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- 1. PREFACE lem is also intended to test students’ understanding of the preceding example. It will reinforce their grasp of the material before moving to the next Features section. In spite of the numerous textbooks on circuit analysis available in the market, students often find the course • In recognition of ABET’s requirement on integrat- difficult to learn. The main objective of this book is ing computer tools, the use of PSpice is encouraged to present circuit analysis in a manner that is clearer, in a student-friendly manner. Since the Windows more interesting, and easier to understand than earlier version of PSpice is becoming popular, it is used texts. This objective is achieved in the following instead of the MS-DOS version. PSpice is covered ways: early so that students can use it throughout the text. Appendix D serves as a tutorial on PSpice for • A course in circuit analysis is perhaps the first Windows. exposure students have to electrical engineering. • The operational amplifier (op amp) as a basic ele- We have included several features to help stu- ment is introduced early in the text. dents feel at home with the subject. Each chapter opens with either a historical profile of some • To ease the transition between the circuit course electrical engineering pioneers to be mentioned in and signals/systems courses, Fourier and Laplace the chapter or a career discussion on a subdisci- transforms are covered lucidly and thoroughly. pline of electrical engineering. An introduction • The last section in each chapter is devoted to appli- links the chapter with the previous chapters and cations of the concepts covered in the chapter. Each states the chapter’s objectives. The chapter ends chapter has at least one or two practical problems or with a summary of the key points and formulas. devices. This helps students apply the concepts to • All principles are presented in a lucid, logical, real-life situations. step-by-step manner. We try to avoid wordiness • Ten multiple-choice review questions are provided and superfluous detail that could hide concepts at the end of each chapter, with answers. These are and impede understanding the material. intended to cover the little “tricks” that the exam- • Important formulas are boxed as a means of ples and end-of-chapter problems may not cover. helping students sort what is essential from what They serve as a self-test device and help students is not; and to ensure that students clearly get the determine how well they have mastered the chapter. gist of the matter, key terms are defined and highlighted. Organization • Marginal notes are used as a pedagogical aid. They This book was written for a two-semester or three-semes- serve multiple uses—hints, cross-references, more ter course in linear circuit analysis. The book may exposition, warnings, reminders, common mis- also be used for a one-semester course by a proper selec- takes, and problem-solving insights. tion of chapters and sections. It is broadly divided into three parts. • Thoroughly worked examples are liberally given at the end of every section. The examples are regard- • Part 1, consisting of Chapters 1 to 8, is devoted to ed as part of the text and are explained clearly, with- dc circuits. It covers the fundamental laws and the- out asking the reader to fill in missing steps. orems, circuit techniques, passive and active ele- Thoroughly worked examples give students a good ments. understanding of the solution and the confidence to • Part 2, consisting of Chapters 9 to 14, deals with ac solve problems themselves. Some of the problems circuits. It introduces phasors, sinusoidal steady- are solved in two or three ways to facilitate an state analysis, ac power, rms values, three-phase understanding and comparison of different systems, and frequency response. approaches. • Part 3, consisting of Chapters 15 to 18, is devoted • To give students practice opportunity, each illus- to advanced techniques for network analysis. trative example is immediately followed by a It provides a solid introduction to the Laplace practice problem with the answer. The students can transform, Fourier series, the Fourier transform, follow the example step-by-step to solve the prac- and two-port network analysis. tice problem without flipping pages or searching The material in three parts is more than suffi- the end of the book for answers. The practice prob- cient for a two-semester course, so that the instructor vv v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 2. vi PREFACEmust select which chapters/sections to cover. Sections will provide access to:marked with the dagger sign (†) may be skipped, 300 test questions—for instructors onlyexplained briefly, or assigned as homework. They can Downloadable figures for overheadbe omitted without loss of continuity. Each chapter has presentations—for instructors onlyplenty of problems, grouped according to the sections Solutions manual—for instructors onlyof the related material, and so diverse that the instruc- Web links to useful sitestor can choose some as examples and assign some as Sample pages from the Problem-Solvinghomework. More difficult problems are marked with a Workbookstar (*). Comprehensive problems appear last; they are PageOut Lite—a service provided to adoptersmostly applications problems that require multiple who want to create their own Web site. Inskills from that particular chapter. just a few minutes, instructors can change The book is as self-contained as possible. At the the course syllabus into a Web site usingend of the book are some appendixes that review PageOut Lite.solutions of linear equations, complex numbers, math- The URL for the web site is www.mhhe.com.alexander.ematical formulas, a tutorial on PSpice for Windows, Although the textbook is meant to be self-explanatoryand answers to odd-numbered problems. Answers to and act as a tutor for the student, the personal contactall the problems are in the solutions manual, which is involved in teaching is not to be forgotten. The bookavailable from the publisher. and supplements are intended to supply the instructor with all the pedagogical tools necessary to effectivelyPrerequisites present the material.As with most introductory circuit courses, the mainprerequisites are physics and calculus. Although famil-iarity with complex numbers is helpful in the later part ACKNOWLEDGMENTSof the book, it is not required. We wish to take the opportunity to thank the staff ofSupplements McGraw-Hill for their commitment and hardSolutions Manual—an Instructor’s Solutions Manual is work: Lynn Cox, Senior Editor; Scott Isenberg,available to instructors who adopt the text. It contains Senior Sponsoring Editor; Kelley Butcher, Seniorcomplete solutions to all the end-of-chapter problems. Developmental Editor; Betsy Jones, ExecutiveTransparency Masters—over 200 important figures Editor; Catherine Fields, Sponsoring Editor;are available as transparency masters for use as over- Kimberly Hooker, Project Manager; and Michelleheads. Flomenhoft, Editorial Assistant. They got numerousStudent CD-ROM—100 circuit files from the book are reviews, kept the book on track, and helped in manypresented as Electronics Workbench (EWB) files; 15–20 ways. We really appreciate their inputs. We areof these files are accessible using the free demo of Elec- greatly in debt to Richard Mickey for taking the paintronics Workbench. The students are able to experiment ofchecking and correcting the entire manuscript. Wewith the files. For those who wish to fully unlock all 100 wish to record our thanks to Steven Durbin at Floridacircuit files, EWB’s full version may be purchased from State University and Daniel Moore at Rose HulmanInteractive Image Technologies for approximately Institute of Technology for serving as accuracy$79.00. The CD-ROM also contains a selection of prob- checkers of examples, practice problems, and end-lem-solving, analysis and design tutorials, designed to of-chapter problems. We also wish to thank the fol-further support important concepts in the text. lowing reviewers for their constructive criticismsProblem-Solving Workbook—a paperback work- and helpful comments.book is for sale to students who wish to practice theirproblem solving techniques. The workbook contains a Promod Vohra, Northern Illinois Universitydiscussion of problem solving strategies and 150 addi- Moe Wasserman, Boston Universitytional problems with complete solutions provided.Online Learning Center (OLC)—the Web site for Robert J. Krueger, University of Wisconsinthe book will serve as an online learning center for stu- Milwaukeedents as a useful resource for instructors. The OLC John O’Malley, University of Florida v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 3. PREFACE vii Aniruddha Datta, Texas A&M University Jack C. Lee, University of Texas at Austin John Bay, Virginia Tech E. L. Gerber, Drexel University Wilhelm Eggimann, Worcester Polytechnic The first author wishes to express his apprecia- Institute tion to his department chair, Dr. Dennis Irwin, for his A. B. Bonds, Vanderbilt University outstanding support. In addition, he is extremely grate- Tommy Williamson, University of Dayton ful to Suzanne Vazzano for her help with the solutions Cynthia Finelli, Kettering University manual. The second author is indebted to Dr. Cynthia John A. Fleming, Texas A&M University Hirtzel, the former dean of the college of engineering Roger Conant, University of Illinois at Temple University, and Drs.. Brian Butz, Richard at Chicago Klafter, and John Helferty, his departmental chairper- Daniel J. Moore, Rose-Hulman Institute of sons at different periods, for their encouragement while Technology working on the manuscript. The secretarial support Ralph A. Kinney, Louisiana State University provided by Michelle Ayers and Carol Dahlberg is gratefully appreciated. Special thanks are due to Ann Cecilia Townsend, North Carolina State Sadiku, Mario Valenti, Raymond Garcia, Leke and University Tolu Efuwape, and Ope Ola for helping in various Charles B. Smith, University of Mississippi ways. Finally, we owe the greatest debt to our wives, H. Roland Zapp, Michigan State University Paulette and Chris, without whose constant support and Stephen M. Phillips, Case Western University cooperation this project would have been impossible. Please address comments and corrections to the Robin N. Strickland, University of Arizona publisher. David N. Cowling, Louisiana Tech University C. K. Alexander and M. N. O. Sadiku Jean-Pierre R. Bayard, California State Universityv v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 4. A NOTE TO THE STUDENT versities. PSpice for Windows is described in Appendix D. Make an effort to learn PSpice,This may be your first course in electrical engineer- because you can check any circuit problem withing. Although electrical engineering is an exciting and PSpice and be sure you are handing in a correctchallenging discipline, the course may intimidate you. problem solution.This book was written to prevent that. A good textbook • Each chapter ends with a section on how theand a good professor are an advantage—but you are material covered in the chapter can be applied tothe one who does the learning. If you keep the follow- real-life situations. The concepts in this sectioning ideas in mind, you will do very well in this course. may be new and advanced to you. No doubt, you • This course is the foundation on which most will learn more of the details in other courses. other courses in the electrical engineering cur- We are mainly interested in gaining a general riculum rest. For this reason, put in as much familiarity with these ideas. effort as you can. Study the course regularly. • Attempt the review questions at the end of each • Problem solving is an essential part of the learn- chapter. They will help you discover some ing process. Solve as many problems as you can. “tricks” not revealed in class or in the textbook. Begin by solving the practice problem following A short review on finding determinants is cov- each example, and then proceed to the end-of- ered in Appendix A, complex numbers in Appendix B, chapter problems. The best way to learn is to and mathematical formulas in Appendix C. Answers to solve a lot of problems. An asterisk in front of a odd-numbered problems are given in Appendix E. problem indicates a challenging problem. Have fun! • Spice, a computer circuit analysis program, is C.K.A. and M.N.O.S. used throughout the textbook. PSpice, the per- sonal computer version of Spice, is the popular standard circuit analysis program at most uni- ix v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 5. Contents Preface v 3.7 Nodal Versus Mesh Analysis 99 Acknowledgments vi 3.8 Circuit Analysis with PSpice 100 †3.9 Applications: DC Transistor Circuits 102 A Note to the Student ix 3.10 Summary 107 Review Questions 107 PART 1 DC CIRCUITS 1 Problems 109 Comprehensive Problems 117 Chapter 1 Basic Concepts 3 1.1 Introduction 4 Chapter 4 Circuit Theorems 119 1.2 Systems of Units 4 4.1 Introduction 120 1.3 Charge and Current 6 4.2 Linearity Property 120 1.4 Voltage 9 4.3 Superposition 122 1.5 Power and Energy 10 4.4 Source Transformation 127 1.6 Circuit Elements 13 †1.7 4.5 Thevenin’s Theorem 131 Applications 15 4.6 Norton’s Theorem 137 1.7.1 TV Picture Tube †4.7 Derivations of Thevenin’s and Norton’s 1.7.2 Electricity Bills Theorems 140 †1.8 Problem Solving 18 4.8 Maximum Power Transfer 142 1.9 Summary 21 4.9 Verifying Circuit Theorems Review Questions 22 with PSpice 144 Problems 23 †4.10 Applications 147 Comprehensive Problems 25 4.10.1 Source Modeling Chapter 2 Basic Laws 27 4.10.2 Resistance Measurement 2.1 Introduction 28 4.11 Summary 153 2.2 Ohm’s Laws 28 Review Questions 153 †2.3 Problems 154 Nodes, Branches, and Loops 33 Comprehensive Problems 162 2.4 Kirchhoff’s Laws 35 2.5 Series Resistors and Voltage Division 41 Chapter 5 Operational Amplifiers 165 2.6 Parallel Resistors and Current Division 42 †2.7 5.1 Introduction 166 Wye-Delta Transformations 50 †2.8 5.2 Operational Amplifiers 166 Applications 54 5.3 Ideal Op Amp 170 2.8.1 Lighting Systems 5.4 Inverting Amplifier 171 2.8.2 Design of DC Meters 5.5 Noninverting Amplifier 174 2.9 Summary 60 5.6 Summing Amplifier 176 Review Questions 61 5.7 Difference Amplifier 177 Problems 63 Comprehensive Problems 72 5.8 Cascaded Op Amp Circuits 181 5.9 Op Amp Circuit Analysis Chapter 3 Methods of Analysis 75 with PSpice 183 †5.10 Applications 185 3.1 Introduction 76 3.2 Nodal Analysis 76 5.10.1 Digital-to Analog Converter 5.10.2 Instrumentation Amplifiers 3.3 Nodal Analysis with Voltage Sources 82 3.4 Mesh Analysis 87 5.11 Summary 188 Review Questions 190 3.5 Mesh Analysis with Current Sources 92 Problems 191 †3.6 Nodal and Mesh Analyses by Inspection 95 Comprehensive Problems 200 xiv v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 6. xii CONTENTS Chapter 6 Capacitors and Inductors 201 8.12 Summary 340 Review Questions 340 6.1 Introduction 202 Problems 341 6.2 Capacitors 202 Comprehensive Problems 350 6.3 Series and Parallel Capacitors 208 6.4 Inductors 211 PART 2 AC CIRCUITS 351 6.5 Series and Parallel Inductors 216 †6.6 Applications 219 Chapter 9 Sinusoids and Phasors 353 6.6.1 Integrator 9.1 Introduction 354 6.6.2 Differentiator 9.2 Sinusoids 355 6.6.3 Analog Computer 9.3 Phasors 359 6.7 Summary 225 9.4 Phasor Relationships for CircuitReview Questions 226 Elements 367Problems 227 9.5 Impedance and Admittance 369Comprehensive Problems 235 9.6 Kirchhoff’s Laws in the Frequency Domain 372 Chapter 7 First-Order Circuits 237 9.7 Impedance Combinations 373 7.1 Introduction 238 †9.8 Applications 379 7.2 The Source-free RC Circuit 238 9.8.1 Phase-Shifters 7.3 The Source-free RL Circuit 243 9.8.2 AC Bridges 7.4 Singularity Functions 249 9.9 Summary 384 7.5 Step Response of an RC Circuit 257 Review Questions 385 7.6 Step Response of an RL Circuit 263 Problems 385 †7.7 Comprehensive Problems 392 First-order Op Amp Circuits 268 7.8 Transient Analysis with PSpice 273 †7.9 Applications 276 Chapter 10 Sinusoidal Steady-State Analysis 393 7.9.1 Delay Circuits 10.1 Introduction 394 7.9.2 Photoflash Unit 7.9.3 Relay Circuits 10.2 Nodal Analysis 394 7.9.4 Automobile Ignition Circuit 10.3 Mesh Analysis 397 10.4 Superposition Theorem 400 7.10 Summary 282 10.5 Source Transformation 404Review Questions 283Problems 284 10.6 Thevenin and Norton EquivalentComprehensive Problems 293 Circuits 406 10.7 Op Amp AC Circuits 411 Chapter 8 Second-Order Circuits 295 10.8 AC Analysis Using PSpice 413 †10.9 Applications 416 8.1 Introduction 296 8.2 Finding Initial and Final Values 296 10.9.1 Capacitance Multiplier 8.3 The Source-Free Series RLC Circuit 301 10.9.2 Oscillators 8.4 The Source-Free Parallel RLC Circuit 308 10.10 Summary 420 8.5 Step Response of a Series RLC Review Questions 421 Problems 422 Circuit 314 8.6 Step Response of a Parallel RLC Circuit 319 Chapter 11 AC Power Analysis 433 8.7 General Second-Order Circuits 322 11.1 Introduction 434 8.8 Second-Order Op Amp Circuits 327 11.2 Instantaneous and Average Power 434 8.9 PSpice Analysis of RLC Circuits 330 11.3 Maximum Average Power Transfer 440 †8.10 Duality 332 11.4 Effective or RMS Value 443 †8.11 Applications 336 11.5 Apparent Power and Power Factor 447 8.11.1 Automobile Ignition System 11.6 Complex Power 449 8.11.2 Smoothing Circuits †11.7 Conservation of AC Power 453 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 7. CONTENTS xiii 11.8 Power Factor Correction 457 14.4 Bode Plots 589 †11.9 Applications 459 14.5 Series Resonance 600 11.9.1 Power Measurement 14.6 Parallel Resonance 605 11.9.2 Electricity Consumption Cost 14.7 Passive Filters 608 11.10 Summary 464 14.7.1 Lowpass Filter Review Questions 465 14.7.2 Highpass Filter Problems 466 14.7.3 Bandpass Filter Comprehensive Problems 474 14.7.4 Bandstop Filter 14.8 Active Filters 613 Chapter 12 Three-Phase Circuits 477 14.8.1 First-Order Lowpass Filter 12.1 Introduction 478 14.8.2 First-Order Highpass Filter 12.2 Balanced Three-Phase Voltages 479 14.8.3 Bandpass Filter 14.8.4 Bandreject (or Notch) Filter 12.3 Balanced Wye-Wye Connection 482 †14.9 Scaling 619 12.4 Balanced Wye-Delta Connection 486 12.5 Balanced Delta-Delta Connection 488 14.9.1 Magnitude Scaling 12.6 Balanced Delta-Wye Connection 490 14.9.2 Frequency Scaling 14.9.3 Magnitude and Frequency Scaling 12.7 Power in a Balanced System 494 †12.8 Unbalanced Three-Phase Systems 500 14.10 Frequency Response Using 12.9 PSpice for Three-Phase Circuits 504 PSpice 622 †14.11 Applications 626 †12.10 Applications 508 12.10.1 Three-Phase Power Measurement 14.11.1 Radio Receiver 12.10.2 Residential Wiring 14.11.2 Touch-Tone Telephone 14.11.3 Crossover Network 12.11 Summary 516 14.12 Summary 631 Review Questions 517 Problems 518 Review Questions 633 Comprehensive Problems 525 Problems 633 Comprehensive Problems 640 Chapter 13 Magnetically Coupled Circuits 527 13.1 Introduction 528 PART 3 ADVANCED CIRCUIT ANALYSIS 643 13.2 Mutual Inductance 528 13.3 Energy in a Coupled Circuit 535 Chapter 15 The Laplace Transform 645 13.4 Linear Transformers 539 15.1 Introduction 646 13.5 Ideal Transformers 545 15.2 Definition of the Laplace 13.6 Ideal Autotransformers 552 Transform 646 †13.7 Three-Phase Transformers 556 15.3 Properties of the Laplace 13.8 PSpice Analysis of Magnetically Coupled Transform 649 Circuits 559 15.4 The Inverse Laplace Transform 659 †13.9 Applications 563 15.4.1 Simple Poles 13.9.1 Transformer as an Isolation Device 15.4.2 Repeated Poles 13.9.2 Transformer as a Matching Device 15.4.3 Complex Poles 13.9.3 Power Distribution 15.5 Applicaton to Circuits 666 13.10 Summary 569 15.6 Transfer Functions 672 Review Questions 570 15.7 The Convolution Integral 677 Problems 571 †15.8 Comprehensive Problems 582 Application to Integrodifferential Equations 685 †15.9 Applications 687 Chapter 14 Frequency Response 583 14.1 Introduction 584 15.9.1 Network Stability 15.9.2 Network Synthesis 14.2 Transfer Function 584 †14.3 The Decibel Scale 588 15.10 Summary 694v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 8. xiv CONTENTSReview Questions 696 17.8 Summary 789Problems 696Comprehensive Problems 705 Review Questions 790 Problems 790 Comprehensive Problems 794 Chapter 16 The Fourier Series 707 16.1 Introduction 708 Chapter 18 Two-Port Networks 795 16.2 Trigonometric Fourier Series 708 18.1 Introduction 796 16.3 Symmetry Considerations 717 18.2 Impedance Parameters 796 16.3.1 Even Symmetry 18.3 Admittance Parameters 801 16.3.2 Odd Symmetry 18.4 Hybrid Parameters 804 16.3.3 Half-Wave Symmetry 18.5 Transmission Parameters 809 16.4 Circuit Applicatons 727 †18.6 Relationships between Parameters 814 16.5 Average Power and RMS Values 730 18.7 Interconnection of Networks 817 16.6 Exponential Fourier Series 734 18.8 Computing Two-Port Parameters Using 16.7 Fourier Analysis with PSpice 740 PSpice 823 †18.9 Applications 826 16.7.1 Discrete Fourier Transform 16.7.2 Fast Fourier Transform 18.9.1 Transistor Circuits †16.8 Applications 746 18.9.2 Ladder Network Synthesis 16.8.1 Spectrum Analyzers 18.10 Summary 833 16.8.2 Filters Review Questions 834 Problems 835 16.9 Summary 749 Comprehensive Problems 844Review Questions 751Problems 751 Appendix A Solution of Simultaneous Equations UsingComprehensive Problems 758 Cramer’s Rule 845 Chapter 17 Fourier Transform 759 Appendix B Complex Numbers 851 17.1 Introduction 760 Appendix C Mathematical Formulas 859 17.2 Definition of the Fourier Transform 760 Appendix D PSpice for Windows 865 17.3 Properties of the Fourier Transform 766 17.4 Circuit Applications 779 Appendix E Answers to Odd-Numbered Problems 893 17.5 Parseval’s Theorem 782 Selected Bibliography 929 17.6 Comparing the Fourier and Laplace Index 933 Transforms 784 †17.7 Applications 785 17.7.1 Amplitude Modulation 17.7.2 Sampling v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 9. P A R T 1 DC CIRCUITS Chapter 1 Basic Concepts Chapter 2 Basic Laws Chapter 3 Methods of Analysis Chapter 4 Circuit Theorems Chapter 5 Operational Amplifier Chapter 6 Capacitors and Inductors Chapter 7 First-Order Circuits Chapter 8 Second-Order Circuits 1v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 10. C H A P T E R 1 BASIC CONCEPTS It is engineering that changes the world. —Isaac Asimov Historical Proﬁles Alessandro Antonio Volta (1745–1827), an Italian physicist, invented the electric battery—which provided the ﬁrst continuous ﬂow of electricity—and the capacitor. Born into a noble family in Como, Italy, Volta was performing electrical experiments at age 18. His invention of the battery in 1796 revolutionized the use of electricity. The publication of his work in 1800 marked the beginning of electric circuit theory. Volta received many honors during his lifetime. The unit of voltage or potential difference, the volt, was named in his honor. Andre-Marie Ampere (1775–1836), a French mathematician and physicist, laid the foundation of electrodynamics. He deﬁned the electric current and developed a way to measure it in the 1820s. Born in Lyons, France, Ampere at age 12 mastered Latin in a few weeks, as he was intensely interested in mathematics and many of the best mathematical works were in Latin. He was a brilliant scientist and a proliﬁc writer. He formulated the laws of electromagnetics. He invented the electromagnet and the ammeter. The unit of electric current, the ampere, was named after him. 3v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 11. 4 PART 1 DC Circuits 1.1 INTRODUCTION Electric circuit theory and electromagnetic theory are the two fundamen- tal theories upon which all branches of electrical engineering are built. Many branches of electrical engineering, such as power, electric ma- chines, control, electronics, communications, and instrumentation, are based on electric circuit theory. Therefore, the basic electric circuit the- ory course is the most important course for an electrical engineering student, and always an excellent starting point for a beginning student in electrical engineering education. Circuit theory is also valuable to students specializing in other branches of the physical sciences because circuits are a good model for the study of energy systems in general, and because of the applied mathematics, physics, and topology involved. In electrical engineering, we are often interested in communicating or transferring energy from one point to another. To do this requires an interconnection of electrical devices. Such interconnection is referred to as an electric circuit, and each component of the circuit is known as an element. An electric circuit is an interconnection of electrical elements. Current A simple electric circuit is shown in Fig. 1.1. It consists of three basic components: a battery, a lamp, and connecting wires. Such a simple + circuit can exist by itself; it has several applications, such as a torch light, − a search light, and so forth. A complicated real circuit is displayed in Fig. 1.2, representing the Battery Lamp schematic diagram for a radio receiver. Although it seems complicated, this circuit can be analyzed using the techniques we cover in this book. Our goal in this text is to learn various analytical techniques and computer software applications for describing the behavior of a circuit like this.Figure 1.1 A simple electric circuit. Electric circuits are used in numerous electrical systems to accom- plish different tasks. Our objective in this book is not the study of various uses and applications of circuits. Rather our major concern is the anal- ysis of the circuits. By the analysis of a circuit, we mean a study of the behavior of the circuit: How does it respond to a given input? How do the interconnected elements and devices in the circuit interact? We commence our study by deﬁning some basic concepts. These concepts include charge, current, voltage, circuit elements, power, and energy. Before deﬁning these concepts, we must ﬁrst establish a system of units that we will use throughout the text. 1.2 SYSTEMS OF UNITS As electrical engineers, we deal with measurable quantities. Our mea- surement, however, must be communicated in a standard language that virtually all professionals can understand, irrespective of the country where the measurement is conducted. Such an international measure- ment language is the International System of Units (SI), adopted by the General Conference on Weights and Measures in 1960. In this system, v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 12. CHAPTER 1 Basic Concepts 5 C3 0.1 R1 47 L1 0.445 mH 8 Oscillator R2 Y1 1 7 C 10 k U1 7 MHz Antenna SBL-1 B C6 5 C1 2200 pF Mixer E R3 Q1 L2 3, 4 2, 5, 6 10 k C4 C2 2N2222A 22.7 mH 910 2200 pF (see text) R4 C5 C7 220 910 532 to R11 U1, Pin 8 47 R6 R10 C8 10 k 100 k C11 + L3 0.1 R5 + GAIN + 100 mF C16 12-V dc 1 mH 100 k 16 V 100 mF Supply U2B 16 V − + C15 1 ⁄2 TL072 U2A 0.47 C9 R9 6 C10 5 1 ⁄2 TL072 16 V + + 1.0 mF 1.0 mF + 7 15 k 3 5 Audio + 3 8 16 V 16 V 6 − C14 + − + Output R8 1 4 R12 C17 0.0022 − 2 10 R7 15 k 4 100 mF 2 U3 16 V C13 0.1 LM386N C18 1M C12 0.0033 Audio power amp 0.1 Figure 1.2 Electric circuit of a radio receiver. (Reproduced with permission from QST, August 1995, p. 23.) there are six principal units from which the units of all other physical quantities can be derived. Table 1.1 shows the six units, their symbols, and the physical quantities they represent. The SI units are used through- out this text. TABLE 1.2 The SI preﬁxes. One great advantage of the SI unit is that it uses preﬁxes based on the power of 10 to relate larger and smaller units to the basic unit. Table Multiplier Preﬁx Symbol 1.2 shows the SI preﬁxes and their symbols. For example, the following 1018 exa E are expressions of the same distance in meters (m): 1015 peta P 600,000,000 mm 600,000 m 600 km 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h TABLE 1.1 The six basic SI units. 10 deka da Quantity Basic unit Symbol 10−1 deci d 10−2 centi c Length meter m 10−3 milli m Mass kilogram kg 10−6 micro µ Time second s 10−9 nano n Electric current ampere A 10−12 pico p Thermodynamic temperature kelvin K 10−15 femto f Luminous intensity candela cd 10−18 atto av v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 13. 6 PART 1 DC Circuits 1.3 CHARGE AND CURRENT The concept of electric charge is the underlying principle for explaining all electrical phenomena. Also, the most basic quantity in an electric circuit is the electric charge. We all experience the effect of electric charge when we try to remove our wool sweater and have it stick to our body or walk across a carpet and receive a shock. Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C). We know from elementary physics that all matter is made of fundamental building blocks known as atoms and that each atom consists of electrons, protons, and neutrons. We also know that the charge e on an electron is negative and equal in magnitude to 1.602×10−19 C, while a proton carries a positive charge of the same magnitude as the electron. The presence of equal numbers of protons and electrons leaves an atom neutrally charged. The following points should be noted about electric charge: 1. The coulomb is a large unit for charges. In 1 C of charge, there are 1/(1.602 × 10−19 ) = 6.24 × 1018 electrons. Thus realistic or laboratory values of charges are on the order of pC, nC, or µC.1 2. According to experimental observations, the only charges that occur in nature are integral multiples of the electronic charge e = −1.602 × 10−19 C. 3. The law of conservation of charge states that charge can neither be created nor destroyed, only transferred. Thus the algebraic sum of the electric charges in a system does not change. We now consider the ﬂow of electric charges. A unique feature of electric charge or electricity is the fact that it is mobile; that is, it can I − − − − be transferred from one place to another, where it can be converted to another form of energy. When a conducting wire (consisting of several atoms) is connected + − to a battery (a source of electromotive force), the charges are compelled Battery to move; positive charges move in one direction while negative charges move in the opposite direction. This motion of charges creates electric Figure 1.3 Electric current due to ﬂow current. It is conventional to take the current ﬂow as the movement of of electronic charge in a conductor. positive charges, that is, opposite to the ﬂow of negative charges, as Fig. 1.3 illustrates. This convention was introduced by Benjamin FranklinA convention is a standard way of describing (1706–1790), the American scientist and inventor. Although we nowsomething so that others in the profession can know that current in metallic conductors is due to negatively chargedunderstand what we mean. We will be using IEEE electrons, we will follow the universally accepted convention that currentconventions throughout this book. is the net ﬂow of positive charges. Thus, 1 However, a large power supply capacitor can store up to 0.5 C of charge. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 14. CHAPTER 1 Basic Concepts 7 Electric current is the time rate of change of charge, measured in amperes (A). Mathematically, the relationship between current i, charge q, and time t is dq i= (1.1) dt I where current is measured in amperes (A), and 1 ampere = 1 coulomb/second The charge transferred between time t0 and t is obtained by integrating both sides of Eq. (1.1). We obtain 0 t t q= i dt (1.2) (a) t0 i The way we deﬁne current as i in Eq. (1.1) suggests that current need not be a constant-valued function. As many of the examples and problems in this chapter and subsequent chapters suggest, there can be several types of current; that is, charge can vary with time in several ways that may be represented by different kinds of mathematical functions. 0 t If the current does not change with time, but remains constant, we call it a direct current (dc). (b) A direct current (dc) is a current that remains constant with time. Figure 1.4 Two common types of current: (a) direct current (dc), (b) alternating current (ac). By convention the symbol I is used to represent such a constant current. A time-varying current is represented by the symbol i. A com- mon form of time-varying current is the sinusoidal current or alternating current (ac). An alternating current (ac) is a current that varies sinusoidally with time. Such current is used in your household, to run the air conditioner, refrig- erator, washing machine, and other electric appliances. Figure 1.4 shows direct current and alternating current; these are the two most common 5A −5 A types of current. We will consider other types later in the book. Once we deﬁne current as the movement of charge, we expect cur- rent to have an associated direction of ﬂow. As mentioned earlier, the direction of current ﬂow is conventionally taken as the direction of positive (a) (b) charge movement. Based on this convention, a current of 5 A may be represented positively or negatively as shown in Fig. 1.5. In other words, a negative current of −5 A ﬂowing in one direction as shown in Fig. Figure 1.5 Conventional current ﬂow: (a) positive current ﬂow, (b) negative current 1.5(b) is the same as a current of +5 A ﬂowing in the opposite direction. ﬂow.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 15. 8 PART 1 DC CircuitsE X A M P L E 1 . 1 How much charge is represented by 4,600 electrons? Solution: Each electron has −1.602 × 10−19 C. Hence 4,600 electrons will have −1.602 × 10−19 C/electron × 4,600 electrons = −7.369 × 10−16 CPRACTICE PROBLEM 1.1 Calculate the amount of charge represented by two million protons. Answer: +3.204 × 10−13 C.E X A M P L E 1 . 2 The total charge entering a terminal is given by q = 5t sin 4π t mC. Cal- culate the current at t = 0.5 s. Solution: dq d i= = (5t sin 4π t) mC/s = (5 sin 4π t + 20π t cos 4π t) mA dt dt At t = 0.5, i = 5 sin 2π + 10π cos 2π = 0 + 10π = 31.42 mAPRACTICE PROBLEM 1.2 If in Example 1.2, q = (10 − 10e−2t ) mC, ﬁnd the current at t = 0.5 s. Answer: 7.36 mA.E X A M P L E 1 . 3 Determine the total charge entering a terminal between t = 1 s and t = 2 s if the current passing the terminal is i = (3t 2 − t) A. Solution: 2 2 q= i dt = (3t 2 − t) dt t=1 1 2 2 t 1 = t3 − = (8 − 2) − 1 − = 5.5 C 2 1 2PRACTICE PROBLEM 1.3 The current ﬂowing through an element is 2 A, 0<t <1 i= 2t 2 A, t >1 Calculate the charge entering the element from t = 0 to t = 2 s. Answer: 6.667 C. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 16. CHAPTER 1 Basic Concepts 9 1.4 VOLTAGE As explained brieﬂy in the previous section, to move the electron in a conductor in a particular direction requires some work or energy transfer. This work is performed by an external electromotive force (emf), typically represented by the battery in Fig. 1.3. This emf is also known as voltage or potential difference. The voltage vab between two points a and b in an electric circuit is the energy (or work) needed to move a unit charge from a to b; mathematically, dw vab = (1.3) dq where w is energy in joules (J) and q is charge in coulombs (C). The voltage vab or simply v is measured in volts (V), named in honor of the Italian physicist Alessandro Antonio Volta (1745–1827), who invented the ﬁrst voltaic battery. From Eq. (1.3), it is evident that 1 volt = 1 joule/coulomb = 1 newton meter/coulomb Thus, a + Voltage (or potential difference) is the energy required to move vab a unit charge through an element, measured in volts (V). − b Figure 1.6 shows the voltage across an element (represented by a Figure 1.6 Polarity rectangular block) connected to points a and b. The plus (+) and minus of voltage vab . (−) signs are used to deﬁne reference direction or voltage polarity. The vab can be interpreted in two ways: (1) point a is at a potential of vab volts higher than point b, or (2) the potential at point a with respect to a a point b is vab . It follows logically that in general + − vab = −vba (1.4) 9V −9 V For example, in Fig. 1.7, we have two representations of the same vol- − + tage. In Fig. 1.7(a), point a is +9 V above point b; in Fig. 1.7(b), point b is b b −9 V above point a. We may say that in Fig. 1.7(a), there is a 9-V voltage (a) (b) drop from a to b or equivalently a 9-V voltage rise from b to a. In other words, a voltage drop from a to b is equivalent to a voltage rise from Figure 1.7 Two equivalent b to a. representations of the same voltage vab : (a) point a is 9 V Current and voltage are the two basic variables in electric circuits. above point b, (b) point b is The common term signal is used for an electric quantity such as a current −9 V above point a. or a voltage (or even electromagnetic wave) when it is used for conveying information. Engineers prefer to call such variables signals rather than mathematical functions of time because of their importance in commu- nications and other disciplines. Like electric current, a constant voltage Keep in mind that electric current is always is called a dc voltage and is represented by V, whereas a sinusoidally through an element and that electric voltage is al- time-varying voltage is called an ac voltage and is represented by v. A ways across the element or between two points. dc voltage is commonly produced by a battery; ac voltage is produced by an electric generator.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 17. 10 PART 1 DC Circuits 1.5 POWER AND ENERGY Although current and voltage are the two basic variables in an electric circuit, they are not sufﬁcient by themselves. For practical purposes, we need to know how much power an electric device can handle. We all know from experience that a 100-watt bulb gives more light than a 60-watt bulb. We also know that when we pay our bills to the electric utility companies, we are paying for the electric energy consumed over a certain period of time. Thus power and energy calculations are important in circuit analysis. To relate power and energy to voltage and current, we recall from physics that: Power is the time rate of expending or absorbing energy, measured in watts (W). We write this relationship as dw p= (1.5) dt where p is power in watts (W), w is energy in joules (J), and t is time in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that dw dw dq p= = · = vi (1.6) dt dq dt or p = vi (1.7) The power p in Eq. (1.7) is a time-varying quantity and is called the instantaneous power. Thus, the power absorbed or supplied by an element is the product of the voltage across the element and the current through it. If the power has a + sign, power is being delivered to or absorbed i i by the element. If, on the other hand, the power has a − sign, power is + + being supplied by the element. But how do we know when the power has a negative or a positive sign? v v Current direction and voltage polarity play a major role in deter- mining the sign of power. It is therefore important that we pay attention − − to the relationship between current i and voltage v in Fig. 1.8(a). The vol- tage polarity and current direction must conform with those shown in Fig. p = +vi p = −vi 1.8(a) in order for the power to have a positive sign. This is known as (a) (b) the passive sign convention. By the passive sign convention, current en- ters through the positive polarity of the voltage. In this case, p = +vi or Figure 1.8 Reference vi > 0 implies that the element is absorbing power. However, if p = −vi polarities for power using the passive sign conven- or vi < 0, as in Fig. 1.8(b), the element is releasing or supplying power. tion: (a) absorbing power, (b) supplying power. Passive sign convention is satisﬁed when the current enters through the positive terminal of an element and p = +vi. If the current enters through the negative terminal, p = −vi. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 18. CHAPTER 1 Basic Concepts 11 Unless otherwise stated, we will follow the passive sign convention throughout this text. For example, the element in both circuits of Fig. 1.9 When the voltage and current directions con- has an absorbing power of +12 W because a positive current enters the form to Fig. 1.8(b), we have the active sign con- vention and p = +vi. positive terminal in both cases. In Fig. 1.10, however, the element is supplying power of −12 W because a positive current enters the negative terminal. Of course, an absorbing power of +12 W is equivalent to a supplying power of −12 W. In general, Power absorbed = −Power supplied 3A 3A 3A 3A + − + − 4V 4V 4V 4V − + − + (a) (b) (a) (b) Figure 1.9 Two cases of an Figure 1.10 Two cases of element with an absorbing an element with a supplying power of 12 W: power of 12 W: (a) p = 4 × 3 = 12 W, (a) p = 4 × (−3) = −12 W, (b) p = 4 × 3 = 12 W. (b) p = 4 × (−3) = −12 W. In fact, the law of conservation of energy must be obeyed in any electric circuit. For this reason, the algebraic sum of power in a circuit, at any instant of time, must be zero: p=0 (1.8) This again conﬁrms the fact that the total power supplied to the circuit must balance the total power absorbed. From Eq. (1.6), the energy absorbed or supplied by an element from time t0 to time t is t t w= p dt = vi dt (1.9) t0 t0 Energy is the capacity to do work, measured in joules ( J). The electric power utility companies measure energy in watt-hours (Wh), where 1 Wh = 3,600 J E X A M P L E 1 . 4 An energy source forces a constant current of 2 A for 10 s to ﬂow through a lightbulb. If 2.3 kJ is given off in the form of light and heat energy, calculate the voltage drop across the bulb.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 19. 12 PART 1 DC Circuits Solution: The total charge is q = i t = 2 × 10 = 20 C The voltage drop is w 2.3 × 103 v= = = 115 V q 20PRACTICE PROBLEM 1.4 To move charge q from point a to point b requires −30 J. Find the voltage drop vab if: (a) q = 2 C, (b) q = −6 C . Answer: (a) −15 V, (b) 5 V.E X A M P L E 1 . 5 Find the power delivered to an element at t = 3 ms if the current entering its positive terminal is i = 5 cos 60π t A and the voltage is: (a) v = 3i, (b) v = 3 di/dt. Solution: (a) The voltage is v = 3i = 15 cos 60π t; hence, the power is p = vi = 75 cos2 60π t W At t = 3 ms, p = 75 cos2 (60π × 3 × 10−3 ) = 75 cos2 0.18π = 53.48 W (b) We ﬁnd the voltage and the power as di v=3 = 3(−60π )5 sin 60π t = −900π sin 60π t V dt p = vi = −4500π sin 60π t cos 60π t W At t = 3 ms, p = −4500π sin 0.18π cos 0.18π W = −14137.167 sin 32.4◦ cos 32.4◦ = −6.396 kWPRACTICE PROBLEM 1.5 Find the power delivered to the element in Example 1.5 at t = 5 ms if the current remains the same but the voltage is: (a) v = 2i V, (b) v = t 10 + 5 i dt V. 0 Answer: (a) 17.27 W, (b) 29.7 W. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 20. CHAPTER 1 Basic Concepts 13 E X A M P L E 1 . 6 How much energy does a 100-W electric bulb consume in two hours? Solution: w = pt = 100 (W) × 2 (h) × 60 (min/h) × 60 (s/min) = 720,000 J = 720 kJ This is the same as w = pt = 100 W × 2 h = 200 Wh @ Network Analysis PRACTICE PROBLEM 1.6 A stove element draws 15 A when connected to a 120-V line. How long does it take to consume 30 kJ? Answer: 16.67 s. 1.6 CIRCUIT ELEMENTS As we discussed in Section 1.1, an element is the basic building block of a circuit. An electric circuit is simply an interconnection of the elements. Circuit analysis is the process of determining voltages across (or the currents through) the elements of the circuit. There are two types of elements found in electric circuits: passive elements and active elements. An active element is capable of generating energy while a passive element is not. Examples of passive elements are resistors, capacitors, and inductors. Typical active elements include generators, batteries, and operational ampliﬁers. Our aim in this section is to gain familiarity with some important active elements. The most important active elements are voltage or current sources that generally deliver power to the circuit connected to them. There are two kinds of sources: independent and dependent sources. An ideal independent source is an active element that provides a speciﬁed voltage or current that is completely independent of other circuit variables. + In other words, an ideal independent voltage source delivers to the circuit + v − V whatever current is necessary to maintain its terminal voltage. Physical − sources such as batteries and generators may be regarded as approxima- tions to ideal voltage sources. Figure 1.11 shows the symbols for inde- pendent voltage sources. Notice that both symbols in Fig. 1.11(a) and (b) (a) (b) can be used to represent a dc voltage source, but only the symbol in Fig. 1.11(a) can be used for a time-varying voltage source. Similarly, an ideal Figure 1.11 Symbols for independent current source is an active element that provides a speciﬁed independent voltage sources: (a) used for constant or current completely independent of the voltage across the source. That is, time-varying voltage, (b) used for the current source delivers to the circuit whatever voltage is necessary to constant voltage (dc).v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 21. 14 PART 1 DC Circuits maintain the designated current. The symbol for an independent current source is displayed in Fig. 1.12, where the arrow indicates the direction i of current i. An ideal dependent (or controlled) source is an active element in which the source quantity is controlled by another voltage or current. Figure 1.12 Symbol for independent current source. Dependent sources are usually designated by diamond-shaped symbols, as shown in Fig. 1.13. Since the control of the dependent source is ac- hieved by a voltage or current of some other element in the circuit, and the source can be voltage or current, it follows that there are four possible types of dependent sources, namely: 1. A voltage-controlled voltage source (VCVS). 2. A current-controlled voltage source (CCVS). v + i − 3. A voltage-controlled current source (VCCS). 4. A current-controlled current source (CCCS). Dependent sources are useful in modeling elements such as transistors, (a) (b) operational ampliﬁers and integrated circuits. An example of a current- controlled voltage source is shown on the right-hand side of Fig. 1.14, Figure 1.13 Symbols for: where the voltage 10i of the voltage source depends on the current i (a) dependent voltage source, (b) dependent current source. through element C. Students might be surprised that the value of the dependent voltage source is 10i V (and not 10i A) because it is a voltage source. The key idea to keep in mind is that a voltage source comes with polarities (+ −) in its symbol, while a current source comes with an arrow, irrespective of what it depends on. A B It should be noted that an ideal voltage source (dependent or in- dependent) will produce any current required to ensure that the terminal i voltage is as stated, whereas an ideal current source will produce the + C + 10i necessary voltage to ensure the stated current ﬂow. Thus an ideal source5V − − could in theory supply an inﬁnite amount of energy. It should also be noted that not only do sources supply power to a circuit, they can absorb power from a circuit too. For a voltage source, we know the voltage butFigure 1.14 The source on the right-hand not the current supplied or drawn by it. By the same token, we know theside is a current-controlled voltage source. current supplied by a current source but not the voltage across it. E X A M P L E 1 . 7 I=5A p2 Calculate the power supplied or absorbed by each element in Fig. 1.15. + − Solution: 6A 12 V We apply the sign convention for power shown in Figs. 1.8 and 1.9. For +20 V + − p1 p3 8V p4 0.2 I p1 , the 5-A current is out of the positive terminal (or into the negative − terminal); hence, p1 = 20(−5) = −100 W Supplied powerFigure 1.15 For Example 1.7. For p2 and p3 , the current ﬂows into the positive terminal of the element in each case. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 22. CHAPTER 1 Basic Concepts 15 p2 = 12(5) = 60 W Absorbed power p3 = 8(6) = 48 W Absorbed power For p4 , we should note that the voltage is 8 V (positive at the top), the same as the voltage for p3 , since both the passive element and the dependent source are connected to the same terminals. (Remember that voltage is always measured across an element in a circuit.) Since the current ﬂows out of the positive terminal, p4 = 8(−0.2I ) = 8(−0.2 × 5) = −8 W Supplied power We should observe that the 20-V independent voltage source and 0.2I dependent current source are supplying power to the rest of the network, while the two passive elements are absorbing power. Also, p1 + p2 + p3 + p4 = −100 + 60 + 48 − 8 = 0 In agreement with Eq. (1.8), the total power supplied equals the total power absorbed. PRACTICE PROBLEM 1.7 Compute the power absorbed or supplied by each component of the circuit 8A 2V I=5A in Fig. 1.16. +− Answer: p1 = −40 W, p2 = 16 W, p3 = 9 W, p4 = 15 W. p2 3A + + − p1 p3 + 0.6I p4 5V − + 3V − − Figure 1.16 For Practice Prob. 1.7. † 1.7 APPLICATIONS 2 In this section, we will consider two practical applications of the concepts developed in this chapter. The ﬁrst one deals with the TV picture tube and the other with how electric utilities determine your electric bill. 1.7.1 TV Picture Tube One important application of the motion of electrons is found in both the transmission and reception of TV signals. At the transmission end, a TV camera reduces a scene from an optical image to an electrical signal. Scanning is accomplished with a thin beam of electrons in an iconoscope camera tube. At the receiving end, the image is reconstructed by using a cath- ode-ray tube (CRT) located in the TV receiver.3 The CRT is depicted in 2 The dagger sign preceding a section heading indicates a section that may be skipped, explained brieﬂy, or assigned as homework. 3 Modern TV tubes use a different technology.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 23. 16 PART 1 DC Circuits Fig. 1.17. Unlike the iconoscope tube, which produces an electron beam of constant intensity, the CRT beam varies in intensity according to the incoming signal. The electron gun, maintained at a high potential, ﬁres the electron beam. The beam passes through two sets of plates for vertical and horizontal deﬂections so that the spot on the screen where the beam strikes can move right and left and up and down. When the electron beam strikes the ﬂuorescent screen, it gives off light at that spot. Thus the beam can be made to “paint” a picture on the TV screen. Horizontal deflection Electron gun plates Bright spot on fluorescent screen Vertical Electron deflection trajectory plates Figure 1.17 Cathode-ray tube. (Source: D. E. Tilley, Contemporary College Physics [Menlo Park, CA: Benjamin/Cummings, 1979], p. 319.)E X A M P L E 1 . 8 The electron beam in a TV picture tube carries 1015 electrons per second. As a design engineer, determine the voltage Vo needed to accelerate the electron beam to achieve 4 W. Solution: The charge on an electron is e = −1.6 × 10−19 C i If the number of electrons is n, then q = ne and dq dn q i= =e = (−1.6 × 10−19 )(1015 ) = −1.6 × 10−4 A dt dt Vo The negative sign indicates that the electron ﬂows in a direction opposite to electron ﬂow as shown in Fig. 1.18, which is a simpliﬁed diagram of Figure 1.18 A simpliﬁed diagram of the the CRT for the case when the vertical deﬂection plates carry no charge. cathode-ray tube; for Example 1.8. The beam power is p 4 p = Vo i or Vo = = = 25,000 V i 1.6 × 10−4 Thus the required voltage is 25 kV.PRACTICE PROBLEM 1.8 If an electron beam in a TV picture tube carries 1013 electrons/second and is passing through plates maintained at a potential difference of 30 kV, calculate the power in the beam. Answer: 48 mW. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 24. CHAPTER 1 Basic Concepts 17 1.7.2 Electricity Bills The second application deals with how an electric utility company charges their customers. The cost of electricity depends upon the amount of energy consumed in kilowatt-hours (kWh). (Other factors that affect the cost include demand and power factors; we will ignore these for now.) However, even if a consumer uses no energy at all, there is a minimum service charge the customer must pay because it costs money to stay connected to the power line. As energy consumption increases, the cost per kWh drops. It is interesting to note the average monthly consumption of household appliances for a family of ﬁve, shown in Table 1.3. TABLE 1.3 Typical average monthly consumption of household appliances. Appliance kWh consumed Appliance kWh consumed Water heater 500 Washing machine 120 Freezer 100 Stove 100 Lighting 100 Dryer 80 Dishwasher 35 Microwave oven 25 Electric iron 15 Personal computer 12 TV 10 Radio 8 Toaster 4 Clock 2 E X A M P L E 1 . 9 A homeowner consumes 3,300 kWh in January. Determine the electricity bill for the month using the following residential rate schedule: Base monthly charge of $12.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 200 kWh per month at 6 cents/kWh. Solution: We calculate the electricity bill as follows. Base monthly charge = $12.00 First 100 kWh @ $0.16/kWh = $16.00 Next 200 kWh @ $0.10/kWh = $20.00 Remaining 100 kWh @ $0.06/kWh = $6.00 Total Charge = $54.00 $54 Average cost = = 13.5 cents/kWh 100 + 200 + 100v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 25. 18 PART 1 DC CircuitsPRACTICE PROBLEM 1.9 Referring to the residential rate schedule in Example 1.9, calculate the average cost per kWh if only 400 kWh are consumed in July when the family is on vacation most of the time. Answer: 13.5 cents/kWh. † 1.8 PROBLEM SOLVING Although the problems to be solved during one’s career will vary in complexity and magnitude, the basic principles to be followed remain the same. The process outlined here is the one developed by the authors over many years of problem solving with students, for the solution of engineering problems in industry, and for problem solving in research. We will list the steps simply and then elaborate on them. 1. Carefully Deﬁne the problem. 2. Present everything you know about the problem. 3. Establish a set of Alternative solutions and determine the one that promises the greatest likelihood of success. 4. Attempt a problem solution. 5. Evaluate the solution and check for accuracy. 6. Has the problem been solved Satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again. 1. Carefully Deﬁne the problem. This may be the most important part of the process, because it becomes the foundation for all the rest of the steps. In general, the presentation of engineering problems is somewhat incomplete. You must do all you can to make sure you understand the problem as thoroughly as the presenter of the problem understands it. Time spent at this point clearly identifying the problem will save you considerable time and frustration later. As a student, you can clarify a problem statement in a textbook by asking your professor to help you understand it better. A problem presented to you in industry may require that you consult several individuals. At this step, it is important to develop questions that need to be addressed before continuing the solution process. If you have such questions, you need to consult with the appropriate individuals or resources to obtain the answers to those questions. With those answers, you can now reﬁne the problem, and use that reﬁnement as the problem statement for the rest of the solution process. 2. Present everything you know about the problem. You are now ready to write down everything you know about the problem and its possible solutions. This important step will save you time and frustration later. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 26. CHAPTER 1 Basic Concepts 19 3. Establish a set of Alternative solutions and determine the one that promises the greatest likelihood of success. Almost every problem will have a number of possible paths that can lead to a solution. It is highly desirable to identify as many of those paths as possible. At this point, you also need to determine what tools are available to you, such as Matlab and other software packages that can greatly reduce effort and increase accuracy. Again, we want to stress that time spent carefully deﬁning the problem and investigating alternative approaches to its solution will pay big dividends later. Evaluating the alternatives and determining which promises the greatest likelihood of success may be difﬁcult but will be well worth the effort. Document this process well since you will want to come back to it if the ﬁrst approach does not work. 4. Attempt a problem solution. Now is the time to actually begin solving the problem. The process you follow must be well documented in order to present a detailed solution if successful, and to evaluate the process if you are not successful. This detailed evaluation may lead to corrections that can then lead to a successful solution. It can also lead to new alternatives to try. Many times, it is wise to fully set up a solution before putting numbers into equations. This will help in checking your results. 5. Evaluate the solution and check for accuracy. You now thor- oughly evaluate what you have accomplished. Decide if you have an acceptable solution, one that you want to present to your team, boss, or professor. 6. Has the problem been solved Satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again. Now you need to present your solution or try another alternative. At this point, presenting your solution may bring closure to the process. Often, however, presentation of a solution leads to further reﬁnement of the problem deﬁnition, and the process continues. Following this process will eventually lead to a satisfactory conclusion. Now let us look at this process for a student taking an electrical and computer engineering foundations course. (The basic process also applies to almost every engineering course.) Keep in mind that although the steps have been simpliﬁed to apply to academic types of problems, the process as stated always needs to be followed. We consider a simple example. Assume that we have been given the following circuit. The instruc- tor asks us to solve for the current ﬂowing through the 8-ohm resistor. 2Ω 4Ω + 5V − 8Ω 3V 1. Carefully Deﬁne the problem. This is only a simple example, but we can already see that we do not know the polarity on the 3-V source. We have the following options. We can ask the professor whatv v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 27. 20 PART 1 DC Circuits the polarity should be. If we cannot ask, then we need to make a decision on what to do next. If we have time to work the problem both ways, we can solve for the current when the 3-V source is plus on top and then plus on the bottom. If we do not have the time to work it both ways, assume a polarity and then carefully document your decision. Let us assume that the professor tells us that the source is plus on the bottom. 2. Present everything you know about the problem. Presenting all that we know about the problem involves labeling the circuit clearly so that we deﬁne what we seek. Given the following circuit, solve for i8 . 2Ω 4Ω i8Ω + − 5V − 8Ω + 3V We now check with the professor, if reasonable, to see if the problem is properly deﬁned. 3. Establish a set of Alternative solutions and determine the one that promises the greatest likelihood of success. There are essentially three techniques that can be used to solve this problem. Later in the text you will see that you can use circuit analysis (using Kirchoff’s laws and Ohm’s law), nodal analysis, and mesh analysis. To solve for i8 using circuit analysis will eventually lead to a solution, but it will likely take more work than either nodal or mesh analysis. To solve for i8 using mesh analysis will require writing two simultaneous equations to ﬁnd the two loop currents indicated in the following circuit. Using nodal analysis requires solving for only one unknown. This is the easiest approach. 2Ω i1 i3 4Ω v1 + v − + v − 1 i2 3 + + − 5V − + 3V Loop 1 v2 8Ω Loop 2 − Therefore, we will solve for i8 using nodal analysis. 4. Attempt a problem solution. We ﬁrst write down all of the equations we will need in order to ﬁnd i8 . v1 v1 i8 = i2 , i2 = , i8 = 8 8 v1 − 5 v1 − 0 v1 + 3 + + =0 2 8 4 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 28. CHAPTER 1 Basic Concepts 21 Now we can solve for v1 . v1 − 5 v1 − 0 v1 + 3 8 + + =0 2 8 4 leads to (4v1 − 20) + (v1 ) + (2v1 + 6) = 0 v1 2 7v1 = +14, v1 = +2 V, i8 = = = 0.25 A 8 8 5. Evaluate the solution and check for accuracy. We can now use Kirchoff’s voltage law to check the results. v1 − 5 2−5 3 i1 = = = − = −1.5 A 2 2 2 i2 = i8 = 0.25 A v1 + 3 2+3 5 i3 = = = = 1.25 A 4 4 4 i1 + i2 + i3 = −1.5 + 0.25 + 1.25 = 0 (Checks.) Applying KVL to loop 1, −5 + v1 + v2 = −5 + (−i1 × 2) + (i2 × 8) = −5 + (−(−1.5)2) + (0.25 × 8) = −5 + 3 + 2 = 0 (Checks.) Applying KVL to loop 2, −v2 + v3 − 3 = −(i2 × 8) + (i3 × 4) − 3 = −(0.25 × 8) + (1.25 × 4) − 3 = −2 + 5 − 3 = 0 (Checks.) So we now have a very high degree of conﬁdence in the accuracy of our answer. 6. Has the problem been solved Satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again. This problem has been solved satisfactorily. The current through the 8-ohm resistor is 0.25 amp ﬂowing down through the 8-ohm resistor. 1.9 SUMMARY 1. An electric circuit consists of electrical elements connected together. 2. The International System of Units (SI) is the international mea- surement language, which enables engineers to communicate their results. From the six principal units, the units of other physical quantities can be derived. 3. Current is the rate of charge ﬂow. dq i= dtv v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 29. 22 PART 1 DC Circuits 4. Voltage is the energy required to move 1 C of charge through an element. dw v= dq 5. Power is the energy supplied or absorbed per unit time. It is also the product of voltage and current. dw p== vi dt 6. According to the passive sign convention, power assumes a positive sign when the current enters the positive polarity of the voltage across an element. 7. An ideal voltage source produces a speciﬁc potential difference across its terminals regardless of what is connected to it. An ideal current source produces a speciﬁc current through its terminals regardless of what is connected to it. 8. Voltage and current sources can be dependent or independent. A dependent source is one whose value depends on some other circuit variable. 9. Two areas of application of the concepts covered in this chapter are the TV picture tube and electricity billing procedure.REVIEW QUESTIONS1.1 One millivolt is one millionth of a volt. 1.8 The voltage across a 1.1 kW toaster that produces a (a) True (b) False current of 10 A is: (a) 11 kV (b) 1100 V (c) 110 V (d) 11 V1.2 The preﬁx micro stands for: (a) 106 (b) 103 (c) 10−3 (d) 10−6 1.9 Which of these is not an electrical quantity? (a) charge (b) time (c) voltage1.3 The voltage 2,000,000 V can be expressed in powers of 10 as: (d) current (e) power (a) 2 mV (b) 2 kV (c) 2 MV (d) 2 GV 1.10 The dependent source in Fig. 1.19 is:1.4 A charge of 2 C ﬂowing past a given point each (a) voltage-controlled current source second is a current of 2 A. (b) voltage-controlled voltage source (a) True (b) False (c) current-controlled voltage source1.5 A 4-A current charging a dielectric material will (d) current-controlled current source accumulate a charge of 24 C after 6 s. io (a) True (b) False vs + 6io1.6 The unit of current is: − (a) Coulomb (b) Ampere (c) Volt (d) Joule1.7 Voltage is measured in: Figure 1.19 For Review Question 1.10. (a) Watts (b) Amperes (c) Volts (d) Joules per second Answers: 1.1b, 1.2d, 1.3c, 1.4a, 1.5a, 1.6b, 1.7c, 1.8c, 1.9b, 1.10d. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 30. CHAPTER 1 Basic Concepts 23 PROBLEMS Section 1.3 Charge and Current q (C) 1.1 How many coulombs are represented by these 50 amounts of electrons: (a) 6.482 × 1017 (b) 1.24 × 1018 0 (c) 2.46 × 1019 (d) 1.628 × 1020 2 4 6 8 t (s) 1.2 Find the current ﬂowing through an element if the −50 charge ﬂow is given by: (a) q(t) = (t + 2) mC Figure 1.21 For Prob. 1.7. (b) q(t) = (5t 2 + 4t − 3) C (c) q(t) = 10e−4t pC 1.8 The current ﬂowing past a point in a device is shown (d) q(t) = 20 cos 50π t nC in Fig. 1.22. Calculate the total charge through the (e) q(t) = 5e−2t sin 100t µC point. 1.3 Find the charge q(t) ﬂowing through a device if the current is: i (mA) (a) i(t) = 3 A, q(0) = 1 C 10 (b) i(t) = (2t + 5) mA, q(0) = 0 (c) i(t) = 20 cos(10t + π/6) µA, q(0) = 2 µC (d) i(t) = 10e−30t sin 40t A, q(0) = 0 0 1 2 t (ms) 1.4 The current ﬂowing through a device is i(t) = 5 sin 6π t A. Calculate the total charge ﬂow Figure 1.22 For Prob. 1.8. through the device from t = 0 to t = 10 ms. 1.5 Determine the total charge ﬂowing into an element 1.9 The current through an element is shown in Fig. for 0 < t < 2 s when the current entering its 1.23. Determine the total charge that passed through positive terminal is i(t) = e−2t mA. the element at: (a) t = 1 s (b) t = 3 s (c) t = 5 s 1.6 The charge entering a certain element is shown in Fig. 1.20. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms i (A) 10 5 q(t) (mC) 0 1 2 3 4 5 t (s) 80 Figure 1.23 For Prob. 1.9. Sections 1.4 and 1.5 Voltage, Power, and Energy 1.10 A certain electrical element draws the current i(t) = 10 cos 4t A at a voltage v(t) = 120 cos 4t V. 0 2 4 6 8 10 12 t (ms) Find the energy absorbed by the element in 2 s. 1.11 The voltage v across a device and the current i Figure 1.20 For Prob. 1.6. through it are v(t) = 5 cos 2t V, i(t) = 10(1 − e−0.5t ) A Calculate: 1.7 The charge ﬂowing in a wire is plotted in Fig. 1.21. (a) the total charge in the device at t = 1 s Sketch the corresponding current. (b) the power consumed by the device at t = 1 s.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 31. 24 PART 1 DC Circuits1.12 The current entering the positive terminal of a 1.16 Determine Io in the circuit of Fig. 1.27. device is i(t) = 3e−2t A and the voltage across the device is v(t) = 5 di/dt V. 12 V (a) Find the charge delivered to the device between + − t = 0 and t = 2 s. Io 3A 5A (b) Calculate the power absorbed. + + (c) Determine the energy absorbed in 3 s. + 20 V − 20 V 8V1.13 Figure 1.24 shows the current through and the − − voltage across a device. Find the total energy absorbed by the device for the period of 0 < t < 4 s. 3A i (mA) 50 Figure 1.27 For Prob. 1.16. 1.17 Find Vo in the circuit of Fig. 1.28. 0 2 4 t (s) Io = 2 A v (V) + − 10 28 V 6A 12 V 1A + − 3A + − 28 V + 0 1 3 4 t (s) + – 30 V − Vo + 5Io − Figure 1.24 For Prob. 1.13. 6A 3ASection 1.6 Circuit Elements1.14 Figure 1.25 shows a circuit with ﬁve elements. If Figure 1.28 For Prob. 1.17. p1 = −205 W, p2 = 60 W, p4 = 45 W, p5 = 30 W, calculate the power p3 received or delivered by Section 1.7 Applications element 3. 1.18 It takes eight photons to strike the surface of a 2 4 photodetector in order to emit one electron. If 4 × 1011 photons/second strike the surface of the 1 3 5 photodetector, calculate the amount of current ﬂow. 1.19 Find the power rating of the following electrical appliances in your household: (a) Lightbulb (b) Radio set Figure 1.25 For Prob. 1.14. (c) TV set (d) Refrigerator1.15 Find the power absorbed by each of the elements in (e) Personal computer (f ) PC printer Fig. 1.26. (g) Microwave oven (h) Blender I = 10 A 10 V 8V 4A 1.20 A 1.5-kW electric heater is connected to a 120-V + − + − source. p2 p4 (a) How much current does the heater draw? 14 A (b) If the heater is on for 45 minutes, how much + + p5 energy is consumed in kilowatt-hours (kWh)? + p3 (c) Calculate the cost of operating the heater for 4530 V − p1 20 V 12 V − − 0.4I minutes if energy costs 10 cents/kWh. 1.21 A 1.2-kW toaster takes roughly 4 minutes to heat four slices of bread. Find the cost of operating the toaster once per day for 1 month (30 days). Assume Figure 1.26 For Prob. 1.15. energy costs 9 cents/kWh. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 32. CHAPTER 1 Basic Concepts 25 1.22 A ﬂashlight battery has a rating of 0.8 ampere-hours otherwise dark staircase. Determine: (Ah) and a lifetime of 10 hours. (a) the current through the lamp, (a) How much current can it deliver? (b) the cost of operating the light for one non-leap (b) How much power can it give if its terminal year if electricity costs 12 cents per kWh. voltage is 6 V? 1.25 An electric stove with four burners and an oven is (c) How much energy is stored in the battery in used in preparing a meal as follows. kWh? Burner 1: 20 minutes Burner 2: 40 minutes 1.23 A constant current of 3 A for 4 hours is required to Burner 3: 15 minutes Burner 4: 45 minutes charge an automotive battery. If the terminal voltage Oven: 30 minutes is 10 + t/2 V, where t is in hours, If each burner is rated at 1.2 kW and the oven at (a) how much charge is transported as a result of the 1.8 kW, and electricity costs 12 cents per kWh, charging? calculate the cost of electricity used in preparing the (b) how much energy is expended? meal. (c) how much does the charging cost? Assume 1.26 PECO (the electric power company in Philadelphia) electricity costs 9 cents/kWh. charged a consumer $34.24 one month for using 1.24 A 30-W incandescent lamp is connected to a 120-V 215 kWh. If the basic service charge is $5.10, how source and is left burning continuously in an much did PECO charge per kWh? COMPREHENSIVE PROBLEMS 1.27 A telephone wire has a current of 20 µA ﬂowing Calculate the total energy in MWh consumed by the through it. How long does it take for a charge of plant. 15 C to pass through the wire? 1.28 A lightning bolt carried a current of 2 kA and lasted p (MW) for 3 ms. How many coulombs of charge were 8 contained in the lightning bolt? 5 1.29 The power consumption for a certain household for 4 a day is shown in Fig. 1.29. Determine: 3 (a) the total energy consumed in kWh (b) the average power per hour. 8.00 8.05 8.10 8.15 8.20 8.25 8.30 t p(t) Figure 1.30 For Prob. 1.30. 1200 W 1000 W 1.31 A battery may be rated in ampere-hours (Ah). An lead-acid battery is rated at 160 Ah. (a) What is the maximum current it can supply for 40 h? 400 W 400 W (b) How many days will it last if it is discharged at 200 W 1 mA? 1.32 How much work is done by a 12-V automobile 12 2 4 6 8 10 12 2 4 6 8 10 12 t battery in moving 5 × 1020 electrons from the noon (hour) positive terminal to the negative terminal? Figure 1.29 For Prob. 1.29. 1.33 How much energy does a 10-hp motor deliver in 30 minutes? Assume that 1 horsepower = 746 W. 1.30 The graph in Fig. 1.30 represents the power drawn 1.34 A 2-kW electric iron is connected to a 120-V line. by an industrial plant between 8:00 and 8:30 A.M. Calculate the current drawn by the iron. | | | Go to the Student OLC |v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 33. C H A P T E R 2 BASIC LAWS The chessboard is the world, the pieces are the phenomena of the universe, the rules of the game are what we call the laws of Nature. The player on the other side is hidden from us, we know that his play is always fair, just, and patient. But also we know, to our cost, that he never overlooks a mistake, or makes the smallest allowance for ignorance. — Thomas Henry Huxley Historical Proﬁles Georg Simon Ohm (1787–1854), a German physicist, in 1826 experimentally deter- mined the most basic law relating voltage and current for a resistor. Ohm’s work was initially denied by critics. Born of humble beginnings in Erlangen, Bavaria, Ohm threw himself into electrical research. His efforts resulted in his famous law. He was awarded the Copley Medal in 1841 by the Royal Society of London. In 1849, he was given the Professor of Physics chair by the University of Munich. To honor him, the unit of resistance was named the ohm. Gustav Robert Kirchhoff (1824–1887), a German physicist, stated two basic laws in 1847 concerning the relationship between the currents and voltages in an electrical network. Kirchhoff’s laws, along with Ohm’s law, form the basis of circuit theory. Born the son of a lawyer in Konigsberg, East Prussia, Kirchhoff entered the University of Konigsberg at age 18 and later became a lecturer in Berlin. His collaborative work in spectroscopy with German chemist Robert Bunsen led to the discovery of cesium in 1860 and rubidium in 1861. Kirchhoff was also credited with the Kirchhoff law of radiation. Thus Kirchhoff is famous among engineers, chemists, and physicists. 27v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 34. 28 PART 1 DC Circuits 2.1 INTRODUCTION Chapter 1 introduced basic concepts such as current, voltage, and power in an electric circuit. To actually determine the values of these variables in a given circuit requires that we understand some fundamental laws that govern electric circuits. These laws, known as Ohm’s law and Kirchhoff’s laws, form the foundation upon which electric circuit analysis is built. In this chapter, in addition to these laws, we shall discuss some techniques commonly applied in circuit design and analysis. These tech- niques include combining resistors in series or parallel, voltage division, current division, and delta-to-wye and wye-to-delta transformations. The application of these laws and techniques will be restricted to resistive cir- cuits in this chapter. We will ﬁnally apply the laws and techniques to real-life problems of electrical lighting and the design of dc meters. 2.2 OHM’S LAW l Materials in general have a characteristic behavior of resisting the ﬂow i of electric charge. This physical property, or ability to resist current, is known as resistance and is represented by the symbol R. The resistance + of any material with a uniform cross-sectional area A depends on A and v R − its length , as shown in Fig. 2.1(a). In mathematical form, Material with resistivity r R=ρ (2.1) Cross-sectional A area A where ρ is known as the resistivity of the material in ohm-meters. Good (a) (b) conductors, such as copper and aluminum, have low resistivities, while Figure 2.1 (a) Resistor, (b) Circuit symbol insulators, such as mica and paper, have high resistivities. Table 2.1 for resistance. presents the values of ρ for some common materials and shows which materials are used for conductors, insulators, and semiconductors. TABLE 2.1 Resistivities of common materials. Material Resistivity ( ·m) Usage Silver 1.64 × 10−8 Conductor Copper 1.72 × 10−8 Conductor Aluminum 2.8 × 10−8 Conductor Gold 2.45 × 10−8 Conductor Carbon 4 × 10−5 Semiconductor Germanium 47 × 10−2 Semiconductor Silicon 6.4 × 102 Semiconductor Paper 1010 Insulator Mica 5 × 1011 Insulator Glass 1012 Insulator Teﬂon 3 × 1012 Insulator The circuit element used to model the current-resisting behavior of a material is the resistor. For the purpose of constructing circuits, resistors are usually made from metallic alloys and carbon compounds. The circuit v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 35. CHAPTER 2 Basic Laws 29 symbol for the resistor is shown in Fig. 2.1(b), where R stands for the resistance of the resistor. The resistor is the simplest passive element. Georg Simon Ohm (1787–1854), a German physicist, is credited with ﬁnding the relationship between current and voltage for a resistor. This relationship is known as Ohm’s law. Ohm’s law states that the voltage v across a resistor is directly proportional to the current i ﬂowing through the resistor. That is, v∝i (2.2) Ohm deﬁned the constant of proportionality for a resistor to be the resis- tance, R. (The resistance is a material property which can change if the internal or external conditions of the element are altered, e.g., if there are changes in the temperature.) Thus, Eq. (2.2) becomes v = iR (2.3) which is the mathematical form of Ohm’s law. R in Eq. (2.3) is measured in the unit of ohms, designated . Thus, The resistance R of an element denotes its ability to resist the ﬂow of electric current; it is measured in ohms ( ). We may deduce from Eq. (2.3) that v + i R= (2.4) i v=0 R=0 so that 1 = 1 V/A − To apply Ohm’s law as stated in Eq. (2.3), we must pay careful attention to the current direction and voltage polarity. The direction of (a) current i and the polarity of voltage v must conform with the passive sign convention, as shown in Fig. 2.1(b). This implies that current ﬂows from a higher potential to a lower potential in order for v = iR. If current + i=0 ﬂows from a lower potential to a higher potential, v = −iR. Since the value of R can range from zero to inﬁnity, it is important v R=∞ that we consider the two extreme possible values of R. An element with R = 0 is called a short circuit, as shown in Fig. 2.2(a). For a short circuit, − v = iR = 0 (2.5) (b) showing that the voltage is zero but the current could be anything. In practice, a short circuit is usually a connecting wire assumed to be a Figure 2.2 (a) Short circuit (R = 0), perfect conductor. Thus, (b) Open circuit (R = ∞).v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 36. 30 PART 1 DC Circuits A short circuit is a circuit element with resistance approaching zero. Similarly, an element with R = ∞ is known as an open circuit, as shown in Fig. 2.2(b). For an open circuit, v (a) i = lim =0 (2.6) R→∞ R indicating that the current is zero though the voltage could be anything. Thus, An open circuit is a circuit element with resistance approaching inﬁnity. (b) A resistor is either ﬁxed or variable. Most resistors are of the ﬁxed type, meaning their resistance remains constant. The two common types Figure 2.3 Fixed resistors: (a) wire- of ﬁxed resistors (wirewound and composition) are shown in Fig. 2.3. wound type, (b) carbon ﬁlm type. The composition resistors are used when large resistance is needed. The (Courtesy of Tech America.) circuit symbol in Fig. 2.1(b) is for a ﬁxed resistor. Variable resistors have adjustable resistance. The symbol for a variable resistor is shown in Fig. 2.4(a). A common variable resistor is known as a potentiometer or pot for short, with the symbol shown in Fig. 2.4(b). The pot is a three-terminal element with a sliding contact or wiper. By sliding the wiper, the resistances between the wiper terminal and the ﬁxed terminals vary. Like ﬁxed resistors, variable resistors can either be of wirewound or composition type, as shown in Fig. 2.5. Although resistors like those in (a) (b) Figs. 2.3 and 2.5 are used in circuit designs, today most circuit components including resistors are either surface mounted or integrated, as typicallyFigure 2.4 Circuit symbol for: (a) a variable shown in Fig. 2.6.resistor in general, (b) a potentiometer. (a) (b) Figure 2.5 Variable resistors: (a) composition type, (b) slider pot. (Courtesy of Tech America.) It should be pointed out that not all resistors obey Ohm’s law. A resistor that obeys Ohm’s law is known as a linear resistor. It has a con- stant resistance and thus its current-voltage characteristic is as illustratedFigure 2.6 Resistors in a thick-ﬁlm circuit. in Fig. 2.7(a): its i-v graph is a straight line passing through the ori-(Source: G. Daryanani, Principles of ActiveNetwork Synthesis and Design [New York: gin. A nonlinear resistor does not obey Ohm’s law. Its resistance variesJohn Wiley, 1976], p. 461c.) with current and its i-v characteristic is typically shown in Fig. 2.7(b). v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 37. CHAPTER 2 Basic Laws 31 Examples of devices with nonlinear resistance are the lightbulb and the v diode. Although all practical resistors may exhibit nonlinear behavior under certain conditions, we will assume in this book that all elements actually designated as resistors are linear. A useful quantity in circuit analysis is the reciprocal of resistance Slope = R R, known as conductance and denoted by G: i 1 i (a) G= = (2.7) R v v The conductance is a measure of how well an element will conduct electric current. The unit of conductance is the mho (ohm spelled back- ward) or reciprocal ohm, with symbol , the inverted omega. Although engineers often use the mhos, in this book we prefer to use the siemens Slope = R (S), the SI unit of conductance: i 1S=1 = 1 A/V (2.8) (b) Thus, Figure 2.7 The i-v characteristic of: (a) a linear resistor, (b) a nonlinear resistor. Conductance is the ability of an element to conduct electric current; it is measured in mhos ( ) or siemens (S). The same resistance can be expressed in ohms or siemens. For example, 10 is the same as 0.1 S. From Eq. (2.7), we may write i = Gv (2.9) The power dissipated by a resistor can be expressed in terms of R. Using Eqs. (1.7) and (2.3), v2 p = vi = i 2 R = (2.10) R The power dissipated by a resistor may also be expressed in terms of G as i2 p = vi = v 2 G = (2.11) G We should note two things from Eqs. (2.10) and (2.11): 1. The power dissipated in a resistor is a nonlinear function of either current or voltage. 2. Since R and G are positive quantities, the power dissipated in a resistor is always positive. Thus, a resistor always absorbs power from the circuit. This conﬁrms the idea that a resistor is a passive element, incapable of generating energy. E X A M P L E 2 . 1 An electric iron draws 2 A at 120 V. Find its resistance.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 38. 32 PART 1 DC Circuits Solution: From Ohm’s law, v 120 R= = = 60 i 2PRACTICE PROBLEM 2.1 The essential component of a toaster is an electrical element (a resistor) that converts electrical energy to heat energy. How much current is drawn by a toaster with resistance 12 at 110 V? Answer: 9.167 A.E X A M P L E 2 . 2 i In the circuit shown in Fig. 2.8, calculate the current i, the conductance + G, and the power p. 30 V + − 5 kΩ v Solution: − The voltage across the resistor is the same as the source voltage (30 V) because the resistor and the voltage source are connected to the same pair of terminals. Hence, the current is Figure 2.8 For Example 2.2. v 30 i= = = 6 mA R 5 × 103 The conductance is 1 1 G= = = 0.2 mS R 5 × 103 We can calculate the power in various ways using either Eqs. (1.7), (2.10), or (2.11). p = vi = 30(6 × 10−3 ) = 180 mW or p = i 2 R = (6 × 10−3 )2 5 × 103 = 180 mW or p = v 2 G = (30)2 0.2 × 10−3 = 180 mWPRACTICE PROBLEM 2.2 i For the circuit shown in Fig. 2.9, calculate the voltage v, the conductance + G, and the power p. 2 mA 10 kΩ v Answer: 20 V, 100 µS, 40 mW. − Figure 2.9 For Practice Prob. 2.2 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 39. CHAPTER 2 Basic Laws 33 E X A M P L E 2 . 3 A voltage source of 20 sin π t V is connected across a 5-k resistor. Find the current through the resistor and the power dissipated. Solution: v 20 sin π t i= = = 4 sin π t mA R 5 × 103 Hence, p = vi = 80 sin2 π t mW PRACTICE PROBLEM 2.3 A resistor absorbs an instantaneous power of 20 cos2 t mW when con- nected to a voltage source v = 10 cos t V. Find i and R. Answer: 2 cos t mA, 5 k . † 2.3 NODES, BRANCHES, AND LOOPS Since the elements of an electric circuit can be interconnected in several ways, we need to understand some basic concepts of network topology. To differentiate between a circuit and a network, we may regard a network as an interconnection of elements or devices, whereas a circuit is a network providing one or more closed paths. The convention, when addressing network topology, is to use the word network rather than circuit. We do this even though the words network and circuit mean the same thing when used in this context. In network topology, we study the properties relating to the placement of elements in the network and the geometric a 5Ω b conﬁguration of the network. Such elements include branches, nodes, and loops. 10 V + − 2Ω 3Ω 2A A branch represents a single element such as a voltage source or a resistor. c In other words, a branch represents any two-terminal element. The circuit in Fig. 2.10 has ﬁve branches, namely, the 10-V voltage source, the 2-A Figure 2.10 Nodes, branches, and loops. current source, and the three resistors. b A node is the point of connection between two or more branches. 5Ω 2Ω A node is usually indicated by a dot in a circuit. If a short circuit (a 3Ω 2A connecting wire) connects two nodes, the two nodes constitute a single a + node. The circuit in Fig. 2.10 has three nodes a, b, and c. Notice that − the three points that form node b are connected by perfectly conducting 10 V c wires and therefore constitute a single point. The same is true of the four points forming node c. We demonstrate that the circuit in Fig. 2.10 has Figure 2.11 The three-node circuit of Fig. 2.10 only three nodes by redrawing the circuit in Fig. 2.11. The two circuits in is redrawn.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 40. 34 PART 1 DC Circuits Figs. 2.10 and 2.11 are identical. However, for the sake of clarity, nodes b and c are spread out with perfect conductors as in Fig. 2.10. A loop is any closed path in a circuit. A loop is a closed path formed by starting at a node, passing through a set of nodes, and returning to the starting node without passing through any node more than once. A loop is said to be independent if it contains a branch which is not in any other loop. Independent loops or paths result in independent sets of equations. For example, the closed path abca containing the 2- resistor in Fig. 2.11 is a loop. Another loop is the closed path bcb containing the 3- resistor and the current source. Although one can identify six loops in Fig. 2.11, only three of them are independent. A network with b branches, n nodes, and l independent loops will satisfy the fundamental theorem of network topology: b =l+n−1 (2.12) As the next two deﬁnitions show, circuit topology is of great value to the study of voltages and currents in an electric circuit. Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current. Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. Elements are in series when they are chain-connected or connected se- quentially, end to end. For example, two elements are in series if they share one common node and no other element is connected to that com- mon node. Elements in parallel are connected to the same pair of termi- nals. Elements may be connected in a way that they are neither in series nor in parallel. In the circuit shown in Fig. 2.10, the voltage source and the 5- resistor are in series because the same current will ﬂow through them. The 2- resistor, the 3- resistor, and the current source are in parallel because they are connected to the same two nodes (b and c) and consequently have the same voltage across them. The 5- and 2- resistors are neither in series nor in parallel with each other.E X A M P L E 2 . 4 Determine the number of branches and nodes in the circuit shown in Fig. 2.12. Identify which elements are in series and which are in parallel. Solution: Since there are four elements in the circuit, the circuit has four branches: 10 V, 5 , 6 , and 2 A. The circuit has three nodes as identiﬁed in v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 41. CHAPTER 2 Basic Laws 35 Fig. 2.13. The 5- resistor is in series with the 10-V voltage source because the same current would ﬂow in both. The 6- resistor is in parallel with the 2-A current source because both are connected to the same nodes 2 and 3. 5Ω 1 5Ω 2 + + 10 V − 6Ω 2A 10 V − 6Ω 2A 3 Figure 2.12 For Example 2.4. Figure 2.13 The three nodes in the circuit of Fig. 2.12. PRACTICE PROBLEM 2.4 How many branches and nodes does the circuit in Fig. 2.14 have? Identify the elements that are in series and in parallel. Answer: Five branches and three nodes are identiﬁed in Fig. 2.15. The 1- and 2- resistors are in parallel. The 4- resistor and 10-V source are also in parallel. 5Ω 1 3Ω 2 + 10 V 1Ω 2Ω + 10 V 4Ω 1Ω 2Ω − 4Ω − 3 Figure 2.14 For Practice Prob. 2.4. Figure 2.15 Answer for Practice Prob. 2.4. 2.4 KIRCHHOFF’S LAWS Ohm’s law by itself is not sufﬁcient to analyze circuits. However, when it is coupled with Kirchhoff’s two laws, we have a sufﬁcient, powerful set of tools for analyzing a large variety of electric circuits. Kirchhoff’s laws were ﬁrst introduced in 1847 by the German physicist Gustav Robert Kirchhoff (1824–1887). These laws are formally known as Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). Kirchhoff’s ﬁrst law is based on the law of conservation of charge, which requires that the algebraic sum of charges within a system cannot change.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 42. 36 PART 1 DC Circuits Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero. Mathematically, KCL implies that N in = 0 (2.13) n=1 where N is the number of branches connected to the node and in is the nth current entering (or leaving) the node. By this law, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa. To prove KCL, assume a set of currents ik (t), k = 1, 2, . . . , ﬂow into a node. The algebraic sum of currents at the node is iT (t) = i1 (t) + i2 (t) + i3 (t) + · · · (2.14) i5 i1 Integrating both sides of Eq. (2.14) gives i4 qT (t) = q1 (t) + q2 (t) + q3 (t) + · · · (2.15) i2 i3 where qk (t) = ik (t) dt and qT (t) = iT (t) dt. But the law of conser- vation of electric charge requires that the algebraic sum of electric charges at the node must not change; that is, the node stores no net charge. Thus Figure 2.16 Currents at qT (t) = 0 → iT (t) = 0, conﬁrming the validity of KCL. a node illustrating KCL. Consider the node in Fig. 2.16. Applying KCL gives i1 + (−i2 ) + i3 + i4 + (−i5 ) = 0 (2.16) since currents i1 , i3 , and i4 are entering the node, while currents i2 and i5 are leaving it. By rearranging the terms, we get Closed boundary i1 + i3 + i4 = i2 + i5 (2.17) Equation (2.17) is an alternative form of KCL: The sum of the currents entering a node is equal to the sum of the currents leaving the node. Note that KCL also applies to a closed boundary. This may be regarded as a generalized case, because a node may be regarded as a closed surface shrunk to a point. In two dimensions, a closed boundaryFigure 2.17 Applying KCL to a closed is the same as a closed path. As typically illustrated in the circuit of boundary. Fig. 2.17, the total current entering the closed surface is equal to the total current leaving the surface. A simple application of KCL is combining current sources in par- allel. The combined current is the algebraic sum of the current supplied by the individual sources. For example, the current sources shown in Fig.Two sources (or circuits in general) are said to be 2.18(a) can be combined as in Fig. 2.18(b). The combined or equivalentequivalent if they have the same i-v relationship current source can be found by applying KCL to node a.at a pair of terminals. IT + I2 = I1 + I3 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 43. CHAPTER 2 Basic Laws 37 or IT IT = I1 − I2 + I3 (2.18) a A circuit cannot contain two different currents, I1 and I2 , in series, unless I1 I2 I3 I1 = I2 ; otherwise KCL will be violated. Kirchhoff’s second law is based on the principle of conservation of b energy: (a) IT Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages a around a closed path (or loop) is zero. IS = I1 – I2 + I3 b Expressed mathematically, KVL states that (b) M Figure 2.18 Current sources in parallel: (a) original circuit, (b) equivalent circuit. vm = 0 (2.19) m=1 where M is the number of voltages in the loop (or the number of branches in the loop) and vm is the mth voltage. To illustrate KVL, consider the circuit in Fig. 2.19. The sign on each voltage is the polarity of the terminal encountered ﬁrst as we travel around the loop. We can start with any branch and go around the loop either clockwise or counterclockwise. Suppose we start with the voltage KVL can be applied in two ways: by taking either a source and go clockwise around the loop as shown; then voltages would clockwise or a counterclockwise trip around the be −v1 , +v2 , +v3 , −v4 , and +v5 , in that order. For example, as we reach loop. Either way, the algebraic sum of voltages branch 3, the positive terminal is met ﬁrst; hence we have +v3 . For branch around the loop is zero. 4, we reach the negative terminal ﬁrst; hence, −v4 . Thus, KVL yields −v1 + v2 + v3 − v4 + v5 = 0 (2.20) + v2 − + v3 − Rearranging terms gives v 2 + v 3 + v 5 = v1 + v 4 (2.21) v1 + − v4 − + which may be interpreted as − + v5 Sum of voltage drops = Sum of voltage rises (2.22) Figure 2.19 A single-loop circuit This is an alternative form of KVL. Notice that if we had traveled coun- illustrating KVL. terclockwise, the result would have been +v1 , −v5 , +v4 , −v3 , and −v2 , which is the same as before except that the signs are reversed. Hence, Eqs. (2.20) and (2.21) remain the same. When voltage sources are connected in series, KVL can be applied to obtain the total voltage. The combined voltage is the algebraic sum of the voltages of the individual sources. For example, for the voltage sources shown in Fig. 2.20(a), the combined or equivalent voltage source in Fig. 2.20(b) is obtained by applying KVL. −Vab + V1 + V2 − V3 = 0v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 44. 38 PART 1 DC Circuits or Vab = V1 + V2 − V3 (2.23) To avoid violating KVL, a circuit cannot contain two different voltages V1 and V2 in parallel unless V1 = V2 . a + + − V1 a Vab + + − V2 + V =V +V −V − Vab − S 1 2 3 + V3 − − b b (a) (b) Figure 2.20 Voltage sources in series: (a) original circuit, (b) equivalent circuit.E X A M P L E 2 . 5 For the circuit in Fig. 2.21(a), ﬁnd voltages v1 and v2 . 2Ω 2Ω + v1 − + v1 − − − + v2 3Ω + v2 3Ω 20 V − + 20 V − i + (a) (b) Figure 2.21 For Example 2.5. Solution: To ﬁnd v1 and v2 , we apply Ohm’s law and Kirchhoff’s voltage law. Assume that current i ﬂows through the loop as shown in Fig. 2.21(b). From Ohm’s law, v1 = 2i, v2 = −3i (2.5.1) Applying KVL around the loop gives −20 + v1 − v2 = 0 (2.5.2) Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain −20 + 2i + 3i = 0 or 5i = 20 ⇒ i =4A Substituting i in Eq. (2.5.1) ﬁnally gives v1 = 8 V, v2 = −12 V v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 45. CHAPTER 2 Basic Laws 39 PRACTICE PROBLEM 2.5 Find v1 and v2 in the circuit of Fig. 2.22. 4Ω Answer: 12 V, −6 V. + v1 − − + 10 V − 8V + + v2 − 2Ω Figure 2.22 For Practice Prob. 2.5 E X A M P L E 2 . 6 Determine vo and i in the circuit shown in Fig. 2.23(a). i 4Ω 2vo 4Ω 2vo +− +− − i − + 12 V + 12 V − 4V + − 4V + 6Ω 6Ω + vo − + vo − (a) (b) Figure 2.23 For Example 2.6. Solution: We apply KVL around the loop as shown in Fig. 2.23(b). The result is −12 + 4i + 2vo − 4 + 6i = 0 (2.6.1) Applying Ohm’s law to the 6- resistor gives vo = −6i (2.6.2) Substituting Eq. (2.6.2) into Eq. (2.6.1) yields −16 + 10i − 12i = 0 ⇒ i = −8 A and vo = 48 V. PRACTICE PROBLEM 2.6 Find vx and vo in the circuit of Fig. 2.24. 10 Ω Answer: 10 V, −5 V. + vx − + 35 V + − − 2vx 5Ω + vo − Figure 2.24 For Practice Prob. 2.6.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 46. 40 PART 1 DC Circuits E X A M P L E 2 . 7 a Find current io and voltage vo in the circuit shown in Fig. 2.25. io Solution: + Applying KCL to node a, we obtain 0.5io vo 4Ω 3A − 3 + 0.5io = io ⇒ io = 6 A For the 4- resistor, Ohm’s law gives Figure 2.25 For Example 2.7. vo = 4io = 24 VPRACTICE PROBLEM 2.7 io Find vo and io in the circuit of Fig. 2.26. + io Answer: 8 V, 4 A.6A 2Ω 8Ω vo 4 −Figure 2.26 For Practice Prob. 2.7. E X A M P L E 2 . 8 Find the currents and voltages in the circuit shown in Fig. 2.27(a). i1 i3 i1 i3 8Ω a 8Ω a + v1 − i2 + v1 − i2 + + + + 30 V + − v2 3Ω v3 6Ω 30 V + − Loop 1 v2 3Ω Loop 2 v3 6Ω − − − − (a) (b) Figure 2.27 For Example 2.8. Solution: We apply Ohm’s law and Kirchhoff’s laws. By Ohm’s law, v1 = 8i1 , v2 = 3i2 , v3 = 6i3 (2.8.1) Since the voltage and current of each resistor are related by Ohm’s law as shown, we are really looking for three things: (v1 , v2 , v3 ) or (i1 , i2 , i3 ). At node a, KCL gives i1 − i2 − i3 = 0 (2.8.2) Applying KVL to loop 1 as in Fig. 2.27(b), −30 + v1 + v2 = 0 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 47. CHAPTER 2 Basic Laws 41 We express this in terms of i1 and i2 as in Eq. (2.8.1) to obtain −30 + 8i1 + 3i2 = 0 or (30 − 3i2 ) i1 = (2.8.3) 8 Applying KVL to loop 2, −v2 + v3 = 0 ⇒ v3 = v2 (2.8.4) as expected since the two resistors are in parallel. We express v1 and v2 in terms of i1 and i2 as in Eq. (2.8.1). Equation (2.8.4) becomes i2 6i3 = 3i2 ⇒ i3 = (2.8.5) 2 Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives 30 − 3i2 i2 − i2 − = 0 8 2 or i2 = 2 A. From the value of i2 , we now use Eqs. (2.8.1) to (2.8.5) to obtain i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 = 6 V, v3 = 6 V @ Network Analysis PRACTICE PROBLEM 2.8 Find the currents and voltages in the circuit shown in Fig. 2.28. 2Ω i1 i3 4Ω Answer: v1 = 3 V, v2 = 2 V, v3 = 5 V, i1 = 1.5 A, i2 = 0.25 A, + v1 − i2 + v3 − i3 =1.25 A. + + − 5V − v2 8Ω + 3V − Figure 2.28 For Practice Prob. 2.8. 2.5 SERIES RESISTORS AND VOLTAGE DIVISION The need to combine resistors in series or in parallel occurs so frequently i a R1 R2 that it warrants special attention. The process of combining the resistors + v1 − + v2 − is facilitated by combining two of them at a time. With this in mind, consider the single-loop circuit of Fig. 2.29. The two resistors are in v + − series, since the same current i ﬂows in both of them. Applying Ohm’s law to each of the resistors, we obtain b v1 = iR1 , v2 = iR2 (2.24) Figure 2.29 A single-loop circuit If we apply KVL to the loop (moving in the clockwise direction), we have with two resistors in series. −v + v1 + v2 = 0 (2.25)v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 48. 42 PART 1 DC Circuits Combining Eqs. (2.24) and (2.25), we get v = v1 + v2 = i(R1 + R2 ) (2.26) i Req or a v + v − i= (2.27) R1 + R 2 v + − Notice that Eq. (2.26) can be written as v = iReq (2.28) b implying that the two resistors can be replaced by an equivalent resistor Req ; that is, Figure 2.30 Equivalent circuit of the Fig. 2.29 circuit. Req = R1 + R2 (2.29) Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30. The two circuits in Figs. 2.29 and 2.30 are equivalent because they exhibit the same voltage-current relationships at the terminals a-b. An equivalent circuit such as the one in Fig. 2.30 is useful in simplifying the analysis of a circuit. In general, The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances.Resistors in series behave as a single resistor For N resistors in series then,whose resistance is equal to the sum of the re- Nsistances of the individual resistors. Req = R1 + R2 + · · · + RN = Rn (2.30) n=1 To determine the voltage across each resistor in Fig. 2.29, we sub- stitute Eq. (2.26) into Eq. (2.24) and obtain R1 R2 v1 = v, v2 = v (2.31) R1 + R 2 R1 + R 2 Notice that the source voltage v is divided among the resistors in direct proportion to their resistances; the larger the resistance, the larger the voltage drop. This is called the principle of voltage division, and the circuit in Fig. 2.29 is called a voltage divider. In general, if a voltage divider has N resistors (R1 , R2 , . . . , RN ) in series with the source voltage v, the nth resistor (Rn ) will have a voltage drop of Rn vn = v (2.32) R1 + R2 + · · · + RN 2.6 PARALLEL RESISTORS AND CURRENT DIVISION Consider the circuit in Fig. 2.31, where two resistors are connected in parallel and therefore have the same voltage across them. From Ohm’s law, v = i1 R1 = i2 R2 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 49. CHAPTER 2 Basic Laws 43 or i Node a v v i1 = , i2 = (2.33) R1 R2 i1 i2 Applying KCL at node a gives the total current i as v + − R1 R2 i = i1 + i2 (2.34) Substituting Eq. (2.33) into Eq. (2.34), we get Node b v v 1 1 v i= + =v + = (2.35) R1 R2 R1 R2 Req Figure 2.31 Two resistors in parallel. where Req is the equivalent resistance of the resistors in parallel: 1 1 1 = + (2.36) Req R1 R2 or 1 R1 + R2 = Req R1 R2 or R1 R 2 Req = (2.37) R1 + R 2 Thus, The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum. It must be emphasized that this applies only to two resistors in parallel. From Eq. (2.37), if R1 = R2 , then Req = R1 /2. We can extend the result in Eq. (2.36) to the general case of a circuit with N resistors in parallel. The equivalent resistance is 1 1 1 1 = + + ··· + (2.38) Req R1 R2 RN Note that Req is always smaller than the resistance of the smallest resistor in the parallel combination. If R1 = R2 = · · · = RN = R, then R Req = (2.39) N For example, if four 100- resistors are connected in parallel, their equiv- alent resistance is 25 . It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equivalent conductance for N resistors in parallel is Conductances in parallel behave as a single con- ductance whose value is equal to the sum of the Geq = G1 + G2 + G3 + · · · + GN (2.40) individual conductances. where Geq = 1/Req , G1 = 1/R1 , G2 = 1/R2 , G3 = 1/R3 , . . . , GN = 1/RN . Equation (2.40) states:v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 50. 44 PART 1 DC Circuits The equivalent conductance of resistors connected in parallel is the sum of their individual conductances. This means that we may replace the circuit in Fig. 2.31 with that in i a Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The equivalent conductance of parallel resistors is obtained the same way as the equivalent resistance of series resistors. In the same manner, the v + v equivalent conductance of resistors in series is obtained just the same way − Req or Geq as the resistance of resistors in parallel. Thus the equivalent conductance Geq of N resistors in series (such as shown in Fig. 2.29) is b 1 1 1 1 1 = + + + ··· + (2.41) Figure 2.32 Equivalent circuit to Geq G1 G2 G3 GN Fig. 2.31. Given the total current i entering node a in Fig. 2.31, how do we obtain current i1 and i2 ? We know that the equivalent resistor has the same voltage, or iR1 R2 v = iReq = (2.42) R1 + R 2 Combining Eqs. (2.33) and (2.42) results in R2 i R1 i i1 = , i2 = (2.43) R1 + R 2 R1 + R 2 which shows that the total current i is shared by the resistors in inverse proportion to their resistances. This is known as the principle of current division, and the circuit in Fig. 2.31 is known as a current divider. Notice that the larger current ﬂows through the smaller resistance. As an extreme case, suppose one of the resistors in Fig. 2.31 is zero, i say R2 = 0; that is, R2 is a short circuit, as shown in Fig. 2.33(a). From Eq. (2.43), R2 = 0 implies that i1 = 0, i2 = i. This means that the i2 = i i1 = 0 entire current i bypasses R1 and ﬂows through the short circuit R2 = 0, R1 R2 = 0 the path of least resistance. Thus when a circuit is short circuited, as shown in Fig. 2.33(a), two things should be kept in mind: 1. The equivalent resistance Req = 0. [See what happens when R2 = 0 in Eq. (2.37).] (a) 2. The entire current ﬂows through the short circuit. i As another extreme case, suppose R2 = ∞, that is, R2 is an open circuit, as shown in Fig. 2.33(b). The current still ﬂows through the path i2 = 0 i1 = i of least resistance, R1 . By taking the limit of Eq. (2.37) as R2 → ∞, we R1 R2 = ∞ obtain Req = R1 in this case. If we divide both the numerator and denominator by R1 R2 , Eq. (2.43) becomes G1 (b) i1 = i (2.44a) G1 + G 2 Figure 2.33 (a) A shorted circuit, G2 (b) an open circuit. i2 = i (2.44b) G1 + G 2 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 51. CHAPTER 2 Basic Laws 45 Thus, in general, if a current divider has N conductors (G1 , G2 , . . . , GN ) in parallel with the source current i, the nth conductor (Gn ) will have current Gn in = i (2.45) G1 + G 2 + · · · + GN In general, it is often convenient and possible to combine resistors in series and parallel and reduce a resistive network to a single equivalent resistance Req . Such an equivalent resistance is the resistance between the designated terminals of the network and must exhibit the same i-v characteristics as the original network at the terminals. E X A M P L E 2 . 9 Find Req for the circuit shown in Fig. 2.34. 4Ω 1Ω Solution: 2Ω To get Req , we combine resistors in series and in parallel. The 6- and Req 5Ω 3- resistors are in parallel, so their equivalent resistance is 6Ω 3Ω 6×3 8Ω 6 3 = =2 6+3 (The symbol is used to indicate a parallel combination.) Also, the 1- Figure 2.34 For Example 2.9. and 5- resistors are in series; hence their equivalent resistance is 1 +5 =6 4Ω Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the two 2- resistors are in series, so the equivalent 2Ω Req resistance is 6Ω 2Ω 2 +2 =4 8Ω This 4- resistor is now in parallel with the 6- resistor in Fig. 2.35(a); (a) their equivalent resistance is 4×6 4Ω 4 6 = = 2.4 4+6 Req The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.4 Ω 2.35(b), the three resistors are in series. Hence, the equivalent resistance 8Ω for the circuit is Req = 4 + 2.4 +8 = 14.4 (b) Figure 2.35 Equivalent circuits for Example 2.9. PRACTICE PROBLEM 2.9 By combining the resistors in Fig. 2.36, ﬁnd Req . 2Ω 3Ω 4Ω Answer: 6 . Req 6Ω 4Ω 5Ω 1Ω 3Ω Figure 2.36 For Practice Prob. 2.9.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 52. 46 PART 1 DC CircuitsE X A M P L E 2 . 1 0 Calculate the equivalent resistance Rab in the circuit in Fig. 2.37. 10 Ω c 1Ω d 1Ω a 6Ω Rab 3Ω 4Ω 5Ω 12 Ω b b b Figure 2.37 For Example 2.10. Solution: The 3- and 6- resistors are in parallel because they are connected to the same two nodes c and b. Their combined resistance is 3×6 10 Ω 3 6 = =2 (2.10.1) c 1Ω d 3+6 a Similarly, the 12- and 4- resistors are in parallel since they are con- 2Ω 3Ω 6Ω nected to the same two nodes d and b. Hence 12 × 4 12 4 = =3 (2.10.2) b 12 + 4 b b b Also the 1- and 5- resistors are in series; hence, their equivalent (a) resistance is 1 +5 =6 (2.10.3) 10 Ω c a With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig. 2.38(a), 3- in parallel with 6- gives 2- , as 2Ω 3Ω calculated in Eq. (2.10.1). This 2- equivalent resistance is now in series with the 1- resistance to give a combined resistance of 1 +2 = 3 . b b b Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In (b) Fig. 2.38(b), we combine the 2- and 3- resistors in parallel to get 2×3 2 3 = = 1.2 Figure 2.38 Equivalent circuits for 2+3 Example 2.10. This 1.2- resistor is in series with the 10- resistor, so that Rab = 10 + 1.2 = 11.2PRACTICE PROBLEM 2.10 20 Ω Find Rab for the circuit in Fig. 2.39. Answer: 11 . 8Ω 5Ω a Rab 18 Ω 20 Ω 1Ω 9Ω 2Ω b Figure 2.39 For Practice Prob. 2.10. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 53. CHAPTER 2 Basic Laws 47 E X A M P L E 2 . 1 1 Find the equivalent conductance Geq for the circuit in Fig. 2.40(a). 5S Solution: Geq The 8-S and 12-S resistors are in parallel, so their conductance is 6S 8S 12 S 8 S + 12 S = 20 S (a) This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b) so that the combined conductance is 5S 20 × 5 =4S Geq 20 + 5 6S 20 S This is in parallel with the 6-S resistor. Hence Geq = 6 + 4 = 10 S (b) 1 We should note that the circuit in Fig. 2.40(a) is the same as that in 5 Ω Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in siemens, they are expressed in ohms in Fig. 2.40(c). To show that the circuits are Req 1 1 1 the same, we ﬁnd Req for the circuit in Fig. 2.40(c). 6 Ω 8 Ω 12 Ω 1 1 1 1 1 1 1 1 1 Req = + = + = (c) 6 5 8 12 6 5 20 6 4 1 × 1 1 Figure 2.40 For Example 2.11: (a) original = 6 4 = circuit, (b) its equivalent circuit, (c) same 1 6 + 1 4 10 circuit as in (a) but resistors are expressed in ohms. 1 Geq = = 10 S Req This is the same as we obtained previously. PRACTICE PROBLEM 2.11 Calculate Geq in the circuit of Fig. 2.41. Answer: 4 S. 8S 4S Geq 2S 6S 12 Ω Figure 2.41 For Practice Prob. 2.11. E X A M P L E 2 . 1 2 Find io and vo in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3- resistor. Solution: The 6- and 3- resistors are in parallel, so their combined resistance is 6×3 6 3 = =2 6+3v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 54. 48 PART 1 DC Circuits i 4Ω io Thus our circuit reduces to that shown in Fig. 2.42(b). Notice that vo is a not affected by the combination of the resistors because the resistors are + in parallel and therefore have the same voltage vo . From Fig. 2.42(b), we 12 V + − 6Ω vo 3Ω can obtain vo in two ways. One way is to apply Ohm’s law to get − 12 i= =2A b 4+2 (a) and hence, vo = 2i = 2 × 2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4- and i 4Ω 2- resistors. Hence, a 2 + vo = (12 V) = 4 V 12 V + vo 2Ω 2+4 − − Similarly, io can be obtained in two ways. One approach is to apply Ohm’s law to the 3- resistor in Fig. 2.42(a) now that we know vo ; thus, b 4 (b) vo = 3io = 4 ⇒ io = A 3 Figure 2.42 For Example 2.12: (a) original Another approach is to apply current division to the circuit in Fig. 2.42(a) circuit, (b) its equivalent circuit. now that we know i, by writing 6 2 4 io = i = (2 A) = A 6+3 3 3 The power dissipated in the 3- resistor is 4 po = vo io = 4 = 5.333 W 3PRACTICE PROBLEM 2.12 i1 12 Ω Find v1 and v2 in the circuit shown in Fig. 2.43. Also calculate i1 and i2 and the power dissipated in the 12- and 40- resistors. + v1 − Answer: v1 = 5 V, i1 = 416.7 mA, p1 = 2.083 W, v2 = 10 V, 6Ω i2 = 250 mA, p2 = 2.5 W. i2 +15 V + − 10 Ω v2 40 Ω −Figure 2.43 For Practice Prob. 2.12.E X A M P L E 2 . 1 3 For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vo , (b) the power supplied by the current source, (c) the power absorbed by each resistor. Solution: (a) The 6-k and 12-k resistors are in series so that their combined value is 6 + 12 = 18 k . Thus the circuit in Fig. 2.44(a) reduces to that v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 55. CHAPTER 2 Basic Laws 49 shown in Fig. 2.44(b). We now apply the current division technique to 6 kΩ ﬁnd i1 and i2 . + 18,000 30 mA vo 12 kΩ i1 = (30 mA) = 20 mA − 9 kΩ 9000 + 18,000 9000 i2 = (30 A) = 10 mA 9000 + 18,000 (a) Notice that the voltage across the 9-k and 18-k resistors is the same, io i2 and vo = 9,000i1 = 18,000i2 = 180 V, as expected. i1 (b) Power supplied by the source is + 30 mA vo 9 kΩ 18 kΩ po = vo io = 180(30) mW = 5.4 W − (c) Power absorbed by the 12-k resistor is (b) −3 2 p = iv = i2 (i2 R) = 2 i2 R = (10 × 10 ) (12,000) = 1.2 W Figure 2.44 For Example 2.13: (a) original circuit, Power absorbed by the 6-k resistor is (b) its equivalent circuit. p = i2 R = (10 × 10−3 )2 (6000) = 0.6 W 2 Power absorbed by the 9-k resistor is 2 vo (180)2 p= = = 3.6 W R 9000 or p = vo i1 = 180(20) mW = 3.6 W Notice that the power supplied (5.4 W) equals the power absorbed (1.2 + 0.6 + 3.6 = 5.4 W). This is one way of checking results. PRACTICE PROBLEM 2.13 For the circuit shown in Fig. 2.45, ﬁnd: (a) v1 and v2 , (b) the power dis- sipated in the 3-k and 20-k resistors, and (c) the power supplied by the current source. 1 kΩ + + 3 kΩ v1 10 mA 5 kΩ v2 20 kΩ − − Figure 2.45 For Practice Prob. 2.13. Answer: (a) 15 V, 20 V, (b) 75 mW, 20 mW, (c) 200 mW. @ Network Analysisv v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 56. 50 PART 1 DC Circuits † 2.7 WYE-DELTA TRANSFORMATIONS R1 Situations often arise in circuit analysis when the resistors are neither in parallel nor in series. For example, consider the bridge circuit in Fig. 2.46. How do we combine resistors R1 through R6 when the resistors R2 R3 are neither in series nor in parallel? Many circuits of the type shown in R4 Fig. 2.46 can be simpliﬁed by using three-terminal equivalent networks. vs + − These are the wye (Y) or tee (T) network shown in Fig. 2.47 and the delta ( ) or pi ( ) network shown in Fig. 2.48. These networks occur by R5 R6 themselves or as part of a larger network. They are used in three-phase networks, electrical ﬁlters, and matching networks. Our main interest Figure 2.46 The bridge network. here is in how to identify them when they occur as part of a network and how to apply wye-delta transformation in the analysis of that network. 1 3 R1 R2 R1 R2 1 3 R3 R3 2 4 2 4 (a) (b) Figure 2.47 Two forms of the same network: (a) Y, (b) T. Delta to Wye Conversion Rc 1 3 Suppose it is more convenient to work with a wye network in a place where the circuit contains a delta conﬁguration. We superimpose a wye Rb Ra network on the existing delta network and ﬁnd the equivalent resistances in the wye network. To obtain the equivalent resistances in the wye network, we compare the two networks and make sure that the resistance 2 4 between each pair of nodes in the (or ) network is the same as the (a) resistance between the same pair of nodes in the Y (or T) network. For terminals 1 and 2 in Figs. 2.47 and 2.48, for example, Rc 1 3 R12 (Y) = R1 + R3 (2.46) R12 ( ) = Rb (Ra + Rc ) Rb Ra Setting R12 (Y)= R12 ( ) gives Rb (Ra + Rc ) 2 4 R12 = R1 + R3 = (2.47a) Ra + R b + R c (b) Similarly, Figure 2.48 Two forms of the Rc (Ra + Rb ) same network: (a) , (b) . R13 = R1 + R2 = (2.47b) Ra + R b + R c Ra (Rb + Rc ) R34 = R2 + R3 = (2.47c) Ra + R b + R c Subtracting Eq. (2.47c) from Eq. (2.47a), we get Rc (Rb − Ra ) R 1 − R2 = (2.48) Ra + R b + R c v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 57. CHAPTER 2 Basic Laws 51 Adding Eqs. (2.47b) and (2.48) gives Rb Rc R1 = (2.49) Ra + R b + R c and subtracting Eq. (2.48) from Eq. (2.47b) yields R c Ra R2 = (2.50) Ra + R b + R c Subtracting Eq. (2.49) from Eq. (2.47a), we obtain R a Rb R3 = (2.51) Ra + R b + R c We do not need to memorize Eqs. (2.49) to (2.51). To transform a network to Y, we create an extra node n as shown in Fig. 2.49 and follow Rc this conversion rule: a b R1 R2 Each resistor in the Y network is the product of the resistors in the two adjacent n branches, divided by the sum of the three resistors. Rb Ra R3 Wye to Delta Conversion To obtain the conversion formulas for transforming a wye network to an c equivalent delta network, we note from Eqs. (2.49) to (2.51) that Ra Rb Rc (Ra + Rb + Rc ) Figure 2.49 Superposition of Y and R1 R 2 + R 2 R 3 + R 3 R 1 = networks as an aid in transforming one to (Ra + Rb + Rc )2 the other. (2.52) Ra R b Rc = Ra + R b + R c Dividing Eq. (2.52) by each of Eqs. (2.49) to (2.51) leads to the following equations: R1 R2 + R2 R3 + R3 R1 Ra = (2.53) R1 R1 R 2 + R 2 R 3 + R 3 R 1 Rb = (2.54) R2 R 1 R 2 + R 2 R 3 + R 3 R1 Rc = (2.55) R3 From Eqs. (2.53) to (2.55) and Fig. 2.49, the conversion rule for Y to is as follows:v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 58. 52 PART 1 DC Circuits Each resistor in the network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor. The Y and networks are said to be balanced when R1 = R2 = R3 = RY , Ra = Rb = Rc = R (2.56) Under these conditions, conversion formulas become R RY = or R = 3RY (2.57) 3 One may wonder why RY is less than R . Well, we notice that the Y- connection is like a “series” connection while the -connection is like a “parallel” connection. Note that in making the transformation, we do not take anything out of the circuit or put in anything new. We are merely substituting different but mathematically equivalent three-terminal network patterns to create a circuit in which resistors are either in series or in parallel, allowing us to calculate Req if necessary.E X A M P L E 2 . 1 4 Convert the network in Fig. 2.50(a) to an equivalent Y network. a b 5Ω 7.5 Ω Rc R1 R2 a b 25 Ω 10 Ω 15 Ω Rb Ra R3 3Ω c c (a) (b) Figure 2.50 For Example 2.14: (a) original network, (b) Y equivalent network. Solution: Using Eqs. (2.49) to (2.51), we obtain v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 59. CHAPTER 2 Basic Laws 53 Rb Rc 25 × 10 250 R1 = = = =5 Ra + R b + R c 25 + 10 + 15 50 R c Ra 25 × 15 R2 = = = 7.5 Ra + R b + R c 50 R a Rb 15 × 10 R3 = = =3 Ra + R b + R c 50 The equivalent Y network is shown in Fig. 2.50(b). PRACTICE PROBLEM 2.14 Transform the wye network in Fig. 2.51 to a delta network. R1 R2 a b Answer: Ra = 140 , Rb = 70 , Rc = 35 . 10 Ω 20 Ω R3 40 Ω c Figure 2.51 For Practice Prob. 2.14. E X A M P L E 2 . 1 5 Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use it i a a to ﬁnd current i. Solution: 12.5 Ω 10 Ω In this circuit, there are two Y networks and one network. Transforming just one of these will simplify the circuit. If we convert the Y network 5Ω 120 V + − c n 30 Ω comprising the 5- , 10- , and 20- resistors, we may select 15 Ω 20 Ω R1 = 10 , R2 = 20 , R3 = 5 Thus from Eqs. (2.53) to (2.55) we have b b R1 R2 + R2 R3 + R3 R1 10 × 20 + 20 × 5 + 5 × 10 Figure 2.52 For Example 2.15. Ra = = R1 10 350 = = 35 10 R1 R2 + R2 R3 + R3 R1 350 Rb = = = 17.5 R2 20 R1 R2 + R2 R3 + R3 R1 350 Rc = = = 70 R3 5 With the Y converted to , the equivalent circuit (with the voltage source removed for now) is shown in Fig. 2.53(a). Combining the three pairs of resistors in parallel, we obtainv v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 60. 54 PART 1 DC Circuits 70 × 30 70 30 = = 21 70 + 30 12.5 × 17.5 12.5 17.5 = = 7.2917 12.5 + 17.5 15 × 35 15 35 = = 10.5 15 + 35 so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we ﬁnd 17.792 × 21 Rab = (7.292 + 10.5) 21 = = 9.632 17.792 + 21 Then vs 120 i= = = 12.458 A Rab 9.632 a 12.5 Ω 17.5 Ω a 70 Ω 30 Ω 7.292 Ω 35 Ω 21 Ω 15 Ω 10.5 Ω b b (a) (b) Figure 2.53 Equivalent circuits to Fig. 2.52, with the voltage removed.PRACTICE PROBLEM 2.15 i a 13 Ω For the bridge network in Fig. 2.54, ﬁnd Rab and i. Answer: 40 , 2.5 A. 24 Ω 10 Ω 20 Ω100 V + − 30 Ω 50 Ω bFigure 2.54 For Practice Prob. 2.15. † 2.8 APPLICATIONS Resistors are often used to model devices that convert electrical energy into heat or other forms of energy. Such devices include conducting wire, lightbulbs, electric heaters, stoves, ovens, and loudspeakers. In this v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 61. CHAPTER 2 Basic Laws 55 section, we will consider two real-life problems that apply the concepts So far, we have assumed that connecting wires developed in this chapter: electrical lighting systems and design of dc are perfect conductors (i.e., conductors of zero meters. resistance). In real physical systems, however, the resistance of the connecting wire may be ap- 2.8.1 Lighting Systems preciably large, and the modeling of the system Lighting systems, such as in a house or on a Christmas tree, often consist must include that resistance. of N lamps connected either in parallel or in series, as shown in Fig. 2.55. Each lamp is modeled as a resistor. Assuming that all the lamps are identical and Vo is the power-line voltage, the voltage across each lamp is Vo for the parallel connection and Vo /N for the series connection. The series connection is easy to manufacture but is seldom used in practice, for at least two reasons. First, it is less reliable; when a lamp fails, all the lamps go out. Second, it is harder to maintain; when a lamp is bad, one must test all the lamps one by one to detect the faulty one. 1 2 + Vo − 1 2 3 N + 3 Vo − Power N plug (a) Lamp (b) Figure 2.55 (a) Parallel connection of lightbulbs, (b) series connection of lightbulbs. E X A M P L E 2 . 1 6 Three lightbulbs are connected to a 9-V battery as shown in Fig. 2.56(a). Calculate: (a) the total current supplied by the battery, (b) the current through each bulb, (c) the resistance of each bulb. I I1 I2 + V2 R2 − + 15 W 9V V1 R1 + − 9V 20 W V3 R3 10 W − (a) (b) Figure 2.56 (a) Lighting system with three bulbs, (b) resistive circuit equivalent model.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 62. 56 PART 1 DC Circuits Solution: (a) The total power supplied by the battery is equal to the total power absorbed by the bulbs, that is, p = 15 + 10 + 20 = 45 W Since p = V I , then the total current supplied by the battery is p 45 I= = =5A V 9 (b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since R1 (20-W bulb) is in parallel with the battery as well as the series com- bination of R2 and R3 , V1 = V2 + V3 = 9 V The current through R1 is p1 20 I1 = = = 2.222 A V1 9 By KCL, the current through the series combination of R2 and R3 is I2 = I − I1 = 5 − 2.222 = 2.778 A (c) Since p = I R,2 p1 20 R1 =2 = = 4.05 I1 2.2222 p2 15 R2 = 2 = = 1.945 I2 2.7772 p3 10 R3 = 2 = = 1.297 I3 2.7772PRACTICE PROBLEM 2.16 Refer to Fig. 2.55 and assume there are 10 lightbulbs, each with a power rating of 40 W. If the voltage at the plug is 110 V for the parallel and series connections, calculate the current through each bulb for both cases. Answer: 0.364 A (parallel), 3.64 A (series). 2.8.2 Design of DC Meters By their nature, resistors are used to control the ﬂow of current. We take a advantage of this property in several applications, such as in a poten- Max tiometer (Fig. 2.57). The word potentiometer, derived from the words b potential and meter, implies that potential can be metered out. The po- Vin + − tentiometer (or pot for short) is a three-terminal device that operates on + Vout the principle of voltage division. It is essentially an adjustable voltage Min − divider. As a voltage regulator, it is used as a volume or level control on c radios, TVs, and other devices. In Fig. 2.57, Rbc Figure 2.57 The potentiometer Vout = Vbc = Vin (2.58) controlling potential levels. Rac where Rac = Rab + Rbc . Thus, Vout decreases or increases as the sliding contact of the pot moves toward c or a, respectively. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 63. CHAPTER 2 Basic Laws 57 Another application where resistors are used to control current ﬂow is in the analog dc meters—the ammeter, voltmeter, and ohmmeter, which measure current, voltage, and resistance, respectively. Each of these me- An instrument capable of measuring voltage, cur- ters employs the d’Arsonval meter movement, shown in Fig. 2.58. The rent, and resistance is called a multimeter or a movement consists essentially of a movable iron-core coil mounted on volt-ohm meter (VOM). a pivot between the poles of a permanent magnet. When current ﬂows through the coil, it creates a torque which causes the pointer to deﬂect. The amount of current through the coil determines the deﬂection of the pointer, which is registered on a scale attached to the meter movement. For example, if the meter movement is rated 1 mA, 50 , it would take 1 mA to cause a full-scale deﬂection of the meter movement. By introduc- ing additional circuitry to the d’Arsonval meter movement, an ammeter, voltmeter, or ohmmeter can be constructed. Consider Fig. 2.59, where an analog voltmeter and ammeter are connected to an element. The voltmeter measures the voltage across a load and is therefore connected in parallel with the element. As shown A load is a component that is receiving energy (an in Fig. 2.60(a), the voltmeter consists of a d’Arsonval movement in par- energy sink), as opposed to a generator supplying allel with a resistor whose resistance Rm is deliberately made very large energy (an energy source). More about loading (theoretically, inﬁnite), to minimize the current drawn from the circuit. will be discussed in Section 4.9.1. To extend the range of voltage that the meter can measure, series multi- plier resistors are often connected with the voltmeters, as shown in Fig. 2.60(b). The multiple-range voltmeter in Fig. 2.60(b) can measure volt- age from 0 to 1 V, 0 to 10 V, or 0 to 100 V, depending on whether the switch is connected to R1 , R2 , or R3 , respectively. Let us calculate the multiplier resistor Rn for the single-range volt- meter in Fig. 2.60(a), or Rn = R1 , R2 , or R3 for the multiple-range voltmeter in Fig. 2.60(b). We need to determine the value of Rn to be connected in series with the internal resistance Rm of the voltmeter. In any design, we consider the worst-case condition. In this case, the worst case occurs when the full-scale current Ifs = Im ﬂows through the meter. This should also correspond to the maximum voltage reading or the full- scale voltage Vfs . Since the multiplier resistance Rn is in series with the Ammeter I scale A spring + Voltmeter V V Element pointer − S N permanent magnet Figure 2.59 Connection of a voltmeter and an ammeter to an element. spring rotating coil stationary iron core Figure 2.58 A d’Arsonval meter movement.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 64. 58 PART 1 DC Circuits Multiplier Meter Rn + Im Rm Probes V − (a) R1 1V Meter R2 10 V Switch + Im Rm 100 V Probes V R3 − (b) Figure 2.60 Voltmeters: (a) single-range type, (b) multiple-range type. In Rn Meter internal resistance Rm , Im Vfs = Ifs (Rn + Rm ) (2.59) Rm I From this, we obtain Probes Vfs Rn = − Rm (2.60) (a) Ifs R1 Similarly, the ammeter measures the current through the load and is connected in series with it. As shown in Fig. 2.61(a), the ammeter con- 10 mA sists of a d’Arsonval movement in parallel with a resistor whose resistance R2 100 mA Switch Rm is deliberately made very small (theoretically, zero) to minimize the voltage drop across it. To allow multiple range, shunt resistors are often 1A connected in parallel with Rm as shown in Fig. 2.61(b). The shunt resis- R3 tors allow the meter to measure in the range 0–10 mA, 0–100 mA, or Meter 0–1 A, depending on whether the switch is connected to R1 , R2 , or R3 , respectively. Im Now our objective is to obtain the multiplier shunt Rn for the single- Rm range ammeter in Fig. 2.61(a), or Rn = R1 , R2 , or R3 for the multiple- I range ammeter in Fig. 2.61(b). We notice that Rm and Rn are in parallel and that at full-scale reading I = Ifs = Im + In , where In is the current Probes through the shunt resistor Rn . Applying the current division principle (b) yieldsFigure 2.61 Ammeters: (a) single-range type, Rn Im = Ifs (b) multiple-range type. Rn + R m v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 65. CHAPTER 2 Basic Laws 59 or Im Rn = Rm (2.61) Ifs − Im The resistance Rx of a linear resistor can be measured in two ways. An indirect way is to measure the current I that ﬂows through it by connecting an ammeter in series with it and the voltage V across it by connecting a voltmeter in parallel with it, as shown in Fig. 2.62(a). Then V Rx = (2.62) I The direct method of measuring resistance is to use an ohmmeter. An ohmmeter consists basically of a d’Arsonval movement, a variable resistor or potentiometer, and a battery, as shown in Fig. 2.62(b). Applying KVL to the circuit in Fig. 2.62(b) gives A E = (R + Rm + Rx )Im I or + E Rx V V Rx = − (R + Rm ) (2.63) − Im The resistor R is selected such that the meter gives a full-scale deﬂection, that is, Im = Ifs when Rx = 0. This implies that (a) E = (R + Rm )Ifs (2.64) Ohmmeter Substituting Eq. (2.64) into Eq. (2.63) leads to Im Ifs Rx = − 1 (R + Rm ) (2.65) Rm R Im As mentioned, the types of meters we have discussed are known as E Rx analog meters and are based on the d’Arsonval meter movement. Another type of meter, called a digital meter, is based on active circuit elements such as op amps. For example, a digital multimeter displays measure- (b) ments of dc or ac voltage, current, and resistance as discrete numbers, instead of using a pointer deﬂection on a continuous scale as in an ana- Figure 2.62 Two ways of measuring log multimeter. Digital meters are what you would most likely use in a resistance: (a) using an ammeter and a modern lab. However, the design of digital meters is beyond the scope voltmeter, (b) using an ohmmeter. of this book. E X A M P L E 2 . 1 7 Following the voltmeter setup of Fig. 2.60, design a voltmeter for the fol- lowing multiple ranges: (a) 0–1 V (b) 0–5 V (c) 0–50 V (d) 0–100 V Assume that the internal resistance Rm = 2 k and the full-scale current Ifs = 100 µA. Solution: We apply Eq. (2.60) and assume that R1 , R2 , R3 , and R4 correspond with ranges 0–1 V, 0–5 V, 0–50 V, and 0–100 V, respectively. (a) For range 0–1 V, 1 R1 = − 2000 = 10,000 − 2000 = 8 k 100 × 10−6v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 66. 60 PART 1 DC Circuits (b) For range 0–5 V, 5 R2 = − 2000 = 50,000 − 2000 = 48 k 100 × 10−6 (c) For range 0–50 V, 50 R3 = − 2000 = 500,000 − 2000 = 498 k 100 × 10−6 (d) For range 0–100 V, 100 V R4 = − 2000 = 1,000,000 − 2000 = 998 k 100 × 10−6 Note that the ratio of the total resistance (Rn +Rm ) to the full-scale voltage Vfs is constant and equal to 1/Ifs for the four ranges. This ratio (given in ohms per volt, or /V) is known as the sensitivity of the voltmeter. The larger the sensitivity, the better the voltmeter.PRACTICE PROBLEM 2.17 Following the ammeter setup of Fig. 2.61, design an ammeter for the fol- lowing multiple ranges: (a) 0–1 A (b) 0–100 mA (c) 0–10 mA Take the full-scale meter current as Im = 1 mA and the internal resistance of the ammeter as Rm = 50 . Answer: Shunt resistors: 0.05 , 0.505 , 5.556 . 2.9 SUMMARY 1. A resistor is a passive element in which the voltage v across it is directly proportional to the current i through it. That is, a resistor is a device that obeys Ohm’s law, v = iR where R is the resistance of the resistor. 2. A short circuit is a resistor (a perfectly conducting wire) with zero resistance (R = 0). An open circuit is a resistor with inﬁnite resis- tance (R = ∞). 3. The conductance G of a resistor is the reciprocal of its resistance: 1 G= R 4. A branch is a single two-terminal element in an electric circuit. A node is the point of connection between two or more branches. A loop is a closed path in a circuit. The number of branches b, the number of nodes n, and the number of independent loops l in a network are related as b =l+n−1 v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 67. CHAPTER 2 Basic Laws 61 5. Kirchhoff’s current law (KCL) states that the currents at any node algebraically sum to zero. In other words, the sum of the currents entering a node equals the sum of currents leaving the node. 6. Kirchhoff’s voltage law (KVL) states that the voltages around a closed path algebraically sum to zero. In other words, the sum of voltage rises equals the sum of voltage drops. 7. Two elements are in series when they are connected sequentially, end to end. When elements are in series, the same current ﬂows through them (i1 = i2 ). They are in parallel if they are connected to the same two nodes. Elements in parallel always have the same voltage across them (v1 = v2 ). 8. When two resistors R1 (= 1/G1 ) and R2 (= 1/G2 ) are in series, their equivalent resistance Req and equivalent conductance Geq are G 1 G2 Req = R1 + R2 , Geq = G1 + G 2 9. When two resistors R1 (= 1/G1 ) and R2 (= 1/G2 ) are in parallel, their equivalent resistance Req and equivalent conductance Geq are R 1 R2 Req = , Geq = G1 + G2 R1 + R 2 10. The voltage division principle for two resistors in series is R1 R2 v1 = v, v2 = v R1 + R 2 R1 + R 2 11. The current division principle for two resistors in parallel is R2 R1 i1 = i, i2 = i R1 + R 2 R1 + R 2 12. The formulas for a delta-to-wye transformation are Rb Rc R c Ra R1 = , R2 = Ra + R b + R c Ra + R b + R c R a Rb R3 = Ra + R b + R c 13. The formulas for a wye-to-delta transformation are R1 R2 + R2 R3 + R3 R1 R1 R2 + R2 R3 + R3 R1 Ra = , Rb = R1 R2 R1 R2 + R2 R3 + R3 R1 Rc = R3 14. The basic laws covered in this chapter can be applied to the prob- lems of electrical lighting and design of dc meters. REVIEW QUESTIONS 2.1 The reciprocal of resistance is: 2.2 An electric heater draws 10 A from a 120-V line. (a) voltage (b) current The resistance of the heater is: (c) conductance (d) coulombs (a) 1200 (b) 120 (c) 12 (d) 1.2v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 68. 62 PART 1 DC Circuits2.3 The voltage drop across a 1.5-kW toaster that draws 2.9 Which of the circuits in Fig. 2.66 will give you 12 A of current is: Vab = 7 V? (a) 18 kV (b) 125 V (c) 120 V (d) 10.42 V2.4 The maximum current that a 2W, 80 k resistor can 5V 5V safely conduct is: +− a −+ a (a) 160 kA (b) 40 kA (c) 5 mA (d) 25 µA 3V + − 3V + −2.5 A network has 12 branches and 8 independent loops. How many nodes are there in the network? (a) 19 (b) 17 (c) 5 (d) 4 +− b +− b2.6 The current I in the circuit in Fig. 2.63 is: 1V 1V (a) −0.8 A (b) −0.2 A (a) (b) (c) 0.2 A (d) 0.8 A 5V 5V 4Ω I +− a −+ a 3V + − + − 5V 3V + − 3V + − 6Ω −+ b −+ b Figure 2.63 For Review Question 2.6. 1V 1V (c) (d)2.7 The current Io in Fig. 2.64 is: (a) −4 A (b) −2 A (c) 4 A (d) 16 A Figure 2.66 For Review Question 2.9. 10 A 2.10 The equivalent resistance of the circuit in Fig. 2.67 is: 2A 4A (a) 4 k (b) 5 k (c) 8 k (d) 14 k 2 kΩ 3 kΩ Io Req 6 kΩ 3 kΩ Figure 2.64 For Review Question 2.7.2.8 In the circuit in Fig. 2.65, V is: (a) 30 V (b) 14 V (c) 10 V (d) 6 V Figure 2.67 For Review Question 2.10. 10 V + − Answers: 2.1c, 2.2c, 2.3b, 2.4c, 2.5c, 2.6b, 2.7a, 2.8d, 2.9d, 2.10a. 12 V + − + − 8V + − V Figure 2.65 For Review Question 2.8. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 69. CHAPTER 2 Basic Laws 63 PROBLEMS Section 2.2 Ohm’s Law 2.7 Determine the number of branches and nodes in the circuit in Fig. 2.71. 2.1 The voltage across a 5-k resistor is 16 V. Find the current through the resistor. 2.2 Find the hot resistance of a lightbulb rated 60 W, 120 V. 2.3 When the voltage across a resistor is 120 V, the 5Ω current through it is 2.5 mA. Calculate its conductance. 2Ω 2.4 (a) Calculate current i in Fig. 2.68 when the switch i 5i + 4Ω − 10 V is in position 1. (b) Find the current when the switch is in position 2. 3Ω 6Ω 1 2 Figure 2.71 For Prob. 2.7. i 100 Ω + 150 Ω − 3V Section 2.4 Kirchhoff’s Laws 2.8 Use KCL to obtain currents i1 , i2 , and i3 in the Figure 2.68 For Prob. 2.4. circuit shown in Fig. 2.72. Section 2.3 Nodes, Branches, and Loops 2.5 For the network graph in Fig. 2.69, ﬁnd the number of nodes, branches, and loops. 12 mA i1 8 mA i2 i3 9 mA Figure 2.72 For Prob. 2.8. Figure 2.69 For Prob. 2.5. 2.9 Find i1 , i2 , and i3 in the circuit in Fig. 2.73. 2.6 In the network graph shown in Fig. 2.70, determine the number of branches and nodes. 1A 2A 10 A i1 i2 3A i3 Figure 2.70 For Prob. 2.6. Figure 2.73 For Prob. 2.9.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 70. 64 PART 1 DC Circuits2.10 Determine i1 and i2 in the circuit in Fig. 2.74. 2.13 Find v1 and v2 in the circuit in Fig. 2.77. + v1 − + 6V + v1 − 4A –2 A 12 V 10 V − i2 +− +− i1 + v2 − 3A Figure 2.77 For Prob. 2.13. Figure 2.74 For Prob. 2.10. 2.14 Obtain v1 through v3 in the circuit of Fig. 2.78. + v1 −2.11 Determine v1 through v4 in the circuit in Fig. 2.75. v2 − + 24 V + v3 + 10 V − + − − − −+ − v2 v1 + 12 V + 12 V + − Figure 2.78 For Prob. 2.14. + 8V − − 6V + 2.15 Find I and Vab in the circuit of Fig. 2.79. − 3Ω 10 V a 5Ω + − +− 10 V v4 v3 I − + + + 30 V + Vab + 8V − − − b Figure 2.75 For Prob. 2.11. Figure 2.79 For Prob. 2.15.2.12 In the circuit in Fig. 2.76, obtain v1 , v2 , and v3 . 2.16 From the circuit in Fig. 2.80, ﬁnd I , the power dissipated by the resistor, and the power supplied by each source. 15 V + − 10 V +− I 25 V 10 V − + + − + v2 − 12 V + 3Ω − + + + 20 V v1 v3 +− − − − −8 V Figure 2.76 For Prob. 2.12. Figure 2.80 For Prob. 2.16. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 71. CHAPTER 2 Basic Laws 65 2.17 Determine io in the circuit of Fig. 2.81. Sections 2.5 and 2.6 Series and Parallel Resistors io 4Ω 2.22 For the circuit in Fig. 2.86, ﬁnd i1 and i2 . 36 V + + 5io − − i1 i2 20 mA 6 kΩ 4 kΩ Figure 2.81 For Prob. 2.17. 2.18 Calculate the power dissipated in the 5- resistor in the circuit of Fig. 2.82. Figure 2.86 For Prob. 2.22. 1Ω + V − 2.23 Find v1 and v2 in the circuit in Fig. 2.87. o + – 3Vo 45 V − + 3 kΩ 5Ω + v1 − + Figure 2.82 For Prob. 2.18. 24 V + v2 9 kΩ − 2.19 Find Vo in the circuit in Fig. 2.83 and the power − dissipated by the controlled source. 4Ω Figure 2.87 For Prob. 2.23. + V − o 6Ω 10 A 2Vo 2.24 Find v1 , v2 , and v3 in the circuit in Fig. 2.88. Figure 2.83 For Prob. 2.19. 14 Ω 2.20 For the circuit in Fig. 2.84, ﬁnd Vo /Vs in terms of + v1 − α, R1 , R2 , R3 , and R4 . If R1 = R2 = R3 = R4 , + + what value of α will produce |Vo /Vs | = 10? 40 V + v2 15 Ω v3 10 Ω − − − Io R1 + Vs + R2 aIo R3 R4 Vo Figure 2.88 For Prob. 2.24. − − 2.25 Calculate v1 , i1 , v2 , and i2 in the circuit of Fig. 2.89. Figure 2.84 For Prob. 2.20. 2.21 For the network in Fig. 2.85, ﬁnd the current, 4Ω 6Ω voltage, and power associated with the 20-k resistor. + v1 − i1 i2 + + + v2 12 V − 3Ω 5 mA 10 kΩ Vo 0.01Vo 5 kΩ 20 kΩ − − Figure 2.85 For Prob. 2.21. Figure 2.89 For Prob. 2.25.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 72. 66 PART 1 DC Circuits2.26 Find i, v, and the power dissipated in the 6- 2.30 Determine i1 , i2 , v1 , and v2 in the ladder network in resistor in Fig. 2.90. Fig. 2.94. Calculate the power dissipated in the 2- resistor. 8Ω i1 8Ω 6Ω 4Ω i2 2Ω i + 9A 6Ω v 4Ω + + − 28 V + − v1 12 Ω 15 Ω 10 Ω v2 13 Ω − − Figure 2.90 For Prob. 2.26. Figure 2.94 For Prob. 2.30.2.27 In the circuit in Fig. 2.91, ﬁnd v, i, and the power 2.31 Calculate Vo and Io in the circuit of Fig. 2.95. absorbed by the 4- resistor. 5Ω 4Ω i 70 Ω 30 Ω Io + + 20 V + − v 10 Ω 6Ω 50 V − + − 20 Ω Vo 5Ω − Figure 2.91 For Prob. 2.27. Figure 2.95 For Prob. 2.31.2.28 Find i1 through i4 in the circuit in Fig. 2.92. 2.32 Find Vo and Io in the circuit of Fig. 2.96. 10 Ω i4 i2 20 Ω 8Ω Io 1Ω 40 Ω i3 i1 30 Ω + 20 A 4V + 3Ω 6Ω Vo − 2Ω − Figure 2.92 For Prob. 2.28. Figure 2.96 For Prob. 2.32.2.29 Obtain v and i in the circuit in Fig. 2.93. 2.33 In the circuit of Fig. 2.97, ﬁnd R if Vo = 4V . i 4S 6S 16 Ω + + 9A v 1S 2S 3S 20 V + − 6Ω R Vo − − Figure 2.93 For Prob. 2.29. Figure 2.97 For Prob. 2.33. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 73. CHAPTER 2 Basic Laws 67 2.34 Find I and Vs in the circuit of Fig. 2.98 if the current 2.37 If Req = 50 in the circuit in Fig. 2.101, ﬁnd R. through the 3- resistor is 2 A. 10 Ω R 2Ω 30 Ω I Req 12 Ω 12 Ω 12 Ω 4Ω 60 Ω + V − s 2A 10 Ω 6Ω 3Ω Figure 2.101 For Prob. 2.37. 2.38 Reduce each of the circuits in Fig. 2.102 to a single resistor at terminals a-b. Figure 2.98 For Prob. 2.34. 5Ω 2.35 Find the equivalent resistance at terminals a-b for a b each of the networks in Fig. 2.99. 8Ω 20 Ω a a 30 Ω R R R R a b (a) R 2Ω 4Ω 5Ω R R R R a b b b 5Ω 3Ω 10 Ω (a) (b) (c) 8Ω 4Ω R a a (b) 3R R R R 2R 3R Figure 2.102 For Prob. 2.38. 2.39 Calculate the equivalent resistance Rab at terminals b b a-b for each of the circuits in Fig. 2.103. (d) (e) 5Ω a Figure 2.99 For Prob. 2.35. 20 Ω 10 Ω 40 Ω 2.36 For the ladder network in Fig. 2.100, ﬁnd I and Req . b (a) I 3Ω 2Ω 1Ω 10 Ω a + 4Ω 6Ω 2Ω 80 Ω 10 V − 60 Ω 20 Ω 30 Ω b Req (b) Figure 2.100 For Prob. 2.36. Figure 2.103 For Prob. 2.39.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 74. 68 PART 1 DC Circuits2.40 Obtain the equivalent resistance at the terminals a-b 2.42 Find the equivalent resistance Rab in the circuit of for each of the circuits in Fig. 2.104. Fig. 2.106. a 10 Ω 60 Ω 30 Ω c b 5Ω 6Ω 20 Ω 10 Ω 8Ω d a b e (a) 5Ω 6Ω 8Ω 9Ωa 20 Ω 3Ω 5Ω f 15 Ω 10 Ω 4Ω 11 Ω 20 Ω 4Ωb Figure 2.106 For Prob. 2.42. (b) Figure 2.104 For Prob. 2.40. Section 2.7 Wye-Delta Transformations2.41 Find Req at terminals a-b for each of the circuits in Fig. 2.105. 2.43 Convert the circuits in Fig. 2.107 from Y to . 70 Ω 10 Ω 10 Ω 30 Ω 20 Ω a a a b b 30 Ω 40 Ω 10 Ω 50 Ω 60 Ω b c c (a) (b) 20 Ω (a) Figure 2.107 For Prob. 2.43. 30 Ω 40 Ω 2.44 Transform the circuits in Fig. 2.108 from to Y. 8Ω a 20 Ω 60 Ω 12 Ω 60 Ω a b a b 6Ω 10 Ω 50 Ω 4Ω 12 Ω 12 Ω 30 Ω 10 Ω b 70 Ω 80 Ω c c (b) (a) (b) Figure 2.105 For Prob. 2.41. Figure 2.108 For Prob. 2.44. v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 75. CHAPTER 2 Basic Laws 69 2.45 What value of R in the circuit of Fig. 2.109 would a cause the current source to deliver 800 mW to the resistors? R R R b 30 mA (b) R R Figure 2.111 For Prob. 2.47. ∗ 2.48 Obtain the equivalent resistance Rab in each of the circuits of Fig. 2.112. In (b), all resistors have a Figure 2.109 For Prob. 2.45. value of 30 . 2.46 Obtain the equivalent resistance at the terminals a-b for each of the circuits in Fig. 2.110. 30 Ω 40 Ω a 20 Ω a 10 Ω 10 Ω 20 Ω 10 Ω 30 Ω 80 Ω 60 Ω 50 Ω 10 Ω 20 Ω b b (a) (a) a 30 Ω 30 Ω 25 Ω 10 Ω 20 Ω a b 5Ω 15 Ω (b) b (b) Figure 2.112 For Prob. 2.48. 2.49 Calculate Io in the circuit of Fig. 2.113. Figure 2.110 For Prob. 2.46.∗ 2.47 Find the equivalent resistance Rab in each of the Io circuits of Fig. 2.111. Each resistor is 100 . a 20 Ω 60 Ω 40 Ω 24 V + − 10 Ω 50 Ω 20 Ω b (a) Figure 2.113 For Prob. 2.49. ∗ An asterisk indicates a challenging problem.v v | | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents
- 76. 70 PART 1 DC Circuits 2.50 Determine V in the circuit of Fig. 2.114. 30 W 40 W 50 W I 30 Ω + 100 V − 16 Ω 15 Ω 10 Ω + 100 V + − V 35 Ω 12 Ω 20 Ω Figure 2.117 For Prob. 2.53. − 2.54 If the three bulbs of Prob. 2.53 are connected in parallel to the 100-V battery, calculate the current Figure 2.114 For Prob. 2.50. through each bulb. 2.55 As a design engineer, you are asked to design a∗ 2.51 Find Req and I in the circuit of Fig. 2.115. lighting system consisting of a 70-W power supply and two lightbulbs as shown in Fig. 2.118. You must select the two bulbs from the following three available bulbs. I 4Ω 2Ω R1 = 80 , cost = $0.60 (standard size) R2 = 90 , cost = $0.90 (standard size) R3 = 100 , cost = $0.75 (nonstandard size) 6Ω 1Ω The system should be designed for minimum cost 12 Ω such that I = 1.2 A ± 5 percent. I + 8Ω 2Ω 20 V − + 4Ω 70 W Power Rx Ry Supply 10 Ω

Full NameComment goes here.abduldadabhai12342 months agoNavotas Polytechnic Collegeat Navotas Polytechnic College 1 year agoVicente Celani, Profesor Tiempo Completo at ITBA - Instituto Tecnológico Buenos Aires 1 year ago