www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`Theory and Problems of
`ELECTRIC
`CIRCUITS
`Fourth Edition
`
`MAHMOOD NAHVI, Ph.D.
`Professor of Electrical Engineering
`California Polytechnic State University
`
`JOSEPH A. EDMINISTER
`Professor Emeritus of Electrical Engineering
`The University of Akron
`
`Schaum’s Outline Series
`McGRAW-HILL
`New York Chicago San Francisco Lisbon
`London Madrid Mexico City Milan New Dehli
`San Juan Seoul Singapore Sydney Toronto
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`Copyright © 2003, 1997, 1986, 1965] by The McGraw-Hill Companies, Inc. All rights reserved. Manufactured in the United
`States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be repro-
`duced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permis-
`sion of the publisher.
`
`0-07-142582-9
`
`The material in this eBook also appears in the print version of this title: 0-07-139307-2.
`
`All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a
`trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention
`of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps.
`
`McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in cor-
`porate training programs. For more information, please contact George Hoare, Special Sales, at george_hoare@mcgraw-
`hill.com or (212) 904-4069.
`
`TERMS OF USE
`This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and its licensors reserve all rights in
`and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the
`right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify,
`create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it
`without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use
`of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms.
`
`THE WORK IS PROVIDED “AS IS”. McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WAR-
`RANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM
`USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA
`HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED,
`INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PAR-
`TICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work
`will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors
`shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any dam-
`ages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work.
`Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, con-
`sequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised
`of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such
`claim or cause arises in contract, tort or otherwise.
`
`DOI: 10.1036/0071425829
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`Sinusoidal Steady-
`State Circuit Analysis
`
`9.1
`
`INTRODUCTION
`
`This chapter will concentrate on the steady-state response of circuits driven by sinusoidal sources.
`For a linear circuit, the assumption of a sinusoidal source
`The response will also be sinusoidal.
`represents no real restriction, since a source that can be described by a periodic function can be replaced
`by an equivalent combination (Fourier series) of sinusoids. This matter will be treated in Chapter 17.
`
`9.2 ELEMENT RESPONSES
`
`The voltage-current relationships for the single elements R, L, and C were examined in Chapter 2
`In this chapter, the functions of v and i will be sines or cosines with the
`and summarized in Table 2-1.
`argument !t. ! is the angular frequency and has the unit rad/s. Also, ! ¼ 2f , where f is the frequency
`with unit cycle/s, or more commonly hertz (Hz).
`Consider an inductance L with i ¼ I cos ð!t þ 458Þ A [see Fig. 9-1(a)]. The voltage is
`vL ¼ L
`¼ !LI½ sin ð!t þ 458ފ ¼ !LI cos ð!t þ 1358Þ
`ðVÞ
`di
`dt
`
`191
`
`
`Copyright 2003, 1997, 1986, 1965 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.1965 by The McGraw Hill Companie1965 by The McGraw Hill Compani
`www.EngineeringEBooksPdf.com
`
`Fig. 9-1
`
`IPR2023-00697
`Theta EX2011
`
`

`

`192
`
`SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
`
`[CHAP. 9
`
`A comparison of vL and i shows that the current lags the voltage by 908 or =2 rad. The functions
`are sketched in Fig. 9-1(b). Note that the current function i is to the right of v, and since the horizontal
`scale is !t, events displaced to the right occur later in time. This illustrates that i lags v. The horizontal
`scale is in radians, but note that it is also marked in degrees (1358; 1808, etc.). This is a case of mixed
`units just as with !t þ 458.
`It is not mathematically correct but is the accepted practice in circuit
`analysis. The vertical scale indicates two different quantities, that is, v and i, so there should be two
`scales rather than one.
`While examining this sketch, it is a good time to point out that a sinusoid is completely defined when
`its magnitude ðV or IÞ, frequency (! or f ), and phase (458 or 1358) are specified.
`In Table 9-1 the responses of the three basic circuit elements are shown for applied current
`i ¼ I cos !t and voltage v ¼ V cos !t.
`If sketches are made of these responses, they will show that
`for a resistance R, v and i are in phase. For an inductance L, i lags v by 908 or =2 rad. And for a
`capacitance C, i leads v by 908 or =2 rad.
`
`Table 9-1
`
`i ¼ I cos !t
`
`v ¼ V cos !t
`
`vr ¼ RI cos !t
`
`iR ¼ V
`R cos !t
`
`vL ¼ !LI cos ð!t þ 908Þ
`
`!L cosð!t 908Þ
`iL ¼ V
`
`vC ¼ I
`!C cos ð!t 908Þ
`
`iC ¼ !CV cosð!t þ 908Þ
`
`EXAMPLE 9.1 The RL series circuit shown in Fig. 9-2 has a current i ¼ I sin !t. Obtain the voltage v across the
`two circuit elements and sketch v and i.
`¼ !LI sin ð!t þ 908Þ
`vL ¼ L
`vR ¼ RI sin !t
`di
`dt
`v ¼ vR þ vL ¼ RI sin !t þ !LI sin ð!t þ 908Þ
`
`Fig. 9-2
`
`Since the current is a sine function and
`v ¼ V sin ð!t þ Þ ¼ V sin !t cos  þ V cos !t sin 
`
`we have from the above
`
`v ¼ RI sin !t þ !LI sin !t cos 908 þ !LI cos !t sin 908
`
`ð1Þ
`
`ð2Þ
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`CHAP. 9]
`
`SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
`
`193
`
`Equating coefficients of like terms in (1) and (2),
`V sin  ¼ !LI
`q
`ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
`R2 þ ð!LÞ2
`v ¼ I
`q
`ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
`R2 þ ð!LÞ2
`V ¼ I
`
`Then
`
`and
`
`V cos  ¼ RI
`and
`sin ½!t þ arctanð!L=Rފ
` ¼ tan
`!L
`1
`R
`
`and
`
`The functions i and v are sketched in Fig. 9-3. The phase angle , the angle by which i lags v, lies within the
`range 08    908, with the limiting values attained for !L  R and !L  R, respectively.
`If the circuit had an
`applied voltage v ¼ V sin !t, the resulting current would be
`q
`Vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
`i ¼
`R2 þ ð!LÞ2
`
`sin ð!t Þ
`
`where, as before,  ¼ tan
`
`1 ð!L=RÞ.
`
`EXAMPLE 9.2 If the current driving a series RC circuit is given by i ¼ I sin !t, obtain the total voltage across the
`two elements.
`
`Fig. 9-3
`
`where
`
`V ¼ I
`
`vC ¼ ð1=!CÞ sin ð!t 908Þ
`vR ¼ RI sin !t
`v ¼ vR þ vC ¼ V sin ð!t Þ
`ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
`q
`R2 þ ð1=!CÞ2
` ¼ tan
`
`and
`
`1 ð1=!CRÞ
`
`The negative phase angle shifts v to the right of the current i. Consequently i leads v for a series RC circuit. The
`phase angle is constrained to the range 08    908. For ð1=!CÞ  R, the angle  ¼ 08, and for ð1=!CÞ  R, the
`angle  ¼ 908. See Fig. 9-4.
`
`Fig. 9-4
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`194
`
`SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
`
`[CHAP. 9
`
`9.3 PHASORS
`
`A brief look at the voltage and current sinusoids in the preceding examples shows that the ampli-
`tudes and phase differences are the two principal concerns. A directed line segment, or phasor, such as
`that shown rotating in a counterclockwise direction at a constant angular velocity ! (rad/s) in Fig. 9-5,
`has a projection on the horizontal which is a cosine function. The length of the phasor or its magnitude
`is the amplitude or maximum value of the cosine function. The angle between two positions of the
`phasor is the phase difference between the corresponding points on the cosine function.
`
`Fig. 9-5
`
`Throughout this book phasors will be defined from the cosine function.
`If a voltage or current is
`expressed as a sine, it will be changed to a cosine by subtracting 908 from the phase.
`Consider the examples shown in Table 9-2. Observe that the phasors, which are directed line
`segments and vectorial in nature, are indicated by boldface capitals, for example, V, I. The phase
`angle of the cosine function is the angle on the phasor. The phasor diagrams here and all that follow
`may be considered as a snapshot of the counterclockwise-rotating directed line segment taken at t ¼ 0.
`The frequency f (Hz) and ! (rad/s) generally do not appear but they should be kept in mind, since they
`are implicit in any sinusoidal steady-state problem.
`
`EXAMPLE 9.3 A series combination of R ¼ 10
` and L ¼ 20 mH has a current i ¼ 5:0 cos ð500t þ 108) (A).
`Obtain the voltages v and V, the phasor current I and sketch the phasor diagram.
`Using the methods of Example 9.1,
`¼ 50:0 cosð500t þ 1008Þ
`vL ¼ L
`vR ¼ 50:0 cos ð500t þ 108Þ
`di
`dt
`v ¼ vR þ vL ¼ 70:7 cos ð500t þ 558Þ
`ðVÞ
`
`The corresponding phasors are
`
`I ¼ 5:0
`
`108 A
`
`and
`
`V ¼ 70:7
`
`558 V
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`CHAP. 9]
`
`SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
`
`195
`
`Table 9-2
`
`Function
`
`Phasor Representation
`
`v ¼ 150 cos ð500t þ 458Þ
`
`ðVÞ
`
`i ¼ 3:0 sin ð2000t þ 308Þ
`¼ 3:0 cos ð2000t 608Þ
`
`ðmAÞ
`ðmAÞ
`
`The phase angle of 458 can be seen in the time-domain graphs of i and v shown in Fig. 9-6(a), and the phasor
`diagram with I and V shown in Fig. 9-6(b).
`
`Fig. 9-6
`
`Phasors can be treated as complex numbers. When the horizontal axis is identified as the real axis
`In view of Euler’s
`of a complex plane, the phasors become complex numbers and the usual rules apply.
`identity, there are three equivalent notations for a phasor.
`V ¼ V 
`V ¼ Vðcos  þ j sin Þ
`V ¼ Ve j
`
`rectangular form
`
`exponential form
`
`polar form
`
`Since
`
`The cosine expression may also be written as
`v ¼ V cos ð!t þ Þ ¼ Re ½Ve jð!tþފ ¼ Re ½Ve j!tŠ
`The exponential
`form suggests how to treat
`the product and quotient of phasors.
`1ÞðV
`2Þ þ V
`ðV
`þ
`Þ
`e jð
`e j
`e j
`V
`,
`1
`2
`2
`
`1
`
`2
`
`1
`
`ðV
`
`ÞðV
`
`
`
`1
`
`1
`
`
`
`2
`
`2
`
`Þ ¼ V
`
`V
`2
`
`
`
`1
`
`1
`
`þ 
`
`2
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`196
`
`SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
`
`[CHAP. 9
`
`and, since ðV
`
`1
`
`e j1Þ=ðV
`
`2
`
`e j2Þ ¼ ðV
`
`1
`
`=V
`
`2
`
`Þe jð12Þ;
`
`V
`
`1
`V
`2
`
`
`
`1
`
`
`¼ V
`
`=V
`2
`
`1
`
` 
`
`2
`
`
`
`1
`
`The rectangular form is used in summing or subtracting phasors.
`¼ 25:0
`¼ 11:2
`143:138 and V2
`26:578, find the ratio V1
`143:138
`¼ 2:23
`¼ 25:0
`116:568 ¼ 1:00 þ j1:99
`26:578
`11:2
`¼ ð20:0 þ j15:0Þ þ ð10:0 þ j5:0Þ ¼ 10:0 þ j20:0 ¼ 23:36
`
`EXAMPLE 9.4 Given V1
`
`=V2
`V1
`þ V2
`
`V1
`
`=V2 and the sum V1
`
`þ V2.
`
`116:578
`
`9.4
`
`IMPEDANCE AND ADMITTANCE
`
`A sinusoidal voltage or current applied to a passive RLC circuit produces a sinusoidal response.
`With time functions, such as vðtÞ and iðtÞ, the circuit is said to be in the time domain, Fig. 9-7(a); and
`when the circuit is analyzed using phasors, it is said to be in the frequency domain, Fig. 9-7(b). The
`voltage and current may be written, respectively,
`vðtÞ ¼ V cos ð!t þ Þ ¼ Re ½Ve j!tŠ
`V ¼ V 
`I ¼ I 
`iðtÞ ¼ I cos ð!t þ Þ ¼ Re ½Ie j!tŠ
`and
`The ratio of phasor voltage V to phasor current I is defined as impedance Z, that is, Z ¼ V=I. The
`reciprocal of impedance is called admittance Y, so that Y ¼ 1=Z (S), where 1 S ¼ 1
`1 ¼ 1 mho. Y and
`Z are complex numbers.
`
`and
`
`Fig. 9-7
`
`When impedance is written in Cartesian form the real part is the resistance R and the imaginary part
`is the reactance X. The sign on the imaginary part may be positive or negative: When positive, X is
`called the inductive reactance, and when negative, X is called the capacitive reactance. When the
`admittance is written in Cartesian form, the real part is admittance G and the imaginary part is suscep-
`tance B. A positive sign on the susceptance indicates a capacitive susceptance, and a negative sign
`indicates an inductive susceptance. Thus,
`Z ¼ R jXC
`Z ¼ R þ jXL
`Y ¼ G þ jBC
`Y ¼ G jBL
`and
`The relationships between these terms follow from Z ¼ 1=Y. Then,
`X ¼ B
`R ¼
`G
`G2 þ B2
`G2 þ B2
`B ¼ X
`G ¼
`R
`R2 þ X 2
`R2 þ X 2
`
`and
`
`and
`
`and
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`CHAP. 9]
`
`SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
`
`197
`
`These expressions are not of much use in a problem where calculations can be carried out with the
`numerical values as in the following example.
`
`EXAMPLE 9.5 The phasor voltage across the terminals of a network such as that shown in Fig. 9-7(b) is
`458 V and the resulting current is 5:0
`158 A. Find the equivalent impedance and admittance.
`100:0
`458
`308 ¼ 17:32 þ j10:0
`
`Z ¼ V
`¼ 20:0
`100:0
`158
`5:0
`I
`¼ 1
`
`Y ¼ I
`Z
`V
`
`2 S.Thus, R ¼ 17:32
`, XL ¼ 10:0
`, G ¼ 4:33  102 S, and BL ¼ 2:50  10
`
`
`¼ 0:05 30 ¼ ð4:33 j2:50Þ  102 S
`
`Combinations of Impedances
`The relation V ¼ IZ (in the frequency domain) is formally identical to Ohm’s law, v ¼ iR, for a
`resistive network (in the time domain). Therefore, impedances combine exactly like resistances:
`þ
`þ
`
`impedances in series
`
`impedances in parallel
`
`Zeq
`
`¼ Z1
`þ Z2
`¼ 1
`þ 1
`1
`Z1
`Zeq
`Z2
`=ðZ1
`þ Z2
`Þ.
`
`In particular, for two parallel impedances, Zeq
`
`¼ Z1Z2
`
`Impedance Diagram
`In an impedance diagram, an impedance Z is represented by a point in the right half of the complex
`plane. Figure 9-8 shows two impedances; Z1, in the first quadrant, exhibits inductive reactance, while
`þ Z2, is obtained
`Z2, in the fourth quadrant, exhibits capacitive reactance. Their series equivalent, Z1
`by vector addition, as shown. Note that the ‘‘vectors’’ are shown without arrowheads, in order to
`distinguish these complex numbers from phasors.
`
`Fig. 9-8
`
`Combinations of Admittances
`Replacing Z by 1/Y in the formulas above gives
`
`admittances in series
`
`admittances in parallel
`
`1
`Yeq
`Yeq
`
`¼ 1
`Y1
`¼ Y1
`
`þ 1
`Y2
`þ Y2
`
`þ
`þ
`
`Thus, series circuits are easiest treated in terms of impedance; parallel circuits, in terms of admittance.
`
`www.EngineeringEBooksPdf.com
`
`IPR2023-00697
`Theta EX2011
`
`

`

`198
`
`SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
`
`[CHAP. 9
`
`Admittance Diagram
`Figure 9-9, an admittance diagram, is analogous to Fig. 9-8 for impedance. Shown are an admit-
`tance Y1 having capacitive susceptance and an admittance Y2 having inductive susceptance, together
`þ Y2, which is the admittance of a parallel combination of Y1 and Y2.
`with their vector sum, Y1
`
`Fig. 9-9
`
`9.5 VOLTAGE AND CURRENT DIVISION IN THE FREQUENCY DOMAIN
`
`In view of the analogy between impedance in the frequency domain and resistance in the time
`domain, Sections 3.6 and 3.7 imply the following results.
`(1) Impedances in series divide the total voltage in the ratio of the impedances:
`¼ Zr
`Vr ¼ Zr
`Zs
`Zeq
`
`or
`
`VT
`
`Vr
`Vs
`
`See Fig. 9-10.
`
`Fig. 9-10
`
`Fig. 9-11
`
`(2) Impedances in parallel (admittances in series) divide the total current in the inverse ratio of the
`impedances (direct ratio of the admittances):
`¼ Zs
`¼ Yr
`Zr
`Ys
`
`Ir
`Is
`
`or
`
`Ir ¼ Zeq
`Zr
`
`IT ¼ Yr
`Yeq
`
`IT
`
`See Fig. 9-11.
`
`9.6 THE MESH CURRENT METHOD
`
`Consider the frequency-domain network of Fig. 9-12. Applying KVL, as in Section 4.3, or simply
`by inspection, we find the matrix equation
`
`IPR2023-00697
`Theta EX2011
`
`