fundamentals of electric circuits

F51-pref.qxd

3/17/00

10:11 AM

Page v

PREFACE
Features
In spite of the numerous textbooks on circuit analysis
available in the market, students often find the course
difficult to learn. The main objective of this book is
to present circuit analysis in a manner that is clearer,
more interesting, and easier to understand than earlier
texts. This objective is achieved in the following
ways:
• A course in circuit analysis is perhaps the first
exposure students have to electrical engineering.
We have included several features to help students feel at home with the subject. Each chapter
opens with either a historical profile of some
electrical engineering pioneers to be mentioned in
the chapter or a career discussion on a subdiscipline of electrical engineering. An introduction
links the chapter with the previous chapters and
states the chapter’s objectives. The chapter ends
with a summary of the key points and formulas.
• All principles are presented in a lucid, logical,
step-by-step manner. We try to avoid wordiness
and superfluous detail that could hide concepts
and impede understanding the material.

• Important formulas are boxed as a means of
helping students sort what is essential from what
is not; and to ensure that students clearly get the
gist of the matter, key terms are defined and
highlighted.
• Marginal notes are used as a pedagogical aid. They
serve multiple uses—hints, cross-references, more
exposition, warnings, reminders, common mistakes, and problem-solving insights.
• Thoroughly worked examples are liberally given at
the end of every section. The examples are regarded as part of the text and are explained clearly, without asking the reader to fill in missing steps.
Thoroughly worked examples give students a good
understanding of the solution and the confidence to
solve problems themselves. Some of the problems
are solved in two or three ways to facilitate an
understanding and comparison of different
approaches.
• To give students practice opportunity, each illustrative example is immediately followed by a
practice problem with the answer. The students can
follow the example step-by-step to solve the practice problem without flipping pages or searching
the end of the book for answers. The practice prob-

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
section.
• In recognition of ABET’s requirement on integrating computer tools, the use of PSpice is encouraged
in a student-friendly manner. Since the Windows
version of PSpice is becoming popular, it is used
instead of the MS-DOS version. PSpice is covered
early so that students can use it throughout the text.

Appendix D serves as a tutorial on PSpice for
Windows.
• The operational amplifier (op amp) as a basic element is introduced early in the text.
• To ease the transition between the circuit course
and signals/systems courses, Fourier and Laplace
transforms are covered lucidly and thoroughly.
• The last section in each chapter is devoted to applications of the concepts covered in the chapter. Each
chapter has at least one or two practical problems or
devices. This helps students apply the concepts to
real-life situations.
• Ten multiple-choice review questions are provided
at the end of each chapter, with answers. These are
intended to cover the little “tricks” that the examples and end-of-chapter problems may not cover.
They serve as a self-test device and help students
determine how well they have mastered the chapter.

Organization
This book was written for a two-semester or three-semester course in linear circuit analysis. The book may
also be used for a one-semester course by a proper selection of chapters and sections. It is broadly divided into
three parts.
• Part 1, consisting of Chapters 1 to 8, is devoted to
dc circuits. It covers the fundamental laws and theorems, circuit techniques, passive and active elements.
• Part 2, consisting of Chapters 9 to 14, deals with ac
circuits. It introduces phasors, sinusoidal steadystate analysis, ac power, rms values, three-phase
systems, and frequency response.
• Part 3, consisting of Chapters 15 to 18, is devoted
to advanced techniques for network analysis.
It provides a solid introduction to the Laplace
transform, Fourier series, the Fourier transform,
and two-port network analysis.

The material in three parts is more than sufficient for a two-semester course, so that the instructor

|

v

v

v

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


F51-pref.qxd

3/17/00

10:11 AM

Page vi

vi

PREFACE

must select which chapters/sections to cover. Sections
marked with the dagger sign (†) may be skipped,
explained briefly, or assigned as homework. They can
be omitted without loss of continuity. Each chapter has
plenty of problems, grouped according to the sections
of the related material, and so diverse that the instructor can choose some as examples and assign some as
homework. More difficult problems are marked with a
star (*). Comprehensive problems appear last; they are
mostly applications problems that require multiple
skills from that particular chapter.
The book is as self-contained as possible. At the
end of the book are some appendixes that review
solutions of linear equations, complex numbers, mathematical formulas, a tutorial on PSpice for Windows,
and answers to odd-numbered problems. Answers to
all the problems are in the solutions manual, which is
available from the publisher.

Prerequisites
As with most introductory circuit courses, the main
prerequisites are physics and calculus. Although familiarity with complex numbers is helpful in the later part
of the book, it is not required.

Supplements

|

v

v

Solutions Manual—an Instructor’s Solutions Manual is
available to instructors who adopt the text. It contains
complete solutions to all the end-of-chapter problems.
Transparency Masters—over 200 important figures
are available as transparency masters for use as overheads.
Student CD-ROM—100 circuit files from the book are
presented as Electronics Workbench (EWB) files; 15–20
of these files are accessible using the free demo of Electronics Workbench. The students are able to experiment
with the files. For those who wish to fully unlock all 100
circuit files, EWB’s full version may be purchased from
Interactive Image Technologies for approximately
$79.00. The CD-ROM also contains a selection of problem-solving, analysis and design tutorials, designed to
further support important concepts in the text.
Problem-Solving Workbook—a paperback workbook is for sale to students who wish to practice their
problem solving techniques. The workbook contains a
discussion of problem solving strategies and 150 additional problems with complete solutions provided.
Online Learning Center (OLC)—the Web site for
the book will serve as an online learning center for students as a useful resource for instructors. The OLC

|

e-Text Main Menu

will provide access to:
300 test questions—for instructors only
Downloadable figures for overhead
presentations—for instructors only
Solutions manual—for instructors only
Web links to useful sites

Sample pages from the Problem-Solving
Workbook
PageOut Lite—a service provided to adopters
who want to create their own Web site. In
just a few minutes, instructors can change
the course syllabus into a Web site using
PageOut Lite.
The URL for the web site is www.mhhe.com.alexander.
Although the textbook is meant to be self-explanatory
and act as a tutor for the student, the personal contact
involved in teaching is not to be forgotten. The book
and supplements are intended to supply the instructor
with all the pedagogical tools necessary to effectively
present the material.

ACKNOWLEDGMENTS
We wish to take the opportunity to thank the staff of
McGraw-Hill for their commitment and hard
work: Lynn Cox, Senior Editor; Scott Isenberg,
Senior Sponsoring Editor; Kelley Butcher, Senior
Developmental Editor; Betsy Jones, Executive
Editor; Catherine Fields, Sponsoring Editor;
Kimberly Hooker, Project Manager; and Michelle
Flomenhoft, Editorial Assistant. They got numerous
reviews, kept the book on track, and helped in many
ways. We really appreciate their inputs. We are
greatly in debt to Richard Mickey for taking the pain
ofchecking and correcting the entire manuscript. We
wish to record our thanks to Steven Durbin at Florida
State University and Daniel Moore at Rose Hulman

Institute of Technology for serving as accuracy
checkers of examples, practice problems, and endof-chapter problems. We also wish to thank the following reviewers for their constructive criticisms
and helpful comments.
Promod Vohra, Northern Illinois University
Moe Wasserman, Boston University
Robert J. Krueger, University of Wisconsin
Milwaukee
John O’Malley, University of Florida

| Textbook Table of Contents |

Problem Solving Workbook Contents


F51-pref.qxd

3/17/00

10:11 AM

Page vii

PREFACE

vii

|

v

v

Aniruddha Datta, Texas A&M University
John Bay, Virginia Tech
Wilhelm Eggimann, Worcester Polytechnic
Institute
A. B. Bonds, Vanderbilt University
Tommy Williamson, University of Dayton
Cynthia Finelli, Kettering University
John A. Fleming, Texas A&M University
Roger Conant, University of Illinois
at Chicago
Daniel J. Moore, Rose-Hulman Institute of
Technology
Ralph A. Kinney, Louisiana State University
Cecilia Townsend, North Carolina State
University
Charles B. Smith, University of Mississippi
H. Roland Zapp, Michigan State University
Stephen M. Phillips, Case Western University
Robin N. Strickland, University of Arizona
David N. Cowling, Louisiana Tech University
Jean-Pierre R. Bayard, California State
University

|

e-Text Main Menu

Jack C. Lee, University of Texas at Austin

E. L. Gerber, Drexel University
The first author wishes to express his appreciation to his department chair, Dr. Dennis Irwin, for his
outstanding support. In addition, he is extremely grateful to Suzanne Vazzano for her help with the solutions
manual.
The second author is indebted to Dr. Cynthia
Hirtzel, the former dean of the college of engineering
at Temple University, and Drs.. Brian Butz, Richard
Klafter, and John Helferty, his departmental chairpersons at different periods, for their encouragement while
working on the manuscript. The secretarial support
provided by Michelle Ayers and Carol Dahlberg is
gratefully appreciated. Special thanks are due to Ann
Sadiku, Mario Valenti, Raymond Garcia, Leke and
Tolu Efuwape, and Ope Ola for helping in various
ways. Finally, we owe the greatest debt to our wives,
Paulette and Chris, without whose constant support and
cooperation this project would have been impossible.
Please address comments and corrections to the
publisher.

| Textbook Table of Contents |

C. K. Alexander and M. N. O. Sadiku

Problem Solving Workbook Contents


F51-pref.qxd

3/17/00

10:11 AM

Page ix

A NOTE TO THE STUDENT
This may be your first course in electrical engineering. Although electrical engineering is an exciting and
challenging discipline, the course may intimidate you.
This book was written to prevent that. A good textbook
and a good professor are an advantage—but you are
the one who does the learning. If you keep the following ideas in mind, you will do very well in this course.
• This course is the foundation on which most
other courses in the electrical engineering curriculum rest. For this reason, put in as much
effort as you can. Study the course regularly.
• Problem solving is an essential part of the learning process. Solve as many problems as you can.
Begin by solving the practice problem following
each example, and then proceed to the end-ofchapter problems. The best way to learn is to
solve a lot of problems. An asterisk in front of a
problem indicates a challenging problem.
• Spice, a computer circuit analysis program, is
used throughout the textbook. PSpice, the personal computer version of Spice, is the popular
standard circuit analysis program at most uni-

versities. PSpice for Windows is described in
Appendix D. Make an effort to learn PSpice,
because you can check any circuit problem with
PSpice and be sure you are handing in a correct
problem solution.
• Each chapter ends with a section on how the
material covered in the chapter can be applied to
real-life situations. The concepts in this section

may be new and advanced to you. No doubt, you
will learn more of the details in other courses.
We are mainly interested in gaining a general
familiarity with these ideas.
• Attempt the review questions at the end of each
chapter. They will help you discover some
“tricks” not revealed in class or in the textbook.
A short review on finding determinants is covered in Appendix A, complex numbers in Appendix B,
and mathematical formulas in Appendix C. Answers to
odd-numbered problems are given in Appendix E.
Have fun!
C.K.A. and M.N.O.S.

|

v

v

ix

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents

f51-cont.qxd

3/16/00

4:22 PM

Page xi

Contents
Preface

3.7
3.8
†3.9
3.10

v

Acknowledgments

vi

A Note to the Student

PART 1 DC CIRCUITS
Chapter 1
1.1
1.2
1.3
1.4

1.5
1.6
†1.7

Review Questions
107
Problems
109
Comprehensive Problems

1

Basic Concepts

3

Introduction
4
Systems of Units
4
Charge and Current
6
Voltage
9
Power and Energy
10
Circuit Elements
13
Applications
15

1.7.1
1.7.2

†1.8

ix

Chapter 4

Chapter 2

18

2.1
2.2
†2.3
2.4
2.5
2.6
†2.7
†2.8

25

Basic Laws

4.11 Summary
Review Questions
153
Problems

154
Comprehensive Problems

41
42

Chapter 5

2.9 Summary

72

Chapter 3

Methods of Analysis

75

Introduction
76
Nodal Analysis
76
Nodal Analysis with Voltage Sources
82
Mesh Analysis
87
Mesh Analysis with Current Sources
92
Nodal and Mesh Analyses by Inspection
95

153
162

Operational Amplifiers

165

Introduction
166
Operational Amplifiers
166
Ideal Op Amp
170
Inverting Amplifier
171
Noninverting Amplifier
174
Summing Amplifier
176
Difference Amplifier
177
Cascaded Op Amp Circuits
181
Op Amp Circuit Analysis
with PSpice
183
†5.10 Applications
185

60

3.1
3.2
3.3
3.4
3.5
†3.6

119

5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9

Lighting Systems
Design of DC Meters

Review Questions
61
Problems
63
Comprehensive Problems

Circuit Theorems

4.10.1 Source Modeling
4.10.2 Resistance Measurement

27

Introduction
28
Ohm’s Laws
28
Nodes, Branches, and Loops
33
Kirchhoff’s Laws
35
Series Resistors and Voltage Division
Parallel Resistors and Current Division
Wye-Delta Transformations
50
Applications
54
2.8.1
2.8.2

117

Introduction
120
Linearity Property
120

Superposition
122
Source Transformation
127
Thevenin’s Theorem
131
Norton’s Theorem
137
Derivations of Thevenin’s and Norton’s
Theorems
140
4.8 Maximum Power Transfer
142
4.9 Verifying Circuit Theorems
with PSpice
144
†4.10 Applications
147

1.9 Summary

Review Questions
22
Problems
23
Comprehensive Problems

102

4.1

4.2
4.3
4.4
4.5
4.6
†4.7

TV Picture Tube
Electricity Bills

Problem Solving
21

Nodal Versus Mesh Analysis
99
Circuit Analysis with PSpice
100
Applications: DC Transistor Circuits
Summary
107

5.10.1 Digital-to Analog Converter
5.10.2 Instrumentation Amplifiers

5.11 Summary
Review Questions
190
Problems
191
Comprehensive Problems

188
200

|

v

v

xi

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


xii

CONTENTS
Chapter 6
6.1
6.2
6.3
6.4
6.5

†6.6

Capacitors and Inductors

Introduction
202
Capacitors
202
Series and Parallel Capacitors
Inductors
211
Series and Parallel Inductors
Applications
219
6.6.1
6.6.2
6.6.3

6.7 Summary

Chapter 7

216

237

Delay Circuits
Photoflash Unit
Relay Circuits
Automobile Ignition Circuit

7.10 Summary
Review Questions
283
Problems
284
Comprehensive Problems

Chapter 8

282

Second-Order Circuits

295

Introduction
296
Finding Initial and Final Values
296
The Source-Free Series RLC Circuit
301
The Source-Free Parallel RLC Circuit
308
Step Response of a Series RLC
Circuit
314
8.6 Step Response of a Parallel RLC
Circuit
319

8.7 General Second-Order Circuits
322
8.8 Second-Order Op Amp Circuits
327
8.9 PSpice Analysis of RLC Circuits
330
†8.10 Duality
332
†8.11 Applications
336

v

v

8.11.1 Automobile Ignition System
8.11.2 Smoothing Circuits

|

351

Chapter 9

9.7
†9.8

9.9

e-Text Main Menu

Sinusoids and Phasors

353

Introduction
354
Sinusoids
355
Phasors
359
Phasor Relationships for Circuit
Elements
367
Impedance and Admittance
369
Kirchhoff’s Laws in the Frequency
Domain
372
Impedance Combinations
373
Applications
379
9.8.1
9.8.2

Phase-Shifters
AC Bridges

Summary

Review Questions
385
Problems
385
Comprehensive Problems

384
392

Chapter 10 Sinusoidal Steady-State Analysis
10.1
10.2
10.3
10.4
10.5
10.6

293

8.1
8.2
8.3
8.4
8.5

|

PART 2 AC CIRCUITS

9.5
9.6

235

Introduction
238
The Source-free RC Circuit
238
The Source-free RL Circuit
243
Singularity Functions
249
Step Response of an RC Circuit
257
Step Response of an RL Circuit
263
First-order Op Amp Circuits
268
Transient Analysis with PSpice
273
Applications
276
7.9.1
7.9.2
7.9.3
7.9.4

340
350

9.1
9.2
9.3
9.4

225

First-Order Circuits

8.12 Summary
Review Questions
340
Problems
341
Comprehensive Problems

208

Integrator
Differentiator
Analog Computer

Review Questions
226
Problems
227
Comprehensive Problems

7.1

7.2
7.3
7.4
7.5
7.6
†7.7
7.8
†7.9

201

10.7
10.8
†10.9

Introduction
394
Nodal Analysis
394
Mesh Analysis
397
Superposition Theorem
400
Source Transformation
404
Thevenin and Norton Equivalent
Circuits
406
Op Amp AC Circuits
411

AC Analysis Using PSpice
413
Applications
416
10.9.1
10.9.2

Capacitance Multiplier
Oscillators

10.10 Summary
Review Questions
Problems
422

420

421

Chapter 11 AC Power Analysis
11.1
11.2
11.3
11.4
11.5
11.6
†11.7

| Textbook Table of Contents |

393

433

Introduction
434
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value
443
Apparent Power and Power Factor
Complex Power
449
Conservation of AC Power
453

434
440
447

Problem Solving Workbook Contents


CONTENTS
11.8
†11.9

xiii
Power Factor Correction
Applications

459
11.9.1
11.9.2

11.10 Summary
Review Questions
465
Problems
466
Comprehensive Problems

Power Measurement
Electricity Consumption Cost

464
474

12.11 Summary
Review Questions
517
Problems
518
Comprehensive Problems

†13.9

13.10 Summary
Review Questions
570
Problems

571
Comprehensive Problems

v

v

|

|

†14.11

14.11.1
14.11.2
14.11.3

619
Magnitude Scaling
Frequency Scaling
Magnitude and Frequency Scaling

622
626

Radio Receiver
Touch-Tone Telephone
Crossover Network

631

640

527

PART 3 ADVANCED CIRCUIT ANALYSIS

643

Chapter 15 The Laplace Transform

645

Transformer as an Isolation Device
Transformer as a Matching Device
Power Distribution

569
582

583

Introduction
584
Transfer Function
584
The Decibel Scale
588

e-Text Main Menu

First-Order Lowpass Filter
First-Order Highpass Filter
Bandpass Filter
Bandreject (or Notch) Filter

PSpice
Applications

14.12 Summary

525

613

14.10 Frequency Response Using

Review Questions
633
Problems
633
Comprehensive Problems

Chapter 14 Frequency Response
14.1
14.2
†14.3

Scaling
14.9.1
14.9.2

14.9.3

516

Introduction
528
Mutual Inductance
528
Energy in a Coupled Circuit
535
Linear Transformers
539
Ideal Transformers
545
Ideal Autotransformers
552
Three-Phase Transformers
556
PSpice Analysis of Magnetically Coupled
Circuits
559
Applications
563
13.9.1
13.9.2
13.9.3

†14.9

Lowpass Filter

Highpass Filter
Bandpass Filter
Bandstop Filter

Active Filters
14.8.1
14.8.2
14.8.3
14.8.4

Three-Phase Power Measurement
Residential Wiring

Chapter 13 Magnetically Coupled Circuits
13.1
13.2
13.3
13.4
13.5
13.6
†13.7
13.8

14.8

477

Introduction
478
Balanced Three-Phase Voltages

479
Balanced Wye-Wye Connection
482
Balanced Wye-Delta Connection
486
Balanced Delta-Delta Connection
488
Balanced Delta-Wye Connection
490
Power in a Balanced System
494
Unbalanced Three-Phase Systems
500
PSpice for Three-Phase Circuits
504
Applications
508
12.10.1
12.10.2

Bode Plots
589
Series Resonance
600
Parallel Resonance
605
Passive Filters
608
14.7.1
14.7.2

14.7.3
14.7.4

Chapter 12 Three-Phase Circuits
12.1
12.2
12.3
12.4
12.5
12.6
12.7
†12.8
12.9
†12.10

14.4
14.5
14.6
14.7

457

| Textbook Table of Contents |

15.1
15.2
15.3
15.4

Introduction

646
Definition of the Laplace
Transform
646
Properties of the Laplace
Transform
649
The Inverse Laplace Transform
15.4.1
15.4.2
15.4.3

15.5
15.6
15.7
†15.8
†15.9

659

Simple Poles
Repeated Poles
Complex Poles

Applicaton to Circuits
666
Transfer Functions
672
The Convolution Integral
677

Application to Integrodifferential
Equations
685
Applications
687
15.9.1
15.9.2

15.10 Summary

Network Stability
Network Synthesis

694

Problem Solving Workbook Contents


xiv

CONTENTS

Review Questions
696
Problems
696
Comprehensive Problems

17.8
705

Chapter 16 The Fourier Series
16.1
16.2
16.3

16.4
16.5
16.6
16.7

†16.8

16.8.1
16.8.2

16.9

Even Symmetry
Odd Symmetry
Half-Wave Symmetry

Summary

Review Questions
751
Problems
751
Comprehensive Problems

746

†17.7

v

v

|

|

18.10 Summary

833
844

Appendix A Solution of Simultaneous Equations Using
758

Cramer’s Rule

845

Appendix B Complex Numbers

759

Introduction
760

Definition of the Fourier Transform
Properties of the Fourier Transform
Circuit Applications
779
Parseval’s Theorem
782
Comparing the Fourier and Laplace
Transforms
784
Applications
785
17.7.1
17.7.2

Transistor Circuits
Ladder Network Synthesis

Review Questions
834
Problems
835
Comprehensive Problems

749

Chapter 17 Fourier Transform
17.1
17.2
17.3
17.4

17.5
17.6

18.9.1
18.9.2

Spectrum Analyzers
Filters

795

Introduction
796
Impedance Parameters
796
Admittance Parameters
801
Hybrid Parameters
804
Transmission Parameters
809
Relationships between Parameters
814
Interconnection of Networks
817
Computing Two-Port Parameters Using
PSpice
823
Applications
826

†18.9

Discrete Fourier Transform
Fast Fourier Transform

Applications

794

Chapter 18 Two-Port Networks
18.1
18.2
18.3
18.4
18.5
†18.6
18.7
18.8

Circuit Applicatons
727
Average Power and RMS Values
730
Exponential Fourier Series
734
Fourier Analysis with PSpice
740
16.7.1
16.7.2

789

707

Introduction
708
Trigonometric Fourier Series
708
Symmetry Considerations
717
16.3.1
16.3.2
16.3.3

Summary

Review Questions
790
Problems
790
Comprehensive Problems

851

Appendix C Mathematical Formulas
760
766

Appendix D PSpice for Windows

859
865

Appendix E Answers to Odd-Numbered Problems
Selected Bibliography
Index

893

929

933

Amplitude Modulation
Sampling

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


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

|

v

v

1

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


C H A P T E R
BASIC CONCEPTS

1

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.

|

v

v

3

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


4

PART 1

DC Circuits

1.1 INTRODUCTION
Electric circuit theory and electromagnetic theory are the two fundamental theories upon which all branches of electrical engineering are built.
Many branches of electrical engineering, such as power, electric machines, control, electronics, communications, and instrumentation, are
based on electric circuit theory. Therefore, the basic electric circuit theory 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
+

−

Battery

Figure 1.1

Lamp

A simple electric circuit.

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
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.
Electric circuits are used in numerous electrical systems to accomplish different tasks. Our objective in this book is not the study of various
uses and applications of circuits. Rather our major concern is the analysis 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

|

v

v

As electrical engineers, we deal with measurable quantities. Our measurement, 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 measurement language is the International System of Units (SI), adopted by the
General Conference on Weights and Measures in 1960. In this system,

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents

CHAPTER 1

L1
0.445 mH

C3

R1 47
Oscillator R2
C
10 k
B

8
7

U1
SBL-1
Mixer

C1
2200 pF
C2
2200 pF

3, 4

R3

Q1
10 k
2N2222A

E
2, 5, 6

R6
100 k

R5
100 k

U2B
1 ⁄ 2 TL072
C10 5
+
7
1.0 mF
16 V 6 −

+

+

R7
1M
C12
0.0033

Figure 1.2

R9
15 k
R8
15 k
C13

C11
100 mF
16 V
C15
U2A
0.47
1 ⁄2 TL072 16 V +
3
8
C14 +
1
0.0022 −
4
2

+

0.1

600,000,000 mm

TABLE 1.1

600,000 m

600 km

The six basic SI units.
Basic unit

Symbol

Length
Mass
Time
Electric current
Thermodynamic temperature
Luminous intensity

v

v

Quantity

|

5

L2
22.7 mH
(see text)

C7
532

R10
10 k
GAIN

3

+

+

C16
100 mF
16 V

+
12-V dc
Supply
−

6

−
2

5
4 R12
10

U3
C18
LM386N
Audio power amp 0.1

Audio
+
Output
C17
100 mF
16 V

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 throughout this text.
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
1.2 shows the SI prefixes and their symbols. For example, the following
are expressions of the same distance in meters (m):

|

C6

C4
910

to
U1, Pin 8

R11
47
C8
0.1

C9
1.0 mF
16 V

Y1
7 MHz

C5
910

R4
220

L3
1 mH

5

0.1
1

Antenna

Basic Concepts

meter
kilogram
second
ampere
kelvin
candela

m
kg
s
A
K
cd

e-Text Main Menu

| Textbook Table of Contents |

TABLE 1.2

The SI prefixes.

Multiplier

Prefix

Symbol

1018
1015
1012
109
106
103
102
10
10−1
10−2
10−3
10−6
10−9
10−12
10−15
10−18

exa
peta
tera
giga
mega
kilo
hecto
deka
deci
centi
milli

micro
nano
pico
femto
atto

E
P
T
G
M
k
h
da
d
c
m
µ
n
p
f
a

Problem Solving Workbook Contents


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

−
−

I

−
−

−

+

Battery

Figure 1.3 Electric current due to flow
of electronic charge in a conductor.
A convention is a standard way of describing
something so that others in the profession can
understand what we mean. We will be using IEEE
conventions throughout this book.

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
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
to move; positive charges move in one direction while negative charges
move in the opposite direction. This motion of charges creates electric
current. It is conventional to take the current flow as the movement of
positive charges, that is, opposite to the flow of negative charges, as Fig.

1.3 illustrates. This convention was introduced by Benjamin Franklin
(1706–1790), the American scientist and inventor. Although we now
know that current in metallic conductors is due to negatively charged
electrons, we will follow the universally accepted convention that current
is the net flow of positive charges. Thus,

|

v

v

1 However,

|

e-Text Main Menu

a large power supply capacitor can store up to 0.5 C of charge.

| Textbook Table of Contents |

Problem Solving Workbook Contents


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
i=

dq
dt

(1.1)

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.
If the current does not change with time, but remains constant, we
call it a direct current (dc).

0

t

(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 common 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.

|

v

v

Such current is used in your household, to run the air conditioner, refrigerator, washing machine, and other electric appliances. Figure 1.4 shows
direct current and alternating current; these are the two most common
types of current. We will consider other types later in the book.
Once we define current as the movement of charge, we expect current to have an associated direction of flow. As mentioned earlier, the
direction of current flow is conventionally taken as the direction of positive
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.
1.5(b) is the same as a current of +5 A flowing in the opposite direction.

|

e-Text Main Menu

| Textbook Table of Contents |

−5 A

5A

(a)

(b)

Figure 1.5 Conventional current flow:
(a) positive current flow, (b) negative current
flow.

Problem Solving Workbook Contents


8

PART 1

DC Circuits

E 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 C

PRACTICE 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. Calculate the current at t = 0.5 s.
Solution:
d

dq
= (5t sin 4π t) mC/s = (5 sin 4π t + 20π t cos 4π t) mA
dt
dt
At t = 0.5,
i=

i = 5 sin 2π + 10π cos 2π = 0 + 10π = 31.42 mA

PRACTICE 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:
q=

2

2

i dt =

=

(3t 2 − t) dt

1

t=1

t3 −

t2
2

2

= (8 − 2) − 1 −
1

1
2

= 5.5 C

PRACTICE PROBLEM 1.3
The current flowing through an element is
i=

2 A,
2t 2 A,

0t >1

Calculate the charge entering the element from t = 0 to t = 2 s.

|

v

v

Answer: 6.667 C.

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


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,
vab =

dw
dq

(1.3)

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,
+

Voltage (or potential difference) is the energy required to move
a unit charge through an element, measured in volts (V).

vab
−

Figure 1.6 shows the voltage across an element (represented by a
rectangular block) connected to points a and b. The plus (+) and minus
(−) 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
point b is vab . It follows logically that in general
vab = −vba

v

v

|

e-Text Main Menu

| Textbook Table of Contents |

b

Figure 1.6

Polarity
of voltage vab .

+

(1.4)

For example, in Fig. 1.7, we have two representations of the same voltage. In Fig. 1.7(a), point a is +9 V above point b; in Fig. 1.7(b), point b is
−9 V above point a. We may say that in Fig. 1.7(a), there is a 9-V voltage
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
b to a.
Current and voltage are the two basic variables in electric circuits.
The common term signal is used for an electric quantity such as a current
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 communications and other disciplines. Like electric current, a constant voltage
is called a dc voltage and is represented by V, whereas a sinusoidally
time-varying voltage is called an ac voltage and is represented by v. A
dc voltage is commonly produced by a battery; ac voltage is produced by
an electric generator.

|

a

a

(a)

a

−9 V

9V
−

−

+
b

b
(b)

Figure 1.7

Two equivalent
representations of the same
voltage vab : (a) point a is 9 V
above point b, (b) point b is
−9 V above point a.

Keep in mind that electric current is always
through an element and that electric voltage is always across the element or between two points.

Problem Solving Workbook Contents


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
p=

dw
dt

(1.5)

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=
(1.6)
=
·
= vi
dt
dq dt
or
p = vi

i

i

+

+

v

v

−

−

p = +vi

p = −vi

(a)

(b)

Figure 1.8

|

v

v

Reference
polarities for power using
the passive sign convention: (a) absorbing power,
(b) supplying power.

|

e-Text Main Menu

(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
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?
Current direction and voltage polarity play a major role in determining 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 voltage polarity and current direction must conform with those shown in Fig.
1.8(a) in order for the power to have a positive sign. This is known as
the passive sign convention. By the passive sign convention, current enters through the positive polarity of the voltage. In this case, p = +vi or
vi > 0 implies that the element is absorbing power. However, if p = −vi
or vi < 0, as in Fig. 1.8(b), the element is releasing or 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.

| Textbook Table of Contents |

Problem Solving Workbook Contents

CHAPTER 1

Basic Concepts

Unless otherwise stated, we will follow the passive sign convention
throughout this text. For example, the element in both circuits of Fig. 1.9
has an absorbing power of +12 W because a positive current enters the
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,

11

When the voltage and current directions conform to Fig. 1.8(b), we have the active sign convention and p = +vi.

Power absorbed = −Power supplied
3A

3A

3A

3A

+

−

+

−

4V

4V

4V

4V

−

+

−

+

(a)

(a)

(b)

Figure 1.9

Two cases of an
element with an absorbing
power of 12 W:

(a) p = 4 × 3 = 12 W,
(b) p = 4 × 3 = 12 W.

(b)

Figure 1.10

Two cases of
an element with a supplying
power of 12 W:
(a) 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

w=
t0

t

p dt =

vi dt

(1.9)

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

|

v

v

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.

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


12

PART 1

DC Circuits

Solution:
The total charge is
q = i t = 2 × 10 = 20 C
The voltage drop is
v=

w
2.3 × 103
=
= 115 V
q
20

PRACTICE 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
v=3

di
= 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 kW

PRACTICE 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

|

v

v

Answer: (a) 17.27 W, (b) 29.7 W.

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


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.

|

v

v

In other words, an ideal independent voltage source delivers to the circuit
whatever current is necessary to maintain its terminal voltage. Physical
sources such as batteries and generators may be regarded as approximations to ideal voltage sources. Figure 1.11 shows the symbols for independent voltage sources. Notice that both symbols in Fig. 1.11(a) and (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
independent current source is an active element that provides a specified
current completely independent of the voltage across the source. That is,
the current source delivers to the circuit whatever voltage is necessary to

|

e-Text Main Menu

| Textbook Table of Contents |

v

+
V
−

+
−

(a)

(b)

Figure 1.11

Symbols for
independent voltage sources:
(a) used for constant or
time-varying voltage, (b) used for
constant voltage (dc).

Problem Solving Workbook Contents


14

PART 1

maintain the designated current. The symbol for an independent current
source is displayed in Fig. 1.12, where the arrow indicates the direction
of current i.

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.

+
−

v

DC Circuits

Dependent sources are usually designated by diamond-shaped symbols,
as shown in Fig. 1.13. Since the control of the dependent source is achieved 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).
3. A voltage-controlled current source (VCCS).
4. A current-controlled current source (CCCS).

i

(a)

(b)

Figure 1.13

Symbols for:
(a) dependent voltage source,
(b) dependent current source.

B

A
i
+
5V
−

+
−

C

10i

Figure 1.14

The source on the right-hand
side is a current-controlled voltage source.

Dependent sources are useful in modeling elements such as transistors,
operational amplifiers and integrated circuits. An example of a currentcontrolled voltage source is shown on the right-hand side of Fig. 1.14,
where the voltage 10i of the voltage source depends on the current i
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.
It should be noted that an ideal voltage source (dependent or independent) will produce any current required to ensure that the terminal
voltage is as stated, whereas an ideal current source will produce the
necessary voltage to ensure the stated current flow. Thus an ideal source
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 but
not the current supplied or drawn by it. By the same token, we know the
current supplied by a current source but not the voltage across it.

E X A M P L E 1 . 7
p2

Calculate the power supplied or absorbed by each element in Fig. 1.15.
−

I=5A
+

12 V
20 V

+
−

p3

p1

6A
+
8V
−

p4

0.2 I

Solution:
We apply the sign convention for power shown in Figs. 1.8 and 1.9. For
p1 , the 5-A current is out of the positive terminal (or into the negative
terminal); hence,
p1 = 20(−5) = −100 W

|

For Example 1.7.

v

v

Figure 1.15

|

e-Text Main Menu

Supplied power

For p2 and p3 , the current flows into the positive terminal of the element
in each case.

| Textbook Table of Contents |

Problem Solving Workbook Contents


CHAPTER 1
p2 = 12(5) = 60 W
p3 = 8(6) = 48 W

Basic Concepts

15

Absorbed power
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
in Fig. 1.16.

I=5A
3A

p3

+
−

p1

+
−

0.6I

p4

3V

−

p2
+
5V
−

Figure 1.16

†

2V
+−

+

Answer: p1 = −40 W, p2 = 16 W, p3 = 9 W, p4 = 15 W.

8A

For Practice Prob. 1.7.

APPLICATIONS 2

1.7

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 cathode-ray tube (CRT) located in the TV receiver.3 The CRT is depicted in
2 The

|

v

v

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.

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents


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.

Electron gun

Horizontal
deflection
plates

Bright spot on
fluorescent screen
Vertical
deflection
plates

Electron
trajectory

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
If the number of electrons is n, then q = ne and
dn
dq
=e
= (−1.6 × 10−19 )(1015 ) = −1.6 × 10−4 A
i=
dt
dt
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
the CRT for the case when the vertical deflection plates carry no charge.
The beam power is
4
p
p = Vo i
or
Vo = =
= 25,000 V
i

1.6 × 10−4
Thus the required voltage is 25 kV.

i
q
Vo

Figure 1.18 A simplified diagram of the
cathode-ray tube; for Example 1.8.

PRACTICE PROBLEM 1.8

|

v

v

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.

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents

CHAPTER 1

Basic Concepts

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

Water heater
Freezer
Lighting
Dishwasher

Electric iron
TV
Toaster

Appliance

kWh consumed

500
100
100
35
15
10
4

Washing machine
Stove
Dryer
Microwave oven
Personal computer
Radio
Clock

120
100
80
25
12
8

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

|

v

v

Total Charge = $54.00
$54
Average cost =
= 13.5 cents/kWh
100 + 200 + 100

|

e-Text Main Menu

| Textbook Table of Contents |

Problem Solving Workbook Contents

17