4415 3000 Solved Problems in Electric Circuits Schaums by 7see.blogspot.com

rJ/ gives you more solved rJ/ includes easy-to-readproblems than any other guide cross-reference index and diagrams rJ/ demonstrates the best rJ/ provides subiect problem-solving strategies coverage so thorough, grad uate students and professionals can use this rJ/ improves guide, too perfonnance on exams rJ/ helps cut study time NASIR .SYED A. Louis San Francisco Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan M ontreal New Delhi San Juan Singapore Sydney Tokyo Toronto .SCHAUM'S SOLVED PROBLEMS SERIES 3000 SOLVED PROBLEMS IN ELECTRIC CIRCUITS by Syed A. Nasar University of Kentucky McGR AW-HILL New York St. Electric circuits-Problems. Ph.D.N36 1988 621. ELEC TRIC MACHINES AND ELECTROMECHANICS and BASIC ELECTRICAL ENGINEERING. Schaum's 3000 solved problems in electric circuits. Library of Congress Cataloging-in-Publication Data Nasar. Index by Hugh C. or stored in a data base or retrieval system... Ph.. Title. Title: Schaum's three thousand solved problems in electric circuits. II. TK454. including two Schaum's Outlines. I.. S. etc. Except as permitted under the United States Copyright Act of 1976. . 6< A Division ofTheMcGraw-HiU Companies .) 12 13 14 15 16 17 18 19 20 VFM VFM 5 4 Copyright © 1988 by The McGraw-Hill Companies. Nasar has written many books.I Syed A. INTRODUCTION TO ELECTRICAL EN GINEERING. Inc. 1. All rights r•!served. and a textbook for McGraw-Hill's College Division. Professor of Electrical Engineerirzg at the University of Kentucky. D.. a power-systems text for Macmillan. McGraw-Hill (.. Maddocks. without lhe prior written permission of the publisher. Project supervision by The Total Book. Dr. A.319'2076 87-25974 ISBN 0-07-045936-3 (Formerly published under ISBN 0-07-045921-5. Nasar. no part of this publication may be reproduced or distributed in any form or by any means. Printed in the United States of America. exercises. WAVEFORMS. CONTENTS Chapter 1 UNITS AND BASIC CONCEPTS 1 Chapter 2 RESISTANCE AND OHM'S LAW 6 Chapter 3 SERIES AND PARALLEL RESISTIVE CIRCUITS 16 Chapter 4 KIRCHHOFF 'S LAWS 43 Chapter 5 NETWORK THEOREMS 72 Chapter 6 CAPACITORS 116 Chapter 7 INDUCTORS 129 Chapter 8 AC SOURCES. RAMP. AND IMPULSE FUNCTI NS 423 Chapter 17 DUALS AND ANALOGS 432 Chapter 18 TRANSIENTS IN AC CIRCUITS 441 Chapter 19 CIRCUITS WITH MULTIFREQUENCY I PUTS 450 Chapter 20 CIRCUITS WITH NONSINUSOIDAL SO 462 Chapter 21 LAPLACE TRANSFORM METHOD 491 Chapter 22 STATE VARIABLES METHOD 579 Chapter 23 TWO-PORT NETWORKS 594 Chapter 24 REVIEW PROBLEMS 620 INDEX 747 iii . AND CIRCUIT RELATIONSHIPS 137 Chapter 9 COMPLEX NUMBERS AND PHASORS 145 Chapter 10 AC CIRCUITS U NDER STEADY STATE 152 Chapter 11 MAGNETICALLY COUPLED CIRCUITS 231 Chapter 12 RESONANCE 268 Chapter 13 FREQUENCY RESPONSE AND FILTERS 291 Chapter 14 THREE-PHASE CIRCUITS 304 I I I Chapter 15 TRANSIENTS IN DC CIRCUITS I 351 Chapter 16 STEP. In using the book. and this book solves 3000 of them for you! Speaking seriously. be sure to sketch out one before undertaking the solution .the Laplace transform or whatever. v . you should . you have here the most careful and complete anthology of examination-type problems on the market today. The heuristic value of a clear circuit diagram need not be stressed: if a problem in this book should carry a diagram but doesn 't. of course. concentrate on the area of your maximum weakness. too. with no other preparation! The reasoning is simple: There are only 4000 possible problems in the field (as you must know). May your success be electric. But do not neglect to work problems involving familiar ma terial. To the Student Think of it!-an expected score of 75% on any exam in Electric Circuits. you might well learn more efficient methods of handling them. 8 Electric current is measured in amperes (A). Express the following values in powers of 10 and write them in their abbreviated forms: 0.000005 F. and 100.000005 F = 5 x 10.1 gigahertz = 0.6 x 10.6 x 10.003 H = 3 x 10.C. These powers of 10 are written in abbreviated forms.1 x 10 Hz = 0. Express the following frequencies in powers of 10 and in their respective abbreviated forms: 1000 Hz.9 Find the current in a conductor through which 2.H = 3 millihenry = 3 mH 1.5 mF = 500 µ.F 3 0.2 Electric capacitance is measured in farads (F). The charge 19 on an electron is approximately 1. I = (no. 0.6 x 10.F = 1. C) = 2.C.000 fl.000 Hz.6 x 10-19 = A 1 5 time. how many electrons pass a given point in 30 s in a conductor carrying 8-A current.4 Electric frequency is measured in hertz (Hz).F = 5 microfarad = 5 µ.H = 10 millihenry = 10 mH 0.F = 0. /7 CHAPTER 1 L/units and Basic Concepts 1.01 H = 10 x 10. If an ampere is expressed as a flow of charge in coulombs per second ( Cl s). I 3 1000 Hz = 1 x 10 Hz = 1 kilohertz = 1kHz 6 5.000. I 3 2000 fl = 2 x 10 fl = 2 kiloohm = 2 kfl 6 3. 5.000000001 F.000 Hz. What is the corresponding current in amperes? I I = 360 = 18 A = s 20 .5 Convert 2 minutes to milliseconds: 3 120 x 10. I 6 0. Express the following values in powers of 10 and write them in their abbreviated forms: 0.5 x 10.000.10 A charge of 360 C passes through a conductor in 20 s.6 x 10. 5 I 2 min = 2 x 60 s = 120 s _3 s = 1. However.1 GHz 1. express the following values in powers of 10 and write them in their abbreviated forms: 2000 fl and 3.000.5 millifarad = 0.003 H. this is rather a large unit.5 x 10 electrons pass in 8 s if the charge on an electron is 19 approximately 1.000.6 Convert 5 kilometers to centimeters: 3 I 3 2 5 km = 5 x 10 m = 5 x 10 x 10 cm = 5 x 10 cm 5 1.01 H and 0.0 picofarad = 1pF 1.2 x 10 ms 10 = 1.000 fl = 3 x 10 fl = 3 megohm = 2 Mfl 1.3 The unit of electric inductance is henry (H).000 Hz = 5 x 10 Hz = 5 megahertz = 5 MHz 9 100.5 x 1020 x 1.C correspond to 1 electron 19 20 240 C correspond to (1 x 240) /(1.000.1 Powers of 10 appear frequently with units of measurements.000.F 9 0. If electric resistance is measured in ohms (fl).0005 F. and 0. ) = 15 x 10 electrons 20 1.000 Hz = 0.7 Convert 15 centimeters to millimeters: I 15 15 cm 102 x 103 = 150 mm = 1.0005 F = 0. I 3 3 0. s 8 1. I Charge = A x s = 8 x 30 = 240 C 19 1.000000001 F = 1 x 10. of electrons)\charge on electron. s) = 6 x 2 x 60 = 720 C 1. Determine the work done in moving a 50-µ.17 s = 7.C charge from one point to t he other? 220 >< 10-" I lV=lJ/C v = 10 x 10-" = 22 v or 1.14 through 50 cm is 10 ms.8 fl I V= = V R . the potential at A). · = 125 w 10 >< 10- 3 1.5 2 5 1.5 < SO x 10. if the force F is given by f = QE. C . V = J /C = (J/s) /(C/s) = W/A. which is also the unit of energy. determine the potential difference bttween point A and infinity (i.35 =- 31) ----.17 From Prob. calculate the ohmic value of the resistance. w 30 = 12 w 30 4. Also. . calculate the corresponding power.11 The current in an electric circuit rises exponentiall) as given by i = 10(1 . C electric charge ( Q) through a distance of 50 cm in the direction of a uniform electric field (E ) of 50 kV/ m. V/ m) = 50 x 10-" x 50 x 10 = 2.e. and is defined as the work done in moving a unit positive charge (from one point to the other).-. (V). Also determine the potential difference between point B and infinity. = 10(0.14 that an electric charge experiences a force in an electric field.16.16 We observed in Prob. J I VAR = charge. What is t he potential difference between two points if it requires 220 µ. charge.14 The unit of force is the newton (N) and work is meas ured in netwon-meters (N · m). Thus.C charge of Prob.25 N · m = 1. 1.13 A current of 6 A flows in a resistor. How much time will be required to pass a 30-C charge through the bulb? I = t ..2 0 CHAPTER 1 1. the unit of power is the joule per second (J /s) which is equal to one watt (W).J to move a 10-µ. c . If the time taken to move the 50-µ.5 A of current.3 mm current.. A 68 < 10- 1. C 2 1.19 If an additional energy of 3 J is required to move t he 2-C charge of Prob.25 J 1. 1. Electric potential difference (between two points) is measured in volt:. I Force = (charge. work or energy..= 441. where 1J = 1N ·m.5 N 3 2 Work done = force x distance = 2. 1.2so I 0 250 q= dt = 21 ) 10 t + -. work or energy.. work clone 1 25 I Power =.= 1.5326 C I i 10(1 - x 0 L () 1. J 12 I V --------- = = -=6V Ax char. A)( tim .e. 1. C)(electric field. Calculate the charge 21 flowing through the circuit in 250 ms.21 ) <J. energy is expressed in joule5 11).12 A 75-W bulb draws a 680-mA current. Ca lculate the potential difference across a resistor dissipat ing 30 W of power while taking 2.18 from point A to another point B. 0. )2 = . Assuming infinity to be at zero potential .5e. calculate the potential difference between points A and B. = Ji = (I 2. D = 0.. J o 2so e. Alternatively.:. dt = ( .time --.18 An energy of 121 is expended in moving a 2-C charge from infinity to a point A.15 Power is defined as the rate of work done or the ra te of energy conversion. How many coulombs of charge pass through the resistor in 2 min? I q = (current.) A.250 + 1e-2 . VRx =. 3 -= 7. 3 cnergy. = 1. .21 Determine the charge that requires 1-kJ energy to be moved from infinity to a point having a 12-V potential.5 V 2 1.--= ..5 v 12 + ..20 The potential difference between two conductors is 110 V. How much work is done in moving a 5-C charge from one conductor to the other? I Work = energy =0 110 x 5 = 550 J 1. V 12 · . J 10 I Charge C = ---.= 83 33 C ' pote 1tial. . so the average value of p. . Determine the charge moved during this period.-2.M. mput mput u 100 x 10 3 t= p= 120 x 20 = 41.07 = $2. At the rate of 7 cents/ kWh. What is the energy consumption in megajoules? I Total kWh = (power.M. . mput 0..25 The heater of Prob.27 A 110-V light bulb takes 0.9 x 10 x 12 x 30 = 35. and 11 P. I Instantaneous power p = ui = ( 100\12 sin t )( 5\12 sin t) = 1000 sin t W = 2 (1- 1000 x cos 2t ) p = 500 .6 x 5 + 1.M.20 A I 1. find the time required to boil the water. A battery is required to supply 0.23 Electric utilities employ as the unit of energy the kilowatt-hour (kWh). 1. kW)( time.67 s 1.5 kW.-1. 99 x 10' I = = = 0.99 or U= = 100 kJ Efficiency -.-0.M.5 A continuously for three days.6 kW. Pav = 500 W.5 x 6 + 0. determine the cost to operate the bulb for 30 days.5.5 kW. to 8 A.24 has an efficiency of 99 percent.22 A car battery supplies 48 J of energy at 12 V over a certain period of time.2 kWh in 30 min at 120 V. .24 An electric heater takes 1.5 x 4 + 2. UNITS AND BASIC CONCEPTS D 3 1.25? v 120 I R= = .s w 2.-!w 1600 vi dt = (200 sin wt)( 8 sin wt) dt = rr J Energy W = 10 () w W 1600rr Average power Pav = 2 rrlw = w( 2 rrlw) = 800 W 1. to 2 P. to 6 P.28 The voltage and current in a circuit element are respectively given by u = 100\12 sin t V and i = 5\12 sin t A Calculate the instantaneous power and the average power delivered to the circuit. If the current taken by the heater is 20 A at 120 V.64 x $0.24 and 1.8 x 10 J = 118. Hence. The power consumed in a household over a 24-h period is as follows: 8 A. determine the average power dissipated in the resistor.M.M.500 cos 2t w The cosine function averages to zero. to 11 P. I U = Pt = 110 x 0. -= I 20 1.26 What is the ohmic value of the resistance of the heating element 6 fl of the heater of Probs.-/w 12.0 x 9 = 33 kWh 3 6 = 33 x 10 x 60 x 60 W ·s = 118. 6 P. I q 48 1 12 = 4C = V 1.. output 99 x 10' .5 x 3 x 24 = 36 Ah .8 MJ 1.99 .9-A current and operates 12 hi day. The heat energy required to boil a certain amount of water is 99 kl ..1. d 2 rr I 0ne per1 0 =.29 A resistor draws a current i = 8 sin wt A at a voltage u = 200 sin wt V. What must be the rating of the battery? I Ah = I x hr = 0.30 The energy capacity or rating of a battery is generally expressed in ampere-hour (Ah).M.M.V 10 / 0. 2 P. .0 kW. 1. h) = 1. 1.50. What is the current input to the heater? 3 Ult -_ 1.64 kWh 3 Cost of operation = 35.2 x120 I -. Calculate the energy consumed by the resistor per cycle (or over one period of the current wave). 1. = . Ah 30 I Time = . = 12 h I 2.5 A? .1.31 A battery is rated at 30 Ah.5 . For how many hours can it continuously supply a current of 2. 6 .. ..-· .. For how long can the batte1y supply a 16-A current at 0°C? I From Fig. at 0°C the rating of the battery reduces to 0. 1-1 1... 1-2. 1-2. 1-2 is not allowed to go below 64 Ah . 1-2. 1-1 and 1-2 to obtain the rating of the battery at 17 A and at 10°C if the battery is rated at 100 percent at 5 A and at 25°C I From Fig... ' µ. me t == -Arh = 5 ' . 1-1...w 8 = 2 .. 1-2 at 64 Ah. T1. 6 300 ' = -15e. Ah 57. Ah 64 h . as shown in Fig... T1me t =' -. at 20 A the rating of the battery becomes 58 Ah.6 Ah. 1..6 h = 3 h 36 mm / 10 ro v -.C.= -= 5 20 mm / 12 1.. for how long can the battery supply the rated current? I From Fig.- = 3. How long will the battery supply 20 A of current? I From Fig.. 300 1.4 D CHAPTER 1 1.32 The capacity of a car battery depends on the ambient temperature as shown in Fig..e.36 The decay of charge in an electric circuit is given by q = 50e.. 0 c Fig. The battery is rated at 70 Ah at a discharge rate of 5 A as shown. -u.33 The capacity of a car battery depends on the current drawn (or discharge) from the battery.-- t 7o bo --<- so. rating at 17 A = 60 Ah which is considered as 100 percent at 25°C. . discharge rate = 12 A.35 Combine the characteristics of Figs. 300 ' mA dt . Time t = I =. <:(90 . Hence... - f.-. rating l0°C = 0... -20 -1 0 c /0 --. Determine the resulting current.34 If the rating of the battery with a discharge characteristic shown in Fig.8 x 72 = 57. From Fig.9 x 60 = 54 Ah. Hence.. 1-1. I i = dq = -50 x 300 x 10.- .:::..9 h = 2 h 54 mm. Therefore.T. A certain battery is rated at 72 Ah at 25°C. 1-1.. I At t = 0: 0 i = -15e = -15 mA 2 300 10 2 t = 10.36 at the following instants: t = 0. 1. x - = -0. and t = ao.s: i = -15e. UNITS AND BASIC CONCEPTS D 5 80 t 1.37 Evaluate the current in Prob.= O . t = 10 ms.7468 mA t = ao: i = -15e. I p = vi = ( 10 sin t)( 2 cos t) = 20 sin t cos t = 10 sin 2t W The instantaeous power pulsates with twice the frequency of the voltage or current.60°) = 68 x ![cos (377t .3 The voltage and current i at the pair of terminals of an electric circuit are given by = 100 sin t V and 9 vi = -5 sin t vEvaluate the average power and state if the circuit absorbs or delivers A. the circuit delivers power. Determine the instantaneous and average powers delivered to the circuit. Determine the instantaneous and average powers.60°)] W Pav = 34 COS 60° = 17 W 1. 1.377t + 60°) . since the average value of sin 2t = 0. and explain your result. I p = vi = (34 sin 377t)[2 sin (377t . .. Zero average power indicates that the circuit is nondissipative or conservative. power. Pav = 0 W.60°) A.1.cos (754t .60°)] = 68 sin 377t sin (377t .cos (377t + 377t . I 2 p = vi = (100 sin t )( -5 sin t ) = -500 sin t W Pav = -500 X ! (since the average value of sin 2 t = !) = -250 W The negative sign indicates that negative power is absorbed by the circuit.38 The voltage v and current i in an ac circuit are respectively given by v = 34 sin 377t V and i = 2 sin (377t .e. i.40 The voltage v and current i in a circuit are given by v = 10 sin t V and i = 2 cos t A.60°)] = 34 [cos 60° .