Basic engineering circuit analysis 9th irwin

BASIC ENGINEERING CI RCU IT ANALYSIS
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Irwin, 1. David Basic engineeri ng circuit analysiS/1. David Irwin. R. Mark Nelms.-9th ed.
p. cm. Includes bibliographical references and index.
ISBN 978·0-470- 12869-5 (cloth) I. Electric ci rcui t analysis. I. Nelms, R. M. II. lit le. TK454.1 7S 2()()7 62l.3 19'2-<1c22
Printell in th~ Un ited State:-- of America
10 9 8 7 6 5 4 3
2007040689
To my loving family: Edie
Geri, Bruno, Andrew and Ryan John, Julie, John David and Abi
Laura
BRIEF CONTENTS
CHAPTER 3 Nodal and Loop Analysis Techniques 95
CHAPTER 4 Operational Amplifiers 149
CHAPTER 5 Additional Analysis Techniques 183
CHAPTER 6 Capacitors 248
CHAPTER 8 AC Steady-State Analysis 375
CHAPTER 9 Steady-State Power Analysis 455
CHAPTER 10 Magnetically Coupled Networks 507
CHAPTER 11 Polyphase Circuits 553
CHAPTER 12 Variable-Frequency Network Performance 587
CHAPTER 13 The Laplace Transform 677
CHAPTER 14 Application of the Laplace Transform to Circuit Analysis 705
CHAPTER 15 Fourier Analysis Techniques 758
CHAPTER 16 Two-Port Networks 809
Preface
CHAPTER 1 BASIC CONCEPTS
J.I System of Units 2 1.2 Basic Quantit ies 2 1.3 Circuit Elements 8
Summary 16 Problems 16
CHAPTER 2 RESISTIVE CIRCUITS
2.1 Ohm's Law 24
2.2 Kirchhoff's Laws 28
2.4 Sing le-Node-Pair Circuits 43
2.5 Series and Para lle l Resistor Combinations 48
2.6 Circuits with Series-Paralle l Combinations of Resistors 53
2.7 Wye ~ Della Transfonmuions 57
2.8 C ircuits with Dependent Sources 60 2.9 Resistor Technologies for
Elec tronic Manufactu ring 64
2.10 Application Examples 67
2.11 Design Examples 71
CHAPTER 3 NODAL AND LOOP ANALYSIS TECHNIQUES 95
3.1 Nodal Analysis 96 3.2 Loop Analys is 11 5 3.3 Application Example 131 3.4 Design Example 133
Summary 133 Problems 134
CHAPTER 4 OPERATIONAL AMPLIFIERS 149
4.1 Introduction 150 4.2 Op-Amp Models 150 4.3 Fundamental Op-Amp Circuits 156 4.4 Comparators 164 4.5 App lication Examples 165 4.6 Design Examples 169
Summary 172
Problems 172
S.I Int roduction 184
xii CON TENT S
5.5 dc SPICE Analys is Using Schemalic Capture 2 12
5.6 Application Example 224
6.1 Capacitors 249
6.2 Inductors 255
7.1 Introduction 292
7.3 Second-Order Circu its 3 14
7.4 Transient PSPICE Analysis Using Schematic Capture 328
8.1 Sinusoids 376
8.3 Phasors 383
8.4 Phasor Relati onshi ps for Circui t Elements 385
8.S Impedance and Adm ittance 389
8.6 Phasor Diagrams 396
8.8 Analys is Techniques 40 I
8.9 AC PSPICE Analysis Using Schematic Capture 4 16
Summary 434
Problems 435
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9.11
Effective or rms Values 466
The Power Factor 469
Complex Power 47 1
Power Factor Correction 476
Single-Phase Three-Wire Circuits 479
Safety Considerat ions 482
Applica tion Examples 490
10.1 Mutuallnductanee 508
10.4 Safety Considerations 529
11.3 SoufcelLoad Connections 560
11.5 Power Factor Correction 572
12.2 Sinusoidal Frequency Analys is 596
12.3 Resonant Circuits 608
13. 1 Defin it ion 678
588
13.3 Transfortn Pairs 681
13.6 Convolution Integral 692
CHAPTER 14 APPLICATION OF THE LAPLACE TRANSFORM TO CIRCUIT ANALYSIS
14.1 Laplace Circuit Solutions 706
14.2 Circuit Element Models 707
705
14.S Pole-Zero PlotlBode Plot Con nect ion 732
14.6 Steady-State Response 735
15.1 Fourier Series 759
IS.2 Fourier Transform 781
IS.3 Application Examples 788
IS.4 Design Examples 795
16.2 Impedance Parameters 8 13
16.3 Hybrid Parameters 8 15
16.4 Transmission Para meters 8 17
16.S Parameter Convers ions 8 18
16.6 Intercon nection of Two~ Port s 8 19
16.8 Design Example 826
INDEX
Circuit analysis is not only fundamental 10 the enti re breadth of e lectrical and computer engineering- the concepts studied here extend far beyond those boundaries. For this reason it remains the starting point for many future engineers who wish to work in this field . The text and all the supplementary materials associated with it will aid you in reaching this goal. We strongly recommend while you are here to read the Preface close ly and view all the resources available to you as a learner. And one last piece of advice, learning requires practice and repetition. take every opportunity to work one more problem or study one more hour than you planned. In the end. you·1I be thankful you did.
The Ninth Edition has been prepared based on a careful examination of feedback received from instructors and students. The revisions and changes made should appeal to a wide vari ety of instructors. We are aware of significant changes taking place in the way this material is being taught and learned . Consequently, the authors and the publisher have created a for midable array of traditional and non-traditional learning resources to meet the needs of students and teachers of modern circuit analysis.
• A new four-color design is employed to enhance and clarify both text and illustrations. This sharply improves the pedagogical presentation, particu larly with complex illustra tions. For example, see Figure 2.5 on page 29.
• New chapte r previews provide moti vation for s tudying the material in the chapter. See page 758 for a chapter preview sample. Learning goals for each chapter have been updated and appear as pan of the new Chapter openers.
• End-of-chapter homework problems have been substantially revised and augmented. There are now over 1200 problems in the Ninth Edition! Multiple-choice Fundamentals of Engineering (FE) Exam problems now appear at the end of each chapter.
• Practical applications have been added for nearly every topic in the tex t. Since these are items sllldenlS will natura lly encounter on a regular basis, they serve 10 answer ques tions such as, "Why is this important?" or " How am I going to use what I learn from this course?" For a typical example app lication, see page 338.
• Problem-Solvi ng Videos have been c reated showing students step-by-step how to solve selected Learning Assessment problems within each chapter and supplemental
PREFACE
xvi PREFACE
end-of-chapter problems. The problem-solving videos (PSVs) are now also ava ilable
for the Apple iPod. Look for the iPod icon ii for the application of this unique feature.
• The end-of-chapter problems noted with the 0 denote problems that are included in OUf WileyPLUS course for this title. In some cases, these problems are presented as a lgo rithmic problems which offer differcllI variables for the problem allowing each student to work a similar problem e nsuring for individual ized work, also some problems have been developed into guided online (GO) problems with tutorials for students, as well as mul ti part problems allowing the student (0 show their work at various steps of the problem.
• The ~ icon next to e nd-of-chapter problems denotes those that are more challenging problems and engage the student to apply multiple concepts to sol ve the problem.
• Problem-Solving Strategies have been retained in the Nin th Edition. They are utili zed as a guide for the solutions contained in the PSVs.
• A new chapter on diodes has been created for the Ninth Edition. This chapter was developed in response to requests for a supplementa l chap ter on this topic. Whi le not included as part o f the printed tex t, it is available through WileyPLUS.
Thi s text is suitable for a one·semester, a two·semester, or a three·quarter course sequence. The first seven chapters are concerned with the anal ysis of dc circuits. An introduction to operat ional amplifiers is presented in Chapter 4. Thi s chapter may be omiued without any loss of continuity; a few examples and homework problems in later chapters must be skipped. Chapters 8-12 are focused on the analys is of ac c ircuits beginning with the analysis of s ingle frequency ci rcuits (single·phase and three·phase) and ending with variable· freq uency circuit operation. Calculat ion of power in single·phase and three-phase ae ci rcuits is also presented. The important topics of the Laplace transform, Fourier transform, and two· port networks are covered in Chapters 13- 16.
The organization of the tex t provides instructors maximum flexibility in designing thei r courses. One instructor may choose to cover the first seven chapters in a single semester, while another may omit Chapter 4 and cover Chapters 1-3 and 5-8. Other instruclOrs have chosen to cover Chapters 1-3, 5-{;, and sections 7. 1 and 7.2 and then cover Chapters 8 and 9. The remai ni ng chapters can be covered in a second semester course.
The pedagogy of this tex t is rich and varied. It includes print and media and much thought has been put into integrating its use. To gain the most from this pedagogy, please review the fo llowing e leme nts commonly available in most chapters of this book.
Learning Goals are provided at the outset of each chapter. This tabular li st te lls the reader what is important and what will be gained fro m studying the material in the chapter.
Examples are the mainstay of any circuit analysis text and numerous examples have always been a trademark of this tex tbook. These examples provide a more graduated level of pres· entation with simple, medium , and challengi ng examples, Besides regular examples, nume r ous Design Examples and Application Examples are fo und throughout the texl. See for
example, page 348.
Hints can often be fo und in the page margins. They faci litate understanding and serve as
reminders of key issues. See for example, page 6.
Learning Assessments are a crilicalleaming tool in this text. These exercises test the cumu lative concepts to that point in a given section or sections. Both a problem and the answer are provided. The student who masters these is ready to move forward. See for example, page 7.
Problem-Solving Strategies are step-by-step problem-solving techniques that many stu dents find particularly useful. They answer the frequentl y asked question, "where do I begin?" Nearly every chapter has one or more of these strategies. which are a kind of sum mation on problem-solving for concepts presented. See for example. page 115.
The Problems have been greatly revised for the Ninth Edi tion, This edition has hundreds of new problems of varying depth and level. Any instructor wi ll find numerous problems appro priate for any level class. There are over 1200 problems in the Ninth Edi tion! Included with the Problems are FE Exam Problems for each chapter. If you plan on taking the FE Exam, these problems closely match problems you will typically fi nd on the FE Exam.
Circuit Simulation and Analysis Software are a fundamental part of engineering circuit design today. Simulation software such as PSpice®, Multisim®, and MATLAB® allow engi neers to design and simulate circuits quickly and efficiently. The use of PSPICE and MATLAB has been integrated throughout the text and the presentation has been lauded as exceptional by past users. These examples are color coded for easy location. After much disclission and consideration, we have continued use of PSPICE 9.1 wi th the Ninth Edition. Although other versions are available, we believe the schemat ic interface with version 9. 1 is simple enough for students to quickly begin solving circuits using a simulation (00 1. A self-extracti ng archive for PSPICE 9. 1 may be downloaded from WileyPLUS or http://www,eng.auburn.edu/ece/download/9 Ipspstu.exe.
New to this edition is NI Multisim with simulated circuit exercises from the text. NI Multisim allows you to design and simulate circuits eas ily with its intuitive schematic capture and interactive simulation environment. The Muhisim software is uniquely available to you for download with this edition. Simulated circuit examples and exercises from different chapters are also available on the book's companion web site. While PSpice is referenced with screen shots in the textbook, either software program may be employed to simulate the circuit simu lation exercises. Use the tool you are comfortable with. If you choose Multisim, we recommend viewing the following Multisim resources. Visit www.ni.com/info and enter IRWIN
• Multisim Circuit Files - 50 Multi sim circuit files provide an interactive solution for improving student learning and comprehension
• SPICE Thtorials - information on SPICE simulation, OrCAD pSPICE simulation, SPICE modeling, and other concepts in circuit simulation
• Multisim Educa tor Edition Eyaluation Download - 30-day evaluations of NI Mu ltisim, NI Ultiboard, and the NI Multisim MCU Module
A rich collection of interactive Multisi m circuit fi les may also be downloaded from the textbook companion page. Try them on your own. This circuit set offers a distinctive and helpful way for explori ng examples and exercises from the text.
PREFACE xvii
The supplements list is extensive and provides instructors and students with a wealth of Supplements traditional and modern resources to match different learning needs.
THE INSTRUCTOR'S OR STUDENT'S COMPANION SITES AT WILEY (WILEYPLUS)
The Student Study Guide contains detai led examples that track the chapter presentation for aiding and checking the student's understanding of the problem-solving process. Many of these involve use of software programs like Excel, MATLAB , PSpice, and Multisim.
The Problem-Solving Companion is freely available for download as a PDF file from the WileyPLUS si te. This companion contai ns over 70 additional problems with detai led solu tions, and a walk through of the entire problem-solving process.
xviii PREFACE
Acknowledg ments
Problem-Solving Videos are conlinued in Ihe N imh Edil ion. A novelty with Ihi s edilion is Ihe ir
availabi li ty fo r use on the Apple iPod. Throughout the text i icons indicate when a video should be viewed. The videos provide step-by-step so lu tions to learning extens ions a nd supplementa l end-of-chapter problems. Videos for learning assessments will follow directly after a chap ter feat ure ca lled Problem-Solving Strategy. Icons with end-oF-chapter proble ms indicate that a video so lut ion is avai lab le for a s imilar problem-not the actua l end-or-chapter problem. Students who use these videos have found them to be very helpful.
The Solutions Manual for the Nimh Edilion has been complelely redone. checked, and double-checked for accuracy. Allhaugh it is handwritte n to avoid typesetti ng errors, it is the 111051 accurate so lutions manual eve r c reated for this textbook. Qualified instructors who adopt the lext for classroom use can download it off' Wiley 's Instructor's Companion Site.
PowerPoint Lecture Slides are an especially valuable supplementary aid for so me instruc to rs. While 1110st publishe rs make on ly fi gures ava ilable, these slides are true lec ture tools that summarize the key learn ing points fo r each chapter and are eas ily editable in PowerPoinl. The s lides are avai lable for download from Wiley 's Instruclor's Companion Sile for qualified adoplers .
Over the two decades (25 years) that this text has been in existence, we estimate more than one thousand instructors have used our book in teaching circuit analysis to hundreds of thou sands of students. As authors there is no greater reward than hav ing your work used by so many. We are grateful for the confidence shown in our text and for the numerous evaluations and suggestions from professors and the ir students over the years. This feedback has helped us continuo ll s ly improve the presentation. For this Ni nth Edition, we especially thank Ji m Rowland from the University of Kansas for his guidance on the pedagogical use of color th roughout the text and Aleck Leedy with Tuskegee Univers ity for assistance in the prepara tion of the FE Exam problems and the solutions manual.
We were fortunate 10 have an outstanding group of reviewers for this editi on. They are:
Charles F. Bunli ng, Oklahoma State University Manha Sloan, Michigan Technological University Thomas M. Sul livan. Carnegie Mellon University Dr. Prasad Enje li , Texas A&M UniversilY Muhammad A. Khaliq , Minneso ta S tate University Hongbi n Li, Slevens Insti lute of Technology
The preparalion of Ihi s book and the maleria ls Ihal supporl il have been handled wi lh both en thusiasm and g reat care . The combined wisdom and leadershi p of Catherine Shultz, our Senior Editor. has resulted in a tremendous team effort lhat has addressed every aspect of the presentat ion. This team included the fo llowing individuals:
Executive Marketing Manager, Christopher Ruel Senior Production Editor, William Murray Senior Designer, Kevi n Murphy Senior 1I1ustralion Editor, Anna Melhorn Projeci Editor, Gladys SOlO Media Editor, Laure n Sapira Editorial Assistant. Caro lyn Weisman
Each member of this team played a v ital role in preparing the package that is the Ninth Edition of Basic Engineerillg Circuit Analysis. We are most appreciative of their many contributions.
As in the past, we arc most pleased to acknowledge the support that has been prov ided by numerous indi viduals to earlier editions of this book. Our Auburn colleagues who have helped are :
Thomas A. Baginski Travis Blalock Henry Cobb Bill Dillard Zhi Ding Kevin Driscoll E. R. Graf L. L. Grigsby Charles A. Gross David C. Hill M. A. Honnell R. C. Jaeger Keith Jones Betty Kelley Ray Kirby Matthew Langford
Aleck Leedy George Li ndsey Jo Ann Loden James L. Lowry David Mack Paulo R. Marino M. S. Morse Sung-Won Park John Purr Monty Rickles C. L Rogers Tom Shumpert Les Simonton James Trivltayakhull1 Susan Williamson Jacinda Woodward
Many of our friends throughout the United States, some of whom are now retired, have also made numerous suggestions for improv ing the book:
Dav id Anderson, University of Iowa Jorge Aravena, Louisiana State University Les Axelrod, lIIi nois Institu te of Technology Richard Baker, UCLA John Choma. University of Southern C;:tli fornia David Conner, University of Alabama at Bi rmingham James L. Dodd, Mississippi State University Kevin Donahue, University of Kentucky John Durkin, University of Akron Earl D. Eyman. University of Iowa Arvin Grabel, Northeastern University Paul Gray. University of Wisconsin-Platteville Ashok Goel, Michigan Technological University Walter Green, University of Tennessee Paul Greil ing, UCLA Mohammad Habli , Uni versity of New Orleans John Hadjilogiou, Florida Institute of Technology Yasser Hegazy, Univers ity of Waterloo Keith Holbert , Arizona State University Aileen Honka, The MOSIS Service- USC Inf. Sciences Institute Ralph Kinney, LSU Marty Kaliski , Cal Poly, San Luis Obispo Robert Krueger, Uni versity of Wisconsin K. S. P. Kumar, University of Minnesota Jung Young Lee, UC Berkeley student Aleck Leedy, Tuskegee University James Luster, Snow College Erik Luther, National Instruments Ian McCausland, University of Toronto Arthur C. Moeller, Marquette University
PREFA C E xix
xx PRE FA C E
Darryl Morrell. Arizona Slate Uni versily M. Paul Murray, Mississ ippi State University Burks Oakley II , University of Illinois al Champaign-Urbana John O'Malley, University of Florida Wi ll iam R. Parkhurst, Wichi la Siale Universily Peylon Peebles, Universily of Florida Clifford Pollock, Cornell UniversilY George Prans. Manhanan College Mark Rabalais. Louisiana State University Tom Robbins, National Instruments Armando Rodriguez. Arizona State University James Rowland, University of Kansas Robert N. Sackelt, Normandale Community College Richard Sanford, Clarkson University Peddapullaiah Sannuti, Rutgers University Ronald Schulz. Cleveland Slale University M. E. Shafeei , Penn Stale University at Harri sburg SCOIt F. Smilh, Boise Siale University Karen M. St. Germaine, University of Nebraska Janusz Strazyk, Ohio University Gene Sluffle, Idaho Siale UniversilY Saad Tabel, Florida Stale UniversilY Val Tareski , North Dakola Siale University Thomas Thomas, University of South Alabama Leonard J. Tung, Florida A&M UniversityfFlorida Siale Universily Darrell Vines. Texas Tech Uni vers ity Carl Wells, Washington State University Selh Wolpert, University of Maine
Finally. Dave Irwin wishes to express his deep appreciation to his wife, Edie, who has been mosl support ive of our efforts in this book. Mark Nelms would like to Ihank his parents, Robert and Elizabeth, for their support and encouragement.
J. David Irwin and R. Mark Nelms
CHAPTER
Review the 51 system of units and standard prefixes
• Know the definitions of basic electrical quantities: voltage, current, and power
• Know the symbols for and definitions of independent and dependent sources
• Be able to calculate the power absorbed by a ci rcuit element using the passive sign convention
Courtesy NA5A/JPL-Caltech
WORLD! In Jan uary 2007. the Mars Rovers
Opportun ity and Spirit-began their fourth
Cameras, antennae. and a computer receive energy from the
electrical system. In order to analyze and design electrical
systems for the rovers or futu re exploratory robots. we must
year of exploring Mars. Solar panels on the rovers collect have a firm understanding of basic electrical concepts such as
energy to power the electrica l systems. Batteries are utilized to voltage. current. and power. The basic introduction provided in
store energy so that the rovers can operate at night. The rovers this chapter wi\( lay the foundation for our study of engineering
are propelled and steered by electric motors on the six wheels. circuit analYSis. < < <
1
1.1 System of Units
(a) positive current flow; (b) negative current flow.
The sys tem of units we employ is the inte rnational system of units, the Syslcme internat ional des Unites, which is normally referred to as the SI standard system. This system, which is composed of the basic units meter (m), kilogram (kg), second (s), ampere (A), kelvi n (K), and candela (cd), is defi ned in all modern physics tex ts and therefore wi ll not be defined here, However, we will discuss the uni ts in some detai l as we encounter them in our subsequent analyses.
The standard prefixes that are employed in SI are shown in Fig. 1.1. Note the decimal rela tionshi p between these pre fi xes. These s tandard pre fi xes are employed throughout O llf slUdy o f elec tric circuits.
C ircuillcchnoiogy has changed drastically over the years. For example, in the early 1960s the space on a c ircuit board occupied by the base o f a s ingle vacuum tube was about the s ize of a quan er (25-ce l11 coin). Today that same space could be occupied by an Intel Penl ium integrated c ircu it chip conta ining 50 mill ion transistors. These c hi ps are the e ngine for a host of e lec tron ic equipment .
10- 12 10- 9 10-6 10- 3 10' 10' 10' 10 12
I I I I I I I I pico (pI nano (n) micro (II.) milli (m) kilo (kl mega (MI giga (GI lera (TI
Before we begin our analys is of e lectric c ircuits, we must define terms that we will employ. However, in th is chapter and throughout the book our de finiti ons and explanations wi ll be as simple as possible to foster an understanding of the use of the material. No attempt wi ll be made to give complete definiti ons of many of the quantities because such defi nitions are not only unnecessary at this level but are often confusing. Although 1110st of us have an intuitive concept of what is meant by <l circuit. we will simply refer to an elee1ric circll i1 as an inter connection of electrical compone nts, each of which we will describe wi th a mathematical model.
The most elementary quantity in an analys is of electric circuits is the e lectric charge. Our interest in e lectric charge is centered around its motion, since charge in motion results in an energy transfer. Of p<1I1icular interest to us are those s ituations in which the motion is confined to a defin ite closed path.
An e lec tric c ircuit is essentiall y a pipe line that facilitat es the transfe r of charge from one point to anothe r. The lime rate o f change of charge constitutes an e lectric current. Mathematically. the re lationship is expressed as
i (l) dq(l)
or '1(1) = ,Li(X) tit 1.1 dl
where i und q represent curre nt and charge, respective ly (lowercase letters represent time depende ncy, and capital le iters are reserved fo r constant quaJ1lities). The basic unit of current is the ampere (A), and I ampere is I coulomb per second.
Although we know that c urrent flow in metallic conductors results from e lectron motion. the convent ional cun·ent flow, which is universally adopted, represents the movement of positive charges. It is important that the reader think of cllrrent tlow as the movement of positive charge rega rdless of the physical phenomcna that take place. The symbolism that will be lIsed to represent currcn t flow is shown in Fig. J .2. I[ = 2 A in Fig. 1.2a indicates that at any poi nt in the wire shown, 2 C of charge pass from left to right each second. I']. = - 3 A in Fig. 1.2b indicates that at any point in the wire shown, 3 C of charge pass from right to left each second. Therefore, it is important to specify not only the magnitude of the variable representing the current, but also its direc tion.
S ECTION 1.2
i(r) i(r)
(a) (b)
The two types of current that we encounter often in our daily lives. alte rnating current (ac) and direct current (dc), are shown as a function of time in Fig. 1.3. A/remming currell! is the C0l111110 11 current found in every household and is used to run the refrigerator. stove, washing machine, and so on. Batteries, which are lIsed in au tomobiles or fl ashlights , are one sourcc of direcr currell!. In addilion 10 these two types of currents, which have a wide variety of uscs. we can generate many other types of currents. We will exami ne some of these other types later in the book . In the meant ime, it is interesti ng to note that the magnitude of currents in e lements fam iliar to us ranges from soup to nuts. as shown in Fig. IA.
We have indicaled lhat charges in motion yield an cncrgy transfcr. Now wc detinc the volrage (also called the elecrromotive force or porenthll) between IWO points in a circuit as the difference in energy level of a unit charge located at each of the two points. Vo ltage is very sim ilar to a gravitational force. Think about a bowling ball being dropped from a ladder into a tank of water. As soon as the ball is re leased, the force of gravity pulls it 100vard the boltom of the tank. The potential energy of the bowling ball decreases as it approaches the bottom. The grav itational force is pushing the bowling ball through the water. Th ink of the bowl ing ball as a charge and the voltage as the force pushing the charge through a circuit. Charges in mOl ion
represent a current, so the motion of the bowling ball could be thought of as a current. The water in the tank will resist the motion of the bowling ball. The motion of charges in an elec tric circuit wi ll be impeded or resisted as wel l. We will introduce the concept of resistance in Chapter 2 to describe this effec\.
Work or energy, w(r ) or W, is measured in jou les (l ); I joule is I newton meter (N · m). Hence, voltage [v(r) or VI is measured in volts (V) and I volt is I joule per coulomb; tha t is, I volt ;:;:: I joule per coulomb = I newton meter per coulomb. If a unit positive charge is moved between two points, the energy req uired to move it is the difference in energy level between the two poi nts and is the defin ed voltage. It is extremely imponant that the variables used to represent voltage between two points be deti ned in such a way that the solution wi ll let us interpret which point is at the higher potential wilh respect to the other.
10' Lightning bolt
10'
00
~ 10- ' ~
0. E Human threshold of sensation ~ 10- 4
.5 C 10-6 ~ 5
10-8 0 Integrated ci rcui t (IC) memory cell current
10- 10
10- 12
.~.- Figure 1.3
current (ae); (b) direct
3
Figure 1.5 ••• ~
Voltage representations.
Figure 1.6 ... ~
+ B
(e)
In Fig. l.5a the variable that represe nts the voltage between points A and 8 has bee n defined as VI. and it is assumed that point A is at a higher potential than point e, as indicated by the + and - signs associated with the variable and deti ned in the tlgurc. The + and - signs define a re ference direction for VI' If ~ = 2 Y, then the difference in potential of points A and B is 2 V and point A is at the higher potential. If a unit positive charge is moved from point A through the c ircuit to point B, it wi ll give up energy to the ci rcuit and have 2 J less energy when it reaches point B. If a unit positive charge is moved from point B to point A. ex tra e nergy must be added to the charge by the circuit, and hence the charge will e nd up with 2 J more energy at poi nt A than it started with at point B.
For Ihe circuil in Fi g. 1.5b, V, ~ - 5 V means Ihat Ihe pOlential between points A and 8 is 5 V and point B is at the higher potential. The voltage in Fig. 1.5b can be expressed as shown in Fig. 1.5c. In this equivalent case , the difference in potential between points A and B is V2 = 5 V, and point B is at the higher pote nti al.
Note that it is important to detine a variable with a reference direction so that the answer can be interpreted to give the physical condition in the circuit. We wi ll find that it is not possible in many cases to define the variable so that the answer is positive. and we will also tind that it is not necessary to do so.
As demonstrated in Figs. 1.5b and c, a negative number for a given variable, for example. V2 in Fig. 1.5b, gives exactly the same information as a positive number, that is. V2 in Fig. 1.5c, except that it has an opposite reference direc tion. Hence, \-vhen we define either current or volt age, it is absolutely necessary that we specify both magnitude and direc tion . Therefore, it is incomplete to say that the voltage between two points is 10 V or the current in a line is 2 A, since only the magnitude and not the direction for the variab les has been defined.
The range of magnitudes for voltage, equivalent to that for currents in Fig. lA , is shown in Fig. 1.6. Once again, note that this range spans many orders of magnitude.
10·
10·
10'
~ 102
Vottage between two points on human scatp (EEG)
Antenna of a radio receiver
SECT I ON 1.2
Light bulb
At this point we have presented the conventions that we employ in our discussions of current and voltage. Ellergy is yet anothe r important tenn of basic significance. Let's investigate the voltage-curren t relationships for energy transfer using the flashlight shown in Fig. 1. 7. The basic e lements of a fl ashlight are a battery, a switch, a ligh t bulb, and connect ing wires. Assuming a good battery, we all know thutthe light bul b will g low when the swi tch is closed. A current now fl ows in this closed circuit as charges fl ow out of the positive ter minal of the battery through the swi tch and light bulb and back into the negative terminal of the battery. The current heats up the filament in the bulb, causing it to glow and emit light. The light bulb converts e lectrical energy to thermal energy; as a result, charges passing through the bulb lose energy. These charges acquire energy as they pass through the battery as chemical energy is converted to e lectrical energy. An energy conversion process is occur ring in the flashlight as the chemical energy in the bane ry is converted to e lec trical energy, which is then converted to thermal energy in the light bulb.
I - BaHery
- Vbauery + + V bu1b -
Let's redraw the fl ashlight as shown in Fig. 1.8. There is a current I fl owing in this dia gram. Since we know that the light bu lb uses energy, the charges coming out of the bu lb have less energy than those entering the light bulb. In other words, the charges expend ene rgy as they move through the bulb. This is indicated by the voltage shown across the bulb. The charges gain energy as they pass through the banery, which is indicated by the voltage across the battery. Note the vo ltage--<:urrent re lationships for the battery and bulb. We know that the bulb is absorb ing energy; the current is entering the positive terminal of the voltage. For the battery, the current is leaving the posit ive temlinal, which indicates that energy is being supplied.
This is further illustrated in Fig . 1.9 where a c ircuit element has been extracted from a larger circuit for examination. In Fig. 1.9a, energy is being supplied to the element by whatever is attached to the terminals. Note that 2 A, that is, 2 C of charge are movi ng from point A to point B through the e lement each second. Each coulomb loses 3 J of energy as it passes th rough the element from point A to point B. Therefore, the element is absorbing 6 ] of energy per second. Note that when the element is absorbing energy. a positive current enters the positive terminal. In Fig. 1.9b energy is being suppl ied by the clement to whatever is connected to terminals A- B. In this case, note that when the element is supplying ene rgy, a positive current enters the negative terminal and leaves via the positive terminal. In thi s con vention, a negat ive current in one direction is equivalent to a positive current in the opposite direction, and vice versa. Similarly, a negative voltage in one direction is equivalent to a pos itive vo ltage in the opposite d irection.
BASIC QUANTITIE S
A 1 ~ 2A
.or" Figure 1.9
•
6 C H APT ER 1 BASIC CONC E PTS
EXAMPLE 1.1
i(l)
is used to determin e whether
power is being absorbed or
supplied.
Suppose that your car will not start. To determine whether the battery is faulty, you turn on the light swi tch and fi nd that the lights are very dim, indicating a weak battery. You borrow a friend's car and a set of jumper cables. However, how do you connect his car's battery to yours? What do you want his batlery to do?
Essentially, his car's battery must supply energy to yours, and therefore it should be connected in the manner shown in Fig. J. I O. Note that the positive current leaves the posi tive termi nal of the good battery (supplyi ng energy) and enters the positive terminal of the weak battery (absorbi ng energy), Note that the same connections are used when charging a battery.
I
I
In practica l appl ications there are often considerations other than simply the e lectrical re lations (e.g., safety). Such is the case with jump-starting an automobile. Automobile batteries produce explosive gases that can be ignited accidental ly, causing severe physical inju ry. Be safe-follow the procedure described in your auto owner's manual.
We have defined voltage in joules per coulomb as the energy required to move a positi ve charge of I C through an e lement . If we assume that we are dea ling with a di ffere ntial alllount of charge and energy, then
li1V v=-
dq
Muhiplying this quan tity by the current in the e lement yields
1.2
1.3
which is the time rale of change of energy or power measured in joules pe r second, or watts (W). Since, in general, both v and i are functions of time, p is a lso a time-vary ing quanti ty. Therefore, the change in energy from time I, to lime 12 can be found by integrati ng Eq. ( 1.3); that is,
1', 1" t.w = P dl = VillI '1 '1
1.4
At this point, leI us summarize our sign convention for power. To determine the sign of any o f the quan tities involved, the variables for the current and voltage shou ld be arranged as shown in Fig. 1. 11 . The variable for the voltage V{ /) is defi ned as the voltage across the e le ment with the posit ive reference at the same termi nal that the current variable i{/ ) is entering. This convention is called the passive sign cOllvelllioll and will be so noted in the remainder of thi s book. The product of V and i , with the ir attendant signs, will determine the magnitude and sign of the powcr. If the sign of the power is posit ive, power is bei ng absorbed by the e le ment; if the sign is ncgative. power is being supplied by the element.
SECTION 1 . 2
Given the two diagrams shown in Fig. 1.12, determine whether the element is absorbing or supplying power and how much.
2V
EXAMPLE 1.2
~ ... Figure 1.12
Elements for Example 1.2 .
• In Fig. 1.l2a the power is P = (2 V)(- 4 A) = -8 W. Therefore, the element is supplying SOLUTION power. In Fig. 1.12b, the power is P = (2 V)(- 2 A) = - 4 W. Therefore, the element is supplying power.
Learning ASS E SSM E N T
+
(b)
We wish to determine the unknown voltage or current in Fig. 1.13.
SA { = ?
B +B
(a) (b)
EXAMPLE 1.3
.~ ... Figure 1.13
Elements for Exa mple 1.3.
• In Fig. 1.13a, a power of -20 W indicates that the element is delivering power. Therefore, SOLUTION the current enters the negat ive terminal (terminal A), and from Eq. (1.3) the voltage is 4 V. Thus, B is the positive terminal , A is the negative terminal, and the voltage between them is 4Y.
In Fig 1.1 3b, a power of +40 \V indicates that the element is absorbing power and. there fore, the current should enter the positive terminal B. The current thus has a value of -8 A, as shown in the figure.
7
+
(0) V, = -20 V; (b) I = -5 A.
Finally, it is imporlum to note that ollr e lec trical ne tworks satisfy the principle of conser vation of energy. Because of the re lat ionship between energy and power, it can be implied that power is also conserved in an e lec trical network. This result was formally stuted in 1952 by B. D. H. Tcllcgcn and is known as Tellegcn's theorem- the slim of the powers absorbed by a ll eleme nts in an c lectricalnclwork is zero. Another statement of this theorem is Ihal the power supplied in a network is exac tl y equal to the power absorbed. Checki ng (0 verify that Te llegcn's theorem is satis fi ed for a particular network is one way to check our calculations when analyzing e lectrical networks.
Thus far we have defined voltage, c urrent , and power. In the remainder of this chapter we will de fine both illllt:pclldent and dependen t curre nt and voltage sources. Although we will assume ideal elements, we will try to indicate the shortcomings of these assumptions as we proceed wi th the discussion.
In general, the e lements we will define are terminal devices that are complete ly charac te ri zed by the current th rough the c lement and/or the voltage across it. These e lements, wh ich we wi ll employ in constructing electric ci rc uits. will be broadly classified as being ei the r active or pass ive. The distinction between these two classifications depends essentially on olle thing-whether they supply OJ' absorb energy. As the words themselves imply, an active e lement is capable of generating energy and a passive element cannot gene rate energy.
However, late r we will show that some passive e lements are capable of storing energy. Typical active e lements are batteries and generators. The three common passive e lements are res istors, capacitors, and inductors.
In Chapter 2 we wi ll launch an examination of passive elements by discussing the resis tor in detail. Before proceeding with that c lement , we first present some very important active e lements.
1. Independent voltage source 3. Two dependent voltage sources
2. Independent c urrent source 4 . Two dependent current sources
INDEPENDENT SOURCES An illdepellde1l1 vo/wge source is a two-Ienninal e lement that maintains a specified vo ltage between its te rminals regardless oj tile CllrreW throllgh it as shown by the v -i plot in Fig. I. 14a. The general symbol for an independent source, a c ircle. is also shown in Fig. 1.14a. As the figure indicates. terminal A is v(t ) volts positive with
respec t to terminal B. In contrast to the independent vo ltage source, the illdepel/{Iem current source is a two
terminaJ e lemeill that maintains a specified c urre nt regardless oj'tlte volwge (lcros.\· its termillals, as illustrated by the '1)-i plot in Fig. 1.14b. The general symbol for an independe nt current source is also shown in Fig. 1.1 4b, where i(t) is the specifi ed Clirreill and the arrow indicates the posit ive direction of c urrent n ow.
} I
'V 'V
A A
"(')~ '(')~ B B
(a) (b)
In their normal mode o f operation, independent sources supply power to the re mainder of the circuit. However, they may also be connected into a circuit in such a way that they absorb power. A s imple example of this latter case is a battery-charg ing circuit such as that shown in Example 1.1.
It is important that we pause here to interjec t a comment concerning a sho rtcoming of the models. 1.11 general, mathematical mode ls approximate actual physical syslems only under a cer tain range of conditions. Rarely does a I1H..x.lel accuntle ly represent a physical system under every set of conditions. To illustrate th is point. consider the model for the voltage source in Fig. I. 14a. We assume thai the vo ltage source deli vers V volts regardless of what is connected 10 its tennina ls. Theoreticall y, we could adjust the ex ternal circuit so that an intinite amount of current would tlow, and therefo re the voltage source would deliver an intinite amount of power. This is, of course, physically impossible, A similar argument could be made for the independ ent current source. Hence, the reader is cautioned to keep in mind that models have limitations and thus are valid representations of physical systems o nly under certain conditions.
For example, can the independent voltage source be utili zed to model the battery in an
automobile under all operating conditions? With the headlights on, turn on the radio. Do the headlights dim wi th the radio on? They probab ly won't if the sound system in your auto mo· bi le was installed at the factory. If you try to crank your car w ith the headlights on, you wi ll not ice that the lights dim. The starter in your car draws considerable current, thus causing the vo ltage at the battery termina ls to drop and dimming the headli ghts . The independent vo lt age source is a good model for the batte ry with the radio turned on; however, an improved model is needed for your battery to predict its performance under cranking conditions.
Determine the power absorbed or supplied by the elements in the network in Fig, 1,1 5 ,
6V
~, •• Figure 1,14
Symbols for (al independent voltage sou rce, (bl independ, ent current source.
EXAMPLE 1.4
~ ••• Figure 1,15
•
10 CHA PTE R 1 BAS I C C ONC EPT S
• SOLUTION
[hin tj Elements that are connected In series have the same current.
The curre nt n ow is out of the positive te rminal of the 24-V source, and therefore thi s e leme nt is supplying (2)(24) ~ 48 W of power. The c urrent is into the positive terminals of e leme nts I and 2, and the refore elements I and 2 are absorbing (2)(6) ~ 12 Wand (2)( 18) ~ 36 W, respectively. Note that the power supplied is equal to the power absorbed.
Learning ASSESSM ENT
E1.3 Find lhe power that is absorbed or supplied by the elements in Fig. E 1.3. ANSWER: Current source supplies 36 W, element
Figure E1.3
Figure 1.16 ... ~
I absorbs 54 W, and element 2 supplies 18 W.
DEPENDENT SOURCES In contras t 10 the indepe ndent sources, which produce a particular voltage or c urrent completely unaffec ted by what is happening in the remainder of [he circui t, dependent sources generate a voltage or c urrent that is detennined by a voltage or current at a speciti ed location in the circui t. These sources are very important because they are an integral purt of the mathematical models used to describe the behavior of many e lec tronic circui t e lements.
For example, mctal-oxide-scmiconduc tor ficld-clfect transistors (MOSFETs) and bipolm transislOrs, both of which are commonly found in a hos t of e lec tJonic equipme nt. are mod e led with dependent sources, and there fore the anal ys is of e lec tronic circuits involves the use of these cont ro lled e lements.
In contrast lO the circle used to represent independent sources, a diamond is used to represent a dependent or controlled source. Figure 1. 16 illustrates the four types of depe nd ent sources. The input termi nals on the left represent the voltage or current that contro ls the dependent source. and the output terminals on the right represent the output current or vo lt age orthe COil tro lled source. Note that in Fi gs. 1.1 6a and d the quan tities ~ and J3 are di men sionless constants because we are transforming voltage to voltage and current to c urrent . This is not the case in Figs. 1. 16b and c; hence, when we employ these e lements a short time later, we must describe the un its of the factors rand g.
E (a) (b)
SECTION 1 . 3 CIRCUIT ELEMEN TS 11
Given the two networks shown in Fig. 1.1 7, we wish to determine the outputs. EXAMPLE 1.5
• [n Fig. 1.17a the output voltage is v" = [.LVs or Va = 20 Vs = (20)(2 V) = 40 V. Note that SOLUTION the output voltage has been amplified from 2 V at the input terminals to 40 V at the output terminals; that is, the circuit is a voltage amplifier with an amplification factor of 20.
IS = 1 rnA
(a) (b)
[n Fig. 1.I7b, the output current is 10 = f3/s = (50)(1 mAl = 50 mA; that is, the circuit has a current gain of 50, meaning that the output current is 50 times greater than the input current.
£1.4 Determine the power supplied by the dependent sources in Fig. E 1.4.
10 = 2A " 15 = 4A
+ B '"'B Vs = 4V
Figure E1.4
Calculate the power absorbed by each element in the network of Fig. I. IB. Also verify that Tellegen 's theorem is satisfied by this network.
24 V
4V
Let's calculate the power absorbed by each element using the sign convemion for power.
PI = ( 16)( 1) = 16 W P, = (4)( 1) =4W P, = ( 12) ( 1) = 12 W
~ .. , Figure 1.17
Circuits for Example 1.5.
ANSWER: (a) Power supplied = BO W; (b) power supplied = 160 W.
EXAMPLE 1.6
~ ... Figure 1.18
• SOLUTION
EXAMPLE 1.7
Figure 1.19 ... ~
•
P, = (8)(2) = 16 W P12V = ( 12)(2) = 24 W P"v = (24)(- 3) = -72 W
Note that to calculate the power absorbed by the 24-V source, the current of 3 A flowing up through the source was changed to a current -3 A flowing down through the 24-V source.
Let 's sum up the power absorbed by all elements: 16 + 4 + 12 + 16 + 24 - 72 = 0
This sum is zero, which verifies that Tellegen's theorem is satisfied .
Use Tellegen's theorem to find the current 10 in the network in Fig. 1.19.
SOLUTION First, we must determine the power absorbed by each element in the network. Using the sign convention for power, we find
P'A = (6)( - 2 ) = - 12 W
P, = (6 )(10) = 6/" W
P, = ( 12)( -9 ) = - 108 W
P, = ( 10)(- 3) = -30 W
PH = (4)(-8) = -32W
Applying Tellegen 's theorem yields,
-12 + 6/" - 108 - 30 - 32 + 176 = 0 or
6/,, + 176 = 12+ 108 +30 + 32 Hence,
10 = IA
E1.5 Find the power that is absorbed or supplied by the circuit elements in the network in Fig. EI.5.
BV
24V
Figure El.5
ANSWER: P" v = 96 W supplied; P, = 32 W absorbed; P4I , = 64 W absorbed.
SECTION 1.3 C IRCUIT ELEMENTS 13
The charge that enters the BOX is shown in Fig. 1.20. Calculate and sketch the current flow- EXA M P L E 1 .8 ing into and the power absorbed by the BOX between 0 and 10 milliseconds.
i (I) ~ ... Figure 1.20
Diagrams for Example 1.8.
- t
- 2
- 3
Recall that current is related to charge by i(l) = dq(l) The current is equal to the slope of SOLUTION dl
the charge waveform.
i(l) = 2 X 10 3 _ I X 10 3 = 2A
i(l) = 0
i(l) = 5 X 10 3 _ 3 X 10-3 = -2.5 A
i(l) = 0
9 X 10 3 - 6 X 10-3
i(l) = 0
5SI s 6ms
t ~ 9ms
•
•
14 C H A P TER 1 BAS I C CO N CEPTS
Figure 1.21 ... ? Charge and current
waveforms for Example 1.8.
Example 1.8.
p(l ) = 12*0 = 0
p(l ) = 12*0 = 0
p(l ) = 12*0 = 0
p( l ) = 12*0 = 0
O StS lms
1 ::51S 2ms
3 ::5 1 ::5 5 ms
5 ::5 t s 6 ill S
6 "1 ,, 9ms
9 10 I (ms)
The power absorbed by the BOX is plotted in Fig. 1.22. For the time intervals, I " I " 2 ms and 6 s 1 ::5 9 ms, the BOX is absorbing power. During the time interval 3 ::5 t ::5 5 ms, the power absorbed by the BOX is negative, which indicates that the BOX is supplying power to the 12-V source.
p(l) (W)
- 12
- 24
9
-36
SECTION 1 . 3 CIRCUIT ELE MEN T S 15
A Universal Serial Bus (US B) port is a common feature on both desklop and nOlebook compulers as well as many handheld devices such as MP3 players. digila l cameras, and cell phones. The new USB 2.0 specification (www.usb.org) permits data transfer between a computer and a peripheral device at rates up to 480 Megabits per second. One important fea ture of USB is the ability to swap peripherals without having to power down a computer. USB ports are also capable of supplying power 10 eXlernal peripherals. Figure 1.23 shows a MOlorola RAZR® and Apple iPod® being charged from the USB ports on a nOlebook computer. A USB cable is a four-conductor cable with two signal conductors and two con ductors for providing power. The amount of current that can be provided over a USB port is defi ned in the USB specification in terms of unit loads, where one unit load is specifi ed 10 be 100 rnA. All USB ports defau illo low-power parIS al one unilload, bUI can be changed under software contro l 10 high-power ports capable of supplying up 10 five unil loads or 500mA.
1. A 680 mAh Lilhium-ion battery is standard in a Motorola RAZR®. If Ihis battery is completely discharged (Le. , 0 mAh), how long will it take to recharge the baltery to its full capacilY of 680 mAh from a low-power USB port? How much charge is stored in the battery at the end of the charging process?
2. A Ihird-generation iPod® wilh a 630 mAh Lithium-ion battery is 10 be recharged from a high-power USB port supplying 150 mA of current. Al the beginning of Ihe recharge, 7.8 C of charge are stored in the battelY, The recharging process halts when the stored charge reaches 35.9 C. How long does illake to recharge Ihe battery?
1. A low-power USB port operates at 100 rnA. Assumjng that the charging current from the US B port remains at 100 rnA throughout the charging process, the time required to recharge Ihe battery is 680 mAh/ IOO mA = 6.8 h. The charge stored in the battery when full y charged is 680mAh • 60 s/h = 40,800 mAs = 40.8 As =
40.8 C. 2. The charge supplied 10 Ihe battery during the recharging process is
35.9 - 7.8 = 28. 1 C. This corresponds 10 28.1 As = 28,100 mAs ' Ih/ 60s =
468.3 mAh. Assuming a constant charging current of 150 rnA from the high-power USB port, the time required 10 recharge the battery is 468.3 mAh/ ISO mA = 3. 12 h.
EXAMPLE 1.9
~". Figure 1.23
Charging a Motorola RAZXR® and Apple iPod® from USB ports. (Courtesy of Mark Nelms and Jo Ann Loden)
• SOLUTION
SUMMARY
n = 10-9 M 106
fJ. 10-6 G 10'
• The relationships between current and charge
dq ( l ) i(l ) =-
• The relationships among power, energy, current, and voltage
dw . p = - = V I
"
"
1.1 If the currem in an electric conductor is 2.4 A, how many coulombs of charge pass any point in a 30-second interval?
1.2 Determine the time interval required for a I2-A battery charger '0 deliver 4800 C.
1.3 A lightning bolt carrying 30,000 A lasts for 50 micro seconds. If the lightning strikes an airplane flying at 20,000 feel, what is the charge deposited on the plane?
1,4 If a 12-V battery delivers 100 J in 5 s, fi nd (a) 'he amount of charge delivered and (b) the current produced.
0 '.5 The current in a conductor is 1.5 A. How many coulombs of charge pass any point in a time interval of 1.5 min?
o 1.6 If 60 C of charge pass through an electric conductor in 30 seconds, determine the currem in the conductor.
0 1.7
0 ,.8
Determine the number of coulombs of charge produced by a 12-A battery charger in an hour.
Five cou lombs of charge pass through the elemem in Fig. PI.8 from point A to point B. If the energy absorbed by the element is 120 J , determine the voltage across the element.
B
Figure P1,S
• The passive sign convention The passive sign convention states that if the voltage and current associated with an element are as shown in Fig. 1.11 . the product of v and i , with their 3t1cndant signs. determines the magnitude and sign of the power. If the sign is positive, power is being absorbed by the element, and if the sign is negative, the clemem is supplying power.
• Independent and dependent sources An
ideal independen t voltage (current) source is a two-terminal element that maintains a specified voltage (current) between its terminals regardless of the current (voltage) through (across) the element. Dependent or controlled sources generate a voltage or current that is detcnnined by a voltage or current at a specified location in the circuit.
• Conservation of energy The electric circuits under investigation satisfy the conservation of energy.
• Tellegen's Theorem The Slim of the powers absorbed by all elements in an electrical network is zero.
1.9 The charge entering an element is shown in Fig. P1.9.
1.10
Find the current in the element in the time interval o ~ I ~ 0.5 s. [Him: The equation for '1(1) is '1 (1) = I + ( 1/ 0.5)1, I ~ 0.1
q (e)
Figure P1.9
The current that enters an element is shown in Fig. P 1.1 O. Find the charge that enters the element in the lime interval 0 < I < 20 s.
i(l) rnA
o
1 .11 The charge entering the positive terminal of an element is q(r) = -30e-41 me. If the vollage across the element is 120e- 21 y , determine the energy delivered to the element in the time interva l 0 < t < 50 ms.
PR O B L E MS 17
1.14 The waveform for the current flowing illlo a circui t element is shown in Fig. P 1. 14. Calculate the amoum of charge which enters the element between(a) 0 and 3 sec onds, (b) 1 and 5 seconds, and (c) 0 and 6 seconds.
£1 1.12 The charge emering the positive terminal of an element is given by the express ion q(t) = - J 2e- 2r me. The power de livered to the element is p(r) = 2.4e-31 W. Compute the currem in the element, the voltage across the element, and the energy del ivered to the element in the time interval 0 < t < 100 InS.
( ) (A) - t t
2 -- -- -- -- --------
1.13 The vo ltage across an element is 12e-2r V. The current entering the positive terminal of the elemem is 2e- z, A.
Find the energy absorbed by the element in 1.5 s starting from t = O.
.5
1
1
Figure Pl.14
2 3
1.15 The current fl owing into a box is given by the waveform shown in Fig. PI . IS . Calcu late the fo llowing quantities: (a) the amount of charge which has entered the box at
2V
t = 1 s, t = 3 s, and t = 4.5 s, (b) the power absorbed by the box at t = I s, 2.5 s, 4.5 s, and 5.5 s and (c) the amount of energy absorbed by the box between 0 and 6 s .
t I . ( ) (A)
2 i (I)
1
18 CHAP T ER 1 BASIC CONC E PTS
1.16 The charge fl owi ng into the box is shown in the graph ill Fig. P 1. 16. Sketch the power absorbed by the box .
q(t) (C)
- 1
Figure P1.16
f} 1 .17 The charge that enters a BOX is shown in Fig. PI .I ? Calculate and sketch the current flowing into and the power www absorbed by the BOX between 0 and 9 mill iseconds. Also calculate the energy absorbed by the BOX between 0 and
~ 9 milliseconds.
q(t) (mC)
Figure Pl.l?
0 1.18 Determine the amoun t of power absorbed or supplied by the e le me nt in Fig. PI.1 8 if
(a) V I = 9 V and I = 2A
(b) V I = 9 V and I = - 3A
(c) V I = - 12 V and I = 2A
(d) V I = - 12 V and I = - 3A
+ /
7 8 9 t (ms)
1.19 Determi ne the missing quanti ty in the circuits in Fig. P1. 19.
I
(a) (b)
Figure Pl.19
1.20 Repeat Problem 1. 19 for the circuits in Fig. P1.20.
1 ~ - 3 A
(b)
1.21 Determ ine the power supplied to the elements in Fig. PI.2t.
Figure Pl.21
1.22 Determine the power supplied to the elements in Fig. P1.22.
Figure Pl.22
(} 1.23 (a) [n Fig. Pt.23 (a), P, = 36 W. [s element 2 absorb ing or supplying power, and how much?
(b) In Fig. PI.23 (b). P, = -48 W. Is element I absorb ing or supplying power, and how much?
(a)
P R OBLEMS 19
1.24 Two elcments are connected in serics, as shown in Fig. P 1.24. Element I supplies 24 W of power. Is element 2 absorbing or supplying power, and how much?
Figure Pl.24
+ 6V
8V +
1.25 Two elements are connected in series, as shown in Fig. P 1.25. Element I suppl ies 24 W of power. Is element 2 absorbing or supplying power, and how much?
Figure P1.25
+ 3V
6V +
1.26 Two elements are connected in series, as shown in Fig. P 1.26. Element I absorbs 36 W of power. Is clement 2 absorbing or supplying power. and how mLlch?
Figure Pl,26
12V + + 4V
1.27 Choose Is such thai the power absorbed by element 2 in Fig. Pl.27 is 7 W.
4V
6V
20 CHAPTER 1 BASIC CONCEPTS
1.28 Delemline the power thaI is absorbed or supplied by the circuit elements in Fig. P1.28 .
10 V
24 V
(a) (b)
Figure P1.2S
1.29 Find the power that is absorbed or supplied by the circuit elements in Fig. P1.29.
8V
(a) (b)
Figure P1.29
0 '.30 Find the power that is absorbed or suppJied by the net work elements in Fig. P1. 30.
' .3' Com pule Ihe power that is absorbed or supplied by Ihe elements in the network in Fig. P 1.31.
9 -
(a)
(b)
2A
+ 28V
o
~ 1·32 Calculate the power absorbed by each e lement in the circuit in Fig. P1.32.
2A 12V
20V +
1.33 Find V, in the network in Fig. PI .33 using Tellegen 's theorem.
1------(- +} -----j
24 V
Figure Pl.33
o 1·34 Find ~ in the network in Fig. PI.34 using Tellegen's theorem.
9 V 12V
Figure Pl.34
o 1·35 Find ~ in the network in Fig. P 1.35 using Tellegen's theorem.
2V +
Figure Pl.35
o 1.36 Find l.x in the circuit in Fig. PI .36 using Tellegen's theorem.
12 V
22 CHAPTER 1 BASIC CONC EPTS
() 1.37 Is the source Vt in the network in Fig. P 1.37 absorbing or supplying power, and how much?
Figure Pl.37
() 1.39 Find I" in the network in Fig. P 1.39 using Tellegen's theorem.
+ 24 V 10V l , ~ 2 A
6V + 4 16 V
Figure Pl.39
1.38 Find Vr in the network in Fig. PI.38 using Tellegen's theorem.
+ 12 V
1 A
Figure Pl.38
+ 16 V
1.40 Find VI" in the network in Fig. PlAO using Tellegen 's theorem.
8V 4V
• Be able to use Ohm's law to solve electric circuits
• Be able to apply Ki rchhoWs current law and KirchhoWs voltage law to solve electric circuits
• Know how to analyze single-loop and single node-pair circuits
• Know how to combine resistors in series and parallel
• Be able to use voltage and current division to solve si mple electric circuits
• Understand whe n and how to apply wye-delta transformat ions to solve electric ci rcu its
• Know how to analyze electric circu its containin g dependent sou rces
YBRID VECHICLES SUCH AS THE
TOYOTA Prius utilize a gasoline engine
and an electric motor to provide propul
sion. The gasoline engine may drive the vehicle, or it may charge
of a hybrid powertrain resu lts in reduced emissions and improved
gas mileage. The electrical system in a Toyota Prius is a de system
with a sealed Nickel-Metal Hydride battery with a rated voltage of
201.6 volts. The analysis and design of the electrical system in
the battery in the electrical system. The electric motor may drive hybrid vehicles require knowledge of fundamental circuit laws
the vehicle by itself, It is possible that the gasoline engine and such as Ohm's law, Kirchhoff's current law, and Kirchhoff's
I electric motor may work together to propel the vehicle, Utili_za_t_io_n __ v_o_lt_a_g_e_la_w_,_T_he_se laws will be introduced in this chapter. < < <
23
2.1 Ohm's Law
[hin tJ The passive sign convention will be employed in conjunction with Ohm's law.
Figure 2 .1 ••• ~
are high·wattage fixed
precision resistor. (7)-(1')
different power ratings. (Photo courtesy of Mark
Nelms and 10 Ann Loden)
i t t)
la)
Ohm's law is named for the German physicist Georg Simon Ohm , who is credited wi th establishing the vo ltage-current re lationship for res istance. As a result of his pioneering work, the unit of resistance bears his name.
Ohm's law stales that the lIo/rage acroH a resistance is directly proporliolla/ro the current flo wing through il. The res istance. measured in ohms. is the constant of proportionali ty between the voltage and current.
A circuit e lement whose electrical characteristic is primarily res istive is called a resistor and is represented by the symbol shown in Fig. 2. 1a. A resistor is a physical device that can be purchased in certain standard va lues in an e lectronic part s store . These res istors, which find use in a variety of elec trical applications, are normall y carbon composition or wire wound. In addition , resistors can be fabricated using thick oxide or thin metal films for usc in hybrid circu its, or they can be diffused in semiconductor in tegrated c ircuits. Some typical discrete resi stors are shown in Fig. 2.1 b.
The mmhematical re lationship of Ohm's law is illustrated by the equation
Ve t ) = R i(t ), where R ~ 0 2.1
or equivalently, by the voltage-<:urrent characteristic shown in Fig. 2.2a. Note carefully the relationship between the polarity of the voltage and the direction of the current. In addition, nOle that we have tacit ly assumed that the resistor has a constant va lue and therefore that the voltage-<:urrent characteristic is linear.
The symbol n is used to represent ohms, and therefore,
In = I V IA
Although in our analysis we will always assume that the resistors are linear and arc thus described by a straigh t-line characteristic that passes through the origin, it is important !.hat readers reali ze that some very useful and practical elements do ex ist that exhibit a nonlinear resis tance charac teristic; that is, the vohage-<:urrent relationship is not a straight line.
(1 ) (2)
(6) ---(4)
(10)
Ib ) -
S E CTIO N 2. 1 O H M ' S LAW 25
V(I) V(I)
i( I)
(a) (b)
The light bulb from the flashlight in Chapter I is an example of an element that exhibits a nonlinear characteristic. A typical characteri stic for a light bulb is shown in Fig. 2.2b.
Since a resistor is a passive clement, the proper current- voltage relationship is illustrated in Fig. 2.l a. The power supplied to the terminals is absorbed by the resistor. Note that the charge moves from the higher to the lower potential as it passes through the resistor and the energy absorbed is dissipated by the resistor in the fonn of heat. As indicated in Chapter J, the rate of energy dissipation is the instantaneous power, and therefore
p(l ) = V( I );(I )
, v' ( I) p(l ) = Ri-(I ) = -
R
2.2
2.3
This equation illustrates that the power is a nonlinear function of either current or voltage and that it is always a positive quanti ty.
Conductance, represented by the symbol G, is another quantity with wide application in circuit analysis. By definiti on, conductance is the reciprocal of resistance; that is,
1 G =
R
The unit of conductance is the siemens, and the relationship between units is
I S = I A/ V
Using Eq. (2.4), we can write two additional expressions,
and
2.4
2.5
2.6
Two specific values of resistance, and therefore conductance, are very important : R = 0 and R = 00.
In examining the two cases, consider the network in Fig. 2.3a. The variable resistance symbol is used to describe a resistor sllch as the volume control on a radio or television set.
~ ••• Figure 2 . 2
Graphical representation of the voltage- current relation
•
Figure 2.3 ... ~
(a) (b) (e)
As the resistallce is decreased and becomes smaller and smaller, we finall y reach a point where the res istance is zero and the c ircuit is reduced to that shown in Fig. 2.3b; that is, the resis tance can be replaced by a short ci rcuit. On the other hand, if the resistance is increased and becomes larger and larger, we finall y reach a point where it is essentially infinite and the res istance can be replaced by an open circuit , as shown in Fig. 2.3c. Note that in the case of a short c ircuit where R = 0,
u(t) = !?i(t )
= 0
Therefore, V( I ) = 0, although the current could theoretically be any value. In the open circuit case where R = 00,
i(l ) = U( I)/ !?
= 0
Therefore. the current is zero regardless of the value of the voltage across the open terminals .
In the ci rcuit in Fig. 2.4a, determine the current and the power absorbed by the res islOr.
SOLUTION Using Eq . (2.1), we find the current to be
Figure 2.4 ... ::.
1 = V/ !? = 12/2k = 6mA
Note that because many of the resislOrs employed in our analysis are in kf1 , we will use k in the equations in place of 1000. The power absorbed by the resistor is given by Eq. (2.2) or (2.3) as
12 V
= I' !? = (6 x lO- l)' (2k) = 0.072 W
= v'/ !? = ( 12 )'/2k = 0.072 W
/
(e) (d)
10 kfl
P = 3.SmW
SECTION 2 . 1 O HM ' S LAW 27
The power absorbed by the 10-kD resistor in Fig. 2.4b is 3.6 mW. Determine the voltage and the current in the circuit.
Using the power relationship, we can determine either of the unknowns.
and
Vs = 6 V
I' R = P
I = 0.6 rnA
Furthermore, once Vs is determined, 1 could be obtained by Ohm 's Jaw, and likewise once I is known, then Ohm's law could be used to derive the value of 11,. Note carefully that the equations for power invol ve the terms 12 and V~. Therefore, I = -0.6 rnA and Vs = - 6 V also satisfy the mathematical equations and, in this case, the direction of both the voltage and current is reversed.
Given the circuit in Fig. 2.4c, we wish to find the value of the voltage source and the power absorbed by the res istance.
The voltage is
Vs = I/ G = (0.5 x 10-3)/( 50 x 10-6 ) = 10 V
The power absorbed is then
P = I' / G = (0.5 x 10-3)' /( 50 x 10-6) = 5 rnW
Or we could simply note that
R = I / G = 20k!1
Vs = I R = (0.5 x 1O-')( 20k) = 10 V
and the power could be determined using P = I ' R = Vl/R = 1',1.
Given the network in Fig. 2.4d, we wish to find Rand Vs.
Using the power re lationship, we find that
R = P/ I ' = (80 x 10-3)/(4 x 10-3) ' = 5 k!1
The voltage can now be derived using Ohm's law as
V, = IR = (4 x IO- J)(5k) = 20V
The voltage could al so be obtained from the remaining power relationships in Eqs. (2.2) and (2.3).
EXAMPLE 2.2
28 CHAPTER 2 RESISTI V E CIRCUITS
Before leaving this initial discussion of circuits containing sources and a single resistor, it is important to note a phenomenon that we will find to be true in circuits containing many sources and resistors. The presence of a voltage source between a pair of tenninals tells us precisely what the voltage is between the two terminals regardless of what is happening in the balance of lhe network. What we do not know is the current in the voltage source. We must apply circuit analy· sis to the entire network to detennine this current. Likewise, the presence of a current source con nected between two tenninals specifies the exact value of the current through the source between the terminals. What we do not know is the value of the voltage across the current source. This value must be calculated by applying circuit analysis to the entire network. Furthennore, it is worth emphasizing that when applying Ohm's law, the relationship V = I R specifies a rela
tionship between the voltage directly across a resistor R and the current that is present in this resistor. Ohm's law does not apply when the vo ltage is present in one part of the network and the current exists in another. This is a common mistake made by students who try to apply V = I R to a resistor R in the middle of the network while using a Vat some other location in the network.
Learning ASS ESS MEN IS
E2.1 Given the circuits in Fig. E2. I. find (a) the current I and the power absorbed by the res is tor in Fig. E2. la, and (b) the voltage across the current source and the power supplied by the source in Fig. E2.J b.
ANSWER: Ca) I = 0.3 rnA, P = 3.6 rnW; Cb) " , = 3.6 Y, P = 2.16 mW.
I
Figure E2.1 (a) (b)
E2.2 Given the circui ts in Fig. E2.2, find Ca) R and Vs in the circuit in Fig. E2.2a, and Cb) find I and R in the circuit in Fig. E2.2b.
ANSWER: Ca) R = 10 kJ1, Vs = 4 Y;
0.4 mA
Figure E2.2
Cb) I = 20.8 rnA, R = 576 it
The previous circuits that we have considered have all contained a single res istor and were analyzed using Ohm's law. At this point we begin to expand our capabilities to handle more complicated networks that result from an interconnection of two or more of these simple elements. We will assume that the interconnect ion is performed by electrical conductors (wires) that have zero resistance-that is , perfect conductors. Because the wires have zero resistance, the energy in the circuit is in essence lumped in each e lement, and we employ the term llimped-parameter circllit to describe the network.
To aid us in our d iscussion, we will define a number of terms that will be employed throughout our analysis. As will be our approach throughout this text, we will use examples to illustrate the concepts and define the appropriate terms. For example . the c ircuit shown
S E C TI O N 2 . 2 K I RCHHO FF ' S L AW S 2 9
(a)
i2(1)
® (b)
Rs
is(I)
in Fig. 2.5a will be used 10 describe Ihe terms 1I0de, loop, and brallch. A node is simply a point of connection of two or more circuit elements. The reader is cautioned to note that, although one node can be spread out wi th perfect conductors. it is still only one node. This is illustrated in Fig. 2.5b where the circuit has been redrawn. Node 5 consists of the entire bottom connector of the circu it.
If we start at some point in the circuit and move along perfect conductors in any direction until we encounter a circuit element. the total path we cover represents a single node. Therefore, we can assume that a node is one end of a circuit element together with all the per fect conductors that are attached to it. Examining the circuit, we note that there are numerous paths through it. A loop is simply any closed parh through Ihe circuil in which no node is encountered more than once. For example, starting from node 1, one loop would contain the elements Rio "-'2. R4 , and i1; another loop would contain R2 , Vio 1h, R4 , and i
J ; and so on.
However, the path R1 , VI . Rs. 1h, R), and i l is not a loop because we have encountered node 3 twice. Finally. a brallch is a portion of a circuit containing only a single element and the nodes at each end of the element. The circuit in Fig. 2.5 contains eighl branches.
Given the previous definitions, we are now in a position to consider Kirchhoff's laws, named after German scientist Gustav Robert Kirchhoff. These two laws are quite simple but extremely important. We will not allempl 10 prove them because the proofs are beyond our current level of understanding. However, we will demonstrate their usefulness and attempt to make the reader proficient in thei r use. The fi rst law is KirchllOjJ's Cll rrelll law (KCL), which states that the algebraic Slim of the currellfs ellferitlg allY node is zero . In mathematical form Ihe law appears as
N
Lij(l ) = 0 2.7 j - I
where i .( I) is the j th current entering the node through branch j and N is the number of branche's connected to the node. To understand the use of this law, consider node 3 shown in Fig. 2.5. Applying Ki rchhoff's currenl law to this node yields
i,(I ) - i.(I) + is(I) - i, (I) = 0
We have assumed that the algebraic signs of the currents entering the node are positive and, therefore, that the signs of the currents leaving the node are negative.
If we multiply the foregoing equation by - I, we obtain the expression
- i,(I) + i.( I) - i5(1) + i,(I) = 0
which simply states that the algebraic Slim of the currents leaving a node is zero. Alternatively, we can write the equation as
~ ... Figure 2.5
Cileuit used to iliustrate KCL.
•
•
EXAMPLE 2.5
•
which siaies that the SIIIll of file Cllrrems entering a node is equal to the sum of the currefl/s leaving the node. Both of these italicized expressions are alternative fOfms of Kirchhoff's current law.
Once again it must be emphasized that the latter statement means that the sum of the variables that have been defined entering the node is equal to the sum of the variables that have been detincd leaving the node, not the actual currents. For example, i) 1) may be defined enter ing the node, but if its actual value is negative. there wi ll be positive charge leaving the node.
Note carefully that Kirchhoff's current law states that the algebraic sum of the currents either entering or leaving a node must be zero. We now begin to see why we stated in Chapter I that it is crit ically important to specify both the magnitude and the direc tion of a cur rent. Recall that c llrrent is charge in motion. Based on our background in physics. charges cannot be stored at a node. In other words, if we have a number of charges entering a node. then an equal number must be leaving that same node. Kirchhoff's current law is based on this principle of conservation of charge .
Let us write KCL for every node in the network in Fig. 2.5 assuming that the currents leaving the node are positive .
SOLUTION The KCL equations for nodes I through 5 are
- i,{I) + i,{ I) + i,{I) = 0
i,{I) - i,{ I) + i.{ I) = 0
-i,{I) + i,{ I) - i,{ I) + i,{ I) = 0
-i,{I) + i,{I) - i8( 1) = 0
-i6(1) - i,{ I) + i,( I) = 0
Note carefully that if we add the first four equations, we obtain the fifth equation. What does this tell us? Recall that this means that this set of equations is not linearly independent. We can show that the first four equations are, however, linearly independent. Store this idea in memory because it will become very important when we learn how to write the equations necessary to solve for all the currents and voltages in a network in the following chapter .
EXAMPLE 2.6 The network in Fig. 2.5 is represented by the topological diagram shown in Fig. 2.6. We wish to find the unknown currents in the network.
R~m2.6~ ID Topological diagram for
the circuit in Fig. 2.5. II
16
14
60mA
@ 15
40mA
S ECTION 2.2 KIRCHHO FF ' S LAWS
Assuming the currents leaving the node are positive, the KCL equations for nodes I th rough SOLUTION 4 are
- I , + 0.06 + 0.02 = 0
I , - I, + I. = 0
- 0.02 + I, - 0.03 = 0
The first equation yields f, and the last equat ion yie lds f, . Knowing f, we can immediately obtain f, from the third equation. Then the values of I, and f, yield the value of I. from the second equation. The results are I , = 80 rnA, f, = 70 rnA, I, = 50 mA, and 16 = - 10 mA.
•
31
Let us write the KCL equations for the circuit shown in Fig. 2.7. EXAMPLE 2.7
:!: v I (I)
The KCL equations for nodes I through 4 fo llow.
i ,(I ) + i,( , ) - i, (,) = 0
-i,(I) + i, (,) - 50i, (, ) = 0
- i,(,) + 50i, (,) + i,( , ) = 0
i,(, ) - i, (, ) - i, (, ) = 0
If we added the first three equations, we wou ld obtain the negative of the fourth. What does this tell us about the set of equations?
Finally, it is possible to generalize Kirchhoff's current law to include a closed surface. By a closed surface we mean some set of elements completely contained within the surface that are interconnected. Si nce the current entering each elemenl within the surface is equal to that leavi ng the element (i.e .. the element stores no nel charge), it fo llows that the current enter ing an il1lerconnection of elements is equal to thil l leaving the interconnection. Therefore, Kirchhoff's current law can also be stated as fo llows: The algebraic SllIll of the Cllrrellls
ellterillg allY closed slIIiace is zero.
~ ••• Figure 2.7
• SOLUTION
•
•
32 C H APTER 2 RE SIS TI V E CI R C U I T S
EXAMPLE 2 .8 Let us find 14 and II in the network represented by the topological diagram in Fig. 2.6.
• SOLUTION This diagram is redrawn in Fig. 2.8; node I is enclosed in surface I, and nodes 3 and 4 are
enclosed in surface 2. A quick review of the previous example indicates that we derived a value for I, from the value of I,. However, I, is now completely enclosed in surface 2. If we apply KCL to surface 2, assuming the currents out of the surface are positive, we obtain
Figure 2.8 ... ~
Diagram used to demon strate KC l for a surface.
I, - 0.06 - 0.Q2 - 0.03 + 0.04 = 0
or I, = 70 mA
which we obtained without any knowledge of 15, Likewise for surface 1, what goes in must come out and, therefore, 11 = 80 rnA. The reader is encouraged to cut the network in Fig. 2.6 into two pieces in any fashion and show that KCL is always satisfied at the boundaries.
Surface 1
60mA 20mA
Surface 2
E2.3 Given the networks in Fig. E2.3, find (a) I, in Fig. E2.3a and (b) IT in Fig. E2.3b.
SOmA
(a) (b)
Figure E2.3
E2.4 Find (a) I , in Ihe nelwork in Fig. E2.4a and (b) I , and I, in the circuit in Fig. E2.4b.
4mA 3mA 12 4mA
ANSWER: (a) I, = - 50 rnA; (b) IT = 70 rnA.
ANSWER: (a) I, = 6 mA; (b) I, = 8 rnA and I , = 5 rnA.
SEC TION 2.2 KIRCHHOFF ' S LAWS 33
E2.5 Find the currelll i x in the circuits in Fig. E2.5.
44 rnA R 120 mA R2
12 rnA
(a) (b)
Figure E2.5
Kirchhoff's second law, called Kin-hilOff's voltage law (KYL), states that the algebraic SII1I1 oj the voltages amwu/ allY loop is zem. As was the case wilh Kirchhoff's current law, we will defer the proof of this law and concentrate on understanding how to apply it. Once again the reader is cautioned to remember that we are dealing only with lumped-parameter circuits. These circuits are conservative, meaning that the work required to move a unit charge around any loop is zero.
In Chapter I , we related voltage to the difference in energy leve ls with in a c ircuit and talked about the energy conversion process in a flashlight. Because o f this relationship between voltage and energy, Kirchhoff's voltage law is based on the conservation of energy.
Recall that in Kirchhoff's current law, the algebraic sign was required to keep track of whether the currents were entering or leaving a node. In Kirchhoff's voltage law, the algebraic sign is used to keep track of the voltage polarity. In other words, as we traverse the circuit , it is necessary to
sum to zero the increases and decreases in energy level. Therefore, it is imp0l1ant we keep track of whether the energy level is increasing or decreasing as we go through each element.
In applying KYL, we mllst traverse any loop in the circuit and sum to zero the increases and decreases in energy level. At thi s point, we have a decision to make. Do we want to con sider a decrease in energy level as positive or negative? We will adopt a policy of consider ing a decrease in energy level as positive and an increase in energy level as negative. As we move around a loop. we encounter the plus sign first for a decrease in energy level and a negative sign first for an increase in energy level.
Consider the circuit shown in Fig. 2.9. If VR! and VR2 are known quantities, let us find VR)'
a Rl b c -+
ANSWER: Ca) i x = 4 mA; (b) i , = 12 rnA.
EXAMPLE 2.9
~ ... Figure 2.9
Circuit used to illustrate KVL.
• Staning at point a in the network and traversing it in a clockwise direction, we obtain the SOLUTION equation
which can be written as
+VR, - 5 + VR, - 15 + VR, - 30 = 0
+ VR, + VR, + VR, = 5 + IS + 30
= 50
•
•
34 CHAPTER 2 RESIS TI V E C IRCUIT S
V V R, + VRL a + RI_ b + -- c
R, R2 R3 +
16 V
lei us demonstrate that only two of the three possible loop equations are linearly independent.
• SOLUTION Note that this nelwork has three closed paths: the left loop, right loop, and outer loop.
Figure 2.11 .J. Equiva lent forms for labeling voitage.
+
(a)
Applying our policy for writing KYL equations and travers ing the left loop staning at point
(I , we obtain VR, + VR• - 16 - 24 = 0
The corresponding equation for the right loop starting at point b is
VR, + VR, + 8 + 16 - VR• = 0
The equat ion for the outer loop start ing at point a is
VR, + VR, + VR, + 8 - 24 = 0
Note that if we add the first two equations, we obtain the third equation. Therefore, as we indicated in Example 2.5, the three equations are not linearly independent. Once again. we will address this issue in the next chapter and demonstrate that we need only the first two equations (0 solve for the voltages in the circuit.
a
Finally, we employ the convention ~'b to indicate the voltage of point a with respect 10 point b: that is. the variable for the voltage between point a and point b, with poi nt (l
+ +
SECTION 2.2 KIRCHHOFF ' S LAWS 35
Consider the network in Fig. 2.12a. Let us apply KVL to determine the voltage between two points. Specifically, in terms of the double-subscript notation, let us find v,,, and v,e'
16 V 12V 16 V 12 V a + b c a + b c
RJ + +
R4 e R3
f - 10 V+ - 6V + d f - 10 V+ - 6V + d
(a) (b)
EXAMPLE 2.11
Network used in Example 2 .11.
• The circuit is redrawn in Fig. 2. 12b. Since points a and e as well as e and c are not physi- SOLUTION cally close, the arrow notation is very useful. Our approach (Q determining the unknown voltage is to apply KVL wi th the unknown voltage in the closed path. Therefore, to determine Vae we can use the path aeJa or abcdea. The equations for the two paths in which Val' is the only unknown are
t-:I~ + 10 - 24 = 0 and
16 - 12 + 4 + 6 - V" = 0
Note that both equations yield v,,, = 14 V. Even before calculating v"" we could calculate V« using the path cdec or cefabc. However, s ince \.r;1l' is now known, we can also use the path ceabc. KVL for each of these paths is
and
4 + 6 + V,e=0 -v'e + IO - 24 + 16 - 12 = 0
-~c - ~It + 16 - 12 = a Each of these equations yields Vrr = - 10 V.
In general , the mathematical representation of Kirchhoff's vo ltage law is
N
2.8
where v .(t ) is the vo ltage across the jth branch (wi th the proper reference direction) in a loop J
containing N voltages. This express ion is analogous to Eq. (2.7 ) for Kirchhoff's current law.
Given the network in Fig. 2.13 containing a dependent source, let us write the KVL equa tions for the two closed paths abda and bcdb.
a V + RI- b c +-
RJ 20 VR, + + Vs VR, R2 R3 VR3
d
[hin tj KVL is an extremely important and useful law.
EXAMPLE 2.12
~ ••• Figure 2.13
•
•
• SOLUTION The two KVL equations are
VR, + VR, - Vs = 0
20VR, + VR, - VR, = 0
E2.6 Find \~ul and V'b in the network in Fig. E2.6. ANSWER: V. d = 26 V. V,b = IOV.
6V
Figure E2.6
E2.7 Find Vbd in the cir