 11.29. If the original inducti ve load has apparent power *S*1, then
-
-$$
-P = S_1 \cos \theta_1
-$$
-,  $Q_1 = S_1 \sin \theta_1 = P \tan \theta_1$  (11.57)
-
-**Figure 11.29** Power triangle illustrating power factor correction.
-
-If we desire to increase the po wer factor from cos *θ*1 to cos *θ*2 without altering the real power (i.e., *P* = *S*2 cos *θ*2), then the new reactive power is
-
-$$
-Q_2 = P \tan \theta_2 \tag{11.58}
-$$
-
-The reduction in the reacti ve power is caused by the shunt capacitor; that is,
-
-$$
-Q_C = Q_1 - Q_2 = P(\tan \theta_1 - \tan \theta_2)
-$$
- (11.59)
-
-But from Eq. (11.46), *QC* = *V*<sup>2</sup> rms∕*XC* = *ωCV* <sup>2</sup> rms. The value of the required shunt capacitance *C* is determined as
-
-$$
-C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{P(\tan \theta_1 - \tan \theta_2)}{\omega V_{\text{rms}}^2}
-$$
- (11.60)
-
-Note that the real po wer *P* dissipated by the load is not af fected by the power factor correction because the a verage power due to the capaci tance is zero.
-
-Although the most common situation in practice is that of an inductive load, it is also possible that the load is capaciti ve; that is, the load is operating at a leading power factor. In this case, an inductor should be connected across the load for po wer factor correction. The required shunt inductance *L* can be calculated from
-
-$$
-Q_L = \frac{V_{\text{rms}}^2}{X_L} = \frac{V_{\text{rms}}^2}{\omega L} \qquad \Rightarrow \qquad L = \frac{V_{\text{rms}}^2}{\omega Q_L} \tag{11.61}
-$$
-
-where *QL* = *Q*<sup>1</sup> − *Q*2, the dif ference between the ne w and old reacti ve powers.
-
-Example 11.15 When connected to a 120-V (rms), 60-Hz po wer line, a load absorbs 4 kW at a lagging po wer factor of 0.8. Find the v alue of capacitance necessary to raise the pf to 0.95.
-
-# **Solution:**
-
-If the pf = 0.8, then
-
-cos *θ*1 = 0.8 ⇒ *θ*1 = 36.87°
-
-where *θ*1 is the phase difference between voltage and current. We obtain the apparent power from the real power and the pf as
-
-$$
-S_1 = \frac{P}{\cos \theta_1} = \frac{4000}{0.8} = 5000 \text{ VA}
-$$
-
-The reactive power is
-
-$$
-Q_1 = S_1 \sin \theta = 5000 \sin 36.87 = 3000
-$$
- VAR
-
-When the pf is raised to 0.95,
-
-$$
-\cos \theta_2 = 0.95 \qquad \Rightarrow \qquad \theta_2 = 18.19^\circ
-$$
-
-<span id="page-503-0"></span>The real power *P* has not changed. But the apparent power has changed; its new value is
-
-$$
-S_2 = \frac{P}{\cos \theta_2} = \frac{4000}{0.95} = 4210.5 \text{ VA}
-$$
-
-The new reactive power is
-
-$$
-Q_2 = S_2 \sin \theta_2 = 1314.4 \text{ VAR}
-$$
-
-The difference between the new and old reactive powers is due to the parallel addition of the capacitor to the load. The reactive power due to the capacitor is
-
-$$
-Q_C = Q_1 - Q_2 = 3000 - 1314.4 = 1685.6 \text{ VAR}
-$$
-
-and
-
-$$
-C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{1685.6}{2\pi \times 60 \times 120^2} = 310.5 \,\mu\text{F}
-$$
-
-*Note:* Capacitors are normally purchased for voltages they expect to see. In this case, the maximum voltage this capacitor will see is about 170 V peak. We would suggest purchasing a capacitor with a voltage rating equal to, say, 200 V.
-
-Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is sup plied by a 220-V (rms), 60-Hz line.
-
-**Answer:** 7.673 mF.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/130_11.9 Applications.md

@@ -1,176 +0,0 @@
-# **11.9** Applications
-
-In this section, we consider two important application areas: how power is measured and how electric utility companies determine the cost of electricity consumption.
-
-# **11.9.1** Power Measurement
-
-The average power absorbed by a load is measured by an instrument called the *wattmeter*.
-
-The wattmeter is the instrument used for measuring the average power.
-
-Figure 11.30 sho ws a w attmeter that consists essentially of tw o coils: the current coil and the voltage coil. A current coil with very low impedance (ideally zero) is connected in series with the load (Fig. 11.31) and responds to the load current. The voltage coil with very high impedance (ideally infinite) is connected in parallel with the load as shown in Fig. 11.31 and responds to the load v oltage. The current coil acts lik e a short circuit because of its low impedance; the voltage coil behaves like
-
-Reactive power is measured by an instrument called the varmeter. The varmeter is often connected to the load in the same way as the wattmeter.
-
- Some wattmeters do not have coils; the wattmeter considered here is the electromagnetic type.
-
-Practice Problem 11.15
-
-A wattmeter.
-
-**Figure 11.31** The wattmeter connected to the load.
-
-an open circuit because of its high impedance. As a result, the presence of the w attmeter does not disturb the circuit or ha ve an ef fect on the power measurement.
-
-When the two coils are energized, the mechanical inertia of the moving system produces a deflection angle that is proportional to the average value of the product *v*(*t*)*i*(*t*). If the current and voltage of the load are *v*(*t*) = *Vm* cos(*ωt* + *θv*) and *i*(*t*) = *Im* cos(*ωt* + *θi*), their corresponding rms phasors are
-
-$$
-\mathbf{V}_{\rm rms} = \frac{V_m}{\sqrt{2}} \underline{\theta_v} \quad \text{and} \quad \mathbf{I}_{\rm rms} = \frac{I_m}{\sqrt{2}} \underline{\theta_i} \quad (11.62)
-$$
-
-and the wattmeter measures the average power given by
-
-$$
-P = |\mathbf{V}_{\text{rms}}||\mathbf{I}_{\text{rms}}| \cos(\theta_{\nu} - \theta_{i}) = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) \qquad (11.63)
-$$
-
-As shown in Fig. 11.31, each wattmeter coil has two terminals with one marked ±. To ensure upscale deflection, the ± terminal of the current coil is toward the source, while the ± terminal of the voltage coil is connected to the same line as the current coil. Re versing both coil connections still results in upscale deflection. However, reversing one coil and not the other results in downscale deflection and no wattmeter reading.
-
-# **Solution:**
-
-1. **Define.** The problem is clearly defined. Interestingly, this is a problem where the student could actually v alidate the results by doing the problem in the laboratory with a real wattmeter.
-
-- 2. **Present.** This problem consists of finding the average power delivered to a load by an external source with a series impedance.
-- 3. **Alternative.** This is a straightforward circuit problem where all we need to do is find the magnitude and phase of the current through the load and the magnitude and the phase of the voltage across the load. These quantities could also be found by using *PSpice*, which we will use as a check.
-- 4. **Attempt.** In Fig. 11.32, the w attmeter reads the average power absorbed by the (8 − *j*6) Ω impedance because the current coil is in series with the impedance while the v oltage coil is in parallel with it. The current through the circuit is
-
-The impedance while the voltage coil is in  
-int through the circuit is  
-$$
-I_{\rm rms} = \frac{150/0^{\circ}}{(12 + j10) + (8 - j6)} = \frac{150}{20 + j4} A
-$$
-
-The voltage across the (8 − *j*6) Ω impedance is
-
-$$
-\mathbf{V}_{\rm rms} = \mathbf{I}_{\rm rms}(8 - j6) = \frac{150(8 - j6)}{20 + j4} \text{ V}
-$$
-
-The complex power is
-
-$$
-\mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* = \frac{150(8 - j6)}{20 + j4} \cdot \frac{150}{20 - j4} = \frac{150^2(8 - j6)}{20^2 + 4^2}
-$$
-$$
-= 423.7 - j324.6 \text{ VA}
-$$
-
-The wattmeter reads
-
-$$
-P = \text{Re}(S) = 432.7 \text{ W}
-$$
-
-5. **Evaluate.** We can check our results by using *PSpice*.
-
-To check our answer , all we need is the magnitude of the current (7.354 A) flowing through the load resistor:
-
-$$
-P = (I_L)^2 R = (7.354)^2 8 = 432.7 \text{ W}
-$$
-
-As expected, the answer does check!
-
-6. **Satisfactory?** We ha ve satisf actorily solv ed the problem and the results can now be presented as a solution to the problem.
-
-For Practice Prob. 11.16.
-
-**Answer:** 1.437 kW.
-
-# **11.9.2** Electricity Consumption Cost
-
-In Section 1.7, we considered a simplified model of the way the cost of electricity consumption is determined. But the concept of po wer factor was not included in the calculations. Now we consider the importance of power factor in electricity consumption cost.
-
-Loads with low power factors are costly to serve because they require large currents, as explained in Section 11.8. The ideal situation would be to draw minimum current from a supply so that *S* = *P*, *Q* = 0, and pf = 1. A load with nonzero *Q* means that energy flows back and forth between the load and the source, gi ving rise to additional po wer losses. In vie w of this, power companies often encourage their customers to have power factors as close to unity as possible and penalize some customers who do not improve their load power factors.
-
-Utility companies divide their customers into categories: as residential (domestic), commercial, and industrial, or as small po wer, medium power, and large power. They have different rate structures for each category. The amount of energy consumed in units of kilowatt-hours (kWh) is measured using a kilo watt-hour meter installed at the customer' s premises.
-
-Although utility companies use different methods for charging customers, the tarif f or char ge to a consumer is often tw o-part. The first part is fixed and corresponds to the cost of generation, transmission, and distribution of electricity to meet the load requirements of the con sumers. This part of the tarif f is generally e xpressed as a certain price
-
-per kW of maximum demand. Or it may be based on kVA of maximum demand, to account for the power factor (pf) of the consumer. A pf penalty charge may be imposed on the consumer whereby a certain percentage of kW or kVA maximum demand is charged for every 0.01 fall in pf below a prescribed value, say 0.85 or 0.9. On the other hand, a pf credit may be given for every 0.01 that the pf exceeds the prescribed value.
-
-The second part is proportional to the ener gy consumed in kWh; i t may be in graded form, for example, the first 100 kWh at 16 cents/kWh, the next 200 kWh at 10 cents/kWh and so forth. Thus, the bill is determined based on the following equation:
-
-Total Cost = Fixed Cost + Cost of Energy **(11.64)**
-
-A manufacturing industry consumes 200 MWh in one month. If the Example 11.17 maximum demand is 1,600 kW, calculate the electricity bill based on the following two-part rate:
-
-Demand charge: \$5.00 per month per kW of billing demand. Energy charge: 8 cents per kWh for the first 50,000 kWh, 5 cents per kWh for the remaining energy.
-
-# **Solution:**
-
-The demand charge is
-
-\$5.00 × 1,600 = \$8,000 **(11.17.1)**
-
-The energy charge for the first 50,000 kWh is
-
-$$
-$0.08 \times 50,000 = $4,000 \tag{11.17.2}
-$$
-
-The remaining energy is 200,000 kWh− 50,000 kWh = 150,000 kWh, and the corresponding energy charge is
-
-\$0.05 × 150,000 = \$7,500 **(11.17.3)**
-
-Adding the results of Eqs. (11.17.1) to (11.17.3) gives
-
-Total bill for the month = \$8,000 + \$4,000 + \$7,500 = \$19,500
-
-It may appear that the cost of electricity is too high. But this is often a small fraction of the overall cost of production of the goods manufactured or the selling price of the finished product.
-
-The monthly reading of a paper mill's meter is as follows:
-
-Maximum demand: 48,000 kW Energy consumed: 750 MWh
-
-Using the two-part rate in Example 11.17, calculate the monthly bill for the paper mill.
-
-**Answer:** \$279,000.
-
-Practice Problem 11.17
-
-<span id="page-508-0"></span>Example 11.18 A 300-kW load supplied at 13 kV (rms) operates 520 hours a month at 80 percent power factor. Calculate the average cost per month based on this simplified tariff:
-
-Energy charge: 6 cents per kWh
-
-Power-factor penalty: 0.1 percent of energy charge for every 0.01 that pf falls below 0.85.
-
-Power-factor credit: 0.1 percent of energy charge for every 0.01 that pf exceeds 0.85.
-
-# **Solution:**
-
-The energy consumed is
-
-$$
-W = 300 \text{ kW} \times 520 \text{ h} = 156,000 \text{ kWh}
-$$
-
-The operating power factor pf = 80% = 0.8 is 5 × 0.01 below the prescribed power factor of 0.85. Since there is 0.1 percent energy charge for every 0.01, there is a power-factor penalty charge of 0.5 percent. This amounts to an energy charge of
-
-$$
-\Delta W = 156,000 \times \frac{5 \times 0.1}{100} = 780 \text{ kWh}
-$$
-
-The total energy is
-
-$$
-W_t = W + \Delta W = 156,000 + 780 = 156,780 \text{ kWh}
-$$
-
-The cost per month is given by
-
-Cost = 6 cents × *Wt* = \$0.06 × 156,780 = \$9,406.80
-
-An 500-kW induction furnace at 0.88 power factor operates 20 hours per day for 26 days in a month. Determine the electricity bill per month based on the tariff in Example 11.18. Practice Problem 11.18
-
-**Answer:** \$15,553.20.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/131_11.10 Summary.md

@@ -1,115 +0,0 @@
-# **11.10** Summary
-
-1. The instantaneous power absorbed by an element is the product of the element's terminal voltage and the current through the element:
-
-$$
-p = vi.
-$$
-
-2. Average or real po wer *P* (in w atts) is the a verage of instantaneous power *p*:
-
-$$
-P = \frac{1}{T} \int_0^T p \, dt
-$$
-
- If *v*(*t*) = *Vm*cos(*ωt* + *θv*) and *i*(*t*) = *Im* cos (*ωt* + *θ<sup>i</sup>* ), then *V*rms = *Vm*∕√ 2 , *I*rms = *Im*∕ √ \_\_ 2 , and
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) = V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i)
-$$
-
- Inductors and capacitors absorb no a verage power, while the a verage power absorbed by a resistor is (1∕2)*Im* <sup>2</sup>*R* = *I*rms <sup>2</sup> *R*.
-
-- 3. Maximum average power is transferred to a load when the load impedance is the comple x conjug ate of the Thevenin impedance as seen from the load terminals, **Z***L* = *Z*Th \* .
-- 4. The effective value of a periodic signal x(*t*) is its root-mean-square (rms) value.
-
-$$
-X_{\rm eff} = X_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T x^2 dt}
-$$
-
- For a sinusoid, the ef fective or rms v alue is its amplitude di vided by √ \_\_ 2 .
-
-5. The power factor is the cosine of the phase difference between voltage and current:
-
-$$
-\mathrm{pf} = \cos(\theta_v - \theta_i)
-$$
-
- It is also the cosine of the angle of the load impedance or the ratio of real power to apparent power. The pf is lagging if the current lags voltage (inductive load) and is leading when the current leads v oltage (capacitive load).
-
-6. Apparent power *S* (in VA) is the product of the rms values of voltage and current:
-
-$$
-S = V_{\rm rms} I_{\rm rms}
-$$
-
- It is also given by *S* = ∣**S**∣ = √ \_\_\_\_\_\_\_ *P*2 + *Q*<sup>2</sup> , where *P* is the real power and *Q* is reactive power.
-
-7. Reactive power (in VAR) is:
-
-$$
-Q = \frac{1}{2} V_m I_m \sin(\theta_v - \theta_i) = V_{\text{rms}} I_{\text{rms}} \sin(\theta_v - \theta_i)
-$$
-
-8. Complex power **S** (in VA) is the product of the rms v oltage phasor and the complex conjugate of the rms current phasor . It is also the complex sum of real power *P* and reactive power *Q*.
-
-$$
-\mathbf{S} = \mathbf{V}_{\rm rms} \, \mathbf{I}_{\rm rms}^* = V_{\rm rms} I_{\rm rms} / \theta_{\rm v} - \theta_{\rm i} = P + jQ
-$$
-
-Also,
-
-$$
-\mathbf{S} = I_{\text{rms}}^2 \, \mathbf{Z} = \frac{V_{\text{rms}}^2}{\mathbf{Z}^*}
-$$
-
-- 9. The total comple x power in a netw ork is the sum of the comple x powers of the individual components. Total real power and reactive power are also, respectively, the sums of the individual real powers and the reactive powers, but the total apparent po wer is not calcu lated by the process.
-- 10. Power factor correction is necessary for economic reasons; it is the process of impro ving the po wer f actor of a load by reducing the overall reactive power.
-- 11. The wattmeter is the instrument for measuring the average power. Energy consumed is measured with a kilowatt-hour meter.
-
-# <span id="page-510-0"></span>Review Questions
-
-**11.1** The average power absorbed by an inductor is zero.
-
-(a) True (b) False
-
-**11.2** The Thevenin impedance of a network seen from the load terminals is 80 + *j*55 Ω. For maximum power transfer, the load impedance must be:
-
-| (a) −80 + j55 Ω | (b) −80 − j55 Ω |  |  |
-|-----------------|-----------------|--|--|
-| (c) 80 − j55 Ω  | (d) 80 + j55 Ω  |  |  |
-
-**11.3** The amplitude of the voltage available in the 60-Hz, 120-V power outlet in your home is:
-
-| (a) 110 V | (b) 120 V |  |
-|-----------|-----------|--|
-| (c) 170 V | (d) 210 V |  |
-
-- **11.4** If the load impedance is 20 − *j*20, the power factor is
-  - (a) ⧸−45*°* (b) 0 (c) 1
-  - (d) 0.7071 (e) none of these
-- **11.5** A quantity that contains all the power information in a given load is the
-  - (a) power factor (b) apparent power (c) average power (d) reactive power
-  - (e) complex power
-- **11.6** Reactive power is measured in:
-  - (a) watts (b) VA
-  - (c) VAR (d) none of these
-- **11.7** In the power triangle shown in Fig. 11.34(a), the reactive power is:
-  - (a) 1000 VAR leading (b) 1000 VAR lagging (c) 866 VAR leading (d) 866 VAR lagging
-
-# **Figure 11.34**
-
-For Review Questions 11.7 and 11.8.
-
-**11.8** For the power triangle in Fig. 11.34(b), the apparent power is: (a) 2000 VA (b) 1000 VAR
-
-| (c) 866 VAR | (d) 500 VAR |
-|-------------|-------------|
-
-**11.9** A source is connected to three loads **Z**1, **Z**2, and **Z**<sup>3</sup> in parallel. Which of these is not true?
-
-| (a) P = P1 + P2 + P3 | (b) Q = Q1 + Q2 + Q3 |
-|----------------------|----------------------|
-| (c) S = S1 + S2 + S3 | (d) S = S1 + S2 + S3 |
-
-**11.10** The instrument for measuring average power is the:

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/132_Review Questions.md

@@ -1,6 +0,0 @@
-| (a) voltmeter           | (b) ammeter  |
-|-------------------------|--------------|
-| (c) wattmeter           | (d) varmeter |
-| (e) kilowatt-hour meter |              |
-
-*Answers: 11.1a, 11.2c, 11.3c, 11.4d, 11.5e, 11.6c, 11.7d, 11.8a, 11.9c, 11.10c.*

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/133_Problems.md

@@ -1,437 +0,0 @@
-# Problems1
-
-# Section 11.2 Instantaneous and Average Power
-
-- **11.1** If *v*(*t*) = 160 cos 50*t* V and *i*(*t*) = −33 sin (50*t* − 30°)A, calculate the instantaneous power and the average power.
-- **11.2** Given the circuit in Fig. 11.35, find the average power supplied or absorbed by each element.
-
-- **11.3** A load consists of a 60-Ω resistor in parallel with a 90-*μ*F capacitor. If the load is connected to a voltage source *vs*(*t*) = 160 cos 2000*t*, find the average power delivered to the load.
-- **11.4** Using Fig. 11.36, design a problem to help other students better understand instantaneous and average power.
-
-<sup>1</sup>Starting with problem 11.22, unless otherwise specified, assume that all values of currents and voltages are rms.
-
-## Problems **489**
-
-**11.5** ssuming that *vs* = 8 cos(2*t* − 40°) V in the circuit of Fig. 11.37, find the average power delivered to each of the passive elements.
-
-**Figure 11.37** For Prob. 11.5.
-
-**11.6** For the circuit in Fig. 11.38, *is* = 6 cos 103 *t* A. Find the average power absorbed by the 50-Ω resistor.
-
-**Figure 11.38** For Prob. 11.6.
-
-**11.7** Given the circuit of Fig. 11.39, find the average power absorbed by the 10-Ω resistor.
-
-For Prob. 11.7.
-
-**11.8** In the circuit of Fig. 11.40, determine the average power absorbed by the 40-Ω resistor.
-
-**11.9** For the op amp circuit in Fig. 11.41, **V***s* = 2⧸30*° V*. Find the average power absorbed by the 20-kΩ resistor.
-
-- For Prob. 11.9.
-- **11.10** In the op amp circuit in Fig. 11.42, find the total average power absorbed by the resistors.
-
-**Figure 11.42** For Prob. 11.10.
-
-**11.11** For the network in Fig. 11.43, assume that the port impedance is
-
-$$
-\mathbf{Z}_{ab} = \frac{R}{\sqrt{1 + \omega^2 R^2 C^2}} \sqrt{-\tan^{-1} \omega RC}
-$$
-
- Find the average power consumed by the network when *R* = 10 kΩ, *C* = 200 nF, and *i* = 33 sin(377*t* + 22°) mA.
-
-# **Figure 11.43** For Prob. 11.11.
-
-# Section 11.3 Maximum Average Power Transfer
-
-**11.12** For the circuit shown in Fig. 11.44, determine the load impedance *ZL* for maximum power transfer (to *ZL*). Calculate the maximum power absorbed by the load.
-
-# **Figure 11.44**
-
-For Prob. 11.12.
-
-- **11.13** The Thevenin impedance of a source is **Z**Th = 120 + *j*60 Ω, while the peak Thevenin voltage is **V**Th = 165 + *j*0 V. Determine the maximum available average power from the source.
-- **11.14** Using Fig. 11.45, design a problem to help other students better understand maximum average power transfer to a load *Z*.
-
-**Figure 11.45** For Prob. 11.14.
-
-**11.15** In the circuit of Fig. 11.46, find the value of **Z***L* that will absorb the maximum power and the value of the maximum power.
-
-For Prob. 11.15.
-
-**11.16** For the circuit in Fig. 11.47, find the value of **Z***L* that will receive the maximum power from the circuit. Then calculate the power delivered to the load **Z***L*.
-
-**11.17** Calculate the value of **Z***L* in the circuit of Fig. 11.48 in order for **Z***L* to receive maximum average power. ‒j3 <sup>Ω</sup> What is the maximum average power received by **Z***L*? <sup>4</sup><sup>Ω</sup>
-
-**Figure 11.48**
-
-For Prob. 11.17.
-
-**11.18** Find the value of **Z***L* in the circuit of Fig. 11.49 for maximum power transfer.
-
-**Figure 11.49**
-
-For Prob. 11.18.
-
-**11.19** The variable resistor *R* in the circuit of Fig. 11.50 is adjusted until it absorbs the maximum average power. Find *R* and the maximum average power absorbed.
-
-**Figure 11.50** For Prob. 11.19.
-
-**11.20** The load resistance *RL* in Fig. 11.51 is adjusted until it absorbs the maximum average power. Calculate the value of *RL* and the maximum average power.
-
-For Prob. 11.20.
-
-**11.21** Assuming that the load impedance is to be purely resistive, what load should be connected to terminals *a*-*b* of the circuits in Fig. 11.52 so that the maximum power is transferred to the load?
-
-**Figure 11.52** For Prob. 11.21.
-
-# Section 11.4 Effective or RMS Value
-
-**11.22** Find the rms value of the offset sine wave shown in Fig. 11.53.
-
-For Prob. 11.22.
-
-**11.23** Using Fig. 11.54, design a problem to help other students better understand how to find the rms value of a waveshape.
-
-- For Prob. 11.23.
-- **11.24** Determine the rms value of the waveform in Fig. 11.55.
-
-**Figure 11.55** For Prob. 11.24.
-
-**Figure 11.56** For Prob. 11.25.
-
-**11.26** Find the effective value of the voltage waveform in Fig. 11.57.
-
-For Prob. 11.26.
-
-**11.27** Calculate the rms value of the current waveform of Fig. 11.58.
-
-**11.28** Find the rms value of the voltage waveform of Fig. 11.59 as well as the average power absorbed by a 2-Ω resistor when the voltage is applied across the resistor.
-
-**11.29** Calculate the effective value of the current waveform in Fig. 11.60 and the average power delivered to a 12-Ω resistor when the current runs through the resistor.
-
-**Figure 11.60** For Prob. 11.29.
-
-**11.30** Compute the rms value of the waveform depicted in Fig. 11.61.
-
-**Figure 11.61**
-
-For Prob. 11.30.
-
-**11.31** Find the rms value of the signal shown in Fig. 11.62.
-
-**Figure 11.62** For Prob. 11.31.
-
-**11.32** Obtain the rms value of the current waveform shown in Fig. 11.63.
-
-**Figure 11.63** For Prob. 11.32.
-
-**11.33** Determine the rms value for the waveform in Fig. 11.64.
-
-**11.34** Find the effective value of *f*(*t*) defined in Fig. 11.65.
-
-**Figure 11.65** For Prob. 11.34.
-
-**11.35** One cycle of a periodic voltage waveform is depicted in Fig. 11.66. Find the effective value of the voltage. Note that the cycle starts at *t* = 0 and ends at *t* = 6 s.
-
-**Figure 11.66** For Prob. 11.35.
-
-**11.36** Calculate the rms value for each of the following functions:
-
-(a) *i*(*t*) = 10 A (b) *v*(*t*) = 4 + 3 cos 5*t* V (c) *i*(*t*) = 8 − 6 sin 2*t* A (d) *v*(*t*) = 5 sin *t* + 4 cos*t* V
-
-**11.37** Design a problem to help other students better understand how to determine the rms value of the sum of multiple currents.
-
-# Section 11.5 Apparent Power and Power Factor
-
-**11.38** For the power system in Fig. 11.67, find: (a) the average power, (b) the reactive power, (c) the power factor. Note that 440 V is an rms value.
-
-**11.39** An ac motor with impedance **Z***L* = 2 + *j*1.2 Ω is supplied by a 220-V, 60-Hz source. (a) Find pf, *P*, and *Q*. (b) Determine the capacitor required to be connected in parallel with the motor so that the power factor is corrected to unity.
-
-**11.40** Design a problem to help other students better understand apparent power and power factor.
-
-**11.41** Obtain the power factor for each of the circuits in Fig. 11.68. Specify each power factor as leading or lagging.
-
-# **Figure 11.68**
-
-For Prob. 11.41.
-
-# Section 11.6 Complex Power
-
-- **11.42** A 110-V rms, 60-Hz source is applied to a load impedance **Z**. The apparent power entering the load is 120 VA at a power factor of 0.707 lagging.
-  - (a) Calculate the complex power.
-  - (b) Find the rms current supplied to the load.
-  - (c) Determine **Z**. (d) Assuming that **Z** = *R* + *jωL*, find the values of *R* and *L*.
-
-**11.43** Design a problem to help other students understand complex power.
-
-**11.44** Find the complex power delivered by *vs* to the network in Fig. 11.69. Let *vs* = 100 cos 2000*t* V.
-
-# **Figure 11.69** For Prob. 11.44.
-
-**11.45** The voltage across a load and the current through it are given by
-
-$$
-v(t) = 20 + 60 \cos 100t
-$$
-  
-$$
-i(t) = 1 - 0.5 \sin 100t
-$$
- A
-
-Find:
-
-(a) the rms values of the voltage and of the current (b) the average power dissipated in the load
-
-**11.46** For the following voltage and current phasors, calculate the complex power, apparent power, real power, and reactive power. Specify whether the pf is leading or lagging.
-
-(a) 
-$$
-V = 220/30^{\circ}
-$$
- V rms,  $I = 0.5/60^{\circ}$  A rms
-
-(b) 
-$$
-V = 250 \div 10^{\circ} \text{ V rms}
-$$
-,
-
-$$
-I = 6.2 \angle -25^{\circ}
-$$
- A rms
-
-(c) 
-$$
-V = 120/0
-$$
-° V rms,  $I = 2.4/15$ ° A rms
-
-- (d) **V** = 160⧸45*°* V rms, **I** = 8.5⧸90*°* A rms
-- **11.47** For each of the following cases, find the complex power, the average power, and the reactive power:
-
-(a) 
-$$
-v(t) = 169.7 \sin(377t + 45^\circ)
-$$
- V,  
- $i(t) = 5.657 \sin(377t)$  A
-
-(b) 
-$$
-v(t) = 339.4 \sin (377t + 90^\circ)
-$$
- V,
-
-$$
-i(t) = 5.657 \sin (377t + 45^{\circ}) \text{ A}
-$$
-
-(c) V = 
-$$
-900/90^{\circ}
-$$
- V rms, Z =  $75/45^{\circ}$   $\Omega$ 
-
-(d) 
-$$
-I = 100/60^{\circ}
-$$
- A rms,  $Z = 50/60^{\circ}$   $\Omega$ 
-
-- **11.48** Determine the complex power for the following cases:
-  - (a) *P* = 269 W, *Q* = 150 VAR (capacitive)
-  - (b) *Q* = 2000 VAR, pf = 0.9 (leading)
-
-(c) 
-$$
-S = 600
-$$
- VA,  $Q = 450$  VAR (inductive)
-
-(d) *V*rms = 220 V, *P* = 1 kW,
-
-∣**Z**∣ = 40 Ω (inductive)
-
-**11.49** Find the complex power for the following cases:
-
-- (a) *P* = 4 kW, pf = 0.86 (lagging) (b) *S* = 2 kVA, *P* = 1.6 kW (capacitive) (c) **V**rms = 208⧸20*°* V, **I**rms = 6.5⧸−50*°* A (d) **V**rms = 120⧸30*°* V, **Z** = 40 + *j*60 Ω
-- **11.50** Obtain the overall impedance for the following cases:
-  - (a) *P* = 1000 W, pf = 0.8 (leading), *V*rms = 220 V
-  - (b) *P* = 1500 W, *Q* = 2000 VAR (inductive), *I*rms = 12 A
-
-(c) 
-$$
-S = 4500/60^{\circ}
-$$
- VA,  $V = 120/45^{\circ}$  V
-
-- **11.51** For the entire circuit in Fig. 11.70, calculate:
-  - (a) the power factor
-  - (b) the average power delivered by the source
-  - (c) the reactive power
-  - (d) the apparent power
-  - (e) the complex power
-
-# **Figure 11.70**
-
-For Prob. 11.51.
-
-- **11.52** In the circuit of Fig. 11.71, device *A* receives 2 kW at 0.8 pf lagging, device *B* receives 3 kVA at 0.4 pf leading, while device *C* is inductive and consumes 1 kW and receives 500 VAR.
-  - (a) Determine the power factor of the entire system.
-  - (b) Find **I** given that **V***s* = 120⧸45*°* V rms.
-
-**Figure 11.71** For Prob. 11.52.
-
-- **11.53** In the circuit of Fig. 11.72, load *A* receives 4 kVA at 0.8 pf leading. Load *B* receives 2.4 kVA at 0.6 pf lagging. Box *C* is an inductive load that consumes 1 kW and receives 500 VAR.
-  - (a) Determine **I**.
-  - (b) Calculate the power factor of the combination.
-
-**Figure 11.72** For Prob. 11.53.
-
-# Section 11.7 Conservation of AC Power
-
-**11.54** For the network in Fig. 11.73, find the complex power absorbed by each element.
-
-# **Figure 11.73**
-
-For Prob. 11.54.
-
-**Figure 11.74**
-
-For Prob. 11.55.
-
-**11.56** Obtain the complex power delivered by the source in the circuit of Fig. 11.75.
-
-**Figure 11.75**
-
-For Prob. 11.56.
-
-For Prob. 11.57.
-
-**11.58** Obtain the complex power delivered to the 10-kΩ resistor in Fig. 11.77 below.
-
-- **11.59** Calculate the reactive power in the inductor and capacitor in the circuit of Fig. 11.78.
-- **Figure 11.78** 100 Ω 100 Ω j100 Ω 100 0° mA ‒j200 Ω 20 0°<sup>V</sup> <sup>+</sup> ‒
-
-For Prob. 11.59.
-
-**11.60** For the circuit in Fig. 11.79, find **V***o* and the input power factor.
-
-**Figure 11.79** For Prob. 11.60.
-
-**11.61** Given the circuit in Fig. 11.80, find *Io* and the overall
-
-**Figure 11.80** For Prob. 11.61.
-
-**11.62** For the circuit in Fig. 11.81, find **V***s*.
-
-complex power supplied.
-
-**11.63** Find **I***o* in the circuit of Fig. 11.82.
-
-**11.64** Determine **I***s* in the circuit of Fig. 11.83, if the voltage source supplies 6 kW and 1.2 kVAR (leading).
-
-# **Figure 11.83**
-
-For Prob. 11.64.
-
-**11.65** In the op amp circuit of Fig. 11.84, *vs* = 4 cos 104 *t* V. Find the average power delivered to the 50-kΩ resistor.
-
-For Prob. 11.65.
-
-**11.66** Obtain the average power absorbed by the 10-Ω resistor in the op amp circuit in Fig. 11.85.
-
-**Figure 11.85**
-
-For Prob. 11.66.
-
-**11.67** For the op amp circuit in Fig. 11.86, calculate:
-
-- (a) the complex power delivered by the voltage source
-- (b) the average power dissipated in the 10-Ω resistor
-
-**11.68** Compute the complex power supplied by the current source in the series *RLC* circuit in Fig. 11.87.
-
-# **Figure 11.87**
-
-For Prob. 11.68.
-
-# Section 11.8 Power Factor Correction
-
-**11.69** Refer to the circuit shown in Fig. 11.88.
-
-- (a) What is the power factor?
-- (b) What is the average power dissipated?
-- (c) What is the value of the capacitance that will give a unity power factor when connected to the load?
-
-# **Figure 11.88**
-
-For Prob. 11.69.
-
-- **11.70** Design a problem to help other students better understand power factor correction.
-  - **11.71** Three loads are connected in parallel to a 120⧸0*°* V rms source. Load 1 absorbs 60 kVAR at pf = 0.85 lagging, load 2 absorbs 90 kW and 50 kVAR leading, and load 3 absorbs 100 kW at pf = 1. (a) Find the equivalent impedance. (b) Calculate the power factor of the parallel combination. (c) Determine the current supplied by the source.
-  - **11.72** Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW at a 0.707 pf lagging. Determine: (a) the pf of the second load, (b) the parallel element required to correct the pf to 0.9 lagging for the two loads.
-  - **11.73** A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find:
-    - (a) the apparent power
-    - (b) the current drawn from the supply
-    - (c) the kVAR rating and capacitance required to improve the power factor to 0.96 lagging
-    - (d) the current drawn from the supply under the new power-factor conditions
-
-For Prob. 11.67.
-
-- **11.74** A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in Fig. 11.89.
-  - (a) Find the power factor of the parallel combination.
-  - (b) Calculate the value of the capacitance connected in parallel that will raise the power factor to unity.
-
-**Figure 11.89** For Prob. 11.74.
-
-- **11.75** Consider the power system shown in Fig. 11.90. Calculate:
-  - (a) the total complex power
-  - (b) the power factor
-  - (c) the parallel capacitance necessary to establish a unity power factor
-
-For Prob. 11.77.
-
-**11.78** Find the wattmeter reading of the circuit shown in Fig. 11.93.
-
-**11.79** Determine the wattmeter reading of the circuit in
-
-# **Figure 11.94**
-
-For Prob. 11.79.
-
-For Prob. 11.80.
-
-- **11.80** The circuit of Fig. 11.95 portrays a wattmeter connected into an ac network.
-  - (a) Find the magnitude of the load current.
-  - (b) Calculate the wattmeter reading.
-
-For Prob. 11.78.
-
-Fig. 11.94.
-
-# Section 11.9 Applications
-
-**11.76** Obtain the wattmeter reading of the circuit in Fig. 11.91.
-
-For Prob. 11.76.
-
-- <span id="page-520-0"></span>**11.81** Design a problem to help other students better understand how to correct power factor to values other than unity.
-- **11.82** A 240-V rms 60-Hz source supplies a parallel combination of a 5-kW heater and a 30-kVA induction motor whose power factor is 0.82. Determine:
-  - (a) the system apparent power
-  - (b) the system reactive power
-  - (c) the kVA rating of a capacitor required to adjust the system power factor to 0.9 lagging
-  - (d) the value of the capacitor required
-- **11.83** Oscilloscope measurements indicate that the peak voltage across a load and the peak current through it are, respectively, 210⧸ 60*°* V and 8⧸25*°* A. Determine:
-  - (a) the real power
-  - (b) the apparent power
-  - (c) the reactive power
-  - (d) the power factor
-
-**11.84** A consumer has an annual consumption of 1200 MWh with a maximum demand of 2.4 MVA. The maximum demand charge is \$30 per kVA per annum, and the energy charge per kWh is 4 cents.
-
-(a) Determine the annual cost of energy.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/134_Comprehensive Problems.md

@@ -1,71 +0,0 @@
-# Comprehensive Problems
-
-- **11.86** A transmitter delivers maximum power to an antenna when the antenna is adjusted to represent a load of 75-Ω resistance in series with an inductance of 4 *μ*H. If the transmitter operates at 4.12 MHz, find its internal impedance.
-- **11.87** In a TV transmitter, a series circuit has an impedance of 3 kΩ and a total current of 50 mA. If the voltage across the resistor is 80 V, what is the power factor of the circuit?
-- **11.88** A certain electronic circuit is connected to a 110-V ac line. The root-mean-square value of the current drawn is 2 A, with a phase angle of 55°.
-  - (a) Find the true power drawn by the circuit.
-  - (b) Calculate the apparent power.
-
-**11.89** An industrial heater has a nameplate that reads:
-
-- 210 V 60 Hz 12 kVA 0.78 pf lagging Determine:
-  - (a) the apparent and the complex power
-  - (b) the impedance of the heater
-- **11.90** A 2000-kW turbine-generator of 0.85 power factor operates at the rated load. An additional load of 300 kW at 0.8 power factor is added.What kVAR \*
-
-- (b) Calculate the charge per kWh with a flat-rate tariff if the revenue to the utility company is to remain the same as for the two-part tariff.
-- **11.85** A regular household system of a single-phase threewire circuit allows the operation of both 120-V and 240-V, 60-Hz appliances. The household circuit is modeled as shown in Fig. 11.96. Calculate:
-  - (a) the currents **I**1, **I**2, and I*<sup>n</sup>*
-  - (b) the total complex power supplied
-  - (c) the overall power factor of the circuit
-
-# **Figure 11.96**
-
-For Prob. 11.85.
-
-of capacitors is required to operate the turbinegenerator but keep it from being overloaded?
-
-**11.91** The nameplate of an electric motor has the following information:
-
-> Line voltage: 220 V rms Line current: 15 A rms Line frequency: 60 Hz Power: 2700 W
-
- Determine the power factor (lagging) of the motor. Find the value of the capacitance *C* that must be connected across the motor to raise the pf to unity.
-
-- **11.92** As shown in Fig. 11.97, a 550-V feeder line supplies an industrial plant consisting of a motor drawing 90 kW at 0.8 pf (inductive), a capacitor with a rating of 20 kVAR, and lighting drawing 10 kW.
-  - (a) Calculate the total reactive power and apparent power absorbed by the plant.
-  - (b) Determine the overall pf.
-  - (c) Find the magnitude of the current in the feeder line.
-
-\* An asterisk indicates a challenging problem.
-
-- **11.93** A factory has the following four major loads:
-  - A motor rated at 5 hp, 0.8 pf lagging (1hp = 0.7457 kW).
-  - A heater rated at 1.2 kW, 1.0 pf.
-  - Ten 120-W lightbulbs.
-  - A synchronous motor rated at 1.6 kVAR, 0.6 pf leading.
-  - (a) Calculate the total real and reactive power.
-  - (b) Find the overall power factor.
-
-- (a) Calculate the cost of capacitors needed.
-- (b) Find the savings in substation capacity released.
-- (c) Are capacitors economical for releasing the amount of substation capacity?
-
-- (a) At what frequency is maximum power transferred to the speaker?
-- (b) If *Vs* = 4.6 V rms, how much power is delivered to the speaker at that frequency?
-
-# **Figure 11.98**
-
-For Prob. 11.95.
-
-- **1.96** A power amplifier has an output impedance of 40 + *j*8 Ω. It produces a no-load output voltage of 146 V at 300 Hz.
-  - (a) Determine the impedance of the load that achieves maximum power transfer.
-  - (b) Calculate the load power under this matching condition.
-- **1.97** A power transmission system is modeled as shown in Fig. 11.99. If **V***s* = 440⧸0*°* rms, find the average power absorbed by the load.
-
-**Figure 11.99** For Prob. 11.97.
-
-*This page intentionally left blank* 
-
-# **chapter**
-
-# 12

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/135_Chapter 12 - Three-Phase Circuits.md

@@ -1,31 +0,0 @@
-# <span id="page-523-0"></span>Three-Phase Circuits
-
-*He who cannot forgive others breaks the bridge over which he must pass himself.*
-
-—G. Herbert
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.e), "an ability to identify, formulate, and solve engineering problems."**
-
-Developing and enhancing your "ability to identify , formulate, and solve engineering problems" is a primary focus of te xtbook. Following our six-step problem-solving process is the best w ay to practice this skill. Our recommendation is that you use this process whene ver possible. You may be pleased to learn that this process w orks well for nonengineering courses.
-
-# **ABET EC 2000 criteria (f), "an understanding of professional and ethical responsibility."**
-
-"An understanding of professional and ethical responsibility" is required of every engineer. To some e xtent, this understanding is v ery personal for each of us. Let us identify some pointers to help you de velop this understanding. One of my f avorite examples is that an engineer has the responsibility to answer what I call the "unasked question." For instance, assume that you own a car that has a problem with the transmission. In the process of selling that car, the prospective buyer asks you if there is a problem in the right-front wheel bearing. You answer no. However, as an engineer, you are required to inform the buyer that there is a problem with the transmission without being asked.
-
-Your responsibility both professionally and ethically is to perform in a manner that does not harm those around you and to whom you are responsible. Clearly, developing this capability will tak e time and ma turity on your part. I recommend practicing this by looking for profes sional and ethical components in your day-to-day activities.
-
-Photo by Charles Alexander
-
-# <span id="page-524-0"></span>Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Understand balanced three-phase voltages.
-- 2. Analyze balanced wye-wye circuits.
-- 3. Understand and analyze balanced wye-delta circuits.
-- 4. Analyze balanced delta-delta circuits.
-- 5. Understand and analyze balanced delta-wye circuits.
-- 6. Explain and analyze power in balanced three-phase circuits.
-- 7. Analyze unbalanced three-phase circuits.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/136_12.1 Introduction.md

@@ -1,23 +0,0 @@
-# **12.1** Introduction
-
-So far in this text, we have dealt with single-phase circuits. A singlephase ac power system consists of a generator connected through a pair of wires (a transmission line) to a load. Figure 12.1(a) depicts a singlephase two-wire system, where *Vp* is the rms magnitude of the source voltage and *ϕ* is the phase. What is more common in practice is a singlephase three-wire system, shown in Fig. 12.1(b). It contains two identical sources (equal magnitude and the same phase) that are connected to two loads by two outer wires and the neutral. For example, the normal household system is a single-phase three-wire system because the terminal voltages have the same magnitude and the same phase. Such a system allows the connection of both 120- and 240-V appliances.
-
-Historical note: Thomas Edison invented a three-wire system, using three wires instead of four.
-
-Circuits or systems in which the ac sources operate at the same fre quency but different phases are known as *polyphase*. Figure 12.2 shows a two-phase three-wire system, and Fig. 12.3 sho ws a three-phase fourwire system. As distinct from a single-phase system, a two-phase system is produced by a generator consisting of tw o coils placed perpendicular to each other so that the voltage generated by one lags the other by 90°. By the same token, a three-phase system is produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120°. Because the three-phase system is by far the most pre valent and most economical polyphase system, discus sion in this chapter is mainly on three-phase systems.
-
-**Figure 12.2** Two-phase three-wire system.
-
-Three-phase systems are important for at least three reasons. First, nearly all electric po wer is generated and distrib uted in three-phase,
-
-# Historical
-
-<span id="page-525-0"></span>**Nikola Tesla** (1856–1943) was a Croatian-American engineer whose inventions—among them the induction motor and the first polyphase ac power system—greatly influenced the settlement of the ac versus dc debate in favor of ac. He was also responsible for the adoption of 60 Hz as the standard for ac power systems in the United States.
-
-Born in Austria-Hungary (now Croatia), to a clergyman, Tesla had an incredible memory and a keen affinity for mathematics. He moved to the United States in 1884 and first worked for Thomas Edison. At that time, the country was in the "battle of the currents" with George Westinghouse (1846–1914) promoting ac and Thomas Edison rigidly leading the dc forces. Tesla left Edison and joined Westinghouse be cause of his interest in ac. Through Westinghouse, Tesla gained the reputation and acceptance of his polyphase ac generation, transmission, and distribution system. He held 700 patents in his lifetime. His other inventions include high-voltage apparatus (the tesla coil) and a wireless transmission system. The unit of magnetic flux density, the tesla, was named in honor of him.
-
-Library of Congress [LC-USZ62-61761]
-
-at the operating frequenc y of 60 Hz (or *ω* = 377 rad/s) in the United States or 50 Hz (or *ω* = 314 rad/s) in some other parts of the w orld. When one-phase or tw o-phase inputs are required, the y are taken from the three-phase system rather than generated independently. Even when more than three phases are needed—such as in the aluminum industry , where 48 phases are required for melting purposes—the y can be pro vided by manipulating the three phases supplied. Second, the instanta neous power in a three-phase system can be constant (not pulsating), as we will see in Section 12.7. This results in uniform power transmission and less vibration of three-phase machines. Third, for the same amount of power, the three-phase system is more economical than the singlephase. The amount of wire required for a three-phase system is less than that required for an equivalent single-phase system.
-
-We begin with a discussion of balanced three-phase v oltages. Then we analyze each of the four possible configurations of balanced threephase systems. We also discuss the analysis of unbalanced three-phase systems. We learn how to use *PSpice for Windows* to analyze a balanced or unbalanced three-phase system. Finally, we apply the concepts developed in this chapter to three-phase po wer measurement and residential electrical wiring.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/137_12.2 Balanced Three-Phase Voltages.md

@@ -1,301 +0,0 @@
-# **12.2** Balanced Three-Phase Voltages
-
-Three-phase voltages are often produced with a three-phase ac genera tor (or alternator) whose cross-sectional view is shown in Fig. 12.4. The generator basically consists of a rotating magnet (called the *rotor*) surrounded by a stationary winding (called the *stator*). Three separate windings or coils with terminals *a*-*a*′, *b*-*b*′, and *c*-*c*′ are physically placed 120° apart around the stator. Terminals *a* and *a*′, for example, stand for one of the ends of coils going into and the other end coming out of the
-
-**Figure 12.3** Three-phase four-wire system.
-
-A three-phase generator.
-
-**Figure 12.5** The generated voltages are 120° apart from each other.
-
-page. As the rotor rotates, its magnetic field "cuts" the flux from the three coils and induces voltages in the coils. Because the coils are placed 120° apart, the induced voltages in the coils are equal in magnitude but out of phase by 120° (Fig. 12.5). Since each coil can be regarded as a singlephase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads.
-
-A typical three-phase system consists of three voltage sources connected to loads by three or four wires (or transmission lines). (Threephase current sources are very scarce.) A three-phase system is equivalent to three single-phase circuits. The voltage sources can be either wyeconnected as shown in Fig. 12.6(a) or delta-connected as in Fig. 12.6(b).
-
-# **Figure 12.6**
-
-Three-phase voltage sources: (a) Y-connected source, (b) ∆-connected source.
-
-> Let us consider the wye-connected voltages in Fig. 12.6(a) for now. The voltages **V***an*, **V***bn*, and **V***cn* are respectively between lines *a*, *b*, and *c*, and the neutral line *n*. These voltages are called *phase voltages*. If the voltage sources have the same amplitude and frequency *ω* and are out of phase with each other by 120°, the voltages are said to be *balanced*. This implies that
-
-$$
-\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = 0 \tag{12.1}
-$$
-
-$$
-|\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.2}
-$$
-
-Thus,
-
-As a common tradition in power systems, voltage and current in this chapter are in rms values unless otherwise stated.
-
-Balanced phase voltages are equal in magnitude and are out of phase with each other by 120°.
-
-Because the three-phase voltages are 120° out of phase with each other, there are tw o possible combinations. One possibility is sho wn in Fig. 12.7(a) and expressed mathematically as
-
-$$
-\mathbf{V}_{an} = V_p / \mathbf{0}^{\circ}
-$$
-\n
-$$
-\mathbf{V}_{bn} = V_p / -120^{\circ}
-$$
-\n
-$$
-\mathbf{V}_{cn} = V_p / -240^{\circ} = V_p / +120^{\circ}
-$$
-\n(12.3)
-
-where *Vp* is the effective or rms value of the phase voltages. This is known as the *abc sequence* or *positive sequence*. In this phase sequence, **V***an* leads **V***bn*, which in turn leads **V***cn*. This sequence is produced when the rotor in Fig. 12.4 rotates counterclockwise. The other possibility is shown in Fig. 12.7(b) and is given by
-
-$$
-\mathbf{V}_{an} = V_p \underline{/0^{\circ}}
-$$
-  
-\n
-$$
-\mathbf{V}_{cn} = V_p \underline{/ -120^{\circ}}
-$$
-  
-\n
-$$
-\mathbf{V}_{bn} = V_p \underline{/ -240^{\circ}} = V_p \underline{/ +120^{\circ}}
-$$
- (12.4)
-
-This is called the *acb sequence* or *negative sequence*. For this phase sequence, **V***an* leads **V***cn*, which in turn leads **V***bn*. The *acb* sequence is produced when the rotor in Fig. 12.4 rotates in the clockwise direction. It is easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy Eqs. (12.1) and (12.2). For example, from Eq. (12.3),
-
-$$
-\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = V_p \underline{\hspace{0.3cm}} \left( 0^{\circ} + V_p \underline{\hspace{0.3cm}} \right) + V_p \underline{\hspace{0.3cm}} \left( 120^{\circ} + V_p \underline{\hspace{0.3cm}} \right)
-$$
-\n
-$$
-= V_p (1.0 - 0.5 - j0.866 - 0.5 + j0.866) \quad (12.5)
-$$
-\n
-$$
-= 0
-$$
-
-The phase sequence is the time order in which the voltages pass through their respective maximum values.
-
-The phase sequence is determined by the order in which the phasors pass through a fixed point in the phase diagram.
-
-In Fig. 12.7(a), as the phasors rotate in the counterclockwise direction with frequenc y *ω*, the y pass through the horizontal axis in a se quence *abcabca* . . . . Thus, the sequence is *abc* or *bca* or *cab*. Similarly, for the phasors in Fig. 12.7(b), as the y rotate in the counterclockwise direction, they pass the horizontal axis in a sequence *acbacba* . . . . This describes the *acb* sequence. The phase sequence is important in threephase power distribution. It determines the direction of the rotation of a motor connected to the power source, for example.
-
-Like the generator connections, a three-phase load can be either wye-connected or delta-connected, depending on the end application. Figure 12.8(a) sho ws a wye-connected load, and Fig. 12.8(b) sho ws a delta-connected load. The neutral line in Fig. 12.8(a) may or may not be there, depending on whether the system is four - or three-wire. (And, of course, a neutral connection is topologically impossible for a delta con nection.) A wye- or delta-connected load is said to be *unbalanced* if the phase impedances are not equal in magnitude or phase.
-
-**Figure 12.7** Phase sequences: (a) *abc* or positive sequence, (b) *acb* or negative sequence.
-
- The phase sequence may also be regarded as the order in which the phase voltages reach their peak (or maximum) values with respect to time.
-
-Reminder: As time increases, each phasor (or sinor) rotates at an angular velocity *ω*.
-
-# **Figure 12.8**
-
-Two possible three-phase load configurations: (a) a Y-connected load, (b) a ∆-connected load.
-
-A balanced load is one in which the phase impedances are equal in magnitude and in phase.
-
-For a *balanced* wye-connected load,
-
-$$
-\mathbf{Z}_1 = \mathbf{Z}_2 = \mathbf{Z}_3 = \mathbf{Z}_Y \tag{12.6}
-$$
-
-where **Z***Y* is the load impedance per phase. For a *balanced* delta-connected
-
-$$
-\mathbf{Z}_a = \mathbf{Z}_b = \mathbf{Z}_c = \mathbf{Z}_{\Delta} \tag{12.7}
-$$
-
-$$
-\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (12.8)
-$$
-
-Because both the three-phase source and the three-phase load can be either wye- or delta-connected, we have four possible connections:
-
-- Y-Y connection (i.e., Y-connected source with a Y-connected load).
-- Y-∆ connection.
-
-load,
-
-Eq. (9.69) that
-
-- ∆-∆ connection.
-- ∆-Y connection.
-
-In subsequent sections, we will consider each of these possible con figurations.
-
-It is appropriate to mention here that a balanced delta-connected load is more common than a balanced wye-connected load. This is due to the ease with which loads may be added or remo ved from each phase of a deltaconnected load. This is very difficult with a wye-connected load because the neutral may not be accessible. On the other hand, delta-connected sources are not common in practice because of the circulating current that will result in the delta-mesh if the three-phase voltages are slightly unbalanced.
-
-Example 12.1 Determine the phase sequence of the set of voltages
-
-$$
-v_{an} = 200 \cos(\omega t + 10^{\circ})
-$$
-
-*vbn* = 200 cos(*ωt* − 230°), *vcn* = 200 cos(*ωt* − 110°)
-
-# **Solution:**
-
-The voltages can be expressed in phasor form as
-
-$$
-\mathbf{V}_{an} = 200 \underline{10^{\circ}} \text{ V}, \qquad \mathbf{V}_{bn} = 200 \underline{100^{\circ}} \text{ V}, \qquad \mathbf{V}_{cn} = 200 \underline{110^{\circ}} \text{ V}
-$$
-
-We notice that **V***an* leads **V***cn* by 120° and **V***cn* in turn leads **V***bn* by 120°. Hence, we have an *acb* sequence.
-
-Practice Problem 12.1 Given that **V***bn* = 220⧸ 30° V, find **V***an* and **V***cn*, assuming a positive (*abc*) sequence.
-
-**Answer:** 220⧸150° V, 220⧸−90° V.
-
-# <span id="page-529-0"></span>**12.3** Balanced Wye-Wye Connection
-
-We begin with the Y-Y system, because any balanced three-phase sys tem can be reduced to an equivalent Y-Y system. Therefore, analysis of this system should be regarded as the key to solving all balanced threephase systems.
-
-A balanced Y-Y system is a three-phase system with a balanced Y-connected source and a balanced Y-connected load.
-
-Consider the balanced four-wire Y-Y system of Fig. 12.9, where a Y-connected load is connected to a Y-connected source. We assume a balanced load so that load impedances are equal. Although the impedance **Z***Y* is the total load impedance per phase, it may also be re garded as the sum of the source impedance **Z***s*, line impedance **Z***ℓ*, and load impedance **Z***L* for each phase, since these impedances are in series. As illustrated in Fig. 12.9, **Z***s* denotes the internal impedance of the phase winding of the generator; **Z***ℓ* is the impedance of the line join ing a phase of the source with a phase of the load; **Z***L* is the impedance of each phase of the load; and **Z***n* is the impedance of the neutral line. Thus, in general
-
-$$
-\mathbf{Z}_{Y} = \mathbf{Z}_{s} + \mathbf{Z}_{\ell} + \mathbf{Z}_{L} \tag{12.9}
-$$
-
-A balanced Y-Y system, showing the source, line, and load impedances.
-
-**Z***s* and **Z***ℓ* are often very small compared with **Z***L*, so one can assume that **Z***Y* = **Z***L* if no source or line impedance is given. In any event, by lump ing the impedances together, the Y-Y system in Fig. 12.9 can be simplified to that shown in Fig. 12.10.
-
-Assuming the positi ve sequence, the *phase* v oltages (or line-toneutral voltages) are
-
-$$
-\mathbf{V}_{an} = V_p / \underline{\mathbf{0}^{\circ}}
-$$
-  
-$$
-\mathbf{V}_{bn} = V_p / \underline{-120^{\circ}}, \qquad \mathbf{V}_{cn} = V_p / \underline{+120^{\circ}}
-$$
-(12.10)
-
-**Figure 12.10** Balanced Y-Y connection.
-
-The *line-to-line* voltages or simply *line* voltages **V***ab*, **V***bc*, and **V***ca* are related to the phase voltages. For example,
-
-$$
-\mathbf{V}_{ab} = \mathbf{V}_{an} + \mathbf{V}_{nb} = \mathbf{V}_{an} - \mathbf{V}_{bn} = V_p \left( \frac{0^\circ}{\rho} - V_p \right) - 120^\circ
-$$
-  
-=  $V_p \left( 1 + \frac{1}{2} + j \frac{\sqrt{3}}{2} \right) = \sqrt{3} V_p / 30^\circ$  (12.11a)
-
-Similarly, we can obtain
-
-$$
-\mathbf{V}_{bc} = \mathbf{V}_{bn} - \mathbf{V}_{cn} = \sqrt{3} V_p \sqrt{-90^\circ}
-$$
- (12.11b)
-
-$$
-\mathbf{V}_{ca} = \mathbf{V}_{cn} - \mathbf{V}_{an} = \sqrt{3} V_p \underline{/-210^\circ}
-$$
- (12.11c)
-
-Thus, the magnitude of the line voltages *VL* is √ 3 times the magnitude of the phase voltages *Vp*, or
-
-$$
-V_L = \sqrt{3} V_P \tag{12.12}
-$$
-
-where
-
-$$
-V_p = |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.13}
-$$
-
-and
-
-$$
-V_L = |\mathbf{V}_{ab}| = |\mathbf{V}_{bc}| = |\mathbf{V}_{ca}| \tag{12.14}
-$$
-
-Also the line voltages lead their corresponding phase voltages by 30°. Figure 12.11(a) illustrates this. Figure 12.11(a) also shows how to determine **V***ab* from the phase voltages, while Fig. 12.11(b) shows the same for the three line voltages. Notice that **V***ab* leads **V***bc* by 120°, and **V***bc* leads **V***ca* by 120°, so that the line voltages sum up to zero as do the phase voltages.
-
-Applying KVL to each phase in Fig. 12.10, we obtain the line cur rents as
-
-$$
-\mathbf{I}_{a} = \frac{\mathbf{V}_{an}}{\mathbf{Z}_{Y}}, \qquad \mathbf{I}_{b} = \frac{\mathbf{V}_{bn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -120^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -120^{\circ} \tag{12.15}
-$$
-\n
-$$
-\mathbf{I}_{c} = \frac{\mathbf{V}_{cn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -240^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -240^{\circ}
-$$
-
-We can readily infer that the line currents add up to zero,
-
-$$
-\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0 \tag{12.16}
-$$
-
-so that
-
-$$
-\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = 0 \tag{12.17a}
-$$
-
-or
-
-$$
-\mathbf{V}_{nN} = \mathbf{Z}_n \mathbf{I}_n = 0 \tag{12.17b}
-$$
-
-that is, the voltage across the neutral wire is zero. The neutral line can thus be removed without affecting the system. In fact, in long distance power transmission, conductors in multiples of three are used with the earth itself acting as the neutral conductor. Power systems designed in this way are well grounded at all critical points to ensure safety.
-
-While the *line* current is the current in each line, the *phase* current is the current in each phase of the source or load. In the Y-Y system, the line current is the same as the phase current. We will use single subscripts
-
-**V**nb **V**ab = **V**an + **V**nb
-
-**V**cn
-
-**V**bn
-
-Phasor diagrams illustrating the relationship between line voltages and phase voltages.
-
-(b)
-
-**V**bc
-
-From **I***a*, we use the phase sequence to obtain other line currents. Thus, as long as the system is balanced, we need only analyze one phase. We may do this even if the neutral line is absent, as in the three-wire system.
-
-Calculate the line currents in the three-wire Y-Y system of Fig. 12.13. Example 12.2
-
-**Figure 12.13**
-
-Three-wire Y-Y system; for Example 12.2.
-
-# **Solution:**
-
-The three-phase circuit in Fig. 12.13 is balanced; we may replace it with its single-phase equivalent circuit such as in Fig. 12.12. We obtain **I***a* from the single-phase analysis as
-
-$$
-\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y}
-$$
-
-where **Z***Y* = (5 − *j*2) + (10 + *j*8) = 15 + *j*6 = 16.155⧸21.8°. Hence,
-
-$$
-\mathbf{I}_a = \frac{110/0^{\circ}}{16.155/21.8^{\circ}} = 6.81/-21.8^{\circ} \text{ A}
-$$
-
-In as much as the source voltages in Fig. 12.13 are in positive sequence, the line currents are also in positive sequence:
-
-$$
-\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 6.81 \underline{/-141.8^\circ} \text{ A}
-$$
-$$
-\mathbf{I}_c = \mathbf{I}_a \underline{/-240^\circ} = 6.81 \underline{/-261.8^\circ} \text{ A} = 6.81 \underline{/-98.2^\circ} \text{ A}
-$$
-
-# <span id="page-532-0"></span>A Y-connected balanced three-phase generator with an impedance of 0.4 + *j*0.3 Ω per phase is connected to a Y-connected balanced load with an impedance of 24 + *j*19 Ω per phase. The line joining the generator and the load has an impedance of 0.6 + *j*0.7 Ω per phase. Assuming a positive sequence for the source voltages and that **V***an* = 120⧸30° V, find: (a) the line voltages, (b) the line currents. Practice Problem 12.2
-
-**Answer:** (a) 207.8⧸60° V, 207.8⧸−60° V, 207.8⧸−180° V, (b) 3.75⧸−8.66° A, 3.75⧸−128.66° A, 3.75⧸ 111.34° A.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/138_12.3 Balanced Wye-Wye Connection.md

@@ -1,246 +0,0 @@
-# **12.4** Balanced Wye-Delta Connection
-
-A balanced Y-∆ system consists of a balanced Y-connected source feeding a balanced ∆-connected load.
-
-This is perhaps the most practical three-phase system, as the three-phase sources are usually Y-connected while the three-phase loads are usually ∆-connected.
-
-The balanced Y-delta system is sho wn in Fig. 12.14, where the source is Y-connected and the load is ∆-connected. There is, of course, no neutral connection from source to load for this case. Assuming the positive sequence, the phase voltages are again
-
-$$
-\mathbf{V}_{an} = V_p \underline{/0^{\circ}} \n\mathbf{V}_{bn} = V_p \underline{/ -120^{\circ}}, \qquad \mathbf{V}_{cn} = V_p \underline{/ +120^{\circ}}.
-$$
-\n(12.19)
-
-As shown in Section 12.3, the line voltages are
-
-$$
-\mathbf{V}_{ab} = \sqrt{3} V_p \frac{\Delta 30^\circ}{\mathbf{V}_{ca}} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \sqrt{3} V_p \frac{\Delta 90^\circ}{\Delta 10^\circ} = \mathbf{V}_{BC}
-$$
-\n
-$$
-\mathbf{V}_{ca} = \sqrt{3} V_p \frac{\Delta 150^\circ}{\Delta 10^\circ} = \mathbf{V}_{CA}
-$$
-\n(12.20)
-
-showing that the line voltages are equal to the voltages across the load impedances for this system configuration. From these voltages, we can obtain the phase currents as
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{\mathbf{Z}_{\Delta}}
-$$
-(12.21)
-
-These currents have the same magnitude but are out of phase with each other by 120°.
-
-**Figure 12.14** Balanced Y-∆ connection.
-
-Another way to get these phase currents is to apply KVL. For example, applying KVL around loop *aABbna* gives
-
-$$
--\mathbf{V}_{an} + \mathbf{Z}_{\Delta} \mathbf{I}_{AB} + \mathbf{V}_{bn} = 0
-$$
-
-or
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{an} - \mathbf{V}_{bn}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{ab}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}}
-$$
-(12.22)
-
-which is the same as Eq. (12.21). This is the more general way of finding the phase currents.
-
-The line currents are obtained from the phase currents by applying KCL at nodes *A*, *B*, and *C*. Thus,
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \quad (12.23)
-$$
-
-Since **I***CA* = **I***AB*<sup>⧸</sup>−240°,
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA} = \mathbf{I}_{AB} (1 - 1/240^\circ)
-$$
-  
-=  $\mathbf{I}_{AB} (1 + 0.5 - j0.866) = \mathbf{I}_{AB} \sqrt{3}/-30^\circ$  (12.24)
-
-showing that the magnitude *IL* of the line current is √ \_\_ 3 times the magnitude *Ip* of the phase current, or
-
-$$
-I^L = \sqrt{3}I_p \tag{12.25}
-$$
-
-where
-
-$$
-I_L = |\mathbf{I}_a| = |\mathbf{I}_b| = |\mathbf{I}_c| \tag{12.26}
-$$
-
-and
-
-$$
-I_p = |\mathbf{I}_{AB}| = |\mathbf{I}_{BC}| = |\mathbf{I}_{CA}| \tag{12.27}
-$$
-
-Also, the line currents lag the corresponding phase currents by 30°, assuming the positive sequence. Figure 12.15 is a phasor diagram illustrating the relationship between the phase and line currents.
-
-An alternative way of analyzing the Y-∆ circuit is to transform the ∆-connected load to an equi valent Y-connected load. Using the ∆-Y transformation formula in Eq. (12.8),
-
-$$
-Z_Y = \frac{Z_{\Delta}}{3}
-$$
- (12.28)
-
-After this transformation, we now have a Y-Y system as in Fig. 12.10. The three-phase Y-∆ system in Fig. 12.14 can be replaced by the singlephase equivalent circuit in Fig. 12.16. This allows us to calculate only the line currents. The phase currents are obtained using Eq. (12.25) and utilizing the fact that each of the phase currents leads the corresponding line current by 30°.
-
-A balanced *abc*-sequence Y-connected source with **V***an* = 100⧸ 10° V is Example 12.3 connected to a ∆-connected balanced load (8 + *j*4) Ω per phase. Calculate the phase and line currents.
-
-Phasor diagram illustrating the relationship between phase and line currents.
-
-**Figure 12.16** A single-phase equivalent circuit of a balanced Y-∆ circuit.
-
-# <span id="page-534-0"></span>**Solution:**
-
-This can be solved in two ways.
-
-■ **METHOD 1** The load impedance is
-
-$$
-\mathbf{Z}_{\Delta} = 8 + j4 = 8.944 / 26.57^{\circ} \,\Omega
-$$
-
-If the phase voltage **V***an* = 100⧸ 10°, then the line voltage is
-
-$$
-\mathbf{V}_{ab} = \mathbf{V}_{an} \sqrt{3} / 30^{\circ} = 100 \sqrt{3} / 10^{\circ} + 30^{\circ} = \mathbf{V}_{AB}
-$$
-
-or
-
-$$
-V_{AB} = 173.2 \angle 40^{\circ}
-$$
- V
-
-The phase currents are
-
-$$
-\begin{aligned}\n\text{I}_{AB} &= \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{173.2/40^{\circ}}{8.944/26.57^{\circ}} = 19.36/13.43^{\circ} \text{ A} \\
-\text{I}_{BC} &= \mathbf{I}_{AB} / -120^{\circ} = 19.36 / -106.57^{\circ} \text{ A} \\
-\text{I}_{CA} &= \mathbf{I}_{AB} / +120^{\circ} = 19.36 / 133.43^{\circ} \text{ A}\n\end{aligned}
-$$
-
-The line currents are
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} \sqrt{3} \underline{/-30^\circ} = \sqrt{3} (19.36) \underline{/13.43^\circ - 30^\circ}
-$$
-  
-= 33.53} \underline{/-16.57^\circ} A  
-$$
-\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 33.53 \underline{/-136.57^\circ} A
-$$
-$$
-\mathbf{I}_c = \mathbf{I}_a \underline{/+120^\circ} = 33.53 \underline{/-103.43^\circ} A
-$$
-
-■ **METHOD 2** Alternatively, using single-phase analysis,
-
-$$
-I_a = \frac{V_{an}}{Z_{\Delta}/3} = \frac{100/10^{\circ}}{2.981/26.57^{\circ}} = 33.54/-16.57^{\circ}
-$$
- A
-
-as above. Other line currents are obtained using the *abc* phase sequence.
-
-Practice Problem 12.3 One line voltage of a balanced Y-connected source is **V***AB* = 120⧸−20° V. If the source is connected to a ∆-connected load of 20 <sup>⧸</sup>40° Ω, find the phase and line currents. Assume the *abc* sequence.
-
-> **Answer:** 6⧸−60° A, 6⧸−180° A, 6⧸60° A, 10.392⧸−90° A, 10.392⧸ 150° A, 10.392⧸30° A.
-
-# **12.5** Balanced Delta-Delta Connection
-
-A balanced ∆-∆ system is one in which both the balanced source and balanced load are ∆-connected.
-
-The source as well as the load may be delta-connected as sho wn in Fig. 12.17. Our goal is to obtain the phase and line currents as usual.
-
-Assuming a positive sequence, the phase voltages for a delta-connected source are
-
-$$
-\mathbf{V}_{ab} = V_p / \underline{\mathbf{0}^{\circ}}
-$$
-  
-$$
-\mathbf{V}_{bc} = V_p / \underline{\mathbf{-120}^{\circ}}, \qquad \mathbf{V}_{ca} = V_p / \underline{\mathbf{+120}^{\circ}}
-$$
-(12.29)
-
-The line voltages are the same as the phase voltages. From Fig. 12.17, assuming there is no line impedances, the phase voltages of the de ltaconnected source are equal to the voltages across the impedances; that is,
-
-$$
-\mathbf{V}_{ab} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \mathbf{V}_{BC}, \qquad \mathbf{V}_{ca} = \mathbf{V}_{CA} \tag{12.30}
-$$
-
-Hence, the phase currents are
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{Z_{\Delta}} = \frac{\mathbf{V}_{ab}}{Z_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{Z_{\Delta}} = \frac{\mathbf{V}_{bc}}{Z_{\Delta}}
-$$
-\n
-$$
-\mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{Z_{\Delta}} = \frac{\mathbf{V}_{ca}}{Z_{\Delta}}
-$$
-\n(12.31)
-
-Because the load is delta-connected just as in the previous section, some of the formulas derived there apply here. The line currents are obtained from the phase currents by applying KCL at nodes *A*, *B*, and *C*, as we did in the previous section:
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \tag{12.32}
-$$
-
-Also, as shown in the last section, each line current lags the corre sponding phase current by 30°; the magnitude *IL* of the line current is √ \_\_ 3 times the magnitude *Ip* of the phase current,
-
-$$
-I_L = \sqrt{3}I_p \tag{12.33}
-$$
-
-An alternative way of analyzing the ∆-∆ circuit is to convert both the source and the load to their Y equivalents. We already kno w that **Z***Y* = **Z**∆∕3. To convert a ∆-connected source to a Y-connected source, see the next section.
-
-A balanced ∆-connected load ha ving an impedance 20 <sup>−</sup> *<sup>j</sup>*15 Ω is Example 12.4 connected to a ∆-connected, positi ve-sequence generator ha ving **V***ab* = 330⧸ 0° V. Calculate the phase currents of the load and the line currents.
-
-# <span id="page-536-0"></span>**Solution:**
-
-The load impedance per phase is
-
-$$
-Z_{\Delta} = 20 - j15 = 25 \sqrt{-36.57^{\circ}} \,\Omega
-$$
-
-Since **V***AB* = **V***ab*, the phase currents are
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{330/0^{\circ}}{25/-36.87^{\circ}} = 13.2/36.87^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_{BC} = \mathbf{I}_{AB}/-120^{\circ} = 13.2/-83.13^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_{CA} = \mathbf{I}_{AB}/+120^{\circ} = 13.2/156.87^{\circ} \text{ A}
-$$
-
-For a delta load, the line current always lags the corresponding phase current by 30° and has a magnitude √ \_\_ 3 times that of the phase current. Hence, the line currents are
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB}\sqrt{3}/-30^\circ = (13.2/36.87^\circ)(\sqrt{3}/-30^\circ)
-$$
-  
-= 22.86/6.87° A  
-$$
-\mathbf{I}_b = \mathbf{I}_a/-120^\circ = 22.86/-113.13^\circ
-$$
- A  
-$$
-\mathbf{I}_c = \mathbf{I}_a/+120^\circ = 22.86/126.87^\circ
-$$
- A
-
-A positive-sequence, balanced ∆-connected source supplies a balanced ∆-connected load. If the impedance per phase of the load is 18 + *j*12 Ω and **I***a* = 9.609⧸35° A, find **I***AB* and **V***AB*. Practice Problem 12.4
-
-**Answer:** 5.548⧸65° A, 120⧸98.69° V.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/139_12.4 Balanced Wye-Delta Connection.md

@@ -1,158 +0,0 @@
-# **12.6** Balanced Delta-Wye Connection
-
-A balanced ∆-Y system consists of a balanced ∆-connected source feeding a balanced Y-connected load.
-
-Consider the ∆-Y circuit in Fig. 12.18. Again, assuming the *abc* sequence, the phase voltages of a delta-connected source are
-
-$$
-\mathbf{V}_{ab} = V_p \underline{\hat{\mathbf{O}}^{\circ}}, \qquad \mathbf{V}_{bc} = V_p \underline{\hat{\mathbf{O}}^{\circ}} = V_p \underline{\hat{\mathbf{O}}^{\circ}}
-$$
-\n
-$$
-\mathbf{V}_{ca} = V_p \underline{\hat{\mathbf{O}}^{\circ}} \tag{12.34}
-$$
-
-These are also the line voltages as well as the phase voltages.
-
-We can obtain the line currents in man y ways. One way is to apply KVL to loop *aANBba* in Fig. 12.18, writing
-
-$$
--\mathbf{V}_{ab} + \mathbf{Z}_{Y}\mathbf{I}_{a} - \mathbf{Z}_{Y}\mathbf{I}_{b} = 0
-$$
-
-or
-
-Thus,
-
-$$
-\mathbf{Z}_{Y}(\mathbf{I}_{a}-\mathbf{I}_{b})=\mathbf{V}_{ab}=V_{p}\underline{\int_{0}^{\circ}}
-$$
-
-**<sup>I</sup>***<sup>a</sup>* <sup>−</sup> **<sup>I</sup>***b* = *Vp*⧸ 0° \_\_\_\_\_ **Z***Y* **(12.35)**
-
-**Figure 12.18** A balanced ∆-Y connection.
-
-But **I***b* lags **I***a* by 120°, since we assumed the *abc* sequence; that is, **I***b* = **I***a*<sup>⧸</sup>−120°. Hence,
-
-$$
-\mathbf{I}_a - \mathbf{I}_b = \mathbf{I}_a (1 - 1/ -120^\circ)
-$$
-  
-=  $\mathbf{I}_a \left( 1 + \frac{1}{2} + j\frac{\sqrt{3}}{2} \right) = \mathbf{I}_a \sqrt{3}/30^\circ$  (12.36)
-
-Substituting Eq. (12.36) into Eq. (12.35) gives
-
-$$
-I_a = \frac{\left(V_p / \sqrt{3}\right) / -30^{\circ}}{Z_Y}
-$$
- (12.37)
-
-From this, we obtain the other line currents **I***b* and **I***c* using the positive phase sequence, i.e., **I***b* = **I***a*<sup>⧸</sup>−120°, **I***c* = **I***a*<sup>⧸</sup>+120°. The phase currents are equal to the line currents.
-
-Another w ay to obtain the line currents is to replace the deltaconnected source with its equi valent wye-connected source, as sho wn
-
-Transforming a ∆-connected source to an equivalent Y-connected source.
-
-per phase, according to Eq. (9.69).
-
-Once the source is transformed to wye, the circuit becomes a wyewye system. Therefore, we can use the equi valent single-phase circuit shown in Fig. 12.20, from which the line current for phase *a* is
-
-$$
-I_a = \frac{V_p / \sqrt{3} / -30^{\circ}}{Z_Y}
-$$
- (12.39)
-
-**Z**Y **I**a **V**<sup>p</sup> ‒30° √3 + ‒
-
-which is the same as Eq. (12.37).
-
-Alternatively, we may transform the wye-connected load to an equivalent delta-connected load. This results in a delta-delta system, which can be analyzed as in Section 12.5. Note that
-
-$$
-\mathbf{V}_{AN} = \mathbf{I}_a \mathbf{Z}_Y = \frac{V_p}{\sqrt{3}} \frac{1 - 30^\circ}{\sqrt{3}} \tag{12.40}
-$$
-\n
-$$
-\mathbf{V}_{BN} = \mathbf{V}_{AN} \frac{1 - 120^\circ}{\sqrt{3}} \qquad \mathbf{V}_{CN} = \mathbf{V}_{AN} \frac{1 + 120^\circ}{\sqrt{3}} \tag{12.40}
-$$
-
-As stated earlier , the delta-connected load is more desirable than the wye-connected load. It is easier to alter the loads in any one phase of the delta-connected loads, as the individual loads are connected directly across the lines. Ho wever, the delta-connected source is hardly used in practice because any slight imbalance in the phase voltages will result in unwanted circulating currents.
-
-Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four connections. Students are advised not to memorize the formulas but to understand ho w
-
-## **TABLE 12.1**
-
-|                               | Summary of phase and line voltages/currents for |  |
-|-------------------------------|-------------------------------------------------|--|
-| balanced three-phase systems. | 1                                               |  |
-
-| Connection  | Phase voltages/currents     | Line voltages/currents                           |
-|-------------|-----------------------------|--------------------------------------------------|
-| Y-Y         | Van =<br>Vp<br>⧸ 0°         | __<br>√<br>Vab =<br>3<br>Vp<br>⧸ 30°             |
-|             | Vbn =<br>Vp<br>−120°<br>⧸   | Vbc =<br>Vab<br>−120°<br>⧸                       |
-|             | Vcn =<br>Vp<br>+120°<br>⧸   | Vca =<br>Vab<br>+120°<br>⧸                       |
-|             | Same as line currents       | I<br>Van<br>Z<br>a =<br>∕<br>Y                   |
-|             |                             | I<br>I<br>−120°<br>b =<br>⧸<br>a                 |
-|             |                             | I<br>I<br>+120°<br>c =<br>⧸<br>a                 |
-| Y-<br>∆     | Van =<br>Vp<br>⧸ 0°         | __<br>√<br>Vab =<br>VAB =<br>3<br>Vp<br>30°<br>⧸ |
-|             | Vbn =<br>Vp<br>−120°<br>⧸   | Vbc =<br>VBC =<br>Vab<br>−120°<br>⧸              |
-|             | Vcn =<br>Vp<br>+120°<br>⧸   | Vca =<br>VCA =<br>Vab<br>+120°<br>⧸<br>__        |
-|             | IAB =<br>VAB<br>Z<br>∕<br>∆ | √<br>I<br>IAB<br>3<br>−30°<br>a =<br>⧸           |
-|             | IBC =<br>VBC<br>Z<br>∕<br>∆ | I<br>I<br>−120°<br>b =<br>⧸<br>a                 |
-|             | ICA =<br>VCA<br>Z<br>∕<br>∆ | I<br>I<br>+120°<br>c =<br>⧸<br>a                 |
-| -<br>∆<br>∆ | Vab =<br>Vp<br>⧸ 0°         | Same as phase voltages                           |
-|             | Vbc =<br>Vp<br>−120°<br>⧸   |                                                  |
-|             | Vca =<br>Vp<br>+120°<br>⧸   | __                                               |
-|             | IAB =<br>Vab<br>Z<br>∕<br>∆ | √<br>I<br>a =<br>IAB<br>3<br>−30°<br>⧸           |
-|             | IBC =<br>Vbc<br>Z<br>∕<br>∆ | I<br>b =<br>I<br>−120°<br>⧸<br>a                 |
-|             | ICA =<br>Vca<br>Z<br>∕<br>∆ | I<br>c =<br>I<br>+120°<br>⧸<br>a                 |
-| ∆-Y         | Vab =<br>Vp<br>⧸ 0°         | Same as phase voltages                           |
-|             | Vbc =<br>Vp<br>−120°<br>⧸   |                                                  |
-|             | Vca =<br>Vp<br>+120°<br>⧸   | −30° ________<br>Vp<br>⧸                         |
-|             | Same as line currents       | I<br>a =<br>__<br>√<br>3<br>Z<br>Y               |
-|             |                             | I<br>I<br>−120°<br>b =<br>⧸<br>a                 |
-|             |                             | I<br>I<br>+120°<br>c =<br>⧸<br>a                 |
-
-1 Positive or *abc* sequence is assumed. <span id="page-539-0"></span>they are derived. The formulas can always be obtained by directly applying KCL and KVL to the appropriate three-phase circuits.
-
-A balanced Y-connected load with a phase impedance of 40 + *j*25 Ω is supplied by a balanced, positive sequence ∆-connected source with a line voltage of 210 V. Calculate the phase currents. Use **V***ab* as a reference.
-
-# **Solution:**
-
-The load impedance is
-
-$$
-\mathbf{Z}_Y = 40 + j25 = 47.17 / 32^{\circ} \,\Omega
-$$
-
-and the source voltage is
-
-$$
-\mathbf{V}_{ab} = 210 \underline{\big/ 0^{\circ}} \, \mathrm{V}
-$$
-
-When the ∆-connected source is transformed to a Y-connected source,
-
-$$
-\mathbf{V}_{an} = \frac{\mathbf{V}_{ab}}{\sqrt{3}} \underline{/-30^{\circ}} = 121.2 \underline{/-30^{\circ}} \text{ V}
-$$
-
-The line currents are
-
-$$
-\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y} = \frac{121.2 \div 30^\circ}{47.12 \div 32^\circ} = 2.57 \div 62^\circ \text{ A}
-$$
-\n
-$$
-\mathbf{I}_b = \mathbf{I}_a \div 120^\circ = 2.57 \div 178^\circ \text{ A}
-$$
-\n
-$$
-\mathbf{I}_c = \mathbf{I}_a \div 120^\circ = 2.57 \div 58^\circ \text{ A}
-$$
-
-which are the same as the phase currents.
-
-In a balanced ∆-Y circuit, **V***ab* = 440⧸ 15° and **Z***Y* = (12 + *j*15) Ω. Practice Problem 12.5 Calculate the line currents.
-
-**Answer:** 13.224⧸−66.34° A, 13.224⧸+173.66° A, 13.224⧸ 53.66° A.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/140_12.7 Power in a Balanced System.md

@@ -1,369 +0,0 @@
-# **12.7** Power in a Balanced System
-
-Let us now consider the power in a balanced three-phase system. We begin by examining the instantaneous power absorbed by the load. This requires that the analysis be done in the time domain. For a Y-connected load, the phase voltages are
-
-$$
-v_{AN} = \sqrt{2} V_p \cos \omega t, \qquad v_{BN} = \sqrt{2} V_p \cos(\omega t - 120^\circ)
-$$
-  
-$$
-v_{CN} = \sqrt{2} V_p \cos(\omega t + 120^\circ)
-$$
- (12.41)
-
-where the factor √ 2is necessary because *Vp* has been defined as the rms value of the phase voltage. If **Z***Y* = *Z*⧸*θ*, the phase currents lag behind their corresponding phase voltages by *θ*. Thus,
-
-$$
-i_a = \sqrt{2} I_p \cos(\omega t - \theta), \qquad i_b = \sqrt{2} I_p \cos(\omega t - \theta - 120^\circ) \quad (12.42)
-$$
-$$
-i_c = \sqrt{2} I_p \cos(\omega t - \theta + 120^\circ)
-$$
-
-Example 12.5
-
-where *Ip* is the rms value of the phase current. The total instantaneous power in the load is the sum of the instantaneous powers in the three phases; that is,
-
-$$
-p = p_a + p_b + p_c = v_{AN}i_a + v_{BN}i_b + v_{CN}i_c
-$$
-  
-=  $2V_pI_p[\cos \omega t \cos(\omega t - \theta)$   
-+  $\cos(\omega t - 120^\circ) \cos(\omega t - \theta - 120^\circ)$   
-+  $\cos(\omega t + 120^\circ) \cos(\omega t - \theta + 120^\circ)$  (12.43)
-
-Applying the trigonometric identity
-
-$$
-\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \quad (12.44)
-$$
-
-gives
-
-$$
-p = V_p I_p [3 \cos \theta + \cos(2\omega t - \theta) + \cos(2\omega t - \theta - 240^\circ) + \cos(2\omega t - \theta + 240^\circ)]
-$$
-  
-=  $V_p I_p [3 \cos \theta + \cos \alpha + \cos \alpha \cos 240^\circ + \sin \alpha \sin 240^\circ + \cos \alpha \cos 240^\circ - \sin \alpha \sin 240^\circ]$  (12.45)  
-where  $\alpha = 2\omega t - \theta$   
-=  $V_p I_p [3 \cos \theta + \cos \alpha + 2(-\frac{1}{2}) \cos \alpha] = 3V_p I_p \cos \theta$ 
-
-Thus the total instantaneous power in a balanced three-phase system is constant—it does not change with time as the instantaneous power of each phase does. This result is true whether the load is Y- or ∆- connected. This is one important reason for using a three-phase system to generate and distribute power. We will look into another reason a little later.
-
-Since the total instantaneous po wer is independent of time, the average po wer per phase *Pp* for either the ∆-connected load or the Y-connected load is *p*∕3, or
-
-$$
-P_p = V_p I_p \cos \theta \tag{12.46}
-$$
-
-and the reactive power per phase is
-
-$$
-Q_p = V_p I_p \sin \theta \tag{12.47}
-$$
-
-The apparent power per phase is
-
-$$
-S_p = V_p I_p \tag{12.48}
-$$
-
-The complex power per phase is
-
-$$
-S_p = P_p + jQ_p = V_p I_p^* \tag{12.49}
-$$
-
-\_\_
-
-where **V***p* and **I***p* are the phase voltage and phase current with magnitudes *Vp* and *Ip*, respectively. The total average power is the sum of the average powers in the phases:
-
-$$
-P = P_a + P_b + P_c = 3P_p = 3V_p I_p \cos \theta = \sqrt{3} V_L I_L \cos \theta \qquad (12.50)
-$$
-
-For a Y-connected load, *IL* = *Ip* but *VL* = √ 3 *Vp*, whereas for a ∆-connected load, *IL* = √ 3 *Ip* but *VL* = *Vp*. Thus, Eq. (12.50) applies for both Y-connected and ∆-connected loads. Similarly, the total reactive power is
-
-$$
-Q = 3V_p I_p \sin \theta = 3Q_p = \sqrt{3} V_L I_L \sin \theta \qquad (12.51)
-$$
-
-and the total complex power is
-
-$$
-\mathbf{S} = 3\mathbf{S}_p = 3\mathbf{V}_p \mathbf{I}_p^* = 3I_p^2 \mathbf{Z}_p = \frac{3V_p^2}{\mathbf{Z}_p^*}
-$$
- (12.52)
-
-where **Z***p* = *Zp*⧸*θ* is the load impedance per phase. (**Z***p* could be **Z***Y* or **Z**∆.) Alternatively, we may write Eq. (12.52) as
-
-$$
-\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta} \tag{12.53}
-$$
-
-Remember that *Vp*, *Ip*, *VL*, and *IL* are all rms values and that *θ* is the angle of the load impedance or the angle between the phase voltage and the phase current.
-
-A second major adv antage of three-phase systems for po wer distribution is that the three-phase system uses a lesser amount of wire than the single-phase system for the same line voltage *VL* and the same absorbed power *PL*. We will compare these cases and assume in both that the wires are of the same material (e.g., copper with resisti vity *ρ*), of the same length *ℓ*, and that the loads are resisti ve (i.e., unity power factor). For the tw o-wire single-phase system in Fig. 12.21(a), *IL* = *PL*∕*VL*, so the power loss in the two wires is
-
-$$
-P_{\text{loss}} = 2I_L^2 R = 2R \frac{P_L^2}{V_L^2}
-$$
- (12.54)
-
-# **Figure 12.21**
-
-Comparing the power loss in (a) a single-phase system, and (b) a three-phase system.
-
-For the three-wire three-phase system in Fig. 12.21(b), *IL*′ = |**I***a*| = |**I***b*| = |**I***c*| = *PL*∕ √ \_\_ 3 *VL* from Eq. (12.50). The power loss in the three wires is
-
-$$
-P'_{\text{loss}} = 3(I'_L)^2 R' = 3R' \frac{P_L^2}{3V_L^2} = R' \frac{P_L^2}{V_L^2}
-$$
- (12.55)
-
-Equations (12.54) and (12.55) show that for the same total power delivered *PL* and same line voltage *VL*,
-
-$$
-\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2R}{R'}
-$$
-\n(12.56)
-
-But from Chapter 2, *R* = *ρℓ*∕*πr* 2 and *R*′ = *ρℓ*∕*πr*′ 2 , where *r* and *r*′ are the radii of the wires. Thus,
-
-$$
-\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2r'^2}{r^2}
-$$
-\n(12.57)
-
-If the same power loss is tolerated in both systems, then *r* 2 = 2r′ 2 . The ratio of material required is determined by the number of wires and their volumes, so \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ Material for three-phase = 2(π*<sup>r</sup>*
-
-$$
-\frac{\text{Material for single-phase}}{\text{Material for three-phase}} = \frac{2(\pi r^2 \ell)}{3(\pi r^2 \ell)} = \frac{2r^2}{3r^2}
-$$
-\n
-$$
-= \frac{2}{3}(2) = 1.333
-$$
-\n(12.58)
-
-since *r* 2 = 2*r*′ 2 . Equation (12.58) shows that the single-phase system uses 33 percent more material than the three-phase system or that the threephase system uses only 75 percent of the material used in the equivalent single-phase system. In other words, considerably less material is needed to deliver the same power with a three-phase system than is required for a single-phase system.
-
-Refer to the circuit in Fig. 12.13 (in Example 12.2). Determine the total average power, reactive power, and complex power at the source and at the load.
-
-# **Solution:**
-
-It is sufficient to consider one phase, as the system is balanced. For phase *a*,
-
-$$
-V_p = 110\frac{0}{0} \text{ V}
-$$
- and  $I_p = 6.81\frac{1}{21.8^{\circ}} \text{ A}$ 
-
-Thus, at the source, the complex power absorbed is
-
-$$
-\mathbf{S}_s = -3\mathbf{V}_p \mathbf{I}_p^* = -3(110/0^\circ)(6.81/21.8^\circ)
-$$
-  
-= -2247/21.8° = -(2087 + j834.6) VA
-
-The real or average power absorbed is −2087 W and the reactive power is −834.6 VAR.
-
-At the load, the complex power absorbed is
-
-$$
-\mathbf{S}_{L} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{p}
-$$
-  
-where  $\mathbf{Z}_{p} = 10 + j8 = 12.81 \underline{ / 38.66^{\circ}}$  and  $\mathbf{I}_{p} = \mathbf{I}_{a} = 6.81 \underline{ / -21.8^{\circ}}$ . Hence,  
-$$
-\mathbf{S}_{L} = 3(6.81)^{2} 12.81 \underline{ / 38.66^{\circ}} = 1782 \underline{ / 38.66^{\circ}}
-$$
-$$
-= (1392 + j1113) \text{ VA}
-$$
-
-The real power absorbed is 1391.7 W and the reactive power absorbed is 1113.3 VAR. The difference between the two complex powers is ab sorbed by the line impedance (5 − *j*2) Ω. To show that this is the case, we find the complex power absorbed by the line as
-
-$$
-\mathbf{S}_{\ell} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{\ell} = 3(6.81)^{2}(5 - j2) = 695.6 - j278.3 \text{ VA}
-$$
-
-which is the difference between **S***s* and **S***L*; that is, **S***s* + **S***ℓ* + **S***L* = 0, as expected.
-
-Example 12.6
-
-For the Y-Y circuit in Practice Prob. 12.2, calculate the complex power at the source and at the load.
-
-**Answer:** −(1054.2 + *j*843.3) VA, (1012 + *j*801.6) VA.
-
-A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.
-
-# **Solution:**
-
-The apparent power is
-
-$$
-S = \sqrt{3}V_L I_L = \sqrt{3}(220)(18.2) = 6935.13 \text{ VA}
-$$
-
-Since the real power is
-
-$$
-P = S \cos \theta = 5600 \text{ W}
-$$
-
-the power factor is
-
-pf = 
-$$
-\cos \theta = \frac{P}{S} = \frac{5600}{6935.13} = 0.8075
-$$
-
-Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 550 V.
-
-**Answer:** 37.05 A.
-
-Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Fig. 12.22(a). Load 1 dra ws 30 kW at a po wer factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging. Assuming the *abc* sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and (c) the kVAR rating of the three capacitors ∆-connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor.
-
-# **Solution:**
-
-(a) For load 1, given that *P*1 = 30 kW and cos *θ*1 = 0.6, then sin *θ*1 = 0.8. Hence,
-
-$$
-S_1 = \frac{P_1}{\cos \theta_1} = \frac{30 \text{ kW}}{0.6} = 50 \text{ kVA}
-$$
-
-and *Q*1 = *S*1 sin *θ*1 = 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is
-
-$$
-S_1 = P_1 + jQ_1 = 30 + j40 \text{ kVA}
-$$
- (12.8.1)
-
-Practice Problem 12.7
-
-Example 12.8
-
-Practice Problem 12.6
-
-Example 12.7
-
-# **Figure 12.22**
-
-For Example 12.8: (a) The original balanced loads, (b) the combined load with improved power factor.
-
-For load 2, if 
-$$
-Q_2 = 45
-$$
- kVAR and  $\cos \theta_2 = 0.8$ , then  $\sin \theta_2 = 0.6$ . We find
-
-$$
-S_2 = \frac{Q_2}{\sin \theta_2} = \frac{45 \text{ kVA}}{0.6} = 75 \text{ kVA}
-$$
-
-and *P*2 = *S*2 cos *θ*2 75(0.8) = 60 kW. Therefore the complex power due to load 2 is
-
-$$
-S_2 = P_2 + jQ_2 = 60 + j45 \text{ kVA}
-$$
- (12.8.2)
-
-From Eqs. (12.8.1) and (12.8.2), the total complex power absorbed by the load is
-
-$$
-S = S_1 + S_2 = 90 + j85 \text{ kVA} = 123.8 \underline{43.36^{\circ}} \text{ kVA} \quad (12.8.3)
-$$
-
-which has a power factor of cos 43.36° = 0.727 lagging. The real power is then 90 kW, while the reactive power is 85 kVAR.
-
-It will help with the calculations to assume that the loads are wye connected and then to work with the phase voltages, i.e. the magnitude of *VAN* = (240∕√ \_\_ 3 ) kV.
-
-(b) Since 
-$$
-S = 3((240 \text{ kV}/\sqrt{3})I_L) = \sqrt{3}(240 \text{ kV})I_L
-$$
-
-the magnitude of the line current is
-
-$$
-I_L = \frac{S}{\sqrt{3}(240,000)}
-$$
-(12.8.4)
-
-We apply this to each load keeping in mind that the magnitude of the phase voltages is equal to (240∕√ \_\_ 3 ) kV. For load 1,
-
-$$
-I_{L1} = \frac{50,000}{\sqrt{3} \, 240,000} = 120.28 \, \text{mA}
-$$
-
-Since the power factor is lagging, the line current lags the line voltage by *θ*1 = cos−1 0.6 = 53.13°. Thus,
-
-$$
-I_{a1} = 120.28 \sqrt{-53.13^{\circ}}
-$$
-
-For load 2,
-
-$$
-I_{L2} = \frac{75,000}{\sqrt{3} \, 240,000} = 180.42 \, \text{mA}
-$$
-
-and the line current lags the line voltage by *θ*2 = cos−1 0.8 = 36.87°. Hence,
-
-$$
-I_{a2} = 180.42 \underline{/-36.87^{\circ}}
-$$
-
-The total line current is
-
-$$
-\mathbf{I}_a = \mathbf{I}_{a1} + \mathbf{I}_{a2} = 120.28 \underline{/ -53.13^\circ} + 180.42 \underline{/ -36.87^\circ}
-$$
-  
-= (72.168 - j96.224) + (144.336 - j108.252)  
-= 216.5 - j204.472 = 297.8 \underline{/ -43.36^\circ} mA
-
-Alternatively, we could obtain the current from the total complex power using Eq. (12.8.4) as
-
-$$
-I_L = \frac{123,800}{\sqrt{3} \, 240,000} = 297.82 \, \text{mA}
-$$
-
-and
-
-$$
-I_a = 297.82 \div 43.36^\circ \text{ mA}
-$$
-
-which is the same as before. The other line currents, **I***b*2 and **I***ca*, can be obtained according to the *abc* sequence (i.e., **I***b* = 297.82⧸−163.36° mA and **I***c* = 297.82⧸76.64° mA).
-
-(c) We can find the reactive power needed to bring the power factor to 0.9 lagging using Eq. (11.59),
-
-$$
-QC = P(\tan \theta_{\text{old}} - \tan \theta_{\text{new}})
-$$
-
-<span id="page-545-0"></span>where *P* = 90 kW, *θ*old = 43.36°, and *θ*new = cos−1 0.9 = 25.84°. Hence,
-
-$$
-Q_C = 90,000 \text{(tan } 43.36^\circ - \text{tan } 25.84^\circ) = 41.4 \text{ kVAR}
-$$
-
-This reactive power is for the three capacitors. For each capacitor, the rating *QC*′ = 13.8 kVAR. From Eq. (11.60), the required capacitance is
-
-$$
-C = \frac{Q'_C}{\omega V_{\text{rms}}^2}
-$$
-
-Since the capacitors are ∆-connected as shown in Fig. 12.22(b), *V*rms in the above formula is the line-to-line or line voltage, which is 240 kV. Thus,
-
-is the line-to-ine of the voltage,  
-$$
-C = \frac{13,800}{(2\pi60)(240,000)^2} = 635.5 \text{ pF}
-$$
-
-Assume that the two balanced loads in Fig. 12.22(a) are supplied by an 840-V rms 60-Hz line. Load 1 is Y-connected with 30 + *j*40 Ω per phase, while load 2 is a balanced three-phase motor drawing 48 kW at a power factor of 0.8 lagging. Assuming the *abc* sequence, calculate: (a) the complex power absorbed by the combined load, (b) the kVAR rating of each of the three capacitors ∆-connected in parallel with the load to raise the power factor to unity, and (c) the current drawn from the supply at unity power factor condition.
-
-**Answer:** (a) 56.47 + *j*47.29 kVA, (b) 15.76 kVAR, (c) 38.81 A.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/141_12.8 Unbalanced Three-Phase Systems.md

@@ -1,216 +0,0 @@
-# **12.8** Unbalanced Three-Phase Systems
-
-This chapter would be incomplete without mentioning unbalanced threephase systems. An unbalanced system is caused by two possible situa tions: (1) The source voltages are not equal in magnitude and/or differ in phase by angles that are unequal, or (2) load impedances are unequal. Thus,
-
-An unbalanced system is due to unbalanced voltage sources or an unbalanced load.
-
-To simplify analysis, we will assume balanced source voltages, but an unbalanced load.
-
-Unbalanced three-phase systems are solved by direct application of mesh and nodal analysis. Figure 12.23 sho ws an e xample of an unbal anced three-phase system that consists of balanced source v oltages (not shown in the figure) and an unbalanced Y-connected load (shown in the figure). Since the load is unbalanced, **Z***A*, **Z***B*, and **Z***C* are not equal. The line currents are determined by Ohm's law as
-
-$$
-\mathbf{I}_a = \frac{\mathbf{V}_{AN}}{\mathbf{Z}_A}, \qquad \mathbf{I}_b = \frac{\mathbf{V}_{BN}}{\mathbf{Z}_B}, \qquad \mathbf{I}_c = \frac{\mathbf{V}_{CN}}{\mathbf{Z}_C}
-$$
-(12.59)
-
-# **Figure 12.23** Unbalanced three-phase Y-connected load.
-
-A special technique for handling unbalanced three-phase systems is the method of symmetrical components, which is beyond the scope of this text.
-
-**I**a
-
-Practice Problem 12.8
-
-This set of unbalanced line currents produces current in the neutral line, which is not zero as in a balanced system. Applying KCL at node *N* gives the neutral line current as
-
-$$
-\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) \tag{12.60}
-$$
-
-In a three-wire system where the neutral line is absent, we can still find the line currents **I***a*, **I***b*, and **I***c* using mesh analysis. At node *N*, KCL must be satisfied so that **I***a* + **I***b* + **I***c* = 0 in this case. The same could be done for an unbalanced ∆-Y, Y-∆, or ∆-∆ three-wire system. As mentioned earlier, in long distance po wer transmission, conductors in mul tiples of three (multiple three-wire systems) are used, with the earth itself acting as the neutral conductor.
-
-To calculate po wer in an unbalanced three-phase system requires that we find the power in each phase using Eqs. (12.46) to (12.49). The total power is not simply three times the power in one phase but the sum of the powers in the three phases.
-
-The unbalanced Y-load of Fig. 12.23 has balanced voltages of 100 V and the *acb* sequence. Calculate the line currents and the neutral current. Take **Z***A* = 15 Ω, **Z***B* = 10 + *j*5 Ω, **Z***C* = 6 − *j*8 Ω.
-
-# **Solution:**
-
-Using Eq. (12.59), the line currents are
-
-$$
-\mathbf{I}_a = \frac{100/0^{\circ}}{15} = 6.67/0^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_b = \frac{100/120^{\circ}}{10 + j5} = \frac{100/120^{\circ}}{11.18/26.56^{\circ}} = 8.94/93.44^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_c = \frac{100/-120^{\circ}}{6 - j8} = \frac{100/-120^{\circ}}{10/-53.13^{\circ}} = 10/-66.87^{\circ} \text{ A}
-$$
-
-Using Eq. (12.60), the current in the neutral line is
-
-$$
-\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = -(6.67 - 0.54 + j8.92 + 3.93 - j9.2)
-$$
-  
-= -10.06 + j0.28 = 10.06/178.4° A
-
-The unbalanced ∆-load of Fig. 12.24 is supplied by balanced line-to-line voltages of 440 V in the positive sequence. Find the line currents. Take **V***ab* as reference.
-
-**Answer:** 
-$$
-39.71 \underline{/-41.06^{\circ}}
-$$
- A,  $64.12 \underline{/-139.8^{\circ}}$  A,  $70.13 \underline{/74.27^{\circ}}$  A.
-
-Example 12.9
-
-Practice Problem 12.9
-
-**Figure 12.24** Unbalanced ∆-load; for Practice Prob. 12.9. For the unbalanced circuit in Fig. 12.25, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source.
-
-For Example 12.10.
-
-# **Solution:**
-
-(a) We use mesh analysis to find the required currents. For mesh 1,
-
-$$
-120 \underline{/- 120^{\circ}} - 120 \underline{/0^{\circ}} + (10 + j5)I_1 - 10I_2 = 0
-$$
-
-or
-
-$$
-(10+j5)\mathbf{I}_1 - 10\mathbf{I}_2 = 120\sqrt{3}/30^{\circ}
-$$
- (12.10.1)
-
-For mesh 2,
-
-$$
-120\underline{\bigg/120^{\circ}} - 120\underline{\bigg/}-120^{\circ} + (10 - j10)\mathbf{I}_2 - 10\mathbf{I}_1 = 0
-$$
-
-or
-
-$$
--10I1 + (10 - j10)I2 = 120\sqrt{3} / -90^{\circ}
-$$
- (12.10.2)
-
-Equations (12.10.1) and (12.10.2) form a matrix equation:
-
-2.10.1) and (12.10.2) form a matrix equation:  
-\n
-$$
-\begin{bmatrix}\n10 + j5 & -10 \\
--10 & 10 - j10\n\end{bmatrix}\n\begin{bmatrix}\n\mathbf{I}_1 \\
-\mathbf{I}_2\n\end{bmatrix} = \n\begin{bmatrix}\n120\sqrt{3}/30^\circ \\
-120\sqrt{3}/-90^\circ\n\end{bmatrix}
-$$
-
-The determinants are
-
-erminants are  
-\n
-$$
-\Delta = \begin{vmatrix} 10 + j5 & -10 \\ -10 & 10 - j10 \end{vmatrix} = 50 - j50 = 70.71 \underline{/-45^{\circ}}
-$$
-\n
-$$
-\Delta_1 = \begin{vmatrix} 120\sqrt{3}/30^{\circ} & -10 \\ 120\sqrt{3}/-90^{\circ} & 10 - j10 \end{vmatrix} = 207.85(13.66 - j13.66)
-$$
-\n
-$$
-= 4015 \underline{/-45^{\circ}}
-$$
-\n
-$$
-\Delta_2 = \begin{vmatrix} 10 + j5 & 120\sqrt{3}/30^{\circ} \\ -10 & 120\sqrt{3}/-90^{\circ} \end{vmatrix} = 207.85(13.66 - j5)
-$$
-\n
-$$
-= 3023.4 \underline{/-20.1^{\circ}}
-$$
-
-The mesh currents are
-
-rents are  
-\n
-$$
-\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{4015.23/-45^{\circ}}{70.71/-45^{\circ}} = 56.78 \text{ A}
-$$
-\n
-$$
-\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{3023.4/-20.1^{\circ}}{70.71/-45^{\circ}} = 42.75/24.9^{\circ} \text{ A}
-$$
-
-The line currents are
-
-$$
-\mathbf{I}_a = \mathbf{I}_1 = 56.78 \text{ A}, \qquad \mathbf{I}_c = -\mathbf{I}_2 = 42.75 \underline{\smash{\big)}\, - 155.1^\circ} \text{ A}
-$$
-\n
-$$
-\mathbf{I}_b = \mathbf{I}_2 - \mathbf{I}_1 = 38.78 + j18 - 56.78 = 25.46 \underline{\smash{\big)}\, 135^\circ} \text{ A}
-$$
-
-(b) We can now calculate the complex power absorbed by the load. For phase A,
-
-$$
-\mathbf{S}_A = \mathbf{I} \mathbf{I}_a \mathbf{I}^2 \mathbf{Z}_A = (56.78)^2 (j5) = j16,120 \text{ VA}
-$$
-
-For phase B,
-
-$$
-\mathbf{S}_B = |\mathbf{I}_b|^2 \mathbf{Z}_B = (25.46)^2 (10) = 6480 \text{ VA}
-$$
-
-For phase C,
-
-$$
-S_C = I I_c^2 Z_C = (42.75)^2(-j10) = -j18,276 \text{ VA}
-$$
-
-The total complex power absorbed by the load is
-
-$$
-S_L = S_A + S_B + S_C = 6480 - j2156
-$$
- VA
-
-(c) We check the result above by finding the power absorbed by the source. For the voltage source in phase *a,*
-
-$$
-S_a = -V_{an}I_a^* = -(120/0^\circ)(56.78) = -6813.6
-$$
- VA
-
-For the source in phase *b,*
-
-$$
-\mathbf{S}_b = -\mathbf{V}_{bn}\mathbf{I}_b^* = -(120/-120°)(25.46/-135°)
-$$
-  
-= -3055.2/105° = 790 - j2951.1 VA
-
-For the source in phase *c*,
-
-$$
-\mathbf{S}_c = -\mathbf{V}_{bn}\mathbf{I}_c^* = -(120/120^\circ)(42.75/155.1^\circ)
-$$
-  
-= -5130/275.1° = -456.03 + j5109.7 VA
-
-The total complex power absorbed by the three-phase source is
-
-$$
-S_s = S_a + S_b + S_c = -6480 + j2156
-$$
- VA
-
-showing that **S***s* + **S***L* = 0 and confirming the conservation principle of ac power.
-
-**Answer:** 128.01⧸ 80.1° A, 76.21⧸−60° A, 85⧸−135° A, 19.36 kW.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/142_12.9 PSpice for Three-Phase Circuits.md

@@ -1,882 +0,0 @@
-# <span id="page-549-0"></span>**12.9** PSpice for Three-Phase Circuits
-
-*PSpice* can be used to analyze three-phase balanced or unbalanced cir cuits in the same way it is used to analyze single-phase ac circuits. However, a delta-connected source presents two major problems to *PSpice*. First, a delta-connected source is a loop of voltage sources—which *PSpice* does not like. To avoid this problem, we insert a resistor of neg ligible resistance (say, 1 *μ*Ω per phase) into each phase of the deltaconnected source. Second, the delta-connected source does not provide a convenient node for the ground node, which is necessary to run *PSpice*. This problem can be eliminated by inserting balanced wye-connected large resistors (say, 1 MΩ per phase) in the delta-connected source so that the neutral node of the wye-connected resistors serves as the ground node 0. Example 12.12 will illustrate this.
-
-For the balanced Y-∆ circuit in Fig. 12.27, use *PSpice* to find the line current **I***aA*, the phase voltage **V***AB*, and the phase current **I***AC*. Assume that the source frequency is 60 Hz.
-
-# **Solution:**
-
-The schematic is shown in Fig. 12.28. The pseudocomponents IPRINT are inserted in the appropriate lines to obtain **I***aA* and **I***AC*, while VPRINT2 is inserted between nodes A and B to print differential voltage **V***AB*. We set the attributes of IPRINT and VPRINT2 each to *AC* = *yes*, *MAG* = *yes*, *PHASE* = *yes*, to print only the magnitude and phase of the currents and voltages. As a single-frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 60, and *Final Freq* = 60. Once the circuit is saved, it is simulated by selecting **Analysis/Simulate**. The output file includes the following:
-
-| FREQ      | V(A,B)       | VP(A,B)      |
-|-----------|--------------|--------------|
-| 6.000E+01 | 1.699E+02    | 3.081E+01    |
-| FREQ      | IM(V_PRINT2) | IP(V_PRINT2) |
-| 6.000E+01 | 2.350E+00    | -3.620E+01   |
-| FREQ      | IM(V_PRINT3) | IP(V_PRINT3) |
-| 6.000E+01 | 1.357E+00    | -6.620E+01   |
-
-# Example 12.11
-
-Schematic for the circuit in Fig. 12.27.
-
-From this, we obtain
-
-$$
-I_{aA} = 2.35 \underline{/-36.2^{\circ}} A
-$$
-  
-$$
-V_{AB} = 169.9 \underline{/30.81^{\circ}} V, \quad I_{AC} = 1.357 \underline{/-66.2^{\circ}} A
-$$
-
-# Practice Problem 12.11
-
-Refer to the balanced Y-Y circuit of Fig. 12.29. Use *PSpice* to find the line current **I***bB* and the phase voltage **V***AN*. Take *f* = 100 Hz.
-
-**Answer:** 100.9⧸ 60.87° V, 8.547⧸−91.27° A.
-
-Consider the unbalanced ∆-∆ circuit in Fig. 12.30. Use *PSpice* t o find the generator current **I***ab*, the line current **I***bB*, and the phase current **I***BC*.
-
-Example 12.12
-
-# **Solution:**
-
-- 1. **Define.** The problem and solution process are clearly defined.
-- 2. **Present.** We are to find the generator current flowing from *a* to *b,* the line current flowing from *b* to *B,* and the phase current flowing from *B* to *C*.
-- 3. **Alternative.** Although there are different approaches to solving this problem, the use of *PSpice* is mandated. Therefore, we will not use another approach.
-- 4. **Attempt.** As mentioned above, we avoid the loop of voltage sources by inserting a 1-*μ*Ω series resistor in the delta-connected source. To provide a ground node 0, we insert balanced wyeconnected resistors (1 MΩ per phase) in the delta-connected source, as shown in the schematic in Fig. 12.31. Three IPRINT pseudocomponents with their attributes are inserted to be able
-
-**Figure 12.31** Schematic for the circuit in Fig. 12.30.
-
-to get the required currents **I***ab*, **I***bB*, and **I***BC*. Since the operating frequency is not given and the inductances and capacitances should be specified instead of impedances, we assume *ω* = 1 rad/s so that *f* = 1∕2*π* = 0.159155 Hz. Thus,
-
-$$
-L = \frac{X_L}{\omega} \quad \text{and} \quad C = \frac{1}{\omega X_C}
-$$
-
- We select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.159155, and *Final Freq* = 0.159155. Once the schematic is saved, we select **Analysis/Simulate** to simulate the circuit. The output file includes:
-
-| FREQ      | IM(V_PRINT1) | IP(V_PRINT1) |
-|-----------|--------------|--------------|
-| 1.592E-01 | 9.106E+00    | 1.685E+02    |
-| FREQ      | IM(V_PRINT2) | IP(V_PRINT2) |
-| 1.592E-01 | 5.959E+00    | -1.772E+02   |
-| FREQ      | IM(V_PRINT3) | IP(V_PRINT3) |
-| 1.592E-01 | 5.500E+00    | 1.725E+02    |
-
-which yields
-
-$$
-I_{ab} = 5.595 \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \
-$$
-
-5. **Evaluate.** We can check our results by using mesh analysis. Let the loop *aABb* be loop 1, the loop *bBCc* be loop 2, and the loop *ACB* be loop 3, with the three loop currents all flowing in the clockwise direction. We then end up with the following loop equations:
-
-Loop 1
-
-(54 + *j*10)*I*<sup>1</sup> − (2 + *j*5)*I*<sup>2</sup> − (50)*I*3 = 208⧸ 10° = 204.8 + *j*36.12
-
-Loop 2
-
-$$
--(2+j5)I1 + (4+j40)I2 - (j30)I3 = 208/ -110o
-$$
-  
-= -71.14 - j195.46
-
-Loop 3
-
-$$
--(50)I_1 - (j30)I_2 + (50 - j10)I_3 = 0
-$$
-
-Using MATLAB to solve this we get,
-
-```
->>Z = [(54+10i),(-2-5i),-50;(-2-5i),(4+40i), 
--30i;-50,-30i,(50-10i)]
-```
-
-Z =
-
-```
-54.0000+10.0000i-2.0000-5.0000i-50.0000
--2.0000-5.0000i 4.0000 + 40.0000i 0-30.0000i
--50.0000 0-30.0000i 50.0000-10.0000i
-```
-
-```
->>V = [(204.8+36.12i);(-71.14-195.46i);0]
-V =
-1.0e+002*
-2.0480+0.3612i
--0.7114-1.9546i
- 0
->>I = inv(Z)*V
-I =
-8.9317+2.6983i
-0.0096+4.5175i
-5.4619+3.7964i
-   IbB = −I1 + I2 = −(8.932 + j2.698) + (0.0096 + j4.518)
-     = −8.922 + j1.82 = 9.106⧸168.47° A Answer checks
-    IBC = I2 − I3 = (0.0096 + j4.518) − (5.462 + j3.796)
-       = −5.452 + j0.722 = 5.5⧸172.46° A Answer checks
-```
-
-Now to solve for *Iab*. If we assume a small internal impedance for each source, we can obtain a reasonably good estimate for *Iab*. Adding in internal resistors of 0.01 Ω, and adding a fourth loop around the source circuit, we now get
-
-Loop 1
-
-$$
-(54.01 + j10)I1 - (2 + j5)I2 - (50)I3 - 0.01I4 = 208/10o
-$$
-  
-= 204.8 + j36.12
-
-Loop 2
-
-$$
--(2+j5)I1 + (4.01+j40)I2 - (j30)I3 - 0.01I4
-$$
-  
-= 208/ $-110^{\circ}$  = -71.14 - j195.46
-
-Loop 3
-
-$$
--(50)I1 - (j30)I2 + (50 - j10)I3 = 0
-$$
-
-Loop 4
-
-$$
--(0.01)I_1 - (0.01)I_2 + (0.03)I_4 = 0
-$$
-
->>Z = [(54.01+10i),(-2-5i),-50,-0.01;(-2-5i), (4.01+40i),-30i,-0.01;-50,-30i,(50-10i), 0;-0.01,-0.01,0,0.03] Z = 54.0100 + 10.0000i -2.0000-5.0000i, -50.0000 -0.0100 -2.0000-5.0000i 4.0100-40.0000i 0-30.0000i 0.0100 -50.0000 0-30.0000i 50.0000-10.0000i 0 -0.0100 -0.0100 0 0.0300 >>V = [(204.8 + 36.12i);(-71.14-195.46i);0;0]
-
-```
-V =
-1.0e+002*
-2.0480+0.3612i
--0.7114-1.9546i
- 0
- 0
->>I = inv(Z)*V
-I =
-8.9309+2.6973i
-0.0093+4.5159i
-5.4623+3.7954i
-2.9801+2.4044i
-   Iab = −I1 + I4 = −(8.931 + j2.697) + (2.98 + j2.404)
-= −5.951 − j0.293 = 5.958⧸−177.18° A. Answer checks.
-```
-
-6. **Satisfactory?** We have a satisfactory solution and an adequate check for the solution. We can now present the results as a solution to the problem.
-
-Practice Problem 12.12 For the unbalanced circuit in Fig. 12.32, use *PSpice* to find the generator current **I***ca*, the line current **I***cC*, and the phase current **I***AB*.
-
-**Answer:** 24.68⧸−90° A, 37.25⧸ 83.79° A, 15.55⧸−75.01° A.
-
-# **12.10** Applications
-
-Both wye and delta source connections have important practical applications. The wye source connection is used for long distance transmission of electric power, where resistive losses ( *I* 2 *R*) should be minimal. This
-
-is due to the fact that the wye connection gives a line voltage that is √ \_\_ 3 greater than the delta connection; hence, for the same power, the line current is √ \_\_ 3smaller. In addition, delta connected are also undesirable due to the potential of having disastrous circulating currents. Sometimes, using transformers, we create the equivalent of delta connect source. This conversion from three-phase to single-phase is required in residential wiring, because household lighting and appliances use single-phase power. Three-phase power is used in industrial wiring where a large power is required. In some applications, it is immaterial whether the load is wye- or delta-connected. For example, both connections are satisfac tory with induction motors. In fact, some manufacturers connect a motor in delta for 220 V and in wye for 440 V so that one line of motors can be readily adapted to two different voltages.
-
-Here we consider two practical applications of those concepts co vered in this chapter: power measurement in three-phase circuits and residential wiring.
-
-# **12.10.1** Three-Phase Power Measurement
-
-Section 11.9 presented the wattmeter as the instrument for measuring the average (or real) power in single-phase circuits. A single wattmeter can also measure the average power in a three-phase system that is bal anced, so that *P*1 = *P*2 = *P*3; the total power is three times the reading of that one wattmeter. However, two or three single-phase wattmeters are necessary to measure power if the system is unbalanced. The *threewattmeter method* of power measurement, shown in Fig. 12.33, will work regardless of whether the load is balanced or unbalanced, wye- or delta-connected. The three-wattmeter method is well suited for power measurement in a three-phase system where the power factor is con stantly changing. The total average power is the algebraic sum of the three wattmeter readings,
-
-$$
-P_T = P_1 + P_2 + P_3 \tag{12.61}
-$$
-
-where *P*1, *P*2, and *P*3 correspond to the readings of wattmeters *W*1, *W*2, and *W*3, respectively. Notice that the common or reference point *o* in Fig. 12.33 is selected arbitrarily. If the load is wye-connected, point *o* can be connected to the neutral point *n*. For a delta-connected load, point *o* can be connected to any point. If point *o* is connected to point *b*, for example, the voltage coil in wattmeter *W*2 reads zero and *P*2 = 0, indicating that wattmeter *W*2 is not necessary. Thus, two wattmeters are sufficient to measure the total power.
-
-The *two-wattmeter method* is the most commonly used method for three-phase power measurement. The two wattmeters must be properly connected to an y two phases, as sho wn typically in Fig. 12.34. Notice that the current coil of each w attmeter measures the line current, while the respective voltage coil is connected between the line and the third line and measures the line v oltage. Also notice that the ± terminal of the voltage coil is connected to the line to which the corresponding current coil is connected. Although the individual wattmeters no longer read the power taken by any particular phase, the algebraic sum of the two wattmeter readings equals the total a verage power absorbed by the load, regardless of whether it is wye- or delta-connected, balanced or
-
-unbalanced. The total real power is equal to the algebraic sum of the two wattmeter readings,
-
-$$
-P_T = P_1 + P_2 \tag{12.62}
-$$
-
-We will show here that the method works for a balanced three-phase system.
-
-Consider the balanced, wye-connected load in Fig. 12.35. Our objective is to apply the two-wattmeter method to find the average power absorbed by the load. Assume the source is in the *abc* sequence and the load impedance **Z***Y* = Z*Y*<sup>⧸</sup>*θ*. Due to the load impedance, each voltage coil leads its current coil by *θ*, so that the po wer factor is cos *θ*. We recall that each line voltage leads the corresponding phase voltage by 30°. Thus, the total phase dif ference between the phase current **I***a* and line v oltage **V***ab* is *θ* + 30°, and the a verage po wer read by wattmeter *W*1 is
-
-$$
-P_1 = \text{Re}[\mathbf{V}_{ab}\mathbf{I}_a^*] = V_{ab}I_a \cos(\theta + 30^\circ) = V_L I_L \cos(\theta + 30^\circ) \qquad (12.63)
-$$
-
-**Figure 12.35** Two-wattmeter method applied to a balanced wye load.
-
-Similarly, we can show that the average power read by wattmeter 2 is
-
-$$
-P_2 = \text{Re}[\mathbf{V}_{cb}\mathbf{I}_c^*] = V_{cb}I_c \cos(\theta - 30^\circ) = V_L I_L \cos(\theta - 30^\circ) \qquad (12.64)
-$$
-
-We now use the trigonometric identities
-
-$$
-\cos(A + B) = \cos A \cos B - \sin A \sin B
-$$
-  
-\n
-$$
-\cos(A - B) = \cos A \cos B + \sin A \sin B
-$$
- (12.65)
-
-to find the sum and the difference of the two wattmeter readings in Eqs. (12.63) and (12.64):
-
-$$
-P_1 + P_2 = V_L I_L [\cos(\theta + 30^\circ) + \cos(\theta - 30^\circ)]
-$$
-  
-=  $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$   
-+  $\cos \theta \cos 30^\circ + \sin \theta \sin 30^\circ)$   
-=  $V_L I_L 2 \cos 30^\circ \cos \theta = \sqrt{3} V_L I_L \cos \theta$  (12.66)
-
-since 2 cos 30° = √ \_\_ 3 . Comparing Eq. (12.66) with Eq. (12.50) shows that the sum of the wattmeter readings gives the total average power,
-
-$$
-P_T = P_1 + P_2 \t\t(12.67)
-$$
-
-Similarly,
-
-$$
-P_1 - P_2 = V_L I_L [\cos(\theta + 30^\circ) - \cos(\theta - 30^\circ)]
-$$
-  
-=  $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$   
- $- \cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ)$  (12.68)  
-=  $-V_L I_L 2 \sin 30^\circ \sin \theta$   
- $P_2 - P_1 = V_L I_L \sin \theta$ 
-
-since 2 sin 30° = 1. Comparing Eq. (12.68) with Eq. (12.51) shows that the difference of the wattmeter readings is proportional to the total reactive power, or
-
-$$
-Q_T = \sqrt{3}(P_2 - P_1)
-$$
- (12.69)
-
-From Eqs. (12.67) and (12.69), the total apparent power can be obtained as
-
-$$
-S_T = \sqrt{P_T^2 + Q_T^2}
-$$
- (12.70)
-
-Dividing Eq. (12.69) by Eq. (12.67) gives the tangent of the power fac tor angle as
-
-$$
-\tan \theta = \frac{Q_T}{P_T} = \sqrt{3} \frac{P_2 - P_1}{P_2 + P_1}
-$$
-\n(12.71)
-
-from which we can obtain the power factor as pf = cos *θ*. Thus, the two-wattmeter method not only provides the total real and reactive powers, it can also be used to compute the power factor. From Eqs. (12.67), (12.69), and (12.71), we conclude that:
-
-- 1. If *P*2 = *P*1, the load is resistive.
-- 2. If *P*2 > *P*1, the load is inductive.
-- 3. If *P*2 < *P*1, the load is capacitive.
-
-Although these results are derived from a balanced wye-connected load, they are equally valid for a balanced delta-connected load. However, the two-wattmeter method cannot be used for power measurement in a three-phase four-wire system unless the current through the neutral line is zero. We use the three-wattmeter method to measure the real power in a three-phase four-wire system.
-
-# Three wattmeters *W*1, *W*2, and *W*3 are connected, respectively, to phases *a, b,* and *c* to measure the total power absorbed by the unbalanced wye-connected load in Example 12.9 (see Fig. 12.23). (a) Predict the wattmeter readings. (b) Find the total power absorbed.
-
-# **Solution:**
-
-# Example 12.13
-
-Part of the problem is already solved in Example 12.9. Assume that the wattmeters are properly connected as in Fig. 12.36.
-
-**Figure 12.36** For Example 12.13.
-
-(a) From Example 12.9,
-
-$$
-\mathbf{V}_{AN} = 100 \underline{\text{/}0^{\circ}}, \qquad \mathbf{V}_{BN} = 100 \underline{\text{/}120^{\circ}}, \qquad \mathbf{V}_{CN} = 100 \underline{\text{/} -120^{\circ}} \text{ V}
-$$
-
-while
-
-$$
-\mathbf{I}_a = 6.67 \underline{\bigcirc}^{\circ}, \qquad \mathbf{I}_b = 8.94 \underline{\bigcirc} 3.44^{\circ}, \qquad \mathbf{I}_c = 10 \underline{\bigcirc} -66.87^{\circ} \text{ A}
-$$
-
-We calculate the wattmeter readings as follows:
-
-$$
-P_1 = \text{Re}(\mathbf{V}_{AN}\mathbf{I}_{a}^{*}) = V_{AN}I_a \cos(\theta_{\mathbf{V}_{AN}} - \theta_{\mathbf{I}_a})
-$$
-  
-= 100 × 6.67 × cos(0° – 0°) = 667 W  
-$$
-P_2 = \text{Re}(\mathbf{V}_{BN}\mathbf{I}_{b}^{*}) = V_{BN}I_b \cos(\theta_{\mathbf{V}_{BN}} - \theta_{\mathbf{I}_b})
-$$
-  
-= 100 × 8.94 × cos(120° – 93.44°) = 800 W  
-$$
-P_3 = \text{Re}(\mathbf{V}_{CN}\mathbf{I}_{c}^{*}) = V_{CN}I_c \cos(\theta_{\mathbf{V}_{CN}} - \theta_{\mathbf{I}_c})
-$$
-  
-= 100 × 10 × cos(-120° + 66.87°) = 600 W
-
-(b) The total power absorbed is
-
-$$
-P_T = P_1 + P_2 + P_3 = 667 + 800 + 600 = 2067
-$$
- W
-
-We can find the power absorbed by the resistors in Fig. 12.36 and use that to check or confirm this result
-
-$$
-P_T = |I_a|^2(15) + |I_b|^2(10) + |I_c|^2(6)
-$$
-  
-= 6.67<sup>2</sup>(15) + 8.94<sup>2</sup>(10) + 10<sup>2</sup>(6)  
-= 667 + 800 + 600 = 2067 W
-
-which is exactly the same thing.
-
-| Practice Problem 12.13 | Repeat Example 12.13 for the network in Fig. 12.24 (see Practice<br>Prob. 12.9). Hint: Connect the reference point o in Fig. 12.33 to point B.<br>Answer: (a) 13.175 kW, 0 W, 29.91 kW, (b) 43.08 kW.                                                                                               |
-|------------------------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
-| Example 12.14          | The two-wattmeter method produces w attmeter readings P1 = 1560 W<br>and P2 = 2100 W when connected to a delta-connected load. If the line<br>voltage is 220 V, calculate: (a) the per-phase average power, (b) the per<br>phase reactive power, (c) the power factor, and (d) the phase impedance. |
-
-# **Solution:**
-
-We can apply the given results to the delta-connected load. (a) The total real or average power is
-
-$$
-P_T = P_1 + P_2 = 1560 + 2100 = 3660
-$$
- W
-
-The per-phase average power is then
-
-$$
-P_p = \frac{1}{3}P_T = 1220 \text{ W}
-$$
-
-(b) The total reactive power is
-
-$$
-Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(2100 - 1560) = 935.3 \text{ VAR}
-$$
-
-so that the per-phase reactive power is
-
-$$
-Q_p = \frac{1}{3}Q_T = 311.77 \text{ VAR}
-$$
-
-(c) The power angle is
-
-$$
-\theta = \tan^{-1} \frac{Q_T}{P_T} = \tan^{-1} \frac{935.3}{3660} = 14.33^{\circ}
-$$
-
-Hence, the power factor is
-
-$$
-\cos \theta = 0.9689(\text{lagging})
-$$
-
-It is a lagging pf because *QT* is positive or *P*2 > *P*1. (d) The phase impedance is **Z***p* = *Zp*⧸*θ*. We know that *θ* is the same as the pf angle; that is, *θ* = 14.33°.
-
-$$
-Z_p = \frac{V_p}{I_p}
-$$
-
-We recall that for a delta-connected load, *Vp* = *VL* = 220 V. From Eq. (12.46),
-
-2.46),  
-\n
-$$
-P_p = V_p I_p \cos \theta \implies I_p = \frac{1220}{220 \times 0.9689} = 5.723 \text{ A}
-$$
-
-Hence,
-
-$$
-Z_p = \frac{V_p}{I_p} = \frac{220}{5.723} = 38.44 \ \Omega
-$$
-
-and
-
-$$
-Z_p = 38.44 / 14.33^{\circ} \,\Omega
-$$
-
-Let the line voltage *VL* = 208 V and the wattmeter readings of the balanced system in Fig. 12.35 be *P*1 = −560 W and *P*2 = 800 W. Determine:
-
-(a) the total average power
-
-- (b) the total reactive power
-- (c) the power factor
-
-(d) the phase impedance
-
-Is the impedance inductive or capacitive?
-
-**Answer:** (a) 240 W, (b) 2.356 kV AR, (c) 0.1014, (d) 18.25 <sup>⧸</sup> 84.18° Ω, inductive.
-
-Practice Problem 12.14
-
-Example 12.15
-
-The three-phase balanced load in Fig. 12.35 has impedance per phase of **Z***Y* = 8 + *j*6 Ω. If the load is connected to 208-V lines, predict the read ings of the wattmeters *W*1 and *W*2. Find *PT* and *QT*.
-
-# **Solution:**
-
-The impedance per phase is
-
-$$
-\mathbf{Z}_{Y} = 8 + j6 = 10/36.87^{\circ} \,\Omega
-$$
-
-so that the pf angle is 36.87°. Since the line voltage *VL* = 208 V, the line current is \_\_
-
-$$
-I_L = \frac{V_p}{|\mathbf{Z}_Y|} = \frac{208/\sqrt{3}}{10} = 12 \text{ A}
-$$
-
-Then
-
-$$
-P_1 = V_L I_L \cos(\theta + 30^\circ) = 208 \times 12 \times \cos(36.87^\circ + 30^\circ)
-$$
-  
-= 980.48 W  
-$$
-P_2 = V_L I_L \cos(\theta - 30^\circ) = 208 \times 12 \times \cos(36.87^\circ - 30^\circ)
-$$
-  
-= 2478.1 W
-
-Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W. Since *P*2 > *P*1, the load is inductive. This is evident from the load **Z***<sup>Y</sup>* itself. Next,
-
-$$
-P_T = P_1 + P_2 = 3.459 \text{ kW}
-$$
-
-and
-
-$$
-Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(1497.6)
-$$
- VAR = 2.594 kVAR
-
-If the load in Fig. 12.35 is delta-connected with impedance per phase of **Z***p* = 30 + *j*40 Ω and *VL* = 220 V, predict the readings of the wattmeters *W*1 and *W*2. Calculate *PT* and *QT*. Practice Problem 12.15
-
-**Answer:** 200.6 W, 1.5418 kW, 1.7424 kW, 2.323 kVAR.
-
-# **12.10.2** Residential Wiring
-
-In the United States, most household lighting and appliances operate on 120-V, 60-Hz, single-phase alternating current. (The electricity may also be supplied at 110, 115, or 117 V, depending on the location.) The local power company supplies the house with a three-wire ac system. Typically, as in Fig. 12.37, the line voltage of, say, 12,000 V is stepped down to 120/240 V with a transformer (more details on transformers in the next chapter). The three wires coming from the transformer are typically colored red (hot), black (hot), and white (neutral). As shown in Fig. 12.38, the two 120-V voltages are opposite in phase and hence add up to zero.
-
-That is, 
-$$
-\mathbf{V}_W = 0/\underline{0}^\circ
-$$
-,  $\mathbf{V}_B = 120/\underline{0}^\circ$ ,  $\mathbf{V}_R = 120/\underline{180}^\circ = -\mathbf{V}_B$ .  
-\n
-$$
-\mathbf{V}_{BR} = \mathbf{V}_B - \mathbf{V}_R = \mathbf{V}_B - (-\mathbf{V}_B) = 2\mathbf{V}_B = 240/\underline{0}^\circ \qquad (12.72)
-$$
-
-# **Figure 12.37**
-
-A 120/240 household power system. Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 324. Pearson Education, Inc., Upper Saddle River, NJ.
-
-# **Figure 12.38**
-
-Single-phase three-wire residential wiring.
-
-Since most appliances are designed to operate with 120 V, the light ing and appliances are connected to the 120-V lines, as illustrated in Fig. 12.39 for a room. Notice in Fig. 12.37 that all appliances are connected in parallel. Hea vy appliances that consume lar ge currents, such as air conditioners, dishwashers, ovens, and laundry machines, are con nected to the 240-V power line.
-
-Because of the dangers of electricity , house wiring is carefully regulated by a code dra wn by local ordinances and by the National Electrical Code (NEC). To avoid trouble, insulation, grounding, fuses, and circuit breakers are used. Modern wiring codes require a third wire for a separate ground. The ground wire does not carry po wer like the neutral wire but enables appliances to have a separate ground connection. Figure 12.40 shows the connection of the receptacle to a 120-V rms line and to the ground. As shown in the figure, the neutral line is connected to the ground (the earth) at man y critical locations. Although the ground line seems redundant, grounding is important for many reasons. First, it is required by NEC. Second, grounding provides
-
-# **Figure 12.39**
-
-A typical wiring diagram of a room. Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 325. Pearson Education, Inc., Upper Saddle River, NJ.
-
-a convenient path to ground for lightning that strik es the po wer line. Third, grounds minimize the risk of electric shock. What causes shock is the passage of current from one part of the body to another . The human body is lik e a big resistor *R*. If *V* is the potential dif ference between the body and the ground, the current through the body is determined by Ohm's law as
-
-$$
-I = \frac{V}{R} \tag{12.73}
-$$
-
-The value of *R* varies from person to person and depends on whether the body is wet or dry. How great or how deadly the shock is depends on the amount of current, the pathway of the current through the body, and the length of time the body is e xposed to the current. Currents less than 1 mA may not be harmful to the body , but currents greater than 10 mA can cause se vere shock. A modern safety de vice is the *ground-fault circuit interrupter* (GFCI), used in outdoor circuits and in bathrooms, where the risk of electric shock is greatest. It is essentially a circuit break er that opens when the sum of the currents *iR*, *iW*, and *iB* through the red, white, and the black lines is not equal to zero, or *iR* + *iW* + *iB* ≠ 0.
-
-The best w ay to a void electric shock is to follo w safety guide lines concerning electrical systems and appliances. Here are some of them:
-
-- Never assume that an electrical circuit is dead. Always check to be sure.
-- Use safety de vices when necessary , and wear suitable clothing (insulated shoes, gloves, etc.).
-- Never use tw o hands when testing high-v oltage circuits, since the current through one hand to the other hand has a direct path through your chest and heart.
-- Do not touch an electrical appliance when you are wet. Remember that water conducts electricity.
-- Be extremely careful when working with electronic appliances such as radio and TV because these appliances ha ve lar ge capacitors in them. The capacitors tak e time to dischar ge after the po wer is disconnected.
-- Always ha ve another person present when w orking on a wiring system, just in case of an accident.
-
-# <span id="page-563-0"></span>**12.11** Summary
-
-- 1. The phase sequence is the order in which the phase v oltages of a three-phase generator occur with respect to time. In an *abc* sequence of balanced source v oltages, **V***an* leads **V***bn* by 120°, which in turn leads **V***cn* by 120°. In an *acb* sequence of balanced v oltages, **V***an* leads **V***cn* by 120°, which in turn leads **V***bn* by 120°.
-- 2. A balanced wye- or delta-connected load is one in which the threephase impedances are equal.
-- 3. The easiest way to analyze a balanced three-phase circuit is to transform both the source and the load to a Y-Y system and then analyze the single-phase equivalent circuit. Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four possible configurations.
-- 4. The line current *IL* is the current flowing from the generator to the load in each transmission line in a three-phase system. The line voltage *VL* is the v oltage between each pair of lines, e xcluding the neutral line if it e xists. The phase current *Ip* is the current flowing through each phase in a three-phase load. The phase voltage *Vp* is the voltage of each phase. For a wye-connected load,
-
-$$
-V_L = \sqrt{3} V_p \qquad \text{and} \qquad I_L = I_p
-$$
-
-For a delta-connected load,
-
-$$
-V_L = V_p \qquad \text{and} \qquad I_L = \sqrt{3}I_p
-$$
-
-- 5. The total instantaneous po wer in a balanced three-phase system is constant and equal to the average power.
-- 6. The total comple x po wer absorbed by a balanced three-phase Y-connected or ∆-connected load is
-
-$$
-\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta}
-$$
-
-where *θ* is the angle of the load impedances.
-
-- 7. An unbalanced three-phase system can be analyzed using nodal or mesh analysis.
-- 8. *PSpice* is used to analyze three-phase circuits in the same w ay as it is used for analyzing single-phase circuits.
-- 9. The total real power is measured in three-phase systems using either the three-wattmeter method or the two-wattmeter method.
-- 10. Residential wiring uses a 120/240-V, single-phase, three-wire system.
-
-# Review Questions
-
-**12.1** What is the phase sequence of a three-phase motor for which **V***AN* = 220⧸−100° V and **V***BN* = 220⧸ 140° V?
-
-(a) *abc* (b) *acb*
-
-**12.2** If in an *acb* phase sequence, *Van* = 100⧸−20°, then **V***cn* is:
-
-(a) 
-$$
-100 \div 140^{\circ}
-$$
- (b)  $100 \div 100^{\circ}$ 
-
-(c) 
-$$
-100\div 50^{\circ}
-$$
- (d)  $100\div 10^{\circ}$ 
-
-**12.3** Which of these is not a required condition for a balanced system:
-
-$$
-(a) |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}|
-$$
-
-(b) 
-$$
-\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0
-$$
-
-(c) 
-$$
-V_{an} + V_{bn} + V_{cn} = 0
-$$
-
-- (d) Source voltages are 120° out of phase with each other.
-- (e) Load impedances for the three phases are equal.
-
-<span id="page-564-0"></span>**12.4** In a Y-connected load, the line current and phase current are equal.
-
-(a) True (b) False
-
-**12.5** In a ∆-connected load, the line current and phase current are equal.
-
-(a) True (b) False
-
-**12.6** In a Y-Y system, a line voltage of 220 V produces a phase voltage of:
-
-(a) 381 V (b) 311 V (c) 220 V (d) 156 V (e) 127 V
-
-**12.7** In a ∆-∆ system, a phase voltage of 100 V produces a line voltage of:
-
-(a) 58 V (b) 71 V (c) 100 V (d) 173 V (e) 141 V
-
-Problems1
-
-# Section 12.2 Balanced Three-Phase Voltages
-
-**12.1** If **V***ab* = 400 V in a balanced Y-connected threephase generator, find the phase voltages, assuming the phase sequence is:
-
-(*a*) *abc* (*b*) *acb*
-
-- **12.2** What is the phase sequence of a balanced threephase circuit for which **V***an* = 120⧸30° V and **V***cn* = 120⧸−90° V? Find **V***bn*.
-- **12.3** Given a balanced Y-connected three-phase generator with a line-to-line voltage of **V***ab* = 100⧸45° V and **V***bc* = 100⧸165° V, determine the phase sequence and the value of **V***ca*.
-- **12.4** A three-phase system with *abc* sequence and *VL* = 440 V feeds a Y-connected load with *ZL* = 40⧸30° Ω. Find the line currents.
-- **12.5** For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at the terminals are:
-
-*vAN* = 120 cos(*ωt* + 32°) V *vBN* = 120 cos(*ωt* – 88°) V *vCN* = 120 cos(*ωt* + 152°) V
-
- Write the time-domain expressions for the line-toline voltages *vAB*, *vBC*, and *vCA*.
-
-# Section 12.3 Balanced Wye-Wye Connection
-
-**12.6** Using Fig. 12.41, design a problem to help other students better understand balanced wye-wye connected circuits.
-
-**12.8** When a Y-connected load is supplied by voltages in *abc* phase sequence, the line voltages lag the corresponding phase voltages by 30°.
-
-(a) True (b) False
-
-**12.9** In a balanced three-phase circuit, the total instantaneous power is equal to the average power.
-
-(a) True (b) False
-
-**12.10** The total power supplied to a balanced ∆-load is found in the same way as for a balanced Y-load.
-
-(a) True (b) False
-
-*Answers: 12.1a, 12.2a, 12.3c, 12.4a, 12.5b, 12.6e, 12.7c, 12.8b, 12.9a, 12.10a.*
-
-# **Figure 12.41**
-
-For Prob. 12.6.
-
-- **12.7** Obtain the line currents in the three-phase circuit of Fig. 12.42 on the next page.
-- **12.8** In a balanced three-phase Y-Y system, the source is an *acb* sequence of voltages and **V***cn* = 120⧸ 35° V rms. The line impedance per phase is (1 + *j*2) Ω, while the per-phase impedance of the load is (11 + *j*14) Ω. Calculate the line currents and the load voltages.
-- **12.9** A balanced Y-Y four-wire system has phase voltages
-
-$$
-\mathbf{V}_{an} = 120 \underline{\text{O}^{\circ}}, \qquad \mathbf{V}_{bn} = 120 \underline{\text{O} - 120^{\circ}}
-$$
-$$
-\mathbf{V}_{cn} = 120 \underline{\text{O} \cdot 120^{\circ}} \text{ V}
-$$
-
- The load impedance per phase is 19 + *j*13 Ω, and the line impedance per phase is 1 + *j*2 Ω. Solve for the line currents and neutral current.
-
-<sup>1</sup> Remember that unless stated otherwise, all given voltages and currents are rms values.
-
-Problems **543**
-
-**12.10** For the circuit in Fig. 12.43, determine the current in the neutral line.
-
-For Prob. 12.10.
-
-# Section 12.4 Balanced Wye-Delta Connection
-
-**12.11** In the Y-∆ system shown in Fig. 12.44, the source is a positive sequence with **V***an* = 440⧸ 0° V and phase impedance **Z***p* = 2 – *j*3 Ω. Calculate the line voltage **V***L* and the line current **I***L*.
-
-**Figure 12.44** For Prob. 12.11.
-
-**Figure 12.46** For Prob. 12.13.
-
-**12.14** Obtain the line currents in the three-phase circuit of Fig. 12.47 on the next page.
-
-For Prob. 12.14.
-
-**12.15** The circuit in Fig. 12.48 is excited by a balanced three-phase source with a line voltage of 210 V. If **Z***l* = 1 + *j*1 Ω, **Z**∆ = 24 − *j*30 Ω, and **Z***Y* = 12 + *j*5 Ω, determine the magnitude of the line current of the combined loads.
-
-For Prob. 12.15.
-
-- **12.16** A balanced delta-connected load has a phase current **I***AC* = 5⧸−30° A.
-  - (a) Determine the three line currents assuming that the circuit operates in the positive phase sequence.
-  - (b) Calculate the load impedance if the line voltage is **V***AB* = 440⧸ 0° V.
-- **12.17** A positive sequence wye-connected source where **V***an* = 120⧸ 90° V, is connected to a delta-connected load where **Z***<sup>L</sup>* = (60 + *j*45) Ω. Determine the line currents.
-- **12.18** If **V***an* = 220⧸ 60° V in the network of Fig. 12.49, find the load phase currents **I***AB*, **I***BC*, and **I***CA*.
-
-**Figure 12.49** For Prob. 12.18.
-
-# Section 12.5 Balanced Delta-Delta Connection
-
-**12.19** For the ∆-∆ circuit of Fig. 12.50, calculate the phase and line currents.
-
-# **Figure 12.50**
-
-For Prob. 12.19.
-
-**12.20** Using Fig. 12.51, design a problem to help other students better understand balanced delta-delta connected circuits.
-
-For Prob. 12.20.
-
-**12.21** Three 440-V generators form a delta-connected source that is connected to a balanced deltaconnected load of *ZL* = (8.66 + *j*5) Ω per phase as shown in Fig. 12.52. Determine the value of *IBC* and *IaA*. What is the pf of the load?
-
-b
-
-+
-
-+ ‒
-
-- **12.22** Find the line currents *IaA*, *IbB*, and *IcC* in the three-phase network of Fig. 12.53. Take **Z***L* = (114 + *j*87) Ω and **Z***<sup>l</sup>* = (2 + *j*) Ω.
-- **12.23** A balanced delta connected source is connected to a balanced delta connected load where *ZL* = (80 + *j*60) Ω and *Zl* = (2 + *j*) Ω. Given that the load voltages are **V***AB* = 100⧸ 0° V, **V***BC* = 100⧸ 120° V, and **V***CA* = 100⧸−120° V. Calculate the source voltages **V***ab*, **V***bc*, and **V***ca*.
-- **12.24** A balanced delta-connected source has phase voltage **V***ab* = 880⧸ 30° V and a positive phase sequence. If this is connected to a balanced delta-connected load, find the line and phase currents. Take the load impedance per phase as 60⧸ 30° Ω and line impedance per phase as 1 + *j*1 Ω.
-- **12.26** Using Fig. 12.55, design a problem to help other students better understand balanced delta connected sources delivering power to balanced wye connected loads.
-
-**Figure 12.55** For Prob. 12.26.
-
-10 ‒ j8 Ω
-
-10 ‒ j8 Ω
-
-# Section 12.6 Balanced Delta-Wye Connection
-
-12.25 In the circuit of Fig. 12.54, if 
-$$
-\mathbf{V}_{ab} = 440/10^{\circ}
-$$
-,  
-\n $V_{bc} = 440/-110^{\circ}$ ,  $V_{ca} = 440/130^{\circ}$  V, find the line currents.
-
-3 + j2 Ω 10 ‒ j8 W
-
-**I**b
-
-**I**a
-
-<sup>a</sup> <sup>3</sup><sup>+</sup> <sup>j</sup>2 W
-
-Vab
-
-- **12.27** A ∆-connected source supplies power to a Yconnected load in a three-phase balanced system. Given that the line impedance is 2 + *j*1 Ω per phase while the load impedance is 6 + *j*4 Ω per phase, find the magnitude of the line voltage at the load. Assume the source phase voltage **V***ab* = 208⧸ 0° V rms.
-- **12.28** The line-to-line voltages in a Y-load have a magnitude of 880 V and are in the positive sequence at 60 Hz. If the loads are balanced with *Z*1 = *Z*2 = *Z*3 = 25⧸30°, find all line currents and phase voltages.
-
-# Section 12.7 Power in a Balanced System
-
-- **12.29** A balanced three-phase Y-∆ system has **V***an* = 240⧸ 0° V rms and **Z**∆ = 51 + *j*45 Ω. If the line impedance per phase is 0.4 + *j*1.2 Ω, find the total complex power delivered to the load.
-- **12.30** In Fig. 12.56, the rms value of the line voltage is 208 V. Find the average power delivered to the load.
-
-**Figure 12.56**
-
-For Prob. 12.30.
-
-- **12.31** A balanced delta-connected load is supplied by a 60-Hz three-phase source with a line voltage of 480 V. Each load phase draws 24 kW at a lagging power factor of 0.8. Find:
-  - (a) the load impedance per phase
-  - (b) the line current
-  - (c) the value of capacitance needed to be connected in parallel with each load phase to minimize the current from the source
-
-- **12.32** Design a problem to help other students better understand power in a balanced three-phase system.
-- **12.33** A three-phase source delivers 4.8 kVA to a wyeconnected load with a phase voltage of 208 V and a power factor of 0.9 lagging. Calculate the source line current and the source line voltage.
-- **12.34** A balanced wye-connected load with a phase impedance of 10 – *j*16 Ω is connected to a balanced three-phase generator with a line voltage of 220 V. Determine the line current and the complex power absorbed by the load.
-- **12.35** Three equal impedances, 60 + *j*30 Ω each, are delta-connected to a 230-V rms, three-phase circuit. Another three equal impedances, 40 + *j*10 Ω each, are wye-connected across the same circuit at the same points. Determine:
-  - (a) the line current
-  - (b) the total complex power supplied to the two loads
-  - (c) the power factor of the two loads combined
-- **12.36** A 4200-V, three-phase transmission line has an impedance of 4 + *j* Ω per phase. If it supplies a load of 1 MVA at 0.75 power factor (lagging), find:
-  - (a) the complex power
-  - (b) the power loss in the line
-  - (c) the voltage at the sending end
-- **12.37** The total power measured in a three-phase system feeding a balanced wye-connected load is 12 kW at a power factor of 0.6 leading. If the line voltage is 440 V, calculate the line current *IL* and the load impedance **Z***Y*.
-- **12.38** Given the circuit in Fig. 12.57 below, find the total complex power absorbed by the load.
-
-**Figure 12.57** For Prob. 12.38.
-
-# For Prob. 12.39.
-
-**12.40** For the three-phase circuit in Fig. 12.59, find the average power absorbed by the delta-connected load with **Z**∆ = 21 + *j*24 Ω.
-
-# **Figure 12.59** For Prob. 12.40.
-
-- **12.41** A balanced delta-connected load draws 5 kW at a power factor of 0.8 lagging. If the three-phase system has an effective line voltage of 400 V, find the line current.
-- **12.42** A balanced three-phase generator delivers 7.2 kW to a wye-connected load with impedance 30 – *j*40 Ω per phase. Find the line current *IL* and the line voltage *VL*.
-- **12.43** Refer to Fig. 12.48. Obtain the complex power absorbed by the combined loads.
-- **12.44** A three-phase line has an impedance of 1 + *j*3 Ω per phase. The line feeds a balanced delta-connected load, which absorbs a total complex power of 12 + *j*5 kVA. If the line voltage at the load end has a magnitude of 240 V, calculate the magnitude of the line voltage at the source end and the source power factor.
-- **12.45** A balanced wye-connected load is connected to the generator by a balanced transmission line with an impedance of 0.5 + *j*2 Ω per phase. If the load is rated at 450 kW, 0.708 power factor lagging, 440-V line voltage, find the line voltage at the generator.
-- **12.46** A three-phase load consists of three 100-Ω resistors that can be wye- or delta-connected. Determine which connection will absorb the most average
-
-power from a three-phase source with a line voltage of 110 V. Assume zero line impedance.
-
-**12.47** The following three parallel-connected three-phase loads are fed by a balanced three-phase source:
-
-> Load 1: 250 kVA, 0.8 pf lagging Load 2: 300 kVA, 0.95 pf leading Load 3: 450 kVA, unity pf
-
- If the line voltage is 13.8 kV, calculate the line current and the power factor of the source. Assume that the line impedance is zero.
-
-- **12.48** A balanced, positive-sequence wye-connected source has **V***an* = 240⧸ 0° V rms and supplies an unbalanced delta-connected load via a transmission line with impedance 2 + *j*3 Ω per phase.
-  - (a) Calculate the line currents if **Z***AB* = 40 + *j*15 Ω, **Z***BC* = 60 Ω, **Z***CA* = 18 – *j*12 Ω.
-  - (b) Find the complex power supplied by the source.
-- **12.49** Each phase load consists of a 20-Ω resistor and a 10-Ω inductive reactance. With a line voltage of 480 V rms, calculate the average power taken by the load if:
-
-(a) the three-phase loads are delta-connected (b) the loads are wye-connected
-
-**12.50** A balanced three-phase source with **V***L* = 240 V rms is supplying 8 kVA at 0.6 power factor lagging to two wye-connected parallel loads. If one load draws 3 kW at unity power factor, calculate the impedance per phase of the second load.
-
-# Section 12.8 Unbalanced Three-Phase Systems
-
-**12.51** Consider the wye-delta system shown in Fig. 12.60. Let **Z**<sup>1</sup> = 100 Ω, **Z**<sup>2</sup> = *j*100 Ω, and **Z**<sup>3</sup> = –*j*100 Ω. Determine the phase currents, **I***AB*, **I***BC*, and **I***CA*, and the line currents, **I***aA*, **I***bB* , and **I***cC*.
-
-**Figure 12.60** For Prob. 12.51.
-
-$$
-\mathbf{V}_{an} = 220/120^{\circ}, \qquad \mathbf{V}_{bn} = 220/0^{\circ}
-$$
-$$
-\mathbf{V}_{cn} = 220/-120^{\circ} \text{ V}
-$$
-
-If the impedances are
-
-$$
-\mathbf{Z}_{AN} = 20/60^\circ, \qquad \mathbf{Z}_{BN} = 30/0^\circ
-$$
-$$
-\mathbf{Z}_{cn} = 40/30^\circ \ \Omega
-$$
-
-find the current in the neutral line.
-
-**12.53** Using Fig. 12.61, design a problem that will help other students better understand unbalanced threephase systems.
-
-- For Prob. 12.53.
-- **12.54** A balanced three-phase Y-source with *VP* = 880 V rms drives a Y-connected three-phase load with phase impedance **Z***A* = 80 Ω, **Z***B* = 60 + *j*90 Ω, and **Z***C* = *j*80 Ω. Calculate the line currents and total complex power delivered to the load. Assume that the neutrals are connected.
-- **12.55** A three-phase supply, with the line-to-line voltage of 240 V rms, has the unbalanced load as shown in Fig. 12.62. Find the line currents and the total complex power delivered to the load.
-
-**Figure 12.62** For Prob. 12.55.
-
-**12.56** Using Fig. 12.63, design a problem to help other students to better understand unbalanced three-phase systems.
-
-# **Figure 12.63**
-
-For Prob. 12.56.
-
-**12.57** Determine the line currents for the three-phase circuit of Fig. 12.64. Let **V***a* = 220⧸ 0°, **V***b* = 220⧸−120°, **V***c* = 220⧸ 120° V.
-
-For Prob. 12.57.
-
-# Section 12.9 PSpice for Three-Phase Circuits
-
-- **12.58** Solve Prob. 12.10 using *PSpice or MultiSim*.
-- **12.59** The source in Fig. 12.65 is balanced and exhibits a positive phase sequence. If *f* = 60 Hz, use *PSpice or MultiSim* to find **V***AN*,**V***BN*, and **V***CN*.
-
-For Prob. 12.59.
-
-**12.60** Use *PSpice or MultiSim* to determine **I***o* in the single-phase, three-wire circuit of Fig. 12.66. Let **Z**1 = 15 – *j*10 Ω, **Z**2 = 30 + *j*20 Ω, and **Z**3 = 12 + *j*5 Ω.
-
-**12.61** Given the circuit in Fig. 12.67, use *PSpice or MultiSim* to determine currents **I***aA* and voltage **V***BN*.
-
-**Figure 12.67**
-
-For Prob. 12.61.
-
-**12.62** Using Fig. 12.68, design a problem to help other students better understand how to use *PSpice or MultiSim* to analyze three-phase circuits.
-
-**12.63** Use *PSpice or MultiSim* to find currents **I***aA* and **I***AC* in the unbal anced three-phase system shown in Fig. 12.69. Let
-
-$$
-Zl = 2 + j, \t Z1 = 40 + j20 Ω,\nZ2 = 50 - j30 Ω, \t Z3 = 25 Ω
-$$
-
-# **Figure 12.69**
-
-For Prob. 12.63.
-
-- **12.64** For the circuit in Fig. 12.58, use *PSpice or MultiSim*  to find the line currents and the phase currents.
-- **12.65** A balanced three-phase circuit is shown in Fig. 12.70 on the next page. Use *PSpice or MultiSim* to find the line currents **I***aA*, **I***bB*, and **I***cC*.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/143_12.10 Applications.md

@@ -1,53 +0,0 @@
-# Section 12.10 Applications
-
-- **12.66** A three-phase, four-wire system operating with a 480-V line voltage is shown in Fig. 12.71. The source voltages are balanced. The power absorbed by the resistive wye-connected load is measured by the three-wattmeter method. Calculate:
-  - (a) the voltage to neutral
-  - (b) the currents **I**1, **I**2, **I**3, and **I***<sup>n</sup>*
-  - (c) the readings of the wattmeters
-  - (d) the total power absorbed by the load
-- **12.67** As shown in Fig. 12.72, a three-phase four-wire line with a phase voltage of 120 V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf lagging. The motor load is connected to the three main lines marked *a, b,* and *c*. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line *c* to the neutral, 15 kW from line b to the neutral, and 9 kW from line *c* to the neutral.
-  - (a) If three wattmeters are arranged to measure the power in each line, calculate the reading of each meter.
-  - (b) Find the magnitude of the current in the neutral line.
-
-\* An asterisk indicates a challenging problem.
-
-**Figure 12.68** For Prob. 12.62.
-
-**Figure 12.70** For Prob. 12.65.
-
-- **12.68** Meter readings for a three-phase wye-connected alternator supplying power to a motor indicate that the line voltages are 330 V, the line currents are 8.4 A, and the total line power is 4.5 kW. Find:
-  - (a) the load in VA
-  - (b) the load pf
-  - (c) the phase current
-  - (d) the phase voltage
-- **12.69** A certain store contains three balanced three-phase loads. The three loads are:
-
-Load 1: 16 kVA at 0.85 pf lagging Load 2: 12 kVA at 0.6 pf lagging Load 3: 8 kW at unity pf
-
- The line voltage at the load is 208 V rms at 60 Hz, and the line impedance is 0.4 + *j*0.8 Ω. Determine the line current and the complex power delivered to the loads.
-
-# **Figure 12.72**
-
-For Prob. 12.67.
-
-- **12.70** The two-wattmeter method gives *P*1 = 1200 W and *P*2 = −400 W for a three-phase motor running on a 240-V line. Assume that the motor load is wye- connected and that it draws a line current of 6 A. Calculate the pf of the motor and its phase impedance.
-- **12.71** In Fig. 12.73, two wattmeters are properly connected to the unbalanced load supplied by a balanced source such that **V***ab* = 208⧸ 0° V with positive phase sequence.
-  - (a) Determine the reading of each wattmeter.
-  - (b) Calculate the total apparent power absorbed by the load.
-
-**Figure 12.73** For Prob. 12.71.
-
-- <span id="page-573-0"></span>**12.72** If wattmeters *W*1 and *W*2 are properly connected respectively between lines *a* and *b* and lines *b* and *c* to measure the power absorbed by the deltaconnected load in Fig. 12.44, predict their readings.
-- **12.73** For the circuit displayed in Fig. 12.74, find the wattmeter readings.
-
-# **Figure 12.74**
-
-For Prob. 12.73.
-
-**12.74** Predict the wattmeter readings for the circuit in Fig. 12.75.
-
-# **Figure 12.75** For Prob. 12.74.
-
-- **12.75** A man has a body resistance of 600 Ω. How much current flows through his ungrounded body:
-  - (a) when he touches the terminals of a 12-V autobattery?
-  - (b) when he sticks his finger into a 120-V light socket?

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/144_Problems.md

@@ -1,39 +0,0 @@
-# Comprehensive Problems
-
-- **12.77** A three-phase generator supplied 10 kVA at a power factor of 0.85 lagging. If 7,500 W are delivered to the load and line losses are 160 W per phase, what are the losses in the generator?
-- **12.78** A three-phase 440-V, 51-kW, 60-kVA inductive load operates at 60 Hz and is wye-connected. It is desired to correct the power factor to 0.95 lagging. What value of capacitor should be placed in parallel with each load impedance?
-- **12.79** A balanced three-phase generator has an *abc* phase sequence with phase voltage **V***an* = 554.3⧸ 0° V. The generator feeds an induction motor which may be represented by a balanced Y-connected load with an impedance of 12 + *j*5 Ω per phase. Find the line currents and the load voltages. Assume a line impedance of 2 Ω per phase.
-- **12.80** A balanced three-phase source furnishes power to the following three loads:
-
-Load 1: 6 kVA at 0.83 pf lagging Load 2: unknown Load 3: 8 kW at 0.7071 pf leading
-
- If the line current is 84.6 A rms, the line voltage at the load is 208 V rms, and the combined load has a 0.8 pf lagging, determine the unknown load.
-
-**12.81** A professional center is supplied by a balanced three-phase source. The center has four balanced three-phase loads as follows:
-
-> Load 1: 150 kVA at 0.8 pf leading Load 2: 100 kW at unity pf Load 3: 200 kVA at 0.6 pf lagging Load 4: 80 kW and 95 kVAR (inductive)
-
- If the line impedance is 0.02 + *j*0.05 Ω per phase and the line voltage at the loads is 480 V, find the magnitude of the line voltage at the source.
-
-- **12.82** A balanced three-phase system has a distribution wire with impedance 2 + *j*6 Ω per phase. The system supplies two three-phase loads that are connected in parallel. The first is a balanced wye-connected load that absorbs 400 kVA at a power factor of 0.8 lagging. The second load is a balanced delta-connected load with impedance of 10 + *j*8 Ω per phase. If the magnitude of the line voltage at the loads is 2400 V rms, calculate the magnitude of the line voltage at the source and the total complex power supplied to the two loads.
-- **12.83** A commercially available three-phase inductive motor operates at a full load of 120 hp (1 hp = 746 W) at 95 percent efficiency at a lagging power
-
-factor of 0.707. The motor is connected in parallel to a 80-kW balanced three-phase heater at unity power factor. If the magnitude of the line voltage is 480 V rms, calculate the line current.
-
-**12.84** Figure 12.76 displays a three-phase delta-connected motor load which is connected to a line voltage of 440 V and draws 4 kVA at a power factor of 72 per cent lagging. In addition, a single 1.8 kVAR capacitor is connected between lines *a* and *b*, while a 800-W lighting load is connected between line *c* and neutral. Assuming the *abc* sequence and taking **V***an* = *Vp*⧸0°, find the magnitude and phase angle of currents **I***a*, **I***b*, **I***c*, and **I***n*.
-
-**Figure 12.76**
-
-For Prob. 12.84.
-
-**12.85** Design a three-phase heater with suitable symmetric loads using wye-connected pure resistance. Assume that the heater is supplied by a 240-V line voltage and is to give 27 kW of heat.
-
-**12.86** For the single-phase three-wire system in Fig. 12.77, find currents **I***aA*, **I***bB*, and **I***nN*.
-
-# **Figure 12.77**
-
-For Prob. 12.86.
-
-**12.87** Consider the single-phase three-wire system shown in Fig. 12.78. Find the current in the neutral wire and the complex power supplied by each source. Take **V***s* as a 220⧸ 0°-V, 60-Hz source.
-
-**Figure 12.78** For Prob. 12.87.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/145_Chapter 13 - Magnetically Coupled Circuits.md

@@ -1,37 +0,0 @@
-# <span id="page-575-0"></span>Magnetically Coupled Circuits
-
-*If you would increase your happiness and prolong your life, forget your neighbor's faults. . . . Forget the peculiarities of your friends, and only remember the good points which make you fond of them. . . . Obliterate everything disagreeable from yesterday; write upon today's clean sheet those things lovely and lovable.*
-
-—Anonymous
-
-# **chapter**
-
-13
-
-# Enhancing Your Career
-
-# **Career in Electromagnetics**
-
-Electromagnetics (EM) is the branch of electrical engineering (or ph ysics) that deals with the analysis and application of electric and magnetic fields. In electromagnetics, electric circuit analysis is applied at low frequencies.
-
-The principles of EM are applied in various allied disciplines, such as electric machines, electromechanical ener gy conversion, radar meteorology, remote sensing, satellite communications, bioelectromagnetics, electromagnetic interference and compatibility, plasmas, and fiber optics. EM devices include electric motors and generators, transformers, electromagnets, magnetic levitation, antennas, radars, microwave ovens, microwave dishes, superconductors, and electrocardiograms. The design of these de vices requires a thorough knowledge of the laws and principles of EM.
-
-EM is regarded as one of the more dif ficult disciplines in electrical engineering. One reason is that EM phenomena are rather abstract. But if one enjoys working with mathematics and can visualize the invisible, one should consider being a specialist in EM, inasmuch as few electrical engineers specialize in this area. Electrical engineers who specialize in EM are needed in micro wave industries, radio/TV broadcasting stations, electromagnetic research laboratories, and se veral communications industries.
-
-Telemetry receiving station for space satellites. © DV169/Getty Images RF
-
-# Historical
-
-<span id="page-576-0"></span>© Bettmann/Corbis
-
-**James Clerk Maxwell** (1831–1879), a graduate in mathematics from Cambridge University, in 1865 wrote a most remarkable paper in which he mathematically unified the laws of Faraday and Ampere. This relationship between the electric field and magnetic field served as the basis for what was later called electromagnetic fields and waves, a major field of study in electrical engineering. The Institute of Electrical and Electron ics Engineers (IEEE) uses a graphical representation of this principle in its logo, in which a straight arrow represents current and a curved arrow represents the electromagnetic field. This relationship is commonly known as *the right-hand rule* . Maxwell was a very active theoretician and scientist. He is best known for the "Maxwell equations." The max well, a unit of magnetic flux, was named after him.
-
-# Learning Objectives
-
-By using the information and exercises in this chapter you will be able to:
-
-- 1. Understand the physics behind mutually coupled circuits and how to analyze circuits containing mutually coupled inductors.
-- 2. Understand how energy is stored in mutually coupled circuits.
-- 3. Understand how linear transformers work and how to analyze circuits containing them.
-- 4. Understand how ideal transformers work and how to analyze circuits containing them.
-- 5. Understand how ideal auto transformers work and know how to analyze them when used in a variety of circuits.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/146_13.1 Introduction.md

@@ -1,67 +0,0 @@
-# **13.1** Introduction
-
-The circuits we have considered so far may be regarded as *conductively coupled*, because one loop affects the neighboring loop through current conduction. When tw o loops with or without contacts between them affect each other through the magnetic field generated by one of them, they are said to be *magnetically coupled*.
-
-The transformer is an electrical de vice designed on the basis of the concept of magnetic coupling. It uses magnetically coupled coils to transfer energy from one circuit to another. Transformers are key circuit elements. They are used in power systems for stepping up or stepping down ac voltages or currents. They are used in electronic circuits such as radio and television receivers for such purposes as impedance matching, isolating one part of a circuit from another, and again for stepping up or down ac voltages and currents.
-
-We will begin with the concept of mutual inductance and introduce the dot convention used for determining the v oltage polarities of inductively coupled components. Based on the notion of mutual inductance,
-
-<span id="page-577-0"></span>we then introduce the circuit element known as the *transformer*. We will consider the linear transformer, the ideal transformer, the ideal autotransformer, and the three-phase transformer. Finally, among their important applications, we look at transformers as isolating and matching devices and their use in power distribution.
-
-# **13.2** Mutual Inductance
-
-When two inductors (or coils) are in a close proximity to each other, the magnetic flux caused by current in one coil links with the other coil, thereby inducing v oltage in the latter . This phenomenon is kno wn as *mutual inductance*.
-
-Let us first consider a single inductor, a coil with *N* turns. When current *i* flows through the coil, a magnetic flux *ϕ* is produced around it (Fig. 13.1). According to Faraday's law, the voltage *v* induced in the coil is proportional to the number of turns *N* and the time rate of change of the magnetic flux *ϕ*; that is,
-
-$$
-v = N \frac{d\phi}{dt} \tag{13.1}
-$$
-
-But the flux *ϕ* is produced by current *i* so that any change in *ϕ* is caused by a change in the current. Hence, Eq. (13.1) can be written as
-
-$$
-v = N \frac{d\phi}{di} \frac{di}{dt}
-$$
- (13.2)
-
-or
-
-$$
-v = L \frac{di}{dt}
-$$
- (13.3)
-
-which is the voltage-current relationship for the inductor. From Eqs. (13.2) and (13.3), the inductance *L* of the inductor is thus given by
-
-$$
-L = N \frac{d\phi}{di}
-$$
- (13.4)
-
-This inductance is commonly called *self-inductance,* because it relates the voltage induced in a coil by a time-varying current in the same coil.
-
-Now consider two coils with self-inductances *L*1 and *L*2 that are in close proximity with each other (Fig. 13.2). Coil 1 has *N*1 turns, while coil 2 has *N*2 turns. F or the sak e of simplicity, assume that the second inductor carries no current. The magnetic flux *ϕ*1 emanating from coil 1 has two components: One component *ϕ*11 links only coil 1, and another component *ϕ*12 links both coils. Hence,
-
-$$
-\phi_1 = \phi_{11} + \phi_{12} \tag{13.5}
-$$
-
-Although the tw o coils are ph ysically separated, the y are said to be *magnetically coupled*. Since the entire flux *ϕ*1 links coil 1, the v oltage induced in coil 1 is
-
-$$
-v_1 = N_1 \frac{d\phi_1}{dt} \tag{13.6}
-$$
-
-Only flux *ϕ*12 links coil 2, so the voltage induced in coil 2 is
-
-$$
-v_2 = N_2 \frac{d\phi_{12}}{dt}
-$$
- (13.7)
-
-**Figure 13.1** Magnetic flux produced by a single coil with *N* turns.
-
-Again, as the fluxes are caused by the current *i*1 flowing in coil 1, Eq. (13.6) can be written as
-
-$$

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/147_13.2 Mutual Inductance.md

@@ -1,398 +0,0 @@
-v_1 = N_1 \frac{d\phi_1}{di_1} \frac{di_1}{dt} = L_1 \frac{di_1}{dt}
-$$
- (13.8)
-
-where *L*1 =  *N*1 *dϕ*1∕*di*1 is the self-inductance of coil 1. Similarly, Eq. (13.7) can be written as
-
-$$
-v_2 = N_2 \frac{d\phi_{12}}{di_1} \frac{di_1}{dt} = M_{21} \frac{di_1}{dt}
-$$
- (13.9)
-
-where
-
-$$
-M_{21} = N_2 \frac{d\phi_{12}}{di_1}
-$$
- (13.10)
-
-*M*21 is known as the *mutual inductance* of coil 2 with respect to coil 1. Subscript 21 indicates that the inductance *M*21 relates the voltage induced in coil 2 to the current in coil 1. Thus, the open-circuit *mutual voltage* (or induced voltage) across coil 2 is
-
-$$
-v_2 = M_{21} \frac{di_1}{dt}
-$$
- (13.11)
-
-Suppose we now let current *i*2 flow in coil 2, while coil 1 carries no current (Fig. 13.3). The magnetic flux *ϕ*2 emanating from coil 2 comprises flux *ϕ*22 that links only coil 2 and flux *ϕ*21 that links both coils. Hence,
-
-$$
-\phi_2 = \phi_{21} + \phi_{22} \tag{13.12}
-$$
-
-The entire flux *ϕ*2 links coil 2, so the voltage induced in coil 2 is
-
-$$
-v_2 = N_2 \frac{d\phi_2}{dt} = N_2 \frac{d\phi_2}{di_2} \frac{di_2}{dt} = L_2 \frac{di_2}{dt}
-$$
- (13.13)
-
-where *L*<sup>2</sup> = *N*<sup>2</sup> *dϕ*2∕*di*2 is the self-inductance of coil 2. Since only flux *ϕ*21 links coil 1, the voltage induced in coil 1 is
-
-$$
-v_1 = N_1 \frac{d\phi_{21}}{dt} = N_1 \frac{d\phi_{21}}{di_2} \frac{di_2}{dt} = M_{12} \frac{di_2}{dt}
-$$
- (13.14)
-
-where
-
-$$
-M_{12} = N_1 \frac{d\phi_{21}}{di_2}
-$$
- (13.15)
-
-which is the *mutual inductance* of coil 1 with respect to coil 2. Thus, the open-circuit *mutual voltage* across coil 1 is
-
-$$
-v_1 = M_{12} \frac{di_2}{dt}
-$$
- (13.16)
-
-We will see in the next section that *M*12 and *M*21 are equal; that is,
-
-$$
-M_{12} = M_{21} = M \tag{13.17}
-$$
-
-and we refer to *M* as the mutual inductance between the tw o coils. Like self-inductance *L*, mutual inductance *M* is measured in henrys (H). Keep in mind that mutual coupling only exists when the inductors or coils are in close proximity, and the circuits are dri ven by time-varying sources. We recall that inductors act like short circuits to dc.
-
-From the two cases in Figs. 13.2 and 13.3, we conclude that mutual inductance results if a v oltage is induced by a time-v arying current in another circuit. It is the property of an inductor to produce a v oltage in reaction to a time-varying current in another inductor near it. Thus,
-
-Mutual inductance *M*12 of coil 1 with respect to coil 2.
-
-Mutual inductance is the ability of one inductor to induce a voltage across a neighboring inductor, measured in henrys (H).
-
-Although mutual inductance *M* is al ways a positi ve quantity, the mutual voltage *M di*∕*dt* may be negative or positive, just like the self-induced voltage *L di*∕*dt*. However, unlike the self-induced *L di*∕*dt*, whose polarity is determined by the reference direction of the current and the reference polarity of the v oltage (according to the passi ve sign convention), the polarity of mutual v oltage *M di*∕*dt* is not easy to determine, because four terminals are involved. The choice of the correct polarity for *M di*∕*dt* is made by examining the orientation or particular way in which both coils are physically wound and applying Lenz's law in conjunction with the right-hand rule. Since it is inconvenient to show the construction details of coils on a circuit schematic, we apply the *dot convention* in circuit analysis. By this convention, a dot is placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil. This is illustrated in Fig. 13.4. Given a circuit, the dots are already placed beside the coils so that we need not bother about how to place them. The dots are used along with the dot con vention to determine the polarity of the mutual voltage. The dot convention is stated as follows:
-
-If a current enters the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is positive at the dotted terminal of the second coil.
-
-# Alternatively,
-
-If a current leaves the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is negative at the dotted terminal of the second coil.
-
-Thus, the reference polarity of the mutual v oltage depends on the ref erence direction of the inducing current and the dots on the coupled coils. Application of the dot con vention is illustrated in the four pairs of mutually coupled coils in Fig. 13.5. Fo r the coupled coils in Fig. 13.5(a), the sign of the mutual voltage *v*2 is determined by the reference polarity for *v*2 and the direction of *i*1. Since *i*1 enters the dotted terminal of coil 1 and *v*2 is positive at the dotted terminal of coil 2, the mutual voltage is +*M di*1∕*dt*. For the coils in Fig. 13.5(b), the current *i*1 enters
-
-Illustration of the dot convention.
-
-**Figure 13.5** Examples illustrating how to apply the dot convention.
-
-the dotted terminal of coil 1 and *v*2 is ne gative at the dotted terminal of coil 2. Hence, the mutual v oltage is −*M di*1∕*dt*. The same reasoning applies to the coils in Figs. 13.5(c) and 13.5(d).
-
-Figure 13.6 shows the dot convention for coupled coils in series. For the coils in Fig. 13.6(a), the total inductance is
-
-$$
-L = L_1 + L_2 + 2M
-$$
- (Series-aiding connection) (13.18)
-
-For the coils in Fig. 13.6(b),
-
-$$
-L = L_1 + L_2 - 2M
-$$
- (Series-opposing connection) (13.19)
-
-Now that we know how to determine the polarity of the mutual voltage, we are prepared to analyze circuits involving mutual inductance. As the first example, consider the circuit in Fig. 13.7(a). Applying KVL to coil 1 gives
-
-$$
-v_1 = i_1 R_1 + L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}
-$$
- (13.20a)
-
-For coil 2, KVL gives
-
-$$
-v_2 = i_2 R_2 + L_2 \frac{di_2}{dt} + M \frac{di_1}{dt}
-$$
- (13.20b)
-
-We can write Eq. (13.20) in the frequency domain as
-
-$$
-\mathbf{V}_1 = (R_1 + j\omega L_1)\mathbf{I}_1 + j\omega M \mathbf{I}_2 \tag{13.21a}
-$$
-
-$$
-\mathbf{V}_2 = j\omega M \mathbf{I}_1 + (R_2 + j\omega L_2) \mathbf{I}_2 \tag{13.21b}
-$$
-
-As a second e xample, consider the circuit in Fig. 13.7(b). We analyze this in the frequency domain. Applying KVL to coil 1, we get
-
-$$
-\mathbf{V} = (\mathbf{Z}_1 + j\omega L_1)\mathbf{I}_1 - j\omega M\mathbf{I}_2
-$$
- (13.22a)
-
-For coil 2, KVL yields
-
-$$
-0 = -j\omega M I_1 + (Z_L + j\omega L_2)I_2 \tag{13.22b}
-$$
-
-Equations (13.21) and (13.22) are solv ed in the usual manner to deter mine the currents.
-
-One of the most important things in making sure one solv es problems accurately is to be able to check each step during the solution pro cess and to mak e sure assumptions can be v erified. Too often, solving mutually coupled circuits requires the problem solv er to track tw o or more steps made at once re garding the sign and v alues of the mutually induced voltages.
-
-# **Figure 13.7**
-
-Time-domain analysis of a circuit containing coupled coils (a) and frequency-domain analysis of a circuit containing coupled coils (b).
-
-# **Figure 13.6**
-
-Dot convention for coils in series; the sign indicates the polarity of the mutual voltage: (a) seriesaiding connection, (b) seriesopposing connection.
-
-Model that makes analysis of mutually coupled easier to solve.
-
-Experience has sho wn that if we break the problem into steps of solving for the value and the sign into separate steps, the decisions made are easier to track. We suggest that model (Figure 13.8 (b)) be used when analyzing circuits containing a mutually c oupled circuit shown in Figure 13.8(a):
-
-Notice that we have not included the signs in the model. The reason for that is that we first determine the value of the induced voltages and then we determine the appropriate signs. Clearly, I1 induces a voltage within the second coil represented by the value *jω*I1 and I2 induces a voltage of *jω*I2 in the first coil. Once we have the values we next use both circuits to find the correct signs for the dependent sources as shown in Figure 13.8(c).
-
-Since I1 enters *L*1 at the dotted end, it induces a voltage in *L*2 that tries to force a current out of the dotted end of *L*2 which means that the source must have a plus on top and a minus on the bottom as sho wn in Figure 13.8(c). I2 leaves the dotted end of *L*2 which means that it induces a voltage in *L*1 which tries to force a current into the dotted end of *L*<sup>1</sup> requiring a dependent source that has a plus on the bottom and a minus on top as shown in Figure 13.8(c). No w all we have to do is to analyze a circuit with two dependent sources. This process allows you to check each of your assumptions.
-
-At this introductory level we are not concerned with the determination of the mutual inductances of the coils and their dot placements. Lik e *R*, *L*, and *C*, calculation of *M* would involve applying the theory of elect romagnetics to the actual ph ysical properties of the coils. In this te xt, we assume that the mutual inductance and the placement of the dots are the "gi vens'' of the circuit problem, like the circuit components *R*, *L*, and *C*.
-
-For Example 13.1.
-
-$$
--12 + (-j4 + j5)\mathbf{I}_1 - j3\mathbf{I}_2 = 0
-$$
-
-$$
-jI_1 - j3I_2 = 12 \tag{13.1.1}
-$$
-
-For loop 2, KVL gives
-
-$$
--j3\mathbf{I}_1 + (12 + j6)\mathbf{I}_2 = 0
-$$
-
-or
-
-$$
-\mathbf{I}_1 = \frac{(12 + j6)\mathbf{I}_2}{j3} = (2 - j4)\mathbf{I}_2
-$$
- (13.1.2)
-
-Substituting this in Eq. (13.1.1), we get
-
-( *j*2 + 4 − *j*3)**I**<sup>2</sup> = (4 − *j*)**I**<sup>2</sup> = 12
-
-or
-
-$$
-\mathbf{I}_2 = \frac{12}{4 - j} = 2.91 \underline{ / 14.04^{\circ}} A \tag{13.1.3}
-$$
-
-From Eqs. (13.1.2) and (13.1.3),
-
-$$
-\mathbf{I}_1 = (2 - j4)\mathbf{I}_2 = (4.472 \angle -63.43^\circ)(2.91 \angle 14.04^\circ)
-$$
-  
-= 13.01 \angle -49.39^\circ A
-
-Practice Problem 13.1 Determine the voltage **V***o* in the circuit of Fig. 13.10.
-
-**Figure 13.10** For Practice Prob. 13.1.
-
-**Answer:** 12⧸ −45° V.
-
-Example 13.2 Calculate the mesh currents in the circuit of Fig. 13.11.
-
-**Figure 13.11** For Example 13.2.
-
-# **Solution:**
-
-The key to analyzing a magnetically coupled circuit is knowing the polarity of the mutual voltage. We need to apply the dot rule. In Fig. 13.11, suppose coil 1 is the one whose reactance is 6 Ω, and coil 2 is the one whose reactance is 8 Ω. To figure out the polarity of the mutual voltage in coil 1 due to current **I**2, we observe that **I**2 leaves the dotted terminal of coil 2. Since we are applying KVL in the clockwise direction, it implies that the mutual voltage is negative, that is, −*j*2**I**2.
-
-Alternatively, it might be best to figure out the mutual voltage by redrawing the rele vant portion of the circuit, as sho wn in Fig. 13.12, where it becomes clear that the mutual voltage is **V**<sup>1</sup> = −2*j* **I**2.
-
-Thus, for mesh 1 in Fig. 13.11, KVL gives
-
-$$
--100 + \mathbf{I}_1(4 - j3 + j6) - j6\mathbf{I}_2 - j2\mathbf{I}_2 = 0
-$$
-
-or
-
-$$
-100 = (4+j3)\mathbf{I}_1 - j8\mathbf{I}_2 \tag{13.2.1}
-$$
-
-Similarly, to figure out the mutual voltage in coil 2 due to current **I**1, consider the relevant portion of the circuit, as shown in Fig. 13.12. Applying the dot convention gives the mutual voltage as **V**<sup>2</sup> = −2*j***I**1. Also, current **I**2 sees the two coupled coils in series in Fig. 13.11; since it leaves the dotted terminals in both coils, Eq. (13.18) applies. Therefore, for mesh 2 in Fig. 13.11, KVL gives
-
-$$
-0 = -2jI_1 - j6I_1 + (j6 + j8 + j2 \times 2 + 5)I_2
-$$
-
-or
-
-$$
-0 = -j8I_1 + (5+j18)I_2 \tag{13.2.2}
-$$
-
-Putting Eqs. (13.2.1) and (13.2.2) in matrix form, we get
-
-$$
-\begin{bmatrix} 100 \\ 0 \end{bmatrix} = \begin{bmatrix} 4+j3 & -j8 \\ -j8 & 5+j18 \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}
-$$
-
-The determinants are
-
-$$
-\Delta = \begin{vmatrix} 4+j3 & -j8 \\ -j8 & 5+j18 \end{vmatrix} = 30 + j87
-$$
-  
-\n
-$$
-\Delta_1 = \begin{vmatrix} 100 & -j8 \\ 0 & 5+j18 \end{vmatrix} = 100(5+j18)
-$$
-  
-\n
-$$
-\Delta_2 = \begin{vmatrix} 4+j3 & 100 \\ -j8 & 0 \end{vmatrix} = j800
-$$
-
-Thus, we obtain the mesh currents as
-
-Thus, we obtain the mesh currents as  
-\n
-$$
-\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{100(5 + j18)}{30 + j87} = \frac{1,868.2/74.5^{\circ}}{92.03/71^{\circ}} = 20.3/3.5^{\circ} \text{ A}
-$$
-\n
-$$
-\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{j800}{30 + j87} = \frac{800/90^{\circ}}{92.03/71^{\circ}} = 8.693/19^{\circ} \text{ A}
-$$
-
-Determine the phasor currents **I**1 and **I**2 in the circuit of Fig. 13.13. Practice Problem 13.2
-
-**Figure 13.13** For Practice Prob. 13.2.
-
-**Answer:** I1 = 17.889⧸86.57° A, I2 = 26.83⧸86.57° A.
-
-**Figure 13.12** Model for Example 13.2 showing the polarity of the induced voltages.
-
-# **13.3** Energy in a Coupled Circuit
-
-In Chapter 6, we saw that the energy stored in an inductor is given by
-
-$$
-w = \frac{1}{2} L i^2
-$$
- (13.23)
-
-We now want to determine the energy stored in magnetically coupled coils.
-
-Consider the circuit in Fig. 13.14. We assume that currents *i*1 and *i*<sup>2</sup> are zero initially, so that the ener gy stored in the coils is zero. If we let *i*<sup>1</sup> increase from zero to *I*1 while maintaining *i*2 = 0, the power in coil 1 is
-
-$$
-p_1(t) = v_1 i_1 = i_1 L_1 \frac{di_1}{dt}
-$$
- (13.24)
-
-and the energy stored in the circuit is
-
-$$
-w_1 = \int p_1 dt = L_1 \int_0^{I_1} i_1 dt_1 = \frac{1}{2} L_1 I_1^2
-$$
- (13.25)
-
-If we now maintain *i*1 = *I*1 and increase *i*2 from zero to *I*2, the mutual voltage induced in coil 1 is *M*<sup>12</sup> *di*2∕*dt*, while the mutual v oltage induced in coil 2 is zero, since *i*1 does not change. The power in the coils is now
-
-$$
-p_2(t) = i_1 M_{12} \frac{di_2}{dt} + i_2 v_2 = I_1 M_{12} \frac{di_2}{dt} + i_2 L_2 \frac{di_2}{dt}
-$$
- (13.26)
-
-and the energy stored in the circuit is
-
-$$
-w_2 = \int p_2 dt = M_{12} I_1 \int_0^{I_2} di_2 + L_2 \int_0^{I_2} i_2 di_2
-$$
-$$
-= M_{12} I_1 I_2 + \frac{1}{2} L_2 I_2^2
-$$
-(13.27)
-
-The total ener gy stored in the coils when both *i*1 and *i*2 have reached constant values is
-
-$$
-w = w_1 + w_2 = \frac{1}{2} L_1 I_1^2 + \frac{1}{2} L_2 I_2^2 + M_{12} I_1 I_2
-$$
- (13.28)
-
-If we reverse the order by which the currents reach their final values, that is, if we first increase *i*2 from zero to *I*2 and later increase *i*1 from zero to *I*1, the total energy stored in the coils is
-
-$$
-w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + M_{21}I_1I_2
-$$
-\n(13.29)
-
-Because the total energy stored should be the same regardless of how we reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us to conclude that
-
-$$
-M_{12} = M_{21} = M \tag{13.30a}
-$$
-
-and
-
-$$
-w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + MI_1I_2
-$$
- (13.30b)
-
-This equation was derived based on the assumption that the coil currents both entered the dotted terminals. If one current enters one dotted
-
-<span id="page-584-0"></span>**Figure 13.14** The circuit for deriving energy stored in a coupled circuit.
-
-terminal while the other current leaves the other dotted terminal, the mutual voltage is negative, so that the mutual energy *MI*1*I*2 is also negative. In that case,
-
-$$
-w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 - MI_1I_2
-$$
- (13.31)
-
-Also, because *I*1 and *I*2 are arbitrary v alues, they may be replaced by *i*1 and *i*2, which gives the instantaneous ener gy stored in the circuit the general expression
-
-$$
-w = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 \pm Mi_1i_2
-$$
- (13.32)
-
-The positive sign is selected for the mutual term if both currents enter or leave the dotted terminals of the coils; the ne gative sign is selected otherwise.
-
-We will now establish an upper limit for the mutual inductance M. The energy stored in the circuit cannot be negative because the circuit is passive. This means that the quantity 1∕2*L*1 *i* 1 <sup>2</sup> + 1∕2*L*2 *i* 2 <sup>2</sup> − *Mi*1*i*2 must be greater than or equal to zero:
-
-$$
-\frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 - Mi_1i_2 \ge 0
-$$
-\n(13.33)
-
-To complete the square, we both add and subtract the term *i*1*i*2( √ *L*1*L*2 on the right-hand side of Eq. (13.33) and obtain
-
-$$
-\frac{1}{2}(i_1\sqrt{L_1} - i_2\sqrt{L_2})^2 + i_1i_2(\sqrt{L_1L_2} - M) \ge 0
-$$
- (13.34)
-
-The squared term is ne ver negative; at its least it is zero. Therefore, the sec ond term on the right-hand side of Eq. (13.34) must be greater than zero; that is,
-
-√
-
-$$
-\overline{L_1 L_2} - M \ge 0
-$$
-  
-$$
-M \le \sqrt{L_1 L_2}

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/148_13.3 Energy in a Coupled Circuit.md

@@ -1,131 +0,0 @@
-$$
- (13.35)
-
-Thus, the mutual inductance cannot be greater than the geometric mean of the self-inductances of the coils. The e xtent to which the mutual inductance *M* approaches the upper limit is specified by the *coefficient of coupling k*, given by
-
-$$
-k = \frac{M}{\sqrt{L_1 L_2}}\tag{13.36}
-$$
-
-$$
-M = k\sqrt{L_1 L_2} \tag{13.37}
-$$
-
-where 0 ≤ *k* ≤ 1 or equivalently 0 ≤ *M* ≤  √ \_\_\_\_ *L*1*L*2 . The coupling coefficient is the fraction of the total flux emanating from one coil that links the other coil. For example, in Fig. 13.2,
-
-$$
-k = \frac{\phi_{12}}{\phi_1} = \frac{\phi_{12}}{\phi_{11} + \phi_{12}}
-$$
- (13.38)
-
-or
-
-or
-
-and in Fig. 13.3,
-
-# **Figure 13.15**
-
-Windings: (a) loosely coupled, (b) tightly coupled; cutaway view demonstrates both windings.
-
-$$
-k = \frac{\phi_{21}}{\phi_2} = \frac{\phi_{21}}{\phi_{21} + \phi_{22}}
-$$
- (13.39)
-
-If the entire flux produced by one coil links another coil, then *k* = 1 and we have 100 percent coupling, or the coils are said to be *perfectly coupled*. For *k* < 0.5, coils are said to be *loosely coupled*; and for *k* > 0.5, they are said to be *tightly coupled*. Thus,
-
-The coupling coefficient k is a measure of the magnetic coupling between two coils; 0 ≤ k ≤ 1.
-
-We expect *k* to depend on the closeness of the two coils, their core, their orientation, and their windings. Figure 13.15 sho ws loosely coupled windings and tightly coupled windings. The air -core trans formers used in radio frequenc y circuits are loosely coupled, whereas iron-core transformers used in po wer systems are tightly coupled. The linear transformers discussed in Section 3.4 are mostly air -core; the ideal transformers discussed in Sections 13.5 and 13.6 are principally iron-core.
-
-**Figure 13.16** For Example 13.3.
-
-Consider the circuit in Fig. 13.16. Determine the coupling coef ficient. Calculate the ener gy stored in the coupled inductors at time *t* = 1 s if *v* = 60 cos(4*t* + 30°) V.
-
-# **Solution:**
-
-The coupling coefficient is
-
-$$
-k = \frac{M}{\sqrt{L_1 L_2}} = \frac{2.5}{\sqrt{20}} = 0.56
-$$
-
-indicating that the inductors are tightly coupled. To find the energy stored, we need to calculate the current. To find the current, we need to obtain the frequency-domain equivalent of the circuit.
-
-60 cos(4*t* + 30°) 
-$$
-\Rightarrow
-$$
- 60/30°,  $\omega = 4$  rad/s  
-\n5 H  $\Rightarrow$   $j\omega L_1 = j20 \Omega$   
-\n2.5 H  $\Rightarrow$   $j\omega M = j10 \Omega$   
-\n4 H  $\Rightarrow$   $j\omega L_2 = j16 \Omega$   
-\n $\frac{1}{16}F \Rightarrow \frac{1}{j\omega C} = -j4 \Omega$ 
-
-The frequency-domain equivalent is shown in Fig. 13.17. We now apply mesh analysis. For mesh 1,
-
-$$
-(10 + j20)\mathbf{I}_1 + j10\mathbf{I}_2 = 60/30^{\circ}
-$$
- (13.3.1)
-
-For mesh 2,
-
-$$
-j10\mathbf{I}_1 + (j16 - j4)\mathbf{I}_2 = 0
-$$
-
-$$
-\overline{a}
-$$
-
-$$
-I_1 = -1.2I_2 \tag{13.3.2}
-$$
-
-<span id="page-587-0"></span>Substituting this into Eq. (13.3.1) yields
-
-$$
-I_2(-12 - j14) = 60/30^{\circ} \qquad \Rightarrow \qquad I_2 = 3.254/160.6^{\circ} \text{ A}
-$$
-
-and
-
-$$
-I_1 = -1.2I_2 = 3.905 \div 19.4^{\circ}
-$$
- A
-
-In the time-domain,
-
-$$
-i_1 = 3.905 \cos(4t - 19.4^\circ),
-$$
-  $i_2 = 3.254 \cos(4t + 160.6^\circ)$ 
-
-At time *t* = 1 s, 4*t* = 4 rad = 229.2°, and
-
-$$
-i_1 = 3.905 \cos(229.2^\circ - 19.4^\circ) = -3.389 \text{ A}
-$$
-  
-$$
-i_2 = 3.254 \cos(229.2^\circ + 160.6^\circ) = 2.824 \text{ A}
-$$
-
-The total energy stored in the coupled inductors is
-
-$$
-w = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 + Mi_1i_2
-$$
-  
-=  $\frac{1}{2}(5)(-3.389)^2 + \frac{1}{2}(4)(2.824)^2 + 2.5(-3.389)(2.824) = 20.73 \text{ J}$ 
-
-**Figure 13.17**
-
-Frequency-domain equivalent of the circuit in Fig. 13.16.
-
-For the circuit in Fig. 13.18, determine the coupling coefficient and the energy stored in the coupled inductors at *t* = 1.5 s. Practice Problem 13.3
-
-**Answer:** 0.7071, 246.2 J.