SCHAUM’S OUTLINE OF THEORY AND PROBLEMS of BASIC CIRCUIT ANALYSIS Second Edition JOHN O’MALLEY, Ph.D Professor of Electrical Engineering University of Florida SCHAUM’S OUTLINE SERIES McGRAW-HILL New York San Francisco Washington, D.C Auckland Bogotci Caracas London Madrid Mexico City Milan Montreal New Dehli San Juan Singapore Sydney Tokyo Toronto Lisbon JOHN R O’MALLEY is a Professor of Electrical Engineering at the University of Florida He received a Ph.D degree from the University of Florida and an LL.B degree from Georgetown University He is the author of two books on circuit analysis and two on the digital computer He has been teaching courses in electric circuit analysis since 1959 Schaum’s Outline of Theory and Problems of BASIC CIRCUIT ANALYSIS Copyright 1992,1982 by The McGraw-Hill Companies Inc All rights reserved Printed in the United States of America Except as permitted under the Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher 10 1 12 13 14 15 16 17 18 19 20 PRS PRS ISBN 0-0?-04?824-4 Sponsoring Editor: John Aliano Product i (I n S u pe rc’i so r : L a u ise K ar a m Editing Supervisors: Meg Tohin, Maureen Walker Library of Congress Cstaloging-in-Publication Data O’Malley John Schaum’s outline of theory and problems of basic circuit analysis ’ John O’Malley 2nd ed p c.m (Schaum’s outline series) Includes index ISBN 0-07-047824-4 Electric circuits Electric circuit analysis I Title TK454.046 1992 62 1.319’2 dc20 McGraw -Hill 1)rrworr o(7ht.McGraw.Hill Cornpanles 90-266I5 Dedicated to the loving memory of my brother Norman Joseph 'Mallej? Lawyer, engineer, and mentor This page intentionally left blank Preface Studying from this book will help both electrical technology and electrical engineering students learn circuit analysis with, it is hoped, less effort and more understanding Since this book begins with the analysis of dc resistive circuits and continues to that of ac circuits, as the popular circuit analysis textbooks, a student can, from the start, use this book as a supplement to a circuit analysis text book The reader does not need a knowledge of differential or integral calculus even though this book has derivatives in the chapters on capacitors, inductors, and transformers, as is required for the voltage-current relations The few problems with derivatives have clear physical explanations of them, and there is not a single integral anywhere in the book Despite its lack of higher mathematics, this book can be very useful to an electrical engineering reader since most material in an electrical engineering circuit analysis course requires only a knowledge of algebra Where there are different definitions in the electrical technology and engineering fields, as for capacitive reactances, phasors, and reactive power, the reader is cautioned and the various definitions are explained One of the special features of this book is the presentation of PSpice, which is a computer circuit analysis or simulation program that is suitable for use on personal computers (PCs) PSpice is similar to SPICE, which has become the standard for analog circuit simulation for the entire electronics industry Another special feature is the presentation of operational-amplifier (op-amp) circuits Both of these topics are new to this second edition Another topic that has been added is the use of advanced scientific calculators to solve the simultaneous equations that arise in circuit analyses Although this use requires placing the equations in matrix form, absolutely no knowledge of matrix algebra is required Finally, there are many more problems involving circuits that contain dependent sources than there were in the first edition I wish to thank Dr R L Sullivan, who, while I was writing this second edition, was Chairman of the Department of Electrical Engineering at the University of Florida He nurtured an environment that made it conducive to the writing of books Thanks are also due to my wife, Lois Anne, and my son Mathew for their constant support and encouragement without which I could not have written this second edition JOHN R O'MALLEY V This page intentionally left blank Contents Chapter BASIC CONCEPTS Digit Grouping International System of Units Electric Charge Electric Current Voltage Dependent Sources Power Energy 5 RESISTANCE Ohm’s Law Resistivity Temperature Effects Resistors Resistor Power Absorption Nominal Values and Tolerances Color Code Open and Short Circuits Internal Resistance 17 17 17 18 19 19 19 20 20 20 SERIES AND PARALLEL DC CIRCUITS 31 31 31 32 32 34 34 1 1 Chapter Chapter Chapter Chapter Branches Nodes Loops Meshes Series- and Parallel-Connected Components Kirchhoffs Voltage Law and Series DC Circuits Voltage Division Kirchhoffs Current Law and Parallel DC Circuits Current Division Kilohm-Milliampere Method DC CIRCUIT ANALYSIS Cramer’s Rule Calculator Solutions Source Transform at io n s Mesh Analysis Loop Analysis Nodal Analysis Dependent Sources and Circuit Analysis DC EQUIVALENT CIRCUITS NETWORK THEOREMS AND BRIDGE CIRCUITS Introduction Thevenin’s and Norton’s Theorems Maximum Power Transfer Theorem Superposition Theorem Millman’s Theorem Y-A and A-Y Transformations Bridge Circuits vii 54 54 55 56 56 57 58 59 82 82 82 84 84 84 85 86 CONTENTS Vlll Chapter Chapter Chapter Chapter Chapter 10 Chapter 11 Chapter 12 OPERATIONAL-AMPLIFIER CIRCUITS 112 112 112 114 116 PSPICE DC CIRCUIT ANALYSIS 136 136 136 138 139 140 CAPACITORS AND CAPACITANCE 153 153 153 153 154 155 155 156 156 157 INDUCTORS INDUCTANCE AND PSPICE TRANSIENT ANALYSIS 174 174 174 175 175 176 177 177 177 Introduction Op-Amp Operation Popular Op-Amp Circuits Circuits with Multiple Operational Amplifiers Introduction Basic Statements Dependent Sources DC and PRINT Contro! Statements Restrictions Introduction Capacitance Capacitor Construction Total Capacitance Energy Storage Time-Varying Voltages and Currents Capacitor Current Single-Capacitor DC-Excited Circuits RC Timers and Oscillators In trod uction Magnetic Flux Inductance and Inductor Construction Inductor Voltage and Current Relation Total Inductance Energy Storage Single-Inductor DC-Excited Circuits PSpice Transient Analysis SINUSOIDAL ALTERNATING VOLTAGE AND CURRENT 194 Introduction Sine and Cosine Waves Phase Relations Average Value Resistor Sinusoidal Response Effective or RMS Values Inductor Sinusoidal Response Capacitor Sinusoidal Response 194 195 197 198 198 198 199 200 COMPLEX ALGEBRA AND PHASORS 217 Introduction Imaginary Numbers Complex Numbers and the Rectangular Form Polar Form Phasors 217 217 218 219 221 BASIC AC CIRCUIT ANALYSIS IMPEDANCE AND ADMITTANCE 232 Introduction Phasor-Domain Circuit Elements AC Series Circuit Analysis 232 232 234 CONTENTS ix Impedance Voltage Division AC Parallel Circuit Analysis Admittance Current Division 234 236 237 238 239 MESH LOOP NODAL AND PSPICE ANALYSES OF AC CIRCUITS Introduction Source Transformations Mesh and Loop Analyses Nodal Analysis PSpice AC Analysis 265 265 265 265 267 268 14 AC EQUIVALENT CIRCUITS NETWORK THEOREMS AND BRIDGE CIRCUITS Introduction Thevenin’s and Norton’s Theorems Maximum Power Transfer Theorem Superposition Theorem AC Y-A and A-Y Transformations AC Bridge Circuits 294 294 294 295 295 296 296 Chapter 15 POWER IN AC CIRCUITS Introduction Circuit Power Absorption Chapter 13 Chapter Wattmeters Reactive Power Complex Power and Apparent Power Power Factor Correction Chapter 16 TRANSFORMERS Introduction Right-Hand Rule Dot Convention 349 349 349 350 350 352 354 356 THREE-PHASE CIRCUITS Introduction Subscript Notation Three-Phase Voltage Generation Generator Winding Connections 384 384 384 384 385 386 387 389 390 391 391 393 393 The Ideal Transformer The Air-Core Transformer The Autotransformer PSpice and Transformers Chapter 17 ~~ 324 324 324 325 326 326 327 Phase Sequence Balanced Y Circuit Balanced A Load Parallel Loads Power Three-Phase Power Measurements Unbalanced Circuits PSpice Analysis of Three-Phase Circuits ~~ INDEX 415 CHAP 11 BASIC CONCEPTS POWER The rute at which something either absorbs or produces energy is the poit'er absorbed or produced A source of energy produces or delivers power and a load absorbs it The SI unit of power is the wutt with unit symbol W The quantity symbol is P for constant power and p for time-varying power If J of work is either absorbed or delivered at a constant rate in s, the corresponding power is W In general, P(watts) = W (joules) [(seconds) The power ubsorbed by an electric component is the product of voltage and current if the current reference arrow is into the positively referenced terminal, as shown in Fig 1-6: P(watts) = V(vo1ts) x I(amperes) Such references are called associated references (The term pussiw skgn convention is often used instead of "associated references.") If the references are not associated (the current arrow is into the negatively referenced terminal), the power absorbed is P = - VZ Fig 1-6 Fig 1-7 If the calculated P is positive with either formula, the component actually uhsorhs power But if P is negative, the component procltrces power it is a source' of electric energy The power output rating of motors is usually expressed in a power unit called the horsepoiwr (hp) even though this is not an SI unit The relation between horsepower and watts is I hp = 745.7 W Electric motors and other systems have an e@cicvq* (17) of operation defined by Efficiency = power output ~ ~~~ ~ power input x 100% or = - P o ~x~ 100% Pin Efficiency can also be based on work output divided by work input In calculations, efficiency is usually expressed as a decimal fraction that is the percentage divided by 100 The overall efficiency of a cascaded system as shown in Fig 1-7 is the product of the individual efficiencies: ENERGY Electric energy used or produced is the product of the electric power input or output and the time over which this input or output occurs: W(joules) = P(watts) x t(seconds) Electric energy is what customers purchase from electric utility companies These companies not use the joule as an energy unit but instead use the much larger and more convenient kilowattltour (kWh) even though it is not an SI unit The number of kilowatthours consumed equals the product of the power absorbed in kilowatts and the time in hours over which it is absorbed: W(ki1owatthours) = P(ki1owatts) x t(hours) BASIC CONCEPTS [CHAP Solved Problems 1.1 Find the charge in coulombs of ( a ) 5.31 x 10" electrons, and ( h ) 2.9 x 10" protons ( a ) Since the charge of a n electron is - 1.602 x 10- l C, the total charge is 5.31 x O ' ms x -1.602 x IO-'"C -1 = c -85.1 (b) Similarly, the total charge is 2.9 x 1022+ret-mKx 1.602 x 10- l C = 4.65 kC -1 1.2 How many protons have a combined charge of 6.8 pC? protons is I C, the number of protons is Because the combined charge of 6.241 x 6.8 x 10-12$?!x 1.3 6.241 x 10'8protons - _ - = I$ 4.24 x 10' protons Find the current flow through a light bulb from a steady movement of in min, and (c) 10" electrons in h (U) 60 C in s, ( h ) 15 C Current is the rate of charge movement in coulombs per second So, Q (a) I = t 60C =- 4s I = 15 C/S l* 60s 15c (b) I = - x - 2& (c) = 1P - x 15 A 0.125 C / S= 0.125 A 1~$ _- 3600 s x - 1.602 x to-'" C -1 = - 0.445 C/S = - 0.445 A The negative sign in the answer indicates that the current flows in a direction opposite that of electron movement But this sign is unimportant here and can be omitted because the problem statement does not specify the direction of electron movement 1.4 Electrons pass to the right through a wire cross section at the rate of 6.4 x 102' electrons per minute What is the current in the wire? Because current is the rate of charge movement in coulombs per second, I = -1 6.4 x 102'hetrun3 1* X 6.241 x c x I& 60s = - 17.1 C s = - 17.1 A The negative sign in the answer indicates that the current is to the left, opposite the direction o f electron movement 1.5 In a liquid, negative ions, each with a single surplus electron, move to the left at a steady rate of 2.1 x to2' ions per minute and positive ions, each with two surplus protons, move to the right at a steady rate of 4.8 x l O I ions per minute Find the current to the right The negative ions moving to the left and the positive ions moving t o the right both produce a current t o the right because current flow is in a direction opposite that of negative charge movement and the same as that of positive charge movement For a current to the right, the movement of electrons to the left is a CHAP 13 BASIC CONCEPTS negative movement Also, each positive ion, being doubly ionized, has double the charge of a proton So, 2.1 x I= x 1* x- 1.6 l* - - lJ?k&VlT = 0.817 60 s -1.602 ~ x 10-19C x I& - - + -2 x 4.8 x - 60 s 10”~ ~ x - 1.602_ _x lO-I9C ~ ~ - - l* -1 A Will a 10-A fuse blow for a steady rate of charge flow through it of 45 000 C/h? The current is 45 000 c x- 3600s = 12.5 A which is more than the 10-A rating So the fuse will blow 1.7 Assuming a steady current flow through a switch, find the time required for (a) 20 C to flow if the current is 15 mA, ( h ) 12 pC to flow if the current is 30 pA, and (c) 2.58 x 10’’ electrons to flow if the current is -64.2 nA Since I (a) t = (h) t = (c) t = 1.8 = Q/t solved for t is t 20 15 - - 10-3 12 x 10-(j 30 x = = Q/I, 1.33 x 103s = 22.2 = x 105 s = 1 h 2.58 1015-64.2 x 10-9A X -1c 6.241 x 1 = * 6.44 x 103s ~ = 1.79 h The total charge that a battery can deliver is usually specified in ampere-hours (Ah) An ampere-hour is the quantity of charge corresponding to a current flow of A for h Find the number of coulombs corresponding to Ah Since from Q = I t , C is equal to one ampere second (As), 3600 s 1.9 - 3600 AS = 3600 C A certain car battery is rated at 700 Ah at 3.5 A, which means that the battery can deliver 3.5 A for approximately 700/3.5 = 200 h However, the larger the current, the less the charge that can be drawn How long can this battery deliver A? The time that the current can flow is approximately equal to the ampere-hour rating divided by the current: Actually, the battery can deliver A for longer than 350 h because the ampere-hour rating for this smaller current is greater than that for 3.5 A 1.10 Find the average drift velocity of electrons in a No 14 AWG copper wire carrying a 10-A current, given that copper has 1.38 x 1024free electrons per cubic inch and that the cross-sectional area of No 14 AWG wire is 3.23 x 10-3 in2 - ~ S The a ~ w - a g drift e ~~clocity ( ' ) cqu:ils the current di\,idcd by the product of the cross-sectional area a n d the electron density: I0 p' 1' 1s = I 3.23 x 10 1j.d 3j€8 1.38 x I o ' e.' 0.0254 111 1)d Ii2lCmim X - 1.603 x 10 q I" -3.56 x I W ' m s The negative sign i n the answer indicates that the electrons rnn\.'e in it direction opposite that o f current f o w Notice the l o w \docity An electron tra\tls only 1.38 111 in h, on the a\wage, e ~ though ~ n the electric impulses produced by the electron inoi~cnienttra\el at near the speed of light (2.998 x 10' m s) 1.1 I Find the work required to lift ii 4500-kg elevator a vertical distance of 50 m The ivork required is the product of the distance moved and the force needed t o oL'crcome the weight of the e l e ~ a t o r Since this weight i n nc\+'tons is 9.8 tinics the 11i;iss in kilograms, 1.I= ' F S = (9.8 x 4500)(50)J= 3.2 1.12 MJ Find the potential energy in joules gained by a 180-lb man in climbing a 6-ft ladder The potential energj' gained by the nian equals the work he had to d o to climb the ladder The force i n ~ ~ o l ~ xisxhis i u ~ i g h t ,and the distance is the height of the ladder The conwrsion factor from ureight in pounds t o ;i force i n newtons is N = 0.225 Ib Thus 11' = IXOJti, x y x 1.13 I 0.22.5fl X 13fi IJY X 0.0254 I l l = I Jd 1.36 x 103 N111 = 1.46 k J How much chemical energy must a 12-V car battery expend in moLing 8.93 x 10'" electrons from its positive terminal to its negative terminal? The appropriate formula is 1.1'- Q I: Although the signs of Q and 1' ;ire important obviously here the product of these quantities must be positive because energq is required to mo\'e the electrons S o , the easiest approach is t o ignore the signs of Q and I : O r , if signs are used, I ' is ncgatiirc because the charge moves to ;i niore negati\ c terminal, and of coiirhe Q is negative bec;iuse electrons h a w ii negative charge Thus, 1.1' = Q I ' 1.14 = 8.03 x 1o2"Am x ( - I2 V ) x - c = 6.34 x IolxLlwhmls If moving 16 C of positive charge from point h to point drop from point I ( to point h (I 1.72 x 10.' VC = 1.72 kJ requires 0.8 J, find 1;,,,, the voltage w,',,0.8 1.15 In mobing from point ( I t o point b, x 10'" electrons J of work Find I;,,,, the voltage drop from point ( I to point 11 Worh done h j * the electron!, cqui\ alcnt to / i c ~ c / t r t i wwork done 1 thc electron\, and \ oltage depends o n u'ork done O I I charge So It,,, = - J but It:,,, = Cl,, = J Thus -3 x I()''- - I c The negative sign indicates that there is a ~ o l t a g crise from bords, point h is more positi\e than point 1 11 to h instead of a ~ o l t a g cdrop In othcr CHAP I] 1.16 BASIC C O N C E P T S Find y,h the voltage drop from point II to point h, if 24 J are required to move charges of ( a ) C, ( h ) -4 C, and (c) 20 x 10" electrons from point N to point h If 24 J are required to motfe the charges from point ( I to point h, then -24 J are required to move them from point h to point (1 In other words it;,, = -34 J So, The negative sign in the answer indicates that point rise from 11 to h ((-1 Vah = 1.17 Wu h Q - -24 J 20 x 10'qsk&mmS X is more ncgative than point h 11 6.241 x 10'H-eketm% -1c = there is a voltage 0.749 V Find the energy stored in a 12-V car battery rated at 650 Ah From U' = QL' and the fact that As W=650A$x- = 3600 s x C V = ~1 " A s x V = M J 1Y 1.18 Find the voltage drop across a light bulb if a 0.5-A current flowing through it for s causes the light bulb to give off 240 J of light and heat energy Since the charge that flotvs is Q = Ir = 0.5 x 1.19 2min- P = Wr I* 60s and from the fact that U'=Pt=60Wx = 305 s = 30 W Ws = J, 3600 s l $ ~ = 216000 WS = 216 kJ 'Y How long does a 100-W light bulb take to consume 13 k J ? From rearranging P = Wt, 1= 1.22 X How many joules does a 60-W light bulb consume in h ? From rearranging 1.21 C, Find the average input power to a radio that consumes 3600 J in 36005 1.20 = w - 0 = 130s _- P 100 How much power does a stove element absorb if it draws 10 A when connected to a 15-V line'? P=C'I=115x 10W=I.l5kW ... books on circuit analysis and two on the digital computer He has been teaching courses in electric circuit analysis since 1959 Schaum’s Outline of Theory and Problems of BASIC CIRCUIT ANALYSIS. ..SCHAUM’S OUTLINE OF THEORY AND PROBLEMS of BASIC CIRCUIT ANALYSIS Second Edition JOHN O’MALLEY, Ph.D Professor of Electrical Engineering University of Florida SCHAUM’S OUTLINE SERIES... Supervisors: Meg Tohin, Maureen Walker Library of Congress Cstaloging-in-Publication Data O’Malley John Schaum’s outline of theory and problems of basic circuit analysis ’ John O’Malley 2nd ed p c.m