Introduction to
Electric
Circuits
Tenth Edition
Introduction to
Electric
Circuits
Herbert W. Jackson
Dale Temple
Brian Kelly
Karen Craigs
Lauren Fuentes
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Library and Archives Canada Cataloguing in Publication
Title: Introduction to electric circuits / Herbert W. Jackson, Dale Temple, Brian Kelly, Karen
Craigs, and Lauren Fuentes.
Names: Jackson, Herbert W., author. | Temple, Dale, author. | Kelly, Brian, 1948– author. | Craigs,
Karen, author. | Fuentes, Lauren, author.
Description: Tenth edition. | Includes index.
Identifiers: Canadiana (print) 20189066806 | Canadiana (ebook) 20190062096 | ISBN 9780199034703
(loose-leaf) | ISBN 9780199031412 (hardcover) | ISBN 9780199031474 (EPUB)
Subjects: LCSH: Electric circuits—Textbooks. | LCGFT: Textbooks.
Classification: LCC TK454 .J28 2019 | DDC 621.319/2—dc23
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Printed and bound in the United States of America
1 2 3 4 — 22 21 20 19
Contents
From the Publisher xix
From the Preface to the First Edition (1959) xxvii
From the Authors of the Tenth Edition xxix
PART I The Basic Electric Circuit
1
1-1
1-2
1-3
1-4
1-5
1-6
1-7
2
2-1
2-2
2-3
2-4
2-5
2-6
2-7
2-8
2-9
Introduction
Key Terms 3
Learning Outcomes 3
Circuit Diagrams 4
The International System of Units 4
Calculators for Circuit Theory 6
Numerical Accuracy 7
Scientific Notation 7
SI Unit Prefixes 10
Conversion of Units 12
Summary 15
Problems 15
Review Questions 17
Integrate the Concepts 18
Practice Quiz 18
Current and Voltage
Key Terms 21
Learning Outcomes 21
The Nature of Charge 22
Free Electrons in Metals 23
Electric Current 24
The Coulomb 26
The Ampere 26
Potential Difference 28
The Volt 31
EMF, Potential Difference, and Voltage 32
Conventional Current and Electron Flow 33
Summary 35
Problems 35
Review Questions 37
Integrate the Concepts 38
Practice Quiz 38
vi
Contents
3
3-1
3-2
3-3
3-4
3-5
4
4-1
4-2
4-3
4-4
4-5
4-6
4-7
4-8
5
5-1
5-2
5-3
5-4
5-5
5-6
5-7
5-8
5-9
Conductors, Insulators, and Semiconductors
Key Terms 41
Learning Outcomes 41
Conductors 42
Electrolytic Conduction 43
Insulators 45
Insulator Breakdown 46
Semiconductors 47
Summary 48
Review Questions 48
Integrate the Concepts 49
Practice Quiz 49
Cells, Batteries, and Other Voltage Sources
Key Terms 51
Learning Outcomes 51
Basic Terminology 52
Simple Primary Cell 52
Carbon-Zinc and Alkaline Cells 53
Other Commercial Primary Cells 55
Secondary Cells 56
Capacity of Cells and Batteries 58
Fuel Cells 60
Other Voltage Sources 60
Summary 65
Problems 65
Review Questions 65
Integrate the Concepts 66
Practice Quiz 66
Resistance and Ohm’s Law
Key Terms 69
Learning Outcomes 69
Ohm’s Law 70
The Nature of Resistance 71
Factors Governing Resistance 72
Resistivity 73
Circular Mils 75
American Wire Gauge 77
Effect of Temperature on Resistance 79
Temperature Coefficient of Resistance 82
Linear Resistors 84
Contents
5-10
5-11
5-12
5-13
5-14
6
6-1
6-2
6-3
6-4
6-5
6-6
Nonlinear Resistors 86
Resistor Colour Code 88
Variable Resistors 91
Voltage-Current Characteristics 91
Applying Ohm’s Law 92
Summary 94
Problems 94
Review Questions 98
Integrate the Concepts 101
Practice Quiz 101
Work and Power
Key Terms 105
Learning Outcomes 105
Energy and Work 106
Power 107
Efficiency 110
The Kilowatt Hour 112
Relationships Among Basic Electric Units 113
Heating Effect of Current 113
Summary 116
Problems 116
Review Questions 120
Integrate the Concepts 121
Practice Quiz 121
PART II Resistance Networks
7
7-1
7-2
7-3
7-4
7-5
7-6
7-7
7-8
7-9
7-10
7-11
Series and Parallel Circuits
Key Terms 125
Learning Outcomes 125
Resistors in Series 126
Voltage Drops in Series Circuits 128
Double-Subscript Notation 130
Kirchhoff’s Voltage Law 130
Characteristics of Series Circuits 131
Internal Resistance 133
Cells in Series 136
Maximum Power Transfer 137
Resistors in Parallel 139
Kirchhoff’s Current Law 141
Conductance and Conductivity 142
vii
viii
Contents
7-12
7-13
7-14
8
8-1
8-2
8-3
8-4
8-5
8-6
8-7
8-8
9
9-1
9-2
9-3
9-4
9-5
9-6
9-7
9-8
9-9
9-10
9-11
Characteristics of Parallel Circuits 145
Cells in Parallel 148
Troubleshooting 150
Summary 154
Problems 154
Review Questions 159
Integrate the Concepts 161
Practice Quiz 161
Series-Parallel Circuits
Key Terms 165
Learning Outcomes 165
Series-Parallel Resistors 166
Equivalent-Circuit Method 167
Kirchhoff’s Laws Method 171
Voltage-Divider Principle 173
Voltage Dividers 175
Current-Divider Principle 181
Cells in Series-Parallel 184
Troubleshooting 186
Summary 188
Problems 188
Review Questions 196
Integrate the Concepts 197
Practice Quiz 198
Resistance Networks
Key Terms 201
Learning Outcomes 201
Network Equations from Kirchhoff’s Laws 202
Constant-Voltage Sources 202
Constant-Current Sources 204
Source Conversion 206
Kirchhoff’s Voltage-Law Equations: Loop Procedure 208
Networks with More Than One Voltage Source 214
Loop Equations in Multisource Networks 216
Mesh Analysis 222
Kirchhoff’s Current-Law Equations 228
Nodal Analysis 231
The Superposition Theorem 237
Summary 242
Problems 242
Review Questions 252
Integrate the Concepts 254
Practice Quiz 255
Contents
10
10-1
10-2
10-3
10-4
10-5
11
11-1
11-2
11-3
11-4
11-5
11-6
11-7
Equivalent-Circuit Theorems
Key Terms 259
Learning Outcomes 259
Thévenin’s Theorem 260
Norton’s Theorem 268
Dependent Sources 271
Delta-Wye Transformation 278
Troubleshooting 283
Summary 284
Problems 284
Review Questions 291
Integrate the Concepts 292
Practice Quiz 292
Electrical Measurement
Key Terms 295
Learning Outcomes 295
Moving-Coil Meters 296
The Ammeter 297
The Voltmeter 300
Voltmeter Loading Effect 302
Resistance Measurement 304
The Electrodynamometer Movement 311
Multimeters 312
Summary 315
Problems 315
Review Questions 318
Integrate the Concepts 319
Practice Quiz 320
PART III C
apacitance and
Inductance
12
12-1
12-2
12-3
12-4
12-5
12-6
12-7
Capacitance
Key Terms 323
Learning Outcomes 323
Electric Fields 324
Dielectrics 327
Capacitance 328
Capacitors 330
Factors Governing Capacitance 333
Dielectric Constant 336
Capacitors in Parallel 338
ix
x
Contents
12-8
Capacitors in Series 338
Summary 342
Problems 342
Review Questions 345
Integrate the Concepts 346
Practice Quiz 346
13
Capacitance in DC Circuits
14
Magnetism
Key Terms 349
Learning Outcomes 349
13-1 Charging a Capacitor 350
13-2 Rate of Change of Voltage 352
13-3 Time Constant 354
13-4 Graphical Solution for Capacitor Voltage 356
13-5 Discharging a Capacitor 357
13-6 Algebraic Solution for Capacitor Voltage 362
13-7 Transient Response 366
13-8 Energy Stored by a Capacitor 370
13-9 Characteristics of Capacitive DC Circuits 372
13-10 Troubleshooting 375
Summary 376
Problems 376
Review Questions 382
Integrate the Concepts 385
Practice Quiz 385
14-1
14-2
14-3
14-4
14-5
14-6
14-7
14-8
14-9
14-10
14-11
14-12
14-13
14-14
Key Terms 389
Learning Outcomes 389
Magnetic Fields 390
Magnetic Field around a Current-Carrying Conductor 393
Magnetic Flux 396
Magnetomotive Force 397
Reluctance 398
Permeance and Permeability 399
Magnetic Flux Density 400
Magnetic Field Strength 401
Diamagnetic, Paramagnetic, and Ferromagnetic Materials 402
Permanent Magnets 404
Magnetization Curves 404
Permeability from the BH Curve 408
Hysteresis 410
Eddy Current 412
Contents
14-15 Magnetic Shielding 413
Summary 414
Problems 414
Review Questions 416
Integrate the Concepts 418
Practice Quiz 418
15
15-1
15-2
15-3
15-4
15-5
15-6
15-7
15-8
15-9
16
16-1
16-2
16-3
16-4
16-5
16-6
16-7
16-8
16-9
16-10
16-11
16-12
16-13
16-14
16-15
Magnetic Circuits
Key Terms 421
Learning Outcomes 421
Practical Magnetic Circuits 422
Long Air-Core Coils 422
Toroidal Coils 425
Linear Magnetic Circuits 425
Nonlinear Magnetic Circuits 426
Leakage Flux 429
Series Magnetic Circuits 430
Air Gaps 433
Parallel Magnetic Circuits 435
Summary 438
Problems 438
Review Questions 442
Integrate the Concepts 443
Practice Quiz 443
Inductance
Key Terms 447
Learning Outcomes 447
Electromagnetic Induction 448
Faraday’s Law 450
Lenz’s Law 451
Self-Induction 453
Self-Inductance 454
Factors Governing Inductance 455
Inductors in Series 458
Inductors in Parallel 458
The DC Generator 459
Simple DC Generator 461
EMF Equation 463
The DC Motor 465
Speed and Torque of a DC Motor 467
Types of DC Motors 469
Speed Characteristics of DC Motors 471
xi
xii
Contents
16-16 Torque Characteristics of DC Motors 474
16-17 Permanent Magnet and Brushless DC Motors 476
Summary 477
Problems 477
Review Questions 479
Integrate the Concepts 482
Practice Quiz 482
17
Inductance in DC Circuits
Key Terms 485
Learning Outcomes 485
17-1 Current in an Ideal Inductor 486
17-2 Rise of Current in a Practical Inductor 487
17-3 Time Constant 490
17-4 Graphical Solution for Inductor Current 491
17-5 Algebraic Solution for Inductor Current 495
17-6 Energy Stored by an Inductor 499
17-7 Fall of Current in an Inductive Circuit 501
17-8 Algebraic Solution for Discharge Current 506
17-9 Transient Response 507
17-10 Characteristics of Inductive DC Circuits 509
17-11 Troubleshooting 510
Summary 512
Problems 512
Review Questions 516
Integrate the Concepts 517
Practice Quiz 518
PART IV Alternating Current
18
18-1
18-2
18-3
18-4
18-5
18-6
18-7
18-8
18-9
Alternating Current
Key Terms 523
Learning Outcomes 523
A Simple Generator 524
The Nature of the Induced Voltage 524
The Sine Wave 526
Peak Value of a Sine Wave 529
Instantaneous Value of a Sine Wave 529
The Radian 532
Instantaneous Current in a Resistor 533
Instantaneous Power in a Resistor 536
Periodic Waves 537
Contents
18-10 Average Value of a Periodic Wave 540
18-11 RMS Value of a Sine Wave 540
Summary 544
Problems 544
Review Questions 546
Integrate the Concepts 548
Practice Quiz 548
19
Reactance
Key Terms 553
Learning Outcomes 553
19-1 Instantaneous Current in an Ideal Inductor 554
19-2 Inductive Reactance 555
19-3 Factors Governing Inductive Reactance 556
19-4 Instantaneous Current in a Capacitor 558
19-5 Capacitive Reactance 559
19-6 Factors Governing Capacitive Reactance 560
19-7Resistance, Inductive Reactance, and Capacitive
Reactance 562
Summary 564
Problems 564
Review Questions 565
Integrate the Concepts 566
Practice Quiz 567
20 Phasors
Key Terms 571
Learning Outcomes 571
20-1 Addition of Sine Waves 572
20-2 Addition of Instantaneous Values 573
20-3 Representing a Sine Wave by a Phasor Diagram 575
20-4 Letter Symbols for Phasor Quantities 576
20-5 Phasor Addition by Geometrical Construction 576
20-6 Addition of Perpendicular Phasors 578
20-7 Expressing Phasors with Complex Numbers 581
20-8 Phasor Addition Using Rectangular Coordinates 585
20-9 Subtraction of Phasor Quantities 587
20-10 Multiplication and Division of Phasor Quantities 589
Summary 591
Problems 591
Review Questions 593
Integrate the Concepts 594
Practice Quiz 595
xiii
xiv
Contents
21
21-1
21-2
21-3
21-4
21-5
21-6
21-7
21-8
21-9
22
22-1
22-2
22-3
22-4
22-5
22-6
22-7
Impedance
Key Terms 599
Learning Outcomes 599
Resistance and Inductance in Series 600
Impedance 601
Practical Inductors 604
Resistance and Capacitance in Series 607
Resistance, Inductance, and Capacitance in Series 608
Resistance, Inductance, and Capacitance in Parallel 610
Conductance, Susceptance, and Admittance 613
Impedance and Admittance 615
Troubleshooting 619
Summary 621
Problems 621
Review Questions 627
Integrate the Concepts 628
Practice Quiz 628
Power in Alternating-Current Circuits
Key Terms 633
Learning Outcomes 633
Power in a Resistor 634
Power in an Ideal Inductor 635
Power in a Capacitor 637
Power in a Circuit Containing Resistance and Reactance 639
The Power Triangle 641
Power Factor 645
Power-Factor Correction 648
Summary 657
Problems 657
Review Questions 660
Integrate the Concepts 662
Practice Quiz 662
PART V Impedance Networks
23 Series and Parallel Impedances
23-1
23-2
23-3
Key Terms 667
Learning Outcomes 667
Resistance and Impedance 668
Impedances in Series 668
Impedances in Parallel 671
Contents
23-4
23-5
Series-Parallel Impedances 678
Source Conversion 682
Summary 684
Problems 684
Review Questions 688
Integrate the Concepts 689
Practice Quiz 690
24 Impedance Networks
24-1
24-2
24-3
24-4
24-5
24-6
24-7
Key Terms 693
Learning Outcomes 693
Loop Equations 694
Mesh Equations 700
Superposition Theorem 702
Thévenin’s Theorem 707
Norton’s Theorem 713
Nodal Analysis 716
Delta-Wye Transformation 724
Summary 729
Problems 729
Review Questions 735
Integrate the Concepts 736
Practice Quiz 736
25 Resonance
25-1
25-2
25-3
25-4
25-5
25-6
25-7
25-8
Key Terms 741
Learning Outcomes 741
Effect of Varying Frequency in a Series RLC Circuit 742
Series Resonance 745
Quality Factor 748
Resonant Rise of Voltage 749
Selectivity 751
Ideal Parallel-Resonant Circuits 753
Practical Parallel-Resonant Circuits 758
Selectivity of Parallel-Resonant Circuits 764
Summary 766
Problems 766
Review Questions 768
Integrate the Concepts 769
Practice Quiz 770
26 Passive Filters
Key Terms 775
Learning Outcomes 775
xv
xvi
Contents
26-1
26-2
26-3
26-4
26-5
26-6
26-7
26-8
26-9
26-10
27
27-1
27-2
27-3
27-4
27-5
27-6
27-7
27-8
27-9
Filters 776
Frequency Response Graphs 778
RC Low-Pass Filters 781
RL Low-Pass Filters 788
RC High-Pass Filters 791
RL High-Pass Filters 797
Band-Pass Filters 799
Band-Stop Filters 804
Practical Application of Filters 809
Troubleshooting 812
Summary 814
Problems 815
Review Questions 819
Integrate the Concepts 821
Practice Quiz 822
Transformers
Key Terms 825
Learning Outcomes 825
Transformer Action 826
Transformation Ratio 828
Impedance Transformation 831
Leakage Reactance 833
Open-Circuit and Short-Circuit Tests 835
Transformer Efficiency 837
Effect of Loading a Transformer 838
Autotransformers 841
Troubleshooting 843
Summary 846
Problems 846
Review Questions 849
Integrate the Concepts 851
Practice Quiz 851
28 Coupled Circuits
28-1
28-2
28-3
28-4
28-5
Key Terms 855
Learning Outcomes 855
Determining Coupling Network Parameters 856
Open-Circuit Impedance Parameters 857
Short-Circuit Admittance Parameters 864
Hybrid Parameters 867
Air-Core Transformers 874
Contents
28-6
28-7
Mutual Inductance 875
Coupled Impedance 878
Summary 882
Problems 882
Review Questions 885
Integrate the Concepts 886
Practice Quiz 886
29 Three-Phase Systems
Key Terms 891
Learning Outcomes 891
29-1 Advantages of Polyphase Systems 892
29-2 Generation of Three-Phase Voltages 895
29-3 Double-Subscript Notation 897
29-4 Four-Wire Wye-Connected System 899
29-5 Delta-Connected Systems 903
29-6 Wye-Delta System 909
29-7 Power in a Balanced Three-Phase System 913
29-8 Phase Sequence 915
29-9 Unbalanced Three-Wire Wye Loads 919
29-10 Power in an Unbalanced Three-Phase System 924
29-11 The AC Generator 927
29-12 Three-Phase Induction Motor 930
29-13 Three-Phase Synchronous Motor 932
29-14 Single-Phase Motors 934
29-15The 30° Difference between Delta-Wye
Configurations 935
Summary 937
Problems 938
Review Questions 940
Integrate the Concepts 943
Practice Quiz 943
30 Harmonics
30-1
30-2
30-3
30-4
30-5
30-6
Key Terms 947
Learning Outcomes 947
Nonsinusoidal Waves 948
Fourier Series 949
Addition of Harmonically Related Sine Waves 951
Generation of Harmonics 954
Harmonics in an Amplifier 956
Harmonics in an Iron-Core Transformer 958
xvii
xviii
Contents
30-7
30-8
30-9
RMS Value of a Nonsinusoidal Wave 960
Square Waves and Sawtooth Waves 961
Nonsinusoidal Waves in Linear Impedance Networks 963
Summary 966
Problems 966
Review Questions 968
Integrate the Concepts 969
Practice Quiz 970
Appendices
1
2
Determinants 974
Calculus Derivations 977
2-1
2-2
2-3
2-4
2-5
2-6
2-7
2-8
2-9
2-10
Maxium Power-Transfer Theorem 977
Instantaneous Voltage in a CR Circuit 978
Energy Stored by a Capacitor 980
Instantaneous Current in an LR Circuit 980
Energy Stored by an Inductor 982
RMS and Average Values of a Sine Wave 982
Inductive Reactance 983
Capacitive Reactance 984
General Transformer Equation 985
Maximum Transformer Efficiency 985
3Multisim Schematic
Capture and Simulation 986
Answers to Selected Problems 988
Glossary 1000
Index 1013
From the Publisher
Since its first appearance in 1959, Introduction to Electric Circuits has been used as a core text by hundreds of
thousands of college and university students enrolled
in introductory circuit analysis courses. Through
its many editions, this classic text helped shape the
way the subject is taught, and it was acclaimed by
instructors and students alike for its accessible writing style, its clear explanations of key concepts, and
its comprehensive end-of-chapter problem sets.
In 2008, Oxford University Press proudly made this
landmark work available once again, thoroughly
updated to reflect the many advances made in the
discipline. With this tenth edition, Oxford is proud
to carry on the pioneering work of founding author
Herbert W. Jackson by recognizing and addressing
the changing needs of twenty-first-century students
and instructors.
Highlights of the Tenth Edition
New!
29
Expanded coverage of breadboards, band
resistors, digital multimeters, nodal analysis, and
three-phase systems. All-new sections on power in unbalanced three-phase systems and Delta-Wye configurations highlights a detailed expansion of key material,
enhancing the book’s coverage of three-phase systems.
Other exciting additions include expanded coverage
of breadboards, colour codes for band resistors, digital
multimeters, and nodal analysis.
s
Three-Phase System
up to this point have
rks we have considered
with
Most of the AC netwo
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had only one source
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phase
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differe
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large amounts of electr
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l systems.
energy. Polyphase circui
contro
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types of electromec
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Chapter Outline
Chapter 29
29-15
s
Three-Phase System
hase Systems 892
924
Advantages of Polyp
895
Three-Phase Voltages
29-2 Generation of
ion 897
Notat
)
29-3 Double-Subscript
) − ( −0.5145 + j0.1392
899
m
= ( 0.9714 + j0.2577
Connected Syste
29-4 Four-Wire WyeSystem 903
= 1.486 + j0.1185 A
29-5 Delta-Connected
m 909
29-6 Wye-Delta Syste
m 913 Therefore, IA = 1.5 A
ced Three-Phase Syste
29-7 Power in a Balan
IB = IBC − IAB
nce 915
Similarly,
29-8 Phase Seque
919
1.005 A ∠+14.86º
Three-Wire Wye Loads
= 0.533 A∠+44.86º −
29-9 Unbalanced
m 924
)
d Three-Phase Syste
lance
Unba
) − ( 0.9714 + j0.2577
an
in
r
29-10 Powe
= ( 0.3778 + j0.3760
927
29-11 The AC Generator
= −0.5936 + j0.1183 A
Induction Motor 930
e
-Phas
Three
29-12
932
IB = 0.61 A
Synchronous Motor
and
29-13 Three-Phase
Lamp B.
Motors 934
935
Lamp A is brighter than
29-14 Single-Phase
gurations, with
a CBA phase sequence,
en Delta-Wye Confi
Hence
betwe
ence
Differ
29-15 The 30°
29-1
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10/31/18 09:05 PM
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use Ohm’s law to solve
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29-60 to 29-65.
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29-29
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29-27
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890
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Circuit Check
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CC 29-4. A 10-hp AC
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60-Hz delta-connected
ate
Calcul
g.
laggin
its power factor is 0.88
(a) the line currents
(b) the phase currents
source powers a wye
4-wire wye-connected
and ZC =
CC 29-5. A 208-V 60-Hz
60º Ω, ZB = 100 ∠20º Ω,
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load that has ZA = 50 ∠
ts and the neutral curren
curren
line
the
ate
Calcul
.
40 ∠90º Ω
of
rise to different types
ing the phase split gives
The method of obtain
art motor uses a startmotors. The resistance-st
g
single-phase induction
r wire than the runnin
thinne
and
turns
fewer
ing winding made with
nce and low reactance
g winding has high resista
winding. Then, the startin
high reactance, giving
g has low resistance and
while the running windin
to start the motor.
with the starting
the necessary phase split
uses a capacitor in series
g
The capacitor-start motor
tor and starting windin
phase split. The capaci
reaches
winding to obtain the
switch when the motor
ugal
centrif
the
by
en
are both switched out
the phase shift betwe
Since
speed.
g
runnin
approximately 75% of
resistance-start motor,
ts is greater than for the
starting and running curren
g torque and can be manudevelops more startin
the
the capacitor-start motor
, the capacitor is left in
In a capacitor-run motor
sizes.
larger
in
d
facture
This configuration has
runs as a two-phase motor.
torcircuit, and the motor
than resistance and capaci
ion
operat
r
quiete
the advantage of much
pulsating torque.
to be noisy due to the
applistart motors, which tend
motors used in small
types of single-phase
There are many other
is essentially a DC series
The universal motor, which
or
ances and power tools.
can operate on either AC
power tools. This motor
many
This
in
fans.
used
is
and
motor,
small appliances
-pole motor is used in
on the stator,
DC power. The shaded
a portion of each pole
around
wire
heavy
of
the
motor has a loop
the unshaded portion of
to lag behind the flux in
causing the magnetic flux
to start the motor. Small
the phase split necessary
pole, and thus producing
devices and clocks.
motors are used for timing
single-phase synchronous
29-76 and 29-77.
See Review Questions
nce bet ween
29-15 The 30° Differe ons
rati
Delta-Wye Configu
onnected load as shown
ire system with a wye-c
es is
Let’s consider a four-w
respective load voltag
phasor diagram of the
in Figure 29-42(a). The
b).
29-42(
Figure
N
shown in
A
B
C
VC
120 V 120°
+
VA
VA
VB
−
− 120 V 0°
− VC
+
120 V 0°
+
alanced
29.1 0 Power in an Unb
Phase System
Three-
in each arm will be differhase system, the power
the
In an unbalanced three-p
will not be able to use
all be the same, and we
ent as the loads may not
load. Instead, the total
hase
three-p
ed
with a balanc
phase.
factor √3 that we used
each
in
power
be the sum of the
power in the load will
10/31/18 09:05 PM
jac31412_ch29_890-945.indd
924
935
rations
n Delta-Wye Configu
The 30° Difference betwee
120 V −120°
VB
120 V 120°
120 V −120°
(b)
(a)
ire system with
Figure 29-42 Four-w
a wye-connected load
10/31/18 09:05 PM
jac31412_ch29_890-945.indd
935
xx
From the Publisher
The best exercise sets available—over 2000
­problems and review questions. From the very
first ­edition of Jackson’s Circuits, instructors and
students alike praised the ­book’s exercises for their
variety and usefulness.
7-6
Internal Resistance
Complete
es in the circuit of Figure
the table listing the voltag
8.0 Ω
A
lems have been added, and every e­ xercise has been
checked for accuracy and relevance. Grounded in
decades of classroom and lab experience, the problems give students the chance to test and refine their
comprehension of the discipline’s key concepts.
133
A
Circuit Check
CC 7-1.
New! For this edition, hundreds of new prob-
2.0 Ω
B
7-6.
“Circuit Check” problem sets, found throughout
each chapter, give students the chance to test and
refine their understanding of material they have
just read.
C
5.0 Ω
D
12 V
5.0 Ω
12 Ω
E
F
Figur e 7-6
VAF
VBF
VEF
VDF
VCF
VCE
VBE
VDE
VAC
VBC
ances in the circuit of
own resist
CC 7-2. Find the unkn
VDC
VEC
Figure 7-7.
20 V
R1
I
R2
+
100 V
−
3.0 Ω
R3
50 V
Figur e 7-7
a 100-Ω resistor will
connected in series with
ination is conCC 7-3. What resistance
of 20 W when the comb
dissipate heat at a rate
e?
sourc
nected to a 120-V
7-6
766
Chapter 25
Internal Resistance
the
e of a source is equal to
that the terminal voltag
depends on
In Section 2-8 we noted
t conditions. The EMF
only under open-circui
e and is independEMF of the source
sourc
the
in
action
g
y-convertin
tial
the nature of the energ
n in Figure 7-8, the poten
nt. However, as show
ent of the circuit curre
Resonance
Summary
• The impedance of a
series-resonant circuit
is capacitive below the
ant frequency and induc
resontive above the resonant
frequency.
• In a series-resonant
circuit at resonance, the
impedance has its minim
value, which equals the
um
resistance.
• At the resonant frequ
ency, the inductive reacta
nce equals the capacitive
reactance.
• The quality factor of
a resonant circuit deter
mines the steepness of
shoulders of the reson
the
ance curve.
• In a series-resonant
circuit with a high Q,
inductance and capac
voltages at resonance
itance
are much greater than
the source voltage.
• The half-power frequ
encies define the band
width of a resonant circui
• Increasing Q increases
t.
the sensitivity and select
ivity of a resonant circui
• The impedance of a
t.
parallel-resonant circui
t is inductive below the
onant frequency and capac
resitive above the resonant
frequency.
• In a parallel-resonan
t circuit with a high Q,
the tank current at reson
is much greater than the
ance
source current.
• In a parallel-resonan
t circuit at resonance,
the impedance has its
imum value, which equal
maxs the resistance.
• The Q of a practical
parallel-resonant circui
t is equal to the Q of the
• The internal resistance
coil.
of the source reduces the
Q and the selectivity of
a parallel-resonant circui
t.
10/31/18 10:23 PM
d 133
jac31412_ch07_123-163.ind
Review exercises and end-of-chapter problem sets,
now ranked for their level of difficulty, encourage
students to reconsider chapter content and apply
their newly learned skills to over 1500 problems.
B = beginner
I = intermediate
A = advanced
circuitS IM
walkthroug h
Problems
B
Section 25-2
25-1.
Series Resonance
Determine the resonant
frequency of a 68-μF
capacitor in series
with a 22-μH coil that
has a Q of 85.
B 25-2. (a) What capac
itance is needed to tune
a 500-μH coil to series
onance at 465 kHz?
res(b) Use Multisim to verify
the capacitance.
B 25-3. What inductance
in series with a 12-pF
capac
itor is resonant at
45 MHz?
I 25-4. A variable capac
itor with a range of 30
pF to 365 pF is conne
in series with an induc
cted
tance. The lowest frequ
ency to which the
circuit can tune is 540
kHz.
(a) Calculate the induc
tance.
(b) Find the highest frequ
ency to which this circui
t can be tuned.
Section 25-3
A
25-5.
Quality Factor
A series RLC resonant
circuit is connected to
a supply voltage of
50 V at a frequency of
455 kHz. At resonance
the maximum current
measured is 100 mA.
Determine the resistance,
capacitance, and
inductance if the quali
ty factor of the circuit
is 80.
jac31412_ch25_740-773.in
dd 766
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xxi
From the Publisher
Practice Quiz
Integrate the Concep
“Integrate the Concepts” exercises draw all the
strands of the chapter together by presenting a
scenario that requires students to make use of the
various concepts that have been covered in order
to solve a real-life problem.
385
ts
A 50-V source, a 20-kΩ
resistor, a 1-μF capacitor,
and an open switch form
series circuit. Calculate
a
the following after the
switch is closed:
(a) the initial value of
the current
(b) the time constant
(c) the capacitor volta
ge 50 ms later
(d) the time for the capac
itor to fully charge
(e) the energy stored
by the capacitor after
it is fully charged
Practice Quiz
1.
Practice quizzes at the end of each chapter help
students test their retention of material and
prepare for exams.
2.
Which of the following
statements are true?
(a) Uppercase letter
symbols represent stead
y-state values that are
dependent on time.
(b) Lowercase letter
symbols represent instan
taneous values that
are dependent on time.
(c) The time constant
for charging a capacitor
is directly proportional to the capacitance
.
(d) A negative rate of
change of voltage indica
tes that the voltage
across the capacitor is
decreasing.
(e) Kirchhoff’s volta
ge law can be applied
to RC circuits to deter
mine the voltage acros
s the capacitor.
What is the initial rate
of change of potential
difference across the
capacitor in Figure 13-35
when the switch is closed
?
(a) 2.4 V/s
(b) 2.4 kV/s
(c) 24 V/s
(d) 0.24 V/s
R1
2.0 GΩ
V1
48 V
C1
0.010 μF
Figur e 13-35
3.
How long will it take
the capacitor of Figur
e 13-35 to charge to the
battery voltage of 48 V?
(a) 20 s
(b) 20 ms
(c) 100 ms
(d) 100 s
jac31412_ch13_348-387.in
dd 385
10/31/18 04:48 PM
10-5
ing
10-5 Troubleshoot
trace
circuit theorems to help
We can use equivalent-
Troubleshooting
283
circuit faults.
Example 10-9
the calculated value
e circuit in Example 10-3,
For the Wheatstone bridg
if the measured value
mine the probable cause
for V3 is 33.33 V. Deter
for V3 is
(c) 66.67 V
(b) 0 V
(a) 100 V
Multisim and exercises. An ­introdu­ction to M
­ ultisim
paves the way to a host of ­Multisim examples and
exercises designed to a­ cclimatize s­ tudents to w
­ orking
with the leading software for ­circuit ­simulation. All
Multisim exercises and examples in the text—
­denoted by the circuitSIM icon in the ­margin—are
tied to the ancillary resource centre packaged with
the book.
Solution
V , so we deduce that
e law gives E = V1 + 3
(a) Kirchhoff’s voltag
(shorted) or I1 = 0 A.
R , then either R1 = 0 Ω
V1 = 0 V. Since V1 = I1 1
However, if R1 is open,
open.
is
R
or
R
3
1
either
For I1 to be 0 A,
ude that R1 is not
concl
we
V,
100
=
Since V3
I3 = 0 A and V3 = 0 V.
open. Remove these
R is shorted or R3 is
open. Therefore, either 1
with an ohmmeter.
them
check
and
t
circui
resistors from the
V , so we deduce that
e law gives E = V1 + 3
(b) Kirchhoff’s voltag
R is shorted or
’s law, V3 = R3I3, so either 3
V1 = 100 V. From Ohm
is open. However, if R3
R
or
R
3
1
either
A,
0
I3 = 0 A. For I1 to be
we conclude
zero,
not
is
V
V = 0 V. Since 1
is open, I1 = 0 A and 1
or R3 is shorted.
fore, either R1 is open
that R3 is not open. There
check them with an
and
t
circui
the
from
Remove these resistors
ohmmeter.
V , so we deduce that
e law gives E = V1 + 3
(c) Kirchhoff’s voltag
and V3 are reversed,
that the values of V1
V1 = 33.33 V. We note
hanged. Checking the
interc
R have been
suggesting that R1 and 3
the 150-Ω rewe discover that R1 is
rs,
resisto
these
on
colour codes
resistor.
sistor and R3 the 300-Ω
Multisim Solution
ite.
EX10-9 from the webs
Download Multisim file
with the given values
as shown in Figure 10-5,
The circuit is the same
terfor the four resistors.
50-Ω resistor between
s of R1 and R3. Insert a
Interchange the value
ect a ground to the
ate a galvanometer. Conn
minals A and B to simul
ure the voltages
meas
to
s
meter
atic. Insert
bottom node of the schem
uacross R1 and R3.
ngs to verify the concl
view the voltage readi
Run the simulation, and
).
sions of Example 10-9(c
circuitS IM
walkthroug h
xamples. Numerous worked examples ­throughout
E
each chapter show students how to perform the
calculations that are essential to circuit analysis.
See Problem 10-42.
10/31/18 11:12 PM
d 283
jac31412_ch10_258-293.ind
xxii
From the Publisher
Chapter 22
650
Power in Alternating
Practical Circuits
-Current Circuits
ion
Power-Factor Correct
raham Hughes
Source: CP Photo/G
measure of how effect
The power factor is a
by a consumer convert
ively devices operated
l
usefu
to
utility
local
electric current from the
heat, light, or mechanpower output, such as
practhe
is
ction
corre
or
ical motion. Power-fact
r factor of an inductive
tice of raising the powe
by inserting a capacitor
load, such as a motor,
,
lower the power factor
in parallel with it. The
required to deliver a
the greater the current
y.
given amount of energ
ly bill commercial
Electricity suppliers usual
a power factor surand industrial customers
of the customer’s
charge if the power factor
The power supplied to
load drops below 90%.
ured with a meter that
these customers is meas
e power and the reactmonitors both the usabl
indicate the power
can
ive power, and hence
factor each month.
Practical applications feature. “­Practical ­Circuits”
boxes show students how circuit t­ heory is used in
everyday situations.
Power meter
14.3 A
A
70% lagging
M power factor
120 V
60 HZ
308 μF
Figur e 22-13
Because
Effect of a series capa
citor on a lagging powe
r-factor load
are in series,
the motor and capacitor
ZT = 8.4 + j8.6 − j8.6 =
8.4 + j0 Ω = 8.4 Ω ∠0º
10/31/18 08:49 PM
300
Chapter 11
dd 650
jac31412_ch22_632-664.in
Electrical Measurem
ent
11-3 The Voltmeter
Moving-coil movement
s can also be used
in voltmeters. If a 1.0-m
movement has a total resist
A
ance of 50 Ω, a voltage
drop of V = IR = 0.0010
× 50 Ω = 50 mV appea
A
rs across the movement
when it is reading full
If we connect a 99.95-kΩ
scale.
resistor in series with
the meter to bring the
resistance up to 100 kΩ,
total
as in Figure 11-5, the voltag
e we must apply to produce a 1.0-mA current
through the meter is
V = IR = 0.0010 A × 100
kΩ = 100 V
99.95 kΩ
Multiplier
Online icons—found throughout the chapters—
direct students to a wide variety of interactive
learning resources available on the mobilefriendly ancillary resource centre.
Figur e 11-5
1.0-mA movement (50 Ω)
V
Simple voltmeter
If we apply a 50-V poten
tial difference to the voltm
eter, the current through
it is
I=
V
50 V
=
= 0.50 mA
R
100 kΩ
Thus, the meter will read
half scale. With the 99.95
-kΩ multiplier resistor,
the scale on the 1.0-mA
movement correspond
s to 0–100 V.
To measure open-circu
it voltage, a perfect voltm
eter should draw no
current. Since current
must flow in a moving-co
il movement to obtain
reading, a 50-μA move
a
ment is much more accur
ate than a 1.0-mA move
for measurements in highment
resistance circuits. The
resistance of a standard
50-μA movement and
its calibrating resistor
is 1000 Ω. To obtain a
full-scale range, the total
100-V
resistance must be
100 V
RT =
= 2.0 MΩ
50 µA
Therefore, the multiplier
resistance must be 2000
kΩ − 1 kΩ = 1999 kΩ.
Similarly, the total resist
ance must be 10 MΩ for
a 500-V scale and 20 kΩ
for a 1.0-V scale. For
any full-scale voltage
with a 50-μA movement
total resistance required
, the
is 20 kΩ for each volt
of the full-scale readi
Therefore, when used
ng.
as a voltmeter, a 50-μA
movement has a voltm
sensitivity of 20 kΩ/V
eter
. We can determine the
total resistance of a movi
coil voltmeter by multi
ngplying the full-scale volta
ge
by
the
movement. For example,
sensitivity of the
when used as a voltm
eter, a 1.0-mA movement
requires a total resistance
of 1000 Ω/V, so total resist
ance for a 500-V scale
is 500 V × 1000 Ω/V =
500 kΩ.
jac31412_ch11_294-320.in
dd 300
10/22/18 05:51 PM
From the Publisher
Updated and revised instructor and student resources. A comprehensive
package of ancillary resources includes a lab manual and an online library
of learning materials for students, and an instructor's manual, a test generator, solutions manuals, and PowerPoint slides for instructors—all thoroughly
updated and revised to reflect new and critical content in the tenth edition.
For Students
• Lab Manual. The perfect accompaniment for courses with a laboratory
component, the revised lab manual is designed to give students hands-on
experience through experiments carefully linked to chapter material.
• Interactive Companion Site. This suite of online learning resources offers
students easy, mobile-friendly access to online videos, quizzes, interactive
animations, MultiSim example files, and other high-interest materials that
expand on key elements from each chapter, allowing students to explore
circuits in new and engaging ways.
For Instructors
• Instructor's Manual. The instructor's manual provides a variety of resources for instructors, including lesson plans, course outlines, suggestions for quizzes and assignments, and helpful hints for labs.
• Expanded! Test Generator. The test generator—now expanded to include fifty percent more questions—offers an exhaustive store of
­multiple-choice, true/false, and short-answer questions in several formats, all suitable for tests and exams.
• Solutions Manuals. The solutions manuals provide worked solutions to
all end-of-chapter problems and questions and all exercises in the student lab manual.
• PowerPoint Slides. These slides will enhance classroom instruction with
lists of learning outcomes, chapter summaries, key equations and terms,
illustrations, problems, and other content drawn from the text.
Details on instructor’s supplements are available from your Oxford
­University Press sales representative or on our website:
www.oupcanada.com/Jackson10e
xxiii
xxiv
From the Publisher
About the Authors
When Herbert W. Jackson published the first edition of Introduction to
Electric Circuits in 1959, he already had nearly two decades of teaching
experience, first as a radar instructor for the RAF and RCAF during World
War II—he was stationed in England during the Battle of Britain—and
later at the veterans’ training institute in Toronto, Ontario, that would become the nucleus of what is today Ryerson University. J­ ackson headed the
electronics and electrical engineering technology department at ­Ryerson
for many years, during which he worked on early editions of this book.
He later joined the Ontario Ministry of Education, where he oversaw the
­creation of that province’s community colleges in the 1960s and 1970s.
So influential was he that some observers labelled him “the father of the
­Ontario college system.” Jackson continued to r­ evise Introduction to Electric Circuits even after retiring from teaching and public service. He and
his wife, Elleda, had two sons, David and Robert. Herb Jackson died in
1999.
Dale Temple taught electronics engineering technology at the College of
the North Atlantic and has also served as coordinator of the electronics
program there. Previous writing credits include work as the co-author of
Canadian editions of Boylestad & Nashelsky’s Electronic Devices and Circuit
Theory and Tocci’s Digital Systems: Principles and Applications. He lives in
St. John’s, Newfoundland and Labrador.
Brian Kelly for many years coordinated the introductory circuit analysis
course at the College of the North Atlantic. Besides his extensive teaching
experience, he is accomplished in numerous computer-aided design (CAD)
software programs and helped prepare the circuit diagrams for previous
editions of Introduction to Electric Circuits.
Lauren Fuentes currently teaches in four different programs at Durham
College in Oshawa, Ontario, and serves as the coordinator of the Electronics Engineering Technician and Technology Programs there. She is always
interested in new challenges, and she has participated in several research
projects at Durham and contributed to revisions of Jackson’s Introduction to
Electrical Circuits textbook, among others.
Karen Craigs is an electronics college professor in Toronto, Ontario. Besides developing courses and tinkering with electronics gadgets, she has
gradually expanded her work over the years to include coaching teams for
technology competitions, mentoring women in technology, and assisting
with various electronics clubs and outreach programs.
From the Publisher
Acknowledgements
Many individuals brought this new edition of Introduction to Electric Circuits to fruition. Cliff Newman, consulting acquisitions editor for Oxford,
knew the book well from his many years as sales manager and editorial
­director at Prentice-Hall and felt it deserved a new lease on life. Elleda
Jackson likewise believed her husband’s lifework should be made available to new generations of students, and she worked tirelessly to secure the
contractual reversions necessary for Oxford to publish the eighth edition.
Development editor Lauren Wing and copy editor Jess Shulman were instrumental in making suggestions for improvement and offering guidance
through the many phases of production for the tenth edition. Various editing teams at Oxford University Press have contributed their outstanding
hard work and dedication to this textbook, ensuring that it continues to be
a relevant resource for students and educators.
The publishers and authors would also like to thank the following reviewers, along with those who chose to remain anonymous, who provided
valuable feedback at various stages in the development of this textbook:
• Laura Curiel, Lakehead University
• Terry Moschandreou, Fanshawe College
• Mingbo Niu, Okanagan College
• Gord Wilkie, Nova Scotia Community College
xxv
From the Preface to
the First Edition (1959)
When a person first attempts to read a book in a foreign language, for
example in French, armed only with a French–English dictionary, it is difficult for him to appreciate the content fully, because of the conscious effort
required in translating the individual words. A prior study of the basic
rules of the grammar of the language, along with sufficient vocabulary
practice, would be a more logical approach to the task. Similarly, basic
electric circuit theory constitutes the language by which the operation of
electric and electronic apparatus is described. Therefore a study of the
basic laws of this language along with sufficient practice to minimize the
conscious effort required in its application are prerequisites to any serious
study of electric and electronic equipment.
To carry the analogy one step further, translation by searching through
the dictionary for the desired word and substituting the equivalent English
xxviii
From the Preface to the First Edition (1959)
word will do little to assist in understanding the material being translated.
Therefore the student is warned not to think of electric circuit theory and
this text as a dictionary of formulas by which one may arrive at the correct numerical solution simply by selecting the right formula and inserting
given data. In electric circuits, a formula is merely a convenient way of expressing a definition of the behaviour of the circuit. Since the task at hand
is to understand circuit behaviour, the student should not apply a formula
unless he is satisfied that he fully understands the definition it represents.
In this respect electric circuit theory has an advantage over language study
in that every definition can be readily and logically developed from a
­preceding step. . . .
H.W. Jackson
1959
From the Authors
of the Tenth Edition
When Oxford University Press told us of their intention to revise and publish the tenth edition of Jackson’s Introduction to Electrical Circuits and that
they wanted two female professors to join the team, we were honoured,
delighted, and thrilled to be the first to start a new trend.
This textbook has been a staple in many institutions for many years and
we hope that our participation in this edition will provide many students,
alumni, and professors a useful tool for teaching and reference.
Karen Craigs
Lauren Fuentes
October 2018
Many thanks to my husband, Jeremy, and my family, John, Lynda,
Theresa, and Mark, for their inspiration and love, always.
—K.E.C
Thanks to my husband, Andy, and my sons, Alberto and Kenny, for always supporting me in whatever new project I embark on. You give me
strength as I continue to push new boundaries in everything I do.
—L.F.
This tenth edition of Introduction to Electric Circuits
is dedicated to the memory of Herbert W. Jackson.
PART
I
The Basic Electric Circuit
Electric circuit theory consists of a few fundamental laws and a series of equations stating relationships that apply to all electric and electronic circuits.
Consequently, much of an introductory electric circuits course deals with the
solution of numerical examples. However, developing an understanding of how
circuits behave involves much more than entering numbers into a calculator or
a computer program. To apply circuit principles to electrical and electronic systems, we also need to understand the physical basis for electric circuit theory.
Part I develops these basic concepts in detail.
1
Introduction
2
Current and Voltage
3
Conductors, Insulators, and Semiconductors
4
Cells, Batteries, and Other Voltage Sources
5
Resistance and Ohm’s Law
6
Work and Power
Photo source: © iStock.com/omada
1
Introduction
Our standard of living depends on an abundant, convenient, and economical
supply of energy and on the relative ease with which we can convert energy
from one form to ­another. We need energy to heat, cool, and light our homes
and to run appliances, vehicles, and industrial machinery. Electricity is an
extremely useful form of energy.
At the beginning of the nineteenth century, electricity was primarily a
scientific curiosity. By the end of that century, the electric telegraph had
revolutionized communications, and by the end of the twentieth century
electricity had become a vital foundation of modern technology.
To work effectively with electrical technology, we must first understand the
basic principles that govern electric ­circuits. We can then develop techniques
for analyzing the performance of practical electric and electronic circuits.
Chapter Outline
1-1
1-2
1-3
Circuit Diagrams 4
The International System of Units
Calculators for Circuit Theory 6
1-4
Numerical Accuracy 7
1-6
SI Unit Prefixes 10
1-5
1-7
Scientific Notation 7
Conversion of Units 12
5
Key Terms
electricity 2
electric circuit 4
conductors 4
switch 4
source 4
load 4
circuit diagrams 4
schematic diagram 4
metric system 5
International System
of Units (SI) 5
metre 5
kilogram 5
second 5
kelvin 5
ampere 5
significant digits 7
scientific notation 8
unit prefixes 10
engineering notation 10
dimensional analysis 12
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• understand the schematic diagram of a
simple electric circuit
• identify the base units of the SI system of units
• use the proper number of significant figures in
a ­calculation
Photo sources: © iStock.com/gwmullis
• use scientific notation to represent quantities
• use engineering notation to represent
quantities
• use dimensional analysis to convert from one
system of units to another
4
Chapter 1 Introduction
1-1 Circuit Diagrams
The term electricity
can mean either a
flow of charged particles or the branch
of physics that deals
with such flows.
In a circuit, a load
is any device that
­converts electric
­energy to some other
form of energy.
Two energy conversions take place in the basic electric circuit illustrated in
Figure 1-1. Chemical energy stored in the battery is converted into electric
­energy. Two copper electric conductors convey the electric energy to the lamp,
which converts it into light and heat. A switch in this circuit allows us to start
and stop the energy conversion at will by interrupting the transmission of
electric energy from the energy source (the battery) to the load (the lamp).
Ba
tte
ry
+
−
Figure 1-1
Pictorial representation of a basic electric circuit
A diagram is usually the best and easiest way to show the interconnection of the components of an electric circuit. For more elaborate circuits,
pictorial representations such as Figure 1-1 can be confusing and difficult
to draw. To keep circuit diagrams simple and clear, we represent the various circuit elements by standard graphic symbols rather than by pictures.
A circuit diagram consisting of lines and symbols is called a schematic
­diagram. Figure 1-2 shows a schematic diagram for the circuit of Figure 1-1.
A table inside the back cover shows the standard schematic symbols for
the most common circuit components.
+
−
Figure 1-2
Circuit diagram for a basic electric circuit
See Review Question 1-26 at the end of the chapter.
1-2 The International System of Units
In 1883, the Scottish physicist William Thomson (Lord Kelvin) said,
“when you can measure what you are speaking about, and express it in
­numbers, you know something about it; but when you cannot measure it,
when you cannot express it in numbers, your knowledge is of a meagre
1-2 The International System of Units
and u
­ nsatisfactory kind.” To apply this philosophy to our study of electric
­circuits, we need a universally recognized system of measuring units.
The English-speaking world in Lord Kelvin’s time still used an awkward
set of units, including inches, feet, yards, rods, chains, fathoms, teaspoons,
cups, pints, bushels, pounds, and several kinds of ounces. Most other
countries used the metric system, a simple decimal system of units, which
originated in France around 1800. As a result of a treaty signed in 1875
by representatives of 17 nations, the International Bureau of Weights and
­Measures (BIPM) was established to provide standards of measurements for
use worldwide. In October 1965, the Institute of Electrical and Electronics
Engineers (IEEE) adopted the International System of Units (SI) proposed
at the eleventh General Conference on Weights and Measures (CGPM) held
in France in 1960. SI unified and rationalized several earlier sets of metric
units, allowing scientists and engineers in all fields to use units based on the
same standards. Today, SI is structured around seven base units that can be
used to derive another 22 named units in physics. In November 2018, after
a unanimous vote of 60 nations at the twenty-sixth CGPM, the definition of
four base units—the kilogram, the ampere, the kelvin, and the mole—will
be redefined by setting exact numerical values for the constants that relate
each of them to fundamental quantities of nature. Once the new definitions
are adopted in May 2019, SI will be wholly derivable from universal natural
phenomena and will finally resolve its objective after 220 years of use.
The base unit of length or distance in the International System of Units
is the metre. Originally calculated as one ten-millionth (1 × 10−7) of the
­distance at sea level from the earth’s equator to the pole, the metre is now
more accurately defined as the distance travelled by light through a vacuum
in 1/299 792 458 of a second. Note that SI uses spaces instead of commas to
separate groups of three digits in long numbers.
The original metric unit of mass, the gram, was defined as the mass of one
cubic centimetre of pure water at 4°C. The gram was too small a unit for
many practical purposes. Hence the SI base unit for mass is 1000 grams—the
kilogram. For over a century, the international standard for the kilogram
was a platinum-iridium cylinder kept at the BIPM in Sèvres, France. As the
last SI unit to be based on a manufactured object, the kilogram officially
changes its definition in 2019 to be based on a fixed value of the Planck constant, h, which links the kilogram to the metre and second.
The SI unit of time is the second. Originally defined as 1/86 400 of a mean
solar day, we now define the second more accurately in terms of the decay
of radioactive cesium in an atomic clock. SI still recognizes minutes and hours
even though these units are not decimal multiples of a second.
The SI unit for temperature is the kelvin (symbol K), which is equal to
one Celsius degree. On the Celsius scale, the freezing point of water is 0°C
(equivalent to 273.15 K) and the boiling point of water is 100°C (equivalent
to 373.15 K). Absolute zero is 0 K, or −273.15°C. In 2019, the new definition
will be based on a fixed value of the Boltzmann constant, k, which links the
kelvin to the kilogram, metre, and second.
5
The abbreviation SI
comes from the French
name, Système Internationale ­d’Unités.
In the 1890s, the
United States changed
their most accurate
objects for fundamental standards of length
and mass from imperial system objects to
metric system objects.
Officials then used
conversion tables to
derive US customary
measurements.
6
A degree symbol
is not used with
­measurements in
­kelvins.
Chapter 1 Introduction
The SI base unit that extends the system of units to electrical measurement is the ampere. In order to relate electrical measurement to mechanical measurement, the ampere was defined in terms of the electromagnetic
force between two current-carrying electric conductors. In 2019, the new
definition of the ampere will be based on a fixed value of the elementary
electric charge, e, which links the ampere to the second. We shall define the
ampere in Section 2-5, with additional details described in Chapter 14.
For equations, we need letter symbols to represent both the quantities
being measured and the SI units of measurement. Table 1-1 shows the letter symbols for the SI base units used in this book. Note that italic (slanted)
letters represent quantities while Roman (upright) letters represent units.
TABLE 1-1 Some letter symbols
Quantity
length or distance
Quantity
Symbol
SI Base Unit
l or d
Unit
Symbol
metre
m
mass
m
kilogram
kg
time
t
second
s
electric current
I
ampere
A
temperature
T
kelvin
K
Occasionally, we may encounter an abbreviation of a unit name. For
­example, electricians often shorten ampere to amp. Equations should always
use the SI unit symbols.
Example 1-1
When the switch is closed in the basic electric circuit of Figure 1-2, the
electric current through the lamp is one quarter of an ampere. Express
this statement as an equation.
Solution
I = 0.25 A
See Review Questions 1-27 and 1-28.
1-3 Calculators for Circuit Theory
Numerical examples using SI units provide the best means of understanding the behaviour of electric and electronic circuits. Although the mathematics required can be kept reasonably simple, the calculations themselves
can often be tedious if done by hand. Solutions of numerical examples in
1-5 Scientific Notation
7
this book assume the use of a suitable scientific or engineering calculator.
For the more complex circuits, students could also use circuit simulation
­software.
For this course, it is convenient to have a calculator with a key that
gives answers in powers of 10 corresponding to SI unit prefixes. This key
is often labelled Eng . Programmable features are not necessary, but the
­calculator must be able to evaluate exponentials and convert between rectangular and polar coordinates. Keys for these functions are often marked
ex , →P , and →R . Some so-called “scientific” calculators do not have
these functions.
1-4 Numerical Accuracy
At first glance, it might seem that an electric current of 12 A is exactly the
same as a current of 12.0 A. However, stating a current as 12 A means that
the exact value is closer to 12 than it is to either 11 or 13, while a current
of 12.0 amperes is closer to 12.0 than it is to 11.9 or to 12.1. Thus, 12.0 A is
a more precise measurement than 12 A. We say that the figure 12 has two
­significant digits, whereas the figure 12.0 has three significant digits.
A calculator treats any number we enter as an exact number. If we enter 34
÷ 2.3, the calculator will show 14.782608. But, if the original measurements
are each accurate to only two digits, the calculated result is also a­ ccurate to
only two digits. We know only that the answer is closer to 15 than it is to 14,
no matter how many digits are displayed by the calculator. When performing chain calculations with a calculator, we can leave the strings of digits
contained in the calculator until the final answer. However, the final answer
should not have more significant digits than the least accurate of the measurements used in the calculations. For most examples in this book, we do
not need to record more than four significant digits in the intermediate steps.
See Problem 1-1.
1-5 Scientific Notation
Electrical measurements can involve very large quantities, such as a radio
frequency of 455 000 Hz, and very small quantities, such as an inductance
of 0.000 75 H. Scientific notation uses powers of 10 to eliminate long strings
of zeros and to avoid confusion over whether the zeros at the end of a number are significant digits.
If we shift the decimal point five places to the left in the frequency
455 000 Hz, we divide it by 100 000. We can restore the original value by
then multiplying by 100 000 or 105. Hence,
455 000 Hz = 4.55 × 100 000 Hz = 4.55 × 105 Hz
The hertz (Hz) is the
SI unit for frequency
and is equal to one
cycle per second.
Chapter 16 describes
inductance and the
henry (H), the unit for
inductance.
8
Chapter 1 Introduction
In the preceding calculation, we assumed that the trailing zeros are not
significant digits. If they are significant, we include them after the decimal
in order to show the precision of the measurement:
f = 455 000 Hz = 4.550 00 × 105 Hz
If we shift the decimal point four places to the right in the inductance
0.000 75 H, we multiply it by 10 000. Consequently, to restore the original
value, we must divide 7.5 by 10 000 or multiply by 10−4. Hence,
0.000 75 H =
7.5 H
= 7.5 × 10 − 4 H
10 000
To express a quantity in scientific notation, we shift the decimal point
until only one significant digit lies to the left of the decimal point, and then
we multiply by the appropriate power of 10 to restore the quantity to its
original value.
Example 1-2
Express 839 000 m in scientific notation, given that the trailing zeros are
not significant.
Solution
To obtain one digit to the left of the decimal point, we must shift the
decimal point five places to the left. To restore the quantity to its original value, we then multiply by 105:
839 000 m = 8.39 × 105 m
Example 1-3
Express 1/200 of a second in scientific notation.
Solution
The first step is to convert the fraction to decimal form. Here we treat
the 1 in the numerator as an exact number:
1
s = 0.005 00 s
200
We then shift the decimal point three places to the right and multiply
the result by 10−3:
0.005 00 s = 5.00 × 10−3 s
1-5 Scientific Notation
9
The following example illustrates the advantage of scientific notation.
Example 1-4
In Chapter 19, we shall derive this formula for inductive reactance:
XL = 2πfL, where f is measured in hertz, L is in henries, and XL is in ohms
­(symbol Ω). Given f = 455 000 Hz and L = 0.000 750 H, calculate XL.
Solution
Substituting the known values into the formula gives
XL = 2 × 3.1416 × 455 000 × 0.000 750 Ω
If we enter the quantities in this form into a calculator, we may have
trouble keeping track of the number of zeros. However, if we express f
and L in scientific notation, we have
XL = 2 × π × 4.55 × 105 × 7.50 × 10−4 Ω
Using the scientific notation feature of the calculator, we can enter this
equation as follows:
2×
π
× 4.55
EE
5 × 7.5
EE
+/−
4 = 2144 = 2.14 × 103
Even in a longhand solution, scientific notation simplifies the compu­
tation. To multiply powers of 10, we simply add their indices algebraically. Hence,
XL = 6.2832 × 4.55 × 7.50 × 10 (5+−4) = 2.14 × 103 Ω
See Problems 1-2 to 1-7 and Review Questions 1-29 and 1-30.
Circuit Check
CC 1-1.
How many significant digits does each quantity have?
(a) 435 m
(b) 0.16 kg
(c) 7.140 A
CC 1-2. Perform the indicated operations, and express your answer in
scientific notation with the correct number of significant digits.
651 × 0.00197 s
50 600 × 0.003 66 × 102 m
(a)
(b)
53.4 × 0.784
39 × 12 × 106
1 kg
1A
(c)
(d)
1
1
1
1
1
1
+
+
6 +
5 +
720 180 500
18 × 10
39 × 10
56 × 103
CC 1-3. Simplify each expression to a power of 10:
(a) (103)9
(b) (1/10−4)−2
(c) (10−2)−3
4 1/2
2
−4 1/2
(d) (10 )
(e) (10 /10 )
(f) (108)−1/2
stands for Enter
Exponent. The π
key retrieves the
­numerical value for
π from the calculator’s permanent
memory. The +/−
key changes the sign.
Some ­calculators have
­different labels for
these keys.
EE
A
10
Chapter 1 Introduction
1-6 SI Unit Prefixes
Root units have no
prefixes. All of the
SI base units except
the kilogram are root
units.
For engineering applications, we generally use a variation of scientific
­notation in which a prefix added to the root unit indicates the appropriate
power of 10. Table 1-2 lists the SI unit prefixes. This system is often called
engineering notation.
SI recommends using prefixes that give the numerical values between
0.1 and 1000. For electrical quantities, we usually shift the decimal point in
multiples of three places. So, we seldom use the prefixes hecto, deka, deci, and
centi when working with electric circuits. The other unit prefixes shown in
lightface type in Table 1-2 are used with quantities much larger or much
smaller than we shall encounter in this book.
TABLE 1-2 SI unit prefixes
Prefix
exa
peta
tera
giga
mega
kilo
hecto
deka
deci
centi
milli
micro
The symbol for micro
is the Greek letter
μ (mu).
nano
pico
femto
atto
Multiplication Factor
Letter Symbol
1 000 000 000 000 000 000 = 10
18
E
1 000 000 000 000 = 1012
P
1 000 000 000 000 000 = 1015
1 000 000 000 = 109
T
G
1 000 000 = 106
M
2
10 = 10
h
1
0.1 = 10−1
da
0.001 = 10−3
0.000 001 = 10−6
m
0.000 000 000 000 001 = 10−15
p
1 000 = 103
100 = 10
0.01 = 10
−2
0.000 000 001 = 10−9
0.000 000 000 001 = 10
−12
0.000 000 000 000 000 001 = 10
−18
k
d
c
μ
n
f
a
Example 1-5
Express 839 000 m in engineering notation. Assume that the trailing
zeros are not significant digits.
Solution
Shifting the decimal point three places to the left gives 839, which we
must multiply by 1000 or 103 to restore its proper magnitude. Hence,
839 000 m = 839 × 103 m = 839 km
The kilometre (km) is 1000 times as large as the root unit, the metre.
1-6 SI Unit Prefixes
Example 1-6
If 200 cycles occur in 1.0 s, how long does one cycle take?
Solution
The first step is to get the answer in decimal form.
1.0 s
= 0.0050 s
200
If we shift the decimal point three places to the right to obtain the figure 5.0,
we must multiply by 0.001 or 10−3 to restore the original magnitude.
From Table 1-2, we see that we can express the answer in milliseconds:
5.0 × 10−3 s = 5.0 ms
When we make calculations involving quantities expressed in units with
prefixes, we must remember that the prefixes represent powers of 10. These
powers of 10 must be included in the computation for the answer to have
the proper magnitude. We can demonstrate this point by using unit prefixes for the quantities in Example 1-4.
Example 1-7
Given XL = 2πf L, f = 455 kHz, and L = 750 μH, calculate XL.
Incorrect Solution
If we forget the prefixes and just carry out the arithmetic we get
2 × π × 455 × 750 = 2.14 × 106
This answer appears to be much larger than the answer in Example 1-4.
By omitting the prefixes, we have calculated the answer in milliohms
(thousandths of an ohm).
Correct Solution
To make sure that the answer is in root units, we include the unit prefixes in the calculation by entering the corresponding exponents:
XL = 2 ×
π
× 455
EE
3 × 750
↑
kilo
EE
+/−
6 = 2144 Ω
↑
micro
↑
root unit
Since it is not good form in SI to leave a numerical value greater than
1000, we now make the conversion
2144 Ω = 2.144 × 103 Ω = 2.14 kΩ
Note that we round to three significant digits to match the precision of
the given measurements.
11
12
Chapter 1 Introduction
Example 1-8
A calculator display reads 7.50 × 10−4 H. Express this answer in engineering notation.
Solution
Table 1-2 has no unit prefix for 10−4. Hence, we must start by simultaneously shifting the decimal point and changing the power of 10 (the
exponent) until it is a positive or negative multiple of 3 and the numerical quantity (the significand) is between 0.1 and 1000. In this particular
example, we can shift in either direction. We can lower the exponent
from −4 to −6 and compensate by shifting the decimal point two places
to the right, or we can raise the exponent from −4 to −3 and shift the
decimal point of the significand one place to the left:
7.50 × 10−4 H = 750 × 10−6 H = 750 μH
or
7.50 × 10−4 H = 0.750 × 10−3 H = 0.750 mH
Engineering calculators have an Eng key that automatically converts
a scientific notation answer to display an exponent that is a ­multiple
of 3 with a significand between 0.1 and 1000. Some scientific calculators have EE↑ and EE↓ keys that adjust the exponent while shifting the
decimal point one place at a time. In this example we can press the EE↓
key twice to lower the exponent from −4 to −6, or we can press the EE↑
key once to raise the exponent from −4 to −3.
See Problems 1-8 to 1-18 and Review Question 1-31.
1-7 Conversion of Units
When only a single conversion from a root unit to a prefixed unit is
­involved, as in the last step of Example 1-7, we can usually convert the
­calculator readout mentally to find an appropriate prefix. But some chain
calculations may require several conversions before the final numerical answer is displayed. To determine the correct unit for the final answer, we can
apply dimensional analysis. With this technique, we write the units in an
equation along with the numerical quantities, and multiply by the required
conversion factors. We cancel out the units that appear in both a numerator
and a denominator, and then ­collect the remaining units to establish the
proper units, or dimensions, of the answer.
1-7 Conversion of Units
13
Example 1-9
Convert 0.250 h to seconds.
Solution
We require two conversion factors—hours to minutes and minutes to
­seconds. We can write the first as
60 min = 1 h
We can rearrange this conversion factor by dividing both sides of the
equation by 1 h:
60 min
1h
=
=1
1h
1h
Thus a conversion factor expressed as a ratio is a dimensionless number
equal to unity. We can, therefore, multiply any quantity by 60 min/1 h,
or its reciprocal, without altering the value of the original quantity. Similarly, we can also multiply by the conversion ratio for minutes to seconds:
0.250 h = 0.250 h ×
60 min
60 s
×
= 900 s
1h
1 min
We can also use dimensional analysis to convert from one system of
units to another. Since electrical engineering in North America has already
adopted the International System of Units (SI), all calculations in this book
are made in SI units. Still, we occasionally encounter quantities in the obsolescent foot-pound system of units. To make the necessary conversion into
SI units for examples in this book, we need two basic conversion factors:
1 inch = 2.54 cm and 1 pound = 0.4536 kg
The conversion from inches to centimetres is exact. The conversion from
pounds to kilograms is an approximation, accurate to four digits.
Example 1-10
An electric conductor is 25 ft long. Express its length in SI units.
Solution
25 ft = 25 ft ×
12 in
2.54 cm
1m
×
×
= 7.6 m
1 ft
1 in
100 cm
See Problems 1-19 to 1-25 and Review Questions 1-32 to 1-36.
The “Weights and
Measures” table at
the back of the book
provides a list of common metric and U.S.
equivalents when
converting a variety
of measuring units.
14
Chapter 1 Introduction
Circuit Check
B
CC 1-4. Express the following as quantities having a metric prefix.
(a) 3 × 10−6 m
(b) 2 million kelvins
(c) 5.6 × 102 V
(d) 256 × 1012 Hz
(e) 55 × 10−9 A
CC 1-5. The speed of sound in dry air at 20°C is 343.2 m/s. Express this
speed in kilometres per hour.
CC 1-6. Derive the conversion factor between metric tonnes (1000 kg)
and imperial short tons (2000 pounds).
15
Problems
Summary
• Schematic diagrams use lines and standard graphic symbols to represent electric circuits.
• The International System of Units (SI) is a unified set of measurement
standards used worldwide. The key base units for electrical work are
the metre for distance, the kilogram for mass, the ­second for time, the
kelvin for temperature, and the ampere for current.
• The values of quantities may be expressed with scientific notation or
with unit prefixes (engineering notation).
• The value of a quantity expressed in one unit may be converted to
­another unit using dimensional analysis.
Problems
B
B
Section 1-4 Numerical Accuracy
1-1.
Assuming that the numbers represent measured quantities, express
the solutions to the following computations with the appropriate
number of significant digits.
(a) 1.47 + 120.6 + 9.581 =
(b) 190 − 52.33 + 4.3 =
(c) 567 × 0.0050 =
(d) 2242 ÷ 37 =
(e) 73.2 + 42 × 9.78 =
73.2 − 42
(f)
=
9.78
Section 1-5
1-2.
B
1-3.
B
1-4.
Scientific Notation
Express each of the following numbers in scientific notation.
(a) 455
(b) 73 000
(c) 10 000
(d) 865
(e) 0.000 21
(f) 0.963
Express each of the following numbers in conventional form.
(a) 1.4 × 106
(b) 4.97 × 102
(c) 6.28 × 10
(d) 3.45 × 10−3
(e) 7.7 × 10−6
(f) 8.672 × 10−1
The distance between two towns is 25 km. Express this distance in
metres, using scientific notation.
B = beginner
I = intermediate
A = advanced
16
Chapter 1 Introduction
B
1-5.
B
1-6.
B
1-7.
B
B
I
A coulomb is equivalent to the amount of charge on
6 241 510 000 000 000 000 electrons. Write this number in scientific
notation with four significant digits.
The mass of an electron is 0.000 000 000 000 000 000 000 000 000 899 9 g.
Express this mass in scientific notation.
The mass of the earth is about 5.98 million million million million
kilograms. Express this mass in scientific notation.
Section 1-6 SI Unit Prefixes
Express each of the following quantities in engineering notation.
(a) 68 000 Ω
(b) 45 000 000 Hz
(c) 1500 W
(d) 0.0505 s
(e) 0.0008 V
(f) 0.000 000 000 39 F
1-9. Express the following quantities with root units, first in scientific
­notation and then in words.
(a) 5.6 MΩ
(b) 75 mm
(c) 0.218 kW
(d) 37 μV
(e) 37 km
(f) 37 kg
1-10. Perform the following calculations using a calculator, first in the
form given, then in scientific notation, and finally in engineering
­notation.
(a) 0.005 × 47 000 − 958 =
(b) 0.005(47 000 − 958) =
(c) 54 000 − 75 × 634 =
(d) 0.00054 =
423 000 + 17 400
=
(e)
560 − 0.04
1-8.
B
B
B
B
B
B
B
1-11.
1-12.
1-13.
1-14.
1-15.
1-16.
1-17.
B
1-18.
(f) √ 1252 + 4 × 5600 =
Express 2250 g in kilograms.
Express 850 mm in metres.
Express 1/60 of a second in milliseconds.
Express 0.000067 km in millimetres.
Express 2.2 × 10−2 kg in grams.
Express 0.044 m2 in square centimetres.
Calculate the cross-sectional area, in square millimetres, of a conductor having a diameter of 3.0 cm.
Express 4.08 × 107 cm3 in cubic metres.
Review Questions
B
I
I
I
I
I
I
Section 1-7
Conversion of Units
1-19. A steak weighs 8 ounces. Using dimensional analysis, determine the
steak’s mass in grams.
1-20. Derive the conversion factor between miles and kilometres.
1-21. Derive the conversion factor between square feet and square
metres.
1-22. Derive the conversion factor between ounces and grams.
1-23. Express a velocity of 50 miles per hour in metres per second.
1-24. Express a velocity of 50 ft/s in kilometres per hour.
1-25. The No. 14 AWG (American Wire Gauge) conductors used in house
wiring have a diameter of 64.08 mils. (A mil is one thousandth
of an inch.) What is the cross-sectional area of this wire in square
­millimetres?
Review Questions
Section 1-1
Circuit Diagrams
1-26. A flashlight is one of the simplest complete electric circuits. It consists of a battery, electric conductors, a switch, and a small lamp.
Using standard schematic symbols, draw a circuit diagram for a
flashlight.
Section 1-2
The International System of Units
1-27. What do the letters SI stand for?
1-28. What is the advantage of a system of units that are all derived from
a few base units?
Section 1-5
Scientific Notation
1-29. What is meant by scientific notation for expressing quantities? What
is the advantage of expressing quantities in this manner?
1-30. How does the scientific notation for expressing quantities differ from
using SI unit prefixes?
Section 1-6
SI Unit Prefixes
1-31. Before the adoption of SI units, wavelengths of light were often
expressed in millimicrons. A micron is one millionth of a metre.
What SI unit is equal to a millimicron?
Section 1-7
Conversion of Units
1-32. Why is it important to write down the proper units when recording
a calculator solution to a numerical example?
1-33. What is the meaning of the term dimensional analysis?
1-34. How would you apply dimensional analysis when you are doing a
chain calculation with a calculator?
17
18
Chapter 1 Introduction
1-35. One butcher shop advertises sirloin steak at $5.99 a pound; another
sells sirloin steak at $12.59 a kilogram. Assuming equal quality,
which is the better price?
1-36. How would you convert the fuel consumption for a vehicle from
miles per gallon to litres per 100 km?
Integrate the Concepts
The hydroelectric power plant at Bay d’Espoir in Newfoundland and Labrador generates 604 million watts.
(a) Express this power using scientific notation.
(b) Express this power using engineering notation.
(c) Convert 604 million watts to horsepower (hp) given that 1 hp ≈ 746 W.
Give the answer to the appropriate number of significant digits.
Practice Quiz
1.
Express each of the following numbers in scientific notation:
(a) 8263
(b) 0.9
(c) 0.000 070
(d) 0.2 × 105
1.0
(e) 2 000 000
2.
Express the following quantities using metric prefixes.
(a) 50 thousand metres
(b) 1.666 × 105 Hz
(c) 4.70 × 10−7 A
(d) 0.000 800 V
3.
Perform the indicated operations:
(a) 46.7 μA = ______________ mA = _____________ A
(b) 0.000 652 A = _______________ mA = _____________ μA
(c) 825 × 103 m =_______________ km = _____________ Mm
(d) 403 μs = _______________ s = _____________ ms
(e) 0.050 ng = _______________ μg = ______________ pg
(f) 25 mH = _______________ H = _____________ μH
(g) 31.4 kHz = ______________ MHz = _____________ Hz
(h) 176 mV = _______________V = _____________ μV
Practice Quiz
4.
Perform the indicated operations:
(a) 633 μA + 2.61 mA = _________________mA
(b) 0.477 mA + 630 μA = ________________ mA
1
(c) = __________ Ω = __________ kΩ
1
1
1
+
+
150 kΩ
220 kΩ
1.0 MΩ
24 V
(d) = ___________________ V = ____________________kV
2.5 × 10−4
100 V
(e) = ___________________ mV = _________________ μV
1.2 × 106
(f) 68.0 km × 145 μm = _____________ m2
19
2
Current and Voltage
In the electric circuit of Figure 1-1, a battery converts chemical energy into
electric energy and a lamp converts that electric energy into light and
heat. Energy conversions are the primary function of any electric circuit.
The concepts of current and voltage are essential for understanding how
­circuits convert energy.
Chapter Outline
2-1
2-2
The Nature of Charge 22
Free Electrons in Metals 23
2-3
Electric Current
2-5
The Ampere 26
2-4
2-6
2-7
2-8
2-9
24
The Coulomb 26
Potential Difference
The Volt
31
28
EMF, Potential Difference, and Voltage
32
Conventional Current and Electron Flow 33
Key Terms
protons 22
neutrons 22
electrostatic force 22
Coulomb’s law 22
free electrons 23
electric current 24
coulomb 26
ampere 26
electromotive force (EMF) 28
generator 28
dynamo 28
potential difference 30
potential rise 30
potential fall
(potential drop) 30
work 31
joule 31
volt 31
voltage 33
voltage drop 33
conventional current 34
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• describe the interactions between like and
unlike electric charges
• describe the behaviour of free electrons in an
electric ­conductor
• calculate the quantity of electric charge
associated with a group of electrons
• perform calculations using the relationship
between charge, current, and time
Photo sources: © iStock.com/IanChrisGraham
• describe potential difference in terms of
potential rise and potential drop
• perform calculations using the relationship
between ­voltage, work, and charge
• differentiate between a source voltage and a
voltage drop
• differentiate between conventional current
and electron flow
22
Chapter 2 Current and Voltage
2-1 The Nature of Charge
The American
­scientist and
­statesman Benjamin
Franklin ­introduced
the terms positive
charge and negative
charge.
An atom consists of a nucleus with electrons orbiting around it. The n
­ ucleus
contains protons bonded with neutrons. Protons have a positive charge,
while neutrons have no charge. An electron is much lighter and smaller
than a proton, and has a negative charge with exactly the same magnitude
as the positive charge on a proton. All electrons have the same charge. In
fact, all electrons are identical, as are all protons.
Normally, the number of electrons orbiting the nucleus of an atom is
equal to the number of protons in the nucleus, making the atom electrically neutral. However, when two dissimilar materials are rubbed together,
some electrons from the surface of one material can transfer to the other.
The material that gains the electrons has a net negative charge, while the
material that loses electrons has a net positive charge. Early researchers
used such transfers to demonstrate that there are two kinds of charge.
­Simple experiments like the one shown in Figure 2-1 reveal a key property
of electric charges.
Like charges repel, unlike charges attract.
Silk
threads
+
+
(a) Like charges
Figure 2-1
Light
balls
+
−
(b) Unlike charges
Interactions of (a) like charges and (b) unlike charges
Such experiments also show that the electrostatic force acts on the
charges without contact between the charges. Therefore, the electrostatic
force is a field force, similar to the gravitational force. However, the electrostatic force can either attract or repel, while the gravitational force
always attracts.
By 1785, a French physicist, Charles Augustin de Coulomb, had shown
that the force between two electrically charged bodies is directly proportional to the product of the magnitudes of the two charges and inversely
proportional to the square of the distance between them. This relationship
is called Coulomb’s law. We can express this law with an equation:
2-2 Free Electrons in Metals
F= k
Q 1Q 2
d2
(2-1)
where F is the force between the two charges, Q1 is the charge on the first
body, Q2 is the charge on the second body, d is the distance between the
charged bodies, and k is Coulomb’s constant, about 8.99 × 109 N⋅m2/C2.
See Review Question 2-27 at the end of the chapter.
2-2 Free Electrons in Metals
In the atoms of some metals, such as copper and silver, the configuration
of electrons is such that the repulsion between the outermost electron and
the other electrons largely offsets the attraction between it and the positive
nucleus. The atoms of such metals form a crystal lattice. Within this ­lattice,
the outermost electron can wind up midway between its parent atom
and a neighboring atom. Since the attractive forces balance at this point,
the electron can leave its atom to orbit the adjacent n
­ ucleus. The thermal
energy of the atoms also contributes to freeing the outermost electrons.
Thus, the lattice consists of positive ions in fixed positions with a cloud of
free electrons that can drift from ion to ion. The lattice is held together by
the attraction between the positive metal ions and the common cloud of
free electrons.
For example, a copper atom has 29 electrons. The first 28 are tightly
bound to the nucleus, but only a very weak attraction holds the 29th
electron in orbit. The copper atoms form a cubical lattice with a metal
ion in each of the eight corners of the cube and also in the centre of each
of the six faces of the cube. Figure 2-2 shows a cross section of such a lattice structure with the outline of one crystal unit sketched in. There are
­roughly 8.5 × 1022 free electrons in a cubic centimetre of copper.
+
+
+
+
+
+
+
+
+
+
+
+ Copper ions
+
+
+
+
Figure 2-2
+
+
+
+
+
+
+
+
Lattice structure of copper
See Review Questions 2-28 to 2-31.
+
+
+
+
23
+
Free electrons
In the units for
Coulomb’s constant,
N stands for newtons
(the SI unit for force)
and C stands for coulomb (the SI unit for
charge), which is described in Section 2-4.
A positive ion is an
atom or group of
atoms that has lost
one or more electrons,
while a negative ion
is an atom or group of
atoms that has gained
one or more electrons.
24
Chapter 2 Current and Voltage
2-3 Electric Current
The moving charged
particles in electric
current will create
a magnetic field,
which is described in
Chapter 14-2.
An electric current is a flow of charged particles. In most circuits these
charged particles are free electrons.
Although there is a free electron for each copper ion, Figure 2-3 shows only
one of these free electrons so that we may trace its motion through the lattice. In Figure 2-3(a), this free electron moves randomly from atom to atom.
The random motion of all of the free electrons in the conductor averages out,
so there is no net flow of electrons in any direction. Figure 2-3(b) shows the
same conductor when connected in an electric circuit that ­conveys energy to
the free electrons. This external source of energy superimposes a net electron
drift to the right on the random motion of the free electrons.
+
+
+
+
+
+
+
+
+
+
(a) Random motion
+
+
+
+
+
+
Figure 2-3
+
+
+
+
+
+
+
+
+
+
(b) Electron drift due to an
external energy source
Motion of a free electron in a copper conductor
The battery in Figure 2-4 adds electrons to the left end of the conductor
and, at the same time, removes the same number of electrons from the
right end of the conductor. The battery uses chemical energy to maintain
a ­surplus of electrons (a net negative charge) at its negative terminal, and
−
−
−
+
Free electrons −
−
+
Copper conductor
−
+
−
+
−
+
− Terminal
Terminal +
Battery
Figure 2-4
Motion of free electrons due to energy from a battery
2-3 Electric Current
a deficiency of electrons (a net positive charge) at its positive terminal. The
free electrons in the conductor are repelled by the negative charge at the
negative terminal of the battery and are attracted by the positive charge at
the positive terminal. Since these electrons are free to move from atom to
atom in the lattice of copper ions, the result is a net electron drift from left
to right in the conductor.
To determine the net movement of free electrons in an electric conductor,
we can consider an imaginary plane cutting across the conductor at right
angles to its length. Figure 2-5(a) shows the conductor with no external
source of energy applied to it. In any given time, the number of electrons
crossing the plane in one direction equals the number crossing in the opposite direction. There is no net electron drift, and, hence, no electric current.
In Figure 2-5(b), an external energy source causes more electrons to cross
the imaginary plane from left to right than from right to left. Consequently,
there is now a net electron drift to the right. This net drift of charge carriers
is an electric current.
(a) No net electron drift
Figure 2-5
(b) Net electron drift
to the right
Net movement of electrons in a conductor
Electric current is the net flow of charge carriers past a given point
in an electric circuit in a given period of time.
In an electric circuit, the current from the energy source to the load is
equal to the current from the load back to the energy source.
See Review Questions 2-32 to 2-37.
Circuit Check
CC 2-1. How is the force between two electric charges affected by
doubling the distance between them?
CC 2-2. What is the net current produced by the random motion
of 1 C of free electrons in a copper wire?
A
25
26
Chapter 2 Current and Voltage
2-4 The Coulomb
Since each electron possesses the same elemental quantity of charge, we
may express the net electron drift across the imaginary plane in Figure 2-5
in terms either of the number of electrons or of the total charge possessed
by this number of electrons. A practical unit for expressing quantity of electric charge must represent the charge carried by many billions of ­electrons.
The coulomb is
named in honour of
Charles Augustin de
Coulomb (1736–1806).
In 2019, the definition
of one elementary
charge, e, will be set
to exactly 1.602 176 634
× 10−19 C. The total
electric charge carried in one coulomb
is the reciprocal of
this value.
The letter symbol
for current, I, comes
from the French
word ­intensité, which
was used to denote
the rate of current
flow. As a unit of
measurement, the
letter C ­represents
coulombs. As a letter
symbol, C is used to
represent the quantity ­capacitance,
which is ­described in
Chapter 12.
The coulomb (symbol C) is the SI unit of quantity of electric charge.
A coulomb is the quantity of electric charge carried by 6.24 × 1018
electrons.
The letter symbol for quantity of electric charge is Q, while the ­letter
e represents the charge on one electron.
Example 2-1
How much charge is carried by 2.40 × 1019 free electrons?
Solution
Q=
2.40 × 1019e
= 3.85 C
6.24 × 1018e/C
See Problems 2-1 and 2-2 and Review Question 2-38.
2-5 The Ampere
Electric current is measured in terms of the rate of charge flow. The SI unit
of electric current, the ampere, is named after a French pioneer of electrical
physics, André-Marie Ampère (1775–1836).
The ampere (symbol A) is the SI unit of electric current.
One ampere is equal to a flow of one coulomb of charge per second:
1 A = 1 C/s.
The letter symbol for electric current is I.
The relationship between current and charge is
I=
Q
t
(2-2)
where I is current in amperes, Q is charge in coulombs, and t is time in
­seconds.
2-5 The Ampere
Example 2-2
Find the current in an electric heater when 75 C of charge pass through
the heater in half a minute.
Solution
I=
Q
75 C
=
= 2.5 C/s = 2.5 A
t
30 s
↑
↑
↑
↑
Step 1
Step 2 Step 3
Step 4
When using a calculator, we tend to simply enter the data to get a numerical answer in a single step. However, a systematic approach is often
helpful, especially for more complex calculations. Here is the sequence
of steps for solving Example 2-2.
Step 1
Note that the problem states two pieces of information and asks for one.
To express this information in equation form, write the symbol for the
unknown quantity on the left of the equals sign and the symbols for the
given data in their proper relationship on the right of the equals sign.
Step 2
Substitute the given data into the equation, making sure that powers
of 10 or unit prefixes are included to preserve the proper magnitude.
Step 3
Perform the numerical computation.
Step 4
The calculator displays the magnitude of the answer in root units.
Express the answer in appropriate units, using unit prefixes if necessary.
Example 2-3
How long will it take 4.0 mC of electric charge to pass through a fuse if
the current is 50 A?
Solution
Since I =
Q
,
t
Q
4.0 mC
t=
=
= 8.0 × 10 − 5 s = 80 × 10 − 6 s = 80 μs
I
50 C/s
See Problems 2-3 to 2-11 and Review Question 2-39.
27
28
Chapter 2 Current and Voltage
2-6 Potential Difference
In Section 2-3, we noted that no current flows in a conductor unless a device
(such as a battery) imparts energy to the free electrons. We say that the
battery is the source of an electron-moving force or electromotive force,
usually abbreviated to EMF. Electromotive force is a property that distinguishes an energy source from the rest of a circuit.
To help understand the relationship between energy and flow, consider
the operation of the hydroelectric generating station shown in Figure 2-6.
Water enters a tunnel at the base of the dam, flows down through a turbine,
and then discharges into the river below. In falling 100 m, the water loses
some of its gravitational potential energy while gaining kinetic energy. The
turbine transfers some of this kinetic energy to the generator or dynamo,
which converts most of it into electric energy.
g
Dam
Generator
100 m
Turbine
g
Figure 2-6
Simplified cross section of a hydroelectric generating station
Since objects at the surface of the earth are 6400 km away from the
earth’s centre of gravity, the difference in gravitational force acting on a
cubic metre of water above and below the generating station is negligible.
But there is an appreciable difference in the potential energy of a cubic
2-6 Potential Difference
29
Source: © iStock.com/IanChrisGraham
Seven Sisters
Generating
Station, the
largest producer
of electricity on
the Winnipeg
River, in Manitoba
metre of water above and below the station. The law of conservation of
energy requires the difference in potential energy between a unit quantity
of water above and below the generating station to be equal to the energy
expended in raising the unit quantity of water the 100 m against the force
of gravity.
Figure 2-7 shows the electric circuit of Figure 2-4 with the battery turned
on its side to parallel the water flow in Figure 2-6.
Battery
−
−
−
Figure 2-7
+
−
−
Conductor
Electrons
−
gain potential
energy as they move
away from + charge
and toward − charge
+
−
−
Electrons lose
potential energy
as they flow
through conductor
Potential
difference
+
+
+
+
Electric potential difference
30
Chapter 2 Current and Voltage
The free electrons in the conductor flow away from the negative ter­
minal of the battery and toward the positive terminal. To maintain the
negative and positive charges at the two battery terminals, an equivalent number of electrons must move inside the battery from the ­positive
­terminal to the ­negative terminal. These electrons move away from the
positive terminal and toward the negative terminal. So, the electrons inside
the ­battery move against the electric forces acting on them, just as water
moves against gravitational force when it is pumped uphill. The electrons
acquire potential e­ nergy at the expense of the chemical energy stored
in the battery. Consequently, an electron moving inside the battery has a
greater potential energy when it arrives at the negative terminal than when
it leaves the positive terminal. There is an electric potential ­difference
­between the negative and positive terminals of the battery, with electrons
at the negative t­ erminal being at a higher potential than those at the positive terminal.
Under the influence of gravity, the water in the hydroelectric generating
station of Figure 2-6 always tends to fall to a lower potential energy level.
Similarly, electrons at the negative terminal of an energy source tend to
“fall” to a lower potential energy level. They can move to a lower potential
via the external conductor connected between the battery terminals. When
flowing from the negative terminal to the positive terminal through the
­external circuit, electrons lose as much potential energy as they gained in
being moved inside the battery from the positive terminal to the negative
terminal. This energy “lost” in the external circuit is converted into light,
heat, or some other form of energy, depending on the nature of the circuit. In travelling the complete circuit around the closed loop of Figure 2-7,
electrons experience a potential rise within the battery and a matching
­potential fall or potential drop in the circuit.
The electric potential difference between any two points in a c­ ircuit
is the rise or fall in potential energy involved in moving a unit
quantity of charge from one point to the other.
The letter symbol for potential difference is E or V.
EMF is the energy per unit of charge that a battery or other energy source
converts in the process of creating a potential difference between its terminals. Since a force has dimensions of mass times acceleration, EMF is not
­actually a force. Electromotive force is a traditional term that is gradually
­disappearing from common use.
See Review Questions 2-40 to 2-42.
2-7 The Volt
31
2-7 The Volt
Work is energy transferred to a body or system. For example, the force of
gravity does work on a falling body. This work increases the kinetic energy
of the body and decreases its potential energy.
The letter symbol for work and energy is W.
The joule (symbol J) is the SI unit of work and energy.
One joule is equal to one newton metre: 1 J = 1 N⋅m = 1 kg⋅m2/s2.
When a force moves a body, the work, W, done is equal to the magnitude, F, of the force times the distance, d, the body moves:
W = Fd
(2-3)
We can express electric potential difference in terms of joules per coulomb,
or volts.
The volt (symbol V) is the SI unit of potential difference.
The potential difference between two points is one volt if one coulomb of charge gains or loses one joule of energy when moving from
one point to the other: 1 V = 1 J/C.
This equation relates potential difference to charge and energy:
E=
W
Q
or
V=
W
Q
(2-4)
where E (or V) is potential difference in volts, W is energy in joules, and
Q is charge in coulombs.
Note that EMF is measured in volts, while a force is measured in newtons.
Example 2-4
A power supply delivers 55 J when 50 C of electrons move from its
negative terminal to its positive terminal. Find the potential difference
between the terminals.
Solution
E=
The joule is named
in honour of the
­English physicist
James Prescott Joule
(1818–89).
55 J
W
=
= 1.1 V
Q
50 C
The volt is named in
honour of the Italian
physicist Alessandro
Volta (1745–1827),
who invented the first
electric battery.
32
Chapter 2 Current and Voltage
Example 2-5
A current of 0.30 A flowing through the filament of a cathode-ray tube
­produces 9.45 J of heat in 5.0 s. What is the potential difference across
the filament?
Solution
Since I = Q/t,
Q = It = 0.30 A × 5.0 s = 1.5 C
V=
9.45 J
W
=
= 6.3 V
Q
1.5 C
See Problems 2-12 to 2-26 and Review Question 2-43.
2-8 EMF, Potential
Difference, and Voltage
At the beginning of Section 2-6, we used the term electromotive force to
­describe the property that an energy source must have to force charge
­carriers to flow in a circuit. EMF is the energy converted per unit quantity
of electric charge moved from one terminal to the other inside the source.
However, the EMF is equal to the potential rise between the terminals of the
source when the circuit is open and no current flows (I = 0). When current
flows in the circuit, inefficiencies in the energy-conversion process make
the potential rise less than the internal EMF of the source.
In the circuit shown in Figure 2-8, the left voltmeter measures the potential difference between the battery terminals. Even with the switch open,
this potential difference is not the same as the EMF of the battery since the
voltmeter itself draws some current from the battery.
Switch
Battery
+
E
V
Lamp
−
Voltmeters
Figure 2-8
circuit
Measuring source voltage and voltage drop in a basic electric
For many circuit calculations, the energy source is assumed to be perfectly
efficient, making its internal EMF and the potential difference between its
terminals equal. Nonetheless, we should not use the term EMF to refer to
the potential rise between the terminals of a generating device.
2-9 Conventional Current and Electron Flow
The term voltage is now commonly used as a simpler expression for
p­ otential difference. We can maintain the distinction between rise and fall
in potential by using the terms source voltage or applied voltage for potential
rise between the terminals of a source and the term voltage drop for the
­potential fall across the circuit load.
The letter symbol for source voltage or applied voltage is E.
The letter symbol for voltage drop is V.
The difference between a source voltage and a voltage drop is illustrated in
Figure 2-8. When the switch is closed, both voltmeters show the same reading since the current through the lamp creates a voltage drop across the lamp
equal to the source voltage (or applied voltage). When the switch is open, the
voltmeter connected to the battery terminals still registers the source voltage,
but the voltmeter connected to the lamp terminals reads zero. A voltage drop
can appear across the lamp only when electrons are flowing through it.
A voltage or potential difference must be measured between two points,
from one point with respect to another, or across a circuit element. There is no
such thing as a “voltage at a point.” On the other hand, we speak of current
in or through a conductor or other component.
The term amperage is sometimes used by electricians for the current normally carried by an electric device. Electrical codes often use ampacity to
mean the maximum allowable current for a given size of conductor.
See Review Questions 2-44 to 2-47.
Circuit Check
B
CC 2-3. How is the ampere related to the coulomb?
CC 2-4. Consider a 4.0-V lithium-ion battery that can maintain a current of 1.0 A for 7.0 h. Assuming that the voltage of the battery
remains constant while it discharges, determine the energy produced by the battery while supplying this current.
CC 2-5. Calculate the work done when 15 C of charge moves between
the terminals of a 12-V battery.
2-9 Conventional Current
and Electron Flow
The most common form of electric conductor is the metallic conductor
­described in Section 2-3. The charge carriers in metallic conductors are free
electrons that flow from the negative terminal of the voltage source t­ oward
the positive terminal. However, as we shall discover in Chapters 3 and 4,
electric current can also result from the flow of positive and negative ions.
33
34
Chapter 2 Current and Voltage
For calculating the algebraic sum of the currents at a junction in a circuit,
we need to define the direction of current flow in a way that we can apply to
all types of charge carriers. Since most circuit conductors are metal, it would
be logical to define current direction in terms of electron flow. ­However,
experimenters established the convention for current direction long before
the electron was discovered. Since Michael Faraday (1791–1867) showed
that metals transfer from the positive terminal to the negative t­erminal in
electrolytic cells, the direction of conventional current was taken to be from
positive to negative. Positive charge carriers flow in the ­opposite direction to
negative charge carriers such as free electrons and ­negative ions.
Therefore, the laws and calculations for electric circuits were based
on current flowing from the positive terminal of the source, through the
circuit, and back to the negative terminal. So, we need to keep in mind that
conventional current is simply a mathematical convention for describing
current and that electrons flow in the opposite direction. The circuit diagrams in this text show the direction of conventional c­ urrent.
Some textbooks now define current as electron flow, and show currents leaving the negative terminal of a voltage source and flowing
into its p
­ ositive terminal. However, the equations and calculations are
the same as for conventional current. In fact, some authors of electrical
and ­electronics textbooks produce two versions, one using conventional
­current and the other using electron flow. The main difference between
the two versions is the direction of the current arrow in the schematic
­diagrams.
Problems
35
Summary
• Like charges repel each other and unlike charges attract each other.
• The electrical properties of atoms depend on the configuration of negatively charged electrons that orbit positively charged nuclei.
• The weak bonding of the outermost electron in the atoms of some
­metals produces free electrons that allow the metal to conduct current
readily.
• Electric current is produced when free electrons move through a conductor in response to the application of an electromotive force.
• Electromotive force is an internal property of a voltage source, and it
causes a potential difference to form across the source’s terminals.
• An electric circuit is the configuration of a continuous path of conductors or devices that connect one terminal of a voltage source to its other
terminal, creating a loop for current to flow.
• Current (I) is the rate of charge flowing past one point in a circuit, and is
measured in amperes (A).
• Voltage (E or V) is the electric potential difference between two points
in a circuit, and is measured in volts (V). E is a voltage source or applied
voltage; V is a voltage drop.
• Conventional current travels from positive to negative. Electron flow
travels from negative to positive.
Problems
B
B
B
Section 2-4
2-1.
2-2.
How many electrons must accumulate to produce a charge of 3.5 C?
How much charge is carried by a trillion free electrons?
Section 2-5
2-3.
B
2-4.
B
2-5.
B
B
2-6.
2-7.
B
2-8.
B
2-9.
The Coulomb
The Ampere
It takes 7.5 s for 30 C of electric charge to pass a switch. Find the
c­ urrent through this switch.
In 1 min, 1 C of charge carriers arrive at the collector (output terminal) of a transistor. Find the current through the transistor.
If the net drift of electrons across the imaginary plane in Figure
2-5(b) is 1014 electrons per second, what is the current in amperes?
Express an electric current of 4.0 mC/min in appropriate SI units.
With the switch closed in Figure 2-8, what is the current through the
lamp if 18 C passes through it in 1.0 min?
With the switch closed in Figure 2-8, how many coulombs of charge
flow through each of the terminals of the battery in 15 ms if the current drain is a steady 0.50 A?
With the switch closed in Figure 2-8, how long will it take for 9.0 C of
electric charge to flow through the switch if the current is 750 mA?
B = beginner
I = intermediate
A = advanced
36
Chapter 2 Current and Voltage
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
2-10. How much charge has passed through a conductor if 20 A flows for
30 min?
2-11. How much current is flowing if 5.5 × 1020 electrons pass a point in
half a minute?
Section 2-7
The Volt
2-12. If it takes 5 J of chemical energy to move 20 C of electric charge
­between the positive and negative terminals of a battery, find the
­potential difference between its terminals.
2-13. If a flow of 40 mC from one terminal of a lamp to the other releases
4.8 J of energy, determine the potential difference developed across
the terminals.
2-14. What electric charge moves through a load that releases 54 J of
energy and creates a potential drop of 120 V across its terminals?
2-15. If the voltage between the terminals of a battery is 1.5 V, how many
electrons move from the positive terminal to the negative terminal
during the conversion of 300 mJ of chemical energy?
2-16. How much energy is required to move 5 mC of electric charge
through a potential rise of 60 V?
2-17. Determine the charge moving through a lamp that converts 2.1 J of
electric energy into light and heat and creates a potential drop of 3.0
V between its terminals.
2-18. How much energy is required to maintain a 0.5-A current through a
battery for half a minute if the potential difference between the battery terminals is 1.1 V?
2-19. How long does a 225-mA current have to flow in order to transfer
5.4 J of energy to a flashlight lamp with a 3.0-V drop between its
­terminals?
2-20. Find the applied voltage of a battery in a transistor radio if a
steady current of 24 mA transfers 13 J of energy to the radio circuit
in 1.0 min.
2-21. If a fully charged 12-V automobile battery can produce 3.5 × 106 J of
electric energy, how long, in hours, can this battery maintain a current
of 2 A?
2-22. Find the current through a load with a 120-V drop across its terminals when energy is delivered to the load at a rate of 18 J/s.
2-23. Assuming perfect efficiency, at what rate is chemical energy being
converted in a battery with a source voltage of 3.0 V while a lamp
connected to the battery draws a steady 120-mA current?
2-24. What is the voltage between the terminals of a 15-A electric heater
that runs for 10 min and produces 1.98 MJ of heat?
2-25. How much current is flowing if 50 J of energy is transferred in 40 s
at a potential of 30 V?
2-26. What is the potential difference when 1.25 J of energy is developed
by a flow of 8.2 × 1017 electrons?
Review Questions
Review Questions
Section 2-1
The Nature of Charge
2-27. What effect would be observed if both balls in Figure 2-1 had a negative charge?
Section 2-2
2-28.
2-29.
2-30.
2-31.
Free Electrons in Metals
Distinguish among an atom, a positive ion, and a negative ion.
Why is silver an excellent conductor of electricity?
Define the term free electron.
How are free electrons related to the ability of metals to act as ­electric
conductors?
Section 2-3
Electric Current
2-32. What causes the random motion of free electrons shown in Figure
2-3(a)?
2-33. Why is the term drift appropriate for describing electric current in a
metallic conductor?
2-34. Explain the forces acting on a free electron at the centre of the
­conductor shown in Figure 2-4.
2-35. Figure 2-5(b) shows the motion of electrons in current-carrying
conductors. Why are some of the electrons moving in opposite
­
­directions?
2-36. Describe the motion of free electrons in the conductors of Figure 2-8
when the switch is open.
2-37. What change in electron motion occurs in the circuit of Figure 2-8
when the switch is closed?
Section 2-4
The Coulomb
2-38. Describe the numerical relationship between the charge of a single
electron and the unit of electric charge, the coulomb.
Section 2-5
The Ampere
2-39. What quantity could be measured in ampere seconds?
Section 2-6
Potential Difference
2-40. The water in Figure 2-6 loses potential energy as it flows through the
turbine of the generating station. How is the energy replenished to
maintain the operation of the station over the years?
2-41. In the hydroelectric generating station of Figure 2-6, gravity c­ reates
a potential difference between the top of the intake pipe and the
end of the discharge pipe below the turbine. What is the equivalent
force that creates a potential difference in an electric ­circuit?
37
38
Chapter 2 Current and Voltage
2-42. Both gravitational force and electric force are measured in newtons.
Explain why electromotive force cannot be measured in the same
units.
Section 2-7
The Volt
Section 2-8
EMF, Potential Difference, and Voltage
2-43. Explain the relationship among the quantities in the equation
V = W/Q.
2-44. Although charge carriers flow from the load back to the source in
one of the conductors of an electric circuit, energy is not transferred
from the load back to the source. Explain why not.
2-45. Explain what happens to the chemical reactions in the battery in
­Figure 2-8 when the switch is opened.
2-46. Why is the letter symbol E used in Example 2-4 and the letter symbol
V in Example 2-5?
2-47. If we connect a voltmeter across the open switch terminals in the
­circuit of Figure 2-8, it will show the same reading as the voltmeter
connected across the battery. Explain why.
Integrate the Concepts
A length of copper wire contains 1.5 × 1020 free electrons.
(a) Find the total charge carried by these free electrons.
(b) Calculate the current when 1.5 × 1020 electrons flow through the wire
in 5.0 s.
(c) Calculate the potential difference across the wire, given that it takes
48 J of energy to produce the current in part (b).
Practice Quiz
1.
Which of the following statements is true?
(a) The force between two electrically charged bodies is inversely
proportional to the product of the magnitudes of the two charges
and directly proportional to the square of the distance between
them.
(b) One ampere is equal to a flow of one coulomb per minute.
(c) Voltage is a measure of the difference in electric potential between
two points.
2.
What amount of charge is carried by 10 mA of current in 0.5 s?
3.
Determine the equivalent current for a flow of
(a) 25 C in 1.0 s (b) 30 C in 10 s
Practice Quiz
4.
5.
If current flowing through a conductor is 10 mA, how many electrons
travel through it in 10 s?
How long will it take a 6.0-A fuse to safely transfer 4.0 × 1020 electrons
between its two terminals?
6.
What applied voltage will allow 17 C of charge to pass through a TV
remote control while it consumes 51 J of energy?
7.
If the difference of potential between the two terminals of a device is 65 V,
how much energy is used to move 10 μC from one terminal to the other?
8.
Determine the current flowing through a lamp if 10.8 kJ of energy is transferred in 1 min at a potential of 120 V.
39
3
Conductors, Insulators,
and Semiconductors
As we noted in Chapter 2, we can proceed with numerical analysis of
electric circuits without reference to the physical process of electrical
conduction. However, to develop a ­better understanding of the behaviour
of circuit components, we shall pause in our numerical analysis to consider
qualitatively the electrical properties of several important non-metals.
Chapter Outline
3-1
Conductors
3-3
Insulators
3-5
Semiconductors
3-2
3-4
42
Electrolytic Conduction
45
Insulator Breakdown
47
46
43
Key Terms
printed circuit boards
breadboard 43
electrolytic cell 43
electrode 43
42
electrolyte 43
anode 43
cathode 44
electrolytic conduction 44
insulator 45
semiconductors 47
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• differentiate between conductors and
insulators
• explain how molecular bonds and crystal
­structure d
­ etermine whether a material is a
­conductor, an i­nsulator, or a semiconductor
Photo sources: © iStock.com/eldadcarin
• explain the process of electrolytic conduction
• explain how a high potential difference can
cause i­nsulator breakdown
42
Chapter 3 Conductors, Insulators, and Semiconductors
3-1 Conductors
As discussed in Section 2-3, electric current is a net drift or flow of charged
particles. It follows that the greater the number of charge carriers per unit
volume, the better a material will conduct electricity. All metals can be used
as electric conductors, although some metals have more free electrons per
unit volume than others.
The material most commonly used for conductors is copper. It is reasonably inexpensive, stable, and easy to shape. Copper has only 5% fewer free
electrons per unit volume than silver, the most conductive element.
Despite having only 60% as many free electrons per unit volume as
copper, aluminum is used for some applications because it is much lighter
and cheaper than either copper or silver. Some of the metal alloys used as
heater elements have less than 1% of the number of free electrons per unit
volume that copper has. The larger the diameter of a wire, the more current
it can carry without overheating.
For electronics, printed circuit boards have largely replaced wiring with
individual conductors. A printed circuit board is a sheet of rigid insulating
material coated with a film of copper. The circuit layout is printed onto the
copper surface with a special ink, then any copper not covered by the ink
is etched away with acid. Alternatively, the unwanted copper can be cut
away by a computerized machine tool. The remaining copper is a n
­ etwork
Source: © iStock.com/eldadcarin
A printed circuit board
3-2 Electrolytic Conduction
43
of traces, conducting paths that link the points where components are
­soldered to the circuit board.
Students and electronic hobbyists have been experimenting with electric
circuits for over 100 years. Since the 1970s, temporary circuit designs have
typically been constructed on a solderless breadboard. Breadboarding
allows a circuit prototype to be built relatively easily using actual throughhole components (which have long metal leads that are pushed through the
holes of breadboards or printed circuit boards), and permits the user to observe and record electrical measurements around the circuit. Many issues
can be addressed during the prototyping phase of a project, even before
starting the design of a printed circuit board.
Breadboards still appear to be popular among electronics learners but
the through-hole component industry has rapidly declined in recent years.
Manufacturers are instead replacing their stock with surface-mount technology (which uses short flat pins, rather than long metal leads, and is designed for use with printed circuit boards), offering improved features in
smaller, less expensive packaging. In the near future, learners will experiment with different circuit designs exclusively through the use of powerful
simulation tools, rather than breadboards.
The term breadboard
emerged about
100 years ago, when
enthusiasts built
circuits on actual
wooden boards,
and the connections
between the
components were
soldered to nails or
thumbtacks pressed
into the board.
Source: Karen Craigs
Removing the protective
backing of a breadboard
reveals its hidden pattern of
electrical connections.
See Review Questions 3-1 and 3-2 at the end of the chapter.
3-2 Electrolytic Conduction
Free electrons are not the only charged particles that can act as charge
­carriers in an electric circuit. In solutions and in gases, current can result
from a flow of ions.
Figure 3-1 shows an electrolytic cell or electroplating bath, consisting
of a silver electrode and an iron electrode immersed in a solution of silver
­nitrate (AgNO3). This current-carrying solution is an electrolyte. The electrode connected to the positive terminal of the battery is an anode, and the
Faraday originated
the terms electrode,
electrolyte, and
­electrolysis.
44
Chapter 3 Conductors, Insulators, and Semiconductors
electrode connected to the negative terminal of the battery is a cathode.
Note that the symbol used for the battery can represent a cell or battery
with any voltage.
Battery
+
Silver anode
Ag+
Electron flow
−
Electron flow
Cathode
(steel spoon)
NO3−
Silver nitrate solution
Figure 3-1
Svante Arrhenius
was the first scientist
to calculate a model
of global warming,
in 1896, due to the
greenhouse effect.
In 1903, he received
the Nobel Prize for
Chemistry, in recognition for his work
with electrolytic
dissociation.
Simple electrolytic cell
From experiments with similar electrolytic cells in the 1830s, Michael
Faraday reasoned that current must be flowing through the silver nitrate
­solution since he could measure an electric current in the copper conductors. Faraday also observed that, as current flowed through the cell, silver
disappeared from the anode and the same amount of silver was deposited
on the cathode.
In 1887 the Swedish scientist Svante August Arrhenius (1859–1927)
­explained the nature of electrolytic conduction. Silver nitrate is an ionic
compound consisting of a positive silver ion bonded to a negative nitrate ion.
Arrhenius suggested that these ions tend to dissociate when the c­ ompound
dissolves in water. The silver and nitrate ions in the electrolyte can then
move independently.
In the cell shown in Figure 3-1, the battery connected to the electrodes
creates a potential difference between them. The positive silver ions are
­attracted to the negatively charged cathode. When a silver ion touches
the cathode, an electron transfers from it to the ion and the resulting
­neutral atom is deposited on the cathode. At the same time, an equal
number of negative nitrate ions are attracted to the positively charged
anode. On reaching the anode, a nitrate ion gives up its surplus electron
to the anode and takes an atom of silver from the anode into the electrolyte solution, thus maintaining the strength of the electrolyte. Since
equal amounts of s­ ilver go into the solution at the anode and deposit
onto the cathode, no chemical energy is used in the cell. The energy required to transport the ­silver from the anode to the cathode comes from
the ­battery.
3-3 Insulators
Although the current in the copper conductors connecting the battery to the electrolytic cell is a unidirectional flow of free electrons, the
­current in the electrolyte results from positive and negative ions flowing
simultaneously in opposite directions. However, Faraday had no way
to observe these ions. All he could measure was the transfer of silver
from the anode to the cathode. Since it was thought that the electric current carried the ­silver, the direction of electric current was assumed to
be from the anode to the cathode. As noted in Section 2-9, positive to
­negative then became ­established as the conventional direction for electric ­current.
See Review Questions 3-3 to 3-5.
3-3 Insulators
A material that does not conduct electricity is an electric insulator. Dry
air is an insulator since it consists mainly of oxygen molecules and nitrogen molecules, in which all the orbital electrons are firmly bound. Free
­electrons cannot readily drift from atom to atom in dry air the way they
can in a metal conductor, and current will not leak from household circuits to the surrounding air. Even high-voltage transmission lines do not
have to be ­insulated from the surrounding air, but they do need long
glass or p
­ orcelain spacers to prevent current from leaking into the steel
support towers.
Some of the most effective insulators are polymers, compounds that
have identical groups of atoms joined into long chains by covalent bonds
in which electrons are shared between adjacent atoms. In some polymers,
these covalent bonds hold the electrons so tightly that there is virtually no
electron drift.
Figure 3-2 shows the structure of polyethylene. Each carbon atom has
six electrons, of which the inner two are tightly bonded to the carbon
nucleus. Two of the outer electrons are shared with hydrogen atoms,
­
which also ­supply an electron each to form strong covalent bonds with
the c­ arbon atom. The other two electrons are shared with the two adjacent
­carbon atoms, forming two more covalent bonds. This form of covalent
bonding creates long, filament-like molecules that are excellent electric insulators. Many plastics and some types of synthetic rubber and varnish are
polymers.
Totally pure polyethylene would have no free electrons, making it an
ideal insulator. However, mass-manufactured polymers usually contain
some impurities with a few free electrons. For example, polystyrene used
in electronic components typically has roughly 6 × 1010 free electrons per
cubic centimetre. Although this number seems large at first glance, it is
­extremely small compared with the number of free electrons found in copper.
45
46
Chapter 3 Conductors, Insulators, and Semiconductors
H
H
H
C
C
C
H
H
H
Electron
H Hydrogen nucleus
Figure 3-2
C Carbon nucleus
Structure of a polyethylene molecule
For practical purposes, the leakage of the charge carriers through a layer of
polystyrene is negligible.
See Review Questions 3-6 and 3-7.
3-4 Insulator Breakdown
Figure 3-3 shows a voltage source connected between the two conductors
of a lamp cord. The potential difference gives rise to an electric force that
acts on the electrons in the molecular bonds of the insulation between the
conductors. If the potential difference is great enough, the force will produce free electrons by breaking some of the insulator’s molecular bonds,
resulting in the rupture or breakdown of the insulator.
The potential difference required to rupture a particular insulator will
depend on the spacing between the two conductors and the strength of the
bonds in the insulating material. Some insulating materials can withstand
a much greater potential difference for a given thickness than others (see
Table 12-1).
+
F
+
−
Electron in
molecular bond
Parallel
conductors
Insulating sheath
−
Figure 3-3
Electric stress on molecular bonds in an insulator
See Review Question 3-8.
3-5 Semiconductors
3-5 Semiconductors
Column IV of the periodic table of elements includes carbon, silicon, and
germanium. These elements each have four outer electrons that can form
covalent bonds. These elements tend to form a crystal structure in which
each atom shares an electron with each of four adjacent atoms. The resulting covalent bonds are not as strong as those in polymers such as
polyethylene or polystyrene. At room temperature, crystals of carbon, silicon, and germanium have too few free electrons to be good conductors
and too many to be good insulators. Materials that conduct more than
insulators but less than conductors are called semiconductors. Crystals
of highly purified silicon are the basis of most diodes and transistors
(see Figure 3-4).
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si Silicon nucleus with 10 inner electrons
Outer electron
Figure 3-4
Covalent bonds in a crystal of pure silicon
See Review Question 3-9.
Circuit Check
CC 3-1.
CC 3-2.
CC 3-3.
List an advantage and a drawback of aluminum as an electric
conductor.
How does the flow of charge in electrolytic conduction differ
from that in metals?
What factor determines whether a material is a conductor, a
semiconductor, or an insulator?
A
47
48
Chapter 3 Conductors, Insulators, and Semiconductors
Summary
• Metals such as copper have a relatively large number of free electrons
and therefore make excellent conductors in electric circuits.
• If a voltage source is connected to the electrodes of an electrolytic cell,
the movement of positive and negative ions through the electrolyte results in a current flow.
• Insulators (such as polymers) have very few free electrons.
• A high potential difference across an insulator can cause it to break down.
• Semiconductors have more free electrons per unit volume than insulators have but fewer than conductors have.
Review Questions
Section 3-1
3-1.
3.2
Calculate the total charge on the free electrons in a cubic centimetre
of copper. (Hint: Refer to Section 2-2.)
Explain why printed circuit boards are not constructed with a rigid
metal base.
Section 3-2
3-3.
3-4.
3-5.
3.7.
Insulator Breakdown
How do the properties of an insulator change when molecular bonds
within the material are broken by a large potential difference?
Section 3-5
3-9.
Insulators
Both copper and sodium chloride (common table salt) are crystalline
solids. Explain why copper is a good conductor of electricity while
sodium chloride is an insulator.
Calculate the total charge on the free electrons in a cubic centimetre
of polystyrene.
Section 3-4
3-8.
Electrolytic Conductors
Explain how the silver anode in the electrolytic cell shown
in ­Figure 3-1 obtains the positive charge required to attract nitrate
ions.
In a solution of an ionic compound, why is it important for the ions
to dissociate before electric current is able to flow?
Describe the major difference between current flow in an electrolyte
and in a metallic conductor.
Section 3-3
3-6.
Conductors
Semiconductors
Why are germanium and silicon neither good conductors nor good
insulators?
Practice Quiz
Integrate the Concepts
(a) Why is copper the most commonly used conductor in electric circuits?
(b)Why does a current flow when a battery is connected to electrodes in
an electrolyte?
(c)
What purpose do insulators serve in electric circuits?
Practice Quiz
1.
Rank the following materials from most conductive to least conductive: aluminum, carbon, copper, glass, silver.
2.
Why are polymers often excellent insulators?
3.
What factors will affect the potential difference required to cause a
breakdown in the insulating sheath shown in Figure 3-3?
4.
What properties do the bonds in crystals of silicon and germanium
have in common?
49
4
Cells, Batteries,
and Other Voltage
Sources
A cell or battery is a chemical source of electric energy. We use these
devices to power portable appliances such as laptop computers, cell
phones, calculators, and TV remote controllers. Cells and batteries come
in a wide range of sizes, types, and voltages, from miniature button cells
for hearing aids and watches to large rechargeable batteries for trucks
and ships.
Chapter Outline
4-1
Basic Terminology 52
4-3
Carbon-Zinc and Alkaline Cells 53
4-2
Simple Primary Cell
52
4-4
Other Commercial Primary Cells 55
4-6
Capacity of Cells and Batteries
4-5
4-7
4-8
Secondary Cells
Fuel Cells 60
56
Other Voltage Sources 60
58
Key Terms
cell 52
battery 52
primary cell 52
secondary cell 52
cathode 52
anode 52
dry cells 53
carbon-zinc cell 53
polarization 53
local action 53
manganese-alkaline cells
(alkaline cells) 53
zinc-chloride cell 55
heavy-duty cell 55
silver-oxide cell 55
zinc-air cell 55
lithium cell 55
lead-acid cell 57
nickel-cadmium (ni-cad)
cell 57
nickel-metalhydride (ni-mh)
cells 57
lithium-ion (li-ion) cell 57
rechargeable alkalinemanganese (RAM) cell 57
capacity 58
ampere-hour rating 58
fuel cell 60
hydroelectricity 61
steam turbine 62
gas turbines 62
wind turbines 62
photovoltaic (solar) cell 62
thermocouple 63
piezoelectricity 64
static electricity 64
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• differentiate between primary and secondary
cells
• explain the basic construction and chemical
action of a simple primary cell
• describe the limitations of simple primary
cells
• describe the basic construction of
commercial primary cells
• describe the chemical actions, advantages,
and ­disadvantages of various types of
primary cells
Photo sources: © iStock.com/hohl
• describe the basic construction of a leadacid secondary cell
• list advantages and disadvantages of
different types of secondary cells
• compare the capacity of various types of
cells
• compare fuel cells to conventional batteries
• discuss how various voltage sources convert
mechanical energy, light, and heat into
electric energy
52
Chapter 4 Cells, Batteries, and Other Voltage Sources
4-1 Basic Terminology
Although not technically correct, most
people, and even
some manufacturers,
refer to single cells as
“batteries.”
A cell is a single unit for converting chemical energy into electric energy.
The 1.5-V, AAA, AA, C, and D cells that we use every day are examples of
cells. A battery is a set of interconnected cells used to provide higher voltage and/or current than is available from a single cell. For example, the
9-V batteries used in devices such as smoke detectors and the 12-V batteries in cars both contain six cells.
In a primary cell the chemical reactions that produce electrical energy
are irreversible, and the cell cannot be recharged. A secondary cell is rechargeable. A power supply connected to the secondary cell uses electric
energy to restore the chemical energy in the cell.
See Review Question 4-3 at the end of the chapter.
4-2 Simple Primary Cell
A cathode is the
­terminal where
­conventional current leaves a device.
­Therefore, the cathode
of a primary cell is
its positive ­terminal
while the cathode of
an ­electrolytic cell
or any other load is
the ­negative terminal.
Similarly, the anode
of a voltage source is
its negative terminal
while the anode of
a load is its positive
­terminal.
In its simplest form, a primary cell consists of two different metal electrodes
immersed in an acidic liquid electrolyte. In solution, the acid molecules
separate into positive and negative ions. For example, in the cell shown in
Figure 4-1, sulfuric acid splits into two positive hydrogen ions and a negative sulfate ion:
H2SO4 → 2H+ + SO42−
When the zinc electrode is immersed in the acid electrolyte, sulfate ions
combine with the zinc to form zinc sulfate (ZnSO4), which precipitates into
the electrolyte. The excess electrons from the sulfate ions remain on the
zinc electrode, making it negative.
Zn + SO42− → ZnSO4 + 2e
Load
Copper rod +
− Zinc rod
H2
+
+
−
Dilute sulfuric acid
−
+
+
Figure 4-1
−
ZnSO4
−
A simple primary cell, the copper-zinc cell
4-3 Carbon-Zinc and Alkaline Cells
At the copper electrode, the hydrogen ions take electrons from the
copper to form neutral hydrogen gas molecules (H2). This reaction leaves
the copper electrode with a deficiency of electrons, making it positive.
2H+ + 2e → H2 ( gas )
These reactions slow as charge builds up on the electrodes, and cease
when the negative charge on the zinc electrode is sufficient to repel sulfate ions, and the positive charge on the copper electrode repels hydrogen ions. At this equilibrium point, the potential difference between
the electrodes is about 1.2 V. When the load is connected between
the electrodes, electrons flow from the negative electrode of the cell
through the external circuit to the positive terminal. This flow of electrons reduces the charge on the electrodes, so the chemical reactions
start again. If the load stays connected, the reactions continue until the
zinc cathode disintegrates or all of the sulfuric acid has become zinc
sulfate.
See Review Questions 4-4 and 4-5.
4-3 Carbon-Zinc and Alkaline Cells
Dry cells use an electrolyte paste or a saturated absorbent spacer instead
of a liquid electrolyte. The first dry cell was the carbon-zinc cell invented
by the German chemist Carl Gassner (1855–1942) in 1886. Variations of
this cell are still being used today. They are usually referred to as general-­
purpose or standard-duty cells.
Negative ions from the electrolyte in the carbon-zinc cell react with the
zinc electrode, making it negative, while hydrogen ions take electrons
from the carbon rod electrode, making it positive. The carbon rod is surrounded by manganese dioxide to prevent polarization, hydrogen bubbles adhering to the electrode and insulating it from the electrolyte. The
hydrogen gas produced at the electrode reacts with the oxygen from the
depolarizer to form water. Impurities in the zinc electrode can cause local
action, reactions that continue without any load connected to the cell.
Local action limits the shelf life of carbon-zinc cells. These cells sometimes leak when nearly used up because the reactions in the cell weaken
the zinc case.
Table 4-1 summarizes a few characteristics of two common types of dry
cells, the carbon-zinc cell and the manganese-alkaline cell, also called
the alkaline cell. Compared to carbon-zinc cells, alkaline cells produce
more energy, have a longer shelf life, and are much less likely to leak.
See Figures 4-2 and 4-3 for a visual representation of their respective
constructions.
53
54
Chapter 4 Cells, Batteries, and Other Voltage Sources
TABLE 4-1 Comparison of common dry cells
Cell Type
Nominal
Voltage
Construction
Usage
Carbon-zinc cell
(see Figure 4-2)
1.5 V
Anode: a zinc can (also the cell case)
Cathode: carbon rod surrounded by a depolarizer
Electrolyte: paste of ammonium chloride and
zinc chloride (acidic)
• First used in flashlights,
radios, and clocks.
• Not as common today as
the alkaline cell.
Manganese-alkaline cell,
also called alkaline cell
(see Figure 4-3)
1.5 V
Anode: zinc powder in a gel containing the
electrolyte
Cathode: manganese dioxide mixed with carbon
powder inside a steel can (the cell case)
Electrolyte: potassium hydroxide (alkaline)
• Most commonly used in
portable devices, such as
toys, flashlights, remote
controls, music players, and
clocks.
Carbon rod (positive electrode)
Sealing compound
Electrolyte paste (ammonium
chloride and zinc chloride)
Depolarizer (manganese dioxide,
ammonium chloride, and graphite)
Zinc container
(negative electrode)
Figure 4-2
Commercial carbon-zinc cell
+
Steel can
(current collector)
Carbon/manganese
dioxide (cathode)
Outer jacket
Powdered zinc, potassium
hydroxide electrolyte/anode
Brass pin
(current collector)
Separator
Insulator seal/vent
−
Figure 4-3
Manganese-alkaline cell
See Review Questions 4-6 to 4-8.
Steel base
4-4 Other Commercial Primary Cells
55
4-4 Other Commercial Primary Cells
Alkaline and carbon-zinc cells are by far the most common types of cells in
use today, but many other types have been developed for special applications.
Table 4-2 summarizes the properties of a few other commercial types of cells.
TABLE 4-2 Comparison of other commercial primary cells
Cell Type
Nominal
Voltage
Construction
Usage
Zinc-chloride
cell, often
called a
heavy-duty
cell
1.5 V
Similar to the carbon-zinc cell.
Anode: a zinc can (also the cell case)
Cathode: manganese dioxide mixed with carbon
Electrolyte: zinc chloride dissolved in water
• Two to four times the operating life of carbon-zinc,
depending on application and continuous use
• Was commonly marketed for “heavy-duty” use
in appliances until alkaline batteries surpassed it
Silver-oxide
cell
1.5 V
Anode: zinc powder in a gel containing the
electrolyte
Cathode: silver oxide
Electrolyte: either potassium hydroxide or
sodium hydroxide
• Voltage remains almost constant over the life
of the cell
• Ideal for limited-space applications such as
watches and calculators
Zinc-air cell
1.4 V
Similar to the alkaline cell.
Anode: zinc powder in a gel containing the
electrolyte
Cathode: a thin layer of carbon with access
to oxygen through venting holes
Electrolyte: potassium hydroxide
• Lightweight with a high energy density
• Shelf life of 3 years or more, if left sealed
• Practical for use in hearing aids, digital cameras,
cell phones, and larger batteries; can power
navigation lights, such as railway signals
• Not suitable for air-tight battery holders
Lithium cell
3.0 V
Anode: lithium foil layer pressed inside a
­stainless steel can
Cathode: manganese dioxide mixed with
carbon
Electrolyte: lithium perchlorate dissolved in
propylene carbonate
• Used in portable electronics and ­implanted
medical devices
• Greater capacity than similarly-sized alkaline cells
• Voltage remains almost constant over the life
of the cell
• Shelf life of 10 years or more
Figure 4-4 shows how the voltages of various types of primary cells vary
as the cells discharge.
Figure 4-4 Discharge
curves for common
primary cells
3.5
Lithium
Terminal voltage (V)
3.0
2.5
2.0
Silver-oxide
1.5
Alkaline
1.0
Carbon-zinc &
zinc-chloride
0.5
20
See Review Question 4-9.
40
60
Operating life (%)
80
100
56
Chapter 4 Cells, Batteries, and Other Voltage Sources
4-5 Secondary Cells
Usually, secondary (rechargeable) cells are preferred for portable devices
that get a lot of use, such as cell phones, laptop computers, and digital cameras. The most common secondary cell is still the lead-acid cell, invented
in 1859 by the French physicist Gaston Planté (1834–89). This type of battery powers the starter motor in most cars. Once running, the car’s engine
drives an alternator, which recharges the battery, reversing the chemical
reactions that produced the electric energy to start the engine. Figure 4-5
shows the construction of a lead-acid secondary cell.
Discharge
Charging supply
A
V
Charge
+
Load
−
Dilute sulfuric acid
(H2SO4)
Lead peroxide (PbO2) on
lead-antimony grid
Figure 4-5
Spongy lead (Pb) on
lead-antimony grid
Lead-acid secondary cell
Several types of rechargeable cells and batteries are made in the same
sizes as primary cells. Some types work best when they are almost fully
discharged before recharging (this is called “deep cycling”); if only partially discharged they exhibit what’s called a “memory effect,” preventing
them from fully recharging. Others can be r­ echarged even if only partially
discharged (“­shallow ­cycling”), which makes them the right choice for
applications that call for shallow c­ ycling or “float” (­continuous) charging. Table 4-3 summarizes the properties of a few commercial types of
secondary cells.
4-5 Secondary Cells
TABLE 4-3 Comparison of secondary cells
Cell Type
Nominal
Voltage
Construction
Usage
Lead-acid cell
(see Figure 4-5)
2.1 V
Anode: a spongy form
of pure lead on a grid of
lead-antimony
Cathode: lead peroxide on
a grid of lead-antimony
Electrolyte: dilute sulfuric
acid
• C
ommon applications
include car starter
motors, emergency
­lighting, ­uninterruptible
power supplies for
­computers, standby
power systems
for ships
Nickel-cadmium
(ni-cad) cell
1.25 V
Anode: cadmium hydroxide
Cathode: nickel hydroxide
Electrolyte: potassium
­hydroxide in water
• S
elf-discharges at a rate
of roughly 20% a month
• Memory effect occurs if
shallow cycled; unsuited
for float charging
• Low cost and can be
recharged up to 1000
times
• Commonly used in
­cordless power tools
Nickel-metal ­
hydride (ni-mh)
cell
1.25 V
Anode: hydrogen storage
alloy such as lanthanumnickel or zirconium-nickel
Cathode: nickel hydroxide
Electrolyte: potassium
hydroxide
• Reduced memory
effect; deep cycling not
required
• More energy density
compared to ni-cads
• Used in cell phones,
digital cameras, laptop
computers
Lithium-ion
(li-ion) cell
3.6 V
Anode: Aluminum
coated with a lithium
compound such as
lithium-cobalt ­dioxide,
lithium-nickel ­dioxide,
or lithium-­manganese
dioxide
Cathode: copper coated
with carbon
Electrolyte: lithium salt such
as lithium-phosphorous
hexafluoride
• Low self-discharge rate
and no memory effect
• Can be deep cycled,
should not be float
charged
• Twice the energy density
compared to ni-cads
• More costly than ni-cad
and ni-mh, thus typically used when size
and weight must be
minimized
• Used in cell phones,
digital cameras, laptop
computers
Rechargeable
alkaline-­
manganese
(RAM) cell
1.1 V
Anode: porous zinc gel
designed to absorb hydrogen
during the charging process
Cathode: manganese
dioxide
Electrolyte: potassium
hydroxide
• L ow self-discharge rate
and no memory effect
• Short operating life,
­especially when deeply
cycled
• Less expensive than
ni-cad
• Suitable for low-cost
consumer applications
with shallow cycling
57
58
Chapter 4 Cells, Batteries, and Other Voltage Sources
Various types of cells and
batteries
Nickel Metal Hydride
Lithium Primary
Lithium Cylindrical
Lithium Rechargeable
Lead Acid
Source: © Panasonic
Lithium Ion
Industrial Alkaline
Carbon Zinc
See Review Questions 4-10 and 4-11.
4-6 Capacity of Cells and Batteries
The capacity of a cell is the total energy it can store. Capacity is determined by the size of the cell and the energy density of the chemicals in it.
Battery capacity is usually expressed in terms of an ampere-hour rating,
which is the product of the discharge current and number of hours to total
discharge.
This relationship can be expressed as an equation:
Capacity = Life × Discharge current
Table 4-4 compares the ampere-hour ratings of several sizes of goodquality alkaline and ni-cad batteries.
The capacity is reduced if the cell is discharged continuously at high current levels. Specifications for cells usually include a maximum discharge
current. Exceeding this maximum can damage the cell.
4-6 Capacity of Cells and Batteries
Example 4-1
A wireless mouse is powered by AA cells. If the average current drawn
from each cell is 50 mA under constant use, determine how long the cells
will last if they are
(a) alkaline
(b) nickel-cadmium
Solution
(a) For alkaline cells,
Life =
Capacity
Discharge current
(b) For ni-cad cells,
Life =
=
2.0 A· h
2.0 A· h
=
= 40 h
50 mA
0.050 A
0.60 A· h
0.60 A· h
=
= 12 h
50 mA
0.050 A
See Problems 4-1 and 4-2 and Review Questions 4-12 and 4-13.
TABLE 4-4 Cell and battery capacities
Capacity (A·h)
Typical
Drain (mA)
Alkaline
Ni-cad
D
200
12.0
4.0
C
100
6.0
2.0
50
2.0
0.60
Battery Type
AA
AAA
10
1.0
0.30
9 Volt
15
0.50
0.15
Circuit Check
CC 4-1.
What is the difference between the cathode and anode of a
­device?
CC 4-2. Why are alkaline cells often used in place of carbon-zinc cells?
CC 4-3. Estimate how long a 9-V alkaline battery will last in a smoke
detector with a continuous current drain of 50 μA.
A
59
60
Chapter 4 Cells, Batteries, and Other Voltage Sources
4-7 Fuel Cells
A fuel cell produces a potential difference by combining hydrogen and
oxygen to form water. This chemical reaction does not consume the
­electrodes. Instead, the chemical energy in the cell is replenished by feeding
more ­hydrogen fuel into the cell (see Figure 4-6). Typical fuel cells produce
less than 1 V at full-rated load, but many cells can be placed in series for higher
voltage outputs, and they can be placed in parallel (or have their surface
area expanded) to allow higher current to flow. Research and development
of fuel-cell technology tends to fall within three main categories: stationary
power generation, portable power generation, and power for transportation. Fuel cells are being developed as an energy source for electric vehicles,
cell-phone towers, and electric grid substations, and as combined heat and
power systems for buildings including homes, schools, hospitals, fire and
police stations, wastewater treatment plants, and airports. The number of
fuel-cell installations around the world is growing, including in the United
States (where California has the most stationary fuel cells and Connecticut
and Delaware have the largest installations [in the 15-30 MW range]) and
South Korea (which boasts one of the world’s largest installations, at close
to 60 MW). Canada is a global leader in hydrogen and fuel-cell research and
development, exporting approximately 90% of this technology through the
supply of parts, testing equipment, and expert services around the world.
−
Hydrogen
(fuel)
in
Electrolytic
barrier
Water
out
Figure 4-6
+
Oxygen
(air)
in
Catalytic electrodes
Simplified cross section of a fuel cell
See Review Question 4-14.
4-8 Other Voltage Sources
All voltage sources create a potential difference by converting some form
of energy into electrical energy.
4-8 Other Voltage Sources
Generators
Source: © iStock.com/leezsnow
In 1831, Michael Faraday discovered that magnetism could produce electricity. Faraday built the first electric generator, a machine that converts
mechanical energy into electrical energy. Large generators produce most of
the electrical energy we use.
Hydroelectricity harnesses the power of falling water to drive turbines
connected to large generators. Figure 2-6 shows a cross s­ ection of a hydroelectric station. The largest hydroelectric stations in the world are installed
in China, Brazil, Paraguay, and Venezuela, with generating capacities in
the range of 10,000 to 22,000 MW.
A row of hydroelectric turbine generators at a
generating station; the one in the foreground has
been taken apart for maintenance.
61
The majority of the
energy consumed in
the world is used for
heat and transportation. About 20% is
used in the form of
electricity.
Chapter 4 Cells, Batteries, and Other Voltage Sources
A steam turbine uses high-pressure steam to drive a generator. The heat
to produce the steam may come from burning a fuel (primarily coal, oil,
and natural gas) or from a nuclear reactor. Coal and natural gas are used in
more than 50% of the world’s electricity generation.
Gas turbines are essentially jet engines that use burning fuel to drive
the turbine directly. Wind turbines use the power of wind to turn a
­propeller connected to a small generator. Denmark generates about
40% of its electricity with wind turbines, and the world’s largest offshore wind farms are found in the United Kingdom, Denmark, and
Germany.
Photoelectricity
A photovoltaic (solar) cell is a semiconductor device that produces
a ­potential difference when light strikes its surface. These devices are
made with two types of silicon crystal, as shown in Figure 4-7. The
n-type s­ ilicon contains a small amount of an element such as antimony,
arsenic, or phosphorus. When added to a silicon crystal lattice, each
atom of these elements has a loosely bound outer electron that can move
within the crystal. The p-type silicon contains a small amount of an element such as boron, gallium, or indium, which can loosely bind a free
electron. At the junction of the two types of silicon, the p-type material
binds the free electrons from the n‑type material, leaving almost no free
charge carriers.
Source: © iStock.com/hohl
62
Panels of Photovoltaic cells at a solar power farm
4-8 Other Voltage Sources
Light energy provides
electron with sufficient
energy to cross
depletion layer
Transparent layer
of p-type silicon
Depletion
layer
−
−
−
−
n-type
silicon
−
−
Load
−
−
+
Figure 4-7
Cross section of a solar cell
Small solar cells power household devices such as calculators and garden
lights. Larger arrays, consisting of hundreds of cells, provide electricity to
pump water, run communications equipment, light homes, and run appliances. Solar cells also power the Hubble Space Telescope and some communications satellites.
Thermocouples
Source: © Omega Engineering Inc.
Reproduced with the permission of Omega
Engineering Inc., Stamford, CT 06907 USA
www.omega.com
A thermocouple is a junction of two dissimilar metals that develops a
small potential difference proportional to the temperature at the junction.
Although not practical for generating large quantities of energy, thermocouples can power small systems such as furnace controls. T
­ hermocouples
are widely used to monitor pilot flames in gas appliances and to measure
temperatures in engines, silos, and industrial ­machinery.
Thermocouples
63
64
Chapter 4 Cells, Batteries, and Other Voltage Sources
Piezoelectricity
Certain crystals, such as quartz and Rochelle salt (potassium sodium
tartrate), produce a potential difference when pressure is applied to them.
This piezoelectric effect converts mechanical energy into electrical energy.
Piezoelectricity has a variety of applications including microphones,
buzzers, strain gauges, and barbecue lighters.
Friction
Friction is another method of producing a potential difference by converting
mechanical energy into electrical energy. When certain dissimilar materials
are rubbed together, the friction between them can transfer electrons from
one of the materials to the other, building up a charge of static electricity.
Friction can generate very high voltages. One application is the Van de
Graaff generator, in which a moving belt builds up a charge on a hollow
metal sphere. Large Van de Graaff generators produce potential differences
of millions of volts. Originally developed for particle accelerators, these
generators are also used to produce radiation for medical treatments and
tests of structural materials.
See Review Questions 4-15 and 4-16.
Review Questions
65
Summary
• Cells produce a voltage as the result of chemical action.
• A battery is a set of cells interconnected to provide a higher voltage and/
or current than is available from a single cell.
• Secondary cells are rechargeable whereas primary cells are not.
• A basic cell has two electrodes in contact with an electrolyte.
• Polarization and local action can limit the output and shelf life of cells.
• Common practical primary cells include carbon-zinc, manganese-­
alkaline, zinc-chloride, silver-oxide, zinc-air, and lithium cells.
• Common secondary cells include lead-acid, nickel-cadmium, nickel-metal
hydride, lithium-ion, and rechargeable alkaline-manganese cells.
• The capacity of a cell is the total energy stored in the cell, usually measured in ampere-hours.
• A fuel cell produces a voltage from the energy released as hydrogen and
oxygen combine to form water.
• Voltage sources convert chemical, mechanical, light, or heat energy to
produce a potential difference.
Problems
B
B
Section 4-6
4-1.
4-2.
Capacity of Cells and Batteries
How long will an alkaline 9 V battery last in a device that draws
5 mA of current for an average of 15 min a day?
How long can a fully charged ni-cad AAA cell maintain a current of
10 mA?
Review Questions
Section 4-1
4-3.
Differentiate between primary and secondary cells.
Section 4-2
4-4.
4-5.
4-8.
Simple Primary Cell
Describe the chemical reaction that takes place at the cathode of a
simple zinc-copper primary cell.
At what point is the simple primary cell considered to be completely
discharged?
Section 4-3
4-6.
4-7.
Basic Terminology
Carbon-Zinc and Alkaline Cells
What is polarization in a cell? How can this effect be prevented?
Compare the terminal voltage and the shelf life of the carbon-zinc
cell and the alkaline cell.
Describe local action in primary cells, and explain why it is
­undesirable.
B = beginner
I = intermediate
A = advanced
66
Chapter 4 Cells, Batteries, and Other Voltage Sources
Section 4-4
4-9.
Other Commercial Primary Cells
Why do most hearing aids use zinc-air cells?
Section 4-5
Secondary Cells
4-10. Describe the various energy conversions that take place from the
time a car’s battery is used to start the engine until the car’s alternator
recharges the battery.
4-11. Why is the ni-mh cell a good choice for powering digital cameras?
Section 4-6
Capacity of Cells and Batteries
4-12. What is the capacity of a cell or battery?
4-13. How is the ampere-hour rating of a battery related to the total energy
stored in the battery?
Section 4-7
Fuel Cells
4-14. Describe one advantage of the fuel cell over a conventional battery.
Section 4-8
Other Voltage Sources
4-15. How does a Van de Graaff generator produce a potential difference?
4-16. What type of voltage source could you use to measure pressure in a
boiler? How does this sensor work?
Integrate the Concepts
(a) What features do all cells have in common?
(b) Choose a suitable type of cell for each application, and explain the
reason for your choice:
i) a flashlight
ii) a cell phone
iii)a memory chip that holds the settings required to start a ­computer
(c)
By what factor would an alkaline D cell outlast a ni-cad D cell powering the same portable device, assuming negligible self-discharge in
the cells?
(d) How might self-discharge affect your answer to part (c)?
Practice Quiz
1.
What is the source of energy in a primary cell?
2.
What is the term for two or more cells connected together?
3.
Why can primary cells not be recharged?
4.
List two advantages of lithium cells.
5.
List three commonly used types of secondary cells.
Practice Quiz
6.
How does continuous discharge at high current levels affect the capacity of a cell?
7.
A wireless computer mouse is powered by AA ni-cad cells that have
a capacity of 0.60 A·h. If the mouse draws an average current of
20 mA while in use, how long can it be used before the cells need to be
­recharged?
8.
Which of the following characteristics apply for fuel cells?
(a) They combine hydrogen and oxygen to form water.
(b) They consume the electrodes.
(c) They are used for electric cars.
(d) None of the above.
9.
Which of the following statements are true?
(a) Every voltage source produces electric energy by converting some
other type of energy.
(b) A photovoltaic cell is a semiconductor device that converts solar
energy into electric energy.
(c) Thermocouples are often used to monitor temperatures.
(d) Piezoelectric crystals convert heat into electric energy.
67
5
Resistance and
Ohm’s Law
Chapter 3 introduced the concepts of voltage and current. Now we shall
explore how these two quantities are related.
Chapter Outline
5-1
Ohm’s Law 70
5-3
Factors Governing Resistance
5-5
Circular Mils
5-2
5-4
The Nature of Resistance
Resistivity
73
75
71
72
5-6
American Wire Gauge
5-8
Temperature Coefficient of Resistance 82
5-7
5-9
77
Effect of Temperature on Resistance 79
Linear Resistors
84
5-10 Nonlinear Resistors 86
5-11
5-12
5-13
Resistor Colour Code 88
Variable Resistors 91
Voltage-Current Characteristics 91
5-14 Applying Ohm’s Law 92
Key Terms
Ohm’s law 70
resistance 70
ohm 70
resistor 72
resistivity 74
specific
resistance 74
ohm metre 74
circular mil 75
mil 75
American Wire
Gauge (AWG) 77
temperature coefficient of
resistance 82
linear resistor 84
wire-wound
resistor 84
precision resistor 85
carbon-composition
resistor 85
nonlinear resistor 86
inrush current 86
thermistor 86
varistor 87
photoresistor 87
photoconductor 87
light dependent
resistor (LDR) 87
nominal value 88
tolerance 88
reliability 88
potentiometer 91
rheostat 91
voltage-current
characteristic 91
IR drop 93
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• state Ohm’s law
• use Ohm’s law to calculate the resistance of,
voltage across, or current through an electric
device
• explain why conductors present an opposition
to current
• state the effect of length, cross-sectional
area, and type of material on the resistance
of a ­conductor
• define resistivity
• calculate the resistance of a material given its
length, cross-sectional area, and resistivity
Photo sources: © iStock.com/David Cannings-Bushell
• calculate the resistance of a length of wire
using the American Wire Gauge table
• calculate the resistance of a material at
different ­temperatures
• describe the construction of different types of
linear, ­nonlinear, and variable resistors
• determine the resistance, tolerance, and
reliability of a ­resistor using the resistor
colour code
• plot the voltage-current characteristic of a
­resistor
70
Chapter 5 Resistance and Ohm’s Law
5-1 Ohm’s Law
In 1827 the German physicist and mathematician Georg Simon Ohm
published a book describing his conclusions from careful measurements
of currents in circuits like the one shown in Figure 5-l. He found the
same current each time he closed the switch in a given circuit. Ohm also
discovered that as long as the temperature of the conductor did not change,
doubling the applied voltage doubled the current, and tripling the applied
voltage tripled the current. Ohm found the same relationships when he
varied the load by using conductors of different lengths and thicknesses.
For a given circuit, the ratio of the applied voltage to the current is
a constant, and
E
=k
I
(5-1)
where E is the applied voltage, I is the current, and the value of k
­depends on the particular circuit.
This relationship is known as Ohm’s law.
+
−
Voltage
source
Figure 5-1
Load
resistance
The basic circuit for Ohm’s resistance experiments
Ohm concluded that the E/I ratio for a given circuit is a property of
that circuit. For a given applied voltage, the current decreases as the ratio
increases. Therefore, this ratio indicates the opposition of the circuit to the
flow of charge carriers. This property is termed resistance, and the unit of
resistance, the ohm, is named for Georg Ohm.
Resistance is the opposition of a circuit to current.
The letter symbol for resistance is R.
The ohm is the SI unit of electric resistance.
The symbol for ohm is the Greek uppercase letter Ω (omega).
One ohm equals one volt per ampere.
We can substitute R for the constant k in Equation 5-1. If we want to
­determine the resistance of the load rather than the resistance of the whole
5-2 The Nature of Resistance
circuit, we must consider the voltage drop across the load, rather than
the voltage applied to the total circuit. The equation for Ohm’s law then
becomes
R=
V
I
(5-2)
where R is the resistance in ohms, V is the voltage drop across the resistance in volts, and I is the current through the resistance in ­amperes.
Example 5-1
Find the resistance of a lamp if a current of 150 mA flows through the
lamp when a voltage of 6.0 V is applied to its terminals.
Solution
R=
V
6.0 V
=
= 40 Ω
I
150 mA
Example 5-2
What value of resistance limits the current through the resistance to
20 μA when the voltage drop across the resistance is 480 mV?
Solution
R=
V
480 mV
=
= 24 000 Ω = 24 kΩ
I
20 μA
See Problems 5-1 to 5-8 and Review Questions 5-70 and 5-71 at the end
of the chapter.
5-2 The Nature of Resistance
As described in Chapter 2, a voltage applied to a conductor causes a net
drift of free electrons along the length of the conductor. Repulsion from electrons entering the conductor from the circuit’s energy source accelerates free
electrons along the conductor. Thus, energy from the source is transferred
to the free electrons as kinetic energy. As the moving electrons collide with
atoms in the conductor, some kinetic energy transfers from the electrons to
the atoms. The transferred energy appears as heat since it increases the vibration of the atoms within the lattice structure of the c­ onductor.
71
72
Chapter 5 Resistance and Ohm’s Law
The collisions between the free electrons and the atoms reduce the speed
at which the electrons drift in response to the applied voltage. The greater
the rate of collisions in a material, the greater its resistance.
Current flowing through a resistance always produces heat. Devices
such as stove elements and incandescent lamps apply this heat. In many
circuits, however, the heat produced is an unavoidable loss of energy from
the system. Some applications require ventilation or other cooling to prevent
the waste heat from damaging the circuit.
A resistor is a component made to have a specific resistance. If we c­ onnect
a resistor between a lamp and a voltage source, as shown in ­Fig­ure 5-2, the
total resistance of the circuit increases and the current is correspondingly
reduced. The rate at which the source transfers energy to the circuit is also
reduced, and that energy is divided between the lamp and the resistor.
Thus, adding the resistor to the circuit dims the lamp.
Resistor
+
−
Figure 5-2
Using a resistor to limit current
See Review Questions 5-72 to 5-74.
5-3 Factors Governing Resistance
Suppose we have two pieces of wire that are identical except that one is
twice the length of the other. For free electrons travelling the full length
of these wires, the average interval between collisions with atoms in
the wire is the same. So, electrons passing through the longer wire have
twice as many collisions as electrons passing through the shorter wire.
Consequently, the opposition of the longer wire to electric current is
twice as great as that of the shorter wire. We can generalize this comparison:
The resistance of a conductor is directly proportional to its length.
Next we compare two pieces of wire that are identical except that
one has twice the cross-sectional area of the other. The thicker wire has
the same cross-sectional area as two pieces of the thinner wire joined
together at both ends (connected in parallel). Each of the two smaller
diameter wires will pass the same current when connected to a given
voltage source. Therefore the thicker wire passes twice as much current
as the thinner wire for a given applied voltage. Since R = V/I, the thicker
5-4 Resistivity
wire has half the ­resistance of the thinner wire. Again, we can generalize
the comparison:
The resistance of a conductor is inversely proportional to its crosssectional area.
We noted in Section 3-1 that some materials possess more free electrons per
unit volume than others. The intervals between collisions of a free electron with
atoms in a material and the energy transferred by the collisions also depend on
the molecular structure of the material. For example, a silver wire has a lower
resistance than a copper wire with the same dimensions, and the copper wire
has a lower resistance than an aluminum wire with the same dimensions.
The resistance of a conductor is dependent on the composition of
the conductor.
Section 5-7 describes how temperature affects resistance.
See Review Questions 5-75 and 5-76.
5-4 Resistivity
The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, we can calculate
the resistance for any dimensions of a conductor if we know the resistance
of a length of the material with a uniform cross-sectional area, assuming no
change in temperature.
R2
l2 A1
= ×
R1
l1 A2
(5-3)
here R is the resistance of the conductor, l is the length, and A is the
w
cross-sectional area.
Example 5-3
A conductor 1.0 m long with a cross-sectional area of 1.0 mm2 has a
resistance of 0.017 Ω. Find the resistance of 50 m of wire of the same
material with a cross-sectional area of 0.25 mm2.
Solution
R2 = 0.017 Ω ×
50 m
1 mm2
×
= 3.4 Ω
1m
0.25 mm2
73
74
Resistivity is
sometimes called
specific resistance.
Chapter 5 Resistance and Ohm’s Law
Resistivity is a convenient quantity for calculating the resistance of a given
conductor.
The resistivity of a material is the resistance of a unit length of the
material with unit cross-sectional area.
The letter symbol for resistivity is the Greek letter ρ (rho).
In SI, the resistivity of a material is the resistance between opposite faces
of a cube of the material measuring 1 m along each side. We can apply
­dimensional analysis to determine the units for resistivity.
ρ = resistance of unit area per unit length
ohms × metre2
metre
= ohm metres
=
The ohm metre is the SI of resistivity.
The unit symbol for ohm metre is Ω·m.
Since temperature affects resistance, values of resistivity are given for a
specified temperature, usually 20°C. Table 5-1 lists the resistivities of the
more common metallic conductor materials at 20°C.
TABLE 5-1 Resistivity of some common conductors at 20°C
Material
Constantan is an alloy
of 50%–60% copper
and 40%–50% nickel.
Nichrome™ II is
an alloy containing
­primarily nickel and
chromium.
Resistivity (nΩ·m)
Silver
16.4
Copper (annealed)
17.2
Gold
24.4
Aluminum
28.3
Tungsten
55
Nickel
about 70
Iron
about 100
Constantan
about 490
Nichrome™ II
about 1100
If we set l1 = 1 m and A1 = 1 m2 in Equation 5-3, R1 becomes ρ. Rearranging the equation gives a formula for the resistance of any conductor:
R=ρ
l
A
(5-4)
where R is the resistance of the conductor in ohms, l is the length of the
­conductor in metres, A is the cross-sectional area in square metres, and ρ
is the resistivity of the conductor material in ohm metres.
5-5 Circular Mils
75
Example 5-4
Find the resistance at 20°C of 200 m of an aluminum conductor with a
cross-sectional area of 4.0 mm2.
Solution
Using the value of ρ for aluminum from Table 5-1,
R=ρ×
l
200 m
= 1.4 Ω
= 2.83 × 10 − 8 Ω · m ×
A
4.0 × 10 − 6 m2
In this solution, converting square millimetres to square metres
allows us the cancel out all the metre units.
Most conductors have a circular cross section. If we know the diameter
of the conductor, we can calculate the cross-sectional area.
Example 5-5
Find the resistance at normal room temperature of 60 m of copper wire
having a diameter of 0.64 mm.
Solution
A = πr2 =
R=ρ×
π 2 π
d = × 0.642 mm2 = 3.217 × 10 −7 m2
4
4
l
60 m
= 1.72 × 10 −8 Ω · m ×
= 3.2 Ω
A
3.217 × 10 −7 m2
See Problems 5-9 to 5-26 and Review Questions 5-77 and 5-78.
Note that CM stands
for circular mil, while
cm is the SI symbol
for centimetre.
5-5 Circular Mils
Many conductors are much less than an inch in diameter. To simplify area
calculations for such conductors, the foot-pound system uses an area unit,
called a circular mil, based on a mil, or one thousandth of an inch (see
Figure 5-3).
One circular mil (CM) is the area of a circle 1 mil in diameter.
0.001"
Figure 5-3
A circle with diameter
of one mil
76
The mathematical
­formula for the area of a
circle is πr2 = ( π/4 ) d2 .
However, with circular
mils the factor of π/4
is included in the units
for the area.
Chapter 5 Resistance and Ohm’s Law
When using circular mils, the formula for the area of a circle becomes
A = d2
(5-5)
here A is the area of the circle in circular mils and d is the diameter in
w
mils.
Example 5-6
Find the cross-sectional area, in circular mils, of a wire 0.0280" in diameter.
d = 0.02800" = 28.0 mils
A = d2 = 28.02 = 784 CM
Solution
Resistivity can be defined in terms of a piece of wire 1 ft long with a crosssectional area of 1 CM, as shown in Figure 5-4. The volume of this wire is
1 circular-mil foot, and the units of ρ become CM-Ω/ft. Table 5-2 lists the
­resistivities for some metals.
1 ft
0.001" = 1 mil
1 Circular mil
Figure 5-4
One circular-mil foot
TABLE 5-2 Resistivity of some metals at 20°C
Material
Silver
9.9
Copper
10.4
Gold
14.7
Aluminum
Brass
Manganin is a ­copper
alloy containing
13%–18% manganese
and 1%–4% nickel.
Resistivity (CM-Ω/ft)
17.0
23.4
Tungsten
33
Nickel
42
Iron
60
Platinum
64
Manganin
290
Constantan
295
Mercury
576
Nichrome™ II
660
5-6 American Wire Gauge
77
Example 5-7
Calculate the resistance of 50.0 miles of copper wire 0.350″ in diameter.
(1 mile = 5280 ft)
Solution
d = 0.350″ = 350 mils
A = d2 = 122.5 × 103 CM
5280 ft
l = 50.0 miles ×
= 2.54 × 105 ft
1 mile
ρl
10.4 ( 2.54 × 105 )
=
R=
= 21.6 Ω
A
122.5 × 103
See Problem 5-27.
Circuit Check
A
CC 5-1. What resistance passes a current of 21.28 mA when the applied
voltage is 10.0 V?
CC 5-2. What length of aluminium wire with a diameter of 0.25 mm has
a resistance of 2.25 Ω at 20ºC?
CC 5-3. An 6.5 m length of wire has a diameter of 0.8 mm and a
­resistance of 0.315 Ω at 20ºC. Of what material is this wire likely
to be made?
5-6 American Wire Gauge
In North America, most wire is produced in the standard American Wire
Gauge (AWG) sizes listed in Table 5-3. The larger the gauge number, the
smaller the wire size. Wires larger than 4/0 (sometimes called 0000) are
usually specified in thousands of circular mils (MCM) instead of having a
gauge number.
The ratio of the diameter of each American Wire Gauge size and the
next one is about 1.1229, and the ratio of their areas is about 1.26. The
area ratio between wire sizes three AWG numbers apart is very close to 2.
For example, #10 wire (10 380 CM) has about twice the area of #13 wire
(5178 CM) and half the area of #7 wire (20 820 CM).
Since resistance is inversely proportional to area, the resistance per unit
length also differs by a factor of about two for wires three AWG numbers
apart. For example, #1 wire (0.1239 Ω/1000′) has half the resistance of #4
wire (0.2485 Ω/1000′) and twice the resistance of #3/0 wire (0.0618 Ω/1000′).
The first M in MCM
is the Roman n
­ umeral
for 1000, not the
­metric prefix mega.
MCM is sometimes
written as kcmil, or
kilo-circular mils.
78
Chapter 5 Resistance and Ohm’s Law
Example 5-8
What size of aluminum wire has a resistance of 1.62 Ω for a length of
2500′?
Solution
Since R =
ρl
,
A
A=
ρl
17 × 2500
=
= 26.2 MCM
R
1.62
From Table 5-3, we see that this area matches AWG #6.
TABLE 5-3 American Wire Gauges
AWG
The resistance of solid
copper is also a function of the copper’s
purity, or the percentage of copper content,
as well as any treatment the copper wire
receives after being
drawn to its AWG
size. These processes
(such as annealing,
bunching, stranding,
tinning, or braiding)
can each affect the
overall wire resistance. The resistances
listed in Table 5-3 for
copper may be used
as a general guideline.
Gauge
mils
circular
mils
211 600
107.22
133 100
67.43
460
11.68
2/0
365
9.27
1
289
3
229
3/0
mm2
10.41
167 800
325
8.25
105 500
53.47
258
6.54
66 370
33.63
4
204
5.19
6
162
0
2
410
Resistance of Solid
Copper at 20°C
Area
mm
4/0
7.35
83 690
85.03
42.41
5.83
52 640
26.67
182
4.62
33 100
16.77
7
144
3.67
20 820
10.55
9
114
13 090
6.63
5
8
4.12
41 740
26 250
128
3.26
102
2.59
10 380
80.8
2.05
6 530
14
64.1
1.63
4 107
16
50.8
18
40.3
10
11
12
13
15
The units Ω/1000 ft
is sometimes listed as
mΩ/ft.
Diameter
17
19
20
21
22
90.7
72.0
2.91
2.31
1.83
16 510
8 234
5 178
57.1
1.45
45.3
1.15
2 048
0.91
1 288
0.72
810
35.9
32.0
28.5
25.3
1.29
1.02
0.81
0.64
3 257
2 583
21.15
13.30
8.37
5.26
4.17
3.31
2.62
2.08
1.65
1.31
1.04
Ω/1000 ft
Ω/km
0.0618
0.202
0.0490
0.0779
0.0983
0.1239
0.1563
0.1970
0.2485
0.3133
0.3951
0.4982
0.405
0.511
0.644
0.812
1.02
1.29
1.63
0.9989
3.27
1.260
1.588
2.003
2.525
2.59
4.12
5.19
6.55
8.25
3.184
10.41
5.064
16.54
4.016
1 022
0.518
10.15
0.326
16.14
642
0.321
2.05
0.7921
0.824
0.411
0.255
0.6282
1 624
0.653
0.160
6.385
8.051
12.80
13.13
20.89
26.32
33.18
41.84
52.63
See Problems 5-28 to 5-33 and Review Questions 5-79 and 5-80.
5-7 Effect of Temperature on Resistance
5-7 Effect of Temperature on Resistance
Resistance
Checking the V/I ratio of conductors at various temperatures shows that
the resistance of most conducting materials increases linearly with temperature except at very hot or very cold temperatures. Temperature has little
effect on the resistance of some alloys, such as constantan. For a few materials, including carbon and other semiconductors, the resistance ­decreases
as the temperature increases.
Figure 5-5 shows the resistance of a conductor increasing with temperature. R1 is the resistance at temperature T1, and R2 is the resistance at
­tem­perature T. The line segment CF shows how the resistance varies for
­temperatures between T1 and T2. The steeper the slope of segment CF, the
more the resistance increases with temperature.
F
R2
x
A
Tx
C
R1
O
B
T1
0°C
E
D
T2
Temperature
Figure 5-5
Effect of temperature on resistance in the majority of materials
Extending CF to the temperature axis produce triangles ABC and ADF.
Since these triangles are similar,
DF
AD
AO + OD
=
=
BC
AB
AO + OB
Substituting the quantities that the sides of the triangles represent gives
R2
x + T2
=
R1
x + T1
(5-6)
where R1 is the resistance of the conductor at temperature T1, R2 is the
­resistance at temperature T2, and x is the difference between 0°C and
the temperature at which the resistance would be zero if it continued to
­decrease linearly at very low temperatures.
Table 5-4 lists values for x for some metals commonly used for conductors. The values for Nichrome™ II and constantan show that x is not the
79
80
Chapter 5 Resistance and Ohm’s Law
a­ ctual temperature at which the resistance becomes zero ohms. However, we
can use x to calculate resistances within the range of temperatures where the
resistance varies linearly with temperature.
TABLE 5-4 Values of x for common conductors
Conductor Material
x (°C)
Silver
243
Copper
234.5
Aluminum
236
Tungsten
202
Nickel
147
Iron
180
Nichrome™ II
6 250
Constantan (55% Cu, 45% Ni)
125 000
Example 5-9
A copper conductor has a resistance of 12.0 Ω at 20°C. Find the resistance at 100°C.
Solution
Since
R2
x + T2
=
,
R1
x + T1
R2 = R1
( 234.5 + 100 ) 12 × 334.5
x + T2
= 12.0 Ω ×
= 15.8 Ω
=
( 234.5 + 20 )
x + T1
254.5
Note that the temperature units, °C, cancel out in this calculation.
Example 5-10
A precision resistor made of constantan wire has a resistance of
10 000 Ω at 20°C. What is its resistance when its temperature rises 20°C?
Solution
R2 = R1
x + T2
125 000 + 40
= 10 000 ×
= 10 002 Ω
x + T1
125 000 + 20
In addition to being able to calculate the effect that temperature has on
r­ esistance, we can use the change in resistance of a certain material to ­calculate
the temperature in locations where it is difficult to place a t­ hermometer.
5-7 Effect of Temperature on Resistance
Example 5-11
The copper winding of an electric motor that has been standing for several hours in a room at 20°C has a resistance of 0.20 Ω. When the motor
has been in use for some hours, the resistance of the winding is found
to be 0.22 Ω. Calculate the temperature rise in the winding.
Solution
R2
x + T2
=
R1
x + T1
0.22
234.5 + T2
=
0.20
234.5 + 20
T2 =
0.22
× ( 234.5 + 20 ) − 234.5 = 45.45ºC
0.20
temperature rise = T2 − T1 = 45.45oC − 20oC = 25oC
Substituting R = ρl/A from Equation 5-4 for R1 in Equation 5-6 gives
R=ρ
l(x + T)
A ( x + 20 )
(5-7)
With values for ρ and x, we can use Equation 5-7 to estimate the resistance
of any metallic conductor at any temperature.
Example 5-12
Find the resistance at 40°C of 300 m of copper wire with a cross-­
sectional area of 1.50 mm2.
Solution
From Tables 5-1 and 5-4, ρ is 1.72 × 10–8 Ω · m and x is 234.5˚C.
R=ρ
l(x + T)
A ( x + 20 )
= 1.72 × 10 − 8 Ω · m ×
= 3.44 ×
= 3.71 Ω
274.5
254.5
( 234.5ºC + 40ºC )
300 m
2 × (
234.5ºC + 20ºC )
1.50 mm
See Review Questions 5-81 to 5-83.
81
82
Chapter 5 Resistance and Ohm’s Law
5-8 Temperature Coefficient of
Resistance
Now we shall derive an alternative method of showing the effect of temperature on the resistance of a conductor. In Figure 5-5, line segment CE
is parallel to the temperature axis. Since resistivity is usually stated for a
temperature of 20°C, we shall let T1 represent 20°C.
Triangle ABC is similar to triangle CEF, so
FE
CE
=
BC
AB
Substituting the quantities that the sides of the triangles represent gives
ΔT
FE
=
, where ΔT is the difference between T2 and 20°C
R1
x + 20
ΔT
Therefore,
FE = R1
x + 20
(
Since CE and BD are parallel,
R2 = ED + FE = R1 + FE = R1 + R1
)
(
ΔT
ΔT
= R1 1 +
x + 20
x + 20
) (
)
The quantity x +1 20 is the temperature coefficient of resistance at 20°C,
the proportion by which the resistance changes per degree of change in
­temperature from 20°C. We represent this coefficient by the Greek letter
α (alpha), so
R2 = R1 ( 1 + αΔT )
(5-8)
where R2 is the resistance of a conductor at any specified temperature,
R1 is its resistance at 20°C, α is the temperature coefficient of resistance
at 20°C for the conductor, and ΔT is the difference between the specified
temperature and 20°C.
Substituting R1 = ρl/A from Equation 5-4 gives a general equation for
the resistance of a conductor:
R=ρ
l
( 1 + αΔT )
A
(5-9)
where R is the resistance of the conductor in ohms, ρ is the resistivity
of the material in ohm metres at 20°C, l is the length in metres, A is the
cross-­sectional area in square metres, α is the temperature coefficient of
resistance of the material at 20°C, and ΔT is the difference between the
temperature of the conductor and 20°C.
Table 5-5 lists temperature coefficients for some materials commonly used
for conductors and resistors.
5-8 Temperature Coefficient of Resistance
TABLE 5-5 Temperature coefficients of resistance at 20°C
Conductor Material
α (K )
−1
Silver
0.003 8
Copper
0.003 93
Aluminum
0.004 3
Tungsten
0.004 5
Nickel
0.006
Iron
0.005 5
Nichrome™ II
0.000 16
Constantan (55% Cu, 45% Ni)
0.000 008
Carbon
−0.000 5
Example 5-12A
What is the resistance of 300 m of copper wire with a cross-sectional
area of 1.5 mm2 at 40°C?
Solution
From Table 5-1 and Table 5-5, ρ is 1.72 × 10−8 Ω · m and α is 0.003 93.
R=ρ
l
( 1 + αΔT )
A
= 1.72 × 10−8 Ω · m ×
= 3.44 × 1.079
= 3.71 Ω
300 m
× ( 1 + 0.003 93 × 20 )
1.5 × 10−6 m2
Example 5-13
A heating element made from Nichrome™ II has a resistance of 16.0 Ω
at 1500°C. Find the resistance at normal room temperature.
Solution
In this example, R2 in Equation 5-8 is given and R1 is unknown. Hence,
R1 =
R2
16.0
16.0
= 12.9 Ω
=
=
( 1 + αΔT )
( 1 + 0.000 16 × 1480 )
1.237
If the required temperature is below 20°C, ΔT will be a negative quantity
and the resulting resistance is less than the resistance at 20°C.
See Problems 5-34 to 5-45 and Review Questions 5-84 and 5-85.
83
K−1 = 1/K , or “per
­Celsius degree”
84
Chapter 5 Resistance and Ohm’s Law
Circuit Check
CC 5-4. What length of iron wire 0.5″ in diameter has a resistance of
3.0 Ω?
CC 5-5. An electric thermometer with a tungsten element has a
­resistance of 10 Ω at 20°C. It is immersed in a fluid and its
­resistance drops to 7.3 Ω. What is the temperature of the fluid?
CC 5-6. In metres, what length of AWG #20 copper wire has a resistance
of 2.5 Ω at 50ºC?
B
5-9 Linear Resistors
For most conductors, a graph of current versus voltage is a straight line,
­indicating a constant resistance (see Figure 5-6). The smaller the resistance,
the steeper the slope of the graph. A resistor that maintains a constant V/I
ratio is a linear resistor. As current through a resistor increases, more heat
is produced in the resistor, raising its temperature. This increase in temperature causes a slight increase in the resistance of most conductor materials. For the common conductor materials such as copper and aluminum, the
change in resistance over the ranges of operating temperatures for most
circuits is so small that these materials are usually considered to be linear
resistors.
Current (mA)
20
5 kΩ
15
10
10 kΩ
5
0
0
Source: © iStock.com/Gueholl
Figure 5-6
Wire-wound resistor
encased in procelain
25
50
Voltage (V)
75
100
Current versus voltage for linear resistors
As the name suggests, wire-wound resistors have a metal wire wound
on a hollow porcelain tube and sealed in position with a porcelain coating. These resistors are usually made with constantan or other alloys with
a temperature coefficient of almost zero. The larger the size of a resistor,
the more readily it can dissipate heat to the surrounding air, and hence
the greater the power rating. (See Chapter 6-2 for more information about
power ratings of resistors.)
Inexpensive, mass-manufactured resistors often have resistances that
vary by 10% or more from their nominal values. While such resistors
Resistance
change (%)
Precision resistors
Carbon-composition
resistor
Source: Photo courtesy of
Ohmite Manufacturing Co.
are adequate for many types of circuits, applications such as measuring
­instruments require resistors that are accurate to within 1% or less of the
nominal resistance. Such precision resistors are often made by depositing
a thin film of metal or carbon on a small ceramic cylinder, to which leads
are ­attached. The film may be etched to adjust the resistance to the specified value before being coated with a layer of insulating material.
Resistors used in electronic devices usually have resistances greater
than a kilohm (a thousand ohms) and pass currents of only a few milliamperes. A carbon-composition resistor is commonly used when the
current through the ­resistor produces less than 2 W of heat. The resistance element consists of finely ground carbon mixed with an insulating
binder such as phenolic and pressed into a cylindrical shape with a wire
lead embedded in each end. The resistance element is then sealed in a
plastic jacket. The length and width of the cylinder, the proportion of
carbon in the mixture, and the way the mixture is compressed determine
the resistance.
Carbon-composition resistors are much cheaper than wire-wound and
film resistors, but the resistance of a carbon-composition resistor increases
if the temperature varies appreciably from 20°C as shown in Figure 5-7.
For moderate temperature variations, the resistance changes by only a
few ­percent, so we can usually treat carbon-composition resistors as linear
­resistors.
85
Source: © iStock.com/catetus
5-9 Linear Resistors
5
4
3
2
1
–50
–25
0
25
50
Temperature (°C)
75
100
Composition resistors are also made with a ceramic consisting of tin
oxide and antimony bound with glass. Ceramic-composition resistors are
particularly useful for circuits where the resistors must withstand voltage
or ­energy surges.
Integrated circuits (ICs) range from simple resistor networks to miroprocessors containing millions of microscopic resistors and transistors. ICs
start as a wafer of highly purified silicon. Pure silicon is a poor conductor
since it has few free charge carriers. The resistors are made by diffusing
tiny regions of the silicon with precisely controlled amounts of elements
that supply free electrons, making the regions more conductive.
Integrated circuits
Source: © iStock.com/alex-mit
Figure 5-7 Resistance-temperature characteristic of carbon-composition
resistors
Chapter 5 Resistance and Ohm’s Law
5-10 Nonlinear Resistors
When an incandescent lamp is switched on, the temperature and the resistance of its tungsten filament increase dramatically. When white-hot, the
filament of an ordinary 60-watt 120-V lamp has a resistance of 240 Ω, but
its resistance at room temperature is about 18 Ω. An incandescent lamp is a
nonlinear resistor. The inrush current at the instant the lamp is turned on
is much greater than its normal operating current:
Normal I =
V
120 V
120 V
V
=
= 0.50 A Inrush I =
=
= 6.6 A
R
240 Ω
R
18 Ω
Fortunately, the mass of the lamp filament is small enough that it gets
white hot in less than a millisecond. Therefore, the current surge is brief,
as shown in Figure 5-8. Nevertheless, switches used with incandescent
lamps have to be designed to withstand the inrush current. To avoid large
inrush currents, heating elements, such as those in stoves, are usually made
from an alloy with a very small temperature coefficient.
7
6
Current (A)
86
5
4
3
2
Switch closes
at t = 0
1
0
1
2
Time (milliseconds)
Figure 5-8
Inrush current of a 60-W incandescent lamp
A resistor with a large negative temperature coefficient, called a thermistor,
can be used to limit inrush currents. Typically, such thermistors have a
­resistance of over 100 Ω at room temperature, but with a current of 1 A
through them, their resistance drops to less than 1 Ω after 10 to 15 s. These
thermistors contain semiconductive metal oxides with a ceramic binder.
Heat produced by current through the resistor breaks covalent bonds in
the metal oxides, creating enough free electrons to reduce the resistance
to a fraction of its value at room temperature. Small thermistors are used
to measure temperatures since a decrease in temperature of less than 20°C
will more than double their resistance.
87
Source: © Omega Engineering Inc.
Reproduced with the permission of Omega
Engineering Inc., Stamford, CT 06907 USA
www.omega.com
5-10 Nonlinear Resistors
Varistors depend on the nonlinear resistance characteristic of zinc oxide
or silicon carbide crystals, which are formed into wafers with a clay binder.
Zinc oxide varistors (also called metal-oxide varistors or MOVs) are used to
protect sensitive electronics from voltage surges. Silicon carbide varistors
(commonly known by the trade name thyrite) can protect high-voltage
systems. They are used as lightning arrestors on power transmission lines.
Temperature has little effect on the resistance of a varistor. Instead a
rapid increase in the number of charge carriers occurs when the potential
difference across the varistor becomes greater than the threshold of the
varistor. As Figure 5-9 indicates, the resulting decrease in resistance is such
that the current through the varistor increases greatly without appreciably increasing the voltage drop across it. Thus, a varistor connected across
the power input to a device can protect it from voltage surges.
Source: Hyper-Sense
Technology Co., Ltd.
Thermistors
Varistors
Voltage surge
supression
Normal operating
voltage
8
6
4
Source: MARTYN F. CHILLMAID/SCIENCE PHOTO LIBRARY
Current (A)
10
2
0
Figure 5-9
30
60
90
Voltage (V)
120
150
Typical varistor characteristic
A photoresistor, photoconductor, or light dependent resistor (LDR)
contains a thin zigzag strip of cadmium sulfide or cadmium selenide. Light
falling on the strip breaks down valence bonds in the cadmium compound,
creating additional charge carriers. The resistance can range from hundreds
of kilohms in the dark to less than 100 Ω in bright daylight. Photoresistors are widely used in light meters, auto-exposure circuits in cameras, and
controllers for outdoor lights.
Photoresistor
88
Chapter 5 Resistance and Ohm’s Law
Thermistors, varistors, and photoresistors all use semiconductors engineered to become more conductive under specific conditions. The schematic
symbols for these nonlinear resistors are shown in Table 5-6.
TABLE 5-6 Schematic symbols for nonlinear ­resistors
General
Thermistor
t°
or
Varistor
V
Photoresistor
See Review Questions 5-86 to 5-87.
5-11 Resistor Colour Code
Many resistors are small enough that any printing on them would be hard
to read. So, most resistors are marked with bands of colour, which have the
further advantage of being visible no matter how the resistor is oriented
(see Figure 5-10).
Some precision
resistors use the first
three bands for digits,
the fourth band for the
multiplier, and the
fifth band for tolerance.
A single-band resistor
only has one black colour band to indicate
its value: zero ohms.
It looks like a resistor
but acts like a wire in
the circuit.
First digit
Second digit
Multiplier
Figure 5-10
Reliability
Tolerance
Colour bands on a resistor
A resistor has from three to six bands, but most common resistors have
only four or five colour bands. The colours are read from the band nearest
to an end of the resistor. On a 4-band resistor, the first three bands indicate
two digits and a multiplier that give the nominal value of the resistor in
ohms. The fourth band indicates the tolerance, which is the maximum percent deviation from the nominal value. A fifth band is sometimes added to
show the reliability, which is the percentage of resistors with a resistance
outside the specified tolerance after 1000 h of use, also known as the fail
rate. Table 5-7 lists the meaning of the colour bands for 4-band resistors
and 5-band reliability resistors. Table 5-8 lists the meaning of the colour
bands for 5-band precision resistors.
5-11 Resistor Colour Code
89
TABLE 5-7 Resistor colour code for 4-band resistors and
5-band reliability resistors
Band
Colour
Black
Brown
Red
Orange
Yellow
Green
First
First Digit
1
2
3
4
5
Second
Second
Digit
Third
Multiplier
Fourth
Tolerance
0
100
2
102
±1%
1
3
4
5
Blue
6
6
Grey
8
8
Violet
White
Gold
7
9
7
9
Silver
101
103
104
1%
0.1%
0.01%
0.001%
105
106
107
108
109
0.1
0.01
None
±2%
Fifth
Fail Rate
±5%
±10%
±20%
TABLE 5-8 Resistor colour code for 5-band precision
resistors
First
First Digit
Second
Second
Digit
Third
Third Digit
Fourth
Multiplier
Black
0
0
0
100
Red
2
2
2
102
Band
Colour
Brown
Orange
Yellow
Green
1
3
4
5
1
3
4
5
1
3
4
5
Blue
6
6
6
Grey
8
8
8
Violet
White
Gold
Silver
None
7
9
7
9
7
9
101
103
104
Fifth
Tolerance
1%
2%
105
0.5%
107
0.1%
106
0.1
0.01
0.25%
5%
10%
20%
Resistors are manufactured with standard nominal values such that any
given resistance is within the tolerance range of at least one of the standard
values. These standardized values for different tolerance ranges are also
known as the E-series, and they can be valid for other types of components,
such as capacitors, inductors, and Zener diodes. The lower the tolerance,
the larger the number of standard values. Table 5-9 lists the two digits for
the standard values for 5% and 10% tolerances. The standard values have
these digits with any multiplier.
Six-band resistors
have colour bands
that indicate three
digits, a multiplier,
the tolerance, and
a temperature coefficient. These highprecision resistors are
used in applications
where temperature is
a critical factor.
90
The most common
E-series are E6, E12,
E24, E48, E92, and
E192.
For example, the E12
series contains 12
two-digit standard
nominal values ranging from 10 to 82,
with a 10% tolerance,
as shown in Table 5-9.
Chapter 5 Resistance and Ohm’s Law
TABLE 5-9 Standard resistor digits
±5%
Tolerance
±10%
Tolerance
10
10
12
12
15
15
18
±5%
Tolerance
±10%
Tolerance
33
33
39
39
47
47
18
56
56
22
22
68
68
27
27
82
82
11
13
16
20
24
30
36
43
51
62
75
91
Example 5-14
Determine the specifications of a reliability resistor coded
brown-black-red-gold-brown.
Solution
First digit: brown = 1
Second digit: black = 0
Multiplier: red = 102
Tolerance: gold = 5%
Reliability: brown = 1%
The nominal resistance is 10 × 102 = 1.0 kΩ. The tolerance is 5%, which
means the actual resistance is between 950 Ω and 1050 Ω. The reliability
of 1% means that one out of every 100 resistors will not lie within the
tolerance range after 1000 h of operation at the resistor’s rated power.
Example 5-15
Interpret these resistor colour codes:
(a) yellow-violet-orange-silver-red, on a precision resistor
(b) yellow-violet-orange-silver-red, on a reliability resistor
(c) green-brown-gold-gold
Solution
(a) Resistance is 473 × 10−2 = 4.73 Ω ±2%.
(b) Resistance is 47 × 103 = 47 kΩ ±10% with a failure rate of 0.1%.
(c) Resistance is 51 × 0.1 = 5.1 Ω ±5% (reliability not specified).
See Problems 5-46 to 5-50 and Review Questions 5-88 and 5-89.
5-13 Voltage-Current Characteristics
91
5-12 Variable Resistors
Source: © Sergpet/Dreamstime/Getstock
Variable resistors have a moving contact, or wiper, that either rotates to
run against a curved resistor (as shown in Figure 5-11) or slides along a
straight resistor. The resistance element is usually carbon composition, cermet (a ceramic and metal mixture), or a wire coil. The resistance between
the wiper and either end of the resistor depends on the position of the
wiper, and varies from zero to the full value of the resistor.
Resistance element
Wiper
Mechanism of a wire-wound variable resistor
A
C
In Figure 5-12, the resistance between terminals A and B is constant.
Since RAC + RCB = RAB, the resistance between terminals A and C increases
as the resistance between terminals C and B decreases.
When all three terminals of a variable resistor are used, it is called a
­potentiometer. Potentiometers provide a simple way of adjusting a v
­ oltage,
and are commonly used as volume and tone controls in audio equipment.
When only the wiper and one end of the resistor are connected, as in
Figure 5-13, the variable resistor is called a rheostat. Rheostats can vary the
current in a circuit, and are sometimes used for motor speed controls.
See Review Question 5-90.
5-13 Voltage-Current Characteristics
Although the resistance of a nonlinear resistor is not constant, the Ohm’s
law relationship R = V/I still applies. It is often convenient to show variations in resistance by plotting a voltage-current characteristic. These
graphs usually have voltage drop on the x-axis and current on the y-axis,
as in Figure 5-14. A low resistance has a graph with a steep slope and a high
resistance has a graph with a shallow slope.
The two straight-line graphs in Figure 5-14 represent linear resistors. For
a linear resistor, the resistance is constant, so we can find the resistance
by calculating the ratio of voltage to current at any point on the graph.
B
Shaft
Figure 5-11 Crosssection of a variable
resistor
A
C
B
Movable arm
(wiper)
Figure 5-12
Schematic of a
potentiometer
C
B
Figure 5-13
Schematic of a rheostat
Chapter 5 Resistance and Ohm’s Law
Low-linear resistance
Nonlinear resistance
ΔI
Current
92
ΔI
High-linear resistance
ΔI
ΔV
Voltage
Figure 5-14
Voltage-current characteristics of resistors
For a given ΔV, ΔI is much greater for a low resistance than it is for a high
resistance.
If the voltage-current characteristic is not linear, we can calculate a
dynamic value of resistance from the ratio of the change in voltage drop to
the associated change in current. Graphs of voltage-current characteristics
are particularly useful for showing the behaviour of nonlinear components
such as varistors.
See Review Question 5-91.
5-14 Applying Ohm’s Law
We can rearrange the equation for Ohm’s law (Equation 5-2) to get an expression for current, I = V/R. The total voltage drop V for a circuit connected between the terminals of a voltage source is equal to the applied
voltage E of the source. Hence, the current is dependent on the applied
voltage and the total resistance of the circuit:
I=
E
R
(5-10)
where I is the current in the circuit (in amperes), E is the applied voltage
(in volts), and R is the total resistance of the circuit (in ohms).
Similarly, we can use Ohm’s law to find an expression for the voltage
drop across a resistor:
V = IR
(5-11)
5-14 Applying Ohm’s Law
where I is the current through a resistor (in amperes), R is the resistance (in ohms), and V is the resulting voltage drop across the resistor
(in volts).
A voltage drop across a resistance is often referred to as an IR drop.
Example 5-16
Calculate the current through a 4.7 kΩ resistor connected across a 9.0-V
source.
Solution
I=
E
9.0 V
=
= 1.9 × 10−3 A = 1.9 mA
R
4.7 kΩ
Example 5-17
Calculate the voltage drop produced by a 15-mA current through a
560-Ω resistor.
Solution
V = IR = 15 mA × 560 Ω = 8.4 V
See Problems 5-51 to 5-69 and Review Question 5-92.
Circuit Check
CC 5-7.
C
Determine the resistance and tolerance for resistors with the
colour bands listed below.
(a) brown-black-yellow-gold
(b) red-red-black-silver
(c) violet-brown-green-black-red, in a precision resistor
CC 5-8. Find the minimum and the maximum acceptable resistances
for the resistors in question CC 5-7.
CC 5-9. How much current will flow if a 120-V source is connected to a
25-kΩ resistance?
93
94
Chapter 5 Resistance and Ohm’s Law
Summary
• Ohm’s law relates the resistance of an electric component to the voltage
across it and the current through it.
• When a voltage is applied to a conductor, it exhibits a resistance or
­opposition to electric current as free electrons collide with atoms in the
conductor.
• The resistance of a conductor depends on its length, its cross-sectional
area, and its composition.
• Resistivity is the resistance of a unit length of a conductor with unit
cross-sectional area.
• The American Wire Gauge specifies diameters for standard sizes of wire
conductors.
• The effect of temperature on the resistance of a material can be calculated using the temperature coefficient of resistance for the material.
• Types of linear resistors include wire-wound, carbon-composition, ceramiccomposition, metal-film, and carbon-film.
• Nonlinear resistors include the thermistor, varistor, and photoresistor.
• Colour bands on resistors indicate their nominal resistance and tolerance. Additional bands can be used to indicate the failure rate and temperature coefficient.
• Variable resistors can be used as either potentiometers or rheostats.
• The voltage-current characteristic is a graph of the current through a
­device versus the voltage across it.
B = beginner
I = intermediate
A = advanced
Problems
B
Section 5-1
5-1.
B
5-2.
B
5-3.
B
5-4.
B
5-5.
B
5-6.
B
B
5-7.
5-8.
Ohm's Law
Find the resistance of a ceiling fan that draws a 0.5-A current from a
120-V source.
What resistance will limit the current in a circuit to 1.5 A when the
applied voltage is 30 V?
A 12.5-A current causes a 110-V voltage drop across a portable electric heater ­element. Find the resistance of this element.
A current of 0.15 A causes a 1.0-V drop across a resistor. Find its
­resistance.
A 15-μA current through a resistor causes a 45-mV drop across it.
Find the resistance of the resistor.
Calculate the resistance of an ammeter that reads 10 A with a 50-mV
drop across it.
What value of resistor has a 7-V drop for a current of 15 mA?
What value of resistor draws 40 mA when connected to a 2.5-kV
source?
Problems
Section 5-4
I
I
I
I
B
B
B
B
B
I
I
I
I
I
I
A
Resistivity
NOTE: Assume a temperature of 20°C for Problems 5-9 to 5-26.
5-9. A conductor with a cross-sectional area of 10 mm2 has a resistance of
1.72 Ω/km. Find the resistance of 500 m of wire of the same material
with a cross-sectional area of 4 mm2.
5-10. A conductor with a cross-sectional area of 2.5 mm2 and a length of
50 m has a resistance of 1.42 Ω. What length of wire of the same
­material with a cross-sectional area of 1.5 mm2 has a resistance of
14.2 Ω?
5-11. A conductor with a diameter of 1.5 mm and a length of 70 m has a
­resistance of 0.65 Ω. Find the resistance of a conductor of the same
material having a length of 40 m and a diameter of 0.8 mm.
5-12. A conductor 40 m in length and 1.2 mm in diameter has a resistance
of 17.3 Ω. What diameter of wire of the same material has a resistance
of 1.0 Ω/m?
5-13. Use Table 5-1 to calculate the resistance of a tungsten wire 250 m
long and with a cross-sectional area of 5.0 mm2.
5-14. Find the resistance of a copper bar with a rectangular cross section
2 cm × 4 cm and a length of 1.5 m.
5-15. Calculate the resistance of 160 m of aluminum wire having a diameter
of 1.6 mm.
5-16. Find the resistance of a tungsten filament 12 cm in length and 0.1 mm
in diameter.
5-17. What length of silver wire 0.5 mm in diameter has a resistance of
0.04 Ω?
5-18. What diameter of copper wire has a resistance of 1 Ω/m?
5-19. Find the diameter of copper wire that will give a relay coil wound
with 85 m of the wire a resistance of 30 Ω.
5-20. Constantan wire 0.4 mm in diameter is wound on a cylindrical core
such that the mean diameter of each turn is 2.0 cm. How many turns
of this wire are needed to make a 24-Ω resistor?
5-21. A carbon rod 0.50 cm in diameter and 10 cm in length has a resistance of 0.153 Ω. Find the resistivity of this carbon rod.
5-22. An electric conductor 40 m in length and 1.8 mm in diameter has a
resistance of 17.3 Ω. What material is used for this conductor?
5-23. A stranded cable consists of eight copper conductors, each having a
diameter of 1.0 mm. Twisting the wires to form the cable makes the
length of each strand 3% longer than the length of the cable. Find the
resistance of 1.0 km of this cable.
5-24. What diameter of solid copper conductor has the same resistance per
kilometre as a stranded cable consisting of six aluminum ­conductors
each having a diameter of 1.3 mm? Assume that the twisted aluminum strands are each 3% longer than the finished cable.
95
96
Chapter 5 Resistance and Ohm’s Law
I
I
5-25. What diameter of copper wire will have a resistance of 10 Ω for a
length of 100 m?
5-26. Calculate the resistivity of manganin (in nanoohm metres) given
that 5.0 m of manganin wire 1.27 mm in diameter has a resistance
of 1.7 Ω.
Sections 5-5 and 5-6
B
B
B
I
I
I
I
B
B
I
I
I
I
Circular Mils and AWG
NOTE: Assume a temperature of 20°C for Problems 5-27 to 5-33.
5-27. Determine the area in circular mils of wires with the following
­diameters:
(a) 0.0063″
(b) 0.035″
(c) 0.565″
5-28. Calculate the resistance of 2500 feet of aluminum wire, 0.150” in
­diameter.
5-29. What length of AWG #22 constantan wire is needed to make a
1.38 Ω resistor?
5-30. What is the diameter, in inches, of 1500 feet of copper wire if its
­resistance is 10 Ω?
5-31. A 1500-foot length of AWG #20 wire has a resistance of 50 Ω. ­Determine
what material the wire is made of by calculating its ­resistivity.
5-32. What length of AWG #10 aluminum wire will have the same resistance as 1500″ of #12 copper wire?
5-33. What is the nearest AWG of Nichrome™ II wire that will have a resistance of 4.5 Ω for an 11′ length?
Sections 5-7
Temperature Effects on Resistance
5-34. A length of copper telephone line has a resistance of 24 Ω at 20°C.
What does the resistance become on a hot summer day with a
­temperature of 36°C?
5-35. Find the resistance at −20°C of an aluminum conductor that has a
­resistance of 1.25 Ω at +30°C.
5-36. What is the resistance at 60°C of 25 m of copper wire having a diameter of 0.50 mm?
5-37. What length of Nichrome™ II wire with a diameter of 0.64 mm has
a resistance of 48 Ω at 200°C?
5-38. If the resistance-temperature graph for brass is extended in a straight
line until the resistance becomes zero, the corresponding temperature would be −480°C. Find the temperature coefficient of brass at
20°C.
5-39. The temperature coefficient of platinum at 20°C is 0.003 per Celsius
degree. Find the temperature at which the resistance of platinum
would become zero if the resistance-temperature graph were extended
as a straight line.
Problems
B
I
I
A
B
A
B
B
B
B
B
5-40. An incandescent lamp draws a 1.0-A current from a 110-V source
to raise the temperature of its tungsten filament to 2800°C. Find the
­resistance of the filament at 20°C.
5-41. At what temperature does the resistance of a nickel rod increase to
105% of the rod’s resistance at 20°C?
5-42. A certain conductor has a resistance of 10.0 Ω at 20°C and 11.35 Ω at
50°C. From which of the materials listed in Table 5-5 is this conductor made?
5-43. An electric motor is set up 250 m from a 120-V source. The current
drawn by the motor is 8.0 A. The temperature of the copper conductors feeding the motor is 40°C. What is the minimum diameter
of wire that can be used without the ­voltage drop in the conductors exceeding 10% of the applied ­voltage?
5-44. The resistance of a length of aluminum wire is 10.5 Ω at 100ºC. What
will its resistance be at −20 ºC?
5-45. A motor with copper windings was tested at an ambient temperature of 20ºC. The voltage across the motor was the 100 V, and the
current through the motor at the start of the test was 5.0 A. D
­ uring
the test the current dropped gradually to a steady-state value of
4.5 A. What was the temperature of the motor at the end of the test?
Section 5-11
Resistor Colour Code
5-46. Determine the value, tolerance, and reliability (if indicated) of resistors with the following colour bands:
(a) orange-white-black-gold-brown, for reliability
(b) red-red-red-red-red, for reliability
(c) green-blue-red-silver-red, for precision
(d) yellow-violet-yellow
5-47. State the colour code for each of the following resistors.
(a) 10 kΩ ±5%, 1% reliability
(b) 2.2 MΩ ±2%, 0.1% reliability
(c) 0.47 Ω ±1%, reliability unspecified
(d) 31.6 Ω ±2%
(e) 825 Ω ±0.5%
(f) 100 kΩ ±5%
5-48. The colour bands on a reliability resistor are yellow-orange-orangegold-red.
(a) What is the minimum resistance for this resistor?
(b) If 56 000 of these resistors are purchased, how many are likely to
be out of tolerance after 1000 h of use?
5-49. What is the colour code for a 2% tolerance resistor with a value of
2.4 Ω?
5-50. A resistor is colour coded brown-green-orange-gold. What is the
mini­mum value it can have?
97
98
Chapter 5 Resistance and Ohm’s Law
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
Section 5-14
Applying Ohm's Law
5-51. Find the current through a 160-Ω resistor connected across a 240-V
source.
5-52. What current through a 150-Ω resistor produces a voltage drop of
75 V across the resistor?
5-53. A 12-Ω heating element has a voltage drop of 95 V across it. Calculate the current flowing through this element.
5-54. Find the current that flows in a circuit with a total resistance of
17 Ω connected to a 117-V source.
5-55. Find the voltage that produces a 300-mA current when applied to a
15-Ω resistor.
5-56. Find the voltage drop across a 125-Ω resistor when a 0.2-A current
flows through it.
5-57. What voltage drop is produced by a 160-mA current flowing through
a 56-Ω resistor?
5-58. What applied voltage causes a current of 0.35 A to flow in a circuit
with a total resistance of 124 Ω?
5-59. A fuse in the power supply of a transistor amplifier has a resistance of
0.02 Ω. Find the current that produces a 500-μV drop across the fuse.
5-60. What current will flow when a 50-μV source is connected to a
25-Ω resistor?
5-61. What current will flow through a 2.7-kΩ resistor connected across a
480-mV source?
5-62. Find the current that produces an 80-V drop across a 6.8-kΩ resistor.
5-63. What applied voltage causes a 200-µA current through a 22-kΩ ­resistor?
5-64. What IR drop is produced by a 5.0-mA current passing through a
15-kΩ load resistor?
5-65. What voltage drop appears across a 2.2-kΩ load resistor when the
current through the resistor is 1.8 mA?
5-66. The contacts of a particular relay close when the current through the
relay’s coil is at least 24 mA. The coil has a resistance of 3.7 kΩ. Find
the minimum applied voltage for operating this relay.
5-67. How much voltage is necessary to send a current of 100 μA through
a 15.0 kΩ resistance?
5-68. A 100 kΩ resistor has a current of 330 μA flowing through it. What is
the voltage drop across this resistor?
5-69. An LED flashlight is powered by a 3.0 V battery. If the effective resistance of the LED is 120 Ω, how much current flow through it?
Review Questions
Section 5-1
Ohm's Law
5-70. Why can Ohm’s law be described in terms of constant ­proportionality?
5-71. Express the ohm in terms of other SI units.
Review Questions
Section 5-2
The Nature of Resistance
5-72. Why does the V/I ratio of a circuit indicate its ability to oppose electric current rather than its ability to permit current?
5-73. Energy must be expended to force current to flow through a resistance. Where does this energy come from? Where does it go?
5-74. How is the resistance of a given conductor related to the number of
free electrons in the conductor?
Section 5-3
Factors Governing Resistance
5-75. Why does shortening a conductor decrease its resistance?
5-76. Why is resistance inversely proportional to the square of the diameter of a conductor?
Section 5-4
Resistivity
5-77. Why does the term resistivity apply to the material of an electric conductor, rather than to a particular conductor?
5-78. Given an accurate resistance-measuring device, how would you
go about determining the resistivity of a sample of an unknown alloy?
Section 5-6
American Wire Gauge
5-79. How are gauge number, diameter, and resistance related for conductors made in American Wire Gauge sizes?
5-80. North American electrical codes specify a limit of 15 A for current
through AWG #14 house wiring. What is the reason for this limit? Is
it possible for a current greater than 15 A to flow through #14 wire?
Section 5-7
Effect of Temperature on Resistance
5-81. A fuse contains a narrow strip of metal and is connected such that
the current in the protected electric circuit flows through this strip
of metal. A current greater than the rating of the fuse will melt the
metal strip. Which has the greater resistance, a 10-A fuse or a 20-A
fuse? Explain your reasoning.
5-82. Explain why the filament of an incandescent lamp is many times
hotter than a Nichrome™ II wire-wound resistor that has the same
resistance and carries the same current.
5-83. Before an electric motor was started, the resistance of its copper
windings at room temperature was 50.0 Ω. After 30 min of operation, the resistance was 53.0 Ω. A half-hour later the resistance was
54.3 Ω, and after a further 40 min it was 54.7 Ω. After a total of 3 h, the
­resistance settled at a steady value of 55 Ω. Plot a graph of temperature against time for the motor. Explain the reasons for the shape of
the graph.
99
100
Chapter 5 Resistance and Ohm’s Law
Section 5-8
Temperature Coefficient of Resistance
5-84. Use the data in Table 5-5 to draw a temperature-resistance graph
for carbon, similar to Figure 5-5. Find the temperature x for
carbon.
5-85. Find the temperature coefficient of copper at 0°C. Explain why the
temperature coefficient at 0°C differs from the temperature coefficient at 20°C.
Section 5-10
Nonlinear Resistors
5-86. Draw a graph similar to Figure 5-5 for a resistor having a fairly pronounced negative temperature coefficient. Would such a resistor be
useful as a current regulator? Explain.
5-87. Resistors made from resistance wire or metal film usually have a
positive temperature coefficient of resistance while resistors made
from semiconductor materials generally have a negative temperature coefficient. Account for this difference.
Section 5-11
Resistor Colour Code
5-88. What information does each of the five colour bands on a resistor
­convey? Provide an example for a precision resistor and a reliability
resistor.
5-89. Alphanumeric markings are sometimes used for surface-mounted
resistors. Look up this labelling system, and determine the resistances indicated by the following labels:
(a) 220
(b) 2R2
(c) 102
(d) 4R7
Section 5-12
Variable Resistors
5-90. Differentiate between a potentiometer and a rheostat.
Section 5-13
Voltage-Current Characteristics
5-91. The voltage-current characteristic of a carbon-composition resistor is a straight line for all values of current (and voltage) within
its ­normal operating limits. If the graph is extended beyond these
limits, the voltage-current characteristic starts to curve slightly. In
which direction will it bend? Why?
Section 5-14
Applying Ohm's Law
5-92. What variables determine the current in a circuit?
Practice Quiz
Integrate the Concepts
When a 120-V, 150-watt incandescent light bulb is switched on, the initial
current through it is in the order of 12.5 A. After the bulb heats up, the current drops to about 1.25 A. How hot does the filament get?
Practice Quiz
1.
Ohm’s law states that
(a) resistance equals voltage times current
(b) voltage equals current divided by resistance
(c) resistance equals voltage divided by current
(d) current equals resistance divided by voltage
2.
When the voltage across a resistor is halved, the current through it is
(a) halved
(b) doubled
(c) quadrupled
(d) unchanged
3.
The resistance that passes a current of 20 mA with an applied voltage
of 1.0 V is
(a) 5.0 Ω
(b) 0.02 Ω
(c) 20 Ω
(d) 50 Ω
4.
5.
6.
The resistance of a conductor is
(a) directly proportional to its length and its cross-sectional area
(b) inversely proportional to its length and its cross-sectional area
(c) directly proportional to its length and inversely proportional to
its cross-sectional area
(d) inversely proportional to its length and directly proportional to
its cross-sectional area
The resistance of an aluminum conductor at room temperature is
100 Ω. If the conductor is 20 m long, its radius is
(a) 13.6 nm
(b) 84.6 μm
(c) 1.8 nm
(d) 42.4 μm
The resistance of 125 miles of copper wire 0.025″ in diameter is
approximately
(a) 50 Ω
(b) 10 kΩ
(c) 11 kΩ
(d) 270 kΩ
101
102
Chapter 5 Resistance and Ohm’s Law
7. What parameter indicates how much effect temperature has on the
resistance of a material?
8. What is the key property of a linear resistor?
9. Give two examples of nonlinear resistors.
10. A resistor colour-coded orange-blue-brown-silver has a resistance of
(a) 360 Ω ±10%
(b) 36 Ω ±10%
(c) 3.60 Ω ±10%
(d) 0.36 Ω ±10%
11. The colour bands for a resistor with a nominal value of 15 kΩ
±20% are
(a) brown-blue-orange-gold
(b) brown-green-orange
(c) brown-blue-orange
(d) brown-green-orange-gold
12. What is the difference between a rheostat and a potentiometer?
13. A resistance is
(a) directly proportional to voltage and directly proportional
­current
(b) directly proportional to voltage and inversely proportional
­current
(c) inversely proportional to voltage and directly proportional
­current
(d) inversely proportional to voltage and inversely proportional
current
14. In the circuit of Figure 5-15, the value of the resistor is
(a) 12.5 Ω
(b) 1.25 Ω
(c) 125 Ω
(d) none of the above
2.0 A
+
E
250 V −
Figure 5-15
R=?
to
to
to
to
Practice Quiz
15. In the circuit of Figure 5-16, the current through the resistor is
(a) 10 mA
(b) 10 A
(c) 100 mA
(d) none of the above
I=?
+
E
330 V –
Figure 5-16
R
33 kΩ
103
6
Work and Power
Thus far we have discussed the relationships among v
­ oltage, current, and
resistance in an electric circuit. Since voltage produces charge in motion
(current) through resistors, the source must be supplying energy and the
circuit is using that energy to do work. The next step is to calculate the
energy and the power associated with the source and the resistors in the
circuit.
Chapter Outline
6-1
6-2
Energy and Work 106
Power
107
6-3
Efficiency
6-5
Relationships Among Basic Electric Units
6-4
6-6
110
The Kilowatt Hour 112
Heating Effect of Current
113
113
Key Terms
power 107
watt 107
efficiency 110
horsepower 111
kilowatt hours 111
British thermal unit (BTU)
114
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• differentiate among work, energy, and power
• perform calculations based on the equations
relating power, voltage, current, and
resistance
• calculate the power rating for a resistor in a
­specific ­application
Photo sources: iStock.com/Peter Englested Jonasen
• define efficiency in terms of energy and power
• calculate the efficiency of a device
• calculate the total energy consumed by an
­electrical ­system
• calculate the heat produced by an electric
­current
106
Chapter 6 Work and Power
6-1 Energy and Work
As explained in Section 2-7, work is energy transferred to a body or system. Since energy conversions involve such transfers, work is done whenever e­ nergy is converted from one form into another. For example, in the
circuit shown in Figure 6-1, the generator does work on free electrons in
its windings while converting mechanical energy into electric energy, and
free electrons do work on atoms in the heater while converting the electric
energy into heat.
Mechanical
energy input
Heat
energy
output
Electrical
conductors
Electric heater
(resistance)
Generator
Figure 6-1
The megagram is
commonly called a
tonne or metric ton (t):
1 Mg = 1 t = 1000 kg
Energy conversions in a simple electric circuit
Example 6-1
How much electric energy must be supplied to an electric motor to
raise an elevator car having a mass of 4.0 t a distance of 40 m? Assume
that 10% of the electric energy supplied to the motor is lost as heat.
Solution
The work, W, done by the motor to raise the elevator car is equal to the
force acting on the car times the distance, d, it travels. The force equals
the car’s mass, m, times the gravitational acceleration, g, which is about
9.8 m/s2.
F = mg = 4.0 t × 9.8 m/s2 = 4.0 × 103 kg × 9.8 m/s2 = 39.2 kN
W = Fd = 39.2 kN × 40 m = 1.568 MJ
The work done by the motor equals 90% of the electric energy supplied to the motor. Therefore,
Win =
1.568 MJ
= 1.7 M J
0.90
See Problems 6-1 to 6-4 and Review Question 6-80 at the end of the chapter.
6-2 Power
107
6-2 Power
If a more powerful motor were installed for the elevator in Example 6-1,
it would transfer energy at a greater rate and raise the car more quickly.
However, the work done would be the same as with the smaller
motor. These comparisons illustrate the relationship between work
and power.
Power is the rate of doing work.
The letter symbol for power is P.
The watt (W) is the SI unit of power.
One watt is equal to one joule per second: 1 W = 1 J/s.
The watt is named for
the Scottish engineer
and inventor James
Watt (1736–1819),
who developed the
steam engine.
Equation 6-1 shows the relationship between power and work:
P=
W
t
(6-1)
where P is power in watts, W is work in joules, and t is time in seconds.
When dealing with work and power, keep in mind that joules are a
measure of energy while watts are a measure of the rate of transferring
­energy.
Example 6-2
At what rate must electric energy be supplied to the electric motor in
Example 6-1 to raise the elevator 40 m in 4.0 min?
Solution
P=
1.742 MJ
Win
=
= 7.3 kW
t
4.0 × 60 s
We can use the equations for potential difference and current to relate
these quantities to power in a circuit.
Since V =
W
(Equation 2-4),
Q
W = QV
108
Chapter 6 Work and Power
Similarly, I =
Q
(Equation 2-2), so
t
t=
Q
I
Substituting for W and t in Equation 6-1 gives
So,
P=
QV
W
I
=
= QV ×
=V×I
t
Q/I
Q
P = VI
(6-2)
where P is power in watts, V is voltage drop in volts, and I is current in
­amperes.
We can also substitute V = IR (Equation 5-11) into Equation 6-2 to get
P = IR × I
P = I2R
(6-3)
where P is power in watts, I is current through the resistance in amperes,
and R is resistance in ohms.
Similarly, substituting I = V/ R into Equation 6-2 gives
P=
V2
R
(6-4)
where P is power in watts, V is voltage drop across the resistance in
volts, and R is resistance in ohms.
Example 6-3
A lamp draws a current of 2.00 A when connected to a 120-V source.
How much power is going to the lamp?
Solution
P = EI = 120 V × 2.00 A = 240 W
6-2 Power
109
When designing circuits, we need to specify the power rating of a resistor as well as its resistance. Since a resistor converts all power it consumes
into heat, the power rating is the rate at which the resistor can dissipate
heat without damage. Heat dissipation depends on the surface area of the
resistor and therefore on its physical size. Standard carbon-composition
resistors are manufactured in sizes ranging from 1/8 W to 2 W, as shown
in Figure 6-2.
2W
1W
1_ W
2
1 cm
1_ W
4
1_ W
8
Figure 6-2 Approximate sizes of carbon-composition resistors
Example 6-4
A 10-kΩ resistor is connected into a circuit where the current through
it is 50 mA. What is the minimum power rating required for this
­resistor?
Solution
P = I 2R = (50 mA)2 × 10 kΩ = 25 W
The resistor should be rated for at least 25 W.
If a resistor exceeds
its power rating, it
can overheat and fail,
often causing irreparable damage to itself.
It may also cause
damage to the circuit
board on which it is
installed, and any
nearby components.
110
Chapter 6 Work and Power
Example 6-5
What is the highest voltage that can be applied to a 3.3-kΩ, 2.0-W
­resistor without exceeding its heat-dissipating capability?
Solution
Since
P=
V2
R
V 2 = PR
V = √ PR
E = V = √ PR = √ 2.0 W × 3.3 kΩ = 81 V
and
See Problems 6-5 to 6-28 and Review Questions 6-81 to 6-85.
Circuit Check
A
CC 6-1. A Miran CO2 surgical laser has an output power of 25 W.
How much energy does its beam transfer in 1.2 s?
CC 6-2. How much power is used by an electric toaster that has a resistance of 10 Ω and operates from a 120 V supply?
CC 6-3. Determine which of the following resistors are likely to be damaged by overheating. Justify your answer.
(a) 560 Ω, ½ W, with 75 V across it
(b) 3 Ω, 20 W, with 4 A through it
(c) ¼ W, with 0.25 mA through it and 40 V across it
6-3 Efficiency
Almost all equipment converts some of its input energy into a form that
does not produce useful work. This wasted energy is usually in the form of
heat. For the motor in Example 6-1, some of this heat comes from friction
and some of it from current heating the conductors in the motor. The wasted
energy increases the cost of making the equipment as well as the cost of
­running it since the equipment has to be designed to dissipate the wasted
energy safely. The efficiency of a device indicates how much of the input
energy, Win, the device converts into useful work, Wout.
Efficiency is the ratio of useful output energy to total input energy.
The letter symbol for efficiency is the Greek letter η (eta).
η=
Wout
Win
(6-5)
We usually express efficiency as a percentage. For example, the efficiency of the system in Example 6-1 is 90%.
6-3 Efficiency
111
Example 6-6
What is the efficiency of an electric hoist if its motor uses 60 KJ to raise
a 300-kg mass through 18 m?
Wout = 300 kg × 9.8 m/s2 × 18 m = 52.9 kJ
Solution
η=
Since P =
52.9 kJ
Wout
=
× 100% = 88%
Win
60 kJ
W
(Equation 6-1),
t
W = Pt
Substituting for Wout and Win in Equation 6-5 gives
η=
Wout
Pout × t
=
Win
Pin × t
η=
So,
Pout
Pin
(6-6)
For electric motors, Pout is mechanical power output, and Pin is electric
power input. All types of power can be measured in watts. However in
North America the mechanical power output of motors is often expressed
in terms of horsepower (hp), a unit invented by James Watt:
1 hp = 550 foot-pounds/second ≈ 746 W
(6-7)
Example 6-7
Find the input power for an electric motor that has an output of 24 hp
and an efficiency of 85%.
Solution
Pout = 24 hp ×
Pin =
746 W
= 17.9 kW
1 hp
17.9 kW
Pout
=
= 21 kW
η
0.85
See Problems 6-29 to 6-38 and Review Questions 6-86 to 6-89.
There are several
­different definitions
for horsepower.
Each gives a slightly
­different conversion factor between
­horsepower and
watts.
112
Chapter 6 Work and Power
6-4 The Kilowatt Hour
A joule is a relatively small quantity of energy. A typical home uses ­millions
of joules of electrical energy each day. Since power × time = energy, we
could use the product of any unit of power with any unit of time as a measure of energy. In fact, a joule is equal to a watt second. If we measure
power in kilowatts and time in hours, we can use kilowatt hours (kW·h)
as a unit for energy:
W = Pt
where W is work or energy in kilowatt hours, P is power in kilowatts, and
t is time in hours.
Since 1 kW = 1000 W and 1 h = 3600 s,
1 kW·h = 1000 W × 3600 s = 3.6 × 106 J = 3.6 MJ
(6-8)
Many North American electric utilities measure their customers’ energy
consumption in kilowatt hours. Most meters that record electric energy
consumption are a form of motor with an aluminum disk that rotates at
a speed directly proportional to both the applied voltage and the current.
Since P = EI, the speed of the disk is proportional to the power used at
any given moment, and the number of rotations indicates the energy consumed (power times time).
Example 6-8
At 7¢ per kilowatt hour, how much will it cost to leave a 60-W lamp
burning for five days?
Solution
W = Pt = 60 W × 24 h × 5 = 7200 Wh = 7.2 kW·h
Cost = 7.2 kW · h ×
7¢
= 50.4¢
kW · h
See Problems 6-39 and 6-40 and Review Questions 6-90 and 6-91.
6-6 Heating Effect of Current
113
6-5 Relationships Among Basic
Electric Units
All of the quantities discussed in this chapter are interrelated. If we measure any three of these quantities for a given circuit, we can calculate values for all of the others. For example, given voltage, current, and elapsed
time, we can determine resistance, work, and power.
As shown for power in Section 6-2, straightforward algebraic manipulation of the defining equations gives formulas for each quantity in terms of
various combinations of the others. Table 6-1 lists some of the more useful
formulas. When dealing with the applied voltage for a circuit, we can substitute the symbol E for V in these formulas.
See Problems 6-41 to 6-70.
TABLE 6-1 Relationships among basic electric units
Quantity
Defining
Equations
Voltage
V = W/Q
Current
Resistance
Power
Work and
energy
Unit and
Unit Symbol
Definition of
Unit
volt, V
joules per
coulomb
I = Q/t
ampere, A
coulombs
per second
R = V/I
ohm, Ω
P = W/t
volts per
ampere
watt, W
W = Fd
joules per
second
joule, J
newton
metres
Useful Derived
Equation
V = IR
V = P/I
V = √PR
transpose Equation 5-2
transpose Equation 6-2
transpose Equation 6-4
R = V 2/P
R = P/I 2
transpose Equation 6-4
transpose Equation 6-3
I = V/R
I = P/V
I = √P/R
P = VI
P = I 2R
P = V 2/R
W = Pt
6-6 Heating Effect of Current
The heat produced by current flowing through a resistor is put to practical
use in devices such as kettles, toasters, stoves, space heaters, and hot-water
tanks. About 4.186 J of electrical energy produces the heat required to raise
the temperature of 1 g of water by 1°C (or 1 K). The equation relating the
quantity of heat to the change in temperature of a mass is
Q ≈ 4186 mCΔT
Derivation
(6-9)
where Q is the heat in joules, m is the mass in kilograms, C is the specific
heat of the material, and ΔT is the change in temperature in kelvins.
­Specific heat is the ratio of the heat required to raise the temperature of
a given mass of a substance by 1 K to that required to raise the temperature of the same mass of water by 1 K.
transpose Equation 5-2
transpose Equation 6-2
transpose Equation 6-3
Equation 6-2
Equation 6-3
Equation 6-4
transpose Equation 6-1
The relationship
­between electrical
­energy and heat was
first measured by
James Joule.
When analyzing problems with heat transfer, the letter symbol
Q can represent the
quantity of heat, measured in joules.
114
Chapter 6 Work and Power
Example 6-9
How long will it take to boil 4.3 kg of water starting from 20°C with a
120-V, 10-A heating element and an efficiency of heat transfer of 80%?
Solution
By definition, the specific heat of water is 1. So, the heat required is
Q = 4186 mCΔT = 4186(4.3)(1)(100 − 20) = 1.44 MJ
Compared to most
other substances,
water has a very high
specific heat. This
means it is easy for
water to store heat
energy, which is an
important factor in
sustaining life on
Earth.
Since heat transfer is only 80% efficient, the heating element must
­produce
1.44 MJ
= 1.80 MJ
0.80
From Table 6-1, W = Pt = VIt.
So,
t=
1.80 MJ
W
1 min
=
= 25 min
= 1500 s ×
( 120 V ) ( 10 A )
VI
60 s
In the system of imperial units, the basic unit of heat is called the British
thermal unit or BTU.
One BTU is the amount of heat required to raise the temperature of
one pound of water by 1°F.
1 BTU ≈ 1055 J.
With foot-pound units, the equation relating the quantity of heat to the
change in temperature of a mass becomes
Q = mCΔT
where Q is the heat in BTUs, m is the mass in pounds, C is the specific heat of
the material, and ΔT is the change in temperature in degrees Fahrenheit.
See Problems 6-71 to 6-79 and Review Questions 6-92 to 6-95.
6-6 Heating Effect of Current
Circuit Check
B
CC 6-4. A 220 V motor draws a current of 51.8 A when operating at full
load. If the efficiency of the motor is 81.6%, determine its fullload output in horsepower.
CC 6-5. Calculate the cost of electricity for a 30-day period for a home
with the energy usage listed below. Assume that electricity costs
6.4¢/kW·h.
Average Daily Use
Purpose
Power
Duration
Lighting
Heating
Dryer
Stove
Water heater
Misc.
0.5 kW
10 kW
4 kW
5 kW
3 kW
5 kW
8h
6h
1h
3h
4h
5h
CC 6-6. A 12-Ω heater is connected to a 220-V supply. How much water
can this heater raise from freezing to boiling in
20 minutes if the efficiency of heat transfer is 90%?
115
116
Chapter 6 Work and Power
Summary
• Work is done whenever one form of energy is converted into another.
• Power is the rate of doing work.
• Power is the product of the voltage across a device and the current
through it.
• A resistor is limited by its ability to dissipate heat quickly.
• The efficiency of a device is the ratio of the useful output energy to the
total input energy and is equal to the ratio of useful output power to the
total input power.
• The kilowatt hour is the most common unit of electrical energy monitored by utility companies.
• The quantity of heat a substance can transfer depends upon its own
mass, its specific heat, and the change in temperature.
B = beginner
I = intermediate
A = advanced
Problems
B
B
Section 6-1
6-1.
6-2.
B
6-3.
B
6-4.
B
How much energy does it take to hoist a 25-kg parcel 13 m?
How much energy is required to pump 1500 kg of water to the
­surface from a well 5.0 m deep?
What distance will a 100-kg pile-driver ram travel if it uses 14.7 kJ of
energy?
Find the mass of an object that requires 81 kJ of energy to lift it 4.5 m.
Section 6-2
6-5.
B
6-6.
B
6-7.
B
6-8.
B
6-9.
B
6-10.
B
6-11.
B
6-12.
I
6-13.
Energy and Work
Power
How much power is used by a toaster that draws a 4.5-A current
from a 120-V source?
How much power is used by a soldering iron that has a 110-V drop
between its terminals while drawing a current of 1.2 A?
If a 500-mA current through the heater of a guitar amplifier tube
produces 3.0 W of heat, what is the voltage drop across the heater?
At what rate is electric energy converted by the heater of a cathoderay tube if the heater draws 0.6 A from a 6.3-V power supply?
What is the maximum current that a 1/2-W resistor can handle if a
voltage drop of 80 V develops across it?
At what rate is energy converted in a resistor that produces 6.0 kJ of
heat in 10 min?
At what rate does a 2.5-A current through a 91-Ω resistor produce
heat?
What power rating must a 12-kΩ resistor have to safely pass a current of 250 mA?
What is the minimum common resistance value that will limit the
current to 24 mA flowing through a 1/8-W resistor?
Problems
B
B
B
B
B
B
B
B
B
B
B
A
B
I
B
I
B
I
B
B
I
6-14. At what rate does a 1.2-Ω motor-starting resistor convert electric
energy while the current through it is 60 A?
6-15. What is the maximum current that a 18-kΩ, 10-W resistor can handle
without overheating?
6-16. What is the maximum current that can flow through a 5-kW, 8-Ω
resistor?
6-17. What power rating would you select for a 470-Ω resistor that has to
develop a voltage across it of 3.3 V?
6-18. The armature winding of an electric motor has a resistance of 0.2 Ω.
How much power is lost as heat when the voltage developed across
the winding is 8 V?
6-19. What is the minimum value of resistance that can be connected
across a 120-V source if the power drawn from the source is not to
exceed 0.45 kW?
6-20. What power is released as heat from a 22-kΩ resistor that develops
400 V across its terminals?
6-21. Find the applied voltage for a 1.2-kW heating element that has a
resistance of 34 Ω when hot.
6-22. Find the voltage drop across a 75-Ω 3.0-W decorative lamp.
6-23. How much energy does a 750-W electric heater use in 24 h?
6-24. How much heat is produced in 1 h in the winding of a motor that
has an I2R loss of 360 W?
6-25. How much power is required to raise a 280-lb object 10 m in 10 s?
6-26. A voltmeter with a resistance of 180 kΩ draws 80 mW of power
when connected across a voltage source. Find the potential difference of the source.
6-27. An electric heater produces heat at the rate of 45 kJ/min when the
current through the heater is 9.4 A. Find the resistance of the heater.
6-28. A fuse element with a resistance of 0.02 Ω is designed to melt when
the current through it produces more than 5 W of heat. What is the
current rating of this fuse?
Section 6-3
Efficiency
6-29. Find the current drawn from a 240-V source by a 6.0-hp electric
motor with an efficiency of 86%.
6-30. How much work can be done by an 80% efficient electric motor
drawing 2 A from a 120-V source for 6 h?
6-31. Calculate the efficiency of an elevator that uses 2.8 × 106 J of electric
energy to raise a 4.0-t mass 55 m.
6-32. What is the efficiency of a 1.8-hp electric motor when the electric
power input is 2.2 kW?
6-33. How much energy does a pumping system with an overall efficiency
of 75% use to raise 1250 kg of water 10.5 m?
6-34. Find the output in horsepower of the pumping system in Problem 6-33 if the system pumps the water at a rate of 1250 kg/h.
117
118
Chapter 6 Work and Power
I
B
I
B
B
B
B
B
B
B
B
B
B
B
B
6-35. The output of a power supply for a transistor amplifier is 25 V at
2.4 A. The efficiency of the power supply is 80%. How much current
does the power supply draw from a 120-V source?
6-36. A central air conditioner draws 12.0 A from a 230-V source. The efficiency of the air conditioner is 80%. How much work does the air
conditioner do when it runs continuously for 1 h?
6-37. The following devices are simultaneously drawing energy from the
12.0-V storage battery in a car:
• two 6-W tail lights
• two parking lamps with a hot resistance of 12 Ω each
• a radio drawing 5.0 A
• the fan motor for the car's heater, which develops 0.05 hp at 75%
­efficiency
How much power is the battery supplying?
6-38. If the overall efficiency of a radio transmitter is 48%, how much input
energy does it need to produce an output of 50 kW from 7:00 A.M. to
midnight?
Section 6-4
The Kilowatt Hour
6-39. How much money can be saved by operating a 12-Ω clothes dryer
from a 240-V power line for 5 hours every week for a year during
off-peak hours instead of on-peak hours? Assume that off-peak
electric energy costs 7.7¢/kW·h, and on-peak electric energy costs
15.7¢/kW·h.
6-40. Find how much electric energy a household used in May given that:
• The household pays 6¢/kW·h for electric energy.
• The meter reading at the end of February was 73 067 kW·h.
• The meter reading at the end of May was 74 267 kW·h.
• The total bill for the energy used in March and April was $48.00.
Section 6-5
Relationships Among Basic Electric Units
6-41. Find the voltage drop produced by a 2.4-A current through a ­42‑Ω
resistor.
6-42. At what rate is electric energy converted into heat in the resistor in
Problem 6-41?
6-43. How long will it take the resistor in Problem 6-42 to convert
2.0 kW·h of energy?
6-44. What voltage must be applied to a 33-kΩ resistor to make it dissipate 180 mW of heat?
6-45. Calculate the power rating for a 72-Ω heater that passes a 6.0-A
­current.
6-46. How long will it take the heater in Problem 6-45 to consume 9400 J
of electric energy?
6-47. How much charge passes through the heater in Problem 6-46?
6-48. Find the current that produces 550 μW of heat in a 1.2-kΩ resistor.
6-49. What resistance draws 12.5 μA from a 220-mV source?
Problems
B
B
B
B
I
B
B
B
B
B
B
B
B
I
B
B
B
B
B
B
B
B
I
B
I
6-50. What current does a 40-W lamp draw from a 117-V source?
6-51. What resistance produces 50 W of heat when connected to a 110-V
source?
6-52. Find the applied voltage that produces an 8-A current through a
650‑W heater.
6-53. Find the current that produces 400 W of heat in a 45-Ω resistor.
6-54. Find the applied voltage that makes a 2.7-kΩ resistor convert 0.15
kW⋅h in 8 h.
6-55. Calculate the efficiency of a motor with a 1-kW input and a 1-hp ­output.
6-56. Find the voltage drop created by a 6.8-mA current flowing through
a 27-Ω resistor.
6-57. Find the applied voltage that delivers 50 μW to a 300-Ω resistance.
6-58. Find the resistance that has a voltage drop of 7.0 V when the current
through the resistor is 18 mA.
6-59. What power rating should the resistor in Problem 6-58 have?
6-60. What current will flow when a 20-μV signal is applied to a 75-Ω load?
6-61. What power is fed to the load in Problem 6-60?
6-62. What current will a 4.5-kW load draw from a 240-V source?
6-63. If a 1.25-hp starter motor has 80% efficiency, what current does the
motor draw from a 12-V battery?
6-64. How much work can a fully charged 1.5-V rechargeable battery do
if its ­capacity is 2500 mA·h?
6-65. Find the resistance of a voltmeter that reads 120 V when a 40-μA
current passes through it.
6-66. How long will it take for 2.8 C of electric charge to pass through the
voltmeter in Problem 6-65?
6-67. What is the resistance of the copper bus bars feeding an aluminum
refining cell if a 4000-A current through them causes a 620-mV drop
across them?
6-68. How much electric energy is lost as heat in 8 h in the bus bars of
Problem 6-67?
6-69. Find the resistance of an ammeter shunt that has a 5-mV drop across
it when the current through it is 9.99 A.
6-70. Calculate the resistance of a 15-A fuse if heat must be developed at
the rate of 4.3 W to melt it.
Section 6-6
Heating Effect of Current
6-71. How much energy is required to raise the temperature of 4.0 kg of
water by 45°C?
6-72. How much heat must an electric stove element produce to increase
the temperature of 1.0 kg of water from 20°C to 85°C if one-third of
the heat is lost to the surroundings?
6-73. If heat is transferred to 1 kg of water at the rate of 1 kW, how long
will it take to raise the temperature of the water by 45°C?
6-74. Find the power rating for a stove element that raises the temperature of 1.0 kg of water from 20°C to 80°C in 4 min, given that only
75% of input energy is transferred to the water.
119
120
Chapter 6 Work and Power
A
A
A
A
A
6-75. An electric kettle takes 15 min to raise the temperature 1.2 kg of
water from 12°C to 95°C. The efficiency of this heating process is
89.4%. Find the heater current if the voltage is 120 V.
6-76. A heating element raises the temperature of 100 kg of water from
20°C to boiling point in 10 min. The efficiency of heat transfer is 85%.
The current through the element is 90.6 A. Find its resistance.
6-77. A heating element operates at 240 V and has a resistance of 20 Ω. The
element heats 15 kg of water for 15 min. The initial temperature of
the water is 20ºC, and the heat transfer is 81% efficient. Find the final
temperature of the water.
6-78. A 10-Ω heating element operating from a 550-V source heats 100 kg
of water from 20°C to boiling point in 20 min. Calculate the efficiency
of the heat transfer.
6-79. The 3.0-kW heater in a vat holding 10 kg of cooking oil takes 27 min
to heat the oil to the desired operating temperature of 240ºC. Given
that the oil has a specific heat of 0.45 and the efficiency of heat
­transfer is 86%, determine the original temperature of the oil.
Review Questions
Section 6-1
Energy and Work
6-80. What is the distinction between work and energy?
Section 6-2
Power
Section 6-3
Efficiency
6-81. A 5-W and a 50-W 1000-Ω resistor are both made with NichromeTM II
wire. Compare the conductor sizes, the overall dimensions, and the
operating temperatures of these two resistors.
6-82. A 35-W soldering iron is used to solder miniature radio components,
whereas a 150-W iron is needed to solder a lead to the radio chassis.
Explain why the greater wattage is needed for soldering to the ­chassis.
6-83. SI defines the volt as the difference in electric potential between two
points of a conductor carrying a constant current of 1 A when the
power dissipated between these points is equal to 1 W. Show that
this definition is consistent with the one used in Section 2-7.
6-84. Why is it common to find larger electric motors on passenger elevators than on freight elevators even though the freight elevators carry
heavier loads?
6-85. What is wrong with the wording of the question: “How much power
is consumed by a toaster drawing 3 A from a 110-V source?”
6-86. Why would a 5-hp motor with a 90% efficiency be physically smaller
than a 5-hp motor with a 60% efficiency?
Practice Quiz
6-87. Why is it possible to express efficiency in terms of the ratio of output
power to input power even though efficiency is defined as the ratio
of output energy to input energy?
6-88. What is wrong with the wording of the question: “How many joules
are there in a horsepower?”
6-89. What is wrong with the wording of the question: “What is the efficiency of an electric motor that has a power input of 5 hp and a
power output of 4 hp?”
Section 6-4
The Kilowatt Hour
6-90. Why is the kilowatt hour often used instead of the joule as a unit for
electric energy?
6-91. What unit is equivalent to a watt second?
Section 6-5
Relationships Among Basic Electric Units
6-92. Why must electric kettles have thermostatic switches to open the
­circuit when the kettle boils dry?
6-93. Some small heaters contain an electric heating coil and a small fan.
The heating element glows brighter when the fan is turned off.
­Explain why.
6-94. Does the heater coil in Question 6-93 draw more current when the
fan is turned off? Explain.
6-95. (a) What is the SI unit for heat?
(b) How is the BTU defined?
Integrate the Concepts
Most building codes require every fire exit in a public building to have
a lighted exit sign that remains on at all times. One of the most common
types of exit signs was designed to use two 15-W incandescent light bulbs.
These two bulbs can now be replaced by a single panel of light-emitting
diodes (LEDs) that uses only 3 W. How long will it take a $20 LED panel to
pay for itself in energy savings? Assume an energy cost of $0.075/kW⋅h.
Practice Quiz
1.
Which of the following statements are true?
(a) Energy is the capacity to do work.
(b) Power is the amount of work done per second.
(c) Efficiency is the ratio of input power to output power.
(d) One BTU is the amount of heat required to raise the temperature
of one pound of water by 1°C.
121
122
Chapter 6 Work and Power
2.
How much power is used by a hair dryer that draws 1.5 A when
­connected to a 120-V wall socket?
(a) 80 W
(b) 112 W
(c) 125 W
(d) 180 W
3.
The highest voltage that can be applied to a 1.5-kΩ, 10-W resistor without risk of damaging it is
(a) 121 V
(b) 12.2 V
(c) 124 V
(d) 122 V
4.
What physical property affects the power rating of a resistor?
5.
If an amplifier has an efficiency of 75% and an input power of 15 W,
the output power is
(a) 5 W
(b) 11.25 W
(c) 1.13 kW
(d) 50 mW
6.
At 10¢ per kilowatt hour, how much will it cost to leave a 100-W lamp
on for one week?
(a) 16.8¢
(b) 24¢
(c) $1.68
(d) $16.8
7.
If 135 kJ of heat is needed to raise 5.1 kg of a substance from 18.0°C to
31.5°C, the specific heat of the substance is about
(a) 0.20
(b) 0.50
(c) 2.0
(d) 26
PART
Resistance Networks
II
Part II explains the principles and techniques we require for the analysis of
any resistance network. We can apply the relationships among current, voltage, resistance, work, and power to all electric circuits, no matter how complex
they may be.
7
Series and Parallel Circuits
8
Series-Parallel Circuits
9
Resistance Networks
10
Equivalent-Circuit Theorems
11
Electrical Measurement
Photo source: © iStock.com/omada
7
Series and Parallel
Circuits
All the electric circuits in residential, automotive, and ­industrial
applications are based on two fundamental types of circuits: series
and parallel. Working with sophisticated circuits requires a good
understanding of series and parallel connections.
Chapter Outline
7-1
Resistors in Series
7-3
Double-Subscript Notation 130
7-5
Characteristics of Series Circuits 131
7-2
7-4
7-6
7-7
126
Voltage Drops in Series Circuits 128
Kirchhoff’s Voltage Law 130
Internal Resistance 133
Cells in Series
136
7-8
Maximum Power Transfer
7-10
Kirchhoff’s Current Law 141
7-9
7-11
7-12
7-13
7-14
Resistors in Parallel
139
137
Conductance and Conductivity 142
Characteristics of Parallel Circuits 145
Cells in Parallel
Troubleshooting
148
150
Key Terms
series circuit 126
equivalent circuit 127
active circuit
element 129
passive circuit
element 129
polarity 129
Kirchhoff’s voltage
law 130
internal resistance 135
open-circuit voltage 135
voltage regulation 135
controlled source 135
maximum power
transfer 138
overload 139
parallel circuit 139
branch 140
duality 141
equivalent resistance 141
Kirchhoff’s current
law 141
conductance 142
siemens 142
conductivity 145
short circuit 152
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• recognize series and parallel circuits
• calculate the total resistance in a series
circuit
• label the polarity of voltages across voltage
sources and resistors in a series circuit
• identify the polarity of a voltage using doublesubscript notation
• apply Kirchhoff’s voltage law to a series circuit
• relate the ratio of resistances to the voltage
drops across resistors in a series circuit
• explain the effect of internal resistance on the
terminal voltage of a practical voltage source
• calculate the terminal voltage of cells in series
and the r­ esulting circuit current
Photo sources: © iStock.com/powerofforever
• state the condition under which a voltage
source delivers maximum power to a load
• apply Kirchhoff’s current law to a parallel circuit
• calculate the total conductance in a parallel
circuit
• calculate the equivalent resistance of two
resistors in ­parallel
• define conductivity
• relate the ratio of conductances to the
currents through two resistors in a parallel
circuit
• calculate the terminal voltage of cells in
parallel and the resulting circuit current
• troubleshoot series and parallel circuits
126
Chapter 7 Series and Parallel Circuits
7-1 Resistors in Series
A series circuit can be identified by the connection between components
or by the current through them. For example, in the circuit of Figure 7-1,
R1 and R2 are connected in series because no other component or branch is
connected to the junction of R1 and R2. None of the junctions in this circuit
have a second branch, so all of the components are in series. Thus, a series
circuit has only one path for current.
Conventional current
I
R1
+
E
R2
−
R3
Electron f low
Figure 7-1
Simple series circuit
Since resistors cannot store charge, the flow of electrons into R1 is
equal to the flow of electrons out of R1, which in turn is equal to the flow
of electrons into R2. Similarly, the flow of electrons through the source is
equal to the flow of electrons through any other part of the series circuit.
Georg Ohm was the first person to recognize this key property of series
circuits.
The current is the same in all parts of a simple series circuit.
Conversely, two or more electric components are in series if a common
current flows through them.
Since the current in a series circuit is common to all components, it is not
necessary to use a subscript with I to distinguish the current through the
various components. In the circuit of Figure 7-1, I represents the current
through R1, R2, and R3, as well as the current through the source and the
connecting wires.
Suppose that resistor Rl is made with 2 m of Nichrome™ wire, while
­resistor R2 contains 1 m and R3 contains 3 m of the same wire. An electron
flowing through the circuit passes through a total of 6 m of Nichrome™
wire. Since the resistance of an electric conductor is directly proportional
to its length, the total resistance of this circuit is the sum of the individual
­resistances. Generalizing,
7-1 Resistors in Series
The total resistance of a series circuit equals the sum of all the individual resistances in the circuit:
RT = R1 + R2 + R3 + . . .
(7-1)
Once we know the total resistance of a series circuit, we can use Ohm’s
law to find the common current:
I = E/RT
This current produces a total voltage drop VT, equal to the applied voltage E. Since the current through all the resistors is equal, the voltage drops
across equal resistances are also equal.
Example 7-1
What current will flow in a series circuit consisting of a 45-V source,
a 20-Ω, a 10-Ω, and a 30-Ω resistor?
Solution
RT = R1 + R2 + R3 = 20 Ω + 10 Ω + 30 Ω = 60 Ω
I=
E
45 V
=
= 750 mA
RT
60 Ω
The current drawn from the source is exactly the same for a single
60-Ω resistor connected to its terminals as for the 20-Ω, 10-Ω, and 30-Ω
resistors connected in series. Therefore, we can think of the resistances
in series as being replaced by a single equivalent resistance Req. Figure 7-2
shows the equivalent circuit for the series circuit in Figure 7-1. When we
analyze more elaborate circuits in later chapters, we simplify a circuit
by replacing two or more components in series with a single equivalent
component.
+
E
Req
−
Figure 7-2
Equivalent circuit of Figure 7-1
127
128
Chapter 7 Series and Parallel Circuits
Example 7-2
Three resistors are connected in series as shown in Figure 7-1. R1 and
R3 are 10 kΩ and 22 kΩ, respectively, and the potential difference of the
source is 16 V. What resistance for R2 gives a current of 340 μA?
Solution
RT =
circuitSIM
walkthrough
E
16 V
=
= 47 kΩ
I
340μA
R2 = RT − ( R1 + R3 ) = 47 kΩ − ( 10 kΩ + 22 kΩ ) = 15 kΩ
Multisim Solution
Download Multisim file EX7-2 from the website.
The circuit is Figure 7-1 with an ammeter inserted and with the component values as shown below:
E = 20 V R1 = 8.2 kΩ R3 = 12 kΩ
An ohmmeter indicates a total resistance of 35.2 kΩ.
Calculate the value of R2. Double-click on R2 in the circuit to reveal the
value of R2.
Calculate the value of the current in the circuit.
Run the simulation for the circuit, and check the value of the current as
displayed on the ammeter.
See Problems 7-1 to 7-4 and Review Questions 7-48 and 7-49 at the end
of the chapter.
7-2 Voltage Drops in Series Circuits
Ammeter is short for
ampere meter.
Figure 7-3 shows the simple series circuit of Figure 7-1 with a voltmeter
and an ammeter added. Although the positions of R1 and R3 have changed
in the circuit diagram, the connections between the components are the
same as in Figure 7-1. The ammeter must be connected in series with the
­circuit in order to measure the current through it. Since the voltmeter is connected between points B and C, it measures the potential difference across
R2. The ideal voltmeter draws negligible current and thus does affect the
behaviour of the circuit.
In the circuit of Figure 7-3, electrons experience a potential rise inside the
voltage source as it uses energy to move them from the positive terminal
to the negative terminal. As electrons flow around the external circuit, they
7-2 Voltage Drops in Series Circuits
Conventional current
A
+
Electron flow
V1
B
+
+
+
V2
V
−
E
−
R1
−
R2
−
C
+
V3
R3
−
−
Figure 7-3
A
+
D
Polarity of voltage drops in a series circuit
lose the potential energy they gained while moving through the source.
Hence, the electrons at point D are at a higher potential than those at point
C, and electrons at point B are at a higher potential than those at point A.
Similarly, electrons at point B are at a lower potential than those at point
C, and so on.
If negative ions in the voltage source are free to move, they flow in
the same direction as the electrons. Any positive charge carriers flow in
the ­op­posite direction. Positive charge carriers inside a voltage source
experience a ­potential rise as they move from the negative to the positive
terminal.
A voltage source is an active circuit element since it generates electric
­energy (at the expense of some other form of energy). Passive circuit
elements consume electric energy. Electrons flow from the negative terminal
to the positive terminal of a passive component. Thus, the polarity of the
voltage drops indicates the direction of current flows in these components.
Note that an ammeter is a passive circuit element. Electrons enter the
­ammeter through its negative terminal and leave through its positive
terminal.
In the circuit of Figure 7-3, electrons flow counterclockwise around
the circuit. We can, therefore, mark the polarities of all voltage drops with
− and + signs. The end of the resistor at which electrons enter is marked −
and the end of the resistor from which electrons leave is marked +. In the
circuit of Figure 7-3, the positive end of R3 is connected to the negative end
of R2, and so on. So, point C is positive with respect to point D, and point B
is even more positive with respect to point D. Similarly, point C is negative
with respect to point A, and so on.
For circuit diagrams, the nature of the actual charge carriers is not particularly important, but we often need to keep track of current direction.
129
130
Chapter 7 Series and Parallel Circuits
The direction of both conventional current and electron flow are shown in
Figure 7-3. The polarities of the voltage drops are exactly the same using
­either approach, but conventional current “flows” from the positive terminal
of a passive circuit element to its negative terminal. The remainder of the
text will show only the direction of conventional current unless the nature
of the actual charge carriers is important.
See Review Question 7-50.
7-3 Double-Subscript Notation
In Figure 7-3, V1 represents the voltage drop across R1. The polarity of V1 is
obvious since terminal A is positive with respect to terminal B.
Double-subscript notation lets us indicate the polarity of voltage drops.
For example, VAB represents the potential at point A with respect to point B.
The second subscript normally designates the reference point. In Figure 7-3,
VAB is a positive voltage, VBA is negative, and VAB = −VBA.
Double-subscript notation is essential for keeping track of voltage
­polarities and current directions in three-phase circuits such as those in
Chapter 29. For most of the dc circuits in this book, single-subscript notation is adequate.
7-4 Kirchhoff’s Voltage Law
Multiplying both sides of Equation 7-1 by I gives
Since
IRT = IR1 + IR2 + IR3 + . . .
E = IRT and V = IR,
E = V1 + V2 + V3 + . . .
(7-2)
In a series circuit, the sum of all the voltage drops across the individual resistances equals the applied voltage.
The German physicist Gustav Robert Kirchhoff (1824–1887) discovered
that a similar relationship applies to any complete electric circuit.
Kirchhoff’s voltage law: In any complete electric circuit, the algebraic
sum of applied voltages equals the algebraic sum of the voltage drops.
Kirchhoff’s voltage law provides us with another method for solving
­ xample 7-1.
E
7-5 Characteristics of Series Circuits
Example 7-1A
What current will flow in a series circuit consisting of a 45-V source, a
20-Ω, a 10-Ω, and a 30-Ω resistor?
E = IR1 + IR2 + IR3
Solution
45 = 20I + 10I + 30I = 60I
45 V
I=
= 750 mA
60 Ω
See Review Question 7-51.
7-5 Characteristics of Series Circuits
Since the current is the same through all components of a series circuit,
­applying Ohm’s law to the voltage drops gives
I=
V1
V2
V3
VT
=
=
=
R1
R2
R3
RT
By transposing variables in each pair of equal terms, we get
V1
R1
=
,
V2
R2
V2
R2
=
,
V3
R3
and so on.
In a series circuit, the ratio of the voltage drops across two resistances equals the ratio of the resistances.
We can use these ratios to save a step when solving for a specific para­
meter of a series circuit.
Example 7-3
A 20-kΩ resistor and a 15-kΩ resistor are connected in series to a 140-V
source. Find the voltage drop across the 15-kΩ resistor.
Solutions
Often, drawing and labelling a schematic diagram, such as Figure 7-4,
can help us see the relationships among the circuit parameters.
R1
20 kΩ
+
140 V
−
Figure 7-4
15 kΩ
R2
Circuit diagram for Example 7-3
131
132
Chapter 7 Series and Parallel Circuits
Long Method Use the current to find the voltage drop.
Source: © iStock.com/powerofforever
RT = R1 + R2 = 20 kΩ + 15 kΩ = 35 kΩ
I=
E
140 V
=
= 4.0 × 10−3 A = 4.0 mA
RT
35 kΩ
V2 = IR2 = 4.0 mA × 15 kΩ = 60 V
Short Method Use ratios to find the voltage drop.
Christmas lights are
connected in series: if one
bulb fails, the whole string will
go out.
Since
R2
15 kΩ
V2
R2
= 60 V
=
, V2 = E
= 140 V ×
RT
35 kΩ
E
RT
Since the same current flows through all components of a series circuit,
changing any one component affects the current through all the components. Therefore, control components such as switches, fuses, and rheostats
are connected in series with a load, as shown in Figure 7-5.
Switch
+
E Source
−
Figure 7-5
Fuse
Rheostat
Lamp
Control components in series with a load
The following characteristics can help us recognize a series circuit:
• The current is the same in all parts of a series circuit.
• The total resistance is the sum of all the individual resistances:
RT = R1 + R2 + R3 + . . . .
• The applied voltage is equal to the sum of all the individual voltage drops:
E = V1 + V2 + V3 + . . . .
• The ratios of voltage drops equal the ratios of resistances.
• A change to any component of a series circuit affects the current through
all components.
See Problems 7-5 to 7-14 and Review Questions 7-52 to 7-54.
7-6 Internal Resistance
Circuit Check
CC 7-1.
A
Complete the table listing the voltages in the circuit of Figure 7-6.
8.0 Ω
A
B
2.0 Ω
C
5.0 Ω
12 V
D
5.0 Ω
12 Ω
F
E
Figure 7-6
VAF
VBF
VCF
VDF
VEF
VBE
VCE
VDE
VAC
VBC
VDC
VEC
CC 7-2. Find the unknown resistances in the circuit of Figure 7-7.
20 V
R1
I
+
100 V
−
R2
3.0 Ω
R3
50 V
Figure 7-7
CC 7-3. What resistance connected in series with a 100-Ω resistor will
dissipate heat at a rate of 20 W when the combination is connected to a 120-V source?
7-6 Internal Resistance
In Section 2-8 we noted that the terminal voltage of a source is equal to the
EMF of the source only under open-circuit conditions. The EMF depends on
the nature of the energy-converting action in the source and is independent of the circuit current. However, as shown in Figure 7-8, the potential
133
Chapter 7 Series and Parallel Circuits
Open-circuit voltage
Terminal
voltage
depends
on load
current
Rint
E
+
−
Constant
potential
difference
Termin
a
l volta
VT
0
Figure 7-8
ge
Vint
E
Voltage
134
VT
Current
Equivalent circuit for a practical voltage source
difference between the two terminals of a practical voltage source decreases
as the current drawn from the source increases. For example, a drop in
terminal voltage causes a car’s headlights to dim when the starter motor is
drawing a large current from the car’s battery.
In some circuits, the change in terminal voltage of the source is small
enough to ignore. However, in many circuits, particularly electronic applications, a decrease in the terminal voltage of the source can significantly
­affect the operation of the circuit. For such circuits we can treat the practical
voltage source as consisting of an internal resistance connected in series
with an ideal source that has a constant potential difference (see Figure 7-8).
We can use this equivalent circuit to calculate the terminal voltage and
­energy losses of the actual source.
When no current is drawn from the practical source, the voltage drop
across the internal resistance is Vint = IRint = 0 since I = 0. Therefore, the opencircuit voltage of the practical source is equal to E, the constant ­potential
difference generated by the ideal source. The following example illustrates
how we can determine the internal resistance of a practical ­voltage source.
Example 7-4
A voltmeter that draws negligible current reads 6.0 V when connected
to the terminals of a battery with no other load. When a 5-Ω resistor
is added, as shown in Figure 7-9, the voltmeter reads 5.0 V. Find the
internal resistance of the battery.
Rint
V 5.0 V
Ideal
source
RL = 5 Ω
+
E 6.0 V
−
Figure 7-9
Circuit diagram for Example 7-4
7-6 Internal Resistance
Solution
Since the open-circuit voltage is 6.0 V, the potential difference of the
ideal source is 6.0 V. Applying Kirchhoff’s voltage law gives
E = Vint + VL
Vint = E − VL = 6.0 V − 5.0 V = 1.0 V
From Ohm’s law, the current in the circuit with the 5-Ω resistor connected is
VL
5.0 V
I=
=
= 1A
RL
5Ω
Vint
1.0 V
Rint =
=
= 1Ω
I
1A
Alternative Solution
Vint
Rint
=
RL
VL
1.0 V
Vint
Rint = RL
= 5Ω ×
= 1Ω
VL
5.0 V
The terminal voltage of the voltage source is equal to the total of the voltage drops in the rest of the circuit, VT. We can use Kirchhoff’s voltage law
to write an equation for the terminal voltage:
VT = E − IRint
(7-3)
where VT is the terminal voltage of a practical voltage source, E is the
open-circuit voltage of the source, I is the current drawn from the source,
and Rint is its internal resistance.
If we can increase the EMF, E, of the voltage source to match the increase
in the voltage drop IRint, the terminal voltage VT will remain constant.
This arrangement is one type of voltage regulation. Since the EMF of a battery depends on the chemical reaction in the cells, we cannot apply this
type of voltage regulation to batteries. However, the operators of powergenerating stations can increase the voltage generated as the current demand
increases. In electronic power supplies, the source voltage can be regulated
by a feedback circuit that monitors the output terminal voltage. A voltage
source with a regulation circuit is called a controlled source. Section 10-3
discusses controlled sources in more detail.
For heavy-duty sources, such as large storage batteries or power generators, the terminal voltage under normal load is essentially the same as
the open-circuit voltage. Some smaller voltage sources do have relatively
high internal resistances. For the circuits in this text, we shall consider the
internal resistance of the sources to be zero, unless otherwise stated.
See Problems 7-15 to 7-18 and Review Question 7-55.
135
136
Chapter 7 Series and Parallel Circuits
7-7 Cells in Series
Most cells have a nominal voltage between 1.1 V and 1.5 V. The current
rating of a cell depends on its size and type. Identical cells connected in
series provide higher voltages than a single cell. The current rating will be
the same as for each individual cell since the current in a series circuit is
the same in all parts of the circuit. If cells with different capacities are connected in series, the lower capacity cells will be used up first. Most batteries for portable ­devices have cells connected in series, like those shown in
Figure 7-10.
S cells in series
Ecell
Rcell
Rcell
Ecell
Ecell
Ebat
+
Figure 7-10
Rcell
−
Equivalent circuit of a series-connected battery
If S is the number of cells connected in series, Ecell is the EMF of each cell,
and Rcell is the internal resistance of each cell, then
Ebat = SEcell
(7-4)
total battery internal resistance is Rbat = SRcell
(7-5)
total battery EMF is and
Example 7-5
How much current flows through an external resistance of 4.0 Ω connected to a battery consisting of ten cells connected in series, with each
cell having an EMF of 1.5 V and an internal resistance of 0.20 Ω? Find
the terminal voltage of the battery.
Solution
Ebat = SEcell = 10 × 1.5 V = 15 V
Rbat = SRcell = 10 × 0.20 Ω = 2.0 Ω
The total resistance in the circuit, RT = Rbat + RL = 2.0 Ω + 4.0 Ω = 6.0 Ω
Applying Ohm’s law to the whole circuit gives
7-8 Maximum Power Transfer
I=
137
Ebat
15 V
= 2.5 A
=
RT
6.0 Ω
VT = IRL = 2.5 A × 4.0 Ω = 10 V
VT = Ebat − IRbat = 15 V − ( 2.5 A × 2.0 Ω ) = 10 V
or
Figure 7-11 shows the equivalent circuit for the battery.
IL
R bat
2.0 Ω
RL = 4.0 Ω
VT
ET
15 V
Figure 7-11
Equivalent circuit for Example 7-5
Multisim Solution
Download Multisim file EX7-5 from the website.
The battery consists of ten cells, each having an EMF of 1.5 V and an internal resistance of 0.2 Ω. These cells and an external resistance of 4.0 Ω
are all connected in series.
Run the simulation and check the values of VT and IL displayed on the
meters.
Note that these values match those calculated using Equations 7-4
and 7-5.
See Problems 7-19 to 7-22.
7-8 Maximum Power Transfer
To determine how the internal resistance of the voltage source and the
­resistance of the load affect the behaviour of a circuit, we examine how
­various parameters of a circuit change as the load resistance increases. The
­internal resistance and voltage of a voltage source are independent of the
external circuit. For the circuit of Figure 7-12, these parameters remain constant at Rint = 6.0 Ω and E = 120 V, so we can calculate other parameters of
the circuit using the following equations:
current:
I=
E
120
=
RL + Rint
RL + 6.0
circuitSIM
walkthrough
Chapter 7 Series and Parallel Circuits
VL = IRL
load voltage:
PL = I2 RL
load power:
internal resistance power: Pint = I2Rint = 6.0 I2
η=
efficiency:
Rint
PL
× 100%
PL + Pint
6.0 Ω
RL
VT
+
Constant
120 V
voltage
−
Figure 7-12
Circuit for observing the effects of changing RL
15
90
10
60
5
30
0
0
Load current
0
8
4
6Ω
Figure 7-13
Load voltage
and efficiency
Load power
12 16 20 24 28 32
Load Resistance, RL (Ω)
36
40
600
100
450
75
300
50
150
25
0
Efficiency, η (%)
120
Load Voltage, VL (V)
20
Load Power, PL (W)
We can use a spreadsheet, a graphing calculator, or circuit simulation
software to graph these parameters for values of RL from 0 Ω to 40 Ω, as
shown in Figure 7-13.
Load Current, IL (A)
138
0
Effects of varying load resistance
Although none of the graphs in Figure 7-13 are linear, we can see that as
the load resistance increases, the current decreases while the load voltage
and the efficiency increase. In fact, the graphs for the load voltage and the
efficiency have exactly the same shape. The load power rises to a maximum when RL reaches 6.0 Ω and then tapers off gradually as RL increases
further. So, the maximum power transfer occurs when the load resistance
has the same value as the internal resistance of the source.
From graphs like Figure 7-13, we find that for all circuits the condition
for maximum power transfer to the load is
RL = Rint
(7-6)
7-9 Resistors in Parallel
139
Further, when the load resistance equals the internal resistance of the source,
1
I = Isc
2
where Isc is the short-circuit current (the current when RL = 0).
VL =
1
E
2
(7-7)
(7-8)
η = 50%
(7-9)
The maximum power transfer does not coincide with maximum efficiency.
When RL = Rint, the power lost as heat in the internal resistance of the source
is equal to the power transferred to the load. A load resistance that is two to
three times the internal resistance of the source results in appreciable reduction in wasted power with only a small reduction in power output.
For a circuit where the output voltage is more critical than the power
output (as in transistor voltage amplifiers), the load resistance should be
roughly five times the internal resistance of the source. A load resistance
much less than the internal resistance of the source reduces the power
­output and the efficiency while increasing the heat produced in the internal
resistance of the source. Such overloads can ruin a voltage source.
As shown in Figure 7-13, the slope of the load power graph is zero at maximum power. Appendix 2-1 derives Equation 7-6 using basic differential calculus.
See Problems 7-23 to 7-29 and Review Questions 7-56 to 7-61.
7-9 Resistors in Parallel
Figure 7-14 shows two different ways of drawing a circuit diagram for
the same simple parallel circuit. Circuit diagrams usually show interconnecting conductors as either horizontal or vertical lines, as in Figure 7-14(a).
However, to illustrate the nature of a parallel circuit, we redraw the circuit
diagram by combining the junctions that are directly connected to each
other, producing Figure 7-14(b).
IT
+
E
IT
I1
I2
I3
R1
R2
R3
−
+
E
A
I1
I2
I3
R1
R2
R3
−
(a)
Figure 7-14
B
IT
IT
(b)
imple parallel circuit: (a) Customary configuration,
S
(b) Equivalent configuration
Using arrows to mark
the direction of the
current in each branch
of a circuit makes it
easier to see how these
currents combine into
the total current.
140
Chapter 7 Series and Parallel Circuits
We can identify a parallel circuit by connections among the components. In Figure 7-14, E, R1, R2, and R3 are all in parallel because they are all
connected between the same two points, A and B. Since each of the resistors is connected directly across the voltage source, V1 = V2 = V3 = E. We
can omit the subscripts since the voltage is common to all the components
­connected in parallel.
Two or more electric components are in parallel if a common voltage
appears across all of the components.
For a series circuit, we determined the total resistance in order to find
the current in the circuit. For a parallel circuit, we find the total current
first, and use it to determine the resistance of the circuit. Given the applied
­voltage and the values of each resistance in the circuit of Figure 7-14, we
can solve for the current in each branch by using Ohm’s law. If we think
of the current in each branch in terms of electrons flowing through the
branch, it is apparent that the current through the source must be the sum
of the branch currents.
In a simple parallel circuit, the total current is the sum of all the
branch currents:
IT = I1 + I2 + I3 + . . .
Example 7-6
(7-10)
For the circuit in Figure 7-14, assume that R1 is 40 Ω, R2 is 30 Ω, R3 is
20 Ω, and E is 120 V. What single resistance would draw the same current from the source?
Solution
I1 =
V1
120 V
=
= 3.0 A
R1
40 Ω
I3 =
V3
120 V
=
= 6.0 A
R3
20 Ω
I2 =
V2
120 V
=
= 4.0 A
R2
30 Ω
7-10 Kirchhoff’s Current Law
141
IT = I1 + I2 + I3 = 3.0 + 4.0 + 6.0 = 13.0 A
Req =
E
120 V
=
= 9.23 Ω
IT
13.0 A
Multisim Solution
circuitSIM
Download Multisim file EX7-6 from the website. The circuit is
Figure 7-14 with the component values as indicated below:
E = 20 V R1 = 80 Ω R2 = 50 Ω R3 = 40 Ω
walkthrough
Calculate IT, I1, I2, and I3.
Run the simulation and check the values of IT, I1, I2, and I3 as displayed
on the ammeters.
For a series circuit, the total resistance is greater than the resistance of any
individual resistor. However, the resistance in a parallel circuit (e.g., the
combination of resistors in Example 7-6) is less than the resistance of any
­individual resistor. Hence, when dealing with parallel circuits, we do not
speak of total resistance. Instead, we speak of the equivalent resistance,
Req, of two or more resistors in parallel.
Parallel circuits have characteristics that, in many respects, are similar
but opposite to those of series circuits.
See Problems 7-30 to 7-34.
Circuit Check
B
CC 7-4. If a battery has an open-circuit terminal voltage of 6 V and an internal resistance of 1 Ω, what will the terminal voltage be when
the load current is 250 mA?
CC 7-5. Calculate the internal resistance of a source that has an efficiency
of 80% when connected to a 10-Ω load.
7-10 Kirchhoff’s Current Law
Kirchhoff extended the total-current principle of Equation 7-10 to apply to
all electric circuits.
Kirchhoff’s current law: At any junction in a circuit, the algebraic
sum of the currents entering the junction equals the algebraic sum
of the currents leaving the junction.
The “similarbut-­opposite”
characteristic of series
and parallel electric
circuits is sometimes
referred to as the
principle of ­duality.
142
Chapter 7 Series and Parallel Circuits
Example 7-7
Find the current in the R2 branch of the circuit of Figure 7-15.
IT = 12 A
Node
X
I2
I1
+
50 V
−
Figure 7-15
R1
R2
10 Ω
Circuit diagram for Example 7-7
Solution
I1 =
V
50 V
=
= 5.0 A
R1
10 Ω
The current flowing into junction X is IT and the currents flowing away
from junction X are I1 and I2. Therefore,
IT = I1 + I2
I2 = IT − I1 = 12 A − 5.0 A = 7 A
See Problem 7-35.
7-11 Conductance and Conductivity
Before the adoption of
SI units, the common
name for the unit of
conductance was the
mho (ohm spelled
backward). The siemens was named in
honor of the G
­ erman
engineer Ernst
Werner von Siemens
(1816–1892).
As Examples 7-6 and 7-7 demonstrate, the total current is always greater
than the current through any branch of a parallel circuit. Therefore, the
equivalent resistance is always less than the smallest of the branch resistances. The more resistors we connect in parallel, the smaller the equivalent
resistance becomes. In other words, the more resistors we connect in parallel, the more readily the circuit can pass current since there are more parallel
branches for current to flow through.
Conductance is a measure of the ability of an electric circuit to pass
current.
The letter symbol for conductance is G.
The siemens (S) is the SI unit of conductance: 1 S = 11Ω .
Conductance is the reciprocal of resistance:
G=
1
R
(7-11)
7-11 Conductance and Conductivity
where G is the conductance of a circuit in siemens and R is the resistance
of the same circuit in ohms.
Dividing both sides of Equation 7-10 by E (or V, since they are the same
for simple parallel circuits), we get
IT
I1
I2
I3 . . .
=
+
+
+
E
V
V
V
Substituting Ohm’s law into Equation 7-11 gives
G=
Therefore,
1
1
I
=
=
R
V/I
V
GT = G1 + G2 + G3 + . . .
(7-12)
In parallel circuits, the total conductance is equal to the sum of the
conductances of all the individual branches.
The equivalent resistance is simply Req = 1/GT .
We can use conductances to find the equivalent resistance for a
parallel circuit without calculating the total current. Try Example 7-6 using
conductance:
Example 7-6A
For the circuit in Figure 7-14, assume that R1 is 40 Ω, R2 is 30 Ω, R3 is
20 Ω, and E is 120 V. What single resistance would draw the same current from the source?
Solution
1
1
1
+
+
= 0.1083 S
40 Ω
30 Ω
20 Ω
1
1
=
=
= 9.23 Ω
GT
0.1083 S
GT = G1 + G2 + G3 =
Req
When only two resistors are connected in parallel, Equation 7-12 becomes
GT = G1 + G2 =
1
1
R1 + R2
+
=
R1
R2
R1R2
143
144
Chapter 7 Series and Parallel Circuits
Req =
and
R1R2
R1 + R2
(7-13)
For two resistors in parallel, the equivalent resistance equals their
product over their sum.
We can rearrange Equation 7-13 to find the resistance R2 that we must
connect in parallel with a given resistance R1 to obtain a desired equivalent
resistance:
R2 =
R1 − Req
R1Req
(7-14)
Now we can use Ohm’s law and Kirchhoff’s current law to find the
­current through resistor R2:
V1 = V2 = I1R1 = I2R2
I1R1 = (IT − I2)R1
I2 = IT
(
R1
R1 + R2
)
(7-15)
When N equal resistors are connected in parallel, we can collect the terms
in Equation 7-12 and show that
Req =
R
N
(7-16)
where R is the resistance of each of the parallel resistors and N is the number
of resistors.
Example 7-8
What is the equivalent resistance of a 1-kΩ and a 4-kΩ resistor in
­parallel?
Solution
Req = =
R1R2
1 kΩ × 4 kΩ
4 000 000
=
= 800 Ω
=
( 1 kΩ + 4 kΩ )
R1 + R2
5000
7-12 Characteristics of Parallel Circuits
We defined resistivity as the resistance of a unit length and cross section
of a material. Conductivity is defined in a similar way.
The conductivity of a material is the conductance of a unit length
and cross section of that material.
The letter symbol for conductivity is the Greek letter σ (sigma).
Conductivity is measured in siemens per metre.
Since conductance is the reciprocal of resistance, conductivity is the reciprocal of resistivity:
σ=
1
ρ
(7-17)
See Problems 7-36 to 7-38 and Review Questions 7-62 to 7-64.
7-12 Characteristics of Parallel Circuits
Since V = IR and R = 1/G, V = I/G. The voltage is the same across all components connected in parallel, so
V=
IT
I1
I2
I3
=
=
=
= ...
GT
G1
G2
G3
(7-18)
By transposing variables in each pair of equal terms, we get
I1
G1
R2
=
= ,
I2
G2
R1
I2
G2
R3
=
= , and so on.
I3
G3
R2
In a parallel circuit, the ratio between any two branch currents equals the
ratio of their conductances or the inverse of the ratio of their ­resistances.
Example 7-9
The total current drawn by a 12.5-kΩ resistor and a 50-kΩ resistor in
parallel is 15 mA. Find the current through the 50-kΩ resistor.
Solution I
From the circuit diagram of Figure 7-16,
I1 + I2 = 15 mA
I1
R2
50 kΩ
=
=
= 4.0
I2
R1
12.5 kΩ
I1 = 4.0 I2
145
146
Chapter 7 Series and Parallel Circuits
Substituting for I1 in the first equation gives
4.0I2 + I2 = 15 mA
15 mA
I2 =
= 3.0 mA
5.0
There is more than
one way to solve most
circuit problems.
Generally, it is best
to select the method
in which we can
­visualize what each
step represents.
+
R1
E
−
Figure 7-16
Solution II
12.5 kΩ
R2
50 kΩ
Circuit diagram for Example 7-9
Req =
R1R2
12.5 kΩ × 50 kΩ
=
= 10 kΩ
( 12.5 kΩ + 50 kΩ )
R1 + R2
V = ITReq = 15 mA × 10 kΩ = 150 V
I2 =
V
150 V
=
= 3.0 mA
R2
50 kΩ
When the internal resistance of the source is negligible, altering the
­resistance of one branch of a parallel circuit does not affect the voltage
across, or the current through, the other branches. Therefore, changes
in one branch of a parallel circuit have a negligible effect on the other
branches. In house wiring, lighting circuits are connected in parallel so
that switching one circuit on or off does not affect the operation of the other
circuits (see Figure 7-17).
15-A fuse
Switches
Power
outlets
Voltage
source
Lighting fixtures
Figure 7-17
Parallel connection of loads in house wiring
Source: © iStock.com/jokerproporduction
7-12 Characteristics of Parallel Circuits
Streetlights connected in parallel
The following characteristics can help us recognize parallel circuits:
• The voltage is the same across all components.
•The total conductance is the sum of all the individual branch conductances:
GT = G1 + G2 + . . ..
• The total current is the sum of all the individual branch currents:
IT = I1 + I2 + I3 + . . . .
•The ratio between branch currents is the same as the conductance ratio
and the inverse of the resistance ratio.
•Each branch is independent of any changes in the other branches,
providing the voltage across the parallel circuit is constant.
See Problems 7-39 to 7-45 and Review Questions 7-65 to 7-70.
147
148
Chapter 7 Series and Parallel Circuits
Circuit Check
C
CC 7-6. The switches S1, S2, and S3 shown in Figure 7-18 are considered closed in the configuration
shown. Determine the conditions of each switch in order to turn on lamps L1, L2, and L3,
respectively.
S2
S1
S3
E
L1
L2
L3
Figure 7-18
CC 7-7. Which resistors of the networks shown in Figure 7-19 are connected in series and which
are connected in parallel?
R2
R1
R1
R4
R3
V1
R2
R3
V1
(a)
R4
(b)
Figure 7-19
CC 7-8. A 33-Ω, a 56-Ω, and a 100-Ω resistor are connected in parallel. If all three resistors are rated at ¼ W,
what is the maximum voltage that can be applied without exceeding the power rating of one of
the resistors? Which resistor will burn out first if the voltage increases above this maximum?
7-13 Cells in Parallel
A battery of identical cells connected in parallel (see Figure 7-20) has the
same EMF as a single cell, but can deliver a greater maximum current. The
total capacity of the battery is proportional to the number of cells. If the cells
do not all have the same EMF, currents will flow between them since the cells
7-13 Cells in Parallel
with lower EMFs will act as loads for the cells with higher EMFs. Standard
AA-size cells are sometimes connected in parallel to make a compact battery
with enough current capacity to run motors in devices such as cordless electric razors.
E
Rcell
E
Rcell
E
Rcell
E
Rcell
+
Figure 7-20
Ebat
P cells in parallel
−
Equivalent circuit of a parallel connected battery
In Figure 7-20, P is the number of cells connected in parallel, Ecell is the
EMF of each cell, and Rcell is the internal resistance of each cell. The battery
EMF is
Ebat = E
and the total internal resistance of the battery is
Rbat =
Rcell
P
Example 7-10
(7-19)
How much current flows through an external resistance of 4.00 Ω connected to a battery consisting of ten cells connected in parallel, with
each cell having an EMF of 1.50 V and an internal resistance of 0.20 Ω?
Find the ­terminal voltage of the battery.
Solution
Ebat = 1.50 V
Rbat =
0.20 Ω
Rcell
=
= 0.020 Ω
P
10
149
150
Chapter 7 Series and Parallel Circuits
I=
E
1.50 V
= 0.373 A
=
RT
4.00 Ω + 0.020 Ω
VT = IRL = 0.373 A × 4.00 Ω = 1.49 V
Figure 7-21 shows an equivalent circuit for the battery.
IL
R bat
20 mΩ
VT
Ebat
1.5 V
Figure 7-21
circuitSIM
walkthrough
RL = 4.0 Ω
Equivalent circuit for Example 7-10
Multisim Solution
Download Multisim file EX7-10 from the website.
The simulated circuit corresponds to the one shown in Figure 7-21, with
a battery that consists of ten cells connected in parallel, each having an
EMF of 1.5 V and an internal resistance of 0.20 Ω.
Run the simulation and check the values of VT and IL displayed on the
meters.
Note that these values match those calculated using Equation 7-19.
See Problems 7-46 and 7-47.
7-14 Troubleshooting
The basic instruments used to measure circuit variables are the ammeter
for current, the voltmeter for voltage, and the ohmmeter for resistance.
Most analog meters have a pointer that moves along a graduated scale.
Digital meters display a numerical value for the measurement. Chapter 11
discusses meters in more detail.
The Ammeter
The ammeter is a two-terminal device that measures the current flowing
through it. Therefore, the ammeter must be connected in series with the component. Since the current through components in series is the same, the current
indicated by the ammeter is also the current flowing through the component.
7-14 Troubleshooting
151
The Voltmeter
The voltmeter is a two-terminal device that measures the voltage across it.
The voltmeter must be connected in parallel with a component. Since the
voltage across parallel components is the same, the voltage indicated by
the voltmeter is also the voltage across the component.
The Ohmmeter
Source: TREVOR CLIFFORD PHOTOGRAPHY/SCIENCE PHOTO LIBRARY
The ohmmeter is a two-terminal device that indicates the resistance of a
resistor connected across its terminals. A circuit should be turned off before
an ohmmeter is connected to it. An ohmmeter connected across a network
of resistors in series or parallel measures the equivalent resistance of the
network. To measure the resistance of an individual resistor in a network,
we first disconnect the resistor from the network.
An ammeter is connected in series with a component while a voltmeter is
connected in parallel (across it).
Troubleshooting Techniques
Malfunctions of a circuit can be caused by an open circuit, a short circuit, or
a wrong value for a component. Troubleshooting is the process of locating
and correcting such problems.
An open circuit can occur if one of the components, such as a resistor, is
burnt out (usually the result of an excessive current through it) or if a wire
or circuit-board trace is not making a good connection. These faults ­prevent
current from flowing in that part of the circuit. An ammeter will indicate
zero current, suggesting an open circuit.
152
Chapter 7 Series and Parallel Circuits
A short circuit is a fault that creates a low-resistance connection between
two points in a circuit. Short circuits are often caused by accidental contact
between wires, component leads, or traces on a printed circuit board. Short
circuits also sometimes occur within a component. Since V = IR, the voltage is zero when the resistance is zero. Thus, a voltmeter connected across
a shorted component indicates zero voltage.
If a wrong component is installed in a circuit, measurements of current
and voltage can help to locate the error.
Example 7-11
The series circuit in Example 7-2 should have a current of 340 μA, as
shown in Figure 7-22. However, the ammeter reads 640 μA instead.
Determine the most likely fault.
340.4
+
DCμA
10 kΩ
−
R1
+
16 V E
−
R2
15 kΩ
22 kΩ
R3
Figure 7-22
Measuring current in the circuit for Example 7-11
Solution
Step 1
Use a voltmeter to check the applied voltage as shown in Figure 7-23.
The voltmeter reads 16 V, verifying proper operation of the source.
16.00
+
DC V
10 kΩ
−
R1
+
16 V E
−
R2
15 kΩ
22 kΩ
R3
Figure 7-23
Checking the applied voltage in the circuit for Example 7-11
Step 2
Check the voltage across R1 with the voltmeter. Ohm’s law tells us that
VR1 should be 10 kΩ × 640 μA = 6.40 V. The voltmeter reads 6.40 V, so
we conclude that there is no problem with R1.
7-14 Troubleshooting
Step 3
Check the voltage across R2. VR2 should be 15 kΩ × 640 μA = 9.60 V. The
voltmeter reads 9.60 V, and we conclude that there is no problem with R2.
Step 4
Check the voltage across R3. VR3 should be 22 kΩ × 640 μA = 14.1 V.
The voltmeter reads 0 V. Since VR3 = IR3 and I is not zero, R3 must be
zero. So, we conclude that R3 is shorted.
Example 7-12
Calculation of the total current for the circuit shown in Figure 7-24 gives
IT = 1.8 A, but the ammeter in the circuit reads only 1.2 A. Determine
the most probable cause.
1.20
+
DC A
+
120 V E
−
Figure 7-24
−
R1
150 Ω
R2
300 Ω
R3
200 Ω
Diagram for Example 7-12
Solution
Step 1
Use a voltmeter to check the applied voltage. The voltmeter reads
120 V, indicating proper operation of the source. We conclude the problem must be in the resistor network.
Step 2
For safety, turn off the power supply at this point before you start removing resistors from the circuit in the next few steps.
Step 3
Remove R1 from the circuit and check its resistance using an ohmmeter.
Its value is 150 Ω, so we conclude that there is no problem with R1.
Step 4
Remove R2 from the circuit and measure its resistance. Its value is
300 Ω, so there is no problem with R2.
Step 5
Remove R3 from the circuit and measure its resistance. Its resistance cannot be measured even on the highest scale of the ohmmeter.
Therefore, R3 is open and must be replaced.
See Review Questions 7-71 to 7-75.
153
154
Chapter 7 Series and Parallel Circuits
Summary
• Two or more components are connected in series if a common current
flows through them.
• The total resistance in a series circuit is the sum of the resistances of all
the resistors in the circuit.
• Either plus and minus signs or double-subscript notation may be used to
show the polarity of applied voltages and voltage drops in a circuit.
• In a complete electric circuit, the sum of the voltage drops equals the
sum of the applied voltages.
• In a series circuit, the ratio of the resistances of two resistors equals the
ratio of their voltage drops.
• Any change to any component in a series circuit affects the current
though all components.
• The internal resistance of a practical voltage source causes the terminal
voltage to drop when a load draws current from the source.
• Cells connected in series provide a higher voltage than a single cell.
• A practical voltage source delivers maximum power to a load when the
resistance of the load equals the internal resistance of the source.
• Two or more electric components are connected in parallel if a common
voltage appears across them.
• In a parallel circuit, the total current is the sum of all the branch c­ urrents.
• The algebraic sum of the currents entering a junction equals the algebraic sum of the currents leaving the junction.
• In a parallel circuit, the total conductance is the sum of the conductances
of all the branches.
• For two resistors in parallel, the equivalent resistance equals the product
of their resistances divided by their sum.
• In a parallel circuit, the ratio of any two branch currents equals the ratio
of their conductances.
• Cells connected in parallel can provide a greater current than a sin­gle cell.
• The ammeter, voltmeter, and ohmmeter are basic instruments used for
troubleshooting circuits.
B = beginner
Problems
I = intermediate
A = advanced
Draw a fully labelled schematic diagram for each problem.
B
Section 7-1
7-1.
B
7-2.
B
7-3.
Resistors in Series
Find the total resistance of a 10-Ω resistor, a 20-Ω resistor, and a 30-Ω
resistor connected in series.
Find the total resistance of a series circuit consisting of R1 = 47 kΩ,
R2 = 220 kΩ, and R3 = 3.3 kΩ.
A string of Christmas tree lights consists of 50 identical mini lamps
connected in series to a 120-V source. Given that the hot resistance
of each lamp is 8 Ω, determine the current.
Problems
B
B
7-4.
A 1.2-kΩ, a 2.0-kΩ, an 850-Ω, and a 500-Ω resistor are connected in
series to a voltage source. Determine the applied voltage that will
cause a current of 26.4 mA to flow in this circuit.
Section 7-5
7-5.
I
7-6.
I
7-7.
I
7-8.
I
7-9.
I
7-10.
A
7-11.
I
7-12.
I
7-13.
B
7-14.
Characteristics of Series Circuits
If the series circuit of Problem 7-1 is connected to a 240-V source,
what is the voltage drop across each resistor?
What resistance must be connected in series with a heater rated at
200 mA with a hot resistance of 270 Ω in order to operate it safely from
a 120-V source? What power rating must this resistor have?
A 5-V electric cooler draws a current of 2.5 A. What resistance must
be connected in series with it in order to operate it in a car with a
12-V battery? What power rating should the resistor have?
An electric stove element is rated at 300 W when connected to a
110-V source. Assuming negligible change in resistance for any
change in temperature, determine the total rate of energy conversion when two of these elements are connected in series.
Three resistors are connected in series to a 120-V generator. The first
has a resistance of 200 Ω, the second passes a current of 320 mA, and
the third has a voltage drop of 24 V. Calculate the resistance of the
second and third resistors.
Three resistors are connected in series to a 120-V source. The voltage
drop across R1 and R2 together is 57 V, and the voltage drop across R2
and R3 together is 78 V. The total resistance is 12 kΩ. Find the resistance of each resistor.
A 50-W 22-kΩ resistor, a 100-W 82-kΩ resistor, and a 150-W 36-kΩ
resistor are connected in series. What is the maximum voltage that
can be applied to this network without exceeding the power rating
of any of the resistors?
A 110-V Christmas tree light set consists of twelve 3-W lamps in
series. Find the hot resistance of each lamp.
What resistance must be connected in series with a 100-Ω resistor
for the 100-Ω resistor to produce heat at the rate of 30 W when the
combination is connected to a 120-V source?
Calculate the voltage drops across each resistor in Figure 7-25.
10 Ω
5.0 Ω
+
150 V
−
15 Ω
20 Ω
Figure 7-25
155
156
Chapter 7 Series and Parallel Circuits
I
I
A
I
I
A
I
A
A
Section 7-6
Internal Resistance
7-15. The high-voltage power supply for a television’s picture tube has an
open-­circuit voltage of 18 kV and an internal resistance of 1.5 MΩ.
(a) Find the voltage at the anode of the picture tube when the anode
current is 800 μA.
(b) What will the anode voltage be if the anode current is increased
by 50%?
(c) Find the short-circuit current of the power supply.
7-16. A certain generator has a terminal voltage of 110 V when a 5.5-Ω load is
connected to its terminals. The terminal voltage becomes 105 V when
the load is 3.5 Ω. Calculate the internal resistance of the generator.
7-17. A 20-Ω resistor dissipates heat at the rate of 10 W when connected
to a certain generator. If a 30-Ω resistor is connected in series with
the 20-Ω resistor, the heat dissipation of the 20-Ω resistor becomes
200 mW. Find the open-circuit voltage of the generator.
7-18. A primary D cell’s open-circuit voltage is measured to be 1.51 V.
Touching the leads of an ammeter to the cell’s terminals for a moment
gives a reading of 5.0 A. Find the terminal voltage of the cell when it
is connected to a 2.5-Ω load.
Section 7-7
Cells in Series
7-19. A 12-Ω load is connected to the terminals of a battery consisting
of 12 cells connected in series. Each cell has an EMF of 1.5 V and
an ­internal resistance of 0.20 Ω. Calculate the load current and the
­terminal voltage.
7-20. A battery consisting of five identical cells in series produces a current of 1.5 A when the load resistance is 4.0 Ω. When the load resistance is 9.0 Ω, the load current is 0.75 A. Calculate:
(a) the internal EMF of each cell
(b) the internal resistance of each cell
7-21. A battery is made from six identical 1.5-V cells connected in series.
When a 12-Ω load resistor is connected to this battery, the terminal
voltage is 8 V. Find the internal resistance of each cell.
7-22. A marine emergency battery room contains an array of ten 24-V
marine batteries, each having an internal resistance of 0.05 Ω, connected in series. At what load power level will this array deliver a
load voltage of 220 V?
Section 7-8
Maximum Power Transfer
7-23. A storage battery has an open-circuit voltage of 4.5 V and an internal
resistance of 0.05 Ω.
(a)Find the terminal voltage of the battery when a 0.4-Ω load is connected to it.
(b) What power will be dissipated by a 0.7-Ω load?
157
Problems
A
7-24.
I
7-25.
I
7-26.
B
7-27.
I
7-28.
I
7-29.
B
B
I
I
(c) Find the efficiency of the system when a 0.35-Ω load is connected
to the battery.
(d) Find the maximum power that a load can draw from the b
­ attery.
A generator has an open-circuit voltage of 32 V and an internal resistance of 0.2 Ω. The generator is designed for a constant-duty power
output of 300 W.
(a)What resistance will draw energy from the generator at the rate
of 300 W?
(b)Find the efficiency of the generator with this load.
The magnetic cartridge on a turntable develops an open-circuit
voltage of 100 mV and has an internal resistance of 1200 Ω.
(a)Find the output voltage when the cartridge is connected to a
5.2‑kΩ load.
(b)What value of load resistor must be used to obtain 75% ­efficiency?
(c)What is the maximum power output of the cartridge?
A generator with an internal resistance of 1.0 Ω feeds a load at
the end of a line, each wire of which has a resistance of 2.0 Ω. The
generator is adjusted to produce 120 V across the load at full load.
If the total heat dissipation of the two wires is not to exceed 10% of
the total generated power, what is the maximum power that can be
delivered to the load?
An audio amplifier has an internal resistance at its output terminals
of 4 Ω. The speaker is connected by a cable, each wire of which has
a resistance of 2 Ω. What condition must be met to obtain maximum
power in the speaker?
A 30-V source with an internal resistance of 1.0 Ω is connected to a
7.0-Ω load. Calculate
(a) power delivered to the load
(b) efficiency of power transfer
(c) maximum power that this source can deliver
(d) current flowing at maximum power point
Download Multisim file P7-29 from the website. Run the simulation,
and adjust the value of RL until VT = E/2. Then compare the values
of RL and Rint.
Section 7-9
Resistors in Parallel
7-30. Find the total current through a 10-Ω resistor, a 20-Ω resistor, and a
30-Ω resistor connected in parallel to a 120-V source.
7-31. Find the equivalent resistance for a parallel circuit consisting of
R1 = 47 kΩ, R2 = 220 kΩ, and R3 = 3.3 kΩ.
7-32. The equivalent resistance of three resistors in parallel is 6 kΩ. If R1 is
10 kΩ and R2 is 20 kΩ, what is the resistance of R3?
7-33. What resistance connected in parallel with a 1-kΩ resistor will result
in a total current of 50 mA drawn from a 10-V source?
circuitSIM
walkthrough
158
Chapter 7 Series and Parallel Circuits
A
I
B
I
I
I
I
A
I
I
B
B
7-34. Two resistors connected in parallel draw a total current of 2 A from
a 90-V supply. One of the resistors is three times the value of the
other. Calculate the value of each resistor.
Section 7-10
Kirchhoff’s Current Law
7-35. A source connected to three resistors in parallel supplies a total
current of 6.5 mA. One resistor is 10 kΩ and has a voltage drop of
4.5 V. The second resistor has a current of 750 µA. How much current
is flowing through the third resistor?
Section 7-11
Conductance and Conductivity
Section 7-12
Characteristics of Parallel Circuits
7-36. A circuit element having a conductance of 150 μS is connected in
parallel with a branch having a conductance of 750 mS. Find the
equivalent resistance of this circuit.
7-37. What resistance must be placed in parallel with a 20-kΩ resistor to
reduce the equivalent resistance to 15 kΩ?
7-38. Find the total conductance of a 10-kΩ, 22-kΩ, and 15-kΩ resistor
connected in parallel. How much current will flow if a 20 V source is
connected to this circuit?
7-39. Three lamps operating in parallel on a 110-V circuit are rated at
40 W, 60 W, and 100 W. What is the equivalent hot resistance of this
load?
7-40. Three resistors in parallel pass a total current of 0.60 A. The first
resistor has a resistance of 400 Ω, the second passes a current of
60 mA, and the third has a voltage drop across it of 150 V. Calculate
the ­resistance of the second and third resistors.
7-41. If the three resistors in Problem 7-31 are each rated at 0.5 W, what
is the maximum total current that the network can handle without
overheating any resistor?
7-42. The total current passed by a 10-kΩ, a 15-kΩ, and a 20-kΩ resistor in
parallel is 20 mA. What is the current through each branch?
7-43. Three resistors connected in parallel have an equivalent resistance of 2.5 kΩ. R1 has a resistance of 15 kΩ, R2 has a voltage drop
of 25 V, and R3 produces heat at a rate of 25 mW. Determine R2
and R3.
7-44. A 500-Ω resistor and a 1-kΩ resistor are connected in parallel. How
much resistance must be connected in parallel with this combination
to draw a current of 100 mA from a 30 V source?
7-45. A 100-Ω resistor and a 50-Ω resistor are connected in parallel
and a total current of 3A flows. What is the value of the source
­voltage?
159
Review Questions
B
B
Section 7-13
Cells in Parallel
7-46. A 0.525-Ω resistor is connected to the terminals of a battery consisting
of four 1.46-V cells connected in parallel. If the load current is 0.80 A,
find the internal resistance of each cell.
7-47. Download Multisim file P7-47 from the website. Run the simulation,
and measure IT with R1 = 150 Ω, R2 = 300 Ω, and R3 = 200 Ω. Then
use a simulation to determine what happens to IT when R3 is open.
circuitSIM
walkthrough
Review Questions
Section 7-1
Resistors in Series
7-48. Justify the statement that the current is the same in all parts of a
simple series circuit.
7-49. Justify the statement that the total resistance of a series circuit is the
sum of all the individual resistances.
Section 7-2
Voltage Drops in Series Circuits
7-50. In the electric circuit of Figure 7-26, one of the “black boxes”
represents a voltage source and the other a resistor. The voltmeter
and ammeter read correctly when their terminals are connected with
the polarity markers as shown. Which “black box” is the source?
Section 7-4
Kirchhoff’s Voltage Law
7-51. Kirchhoff’s voltage law may be stated as “The voltage between any
two points in a network is the same via any path between those two
points.” Verify this statement for two ends of R2 in Figure 7-1, using
the values given in Example 7-1.
Section 7-5
Characteristics of Series Circuits
7-52. Prove that the ratio between any two power dissipations in a series
circuit is the same as the ratio between the two resistances.
7-53. Two 120-V lamps, rated at 100 W and 25 W, are connected in series
to a 120-V source. Which one will glow more brightly? Explain.
7-54. Explain the visible effect when the 25-W lamp in Review Question
7-53 is
(a) open-circuit
(b) short-circuit
Section 7-6
Internal Resistance
7-55. How would you determine the internal resistance of a given flashlight battery?
1
−
V
+
A
−
2
Figure 7-26
+
160
Chapter 7 Series and Parallel Circuits
Section 7-8
Maximum Power Transfer
7-56. Explain the shape of the graph of load power versus load resistance
in Figure 7-13.
7-57. Express the condition for maximum power output in relation to
short-circuit and open-circuit current.
7-58. Although there are two numerically correct answers to Problem 7-24(a),
we select only one as the electrically correct answer. Why?
7-59. Generator B develops three times the open-circuit voltage of generator A but has three times as great an internal resistance. Each generator is developing the same power in its load. Which generator has
the greater efficiency?
7-60. A generator is operated in such a manner that its terminal voltage
is kept constant by increasing its generated potential difference
as the load current is increased. Is this system 100% efficient?­
Explain.
7-61. In electronic equipment, the load resistance is often made equal to
the internal resistance. Why is the load resistance not set this way for
a power-generating station?
Section 7-11
Conductance and Conductivity
7-62. What is the disadvantage of a series-circuit string of decorative lights
as compared to a parallel-circuit string?
7-63. Why must the equivalent resistance of a parallel circuit always be
less than that of the smallest of the branch resistances?
7-64. Show that the statement “the total current in a parallel circuit is the
sum of all the branch currents” confirms Kirchhoff’s current law.
Section 7-12
Characteristics of Parallel Circuits
7-65. When working with parallel circuits, why is it preferable to think in
terms of conductance rather than resistance?
7-66. If the two lamps in Question 7-53 are connected in parallel, explain the visible effect when the 25-W lamp is (a) open-circuit and
(b) short-circuit.
7-67. An electric stove element has a resistance of 50 Ω with a centre-tap
connection. Draw circuit diagrams to show three means of connecting it to a 110-V source to obtain three different rates of conversion of electric energy to heat.
7-68. Most homes have a lighting fixture that can be switched on or
off from two different locations. This is accomplished with two
single-pole, double-throw switches (which electricians call “threeway switches”). Draw a circuit diagram for a light controlled in this
way.
7-69. Figure 7-17 is not a simple parallel circuit because the fuse is in series
with the remainder of the circuit. Why must the fuse be located in
the position shown?
Practice Quiz
7-70. Two 50-kΩ resistors are connected in series across a voltage source. If
a voltmeter with a resistance of 100 kΩ is connected first across one
resistor and then the other and finally across both, the sum of the
first two readings does not equal the third reading. Is this third reading greater or less than the sum of the other two? ­Explain.
Section 7-14
Troubleshooting
7-71. How must an ammeter be connected into a circuit to measure c­ urrent
through a component?
7-72. How must a voltmeter be connected into a circuit to measure voltage
across a component?
7-73. Why is it often necessary to disconnect a resistor from a circuit b
­ efore
measuring the resistance of the resistor?
7-74. List two probable causes for the ammeter in Figure 7-22 reading less
than 340 μA.
7-75. List two probable causes for the ammeter in Figure 7-24 reading
more than 1.8 A.
Integrate the Concepts
One string of patio lights has 20 bulbs connected in series, with each bulb
having a resistance of 30 Ω when hot. Another brand of patio lights has
20 bulbs connected in parallel with each bulb having a resistance of 12 kΩ
when hot. Both strings run on 120 V. For each string, determine
(a) how much current flows through each bulb
(b) how much power each bulb consumes
(c) how much power the whole string uses
(d) what happens to the remaining bulbs when one burns out
Practice Quiz
1.
Four resistors are connected in series, and the current through the
first resistor is 2 A. The current through the third resistor is
(a) 0.67 A
(b) 2 A
(c) 6 A
(d) 0.3 A
2.
When a third resistor is connected in series with two other resistors,
the total resistance of the circuit
(a) decreases
(b) remains the same
(c) increases
(d) may either increase or decrease depending on the relative values
of the resistors
161
162
Chapter 7 Series and Parallel Circuits
3.
Kirchhoff’s voltage law states that
(a) the algebraic sum of applied voltages equals the algebraic sum
of the voltage drops around a closed loop
(b) the algebraic sum of the resistances is equal to the sum of the
voltages
(c) the algebraic sum of the individual currents around a closed
loop is zero
(d) the voltages developed across each element in a series circuit are
identical
4.
Identify the active and passive circuit elements in Figure 7-27.
R1
R1
R2
Vs
R2
Vs
R3
R4
(a)
(b)
Figure 7-27
5.
Show the direction of conventional current through each component
of the circuits in Figure 7-27.
6.
A 1.0-kΩ resistor, a 2.2-kΩ resistor, and a 10-kΩ resistor are in parallel. The equivalent resistance is approximately
(a) 11 kΩ
(b) 13 kΩ
(c) 643 Ω
(d) 1.7 MΩ
7.
If one resistor in a parallel circuit is removed, the total resistance
(a) decreases
(b) increases
(c) remains the same
(d) may either increase or decrease depending on the relative values
of the resistors
8.
In a parallel circuit, the largest resistance has the
(a) most voltage
(b) least voltage
(c) most current
(d) least current
Practice Quiz
9.
Determine the current flowing in each branch of the circuit in
­Figure 7-28.
Vs
20 V
R1
510 Ω
R2
680 Ω
R3
1.0 kΩ
Figure 7-28
10.
Conductance is a measure of the ability of an electric circuit to
(a) pass current
(b) limit current
(c) limit voltage
(d) pass voltage
163
8
Series-Parallel Circuits
In Chapter 7, we considered only simple series and parallel circuits.
­Although electric networks are seldom that simple, they are often
combinations of series and parallel ­circuits. This chapter describes
techniques for analyzing such combinations.
Chapter Outline
8-1
Series-Parallel Resistors 166
8-3
Kirchhoff’s Laws Method
8-2
Equivalent-Circuit Method 167
171
8-4
Voltage-Divider Principle
8-6
Current-Divider Principle 181
8-5
8-7
8-8
Voltage Dividers
175
Cells in Series-Parallel
Troubleshooting
186
173
184
Key Terms
series-parallel circuit 167
ladder network 168
voltage-divider principle 174
bleeder resistor 176
voltage regulation 176
chassis 177
ground 177
current-divider principle
181
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• simplify a series-parallel circuit into a series
or a parallel circuit
• analyze a series-parallel circuit to calculate
­voltage, c
­ urrent, and power
• apply Kirchhoff’s voltage and current laws to
series-­parallel circuits
• state the voltage-divider principle
• calculate voltages across resistors in series
using the ­voltage-divider principle
• calculate an appropriate value for a bleeder
resistor in a voltage divider
Photo sources: © iStock.com/james steidl
• reference a voltage at any point in a circuit
with respect to ground
• state the current-divider principle
• calculate currents through resistors in parallel
using the current-divider principle
• calculate the terminal voltage and total
resistance of a s
­ eries-parallel combination of
cells
• troubleshoot series-parallel circuits
166
Chapter 8 Series-Parallel Circuits
8-1 Series-Parallel Resistors
Considered as a whole, the circuit in Figure 8-1 is neither a series nor a
­parallel circuit. However, R2 and R3 are connected between the same
two points in the circuit and must have the same voltage drop. Therefore
these two resistors are in parallel, and we can calculate a single equivalent
resistance for them. As measured from the terminals of the voltage source,
the simplified circuit of Figure 8-2 is equivalent to the original circuit of
Figure 8-1. In Figure 8-2, R1 is in series with the equivalent resistance of
R2 and R3 in parallel. Therefore, we can solve the circuit of Figure 8-2 as a
simple series circuit.
I1
I2
R1
+
I3
R2
E
R3
−
Figure 8-1
Simple series-parallel circuit
I1
I2 + I3
R1
+
Req =
E
−
Source: © Sundeepgoel/GetStock/Dreamstime
Figure 8-2
R2R3
R2 + R3
Equivalent circuit for Figure 8-1
In the circuit shown in Figure 8-3(a), R2 and R3 have the same current through them and are therefore in series. We can replace them with
an equivalent resistor, as shown in Figure 8-3(b). We can now solve the
­simplified circuit of Figure 8-3(b) as a simple parallel circuit.
R2
+
In this flashlight,
a battery of cells
connected in series
provides power to two
lamps connected in
parallel.
E
R1
R3
−
R1
E
−
(a)
Figure 8-3
+
(b)
(a) Series-parallel circuit; (b) Equivalent circuit
Req = R2 + R3
8-2 Equivalent-Circuit Method
We define a series-parallel circuit as one in which some portions of the
circuit have the characteristics of simple series circuits while the other portions have the characteristics of simple parallel circuits.
See Review Questions 8-40 to 8-42 at the end of the chapter.
8-2 Equivalent-Circuit Method
We can solve series-parallel circuits by substituting the equivalent resistances for various portions of the circuit until the original circuit is reduced
to either a simple series or a simple parallel circuit.
Example 8-1
Find the voltage drop, current, and power for each resistor in the circuit diagram of Figure 8-4.
Solution
Step 1
Draw a fully labelled schematic diagram for this particular circuit.
I1
R1
R1
I2
12 Ω
+
100 V
−
Figure 8-4
R2
I3
10 Ω
R3
40 Ω
12 Ω
+
100 V
−
Veq Req
Circuit diagram for Example 8-1
Step 2
Since R2 and R3 are in parallel,
Req =
R2R3
10 Ω × 40 Ω
=
= 8.0 Ω
( 10 Ω + 40 Ω )
R2 + R3
The total resistance of the circuit is
Step 3
From Ohm’s law,
RT = R1 + Req = 12 + 8.0 = 20 Ω
IT =
E
100 V
=
= 5.0 A
RT
20 Ω
Step 4
Since R1 is in series with the source,
I1 = IT = 5.0 A
8.0 Ω
167
168
Chapter 8 Series-Parallel Circuits
Step 5
Applying Ohm’s law to R1,
V1 = I1R1 = 5.0 A × 12 Ω = 60 V
To verify our
­calcula­tions, we can
check that the currents we calculated
satisfy Kirchhoff’s
current law for this
circuit: I1 = I2 + I3.
Step 6
From Kirchhoff’s voltage law,
Veq = E − V1 = 100 V − 60 V = 40 V
Returning now to the original circuit,
V2 = V3 = Veq = 40 V
Step 7
Applying Ohm’s law to R2 and R3,
V2
I2 =
=
R2
V3
I3 =
=
R3
Since P = VI,
Step 8
We can verify our
­calculations by
­checking that
PT = P1 + P2 + P3.
circuitSIM
walkthrough
40 V
= 4.0 A
10 Ω
40 V
= 1.0 A
40 Ω
P1 = V1I1 = 60 V × 5.0 A = 0.30 kW
P2 = V2I2 = 40 V × 4.0 A = 0.16 kW
P3 = V3I3 = 40 V × 1.0 A = 40 W
PT = VTIT = 100 V × 5.0 A = 0.50 kW
Multisim Solution
Download Multisim file EX8-1 from the website.
Connect a ground to the bottom node of the circuit. Insert current
meters to measure I1, I2, and I3.
Run the simulation, and note the currents displayed on the ammeters.
Confirm that these values for I1, I2, and I3 match those calculated using
equivalent resistances in the first solution for this example.
Example 8-2
Find the resistance at the input terminals of the ladder network in
­Fig­ure 8-5 with
(a) the output terminals open-circuit
(b) a 600-Ω load resistance connected to the output terminals
8-2 Equivalent-Circuit Method
100 Ω
200 Ω
A
Input
A
800 Ω
B
100 Ω
Figure 8-5
200 Ω
100 Ω
Output
800 Ω
B
100 Ω
Circuit diagram for Example 8-2
Solution
(a) Step 1
With the output terminals open-circuit, no current flows in the two
right-hand 100-Ω resistors. Thus they are effectively out of the circuit, which is therefore equivalent to Figure 8-6(a).
100 Ω
200 Ω
A
Input
800 Ω
100 Ω
B
100 Ω
800 Ω
A
Input
200 Ω
800 Ω
100 Ω
(a)
1200 Ω
B
(b)
100 Ω
Input
480 Ω
100 Ω
(c)
Figure 8-6
Input
680 Ω
(d)
Equivalent circuits for Example 8-2(a)
Step 2
The two 200-Ω resistors and the right-hand 800-Ω resistor have the
same current through them. Hence they have the characteristics of
a ­series ­circuit:
Req = 200 Ω + 800 Ω + 200 Ω = 1200 Ω
Substituting Req for the three resistors gives the equivalent circuit
of Figure 8-6(b).
Step 3
Since the 800-Ω and 1200-Ω resistors in the equivalent circuit of
Fig­ure 8-6(b) are connected between the same junction points,
they are in parallel.
169
170
Chapter 8 Series-Parallel Circuits
Req =
800 × 1200
= 480 Ω
( 800 + 1200 )
Step 4
Since the equivalent circuit in Figure 8-6(c) is a simple series circuit,
Rin = 100 Ω + 480 Ω + 100 Ω = 680 Ω
(b) Step 1
A 600-Ω load resistor connected to the output terminals is in series
with the two right-hand 100-Ω resistors. Hence,
Req = 100 Ω + 600 Ω + 100 Ω = 800 Ω
Step 2
We can replace the two 800-Ω resistors in parallel in the equivalent
circuit of Figure 8-7(a) with a single 400-Ω resistance, as in Figure 8-7(b).
Step 3
The two 200-Ω resistors and the 400-Ω resistor of Figure 8-7(b) are
in ­series, so we can replace them with a single 800-Ω equivalent
resistance, as in ­Figure 8-7(c).
Step 4
Once again, two 800-Ω resistors in parallel are equivalent to a single
400-Ω resistance, giving the equivalent circuit in Figure 8-7(d).
Step 5
Finally, the total resistance of the series circuit in Figure 8-7(d) is
100 Ω
200 Ω
A
Input
Rin = 100 Ω + 400 Ω + 100 Ω = 600 Ω
800 Ω
B
100 Ω
100 Ω
A
800 Ω
200 Ω
(a)
B
100 Ω
A
Input
800 Ω
B
100 Ω
200 Ω
100 Ω
800 Ω
800 Ω
Input
B
(c)
Figure 8-7
200 Ω
(b)
Input
100 Ω
800 Ω
A
Equivalent circuits for Example 8-2(b)
See Problems 8-1 to 8-13.
400 Ω
100 Ω
(d)
400 Ω
8-3 Kirchhoff’s Laws Method
8-3 Kirchhoff’s Laws Method
Kirchhoff’s laws provide us with a method for solving series-parallel
­circuits without reducing them to simple series or parallel circuits by substituting equivalent resistances.
Kirchoff’s Laws Solution to Example 8-1
Applying Kirchhoff’s current law to the circuit in Figure 8-4 gives
I1 = I2 + I3
From Ohm’s law, I1 = V1/R1, I2 = V2/R2, and I3 = V3/R3. Therefore,
V1
V2
V3
=
+
R1
R2
R3
and
V1
V2
V3
=
+
12 Ω
10 Ω
40 Ω
Since R2 and R3 are in parallel, V2 = V3. Combining this equation with
Kirchhoff’s voltage law gives
V2 = V3 = E − V1
Substitution in the preceding equation gives
V1
100 V − V1
100 V − V1
=
+
12 Ω
10 Ω
40 Ω
Multiplying by 120 gives
10V1 = 1200 − 12V1 + 300 − 3V1
25V1 = 1500
V1 = 60 V
V2 = V3 = E − V1 = 100 − 60 = 40 V
Therefore,
We can now use Ohm’s law to determine the various currents:
I1 =
I3 =
V1
60 V
V2
40 V
=
= 5.0 A I2 =
=
= 4.0 A
R1
12 Ω
R2
10 Ω
V3
40 V
=
= 1.0 A
R3
40 Ω
If we do not know all the resistances in a circuit, we cannot readily solve
for the equivalent resistances to reduce the circuit to a simple series or
parallel circuit. However, we can use Kirchhoff’s laws to calculate the unknown resistance.
171
172
Chapter 8 Series-Parallel Circuits
Example 8-3
A resistor passing a 20-mA current is in parallel with a 5.0-kΩ resistor. This combination is in series with another 5.0-kΩ resistor, and the
whole network is connected to a 500-V source. (See Figure 8-8.) Find
the resistance of the resistor that is passing the 20-mA current.
I1
R1
5.0 kΩ
+
500 V
−
Figure 8-8
5.0 kΩ
I2
I3 = 20 mA
R2
R3
Circuit diagram for Example 8-3
Solution
Applying Kirchhoff’s voltage law,
V1 + V2 = E
Substituting V = IR,
I1R1 + I2R2 = E
Applying Kirchhoff’s current law,
Therefore,
I2 = I1 − 0.020 A
R1I1 + R2(I1 − 0.020) = E
Substituting given values for R1, R2, and E,
5000I1 + 5000(I1 − 0.020) = 500
10 000I1 = 600
Therefore,
and
I1 = 60 mA
I2 = I1 − 0.020 A = 0.060 − 0.020 = 0.040 A = 40 mA
V2 = I2R2 = 40 mA × 5 kΩ = 200 V
Since R2 and R3 are in parallel, V2 = V3 and
R3 =
V3
200 V
=
= 10 kΩ
I3
20 mA
It is also possible to start the solution by applying Kirchhoff’s
current law first and then substituting I = V/R into I1 = I2 + 0.020 A.
See Problems 8-14 to 8-17 and Review Questions 8-43 and 8-44.
8-4 Voltage-Divider Principle
Circuit Check
CC 8-1. Calculate the equivalent resistance of the network in Figure 8-9.
A
400 Ω
5.0 kΩ
500 Ω
15 kΩ
600 Ω
Figure 8-9
CC 8-2. Calculate the resistance measured from terminal T to ground in
the network in Figure 8-10.
T
10 kΩ
27 kΩ
2.2 kΩ
10 kΩ
2.2 kΩ
Figure 8-10
8-4 Voltage-Divider Principle
For the series circuit of Figure 8-11, Kirchhoff’s voltage law states that
E = V1 + V2 + V3. In other words, the total applied voltage is divided among
the three resistors. We can connect a voltmeter across six possible combinations of the terminals A, B, C, and D, as shown in Figure 8-11. Thus, the
­series combination of R1, R2, and R3 acts as a voltage divider.
A
R1
200 V
20 kΩ
B
+
350 V
−
R2
250 V
50 V
5 kΩ
150 V
C
R3
100 V
10 kΩ
D
Figure 8-11
350 V
Voltage-divider principle
173
174
Chapter 8 Series-Parallel Circuits
To solve for the six voltages in Figure 8-11, we could calculate RT, find I from
Ohm’s law, and then calculate the voltage drop across each r­ esistance from
V1 = IR1, V2 = IR2, and so on. However, as demonstrated in Example 7-3, we
can calculate each voltage in a single step. Section 7-5 described a property
of series circuits that is often called the voltage-­divider principle:
In a series circuit, the ratio between any two voltage drops equals the
ratio of the two resistances across which these voltage drops occur.
Hence, for any resistor Rn in a voltage divider,
Vn
Rn
=
E
RT
Vn = E
and
Rn
RT
(8-1)
Example 8-4
What is the voltage between terminals B and D in the circuit of
­Figure 8-11?
Solution
VBD = E
R2 + R3
R1 + R2 + R3
( 5.0 kΩ + 10.0 kΩ )
( 20.0 kΩ + 5.0 kΩ + 10.0 kΩ )
15.0 kΩ
= 350 V ×
35.0 kΩ
= 150 V
= 350 V ×
Using a variable resistor as a voltage divider provides a continuously
variable terminal voltage, as shown in Figure 8-12.
A
E
RT
C
Vout
B
D
Figure 8-12
Potentiometer circuit
See Review Questions 8-45 and 8-46.
8-5 Voltage Dividers
8-5 Voltage Dividers
Voltage dividers are widely used in power supplies for electronic ­circuits
so that a single voltage source can supply all the various voltages
­required by a piece of equipment, reducing the cost, size, and weight of
the ­equipment. The series-dropping resistor of Figure 8-13 provides the
simplest method of obtaining the required voltage drop across a particular circuit e­ lement.
RS
+
25 V
−
Series-dropping
resistor
Figure 8-13
15 V
RL
Series-dropping resistor
Example 8-5
A portion of an electronic circuit requires an operating voltage of 15 V
and a current of 20 mA. If the supply terminal voltage is 25 V, what value
of series-dropping resistor is required?
Solution
From Kirchhoff’s voltage law, the voltage drop across the series-­
dropping resistor must be
VD = E − VL = 25 V − 15 V = 10 V
ID = IL = 20 mA
Since the dropping resistor and the load are in series,
RS =
VD
10 V
=
= 500Ω
ID
20 mA
To complete the design, we find the minimum power rating for RS:
P = VI = 10 V × 20 mA = 0.20 W
A 0.5-W resistor would be a good choice since it will operate at a lower
temperature than a 0.25-W resistor and thus be less likely to fail.
The advantage of the simple series-dropping resistor is that the only
drain on the power supply is the current through the load. But this circuit
has the disadvantage that any change in load resistance changes the current
175
176
Chapter 8 Series-Parallel Circuits
through the series-dropping resistor, which, in turn, changes the voltage
drop across the resistor and the voltage supplied to the load.
In circuits where the load is a transistor, RL can become very high. Under
such circumstances, the voltage drop across a series-dropping resistor is
­almost zero and the voltage across the load is close to the full applied
­voltage, which may be high enough to damage the transistor. To prevent
such damage, some voltage-divider designs include a bleeder resistor in
parallel with the load, as in Figure 8-14. The bleeder resistor ensures there
is always enough current through the series-dropping resistor to maintain
an appreciable voltage drop across it. Hence, the load voltage cannot rise
to the full applied voltage.
RS
Series-dropping resistor
+
Bleeder
25 V
resistor
−
Figure 8-14
IL = 20 mA
RB
RL
VL = 15 V
Voltage divider with bleeder resistor
In power supplies for electronic equipment, a bleeder current of 10–25%
of the total current drawn from the source provides sufficient protection
against excessive load voltage when the load resistance increases. The
greater the bleeder current, the less variations in load current affect the load
voltage. Thus, improved voltage regulation is achieved at the expense of
extra current drain from the source and extra heat produced in the voltagedivider resistors. In designing voltage dividers for loads consisting of a
­single transistor, voltage regulation is more significant than a few extra
­milliamperes of bleeder current.
Example 8-6
Allowing a bleeder current of 50 mA, design a voltage divider to supply
15 V at 20 mA from a 25-V source. Calculate the open-circuit terminal
voltage of the voltage divider.
Solution
Since the bleeder resistor is in parallel with the load, the voltage drop
across RB is 15 V. Therefore,
RB =
VL
15 V
=
= 300 Ω
IB
50 mA
8-5 Voltage Dividers
The power dissipated by the bleeder resistor is
PB = VLIB = 15 V × 50 mA = 0.75 W
The voltage drop across the series-dropping resistor is
VD = E − VL = 25 V − 15 V = 10 V
ID = IB + IL = 50 mA + 20 mA = 70 mA
and
RS =
VD
10 V
= 143 Ω
=
ID
70 mA
PS = VDID = 10 V × 70 mA = 0.70 W
We would use at least 1-W resistors for this voltage divider.
Under open-circuit conditions, we can use the voltage-divider principle (Equation 8-1):
VT = VB = 25 V ×
300 Ω
= 16.9 V
143 Ω + 300 Ω
In circuits that obtain several different voltages from one power supply,
it is convenient to have a common reference point for all voltage measurements. Often the circuit’s metal frame or chassis is connected to the circuit
and used as the reference point. The chassis may also be connected to the
earth, usually through the ground wire of the electrical system. The symbol
at the bottom of Figure 8-15 indicates that the reference point is grounded.
Hence, we can say that the voltage at point A is +250 V with respect to
ground. The symbol in the middle of Figure 8-16 indicates the point where
the chassis is electrically connected to the circuit. With respect to chassis, the
voltage at point A in this circuit is +12 V and the voltage at point B is −12 V.
+250 V 50 mA
A
R1
+150 V 25 mA
B
+
250 V E
−
R2
+100 V 10 mA
IB = 10 mA
C
RB
Ground
Figure 8-15
D
Voltage divider with multiple output voltages
177
178
Chapter 8 Series-Parallel Circuits
Example 8-7
Design a voltage divider for the specifications shown in Figure 8-15.
Solution
Step 1
Figure 8-15 uses a resistor to represent the load for each specified
output. These resistors make it easier to trace the currents and draw in
arrows showing the direction of the current in each branch of the circuit. None of the 50-mA current drawn from terminal A flows through
the voltage-­divider resistors. Hence, this current does not enter into
our ­calculations.
Step 2
Starting with the bleeder resistor, calculate the current through
each resistor of the voltage divider and mark the circuit diagram of­
Figure 8-15 a­ ccordingly.
The only current through RB is the 10-mA bleeder current. Since R2
passes this bleeder current and the 10-mA current drawn by the 100-V
load,
I2 = 10 mA + 10 mA = 20 mA
Similarly, R1 passes I2 plus the 25-mA current drawn by the 150-V load:
I1 = 20 mA + 25 mA = 45 mA
Step 3
Calculate the resistance and power rating for each of the three resistors
of the voltage divider.
and
RB =
VB
100 V
=
= 10 kΩ
IB
10 mA
PB = VB IB = 100 V × 10 mA = 1.0 W
The voltage drop across R2 is the potential difference between terminals B and C:
V2 = 150 V − 100 V = 50 V
Hence,
and
R2 =
V2
50 V
=
= 2.5 kΩ
I2
20 mA
P2 = V2I2 = 50 V × 20 mA = 1.0 W
For reliability, we would use 2-W resistors or even 5-W resistors for R2
and RB.
8-5 Voltage Dividers
V1 = 250 V − 150 V = 100 V
Similarly,
R1 =
V1
100 V
=
= 2.2 kΩ
I1
45 mA
P1 = V1I1 = 100 V × 45 mA = 4.5 W
For R1, we would use a 10-W resistor.
Transistor circuits can require positive or negative voltages relative to a
common chassis connection, and many circuits require both. We can obtain
these positive and negative voltages from one power supply by connecting
a tap on a voltage divider to the chassis as shown in ­Figure 8-16.
A
IT = 0.50 A
+
R1
+
24 V
−
Load 1
400 mA
12 V
Load 2
200 mA
Chassis
R2
B
Figure 8-16
12 V
−
Power-supply voltage divider for a transistor amplifier
Example 8-8
At full load, the power supply in Figure 8-16 has a terminal voltage of
24 V DC. Design a voltage divider to provide +12 V and −12 V with
respect to the chassis when the current drains are 400 mA at +12 V
and 200 mA at −12 V. The total current drain on the power supply is
500 mA.
Solution
Since R1 is in parallel with load 1, V1 = 12 V. Applying Kirchhoff’s current law to junction A,
I1 = 500 mA − 400 mA = 100 mA
R1 =
V1
12 V
=
= 120 Ω
I1
100 mA
P1 = V1I1 = 12 V × 100 mA = 1.2 W
179
180
Chapter 8 Series-Parallel Circuits
Similarly, V2 = 12 V, and
I2 = 500 mA − 200 mA = 300 mA
R2 =
12 V
= 40 Ω
300 mA
P2 = 12 V × 300 mA = 3.6 W
See Problems 8-18 to 8-27 and Review Questions 8-47 to 8-51.
Circuit Check
B
CC 8-3. Determine values of R1 and R2 for the potentiometer in
Figure 8-17 such that the current I is limited to 250 mA when the
output voltage Vout is 50 V with no load connected.
+
Vin R
10 kV 1
−
I
+
R2 Vout
−
Figure 8-17
CC 8-4. Determine the voltage drop across each resistor shown in
Figure 8-18.
R1
R2
1.0 kΩ
2.2 kΩ
R3
3.3 kΩ
V1
15 V
R4
5.1 kΩ
R5
10 kΩ
Figure 8-18
8-6 Current-Divider Principle
8-6 Current-Divider Principle
In Example 8-6, the total current drawn from the voltage source divides between
the load and the bleeder resistor. In circuits where current ­divides among parallel branches, we can apply the parallel circuit dual of the v
­ oltage-divider principle for series circuits. Section 7-12 described this ­current-divider principle:
In a parallel circuit, the ratio between any two branch currents equals
the ratio of the two conductances through which these ­currents flow.
Since the voltage is common for resistors in parallel,
V=
I1
I2
In
IT
=
=
=
G1
G2
Gn
GT
In = IT
and
Gn
GT
(7-18)
(8-2)
Substituting Gn = 1/Rn, and GT = 1/Req in Equation 8-2 gives
In = IT
Req
Req
In
or
=
Rn
IT
Rn
(8-3)
Hence, we can restate the current-divider principle:
In a parallel circuit, the ratio between any two branch currents is the
inverse of the ratio of the branch resistances.
For two resistors in parallel,
Req =
R1R2
R1 + R2
I1 = IT
and
I1 = IT
Similarly,
I2 = IT
(7-13)
R1 R2
(
R1 R1 + R2 )
R2
R1 + R2
R1
R1 + R2
(8-4)
(8-5)
181
182
Chapter 8 Series-Parallel Circuits
Example 8-9
Calculate the bleeder current in the voltage divider designed in
Example 8-6 if the load resistance increases to 3.0 kΩ, as shown in
­Figure 8-19.
RS
143 Ω
+
25 V
−
RB
Figure 8-19
300 Ω
RL
3000 Ω
Circuit diagram for Example 8-9
Solution
The equivalent resistance of the bleeder resistor and the load resistance
in parallel is
Req =
300 Ω × 3 kΩ
= 273 Ω
( 300 Ω + 3.0 kΩ )
RT = RS + Req = 143 + 273 = 416 Ω
IT =
E
25 V
=
= 60.1 mA
RT
416 Ω
IB = IT
RL
3.0 kΩ
= 60.1 mA ×
= 55 mA
RB + RL
3.3 kΩ
Example 8-10
The ammeter in Figure 8-20 has an internal resistance of 50 Ω and reads
full scale when the current through it is 1.0 mA. What shunt resistance
must be connected in parallel with the meter so that it will read full
scale for a circuit current of 1.0 A?
RM = 50 Ω
A
1.0 mA
1.0 A
Figure 8-20
Shunt
Circuit diagram for Example 8-10
8-6 Current-Divider Principle
Solution
From Kirchhoff’s current law,
Ish = IT − IM = 1000 mA − 1.0 mA = 999 mA
Applying the current-divider principle,
Rsh
IM
=
RM
Ish
Rsh = RM ×
IM
1.0
= 50 mΩ
= 50 Ω ×
Ish
999
In Example 8-2, we calculated the input resistance of a typical ladder
network, but we did not compare the output voltage and current with
the input voltage and current. Now we can use the current-divider principle to complete the circuit analysis without a series of tedious Ohm’s-law
­calculations.
Example 8-11
Find the output voltage for the ladder network in Figure 8-21 with
(a) the output terminals open-circuit
(b) a 600-Ω load connected to the output terminals
Iin 100 Ω
+
A I2 200 Ω
204 mV
100 Ω
C
I1
I3
IL
800 Ω
800 Ω
RL
−
100 Ω
Figure 8-21
B
200 Ω
D
100 Ω
Circuit diagram for Example 8-11
Solution
(a)From the solution of Example 8-2(a), we know that Rin = 680 Ω
when the output terminals are open-circuit. Therefore,
Iin =
E
204 mV
=
= 300 μA
Rin
680 Ω
This current splits into two parallel branches at junction A. Substi­
tuting into Equation 8-5 gives
R1
800 Ω
I2 = IT
= 300 µA
= 120 µA
R1 + R2
800 Ω + 1200 Ω
183
184
Chapter 8 Series-Parallel Circuits
Since the output terminals are open-circuit, IL = 0 and I3 = I2.
Also, there will be zero voltage drop across the two right-hand
100-Ω resistors. Therefore,
VL = VCD = I2RCD = 120 μA × 800 Ω = 96.0 mV
(b)
From the solution of Example 8-2(b), Rin = 600 Ω when
RL = 600 Ω. Then,
E
204 mV
Iin =
=
= 340 µA
Rin
600 Ω
This current splits into two branches at junction A. From
Figure 8-7(c), we note that the two branches between junctions
A and B have the same resistance. Therefore,
I2 = 0.5 × 340 μA = 170 μA
At junction C, I2 splits into two branches. Figure 8-7(a) shows that
these branches have equal resistances, so
IL = 0.5 × 170 μA = 85 μA
VL = ILRL = 85 μA × 600 Ω = 51 mV
See Problems 8-28 to 8-34 and Review Questions 8-52 to 8-54.
8-7 Cells in Series-Parallel
A series-parallel battery connection, as shown in Figure 8-22, provides both
greater voltage and greater capacity than the individual cells. The cells must
be identical to avoid circulating currents within the battery. The total capacity
of the battery is determined by the number of rows connected in parallel.
S cells in series
Ecell Rcell Ecell Rcell Ecell Rcell
Ecell Rcell Ecell Rcell Ecell Rcell
P rows in parallel
Ecell Rcell Ecell Rcell Ecell Rcell
Ecell Rcell Ecell Rcell Ecell Rcell
+
Figure 8-22
Ebat
−
Equivalent circuit of a series-parallel connected battery
8-7 Cells in Series-Parallel
185
For a battery consisting of P parallel rows with S cells in series in each
row, the EMF of the battery is the EMF of one series row, the same as for a
­series battery connection:
Ebat = SEcell
The total internal resistance is the combination of P identical parallel rows
of S resistors in series, with each resistor having a resistance of Rcell, so
Rbat =
resistance of each row
SRcell
=
number of rows
P
(8-6)
Example 8-12
Twenty identical cells are connected in four parallel rows, each with
five cells in series. Each cell has an internal resistance of 0.20 Ω. The
battery delivers a current of 5.0 A to a 1.25-Ω resistor as shown in
­Figure 8-23. Calculate the EMF of each cell.
Solution
Rbat =
SRcell
5 × 0.20 Ω
=
= 0.25 Ω
P
4
Ecell =
Ebat
7.5 V
=
= 1.5 V
S
5
Ebat = ILRT = 5.0 A × ( 0.25 Ω + 1.25 Ω ) = 7.5 V
Rbat
IL = 5.0 A
0.25 Ω
RL = 1.25 Ω
Ebat
Figure 8-23
Equivalent circuit for Example 8-12
Multisim Solution
Download Multisim file EX8-12 from the website.
Connect a ground to the bottom node. Run the simulation, and check
the value of IL displayed on the ammeter. Confirm that this value
matches the one calculated using Equation 8-6.
See Problems 8-35 to 8-39 and Review Question 8-55.
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Chapter 8 Series-Parallel Circuits
Circuit Check
C
CC 8-5. A 1.0-kΩ resistor, a 2.0-kΩ resistor, and a 4.0-kΩ resistor are
­connected in parallel and draw a total of 175 mA from a voltage
source. Find the current through each resistor, and calculate the
applied voltage.
CC 8-6. A battery has three parallel rows, each with four cells ­connected
in series. Given that each cell has an EMF of 1.5 V and an internal
resistance of 0.2 Ω, find the current this battery will deliver to a
4-Ω load.
8-8 Troubleshooting
We can use the voltage and current laws that apply to series-parallel networks to help diagnose faults in circuits.
Example 8-13
Determine the fault most likely to cause each of the following voltages
in the voltage divider designed in Example 8-6:
(a)
VL = 21 V
(b) VL = 0 V
RS shorted:
VL = 25 V
(c) VL = 25 V
Solution
(a) Either the supply voltage is not 25 V or one of the resistors is
shorted or open. Check the source voltage with a voltmeter. If the
source voltage is 25 V, then we calculate the value of VL that results if any of the resistors are either shorted or open.
RL shorted:
RB shorted:
RS open:
RL open:
RB open:
VL = 0 V
VL = 0 V
VL = 0 V
VL =
RB
300
( 25 V ) = 17 V
E=
RS + RB
300 + 143
The circuit is designed for RL =
VL =
VL
15 V
=
= 750 Ω
IL
20 mA
RL
750
( 25 V ) = 21 V
E=
RL + RS
750 + 143
The calculation for RB open yields the same result for VL as the
measured value. Therefore, we conclude that RB is open.
8-8 Troubleshooting
187
(b) As noted in part (a), VL = 0 V if RS is open or either of RB or RL
is shorted. Each resistor should be removed from the circuit and
checked with an ohmmeter to determine which one is defective.
(c)
As noted in part (a), VL = 25 V if RS is shorted. However, VL will
be 25 V if both RB and RL are open (leaving no complete path for
current). It is more likely that we have one defective resistor, rather
than two. Therefore, we first remove RS and check it with an ohmmeter. If it is not defective, then we check the other resistors.
Multisim Solution
Download Multisim file EX8-13(a) from the website.
Run the simulation, and note the meter readings for VL and IL are 15 V
and 20 mA, respectively.
Replace RS with a short circuit. Run the simulation, and note that VL
has become 25 V.
Restore RS and replace RL with a short circuit. Run the simulation, and
note that VL has become 0 V.
Restore RL and replace RB with a short circuit. Run the simulation, and
note that VL has become 0 V.
Restore RB and replace RS with an open circuit. Run the simulation, and
note that VL has become 0 V.
Restore RS and replace RL with an open circuit. Run the simulation, and
note that VL has become 175 V.
Restore RL and replace RB with an open circuit. Run the simulation, and
note that VL has become 21 V.
See Review Question 8-56.
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Chapter 8 Series-Parallel Circuits
Summary
• A series-parallel circuit can be simplified into an equivalent series or a­
parallel circuit.
• Series-parallel circuits can be analyzed using Kirchhoff’s voltage and
current laws.
• Voltages across resistors in series can be calculated using the voltage-­
divider principle.
• Bleeder resistors are used in voltage dividers to improve voltage
­regulation.
• Voltages in a circuit are often referenced with respect to chassis or
ground.
• Currents through resistors in parallel may be calculated using the currentdivider principle.
• Cells connected in series-parallel provide higher voltage and current to
a load.
• A fault in a resistive network is usually the result of an open or shorted
resistor.
B = beginner
I = intermediate
A = advanced
Problems
I
Section 8-2
8-1.
Equivalent-Circuit Method
Calculate the equivalent resistance of each of the resistance networks
in Figure 8-24.
60 Ω
400 Ω
40 Ω
30 Ω
20 Ω
(a)
Figure 8-24
1.0 kΩ
600 Ω
(b)
4.0 kΩ
189
Problems
I
8-2.
Calculate the resistance measured from terminal T to ground in each
of the resistance networks of Figure 8-25.
T
264 Ω
T
240 Ω
600 Ω
900 Ω
300 Ω
10 kΩ
27 kΩ
10 kΩ
2.2 kΩ
2.2 kΩ
(a)
(b)
Figure 8-25
I
8-3.
How much current is drawn from the voltage source in each of the
circuits in Figure 8-26?
Short circuit
136 Ω
+
50 V
−
20 Ω
320 Ω
60 Ω
136 Ω
+
50 V
−
320 Ω
(a)
All resistors 2.2 kΩ
20 Ω
60 Ω
200 mV
(b)
(c)
Figure 8-26
A
8-4.
Determine the source current in each of the circuits in Figure 8-27.
Short circuit
−48 V
16 Ω
12 Ω
18 Ω
20 Ω
+
48 V
−
8.0 Ω
16 Ω
32 Ω
12 Ω
20 Ω
18 Ω
8.0 Ω
4.0 kΩ
32 Ω
2.0 kΩ
+
80 V
−
6.0 kΩ
8.0 kΩ
Open circuit
(a)
Figure 8-27
(b)
(c)
190
Chapter 8 Series-Parallel Circuits
100 Ω
200 Ω
800 Ω
I
8-5.
I
8-6.
By connecting two leads to various terminals of the resistance network in Figure 8-28, six different values of equivalent resistance can
be obtained. Calculate all six.
Calculate RAB, RBC, and RCA in each of the networks in Figure 8-29.
A
400 Ω
Figure 8-28
A
60 Ω
300 Ω
200 Ω
100 Ω
150 Ω
B
C
B
(a)
500 Ω
(b)
C
Figure 8-29
I
8-7.
Determine the equivalent resistance of the circuit in Figure 8-30
(a) with the switch open
(b) with the switch closed
40 Ω
200 Ω
500 Ω
300 Ω
400 Ω
+
120 V
−
Figure 8-30
A
8-8.
I
8-9.
Find the voltage drop across the switch in the circuit of Figure 8-30
when the switch is open.
Find the input resistance of the ladder network in Figure 8-31
(a) with a 75-Ω load resistor connected as shown
(b) with the load resistor disconnected
25 Ω
Input
Figure 8-31
50 Ω
100 Ω
50 Ω
100 Ω
25 Ω
100 Ω
Output
75 Ω
191
Problems
I
I
A
8-10. What voltage applied to the input terminals of the ladder network in
Figure 8-31 produces 200 μW of power in the 75-Ω load resistor?
8-11. For each set of values for the circuit of Figure 8-1, draw fully labelled
diagrams of the equivalent circuits and calculate the voltage drop,
current, and power for each resistor.
(a) E = 40 V; R1 = 48 Ω; R2 = 40 Ω; R3 = 160 Ω
(b) PT = 625 W; R1 = 8 Ω; R2 = 10 Ω; R3 = 40 Ω
8-12. For the circuit shown in Figure 8-32, calculate
(a) the equivalent resistance
(b) the current through the 18-Ω resistor
(c) the voltage drop across the 8.0-Ω resistor
(d) the power dissipated by the 20-Ω resistor
20 Ω
8Ω
11 Ω
30 Ω
3Ω
5Ω
19 Ω
18
Ω
+
40 V
−
12
9Ω
Ω
4Ω
3Ω
10 Ω
Figure 8-32
I
A
A
I
8-13. Use Multisim to verify your answer to Problem 8-3(b).
Section 8-3
Kirchhoff’s Laws Method
8-14. For each set of values for the circuit of Figure 8-1, draw fully labelled
diagrams of the equivalent circuits and calculate the voltage drop,
current, and power for each resistor.
(a) E = 320 V; R1 = 11.0 Ω; I2 = 1.6 A; V3 = 144 V
(b) E = 100 V; PT = 75 W; R2 = 100 Ω; I3 = 0.50 A
(c) E = 150 V; R1 = 5.4 Ω; R2 = 12 Ω; I3 = 2.0 A
(d) E = 250 V; R1 + R2 = 20 kΩ; V2 = 120 V; I3 = 6.0 mA
8-15. A 12-V generator has an internal resistance of 50 mΩ. Two loads are
connected in parallel to its terminals, one drawing a 12-A current
and the other dissipating energy at the rate of 200 W. Find the terminal voltage of the generator with this load.
8-16. A digital voltmeter with an input resistance of 10 MΩ reads 4.5 V
when measuring the automatic-gain-control (AGC) voltage of a radio
receiver. When an analog voltmeter with an input resistance of 10 kΩ
is also connected across the AGC voltage source, both meters read
0.40 V. Calculate the internal resistance of the AGC voltage source.
circuitSIM
walkthrough
192
Chapter 8 Series-Parallel Circuits
A
8-17. In the circuit of Figure 8-33, determine the resistances of R1 and R2 if
(a) RL = 400 Ω and IL = 50 mA
(b) RL = 200 Ω and IL = 75 mA
R1
+
48 V
−
R2
RL
Figure 8-33
B
I
A
A
Section 8-5
Voltage Dividers
8-18. What bleeder resistor is required to complete a voltage divider to
give 250 V at 40 mA from a 500-V source if the series dropping resistor is 5.0 kΩ?
8-19. Design a voltage divider to supply 120 V at 75 mA from a 240-V
source if the total drain on the source is to be 100 mA.
8-20. Design a voltage divider to deliver 100 V at 1.0 mA from a 240-V
source so that a 100% increase in load current causes only a 5% reduction in load voltage.
8-21. Figure 8-34 shows the focus control for a cathode-ray tube. In the
one extreme position, the focus-anode voltage is to be 800 V with a
current of 10.0 μA. At the other extreme, the focus-anode voltage is
to be 1400 V with a current of 14.0 μA. Calculate the resistance of R1
and of the focus control.
Cathode
ray
tube
+
R1
2.0 kV
Source
Focus
control
R2
1.0 MΩ
−
Figure 8-34
I
circuitSIM
walkthrough
8-22. (a)Design a voltage divider to supply 10 V at 10 mA and 25 V
at 25 mA from a 40-V DC supply with a bleeder current (IB in
­Figure 8-35) of 10 mA.
(b)Use Multisim to verify that your circuit design produces the required voltages and currents.
Problems
I1
R1
25 mA
I2
+
40 V E
−
R2
10 mA
25 V
IB
RB
10 V
Figure 8-35
I
A
I
I
I
B
8-23. Assuming that the load 1 in Figure 8-16 acts like a linear resistance,
­determine the voltage distribution (with respect to the chassis) in
Example 8-8 if the load 2 becomes open-circuit.
8-24. A voltage divider consists of a 10-kΩ resistor with an adjustable tap.
The tap is set so that it feeds 80 V at 5 mA to a load when the voltage
divider is connected across a 200-V source. What is the resistance of
the bleeder portion of the voltage divider?
8-25. A transistorized stereo amplifier requires the following DC supplies:
−30 V at 2.0 A, −16 V at 250 mA, and +12 V at 2 A. All voltages are
measured with respect to ground. Design a voltage divider to operate this amplifier from a single DC supply that ­develops a ter­minal
voltage of 46 V with a total current drain of 2.5 A.
8-26. A voltage divider circuit is to supply a 100-mA load at 30 V. If the
power supply is 50 V and the total current drain is to be 150 mA,
design the circuit and sketch the circuit diagram.
8-27. Draw and design a voltage-divider circuit to provide the following
load currents from a 50-V supply:
500 mA at 50 V 200 mA at 30 V 100 mA at 15 V
Design for a bleeder current of 100 mA.
Section 8-6
Current-Divider Principle
8-28. Calculate the current through each resistor in the circuit of
­Figure 8-36(a).
193
194
Chapter 8 Series-Parallel Circuits
I
8-29. What resistance must be connected across the terminals of the
­constant-current source shown in Figure 8-36(b) to reduce the terminal voltage to 75 V?
IT = 15 mA
+
E
−
8.0 kΩ
16 kΩ
40 mA
5.0 kΩ
(a)
(b)
Figure 8-36
I
I
8-30. In the circuit in Figure 8-37, what resistance R will draw a current
equal to 25% of the source current?
8-31. In the circuit in Figure 8-37, what resistance R makes the source current 160 mA?
50 Ω
+
20 V
−
50 Ω
100 Ω
R
Figure 8-37
A
circuitSIM
walkthrough
8-32. (a)For the circuit of Figure 8-38, calculate the equivalent resistance
for the resistor network, the current in the 5.2-Ω resistor, and the
voltage across the 1.0-Ω resistor.
(b) Use Multisim to verify your answers for part (a).
6.0 Ω
3.0 Ω
5.2 Ω
20 Ω
60 Ω
30 Ω
20
Ω
+
30 V
−
8.0 Ω
12 Ω
4.0 Ω
1.0 Ω
Figure 8-38
5.0 Ω
Problems
A
8-33. For the circuit of Figure 8-39, calculate
(a) the total resistance
(b) the current in the 24-Ω resistor
(c) the voltage across the 14-Ω resistor
(d) the power to the 20-Ω resistor
3.0 Ω
5.0 Ω
6.0 Ω
20 Ω
27 Ω
10
24
Ω
Ω
+
80 V
−
15 Ω
18 Ω
22 Ω
8.0 Ω
6.0 Ω
14 Ω
Figure 8-39
A
8-34. For the circuit shown in Figure 8-40, calculate
(a) the total resistance
(b) the current in the 15-Ω resistor
12 Ω
6.
0
Ω
+
50 V
−
30 Ω
12
Ω
15
Ω
2.0 Ω
Figure 8-40
I
Section 8-7
Cells in Series-Parallel
8-35. A battery contains four parallel-connected rows of five identical cells
in series. Each cell has an EMF of 800 mV and an internal resistance of
0.20 Ω. How much current will this battery deliver to a 6.0-Ω load?
195
196
Chapter 8 Series-Parallel Circuits
I
A
I
I
8-36. A battery is made up of four parallel-connected rows, each with six
identical cells connected in series. Each cell has an internal r­ esistance
of 2.0 Ω. The battery sends a current of 1.0 A through an 6.0-Ω load.
Calculate the EMF of a cell.
8-37. The terminal voltage of a battery is 12.0 V when the load current is
1.6 A, and 10.0 V when the current is 2.4 A. Calculate the internal
­resistance of the battery.
8-38. The maximum sustained current for an alkaline C cell is 300 mA.
Specify how to connect alkaline C cells to make a 12-V battery with
a capacity of 1.2 A. Given that the internal resistance of each cell
is 0.150 Ω, find the ­actual terminal voltage of the battery when the
­current drain is 1.2 A.
8-39. A battery is made from 40 identical 1.5-V cells connected eight
in series per row, five rows in parallel. When a 10-Ω load resistor is connected, the terminal voltage is 10 V. Find the internal
resistance/cell.
Review Questions
Section 8-1
Series-Parallel Resistors
8-40. With respect to series-parallel resistance networks, what is meant by
an equivalent resistance?
8-41. What characteristics must an equivalent resistance possess?
8-42. What circuit rules allow us to replace R2 and R3 in Figure 8-41 with
an equivalent resistance?
Rint
R2
+
R1
E
−
R3
Figure 8-41
Section 8-3
Kirchhoff’s Laws Method
8-43. What is the advantage of solving series-parallel resistance networks
by writing Kirchhoff’s law equations rather than using the equivalent-­
resistance method?
8-44. Write one Kirchhoff’s current law equation and two Kirchhoff’s
voltage law equations for the network of Figure 8-41.
Section 8-4
Voltage-Divider Principle
8-45. What circuit condition must exist for the voltage-divider principle to
be applied in network solutions?
Integrate the Concepts
8-46. Explain the operation of the volume-control circuit shown in
­Figure 8-42.
Input
signal
Output to
amplifier
Figure 8-42
Section 8-5
Volume control
Voltage Dividers
8-47. What is the purpose of a bleeder resistor in a practical voltage-­
divider circuit?
8-48. Calculate the load voltage in Examples 8-5 and 8-6 if the load current
decreases to 10 mA. Account for the change in the load voltage.
8-49. Why does a decrease in the bleeder resistance of a given voltage d
­ ivider
improve the voltage regulation of the output of the voltage divider?
8-50. In Chapter 2, we noted that voltage is measured between two points in
a circuit. Yet circuit diagrams often have notations such as “Voltage
at Point A is +12 V.” Explain.
8-51. A DC power supply for an amplifier has only two output terminals.
Describe how to connect this power supply to obtain both positive
and negative potentials with respect to the chassis of the amplifier.
Section 8-6
Current-Divider Principle
8-52. What circuit conditions must exist for the current-divider principle
to be applied in network solutions?
8-53. The current-divider principle is sometimes called the parallel-circuit
dual of the voltage-divider principle. Explain this duality relationship.
8-54. Explain why the ratio of the currents in two parallel branches is the
inverse of the ratio of the resistances of the branches.
Section 8-7
Cells in Series-Parallel
8-55. How is the total internal resistance of a series-parallel combination of
identical cells related to the internal resistance of the individual cells?
Section 8-8
Troubleshooting
8-56. What circuit conditions in Figure 8-14 would lead one to conclude
that the bleeder resistor is open?
Integrate the Concepts
Refer to the circuit of Figure 8-1 and its equivalent circuit in Figure 8-2.
Set R1 = 48 Ω, R2 = 220 Ω, R3 = 330 Ω, and E = 450 V. Calculate Req, RT, I1,
VR1, I2, and I3.
197
198
Chapter 8 Series-Parallel Circuits
Practice Quiz
1.
The equivalent resistance of the circuit in Figure 8-43 is
(a) 260 Ω
(b) 473 Ω
(c) 590 Ω
(d) 143 Ω
R2
R1
220 Ω
50 Ω
R3
220 Ω
R4
100 Ω
V1
Figure 8-43
2.
The equivalent resistance of the ladder network in Figure 8-44 is
(a) 763 Ω
(b) 783 Ω
(c) 842 Ω
(d) 963 Ω
330 Ω
1.0 kΩ
100 Ω
100 Ω
1.0 kΩ
1.0 kΩ
100 Ω
Figure 8-44
3.
The total current in the circuit of Figure 8-45 is
(a) 999 μA
(b) 13.2 mA
(c) 10.8 mA
(d) 99.8 mA
Practice Quiz
R1
IT
IR3
100 Ω
R2
10 kΩ
12 V
R3
1.0 kΩ
R4
100 Ω
Figure 8-45
4.
The current through resistor R3 in Figure 8-45 is
(a) 984 μA
(b) 9.84 mA
(c) 9.75 mA
(d) 97.5 mA
5.
For the battery shown in Figure 8-46, the EMF of each cell is
(a) 3.0 V
(b) 3.5 V
(c) 2.0 V
(d) 2.5 V
XMM1
−
V1
R1
V2
0.15 Ω
+
V4
R4
V3
0.15 Ω
V5
0.15 Ω
R5
0.15 Ω
RL
220 Ω
Figure 8-46
R2
R3
0.15 Ω
V6
R6
0.15 Ω
199
9
Resistance Networks
The techniques developed to this point work well for c
­ ircuits with
resistors in series or in parallel. However, some circuits have components
connected in patterns that are ­neither ­series nor parallel combinations.
This chapter introduces methods of circuit analysis for such circuits.
Chapter Outline
9-1
9-2
9-3
9-4
Network Equations from Kirchhoff’s Laws 202
Constant-Voltage Sources 202
Constant-Current Sources 204
Source Conversion 206
9-5Kirchhoff’s Voltage-Law Equations:
Loop Procedure 208
9-6Networks with More Than One
Voltage Source
214
9-7Loop Equations in Multisource
9-8
9-9
9-10
9-11
Networks
216
Mesh Analysis
222
Nodal Analysis
231
Kirchhoff’s Current-Law Equations 228
The Superposition Theorem
237
Key Terms
source conversion 202
short-circuit current 204
loop equations 208
mesh equations 208
nodal equations 208
mesh analysis 222
mesh 222
planar 222
nodes 228
reference node 229
nodal analysis 231
superposition theorem 237
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• calculate the internal resistance of constantvoltage and constant-current sources from
their open-circuit voltages and short-circuit
currents
• convert a constant-voltage source to an
­equivalent ­constant-current source
• convert a constant-current source to an
­equivalent ­constant-voltage source
• analyze a circuit containing constant-voltage
sources using the loop procedure based on
Kirchhoff’s voltage law
Photo source: © MARK SYKES/SCIENCE PHOTO LIBRARY
• analyze a planar circuit containing constantvoltage sources using mesh equations
• analyze a circuit using Kirchhoff’s current law
• analyze a circuit containing constant-current
sources using nodal analysis
• analyze a circuit containing more than one
source using the superposition theorem
202
Chapter 9 Resistance Networks
9-1 Network Equations
from Kirchhoff’s Laws
The Wheatstone bridge in Figure 9-1 may appear at first glance to be a
­simple series-parallel circuit, but no two resistors in the circuit are actually connected in series or parallel. Hence, we cannot use the equivalent-­
resistance techniques of Chapter 8 to analyze a bridge circuit. However,
there are two general methods we can use to solve such networks.
R1
R5
+
E
−
R3
Figure 9-1
R2
R4
Wheatstone bridge
One method uses various network theorems to simplify the original network so that we can apply equivalent-circuit techniques. The second method
is based on Kirchhoff’s voltage and current laws, which apply to all circuits
no matter how complex. To apply the Kirchhoff’s-laws method, we need
to learn a few basic rules for writing an orderly set of simultaneous equations. This structured approach makes this method particularly suitable for
circuit-analysis software. The disadvantage of Kirchhoff’s-laws techniques
in an ­introductory course is that we can become more involved with plugging numbers into computer programs than with learning the principles of
circuit behaviour. The equivalent-circuit and network-theorem methods let
us analyze the same circuits with less emphasis on mathematical formats.
An understanding of the principle of source conversion enables us to
simplify the final formats for Kirchhoff’s-laws equations for various resistance networks. Hence, we shall discuss equivalent-circuit techniques in
the next three sections of this chapter.
9-2 Constant-Voltage Sources
The terminal voltage of most practical sources falls as the current drawn
from the source increases. In Section 7-6 we explained this effect by assuming that a practical voltage source consists of a constant-voltage source in
series with an internal resistance. In other words, the circuit of Figure 9-2(a)
behaves exactly the same as the equivalent circuit of Figure 9-2(b).
9-2 Constant-Voltage Sources
RL
Battery
Rint
+
RL
E
−
(a)
Figure 9-2
(b)
Equivalent circuit for a practical voltage source
In replacing a practical voltage source in a circuit diagram with an equivalent series constant-voltage source, we treat the actual voltage source as a
“black box” that has only two terminals exposed. For example, consider a
power supply that provides a DC voltage to operate a transistor amplifier.
As shown in Figure 9-3(a), this power supply is essentially a “black box”—
we can make measurements at the terminals but we cannot see how the
­terminals are interconnected inside the box. This black box outputs 16 V
DC when connected to a 120-V 60-Hz power supply. Viewed from the load
side, the power supply is simply a source of direct current. The details of
how the black box converts its input power to direct current do not affect
the operation of the load circuit.
Rint
A
+
0.20 Ω
Power
supply
V
RL
+
16 V
−
RL
−
Terminal voltage (V)
(a)
(b)
16
14
0
10
20
30
40
50
Load current (A)
60
(c)
Figure 9-3
Equivalent circuit for a DC power supply
70
80
203
204
Chapter 9 Resistance Networks
We can measure the open-circuit terminal voltage of the power supply
by replacing the load with a voltmeter that draws negligible current. When
we connect progressively smaller values of RL , we find that the terminal
voltage falls as the current drawn from the power supply increases, as
shown by the graph in Figure 9-3(c). All that we can conclude from the data
in Figure 9-3(c) is that the power supply behaves like a constant-voltage
source with an internal resistance of
Rint =
ΔV
2.0 V
=
= 0.20 Ω
ΔI
10 A
By extending the graph (as shown by the dashed line), we can find the
short-circuit current, that is, the current when RL = 0 Ω and VL = 0 V. When
RL = 0 Ω, the internal resistance is in series with just the voltage source, so
Rint =
open-circuit terminal voltage
short-circuit terminal current
(9-1)
For the graph of Figure 9-3(c), the short-circuit current is 80 A. Hence,
the internal resistance for the constant-voltage source of Figure 9-3(b) is
Rint =
Voc
16 V
=
= 0.20 Ω
Isc
80 A
See Problems 9-1 and 9-2 and Review Questions 9-44 to 9-49 at the end
of the chapter.
9-3 Constant-Current Sources
We can use another approach to account for the properties of the blackbox power supply in Figure 9-3(a). Our measurements tell us that the
sealed power supply has an open-circuit terminal voltage of 16 V and a
short-­circuit terminal current of 80 A. Theoretically, the box could contain
a ­special generator with a variable terminal voltage that always produces
a current equal to the short-circuit terminal current. For this constant current to flow when the terminals of the power supply are open-circuit, the
internal resistance must be in parallel with the constant-current generator,
as shown in Figure 9-4(b). With no load connected, all of the current flows
through the internal resistance, so
Rint =
open-circuit terminal voltage
short-circuit terminal current
(9-1)
Note that the equation for internal resistance is the same for both
c­onstant-voltage and constant-current sources. The arrow in the symbol
for a constant-current source shows the direction of conventional current
­flowing through the source.
9-3 Constant-Current Sources
A
+
Power
supply
V
RL
Rint
0.20 Ω
80 A
RL
−
(a)
Figure 9-4
(b)
Constant-current source
Suppose that RL = 0.60 Ω in the circuits in Figures 9-3(b) and 9-4(b). With
the constant-voltage source of Figure 9-3(b), the voltage-divider principle
tells us that
VL =
and
RL
0.60
× 16 V = 12 V
× Voc =
Rint + RL
0.20 + 0.60
VL
12 V
IL =
=
= 20 A
RL
0.60 Ω
With the constant-current source of Figure 9-4(b), we switch to the
­current-divider principle:
and
IL =
Rint
0.20
× Isc =
× 80 A = 20 A
Rint + RL
0.20 + 0.60
VL = ILRL = 20 A × 0.60 Ω = 12 V
For all values of load resistance, the constant-voltage source of Figure 9-3(b)
and the constant-current source of Figure 9-4(b) are exact equivalents, as “seen”
by RL. The constant-current equivalent source is the dual of the constant-­
voltage source. An American engineer, Edward L. Norton (1898–1983), devised
the method for using constant-current equivalent sources for circuit analysis.
In most practical electric circuits, we open the circuit (disconnect the load) to
switch the circuit off. When we do so, both the current through the source and
the current in the load become zero. The power dissipation in the internal resistance of the source is also zero under open-circuit c­ onditions. Hence, equivalent circuits with constant-voltage sources operate much like the actual circuits.
If we disconnect RL in the current-source circuit of Figure 9-4(b), the current through Rint increases to the full constant-current value and the internal
dissipation of the source becomes Voc × Isc = 1280 W. Theoretically, to switch
off the circuit of Figure 9-4(b), we short-circuit RL so that the terminal voltage
and the energy transfer to the load become zero. Constant-current equivalent sources are primarily just theoretical devices for network analysis.
Although we will not encounter current sources in power-system networks, we do find devices that function as if they were current sources in
low-power electronic circuits. For example, it is usually easier to analyze a
transistor circuit if we show a current source in its equivalent circuit. Also, it
is possible to use electronic feedback circuits to construct a DC power supply
205
206
Chapter 9 Resistance Networks
that maintains a constant load current with a terminal voltage that varies
as the load resistance varies. Such constant-current power supplies are less
common than power supplies that maintain a constant terminal voltage.
See Problem 9-3 and Review Questions 9-50 to 9-52.
9-4 Source Conversion
For a given constant-voltage source, Rint in the equivalent constant-current
source has the same value but appears in parallel with the ideal current
source, as shown by the examples in Figure 9-5. Similarly, for a given
constant-current source, Rint in the equivalent constant-voltage source has
the same value but appears in series with the ideal voltage source.
In order for us to replace a constant-voltage source in a network with
its constant-current equivalent, there must be some resistance Rx in series
with an ideal voltage source. Rx may be an internal resistance of the source,
or it may be one of the network resistors in series with the voltage source.
­Similarly, there must be some form of resistance in parallel with an ideal
current source before we can replace it with its constant-voltage equivalent.
Applying Equation 9-1 to the constant-voltage source of Figure 9-5(a)
shows that Ix for the equivalent constant-current source (Figure 9-5(b)) is
the short-circuit current:
Ix =
Eoc
Rx
(9-2)
Similarly, given the constant-current source of Figure 9-5(c), Ex for the equivalent constant-voltage source of Figure 9-5(d) is the open-circuit voltage:
Ex = IscRx
(9-3)
Rx
6.0 Ω
+
+
+
120 V Eoc
−
20 A
Ix
6.0 Ω
Rx
−
−
(a)
(b)
Rx
−
−
40 A
−
32 V Ex
+
0.80 Ω
Isc Rx
0.80 Ω
+
+
(c)
Figure 9-5
Source conversion
(d)
9-4 Source Conversion
Example 9-1
(a)Determine the equivalent constant-current source for a voltage
source with an open-circuit voltage of 120 V and an internal
resistance of 6.0 Ω.
(b)Determine the equivalent constant-voltage source for a current
source with a constant current of 40 A and an internal resistance of
0.80 Ω.
(c)Check that the equivalent sources in part (a) produce the same
results when connected to a 12-Ω load.
Solution
(a)The internal resistance for the constant-current source is the same
as for the voltage source: Rint = 6.0 Ω
Ix =
Eoc
120 V
=
= 20 A
Rint
6.0 Ω
(b)
Rint for the constant-voltage equivalent source is the same as for
the current source:
Rint = 0.80 Ω
Ex = IscRint = 40 A × 0.80 Ω = 32 V
(c) When we connect a 12-Ω load to the voltage source of Figure 9-5(a),
IL =
and
Ex
120 V
=
= 6.7 A
Rx + RL
6.0 Ω + 12 Ω
VL = 6.7 A × 12 Ω = 80 V
When we connect a 12-Ω load to the current source of Figure 9-5(b),
VL = Ix ×
and
Rx × RL
6.0 × 12
= 80 V
= 20 A ×
Rx + RL
6.0 + 12
VL
80 V
IL =
=
= 6.7 A
RL
12 Ω
Once we are familiar with the concept, we can convert voltage
sources to equivalent current sources (and vice versa) for network analysis ­purposes with just a bit of mental arithmetic. Analysis of transistor
circuits often i­nvolves resistance networks containing both voltage and
current sources. Such networks are usually easier to solve if we convert the sources so that they are either all voltage sources or all current
sources.
See Problems 9-4 to 9-6 and Review Questions 9-53 to 9-55.
207
208
Chapter 9 Resistance Networks
Circuit Check
A
CC 9-1. Find the voltage across a 12-Ω load when it is connected to a
24-V source that has an internal resistance of 3.0 Ω.
CC 9-2. Use an equivalent constant current source to verify the load voltage calculated in question CC 9-1.
9-5 Kirchhoff’s Voltage-Law Equations:
Loop Procedure
Kirchhoff’s laws let us analyze a circuit in several different ways, each with
its own advantages and disadvantages. We can use Kirchhoff’s voltage law
to write a set of either loop equations or mesh equations. With Kirchhoff’s
current law, we can write a set of nodal equations. We shall consider the
loop method first.
In Section 7-4, we stated Kirchhoff’s voltage law:
In any complete electric circuit, the algebraic sum of the source
voltages must equal the algebraic sum of the voltage drops.
This statement leads to equations in the form
E = V1 + V2 + V3 + . . .
(9-4)
Here is another way of stating Kirchhoff’s voltage law:
In any closed loop, the algebraic sum of all voltage rises and voltage
drops is zero.
This statement leads to an equation in which we must carefully record the
sign of voltage drops and voltage rises:
E − V1 − V2 − V3 = 0
(9-5)
V1 + V2 + V3 + . . . = E
(9-6)
To help set up our equations in a form that can be solved by determinants,
we reverse Equation 9-4:
We start with an example that we can readily check with the equivalent
resistance techniques of Chapter 8.
9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure
Example 9-2
Determine the magnitude and polarity of the potential difference
between terminals A and B in the bridge network of Figure 9-6.
10 Ω
Rint
+
80 V
−
Figure 9-6
R2
R1
20 Ω
A
R3
30 Ω
15 Ω
B
R4
10 Ω
Bridge network for Example 9-2
Solution
Step 1
Draw a circuit diagram large enough to mark on tracing loops, voltage
­polarities, and any other data we need. We may change the location of
components in the diagram so long as we do not change the electric
­connections (see Figure 9-7).
Rint
+
C
−
10 Ω
+
80 V
−
+
+
R 1 20 Ω
−
I1
A
15 Ω
−
I2
B
+
+
R3
R2
30 Ω
−
R4
10 Ω
−
D
Figure 9-7
Circuit diagram for Example 9-2
Step 2
Label the positive and negative terminals of the sources. We will draw
current tracing loops on the circuit diagram, beginning at the positive
terminal of the source and proceeding along a path through the network back to the negative terminal of the source. These tracing loops
help us to write Kirchhoff’s voltage-law equations. We need enough
tracing loops to include all components and to provide as many
209
210
Chapter 9 Resistance Networks
s­ eparate equations as there are unknowns. In this example, we require
two tracing loops for the two unknown loop currents, I1 and I2. Note
that reversing the polarity of a source in a network with more than one
source alters all the current and voltage relationships in the network.
Step 3
Mark the polarity of each voltage drop. In Section 7-2, we considered
the polarities of voltage drops and voltage rises in relation to current
direction, as shown in Figure 9-8. We can generalize this concept:
In any passive component (such as a resistor) in an electric net­
work, the tracing direction is from positive to negative through
that ­component.
+
E
−
I
+
V
−
Figure 9-8 Polarity of voltage drop and source
voltage in relation to direction of current tracing
loop
Once we have established the directions of tracing loops for I1 and
I2, we can mark the polarity of the various voltage drops on the circuit
diagram in Figure 9-7.
Step 4
Write a Kirchhoff’s voltage-law equation for each loop. For the I1 loop,
VRint + VR1 + VR3 = E
Since I1 is the current through R1 and R3,
VR1 = I1R1 and
VR3 = I1R3
Current I1 leaves junction C through R1 and R3, and the current I2 leaves
this junction through R2 and R4. Hence, from Kirchhoff’s current law,
the current through Rint is I1 + I2, and
VRint = ( I1 + I2 ) Rint
Substituting for the voltages in the Kirchhoff’s voltage-law equation
for the I1 tracing loop gives an equation in the standard form we shall
use for the loop method:
(I1 + I2)Rint + I1R1 + I1R3 = E
Writing the voltage drops as IR drops minimizes the number of
­ nknowns.
u
9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure
Step 5
Substitute the known quantities into the equations. The equation for
the I1 tracing loop becomes
10 Ω(I1 + I2) + 20 Ω(I1) + 30 Ω(I1) = 80 V
Collecting terms gives
Similarly, for the I2 tracing loop,
60I1 + 10I2 = 80
10 Ω ( I1 + I2 ) + 15 Ω ( I2 ) + 10 Ω ( I2 ) = 80 V
10I1 + 35I2 = 80
(1)
(2)
We then solve Equations 1 and 2 simultaneously.
Elimination Method
6 × Equation 2: 60I1 + 210I2 = 480
subtract Equation 1: 60I1 + 10I2 = 80
200I2 = 400
I2 = 2.0 A
Substituting this value in Equation 1, we obtain
I1 = 1.0 A
Determinant Method
The determinant method provides a standard procedure that we can
use to solve any array of simultaneous equations. Appendix 1 explains
the matrix algebra for calculating determinants.
I1
I2
│8080
=
│6010
│6010
=
│6010
60I1 + 10I2 = 80
10I1 + 35I2 = 80
│ 2800 − 800 2000
=
=
= 1.0 A
2100 − 100
2000
10
35│
10
35
│ 4800 − 800 4000
=
=
= 2.0 A
2000
2000
10
35│
80
80
(1)
(2)
211
212
Chapter 9 Resistance Networks
Simultaneous equations can also be solved using a programmable
calculator. Select the program for solving simultaneous equations and
input the coefficients. Run the program and the display will show the
values of I1 and I2.
Step 6
Find the voltage drops across the components of the network. In this
example, we need to solve for just two voltages:
and
circuitSIM
walkthrough
VR3 = I1R3 = 1.0 A × 30 Ω = 30 V
VR4 = I2R4 = 2.0 A × 10 Ω = 20 V
A voltage drop across a resistance is just another way of expressing a
p­ otential difference between the two terminals of the resistance. Hence,
terminal A in Figure 9-7 is 30 V more positive than junction D, and terminal B is 20 V more positive than junction D. Therefore, terminal A is
10 V more positive than terminal B:
VAB = +10 V
Multisim Solution
Download Multisim file EX9-2 from the website.
The circuit is the one shown in Figure 9-7, except a voltmeter has been
added to measure VAB.
Source: © MARK SYKES/SCIENCE PHOTO LIBRARY
Run the simulation. Note that the voltmeter reading matches the value
for VAB calculated using the loop procedure.
Wheatstone bridge.
A
Invented by Samuel
Hunter Christie in
1833, this circuit
was improved and
popularized by Charles
Wheatstone in 1843.
Yet another approach to the potential difference between terminals A
and B in Figure 9-6 is to use circuit simulation software that can capture
schematics. Once the circuit is drawn on the computer screen, any current
or voltage of interest can be displayed on the screen. Unfortunately, this
method does not allow a student to learn anything about loop equations,
determinants, or network theorems as they are buried within the computer
program. However, circuit simulations are useful for predicting currents
and voltages before a circuit is built.
If we add a meter between terminals A and B of the bridge circuit of
Figure 9-6, the network becomes a Wheatstone bridge, which can be used
for high-precision resistance measurements. We now require one more
tracing loop for our circuit analysis. In Figure 9-9 we have chosen a tracing loop for I3 passing through R1, RM (the resistance of the meter), and R4.
We could choose a tracing loop passing through R2 , RM, and R3 instead. I3
would then pass in the opposite direction through RM, and have the opposite
sign. If a tracing loop current has a negative sign, we simply treat the current as a negative quantity when we add the loop currents algebraically.
Using the I3 loop shown in Figure 9-9, the Kirchhoff’s voltage-law equations for the three tracing loops are
9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure
R1(I1 + I3) + R3 I1 = E
R2I2 + R4(I2 + I3) = E
R1(I1 + I3) + RMI3 + R4(I2 + I3) = E
R1
+
E
−
I1
I2
R3
Figure 9-9
R2
RM
I3
R4
Solving a Wheatstone bridge by loop procedure
Substituting the given values of resistance and source voltage, we can
solve these three simultaneous equations for the three loop currents. We
can then find the current through each component.
Total current:
I1 + I2 + I3
Current through R2:
I1 + I3
Current through R3:
I1
Meter current:
Current through R1:
Current through R4:
I3
I2
I2 + I3
Example 9-3
Find the current through a galvanometer with a resistance of 50 Ω
when the meter is connected between terminals A and B of the bridge
circuit of Figure 9-7.
Solution
For the I1 loop,
10 Ω(I1 + I2 + I3) + 20 Ω(I1 + I3) + 30 Ω(I1) = 80 V
213
214
Chapter 9 Resistance Networks
For the I2 loop,
10 Ω(I1 + I2 + I3) + 15 Ω(I2) + 10 Ω(I2 + I3) = 80 V
For the I3 loop,
10 Ω(I1 + I2 + I3) + 20 Ω(I1 + I3) + 50 Ω(I3) + 10 Ω(I2 + I3) = 80 V
Collecting terms gives
│
│
60I1 + 10I2 + 30I3 = 80
10I1 + 35I2 + 20I3 = 80
30I1 + 20I2 + 90I3 = 80
│
│
(1)
(2)
(3)
Since we are asked only to determine the meter current, we need
only solve for I3. Using the determinant method,
I3 =
=
=
60
10
30
10
35
20
80
80
80
60
10
30
10
35
20
30
20
90
168 000 + 24 000 + 16 000 − 84 000 − 96 000 − 8000
189 000 + 6000 + 6000 − 31 500 − 24 000 − 9000
20 000
136 500
= 0.15 A
See Problems 9-7 to 9-9 and Review Questions 9-56 to 9-61.
9-6 Networks with More
than One Voltage Source
Most flashlights have two or more cells connected in series aiding to provide
the lamp with a potential difference higher than that of a single cell. Since
both voltage sources in Figure 9-10(a) make current flow in the same direction through any load, the total effective source voltage is the sum of the individual source voltages. In circuit diagrams, we can replace the individual
series-connected voltage sources with a single equivalent source.
If we load one cell backward in a flashlight, the lamp will glow dimly, if
at all. With a series opposing connection, like the one shown in Figure 9-10(b),
each source tries to make current flow in opposite directions through the
load. The net terminal voltage for the equivalent source is the difference
9-6 Networks with More than One Voltage Source
­ etween the individual source voltages. We avoid series opposing con­
b
nection of voltage sources in practical electric circuits other than battery
chargers because the equivalent source voltage is actually reduced and the
higher voltage source causes current to travel the wrong way through the
lower voltage source when a load is connected.
+
+
−
−
+
+
6V
6V
+
14 V
−
−
+
8V
−
−
2V
−
−
8V
+
+
−
−
(a)
Figure 9-10
+
(b)
+
Voltage sources: (a) Series aiding; (b) Series opposing
Figure 9-11(a) shows a circuit in which both a storage battery and a generator driven by an automobile engine provide current to a common load.
The terminals of these two voltage sources (each with its own internal
resistance) are connected in parallel. If the engine stops, the generator EMF
must fall to zero and a heavy current—limited only by the two internal
­resistances in series—flows from the battery through the generator. In an
electrical system for an automobile, the generator is automatically disconnected from the rest of the system when the engine is stopped. Parallel
connection of voltage sources requires special precautions.
Since a single power supply often feeds dozens of stages of an electronic
system, we usually avoid cluttering circuit diagrams with long interconnecting leads by marking the DC supply voltages for individual stages as
shown in the coupling circuit in Figure 9-11(b). Since both networks in
­Figure 9-11 include more than one voltage source, we cannot solve them
by simple Ohm’s-law calculations.
−15 V
2.2 kΩ
Rgen
5.6 kΩ
Rbat
Load
+
+
Egen
−
Ebat
3.3 kΩ
RB
−
(a) Automobile electrical system
Figure 9-11
6.8 kΩ
+10 V
(b) Transistor coupling circuit
Examples of networks with more than one voltage source
215
216
Chapter 9 Resistance Networks
9-7 Loop Equations
in Multisource Networks
The procedure we developed in Section 9-5 for writing Kirchhoff’s voltagelaw equations for closed tracing loops applies equally well to multisource
networks. As our first example, we shall use the automobile electrical system of Figure 9-11(a).
Example 9-4
An automobile generator with an internal resistance of 0.20 Ω develops
an open-circuit voltage of 16.0 V. The storage battery has an internal
resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources
are connected in parallel to a 1.0-Ω load. Determine the generator current, battery current, and load current.
Solution I
We can redraw the circuit diagram of Figure 9-11(a) as shown in
Figure 9-12 to make it easier to draw the tracing loops.
− 0.10 Ω +
+ 0.20 Ω −
Rgen
Rbat
+
Egen
16.0 V
Igen
1.0 Ω
−
Figure 9-12
+
RL
−
+
Ibat
Ebat
12.8 V
−
Circuit diagram for Solution I of Example 9-4
For the generator loop with current Igen,
But from Ohm’s law,
VRgen + VL = Egen
VRgen = Igen Rgen = Igen × 0.20 Ω
Since Figure 9-12 shows both Igen and Ibat loops going in the same direction through the load, the load current is ­Igen + Ibat , and
VL = ILRL = 1.0 Ω × ( Igen + Ibat )
Hence, the Kirchhoff’s voltage-law equation for the generator loop becomes
or
0.20Igen + 1.0 ( Igen + Ibat ) = Egen
1.2Igen + 1.0Ibat = 16.0
(1)
1.0Igen + 1.1Ibat = 12.8
(2)
Similarly, the Kirchhoff’s voltage-law equation for the battery loop becomes
or
0.10Ibat + 1.0(Igen + Ibat) = Ebat
9-7 Loop Equations in Multisource Networks
Now we solve Equations 1 and 2 simultaneously.
Elimination Method
5 × Equation 1: 6Igen + 5.0Ibat = 80
6 × Equation 2: 6Igen + 6.6Ibat = 76.8
− 1.6Ibat = 3.2
subtracting gives
Ibat =−2.0 A
The negative value for Ibat is correct. This value tells us that the battery is not supplying current to the load, as we had supposed when
setting up the direction for the tracing loop for Ibat. In this example, the
storage battery is charging at a rate of 2.0 A. We do not need to change
the direction of the Ibat tracing loop as long as we treat Ibat as a negative
quantity. Substituting for Ibat in Equation 1 gives
1.2Igen + (1.0)(−2.0) = 16
Igen = 15 A
Determinant Method
Igen
Ibat
IL = Igen + Ibat = 15 + (−2.0) = 13 A
│16.0
12.8
=
│1.2
1.0
│1.2
1.0
=
│1.2
1.0
1.2Igen − 2.0 = 16
1.2Igen + 1.0Ibat = 16.0
1.0Igen + 1.1Ibat = 12.8
│ 17.6 − 12.8 4.8
=
=
= 15 A
1.32 − 1.0
0.32
1.0
1.1│
1.0
1.1
│ 15.36 − 16.0 −0.64
=
=
= −2.0 A
0.32
0.32
1.0
1.1│
16.0
12.8
IL = Igen + Ibat = ( 15 − 2 ) = 13 A
Solution II
The tracing loops in Solution I are the normal choices from a circuit-­
operation point of view. However, all other choices lead to the same
solution. To illustrate the loop technique further, we return to the circuit layout given in the original circuit diagram of Figure 9-11(a) and
add tracing loops as shown in Figure 9-13.
217
218
Chapter 9 Resistance Networks
−
Rgen
0.20 Ω
+
Rbat
+
Igen
IL
+
RL
+
Egen
16.0 V
0.10 Ω
Ebat
−
12.8 V
1.0 Ω
−
−
Figure 9-13 Circuit diagram for Solution II of Example 9-4
Igen flows from the positive terminal of the generator through Rgen,
Rbat, and the battery, back to the negative terminal of the generator. The
IL tracing loop follows the same path as the Ibat loop in Solution I. But IL
is not the same current as Ibat. The total load current is IL since the IL tracing loop is the only one that passes through RL. The battery current in
this solution is the discharge current IL minus the charging current Igen.
Note that the voltage drop across Rbat will be Rbat times the difference
of the two loop currents. In writing loop equations, the direction of the
tracing loop is always the positive current direction. Any current going
against the tracing direction is negative. Hence, Kirchhoff’s voltage-law
equation for the IL loop becomes
0.10 ( IL − Igen ) + 1.0IL = 12.8
−0.10Igen + 1.1IL = 12.8
or
(1)
In writing the equation for the Igen loop, we note that Igen enters the
+ terminal of the battery, just as it enters the + terminal for voltage
drops across resistors. Thus, Ebat can appear as a positive term on the
left-hand side of the equation. Alternatively, we can think of Ebat as a
voltage source bucking the generator voltage. Then Ebat appears on the
right-hand side of the equation as a negative quantity. The equation for
the Igen loop becomes
0.20Igen + 0.10 ( Igen − IL ) + 12.8 = 16.0
or
│12.8
3.2
=
│−0.1
0.3
0.30Igen − 0.10IL = 3.2
│ −1.28 − 3.52 −4.80
=
=
= 15 A
+0.01 − 0.33
−0.32
1.1
−0.1│
(2)
Again we can use determinants to solve simultaneous Equations 1
and 2:
Igen
1.1
−0.1
As shown in Appendix 1, the letter D represents the matrix of all the
­coefficients of a system of equations. In this example,
9-7 Loop Equations in Multisource Networks
D=
IL =
│−0.1
0.3
│−0.1
0.3
1.1
0.1
│
12.8
3.2
D
│
=
219
−0.32 − 3.84
−4.16
=
= 13 A
−0.32
−0.32
Battery current = IL − Igen = 13 − 15 = −2 A
Again we find that the battery is charging rather than discharging.
Multisim Solution
Download Multisim file EX9-4 from the website.
The circuit is the same as shown in Figure 9-12, except that ammeters
have been added to measure the generator current, battery current,
and load current.
Run the simulation. Note that the ammeter readings match the c­ urrents
calculated using loop equations.
Our next example illustrates how the Edison three-wire distribution
­system reduces power losses in transmission lines.
Example 9-5
(a)Calculate the total power loss in the three 1.0-Ω conductors in the
three-wire distribution system shown in Figure 9-14.
(b)Calculate the total power loss if the two loads are fed in parallel
from a single 110-V source through a pair of 1.0-Ω conductors.
R1
1.0 Ω
+
110 V E1
−
I1
RL1
10 Ω
RL2
15 Ω
RN
1.0 Ω
+
110 V E2
−
I2
R2
1.0 Ω
Figure 9-14
Three-wire distribution system
circuitSIM
walkthrough
220
Chapter 9 Resistance Networks
Solution
(a)The voltage drop across the resistance of the common or neutral
conductor is 1.0 Ω times the difference of the two loop currents. As
in the second solution for Example 9-4, current in the direction of
a tracing loop direction is positive and any current flowing against
the tracing ­direction is negative. Therefore, the Kirchhoff’s voltagelaw equation for the I1 loop becomes
1.0I1 + 10I1 + 1.0(I1 − I2) = E1
or 12I1 − I2 = 110
(1)
or −I1 + 17I2 = 110
(2)
For the I2 loop,
I1
│110
110
=
│−112
I2 =
12
│−1
1.0(I2 − I1) + 15I2 + 1.0I2 = E2
│ 1870 − ( −110 ) 1980
=
= 9.75 A
=
204 − ( +1 )
203
−1
17│
−1
17
110
110
D
│
=
1320 − ( −110 )
1430
=
= 7.04 A
203
203
P1 = I21R1 = 9.752 × 1.0 = 95 W
Power loss in line 1:
P2 = I22R2 = 7.042 × 1.0 = 50 W
Power loss in line 2:
Neutral current: IN = I1 − I2 = 9.75 − 7.04 = 2.71 A
Neutral power loss:
PN = IN2RN = 2.712 × 1.0 = 7 W
Total power loss: PT = P1 + P2 + PN = 95 + 50 + 7 = 152 W
(b)If the two loads are connected in parallel, the equivalent load
resistance is
Req =
10 × 15
= 6.0 Ω
10 + 15
Adding the resistance of two conductors, RT = 8.0 Ω. Hence, the
circuit ­current becomes
I=
E
110 V
=
= 13.75 A
RT
8.0 Ω
Power loss in each line = I2R = 13.752 × 1.0 = 0.19 kW
Therefore, total power loss in the two conductors is
PT = 2 × 0.19 kW = 0.38 kW
9-7 Loop Equations in Multisource Networks
The three-wire distribution system reduces transmission line losses by
a­ rranging loop currents that largely cancel in the neutral lead. If we arrange
for equal values of RL in Figure 9-14, the load is balanced and the neutral current is zero. Unbalanced load does cause some power loss in the neutral line,
but not as much as operating both loads in parallel on a two-wire system.
Example 9-6
If RB is 1.0 kΩ in the coupling circuit shown in Figure 9-11(b), determine the magnitude and polarity of the voltage drop across RB.
Solution
The original circuit diagram does not show the actual voltage sources.
We draw them on a working diagram (Figure 9-15) to enable us to draw
complete tracing loops.
2.2 kΩ
5.6 kΩ
3.3 kΩ
−
15 V
+
I1
Figure 9-15
6.8 kΩ
I3
I2
RB
1.0 kΩ
+
10 V
−
Circuit diagram for Example 9-6
We require three loops to include all components in this circuit.
Several combinations of loops are possible. The loops drawn in
Figure 9-15 are perhaps the most straightforward choices. Since
volts = ­milliamperes × kilohms, we can write the three Kirchhoff’s
voltage-law ­equations without units as long as we remember that the
currents are measured in milliamperes.
For the I1 loop,
6.8(I1 − I2) + 2.2I1 = 15
For the I2 loop,
3.3(I2 + I3) + 5.6I2 + 6.8(I2 − I1) = 10
For the I3 loop,
3.3(I3 + I2) + 1.0I3 = 10
Collecting like terms in each equation gives
9I1 − 6.8I2 = 15
−6.8I1 + 15.7I2 + 3.3I3 = 10
3.3I2 + 4.3I3 = 10
(1)
(2)
(3)
221
222
Chapter 9 Resistance Networks
│
│
│
│
We need to solve for just I3 since it is the only current through RB.
I3 =
=
=
9
−6.8
0
9
−6.8
0
−6.8
15.7
3.3
−6.8
15.7
3.3
15
10
10
0
3.3
4.3
1413 + 0 − 336.6 − 0 − 297 − 462.4
607.59 + 0 + 0 − 0 − 98.01 − 198.83
317
310.75
= 1.02 mA
Hence, the voltage drop across RB is
VB = 1.02 mA × l.0 kΩ = 1.0 V
Since I3 is a positive quantity, the top end of RB is positive with respect
to the chassis.
circuitSIM
walkthrough
Multisim Solution
Download Multisim file EX9-6 from the website.
The circuit is the same as shown in Figure 9-15. Insert a voltmeter to
measure the voltage across RB.
Run the simulation. Note that the voltmeter reading matches the value
calculated using loop equations.
See Problems 9-10 to 9-26 and Review Questions 9-62 to 9-65.
Mesh equations
were developed
by James Clerk
Maxwell (1831–79),
the British physicist
who also ­developed
electromagnetic
­theory and predicted
the existence of radio
waves before they
were discovered.
9-8 Mesh Analysis
Now that we have a feel for writing equations for Kirchhoff’s voltage law,
we can turn our attention to mesh analysis, a procedure that simplifies
and speeds up writing the simultaneous equations for solving various resistance networks. The format for mesh equations is straightforward, but
it cannot handle some of the networks that we can solve with the loop
procedure.
A mesh is a closed loop that does not enclose any circuit elements. Therefore, the circuit diagram cannot contain any conductors that cross without
being joined at the point of crossing. In other words, the network diagram
9-8 Mesh Analysis
must be strictly two-dimensional or planar. The mesh format requires
all sources to be voltage sources. We must convert any current sources
to equivalent voltage sources before we use the mesh format. Instead of
choosing convenient tracing loops, we draw a current loop for each mesh.
All mesh currents must have the same direction, either clockwise or counterclockwise. The more common choice is clockwise. These rules provide a
standard format for every mesh equation.
In the loop procedure, we write equations for voltage drops around the
current loops, and then rewrite the Kirchhoff’s voltage-law equations to
collect the current terms. In the mesh format, we go directly to this second
step. We can use the network shown in Figure 9-16 to illustrate the mesh
format. The network is clearly planar and the four meshes have no internal
electric components.
10 Ω
+
20 V
−
70 Ω
IA
30 Ω
IC
IB
40 Ω
60 Ω
50 Ω
80 Ω
Figure 9-16
20 Ω
+
30 V
−
ID
90 Ω
+
40 V
−
Mesh network for Example 9-7
For any particular mesh current, all other currents through common
­circuit elements are in the opposite direction. For example, IA and IB flow
through the 30-Ω resistor in opposite directions. The mesh equation for
mesh A has the form
RTIA − RAB IB − RACIC = ΣE
(9-7)
where RT is the total resistance around mesh A, RAB is the resistance
that is common to mesh A and mesh B, RAC is the resistance that is common to mesh A and mesh C, and ΣE is the algebraic sum of the voltage
sources around the mesh.
Where IA passes inside a voltage source from its negative terminal
to its positive terminal, E is a positive. If the direction of IA through a
voltage source is from the positive terminal to the negative terminal, E is
­negative.
223
224
Chapter 9 Resistance Networks
Example 9-7
Find the magnitudes and polarities of the voltage drops across the
30-Ω, 50-Ω, and 60-Ω resistors in the network shown in Figure 9-16.
Solution
Using the mesh format, write a Kirchhoff’s voltage-law equation for
each of the four meshes in Figure 9-16. In this example, the resistances
are in ohms and the potential differences are in volts, so the currents
will be in amperes.
(10 + 30 + 50)IA − 30IB − 50IC = 20
(30 + 20 + 40 + 60)IB − 30IA − 60ID = 0
(80 + 70 + 50)IC − 50IA = −30
(90 + 60)ID − 60IB = 30 − 40
Collecting like terms in each equation gives
90IA − 30IB − 50IC
−30IA − 150IB
−50IA
− 60IB
+ 200IC
=
20
(1)
0
(2)
+ 150ID = −10
(3)
− 60ID =
= −30
(4)
Unfortunately, the calculations for solving fourth-order determinants are not as simple as those for second- and third-order determinants. We can get around this problem by eliminating one of the
unknowns. From Equation 4,
60IB − 10
150
ID =
Substituting for ID in Equation 2 gives
−30IA + 150IB − (24IB – 4) = 0
We now have three simultaneous equations:
IA =
│
│
20
−4
−30
90
−30
−50
90IA − 30IB − 50IC =
−30IA − 126IB
−50IA
−30
126
0
−30
126
0
−50
0
200
│
│
−50
0
200
=
20
(1)
+ 200IC = −30
(5)
= −4
291 000
= 0.164 A
1 773 000
(3)
IB =
IC =
│
│
ID =
90
−30
−50
90
−30
−50
20
−4
−30
D
−30
126
0
D
−50
0
200
20
−4
−30
│
│
9-8 Mesh Analysis
=
13 000
= 0.0073 A
1 773 000
=
−193 200
= −0.109 A
1 773 000
225
60 × 7.33 × 10−3 − 10
= −0.064 A
150
I30 = IA − IB = 0.164 A − 0.0073 A = 157 mA
V30 = 30 Ω × 157 mA = 4.7 V [positive at top]
I50 = IA − IC = 0.164 A + 0.109 A = 273 mA
V50 = 50 Ω × 273 mA = 14 V [positive on the right]
I60 = IB − ID = 0.0073 A − 0.064 A = 71 mA
V60 = 60 Ω × 71 mA = 4.3 V [positive on the right]
Multisim Solution
Download Multisim file EX9-7 from the website.
The circuit is the same as shown in Figure 9-16, except that voltmeters
have been added to measure the voltage drops across the 30-Ω, 50-Ω,
and 60-Ω resistors.
Run the simulation. Note that the voltmeter readings match the values
calculated using mesh analysis.
To demonstrate the application of mesh analysis, we can now repeat
­Examples 9-2 to 9-6 using the mesh equations.
Example 9-2A
The circuit diagram of Figure 9-7 has two meshes. The I1 mesh is the
same as the I1 loop. The second mesh is the closed loop through R3, R1,
R2, and R4. The two mesh equations are
(10 + 20 + 30)I1 − (20 + 30)I2 = 80 V
(30 + 20 + 15 + 10)I2 − (20 + 30)I1 = 0 V
IR3 = I1 − I2 and IR4 = I2
The remainder of the solution is the same as in Example 9-2.
circuitSIM
walkthrough
226
Chapter 9 Resistance Networks
Example 9-3A
The circuit diagram of Figure 9-9 has three meshes. The I1 mesh is the
same as the I1 loop. The second mesh is the closed loop through R1, R2,
and RM. The third mesh consists of R3, RM, and R4. With the like terms
collected, the three mesh equations become
60I1 − 20I2 − 30I3 = 80
−20I1 + 85I2 − 50I3 = 0
−30I1 − 50I2 + 90I3 = 0
We need to solve for both I2 and I3, since IM = I2 − I3.
(1)
(2)
(3)
Example 9-4A
In the circuit diagram of Figure 9-12, we can write mesh-current equations by simply reversing the direction of Ibat. The second solution for
Example 9-4 using the circuit configuration of Figure 9-13 already satisfies the rules for the mesh format and yields the same simultaneous
equations when the terms are collected.
Example 9-5A
The circuit of Figure 9-14 also satisfies the mesh format rules.
Example 9-6A
Before we can write the mesh equations for the circuit shown in
Figure 9-15, all mesh currents must be in the same direction.
Reversing I3 so that all the mesh currents are counterclockwise gives
(2.2 kΩ + 6.8 kΩ) × I1 − 6.8 kΩ × I2 = 15 V
(3.3 kΩ + 5.6 kΩ + 6.8 kΩ) × I2 − 6.8 kΩ × I3 − 3.3 kΩ × I3 = 10 V
(1.0 kΩ + 3.3 kΩ) × I3 − 3.3 kΩ × I2 = −10 V
9-8 Mesh Analysis
Collecting terms gives
9.0 kΩ × I1 − 6.8 kΩ × I2 = 15
−6.8 kΩ × I1 + 15.7 kΩ × I2 − 3.3 kΩ × I3 = 10
−3.3 kΩ × I2 + 4.3 kΩ × I3 = −10
(1)
(2)
(3)
To solve for VB, we need to solve only for I3, which is −1.0 mA since
we reversed the direction of I3.
See Problems 9-27 to 9-31 and Review Questions 9-66 to 9-71.
Circuit Check
B
CC 9-3. Use loop equations to calculate the current in the 4.0-Ω resistor
in circuit shown in Figure 9-17.
5.0 Ω
Ix
4.0 Ω
+
10 V
2.0 Ω
−
−
2.0 Ω
10 V
1.0 Ω
+
Figure 9-17
CC 9-4. Use mesh analysis to calculate the unknown current, Ix, in the
circuit shown in Figure 9-18.
5.0 Ω
6.0 Ω
Ix
−
2.0 Ω
2V
+
2.0 Ω
1.0 Ω
+
10 V
−
4.0 Ω
−
−
3V
+ 6V
10
+
7.0 Ω
−
+ 4V
Figure 9-18
Ω
227
228
Chapter 9 Resistance Networks
9-9 Kirchhoff’s Current-Law Equations
The loop-current and mesh-equation methods of solving resistance networks are based on Kirchhoff’s voltage law. With Kirchhoff’s current law
we can write a different set of equations by considering currents entering
and leaving junction points or nodes in a circuit. First we must identify
the nodes in a given network. In the network shown in Figure 9-19, we do
not consider the junction of the voltage source and R1 to be a node because
the current does not branch at this point. The current law equation for this
junction is simply I1 = I1, which is redundant. Similarly, the junction of R3
and R6 is not a node. As shown in Figure 9-19(b), we treat all connections
to a common grounded conductor as a single node. Hence, the network in
Figure 9-19 has only three nodes.
R1
I1
Node
A
I4
I
2
R2
Node
B
I5 I
3
Node
A
R3
I6
+
E
R4
R5
R6
R1
+
Node
B
R3
E
−
R2
R4
R5
R6
−
Reference node
(a)
Figure 9-19
Reference node
(b)
Identifying nodes in a network
The simplest statement of Kirchhoff’s current law is that the algebraic
sum of currents at a node is zero. To use this statement to write equations,
we have to remember which currents are positive and which are negative.
For nodal circuit analysis, it is more convenient to state Kirchhoff’s current
law in this form:
The sum of the currents leaving a node in a circuit equals the sum
of the currents entering the node.
For the three nodes in the network of Figure 9-19, this statement leads to
the following equations:
Node A:
Node B:
Reference node:
I4 + I2 = I1
(1)
I1 = I4 + I5 + I3
(3)
I5 + I3 = I2
(2)
9-9 Kirchhoff’s Current-Law Equations
Note that if we add Equations 1 and 2, we obtain Equation 3. Hence,
Kirchhoff’s current-law equation for the third node is redundant. The
number of Kirchhoff’s current-law equations we need to solve any network
is ­always one less than the total number of nodes. We label the redundant
node as the reference node. Choosing a reference node that is common to
as many branch currents as possible usually gives simpler equations. The
grounded conductor or chassis connection is a good choice for the reference node.
The equations for nodes A and B in Figure 9-19 have five unknown
currents. However, there are only two node voltages, VA and VB, with
respect to the reference node in the network. To reduce the unknowns in
the current-law equations to two, we use I = V/R to substitute V/R ratios
for each current, so that we can solve for VA and VB. Once we know the
node voltages, we can use Ohm’s law to find any other unknown in
the network. In the network shown in Figure 9-19, V4 = VA and V5 = VB.
For node A:
I4 =
VA
R4
I1 =
V1
E − VA
=
R1
R1
I2 =
and
V2
VA − VB
=
R2
R2
Substituting these expressions in Equation 1 gives
VA
VA − VB
E − VA
+
=
R4
R2
R1
(4)
For node B:
I5 =
I3 =
and
I2 =
VB
R5
VB
R3 + R6
VA − VB
R2
VB
VB
VA − VB
+
=
R5
R3 + R 6
R2
Given values for E and all the resistors, we can solve for VA and VB.
(5)
229
230
Chapter 9 Resistance Networks
Example 9-4B
An automobile generator with an internal resistance of 0.20 Ω develops
an open-circuit voltage of 16.0 V. The storage battery has an internal
resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources
are connected in parallel to a 1.0-Ω load. Determine the load current.
Solution
The Kirchhoff’s current-law equations for the two nodes in Figure 9-20
are the same except that all the signs are reversed. Hence, we need to
write only one Kirchhoff’s current-law equation.
IL = Igen + Ibat
0.20 Ω
0.10 Ω
Node
Rgen
+
16.0 V
−
Rbat
+
Egen
1.0 Ω
RL
Ebat
12.8 V
−
Reference node
Figure 9-20
Circuit diagram for Example 9-4B
Note that we again assume that the battery is discharging. Applying
Ohm’s law gives
IL =
Egen − VL
VL
Ebat − VL
, Igen =
, and Ibat =
RL
Rgen
Rbat
Substituting these V/R ratios for the currents in the Kirchhoff’s currentlaw equation, we obtain
VL
16.0 V − VL
12.8 V − VL
=
+
1.0 Ω
0.20 Ω
0.10 Ω
Multiplying the equation by 0.1 gives
0.10VL = 8.0 − 0.5VL + 12.8 − VL
VL =
and
IL =
20.9
= 13 V
1.6
VL
13 V
=
= 13 A
RL
1.0 Ω
9-10 Nodal Analysis
Since we only required one nodal equation in the solution of Example 9-4B,
we avoided the simultaneous equations of Example 9-4, which required
two tracing loops. However, loop and mesh equations are free from algebraic fractions since we substitute a product—the IR drop—for each voltage
drop. In the nodal equation of Example 9-4B, we substitute a V/ R ratio for
each current term, leaving us with an equation in a more complex form. We
can simplify such equations by using conductance instead of resistance.
We then substitute I = VG for the currents.
See Review Questions 9-72 to 9-74.
9-10 Nodal Analysis
Nodal analysis is a circuit-analysis format that combines Kirchhoff’s currentlaw equations with the source conversions we developed in Section 9-4.
Converting all voltage sources to equivalent constant-current sources allows
us to standardize the way we write the Kirchhoff’s current-law equations.
For nodal analysis, we consider source currents to flow into a node. If the
arrows in a circuit diagram show that a source current actually flows out of
a particular node, the current is a negative quantity for that node. Similarly,
we consider all resistor currents to flow out of a node. We are thus restating
Kirchhoff’s current law in the form:
At any independent node, the algebraic sum of the resistor currents
leaving the node equals the algebraic sum of the source currents
­entering the node.
The Kirchhoff’s current-law equations then have the format
IR1 + IR2 + . . . = IS1 + IS2 + . . .
(9-8)
The next step is to replace the resistor currents using I = VG. The voltage
across each resistor equals the difference between the voltages, relative to the
reference node, of the nodes at each end of the resistor. We always subtract
the voltage of the adjacent node from the voltage of the node for which we
are writing the equation. For node 1 in Figure 9-21, the current through R1
Node 1
IS1
Node 2
R3
R1
R2
IS2
Reference node
Figure 9-21
Circuit diagram for writing nodal equations
231
232
Chapter 9 Resistance Networks
equals V1G1. The voltage across R3 is V1 – V2. Hence the Kirchhoff’s currentlaw equation for node 1 becomes
V1G1 + ( V1 − V2 ) G3 = IS1
Collecting the voltage terms gives
( G1 + G3 ) V1 − G3 V2 = IS1
(9-9)
Equation 9-9 illustrates the standard format for writing a nodal equation. We can use determinants to solve for the voltages in the same way
that we solved for currents in loop and mesh equations.
We can write Equation 9-9 directly by noting that the positive voltage
term on the left-hand side of the equation is the unknown voltage for the
node in question multiplied by the sum of the conductances connected to
that node. From this positive term, we must subtract a term for the node
voltage at every adjacent node connected by a resistor to the node in question. This term is the adjacent node voltage multiplied by the conductance
between the two nodes. Using this procedure to write the nodal equation
for node 2 gives
( G2 + G3 ) V2 − G3V1 = −IS2
Note the duality between this nodal equation and the mesh format equation Equation 9-7. Nodal analysis is particularly useful for networks where
a common portion of the network is fed from several sources in parallel.
Example 9-4C
An automobile generator with an internal resistance of 0.20 Ω develops
an open-circuit voltage of 16.0 V. The storage battery has an internal
resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources
are connected in parallel to a 1.0-Ω load. Determine the load current.
Solution
Step 1
First we draw the equivalent circuit with the voltage sources converted
to constant-current sources, as shown in Figure 9-22.
80 A
Igen 0.20 Ω
Figure 9-22
Rgen
1.0 Ω
RL
0.10 Ω
Circuit diagram for Example 9-4C
Rbat
Ibat
128 A
9-10 Nodal Analysis
The short-circuit current for the battery is
Ibat = =
Ebat
12.8 V
=
= 128 A
Rbat
0.10 Ω
Similarly, the short-circuit current of the generator is
Igen =
Egen
Rgen
=
16.0 V
= 80 A
0.20 Ω
With a bit of mental arithmetic we can fill in the required data on the
c­ ircuit diagram of Figure 9-22. Note that the generator and battery currents in this equivalent circuit are not the same as the real currents in the
original circuit of Figure 9-20.
Step 2
Next we find the conductance of each branch:
Ggen =
1
1
=
= 5.0 S
Rgen
0.20 Ω
GL =
and
Gbat =
1
= 1.0 S
RL
1
1
=
= 10 S
Rbat
0.10 Ω
Step 3
At first glance, Figure 9-22 appears to show six junction points, but
there are only two because the same voltage appears across all parallel branches. For our nodal analysis, we can redraw the circuit in the
form shown in F
­ igure 9-23. If we label the lower node as the reference
node, the upper node is the only independent node. Consequently,
we require only one Kirchhoff’s current-law equation for the circuit of
­Figure 9-23.
Node
80 A
Igen 5.0 S
Ggen
1.0 S
GL
10 S
Gbat
Reference node
Figure 9-23
Equivalent circuit for Example 9-4C
Ibat
128 A
233
234
Chapter 9 Resistance Networks
At the single independent node, Equation 9-8 becomes
I5 + I1 + I10 = Igen + Ibat
Or, we can go directly to the format of Equation 9-9,
(G5 + G1 + G10)V = Igen + Ibat
V=
circuitSIM
walkthrough
80 + 128
208 A
=
= 13 V
5.0 + 1.0 + 10
16 S
IL = VGL = 13 V × 1.0 S = 13 A
Multisim Solution
Download Multisim file EX9-4C from the website.
The circuit is the same as shown in Figure 9-20. Insert an ammeter to
measure the current through RL.
Run the simulation. Note that the ammeter reading matches the value
calculated using nodal analysis.
Nodal analysis allowed us to solve Example 9-4C with not much more
than mental arithmetic since the circuit has only a single independent node.
However, we do need to solve simultaneous equations when the network
has more than one independent node.
Example 9-6B
If RB is 1.0 kΩ in the coupling circuit shown in Figure 9-24, determine
the magnitude and polarity of the voltage drop across RB.
2.2 kΩ
5.6 kΩ
3.3 kΩ
−
15 V
+
Figure 9-24
RB
6.8 kΩ
+
10 V
−
Circuit diagram for Example 9-6B
Solution
Step 1
In the circuit diagram of Figure 9-24, we can consider the 2.2-kΩ
resistor to be the internal resistance of a 15-V constant-voltage
9-10 Nodal Analysis
235
source c­onnected across the 6.8-kΩ resistor. Similarly, we treat the
3.3-kΩ resistor as the internal resistance of a 10-V constant-voltage
source connected across RB. Therefore, when we convert the voltage
sources to constant-current sources, we obtain the equivalent circuit in
Figure 9-25 with
I1 =
15 V
10 V
= 6.82 mA and I2 =
= 3.03 mA
2.2 kΩ
3.3 kΩ
We must make sure that the directions of I1 and I2 are consistent with
the polarities of the voltage sources they replace.
Step 2
By converting each of the five resistances into its equivalent conductance, we obtain the values shown in Figure 9-25.
V1 – V 2
5.6 kΩ
Node 1
Node 2
179 μS
−
V1
−
I1
6820 μA
+
2.2 kΩ
454 μS
−
−
+
+
6.8 kΩ
147 μS
+
+
I2
+
3030 μA
3.3 kΩ R
bat
303 μS
−
−
+
1.0 kΩ
1000 μS
V2
−
Reference node
Figure 9-25
Equivalent circuit for Example 9-6B
Step 3
Choosing the bottom conductor (the chassis) as the reference node
leaves two independent nodes in Figure 9-25. We need a nodal equation for each independent node. At node 1, all the currents are negative
quantities since the source current actually leaves the node and resistor
currents enter the node. Since volts × microsiemens = microamperes,
the voltages will be in volts if we express the currents in microamperes
and the conductances in microsiemens.
For node 1,
(454 + 147 + 179)V1 − 179V2 = −6820
For node 2, (179 + 303 + 1000)V2 − 179V1 = 3030
Collecting terms gives
780V1 − 179V2 = −6820 −179V1 + 1482V2 =
3030 (1)
(2)
If we solve for V1, we
find that it is negative, indicating that
node 1 has a negative
­potential difference
with ­respect to the
chassis.
236
Chapter 9 Resistance Networks
780
│−179
=
780
│−179
│ 2 363 400 − 1 220 780 1 142 620
=
= 1.02 V
=
1 155 960 − 32 041
1 123 919
−179
1482│
Step 4
To find the voltage drop across RB, we need solve only for V2.
V2
−6820
3030
Hence, the voltage drop across RB is 1.02 V. Since V2 is positive, node 2
is positive with respect to the chassis.
circuitSIM
Multisim Solution
Download Multisim file EX9-6B from the website.
walkthrough
The circuit is the same as shown in Figure 9-23. Insert a voltmeter to
measure the voltage across RB.
Run the simulation. Note that the voltmeter reading matches the value
calculated using nodal analysis.
Although we had to solve simultaneous equations in Example 9-6B, they
involved only two unknowns rather than three as in the loop and mesh
­solutions for this example.
The Wheatstone bridge circuit of Figure 9-26(a) has four nodes, so we
need three nodal equations to solve it. Since the loop or mesh procedure
also needs three simultaneous equations, we would probably not use nodal
analysis to solve the bridge circuit. However, using loop or mesh format
to solve the circuit in Figure 9-26(b) requires four simultaneous equations
while nodal analysis requires only one nodal equation to solve the whole
network.
Node
Node 1
Node 2
Node 3
RL
+
E
+
+
E1
−
Reference node
(a)
−
+
E2
−
+
E3
−
E4
−
Reference node
(b)
Figure 9-26 (a) Wheatstone bridge; (b) Four parallel sources supplying
a common load
See Problems 9-32 to 9-35 and Review Questions 9-75 and 9-76.
9-11 The Superposition Theorem
9-11 The Superposition Theorem
The Kirchhoff’s-laws methods for analyzing resistance networks usually
­require solving a set of simultaneous linear equations. If there are no more
than two tracing loops or independent nodes, the computation is not particularly onerous. But when we need three or more simultaneous equations,
we start looking for alternative solutions. We can often avoid simultaneous
equations in multi-source networks by applying a fundamental principle of
linear networks known as the superposition theorem.
In a network of resistors with multiple voltage sources, the current
in any branch is the algebraic sum of the component currents that
would be caused in that branch by each source acting independently
with the others replaced by their respective internal resistances.
This theorem is useful only for determining the current through or
voltage drop across one branch of a network containing more than one
source. However, in many cases, once we know the current through one
branch of a network, we can solve the remainder of the network quite
readily.
In both the loop-current and nodal methods of solving resistance networks, the currents appearing in the simultaneous equations are not
­necessarily actual currents that we could measure at various points in the
network. To obtain actual branch currents, we have to add the appropriate loop or mesh currents algebraically. Similarly, component currents calculated by the superposition theorem are only a means of calculating the
­actual current in a certain branch and do not represent currents that we can
measure directly.
Example 9-4D
An automobile generator with an internal resistance of 0.20 Ω develops
an open-circuit voltage of 16.0 V. The storage battery has an internal
resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources
are connected in parallel to a 1.0-Ω load. Determine the load current.
Solution
Because there are two sources, there are two component currents
through the load resistance. To apply the superposition theorem, we
redraw the circuit of Figures 9-11(a) and 9-12 as the two equivalent circuits in Figure 9-27.
237
238
Chapter 9 Resistance Networks
IT
0.20 Ω
16.0 V
−
I L1
Rgen
+
Egen
1.0 Ω
RL 0.10 Ω
Rbat
(a)
0.10 Ω
IL2
0.20 Ω
Rgen 1.0 Ω
IT
Rbat
+
RL
Ebat
12.8 V
−
(b)
Figure 9-27
Equivalent circuits for Example 9-4D
The first equivalent circuit of Figure 9-27(a) is a series-parallel network.
Therefore,
RT = Rgen +
RL × Rbat
1.0 × 0.10
= 0.20 +
= 0.29 Ω
RL + Rbat
1.0 + 0.10
Egen
16.0 V
IT =
=
= 55 A
RT
0.29 Ω
From the current divider principle, the first component of the load current is
­IL1 = IT ×
Rbat
0.10
= 55 ×
= 5.0 A
RL + Rbat
1.1
Similarly, for the circuit of Figure 9-27(b),
RT = Rbat +
IT =
Rgen × RL
Rgen + RL
= 0.10 +
Ebat
12.8 V
= 48 A
=
RT
0.267 Ω
0.20 × 1.0
= 0.267 Ω
0.20 + 1.0
The second component of the load current is
IL2 = IT ×
0.20
= 8.0 A
Rgen + RL
1.2
Since the component currents have the same direction in Figures 9-27(a)
and (b), the actual load current is the sum of the component currents.
Rgen
= 48 ×
IL = IL1 + IL2 = 5 + 8 = 13 A
9-11 The Superposition Theorem
Example 9-6C
If RB is 1.0 kΩ in the coupling circuit in Figure 9-11(b), determine the
magnitude and polarity of the voltage drop across RB.
Solution
Again there are two sources, so the superposition theorem yields two
component currents. In the equivalent circuit of Figure 9-28(a), the
10-V source has been replaced by its zero-ohm internal resistance so
that only the 15-V source contributes to the current through RB. Again
we can solve the circuit as a series-parallel network. First we replace
the 5.6-kΩ and 3.3-kΩ resistors and RB with their equivalent resistance,
Req, then we find the total resistance for the equivalent circuit shown
in Figure 9-28(b):
Req = 5.6 kΩ +
RT = 2.2 kΩ +
IT =
2.2 kΩ
−
15 V 6.8 kΩ
+
3.3 kΩ × 1 kΩ
= 6.4 kΩ
3.3 kΩ + 1 kΩ
6.8 kΩ × 6.4 kΩ
= 5.5 kΩ
6.8 kΩ + 6.4 kΩ
E
15 V
=
= 2.7 mA
RT
5.49 kΩ
5.6 kΩ
2.2 kΩ
−
Rbat
1.0 kΩ
+
3.3 kΩ
IT
−
15 V 6.8 kΩ
+
6.4 kΩ
IT
I B1
Req
Ieq
(b)
(a)
IB2
IB2
5.6 kΩ
3.3 kΩ
2.2 kΩ
6.8 kΩ
IT
+
10 V
−
3.3 kΩ
1.0 kΩ
Rbat
Req
(c)
Figure 9-28
Equivalent circuit for Example 9-6C
7.26 kΩ
IT
+
10 V
−
(d)
1.0 kΩ
Rbat
239
240
Chapter 9 Resistance Networks
Now we use the current-divider principle twice to find the first component of the current through RB. In Figure 9-28(b), the current through
Req is
Ieq = 2.73 mA ×
6.8 kΩ
= 1.41 mA
6.8 kΩ + 6.4 kΩ
IB1 = 1.41 mA ×
3.3 kΩ
= 1.08 mA
3.3 kΩ + 1.0 kΩ
In the circuit of Figure 9-28(a),
Similarly, when the 10-V source operates alone, as in Figure 9-28(c), we
start by replacing the 5.6-kΩ, 2.2-kΩ, and 6.8-kΩ resistors with an equivalent ­resistance.
Req = 5.6 kΩ +
RT = 3.3 kΩ +
IT =
2.2 kΩ × 6.8 kΩ
= 7.26 kΩ
2.2 kΩ + 6.8 kΩ
7.26 kΩ × 1.0 kΩ
= 4.17 kΩ
7.26 kΩ + 1.0 kΩ
E
10 V
=
= 2.39 mA
RT
4.17 kΩ
Applying the current-divider principle to the equivalent circuit in
Figure 9-28(d) gives the second component of the current through RB:
IB2 = 2.39 mA ×
7.26 kΩ
= 2.10 mA
7.26 kΩ + 1 kΩ
In Figures 9-28(a) and (c) the two component currents flow in o­ pposite
­directions through RB. Hence, the real current through RB is
and
IB = 2.10 mA – 1.08 mA = 1.02 mA
VB = 1.02 mA × 1.0 kΩ = 1.02 V
Because the component current IB2 is greater than IB1, the real current
has the same direction through RB as IB2 in Figure 9-28(c). Thus, the top
end of RB is 1.02 V positive with respect to the chassis.
See Problems 9-36 to 9-43 and Review Questions 9-77 to 9-81.
9-11 The Superposition Theorem
Circuit Check
C
CC 9-5. Use nodal analysis to calculate the voltage across the 4-Ω ­resistor
in the circuit in Figure 9-29.
2.0 Ω
5A
1.0 Ω
1.0 Ω
5.0 Ω
2A
4.0 Ω
2.0 Ω
Figure 9-29
CC 9-6. Use the superposition theorem to calculate the unknown current
in the circuit in Figure 9-30.
Ix
+
90 V
−
36 Ω
Figure 9-30
12 Ω
2A
+
60 V
−
8.0 Ω
241
242
Chapter 9 Resistance Networks
Summary
• A constant-voltage source can be represented by a voltage source in series
with a resistor.
• A constant-current source can be represented by a current source in parallel with a resistor.
• A constant-voltage source may be converted to an equivalent constantcurrent source and vice versa.
• A circuit containing constant-voltage sources may be analyzed using the
loop procedure based on Kirchhoff’s voltage-law equations.
• A planar circuit containing constant-voltage sources may be analyzed
using mesh equations.
• A circuit containing constant-current sources can be analyzed using
nodal analysis, which is based on Kirchhoff’s current law.
• A circuit containing more than one source may be analyzed using the
­superposition theorem.
B = beginner
Problems
I = intermediate
A = advanced
Solve each network problem with one method and use a different method to check
the solution.
B
B
B
B
Section 9-2
9-1.
9-2.
As the current drawn from a DC power supply is increased from 600
mA to 800 mA, the terminal voltage decreases from 25 V to 23 V.
Determine the equivalent constant-voltage source.
Find the voltage across a load resistance of 21 Ω connected to a
constant-voltage source that has an open-circuit voltage of 12 V and
a short-circuit current of 8 A.
Section 9-3
9-3.
Constant-Current Sources
A 15-A constant current source has an internal resistance of 400 Ω.
How much current will this source deliver to a 10-Ω load? Find the
voltage across the load.
Section 9-4
9-4.
Constant-Voltage Sources
Source Conversion
A voltage source has an open-circuit terminal voltage of 120 V and
an internal resistance of 0.8 Ω. Determine the equivalent constantcurrent source.
243
Problems
B
9-5.
Convert the voltage sources of Figure 9-31 into the equivalent
­constant-current sources.
180 Ω
+
160 V
−
68 kΩ
+
20 kΩ
E 25 mV
25 kΩ
100 kΩ
+
100 V
−
45 Ω
−
85 V
+
−
(b)
(a)
(c)
Figure 9-31
B
9-6.
Replace the current sources in Figure 9-32 with equivalent constantvoltage sources.
33 kΩ
60 A
2.5 Ω
45 μA
(a)
22 kΩ
16 A
(b)
4.0 Ω
(c)
Figure 9-32
I
Section 9-5 Kirchhoff’s Voltage-Law Equations:
Loop Procedure
9-7.
Determine the current through the 400-Ω resistor in Figure 9-33.
120 Ω
240 Ω
+
−
6.0 V
400 Ω
360 Ω
Figure 9-33
80 Ω
24 A
244
Chapter 9 Resistance Networks
I
9-8.
When the input voltage is 3.0 V, what is the current through a 250-Ω
load connected to the output terminals of the bridged-T attenuator
of Figure 9-34?
400 Ω
100 Ω
Input
800 Ω
Figure 9-34
A
9-9.
100 Ω
Output
Bridged-T network
If 320 V DC is applied to the input terminals of the lattice network of
Figure 9-35, determine the current through a 40-Ω resistor connected
to the output terminals. (Hint: Redraw the circuit to eliminate the
crossing wires.)
20 Ω
30
Ω
Output
30
Ω
Input
20 Ω
Figure 9-35
A
Section 9-7
Lattice network
Loop Equations in Multisource Networks
9-10. Determine the magnitude and polarity of the voltage drop across
the 50-Ω resistor in Figure 9-36.
10 Ω
+
20 V
−
20 Ω
50 Ω
40 Ω
Figure 9-36
30 Ω
−
40 V
+
Problems
A
245
9-11. Find the current drawn from each of the sources in Figure 9-37.
5.0
kΩ
20
kΩ
10 kΩ
+
100 V
−
+
150 V
−
+
200 V
−
Figure 9-37
I
A
I
I
I
I
A
9-12. (a)Find the battery current when the open-circuit terminal voltage
of the generator in Example 9-4 drops to 15.0 V.
(b) Use Multisim to verify the battery current in part (a).
9-13. For the circuit of Example 9-4, find the open-circuit generator ­voltage that
(a) makes the battery current zero
(b) causes a 10-A charging current to flow into the battery
9-14. Find the neutral current in the circuit of Example 9-5 if a second
15-Ω load is connected in parallel with the existing 15-Ω load.
9-15. Calculate the total power loss in the three 1.0-Ω conductors in the
three-wire distribution system shown in Figure 9-14 if the polarity
of source E2 is reversed.
9-16. (a)Two batteries are connected in parallel to feed a 12-Ω load. Battery
A has an open-circuit voltage of 6.3 V and an internal resistance of
1.5 Ω. Battery B has an open-circuit voltage of 6.0 V and an internal resistance of 2.1 Ω. Find the current drain from each battery.
(b) Use Multisim to verify the current drains calculated in part (a).
9-17. If the 12-Ω load in Problem 9-16 is replaced by a battery charger with
an open-circuit voltage of 7.2 V and an internal resistance of 0.20 Ω,
find the charging current of each battery.
9-18. (a)Find the current through each voltage source in the circuit in
­Fig­ure 9-38.
(b) Use Multisim to verify the currents calculated in part (a).
+
120 V
−
Figure 9-38
2.2 kΩ
walkthrough
circuitSIM
walkthrough
circuitSIM
walkthrough
3.3 kΩ
4.7 kΩ
circuitSIM
6.8 kΩ
−
90 V
+
246
Chapter 9 Resistance Networks
B
circuitSIM
A
walkthrough
9-19. Determine the magnitude and direction of the current through the
2.2-kΩ resistor in the circuit of Figure 9-38.
9-20. (a)Determine the current through each of the five resistors in the
circuit of Figure 9-39.
(b) Use Multisim to verify the voltage calculated in part (a).
100 Ω
+
50 V
−
50 kΩ
300 Ω
Egen
A
90 V
50 kΩ
I
Figure 9-40
I
circuitSIM
I
walkthrough
−
+
25 V
−
Figure 9-39
V
+
Reference
battery
−
400 Ω
500 Ω
+
80 V
−
200 Ω
+
100 V
9-21. The 5.6-kΩ resistor in the circuit of Figure 9-11(b) is replaced with a
variable resistor that is adjusted until the voltage across RB is exactly
zero. What is the resistance of the variable resistor under these
­circumstances?
9-22. In the balance-detecting circuit of Figure 9-40, the voltmeter has a
­resistance of 100 kΩ. Find the voltmeter reading.
9-23. The current through the unknown resistance R in Figure 9-41 is
400 mA. Find this resistance.
9-24. (a)In Figure 9-42, a pulse generator circuit acts as a switch in series
with a 1.0-kΩ resistor. Calculate the output voltage across a
42-kΩ resistor connected from the output terminal to ground
first with the switch open and then with the switch closed.
(b) Use Multisim to verify the voltage across the 42-kΩ resistor with
the switch closed.
100 Ω
22 kΩ
R
Out
1.0 kΩ
4.7 kΩ
100 Ω
+
10 kΩ
−
200 V
+25 V
Figure 9-41
−15 V
Figure 9-42
I
9-25. The circuit shown in Figure 9-42 is modified as shown in Figure 9-43.
Calculate the voltage drop across a 42-kΩ resistor connected from
the output terminal to ground
(a) with the switch open
(b) with the switch closed
Problems
22 kΩ
247
Out
1.0 kΩ
4.7 kΩ
+25 V
10 kΩ
−15 V
Figure 9-43
I
A
9-26. A trolleybus is halfway along a section of trolley wire that is fed at
one end from a 600-V source and at the other end from a 596-V source.
The internal resistances of the sources are negligible, but each trolley
wire in the section has a resistance of 1.0 Ω. The trolleybus is the only
one in the section at the moment, and draws a 50-A current from the
trolley wires. Find the voltage drop across the trolleybus.
Section 9-8
Mesh Analysis
9-27. In the network shown in Figure 9-44, there is no electric connection
where the leads of the 20-Ω and 40-Ω resistors cross. Redraw the
network to show that it is planar. Determine the voltage drop across
the 16-Ω resistor.
16 Ω
Ω
+
12 V
40
−
Ω
20
+
12 V
−
32 Ω
Figure 9-44
I
A
9-28. (a)Find the current drawn from each of the sources in the network
shown in Figure 9-37 when a 25-kΩ resistor is connected across
the top of the network (from the + terminal of the 100-V source
to the + terminal of the 200-V source).
(b) Use Multisim to verify the current calculated in part (a).
9-29. Use loop or mesh equations to determine the current drawn from
each source in the network of Figure 9-54.
circuitSIM
walkthrough
248
circuitSIM
Chapter 9 Resistance Networks
I
walkthrough
9-30. (a)For the circuit shown in Figure 9-45, write the mesh equations
and solve to find Ix.
(b) Use Multisim to verify the current Ix calculated in part (a).
2.0 Ω
9.0 Ω
4.0 Ω
5.0 Ω
1.0 Ω
50 V
Ix
20 V
15 Ω
30 V
20 V
3.0 Ω
Figure 9-45
I
9-31. For the circuit shown in Figure 9-46, find Ix by mesh analysis.
8.0 Ω
7.0 Ω
3.0 Ω
4.0 Ω
5.0 Ω
20 V
15 V
Ix
20 V
5V
5.0 Ω
1.0 Ω
7.0 Ω
35 V
4.0 Ω
6.0 Ω
Figure 9-46
B
Section 9-10
Nodal Analysis
9-32. Determine the voltage drop across each of the three resistors in the
network shown in Figure 9-47.
Problems
10 Ω
3.0 A
5.0 A
30
20
Ω
Ω
6.0 A
Figure 9-47
I
9-33. Determine the voltage drop across the 200-Ω resistor in the network
shown in Figure 9-48.
300 Ω
20 mA
200 Ω
40 mA
400 Ω
Figure 9-48
I
9-34. Identify the nodes in Figure 9-49. Using the reference node shown,
write the nodal equations.
10 A
2S
1S
5A
5A
4S
2S
2S
5A
1A
6S
4S
1S
3S
2A
1S
5S
Reference node
Figure 9-49
5A
3S
15 A
249
250
Chapter 9 Resistance Networks
I
9-35. For the circuit shown in Figure 9-50, find Ex by nodal analysis.
3S
5A
1S
Ex
4S
2S
5A
3S
4
5S
S
1S
15 A
5A
Figure 9-50
circuitSIM
I
walkthrough
Section 9-11
The Superposition Theorem
9-36. (a)In the network shown in Figure 9-51, find the voltage drops
across the 2.2-kΩ resistor and the 10-kΩ resistor
(b) Use Multisim to verify the voltage drops calculated in part (a).
4.0 mA
10 kΩ
+
4.0 V
4.7 kΩ
2.2 kΩ
−
Figure 9-51
I
9-37. In the network shown in Figure 9-52, determine the current drawn
from each of the voltage sources.
50 Ω
+
200 V
−
40 Ω
5.0 A
20 Ω
−
80 V
+
Figure 9-52
I
9-38. Determine the voltage drop across the 30-Ω resistor in the network
shown in Figure 9-53.
30 Ω
4.0 A
20 Ω
50
Ω
+ 100 V
−
10 Ω
Figure 9-53
40 Ω
3.0 A
251
Problems
9-39. Use the superposition theorem to determine the voltage drop across
the 30-kΩ resistor in Figure 9-54.
10
+
150 V
−
kΩ
kΩ
40 kΩ
25
15
30 kΩ
kΩ
+
120 V
−
20
kΩ
I
Figure 9-54
I
A
9-40. Use the superposition theorem to determine the voltage drop across
the 4.7-kΩ resistor in Figure 9-51.
9-41. Use the superposition theorem to find Ix in Figure 9-55.
15 Ω
Ix
10 A
5.0 Ω
2.0 Ω
30 V
12 Ω
25 V
50 V
7.0 Ω
10 Ω
5.0 Ω
Figure 9-55
A
9-42. (a) Use the superposition theorem to find Ix in Figure 9-56.
(b) Use Multisim to verify the current calculated in part (a).
40 V
8.0 Ω
50 V
10 Ω
2.0 A
6.0 Ω
Ix
Figure 9-56
9.0 Ω
4.0 Ω
4.0 A
circuitSIM
walkthrough
252
Chapter 9 Resistance Networks
A
9-43. Calculate Ix in the following circuit by superposition.
Ix
8.0 Ω
+
20 V
−
2A
2.0 Ω
13.0 Ω
5.0 Ω
10.0 Ω
3.0 Ω
7.0 Ω
−
+
32 V
Figure 9-57
Review Questions
Section 9-2
Constant-Voltage Sources
9-44. Define a constant-voltage source.
9-45. Why is the terminal voltage of a constant-voltage source not necessarily constant as the load resistance varies?
9-46. Describe a laboratory procedure for determining the internal resistance of an automobile storage battery.
9-47. What is the significance of the term short-circuit current?
9-48. How would you determine the short-circuit current of a car
­battery?
9-49. What is the significance of the term open-circuit voltage?
Section 9-3
Constant-Current Sources
9-50. Define a constant-current source.
9-51. Why is the current drawn from a constant-current source not necessarily constant as the load resistance varies?
9-52. Explain what happens to the terminal voltage of a constant-current
source when a load resistor is connected to it.
Section 9-4
Source Conversion
9-53. Compare equivalent constant-voltage and constant-current sources
in terms of their open-circuit voltage and short-circuit current.
9-54. Describe the two methods for determining the internal resistance of
constant-voltage and constant-current sources.
9-55. Why can we not replace the source shown in Figure 9-1 with an
equivalent constant-current source?
Review Questions
Section 9-5 Kirchhoff’s Voltage-Law Equations:
Loop Procedure
9-56. Show that Equations 9-5 and 9-6 are algebraically the same and, therefore, represent two different ways of stating Kirchhoff’s voltage law.
9-57. Explain the procedure for determining the polarity of voltage drops
around a current tracing loop, using Figure 9-7 as an example.
9-58. How does the loop procedure apply when determining the voltage
drop across R in the circuit in Figure 9-41?
9-59. If, in solving Example 9-3, we had chosen the I3 tracing loop from the
+ terminal of the source through R2, RM, and R3 back to the − terminal
of the source, the solution for I3 would have been −0.15 A. What is
the significance of a negative current in solving loop equations?
9-60. Is it likely that I1 or I2 in Figure 9-9 would ever work out to be negative quantities? Explain.
9-61. If R1 = 40 Ω, R2 = 10 Ω, and R3 = R4 = 20 Ω in the Wheatstone bridge
of Figure 9-9, what path would you select for the I3 tracing loop in
order to avoid a negative value for I3?
Section 9-7 Loop Equations in Multisource Networks
9-62. How do you write loop equations for the circuit of Figure 9-11(b)
when no complete loops appear in the circuit diagram?
9-63. Loop equations are based on Kirchhoff’s voltage law but the unknown
quantities in the loop equations are currents. Explain the presence of
current terms in voltage-law equations.
9-64. In the second solution to Example 9-4, how do we know that the battery current should be IL − Igen rather than Igen − IL?
9-65. In distributing loads on a three-wire residential service, why is it d
­ esirable
to have the two values of RL in Figure 9-14 as nearly equal as possible?
Section 9-8
Mesh Analysis
9-66. Explain why the circuit in Figure 9-44 is actually planar even though
this circuit diagram shows conductors crossing with no electrical
connection.
9-67. Why are we more likely to encounter negative current terms with
the mesh format for writing Kirchhoff’s voltage-law equations than
with the loop procedure?
9-68. Explain why many of the currents calculated with either the loop or
mesh formats cannot be measured in the actual network.
9-69. How do we determine whether the source voltage in a mesh equation
is a positive or negative quantity?
9-70. How do we determine the polarity of the voltage drop across a resistor in a network where the actual current is made up of two different
mesh currents?
9-71. In writing loop equations, we insert branch currents in the basic
Kirchhoff’s voltage-law equation before we collect the loop-current
253
254
Chapter 9 Resistance Networks
terms. With the mesh format, we simply write the equations in terms
of the mesh currents. What features of the mesh format permit us to
take this shortcut?
Section 9-9
Kirchhoff’s Current-Law Equations
9-72. Nodal analysis is based on Kirchhoff’s current law, but the unknown
quantities in the equations are voltages. Explain the presence of voltage terms in current-law equations.
9-73. When we write equations for node voltages, we require one less
equation than the total number of nodes. Explain why.
9-74. What do we mean when we refer to “the voltage at node X”?
Section 9-10
Nodal Analysis
9-75. What factors would lead us to select nodal analysis in preference to
loop or mesh analysis when solving a given network?
9-76. Under what circumstances are loop equations preferable to nodal
analysis?
Section 9-11
The Superposition Theorem
9-77. What advantage does the superposition theorem have over
­Kirchhoff’s-law methods for analyzing networks?
9-78. What factors tend to limit the application of the superposition theorem in network analysis?
9-79. Are we likely to encounter negative values for component currents when
applying the superposition theorem to a network problem? Explain.
9-80. Is it possible that we have to subtract component currents when we
apply the superposition theorem to a branch current in a network?
­Explain.
9-81. To use mesh equations, we convert any current sources to equivalent constant-voltage sources. To use nodal analysis, we convert any
voltage sources to equivalent constant-current sources. Do we have
to make such conversions when using the superposition theorem?
Integrate the Concepts
In Example 8-1 we used the equivalent-circuit method to determine the
values of I1, I2 and I3 in the circuit shown in Figure 9-58. Solve for these
­currents using four other techniques.
I1
+
100 V
−
R1
I2
12 Ω
Figure 9-58
R2
10 Ω
I3
R3
40 Ω
Practice Quiz
Practice Quiz
1.
When a load resistor is connected across a voltage divider output, the
output voltage
(a) decreases
(b) increases
(c) remains the same
(d) will decrease only if the load resistance is less than the resistances in the voltage divider
2.
A practical voltage source can be represented by
(a) a constant-voltage source in parallel with an internal resistance
(b) a constant-voltage source in series with an internal resistance
(c) a constant-voltage source in parallel with the load resistor
(d) a constant-voltage source in series with the load resistor
3.
A constant-current source can be represented by a
(a) current source in parallel with its internal resistor
(b) current source in series with its internal resistor
(c) current source in parallel with the load resistor
(d) current source in series with the load resistor
4.
To convert a constant current source into a constant-voltage source
you should use Ohm’s law to determine a constant-voltage source
connected
(a) in series with the current source internal resistor
(b) in series with the load resistor
(c) in parallel with the current source internal resistor
(d) in parallel with the load resistor
5.
The equivalent constant-current source for the constant-voltage
source of Figure 9-59 is
(a) a 1.8-A current source in series with a 0.15-Ω resistor
(b) an 80-A current source in parallel with a 0.15-Ω resistor
(c) an 80-A current source in parallel with a 100-Ω resistor
(d) a 1.8-A current source in parallel with a 100-Ω resistor
R1
0.15 Ω
V1
12 V
Figure 9-59
RL
100 Ω
255
256
Chapter 9 Resistance Networks
6.
The current in the circuit of Figure 9-60 is
(a) 151 mA
(b) 86 mA
(c) 22 mA
(d) 11 mA
R1
100 Ω
V1
12 V
V2
20 V
R2
50 Ω
V3
24 V
R3
220 Ω
Figure 9-60
7.
The current through resistor R2 in Figure 9-61 is
(a) 0.75 A
(b) 0.66 A
(c) 0.50 A
(d) 85 mA
R1
R4
10 Ω
22 Ω
R2
50 Ω
V1
20 V
R3
R5
33 Ω
10 Ω
R6
50 Ω
Figure 9-61
V2
12 V
V3
15 V
Practice Quiz
8.
The voltage drop across resistor R5 in Figure 9-61 is
(a) 37 V
(b) 33 V
(c) 25 V
(d) 4.3 V
9.
Which method of circuit analysis uses simultaneous equations to
­ etermine the current flowing through each component?
d
(a) Loop equations
(b) Mesh analysis
(c) Superposition
(d) (a) and (b)
10.
When applying Kirchhoff’s current law to a parallel circuit,
(a) the maximum branch current corresponds to the smallest
branch resistor
(b) the sum of currents entering a node is equal to the difference of
all the branch currents
(c) the sum of the currents leaving a node equals the sum of the
­currents entering the node
(d) all of the above
11.
The current flowing through the resistor R3 in Figure 9-62 is
(a) 667 mA
(b) 674 mA
(c) 818 mA
(d) 1.0 A
R1
100 Ω
10.0 V
Figure 9-62
R3
330 Ω
R2
220 Ω
2.00 A
R4
2.7 kΩ
257
10
Equivalent-Circuit
Theorems
With the Kirchhoff’s-laws methods for solving resistance
networks, we can solve almost any network without having to
tamper with the original configuration of the network. However,
these methods can involve many tedious computations with
possibilities for errors in either the equations or the calculations.
This chapter introduces more methods for using equivalent
circuits to simplify network calculations.
Chapter Outline
10-1
10-2
Thévenin’s Theorem
Norton’s Theorem
10-3
Dependent Sources
10-5
Troubleshooting
260
268
271
10-4 Delta-Wye Transformation
283
278
Key Terms
Thévenin’s theorem 261
Norton’s theorem 268
independent sources 271
dependent or controlled
sources 271
delta-to-wye
transformation 279
wye-to-delta
transformation
280
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• state Thévenin’s theorem
• determine the Thévenin-equivalent source of
a network consisting of resistors and one or
more independent sources
• state Norton’s theorem
• determine the Norton-equivalent source of
a network consisting of resistors and one or
more independent sources
• differentiate between independent and
dependent sources
• analyze circuits consisting of resistors,
independent sources, and dependent
sources
Photo sources: © iStock.com/calvio
• determine the Thévenin-equivalent source of
a network consisting of resistors, independent
sources, and dependent sources
• convert a resistive Y-network to its equivalent
Δ-network
• convert a resistive Δ-network to its equivalent
Y-network
• determine if a fault exists in a circuit by
comparing ­measured and calculated
voltages
260
Chapter 10 Equivalent-Circuit Theorems
10-1 Thévenin’s Theorem
In Sections 9-2 and 9-3, we developed a technique for replacing any prac­
tical voltage source in a network with an equivalent constant-voltage or
constant-current source. In this section, we shall expand this principle into
one of the most useful theorems for simplifying circuit calculations.
Suppose that the circuit of Figure 10-1(a) is sealed in a black box
­(represented by the thick line) with only the terminals A and B exposed.
A high-resistance voltmeter connected across these terminals shows that
the open-circuit output voltage of the circuit is 80 V. Similarly, connecting
a very low resistance ammeter across terminals A and B shows that the
short-circuit current is about 2.0 mA. Hence, the black box appears to have
an internal resistance of
RTh =
25 kΩ
20 kΩ
Eoc
80 V
= 40 kΩ
=
Isc
2.0 mA
RTh
A
A
40 kΩ
+
100 V
−
+
ETh 80 V
−
100 kΩ
B
(a)
Figure 10-1
B
(b)
(a) Original voltage source; (b) Thévenin-equivalent source
As far as we can tell without opening it, the box contains an 80-V source,
as shown in Figure 10-1(b). Any load connected to terminals A and B of the
equivalent circuit of Figure 10-1(b) draws exactly the same current with
­exactly the same voltage drop as the load would if it were connected to the
network of Figure 10-1(a).
In fact, all that we can determine from measurements at the output
­terminals of any network containing one or more voltage sources is that
the network is equivalent to a simple constant-voltage source with a single
­internal resistance in series with it. The French engineer Léon Charles
Thévenin (1857–1926) stated the principle of this equivalence as a theorem:
Any two-terminal network of fixed resistances and voltage sources
may be replaced by a single voltage source that has
•an equivalent voltage equal to the open-circuit voltage at the
­terminals of the original network, and
•an internal resistance equal to the resistance looking back into
the network from the two terminals with all the voltage sources
replaced by their internal resistances.
10-1 Thévenin’s Theorem
We can apply Thévenin’s theorem to any of the resistance networks we
have solved so far by treating one branch of the network as a load and
the remainder of the network as a two-terminal network containing one or
more voltage sources. Having decided which branch of the original network to treat as a load, we remove it from the original network and place
it in a Thévenin-equivalent circuit. Then, we apply Thévenin’s theorem to
­determine the rest of the equivalent circuit. Note that Thévenin uses an
“ohmmeter” approach to determine the internal resistance “looking back
into” the open-circuit terminals of the source network, rather than the
short-circuit current method we used with our black-box technique. Both
approaches lead to the same value for the internal resistance.
Example 10-1
What current does a 10-kΩ resistor draw when it is connected to a
100-V source through the T-network shown in Figure 10-1(a)?
Solution
Step 1
Connect the 10-kΩ resistor to a Thévenin-equivalent source consisting of a constant voltage source ETh with a series internal resistance RTh, as shown in Figure 10-2(b). According to Thévenin’s theorem,
ETh equals the open-circuit voltage between terminals A and B of
Figure 10-1(a). Since an open circuit draws no current, there is no voltage drop across the 20-kΩ resistor. Hence, the open-circuit terminal
voltage is the same as the voltage drop across the 100-kΩ resistor. When
terminal A is open-circuit, the 25-kΩ and the 100-kΩ resistors form a
simple series circuit. Then, the voltage-divider principle provides us
with the voltage drop across the 100-kΩ resistor. This voltage drop is
the open-circuit voltage between terminals A and B, which is ETh in the
Thévenin-equivalent circuit.
ETh =
100
× 100 V = 80 V
25 + 100
A
20 kΩ
40 kΩ
+
25 kΩ
RTh
10 kΩ
ETh 80 V
−
100 kΩ
B
(a)
Figure 10-2
A
B
(b)
Equivalent circuit for Step 2 of Example 10-1
261
262
Chapter 10 Equivalent-Circuit Theorems
Step 2
Replace the actual source in Figure 10-1(a) by its internal resistance,
which is zero ohms. The original circuit of Figure 10-1(a) then becomes
the series-parallel resistance network of Figure 10-2(a). The equivalent
resistance for this network is
RTh = 20 kΩ +
25 kΩ × 100 kΩ
= 20 kΩ + 20 kΩ = 40 kΩ
25 kΩ + 100 kΩ
Step 3
Solve for the current through the 10-kΩ resistor in the Théveninequivalent circuit of Figure 10-2(b).
I=
circuitSIM
walkthrough
ETh
80 V
=
= 1.6 mA
RT
40 kΩ + 10 kΩ
Multisim Solution
We can use a circuit simulation to show that the circuits in Figure 10-1
are equivalent.
Download Multisim file EX10-1(a) from the website.
The circuit is the same as shown in Figure 10-1(a). Connect a 10-kΩ
load resistor across terminals A and B. Then, add a meter to measure
the current through this load resistor.
Connect a ground to the bottom node of the schematic.
Run the simulation, and record the meter reading for the current
through the load.
Repeat the simulation with a 50-kΩ load and with a 100-kΩ load, and
record the load currents.
Download Multisim file EX10-1(b) from the website.
The circuit is the same as shown in Figure 10-1(b). Using the same
procedure as for the first circuit, run simulations with a 10-kΩ load, a
50-kΩ load, and a 100-k Ω load. Record the load currents.
Comparing the load current values for the two circuits, we find that
both circuits produce the same current for a given load. If we run simulations with very low and very high load resistances, we still find that
the load currents are the same for both circuits. Therefore, the circuits
are equivalent.
To further demonstrate the usefulness of Thévenin’s theorem, we apply
it to Example 8-3, which we solved originally using Kirchhoff’s-laws
­equations.
10-1 Thévenin’s Theorem
Example 10-2
A resistor passing a 20-mA current is in parallel with a 5.0-kΩ resistor. This combination is in series with another 5.0-kΩ resistor, and the
whole network is connected to a 500-V source. Find the resistance of
the resistor that is passing the 20-mA current.
Solution
Step 1
We select the unknown resistance passing the 20-mA current as the
load, r­ emove it from the original circuit in Figure 10-3(a), and place it in
the Thévenin-equivalent circuit of Figure 10-3(b). The voltage-divider
principle then gives the open-circuit terminal voltage of the remaining
circuit in ­Figure 10-3(a):
ETh =
5.0 kΩ
A
+
500 V
−
5.0
× 500 V = 250 V
5.0 + 5.0
RTh
IL = 20 mA
A
IL = 20 mA
RL
5.0 kΩ
+
ETh 250 V
−
RL
B
B
(a)
Figure 10-3
(b)
Circuit diagram for Step 1 of Example 10-2
Step 2
If we replace the original voltage source with its internal resistance of
zero ohms, the circuit of Figure 10-3(a) becomes the simple parallel resistance network of Figure 10-4(a), and
RTh =
5.0 kΩ × 5.0 kΩ
= 2.5 kΩ
5.0 kΩ + 5.0 kΩ
RTh
A
5.0 kΩ
+
ETh 250 V
−
5.0 kΩ
Figure 10-4
IL = 20 mA
2.5 kΩ
RL
B
(a)
A
B
(b)
Equivalent circuit for Step 2 of Example 10-2
263
264
Chapter 10 Equivalent-Circuit Theorems
Step 3
We can now use Ohm’s law to solve the Thévenin-equivalent circuit of
­Fig­ure 10-4(b):
RT =
circuitSIM
walkthrough
ETh
250 V
=
= 12.5 kΩ
I
20 mA
RL = RT − RTh = 12.5 kΩ − 2.5 kΩ = 10 kΩ
Multisim Solution
Download Multisim file EX10-2 from the website.
The circuit is the same as shown in Figure 10-3(a) with RL = 10 kΩ.
Insert a meter to measure the current through RL. Connect a ground to
the bottom node of the schematic.
Run the simulation, and note that the load current is 20 mA. If we
run the simulation with smaller load resistances, the load current increases. Similarly, if we use larger resistances, the load current decreases. Therefore, 10 kΩ is the only load resistance that passes a
current of 20 mA.
Thévenin’s theorem allows us to solve a Wheatstone bridge network without having to solve simultaneous equations as required for the Kirchhoff’slaws solution in Chapter 9.
Example 10-3
In the Wheatstone bridge circuit of Figure 10-5, R1 = R4 = 300 Ω and
R2 = R3 = 150 Ω. Find the current through a 50-Ω galvanometer connected across terminals A and B when the source voltage is 100 V.
Solution
Step 1
Since we are asked to find the current through only one branch of the
Wheatstone bridge, we can treat the meter as if it were connected to the
Thévenin-equivalent circuit, as shown in Figure 10-5(b).
Step 2
Applying the voltage-divider principle to Figure 10-5(a) gives
V3 = 100 V ×
and
V4 = 100 V ×
150
= 33.33 V
300 + 150
300
= 66.67 V
300 + 150
10-1 Thévenin’s Theorem
R1
R2
A
+
E
RTh
265
B
B
+
ETh
−
R4
R3
RM G
−
A
(a)
(b)
R1
R3
R2
RTh
R4
(c)
Figure 10-5
Solving Wheatstone bridge by Thévenin’s theorem
ETh = VAB = V3 − V4 = 33.33 V − 66.67 V = −33.34 V
From Kirchhoff’s voltage law,
Since VAB is negative, terminal A is negative with respect to terminal B.
Step 3
Assuming that the voltage source in Figure 10-5(a) has negligible internal resistance, we short it out to determine RTh. Then, R1 is in parallel
with R3, R2 is in parallel with R4, and the two pairs of parallel resistors
are connected in series between terminals A and B. Therefore,
RTh =
R1R3
R2R4
300 Ω × 150 Ω
150 Ω × 300 Ω
+
=
+
= 200 Ω
R1 + R3
R2 + R4
300 Ω + 150 Ω
150 Ω + 300 Ω
Step 4
Applying Ohm’s law to the Thévenin-equivalent circuit gives the meter
current:
ETh
33.34 V
IM =
=
= 133 mA
RTh + RM
250 Ω
Multisim Solution
Download Multisim file EX10-3 from the website.
The circuit is the Wheatstone bridge shown in Figure 10-5(a) with the
given values for the four resistors.
Insert a 50-Ω resistor between terminals A and B to simulate a galvano­
meter. Add a meter in series with the 50-Ω resistor to measure the current.
Connect a ground to the bottom node of the schematic.
Run the simulation. Note that the meter reading of 133 mA matches the
current calculated using Thévenin’s theorem.
circuitSIM
walkthrough
266
Chapter 10 Equivalent-Circuit Theorems
6.0 Ω
+
6.0 V
−
4.8 Ω
4.0 Ω
+
8.0 V
−
Figure 10-6 Network
with two voltage sources
We can use the same method to determine the current through any
other branch of the bridge circuit. Thévenin’s theorem also applies to networks with two or more voltage sources. The theorem’s usefulness for
such ­networks depends on whether we can determine ETh without having
to solve simultaneous equations. With only two sources, finding ETh is
­usually straightforward, as we can show with the network of Figure 10-6,
which is similar to the automobile generator example in Chapter 9.
Example 10-4
Determine the current in the 4.8-Ω resistor in the network of Figure 10-6.
Solution
Step 1
Remove the 4.8-Ω resistor from the original circuit of Figure 10-6 and
place it in the Thévenin-equivalent circuit of Figure 10-7(b).
1.2 V
−
0.80 V
+
6.0 Ω
+
6.0 V
−
−
A
+
2.4 Ω
C
RTh
4.0 Ω
+
8.0 V
−
D
B
+
ETh 7.2 V
−
(a)
Figure 10-7
A
4.8 Ω
B
(b)
Circuit diagram for Example 10-4
Step 2
The sum of the potential differences must be the same along both
paths from point C to point D in Figure 10-7(a). One path is the
8.0-V source, and the other path consists of the 6.0-V source and
the two resistors. Therefore, there must be a 2.0-V drop across the
6.0-Ω and 4.0-Ω resistors in series. From the voltage-divider principle, these voltage drops are 1.2 V and 0.8 V, respectively, as shown in
Figure 10-7(a). Tracing either path from point A to point B shows that
the open-circuit voltage ETh is 7.2 V.
Step 3
Replace the sources by their internal resistances (zero ohms). The 6.0-Ω and
4.0-Ω resistors then appear in parallel between points A and B. Therefore,
RTh =
6Ω × 4Ω
= 2.4 Ω
6Ω + 4Ω
Step 4
Apply Ohm’s law to the Thévenin-equivalent circuit of Figure 10-7(b):
IL =
ETh
7.2 V
=
= 1A
RTh + RL
2.4 Ω + 4.8 Ω
10-1 Thévenin’s Theorem
267
Multisim Solution
Download Multisim file EX10-4 from the website.
circuitSIM
walkthrough
The circuit is the same as shown in Figure 10-6.
Insert a meter to measure the current through the 4.8-Ω resistor. ­Connect
a ground to the bottom node of the schematic.
Run the simulation. Note that the meter reading of 1 A matches the current calculated using Thévenin’s theorem.
Practical Circuits
Thévenin’s Theorem Applications
I
A
Linear two-terminal
circuit
B
+
V
−
Load
+
V
−
Load
(a)
Source: © iStock.com/gwmullis
I
A
RTh
+
VTh
B
−
(b)
Figure 10-8
An electrical outlet
Often the load in a circuit varies while the other
circuit elements are fixed. For example, various
appliances may be connected to a household
outlet, thus changing the load on that branch
of the house wiring. When we are analyzing
a circuit that has a varying load, Thévenin’s
theorem can save us some tedious calculations
by letting us replace the unchanged portion of
the circuit with a simple equivalent circuit, as
shown in Figure 10-8. However, we can apply
Thévenin’s theorem only to linear components
(components with values that do not vary with
voltage or current).
See Problems 10-1 to 10-18 and Review Questions 10-43 to 10-48.
268
Chapter 10 Equivalent-Circuit Theorems
10-2 Norton’s Theorem
Figure 10-1 demonstrates how we can reduce a network containing one or
more voltage sources into a simple equivalent constant-voltage source. In
Chapter 9, we found that we can readily convert a constant-voltage source
into an exactly equivalent constant-current source. Therefore, we should
be able to reduce any network containing one or more voltage sources
to a simple equivalent constant-current source, as shown in Figure 9-5.
­Norton’s theorem combines Thévenin’s theorem and source conversion
into a single procedure:
Any two-terminal network of fixed resistances and voltage sources
may be replaced by a single constant-current source that has
•a current equal to the current drawn by a short circuit across the
terminals of the original network, and
•a resistance in parallel equal to the resistance that would be ­measured
from the two terminals if each voltage source were r­eplaced by its
­internal resistance.
Example 10-5
Determine the Norton-equivalent constant-current source for the network of Figure 10-9(a).
25 kΩ
+
100 V
−
20 kΩ
A
A
IN
100 kΩ
2.0 mA
RN
40 kΩ
B
B
(b)
(a)
Figure 10-9
source
(a) Original voltage source; (b) Equivalent constant-current
Solution I
Step 1
Find the short-circuit current. When terminals A and B are shorted, the
total resistance connected across the 100-V source is
RT = 25 kΩ +
100 kΩ × 20 kΩ
= 41.67 kΩ
100 kΩ + 20 kΩ
The current drawn from the 100-V source is
10-2 Norton’s Theorem
IT =
E
100 V
=
= 2.4 mA
RT
41.67 kΩ
From the current-divider principle,
IN = Isc =
100 kΩ
× 2.4 mA = 2.0 mA
100 kΩ + 20 kΩ
Step 2
Replace the 100-V source with its internal resistance of zero ohms. Then
the equivalent resistance connected between terminals A and B is
RN = 20 kΩ +
25 kΩ × 100 kΩ
= 40 kΩ
25 kΩ + 100 kΩ
IN and RN are the required values for the Norton-equivalent constantcurrent source of Figure 10-9(b).
Solution II
Step 1
We can avoid the current-divider calculations of Solution 1 by first
finding the Thévenin-equivalent constant-voltage source as shown in
Figure 10-10(a).
ETh = Eoc =
40 kΩ
100 kΩ
× 100 V = 80 V
100 kΩ + 25 kΩ
A
A
RTh
+
80 V ETh
−
RL
IN
2.0 mA
RN
40 kΩ
V
RL
B
B
(a)
(b)
Figure 10-10 (a) Thévenin-equivalent constant-voltage source;
(b) Norton-equivalent constant-current source
Step 2
Find RTh with the same method we used to find RN in Solution 1.
RTh = 40 kΩ
Step 3
Convert the Thévenin-equivalent voltage source of Figure 10-10(a) into
a constant-current source, obtaining the Norton-equivalent source of
Fig­ure 10-10(b).
IN = Isc =
ETh
80 V
=
= 2.0 mA
RTh
40 kΩ
269
270
Chapter 10 Equivalent-Circuit Theorems
Example 10-4A
Determine the current in the 4.8-Ω resistor in the circuit shown in
­Figure 10-6.
Solution
Step 1
Remove the 4.8-Ω resistor from the original circuit and place it in the
­Norton-equivalent circuit of Figure 10-11(b).
6.0 Ω
4.0 Ω
A
+
6.0 V
−
A
+
8.0 V
−
Isc
B
IN
3.0 A
RN
4.8 Ω
B
(a)
Figure 10-11
(b)
Circuit diagram for Step 2 of Example 10-4A
Step 2
Replace the 4.8-Ω resistor in the original circuit by a short circuit, as
shown in Figure 10-11(a). Drawing arrows for current directions shows
that the short-circuit current is the sum of the currents resulting from
connecting the 6-Ω resistor directly across the 6.0-V source and connecting the 4.0-Ω resistor directly across the 8.0-V source. Therefore, the
short-circuit current is
IN =
E1
E2
6V
8V
+
=
+
= 3A
R1
R2
6Ω
4Ω
A
A
6.0 Ω
B
4.0 Ω
3.0 A
IN RN
2.4 Ω
4.8 Ω
B
(a)
Figure 10-12
(b)
Circuit diagram for Step 3 of Example 10-4A
Step 3
Replace the sources with their negligible internal resistances. In the resulting circuit, shown in Figure 10-12(a), the 6.0-Ω and 4.0-Ω resistors are
connected in parallel across terminals A and B. Hence,
RN =
6Ω × 4Ω
= 2.4 Ω
6Ω + 4Ω
10-3 Dependent Sources
Step 4
Apply the current-divider principle to the Norton-equivalent circuit of
­Figure 10-12(b):
2.4 Ω
× 3A = 1A
IL =
2.4 Ω + 4.8 Ω
See Problems 10-19 to 10-26 and Review Questions 10-49 and 10-50.
10-3 Dependent Sources
The EMF generated by the voltage sources we have encountered so far has
been determined entirely by the energy-conversion process within the source.
The open-circuit terminal voltage is, therefore, independent of what happens
elsewhere in a network containing such constant-voltage or ­independent
sources. The terminal voltage of a practical voltage source ­decreases as load
current increases, but we can account for this effect by including an internal resistance in series with the ideal constant-voltage source in network
­diagrams. Constant-current sources are also independent sources.
Transistors or other electronic devices can be connected such that they
behave like voltage or current sources with an output that is dependent on
or controlled by a voltage or current appearing in some other branch of the
circuit. To distinguish dependent or controlled sources from independent
sources in circuit diagrams, we use a diamond-shaped symbol with + and −
signs for a dependent voltage source and a diamond with a current arrow
for a dependent current source. As shown in Table 10-1, there are four types
of dependent sources.
TABLE 10-1 Dependent sources
Type of Source
Voltage-controlled voltage source
(VCVS)
Current-controlled voltage source
(CCVS)
Current-controlled current source
(CCCS)
Voltage-controlled current source
(VCCS)
Magnitude
V = kV1
Symbol
V = kI1
I = kI1
I = kV1
Figure 10-13 shows a simple network containing a current-controlled
voltage source (CCVS) in series with a 120-V independent voltage source
and its internal resistance. This dependent source is designed so that
271
272
Chapter 10 Equivalent-Circuit Theorems
Rint
+
−
5.0 Ω
+
+
120 V
−
I1
RL
5I1
−
Figure 10-13
Network with a dependent source
its ­terminal voltage is V = 5 Ω × I1. Equations for dependent sources
are commonly written without showing the units for the constant. For
­example, the equation for the CCVS in Figure 10-13 would be simplified
to V = 5I1. The necessary units are understood to be included in the
­constant.
As we increase RL in the circuit of Figure 10-13, I1 increases and so does
the IR drop across Rint. The dependent source voltage also increases since
its output is V = 5I1. Thus, the dependent source adds a voltage that ­exactly
offsets the IR drop across Rint. The series combination of the two voltage
sources acts as a voltage-regulated power supply.
We think of the CCVS in Figure 10-13 as a source because the relationship
between the polarity of the voltage between its terminals and the current
through it is that of a source. However, dependent sources differ from real
energy sources. If we disconnect RL in the circuit of Figure 10-13, I1 = 0
and the voltage across the CCVS disappears while the independent source
maintains its voltage. Hence, we use the symbol V rather than E for dependent sources. Although Kirchhoff’s laws and network theorems hold
true for networks containing dependent sources, we cannot always use
these methods to solve such networks.
Example 10-6
Determine the load voltage VL in the network shown in Figure 10-14.
200 Ω
+
−
+
5.0 V
−
800 Ω
−
+
Figure 10-14
V1
0.010V1
+
RL
500 Ω
−
Circuit diagram for Example 10-6
10-3 Dependent Sources
Solution I
At first glance, it might seem that we could write two mesh equations
to solve the network shown in Figure 10-14. But the rules for mesh
equations require us to convert all current sources to equivalent voltage sources. We cannot convert the voltage-controlled current source
(VCCS) in Figure 10-14 to an equivalent voltage-controlled voltage
source (VCVS) because there is no resistor that we can use as Rint in
parallel with the original current source. We can, however, apply
Thévenin’s theorem to the left loop of the network without affecting V1.
Figure 10-15 shows the resulting single-loop network.
0.010V1
+
−
160 Ω
+
+
4.0 V
−
+
RL
V1
500 Ω
−
−
Figure 10-15
Equivalent circuit for Solution I for Example 10-6
The current in the equivalent circuit in Figure 10-15 is the output
of the VCCS,
I = 0.01V1
V1 = 4.0 V − 160I = 4.0 − 160(0.01V1) = 4.0 − 1.6V1
2.6V1 = 4.0
V1 = 1.54 V
I = 0.01V1 = 15.4 mA
VL = IRL = 15.4 mA × 500 Ω = 7.7 V
Applying Ohm’s law and substituting for I gives
Solution II
We cannot use the superposition theorem for this example since removing the voltage source will drop V1 to zero, making the output
of the VCCS zero as well. But we can convert the independent voltage
source to its equivalent constant-current source and use nodal analysis.
0.010V1
Node
+
+
25 mA
200 Ω
800 Ω
V1
RL
500 Ω
−
−
Figure 10-16
Equivalent circuit for Solution II for Example 10-6
273
274
Chapter 10 Equivalent-Circuit Theorems
In the equivalent circuit of Figure 10-16, the single nodal equation is
1
1
+
V1 = 25 mA − 0.01V1
200 Ω
800 Ω
25 mA
= 1.54 V
This equation reduces to
V1 =
16.25 mS
(
)
IL = 0.01V1 = 0.01 × 1.54 = 15.4 mA
VL = ILRL = 15.4 mA × 500 Ω = 7.7 V
Then
and
In Figure 10-17, a current-controlled voltage source has replaced the
voltage-controlled current source of Figure 10-14. The superposition
theorem still does not work, and we cannot replace the left-hand part of
the ­network with its Thévenin-equivalent source because we will lose the
I1 current, which governs the voltage produced by the CCVS. However, we
can use either mesh or nodal analysis techniques.
200 Ω
+
Node N
100I1
−
VB
I1
+
5.0 V
−
IA
Figure 10-17
+
800 Ω
−
IB
+
RL
500 Ω
−
Circuit diagram for Example 10-7
Example 10-7
Determine VL in the network shown in Figure 10-17.
Solution I
From Kirchhoff’s voltage law,
(200 + 800)IA − 800IB = 5.0
(500 + 800)IB − 800IA = 100I1 = 100(IA − IB)
Collecting terms gives
1000IA − 800IB = 5.0
900IA − 1400IB = 0
(1)
(2)
10-3 Dependent Sources
IB
│1000
900
=
│1000
900
│ −4500
=
= 6.62 mA
−680 000
−800
−1400│
5.0
0
VL = IBRL = 6.62 mA × 500 Ω = 3.31 V
Solution II
Figure 10-18 shows the Norton-equivalents for both voltage sources in
the original network.
Node N
+
25 mA
IA 200 Ω
Figure 10-18
IB
+
800 Ω
−
500 Ω
−
RL
0.20I1
Equivalent circuit for Solution II for Example 10-7
At the independent current source,
IA =
5.0 V
= 25 mA
200 Ω
At the dependent voltage source, VB = 100I1 and
Since I1 =
IB =
VN
VN
, IB =
800
4000
VB
100I1
=
= 0.200I1
500
500
For the single independent node, the nodal equation is
(
VN
1
1
1
VN = 25 mA −
+
+
200
800
500
4000
)
VN = 25 mA × 117.6 Ω = 2.94 V
Note that VN is not the same as VL in the original network of ­Figure 10-17.
Substituting for VN in the equation for I1 gives
I1 =
2.94 V
VN
=
= 3.676 mA
800
800 Ω
VB = 100I1 = 0.368 V
VL = VN + VB = 2.94 V + 0.37 V = 3.31 V
275
276
circuitSIM
walkthrough
Chapter 10 Equivalent-Circuit Theorems
Multisim Solution
Download Multisim file EX10-7 from the website.
The circuit is the same as shown in Figure 10-17.
Insert a meter to measure the voltage across the 500-Ω load ­resistor.
Run the simulation. Note that the meter reading of 3.31 V matches
the voltage calculated using either Kirchhoff’s voltage law or Nortonequivalent sources.
We can also use a simulation to check the output of the dependent voltage source. Place two meters in the circuit, one to measure the ­current,
I1, and the other to measure the voltage, VB.
Run the simulation. The meter readings confirm that VB equals
100I1.
We can include dependent sources in a Thévenin transformation as long
as we are careful how we determine RTh. If the controlling variable for
the dependent source is within the portion of the network included in the
Thévenin-equivalent source, we cannot find RTh by simply calculating the
equivalent resistance that would appear across the open-circuit terminals
when the voltage sources are replaced by the internal resistances. ­Instead,
we must derive the equivalent resistance from the short-circuit current of
the network, as we did in Section 9-2:
RTh =
open-circuit terminal voltage
short-circuit terminal current
(9-1)
Example 10-8
Remove RL from the network in Figure 10-17, and determine the
Thévenin-equivalent source that can replace the remainder of the
­network.
Solution
For the open-circuit condition shown in Figure 10-19(a),
I1 =
5.0 V
= 5.0 mA
200 Ω + 800 Ω
10-3 Dependent Sources
100I1
200 Ω
100I1
200 Ω
I1
+
5.0 V
−
I1
+
+
5.0 V
−
Voc = ETh
800 Ω
−
277
+
800 Ω
−
(a)
Isc
(b)
180 Ω
RTh
+
4.5 V ETh
−
(c)
Figure 10-19
Dependent source in a Thévenin transformation
The dependent source voltage is
V = 100I1 = 0.50 V
Voc = ETh = 5.0 mA × 800 Ω + 0.50 V = 4.5 V
To find the short-circuit current, we can write two mesh equations for
the circuit of Figure 10-19(b).
(200 + 800)IA − 800IB = 5.0 and 800IB − 800IA = 100I1 = 100(IA − IB)
Collecting terms gives
1000IA − 800IB = 5.0
900IA − 900IB = 0
Isc = IB
│1000
900
=
│1000
900
RTh =
│ −4500
= 25 mA
=
−180 000
−800
−900│
(1)
(2)
5
0
Eoc
4.5 V
=
= 180 Ω
Isc
25 mA
Note that RTh is not the same as the resistance “seen” looking back
into the open circuit terminals of Figure 10-19(a).
See Problems 10-27 to 10-35 and Review Questions 10-51 to 10-53.
We can check ETh
and RTh by comparing
the load voltage with
the value found in
­Example 10-7:
VL =
500 × 4.5 V
180 + 500
= 3.31 V
278
Chapter 10 Equivalent-Circuit Theorems
Circuit Check
CC 10-1.
A
Use Thévenin’s theorem to calculate the voltage across the
8.0-Ω resistor in the circuit shown in Figure 10-20.
120 Ω
+
36 V
−
40 Ω
60 Ω
12 Ω
20 Ω
18 Ω
8.0 Ω
Figure 10-20
CC 10-2. Use a Norton-equivalent source to calculate the voltage across
the 40-Ω resistor in Figure 10-20.
CC 10-3. Determine the voltage across the 15-Ω resistor in Figure 10-21.
1.5V1
V1
−
+
25 V
−
+
10 Ω
15 Ω
20 Ω
Figure 10-21
10-4 Delta-Wye Transformation
Another useful type of circuit transformation applies to three-terminal
­resistance networks like the one in Figure 10-22. If we can state the conditions under which the wye (Y) circuit of Figure 10-22(a) is equivalent to the
delta (Δ) circuit of Figure 10-22(b), we can substitute the wye connection
for the delta, and vice versa.
A
RA
RB
A
RZ
RC
C
B
(a)
Figure 10-22
B
RY
RX
(b)
(a) Y-network; (b) Δ-network
C
10-4 Delta-Wye Transformation
For the circuits to be equivalent, the resistance between terminals A and B
must be the same for both networks. For the wye circuit of Figure 10-22(a),
the circuit between A and B is a simple series circuit, so RAB = RA + RB.
The delta circuit of Figure 10-22(b) has two branches in parallel between
terminals A and B, so
RAB =
RA + RB =
Similarly,
RB + RC =
and
RC + RA =
RZ ( RX + R Y )
RZ + ( RX + RY )
RXRZ + RYRZ
RX + RY + RZ
RXRY + RXRZ
RX + RY + RZ
RYRZ + RXRY
RX + RY + RZ
(1)
(2)
(3)
Subtracting Equation 2 from Equation 1 gives
RA − RC =
RYRZ − RXRY
RX + RY + RZ
(4)
Adding Equation 3 and Equation 4 gives
2RA =
RA =
Likewise,
RB =
and
RC =
2RYRZ
RX + RY + RZ
RYRZ
RX + RY + RZ
RXRZ
RX + RY + RZ
RXRY
RX + RY + RZ
(10-1)
(10-2)
(10-3)
Equations 10-1 to 10-3 are the delta-to-wye transformation equations.
Note that there is a simple pattern to the three equations:
RA =
product of the two Δ sides connected to terminal A
sum of all three Δ sides
279
280
Chapter 10 Equivalent-Circuit Theorems
We can solve Equations 1, 2, and 3 for the equivalent delta circuit for a
given wye circuit. The wye-to-delta transformation is usually stated as
RX =
RY =
RZ =
RARB + RBRC + RCRA
RA
RARB + RBRC + RCRA
RB
RARB + RBRC + RCRA
RC
(10-4)
(10-5)
(10-6)
Again the three equations have a pattern:
RX =
sum of products of each pair of Y-arms
opposite Y-arm
Inverting Equation 10-4 and substituting R = 1/G for each resistance gives
1
1
1
GA
GA
= GX =
=
RX
1
1
1
GA + GB + GC
+
+
GAGB
GBGC
GCGA
GAGBGC
So,
GX =
Similarly,
GY =
and
GZ =
GBGC
GA + GB + GC
GAGC
GA + GB + GC
GAGB
GA + GB + GC
(10-7)
(10-8)
(10-9)
Equations 10-7 to 10-9 are the duals of Equations 10-1 to 10-3.
Inverting Equation 10-7 gives
RX = RBRC ( GA + GB + GC )
(10-10)
10-4 Delta-Wye Transformation
RY = RARC ( GA + GB + GC )
Similarly,
(10-11)
RZ = RARB ( GA + GB + GC )
and
(10-12)
The conductance forms of the wye-to-delta transformation equations are
particularly useful for analyzing alternating-current circuits.
Example 10-3A
In the Wheatstone bridge circuit of Figure 10-23(a), R1 = R4 = 300 Ω
and R2 = R3 = 150 Ω. Find the current through a 50-Ω galvanometer
connected across terminals A and B when the source voltage is 100 V.
RA
R1
+
RM
R2
RC
R3
R4
E
G
E
RB
+
−
−
R3
R4
(a)
Figure 10-23
(b)
Solving Wheatstone bridge by delta-wye transformation
Solution
Step 1
Replace the delta connection of R1, R2, and RM with its equivalent wye
­circuit, as in Figure 10-23(b).
RA =
RB =
RC =
300 Ω × 150 Ω
= 90 Ω
300 Ω + 150 Ω + 50 Ω
300 Ω × 50 Ω
= 30 Ω
500 Ω
50 Ω × 150 Ω
= 15 Ω
500 Ω
281
282
Chapter 10 Equivalent-Circuit Theorems
Step 2
The load on the source is now a simple series-parallel circuit with an
equivalent resistance of
Req = RA +
Therefore,
( R B + R3 ) × ( RC + R4 )
180 Ω × 315 Ω
= 204.55 Ω
= 90 Ω +
( RB + R3 + RC + R4 )
495 Ω
IT =
E
100 V
=
= 489 mA
Req
204.55 Ω
Step 3
From the current-divider principle,
I3 = 489 mA ×
180 Ω
= 178 mA
495 Ω
Since R3 and R4 were not altered by the transformation, I3 and I4 are
the same as for the original circuit. Therefore, in Figure 10-23(a),
and
The right terminal of
the meter is positive
with respect to the left
terminal.
and
I4 = 489 mA ×
315 Ω
= 311 mA
495 Ω
V3 = I3R3 = 311 mA × 150 Ω = 46.65 V
V4 = I4R4 = 178 mA × 300 Ω = 53.4 V
Step 4
From Kirchhoff’s voltage law,
Therefore,
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VM = V4 − V3 = 53.4 − 46.65 = 6.75 V
IM =
6.75 V
= 0.14 A
50 Ω
Multisim Solution
We can use simulations to confirm that the two circuits shown in
­Figure 10-23 are indeed equivalent. If two circuits are equivalent, then
the voltages across R3 in both circuits must be identical, and the voltages across R4 must also match.
Download Multisim file EX10-3A(b) from the website. The circuit is same
as shown in Figure 10-23(b). Insert multimeters to measure the voltages
across R3 and R4. Connect a ground to the bottom node of the s­ chematic.
Run the simulation, and record the voltage readings.
Download Multisim file EX10-3A(a) from the website. The circuit is the
same as shown in Figure 10-23(a). Insert a 50-Ω resistor between terminals
A and B to simulate a galvanometer. Insert meters to measure the voltages
across R3 and R4. Connect a ground to the bottom node of the schematic.
Run the simulation and record the voltage readings. Comparing the
two sets of voltage measurements confirms that the delta and wye circuits are equivalent.
See Problems 10-36 to 10-41 and Review Questions 10-54 and 10-55.
10-5 Troubleshooting
283
10-5 Troubleshooting
We can use equivalent-circuit theorems to help trace circuit faults.
Example 10-9
For the Wheatstone bridge circuit in Example 10-3, the calculated value
for V3 is 33.33 V. Determine the probable cause if the measured value
for V3 is
(a) 100 V
(b) 0 V
(c) 66.67 V
Solution
(a) Kirchhoff’s voltage law gives E = V1 + V3, so we deduce that
V1 = 0 V. Since V1 = I1R1, then either R1 = 0 Ω (shorted) or I1 = 0 A.
For I1 to be 0 A, either R1 or R3 is open. However, if R1 is open,
I3 = 0 A and V3 = 0 V. Since V3 = 100 V, we conclude that R1 is not
open. Therefore, either R1 is shorted or R3 is open. Remove these
resistors from the circuit and check them with an ­ohmmeter.
(b) Kirchhoff’s voltage law gives E = V1 + V3, so we deduce that
V1 = 100 V. From Ohm’s law, V3 = R3I3, so either R3 is shorted or
I3 = 0 A. For I1 to be 0 A, either R1 or R3 is open. However, if R3
is open, I1 = 0 A and V1 = 0 V. Since V1 is not zero, we conclude
that R3 is not open. Therefore, either R1 is open or R3 is shorted.
Remove these resistors from the circuit and check them with an
ohmmeter.
(c) Kirchhoff’s voltage law gives E = V1 + V3, so we deduce that
V1 = 33.33 V. We note that the values of V1 and V3 are reversed,
suggesting that R1 and R3 have been interchanged. Checking the
colour codes on these resistors, we discover that R1 is the 150-Ω resistor and R3 the 300-Ω resistor.
Multisim Solution
Download Multisim file EX10-9 from the website.
The circuit is the same as shown in Figure 10-5, with the given values
for the four resistors.
Interchange the values of R1 and R3. Insert a 50-Ω resistor between terminals A and B to simulate a galvanometer. Connect a ground to the
bottom node of the schematic. Insert meters to measure the voltages
across R1 and R3.
Run the simulation, and view the voltage readings to verify the conclusions of Example 10-9(c).
See Problem 10-42.
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284
Chapter 10 Equivalent-Circuit Theorems
Summary
• Any two-terminal network of fixed resistances and voltage sources can
be replaced by a Thévenin constant-voltage source consisting of a single
voltage source in series with a resistor.
• Any two-terminal network of fixed resistances and voltage sources can
be replaced by a Norton constant-current source consisting of a single
current source in parallel with a resistor.
• A dependent source has an output that depends on a voltage or current
in some other branch of the circuit.
• The Thévenin constant-voltage source may be determined for a network
containing a dependent source provided the open-circuit v
­ oltage and
short-circuit current are used to determine the equivalent resistance.
• A resistive Y-network can be converted to an equivalent Δ-network, and
vice versa.
• Comparing measured and calculated voltages enables us to identify a
fault in a circuit.
B = beginner
I = intermediate
A = advanced
Problems
B
Section 10-1
10-1.
Thévenin's Theorem
Determine the Thévenin-equivalent constant-voltage source for the
two-terminal T-network of Figure 10-24.
100 Ω
+
200 V
−
Figure 10-24
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I
10-2.
100 Ω
400 Ω
T-network
(a)Determine the Thévenin-equivalent source for the π-network in
Figure 10-25.
(b)Use Multisim to verify the voltage of the Thévenin-equivalent
source in part (a).
500 Ω
40 mA
Figure 10-25
I
10-3.
I
10-4.
2.0 kΩ
π-network
2.0 kΩ
Use Thévenin’s theorem to determine the current through the
300-Ω resistor in the circuit of Figure 10-26.
Determine the current in the 100-Ω resistor in the circuit of Figure 10-26.
285
Problems
200 Ω
1.5 kΩ
100 Ω
300 Ω
500 Ω
−
200 V
+
Figure 10-26
B
B
I
I
I
A
10-5.
Use Thévenin’s theorem to find the load resistance that draws
400 mA from the network shown in Figure 10-24.
10-6. (a)Use Thévenin’s theorem to determine the current through a
3.3-kΩ load connected to the network shown in Figure 10-25.
(b) Use Multisim to verify the current calculated in part (a).
10-7. Find the maximum power that can be developed in a load connected to the network shown in Figure 10-24.
10-8. Determine the load resistance that can transfer the maximum
power when connected to the network shown in Figure 10-25.
10-9. Solve for the current drawn from the 20-V source in Figure 10-27 by
replacing the portion of the circuit to the left of the dashed line with
its Thévenin-equivalent source.
10-10. Find the two load resistances that dissipate energy at the rate of
200 mW when connected to the network in Figure 10-25.
−
30 V
+
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15 Ω
50 Ω
+
20 V
−
50 Ω
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Figure 10-27
A
I
10-11. (a)Determine the Thévenin-equivalent source as “seen” by the
10-kΩ resistor in Figure 10-28.
(b)Use Multisim to verify the voltage of the Thévenin-equivalent
source in part (a).
10-12. Find the voltage drop across the 25-Ω resistor in Figure 10-29 by replacing the current source with an equivalent constant-voltage source.
100 Ω
+
100 V
−
+300 V
−150 V
22 kΩ
47 kΩ
10 kΩ
33 kΩ
25 Ω
5.0 A
Figure 10-28
Figure 10-29
68 kΩ
286
Chapter 10 Equivalent-Circuit Theorems
I
A
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10-13. Given E = 200 V, R1 = 8 kΩ, R2 = 2 kΩ, R3 = 2 kΩ, R4 = 14 kΩ,
and RM = 25 kΩ for the Wheatstone bridge of Figure 10-23(a), use
Thévenin’s theorem to determine the meter current.
10-14. (a)Use Thévenin’s theorem to determine the current through a
250-Ω load connected to the output terminals of the bridged-T
attenuator of Figure 10-30 when the input voltage is 3.0 V.
(b) Use Multisim to verify the current calculated in part (a).
400 Ω
100 Ω
100 Ω
Input
Figure 10-30
A
Output
800 Ω
Bridged-T network
10-15. Use Thévenin’s theorem to find the load resistance that draws
1.0 A from the output terminals of the lattice network in Figure 10-31
when 500 V is applied to its input terminals.
20 Ω
30
Ω
Output
30
Ω
Input
20 Ω
Figure 10-31
A
Lattice network
10-16. The switch in the pulse generator circuit of Figure 10-32 alternately
closes for 1 ms and opens for 4 ms. Draw the open-circuit output-­voltage
waveform (with respect to chassis) and determine its a­ mplitude.
−24 V
5.6 kΩ
22 kΩ
500 Ω
+12 V
Figure 10-32
Out
27 kΩ
+24 V
287
Problems
A
10-17
Use Thévenin’s theorem to calculate the current through the 10-Ω
load resistor in Figure 10-33.
7.0 Ω
8.0 Ω
5.0 Ω
10 Ω
4.0 Ω
RL
+
50 V
−
25 Ω
55 Ω
Figure 10-33
I
B
B
B
B
I
I
A
I
10-18. Use another method of analysis to verify your answer for
Problem 10-17.
Section 10-2
Norton's Theorem
10-19. Determine the Norton-equivalent constant-current source for the
two-terminal network of Figure 10-24.
10-20. (a)Determine the Norton-equivalent source for the network of
­Fig­ure 10-25.
(b) Use Multisim to verify the current for the Norton-equivalent
source in part (a).
10-21. Use Norton’s theorem to find the current that a 400-Ω load draws
from the source shown in Figure 10-24.
10-22. (a)Use Norton’s theorem to determine the load resistance that
draws a 950-μA current from the source shown in Figure 10-25.
(b) Use Multisim to verify the load resistance calculated in part (a).
10-23. Replace each voltage source in Figure 10-27 with a Norton-equivalent
source. Then use nodal analysis to determine the current through
the centre 50-Ω resistor.
10-24. (a)Find the current through the 100-Ω resistor in Figure 10-29 by
­leaving the sources in their original form and using the super­
position theorem.
(b) Use Multisim to verify the current calculated in part (a).
10-25. Use Norton’s theorem to calculate the current through the 10-Ω
load resistor in Figure 10-33.
10-26. Compare the internal resistance of the Norton-equivalent source
for the circuit in Figure 10-33 with the internal resistance of the
Thévenin-equivalent source for the circuit.
Section 10-3
Dependent Sources
10-27. Figure 10-34 shows the equivalent output circuit for a typical transistor amplifier. Calculate the output voltage if the input current I1
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288
Chapter 10 Equivalent-Circuit Theorems
is 20 μA, the forward current ratio hfe is 45, the output conductance
hoe is 3.0 × 10−5 S, and the load resistance RL is 20 kΩ.
hoe
I = hfeI1
Figure 10-34
B
I
RL
Equivalent output circuit for a transistor amplifier
10-28. Determine the Thévenin-equivalent source for the transistor circuit
of Problem 10-27.
10-29. Determine the load voltage VL in the circuit of Figure 10-35.
I1
2.2 kΩ
+
40 V
−
2.0I1
RL
6.8 kΩ
10 kΩ
Figure 10-35
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10-30. (a) Determine V1 in the circuit of Figure 10-36.
(b) Use Multisim to verify the voltage calculated in part (a).
2.0V1
2.2 kΩ
−
40 V
+
RL
V1
10 kΩ
Figure 10-36
A
10-31. Find the Thévenin-equivalent voltage source for the network of
­Figure 10-37.
0.50V1
1.0 kΩ
2.0 kΩ
+
36 V
−
Figure 10-37
V1
289
Problems
A
10-32. (a)Find the Thévenin-equivalent voltage source for the network of
­Figure 10-38.
(b) Use Multisim to verify the voltage calculated in part (a).
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50 Ω
0.020V1
80 Ω
V1
−
20 V
+
Figure 10-38
I
10-33. Determine the load voltage VL in the amplifier equivalent circuit of
Figure 10-39.
200 Ω
+
5.0 V
−
V1 0.010V1
800 Ω
1000 Ω
RL
500 Ω
Figure 10-39
A
I
B
10-34. (a)Determine VL when a 2500-Ω feedback resistor is connected between the junction of the 200-Ω and 800-Ω resistors and the top
of the VCCS in Figure 10-39.
(b) Use Multisim to verify the load voltage calculated in part (a).
10-35. What value of RTh is “seen” looking back into the terminals of the
dependent voltage source of Figure 10-40? (Hint: Connect a fixed
voltage source to the terminals and calculate the terminal voltage
and current.)
Section 10-4
Delta-Wye Transformation
10-36. Use a delta-to-wye transformation to determine the equivalent
T-network for the π-network of Figure 10-41.
480 Ω
Input
1.2 kΩ
Figure 10-41
1.2 kΩ
π-network
Output
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100 Ω
20I1
RTh
I1
Figure 10-40
290
Chapter 10 Equivalent-Circuit Theorems
B
10-37. Use a wye-to-delta transformation to determine the equivalent
π-network for the T-network of Figure 10-42.
3.3 kΩ
6.8 kΩ
Input
22 kΩ
Figure 10-42
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I
I
I
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I
Output
T-network
10-38. (a)Given E = 50 V, R1 = 5 Ω, R2 = 25 Ω, R3 = 10 Ω, R4 = 15 Ω, and
RM = 50 Ω for the Wheatstone bridge of Figure 10-23(a), use a
delta-to-wye transformation to determine the meter ­current.
(b) Use Multisim to verify the meter current calculated in part (a).
10-39. Use a delta-to-wye transformation to determine the current
through a 400-Ω load connected to the output terminals of the
bridged-T ­
attenuator of Figure 10-30 when the input voltage
is 5.0 V.
10-40. (a)
Use a delta-to-wye transformation to determine what voltage ­applied to the input terminals of the lattice network of
Figure 10-31 will produce a 2.0-A current through a 20-Ω load
connected to the output terminals.
(b) Use Multisim to verify the input voltage calculated in part (a).
10-41. Use a delta-to-wye transformation to calculate the current delivered by the source in Figure 10-43.
20 Ω
+
50 V
−
25 Ω
15 Ω
20 Ω
30 Ω
5.0 Ω
Figure 10-43
A
Section 10-5
Troubleshooting
10-42. A voltmeter is used to measure the voltage across the 300-Ω resistor
in Figure 10-26. Determine the most probable cause if the voltmeter
reads
(a) 0 V
(b) −200 V
(c) −64 V
Review Questions
Review Questions
Section 10-1
Thévenin's Theorem
10-43. What is a Thévenin-equivalent source?
10-44. What is the difference between the black-box method and the
Thévenin’s-theorem method of determining the internal resistance
of a Thévenin-equivalent source?
10-45. Show that the Thévenin-equivalent resistance found by using the
short-circuit current is the same as that found by the procedure in
the statement of Thévenin’s theorem.
10-46. Draw a Thévenin-equivalent circuit for the source shown in
Figure 10-44 and express the equivalent voltage, ETh, and the
­equivalent internal resistance, RTh, in terms of the parameters of
the original circuit.
R1
+
R2
E
R3
−
Figure 10-44
10-47. Derive a single equation for IL when a load RL is connected to the
­circuit of Figure 10-44.
10-48. In a Thévenin-equivalent circuit, IL = ETh/(RTh + RL). Show that
this expression is equal to the equation for IL in Question 10-47,
thereby verifying Thévenin’s theorem.
Section 10-2
Norton's Theorem
10-49. What is a Norton-equivalent source?
10-50. Explain what happens to the terminal voltage of a Norton-­equivalent
source when a load resistor is connected to its terminals.
Section 10-3
Dependent Sources
10-51. The constant for the current-controlled voltage source in Figure 10-13
is understood to include the unit ohm. What unit is included in the
constant for
(a) a voltage-controlled voltage source
(b) a voltage-controlled current source
10-52. What are the restrictions on using the superposition theorem in a
network containing dependent sources?
10-53. What are the restrictions on using Thévenin’s and Norton’s theorems in a network containing dependent sources?
Section 10-4
Delta-Wye Transformation
10-54. Derive the wye-to-delta transformation Equation 10-4.
291
292
Chapter 10 Equivalent-Circuit Theorems
10-55. Figure 10-23(b) shows a method for solving a Wheatstone bridge using
a delta-to-wye transformation. Draw a diagram showing how we can
solve the same bridge circuit using a wye-to-delta ­transformation.
Integrate the Concepts
In Example 8-1 of Chapter 8, we used the equivalent-circuit method to
­determine I1, I2, and I3 in the circuit shown in Figure 10-45. In Chapter 9 we
used four other techniques to analyze this circuit. Solve for I3 using another
two different techniques.
I1
R1
I2
12 Ω
+
100 V
−
R2
I3
10 Ω
R3
40 Ω
Figure 10-45
Practice Quiz
1.
2.
A Thévenin-equivalent circuit consists of
(a) a constant-voltage source in series with a
­resistance
(b) a constant-current source in series with a
­resistance
(c) a constant-voltage source in parallel with a
­resistance
(d) a constant-current source in parallel with a
­resistance
single internal
single internal
single internal
single internal
The Thévenin internal resistance for the network in Figure 10-46 is
(a) 10 kΩ
(b) 1.8 kΩ
(c) 2.07 kΩ
(d) 2.11 kΩ
I2
10 mA
R3
1.2 kΩ
I1
15 mA
R1
1.0 kΩ
A
R2
10 kΩ
RL
B
Figure 10-46
Practice Quiz
3.
The Thévenin-equivalent constant-voltage source for the network in
Figure 10-46 is
(a) 15 V
(b) 4.9 V
(c) 22.1 V
(d) 12 V
4.
A Norton-equivalent circuit consists of
(a) a constant-voltage source in parallel with a single internal
­resistance
(b) a constant-current source in series with a single internal ­resistance
(c) a constant-voltage source in series with a single internal
­resistance
(d) a constant-current source in parallel with a single internal r­ esistance
5.
The Norton-equivalent internal resistance viewed between the terminals A and B of Figure 10-47 is
(a) 13.64 Ω
(b) 60.76 Ω
(c) 395 Ω
(d) 143.8 Ω
R1
100 Ω
A
R2
150 Ω
V1
12 V
R3
150 Ω
B
R4
220 Ω
22 V
V2
Figure 10-47
6.
7.
The Norton’s constant-current source viewed between the terminals
A and B of Figure 10-47 is
(a) −31.25 mA
(b) −25.32 mA
(c) +31.25 mA
(d) +25.32 mA
The output of a dependent source depends on some other variable in
the circuit. Which of the following is a dependent source?
(a) voltage-controlled pressure source
(b) current-controlled light source
(c) voltage-controlled current source
(d) temperature-controlled voltage source
293
11
Electrical Measurement
To understand electric circuit behaviour, we must be able to measure the circuit
parameters. In this chapter, we consider how commonly used instruments
measure current, voltage, and resistance.
Chapter Outline
11-1
11-2
Moving-Coil Meters
The Ammeter
297
296
11-3
The Voltmeter
11-5
Resistance Measurement
11-7
Multimeters
11-4
11-6
300
Voltmeter Loading Effect 302
304
The Electrodynamometer Movement
312
311
Key Terms
moving-coil movement 296
d’Arsonval movement 296
motor principle 296
diodes 297
sensitivity 297
resistance 297
shunt resistor 297
swamping or calibrating
resistor 297
ammeter 299
voltmeters 300
multiplier resistor 300
voltmeter sensitivity 300
ohmmeter 305
galvanometer 309
electrodynamometer
movement 311
wattmeter 311
volt-ohm-milliammeter (VOM)
or multimeter 312
digital multimeter (DMM)
312
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• explain how a moving-coil movement
measures current
• calculate the calibrating and shunt resistors
required for a multirange ammeter
• calculate the multiplier resistors required for a
­multirange voltmeter
• calculate the internal resistance of a movingcoil ­voltmeter
• calculate the loading effect of a voltmeter
Photo sources: iStock.com/Zmiy
• design an ohmmeter using a moving-coil
movement
• use a Wheatstone bridge for accurate
resistance ­measurements
• explain the operation of the
electrodynamometer ­movement
• differentiate between analog and digital
multimeters
296
Chapter 11 Electrical Measurement
11-1 Moving-Coil Meters
The most common mechanism for meters that have a pointer that moves
across a scale is the moving-coil or d’Arsonval movement. This movement
can be used to measure current, voltage, and resistance. The operation of
such analog meters is based on the motor principle, the fact that a magnetic
field exerts a force on a current flowing perpendicular to it.
Figure 11-1 shows the basic parts of a D’Arsonval movement. When current flows through the coil, it becomes an electromagnet with a north and
a south pole. These poles are repelled by the permanent magnet, rotating
the coil until the magnetic force is balanced by the spiral spring. As the coil
rotates around the stationary iron core, the pointer moves to the right. The
iron core increases the strength of the magnetic field in the air gap between
the core and the permanent magnet. Often the spiral spring also provides
an electric connection to the coil.
The French physicist
Jacques-Arsène
d’Arsonval (1851–1940)
developed the
moving-coil movement
in 1885.
Scale
Volts
Pointer
Permanent magnet
Air gap
Spiral spring
N
Moving coil
Moving-coil meter
Source: ANDREW LAMBERT PHOTOGRAPHY/SCIENCE PHOTO LIBRARY
Figure 11-1
A moving-coil mechanism
S
Soft-iron core
Pole face
11-2 The Ammeter
If the current in the coil increases, the electromagnetic force also increases,
moving the pointer farther across the scale. The deflection of the pointer is
directly proportional to the current, so the scale is linear.
If the current in the coil is reversed, the direction of its magnetic field
­reverses and the pointer hits a stop at the left end of the scale. Since reverse
deflection can damage some meters, it is important to ensure proper polarity on the meter connections.
Moving-coil movements cannot measure alternating current directly
since the alternating polarity makes the pointer vibrate back and forth.
However, a meter can include diodes connected such that current always
flows in the same direction through the coil. The pointer then shows the
­average magnitude of the alternating current.
A moving-coil meter movement has two key characteristics:
Sensitivity Sensitivity is usually given in terms of the current that causes
a full-scale deflection. This current is determined by the strength of the
permanent magnet and the number of turns on the moving coil. Most
moving-coil movements have a sensitivity of 1 mA or less.
Resistance The resistance of the movement comes from the resistance of
the wire in the coil. Generally, a more sensitive meter has more turns on the
moving coil and consequently a higher resistance.
See Review Questions 11-23 and 11-24 at the end of the chapter.
11-2 The Ammeter
Commercial movements with strong magnets, delicate springs, and lowfriction pivots can have a sensitivity of 20 μA or less. Using a stronger spring
and fewer turns with a thicker wire on the moving coil gives a ­sensitivity of
several milliamperes. For meters up to 8 cm in diameter, the maximum wire
size that can be conveniently used for the coil limits the current the meter can
pass without damage to about 30 mA. To measure larger currents with these
movements, we need to connect a shunt resistor in parallel.
Sensitive 50-μA movements are commonly used for measurements in electronic circuits while more rugged 5.0-mA movements are used with electric
power circuits. Usually a swamping or calibrating resistor is connected in
series with the moving coil to bring the total resistance of the movement up
to some convenient number, as shown in Figure 11-2. It is common practice to
select a calibrating resistor such that the full-scale c­ urrent produces a 50-mV
drop across the movement and its calibrating ­resistor. For a moving-coil
movement with a sensitivity of 1.0 mA and a r­ esistance of 27 Ω,
RT =
V
50 mV
=
= 50 Ω
I
1.0 mA
297
A diode has a high
­resistance for current
flowing through it
in one direction and
a low resistance for
­current flowing in the
opposite direction.
298
Chapter 11 Electrical Measurement
Moving coil
(27 Ω)
Calibrating
resistance
(23 Ω)
Shunt
Figure 11-2
Moving-coil ammeter with shunt resistor
Therefore, the calibrating resistance is
R c = 50 − 27 = 23 Ω
The calibrating resistor is made from precision resistance wire having a
small negative temperature coefficient to offset the positive temperature
coefficient of the moving coil. Since accurate meters require precision resistors at key points in their circuits, we calculate the resistances here to four
significant digits.
To use the 1.0-mA movement to measure a current of up to 1 A, we
arrange the meter circuit as shown in Figure 11-3, forming two parallel
branches with 999 times as much current flowing through the shunt resistor as through the meter movement. Since the shunt and the movement
are in parallel, the same voltage drop appears across both and their current
ratio is inversely proportional to their resistance ratio. Therefore,
Rsh =
and
Vsh = VM
Ish × Rsh = IM × RM
1
50
× RM =
= 50.05 mΩ
999
999
1.0-mA movingcoil movement and
calibrating resistor
(RT = 50 Ω)
V
Shunt
50.05 mΩ
+
E
−
Figure 11-3
Measuring current
Lamp
11-2 The Ammeter
If the current flowing through the lamp in Figure 11-3 is sufficient to
make the movement read full scale, the current in the moving-coil branch
is 1 mA. Therefore the current through the shunt resistor is 999 mA and the
current through the lamp is 1 mA + 999 mA = 1 A. As long as this movement and shunt are used together, we can interpret the meter scale as being
calibrated for 1 A full scale.
If the lamp current is 0.6 A, this total current splits so that one part
(0.6 mA) flows through the moving coil and 999 parts (0.5994 A) flows
through the shunt. Therefore, we can interpret the 0.6 scale marking of the
1-mA movement as indicating 0.6 A. For currents up to about 25 A, the
shunt is physically small enough to be placed inside the meter case. For
larger currents, external shunt resistors are required.
An ideal ammeter would have no internal resistance so that connecting
the meter would not reduce the current being measured. For most circuits,
connecting an ammeter has only a slight effect since we are adding just a
­little extra resistance to the circuit. For example, the total resistance of
the meter and its shunt in Figure 11-3 is 0.05 Ω, while the hot resistance
of the lamp is in the order of 200 Ω. Therefore, the change in circuit current due to adding the ammeter to the circuit is only 0.05/200, or 0.025%.
In practice, the main problem with making current measurements is that
we must place the ammeter in series with a circuit in order to measure
its ­current. ­Unless the meter is left in the circuit permanently, current
measurements ­require opening the circuit to connect the meter and then
restoring the original ­connections when we remove the meter. Because
an ammeter has low r­esistance, we must guard against connecting an
ammeter across a source.
To make a test meter that can measure a wide range of currents, we arrange the meter circuit so that we can easily change the shunt resistance,
as shown in Figure 11-4. The switch must be a make-before-break type so
that the meter is not damaged by passing the entire current as we change
the shunt.
V
Figure 11-4
Multirange shunt
See Problems 11-1 to 11-4 and Review Questions 11-25 to 11-29.
299
300
Chapter 11 Electrical Measurement
11-3 The Voltmeter
Moving-coil movements can also be used in voltmeters. If a 1.0-mA
­move­ment has a total resistance of 50 Ω, a voltage drop of V = IR = 0.0010 A
× 50 Ω = 50 mV appears across the movement when it is reading full scale.
If we connect a 99.95-kΩ resistor in series with the meter to bring the total
­resistance up to 100 kΩ, as in Figure 11-5, the voltage we must apply to produce a 1.0-mA current through the meter is
V = IR = 0.0010 A × 100 kΩ = 100 V
99.95 kΩ
Multiplier
Figure 11-5
1.0-mA movement (50 Ω)
V
Simple voltmeter
If we apply a 50-V potential difference to the voltmeter, the current through
it is
V
50 V
I=
=
= 0.50 mA
R
100 kΩ
Thus, the meter will read half scale. With the 99.95-kΩ multiplier resistor,
the scale on the 1.0-mA movement corresponds to 0–100 V.
To measure open-circuit voltage, a perfect voltmeter should draw no
current. Since current must flow in a moving-coil movement to obtain a
reading, a 50-μA movement is much more accurate than a 1.0-mA movement
for measurements in high-resistance circuits. The resistance of a standard
50-μA movement and its calibrating resistor is 1000 Ω. To obtain a 100-V
full-scale range, the total resistance must be
RT =
100 V
= 2.0 MΩ
50 µA
Therefore, the multiplier resistance must be 2000 kΩ − 1 kΩ = 1999 kΩ.
Similarly, the total resistance must be 10 MΩ for a 500-V scale and 20 kΩ
for a 1.0-V scale. For any full-scale voltage with a 50-μA movement, the
total resistance required is 20 kΩ for each volt of the full-scale reading.
Therefore, when used as a voltmeter, a 50-μA movement has a voltmeter
sensitivity of 20 kΩ/V. We can determine the total resistance of a movingcoil voltmeter by multiplying the full-scale voltage by the sensitivity of the
movement. For example, when used as a voltmeter, a 1.0-mA movement
requires a total resistance of 1000 Ω/V, so total resistance for a 500-V scale
is 500 V × 1000 Ω/V = 500 kΩ.
11-3 The Voltmeter
301
Example 11-1
The meter in Figure 11-6 uses a 1.0-mA moving-coil movement with a
resistance of 50 Ω, and has scales of 5 V, 50 V, 150 V, and 500 V. Calculate
RA, RB, RC, and RD.
V
RA
RB
5V
−
Figure 11-6
RC
50 V
RD
150 V
500 V
Multirange voltmeter
Solution
For the 5-V range,
RT =
5.0 V
= 5000 Ω and RA = 5000 Ω − 50 Ω = 4950 Ω
1.0 mA
Similarly,
for the 50-V range, RT = 50 kΩ and RB = 50 000 Ω − 5000 Ω = 45 kΩ
for the 150-V range, RT = 150 kΩ and RC = 150 kΩ − 50 kΩ = 100 kΩ
for the 500-V range, RT = 500 kΩ and RD = 500 kΩ − 150 kΩ = 350 kΩ
Multisim Solution
Download Multisim file EX11-1 from the website.
The circuit is the same as shown in Figure 11-6 with the calculated
values for RA, RB, RC, and RD. A 50-Ω resistor, Rm, represents the resistance of the moving-coil movement.
Insert a meter to measure the current through the movement. Run the
simulation, and check that the ammeter reading is 1 mA, the full-scale
current for the meter movement. This reading confirms that the value
of RA is correct.
Move the positive lead of the voltage supply to the connection between
RB and RC, and increase the voltage to 50 V. Run the simulation. Now a
meter reading of 1 mA confirms that the value of RB is correct.
Verify the calculations for RC and RD by running a simulation first with
a lead from a 150-V supply connected between RC and RD and then
with a lead from a 500-V supply connected to the right terminal of RD.
The current readings should be 1 mA.
circuitSIM
walkthrough
302
Chapter 11 Electrical Measurement
The selector switch and resistance network in a multirange meter
The voltmeter is a useful instrument for checking the operation of a circuit. With suitable clips on the test leads, we can connect a voltmeter across
any part of a circuit without having to disconnect any of the components
in the circuit.
See Problems 11-5 to 11-8 and Review Questions 11-30 to 11-35.
Circuit Check
CC 11-1.
CC 11-2.
CC 11-3.
A
Determine the sensitivity of a voltmeter that uses a 50-µA
meter movement.
Determine the total resistance for the 100-V scale range of a
voltmeter with a 10-mA moving-coil movement.
Find the total resistance for the 30-V scale of a meter that has a
sensitivity of 30 kΩ/V.
11-4 Voltmeter Loading Effect
Since a voltmeter is connected in parallel with the portion of the circuit in
which the voltage drop is measured, an ideal voltmeter should draw zero
current and should, therefore, have an infinitely high internal resistance.
However, the moving-coil type of voltmeter must draw some current in
order to deflect the pointer. The smaller the current required for full-scale
deflection, the more sensitive the voltmeter and the smaller the loading
­effect it has on the circuit being tested.
When measuring voltage in circuits with fairly large currents, an extra
5 mA does not alter the circuit conditions appreciably. For such circuits,
the more rugged and less sensitive voltmeters give adequate accuracy.
11-4 Voltmeter Loading Effect
Example 11-2
(a) Find the voltage drop across the 200-kΩ resistor in Figure 11-7.
(b) What will a 20-kΩ/V meter set to a 150-V scale read when connected across the 200-kΩ resistor?
(c) What will a 1000-Ω/V meter set to a 150-V scale read when connected across the 200-kΩ resistor?
100 kΩ
+
200 V
−
200 kΩ
Figure 11-7
V
Effect of voltmeter sensitivity
Solution
(a) From the voltage-divider principle,
V=
200 kΩ
× 200 V = 133.3 V
300 kΩ
(b) A 20-kΩ/V meter set to a 150-V range has a total resistance of
20 kΩ × 150 = 3.0 MΩ. This resistance is in parallel with the 200-kΩ
resistor, thus ­reducing the resistance of the parallel combination to
R=
Therefore,
200 kΩ × 3000 kΩ
= 187.5 kΩ
200 kΩ + 3000 kΩ
V=
187.5 kΩ
× 200 V = 130.4 V
287.5 kΩ
R=
200 kΩ × 150 kΩ
= 85.7 kΩ
200 kΩ + 150 kΩ
(c) A 1000-Ω/V meter set to a 150-V range has a total resistance of
1000 Ω × 150 = 150 kΩ. Therefore, the 200-kΩ resistor and the
voltmeter in parallel have an equivalent resistance of
and
V=
85.7 kΩ
× 200 V = 92.3 V
185.7 kΩ
303
304
Chapter 11 Electrical Measurement
Multisim Solution
Download Multisim file EX11-2 from the website. The circuit is the
same as shown in Figure 11-7. Connect a ground to the bottom node of
the schematic.
circuitSIM
walkthrough
(a) Double-click on the voltmeter and view the value of the voltmeter
resistance. Set it to a value of at least 1 GΩ.
Run the simulation. The voltmeter reading of 133.3 V matches the
value calculated using the voltage divider principle.
(b) Set the simulated voltmeter resistance to 3.0 MΩ, the total resistance
for a 150-V scale on an actual 20-kΩ/V meter.
Run the simulation. The voltmeter now reads 130.4 V.
(c) Set the voltmeter resistance to 150 kΩ, the total resistance for a 150-V
scale on a 1000-Ω/V meter.
Run the simulation. The voltmeter now reads 92.3 V.
However, in electronic circuits with small currents, the resistance of a
­voltmeter can alter circuit conditions considerably. Even the 20-kΩ/V
meter in part (b) of Example 11-2 introduced an error in excess of 2%.
Consequently, the simple moving-coil type of voltmeter is not suitable
for voltage ­measurement in high-resistance circuits. However, such circuits can be measured accurately with an electronic voltmeter, like those
described in Section 11-7. These meters have internal resistances much
higher than those of moving-coil meters.
See Problems 11-9 to 11-11 and Review Questions 11-36 and 11-37.
11-5 Resistance Measurement
We can determine an unknown resistance in Figure 11-8 by applying
Ohm’s law to the readings obtained from the voltmeter and ammeter. This
method requires that the unknown resistance be connected into a special
circuit with two separate meters. This method is useful for measuring very
low resistances, such as those of motors, and for measuring the resistance
of a component while it carries its normal operating current.
Current-adjusting
resistor
A
+
E
V
Rx
−
Figure 11-8
Measuring resistance with a voltmeter and an ammeter
11-5 Resistance Measurement
With the voltmeter connected as shown in Figure 11-8, the ammeter
indicates the sum of the currents through Rx and the voltmeter. Thus,
the V/I ratio calculated from the meter readings equals the equivalent
resistance of the two parallel branches. This ratio is less than the actual
value of Rx unless the voltmeter current is negligible compared to the
current through Rx. We can check for voltmeter loading by watching
the ammeter reading as we disconnect the voltmeter. If there is any noticeable decrease in the current, we then connect the voltmeter on the
source side of the ammeter. The a­ mmeter now reads only the current
through Rx while the voltmeter now shows the sum of the voltage drops
across the ammeter and Rx. However, if Rx is large enough to show a
noticeable voltmeter loading effect, the voltage drop across the
ammeter is insignificant compared to Vx and we can ­ignore the ammeter
loading effect.
For quick checks of circuit resistance, we can use an ohmmeter, a meter
designed to measure resistance directly. The simple ohmmeter shown in
­Figure 11-9(a) consists of a 1.0-mA movement, a 4.5-V battery, and a resistance that passes a 1.0-mA current when we short-circuit the ohmmeter
­terminals. A portion of the total resistance is adjustable so that we can
­calibrate the meter to read exactly full scale when we connect the two test
leads together. In this example, the total resistance, including the meter
movement, is
RT =
E
4.5 V
=
= 4.5 kΩ
I
1.0 mA
As shown in Figure 11-9(b), the scale of an ohmmeter is nonlinear. The
nonlinear scale makes low resistances easier to read accurately, but high
­resistances are crammed together at the left end of the scale. For circuit
testing where we need a high-range ohmmeter, we could use a 50-μA
1-mA
movement
V
+
4.5 V
−
40.5 k
13.5 k 4.5 k
Unknown
resistor
Zero
adjust
Figure 11-9
0
∞
(a)
OHMS
(b)
Simple ohmmeter
1.5 k
305
306
Chapter 11 Electrical Measurement
Example 11-3
Determine the resistance values that should be marked at full scale,
centre scale, one-quarter of full scale, and one-tenth of full scale for the
ohmmeter in Figure 11-9.
Solution
As we have already noted, the total internal resistance of the ohmmeter
is adjusted so that the meter reads exactly full scale when the test leads
are short-circuited. Therefore, the end mark on the scale represents 0 Ω.
When the meter reads half scale, the current is 0.5 mA and the total
resistance in the series loop is double the total ohmmeter resistance.
Therefore, a centre-scale reading represents
Rx = RM =
4.5 V
= 4.5 kΩ
1.0 mA
For one-quarter of full scale, the current is 0.25 mA. Then, the total
resistance in the loop is
RT =
4.5 V
= 18 kΩ
0.25 mA
Rx = 18 kΩ − 4.5 kΩ = 13.5 kΩ
and
Similarly, one-tenth of full scale represents
Rx =
4.5 V
− 4.5 kΩ = 40.5 kΩ
0.10 mA
1.0-mA movement
(50 Ω)
V
0.505-Ω
shunt
movement rather than a 1.0-mA movement. With a 50-μA movement the
total i­ nternal resistance of the ohmmeter in Figure 11-9(a) is
4.5 V
44.5 Ω
Zero adjust
Figure 11-10
Low-range ohmmeter
RT =
E
4.5 V
=
= 90 kΩ
I
50 µA
The centre-scale reading of the meter will also be 90 kΩ.
These examples demonstrate that the resistance represented by a
centre-scale reading is inversely proportional to the full-scale current
for the meter. Therefore, we can convert the basic ohmmeter into a lowrange ohmmeter by placing a shunt across the moving coil, as shown in
Figure 11-10.
11-5 Resistance Measurement
307
Example 11-4
Using a 1.0-mA movement with an internal resistance of 50 Ω and a
4.5-V battery, design an ohmmeter that reads 45 Ω at centre scale.
Solution
For a centre-scale reading, Rx = RM, so the total internal resistance of
the ohmmeter is 45 Ω. Therefore, full-scale current is
I=
E
4.5 V
=
= 0.10 A
R
45 Ω
Since the meter movement passes 1.0 mA at full scale, the shunt current
must be 99 mA. The shunt resistance is
Rsh =
1 mA
× 50 Ω = 0.505 Ω
99 mA
The equivalent resistance of the meter movement and shunt in
­parallel is
Req =
50 Ω × 0.505 Ω
= 0.500 Ω
50 Ω + 0.505 Ω
Therefore, the total resistance of the series resistor and the “ohms
adjust” rheostat is
RS = 45 Ω − 0.5 Ω = 44.5 Ω
Multisim Solution
Download Multisim file EX11-4 from the website.
The circuit is the same as shown in Figure 11-10, with a 50-Ω resistor
representing the resistance of the moving-coil movement.
Place a short across the ohmmeter terminals. Insert a meter to measure
the current flowing through the short. Connect a ground to the bottom
left terminal of the circuit.
Run the simulation, and check that the ammeter shows 100 mA.
This reading confirms the calculation of the full-scale current for the
ohmmeter.
Replace the short across the ohmmeter terminals with a 45-Ω resistor.
Run the simulation, and check that the ammeter shows 50 mA, which
is the half-scale current for the ohmmeter. This reading verifies that the
ohmmmeter design is correct.
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walkthrough
308
Chapter 11 Electrical Measurement
As with a voltmeter, we can readily connect an ohmmeter across a portion
of a circuit. However, we must make sure that the circuit being measured
is switched off. Current from sources in the circuit can cause an inaccurate
reading or damage the ohmmeter.
When we wish to measure a very high resistance, such as the leakage
­between the two conductors of a cable as in Figure 11-11, we use a voltmeter and a separate voltage source. This system has the added advantage
of testing the resistance with the normal operating voltage applied.
V
+
120 V
−
Figure 11-11
Two-conductor cable
Using a voltmeter as a high-resistance ohmmeter
Example 11-5
The voltmeter in Figure 11-11 has a 20-kΩ/ V movement. The meter
reads 120 V on its 150-V scale when the switch is closed, and 10 V
when the switch is open. Calculate the leakage resistance of the cable
­insulation.
Solution
The resistance of the voltmeter is
RM = 20 kΩ/ V × 150 V = 3.0 MΩ
The voltage drop across the leakage resistance of the cable is
Vleak = 120 V − 10 V = 110 V
Since the resistance of the voltmeter and the leakage resistance of the
cable form a simple series circuit,
Rleak
Vleak
=
RM
VM
Rleak = 3.0 MΩ ×
110 V
= 33 MΩ
10 V
11-5 Resistance Measurement
309
For precision resistance measurements, we can use a Wheatstone bridge
as shown in Figure 11-12. If we adjust Ry such that there is no deflection
of the galvanometer G when we close the switch, the voltage drops across
Rx and RA must be exactly the same since a potential difference across the
galvanometer would cause current to flow through the galvanometer. With
no current through the meter movement, Ix = Iy. Hence,
Galvanometer is a term
for a sensitive ammeter
that can measure
­current flowing
through it in either
­direction.
Vx = IxRx and Ix =
Vx =
Therefore,
VA =
Similarly,
E
Rx + Ry
ERx
Rx + Ry
ERA
RA + RB
Rx
RA
G
+
E
−
RB
Figure 11-12
Ry
Measuring resistance with a Wheatstone bridge
Therefore, for perfect balance,
ERx
ERA
=
Rx + Ry
RA + RB
RxRA + RxRB = RxRA + RARy
RxRB = RyRA
and
Rx =
RARy
RB
(11-1)
For the bridge to be balanced, the product of the resistances in one pair
of opposite arms of the bridge must equal the product of the resistances in
other pair of opposite arms. Since E does not appear in Equation 11-1, the
magnitude of the source voltage used with a bridge circuit has no effect on
the accuracy of the measurement. The source merely causes a deflection of
310
Chapter 11 Electrical Measurement
the galvanometer pointer if the bridge is not properly balanced. With precision resistors for RA, RB, and Ry, we can use Equation 11-1 to determine an
accurate value for Rx.
Example 11-6
The Wheatstone bridge circuit of Figure 11-12 is balanced when RA =
1.0 Ω, RB = 50 Ω, and Ry = 17 Ω. Calculate Rx.
Solution
Rx =
circuitSIM
walkthrough
RARy
RB
=
1.0 Ω × 17 Ω
= 0.34 Ω
50 Ω
Multisim Solution
Download Multisim file EX11-6 from the website.
The circuit is a Wheatstone bridge as shown in Figure 11-12, with the
given values for RA, RB, and Ry, and the calculated value for Rx.
Set E = 10 V. Connect meters to measure the voltages across RA and Rx.
Connect a ground to the bottom node of the circuit.
Run the simulation, and note that voltage readings show that
VA = Vx. These readings verify that the bridge is balanced.
See Problems 11-12 to 11-22 and Review Questions 11-38 to 11-44.
Circuit Check
CC 11-4.
B
Find the voltmeter reading when the terminal voltage of a 12-V
source with an internal resistance of 100 kΩ is checked with a
20-kΩ/V voltmeter set to its 50-V range.
CC 11-5. Design a simple series ohmmeter that uses a 50-μA, 300-Ω
movement, and a 1.5-V battery. Calibrate the meter for the
quarter, half, and three-quarter scale positions. What deflection will show on the meter if an external resistance of 50 Ω is
­connected?
CC 11-6. In a commercial Wheatstone bridge, the values of RA and RB
are normally set as a variable ratio for determining low or high
value resistances. Given that the RA/RB ratio is 0.01 and the
bridge balances with an external resistance of 0.37 Ω,
determine the value of the variable resistance.
11-6 The Electrodynamometer Movement
11-6 The Electrodynamometer Movement
We can replace the permanent magnet in a meter movement with an electromagnet consisting of two stationary coils, as shown in Figure 11-13. Such
electrodynamometer movements can measure alternating currents if the
stationary and moving coils are connected in series. Whenever the current in the moving coil reverses direction, the magnetic field produced by
the stationary coil also reverses. Therefore, the magnetic force moves the
pointer clockwise, regardless of the current direction in the coils.
Spring
Pointer
Moving
coil
Stationary
coils
Figure 11-13
Electrodynamometer movement
The magnetic force acting on the moving coil is proportional to the product
of the current in the moving coil and strength of the magnetic field from stationary coils. Since the magnetic field strength for a coil is proportional to the
current through the coil, the reading of an electrodynamometer movement is
proportional to the product of the stationary-coil and moving-coil currents.
If the stationary coils are connected in series with a load as in Figure 11-14,
the stationary-coil current is the same as the load current. If the moving
coil in series with a multiplier resistor is then connected across the load, the
moving-coil current is directly proportional to the load voltage. Therefore,
the scale reading of the electrodynamometer movement is proportional to
the product of the load current and load voltage. Since P = IV, the scale
shows the power in the load. Thus, we can use an electrodynamometer as
a wattmeter in both DC and AC circuits.
±
+
Load
E
−
Multiplier
±
Figure 11-14
Electrodynamometer movement as a wattmeter
311
312
Chapter 11 Electrical Measurement
If we reverse the leads to either the current coil or the voltage coil when
we connect a wattmeter into a circuit, the pointer is deflected counterclockwise instead of clockwise. Wattmeters usually identify one end of each coil
with a ± mark, as shown in Figure 11-14. The marked terminal of the current coil (the stationary coil) should be connected to the voltage source and
the marked terminal of the voltage coil (the moving coil) should be connected to the line containing the current coil.
See Review Question 11-45.
11-7 Multimeters
Source: © iStock.com/EricFerguson
The most common electrical test instrument is the volt-ohm-milliammeter
(VOM), or multimeter. Analog multimeters have a single moving-coil movement with a multiposition switch for selecting various DC current ranges,
DC and AC voltage ranges, and resistance ranges. The multimeter’s dial has
a scale for each range. All these scales can be difficult to read accurately.
An analog multimeter
A digital instrument, on the other hand, displays data as discrete numbers. Since improved technology has reduced the size and cost of integrated
circuits, digital multimeters (DMMs) have virtually replaced analog VOMs.
DMMs have high input resistances (typically 20 MΩ), greater accuracy than
moving-coil meters, and clear numerical readouts. DMMs are generally
cheaper and more rugged than equivalent analog multimeters. In a­ ddition,
many DMMs have extra features such as functions for testing diodes and
transistors. The photographs show two types of digital multimeters.
Source: Photo © AEMC Instruments, a leader in electrical test and
measurement instruments
11-7 Multimeters
Source: Copyright © Tektronix.
Reprinted with permission. All
Rights Reserved.
Handlheld digital multimeter
Benchtop digital multimeter
Most electronic meters use a field-effect transistor (FET) to give them a very
high input resistance, as shown in Figure 11-15. The voltage between the
gate and source terminals of the FET controls the current from the source
to the drain. Since the resistance between the gate and the source is extremely high, the FET draws only a tiny current from the circuit connected
to the input terminals of the meter. Almost all of the current that goes to the
measurement circuit comes from the meter’s battery.
Gate
−
FET
Drain Measurement
circuit
Source
20 MΩ
+
Figure 11-15
Input stage of an electronic meter
313
314
Chapter 11 Electrical Measurement
The measurement circuit inside a digital multimeter can vary between
manufacturers but they typically follow similar logic principles.
An example measurement process that converts an analog input into a
digital output is shown in Figure 11-16. The red and black terminal leads of
a DMM are attached to an analog input, such as a DC voltage. Many samples
of the analog signal will be read by the circuitry in a very short amount of
time (100 000 samples per second or more) but only a few of the samples
will be saved into a buffer and averaged into a single, steady reading. This
method of acquiring a steady reading can help to reduce fluctuations from
noise and other small variations, thus improving the overall accuracy of
the reading.
The analog reading is sent to a comparator to determine if it is above or
below the last reading. (If it is the first reading, it usually compares itself
to half the maximum scale value on the DMM.) The output of the comparator drives either a high or a low signal into the Successive Approximation
Register (SAR), which establishes the most significant digit of the reading.
The digital output of the SAR then feeds back into the comparator through
a digital-to-analog converter (DAC), where it is compared again so the next
digit can be established, and so on. This iterative process continues until
all the significant digits of the analog signal have been identified, perhaps
requiring only a few microseconds to complete. After the analog input has
transformed into a digital output, the DMM can then display the numerical
value on the meter’s digital screen.
Analog
input
Sample
acquisition
Comparator
High or low
output
SAR
Digital
output
DAC
Figure 11-16 A DMM’s measurement circuit processes an analog input into a
digital output
See Review Question 11-46.
315
Problems
Summary
• Since the deflection of the pointer in a moving-coil movement is proportional to the current through the coil, the movement can be used as the
measuring device in an ammeter.
• A voltmeter with a number of voltage ranges can be constructed from a
moving-coil movement by placing multiplier resistors in series with it.
• An analog voltmeter can reduce the voltage being measured.
• An ohmmeter with a number of resistance ranges can be constructed using
a moving-coil movement, a battery, shunt resistors, and series ­resistors.
• The Wheatstone bridge can be used for accurate measurements of
­resistances.
• The electrodynamometer movement is a practical means of measuring
power in both DC and AC circuits.
• Digital multimeters typically have higher input resistances and greater
accuracy than moving-coil meters.
Problems
B
Section 11-2
11-1.
B
11-2.
B
11-3.
I
11-4.
B
B
The full-scale voltage drop across a 5.0-mA movement is 75 mV.
­ alculate the shunt resistance required to use the movement to
C
measure currents up to 1.0 A.
If the shunt resistance calculated in Problem 11-1 is used with a
1.0-mA movement having a total resistance of 50 Ω, what current
would its full-scale reading represent?
Design an ammeter with a range of 50 mA from a 250-μA, 500-Ω
movement.
A milliammeter has a range of 20 mA and a shunt resistor of 2.63 Ω. If
the movement has a full-scale sensitivity of 1 mA, what is its r­ esistance?
Section 11-3
11-5.
11-6
B
11-7.
B
11-8.
The Ammeter
B = beginner
I = intermediate
A = advanced
The Voltmeter
Calculate the multiplier resistors required for 15-V, 50-V, 150-V, and
500-V ranges in a voltmeter using a 50-μA, 1000-Ω meter movement.
(a) The resistance of the movement for a 200 Ω/V voltmeter is
10. Calculate the multiplier resistors required for 30-V, 150-V, and
1500‑V ranges.
(b) Use Multisim to verify the resistor values calculated in part (a).
A voltmeter has a 2000-Ω, 75-μA movement and multiplier resistors of 38 kΩ, 160 kΩ, and 600 kΩ. What are the voltage ranges for
this voltmeter?
Compare your answers to Problems 11-5 and 11-7. How does using
a less sensitive meter movement affect the multiplier resistors
required?
circuitSIM
walkthrough
316
Chapter 11 Electrical Measurement
I
Section 11-4
11-9.
Voltmeter Loading Effect
A voltmeter is constructed by using a 50-μA, 1000-Ω movement. It is
used on the 20-V range to measure the voltage across R2 in the circuit
shown in Figure 11-17. Find the percent error caused by the loading
effect of the voltmeter.
100 kΩ
R1
+
R2
10 V
250 kΩ
−
Figure 11-17
I
circuitSIM
walkthrough
11-10. (a) Resistor R2 in the voltmeter shown in Figure 11-18 has burned
out, and its original value is not known. R1 and R3 measure 122 kΩ
and 375 kΩ, respectively. What value of resistance for R2 is required
to repair the voltmeter?
(b) Use Multisim to verify the resistance calculated in part (a).
V
R1
50 V
−
Figure 11-18
I
circuitSIM
walkthrough
circuitSIM
walkthrough
B
B
B
I
R2
R3
100 V
250 V
Multirange voltmeter
11-11. Using the data from Problem 11-10, calculate the meter resistance,
full-scale current, and ohms per volt rating for the voltmeter shown
in Figure 11-18.
Section 11-5
Resistance Measurement
11-12. (a) What resistance corresponds to the centre-scale mark on a basic
ohmmeter constructed with a 50-μA movement and a 1.5-V battery?
(b) Use Multisim to verify the resistance calculated in part (a).
11-13. Find the battery voltage required for an ohmmeter with a 1.0-mA
movement to use the same scale as the meter in Problem 11-12.
11-14. (a) What resistance must be connected to the terminals of the ohm­meter
in Problem 11-12 for the pointer to read 20% of full-scale d
­ eflection?
(b) Use Multisim to verify the resistance calculated in part (a).
11-15. Design an ohmmeter with 500 Ω at centre scale, using a 2.5-V battery and a 50-μA, 500-Ω movement.
Problems
I
B
B
B
I
B
11-16. The insulation resistance of a solenoid is checked by connecting
one terminal of a 1.0-kV source to the solenoid and connecting a
20-kΩ/V voltmeter between the other terminal and the iron core
of the solenoid. Calculate the insulation resistance given that the
meter reads 360 V when set to 1500-V range.
11-17. What would the meter read in Problem 11-16 if the leakage
resistance between the solenoid winding and the core dropped to
0.5 MΩ?
11-18. (a) The Wheatstone bridge in Figure 11-12 balances when RA is
5.0 kΩ, RB is 1000 Ω, and Ry is 42 kΩ. Calculate Rx.
(b) Use Multisim to verify the resistance calculated in part (a).
11-19. Find the value of Ry that balances the Wheatstone bridge in
Figure 11-12 when RA is 1200 Ω, RB is 1500 Ω, and Rx is 500 Ω.
11-20. The Wheatstone bridge in Figure 11-12 balances when VA is 2.4 V,
IB is 0.24 A, and the total current drawn from the battery is 300 mA.
Calculate Rx.
11-21. Figure 11-19 shows a variation of the Wheatstone bridge known
as the slide-wire bridge. The slide wire is a length of uniform
­resistance wire 1.0 m long. The standard resistor, RS, is 37 Ω. Find
Rx given that the bridge balances when the slider is 45 cm from the
end connected to the standard resistor.
Rx
G
RS
Slider
Resistance
wire
1 metre
Figure 11-19
I
Slide-wire bridge
11-22. Figure 11-20 shows a variation of the Wheatstone bridge called
the Varley loop, used for ­locating ground faults on long transmission lines. To trace the fault, the far ends of the two identical
conductors are connected together. RA and RB have fixed values
of 100 Ω and 1000 Ω, respectively. When the switch is in the line
(L) position, the bridge balances with RC = 200 Ω. When the switch
is thrown to the ground (G) position, the bridge balances with
RC = 150 Ω. How far is the ground fault from the end of the cable
at which the apparatus is located? Assume negligible resistance
317
circuitSIM
walkthrough
318
Chapter 11 Electrical Measurement
between the ground connection at the switch and the fault ground
on the transmission line.
RA
20 km
G
Transmission line
RB
RC
Figure 11-20
L
Fault
ground
G
Varley loop
Review Questions
Section 11-1
Moving-Coil Meters
11-23. Describe how current passing through the coil in Figure 11-1 affects
the pointer.
11-24. How would leaving out the soft-iron core affect the operation of a
moving-coil movement in Figure 11-1?
Section 11-2
The Ammeter
11-25. What is the purpose of the calibrating resistor used with movingcoil movements?
11-26. What is the advantage of arranging that moving-coil movements
have a 50-mV drop across their terminals when full-scale current
flows through them? (Refer to Problem 11-2.)
11-27. Show that
IM
RS = RM
IT − IM
where RS is the resistance of an ammeter shunt, RM is the resistance
of the moving-coil movement with its calibrating resistor, IM is the
coil current required for full-scale deflection, and IT is the desired
full-scale ammeter reading.
11-28. What undesired effect could result from connecting an external
shunt to an ammeter as shown in Figure 11-21(a)?
(
)
Moving-coil movement
Shunt
E
Moving-coil
movement
(a) Wrong
Figure 11-21
Shunt
Load
Load
E
(b) Right
Connecting an external ammeter shunt
11-29. What is the effect of connecting an ammeter across a load?
319
Integrate the Concepts
Section 11-3
The Voltmeter
11-30. The basic moving-coil movement is a current-indicating device.
How is it possible to calibrate it as a voltmeter?
11-31. What is the significance of voltmeter sensitivity?
11-32. Why is a 20-kΩ/V meter more suitable than a 1000-Ω/V meter for
checking electronic circuitry?
11-33. Why is a 200-Ω/V meter satisfactory for use with electric machinery?
11-34. What would be the effect of connecting a voltmeter in series with a load?
11-35. What circuit information could be obtained from the voltmeter
reading in Question 11-34?
Section 11-4
Voltmeter Loading Effect
11-36. Why is voltmeter loading more of a problem than ammeter resistance in practical measurements?
11-37. Why is the loading effect of a DMM considerably less than that of a
20-kΩ meter?
Section 11-5
Resistance Measurement
11-38. Why is a galvanometer used in the circuit of Figure 11-12 rather
than a microammeter?
11-39. Draw a circuit diagram showing how to use a voltmeter and an
ammeter to determine a resistance. Discuss the possible errors that
might occur when measuring a 50-kΩ resistor with this method.
11-40. What precaution must be observed when checking the resistance of
a circuit with an ohmmeter?
11-41. Lay out a scale for a simple ohmmeter that uses a 50-μA movement
and a 1.5-V battery.
11-42. Lay out a scale for the low-resistance ohmmeter shown in Figure 11-22.
11-43. Would you select a 50-μA movement or a 5-mA movement for the
Wheatstone bridge circuit of Figure 11-12? Explain your choice.
11-44. How does the accuracy of the galvanometer scale calibration
affect the accuracy of the resistance measurements made with a
Wheatstone bridge?
Section 11-6
The Electrodynamometer Movement
11-45. Why is it possible to use an electrodynamometer wattmeter in both
DC and AC circuits?
Section 11-7
Multimeters
11-46. List the advantages of the DMM over an analog VOM. Describe a
­situation where an analog instrument might be more convenient.
Integrate the Concepts
Design a VOM that will accurately measure
(a) direct current up to 50 mA
(b) DC voltages up to 40 V
(c) 100-Ω resistors
Ohms adjust
1.5 V
V
1.0-mA
movement
(50 Ω)
Figure 11-22
Low-resistance
ohmmeter
320
Chapter 11 Electrical Measurement
Practice Quiz
1.
What device can enable a moving-coil movement to measure alternating current?
(a) transistor
(b) resistor
(c) diode
(d) fuse
2.
List two key characteristics of a moving-coil meter movement.
3.
Why would an ideal ammeter have zero internal resistance?
4.
Which of the following statements is true?
(a) A galvanometer is a sensitive ammeter that can measure current
flowing through it in only one direction.
(b) An electrodynamometer is an instrument that can measure electric power in any DC and AC circuit.
PART
III
Capacitance and Inductance
In all the circuits we have considered so far, the passive components have only
one basic property: resistance. Part III introduces the other two basic properties of circuits: capacitance and inductance. While resistance is opposition to the
flow of current in a circuit, inductance is ­opposition to any change in the current.
Similarly, capacitance is opposition to any change in the voltage between two
points in a circuit.
12
Capacitance
13
Capacitance in DC Circuits
14
Magnetism
15
Magnetic Circuits
16
Inductance
17
Inductance in DC Circuits
Photo source: © iStock.com/omada
12
Capacitance
Suppose we have two metal plates that are close together
but insulated from each other. If we connect one plate to each
terminal of a voltage source, electrons are removed from one
plate and deposited on the other. The first plate now has a net
positive charge while the second has a net negative charge. If we
remove the plates from the circuit and leave them isolated from
each other, they will retain their net charge indefinitely. In this
chapter we shall investigate how these plates—and other forms of
capacitors—store charge.
Chapter Outline
12-1
Electric Fields
324
12-3
Capacitance
328
12-5
Factors Governing Capacitance 333
12-2
12-4
12-6
12-7
12-8
Dielectrics
Capacitors
327
330
Dielectric Constant
336
Capacitors in Parallel
338
Capacitors in Series 338
Key Terms
electric field 324
strength (intensity)
of an electric field 324
vector 324
electric lines of force
(electric flux lines) 324
electric flux 325
electric flux density 325
voltage gradient 326
dielectric 327
polarized 327
dielectric strength 327
dielectric absorption 327
capacitor 328
capacitance 329
farad 329
electrolytic capacitor 331
working voltage 332
permittivity (absolute
permittivity) 334
permittivity of
free space 335
relative permittivity
(dielectric constant) 336
ferroelectric
dielectric 337
electrostatic
induction 338
elastance 340
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• describe the nature of an electric field
• explain the relationships among the strength
of an electric field, electric flux, electric flux
density, and potential ­difference
• explain how the electric charges in a
dielectric behave in the presence of the
electric field
• describe the construction of various types of
capacitors
• define capacitance in terms of the charge
on the plates of a capacitor and the voltage
across it
Photo sources: © sciencephotos/Alamy Stock Photo
• state the effect that the area of the plates,
the spacing of the plates, and the type of
dielectric have on the capacitance of a
capacitor
• define dielectric constant
• calculate the capacitance of a capacitor
given the dielectric and the area and spacing
of the plates
• calculate the total capacitance of capacitors
in parallel
• calculate the equivalent capacitance of
capacitors in series
324
Chapter 12 Capacitance
12-1 Electric Fields
In Section 2-1, we noted that Coulomb developed an equation to describe
the force that any two electric charges exert on each other:
F=k
Q1Q2
d2
(2-1)
where F is the magnitude of the force in newtons, Q1 and Q2 are the two
charges in coulombs, d is the distance between the charges in metres,
and k is Coulomb’s constant, 8.99 × 109 N · m2/C2.
We can think of this force as resulting from an electric field.
Some books use the
script letter ℰ for
­electric field strength
to avoid confusion
with the letter symbol
for EMF, E.
An electric field is that region in which an electric charge is acted
upon by an electric force.
The strength (or intensity) of an electric field is the force that the
field exerts on a unit of charge.
The letter symbol for electric field strength is the Greek letter E
(capital epsilon).
The units for electric field strength are newtons per coulomb, which
are equivalent to volts per metre.
Thus,
E=
F
Q
(12-1)
where E is the strength of the electric field in newtons per coulomb, F is
the force in newtons, and Q is the charge in coulombs.
Vectors are sometimes
indicated by putting
an arrow over the
­letter representing
the variable. Thus, E
→
could be written as E.
This alternative notation can prevent confusion in handwritten
material.
A quantity that has direction as well as magnitude is called a vector.
Such quantities are set in boldface, italic type. To refer to just the magnitude of a vector, we use the same letter symbol in lightface, italic type. For
example, to fully specify a 10-N force directed upward, we write F = 10 N
[up], but if we are concerned only with the magnitude of the force, we can
write just F = 10 N. Electric field strength and many of the other quantities
relating to forces and fields are vectors. Section 20-4 describes vectors and
other types of variables in more detail.
In diagrams, we represent an electric field with electric lines of force (or
electric flux lines), which show the path that a tiny positive charge would
follow because of the force exerted by the electric field. Figure 12-1 shows
the lines of force around positive (a) and negative (b) point charges. The
closer ­together the lines of force are, the greater the strength of the electric
field. Note that the lines of force are directed away from the positive charge
in (a), and directed toward the negative charge in (b). At every point, the
electric field has both a magnitude and a direction.
Figure 12-2 shows the electric fields around two parallel conductors (a)
and between two concentric (or coaxial) conductors (b). These conductor
12-1 Electric Fields
+
−
(a)
(b)
Figure 12-1
325
Electric fields surrounding point electric charges
+
+
−
(a)
Figure 12-2
+
−
+
+
(b)
Electric field between (a) parallel conductors and (b) coaxial conductors
c­onfigurations have numerous applications in telecommunications,
including radio, television, and computer networks.
The electric field passing through a given surface is called electric flux,
Ψ, and the flux per unit area is called the electric flux density, D. (Ψ is the
Greek uppercase letter psi.) The greater the charge producing the electric
field, the greater the flux. For example, doubling the charge in Figure 12-1(a)
doubles the flux through a plane beside the charge. Similarly, increasing
the charge by a factor of 10 increases the flux by the same factor. Thus, we
can define electric flux and charge as equivalent:
ψ≡Q
(12-2)
If an electric field is uniform over a given area, A, then the flux density
for that area is simply
D=
ψ
Q
=
A
A
(12-3)
In SI, the units for electric flux are coulombs and the units for flux density are coulombs per square metre.
When two parallel conducting plates are connected to a voltage source,
as shown in Figure 12-3, electrons flow into the plate connected to the
negative terminal and an equal number of electrons flow out of the plate
connected to the positive terminal. On each plate, the force of repulsion
between like charges causes the charge to be distributed evenly, except at
the edges. As a result, the electric field between the parallel conducting
plates is uniform. Therefore, a constant force is exerted on a charged particle as it moves from one plate to the other. The decrease in repulsive force
from the charge on the plate that the particle leaves is balanced by the
increase in attractive force from the charge on the plate that the particle
is approaching.
One of Maxwell’s
equations, also
known as Gauss’s
Law, describes how
an electric field
behaves around an
electric charge: If an
electric charge exists
at a point in space, as
in Figure 12-1, then
the measure of the
electric flux entering
or exiting a surface
surrounding that
point is non-zero.
Because of the attraction between unlike
charges, the charges
are concentrated at
the surface of each
plate facing the other
plate. Therefore, we
speak of the charge
on the plate rather
than the charge in
the plate.
326
Chapter 12 Capacitance
Uniform voltage
gradient
+
−
+
−
+
−
+
−
Figure 12-3
Electric field between parallel plates
The work done on the charge when it moves from one plate to the other is
W = Fd
(2-3)
W = QV
(2-4)
This work can also be expressed in terms of the potential difference between
the plates
Therefore,
Fd = QV
Substituting for F from Equation 12-1 gives
EQd = QV
and
E=
V
d
(12-4)
where E is the magnitude of the electric field strength in newtons per
coulomb or volts per metre, V is the potential difference between the
parallel plates in volts, and d is the distance between the plates in metres.
This expression for electric field intensity leads to an interesting characteristic of electric fields. According to Kirchhoff’s voltage law, when we
apply 400 V to the two parallel plates of Figure 12-3, there must be a 400-V
potential difference between them. Since the electric lines of force begin at
the positive plate and end at the negative plate, this potential difference
appears across each line of force. Because the electric field between parallel
plates is uniform, Equation 12-4 shows that the voltage gradient between
the plates is also uniform. In other words, the potential difference is distributed evenly along the length of the electric line of force, somewhat like the
way voltage is distributed along a linear potentiometer.
Later in this chapter we shall see how the behaviour of charge on parallel conducting plates is a key to understanding capacitance.
See Problems 12-1 to 12-6 and Review Questions 12-27 to 12-30.
12-2 Dielectrics
12-2 Dielectrics
Figure 12-4 shows an electrical insulator, or dielectric, placed between
the parallel charged plates. Although tightly bound to the molecules in the
­dielectric, the electrons orbiting each nucleus are attracted by the positive
plate and repelled by the negative plate. As a result, the orbits of the electrons in each atom are displaced toward the positive plate and are no longer
centred on the nucleus. Thus, the atoms in the dielectric become polarized.
Nucleus
+
−
Electron
orbit
+
−
+
−
+
−
+
−
+
−
+
−
Dielectric
Figure 12-4 Effect of an electric field on a dielectric (The displaced orbit is
not drawn to scale.)
The extent of the polarization depends on the strength of the electric field.
If we keep increasing the potential difference between the two parallel plates,
we increase the electric field strength to the point where the ­resulting force
tears the outer electrons free of their orbits, causing the ­dielectric to break
down. The dielectric then becomes conductive and short-circuits the plates.
The field strength required to break down a dielectric is called its d
­ ielectric
strength. Some dielectrics are able to withstand a much stronger electric field
than others. Table 12-1 lists typical values for the dielectric strength of some
common dielectrics. The dielectric strength of a material sometimes depends
on the manufacturing process and can vary considerably.
After exposing a dielectric to a strong electric field between two parallel plates, we could remove the voltage source and temporarily short the
plates together. We would expect to completely neutralize the charge on
the plates. However, with some dielectrics there is still a small potential
difference between the two plates, indicating that the electron orbits in
the dielectric did not return fully to their original positions. The dielectric
­remains slightly polarized even though there is no electric field acting on
it. This effect is called dielectric absorption.
327
328
Chapter 12 Capacitance
TABLE 12-1 Dielectric strength
Typical Dielectric
Strength (kV/mm)
Material
Air
3
Barium-strontium titanate
3
Porcelain
8
Transformer oil
16
Bakelite
16
Polystyrene
16
Paper
20
Mylar
24
Rubber
28
Teflon
60
Glass
120
Mica
200
See Problems 12-7 and 12-8 and Review Questions 12-31 to 12-34.
12-3 Capacitance
The symbol at the right side of Figure 12-5 represents a capacitor that has
two conducting plates separated by an insulator. Since the current is the
same in all parts of a series circuit, no charge can build up on either plate
of the capacitor until the switch is closed. When the plates have no charge,
there is no electric field and no potential difference between them.
When we close the switch, the galvanometer shows a surge of current
with electrons flowing counterclockwise around the circuit. Since electrons cannot pass through the insulation between the plates of the capacitor, a negative charge builds up on the bottom plate and a positive charge
builds up on the top plate. As time passes with the switch closed and these
charges build up, the electric field between the plates gets stronger and the
potential difference between them increases. According to Kirchhoff’s voltage law, the potential difference between the two plates cannot be greater
than the potential difference of the source. The galvanometer reads zero
Electrons flow
counterclockwise
as capacitor charges
+
E
V
−
+
−
G
Figure 12-5
Charging a capacitor
12-3 Capacitance
329
as soon as enough charge has built up on the c­ apacitor to make its voltage
drop equal to the applied voltage.
If we suddenly double the applied voltage after the capacitor has finished charging, there will be another surge of current until the capacitor
plates charge sufficiently to raise the potential difference between them
to equal the increased applied voltage. There is a fixed ratio between the
potential difference between any given pair of insulated conducting plates
and the charge required to establish this potential difference:
Q
= a constant
V
(12-5)
If we now open the switch in Figure 12-5, there is no longer any conductive path between the two plates, and the surplus electrons on the bottom
plate cannot flow to the top plate. Thus, the parallel plates store electric
charge when a potential difference is applied between the plates. Since
Q/V is a constant, the stored charge produces a potential difference between
the plates.
A capacitor is a component that can store electric charge.
To determine just how much charge a capacitor can store when it is
­connected to a given voltage source, we must determine the factors that
govern the value of the constant in Equation 12-5. This constant, which
­represents the ability of a capacitor to store an electric charge, is called its
capacitance.
The letter symbol for capacitance is C.
The SI unit of capacitance is the farad, which is equal to one coulomb per volt.
The unit symbol for farad is F.
A circuit has a capacitance of 1 F when a charge of 1 C raises the potential difference by 1 V: 1 F = 1 C/V.
C=
Q
V
(12-6)
where C is the capacitance in farads, Q is the charge in coulombs, and V
is the potential difference in volts.
One farad is a much greater capacitance than appears in most circuits.
Consequently, capacitances are often expressed in microfarads, nanofarads,
or even picofarads. However, when substituting for C in equations, we
must remember that these equations were set up in terms of the basic unit
of capacitance, the farad.
An older term for
­capacitance is capacity.
However, the ability
of a capacitor to store
a charge at a certain
potential difference
is sometimes referred
to as its capacity. In
present-day usage,
the term capacitance
(which rhymes with
resistance and inductance) has replaced the
older form.
330
The farad is named
in honour of Michael
Faraday.
Chapter 12 Capacitance
Note that in defining the farad we spoke of the capacitance of a circuit.
Capacitors are not the only type of component that has capacitance,
although capacitors generally have much more capacitance for a given
size. In fact, any pair of conductors that are insulated from each other have
some capacitance. For example, an open-circuit transmission line, like the
one in ­Figure 12-6, can have significant capacitance.
+
+
+
+
+
Transmission line
E
−
−
−
−
−
G
Figure 12-6
Capacitance between parallel conductors
We have already noted that a capacitor cannot charge instantly. It takes
time, even if only a fraction of a second, for the charge to build up. Consequently, the full potential difference across a capacitor does not appear
at the instant the switch is closed. This potential difference remains until
there is a conducting path for electrons to flow from the negative plate to
the positive plate. Since time is involved in any change in the p
­ otential difference between the plates of a capacitor, we can think of the capacitance as
opposing any change in the potential difference across the capacitor.
Capacitance is that property of a circuit that opposes any change in
the voltage across the circuit.
Note how this definition somewhat parallels the definition of resistance as
that property of a circuit that opposes electric current through the circuit.
See Problems 12-9 and 12-10 and Review Questions 12-35 to 12-38.
Circuit Check
A
CC 12-1.
Calculate the force of attraction that is exerted on an electron
located 0.03 mm from a point charge of 5 mC.
CC 12-2. How much capacitance is present if a voltage of 50 V produces
a charge of 11 µC?
CC 12-3. How much voltage will produce a charge of 4 µC on a 50-nF
capacitor?
12-4 Capacitors
Parallel-plate capacitors can be constructed by supporting two metal plates
with air between them, or by coating the two sides of a ceramic disk with
metal, as in Figure 12-7(a). Some capacitors consist of interleaved parallel
12-4 Capacitors
331
Plastic film
Metal
Ceramic
Aluminum foil
(a)
Figure 12-7
(b)
(c)
Construction of capacitors
Capacitors are sometimes referred to by
the now obsolete term
condenser.
Sources: © sciencephotos/Alamy Stock Photo
Sources: GIPHOTOSTOCK/SCIENCE PHOTO LIBRARY
plates with either air or mica dielectric, as in Figure 12-7(b). Others consist
of two long strips of aluminum foil interleaved with strips of a dielectric
and rolled up, as shown in Figure 12-7(c). Originally, waxed paper was
used as the dielectric for such tubular capacitors. Modern tubular capacitors use thin films of a plastic such as polyethylene. Capacitors constructed
by the methods shown in Figure 12-7 range in size from smaller than the
head of a match to larger than a desktop computer. These types of capacitors usually have capacitances of less than 1 μF.
An ingenious means of obtaining large capacitances in a compact package employs the electrolysis principle described in Chapter 3. Electrolytic
­capacitors have a tubular construction like that in Figure 12-7(c) but with
thin gauze soaked in an aluminum hydroxide solution instead of the plastic
dielectric. When first assembled, an electrolytic capacitor acts like a short
circuit since the aluminum hydroxide electrolyte dissociates into positive
and negative ions that act as charge carriers. However, passing a direct
current through the electrolyte attracts the negative hydroxide ions to the
Capacitors: (top row, l–r) polypropylene plastic film,
polyester film, ceramic, mica; (bottom row, l–r) silver
mica, metallized polyester film, tantalum electrolytic,
aluminum electrolytic
One set of the interleaved vanes can
be rotated to adjust the capacitance of
this ­air-dielectric tuning capacitor.
332
Chapter 12 Capacitance
Figure 12-8 Schematic
symbols for capacitors
positive foil, where they react with the aluminum to form a very thin layer
of insulating aluminum oxide. We now have a capacitor where one plate is
the positive aluminum foil, the other is the electrolyte in contact with the
negative foil, and the dielectric is the thin oxide layer. Electrolytic capacitors with capacitances from 1 to 5000 μF are readily available.
Electrolytic capacitors must always be connected so that the polarity of
the potential difference between their terminals is the same as the polarity of the forming voltage. Reversing the polarity of the operating voltage r­ everses the electrolytic action, which can then short out the capacitor
by r­emoving the oxide film. The terminals of electrolytic capacitors
have + and − signs or colour-coded wires to indicate the correct polarity.
This ­polarity restriction greatly limits the use of electrolytic capacitors
in alternating-­current circuits. However, two electrolytic capacitors connected in series opposing can be used as a starting capacitor for AC motors.
The thinness of the oxide film enables electrolytic capacitors to have
a large capacitance compared to other types of capacitor of similar size.
However, there is always some leakage of current through the dielectric
in an electrolytic capacitor. The thin dielectric also makes the breakdown
voltage of electrolytic capacitors relatively low. Capacitors are rated with
a working voltage, which is the maximum DC voltage that can be applied
continuously without breaking down the dielectric.
In Figure 12-8 we can see that the two symbols most commonly used to
show capacitors on circuit diagrams are both stylized representations of
the plates in a capacitor. The version with parallel plates is an international
standard, but North America generally uses the version with one curved
plate. For polarized capacitors, the curved line indicates the negative plate.
See Review Question 12-39.
Sources: GIPHOTOSTOCK/SCIENCE PHOTO LIBRARY
Electrolytic capacitors
12-5 Factors Governing Capacitance
Practical Circuits
Capacitors
As you will see in later chapters, capacitors
have a variety of applications in electronics and electrical engineering. Capacitors can
store energy and pass alternating current while
blocking direct current. Here are some common
applications that make use of these properties.
Power Filtering
Most power supplies for electronic devices
use diodes to convert AC into a pulsating DC.
Capacitors can smooth out these DC pulses by
storing energy as the voltage peaks and then
releasing it as the voltage drops.
Sources: DAVID HAY JONES/SCIENCE PHOTO LIBRARY
Circuit Coupling
Capacitors are often used to link two circuits,
such as the stages of an amplifier. A coupling
capacitor allows AC signals to pass from one
circuit to the other but prevents DC from flowing between the two circuits. With this type of
coupling, the two circuits can be powered by
different DC voltages.
Bypassing
When a capacitor is connected in parallel with
a resistor or other component, AC will flow
through the capacitor quite readily, effectively
bypassing the other component. Thus, such
bypass capacitors can provide a path for AC
signals without significantly changing the DC
voltage across the resistor or other components.
The blue and orange structure is a capacitor array at Fermilab, the US National
Accelerator Laboratory at Batavia, Illinois.
12-5 Factors Governing Capacitance
We can consider all the capacitors shown in Figure 12-7 to be parallel-plate
capacitors. In Section 12-1, we noted that the strength of the electric field
between two parallel plates depends on the quantity of charge per unit
333
334
Chapter 12 Capacitance
area on the surface of the plates. If we double the area of the plates, it takes
twice as much charge to produce an electric field of a given strength. From
Equation 12-4, V = Ed, so the enlarged plates hold twice as much charge at
a given potential difference. Since C = Q/V (Equation 12-6), doubling the
area of the plates doubles their capacitance. We can apply the same logic to
any change in the area of the plates.
The capacitance of parallel plates is directly proportional to their area.
The positively charged plate in a capacitor attracts free electrons to
the closer surface of the negatively charged plate. This attraction aids the
applied voltage in transferring electrons to the negative plate. Similarly,
the negatively charged plate helps to drive electrons from the positively
charged plate into the positive terminal of the source. Moving the plates
closer together increases these forces of attraction and repulsion, enabling
the capacitor to store a greater charge for a given voltage source.
The capacitance of parallel plates is inversely proportional to their
spacing.
Suppose we leave the switch closed after the charging current has ceased
flowing in the circuit of Figure 12-5. If we then slide a sheet of glass or mica
between the plates of the capacitor, the galvanometer shows an additional
surge of charging current. Since the charge on the plates increased but the
potential difference did not change, the capacitance must have ­increased.
This simple experiment demonstrates that
The capacitance of parallel plates depends on the type of dielectric
between plates.
To compare conductive materials, we defined the conductivity of a
material as the conductance between parallel faces of a one-metre cube of the
material (refer to the sections on resistivity [5-4] and conductivity [7-11] for
details). We can define a similar quantity for comparing dielectric m
­ aterials.
The permittivity (or absolute permittivity) of a material is the capacitance between opposite faces of a unit length and cross section of
the material.
The letter symbol for permittivity is the Greek letter ε (lowercase
epsilon).
The units for permittivity are farads per metre.
Permittivity is the ability of a material to permit an electric field within it.
Permittivity measures the extent to which polarization caused by an electric field reduces the strength of the field inside the material.
12-5 Factors Governing Capacitance
For parallel plates,
C=ε
A
d
(12-7)
where C is the capacitance in farads, ε is the permittivity in farads per
metre, A is the area of each plate in square metres, and d is the distance
between the plates in metres.
Note the similarity between Equation 12-7 and the formula for the resistance of a conductor
R=ρ
Just as
l
A
(5-4)
Substituting for C = Q/V into Equation 12-7 and solving for ε gives
ε=
Q
Q
d
d
×
=
×
V A
A V
For parallel plates, Equation 12-3 gives D = Q/A and inverting
­Equation 12-4 gives d/V = 1/E. Therefore,
ε=
D
E
(12-8)
where ε is the permittivity of a given dielectric in farads per metre, D is
the electric flux density in coulombs per square metre, and E is the electric field strength in volts per metre.
Parallel plates have capacitance even without a dielectric material
between them. Applying a voltage to two parallel plates in a vacuum still
produces an electric field between them. The ratio of electric flux density to electric field strength in a vacuum is a physical constant called the
­permittivity of free space, ε0
ε0 =
D
= 8.85 × 10 − 12 F/m
E
(12-9)
When we do place a dielectric between the plates of a capacitor, the electron orbits of atoms are forced off-centre, as shown in Figure 12-4. As a
result, the negative charge of each atom of the dielectric is a little closer to
the positive plate. This displacement has the same effect on the capacitance
335
336
Formulas for the
­capacitance of other
configurations,
such as parallel and
­concentric conductors, are given in electrical and electronics
­handbooks.
Chapter 12 Capacitance
as moving the parallel plates a little closer together. The permittivity of
most solid and liquid dielectrics is considerably greater than that of free
space, while the permittivity of air is very close to ε0. Therefore we can
substitute ε0 into Equation 12-7 to find the capacitance of a parallel-plate
air-dielectric capacitor
C = 8.85 × 10 −12
A
d
(12-10)
where C is the capacitance in farads, A is the area of each plate in square
metres, and d is the distance between the plates in metres.
See Problems 12-11 and 12-12 and Review Questions 12-40 to 12-43.
12-6 Dielectric Constant
Data tables for dielectrics commonly list the ratio between the permittivity
of a given material and the permittivity of free space. This ratio is called the
relative permittivity or dielectric constant of the material.
k or εr =
ε
ε0
(12-11)
where k (or εr) is the dielectric constant of a material, ε is its permittivity,
and ε0 is the permittivity of free space (8.85 × 10−12 F/m).
Therefore, the equation for the capacitance of any parallel-plate capacitor
becomes
C=
k ε0 A
d
(12-12)
where C is the capacitance in farads, k is the dielectric constant, A is the
area of each plate in square metres, and d is the distance between the
plates in metres.
Like dielectric strength, the dielectric constant of a given type of material
sometimes depends on the manufacturing process and can vary ­considerably.
Table 12-2 gives typical values of dielectric constant for the dielectric materials listed in Table 12-1. Comparing Table 12-1 and 12-2, we note that there
is no correlation between dielectric strength and dielectric constant. Many
­materials with a high dielectric constant cannot be used in a capacitor for
high-voltage circuits because their dielectric strengths are too low. For example, distilled water has a dielectric constant of 80, but the dielectric strength
of water is so low that it is not practical to use it as a capacitor d
­ ielectric.
12-6 Dielectric Constant
TABLE 12-2 Dielectric constants
Typical Dielectric
Constant
Material
Air
1.0006
Barium-strontium titanate
7500
Porcelain
6
Transformer oil
4
Bakelite
7
Polystyrene
2.6
Paper
2.5
Mylar
3.0
Rubber
3
Teflon
2
Glass
6
Mica
5
Ceramics called ferroelectric dielectrics (such as barium-strontium
t­ itanate) have molecules that are very easily polarized by an electric field.
As a result, ferroelectric dielectrics have extremely high dielectric constants. The development of these ceramics made it possible to manufacture
capacitors that are much more compact for a given capacitance and voltage rating. The term ferroelectric does not mean that these materials contain
iron, but rather that they permit electric fields in a similar way to how
iron permits magnetic fields. (Ferromagnetic materials are described in
Chapter 14.) The dielectric constant of ferroelectric materials varies slightly
with changes in the strength of the electric field.
Example 12-1
Use Tables 12-1 and 12-2 to calculate the capacitance and the breakdown voltage for a capacitor made by plating silver on each side of a
barium-strontium titanate disk 1.0 cm in diameter and 0.20 mm thick.
Solution
From Table 12-2, the dielectric constant, k, is about 7500.
C=
kε0A
7500 × 8.85 × 10−12 F/m × π ( 5.0 mm ) 2
=
d
0.20 mm
= 2.6 × 10−8 F = 26 nF
From Table 12-1, the dielectric strength of barium-strontium titanate is
about 3 kV/mm. Therefore, the breakdown voltage is approximately
3 kV/mm × 0.20 mm = 0.6 kV.
See Problems 12-13 to 12-16 and Review Questions 12-44 and 12-45.
337
338
Chapter 12 Capacitance
12-7 Capacitors in Parallel
+
E
C1
+
−
C2
+
−
−
Figure 12-9
in parallel
Connecting capacitors in parallel is like increasing the area of the plates of
a single capacitor. Therefore, the total capacitance is greater than that of
any individual one. Since both capacitors in Figure 12-9 are connected to
the same battery, the total charge drawn from the battery is
QT = Q1 + Q2
Capacitors
(12-13)
Since Figure 12-9 is a simple parallel circuit,
E = V1 = V2
QT
Q2
Q1
+
=
E
V1
V2
CT = C1 + C2
and
Therefore,
The total capacitance of capacitors connected in parallel is the sum
of all the individual capacitances.
CT = C1 + C2 + C3 + . . .
(12-14)
Example 12-2
What single capacitance can be used to replace a 10-nF and a 0.05- μF
­capacitor connected in parallel?
Solution
CT = C1 + C2 = l0 nF + 50 nF = 60 nF
See Problems 12-17 and 12-18.
12-8 Capacitors in Series
When we connect capacitors in series, as shown in Figure 12-10, the plates
connected directly to each other are charged by electrostatic ­induction. The
positive charge on the top plate of C1 attracts free electrons from the top
plate of C2 onto the bottom plate of C1. Similarly, the negative charge on the
bottom plate of C2 repels electrons from the top plate of C2 onto the bottom
plate of C1. The effect is much the same as increasing the spacing between
12-8 Capacitors in Series
C1
+
E
C2
−
+
−
+
−
Figure 12-10
Capacitors in series
the plates of a single capacitor. The capacitance of the s­ eries combination is
less than that of either capacitor, but the combination can withstand a higher
­potential difference than either capacitor can by ­itself.
Since the capacitors and the voltage source are a simple series circuit,
E = V1 + V2
Since the current is the same in all parts of a simple series circuit,
QT = Q1 = Q2
E
V1
V2
=
+
QT
Q1
Q2
1
1
1
=
+
Ceq
C1
C2
or
Ceq =
1
C1C2
=
1
1
C1+ C2
+
C1
C2
Therefore,
The equivalent capacitance of series capacitors is
Ceq =
1
1/C1 + 1/C2 + 1/C3 + . . .
(12-15)
Example 12-3
What single capacitance can be used to replace a 0.01-μF and a 0.05-μF
­capacitor connected in series?
Solution
Ceq =
( 10 × 10−9 ) × ( 50 × 10−9 )
C1C2
=
= 8.3 × 10−9 F = 8.3 nF
( 10 + 50 ) × 10−9
C1 + C2
339
340
Chapter 12 Capacitance
Equation 12-15 has the same form as the equation for the equivalent
r­esistance of resistors connected in parallel. To simplify calculations for
parallel resistors, we considered conductance, which is the reciprocal of
­resistance. Similarly, for capacitors in series, we can use elastance, the
reciprocal of capacitance.
The reciprocal of a
farad is sometimes
called a daraf ( farad
spelled backwards).
This term is not
­approved by SI.
Elastance is the opposition of material to the setting up of electric
lines of force in an electric insulator or dielectric.
The letter symbol for elastance is S.
The units for elastance are reciprocal farads (F−1).
S=
1
C
(12-16)
where S is elastance in reciprocal farads and C is capacitance in farads.
With elastance, Equation 12-15 for capacitors in series becomes
ST = S1 + S2 + S3 + . . .
Since Q1 = Q2 in a series circuit and Q = CV,
(12-17)
Q1 = Q2 = C1V1 = C2V2
V1
C2
=
V2
C1
and
(12-18)
When capacitors are connected in series, the ratio between any two ­potential
differences across the capacitors is the inverse of the ratio of their capacitances.
Example 12-4
Compare the following quantities for a 10-nF capacitor and a 40-nF
capacitor connected in parallel to a 500-V source and the same capacitors connected in series to the source:
(a) the total capacitance
(b) the total charge transferred by the source
(c)the charge on each capacitor and the potential difference across
each capacitor
Solution
(a) Parallel:
Series:
CT = C1 + C2 = 10 nF + 40 nF = 50 nF
Ceq =
C1C2
10 nF × 40 nF
=
= 8 nF
C1 + C2
10 nF + 40 nF
12-8 Capacitors in Series
(b) Parallel:
Series:
(c) Parallel:
Series:
QT = CTVT = 50 nF × 500 V = 25 μC
QT = CeqVT = 8.0 nF × 500 V = 4.0 μC
V1 = V2 = E = 500 V
Q1 = C1V1 = 10 nF × 500 V = 5.0 μC
Q2 = C2V2 = 40 nF × 500 V = 20 μC
Q1 = Q2 = QT = 4.0 μC
V1 =
V2 =
4.0 μC
Q1
=
= 400 V
C1
10 nf
4.0 μC
Q2
=
= 100 V
C2
40 nF
See Problems 12-19 to 12-26 and Review Questions 12-46 to 12-48.
Circuit Check
B
CC 12-4. What spacing between two 3 cm × 5 cm parallel plates will produce a capacitance of 47 pF if air is the only material between
the plates? Find the maximum working voltage of this capacitor.
CC 12-5. (a)Calculate the total capacitance of a 1.0-µF capacitor, a 250-nF
capacitor, and a 0.50-µF capacitor connected in parallel.
(b)Calculate the charge on each capacitor when the voltage
across the combination is 50 V.
CC 12-6. A 50-µF capacitor and a 100-µF capacitor are connected in series
to a 150 V source. Calculate the total capacitance and the voltage drop across each capacitor.
341
342
Chapter 12 Capacitance
Summary
• An electric field is a region in which an electric charge is acted upon by
an electric force.
• The strength of an electric field at a certain point is the electric force
­acting on a charge at that point divided by the quantity of the charge.
• The flux density of an electric field is the electric flux per unit area.
• The electric field between parallel conductive plates is uniform except
near the edges.
• Dielectric strength indicates the electric field strength that a dielectric
can withstand.
• Capacitance opposes change in voltage and can store charge.
• The maximum voltage that a capacitor can withstand is expressed as its
working voltage.
• Capacitance is the ratio of the charge on the plates of a capacitor to the
voltage across it.
• Capacitance is proportional to the area of the plates of a capacitor and
inversely proportional to the spacing of the plates.
• Capacitance is dependent on the type of dielectric in a capacitor.
• Permittivity is the capacitance of a unit length and cross section of a ­material.
• The total capacitance of capacitors connected in parallel is the sum of all
the individual capacitances.
• The equivalent capacitance of capacitors connected in series is the reciprocal of the sum of the reciprocals of the individual capacitances.
• Elastance is the reciprocal of capacitance.
B = beginner
I = intermediate
A = advanced
Problems
B
Section 12-1
12-1.
I
12-2.
I
12-3.
I
12-4.
B
12-5.
B
12-6.
B
B
Calculate the force of attraction that acts on a single electron located
0.5 cm from a point charge of +3 μC.
Two electric charges, one of which is four times as large as the other,
exert a mutual force of repulsion of 24 N when they are 15 cm apart.
Determine the magnitude of each charge.
Two parallel plates 1.0 cm apart are connected to a 250-V source.
Find the force exerted on a free electron between the plates.
If 5 × 109 electrons are removed from one parallel plate and added
to the other, what is the total electric flux between the plates?
Determine the electric flux density between two parallel plates
measuring 25 cm by 50 cm and carrying an electric charge of 600 nC.
Calculate the electric field produced by a capacitor when 500 V is
connected across plates with a spacing of 8 mm.
Section 12-2
12-7.
12-8.
Electric Fields
Dielectrics
What is the maximum voltage that can be applied to a parallel plate
capacitor if 0.4 mm of polystyrene is used as the dielectric?
Find the minimum thickness of rubber required to withstand an
applied voltage of 21 kV.
Problems
B
B
B
B
B
I
B
I
B
A
Section 12-3
Capacitance
If moving 3.0 × 1010 electrons from one parallel plate to the other
produces a potential difference between the plates of 220 V, what is
the capacitance of the plates?
12-10. How many electrons must be removed from one plate of a 270-pF
capacitor and added to the other to raise the voltage between the
plates to 420 V?
12-9.
Section 12-5
Factors Governing Capacitance
12-11. A neutralizing capacitor in a radio transmitter consists of two
­aluminum disks, each 10 cm in diameter and 0.50 cm apart with air
between them. What is its capacitance?
12-12. If the electric flux density between the parallel plates of the capacitor
in Problem 12-11 is 1.0 μC/m2, what is the voltage between the plates?
Section 12-6
Dielectric Constant
Section 12-7
Capacitors in Parallel
12-13. Find the capacitance of a parallel-plate capacitor that consists of eleven
2 cm × 0.5 cm sheets of aluminum foil interleaved with ten sheets of
mica 0.02 mm thick and connected as shown in Figure 12-7(b).
12-14. Two sheets of aluminum foil 2.5 cm wide and 1.0 m long and two sheets
of 0.1-mm-thick polystyrene 3.0 cm wide and 1.0 m long are rolled as
shown in Figure 12-7(c) to form a tubular capacitor. The polystyrene
has a dielectric strength of 16 kV/mm and a dielectric constant of 3.
­Calculate the capacitance and voltage rating of this c­ apacitor.
12-15. Calculate the capacitance of a parallel plate capacitor with plates
that are 6 cm × 3 cm, separated by 0.9 mm of polystyrene.
12-16. Two sheets of aluminum 4 cm × 10 m long are rolled up with two
sheets of 0.5-mm mylar separating them. Calculate the capacitance
and maximum working voltage.
12-17. Find the total capacitance and the charge stored by each capacitor when
a 12-μF and a 68-μF capacitor are connected in parallel to a 250-V source.
12-18. Initially, the capacitors in Figure 12-11 are completely discharged.
The switch is first moved to connect its left terminal. When the
charging current ceases, the switch is moved to connect its right terminal. When the current ceases again, the switch is moved back to
its left position, and the switching cycle is repeated. Sketch a graph
of the voltage across the 10-μF capacitor as the switch is operated
for five complete cycles. Label the exact voltage after each cycle of
switch operation. Assume no leakage in either capacitor.
+
300 V
−
Figure 12-11
1 μF
10 μF
343
344
Chapter 12 Capacitance
I
I
B
I
Section 12-8
Capacitors in Series
12-19. A 15-μF, a 22-μF, and a 47-μF capacitor are connected in series to a
240-V source.
(a) Calculate the equivalent capacitance.
(b) Calculate the total charge stored by this capacitor network.
(c) Calculate the voltage across each capacitor.
12-20. After charging, the three capacitors in Problem 12-19 are disconnected from the source and from each other. They are then connected in parallel, with the three positive plates connected to each
other. Find the voltage across each capacitor.
12-21. Two aluminum disks, each 6 cm in diameter, have a sheet of glass
0.3 cm thick separating them. Determine the capacitance.
12-22. Find the equivalent capacitance and the voltage across each of the
capacitors in the circuit of Figure 12-12.
5 μF
+
240 V
−
8 μF
20 μF
Figure 12-12
I
12-23. Find the equivalent capacitance and the voltage across each of the
capacitors in the circuit of Figure 12-13.
20 nF
30 V
Figure 12-13
I
I
330 pF
470 pF
12-24. Three capacitors have capacitances of 0.01 μF, 0.02 μF, and 0.05 μF,
respectively, and a rated working voltage of 400 V for each capacitor. Find the highest voltage that can safely be applied to the series
combination of the three capacitors.
12-25. Find the total elastance of the three capacitors of Problem 12-24
when connected
(a) in series
(b) in parallel
Review Questions
I
12-26. For the circuit shown in Figure 12-14, calculate the total capacitance
and voltage across each capacitor.
20 μF
40 μF
+
100 V
−
10 μF
12 μF
Figure 12-14
Review Questions
Section 12-1
Electric Fields
12-27. Coulomb’s law of electrostatic force states that electric force is proportional to the product of two electric charges. Yet Figure 12-1
shows an electric field for a single isolated electric charge. Explain
the purpose of sketches such as those in Figure 12-1.
12-28. Why does a potential difference exist between two electric conductors possessing unlike charges?
12-29. What is the significance of the arrows on the electric lines of force
­between the two parallel plates in Figure 12-3?
12-30. What characteristic of an electric field is referred to by the term
electric field strength?
Section 12-2
Dielectrics
12-31. What is meant by a polarized atom?
12-32. What type of atoms can become polarized?
12-33. Why is the strength of a dielectric expressed in terms of the k
­ ilovolts
per millimetre rather than the total potential difference?
12-34. Define dielectric absorption.
Section 12-3
Capacitance
12-35. Why must the ratio between the charge on a given pair of insulated
conductors and the potential difference between them be a constant?
12-36. Define capacitance in terms of charge and voltage.
12-37. Why can we define capacitance as that property of an electric ­circuit
that opposes any change in voltage across that circuit?
12-38. Define the farad in terms of flux density and strength of the electric
field in the dielectric between the plates of a capacitor.
Section 12-4
Capacitors
12-39. Explain why a motor-starting capacitor consisting of two electrolytic
capacitors connected in series opposing can be used safely in an AC
circuit.
345
346
Chapter 12 Capacitance
Section 12-5
Factors Governing Capacitance
12-40. Why does the spacing between two parallel plates have an effect on
their capacitance?
12-41. Why does the voltage applied to two parallel plates not affect their
capacitance?
12-42. Why is it possible to have capacitance between two parallel plates
when the space between them is evacuated?
12-43. Why does the presence of an insulating material between parallel
plates increase their capacitance compared with that of the same
plates in a vacuum?
Section 12-6
Dielectric Constant
12-44. Why is it possible to define the dielectric constant of a given insulating
material as the ratio between the capacitance of a pair of parallel
plates with that material as a dielectric and the same parallel plates
with air as a dielectric?
12-45. Distinguish between the (absolute) permittivity and the relative
permittivity of a given dielectric.
Section 12-8
Capacitors in Series
12-46. When two capacitors are connected in series, one plate of each has
no conductive path to any voltage source. How is it possible for
these capacitors to become charged?
12-47. Why must capacitors in series all store the same charge?
12-48. A 0.1-μF 400-V capacitor and a 0.05-μF 200-V capacitor are c­ onnected
in series. What is the maximum voltage that can be safely applied
to this combination?
Integrate the Concepts
Three capacitors, C1, C2, and C3, appear in a network such that C1 and C2 are
in parallel and that combination is in series with C3.
• C1 is a TeflonTM capacitor with square plates whose sides measure 1 cm
and are 0.05 mm apart.
• C2 is a ceramic-disc capacitor whose plates have a diameter of 1 cm and
are 0.1 mm apart.
• C3 is a mica capacitor with rectangular plates that measure 1 cm × 0.5 cm
and that are 0.025 mm apart.
Determine the total capacitance of the series-parallel combination.
Practice Quiz
1.
Why is electric field strength described as a vector quantity?
2.
What does the concentration of lines of electric force on a diagram
represent?
Practice Quiz
3.
Which of the following statements are true?
(a) An electric field is directly proportional to the difference of
potential between two points and inversely proportional to the
distance between those two points.
(b) A capacitor blocks alternating current and passes direct current.
(c) The larger the capacitance, the more charge the capacitor can store.
4.
How are the permittivity and the dielectric constant of a material related?
5.
The capacitance of parallel plates is
(a) inversely proportional to their area and the distance between the plates
(b) directly proportional to their area and the distance between the plates
(c) directly proportional to their area and inversely proportional to
the distance between the plates
(d) inversely proportional to their area and directly proportional to
the distance between the plates
6.
The equivalent capacitance for the circuit of Figure 12-15 is
(a) 46.87 nF
(b) 5.87 nF
(c) 30 nF
(d) 72 nF
C1
10 nF
C2
22 nF
C3
40 nF
Figure 12-15
7.
The equivalent capacitance for the circuit of Figure 12-16 is
(a) 4.07 nF
(b) 2.45 nF
(c) 27 nF
(d) 42 nF
C1
10 nF
C2
10 nF
C3
22 nF
Figure 12-16
347
13
Capacitance in DC
Circuits
Now that we know something about capacitance as a basic
property of electric circuits, we can examine the behaviour
of practical electric circuits containing capacitance, either
­intentionally from capacitors, or as a byproduct of other
components such as parallel traces on a printed circuit board.
Chapter Outline
13-1
13-2
Charging a Capacitor 350
Rate of Change of Voltage 352
13-3
Time Constant
13-5
Discharging a Capacitor 357
13-4
13-6
13-7
13-8
13-9
354
Graphical Solution for Capacitor Voltage 356
Algebraic Solution for Capacitor Voltage 362
Transient Response
366
Energy Stored by a Capacitor 370
Characteristics of Capacitive DC Circuits
13-10 Troubleshooting
375
372
Key Terms
steady-state value 350
instantaneous value 351
time constant 354
transient response 368
transient phase 368
stray capacitance 374
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• calculate the initial current through a
capacitor when it is charging or discharging
• calculate the initial rate of change of v
­ oltage
across a ­capacitor when it is charging or
­discharging
• calculate the time constant of a CR circuit
• analyze a CR circuit by using universal
­exponential curves
• calculate the capacitor voltage for any
instant of time by using the charging and
discharging equations
Photo sources: Tomáš Hašlar/Alamy Stock Photo
• compute the transient response of the
capacitor voltage in a CR circuit when the
capacitor has an initial voltage across it
• explain how energy can be stored by a
capacitor
• calculate the energy stored by a charged
­capacitor
• describe applications of capacitors that use
their energy-storage capabilities
• describe the undesirable effects of stray
capacitance on a circuit
• troubleshoot a capacitor for defects
350
Chapter 13 Capacitance in DC Circuits
13-1 Charging a Capacitor
We can think of charging a capacitor in terms of the quantity of charge transferred to its plates. However, Q = CV and C is a constant for a given capacitor, so the charge is directly proportional to the potential difference b
­ etween
the plates of the capacitor. Since we can easily measure this ­potential difference with a voltmeter, it is more convenient to speak of the charging of
a c­ apacitor in terms of the potential difference built up between its plates.
At the instant we close the switch in the circuit of Figure 13-1, no e­ lectrons
have had enough time to flow into the bottom plate and out of the top plate,
so no potential difference has yet built up between the plates. Kirchhoff’s
voltage law states that there must be a voltage drop in the c­ ircuit equal to
the applied voltage. Since there is no potential difference across the capacitor, there must be an IR drop in the circuit. Although no r­esistance is
shown, we shall assume that the resistance of the wiring and the internal
resistance of the voltage source battery total 0.5 Ω. Therefore, at the instant
we close the switch, the current is
I0 =
+
100 V E
−
Figure 13-1
E
100 V
=
= 200 A
R
0.5 Ω
v
1.0 μF
Charging a capacitor without a current-limiting resistor
A 200-A current charges the plates of the capacitor very rapidly, and the
potential difference across the capacitor rises to 100 V in a few microseconds. When the potential difference across the capacitor equals the applied
voltage, current ceases. Therefore, if we charge a capacitor by connecting it
directly across a voltage source, there is a very short-duration pulse of very
high current. Since such pulses can damage a source, we usually charge
­capacitors through a resistance that limits the initial current surge.
When analyzing circuits that contain capacitors, we have to deal with
voltages and currents that vary with time. It is conventional practice to
use lowercase letters to represent quantities that vary with time. The resistance and capacitance of most circuits and components depend on physical
­factors (such as length, area, and material type) that are independent of time.
Consequently, we represent resistance and capacitance with the ­uppercase
letters R and C. In the preceding chapters, the source voltage and the resulting currents, voltage drops, and power were also independent of time.
­Accordingly, we represented all these steady-state values with ­uppercase
letters. In a circuit like the one in Figure 13-2, the charging ­current and the
13-1 Charging a Capacitor
351
i
+
100 V E
−
vR
vC
+
−
+
2.0 MΩ
1.0 μF
−
Figure 13-2
Charging a capacitor through a current-limiting resistor
potential difference across the capacitor change from instant to instant and
depend on how long the switch has been closed. Therefore, we represent
the instantaneous values of the voltage and current with the lowercase letters i and v. Note that we represent time itself by lowercase t.
Lowercase letter symbols represent instantaneous values that are
dependent on time.
Uppercase letter symbols represent steady-state values that are not dependent on time.
As long as the switch in Figure 13-2 is closed, the sum of the potential
difference across the capacitor and the IR drop across the resistor must
equal the applied voltage. Therefore,
E = vC + iR
(13-1)
At the instant we close the switch, there is no potential difference across
the capacitor. Hence, the initial current, I0, must make the IR drop in the
­circuit equal to the applied voltage:
I0 =
E
R
(13-2)
The initial current in the circuit of Figure 13-2 is only 100 V/2.0 MΩ =
50 μA. This current transfers charge to the capacitor. Thus, a potential difference starts to build up across the capacitor. To satisfy Kirchhoff’s voltage
law, the IR drop across the resistor must decrease. The instantaneous ­current
decreases correspondingly, and the potential difference across the capacitor rises more slowly. Therefore, the rate of rise of the instantaneous voltage
across the capacitor depends on the magnitude of the instantaneous current,
which depends on the IR drop across the resistor, which, in turn, depends on
the instantaneous voltage across the capacitor, which ­depends on how long
the switch has been closed. Before we can sort out these interdependences,
we must consider how we can express rate of change of voltage.
See Review Question 13-27 at the end of the chapter.
A circuit like the
one in Figure 13-2
is often referred to
as a CR ­circuit since
the only components other than the
source and the switch
are a ­resistor and a
­capacitor.
Chapter 13 Capacitance in DC Circuits
13-2 Rate of Change of Voltage
4
2
0
1
Time (seconds)
2
C
it
6
ΔV
4
Δt
2
0
(a)
Figure 13-3
rcu
6
8
Ci
rcu
Output Voltage (volts)
it A
10
Ci
8
Circ
10
uit B
Figure 13-3 shows the output potential difference for three circuits that are
switched on at t = 0. In Figure 13-3(a), the potential difference increases as
a linear function of time rather than immediately jumping to a steady-state
value, as in a simple resistance network. We can read the rate of change
of voltage directly from the straight-line graphs by noting the increase in
­potential difference during 1 s. For circuit A, the voltage rises from 0 to 5 V
in the first second and from 5 V to 10 V during the next second. For both
of these intervals, the rate of change of voltage is 5 V/s. Similarly, the rate
of change of voltage in circuit B is 10 V/s. Note that the graph for circuit B
is twice as steep as the graph for circuit A. The slope of a graph of voltage
­versus time represents the rate of change of the voltage.
Output Voltage (volts)
352
t1
1
2
Time (seconds)
3
(b)
Graphical determination of rate of change of voltage
In circuit C the output potential difference is a function of the square of
the elapsed time. Since this graph is not a straight line, the rate of change
of voltage is not the same for all time intervals. By drawing a tangent to the
graph at time t1 in Figure 13-3(b), we can read the rate of change of voltage at that instant. Since the tangent is a straight line, we simply note the
change in voltage, ΔV, over any convenient interval, Δt. Hence, the rate of
change of voltage at time t1 is ΔV/Δt. To represent the change in voltage
at any given instant, we use calculus notation.
The symbol for rate of change of voltage with time is
dv
dt
With differential calculus we find an equation for the rate of change of
voltage at any instant:
For circuit A,
v = 5t and dv
= 5 V/s
dt
13-2 Rate of Change of Voltage
v = 10t
For circuit B,
v = t2,
For circuit C,
and
dv
= 10 V/s
dt
dv
= 2t V/s
dt
When we close the switch in the circuit of Figure 13-2, current flows
i­nstantly at its initial value and then decreases as the voltage across the
­capacitor rises (see Figure 13-4). From Ohm’s law we know that the voltage
drop across the resistance is always proportional to the current through it.
Since the resistance is constant, the graph of the IR drop across the resistor
has the same shape as the graph of the current.
I0
Current (mA)
50
40
30
i
20
10
0
0
Figure 13-4
2
4 6 8
Time (s)
10
Instantaneous charging current
Since the initial value of capacitor voltage, v, is 0 V, the initial value of
­resistor voltage is E = I0R. Therefore
I0 =
E
R
(13-3)
From Equation 12-6, we get v = q/C. The capacitance C does not change
with time, so
dq
dv
1
= ×
dt
C
dt
Since the current transfers charge to the capacitor, the rate of change of the
charge on the capacitor is dq/dt = i . Therefore, dv/dt = i/C.
When the switch is closed at t = 0,
dv
I0
E/R
= =
dt
C
C
Therefore,
initial
E
dv
=
dt
CR
(13-4)
353
Chapter 13 Capacitance in DC Circuits
This initial rate of change of the capacitor voltage is represented by
the slope of the tangent at t = 0 in Figure 13-5. Since the initial current, I0, in
the CR circuit of Figure 13-2 is only 50 μA, it takes about 10 s for the voltage
across the capacitor to rise to a value close to the applied voltage.
Capacitor Voltage, vC
354
τ
Slope =
100
dv
E
=
dt
CR
E
vC
80
60
40
20
0
63% of E
4
6
8
2
Time after switch is closed (seconds)
10
Figure 13-5 Rise of potential difference across a capacitor charging through
a resistance
See Review Questions 13-28 to 13-30.
13-3 Time Constant
To help analyze the charging and discharging of capacitors, we define the
time constant of a CR circuit:
The time constant of a CR circuit is the time it would take the potential difference across the capacitor to rise to the same value as the
applied voltage if the potential difference increased at a constant
rate equal to its initial rate of change. The letter symbol for time
constant is the Greek lowercase letter τ (tau).
As shown in Figure 13-5, if the rate of change of the capacitor voltage
­remains equal to the initial rate of change, the voltage across the capacitor
when it is fully charged is
V=E=τ×
Therefore,
τ = CR
E
CR
(13-5)
where τ is the time constant of the CR circuit in seconds, C is the capacitance in farads, and R is the resistance in ohms. (Equation 13-5 also applies
with the capacitance in microfarads and the resistance in megohms.)
The time constant of a CR circuit is directly proportional to the capacitance. If we double the capacitance, twice as much charge must flow
13-3 Time Constant
355
into the capacitor in order to raise the potential difference across it to the
­applied voltage. But changing the capacitance does not affect the initial
current, so the capacitor takes twice as long to charge. If we double the
­resistance instead, the initial current is cut in half and the capacitor again
takes twice as long to charge to a given voltage.
The applied voltage has no effect on the time constant. If we double the
applied voltage, the capacitor must store twice the charge but the initial
current is doubled. Therefore, it will take exactly the same time for the
­potential difference across the capacitor to reach the applied voltage.
For every CR circuit, the relationship between the actual charging time
and the time constant is the same. The instantaneous potential difference
across a capacitor is always about 63.2% of the applied voltage one time
constant after the switch is closed. For practical purposes, the potential
­differ­ence across the capacitor reaches a steady-state value equal to the
­applied voltage after a time interval equal to five time constants.
Example 13-1
(a) Find the initial rate of change of potential difference across the
capacitor in Figure 13-2.
(b) How long will it take the capacitor in Figure 13-2 to charge to a
potential difference of 100 V?
Solution
(a)
(b)
Initial
dv
E
100 V
= 50 V/s
=
=
dt
CR
1.0 μF × 2.0 MΩ
τ = CR = 1.0 μF × 2.0 MΩ = 2.0 s
Since the applied voltage is 100 V and V ≈ E after 5τ, the charging time is
t = 5τ = 5 × 2.0 s = 10 s
Multisim Solution
(b) Download Multisim file EX13-1 from the website.
The circuit is the same as shown in Figure 13-2. Connect a ground to
the bottom node of the schematic.
On the Menu bar, select Simulate/Analyses/Transient Analysis. Under
Analysis Parameters, set the initial conditions to zero and the end time
to 15 s. Under Output, add the voltage across the capacitor as the variable for analysis.
At the bottom of the Transient Analysis dialog box, click Simulate. Use
the graph to determine that the voltage reaches 100 V after about 10 s.
See Problem 13-1 and Review Question 13-31.
circuitSIM
walkthrough
Chapter 13 Capacitance in DC Circuits
13-4 Graphical Solution
for Capacitor Voltage
Because of the fixed relationship between the time constant and the charging time for a CR circuit, we can use an exponential curve to find the potential differences at any instant in any CR circuit. If we replot the graph of
vC in Figure 13-5 with the horizontal (time) axis marked in time constants
rather than seconds and with a vertical (voltage) axis marked in percentage
of the applied voltage, we get the charging curve in Figure 13-6.
100
Potential Difference (percentage of applied voltage)
356
y = 1 − e−x
90
Charging
80
70
60
50
40
30
Discharging
20
10
0
Figure 13-6
capacitors
y = e−x
1
2
3
Time (time constants)
4
5
Universal exponential curves of the charge and discharge of
Example 13-2
(a) Find the potential difference across the capacitor in the circuit of
­Figure 13-2 3.0 s after the switch is closed.
(b) How long will it take the potential difference across the capacitor
to rise from 0 V to 55 V?
Solution
(a) Since τ = CR = 1.0 μF × 2.0 MΩ = 2.0 s,
3.0 s =
3.0 s
× τ = 1.5τ
2.0 s
13-5 Discharging a Capacitor
357
On the charging curve in Figure 13-6, y = 77% when t = 1.5τ, as
shown in Figure 13-7(a). Therefore,
vC = 77% of E = 0.77 × 100 V = 77 V
55 V
× 100% = 55% of E
(b)
55 V is
100 V
On the charging curve in Figure 13-6, t = 0.80τ when y = 55 as
shown in ­Figure 13-7(b).
t = 0.80τ = 0.80 × 2.0 s = 1.6 s
100
Voltage (% of E)
Voltage (% of E)
Therefore,
77
0
1.5τ
100
55
0
0.80τ
Time
(a)
Figure 13-7
Time
(b)
Using the graph of Figure 13-6 to solve Example 13-2
Multisim Solution
Download Multisim file EX13-2 from the website.
The circuit is the same as shown in Figure 13-2. Check that the switch is
closed. Connect a ground to the bottom node of the schematic.
(a) On the Menu bar, select Simulate/Analyses/Transient Analysis.
Under Analysis Parameters, set the initial conditions to zero and the
end time to 10 s. Under Output, add the voltage across the capacitor
as the variable for analysis.
At the bottom of the Transient Analysis dialog box, click Simulate.
Use the graph to determine that the capacitor voltage at t = 3 s is 77 V.
(b) The graph also shows that the capacitor voltage reaches 55 V at
t = 1.6 s.
See Problems 13-2 to 13-6 and Review Questions 13-32 and 13-33.
13-5 Discharging a Capacitor
When the capacitor in Figure 13-8(a) is fully charged, the potential difference across the capacitor is equal to the applied voltage, 200 V. If we then
turn the switch to position 2, the potential difference across the capacitor
remains at 200 V for some considerable time since there is no c­ onducting
circuitSIM
walkthrough
358
Chapter 13 Capacitance in DC Circuits
path between the plates except a bit of leakage through the dielectric. At
the instant we turn the switch to position 3, the potential difference across
the capacitor is still 200 V since it takes time for the surplus electrons to
flow from the bottom plate to the top plate of the capacitor. To satisfy
Kirchhoff’s voltage law, the initial current has to be such that a 200-V IR
drop develops across the very low resistance of the conductor connecting
the two plates together. Therefore, the initial current is very large and the
capacitor discharges in a fraction of a second.
100 kΩ
1
1
2
2
3
+
200 V
−
50 nF
+
−
(a)
Figure 13-8
3
+
200 V
−
100 kΩ
50 nF
+
−
(b)
Discharging a capacitor
To reduce the initial discharge current, we can add a 100-kΩ resistor as
shown in Figure 13-8(b). The initial current is now limited to
I0 =
V0
200 V
=
= 2.00 mA
R
100 kΩ
(13-6)
The reduced current takes much longer to discharge the capacitor fully.
As the charge on the capacitor decreases, the potential difference across the
capacitor must also decrease. Therefore, the initial current is the maximum
value that the instantaneous discharge current can have. According to
Kirchhoff’s voltage law, the IR drop must have the same magnitude as the
potential difference of the capacitor. As this potential difference decreases,
the current decreases and the capacitor discharges more slowly. This interdependence among the potential differences, the IR drop, the current, and
the rate of decrease of the potential differences results in the exponential
discharge curve shown in Figure 13-9(a).
When the capacitor is discharging, the current through the 100-kΩ
­resistor in Figure 13-8(b) flows in the opposite direction to the current
when the capacitor is charging. Consequently, the polarity of the IR drop
across the resistance reverses when the capacitor is switched from charging
to discharging. Therefore, the discharge current and the resistor voltage are
shown as negative quantities in Figure 13-9(b).
Capacitor Voltage
(% of E)
V0
100
80
60
40
vC
20
0
Figure 13-9
0
37%
1τ 2τ 3τ
Time
(a)
4τ
5τ
Current/Resistor Voltage
(% of maximum)
13-5 Discharging a Capacitor
0
1τ
0
Time
2τ 3τ
4τ
5τ
20
i and vR
40
60
80
100
V
I0 = R0
(b)
Instantaneous voltage and current as a capacitor discharges
With the same procedure we used for deriving Equation 13-4, we can
show that the initial rate of change of the potential difference across the
­capacitor as it starts to discharge is
dv
V0
=−
dt
CR
(13-7)
where V0 is the initial voltage across the capacitor.
The negative sign results from the discharge current flowing through
the resistance in the direction opposite to that of the charging current.
The ­negative rate of change of voltage indicates that the voltage across the
capacitor is decreasing.
We define the discharge time constant, τ, as the time it would take for
the capacitor to completely discharge if it were to continue to discharge
at the initial rate. Since the final voltage is 0 V, the total change in voltage,
Δv, after the capacitor has discharged is −Vinit. If the capacitor continues
to ­discharge at the initial rate, the time to discharge fully is Δt = τ and the
­discharge rate is dv/di = Δv/Δt.
Equation 13-7 then gives
−V0
−V0
dv
Δv
=
=
=
τ
di
Δt
CR
and
τ = CR
(13-5)
Therefore, we can use the same formula for calculating the time constant for charging and for discharging. Again, we can use the exponential
curve in Figure 13-6 to find the potential differences and currents at any
instant in any CR circuit. After a time interval equal to one time constant,
the ­potential difference of the capacitor drops to about 37% of its initial
value—in other words, the capacitor loses about 63% of its initial potential
359
Chapter 13 Capacitance in DC Circuits
difference. For practical purposes, the charge on a capacitor is negligible
after 5τ. For the CR circuit of Figure 13-8(b),
τ = CR = 50 nF × 100 kΩ = 5.0 ms
The capacitor discharges fully in about 5 × 5.0 ms = 25 ms.
Example 13-3
When the capacitor in Figure 13-8(b) is fully charged, the switch is
turned to position 3.
(a)Find the initial rate of change of potential difference across the
­capacitor.
(b) Find the potential difference across the capacitor after 7.0 ms.
(c)How long does it take for the voltage drop across the 100-kΩ
resistor to reach 110 V?
Solution
(a)
Initial
(b)
dv
V0
200 V
=−
=−
= −40 kV/s
dt
CR
50 nF × 100 kΩ
τ = CR = 0.05 μF × 100 kΩ = 5.0 ms
7.0 ms is
7.0 ms
= 1.4 τ
5.0 ms
On the discharging curve in Figure 13-6, y = 25% when t = 1.4τ, as
shown in Figure 13-10(a).
Capacitor Voltage
(% of V 0 )
vC = 25% of V0 = 0.25 × 200 V = 50 V
Capacitor Voltage
(% of V 0 )
360
100
25
0
3τ
1.4τ
Time
(a)
Figure 13-10
100
55
0
3τ
0.6τ
Time
(b)
Using the graph of Figure 13-6 to solve Example 13-3
13-5 Discharging a Capacitor
361
(c) When the switch is in position 3, the resistor is connected directly
across the capacitor. From Kirchhoff’s voltage law,
vC + vR = 0 and vR = −vC
110 V is
110
= 55% of Vinit
200
On the discharging curve in Figure 13-6, t = 0.6τ when y = 55%,
as shown in Figure 13-10(b).
t = 0.6τ = 0.6 × 5.0 ms = 3 ms
circuitSIM
Multisim Solution
Download Multisim file EX13-3 from the website.
The circuit is as shown in Figure 13-8(b) with the switch in position 3.
Connect a ground to the bottom node of the schematic.
(a) Double-click on the capacitor in the circuit. Under the Values tab,
set the initial condition to 200 V.
On the Menu bar, select Simulate/Analyses/Transient Analysis.
Under Analysis Parameters, set the initial conditions to Userdefined and the end time to 25 ms. Under Output, add the capacitor
voltage as the variable for analysis.
At the bottom of the Transient Analysis dialog box, click Simulate.
Use the graph to determine that the capacitor voltage at t = 7.0 ms
is 50 V.
(b) The graph also shows that the capacitor voltage reaches 110 V
after about 3 ms. Since the resistor voltage and the capacitor voltage
have the same magnitude, the resistor voltage also reaches 110 V
when t = 3 ms.
See Problems 13-7 to 13-13 and Review Question 13-34.
Circuit Check
CC 13-1.
A
A capacitor is charged through a 22-kΩ resistor and reaches full
charge in 5.2 s. Determine the capacitance.
CC 13-2. A 5.0-μF capacitor is connected in series with a 30-kΩ resistor
and a 60-V source. What current will be flowing in this circuit
0.27 s after the capacitor starts charging?
walkthrough
362
Chapter 13 Capacitance in DC Circuits
13-6 Algebraic Solution
for Capacitor Voltage
When we require greater accuracy than we can read from the graphs
of Figure 13-6, we can use the equations for the exponential curves. We
start with the Kirchhoff’s voltage-law equation for the basic circuit of
­Figure 13-2.
E = iR + vC
(13-1)
At every instant, the current is the rate of change of the charge on the
­capacitor:
Since q = CvC ,
i=
i=C
dq
dt
dvC
dt
(13-8)
Substituting for i in Equation 13-1,
E = CR
dvC
+ vC
dt
(13-9)
Equation 13-9 is a differential equation for an exponential function.
­ ppendix 2-2 shows how to solve this equation with basic integral calcuA
lus. This solution gives an equation for the potential difference that builds
up on an uncharged capacitor that begins charging at t = 0:
vC = E ( 1 − e−x )
(13-10)
where e = 2.7182818 . . . (the base of natural logarithms) and x = t/CR.
Since
τ = CR, x =
t
τ
Thus, x represents elapsed time measured in time constants.
Substituting for vC in Equation 13-1 gives
E = iCR + E(1 − e−x)
0 = iCR − Ee−x
13-6 Algebraic Solution for Capacitor Voltage
iC =
E −x
e
R
363
(13-11)
When the switch in the circuit of Figure 13-8(b) is in position 3, there is
no voltage source in the discharge loop. Hence, the applied voltage E is
zero. To satisfy Kirchhoff’s voltage law, the sum of the IR drop across the
resistor and the potential difference across the capacitor must be zero:
0 = iR + vC = CR
dvC
+ vC
dt
As shown in Appendix 2-2, this equation is readily solved with integral calculus. The resulting equation for the voltage across a discharging
­capacitor is
vC = V0e−x
(13-12)
where V0 is the initial voltage across the capacitor and x = t/CR.
Equations 13-10 and 13-12 are the equations for the exponential curves
in Figure 13-6.
Example 13-2A
(a) Calculate the voltage across the capacitor in Figure 13-2 when the
switch has been closed for 3.0 s.
(b) How long will it take the voltage across the capacitor to rise from
0 V to 55 V?
Solution
(a)
x=
t
3.0 s
=
= 1.5
CR
1.0 μF × 2.0 MΩ
vC = E ( 1 − e−x ) = 100 V × ( 1 − e−1.5 ) = 78 V
55 V = 100 V(1 − e−x)
(b)
e−x = 0.45
−x = ln 0.45 = −0.7985
Hence,
x = 0.80 =
t
t
=
τ
CR
t = 0.80 × 2.0 s = 1.6 s
The symbol ln is
­standard notation for
a natural logarithm,
based on powers of
the mathematical
­constant e. We can
calculate natural
­logarithms with the
ln key on a scientific
calculator.
364
Chapter 13 Capacitance in DC Circuits
Example 13-4
Starting from position 1, the switch in Figure 13-11 stays in each position for 1.0 ms. Calculate the voltage across the capacitor 3.0 ms after
the switch first moves to position 2.
1
100 kΩ
Figure 13-11
100 kΩ
2
+
500 V
−
10 nF
Circuit diagram for Example 13-4
Solution
Step 1
When the switch first moves to position 2, the capacitor charges from
0 V toward 500 V. At any instant,
At 1.0 ms,
vC = E(1 − e−x) = 500(1 − e−x)
x=
t
1 ms
1.0 ms
=
=
= 1.0
CR
10 nF × 100 kΩ
1.0 ms
vC = 500(1 − e−1) = 500 × (1 − 0.3679) = 316 V
Step 2
As the capacitor discharges in position 1,
After 1.0 ms,
vC = V0e−x = 316 × e−x
x=
1.0 ms
t
=
= 0.5
CR
10 nF × 200 kΩ
vC = 316 × e−0.5 = 316 × 0.6065 = 192 V
Step 3
When the switch returns to position 2, vC starts at 192 V. Now we
need to calculate how long it would take the capacitor to charge from
0 V to 192 V.
192 = 500(1 − e−x)
13-6 Algebraic Solution for Capacitor Voltage
e− x =
500 − 192
= 0.616
500
We can use
­approximate
values read from the
­exponential curves
to check ­answers
calculated by the
­algebraic method.
x = −ln 0.616 = 0.4845
x=
Since
t
CR
t = 0.4845 × 1.0 ms = 0.4845 ms
One millisecond later, the capacitor will reach a voltage equivalent to
having charged for a total time interval of 0.4845 + 1.0 = 1.4845 ms.
Hence,
1.4845 ms
t
=
= 1.4845
τ
1.0 ms
x=
vC = 500(1 − e−1.4845) = 500 × (1 – 0.2266) = 387 V
Example 13-5
The switch in Figure 13-12(a) is closed at t = 0 s and opened again at
t = 1.0 s. Find the voltage across the capacitor at t = 2.0 s.
Solution
The discharge circuit when the switch is open is quite straightforward.
However, to calculate the charge time constant when the switch is
closed, we need Thévenin’s theorem.
Step 1
Considering the capacitor to be the load, we remove it from the original
­circuit of Figure 13-12(a) and place it in the Thévenin-equivalent circuit
of Figure 13-12(b). We can calculate mentally that ETh = 400 V and RTh =
160 kΩ for the equivalent circuit of Figure 13-12(b).
200 kΩ
RTh
+
500 V
−
800 kΩ
(a)
2.0 μF
160 kΩ
+
E
400 V Th
−
(b)
Figure 13-12 (a) Capacitor-charging circuit and (b) its Théveninequivalent circuit
365
2.0 μF
366
Chapter 13 Capacitance in DC Circuits
Step 2
The charging time constant is
and
τch = CRTh = 2.0 μF × 0.16 MΩ = 0.32 s
1.0 s =
1.0
= 3.125 τ
0.32
Since the switch was originally open, initial voltage across the capacitor is 0 V. We substitute into Equation 13-10 to find the voltage after
1.0 s of charging:
vC = E(1 − e−x) = 400(1 − e −3.125) = 400(1 − 0.044) = 382.4 V
Step 3
The discharging time constant is simply
τdis = CR = 2.0 μF × 800 kΩ = 1.6 s
1.0
= 0.625 τ
1.6
From Equation 13-12, at t = 2.0 s,
and
circuitSIM
walkthrough
1.0 s =
vC = 382.4 × e−0.625 = 382.4 × 0.535 = 205 V
Multisim Solution
Download Multisim file EX13-5 from the website.
The circuit is the same as shown in Figure 13-12(a). Connect a ground
to the bottom node of the schematic, and close the switch.
On the Menu bar, select Simulate/Analyses/Transient Analysis. Under
Analysis Parameters, set the initial conditions to zero and the end time to
1 s. Under Output, add the capacitor voltage as the variable for analysis.
At the bottom of the Transient Analysis dialog box, click Simulate.
Use the graph to determine that the capacitor voltage at t = 1 sec is
about 382 V.
Open the switch. Double-click on the capacitor in the circuit. Under the
Values tab, set the initial condition to 382 V. On the Menu bar, select
Simulate/Analyses/Transient Analysis. Under Analysis Parameters,
set the initial conditions to User-defined and the end time to 1 s.
At the bottom of the Transient Analysis dialog box, click Simulate. Use
the graph to determine that 1.0 s after the switch was closed the capacitor voltage is about 205 V.
See Problems 13-14 and 13-15.
13-7 Transient Response
We can now consider another method for analyzing the response of CR
networks to changes in applied voltage. If we charge a capacitor that is
­already partially charged, the voltage across it equals the initial voltage
13-7 Transient Response
367
plus an additional voltage that depends on the elapsed time. Using this
­approach to adapt Equation 13-10 gives
vC = V0 + ( VF − V0 ) ( 1 − e−x )
(13-13)
where V0 is the initial capacitor voltage and VF is the steady-state voltage to which the capacitor can charge, and x is the charging time measured in time constants.
Simplifying gives us a universal equation for instantaneous capacitor voltage
in a DC CR network:
vC = V0 + VF – V0 − VF e−x + V0 e−x
vC = VF + ( V0 − VF ) e − x
(13-14)
where vC is the instantaneous voltage across the capacitor, V0 is the i­ nitial
voltage across the capacitor, and VF is the final steady-state voltage.
When we close the switch at t = 0 in the purely resistive network of
­ igure 13-13(a), the output voltage of the two-terminal network instantly
F
drops from 500 V to its new steady-state value of 100 V, as shown in
Figure 13-13(b). But in the CR network of Figure 13-13(c), the output voltage is the voltage across the capacitor, which takes five time constants to
­discharge from its initial voltage of 500 V to its final steady-state voltage of
100 V, as shown in Figure 13-13(d).
800 kΩ
Output Voltage, vout (V)
vout
200 kΩ
+
500 V
−
500
400
300
200
100
0
vout
800 kΩ
200 kΩ
2.0 μF
+
500 V
−
Output Voltage, vout (V)
(a)
500
400
300
Transient
Steady
state
200
100
0
(c)
Figure 13-13
0 1τ 2τ 3τ 4τ 5τ
Time
(b)
Transient response of a DC CR network
0 1τ 2τ 3τ 4τ 5τ
Time
(d)
We can also derive
Equation 13-14 by
considering a charged
capacitor discharging
toward a given
steady-state voltage.
368
Chapter 13 Capacitance in DC Circuits
Equation 13-14 represents the output voltage as consisting of two
­components: the steady-state response, VF, and the transient response
(V0 − VF)e−x, which is shown in red in Figure 13-14. This transient response
is negligible when t > 5τ.
Output Voltage (volts)
Transient response is the manner in which a voltage or current
changes from one steady-state value to another.
A transient phase is the time interval during which a network’s
voltages and currents adjust from one steady state to another.
500
V0
400
Transient response
300
Steady state
200
100
0
VF
0
Figure 13-14
1τ
2τ 3τ
Time
4τ
5τ
Separating Vout into steady-state and transient components
Example 13-6
If the switch in the two-terminal network of Figure 13-13(c) has been
open for several minutes, what is the instantaneous output voltage
500 ms after the switch is closed?
Solution
From inspection of the network, V0 = 500 V and VF = 100 V. The discharge time constant is governed by the Thévenin equivalent resistance.
x=
0.500 s
= 1.5625
2.0 μF × 0.16 MΩ
vout = VF + ( V0 − VF ) e − x
= 100 + ( 500 − 100 ) e − 1.5625
= 100 + 400 × 0.2096
circuitSIM
walkthrough
= 184 V
Multisim Solution
Download Multisim file EX13-6 from the website.
The circuit is the same as shown in Figure 13-13(c). Close the switch.
Double-click on the capacitor in the circuit. Under the Values tab,
set the initial condition to 500 V. On the Menu bar, select Simulate/­
Analyses/Transient Analysis. Under Analysis Parameters, set the
13-7 Transient Response
369
i­nitial conditions to User-defined and the end time to 500 ms. Under
Output, add the capacitor voltage as the variable for analysis.
At the bottom of the Transient Analysis dialog box, click Simulate. Use
the graph to determine that 500 ms after the switch is closed the instantaneous output voltages is about 184 V.
In the simple CR charging circuit of Figure 13-2, V0 = 0 and VF = 100 V for
the capacitor. Since VR = E − vC, then V0 = 100 V and VF = 0 for the resistor.
As long as we are careful to specify the correct V0 and VF, Equation 13-14
also ­applies to the instantaneous voltage drop across series charging resistors and shunt discharging resistors. The transient response approach lets
us use a s­ ingle equation to determine the voltage across a partially charged
capacitor.
Example 13-2B
Given that the switch in Figure 13-2 has been closed for 3.0 s, find the
­voltage across
(a) the capacitor
(b) the series charging resistor
Solution
x=
t
3.0 s
=
= 1.5
CR
1.0 μF × 2.0 MΩ
(a) For the capacitor, V0 = 0 and VF = 100 V.
VC = VF + (V0 − VF)e−x
= 100 V + (0 − 100 V)e−1.5
= 100 V − 100 V × 0.2231
= 77.7 V
(b) For the resistor, V0 = 100 V and VF = 0.
vR = VF + (V0 − VF)e−x
= 0 + (100 V − 0)e−1.5
= 100 V × 0.2231
= 22.3 V
See Problems 13-16 to 13-22 and Review Questions 13-35 to 13-39.
We can check our
­calculations by
­ver­ify­ing that
E = vC + vR
= 77.7 V + 22.3 V
= 100 V.
370
Chapter 13 Capacitance in DC Circuits
13-8 Energy Stored by a Capacitor
The current that flows while a capacitor is charging transfers energy from
the voltage source to the rest of the circuit. Some of this energy is converted
into heat as the charging current flows through the series current-­limiting
resistance in the circuit. This heat accounts for only some of the energy
drawn from the source. The remainder is stored by the capacitor. Work is
done on the charge that moves into a capacitor against the opposing potential difference building up between the plates.
When we disconnect a charged capacitor from the source, the potential
difference across the capacitor remains constant unless there is leakage
of the charge through the dielectric. Since no current is required to maintain the potential difference between the plates of a charged capacitor, the
­capacitance stores electric energy. In fact, this energy is stored in the electric
field between the plates of the capacitor. This electric field alters the orbits of
electrons bound to the molecules of the dielectric, as shown in Figure 12-4.
We might compare the electric energy stored by a capacitor to the mechanical energy stored in a cylinder of compressed air. Energy is used to force
the air into the cylinder against the increasing pressure. This stored energy
is ­released as kinetic energy when the compressed air is used to operate a
paint sprayer or a pneumatic tool. Similarly, we can use the electric energy
stored by a charged capacitor, as we shall discover in Section 13-9.
At any instant, the rate at which a capacitor stores energy is power,
p = vi. We can plot a graph of power versus time by calculating the ­product
of the potential difference across the capacitance and the charging current at ­various times from 0 to 5τ. As shown in Figure 13-15, the power
input to the capacitor peaks when the graphs of the potential difference
and the charging current cross. At this instant, the instantaneous voltage
has risen to half of its final value and the instantaneous charging current
has dropped to half of its initial value. The peak power input occurs about
0.693 time constants after the capacitor begins charging.
iC
vC
Area = WC
PC
0
1
2
3
4
Time (time constants)
Figure 13-15
5
Energy stored by a capacitor
13-8 Energy Stored by a Capacitor
After five time constants, the charging current is negligible and therefore
the power input is also negligible. Thus, when a capacitor is fully charged,
it has stored all the energy that it can at the particular voltage applied to
it. This total energy is represented by the area under the power curve in
Figure 13-15. Appendix 2-3 describes how to use integral calculus to calculate energy stored by a capacitor.
However, we can simplify the calculation by considering a capacitor
that charges at a constant rate equal to the initial rate of change of voltage.
­Suppose we connect the capacitor to a constant-current source with a terminal voltage that can vary while the current remains I = E/R, as shown in
Figure 13-16(a). Since the voltage across the capacitor now rises at a ­constant
rate from zero to its final value of V = E, as shown in Figure 13-16(b),
the a­ verage voltage during this time interval must be 1⁄2V. Therefore,
1
W = Pt = Vav × I × t = V × It
2
Since It = q,
1
W = Vq
2
When the capacitor is fully charged, q = CV and
1
1
W = V × CV = CV 2
2
2
(13-15)
where W is the energy stored by the capacitance in joules, C is the
­capacitance in farads, and V is the potential difference across the capacitance in volts.
The total energy stored by a fully charged capacitor is not affected by the
rate at which it is charged.
Capacitor Voltage
I = E/R
Constant
current
source
(a)
Figure 13-16
V =E
Vav
vC
0
t
Time
(b)
Determining the energy stored by a capacitor
371
Chapter 13 Capacitance in DC Circuits
Example 13-7
How much energy is stored in a 500-pF high-voltage filter capacitor
when it is charged to 15.0 kV?
Solution
W=
1
1
CV2 = × 500 pF × ( 15 kV ) 2 = 5.63 × 10 − 2 J = 56 mJ
2
2
See Problems 13-23 to 13-26 and Review Questions 13-40 to 13-42.
Circuit Check
B
CC 13-3. A capacitor is connected to a 300-V source until it is fully
charged. It is then discharged through a 2.2-MΩ resistor. After
20 min the capacitor voltage has fallen to 94 V. Calculate the
­capacitance of the capacitor.
CC 13-4. Find the voltage across a 3000-µF filter capacitor when it is
­storing 750 mJ of energy.
13-9 Characteristics
of Capacitive DC Circuits
The ability of a capacitor to store electric charge and, as a consequence, to
oppose a rapid change in potential difference between its terminals makes
the capacitor a particularly useful component in DC circuits. Its ability to
oppose any instant change in the voltage across it enables the capacitor to
filter or smooth the output from a pulsating voltage source such as the rectifier in Figure 13-17(a). When the source voltage drops below the voltage
across the capacitor, the capacitor maintains the load voltage by discharging
some of its stored energy into the load. The c­ apacitor recharges the next time
that the source voltage rises to its peak value, as shown in Figure 13-17(b).
+
Pulsating E
source g
−
DC
C
(a)
Figure 13-17
Load
Voltage
372
vC
e
Time
(b)
Filtering voltage fluctuations with a capacitor
13-9 Characteristics of Capacitive DC Circuits
This application of a capacitor is somewhat like the use of a reservoir
tank with an air compressor for a paint sprayer. The tank stores enough
compressed air to maintain a fairly constant pressure to the paint sprayer
in spite of the fluctuations in the pressure developed by the piston action
of the compressor. (The compressed-air-tank analogy provides a means of
­visualizing the charge and discharge action of capacitance, with the friction of the air hose representing resistance.)
Energy storage in a capacitor is also useful when we wish to obtain a
very high current for a short period of time without subjecting the source
to a severe current surge. With a circuit patterned after the basic circuit
of Figure 13-18, the capacitor charges slowly from the source and then
­discharges rapidly when we connect the low-resistance load across it.
The peak current drawn from the source is limited by the series resistor.
This system is used in radar transmitters to generate micro­second pulses
with a peak power of over a million watts. Capacitors can also produce
short surges of very high current for spot-welding ­aluminum.
High resistance
+
E
C
Low-resistance
load
−
Figure 13-18 Obtaining short-duration, high-current pulses from a reservoir
capacitor
The potential difference across the capacitor of a given CR circuit ­always
rises along exactly the same exponential curve. Therefore, we can use a
­series CR circuit as the basis for a timing device. In the simple circuit of
­Figure 13-19, the neon lamp “strikes” (starts glowing) when the potential difference across it reaches 100 V. The lamp stops glowing when the
­voltage drops below 20 V. When a neon lamp is not struck, the gas is an
­insulator and the lamp acts like an open circuit. When the lamp strikes,
the neon gas becomes ionized with the positive gas ions flowing in one
­direction and free electrons moving in the other direction. The resistance
of the glowing lamp is about 100 Ω. It takes about 1.8 time constants for
the potential difference across the capacitor in Figure 13-19 to rise to the
striking voltage of the neon lamp. The low resistance of the ionized neon
then quickly discharges the capacitor down to the extinguishing ­voltage
of the lamp, whereupon the resistance of the neon lamp again becomes
­extremely high and the capacitor starts to recharge. The intervals b
­ etween
the short flashes from the neon lamp, therefore, depend on the time
­constant CR.
373
374
Chapter 13 Capacitance in DC Circuits
R
+
120 V
Neon
lamp
C
−
Figure 13-19
C
Leakage
resistance
Figure 13-20
Equivalent circuit of a
“leaky” capacitor
A CR timing circuit
In some instances, capacitors can have undesirable characteristics. So
far, we have assumed that there is negligible leakage of charge through
the dielectric of capacitors. This assumption is reasonable for ceramic,
mica, and polymer dielectrics. However, electrolytic capacitors often have
enough leakage to affect the operation of a circuit. We can represent capacitor leakage in practical circuits by an appropriate high resistance in parallel
with the capacitance, as shown in Figure 13-20.
Capacitance in circuits is not limited to capacitors. As shown in
­Figure 12-6, there is some capacitance between any pair of electric conductors that are insulated from each other. In many circuit applications, this
stray capacitance is small enough that its effect on the circuit is negligible.
But in computers, where timing pulse voltages must rise and fall in a fraction of a nanosecond, stray capacitance can degrade circuit performance.
In Figure 13-21, the stray capacitance of the conductors is directly in parallel with the input terminals of the transistor (even though this capacitance
would not appear in the actual circuit diagram). Hence, the input signal to
the transistor is the potential difference across this stray capacitance, which
charges and discharges through the internal resistance of the pulse generator. If the time constant for the internal resistance and stray capacitance is
more than a small fraction of the pulse duration, the capacitance will cause
appreciable distortion by delaying the rise and fall of the voltage pulses.
Internal
resistance
Stray
capacitance
Pulse
generator
Pulse produced by
pulse generator
Figure 13-21
Pulse appearing across
stray capacitance
Effect of stray capacitance
See Review Question 13-43.
13-10 Troubleshooting
13-10 Troubleshooting
Capacitors can fail in several different ways.
• If a capacitor is subjected to a voltage higher than its working voltage,
the dielectric often breaks down, allowing a high current to pass through
the capacitor. Usually the capacitor burns out and the circuit b
­ ecomes an
open circuit.
• As capacitors age, the leakage current flowing through them tends to
­increase. Electrolytic capacitors in particular are prone to this problem
because the electrolyte dries out somewhat over time. The increasing
leakage current results in a lower leakage resistance for the capacitor.
• A capacitor can short internally.
A capacitor can be checked for faults with an ohmmeter or a capacitance
meter. The capacitor should first be removed from the circuit and then
fully discharged by shorting its leads together. To test the capacitor with
an ohmmeter, connect it across the ohmmeter terminals. If the capacitor is
shorted, the ohmmeter will give a steady reading very close to 0 Ω. If the
capacitor is open, the meter will read infinity or over range.
If the capacitor is not defective, it will be charged by the internal battery in an analog ohmmeter. When the meter is connected, the pointer immediately goes toward the zero ohms position and then moves toward the
infinity position as the current decreases. Electrolytic capacitors usually
have a resistance in the megohm range, while most other types should
read a resistance of hundreds of megohms. If the capacitor has a significant
leakage current, the pointer will stop at a fairly low value of resistance.
Most digital ohmmeters cannot be used to measure the leakage resistance
of ­capacitors.
To test a capacitor with a capacitance meter, simply connect the capacitor
leads to the terminals of the meter and read the capacitance from the dial
or digital display of the meter. If no capacitance is displayed, the ­capacitor
is shorted or open.
See Review Questions 13-44 to 13-46.
375
376
Chapter 13 Capacitance in DC Circuits
Summary
• When a capacitor is charging or discharging, the initial value of current
depends on the values of V0 and R.
• When a capacitor is charging or discharging, the initial rate of change of
capacitor voltage depends on the values of V0, C, and R.
• The time constant for the charging and discharging of a series CR circuit
is τ = CR.
• The universal exponential curves provide a graphical method for analyzing a CR circuit when the capacitor is being charged or discharged.
• The charge and discharge equations for a capacitor provide an algebraic
method for analyzing a CR circuit.
• The transient response of a CR circuit with an initial voltage across the
capacitor can be determined with the universal equation for capacitor
voltage.
• A capacitor stores energy when it is charged.
• Stray capacitance in circuits can have undesirable effects.
B = beginner
I = intermediate
A = advanced
Problems
B
Section 13-3
Time Constant
13-1. Find the following quantities when the switch in Figure 13-22 is first
closed:
(a) the initial current
(b) the initial voltage across the capacitor
(c) the initial rate of rise of voltage across the capacitor
(d) the charging time constant
(e) the time it will take the capacitor to charge
+
250 V
−
200 kΩ
20 μF
Figure 13-22
B
Section 13-4 Graphical Solution for Capacitor
Voltage
13-2. (a) Given that vC = 0 until the switch in Figure 13-23 is closed, find:
(i) the initial charging current
(ii) the initial rate of change of voltage across the capacitor
(iii) the charging time constant
Problems
(iv) the voltage across the capacitor at t = τ
(v) the charge on the capacitor at t = τ
(b) Use Multisim to determine the initial charging current and the
capacitor voltage one time constant after the switch is closed.
377
circuitSIM
walkthrough
10 nF
+
16 V
−
33 kΩ
Figure 13-23
B
I
I
A
13-3. For the circuit of Figure 13-22, use the exponential curves of
­Figure 13-6 to find:
(a) the voltage across the capacitor 3.0 s after the switch is closed
(b) the charging current 4.0 s after the switch is closed
(c) the time taken for the voltage across the capacitor to reach 100 V
(d) the time taken for the voltage across the capacitor to rise from
50 V to 200 V
13-4. (a)For the circuit of Figure 13-23, use the exponential curves of
­Figure 13-6 to find:
(i) how long it will take the capacitor to charge to 16 V
(ii) how long it will take the capacitor to charge to 10 V
(iii) the instantaneous voltage across the capacitor at t = 0.50 ms
(iv) the instantaneous charging current at t = 0.25 ms
(v)the time taken for the voltage across the capacitor to rise
from 4 V to 12 V
(b) Use Multisim to verify your calculations in part (a).
13-5.An unknown capacitor is charged by a 200-V source through a 1.5-MΩ
resistor. Given that it takes 45 s for the voltage across the capacitor to
reach 100 V, determine the capacitance of the ­capacitor.
13-6. For the circuit shown in Figure 13-24, determine the voltage across
the capacitor 5.0 s after switch S1 is closed.
t=0s
t=3s
S1
S2
10 kΩ
+
50 V
−
Figure 13-24
+
100 V
−
100 μF
circuitSIM
walkthrough
378
Chapter 13 Capacitance in DC Circuits
B
circuitSIM
walkthrough
Section 13-5
13-7.
B
13-8.
I
13-9.
A
13-10.
A
13-11.
I
13-12.
I
13-13.
Discharging a Capacitor
A 33-μF capacitor has been charged to a voltage of 450 V.
(a) How long will it take to discharge the capacitor through a
220‑kΩ resistor?
(b) Find the maximum discharge current.
The capacitor of Problem 13-7 is discharged by shorting its terminals with an electric conductor having a resistance of 0.4 Ω.
(a) Find the peak discharge current.
(b) How long will it take to discharge the capacitor?
(c) Use Multisim to verify your calculations for parts (a) and (b).
A 33-μF capacitor, which has been charged to a voltage of 450 V, is
discharged through a 220-kΩ resistor. Using the universal exponential curves, determine:
(a) how long it will take the voltage across the capacitor to drop to
75 V
(b) the voltage across the capacitor 15 s after it starts to d
­ ischarge
(c) the instantaneous discharge current 10 s after it starts to
­discharge
(d) how long it will take the voltage across the capacitor to drop
from 350 V to 50 V
If the resistor in the circuit of Figure 13-19 is 5.0 MΩ and the capacitor has a capacitance of 8.0 μF, calculate the time interval between
flashes of the neon lamp. The lamp strikes at 100 V, and its resistance then becomes 100 Ω. It extinguishes when the voltage across
it drops to 20 V, and its resistance then becomes almost infinite.
A capacitor is connected to a 100-V source until it is fully charged.
It is then discharged through a 3.3-MΩ resistor. After 11 min the
capacitor voltage has fallen to 40 V. Determine the probable value
of the capacitor.
A 10-kΩ resistor is connected in series with a 10-μF capacitor, a
100-V source, and a switch. Calculate:
(a) the time constant
(b) the voltage on the capacitor 50 ms after the switch is closed
(c) the charging current 100 ms after the switch is closed
In the circuit shown in Figure 13-25, S1 is closed at t = 0 s and S2 is
moved to the lower contact at t = 15 s, disconnecting the voltage
source from the capacitor.
(a) How long it will take the capacitor to fully discharge after S2 is
moved?
(b) Calculate the voltage across the capacitor at t = 4.0 s.
Problems
379
(c) Calculate the voltage across the capacitor at t = 24 s.
(d) Sketch a graph of the capacitor voltage from t = 0 s until the
capacitor is fully discharged.
t=0s
t = 15 s
S1
S2
+
100 V
−
10 kΩ
200 μF
10 kΩ
Figure 13-25
A
A
A
Section 13-6
Algebraic Solution for Capacitor Voltage
13-14. Repeat Problem 13-3 using algebraic expressions to solve for instantaneous values of voltage and current.
13-15. Repeat Problem 13-4 using algebraic equations to solve for instantaneous values of voltage and current.
Section 13-7
Transient Response
13-16. After a considerable period of-time in position 2, the switch in the
circuit of Figure 13-26 is thrown to position 1 at t = 0.
(a) Calculate iC and dvC/dt at t = 0.
(b) How long must the switch remain in position 1 for vC to rise to
100 V?
(c) The switch is thrown to position 2 at t = 5.0 min. Calculate iC and
dvC/dt as the switch makes contact in position 2.
(d) If the switch returns to position 1 at t = 5.5 min, calculate vC at
t = 10 min.
(e) Use Multisim to verify your calculations for parts (a) and (b).
5.0 MΩ
470 kΩ
1
+
200 V
−
Figure 13-26
2
50 μF
circuitSIM
walkthrough
380
Chapter 13 Capacitance in DC Circuits
A
13-17. After a considerable period of time in position 1, the switch in the
circuit of Figure 13-27 is thrown to position 2 for 2.0 ms and is then
returned to position 1.
(a) Calculate vC 2.0 ms after the switch moves to position 2.
(b) Calculate vC 2.0 ms after the switch returns to position 1.
(c) Draw a graph of the output voltage at terminal A with respect
to ground for this 4-ms interval.
+
48 V
−
2
68 kΩ
A
1
+
12 V
−
10 nF
22 kΩ
Figure 13-27
A
circuitSIM
walkthrough
13-18. (a)Commencing from the open position, the switch in the circuit of
Figure 13-28 closes for 1.0 ms and then opens. Calculate the voltage across the capacitor 1.0 ms after the switch reopens.
(b) Use Multisim to verify the capacitor voltage calculated in part (a).
68 kΩ
+
300 V
−
A
470 kΩ
10 nF
Figure 13-28
13-19. Repeat Problem 13-18 by using the circuit in Figure 13-29.
220 kΩ
+
15 kΩ
250 V
−
Figure 13-29
5.0 nF
Problems
A
381
13-20. For the circuit shown in Figure 13-30, S1 is closed at t = 0 s and S2 at
t = 10 s.
(a) Calculate the time for the capacitor to be fully charged.
(b) Calculate the capacitor voltage at t = 4.0 s.
10 kΩ
t=0
2.0 kΩ
S1
+
50 V
−
S2
t = 10 s
100 μF
Figure 13-30
A
(c) Calculate the capacitor voltage at t = 11 s.
(d) Sketch the waveform of the capacitor voltage for the interval
from 0 s to 15 s.
(e) Use Multisim to verify your calculations for parts (a), (b), and (c).
13-21. The switch shown in Figure 13-31 is thrown to position 1 at t = 0 s.
After 6.0 s, the switch is thrown to position 2.
(a) Sketch the capacitor voltage waveform for the time period from
0 s to 50 s.
(b) Calculate capacitor voltage at t = 4.0 s.
(c) Calculate capacitor voltage at t = 20 s.
(d) How long after the switch is closed to position 1 will the capacitor voltage reach 75 V?
5.0 kΩ
circuitSIM
walkthrough
1
2
5.0 kΩ
+
100 V
−
25 kΩ
200 μF
Figure 13-31
A
13-22. (a)The switch in the network of Figure 13-32 has been in position
1 long enough for the circuit to reach a steady state. Find the
output voltage with respect to ground (i) 2.0 s and (ii) 4.0 s after
the switch is thrown to position 2.
(b) Use Multisim to verify the output voltage calculated in part (a).
circuitSIM
walkthrough
382
Chapter 13 Capacitance in DC Circuits
2.0 MΩ
5.0 μF
1
+
500 kΩ
2
−
200 V
Figure 13-32
B
B
B
B
Section 13-8
Energy Stored by a Capacitor
13-23. Calculate the energy stored by the capacitors in the circuits of
­Figures 13-22 and 13-23 when they are fully charged.
13-24. Calculate the energy stored by the capacitor in Problem 13-7
(a) before it starts to discharge
(b) after it has been discharging for one time constant
13-25. A 100-μF capacitor, charged to a potential difference of 36 V, is discharged through a 15-kΩ resistor. How much energy is dissipated
by the resistor as the capacitor is completely discharged? If the
capacitor had been discharged through a 680-Ω resistor, how much
energy would have been dissipated by the resistor?
13-26. In Problem 13-10, how much energy is transferred to the neon lamp
with each flash?
Review Questions
Section 13-1
Charging a Capacitor
13-27. What factors determine the initial charging current of a capacitor?
Section 13-2
Rate of Change of Voltage
13-28. What factors determine the rate of change of potential difference
­between the plates of a capacitor?
13-29. When a discharged capacitor is connected to a voltage source, the
current changes instantly from zero to a value of E/R. Why then
can the voltage across the capacitor not change instantly?
13-30. Why must the rate of change of voltage across a capacitor diminish
as the charge on its plates increases?
Section 13-3
Time Constant
13-31. Account for the exponential shape of the graph of vC in Figure 13-5.
Review Questions
Section 13-4
Graphical Solution for Capacitor Voltage
13-32. What is the advantage of using the time constant of a CR circuit as
a unit of time in determining the instantaneous voltage?
13-33. Figure 13-6 shows a graph of the algebraic equation y = 1 − e−x.
Why do we use this equation to solve for vC rather than the simple
exponential equation y = e−x?
Section 13-5
Discharging a Capacitor
13-34. Why does the initial charge on a capacitor have no effect on the
length of time required for the capacitor to discharge through a
given resistance?
Section 13-7
Transient Response
13-35. In the circuit of Figure 13-28, we use the Thévenin-equivalent resistance to calculate the charge time constant, but not the discharge
time constant. In the circuit of Figure 13-29, we use the Théveninequivalent resistance to calculate the discharge time constant, but
not the charge time constant. Explain the difference.
13-36. Why are the charge and discharge time constants the same in the
­circuit of Figure 13-33?
13-37. Draw an accurately labelled graph of the instantaneous voltage
across the capacitor in the circuit of Figure 13-33 plotted against
time as the double-pole, double-throw switch is thrown to the
opposite position.
500 kΩ
+
120 V
−
1.0 μF
Figure 13-33
13-38. When steady-state voltages are suddenly switched in a DC CR
­network, the transient current always starts from zero and returns
to zero, having either a positive or negative direction, as shown
by the graph of Figure 13-34. Does this mean that the current in
­resistors of the network must be zero except during the transient
phase? ­Explain.
383
Chapter 13 Capacitance in DC Circuits
+
Current
384
0
Time
−
Figure 13-34
13-39. Draw superimposed graphs (in different colours) to show how a
graph of vC is obtained by combining a sudden change in steadystate voltage with a transient having the form of either the positive
or negative transient shown in Figure 13-34.
Section 13-8
Energy Stored by a Capacitor
13-40. Using the techniques we employed in producing Figure 13-15, plot
a graph of the instantaneous power in a resistor when a charged
­capacitor is connected across it.
13-41. An air-dielectric capacitor is charged from a voltage source and
then disconnected. The plates are then moved twice as far apart.
Assuming no leakage of the charge, account for the difference in
the energy stored by the capacitor before and after the plates are
moved.
13-42. Instead of changing the spacing between the plates in Question
13-41, a dielectric with a dielectric constant of 2.0 is placed
between the plates after the capacitor is disconnected from the
source. Account for any change in the energy stored by the
­capacitor.
Section 13-9
Characteristics of Capacitive DC Circuits
13-43. Briefly describe three applications of the capacitor.
Section 13-10
Troubleshooting
13-44. Name three problems that one may encounter with capacitors.
13-45. You use an analog ohmmeter to test a capacitor and determine
that it is not defective. Describe the action of the pointer on the
meter.
13-46. What would you conclude about a capacitor that is tested with an
ohmmeter if
(a) it indicates a leakage resistance of 10 kΩ
(b) the pointer goes to the extreme right and stays there
(c) the pointer does not move
Practice Quiz
Integrate the Concepts
A 50-V source, a 20-kΩ resistor, a 1-μF capacitor, and an open switch form a
series circuit. Calculate the following after the switch is closed:
(a) the initial value of the current
(b) the time constant
(c) the capacitor voltage 50 ms later
(d) the time for the capacitor to fully charge
(e) the energy stored by the capacitor after it is fully charged
Practice Quiz
1.
Which of the following statements are true?
(a) Uppercase letter symbols represent steady-state values that are
dependent on time.
(b) Lowercase letter symbols represent instantaneous values that
are dependent on time.
(c) The time constant for charging a capacitor is directly proportional to the capacitance.
(d) A negative rate of change of voltage indicates that the voltage
across the capacitor is decreasing.
(e) Kirchhoff’s voltage law can be applied to RC circuits to determine the voltage across the capacitor.
2.
What is the initial rate of change of potential difference across the
capacitor in Figure 13-35 when the switch is closed?
(a) 2.4 V/s
(b) 2.4 kV/s
(c) 24 V/s
(d) 0.24 V/s
R1
2.0 GΩ
V1
48 V
C1
0.010 μF
Figure 13-35
3.
How long will it take the capacitor of Figure 13-35 to charge to the
battery voltage of 48 V?
(a) 20 s
(b) 20 ms
(c) 100 ms
(d) 100 s
385
386
Chapter 13 Capacitance in DC Circuits
4.
How long after the switch is closed will the voltage across the capacitor of Figure 13-35 increase to 30 V?
(a) 20 s
(b) 19.8 s
(c) 19.6 s
(d) 19 s
5.
What is the rate of change of potential difference across the capacitor of Figure 13-36 when the switch is moved to position 2 with the
capacitor fully charged?
(a) 125 kV/s
(b) −125 V/s
(c) −125 kV/s
(d) 125 V/s
1
R1
2
2.0 MΩ
C1
1.0 μF
V1
250 V
Figure 13-36
6.
How long after the switch is moved to position 2 will the voltage
across the capacitor of Figure 13-36 drop to 100 V?
(a) 1.8 ms
(b) 1.0 ms
(c) 1.9 ms
(d) 1.7 ms
7.
Which equation describes the voltage drop across the resistor during
the discharge of the capacitor in a simple RC circuit?
(a)
VR = E + Ee RC
(c)
VR = E 1 − e RC
−t
(b) VR = E 1 − e RC
(
(
−t
t
(d) VR = Ee RC
−t
)
)
Practice Quiz
8.
The energy stored by a capacitor is
(a) directly proportional to the potential difference across its­
terminals
(b) inversely proportional to the potential difference across its
terminals
(c) directly proportional to the square of the potential difference
across its terminals
(d) inversely proportional to the square of the potential difference
across its terminals
9.
How much energy is stored by a 100-μF capacitor when it is charged
to 1000 V?
(a) 50 mJ
(b) 100 mJ
(c) 5 J
(d) 50 J
10.
How much energy is stored by the capacitor of Figure 13-36 when it
is fully charged?
(a) 125 μJ
(b) 31.3 mJ
(c) 15.6 mJ
(d) 62.5 mJ
387
14
Magnetism
We now turn our attention to the third basic characteristic
of electric ­circuits: inductance. Inductance depends on the
relationship between magnetic fields and electric current just
as capacitance depends on the relationship between electric
fields and potential difference. An understanding of inductance
requires an understanding of the nature of magnetism and its
interactions with electricity.
Chapter Outline
14-1
Magnetic Fields
390
14-2Magnetic Field around a Current-Carrying
14-3
Conductor
393
Magnetic Flux
396
14-4 Magnetomotive Force 397
14-5 Reluctance
14-6
14-7
398
Permeance and Permeability
Magnetic Flux Density
14-8 Magnetic Field Strength
400
399
401
14-9Diamagnetic, Paramagnetic, and Ferromagnetic
Materials
402
14-10 Permanent Magnets
14-11 Magnetization Curves
404
404
14-12 Permeability from the BH Curve 408
14-13 Hysteresis
410
14-14 Eddy Current
412
14-15 Magnetic Shielding
413
Key Terms
magnetic field 390
magnetic line of force (flux) 390
north pole 390
south pole 390
solenoid 394
electromagnet 394
magnetic flux 396
magnetic circuit 397
weber 397
magnetomotive force (MMF) 397
reluctance 398
reciprocal henry 398
permeance 399
permeability 399
relative permeability (μr) 399
magnetic flux density 400
tesla 400
magnetic field strength 401
nonmagnetic 402
diamagnetic 402
paramagnetic 402
ferromagnetic 402
magnetic domains 403
magnetized 403
saturated 404
retentivity 404
residual magnetism
(remanence) 404
temporary magnet 404
permanent magnet 404
ferrites 404
magnetization curve
(BH curve) 405
normal permeability 408
incremental permeability 409
differential permeability 409
initial permeability (μi) 409
average permeability 409
coercive force 410
hysteresis 411
hysteresis loop 411
eddy current 413
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• state the properties of magnetic fields
• use the right-hand rule to determine the
direction of a magnetic field around a straight
current-­carrying ­conductor or a solenoid
• illustrate electromagnetic induction by means
of a ­magnet, solenoid, and galvanometer
• explain the relationship between magnetomotive
force and the number of turns of wire in a solenoid
and the current flowing through it
• state the role of reluctance in a magnetic circuit
• state the role of permeability in a magnetic circuit
• state the relationship between flux density,
magnetic field strength, and permeability
Photo sources: © Tim@awe/Dreamstime/Get-stock
• differentiate among diamagnetic,
paramagnetic, and ­ferromagnetic materials
• differentiate between permanent and
temporary magnets in terms of residual
magnetism
• determine the permeability of a magnetic
material from its BH curve
• explain the hysteresis characteristic of a
magnetic ­material
• explain how an eddy current is established in
a solenoid
• explain how electric devices may be
protected from ­magnetic fields
390
Chapter 14 Magnetism
14-1 Magnetic Fields
Two thousand years ago, Chinese mariners used a crude form of magnetic compass for navigation. In ancient compasses and their modern
counterparts, the pointer is a magnet mounted so that it is free to rotate
around its centre. This pointer will deflect if another magnet, or any iron
object, is brought close to it, demonstrating that magnetic force can act
without physical contact. Like electric and gravitational force, magnetic
force is a field force. The definition of a magnetic field is similar to that of
an electric field.
A magnetic field is that region in which a magnetic material is acted
upon by a magnetic force.
Some theories of
­subatomic particles
predict the existence of a magnetic
­monopole, a tiny form
of magnet with only
one pole. To date,
there has been no
­direct experimental
evidence of magnetic
monopoles.
As shown in Figure 14-1, the strength and direction of a magnetic field
can be represented by magnetic lines of force (or magnetic flux lines).
Every magnet has two areas, called a north pole and a south pole, where
the lines of force are concentrated. Like poles repel each other and oppo­
site poles attract. One pole of Earth’s magnetic field is in the Canadian
arctic near 83° north latitude. The north pole of a magnet is the one that is
­attracted to this northern pole of Earth’s magnetic field.
N
Figure 14-1
S
The magnetic field around a bar magnet
Like electric lines of force, magnetic lines of force are imaginary. They
are simply a way of representing a magnetic field. The strength and direction of magnetic fields have been measured by means ranging from
­observing patterns in iron filings or the deflections of compass needles
to readings from sophisticated electronic sensors. The nature of magnetic
fields is such that the lines of force must have the five characteristics
­described below.
At any point in a magnetic field, the direction of a line of force is the
direction of the magnetic force that the field would exert on a north
pole at that point.
14-1 Magnetic Fields
If we bring one end of a bar magnet close to a compass needle, the compass
needle will align itself with the magnetic lines of force. If we now bring
the opposite pole of the bar magnet near the compass, the other end of
its n
­ eedle will move toward the bar magnet. Hence, the direction of the
­magnetic field at one pole of the bar magnet must be opposite to that at
the other pole. By convention, the positive direction of magnetic lines of
force is defined so that the lines leave the north pole of a magnet and enter
the south pole.
Magnetic lines of force always form complete loops.
Magnetic lines of force do not begin at the north pole of a magnet and end
at the south pole. They continue between south pole and north pole ­inside the
magnet to form complete closed loops. If we cut the magnet of Figure 14-1 in
two and pull the two sections apart, the magnetic lines of force cross the gap
and still form continuous loops. New north and south poles are formed on
each side of the gap, as shown in Figure 14-2(a). No matter how many times
we cut the magnet, each section has its own north pole and south pole.
N
S
N
S
(a)
N
S
S
N
(b)
Figure 14-2
Magnetic lines of force between unlike and like poles
391
Chapter 14 Magnetism
Magnetic lines of force tend to be as short as possible.
When we separate the two magnets in Figure 14-2(a), we lengthen the magnetic lines of force, and the magnetic force opposes this motion. We can think
of the magnetic lines of force as trying to pull the two parts back together in
order shorten the loops. In this respect, the magnetic lines of force act somewhat like elastic bands. The tendency for magnetic lines of force to become as
short as possible accounts for the attraction between north and south poles.
Magnetic lines of force repel one another and cannot intersect.
This characteristic accounts for the mutual repulsion of like poles and helps
explain the pattern of the magnetic fields in Figure 14-1 and Figure 14-2. The
magnetic lines of force diverge as they get farther and farther away from the
poles. The magnetic lines of force are kept in equilibrium by each line tending
to shrink its loop length and at the same time being p
­ revented from doing so
by the repulsion of the next smaller loop. The magnetic field around a uniform
magnet is symmetrical unless there are other magnetic materials in the field.
The magnetic field is distributed so as to maximize the number of
magnetic lines of force.
S
When a piece of soft iron enters the field of a magnet, the symmetrical
pattern of the magnetic lines of force is altered as shown in Figure 14-3.
Some of the lines of force detour to pass through the iron because it permits
a greater concentration of lines of force than air does. As a result the number
of lines of force increases. The magnetic lines of force that pass through the
iron object will become shorter if it moves closer to the magnet. Therefore,
the magnetic field attracts iron objects to the closer pole of a magnet.
N
392
Soft iron
Bar magnet
Figure 14-3
N
S
Soft-iron bar in a magnetic field
Note that some characteristics of magnetic lines of force differ from those
of electric lines of force.
See Review Questions 14-27 to 14-30 at the end of the chapter.
14-2 Magnetic Field around a Current-Carrying Conductor
393
14-2 Magnetic Field around a
Current-Carrying Conductor
In 1819, the Danish physicist Hans Christian Oersted (1777–1851) discovered that a current-carrying conductor deflected the needle of a nearby
magnetic ­compass, as shown in Figure 14-4. Oersted thus demonstrated
that a moving charge ­produces a magnetic field. The magnetic lines of
force form concentric circles perpendicular to the axis of the conductor. As
we move outward from the centre of the conductor, these circular lines of
force are spaced farther and farther apart, and the strength of the magnetic
field decreases. When the current is turned off, the lines of force collapse
back into the conductor and the magnetic field disappears.
Current-carrying
conductor passing
through sheet of
cardboard
Compass needle
+
+
−
−
(a)
Figure 14-4
(b)
Detecting the magnetic field around a current-carrying conductor
Although the magnetic field around a straight current-carrying conductor
has no north and south poles, the lines of force still have direction, as shown
by the compasses in Figure 14-4(b). Reversing the direction of the current
through the conductor causes all the compass needles to reverse their positions. We can determine the direction of the magnetic field around a straight
current-carrying conductor by the right-hand rule illustrated in Figure 14-5.
When the thumb of a right hand points in the direction of conventional current through a conductor, the fingers point in the direction
of the magnetic lines of force when grasping the conductor.
Right hand
Direction
of flux
Conventional current
−
Figure 14-5
+
Right-hand rule for a straight conductor
Modern atomic theory
suggests that all magnetic properties of
matter stem from the
motion of electrons
within atoms.
394
Chapter 14 Magnetism
If we form the current-carrying conductor into a loop, as in Figure 14-6,
the magnetic lines of force all pass through the centre of the loop in the
same direction. For a given current, the total number of lines of force has
not changed; therefore, we have strengthened the magnetic field by concentrating it into a smaller area.
−
+
Figure 14-6
a loop
Concentrating the magnetic field by forming the conductor into
We can concentrate the magnetic field of the conductor even more by
winding the conductor around a cylindrical core to form a solenoid, as in
Figure 14-7. Because the current in the adjacent turns of the solenoid travels in the same direction around the coil, the solenoid acts like a single loop
of a stranded conductor, with each strand carrying a current equal to the
S
−
+
N
Figure 14-7
Magnetic field around a current-carrying solenoid
actual current through the solenoid. As shown in Figure 14-7, the magnetic
field around the solenoid has a pattern similar to that around a bar magnet.
This solenoid is a simple type of electromagnet.
We can again apply the right-hand rule to find the direction of the magnetic lines of force around a solenoid, as illustrated in Figure 14-8.
When the fingers of a right hand circling a solenoid point in the
­direction of conventional current, the thumb points in the direction
of the magnetic lines of force through the centre of the solenoid.
Source: © Tim@awe/Dreamstime/Get-stock
14-2 Magnetic Field around a Current-Carrying Conductor
Iron filings on a sheet of paper resting on top of a
solenoid align with the lines of magnetic force around
the solenoid.
Direction
of flux
Right hand
N
S
+
−
Conventional
current
Figure 14-8
Right-hand rule for a solenoid
Because the solenoid acts as an electromagnet, we can treat the end of
the solenoid where the direction of the magnetic lines of force is away from
the coil (Figure 14-8) as its north pole. Therefore, the thumb in the righthand rule also indicates the north pole end of an electromagnet.
395
396
Chapter 14 Magnetism
A magnetic force exists between two parallel current-carrying conductors. Figure 14-9(a) shows a cross section of equal currents flowing in opposite directions through two parallel conductors. The cross on the conductor
on the left represents the tail feathers of an arrow, indicating that conventional current is flowing into the page. Similarly, the dot on the conductor
on the right represents the point of the arrow, indicating that conventional
current is flowing out of the page. If we imagine a right hand with the
thumb pointing down into the left conductor, we can verify the direction of
the magnetic lines of force. Between the two conductors, the magnetic lines
of force all have the same direction. Since they repel one another, there is a
resultant magnetic force of repulsion between the two conductors.
×

×
(a)
Figure 14-9
Starting in 2019,
the definition of the
ampere will be based
on the fixed value
of the elementary
charge, e, rather than
the magnetic force
described in this
section.
While no longer
­considered the
­defining terms of the
ampere, the magnetic
force described in this
section is still measurable and real.
×
(b)
Magnetic field around parallel current-carrying conductors
In Figure 14-9(b), the conventional current in both conductors flows into
the page. The magnetic lines of force between the conductors have opposite
directions. The outer magnetic lines of force around each conductor have
combined to form a shorter single loop that encompasses both conductors.
The tendency of these lines of force to become even shorter results in a
magnetic force of attraction between the two conductors. The magnitudes
of these magnetic forces depend on the length and spacing of the two conductors and on the magnitude of the currents through them.
In the International System of Units prior to 2019, the ampere was
­defined in terms of the magnetic force between two current-carrying electric conductors:
An ampere is the constant current that produces a force of 2 × 10−7 N/m
when flowing through two straight parallel conductors of ­infinite
length and negligible cross section 1 m apart in a v
­ acuum.
See Problems 14-1 to 14-4 and Review Question 14-31.
14-3 Magnetic Flux
We have been using magnetic lines of force to represent magnetic flux
in diagrams of magnetic fields. We can also think of magnetic flux as the
average magnetic field passing through a perpendicular surface times the
area of the surface.
14-4 Magnetomotive Force
397
The letter symbol for magnetic flux is the Greek letter Φ (phi).
For current to flow in an electric circuit, the circuit must form a closed
loop. Since magnetic lines of force always form closed loops, we can refer
to the path we can trace around these loops as a magnetic circuit. Although
there is no flow of magnetic particles comparable to the flow of electrons
in an electric circuit, there are many similarities between the properties of
magnetic circuits and those of electric circuits.
Faraday connected a galvanometer across a solenoid, as shown in
Figure 14-10. When he thrust a permanent magnet into the solenoid, he
noticed a momentary deflection of the galvanometer’s pointer, indicating
that the moving magnetic field induced a current in the solenoid. Faraday
discovered that the induced voltage appeared only when he moved the
magnet with respect to the electric circuit, and that the faster he moved
the magnet, the greater the deflection of the meter pointer. Thus, the magnetic field links the magnetic circuit to the electric circuit and produces
an EMF in the solenoid. Faraday’s discovery provided a means of specifying the basic unit of magnetic flux, the weber.
The SI unit of magnetic
flux was named in
honor of the German
physicist Wilhelm
­Eduard Weber
(1804–91), who
­established many
of the fundamental
theories of magnetism.
Solenoid
N
S
Motion
Bar
magnet
G
Galvanometer
Figure 14-10
Faraday’s demonstration of electromagnetic induction
A change of one weber per second in the magnetic flux linking a
single-turn coil induces an average EMF of one volt in the coil.
The unit symbol for weber is Wb.
A weber is equivalent to a volt second (1 Wb = 1 V·s).
See Review Question 14-32.
14-4 Magnetomotive Force
Just as an electric current cannot flow in an electric circuit until the circuit
is connected to a voltage source, a magnetomotive force (MMF) is necessary to produce magnetic flux. Magnetomotive force in a magnetic circuit
is the counterpart of electromotive force in an electric circuit. Like EMF,
MMF is not actually a force that could be measured in newtons. Since magnetic flux in a solenoid appears only when current flows through the coil,
Although current
­standards avoid the
use of script letters,
the script letter 𝔉 is
sometimes used to
­distinguish magnetomotive force from
magnetic force,
which also has the
symbol Fm.
398
Chapter 14 Magnetism
­ agnetomotive force must be a direct result of the current. Therefore, the
m
unit of MMF is based on the current in a single-turn coil of wire.
The letter symbol for magnetomotive force is Fm.
The ampere is the SI unit of magnetomotive force.
Since each turn of wire added to a solenoid increases the strength of the
magnetic field in the solenoid by the same amount, the magnetomotive
force is the product of the coil current and the number of turns in the coil.
Fm = NI
The ampere-turn is
not an SI unit and is
no longer used by the
Institute of Electrical
and Electronics
­Engineers (IEEE).
(14-1)
where Fm is the magnetomotive force in amperes (effective), N is the
number of turns of wire in the coil, and I is the actual current through
the coil in amperes.
To avoid confusion between effective amperes of MMF and actual amperes
of current, this book uses the ampere-turn (At) as the unit for MMF. For
­example, the MMF of a 10-turn coil carrying a current of 2.0 A is 20 At.
See Problem 14-5.
14-5 Reluctance
For a given magnetic circuit of nonmagnetic materials, the magnetic flux is
directly proportional to the applied magnetomotive force. This constant is
called the reluctance of the magnetic circuit. Reluctance in a magnetic circuit is the counterpart of resistance in an electric circuit.
Some books use the
script letter ℜ to
­distinguish reluctance
from resistance, R.
Reluctance is the opposition of a magnetic circuit to the establishing
of magnetic flux.
The letter symbol for reluctance is Rm.
Rm =
Fm
Φ
(14-2)
where Rm is reluctance, Fm is magnetomotive force in amperes, and Φ is
the magnetic flux in webers.
Section 16-5 describes
the henry, which
is the SI unit for
­inductance.
Note how Equation 14-2 closely resembles the equation for Ohm’s law,
R = V/I, with reluctance corresponding to resistance, magnetomotive
force corresponding to voltage (or electromotive force), and magnetic flux
corresponding to current.
Reluctance can be expressed in ampere-turns per weber. The SI unit
of ­reluctance is the reciprocal henry (symbol H−1), which is equal to one
­ampere per weber: 1 H−1 = 1 A/Wb.
See Problems 14-6 to 14-8 and Review Question 14-33.
14-6 Permeance and Permeability
399
14-6 Permeance and Permeability
When dealing with parallel electric circuits, we found it more convenient
to think in terms of the conductances of the circuit elements rather than in
terms of their resistances. Similarly, when dealing with magnetic circuits, we
find it more convenient to think in terms of the ability of magnetic circuits to
permit magnetic flux rather than in terms of their opposition to it. Permeance
in a magnetic circuit is the counterpart of conductance in an electric circuit.
Permeance is a measure of the ability of a magnetic circuit to permit
magnetic flux.
The letter symbol for permeance is Pm.
Pm =
1
Rm
(14-3)
We can express permeance in webers per ampere-turn. The SI unit of
permeance is the henry.
To compare the magnetic properties of various materials, we calculate the
permeance per unit length and cross section, which is called ­permeability.
Permeability indicates the ability of a material to permit magnetic flux,
whereas permeance is a measure of the ability of a given magnetic circuit to
permit magnetic flux. Permeability is comparable to permittivity, which is
a measure of the ability of a material to permit electric flux.
The letter symbol for permeability is the italic lowercase Greek
­letter μ (mu).
μ = Pm
l
l
=
A
RmA
(14-4)
where μ is the permeability in ampere-turns per metre, Pm is permeance in
webers per ampere-turn, Rm is reluctance in ampere-turns per weber, l is
the length of the magnetic circuit in metres, and A is the cross-sectional area
in square metres. In SI, permeability is expressed in henrys per metre.
Note that the Greek letter μ (not italic) is also used for the prefix micro.
Fortunately, this prefix is rarely needed in calculations for magnetic circuits.
As with permittivity, it is often convenient to express permeability in
terms of the ratio of the permeability of a material to the permeability of
free space, μ0, which is a physical constant. This ratio is referred to as relative permeability, μr. Since it is a ratio, relative permeability has no units.
μ
μr =
(14-5)
μ0
where μ0 = 4π × 10−7 H/m.
For all non-magnetic materials, μr = 1. For magnetic materials, μr ranges
from 50 to 25 000.
Some books use the
script letter 𝔓 to distinguish permeance
from power, P.
400
Chapter 14 Magnetism
Example 14-1
A solenoid of 1000 turns is energized by a current of 250 mA. A fluxmeter measures a total flux of 1.0 mWb in this magnetic circuit. The core
has an average path length of 25 cm and a cross-sectional area of 6.5 cm2.
Calculate
(a) the reluctance of the magnetic circuit
(b) the permeability of the core
(c) the relative permeability of the core
Solution
(a) Fm = NI = 1000 × 0.25 A = 250 At
Rm =
(b) μ =
(c)
Fm
250 At
= 2.5 × 105 At/ Wb
=
Φ
1.0 × 10 − 3 Wb
1
RmA
μr =
=
0.25 m
= 1.5 × 10 − 3H/m
( 2.5 × 10 At/Wb ) ( 6.5 × 10 −4 m2 )
5
μ
1.54 × 10 − 3
=
= 1.2 × 103
μ0
4π × 10 − 7
See Problems 14-9 to 14-16 and Review Question 14-34.
Circuit Check
A
CC 14-1.
Find the current that will develop a magnetomotive force of
180 At when flowing through a 1500-turn coil.
CC 14-2. How much flux will be produced if a current of 4.0 A flows in a
1600-turn coil wound on a magnetic circuit having a reluctance
of 5.8 MAt/Wb?
14-7 Magnetic Flux Density
The tesla is named in
honor of the inventor
Nikola Tesla (1856–1943),
who ­discovered key
­principles of alternating
current and ­developed
the first AC induction
motor.
For calculations involving permeability we are not dealing with total flux
in a magnetic circuit. Rather, we are interested in the magnetic flux density.
Magnetic flux density is the magnetic flux per unit area.
The letter symbol for flux density is B.
The tesla is the SI unit of flux density.
The unit symbol for tesla is T.
One tesla equals one weber per square metre: 1 T = 1 Wb/m2
14-8 Magnetic Field Strength
B=
Φ
A
401
(14-6)
where B is the magnitude of the magnetic flux density in teslas, Φ is the
magnitude of the total magnetic flux in a magnetic circuit, and A is the
cross-sectional area of the circuit in square metres.
See Problem 14-17 and Review Question 14-35.
Circuit Check
B
CC 14-3. Determine the reluctance for the coil in Example 14-1 if the flux
is 4 mWb.
CC 14-4. Determine the reluctance of an iron rod 3.0 cm in diameter and
8.0 cm long. The relative permeability of the iron is 3500.
CC 14-5. Calculate the flux density for the magnetic circuit in Example 14-1.
14-8 Magnetic Field Strength
Since permeability is based on a unit cube of the material, we are not concerned with the magnetomotive force required for the full length of the magnetic circuit. Rather, we are interested in the magnetomotive force required
to create a certain flux density in a unit length of the magnetic circuit.
Magnetic field strength is magnetomotive force per unit length.
The letter symbol for magnetic field strength is H.
H=
Fm
l
(14-7)
where H is the magnitude of the magnetic field strength in ampereturns per metre, Fm is the magnetomotive force in amperes, and l is the
length of the magnetic circuit in metres. In SI, magnetic field strength is
expressed in effective amperes per metre.
Now that we have defined permeability, flux density, and magnetic field
strength, we can derive a very useful relationship among these three magnetic quantities.
Since Rm =
Fm
,
Φ
Substituting in Equation 14-4 gives
μ=
Pm =
Φ
Fm
Φ
l
Φ
l
×
=
×
Fm A
A Fm
Magnetic field
strength is also
called magnetic field
­intensity, magnetizing
force, and magnetomotive force gradient.
402
Chapter 14 Magnetism
But
Therefore,
Φ
=B
A
Fm
=H
l
and
μ=
B
H
(14-8)
where μ is permeability in henrys per metre, B is the magnitude of the
magnetic flux density in teslas, and H is the magnitude of the magnetic
field strength in ampere-turns per metre (or amperes per metre).
Comparing Equation 14-8 with Equation 12-8, we note that:
• magnetic flux density in a magnetic circuit is the counterpart of electric
flux density in a dielectric in an electric circuit
• magnetic field strength in a magnetic circuit is the counterpart of electric
field strength in an electric circuit
• permeability in a magnetic circuit is the counterpart of permittivity in an
electric circuit
See Problem 14-18 and Review Question 14-36.
14-9 Diamagnetic, Paramagnetic,
and Ferromagnetic Materials
The way atoms in a material respond to a magnetic field determines the
­permeability of the material, which, in turn, governs the magnetic flux that
can be established by a given magnetomotive force. However, the behaviour of materials in magnetic circuits is not as simple as the polarization of
atoms of a dielectric shown in Figure 12-3.
Most substances have no more effect on a magnetic circuit than free
space does. These nonmagnetic materials have the same permeability as free space. A few materials have a permeability slightly less than
that of free space. Such diamagnetic materials oppose magnetic flux
slightly more than free space does. Silver, copper, and hydrogen are diamagnetic. A few materials have a permeability just slightly greater than
that of free space. These paramagnetic materials are, therefore, slightly
magnetic. Platinum, aluminum, and oxygen are paramagnetic. For most
­applications, we can treat diamagnetic and paramagnetic materials as if
they were nonmagnetic.
Some substances, including iron and its compounds (and, to a lesser
extent, nickel and cobalt), have permeabilities many hundreds of times
greater than that of free space. These substances are called ferromagnetic
materials since iron is the significant element in this group.
We can explain the magnetic properties of materials in terms of the ­magnetic
fields produced by moving charges within the atoms in the materials. Every
electron travelling in an orbit around the nucleus of an atom is somewhat
14-9 Diamagnetic, Paramagnetic, and Ferromagnetic Materials
like a tiny electric current flowing in a single-turn loop. Consequently,
every orbital electron generates a tiny magnetic field (and magnetic flux)
­perpendicular to the plane of its orbit. Each electron also behaves as if it were
spinning on an axis through its centre. Thus the charge on the electron circulates around the axis through the electron, generating an additional magnetic
field. The ­positively charged nucleus can have a similar magnetic field.
In most materials the orbital and spin magnetic fields all cancel out, so
the atoms have no net magnetic field. For this reason, nonmagnetic materials have virtually the same permeability as a vacuum. However, an external magnetic field can shift the orbitals of the electrons, causing the atoms
to have a slight net magnetic field that opposes the external field. Hence,
the diamagnetism reduces the net strength of the magnetic field in a material. All materials are diamagnetic to some degree.
The atoms in paramagnetic materials have a small net magnetic field.
These atoms tend to behave individually like permanent magnets. When
we expose paramagnetic materials to an external magnetic field, their atoms
tend to align their magnetic fields with the external field, thus slightly increasing the total magnetic field and the resulting flux. In paramagnetic
materials, this effect is stronger than the opposing diamagnetism.
The atoms in ferromagnetic materials have appreciable net magnetic
fields. As a result, these atoms tend to form magnetic domains, groups of
atoms that have aligned their magnetic fields with each other. A domain
usually contains from 1012 to 1018 atoms. Ferromagnetic materials are made
up of crystals with the atoms arranged in a regular lattice. Initially, the magnetic fields of the domains within each crystal are aligned with the edges of
the faces of the crystal lattice. Since half of the fields aligned with each of
these axes are directed one way and the other half are directed the opposite
way, the ferromagnetic material has no net magnetic field.
When we place a ferromagnetic material inside a solenoid, as in
Fig­ure 14-11, and gradually increase the current in the coil, the material
­becomes magnetized. At first the domains that have magnetic fields oriented close to the direction of the magnetic field in the solenoid expand
as adjacent atoms join them. As the solenoid current increases, the other
domains shift to align with the lattice edges that are closest to being parallel to the magnetic field. This shift greatly increases the total field strength
Iron
sample
−
+
Rheostat
Figure 14-11
Magnetizing an iron sample
403
404
Chapter 14 Magnetism
in the domains, making the permeability appear very high. Then, as the
­magnetizing force increases even further, the orientation of the domains
finally shifts to align with the magnetic flux rather than with the lattice
edges of the crystals. If we increase the current to a point where all the domains are aligned with the magnetic field, any further increase in ­current
can only increase the total flux by the same amount as if the iron were not
present. The iron is now saturated and its permeability has dropped to
that of any nonmagnetic material. The current that is needed to saturate
a ­sample of iron within the solenoid depends on the type and grade of
iron used.
See Review Questions 14-37 and 14-38.
14-10 Permanent Magnets
When the current in the magnetizing solenoid of Figure 14-11 is switched
off, the magnetic domains tend to return to their original orientations.
In materials that have a low magnetic retentivity, such as soft iron, the
­residual magnetism or remanence is practically nil. Such materials are
classified as temporary magnets. Residual magnetism in ferromagnetic
materials is the magnetic counterpart of dielectric absorption (described
in Section 12-2).
In permanent magnets, a large proportion of the domains retain the
orientation they were forced to take in the strong magnetic field produced
by the energizing coil. The permanent magnets for meter movements,
loudspeakers, and similar devices are often made of alnico, a very hard
alloy containing iron, aluminum, nickel, and cobalt. Because of its hardness, an alnico magnet is usually cast in the desired shape and subjected
to a very high magnetic flux density produced by an electromagnet while
it is cooling.
Ceramics containing iron oxides can have strong ferromagnetic properties. Among these ferrites are some that make excellent permanent
magnets. Such ceramic magnets can be formed by grinding barium carbonate and ferric oxide into particles the size of a magnetic domain,
pressing the powder into the desired shape, and firing it in an oxygen
atmosphere. Unlike iron magnetic materials, ferrites are poor electric
conductors.
See Review Question 14-39.
14-11 Magnetization Curves
The permeability of a sample of iron varies considerably as the current in
the coil is increased. The permeability depends on what percentage of the
magnetic domains has aligned with the magnetic field. This percentage, in
14-11 Magnetization Curves
turn, is dependent on the flux density. Even though we can draw a graph,
as in Figure 14-12, showing how the permeability of a sample of iron varies
with flux density, we are faced with the interdependence that becomes
­evident when we rearrange Equation 14-8:
B = μH
Permeability, µ
For a given magnetic field strength H, the flux density B depends on the
permeability μ, which depends on the flux density B. However, we can disregard permeability and concern ourselves only with the manner in which
the flux density of a given material varies with the magnetic field strength.
Since B and H are both defined in terms of a cube with unit dimensions, the
magnetization curve, or BH curve, (produced by plotting a graph of the
manner in which B varies with H) is dependent only on the type of material
and not on its dimensions.
Magnetic Flux Density, B
Figure 14-12
Variation of the permeability of cast steel with flux density
Magnetic Flux Density, B
The magnetization curve of Figure 14-13 can be divided into three
­sections. The magnetic flux density increases slowly at first. The lower
knee of the BH curve corresponds to the conditions where the domains
that are most nearly aligned with the applied magnetic field grow at the
expense of neighbouring domains. In the steep portion of the BH curve,
the domains are aligning with whichever axes of the crystals are closest to the direction of the applied magnetic field. When the magnetic
field is great enough to i­ncrease the flux density beyond the upper knee
of the BH curve, the magnetic field strength is great enough to force
Saturation
Upper knee
Lower knee
Magnetic Field Strength, H
Figure 14-13
Typical BH curve for cast steel
405
Chapter 14 Magnetism
the domains to align themselves with the external magnetic field rather
than with the axes of the crystals. F
­ inally, the BH curve becomes a straight
line with a gradual slope when all the domains are aligned with the magnetic field. This slope shows the same increase in B for a given increase in
H as for any nonmagnetic material.
It is difficult to mark the exact point on the BH curve of Figure 14-13
where saturation occurs. Theoretical saturation occurs when all the domains are finally aligned with the external magnetic field. For some types
of iron, this saturation requires considerable magnetic field strength.
Practical saturation may be defined as the flux density beyond which it
is impractical to magnetize a certain magnetic material. For some applications, this point is the upper knee in the BH curve where the steep,
linear part of the slope ends. In high-quality audio transformers, magnetic field strength must be kept below this point, since any swing in H
past this point causes distortion in the output signal. However, we can
reduce the amount of iron needed for a power transformer if we design
it so that the magnetic cores are magnetized to a point slightly beyond
the upper knee of the BH curve.
The gradual saturation of most types of iron, as shown in Figure 14-15, is
due to the random orientation of the iron crystals. If iron can be manufactured so that all crystals have the same orientation, the flux density will rise
very steeply with an increase in magnetic field strength. At a certain value of
magnetic field strength, the iron will suddenly become saturated (as shown
in Figure 14-14) because the same magnetic field strength is needed to force
all of the domains to align themselves with the magnetic field. This type of
iron was typically used for magnetic recording of audio and video signals
as well as computer data and measurements in videotapes, floppy disks,
scientific and medical research equipment, and hard drives.
Magnetic Flux Density, B
406
Saturation
Magnetic Field Strength, H
Figure 14-14
BH curve for iron with all crystals having the same orientation
Figure 14-15 shows typical magnetization curves for a number of ferromagnetic materials used in practical applications.
See Problems 14-19 and 14-20 and Review Questions 14-40 and 14-41.
14-11 Magnetization Curves
407
1.8
1.7
Sheet
1.6
steel
el
t ste
Cas
1.5
1.4
1.3
Magnetic Flux Density, B (teslas)
1.2
1.1
1.0
0.9
0.8
0.7
n
t iro
Cas
0.6
0.5
0.4
0.3
0.2
0.1
0
0
400
Figure 14-15
800
1200
1600
2000 2400 2800 3200 3600 4000 4400 4800
Magnetic Field Strength, H (ampere-turns per metre)
Typical magnetization curves
5200
5600
6000
6400
Chapter 14 Magnetism
14-12 Permeability from the BH Curve
As shown in Figure 14-12, the permeability of a ferromagnetic material
is far from constant since it is quite dependent on flux density. However,
we can determine permeability at a given magnetic flux density by applying the relationship μ = B/H (Equation 14-8) to the BH curve for a given
type of magnetic material. The manner in which we apply this relationship depends on the conditions under which the magnetic circuit is to be
operated.
If the current in the energizing solenoid is to be kept constant, we can locate
the appropriate value for magnetic field strength and the resulting value
of flux density on the BH curve, as shown in Figure 14-16(a). The normal
permeability for this flux density is simply
μ=
B
H
(14-8)
Magnetic Flux Density, B
Magnetic Flux Density, B
In some magnetic circuits, the magnetic field strength varies between
two limits, as shown by ΔH in Figure 14-16(b). This situation occurs when a
fluctuating direct current flows in the coil, as in many audio transformers.
μ= B
H
ΔB
μΔ = ΔB
ΔH
ΔH
Magnetic Field Strength, H
(b) Incremental permeability
Magnetic Field Strength, H
dH
dB
X
μd = dB
dH
Magnetic Field Strength, H
(c) Differential permeability
Figure 14-16
Magnetic Flux Density, B
(a) Normal permeability
Magnetic Flux Density, B
408
Knee
μav = B
H
Magnetic Field Strength, H
(d) Average permeability
Determining permeability from a BH curve
14-12 Permeability from the BH Curve
In this case, we are interested in the permeability over a limited operating
range. This incremental permeability is
μΔ =
ΔB
ΔH
(14-9)
where μΔ is the incremental permeability, ΔB is the difference between
the maximum and minimum magnitudes of magnetic flux density, and
ΔH is the difference between the maximum and minimum magnitudes
of magnetic field strength.
If alternating current flows in the solenoid, the magnetic field strength
is constantly changing. In this case, we are likely to be concerned with the
permeability as the magnetic field strength swings through a certain value,
rather than as it maintains that value. This particular value of permeability is called the differential permeability for a particular point on the BH
curve. We determine differential permeability from the slope of the magnetization curve at the point in question. To calculate the slope of the BH
curve at point X in Figure 14-16(c), we draw a tangent to the curve at point
X and then determine dB/dH from the slope of the tangent. The differential permeability at point X is
μd =
dB
dH
(14-10)
In electronic circuits where the current is in the order of microamperes,
the magnetic field strength is so small that the B/H ratio is difficult to read
from the magnetization curve. However, we can use a tangent to calculate
the permeability for very small values of H and B. Thus, initial permeability (μi) is the differential permeability when the magnetic field strength is
zero or very small.
We can simplify the problem of solving magnetic circuits in which the
magnetic field strength and, consequently, the permeability are continually changing by determining an average permeability. The average permeability is obtained by drawing a straight line through the origin of the
BH graph and the upper knee of the curve as in Figure 14-16(d). Because
this line is straight, we can pick any point on this line to read off the corresponding values of B and H for determining μav. As long as the flux density
in a magnetic circuit is kept below saturation, this average permeability is
reasonably accurate.
See Problems 14-21 to 14-26 and Review Question 14-42.
409
410
Chapter 14 Magnetism
Circuit Check
C
CC 14-6. For the magnetic circuit shown in Figure 14-17, the total flux
­produced is 2.0 mWb. Given that the coil has 1250 turns and
the ring has a circular cross section with a diameter of 2.0 cm,
calculate the magnetic field strength and the permeability.
2.5 A
+
E
R12.5 cm
−
Figure 14-17
CC 14-7. Use the magnetization curves in Figure 14-15 to determine the
normal permeability of sheet steel at a flux density of 1.15 T.
CC 14-8. Determine the average permeability of cast iron.
Magnetic Flux
Density, B
14-13 Hysteresis
+
A
B
−
C
F
O
E
+
Magnetic
Field
Strength, H
D
−
Figure 14-18
Typical hysteresis loop
The BH curves of Figures 14-12 to 14-16 represent the manner in which
the flux density of an unmagnetized sample of iron rises as the value of H
is increased. However, they do not represent the manner in which the flux
density drops as the magnetic field strength is decreased. When the magnetizing force has returned to zero, the residual magnetism of the iron
will still produce an appreciable magnetic flux density. This is referred
to as r­ etentivity. To get rid of this residual flux, it is necessary to pass
some current through the solenoid in the opposite direction. The amount
of ­reverse magnetic field strength or magnetizing force ­required to demagnetize a particular sample of material is called the coercive force.
Figure 14-18 shows the complete cycle of magnetization of a piece of
iron as we pass an alternating current through the solenoid. If we start
with an unmagnetized sample and increase the magnetic field strength
H from zero to the maximum positive value, the magnetic flux density B
increases along the magnetization curve OA. As we reduce H to zero, the
flux density decreases from saturation to a residual flux density at point B.
To bring the flux density in the iron to zero, we must reverse the current
through the coil and increase it until H reaches the value represented
14-13 Hysteresis
by OC. As we increase the reverse current, the iron saturates at point D
(with the magnetic domains all oriented in the direction opposite to the
orientation they had at point A). When we reduce the reverse current to
zero, the flux density in the iron ­decreases only to the value at point E.
We run current through the solenoid in the original direction to bring
the flux density to zero at point F. If we again ­increase H to its maximum
positive value, the iron again becomes saturated and the flux density returns to its value at point A.
As shown in Figure 14-18, when the magnetic field strength is due to
an alternating current, the flux density of ferromagnetic materials tends
to lag behind the magnetic field strength that creates it. This lagging of the
magnetization of the iron behind the magnetic field strength is called hysteresis. The graph that shows this lag is called a hysteresis loop. Since
only ferromagnetic materials have residual magnetism, nonmagnetic materials show no hysteresis effect.
Hysteresis occurs because some of the magnetic domains in a ferromagnetic material do not return to their original orientation. They have to be
forced to do so by coercive force from a reversed magnetic field. Whenever
motion is accomplished against an opposing force, an energy transfer must
take place. This energy comes from the source of alternating voltage and
is transferred to the molecules of the ferromagnetic material in the form of
heat. The higher the frequency of the alternating current in the solenoid,
the more rapidly the magnetic domains have to change their alignment
and, therefore, the greater the hysteresis loss.
The greater the retentivity of a particular type of iron, the greater the
­coercive force needed to demagnetize it. The greater the opposition by
the magnetic domains to reorientation, the greater the heating of the
­ferromagnetic material. Therefore, hysteresis loss is also proportional
to the retentivity of the iron. The iron used in the magnetic circuits of
transformers for AC circuits should have as little retentivity as possible in
order to minimize hysteresis losses. In graphs like Figure 14-18, the area
within the hysteresis loop increases as the residual flux density increases.
Therefore, the area within a hysteresis loop is a useful indication of
hysteresis loss.
Hysteresis can be used to advantage in some applications. Certain
f­errites composed of oxides of magnesium, manganese, and iron have an
almost rectangular hysteresis loop, as shown in Figure 14-19. As the magnetic field strength H is gradually increased in a positive direction, the flux
density in the core remains unchanged until the magnetic field strength
reaches a ­critical level, at which the core suddenly magnetizes and ­saturates
in the positive direction. A sudden demagnetization also occurs when the
magnetic field strength is gradually increased in the negative direction.
Consequently, these ferrite cores are always saturated in either the positive
or negative ­direction. These two stable states of ferrite cores make them
very useful for switching circuits.
411
Chapter 14 Magnetism
Magnetic Flux
Density, B
412
+
−
+
Magnetic
Field
Strength, H
−
Figure 14-19
Rectangular hysteresis loop of a ferrite switching core
See Review Questions 14-43 and 14-44.
14-14 Eddy Current
In Section 14-3, we found that moving a magnetic field in relation to an
electric conductor induced a voltage in the conductor. The direction of the
magnetic lines of force, the motion of the magnetic lines of force with respect to the conductor, and the flow of current are all at right angles to one
another. If we place a metal core inside a solenoid, any motion or change in
the magnetic field will induce a voltage in the metal.
As we increase the current in the solenoid, more magnetic lines of force
(or flux) must be set up. Crowding more lines of force into the core causes
the lines of force already present to move outward along a radius from the
axis of the solenoid. The direction of the lines of force themselves is parallel
to the axis of the solenoid. When an alternating current flows in the ­solenoid,
the lines of force move alternately away from the axis of the core and then
back toward it. This changing magnetic field induces a voltage in the metal
core of the solenoid. The resulting current must be perpendicular to the
radius and the axis of the solenoid. Therefore, the current flows in a circle
around the circumference of the metallic core, as shown in Figure 14-20.
This circular current is an example of an eddy current. Eddy current
can be produced in any conductive core material, including nonmagnetic
­materials. As the eddy current flows through the resistance of the core
­material, the resulting I2R power loss heats the core.
Solenoid winding
Eddy current flows around
circumference of core
Alternator
Motion of magnetic
lines of flux along
radii of solenoid
Solid metal core
Figure 14-20
Eddy current in a solid metal core in a solenoid (cross section)
14-15 Magnetic Shielding
Eddy-current losses can be greatly reduced by using a core made of thin
laminations that are insulated from each other with a thin coat of varnish.
As shown in Figure 14-21, the laminations are oriented so that the eddy current would have to flow from lamination to lamination through the varnish.
Because of the high resistance of the varnish, the eddy current in a laminated
core is very small and the resulting power loss is, in many cases, negligible.
Coil
Alternator
Laminated steel core
Figure 14-21
Cross section of solenoid with a laminated core
Eddy-current losses in alternating current machinery such as ironcore transformers and electric motors often reduce the efficiency of these
devices. However, eddy currents may also be transformed into other types
of energy, such as heat and electromagnetic forces. Coreless induction furnaces use eddy currents to melt steel and other ferrous metals while at
the same time stirring the molten materials inside their crucibles. Some
trains use eddy-current brakes, whose magnetic forces are proportional to
the velocity of the train, since they typically produce a smoother stopping
motion compared to air brakes.
See Review Questions 14-45 and 14-46.
14-15 Magnetic Shielding
Although there is no “insulator” for magnetic lines of force, it is possible
to shield sensitive equipment from stray magnetic fields by surrounding it
with a high-permeability material. As shown in Figure 14-22, most of the
magnetic flux follows the lower reluctance path, greatly reducing the flux
within the enclosure.
Shielded space
Stray
magnetic
field
Soft iron cylinder (not to scale)
Figure 14-22
Magnetic shielding
See Review Question 14-47.
413
414
Chapter 14 Magnetism
Summary
• A magnetic line of force (or flux) represents the path along which a theoretical
isolated magnetic pole would move from one pole of a magnet to another.
• Magnetic lines of force form complete loops extending from the north
pole to the south pole of a magnet and then through the magnet back to
the north pole.
• Like poles of magnets repel one another.
• A magnetic field exists around a current-carrying conductor.
• The right-hand rule may be used to find the direction of the magnetic
field around a current-carrying conductor or solenoid.
• A moving magnetic field whose flux cuts across a solenoid induces a
voltage across the solenoid.
• Magnetomotive force is proportional to both the number of turns of
wire and the current flowing through a solenoid.
• Reluctance is the opposition of a magnetic circuit to the establishing of
magnetic flux.
• Permeability, a measure of the ability of a magnetic circuit to establish magnetic
flux, is equal to the ratio of magnetic flux density to magnetic field strength.
• A material is classified as diamagnetic, paramagnetic, or ferromagnetic
depending on its permeability compared to that of free space.
• Iron and steel alloys may be magnetized either temporarily or permanently depending on their retentivities.
• The permeability of a material may be determined from its BH curve.
• A magnetic material exhibits hysteresis as it is magnetized in one direction and then magnetized in the opposite direction.
• An eddy current is established in a solenoid when it is energized by an
alternating current.
• Magnetic shielding may be used to protect electric devices from magnetic fields.
+
−
Figure 14-23
B = beginner
Problems
I = intermediate
A = advanced
GEN
Figure 14-24
B
Section 14-2
Conductor
14-1.
B
14-2.
B
14-3.
B
14-4.
B
Sketch a diagram of the magnetic lines of force around the ­conductor
in Figure 14-23 and indicate their direction.
What is the direction of conventional current through the ­conductor
in Figure 14-24?
Sketch a diagram of the magnetic field around the coil in Figure 14-25
and indicate its direction.
What is the direction of conventional current through the coil of
Fig­­­ure 14-26?
Section 14-4
14-5.
Magnetic Field around a Current-Carrying
Magnetomotive Force
A current of 5 A is flowing through the conductor in Figure 14-25.
Find the resulting magnetomotive force.
415
Problems
B
Section 14-5
14-6.
B
14-7.
B
14-8.
B
B
14-10.
B
14-11.
B
14-12.
I
14-13.
I
14-14.
B
14-15.
I
14-16.
I
B
Find the reluctance of a magnetic circuit in which a total flux of
2.0 × 10–3 Wb is set up by a 4.0-A current flowing in a solenoid
­consisting of 300 turns of wire.
Calculate the current that must pass through a 400-turn solenoid to
produce a total flux of 1.5 mWb in a magnetic circuit whose reluctance is 1.2 × 106 At/Wb.
A 250-mA current through a solenoid produces a total flux of 6.0 ×
10–4 Wb in a magnetic circuit whose reluctance is 2.5 × 105 At/Wb.
How many turns of wire are in the solenoid?
Section 14-6
14-9.
Reluctance
Permeance and Permeability
The core of a solenoid consists of a brass cylinder 7 cm in length
and 1.5 cm in diameter. Find its reluctance.
The magnetic circuit in Problem 14-6 has a uniform cross-sectional
area of 6.0 cm2 and an average path length of 20 cm. What is the
permeability of the core material?
If the magnets in Figure 14-2(a) have a rectangular cross section of
1 × 2 cm and the spacing between them is 2.5 mm, find the reluctance of the air gap.
A piece of iron with a length of 8 cm and a rectangular cross section of 1.5 × 2.5 cm has a reluctance of 5.3 × 105 At/Wb. Find the
permeability of the iron.
An iron rod is 50 cm long and has a diameter of 4.0 cm. A 250-turn
coil is wound on the rod. When 2.0 A of current is supplied, a flux of
5 mWb is measured. Find the permeability of the iron.
Calculate the reluctance of an iron rod 100 cm long and 4.0 cm in
­diameter. The relative permeability of the iron is 3000.
Determine the flux produced if the solenoid in Example 14-1 has
2500 turns.
A flux of 1.5 mWb is produced when a current of 400 mA flows
through a 300-turn coil. The magnetic core is a cylindrical rod of
diameter 4.0 cm and length 25 cm. Calculate the
(a) reluctance
(b) permeability
(c) relative permeability
Section 14-7
Magnetic Flux Density
14-17. If 150 turns of wire are wrapped around the iron bar in Problem 14-12
and a current of 1.6 A is supplied, what is the flux density in the iron?
Section 14-8
Magnetic Field Strength
14-18. A magnetic field strength of 2000 At/m produces a flux density of
1.0 T in a certain type of iron. What is its permeability at this flux
density?
+
−
Figure 14-25
S
N
GEN
Figure 14-26
416
Chapter 14 Magnetism
Refer to the typical magnetization curves of Figure 14-15 in solving the r­ emaining
problems.
B
B
B
B
B
B
B
B
Section 14-11
Magnetization Curves
14-19. What flux density is created in sheet cast by a magnetizing force of
1400 At/m?
14-20. What magnetic field strength is required in order to maintain a
­constant flux density of 1.4 T in sheet steel?
Section 14-12
Permeability from the BH Curve
14-21. Find the normal permeability of cast iron when the flux density
­remains at 0.625 T.
14-22 Find the average permeability of sheet steel.
14-23. Find the incremental permeability of cast iron between flux
­densities of 0.30 T and 0.35 T.
14-24. Determine the differential permeability of cast steel for a magnetic
field strength of 1600 At/m.
14-25. Calculate the incremental permeability of cast steel between a
­magnetic field strength of 2000 and 3600 At/m.
14-26. Calculate the differential permeability of sheet steel at a flux
­density of 1.3 T.
Review Questions
Section 14-1
Magnetic Fields
14-27. In what way are gravity, electricity, and magnetism similar?
14-28. Sketch the magnetic field around a horseshoe-shaped perma­nent
magnet.
14-29. With reference to the characteristics of magnetic lines of force,
show why a nail is attracted to either pole of a horseshoe-shaped
permanent magnet.
14-30. State six characteristics of magnetic lines of force.
Section 14-2
Conductor
Magnetic Field around a Current-Carrying
14-31. Show with sketches how it is possible to determine the direction of
the current in an electric conductor with the aid of a compass needle.
Section 14-3
Magnetic Flux
14-32. How is the magnitude of the weber established?
Section 14-5
Reluctance
14-33. Define the term reluctance.
Section 14-6
Permeance and Permeability
14-34. Define the term permeability.
Review Questions
Section 14-7
Magnetic Flux Density
14-35. Why is the flux density of a magnetic field greater in the centre of a
current-carrying solenoid than around a straight wire carrying the
same current?
Section 14-8
Magnetic Field Strength
14-36. What is the distinction between magnetomotive force and magnetizing force?
Section 14-9 Diamagnetic, Paramagnetic, and
Ferromagnetic Materials
14-37. Compare the permeabilities of nonmagnetic, diamagnetic, and para­
magnetic materials.
14-38. Why can ferromagnetic materials become magnetized whereas para­
magnetic materials cannot?
Section 14-10
Permanent Magnets
14-39. Compare temporary and permanent magnets in terms of residual
magnetism.
Section 14-11
Magnetization Curves
14-40. Account for the upper knee on the BH curve of ferromagnetic materials.
14-41. Sketch a BH curve for aluminum.
Section 14-12
Permeability from the BH Curve
14-42. What is the distinction between differential and incremental
­permeability?
Section 14-13
Hysteresis
14-43. Compare permanent and temporary magnets in terms of coercive
force.
14-44. Define the term hysteresis.
Section 14-14
Eddy Current
14-45. Laminating the iron core of a transformer greatly reduces the
eddy-current losses. What effect does the laminating process have
on ­hysteresis losses? Explain.
14-46. Some magnetic circuits consist of powdered iron mixed with a ceramic binder. Compare such a core with a solid core of the same
type of iron in terms of
(a) permeability
(b) hysteresis
(c) eddy current
Section 14-15
Magnetic Shielding
14-47. How may a sensitive meter movement be protected from a nearby
magnetic field?
417
418
Chapter 14 Magnetism
Integrate the Concepts
Figure 14-27 shows a solenoid of 1000 turns energized by a current of 250 mA.
A flux meter measures a total flux of 1.0 mWb through the solenoid. The core
has an average path length of 25 cm and a cross-sectional area of 6.5 cm2.
250 mA
+
1000-turn
solenoid
E
−
Figure 14-27
(a)Draw the magnetic lines of force, given that the wire is wrapped
around the solenoid so that the upper lead from the voltage source
goes behind the solenoid.
(b) Calculate the magnetomotive force.
(c) Calculate the reluctance of the solenoid.
(d) Calculate the permeance of the solenoid.
(e) Calculate the permeability of the solenoid.
(f) Calculate the flux density in the solenoid.
(g) Calculate the magnetic field strength.
Practice Quiz
1.
Which of the following statements are true?
(a) Ferromagnetic materials cannot be magnetized.
(b) Magnetic lines of force rarely form complete loops.
(c) Magnetic lines of force repel each other and cannot intersect.
(d)The right hand rule is used to determine the direction of the magnetic field around a current-carrying conductor.
2.
The letter symbol for magnetic flux is
(a)B
(b)A
(c)Φ
(d)T
3.
Magnetomotive force is
(a)directly proportional to the number of turns of wire in a coil and
directly proportional to the current through the coil
(b)inversely proportional to the number of turns of wire in a coil
and directly proportional to the current through the coil
(c)directly proportional to the number of turns of wire in a coil and
inversely proportional to the current through the coil
(d)inversely proportional to the number of turns of wire in a coil
and inversely proportional to the current through the coil
Practice Quiz
4. Magnetomotive force can be measured in
(a)milliampere-turns
(b)ampere-turns
(c)microampere-turns
(d) all of the above
5. The magnetomotive force produced by a current of 250 mA flowing
through in a 20-turn coil is
(a) 200 mAt
(b) 5.0 At
(c) 12.5 mAt
(d) 80 At
6. The opposition to magnetic flux is called
(a) relative permeability
(b)permeability
(c)permeance
(d)reluctance
7. The current that flows through a 600-turn solenoid that generates
a total flux of 7.5 mWb in a magnetic circuit that has a reluctance
1.25 × 105 At/Wb is
(a) 15.6 mA
(b) 15.6 A
(c) 1.56 A
(d) 156 mA
8. What property of a magnetic circuit corresponds to resistance in an
electric circuit?
9. Permeability is expressed in units of
(a) henrys per turn
(b) henrys per second
(c) henrys per metre
(d) henrys per weber
10. The tesla (T) is a unit of magnetic flux density that is equivalent to
(a) 1 Wb/m2
(b) 1 Wb/cm2
(c) 1 Wb/cm
(d) 1 Wb/m
11. The letter symbol for magnetic field strength is
(a)Φ
(b)H
(c)B
(d)T
12. Why should eddy currents in a core be minimized?
419
15
Magnetic Circuits
Magnetic circuits are a crucial part of equipment ranging from
speakers and relays to large industrial machines and power
transmission systems. The principles of magnetic circuits
described in this chapter will be of particular interest to students
considering further study of motors and power systems.
Chapter Outline
15-1
15-2
Practical Magnetic Circuits 422
Long Air-Core Coils
422
15-3
Toroidal Coils 425
15-5
Nonlinear Magnetic Circuits 426
15-7
Series Magnetic Circuits 430
15-9
Parallel Magnetic Circuits 435
15-4 Linear Magnetic Circuits 425
15-6
15-8
Leakage Flux 429
Air Gaps
433
Key Terms
magnetic circuit 422
long air-core coil 422
air-core coil 422
toroidal coil 425
linear magnetic
circuits 425
nonlinear
magnetic circuit 425
leakage flux 429
leakage factor 429
air gap 433
fringing 433
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• analyze the magnetic field of a long aircore coil
• determine magnetic field variables of a
toroidal winding with a nonmagnetic core
• use magnetization curves to analyze a
nonlinear magnetic circuit
Photo sources: © iStock.com/thiel_andrzej
• calculate magnetic field characteristics of a
series magnetic circuit
• calculate the effect of an air gap in a
magnetic circuit
• analyze the magnetic field of a parallel
magnetic circuit
422
Chapter 15 Magnetic Circuits
15-1 Practical Magnetic Circuits
For devices such as motors, generators, transformers, speakers, and relays,
it is just as important to control the path and the strength of the magnetic
fields as it is to control the current. With ferromagnetic materials, we can
construct magnetic circuits (Figure 15-1) to control the path of the magnetic flux.
(a) Relay
(b) Meter movement
(c) Two-pole motor
(d) Loudspeaker
Figure 15-1
Common magnetic circuits
In most electric machines, the magnetomotive force is produced by current flowing through coils of wire. The shape and the location of the resulting magnetic field depends, in turn, on the reluctance of the magnetic
circuit and the available magnetomotive force. We can determine the MMF
required in much the same manner that we determined the applied voltage
(resulting from an electromotive force) in a desired current in a circuit with
a given resistance.
See Review Question 15-16 at the end of the chapter.
15-2 Long Air-Core Coils
A long air-core coil is a solenoid that has air or any other nonmagnetic material as a core (an air-core coil) whose core is at least 10 times as long as its
diameter. We can calculate fairly accurately the relationship between the
flux at the centre of this coil and the current through it.
15-2 Long Air-Core Coils
As shown in Figure 15-2, all the loops of the magnetic lines of force pass
through the central part of the coil and through the space outside the coil.
Since the permeability of the air outside the coil is the same as that of the
nonmagnetic core, about half the lines of force do not travel the whole
length of the core. Instead they leave and enter the core through the sides
of the coil because magnetic lines repel one another and follow the shortest available path. Therefore, our calculations apply only to the centre of
the coil.
+
Figure 15-2
−
Magnetic field around a long air-core coil
Figure 15-2 also shows that all the lines of force that are crowded
into the small cross section at the centre of the coil complete their
loops through a very large cross-sectional area outside the coil. Since
the reluctance of a magnetic circuit is inversely proportional to its
­cross-sectional area, the r­ eluctance of the portion of the magnetic path
within the coil is much greater than that of the return portion outside
the coil. Therefore,
For long air-core coils, the total reluctance is approximately equal to
the reluctance of the nonmagnetic core of the coil itself.
Transposing Equation 14-4 and substituting μ = μ0 gives
Rm =
l
μ0A
(15-1)
where Rm is the reluctance of the nonmagnetic core in ampere-turns
per weber, l is the length of the coil in metres, μ0 is the permeability of
the core material (nonmagnetic materials, 4π × 10−7 H/m), and A is the
inside cross-sectional area of the coil in square metres.
423
424
Chapter 15 Magnetic Circuits
For air-core coils, μ = μ0, the permeability of free space.
Equation 15-1 is the magnetic counterpart of the electric circuit equation R = l/σA, where R is the resistance of the conductor, l is the length of
the conductor, σ is the conductivity of the conductor material, and A is the
cross-sectional area of the conductor.
Having calculated the reluctance of a magnetic circuit, we can readily
determine the magnetomotive force for a required magnetic flux by transposing equation 14-2 to get
Fm = ΦRm
(15-2)
Example 15-1
An air-core coil 20 cm in length with an inside diameter of 1.0 cm has
1000 turns. What current must flow in the coil to develop a total flux of
10−6 Wb at the centre of the coil?
Solution
Step 1
Solve for the reluctance of the core.
Rm =
=
l
μ0A
4π × 10
−7
20 cm
× π × ( 0.50 cm ) 2
= 2.03 × 109 At/Wb
Step 2
Solve for the magnetomotive force with Equation 15-2.
Fm = ΦRm = 10−6 Wb × 2.03 × 109 At/Wb = 2.03 × 103 At
Step 3
Solve for the current in the coil with Equation 14-1.
I=
Fm
2.03 × 103 At
=
= 2.0 A
N
1000 t
See Problems 15-1 to 15-3 and Review Question 15-17.
15-4 Linear Magnetic Circuits
425
15-3 Toroidal Coils
If we place an iron core in the coil of wire in Figure 15-2, the high permeability of the iron will reduce the reluctance of the core well below that of
the return path outside the core. Now the reluctance of the return circuit
outside the core is the major factor in determining the total reluctance of the
magnetic circuit. The reluctance outside the coil is considerably more difficult to calculate than the reluctance of the core with its finite dimensions. We
can make a magnetic circuit with a much lower reluctance by winding the
coil completely around a ring-shaped core, as shown in Figure 15-3. Such
toroidal coils confine the magnetic flux to a path completely within the coil.
Source: © iStock.com/thiel_andrzej
Figure 15-3
A toroidal winding
A toroidal coil mounted on a printed circuit board.
See Review Question 15-18.
15-4 Linear Magnetic Circuits
The calculations in Example 15-1 were straightforward because the reluctance of the magnetic circuit depended only on the physical properties of
the circuit and was not affected by the applied magnetomotive force or the
­resulting magnetic flux. In such linear magnetic circuits the permeability,
μ, is independent of the strength of the magnetic field.
As shown in Figure 14-12, the permeability of ferromagnetic materials
varies considerably with the magnetic flux density, except for a limited
range at the hump of the graph. Hence, the reluctance of practical ferromagnetic circuits changes with variations in magnetomotive force and magnetic
flux. Accordingly, ferromagnetic circuits are nonlinear magnetic circuits.
426
Chapter 15 Magnetic Circuits
Linear magnetic circuits are, therefore, limited to nonferromagnetic
­materials such as air, brass, aluminum, plastic, and wood.
Example 15-2
The brass core in the toroidal coil in Figure 15-3 has a circular cross section. The inside diameter of the core is 9.0 cm, and the outside diameter
is 11.0 cm. If there are 2000 turns in the winding, what total flux will be
produced by a 1.0-A current in the coil?
Solution
Step 1
From the dimensions given, it follows that the diameter of the cross
section of the brass core is 2.0 cm, and
A = π × (1.0 cm)2 = π × 10−4 m2
The effective length of the magnetic circuit is the average path length. In
this example, the inside circumference is π × 9 cm and the outside circumference is π × 11 cm. Hence, the average path length is π × 10 cm.
Step 2
Rm =
Step 3
Step 4
l
π × 10 cm
= 7.958 × 108 At/Wb
=
μ0A
4π × 10 − 7 × π × 10 − 4 m2
Fm = NI = 2000 t × 1.0 A = 2000 At
Φ=
Fm
2000 At
= 2.5 × 10−6 Wb
=
Rm
7.958 × 108 At/Wb
See Problems 15-4 and 15-5.
15-5 Nonlinear Magnetic Circuits
To work with nonlinear resistors, we require data such as voltage-current
graphs for the resistors. To solve nonlinear magnetic circuits, we need comparable data on the properties of the ferromagnetic materials we wish to
use in our magnetic circuit. This information is usually published in the
form of detailed BH graphs like those in Figure 14-15.
15-5 Nonlinear Magnetic Circuits
Example 15-3
A toroidal coil has a circular cast-steel core with a square cross section.
The inside diameter of the core is 12 cm and the outside diameter is
15 cm. If there are 2000 turns in the coil, what current must flow through
it to develop a total flux of 3.0 × 10−4 Wb?
Solution I
Step 1
Determine the dimensions of the magnetic circuit:
Cross-sectional area: A = (1.5 cm)2 = 2.25 cm2 = 2.25 × 10−4 m2
Average path length: l = π × 13.5 cm = 42.4 cm = 0.424 m
Step 2
Determine the magnetic flux density:
B=
Φ
3.0 × 10 − 4 Wb
= 1.33 T
=
A
2.25 × 10 − 4 m2
Step 3
Determine the permeability. From the magnetization curve for cast
steel in Figure 14-15, a magnetic field strength of 1750 At/m is needed
to produce a flux density of 1.33 T.
μ0 =
Step 4
Rm =
Step 5
Step 6
B
1.33 T
=
= 7.60 × 10 − 4 H/m
H
1750 At/m
l
0.424 m
= 2.48 × 106 At/Wb
=
−4
μ0A
7.60 × 10 × 2.25 × 10 − 4 m2
Fm = ΦRm = 3.0 × 10−4 Wb × 2.48 × 106 At/Wb = 744 At
I=
Fm
744 At
=
= 0.37 A
N
2000 t
Actually, the above procedure is a rather roundabout way of solving
magnetic circuits if reference to BH curves is involved. Once we have determined the flux density, the magnetization curve tells us how much magnetic field strength we require for each unit length of the magnetic c­ ircuit.
We can, therefore, solve for the MMF directly without calculating the permeability and reluctance.
427
428
Chapter 15 Magnetic Circuits
Solution II
Step 1
Determine the dimensions of the magnetic circuit. As in Solution 1,
A = 2.25 × 10−4 m2 and l = 0.424 m
Step 2
B=
Φ
3.0 × 10 − 4 Wb
=
= 1.33 T
A
2.25 × 10 − 4 m2
Step 3
From the BH curve for cast steel in Figure 14-15,
Step 4
Step 5
H = 1750 At/m
Fm = H l = 1750 At/m × 0.424 m = 742 At
I=
Fm
742 At
=
= 0.37 A
N
2000 t
In a simple magnetic circuit where all the magnetomotive force produces
flux in a ferromagnetic core with a uniform cross section, solving for the
flux is straightforward when we know the magnetomotive force.
Example 15-4
A current of 0.50 A is passing through a toroidal coil consisting of
960 turns wound on a cast-steel ring with a cross-sectional area of 12 cm2
and an a­ verage path length of 30 cm. Find the total flux in the core.
Solution
Fm = NI = 960 × 0.50 A = 480 At
H=
Fm
480 At
=
= 1600 At/m
l
30 cm
On the BH curve for cast steel (Figure 14-15), B = 1.3 T when
H = 1600 At/m.
Therefore,
Φ = BA = 1.3 T × 12 cm2 = 1.6 × 10−3 Wb
See Problem 15-6.
15-6 Leakage Flux
Circuit Check
A
CC 15-1.
An air-core coil is wound with 1750 turns and is 25 cm long
by 3 cm in diameter. Calculate the flux produced when a
current of 600 mA flows through the winding.
CC 15-2. Calculate the flux produced by a coil wound on a sheet-steel
core but otherwise identical to the coil in question CC 15-1.
15-6 Leakage Flux
Toroidal windings are expensive because the wire has to be threaded
through the centre of the core for each turn. Such windings are normally
used only in circuits where the flux outside the core must be kept to a minimum. For other applications, the wire can be wound on a bobbin before
the parts of ferromagnetic core are assembled to form a loop with one side
passing through the bobbin, as shown in Figure 15-4.
Iron core
Leakage
flux
Coil
Figure 15-4
Leakage flux in a magnetic circuit
Sections 7-9 to 7-12 described how the current through parallel resistors
divides in proportion to the conductance of the parallel branches. Similarly,
magnetic flux divides in proportion to the permeance of parallel paths.
When the core loop is made of high-permeability iron, most of the magnetic
flux flows through the core. However, some leakage flux flows through the
air around the core. In the magnetic circuit of Figure 15-4, the total flux
through the top section of the circuit is less than that through the centre of
the coil ­because the leakage flux bypasses the top section. The ratio between
the total developed flux and the useful flux is known as the leakage factor.
Keeping a magnetic circuit compact helps to minimize the leakage flux.
The cross-sectional area of the core should be large enough to ensure that the
iron does not become saturated. If the iron does become saturated, its permeabilty drops drastically, greatly increasing the reluctance of the core. Air gaps
in the magnetic circuit also greatly increase its total reluctance and hence the
­proportion of leakage flux. In well-designed magnetic circuits, the leakage flux
429
430
Chapter 15 Magnetic Circuits
is often only a very small percentage of the total flux. For the calculations in
this chapter, we can ignore the leakage flux unless the leakage factor is given.
See Review Question 15-19.
15-7 Series Magnetic Circuits
In Figure 15-4, the straight section of the core and the U-shaped section are
in series since the magnetic flux flows in a loop first through one section
and then through the other. The total reluctance of a series magnetic circuit
is similar to the total resistance of a series electric circuit.
The total reluctance of a series magnetic circuit is the sum of all the
individual reluctances.
RmT = Rm1 + Rm2 + Rm3 + . . .
(15-3)
Equation 15-3 has exactly the same form as the equation for the total
­resistance of a series electric circuit:
RT = R1 + R2 + R3 + . . .
Reluctance of U-shaped
cast-steel section
Rmc
Rms
Fm
Reluctance of I-shaped
sheet-steel section
f
RmT = Rmc + Rms
Figure 15-5
Schematic diagram of a series magnetic circuit
4.0 cm
15 cm
2.5 cm
10 cm
12.5 cm
X
Cast
steel
Y
Average path
2.5 cm
Laminated sheet steel
Figure 15-6
Magnetic circuit for Example 15-5
(7-1)
15-7 Series Magnetic Circuits
There are no standard graphic symbols for drawing schematic diagrams
of magnetic circuits. However, since electric and magnetic series circuits
are similar, we could use Figure 15-5 to represent the magnetic circuit of
Figure 15-6.
Example 15-5
Find the magnetomotive force that produces a total flux of 1.2 × 10−3 Wb
in the magnetic circuit of Figure 15-6. The varnish on the laminations
­accounts for 10% of the thickness of the laminations.
Solution I
Step 1
Determine the reluctance, Rmc, of the U-shaped section.
The dimensions in Figure 15-6 show that all three arms of the cast-steel
­section are 2.5 cm wide.
Therefore,
Ac = 2.5 cm × 4.0 cm = 10 cm2 = 1.0 × 10−3 m2
The average path length, lc, is equal to the average of the outside length
of the U-shaped section (12.5 cm + 15 cm + 12.5 cm) and the inside
length (10 cm + 10 cm + 10 cm).
lc =
Bc =
1
( 40 cm + 30 cm ) = 35 cm = 0.35 m
2
Φ
1.2 × 10 − 3 Wb
=
= 1.2 T
Ac
1.0 × 10 − 3 m2
On the BH curve for cast steel in Figure 14-15, Bc = 1.2 T when
Hc = 1220 At/m.
Bc
1.2 T
μc =
=
= 9.84 × 10 − 4 H/m
Hc
1220 At/m
Rmc =
l
35 cm
=
= 3.557 × 105 At/Wb
−4
μcAc
9.84 × 10 H/m × 10 − 3 m2
Step 2
Determine the reluctance, Rms, of the laminated section.
As = 2.5 cm × 4.0 cm × 90% = 9.0 cm2 = 9.0 × 10−4 m2
ls =
Bs =
1
( 20 cm + 10 cm ) = 15 cm = 0.15 m
2
Φ
1.2 × 10 − 3 Wb
=
= 1.33 T
As
9.0 × 10 − 4 m2
431
432
Chapter 15 Magnetic Circuits
On the BH curve for sheet steel, Bs = 1.33 T when Hs = 900 At/m
Rms =
Step 3
μs =
Bs
1.33 T
=
= 1.48 × 10 − 3 H/m
Hs
900 At/m
ls
15 cm
=
= 1.126 × 105 At/Wb
−3
μsAs
1.48 × 10 H/m × 9.0 × 10 − 4 m2
RmT = Rmc + Rms = (3.557 × 105) + (1.126 × 105) = 4.683 × 105 At/Wb
Fm = ΦRmT = (1.2 × 10−3 Wb)(4.683 × 105 At/Wb) = 5.6 × 102 At
In many series magnetic circuits, the material and the cross-sectional
area of the various sections are not the same, so we must calculate the flux
density or the magnetomotive force for each section.
The total voltage drop of a series electric circuit is the sum of all the individual voltage drops. Similarly,
The total magnetomotive force of a series magnetic circuit is the
sum of the MMFs for all of the sections of the circuit.
FmT = Fm1 + Fm2 + Fm3 + . . .
(15-4)
Equation 15-4 has exactly the same form as the equation for the total
voltage in a series electric circuit:
VT = V1 + V2 + V3 + . . .
(7-2)
We can apply Equation 15-4 to solve Example 15-5 without calculating
the permeability and reluctance of each section.
Solution II
As in Solution I, HC = 1220 At/m for the U-shaped section. Therefore,
Fmc = Hclc = 1220 At/m × 35 cm = 427 At
Similarly, for the laminated section, Hs = 900 At/m, and
Therefore,
Fms = Hsls = 900 At/m × 15 cm = 135 At
FmT = Fmc + Fms = 427 + 135 = 5.6 × 102 At
See Problems 15-7 and 15-8 and Review Question 15-20.
15-8 Air Gaps
15-8 Air Gaps
In the magnetic circuits shown in Figure 15-1, each line of force crosses an
air gap. In motors, meters, and speakers, such air gaps provide space for
the electric conductors to move. Even in equipment with no moving parts,
a small gap may be left between sections of the magnetic circuit. Such
gaps increase the total reluctance of the circuit, which reduces the total
magnetic flux and can prevent saturation of the iron in the circuit. Often,
a sheet of a nonmagnetic material such as fibreboard or brass is inserted
at a joint between sections to produce an “air” gap with an accurate and
uniform width.
In magnetic circuits like those in Figure 15-1, all the magnetic lines of
force must pass through the air gaps. Thus, the gaps in each circuit are in
series with the remainder of the circuit. However, since the air adjacent to
a gap has the same permeability as the air in the gap, the magnetic lines of
force tend to spread out as they cross the gap. This fringing makes the flux
density in the air gap slightly less than that in the adjoining iron sections of
the magnetic circuit (see Figure 15-7).
Iron
polepiece
Iron
polepiece
(a)
Figure 15-7
Iron field pole
Air gap
Iron rotor
(b)
Fringing of magnetic lines of force in an air gap
If the distance across the air gap is small, we can make reasonably accurate calculations by assuming that the cross-sectional dimensions of the air
gap are greater than the cross-sectional dimensions of the adjacent iron by
an amount equal to the distance across the gap.
With series magnetic circuits, it is much more difficult to use the applied
magnetomotive force to calculate the total flux. Since the ferromagnetic
433
434
Chapter 15 Magnetic Circuits
Example 15-6
Find the magnetomotive force that produces a total flux of 1.2 ×
10−3 Wb in the magnetic circuit of Figure 15-6 with pieces of fibreboard
0.1 mm thick inserted at points X and Y. Assume that varnish accounts
for 10% of the thickness of the laminations.
Solution I
Since the total flux is the same as in Example 15-5, the reluctance of the
two steel sections is the same as we have already calculated, 3.557 × 105
At/Wb for the cast-steel section and 1.126 × 105 At/Wb for the laminated sheet-steel section. When we allow for fringing, the cross-sectional
area of the gap is
Ag = 2.51 cm × 4.01 cm = 10.07 cm2 = 1.007 × 10−3 m2
Rmg =
lg
μgAg
=
4π × 10
−7
0.1 mm
4
−3
2 = 7.9 × 10 At/Wb
H/m × 1.007 × 10 m
Since each magnetic line of force has to pass through both gaps, they
are in series. Since the two gaps are identical, each has a reluctance of
7.9 × 104 At/Wb.
RmT = (3.557 × 105) + (1.126 × 105) + 2(7.9 × 104) = 6.26 × 105 At/Wb
Fm = ΦRmT = (1.2 × 10−3 Wb)(6.26 × 105 At/Wb) = 7.5 × 102 At
Solution II
The MMF required for the iron will be the same as in Example 15-5,
562 At. For each air gap,
Bg =
Hg =
Φ
1.2 × 10 − 3 Wb
=
= 1.19 T
Ag
1.007 × 10 − 3 m2
Bg
1.19 T
= 9.47 × 105 At/m
=
−7
μg
4π × 10 H/m
Fmg = Hg lg = 9.47 × 105 At/m × 0.1 mm = 94.7 At
FmT = 562 + ( 2 × 94.7 ) = 7.5 × 102 At
sections are nonlinear, we cannot easily determine how much of the total
MMF applies to each section. Example 15-7 shows one method of tackling
the problem in a series magnetic circuit. The simplest method for solving
such problems is to make an educated guess at the answer, solve for the
other parameters of the circuit, and check if these calculated parameters
agree with the known values. If the known and the calculated values are
not reasonably close, we adjust our estimate and repeat the process.
15-9 Parallel Magnetic Circuits
Example 15-7
A current of 0.50 A is passing through a coil consisting of 960 turns
wound on a cast-steel core with a cross section 5.0 cm × 2.5 cm and an
average path length of 30 cm. This core has a single air gap 0.25 mm
wide. Find the total flux in the core.
Solution
The area of the air gap, with allowance for fringing, is:
Ag = 5.025 cm × 2.525 cm = 12.7 cm2
FmT = NI = 960 t × 0.50 A = 480 At
Assume that the MMF for the air gap is 210 At. Then
Hg =
Fmg
lg
=
210 At
= 8.4 × 105 At/m
0.25 mm
Bg = μgHg = 4π × 10 − 7 H/m × 8.4 × 105 At/m = 1.056 T
Φ = BgAg = 1.056 T × 12.7 cm2 = 1.34 × 10 − 3 Wb
If this is so, the flux density in the iron must be
Bc =
ϕ
1.34 × 10 − 3 Wb
=
= 1.07 T
Ac
12.5 cm2
On the BH curve for cast steel in Figure 14-15, Bc = 1.07 T when
Hc = 885 At/m.
Therefore,
and
Fmc = Hclc = 885 At/m × 30 cm = 266 At
FmT = 210 + 266 = 476 At
Since this value is just slightly less than the MMF calculated from
known quantities, the total flux must be just slightly greater than
1.34 × 10−3 Wb.
See Problems 15-9 to 15-12 and Review Questions 15-21 to 15-23.
15-9 Parallel Magnetic Circuits
In the magnetic circuits of Figure 15-1(c) and 15-1(d), all of the magnetic
flux passes through some portions of the magnetic circuit, but in other
­portions the flux divides between two symmetrical paths. Designers have
found that such multiple paths can appreciably reduce the leakage flux.
These paths are in parallel. When a magnetic circuit has two identical paths
in parallel, the total flux splits equally between the two paths.
435
436
Chapter 15 Magnetic Circuits
Example 15-8
The core in Figure 15-8 is made of sheet-steel laminations stacked to
a total thickness of 4 cm with varnish making up 5% of this thickness.
The width of each outside leg of the core is 2.5 cm, and the width of
the centre leg is 5.0 cm. The average path length is 25 cm. What is the
maximum current that can pass through a 1000-turn coil on the centre
leg of the core without the total flux in the core exceeding 2 × 10−3 Wb?
Average
path
Figure 15-8
CL
Magnetic circuit for Example 15-8
Solution I
Since the core is symmetrical, we could divide the magnetic circuit
down the centreline shown in Figure 15-8 and stack one section on top
of the other without changing the total reluctance of the magnetic circuit. The cross-sectional area of the core then becomes
A = 2 × 2.5 cm × 4.0 cm × 95% = 19 cm2
Therefore, maximum B =
Φ
2 × 10 − 3 Wb
=
= 1.05 T
A
19 cm2
On the BH curve for sheet steel (Figure 14-15), B = 1.05 T when
H = 460 At/m. Therefore,
Fm = Hl = 460 At/m × 25 cm = 115 At
and maximum I =
115 At
Fm
=
= 115 mA
N
1000 t
Solution II
Each of the parallel halves of the magnetic circuit passes half of the total
flux. The flux density in the iron is
B=
Φ
1 × 10 − 3 Wb
=
= 1.05 T
A
9.5 cm2
15-9 Parallel Magnetic Circuits
Fm = Hl = 460 At/m × 25 cm = 115 At
As in Solution I,
and
maximum I = 115 mA
The same MMF in the other half of the magnetic circuit produces
1 × 10−3 Wb of flux, for a total of 2 × 10−3 Wb.
See Problems 15-13 to 15-15 and Review Question 15-24.
Circuit Check
B
CC 15-3. A coil of 2000 turns is wound on the cast iron portion of the
core shown in Figure 15-9. Calculate the current required to
produce a flux of 3 mWb.
40 cm
5 cm
Cast steel
5 mm
20 cm
Cast iron
10 cm
10 cm
10 cm
10 cm
Figure 15-9
CC 15-4. A coil is wound on the centre leg of the transformer core shown
in Figure 15-10. How many turns are required if a current of
1.0 A through the coil is to produce a flux density of 0.31 T in
the outside leg?
2.5 cm
2.5 cm
2.5 cm
Cast iron
Copper
15 cm
2.5 mm
2.5 cm
15 cm
Figure 15-10
15 cm
437
438
Chapter 15 Magnetic Circuits
Summary
• For long air-core coils, the total reluctance is approximately equal to the
reluctance of the nonmagnetic core itself.
• In a linear magnetic circuit, the permeability is constant.
• In a nonlinear magnetic circuit, the permeability is dependent on the
magnetic field strength.
• Some leakage magnetic flux passes through the air in the area surrounding the ferromagnetic core of a coil.
• The total reluctance of a series magnetic circuit is the sum of all the individual reluctances.
• An air gap can substantially increase the total reluctance of a magnetic
circuit.
• The magnetic flux density in an air gap is somewhat less than that in
­adjacent ferromagnetic sections of a magnetic circuit.
• The total magnetic flux divides among parallel paths in a magnetic
­circuit.
B = beginner
I = intermediate
A = advanced
Problems
I
Section 15-2
15-1.
I
15-2.
I
15-3.
I
I
An air-core coil passes a 500-mA current. Its 1000 turns occupy a
length of 12 cm on a 1.0-cm diameter form. What flux is developed
at the centre of the coil?
An 10-cm-long coil of wire is wound on a hardwood dowel
5.0 mm in diameter. If there are 750 turns in the coil, what current
must it pass to develop a flux of 6.0 × 10−7 Wb at the centre of the
coil?
How many turns of wire must be wound on a 1.5-cm-diameter plastic rod to produce 4.0 × 10-7−7 Wb at the centre of the coil from a
450-mA current through the coil? The length of the coil is 15 cm.
Section 15-4
15-4.
15-5.
Long Air-Core Coils
Linear Magnetic Circuits
The core for a low-loss coil is made from a piece of flat plastic stock
8 mm thick by 25 mm wide by 20 cm long, bent into a circular
loop with the ends joined together. The coil has 1000 turns of wire
wound evenly around the circular form. What current must flow in
the coil to develop a flux of 10−5 Wb in the core?
The toroidal coil shown in Figure 15-11 has 350 turns wound on a
steel ring with an inside diameter of 18 cm and an outside diameter of 22 cm. Find the flux density in the steel when the current
through the coil is 2.5 A.
439
Problems
2.5 A
μr = 3000
+
E
−
Figure 15-11
I
I
Section 15-5
15-6.
If the core in Problem 15-4 were a cast-iron ring with the same dimensions, what current would produce a flux of 105 μWb?
Section 15-7
15-7.
Nonlinear Magnetic Circuits
Series Magnetic Circuits
The horseshoe-shaped cast-steel electromagnet of Figure 15-12 is
2.3 cm in diameter and has an average path length of 25 cm. The
­dimensions of the sheet-steel bar are 12.0 cm × 4.0 cm × 1.0 cm.
What current through the windings of the electromagnet develops
a total flux of 0.5 mWb in the bar held by the magnet?
Cast-steel
electromagnet
core
1000 turns
1000 turns
Sheet-steel bar
Figure 15-12
I
A
A
15-8.
Find the total flux in the sheet-steel bar in Figure 15-12 if the current
in the coils of the electromagnet is 0.5 A.
Section 15-8
Air Gaps
The magnetic circuit of Figure 15-13 is made from a sheet-steel bar
15 cm long with a 12 mm × 20 mm rectangular cross section. The
bar is bent into a ring, leaving an air gap 0.25 mm across. What
current must flow through the 800-turn coil wound on this core in
order to produce a total flux of 3 × 10−4 Wb in the air gap?
15-10. If a sheet of aluminum 1.5 mm thick is placed between the sheetsteel bar and the electromagnet of Figure 15-12, what current must
flow in the coils to develop a total flux of 7.5 × 10−4 Wb in the sheetsteel bar if the leakage factor is now 1.15?
15-9.
Air
gap
Sheet
steel
800 turns
Figure 15-13
440
Chapter 15 Magnetic Circuits
A
15-11. In the magnetic circuit shown in Figure 15-14, the coil has 1000
turns. Find the current required to produce a total flux of 1.0 mWb.
Plastic 2.5 mm
Cast steel
5.0 cm
Plastic
2.5 mm
Cast
iron
13.0 cm
2.5 cm
Sheet steel
2.5 cm
2.5 cm
10.0 cm
5.0 cm
Figure 15-14
A
15-12. When a 4.0-A current flows through a coil wound on the core
shown in Figure 15-15, the total flux is 0.80 mWb. How many turns
does this coil have?
5.0 cm
Cast iron
Cast
steel
Sheet
steel
Aluminum
2.0 mm
thick
15.0 cm
2.0 cm
2.0 cm
10.0 cm
8.0 cm
3.0 cm
Figure 15-15
A
Section 15-9
Parallel Magnetic Circuits
15-13. If the sheet-steel laminations of Figure 15-16 are stacked to a total
thickness of 2.5 cm, how many turns of wire must be wound on the
Problems
Cast steel
2.5 cm in diameter
11.5 cm
2.0 cm
7.5 cm
2.0
cm
4.0
cm
1000
turns
2.0
cm
25 cm
2.5 cm
Figure 15-17
Figure 15-16
A
A
centre leg in order for a current of 100 mA to develop a total flux
of 1.2 × 10−3 Wb in the centre leg? Varnish ­accounts for 5% of the
thickness of each lamination. Assume no air gaps and negligible
leakage flux.
15-14. For the magnetic circuit shown in Figure 15-17, calculate the current required to develop 0.80 mWb of flux in the centre leg.
15-15. For the magnetic circuit shown in Figure 15-18, calculate the current needed to develop a total flux of 1.0 mWb.
Plastic 1.0 mm thick
Sheet steel
2.5 cm
1000
turns
20 cm
Cast
iron
Cast
iron
2.5 cm
2.5 cm
5.0 cm
Figure 15-18
15 cm
2.5 cm
5.0 cm
441
442
Chapter 15 Magnetic Circuits
Review Questions
Section 15-1
Practical Magnetic Circuits
15-16. State two methods of supplying a magnetomotive force to a magnetic circuit.
Section 15-2
Long Air-Core Coils
15-17. On what basis can we assume that the total reluctance of a long
air-core coil is approximately equal to the reluctance of the nonmagnetic core of the coil itself?
Section 15-3
Toroidal Coils
15-18. Describe the effect of a toroidal winding on a magnetic circuit.
What is the advantage of this type of winding?
Section 15-6
Leakage Flux
15-19. What causes leakage flux to occur in magnetic circuits? State two
means of reducing leakage flux in magnetic circuit design.
Section 15-7
Series Magnetic Circuits
15-20. What characteristic of magnetic circuits allows us to calculate the
total magnetomotive force needed in a circuit made up of several
different sections?
Section 15-8
Air Gaps
15-21. Many practical magnetic circuits contain a gap in the form of a
sheet of plastic or brass. Why can we treat such materials as air
gaps?
15-22. In an all-iron magnetic circuit, the relationship between the applied
magnetomotive force and the resulting total magnetic flux is quite
nonlinear, as indicated by the graphs of Figure 14-15. Adding an air
gap makes the relationship between applied MMF and total flux in
the circuit more linear. Explain why.
15-23. What causes fringing of the magnetic lines of force in an air gap?
Section 15-9
Parallel Magnetic Circuits
15-24. If the two parallel paths for magnetic lines of force in the magnetic circuit of Figure 15-8 were not symmetrical, what complication would be introduced into our solution of a problem such as
Example 15-8?
Practice Quiz
Integrate the Concepts
Figure 15-19 shows a series magnetic circuit. Given that the current produces a total magnetic flux of 0.40 mWb, calculate:
Cast steel
I
2.0 cm
2000
turns
3.0 cm
Plastic
1.0 mm
thick
Cast
iron
2.0 cm
2.0 cm
6.0 cm
2.0 cm
3.0 cm
Figure 15-19
(a)
(b)
(c)
(d)
(e)
the magnetomotive force in the cast-iron section
the magnetomotive force in the sheet-steel section
the magnetomotive force in the plastic sections
the total magnetomotive force
the current through the coils
Practice Quiz
1.
An example of a magnetic circuit is
(a) a loudspeaker
(b) a hard-drive read/write head
(c) a relay
(d) all of the above
2.
The reluctance of a nonmagnetic core is
(a)
directly proportional to the length of the coil and inversely
­proportional to the cross-sectional area of the coil
(b)directly proportional to the length of the coil and to the crosssectional area of the coil
(c)inversely proportional to the length of the coil and to the crosssectional area of the coil
(d)
inversely proportional to the length of the coil and directly
­proportional to the cross-sectional area of the coil
3.
The reluctance of a 400-turn air-core coil with a length of 50 cm and a
radius of 5.0 cm is
(a) 5.1 × 107 At/Wb
(b) 1.3 × 107 At/Wb
(c) 2.0 × 108 At/Wb
(d) 5.1 × 108 At/Wb
443
444
Chapter 15 Magnetic Circuits
4
Which of the following statements are true?
(a)The principal advantage of toroidal coils is that they are easy to
make.
(b)The ratio between the total developed flux and the useful flux is
known as leakage factor.
(c)The total reluctance of a series magnetic circuit is the sum of all
individual reluctances.
(d)The total magnetic flux divides between parallel paths in a magnetic circuit.
5.
The toroidal coil of Figure 15-20 has a brass core with an inside diameter
of 12 cm and an outside diameter of 16 cm. The coil has 1000 turns. The
total flux produced when a current of 1.5 A flows through the coil is
(a) 54 μWb
(b) 2.5 μWb
(c) 5.4 μWb
(d) 6.3 μWb
I
I
Figure 15-20
6.
If the core of toroidal coil in Question 5 were made of cast steel, the
magnetic flux density necessary to develop a total flux of 880 μWb
would be
(a) 1.4 T
(b) 600 mT
(c) 400 mT
(d) 700 mT
7.
The current through the coil in Question 6 would be
(a) 176 mA
(b) 352 mA
(c) 264 mA
(d) 88 mA
Practice Quiz
8. The total magnetomotive force of a series magnetic circuit is
(a)zero
(b) the product of all MMFs for all the sections of the circuit
(c) the sum of all MMFs for all the sections of the circuit
(d) the difference of all MMFs for all the sections of the circuit
9. When an air gap is inserted a magnetic circuit the total reluctance
(a) becomes zero
(b) does not change
(c) decreases
(d) increases
10. What is the advantage of symmetrical parallel paths in a magnetic
circuit?
445
16
Inductance
Chapter 12 described how capacitance in an electric circuit
depends on electric fields. The third basic property of electric
circuits, inductance, stems from the interaction of magnetic
fields with electric circuits.
Chapter Outline
16-1
Electromagnetic Induction 448
16-3
Lenz’s Law
16-2
Faraday’s Law 450
451
16-4
Self-Induction
16-6
Factors Governing Inductance 455
16-5
453
Self-Inductance 454
16-7
Inductors in Series
458
16-9
The DC Generator
459
16-8
Inductors in Parallel 458
16-10 Simple DC Generator
16-11 EMF Equation 463
461
16-12 The DC Motor 465
16-13 Speed and Torque of a DC Motor 467
16-14 Types of DC Motors
469
16-15 Speed Characteristics of DC Motors 471
16-16 Torque Characteristics of DC Motors 474
16-17 Permanent Magnet and Brushless DC Motors
476
Key Terms
electromagnetic
induction 448
induced voltage 448
primary winding 448
secondary winding 448
mutual induction 450
Faraday’s law 450
Lenz’s law 451
self-induction 453
counter EMF (CEMF) 453
self-inductance 454
inductance 454
henry 454
inductor 456
field system 460
armature 460
rotor 460
stator 460
yoke 460
shunt coil 460
series coil 460
commutator 460
brush 461
pulsating DC 463
wave winding 464
lap winding 464
torque 466
back or counter EMF 467
field excitation 469
shunt motor 470
load current 470
series motor 470
compound motor 470
cumulatively compounded 470
differentially
compounded 470
universal motor 472
compound motor 472
permanent magnet
motor 476
brushless motor 476
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• demonstrate electromagnetic induction with
a magnet, a conductor, and a galvanometer
• explain electromagnetic induction in an iron
core with primary and secondary windings
• explain Faraday’s law relating induced voltage
in an electric circuit to the rate of change of
­magnetic flux
• use Lenz’s law to determine the polarity of an
induced voltage
• use Lenz’s law to determine the direction of
current ­produced by an induced voltage
• explain how self-induction occurs in a coil
• express inductance in terms of induced
voltage and the rate of change of current
• express the inductance of a coil in terms of the
number of turns in the coil and its reluctance
• calculate the inductance of a coil given
the number of turns and the permeability,
Photo sources: © iStock.com/albin
•
•
•
•
•
•
•
•
•
•
cross-sectional area, and length of the
magnetic circuit
calculate the total inductance of inductors in
series when there is no mutual inductance
calculate the equivalent inductance of
inductors in parallel when there is no mutual
inductance
list the components of a DC machine
graph the voltage produced by a rotating loop
explain commutator action
calculate generated voltage
calculate the force on a current-carrying
conductor
calculate the speed and torque of a DC motor
describe and draw circuits of different types
of DC motors
graph speed and torque characteristics
448
Chapter 16 Inductance
16-1 Electromagnetic Induction
In Section 14-3, Figure 14-10 illustrates how Faraday discovered that
thrusting a permanent magnet into a solenoid induces a current in the
solenoid. Figure 16-1 shows another arrangement for demonstrating such
­electromagnetic induction. As the conductor is moved downward between the poles of the magnet, the galvanometer pointer swings one way
from the zero mark at the centre of the meter’s scale. As the conductor is
moved upward, the galvanometer pointer swings in the opposite direction. When the conductor is stationary, the pointer does not deflect.
Motion
N
S
G
Figure 16-1
Demonstrating electromagnetic induction
For current to flow through the galvanometer, there must be some voltage source in the circuit. Experiments show that this voltage appears only
when the conductor is cutting across the magnetic lines of force (and the
magnetic flux they represent). Moving the conductor parallel to the lines
of force produces no deflection of the galvanometer pointer. In Figure 16-1,
the conductor is moved while the magnetic field remains stationary,
whereas the magnetic field in Figure 14-10 is moved while the electric conductor remains stationary. In both cases, an induced voltage appears in the
conductor when the conductor and the magnetic field are moving perpendicular to each other.
Figure 16-2 shows electromagnetic induction produced by a changing
magnetic field without mechanical motion of either a magnet or a conductor.
As the resistance of the rheostat decreases, the current in the primary winding (the winding connected to the source) increases. The magnetic flux in
the iron core increases correspondingly. Since the secondary winding (the
winding connected to the load) is wound on the same core as the primary
winding, the magnetic flux through the secondary winding also increases.
The pointer of the galvanometer deflects, indicating an induced voltage in
16-1 Electromagnetic Induction
Iron core
E
449
Magnetic field
+
G
−
Figure 16-2
Demonstrating mutual induction
Practical Circuits
Source: © iStock.com/tbradford
Induction and Metal Detectors
Airport security
The operation of metal detectors is based upon
the principles of electro­
magnetic induction.
Figure 16-3 shows a metal-detecting security gate
commonly used in airports. Current through
the transmitting coils produces a magnetic field
that induces a current in the receiving coils.
When a metal object passes through the gate,
the magnetic field induces eddy currents in the
object. The resulting change in the magnetic flux
through the receiving coil changes the induced
current. Security gates can detect the location of a
metallic object as small as a paper clip.
Transmitting
coils
Receiving
coils
Lines of magnetic force
Figure 16-3
Metal-detecting gate
450
Chapter 16 Inductance
the secondary winding. Since most of the magnetic flux is ­confined to the
iron core, it is more usual to think of the flux as linking the secondary winding rather than cutting across the turns of the secondary winding.
When the resistance of the rheostat increases, the current in the ­primary
winding and the magnetic flux linking the secondary winding both
­decrease. At the same time, the pointer of the galvanometer swings in
the opposite direction, indicating that the induced voltage now has the
­opposite polarity. When the current in the primary winding is steady, there
is no deflection of the galvanometer pointer. The generation of a voltage
in a secondary winding by a changing current in a primary winding is an
example of mutual induction.
See Review Question 16-29 at the end of the chapter.
16-2 Faraday’s Law
With apparatus like that shown in Figure 16-2, Faraday noted that the faster
the flux in the core builds up or collapses, the greater the deflection of the
galvanometer pointer. This relationship is now known as Faraday’s law:
The voltage induced in an electric circuit is proportional to the rate
of change of the magnetic flux linking the circuit.
To express rate of change of flux at any given moment, we use the same
calculus derivative notation as we used for the rate of change of voltage in
Section 13-2. From the definition of the weber in Section 14-3,
V=
dϕ
dt
where V is the voltage in volts and dϕ/dt is the rate of change of magnetic
flux in webers per second. Note the use of lowercase ϕ to indicate changing
flux.
In the secondary winding in Figure 16-2, the same changing flux links
all the turns and thus induces identical voltages in each turn. Since these
voltages are all in series, the total induced voltage is the number of turns
times the voltage per turn:
eT = N
dϕ
dt
(16-1)
where eT is the instantaneous value of the induced voltage between
the terminals of the winding in volts, N is the number of turns in the
winding, and dϕ/dt is the rate of change of magnetic flux in webers per
second.
16-3 Lenz’s Law
Combining Equations 14-1 and 14-2 gives
Φ=
Fm
NI
=
Rm
Rm
The number of turns in the primary winding is fixed. If the flux density
stays below the saturation point of the iron core, the reluctance of the magnetic circuit is reasonably constant. Therefore, the rate of change of flux is
­directly proportional to the rate of change of current. When dealing with electromagnetic induction due to a changing current, we can restate Faraday’s
law in this form:
The magnitude of the induced voltage is directly proportional to
the rate of change of current.
See Review Question 16-30.
16-3 Lenz’s Law
The Russian physicist Heinrich F. E. Lenz (1804–65) reasoned that electromagnetic induction must conform to the physics principle that reaction is
opposite to action. We can use this statement of Lenz’s law to determine
the polarity of an induced voltage:
Any current resulting from an induced voltage opposes the change
in the original magnetic flux.
When we first close the switch in Figure 16-4(a), the magnetic flux through
the primary winding increases from zero, inducing a voltage across the
secondary winding. As a result, current flows through the closed circuit
formed by the resistor and the secondary winding. According to Lenz’s
law, the magnetic flux produced by this secondary current must o
­ ppose
the increase in primary flux. Applying the right-hand rule to the primary
winding, we find that the primary flux flows clockwise around the core.
Rising
primary
flux
+
E
−
(a)
Figure 16-4
Collapsing
primary
flux
Opposing
secondary
flux
Sustaining
secondary
flux
+
−
+
+
−
E
−
(b)
Determining the direction of an induced voltage
451
452
Chapter 16 Inductance
To oppose the increase in the primary flux, the flux produced by the secondary current must flow counterclockwise. Applying the right-hand rule
to the secondary winding, we find the direction in which the secondary
current flows through the resistor and can thus determine the polarity of
the induced voltage.
When we open the switch, the primary current stops flowing and the primary magnetic flux collapses, as shown in Figure 16-4(b). This r­ eduction in
flux induces a voltage across the secondary winding. According to Lenz’s
law, the secondary current now opposes the collapse of the primary flux.
Thus, the secondary current now produces a clockwise flux. Applying the
right-hand rule gives the polarity shown in Figure 16-4(b). The polarity of
the induced voltage when the primary current is ­decreasing is, therefore,
opposite to the polarity when the primary current is increasing.
We can also use Lenz’s law to determine the direction of the induced
voltage in Figure 16-1. Since the galvanometer completes the circuit, current flows through the conductor as it moves through the magnetic field.
This current produces a magnetic field around the conductor. If we consider the two magnetic fields separately, the field of the stationary magnet
is uniform between the poles of the magnet, as shown in Figure 16-5(a),
while the conductor is surrounded by concentric circular lines of force,
which are directed clockwise if the current is flowing out of the page, as
shown in Figure 16-5(b). The two component fields combine to form the
magnetic field shown in Figure 16-6.
In Figure 16-5(b), the direction of the magnetic lines of force above the
conductor is opposite to the direction of the lines between the poles of the
permanent magnet field in Figure 16-5(a). Consequently, the flux from
the conductor cancels some of the flux from the magnet and the resulting
lines of force bend away from the conductor (see Figure 16-6). Below the
­conductor, the magnetic fields from the conductor and the magnet have
the same direction. The resulting flux density is greater than that above the
conductor, as shown by the spacing of the lines of force. Thus the induced
current in the conductor bends some of the magnetic lines of force between
the poles of the magnet to pass below the conductor.
N
S
Stationary permanent
magnet field
Magnetic field around
current-carrying conductor
(a)
(b)
Figure 16-5
Component magnetic fields for the demonstration of Figure 16-1
16-4 Self-Induction
Motion of
conductor
453
Direction of
magnetic force
N
S
Figure 16-6 Using Lenz’s law to determine the direction of the current
produced by an induced voltage
Since magnetic lines of force tend to become as short as possible, the
magnetic field exerts an upward force on the current-carrying conductor.
In keeping with Lenz’s law, this force opposes the motion of the conductor
since that motion produces a current that changes the original flux from the
magnet. Consequently, when the conductor in Figure 16-6 moves downward through the stationary magnetic field, the induced voltage has a
­polarity that makes current flow out of the page.
See Review Questions 16-31 to 16-33.
16-4 Self-Induction
When we first close the switch in Figure 16-7, the increasing flux produced
by the rising current in turn A induces a voltage into turn B. The same
­rising current in turn B induces a voltage into turn A. These voltages are in
series, and according to Lenz’s law, both have a polarity such as to oppose
the increase in current and magnetic flux. For a coil with N turns, Faraday’s
law shows the total induced voltage is Ndϕ/dt, as in Equation 16-1. The
generation of a voltage in a circuit by a changing current in the same circuit
is called self-induction.
As current flows from the voltage source in Figure 16-7, the voltage
source provides energy to build up a magnetic field around the coil. The
rising magnetic field induces an EMF in the turns of the coil. This EMF separates charge within the coil, causing a surplus of electrons at one end of the
coil and a deficiency at the other. This process transfers energy from the
magnetic field to the coil and produces a potential difference across the coil.
Since the induced voltage opposes the change in current, the self-induced
voltage is sometimes called a counter EMF or CEMF.
When we open the switch in Figure 16-7, the magnetic field collapses
and again induces an EMF in the coil. The polarity of the self-induced voltage is reversed so that it opposes the decrease in current and magnetic flux.
See Review Question 16-34.
E
+
A
B
−
Figure 16-7 Self-induction
454
The henry is named in
honour of the ­American
physicist Joseph Henry
(1797– 1878),
who made an
­extensive study of
electromagnetism and
constructed the first
electromagnetic motor.
The definition of the
henry in terms of
­webers per ampere
underlies the units for
magnetic permeance
(henrys or webers
per effective ampere),
magnetic reluctance
(reciprocal henrys or
effective amperes per
weber), and magnetic
permeability (henrys
per metre) introduced
in Section 14-6.
Chapter 16 Inductance
16-5 Self-Inductance
Section 16-4 showed that the simple electric circuit of Figure 16-7 tends
to oppose any change in current through it or in the magnetic flux linking
it. This property is called its self-inductance Because it is usually apparent whether the electromagnetic induction in a circuit is self-induction or
­mutual induction, it is customary to refer to self-inductance as simply the
inductance of the circuit.
The letter symbol for inductance is L.
The henry (symbol H) is the SI unit of inductance.
We can define the unit of inductance either in terms of the change in flux
­resulting from a given change in current or in terms of the induced voltage resulting from a given change in current. The general definition of the
henry is based on magnetic flux linkage:
An electric circuit has an inductance of one henry when a change in
current of one ampere produces a change in total linkage flux of one
weber: 1 H = 1 Wb/A.
The total linkage flux of a coil with N turns is N times as great as for a
single turn carrying the same current. Therefore,
L=N
dϕ
di
(16-2)
where L is the inductance of a circuit in henrys, N is the number of turns
linked by the magnetic flux, and dϕ/di is the change in flux for a given
change in current in webers per ampere.
Rearranging Equation 16-1 and Equation 16-2 gives
N=
N=
and
Therefore,
and
or
eL
dΦ/dt
L
dϕ/di
eL
L
=
dΦ/di
dΦ/dt
L = eL ×
dt
dΦ
dt
×
= eL ×
di
dΦ
di
L=
eL
di/dt
(16-3)
16-6 Factors Governing Inductance
where L is the inductance of a circuit in henrys, eL is the induced voltage in volts, and di/dt is the rate of change of the current in amperes
per second.
Thus, we can also define the henry in terms of just electrical units:
An electric circuit has an inductance of one henry when current
changing at the rate of one ampere per second induces an EMF of one
volt into the circuit.
See Problems 16-1 to 16-5 and Review Question 16-35.
16-6 Factors Governing Inductance
Even a straight wire has some inductance. As the current through a conductor increases, magnetic lines of force appear as tiny loops at the centre
of the conductor and then expand outward. As they expand, they cut across
the conductor, inducing a voltage into the conductor. However, the inductance of a straight wire is so small that we can ignore it, except at very high
radio frequencies.
The inductance of a coil can be quite large since it depends on the number
of turns. Combining Equations 14-1 and 14-2 gives
Φ=
Dividing both sides by t gives
from which
Fm
NI
=
Rm
Rm
Φ
N
I
=
×
t
Rm t
dϕ
N
di
=
×
dt
Rm dt
Substituting for dϕ/dt in Equation 16-2, we get
eT =
N2
di
×
Rm dt
Now we substitute for the induced voltage in Equation 16-3:
L=
N2
Rm
(16-4)
455
Chapter 16 Inductance
This equation shows that the inductance of a coil is dependent on the
­ umber of turns in the coil and the reluctance of the magnetic circuit on
n
which the coil is wound. Inductors are circuit components constructed for
the express purpose of adding inductance to a circuit. An iron core greatly
reduces the reluctance of the magnetic circuit and correspondingly increases
the inductance of a coil. Doubling the number of turns of wire in the coil of
Figure 16-7 not only doubles the flux linking the coil for a given current (thus
doubling the induced voltage per turn), but also doubles the number of turns
that this flux links. Therefore, the total induced voltage (and consequently
the inductance) is proportional to the square of the number of turns.
Since Rm = L/μA (Equation 15-1),
L=
N2μA
l
(16-5)
where L is the inductance of the inductor in henrys, N is the number of
turns in the coil, μ is the permeability of the magnetic circuit in henrys
per metre, A is the cross-sectional area of the magnetic circuit in square
metres, and l is the length of the magnetic circuit (or the length of the coil
for air-core coils) in metres.
Equation 16-5 is based on the assumption that all of the magnetic flux
links all of the turns. For toroidal and iron-core coils, Equation 16-5 is
reasonably accurate. However, where there is appreciable leakage flux,
as in long air-core coils, we must use empirical formulas to calculate
inductance. Electrical and radio handbooks list formulas for various
­
types of inductors.
Source: Clive Streeter/Getty Images
456
Inductor with a ferite core
16-6 Factors Governing Inductance
Example 16-1
Find the inductance of a toroidal coil with 2000 turns of wire wound on
a cast-steel ring, given that the ring has a cross-sectional area of 2.25 cm2,
an average path length of 42.4 cm, and a permeability of 7.60 × 10−4 H/m.
(See Example 15-3.)
Solution
L=
N2μA
20002 × 7.60 × 10 − 4 H/m × 2.25 × 10 − 4 m2
= 1.61 H
=
l
0.424 m
See Problems 16-6 to 16-13 and Review Questions 16-36 to 16-38.
Circuit Check
A
CC 16-1.
A coil carrying a current of 8.0 A develops a voltage of 150 V
when the current is switched over 0.5 s. Find the inductance of
the coil.
CC 16-2. How much voltage will be induced in a 6-H inductor carrying
4 A if the current is reversed over 20 ms?
CC 16-3. The inductor shown in Figure 16-8 has a 600-turn coil wound
on a sheet-steel core with a 5.0 cm × 5.0 cm cross section.
(a)Given that the relative permeability of the steel is 4500,
­determine the inductance.
(b)How many extra turns are required on the coil to raise the
inductance by 25%?
25 cm
20 cm
Figure 16-8
457
458
Chapter 16 Inductance
16-7 Inductors in Series
L1
+
E
L2
−
Figure 16-9
Air-core inductors in
series
The two inductors connected in series in Figure 16-9 have the same current flowing through them and experience the same rate of change of current. For the present, we shall assume that the two inductors are physically
­located so that the magnetic field of one cannot induce a voltage into the
other. A changing current in the circuit of Figure 16-9 induces a voltage of
e1 in L1 and e2 in L2. The total induced voltage is
From Equation 16-3,
LT =
Generalizing,
eT = e1 + e2
eT
e1 + e2
e1
e2
=
=
+
= L1 + L2
di/dt
di/dt
di/dt
di/dt
When there is no mutual coupling between inductors in series, the
total inductance is the sum of the individual inductances:
LT = L1 + L2 + L3 + . . .
(16-6)
Note that the equation for inductors in series is similar to the equation for
the total resistance of resistors in series.
See Problem 16-14 and Review Questions 16-39 and 16-40.
16-8 Inductors in Parallel
The two inductors connected in parallel in Figure 16-10 have the same
­potential difference between their terminals. Since this voltage is produced
by self-induction, the current in Ll must change at the rate of di1/dt, and
the current in L2 must change at the rate of di2/dt. Therefore, the total current must change at the rate of (di1 + di2)/dt, and
Leq =
eL
di1 + di2
dt
Inverting both sides of this equation gives
di1 + di2
di1
di2
di1
di2
+
l
dt
dt
dt
dt
dt
=
=
=
+
eL
eL
eL
eL
Leq
l
l
l
=
+
Leq
L1
L2
16-9 The DC Generator
+
E
L1
i1
L2
i2
−
Figure 16-10
Iron-core inductors in parallel
Generalizing,
The equivalent inductances of inductors in parallel is
Leq =
1
1
1
1
+
+
+ ...
L1
L2
L3
(16-7)
For just two inductors in parallel, Equation 16-7 reduces to
Leq =
L1 × L2
L1 + L2
(16-8)
Note that Equations 16-6, 16-7, and 16-8 are valid only if there is no ­mutual
induction between the individual inductors. Section 27-6 deals with
­combinations of inductors that have both self and mutual i­ nductions.
See Problems 16-15 and 16-16 and Review Question 16-41.
Circuit Check
B
CC 16-4. What inductance in series with a 100-µH and a 0.25-mH inductor produces a total inductance of 0.5 mH?
CC 16-5. Two inductors in parallel develop a voltage of 6.0 V when the
current is changed at a rate of 32 A/s. Given that one of the inductors is 200 mH, determine the value of the other inductor.
16-9 The DC Generator
A generator is a device that converts mechanical energy into electric energy.
The DC generator is an application of Faraday’s law: a mechanical force (the
prime mover) rotates coils in a stationary magnetic field, inducing a voltage in the coils.
459
460
Chapter 16 Inductance
All motors and generators include two key components:
• a field system of magnets or coils
• an armature, which has conductors that cut the magnetic flux
One of these key components is located on the rotor (the moving part
of the machine) and the other is on the stator (the stationary part). For DC
machines and small AC machines, the armature is on the rotor and the field
system on the stator, as shown in Figure 16-11. For large AC machines, the
field system is on the rotor and the armature is on the stator.
Brush
Rotor
Frame (yoke)
Commutator
Field winding
Field pole core
Armature winding
Shaft
Field pole shoe
Figure 16-11
Construction of a DC generator
The frame or yoke is normally made of annealed steel. It provides the
mechanical support for the poles and also serves as a return path for the
magnetic flux created by the field coils. The poles are made of soft iron,
which has good magnetic characteristics for developing a strong magnetic
field. The poles have shaped pieces called shoes, which minimize the air
gap between the poles and the rotating armature. The field poles are built of
thin laminations insulated from each other by a layer of varnish. This construction is used in magnetic circuits to minimize the power loss through
eddy currents. The field coils are wound on the poles and are either shunt
coils or series coils. Shunt coils have a large number of turns of fine wire,
and series coils have a small number of turns of heavy wire. The commutator consists of copper segments insulated from each other by plastic
spacers. The armature coils ends are connected to the commutator segments. The commutator provides mechanical rectification for the voltage
16-10 Simple DC Generator
­ eveloped by the armature. The brushes are usually made with a mixture
d
of powdered carbon and graphite along with binders and other additives.
The brushes are held against the commutator by a spring arrangement,
and make an electrical connection to the moving armature.
See Review Question 16-42.
16-10 Simple DC Generator
In its simplest form a generator consists of a single loop of wire rotating in
a magnetic field between two poles, as shown in Figure 16-12.
Wire loop
S
N
Slip ring
Brush
V
Figure 16-12
Load
Single loop AC generator
Voltage is induced in the coil sides as they rotate through the magnetic
flux. Since the sides of the coil move in opposite directions, the induced
voltages in the two sides have opposite polarities, which add in series
around the loop to give the total generated voltage.
The induced voltage is determined by the rate of cutting the flux. Since
the motion is rotational, the generated voltage will depend on the angle at
which the flux is cut. In the vertical position, the conductors move parallel to the flux, and therefore no voltage is induced. As the loop rotates, the
voltage increases and will be maximum when the coil is moving perpendicular to the flux, after 90º of rotation. Beyond this point the voltage decreases again and will reach zero after 180º of rotation when the conductors
are again moving parallel to the flux. As the coil rotates farther, the side
that was originally at the top and moving in the direction of the magnetic
field is now at the bottom and moving in the opposite direction. Similarly,
the side that was at the bottom is now at the top and also reverses its direction relative to the magnetic field. Thus, the polarity of the induced voltage
461
Chapter 16 Inductance
reverses as it begins to increase again. The cycle is completed as the coil
rotates back to its initial position.
If the speed of rotation is constant, the induced voltage has the shape of
a sine wave, as shown in Figure 16-13. This alternating voltage is described
in more detail in Section 18-2.
Voltage
462
Angle
90º 180º 270º 360º
Figure 16-13
Output voltage of a simple AC generator
To produce a DC output from the simple generator, the slip rings are replaced with a commutator, which in its simplest form consists of two copper
segments insulated from each other, as shown in Figure 16-14.
S
N
Commutator
V
Figure 16-14
Load
Simple DC generator
The commutator provides mechanical rectification by switching the connections to the armature just as the voltage induced in the coils reverses
polarity. For the first 180º of rotation, brush 1 is connected to commutator segment A, and coil side A is positive (see Figure 16-15). Brush B is
connected to commutator segment B and is negative. After 180º of rotation, sides A and B reverse polarity. At the same time the brushes move to
different commutator segments. Brush 1 is now connected to coil side B
and therefore remains positive, while brush 2 is now connected to coil side
A and remains negative (see Figure 16-16).
16-11 EMF Equation
−
A
+
1
A
1
+
2
B
A
B
B
2
−
+ −
+
−
1
2
Figure 16-15
Commutator connections at start of cycle
−
B
+
1
B
1
+
A
A
2
B
A
2
−
+ −
+
−
1
2
Figure 16-16
Commutator connections after 180º of rotation
Voltage
Mechanical rectification is accomplished by physically switching the
connections to the coil sides as the voltage polarity reverses. The resulting
voltage is a rectified sine wave, referred to as pulsating DC (Figure 16-17).
90º
Figure 16-17
180º
270º
360º
Angle
Output voltage
See Review Questions 16-43 to 16-45.
16-11 EMF Equation
When the armature of a generator is wound with more than one coil, the
output voltage is the sum of the voltages in the individual coils. If the armature has two coils positioned at right angles to each other, the second
463
464
Chapter 16 Inductance
coil is a quarter of a revolution behind the first coil, and the pulsating voltage in the second coil lags 90º behind the voltage in the first coil, as shown
in Figure 16-18. When these two voltages are added together, the fluctuations in total voltage are much less than those in the individual coils.
et
1
N
e1
e2
S
2
Figure 16-18
Output voltage from a two-coil armature
A practical generator has many coils wound in slots in the armature core.
The output voltage is the sum of the coil voltages and has greatly reduced
fluctuations. The commutator will have as many segments as there are
coils on the armature. The output voltage can also be smoothed by using
more than one pair of poles to produce the magnetic field. The voltage produced by a DC generator depends on the speed of rotation, the strength of
the magnetic field, and the number of turns in each armature winding. The
equation for generated voltage is given below:
Eg =
ZΦN
P
×
60
A
(16-9)
where Z is the total number of armature conductors, Φ is the flux per
pole in webers, N is the speed in revolutions per minute (r/min or RPM),
P is the total number of poles in the field system, and A is the number of
parallel paths through the armature.
A depends on the way the armature coils are connected to the commutator. The two basic types of armature connections are wave winding, which
results in two parallel paths, and lap winding, which results in as many
paths as there are poles.
Example 16-2
An 8-pole, lap-wound armature has 96 slots with 10 conductors in each
slot. The flux per pole is 20 mWb. Calculate the EMF generated when the
armature is driven at 500 RPM.
Solution
Z = 96 slots × 10 conductors/slot = 960 conductors
A = P = 8 parallel paths
Eg =
ZΦN
P
960 × 20 × 10 − 3 Wb × 500 RPM
8
×
=
× = 160 V
60
A
60
8
16-12 The DC Motor
465
Source: © iStock.com/albin
Modern day applications for DC generators are limited to standalone battery charging units, welding generators, and some specialty manufacturing processes. Even in these applications, an AC generator with solid-state
rectification is often used instead of a DC generator.
A field system with eight poles
See Problems 16-17 and 16-18 and Review Questions 16-46 and 16-47.
16-12 The DC Motor
A motor is a device that converts electrical energy into mechanical energy.
A DC motor has the same construction as the DC generator and in fact the
two are interchangeable and require only an adjustment in brush position
to reduce sparking.
Motor action is based on the principle that a conductor in a magnetic
field will experience a force when current flows through it (Figure 16-19).
F
I
N
Figure 16-19
S
Force on a current-carrying conductor
466
John Ambrose
Fleming (1849–1945)
was an English
electrical engineer
and physicist. In 1904,
he invented the first
vacuum tube that was
able to detect radio
signals. The vacuum
tube was used in
radios and radars
for the next 50 years,
until semiconductor
electronics eventually
replaced it.
Chapter 16 Inductance
The direction of force is given by Fleming’s left-hand (motor) rule.
Extend the thumb, forefinger and middle finger of the left hand at right
angles to each other. Orient the hand so that the forefinger is in the ­direction
of the magnetic field and the middle finger in the direction of current flow.
The thumb then indicates the direction of developed force.
The magnitude of force created depends on three variables:
• the strength of the magnetic field
• the magnitude of the current
• the length of the conductor in the magnetic field
The magnitude of force, in newtons, is given by
F = BlI (16-10)
where B is the flux density in teslas, l is the length of the conductor in
metres, and I is the current in amperes.
Example 16-3
A rectangular coil has 100 turns and is in a magnetic field of flux density
0.800 T. Given that the length of the coil is 30.0 cm and the current flowing in the conductors is 15.0 A, calculate the force acting on each side
of the coil.
Solution
F = BlI = 0.800 T × (100 turns × 0.300 m) × 15.0 A = 360 N
A practical DC motor has many armature coils in slots on the rotor. The
force on each conductor is offset from the centre of the rotor, so these forces
each produce a torque on the rotor (Figure 16-20). The total torque turns
the rotor.
N
Figure 16-20
S
x Current in
Current out
Torque created in a DC armature
Source: JEFF DALY, VISUALS UNLIMITED /SCIENCE PHOTO LIBRARY
16-13 Speed and Torque of a DC Motor
Multiple windings on a motor armature
A DC motor has a commutator that reverses the current direction in the
armature coils as they pass from one field pole to the next, which has the
opposite direction. As a result torque continues to turn the rotor in the
same direction.
See Problem 16-19 and Review Questions 16-48 and 16-49.
16-13 Speed and Torque of a DC Motor
When a motor armature rotates, the conductors on the armature cut the
flux provided by the field poles. Thus by Faraday’s law, an EMF will be
induced in the armature conductors. This EMF opposes the applied voltage
as per Lenz’s law and is therefore referred to as back or counter EMF (Eb).
Since the back EMF is a generated EMF, it can be calculated using the generated EMF equation.
467
468
Chapter 16 Inductance
Eg =
ZΦN
P
×
60
A
(16-11)
Solving this equation for N gives the speed equation for a DC motor:
N=
60Eb A
×
ZΦ
P
(16-12)
Example 16-4
A 440-V DC motor has four poles with a flux per pole of 20 mWb. The
armature is wave wound with 44 slots and 12 conductors in each slot.
The back EMF is 400 V. Calculate the motor speed.
Solution
Z = 44 slots × 12 conductors/slot = 528 conductors
A=2
N=
60Eb A
60 × 400 V
2
×
=
× = 1.1 × 103 RPM
ZΦ
P
528 × 0.02 Wb 4
Torque is the turning or twisting moment of a force about an axis, and is
often used as a measure of the strength of the motor. The torque in a motor
is created by the forces developed in the armature conductors, and consequently is proportional to the current flowing in the armature (IA) and
strength of the magnetic field. The equation for torque in a motor is
T = KΦIA
(16-13)
K=
(16-14)
where T is the torque in newton metres (N⋅m), and K is the torque constant, which depends on the physical characteristics of the machine and
is given by the following equation:
0.159Z P
A
Combining for K in Equation 16-13 gives
T=
0.159Z P
ΦIA
A
(16-15)
16-14 Types of DC Motors
Example 16-5
Given that the full-load armature current for the motor in Example 16-4
is 60.0 A, calculate the full-load torque developed.
Solution
T=
0.159Z P
0.159 × 528 × 4
ΦIA =
× 0.02 Wb × 60 A = 197 N . m
A
2
As the motor rotates because of the torque developed in the armature,
the motor shaft can do work on a mechanical load. The power produced
by a motor depends on the torque and speed. In North America, the old
imperial unit of power, the horsepower (hp), is still commonly used to rate
the power output of electric motors:
power ( in hp ) =
2πNT
33 000
(16-14)
where T is the torque in pound-feet (1 lb-ft = 1.356 N.m).
Example 16-6
Calculate the full-load horsepower output of the motor in Example 16-5.
Solution
T = 197 N. m ×
power =
1 1b-ft
= 145 1b-ft
1.356 N. m
2π × 1.1 × 103 RPM × 145 1b-ft
2πNT
= 30 hp
×
33 000
33 000
See Problems 16-20 to 16-22 and Review Questions 16-50 and 16-51.
16-14 Types of DC Motors
motors are classified according to the method of field excitation. The
field may be produced electromagnetically by current flow in the field coils
or by permanent magnets.
DC
469
470
Chapter 16 Inductance
The shunt motor has field coils connected in parallel with the armature
(Figure 16-21). Shunt coils have many turns of fine wire and require a relatively small current to produce a strong magnetic field.
Ish
Il
Ia
Ra
+
E
+
−
Rsh
Eb
−
Figure 16-21
Shunt motor circuit diagram
The total current is called the load current and is the sum of the armature
and field currents, IL = IA + Ish. Since the shunt field is connected in parallel with the source, Ish can be calculated from Ohm’s law, Ish = E/Rsh.
The back EMF can be found using Kirchhoff’s voltage law, Eb = E − IARA.
The series motor has field coils connected in series with the armature
(Figure 16-22). Since the field is now developed by the large armature current, series coils require only a few turns of heavy wire.
Il
Ise
Rse
Ia
Ra
+
E
+
−
Eb
−
Figure 16-22
Series motor circuit diagram
Since the same current flows through all components in a series circuit,
IL = Ise = IA. The equation for back EMF now includes the voltage drop in
the resistance of the series coil: Eb = E− IA (RA+ Rse).
The compound motor has both shunt and series coils. The series coil may
be connected to either aid or oppose the shunt field. With the aiding configuration, the motor is cumulatively compounded, and with the opposing
configuration the motor is differentially compounded. The differentially
compounded motor is rarely used because it tends to become unstable under
heavy load and may even reverse its direction of rotation. The series coil
16-15 Speed Characteristics of DC Motors
may be connected in two positions, in series with the source (short shunt) or
in series with the armature (long shunt), as shown in Figure 16-23. There is
very little operational difference between these two series connections.
Ish
Il
Ish
Il
Rse
Rse
Ia
Ia
Ra
E
+
E
Rsh
+
−
Rsh
−
+
Eb
−
Eb
−
(a)
Figure 16-23
Ra
+
(b)
Compound motor circuits: (a) Short shunt; (b) Long shunt
Example 16-7
A 440-V DC compound motor is connected long shunt and has shunt
field, series field, and armature resistances of 110 Ω, 0.25 Ω, and 0.40 Ω,
respectively. Calculate the back EMF when the motor operates with a
load current of 60 A.
Solution
Eb = E − IA ( Rse + RA )
IA = IL − Ish and Ish =
E
440 V
=
= 4.00 A
Rsh
110 Ω
IA = 60 A − 4.00 A = 56 A
Eb = 440 V − 56 A × ( 0.65 Ω ) = 440 V − 36.4 V = 404 V
See Problems 16-23 to 16-24 and Review Questions 16-52 and 16-53.
16-15 Speed Characteristics of DC Motors
The speed characteristic is the variation of the speed of a motor as a mechanical load is applied to the shaft. When a load is applied, the motor draws
more current from its power supply and produces more torque.
471
Chapter 16 Inductance
As shown in Section 16-13, the speed of a DC motor is given by
60Eb A
N=
×
ZΦ
P
For any given machine, Z, P, and A are constant, and therefore
Eb
N∝
Φ
We can examine the speed characteristic for each type of DC motor by
observing the variation of Eb and Φ as the load current increases.
For the shunt motor,
E
Φ ∝ Ish =
Rsh
If E and Rsh are kept constant, the flux will remain constant. The back EMF is
Eb = E − IARA, and will decrease as load current and therefore IA increase.
Since RA is small, this decrease is small and the speed of the shunt motor
remains relatively constant as a load is applied (Figure 16-24).
For the series motor, the field is produced by the armature current,
which is also the load current. Therefore, the flux will increase with increasing load, and the speed will decrease. The back EMF, Eb, will again
decrease slightly as the load current increases, contributing to the reduction in speed. At light loads, the flux is small, and the motor speed is high.
A series motor can never be operated at no load since the excessive speed
would damage the motor. A series motor is usually coupled to its load by
gears rather than a pulley system. Because a series motor has only a few
turns on its field coil, the inductance is small and the flux produced stays
synchronized with the armature flux even when alternating current flows.
This motor will operate on AC and on DC, and is often referred to as the
universal motor.
The compound motor combines the characteristics of shunt and series
motors. The series field will increase as a load is applied, and therefore the
total flux will increase. The speed will drop more than in the shunt motor
but not as much as in the series motor.
Shunt
Compound (cumulative)
Speed, N
472
Series
Load Current (IL)
Figure 16-24
Speed characteristics of DC motors
16-15 Speed Characteristics of DC Motors
Example 16-8
A 220-V DC shunt motor has a shunt field resistance of 110 Ω and an
armature resistance of 0.10 Ω. When unloaded, the motor draws a current of 7.0 A and rotates at 1250 RPM. A series winding of resistance
0.05 Ω is cumulatively connected long shunt. This winding increases
the flux per pole by 20% when the motor is fully loaded and taking a
current of 62 A. Calculate the full-load speed of this compound motor.
Solution
When the shunt motor is unloaded (Figure 16-25),
Ish =
N∝
E
220 V
= 2.00 A
=
Rsh
110 Ω
Eb
Φ
IA = IL − Ish = 7.0 A − 2 A = 5 A
Eb1 = E − IARA = 220 V − ( 5 A × 0.10 Ω ) = 219.5 V
Ish = 2 A
Il = 7 A
Ia
Ra = 0.10 Ω
+
Rsh = 110 Ω
E
+
−
Eb
−
Figure 16-25
When the compound motor is loaded (Figure 16-26),
IA = IL − Ish = 62 A − 2 A = 60 A
Eb2 = E − IA ( RA + Rse ) = 220 V − ( 60 A × 0.15 Ω ) = 211 V
N1 ∝
Eb2╱Φ2
Eb2 Φ1
Eb1
Eb2
N
→ 2=
×
and N2 ∝
=
Eb1 Φ2
Φ1
Φ2
N1
Eb1╱Φ1
Φ2 is increased by 20% because of the series field; therefore, Φ2 = 1.2Φ1
Φ1
1
=
= 0.833
Φ2
1.2
N2 = 1250 RPM ×
211 V
× 0.833 = 1001
219.5 V
RPM
473
474
Chapter 16 Inductance
Ish = 2 A
Il = 7 A
Rse = 0.05 Ω
Ia
Ra = 0.10 Ω
+
E
Rsh = 110 Ω
−
+
Eb
−
Figure 16-26
See Problems 16-25 to 16-26 and Review Question 16-54.
Circuit Check
C
CC 16-6. How much flux per pole is required for a 12-pole lap-wound
motor to rotate at 500 RPM if the back is 224 V? The motor
has a total of 480 conductors.
CC 16-7. A 550-V long-shunt compound motor has shunt field, series
field, and armature resistances of 100 Ω, 0.24 Ω, and 0.32 Ω,
respectively. What load current will produce a back of
500 V?
16-16 Torque Characteristics of DC Motors
DC motors are used to drive a variety of mechanical loads, such as compres-
sors, fans, elevators, and electric vehicles. It is important to understand
how the torque developed by the motor varies as the mechanical load is
increased. From Equation 16-13, T ∝ ΦIA, so we can predict how torque
varies as load is applied and IA and therefore IL increase.
The shunt motor has a constant flux. Therefore, the torque produced will
be directly proportional to the load current, and the torque graph is linear
as load is applied (Figure 16-27).
In the series motor, Φ ∝ IA, and consequently T ∝ I2A. The resulting graph
is a parabola, and torque increases rapidly as load is applied.
The cumulatively compounded motor has a combination of series and
shunt characteristics with the series torque adding to the shunt torque as
load is increased.
16-16 Torque Characteristics of DC Motors
The small torque developed at no load is necessary to overcome friction
and other rotational losses, allowing the motor to turn. At starting, the back
EMF is zero and a large current will flow.
For a shunt motor:
E − Eb
E
Eb = E − IARA ⟶ IA =
; therefore, at starting IA =
RA
RA
Compound (cumulative)
Torque, T
Shunt
Series
No load
torque
Load Current (IL)
Figure 16-27
Torque Characteristics of DC motors
Since RA is small, the starting current will be high. This current creates
a large torque at starting which will help accelerate a heavy load. For this
reason, the DC motor is well suited to traction and lifting applications.
Example 16-9
A 4.0-hp series motor when running at full load draws a current of 40
A and turns at 600 RPM. If the starting current is 70 A, calculate the fullload and starting torque.
Solution
At full load, IA = 40 A
Since hp =
T=
2πNT
33 000
hp × 33 000
4.0 hp × 33 000
= 35 lb- ft
=
2πN
2π × 600 RPM
At starting, IA = 70A
For a series motor, T ∝ IA2 and
Tst = Trun ×
I2Ast
I2Arun
I2Ast
Tst
= 2
Trun
IArun
2
70
= 35 l b-ft × 2 = 107 lb-ft
40
See Review Problems 16-27 to 16-28 and Review Question 16-55.
475
476
Chapter 16 Inductance
16-17 Permanent Magnet and Brushless
DC Motors
Many of the traditional DC motors with field windings have been replaced
by permanent magnet motors. This design is typically used in fractionalhorsepower motors and even up to 3 hp sizes. Advances in magnetic
materials enable the construction of motors with strong magnetic fields
provided by permanent magnets. Such motors have a higher efficiency
because they do not have any energy losses due to resistance in field
windings.
A brushless motor operates much in the same way as a traditional
motor with brushes but uses electronic switching instead of brushes and
a commutator. In a typical brushless motor, the field is on the rotor and is
created by permanent magnets. Power is supplied directly to the armature,
which is on the stator and therefore can be directly connected. A rotating
magnetic field is produced by the armature as current flows and commutation is effected with electronic switches, often by using transistors. The
rotor is pulled around in synchronism with this rotating magnetic field.
Brushless motors have excellent reliability and long lives, and they can
operate at high speeds. Brushless motors are used in computer fans, hard
drives, and biomedical and aerospace applications—industries in which
reliability is critical.
See Review Question 16-56.
Problems
477
Summary
• Perpendicular motion of a conductor relative to a magnetic field p
­ roduces
electromagnetic induction.
• Changing current in a primary winding induces a voltage in a secondary
winding on the same core.
• Faraday’s law relates the induced voltage in an electric circuit to the rate
of change of the magnetic flux.
• The polarity of an induced voltage is always such that any resulting
­current opposes the change in the original magnetic flux.
• Self-induction produces an induced voltage across a coil when the
­current through it is changing.
• Inductance opposes any change in current in an electric circuit.
• Inductance is proportional to the number of turns in a coil and the rate
of change of flux with respect to current.
• The inductance of a coil is proportional to the square of the number of
turns, proportional to the permeability and cross-sectional area of the
magnetic circuit, and inversely proportional to the reluctance and the
length of the magnetic circuit.
• The total inductance of inductors in series (when there is no mutual
­coupling) is the sum of the individual inductances.
• The reciprocal of the equivalent inductance of inductors in parallel equals
the sum of the reciprocals of the individual inductances, provided there
is no mutual inductance.
• All rotating electrical machines have a field system and an armature ­system.
• A rotating coil in a magnetic field produces an alternating voltage.
• A commutator can convert a generated voltage to DC.
• Generated voltage depends on the strength of the field flux and the
speed of rotation.
• A DC motor has the same construction as a DC generator.
• The force produced when a current flows through a conductor in a m
­ agnetic
field depends on the strength of the field and the magnitude of the current.
• The speed of a motor is directly proportional to back EMF and inversely
proportional to magnetic flux.
• The torque of a motor is directly proportional to the armature current
and the magnetic flux.
• DC motors are classified based on the configuration of the field s­ ystem.
Problems
B
Section 16-5
16-1.
B
16-2.
B
16-3.
B
16-4.
Self-Inductance
Current changing at the rate of 250 mA/s in an inductor induces a
voltage of 75 mV. Find the inductance.
A current rising at a uniform rate from 2.0 mA to 10 mA in 100 μs ­induces
a voltage of 20 mV in a load. What is the inductance of the load?
Find the voltage induced into an 8-H inductor by a current that
changes from 12 A to 6 A in 75 ms.
How long will it take the current in a 1.5-H inductor to rise from
zero to 5 A if the voltage induced into the inductor is constant at 6 V?
B = beginner
I = intermediate
A = advanced
478
Chapter 16 Inductance
B
B
16-5.
Section 16-6
16-6.
I
16-7.
B
B
I
16-8.
16-9.
16-10.
I
16-11.
I
16-12.
I
16-13.
B
I
B
B
B
B
Find the inductance of a coil if 250 V is induced across it by an 8-A
current that reverses direction in 20 ms.
Factors Governing Inductance
Find the approximate inductance of a 15-turn air-core coil with an
inside diameter of 3.0 cm and a length of 2.0 cm, assuming that the
leakage flux is negligible.
How many turns must an 80 μH inductor have if it has an air-core
coil 3.0 cm long and 4.0 cm in diameter?
Find the inductance of the inductor in Problem 15-9.
Find the inductance of the coil in Problem 15-13.
A 1-H inductor has 1400 turns. How many turns must be added to
raise its inductance to 2 H?
An 800-mH inductor has 600 turns. Where would you place a tap
to obtain an inductance of 400 mH between one end and the tap?
What is the inductance between the other end and the tap?
An iron core has a relative permeability of 3000, a diameter of 4 cm,
and a length of 15 cm. How many turns do you have to wind on
this core to make a 10-H inductor?
A 750-turn coil is wound on a magnetic core with a relative permeability of 2000, a length of 18 cm, and a cross-sectional area of
6.0 cm2. Calculate:
(a) the inductance of the coil
(b)the voltage induced when the current through the coil decreases
steadily from 8 A to zero in 35 ms
Section 16-7
Inductors in Series
16-14. Three inductors with inductances of 10 mH, 20 mH, and 30 mH are
positioned so that there is no mutual induction between them. Find
the effective inductance when the three coils are connected in series.
Section 16-8
Inductors in Parallel
Section 16-11
EMF Equation
16-15. What inductance must be connected in parallel with an inductance
of 40 μH to reduce the net inductance to 15 μH?
16-16. Find the equivalent inductance when the coils of Problem 16-14 are
connected in parallel.
16-17. An 8-pole generator has a wave-wound armature with 52 slots,
each containing 10 conductors. If the total flux produced by the
field system is 0.32 Wb, calculate the EMF generated when the armature is driven at 600 RPM.
16-18. A 4-pole generator with a lap-wound armature has 120 slots with
4 conductors in each slot. If the flux per pole is 50 mWb and the generated EMF is to be 240 V, at what speed must this generator be driven?
Section 16-12
The DC Motor
16-19. What current will develop a force of 270 N on a 300-turn coil in a
magnetic field of 1.8 T? The length of the coil in the field is 20 cm.
Review Questions
I
B
I
I
A
I
A
B
A
Section 16-13
Speed and Torque of a DC Motor
16-20. A 6-pole lap-wound DC motor has 24 coils on the armature, and each
coil has 10 turns. When the flux per pole is 80 mWb, the motor turns
at 1000 RPM. Calculate the back EMF generated in the motor armature.
16-21. Calculate the starting torque for the motor of Problem 16-20, given
that the starting current is 55 A.
16-22. A 5.0-hp DC motor has a 4-pole wave-wound armature containing
16 coils with 20 turns each. The field system produces a flux per
pole of 10 mWb. Given that the full-load speed is 450 RPM, calculate
the full-load armature current.
Section 16-14
Types of DC Motors
Section 16-15
Speed Characteristics of DC Motors
Section 16-16
Torque Characteristics of DC Motors
16-23. A 220-V DC compound motor is connected short shunt and has
shunt field, series field, and armature resistances of 55 Ω, 0.12 Ω,
and 0.18 Ω, respectively. Calculate the back EMF when the motor has
a load current of 72 A.
16-24. A 440-V DC compound motor is connected short shunt and has
shunt field resistance of 75 Ω and series field resistance of 0.2 Ω.
When the motor has a load current of 40 A, the back EMF is 420 V.
Calculate the resistance of the armature.
16-25. Redo Example 16-8 with the series field connected differentially.
16-26. A 220-V DC 6-pole series motor has a field resistance of 0.15 Ω and
an armature resistance of 0.20 Ω. The armature is lap-wound with
a total of 248 conductors. When the motor delivers its full-load
output, the load current is 52 A and the flux per pole is 35 mWb.
When the motor is unloaded, the current drops to 7 A. Assuming
the field core is unsaturated and has a linear magnetization curve,
calculate the full-load and no-load speeds.
16-27. Calculate the full-load and no-load torque developed by the motor
in Problem 16-26.
16-28. A 35-hp 440-V DC shunt motor has a field resistance of 80 Ω. When
delivering its rated full-load horsepower, this motor rotates at
1150 RPM and draws a current of 75 A. When starting, the motor
draws a current of 120 A. Calculate:
(a) the full-load and starting torque
(b) the full-load efficiency
Review Questions
Section 16-1
Electromagnetic Induction
16-29. What is meant by the term electromagnetic induction?
Section 16-2
Faraday’s Law
16-30. What factors govern the magnitude of the EMF induced into a c­ onductor?
479
480
Chapter 16 Inductance
Section 16-3
Lenz’s Law
16-31. What factors govern the polarity of the EMF induced into an electric
conductor?
16-32. State the direction of conventional current in Figure 16-28. The
arrow in the figure indicates motion.
S
N
G
Figure16-28
16-33. State the direction of conventional current in both the primary and
secondary windings of Figure 16-29 when
(a) the switch is first closed
(b) the switch is opened
E
+
−
Figure16-29
Section 16-4
Self-Induction
16-34. What is meant by self-induction?
Section 16-5
Self-Inductance
16-35. How is the unit of inductance derived from other SI units?
Section 16-6
Factors Governing Inductance
16-36. Why is the self-inductance of a coil proportional to the square of the
number of turns?
16-37. Why does the inductance of an iron-core coil depend on the current
through it?
16-38. What effect does the addition of an air gap to the magnetic circuit
of a coil have on its inductance?
Section 16-7
Inductors in Series
16-39. Why is it necessary to stipulate that there must be no mutual induction when one is deriving the equation for the total inductance of
inductors in series?
16-40. Why is the total inductance of series inductors equal to the sum of
the individual inductances?
Review Questions
Section 16-8
Inductors in Parallel
16-41. In what way is the equation for the equivalent inductance of inductors in parallel similar to the equation for the equivalent resistance
of resistors in parallel? Compare these equations to the equation for
the equivalent capacitance of capacitors in parallel.
Section 16-9
The DC Generator
16-42. Why is it important for the frame or yoke of a
made from ferromagnetic material?
Section 16-10
DC
generator to be
Simple DC Generator
16-43. Explain why the voltage generated in a rotating coil is alternating.
16-44. Explain how a commutator converts the alternating voltage produced in the armature to DC voltage.
16-45. Why is the output voltage of a simple DC generator referred to a
pulsating DC?
Section 16-11
EMF Equation
16-46. What can be done to smooth the output voltage of a DC generator?
16-47. For a particular generator, which factors determine the magnitude
of generated voltage?
Section 16-12
The DC Motor
16-48. How does a DC motor differ in construction from a DC generator?
16-49. Why does a current-carrying conductor in a magnetic field experience a force?
Section 16-13
Speed and Torque of a DC Motor
16-50. What quantities determine the speed of a DC motor?
16-51. Why is torque an important characteristic of a DC motor?
Section 16-14
Types of DC Motors
16-52. Explain the difference between a short-shunt and a long-shunt
compound motor.
16-53. Why is a differentially connected compound motor rarely used?
Section 16-15
Speed Characteristics of DC Motors
16-54. Which DC motor is best suited for driving a machine that uses
magnetic tape for storage? Explain your answer.
Section 16-16
Torque Characteristics of DC Motors
16-55. Which DC motor is best suited for an electric golf cart? Explain your
answer.
Section 16-17
Permanent Magnet and Brushless DC Motors
16-56. Describe how a brushless motor operates.
481
482
Chapter 16 Inductance
Integrate the Concepts
(a) Calculate the inductance of the air-core coil of Example 15-1.
(b) Calculate the inductance of the toroidal coil of Example 15-2.
(c)Calculate the total inductance of these two inductors connected in series
(assume no mutual inductance).
(d)Calculate the equivalent inductance of these two inductors connected
in parallel (assume no mutual inductance).
Practice Quiz
1.
Which of the following statements is true?
(a) An induced voltage will appear in the conductor if the conductor
and the magnetic field are moving parallel to each other.
(b) The polarity of an induced voltage can be determined using Lenz’s law.
2.
Faraday’s law states that
(a) the voltage induced in an electric circuit is inversely proportional
to the rate of change of the magnetic flux linking the circuit
(b) the current induced in an electric circuit is directly proportional
to the rate of change of the magnetic flux linking the circuit
(c) the voltage induced in an electric circuit is directly proportional to
the rate of change of the magnetic flux linking the circuit
(d) the current induced in an electric circuit is inversely proportional
to the rate of change of the magnetic flux linking the circuit
3.
Self-inductance is
(a) directly proportional to the number of turns linked by the magnetic flux and to the current flowing through the circuit
(b) inversely proportional to the number of turns linked by the magnetic flux and to the current flowing through the circuit
(c) directly proportional to the number of turns linked by the magnetic flux
and inversely proportional to the current flowing through the circuit
(d) inversely proportional to the number of turns linked by the magnetic
flux and directly proportional to the current flowing through the circuit
4.If the permeability of the magnetic circuit of a coil decreases, the ­inductance
(a) decreases
(b) stays the same
(c) increases
5.
If the length of the magnetic circuit of a coil decreases, the inductance
(a) decreases
(b) stays the same
(c) increases
6.
If the number of turns in a coil is increased, the inductance
(a) decreases
(b) stays the same
(c) increases
Practice Quiz
7.
The value of the equivalent inductor for the circuit of Figure 16-30 is:
(a) 99 nH
(b) 20 nH
(c) 190 nH
(d) 19 mH
L1
0.0745 μH
L3
0.068 μH
L2
0.047 μH
Figure 16-30
8. The value of the equivalent inductor for the circuit of Figure 16-31 is
(a) 0.48 mH
(b) 12 mH
(c) 0.91 mH
(d) 1.2 mH
L1
1.0 mH
L2
10 mH
L3
1.0 mH
Figure 16-31
9.A 250-turn coil with a length of 25 cm is in a magnetic field with a flux
density of 2.0 T. If the force is 300 N, the current through the coil is
(a) 4.8 mA
(b) 24 mA
(c) 1.67 A
(d) 2.4 A
10. The torque produced by a 2.5-hp DC motor operating at 1500 RPM is
(a) 0.11 lb-ft
(b) 8.8 lb-ft
(c) 4.4 lb-ft
(d) 18 lb-ft
483
17
Inductance in DC Circuits
We now examine the behaviour of practical electric circuits when
inductance is present. Often an inductor is used to give the circuit
inductance for a specific purpose, as in the tuned and filtering
circuits descibed in Chapter 25. However, some other components,
such as wire-wound p
­ otentiometers, have some inductance in
addition to their primary property.
Chapter Outline
17-1
17-2
Current in an Ideal Inductor 486
Rise of Current in a Practical Inductor 487
17-3
Time Constant
17-5
Algebraic Solution for Inductor Current 495
17-4
17-6
490
Graphical Solution for Inductor Current 491
Energy Stored by an Inductor 499
17-7
Fall of Current in an Inductive Circuit 501
17-9
Transient Response
17-8
Algebraic Solution for Discharge Current 506
507
17-10 Characteristics of Inductive DC Circuits
17-11
Troubleshooting
510
509
Key Terms
LR circuit 490
discharge resistor
503
transient response 507
filter choke 510
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• draw a graph of the rise of current in an ideal
inductor
• determine the time constant of a circuit
­containing ­resistance and inductance
• use universal exponential curves to find
­instantaneous currents and voltages in a DC
­circuit containing ­resistance and inductance
Photo sources: iStock.com/Marco Hegner
• calculate exact values of instantaneous
currents and v
­ oltages in a DC circuit
containing resistance and ­inductance
• calculate the energy stored in an inductor
• detect faults in an inductor with an ohmmeter
or an LC meter
486
Chapter 17 Inductance in DC Circuits
17-1 Current in an Ideal Inductor
An ideal inductor would have no resistance. However, all known materials
have some resistance at room temperature. For the moment, we shall assume
that the inductor in Figure 17-1 has negligible resistance.
+
10 V
−
Figure 17-1
E = VL
2.0 H
Voltage drop across an ideal inductor
According to Kirchhoff’s voltage law, a voltage drop exactly equal to the
applied voltage appears across an external circuit connected to a voltage
source. If this external circuit contains only resistance, current flowing
through the resistance produces this voltage drop. When the circuit is
switched on, the current instantly rises to a steady value of I = V/R = E/R.
However, the voltage drop across an ideal inductor is produced by a changing current.
Section 16-4 described how a changing current can produce a voltage
across a coil by inducing an EMF in the coil. In Equation 16-3, eL represented
the induced voltage. The lowercase e indicates a changing voltage resulting
from an EMF. However, we do not think of the inductor in Figure 17-1 as a
voltage source. In this circuit we treat the induced voltage as the ­voltage
drop required by Kirchhoff’s voltage law. Thus, we can rewrite Equation 16-3 as
vL = L ×
di
dt
(17-1)
Rearranging Equation 17-1 gives
di
vL
=
dt
L
Since the applied voltage is constant, the voltage across the inductor
must also be constant. Therefore, we can represent the voltage drop with
the uppercase letter V. The inductance of a coil depends on the number
of turns and the reluctance of the magnetic circuit, so the inductance of a
given air-core coil is constant. For such inductors, the ratio V/L is constant.
Hence, the rate of change of current must also be constant, and
di
Δi
V
=
=
dt
Δt
L
where t is the number of seconds elapsed since the switch was closed.
17-2 Rise of Current in a Practical Inductor
Since the initial current is zero,
Δi
i −0
i
V
=
= =
Δt
t−0
t
L
i=
V
×t
L
For the circuit in Figure 17-1,
i=
10 V
× t = 5tA
2.0 H
The graph of current versus time is a straight line, as shown in Figure 17-2.
The slope of this graph represents the rate of change of current. As shown
above, this rate of change equals V/L. Increasing the inductance slows the
rise of the current. This relationship shows how inductance opposes the
change in current in a circuit.
Current (amperes)
10
8
6
4
2
0
Figure 17-2
1
Time (seconds)
2
Current in a DC circuit containing an ideal inductor
See Review Questions 17-27 to 17-29 at the end of the chapter.
17-2 Rise of Current
in a Practical Inductor
At room temperature, all practical inductors possess some resistance as
well as inductance. Since there is only one path for current through an
inductor, its inductance and resistance are effectively in series, as shown in
Figure 17-3.
487
488
Chapter 17 Inductance in DC Circuits
vR
+
10 V E
−
Figure 17-3
vL
+
1.0 Ω
−
+
2.0 H
−
Equivalent circuit for a practical inductor
When we close the switch in the circuit of Figure 17-3, the voltage drop
across the resistance of the coil plus the voltage drop across the inductance
of the coil must equal the applied voltage in order to satisfy Kirchhoff’s
voltage law:
E = vR + vL
Applying Ohm’s law and substituting the expression for vL from Equation 17-1 gives
E = iR + L
di
dt
(17-2)
If we can solve for the instantaneous current in the circuit, we can then
readily determine the voltage drop across the inductor. However, solving
Equation 17-2 for i requires integral calculus. Therefore, we shall start by
using our understanding of the current in an ideal inductor to determine
the overall behaviour of a practical inductor in a circuit.
The inductance of the practical inductor and the applied voltage in
­Figure 17-3 are the same as for the ideal inductor in Figure 17-1. Therefore, the current through the practical inductor cannot rise more rapidly
than shown in the graph of Figure 17-2. Otherwise the induced voltage
across the inductance would be greater than the applied voltage. Similarly, the current cannot rise to a value such that the IR drop across the
resistance ­exceeds the applied voltage. The greatest value that the instantaneous ­current can have is such that vR = E. Then, vL is zero, so di/dt
is also zero. Therefore, the current has reached a steady-state maximum
value, which is
Im =
E
R
(17-3)
The dashed black lines in Figure 17-4 show the two limits for the current.
17-2 Rise of Current in a Practical Inductor
Current (amperes)
Slope =
10
τ
di
E
=
dt
L
Im = E
R
i
8
6
4
63% of Im
2
0
Figure 17-4
2
4
6
Time (seconds)
8
10
Current in practical inductor
As shown in Section 17-1, the inductance limits the rate at which the
c­ urrent can rise. At the instant we close the switch in Figure 17-3, the current through the resistance is zero and hence the voltage drop across it is
also zero. Therefore, voltage across the inductor at this instant equals the
applied voltage. Since di/dt = vL/L, the initial rate of change of current is
the same as for the ideal inductor in Figure 17-1:
initial
di
E
=
dt
L
(17-4)
However, since the current is rising in order to induce a voltage in the
i­ nductance, the voltage drop across the resistance must be rising. Because
vR + vL = E, the induced voltage across the inductor decreases correspondingly. Consequently, the rate of change of current also decreases
and the slope of the current graph curve in Figure 17-4 becomes more
gradual. ­Although the current is now rising less rapidly, nevertheless
it is still rising. As a result vL and, therefore, the rate of change of current decrease further, and the current graph becomes still more gradual.
Eventually, the current reaches the steady-state maximum value given
by Equation 17-3.
When we close the switch in a DC circuit containing inductance and
­resistance in series, the current cannot instantly rise to E/R as it would if
there were no inductance in the circuit. The effect of the inductance is to
slow the rise of current, as shown by the red curve in Figure 17-4. Once
the current reaches its maximum value, the effect of the inductance in the
circuit disappears.
Since vR = iR and R is a constant, the graph of the voltage drop across
the resistance of the circuit must have the same shape as the graph of the
current (see Figure 17-5). We can graph the instantaneous voltage across
the inductance by subtracting the instantaneous IR drop from the constant
applied voltage:
vL = E − vR
489
Chapter 17 Inductance in DC Circuits
E
10
Instantaneous
Voltage (volts)
490
τ
vR
8
6
4
2
0
Figure 17-5
vL
2
4
6
Time (seconds)
8
10
Instantaneous voltage across the resistance and inductance
See Review Question 17-30.
17-3 Time Constant
As with a capacitor, it is convenient to describe the current and voltage
curves for an inductor in terms of a time constant related to the parameters
of the circuit.
The time constant, τ, of an LR circuit (a circuit containing inductance
and resistance, but not capacitance) is the time it would take the current to
rise to its steady-state value if the current increased at a constant rate equal
to its initial rate of change.
If the rate of change of current remains constant (as shown by the straight
line in Figure 17-4), the steady-state current would be simply
Im = τ × initial
di
dt
But Im = E/R (Equation 17-3) and initial di/dt = E/L (Equation 17-4).
Therefore,
and
Im =
E
E
=τ
R
L
τ
L
R
(17-5)
where τ is the time constant in seconds, L is the inductance of the circuit
in henrys, and R is the resistance of the circuit in ohms.
Note that the time constant of an LR circuit, such as a practical inductor, is directly proportional to the inductance. Doubling the inductance
reduces the initial rate of rise of current by half, and the current takes
twice as long to reach the steady-state value. Since the time constant is
inversely proportional to the resistance, doubling the resistance reduces
the steady-state current by half. The current then takes half as long to
17-4 Graphical Solution for Inductor Current
491
reach the steady-state value. However, the applied voltage has no effect
on the time constant. Doubling the applied voltage doubles both the
­initial rate of rise of current (E/L) and the steady-state current (E/R).
Therefore, the current takes the same time to reach the steady-state value.
The current in an LR circuit does not actually continue to rise at its initial rate. Instead, the current follows an exponential curve, as shown in
Fig­ure 17-4. This exponential curve starts with a slope of E/L, and then
levels off more and more slowly to finally merge with the horizontal line
representing the steady-state current. For any values of L and R, the current reaches about 63.21% of the steady-state value one time constant after
the circuit is switched on. Although theoretically the current never quite
reaches the steady-state value, for practical purposes the current is E/R
after five time constants have elapsed.
Example 17-1
(a) Find the steady-state current in the circuit of Figure 17-3.
(b)How long does it take the current to reach this value after the
switch is closed?
Solution
(a)
(b)
Im =
τ=
E
10 V
=
= 10 A
R
1.0 Ω
L
2.0 H
=
= 2.0 s
R
1.0 Ω
Time to reach steady state = 5τ = 5 × 2.0 s = 10 s
To verify the calculation of the current in part (a) and the time in
part (b), download Multisim file EX17-1 from the website and follow
the instructions in the file.
See Problems 17-1 to 17-6 and Review Questions 17-31 to 17-34.
17-4 Graphical Solution
for Inductor Current
As with capacitors, we can use universal exponential graphs to make fairly
accurate estimates of the current through an inductor at a given moment
and the time when the current reaches a particular value. For inductors,
the vertical axis of these graphs shows the percentage of the steady-state
current and the horizontal axis shows time measured in time constants (see
Figure 17-6).
circuitSIM
walkthrough
Chapter 17 Inductance in DC Circuits
100
90
Current (% of maximum)
492
y = 1 − e−x
80
70
60
50
40
30
20
y = e−x
10
0
3
2
Time, x (time constants)
1
4
5
Figure 17-6 Universal exponential curves for inductive DC circuits
Example 17-2
(a)What is the instantaneous current in the circuit of Figure 17-3 1.0 s
after the switch is closed?
(b)How long does the current take to rise from 0 A to 5.0 A?
Solution
(a) We have already determined that the time constant of this circuit is
Therefore,
τ=
L
2.0 H
=
= 2.0 s
R
1.0 Ω
1.0 s = 0.50τ
On the universal exponential graph, y = 39% when x = 0.50 as
shown in Figure 17-7(a). Therefore the current at t = 1.0 s is
i = 39% ×
(b)
E
10 V
= 0.39 ×
= 3.9 A
R
1.0 Ω
Im =
E
10 V
=
= 10 A
R
1.0 Ω
5.0 A
5.0 A
=
= 50%
Im
10 A
On the universal exponential graph, x = 0.70 when y = 50%, as
shown in Figure 17-7(b). Therefore,
t = 0.70τ = 0.70 × 2.0 s = 1.4 s
17-4 Graphical Solution for Inductor Current
100
Current (% of Im )
Current (% of Im )
100
493
50
39
0 0.5τ
50
0
0.7τ
Time
(b)
Time
(a)
Figure 17-7 Using a universal exponential graph to solve Example 17-2
To verify the calculation of the current in part (a) and the time in part (b),
download Multisim file EX17-2 from the website and follow the
instructions in the file.
Example 17-3
In the circuit shown in Figure 17-8, switch S2 is closed 1.0 s after switch
S1 is closed. What is the current 1.0 s after switch S2 is closed?
S2
S1
4.0 Ω
1.0 Ω
+
10 V
−
2.0 Ω
Figure 17-8 Circuit diagram for Example 17-3
Solution
Step 1
While S1 is closed and S2 is open, the total resistance in the circuit is
The time constant is
RT = 4.0 Ω + 1.0 Ω = 5.0 Ω
τ=
L
2.0 H
=
= 0.40 s
R
5.0 Ω
circuitSIM
walkthrough
Chapter 17 Inductance in DC Circuits
1.0 s
= 2.5τ
0.40 s/T
Therefore,
From the curve of Figure 17-6, when t = 2.5τ,
i = 92% of
E
10 V
= 0.92 ×
= 1.84 A
R
5.0 Ω
There are two ways of dealing with the 1-s interval after S2 is closed.
Step 2
Consider the current curve for the circuit, which now has only 1.0 Ω of
­resistance. The steady-state current becomes
Im =
E
10 V
=
= 10 A
R
1.0 Ω
At the instant S2 is closed, i is already 1.84 A. Find the point on the
graph corresponding to this current and mark off an additional 1.0 s, as
shown in Figure 17-9(a).
1.84 A
1.84
=
= 18.4%
Im
10
From the curve of Figure 17-6, when i = 18.4% of Im,
t = 0.2τ = 0.2
L
2.0 H
= 0.2 ×
= 0.4 s
R
1.0 Ω
Therefore, on the curve in Figure 17-9(a), the current at t = 1.4 s is the
current 1.0 s after S2 is closed. In terms of time constants,
1.4 s =
1.4
τ = 0.70τ
2.0
Again, from Figure 17-6, when t = 0.7τ,
10
i
5
50%
1.84
0 0.2τ 0.7τ
1.0 s
Time
(a)
Additional Current (amperes)
i = 50% of Im = 0.50 × 10 A = 5.0 A
Additional Current (amperes)
494
10
8.16
i
5
3.18
100%
39%
0
0.5τ
1.0 s
Time
(b)
Figure 17-9 Graphical solution for Step 2 of Example 17-3
17-5 Algebraic Solution for Inductor Current
495
Alternative Step 2
Consider the additional rise in current after S2 is closed. At the moment
S2 is closed, a current of 1.84 A is flowing in the i­nductance. As calculated in Step 2a, Im is now 10 A. Therefore, the maximum additional rise
in current is 10 − 1.84 = 8.16 A.
The time constant is now
τ=
L
2.0 H
=
= 2.0 s
R
1.0 Ω
Therefore, 1.0 s now corresponds to 0.5τ. During this time, the ­additional
current rises to 39% of its maximum value, as shown in Figure 17-9(b).
Therefore, the change in current is
Δi = 39% of 8.16 A = 3.18 A
Therefore, the total instantaneous current 1.0 s after S2 is closed is
i = 1.84 A + 3.18 A = 5.0 A
To verify the calculation of the current, download Multisim file EX17-3
from the website and follow the instructions in the file.
circuitSIM
walkthrough
See Problems 17-7 to 17-9 and Review Question 17-35.
17-5 Algebraic Solution
for Inductor Current
The graphic solution for an LR circuit is accurate enough for most purposes.
If we need greater accuracy, we must return to Equation 17-2. Appendix 2-4
gives a calculus solution for this differential equation. The solution yields
an exponential equation for the current:
i=
E
(1 − e − x )
R
(17-6)
where e is the base of natural logarithms (2.718 . . .) and x is the elapsed
time measured in time constants (x = tR/L = t/τ).
Example 17-2A
(a) What is the instantaneous current in the circuit of Figure 17-3 1.0 s
after the switch is closed?
(b) How long does it take for the current to rise from 0 A to 5.0 A?
Most scientific calculators have an ex key
for calculating exponentials and a ln key
for calculating natural
logarithms.
496
Chapter 17 Inductance in DC Circuits
Solution
(a)
x=t×
i=
R
1.0 Ω
= 1.0 s ×
= 0.50
L
2.0 H
E
10 V
( 1 − e− x ) =
× ( 1 − e− 0.50 )
R
1.0 Ω
= 10 A × ( 1 − 0.6065 ) = 3.93 A
i=
(b)
5.0 A =
E
( 1 − e− x )
R
10 V
× ( 1 − e− x )
1.0 Ω
5 = 10 − 10e−x
e−x = 0.5
−x = ln 0.5 = −0.693
t=x×
L
2.0 H
= 0.693 ×
= 1.4 s
R
1.0 Ω
Example 17-3A
In the circuit shown in Figure 17-8, switch S2 is closed 1.0 s after switch
S1 is closed. What is the current 1.0 s after switch S2 is closed?
Solution
Step 1
At t = 1.0 s,
x=t×
i=
R
5.0 Ω
= 1.0 s ×
= 2.5
L
2.0 H
E
10 V
( 1 − e− x ) =
× ( 1 − e− 2.5 ) = 2.0 × ( 1 − 0.0821 ) = 1.84 A
R
5.0 Ω
Step 2
Using the method illustrated by Figure 17-9(a) gives
i=
1.84 =
E
( 1 − e−x )
R
10 V
× ( 1 − e−x )
1Ω
e − x = 0.816
17-5 Algebraic Solution for Inductor Current
x = 0.203 = t ×
R
L
−x = ln 0.816 = −0.203
2.0 H
Therefore,
t = 0.203 ×
= 0.406 s
1.0 Ω
One second later,
t = 1.406 s
1.0 Ω
and
x = 1.406 s ×
= 0.703
2.0 H
E
10 V
× ( 1 − e− 0.703 )
i = ( 1 − e−x ) =
R
1.0 Ω
= 10 A × ( 1 − 0.495 ) = 5.05 A
from which
Alternative Step 2
Using the method illustrated by Figure 17-9(b), we can show that the
maximum additional current is 10 − 1.84 = 8.16 A.
Once S2 is closed, the time constant becomes τ = 2.0 s.
For a time interval of 1.0 s,
x=t×
R
1.0 Ω
= 1.0 s ×
= 0.50
L
2.0 H
Δi = 8.16(1 − e−0.50 ) = 8.16(1 − 0.6065) = 3.21 A
The change in current during the first second after S2 is closed is
Therefore, the total instantaneous current 1 s after S2 is closed is
i = 1.84 A + 3.21 A = 5.05 A
The LR network of Figure 17-10(a) includes a 4.0-Ω discharge resistor in
parallel with a practical inductor. (Section 17-7 describes the function of
this resistor.) To determine the time constant for the buildup of current in
the inductor, we consider the 4.0-Ω resistor to be part of the source network
as “seen” by the inductor. We can then use the Thévenin-equivalent circuit
of Figure 17-10(b) to calculate the current through the inductor.
1.0 Ω
1.0 Ω
+
10 V
−
RTh
1.0 Ω
0.80 Ω
4.0 Ω
2.0 H
(a)
+
ETh 8 V
−
2.0 H
(b)
Figure 17-10 (a) Inductor circuit with discharge resistor; (b) Thévenin-equivalent circuit
497
498
Chapter 17 Inductance in DC Circuits
Example 17-4
For the circuit of Figure 17-10(a), calculate the current in the inductor
1.0 s after the switch is first closed.
Solution
Step 1
With the inductor disconnected from the rest of the circuit, the opencircuit voltage across the 4.0-Ω resistor is
ETh = 10 V ×
and
RTh =
4.0 Ω
= 8.0 V
4.0 Ω + 1.0 Ω
4.0 Ω × 1.0 Ω
= 0.80 Ω
4.0 Ω + 1.0 Ω
Hence, the time constant is
τ=
Step 2
and
circuitSIM
walkthrough
L
2.0 H
=
= 1.111
( 1.0 + 0.80 ) Ω
RT
x=
1.0 s
t
=
= 0.90
τ
1.111 s
i=
ETh
8.0 V
( 1 − e−x ) =
( 1 − e − 0.90 )
RT
1.8 Ω
= 4.444 A ( 1 − 0.4066 ) = 2.6 A
To verify the calculation of the current, download Multisim file
EX17-4 from the website and follow the instructions in the file.
See Problems 17-10 to 17-15.
Circuit Check CC 17-1.
A
Calculate the resistance in series with a 240-mH inductor, given
that the current reaches full value in 25 ms.
CC 17-2. An inductor with 20-Ω resistance is connected to an 80-V
source. After 15 ms, the current through the inductor is 1.8 A.
Calculate the inductance.
17-6 Energy Stored by an Inductor
17-6 Energy Stored by an Inductor
Current
If we connect an ideal inductor to a voltage source having no internal
­resistance, the voltage across the inductance must remain equal to the applied voltage. Therefore, the current rises at a constant rate, as shown in
Fig­ure 17-11(b). The source supplies electrical energy to the ideal inductor at the rate of p = Ei. Unlike resistance, inductance cannot convert this
energy into heat or light. Instead, the energy is stored in the magnetic field
as the rising current forces the magnetic lines of force to expand against
their tendency to become as short as possible—somewhat as a rubber band
stores energy when it is stretched.
+
E
Im
i
Iav
−
Time
(b)
(a)
Figure 17-11 Determining the energy stored by an inductor
In resistance circuits where the current and voltage do not change with
a change in time, the energy transferred from the source to the resistance
is W = Pt = VIt. Although the voltage remains constant in the circuit of
Figure 17-11(a), the current steadily increases as time elapses. However,
since the rate of change of current is constant, the average value of the current, I, as it rises from zero to Im is 1⁄2 Im. Therefore, the energy stored by an
inductor as the current rises from zero to Im is
1
W = V × Im × t
2
Since the rate of change of current is constant, Equation 17-1 becomes
V=L
di
LIm
=
dt
t
(17-1)
Substituting for V in the equation for W gives
W=L×
Im 1
× Im × t
t
2
1
W = LI2m
2
(17-7)
499
500
Chapter 17 Inductance in DC Circuits
With a practical inductor, both the instantaneous voltage and current are
changing and their rates of change are not constant. If we find the voltage
across and the current through the inductance for given moment, we can
use relationship p = vi to calculate the rate at which the inductance of the
circuit stores energy at that moment. With a series of such calculations we
can plot the graph in Figure 17-12.
vL
iL
Area = WL
pL
0
1
2
3
4
Time (time constants)
5
Figure 17-12 Energy stored by a practical inductor
When the current in a practical inductor reaches its steady-state value
of Im = E/R, the magnetic field ceases to expand. The voltage across the
­inductance has dropped to zero, so the power p = vi is also zero. Thus, the
energy stored by the inductor increases only while the current is building
up to its steady-state value. When the current remains constant, the energy
stored in the magnetic field is also constant. Although no additional energy
is stored by the inductance of the practical inductor, the resistance of the
­inductor dissipates energy at a steady rate of P = I2mR. Current must continue to flow to maintain the magnetic field.
Thus, an inductor stores energy in a dynamic form, somewhat like the
­kinetic energy stored by the flywheel of an engine. By contrast, no current is required to maintain the energy stored by a capacitor (except to
offset leakage through the dielectric). The electric field of a capacitor
stores ­energy in a static form, somewhat like the energy stored in a tank
of compressed air.
Note the similarity between Figure 17-12 and Figure 13-15. The significant difference is that the voltage curve for one graph becomes the current curve for the other. Since p = vi, the power curve is the same in both
figures. The rate of storage of energy, p, reaches its peak value at the same
instant that iL rises to 1⁄2Im and vL drops to 1⁄2 E.
The area under the power curve represents the energy stored by the
inductance and is equal to the product of the average power and the elapsed
time. Appendix 2-5 uses basic integral calculus to show that energy stored
in the magnetic field of an inductor is
17-7 Fall of Current in an Inductive Circuit
1
w = Li2
2
(17-8)
where w is the stored energy in joules, L is the inductance in henrys, and
i is the current in amperes.
Example 17-5
Find the maximum energy stored by an inductor with an inductance
of 5.0 H and a resistance of 2.0 Ω when the inductor is connected to a
24-V source.
Solution
Im =
W=
E
24 V
=
= 12 A
R
2.0 Ω
1 2
1
LIm = × 5.0 H × (12 A)2 = 360 J
2
2
See Problems 17-16 and 17-17 and Review Question 17-36.
17-7 Fall of Current
in an Inductive Circuit
When the current through an inductor stops, the magnetic field collapses,
releasing the stored energy. Since the inductance opposes any change in
current, the current cannot cease instantly. Any change in the current takes
time, even if only a few microseconds.
At the instant we open the switch in Figure 17-13, the current in the
­inductor is the same as just before the switch was opened. The steady-state
current before the switch is opened is
I=
500 kΩ
+
12 V
−
E
12 V
=
= 6.0 A
R
2.0 Ω
2.0 Ω
4.0 H
Figure 17-13 Interrupting current in an inductor
501
502
Chapter 17 Inductance in DC Circuits
At the instant the switch opens, the current in the inductor is still 6.0 A.
To demonstrate the effect of interrupting this current, we assume that the
leakage resistance of the open switch is 500 kΩ. Since this leakage resistance is
in series with the source and the inductor, the 6.0-A current flows through
the leakage resistance, and the voltage drop between the switch contacts is
V = IR = 6.0 A × 500 kΩ = 3.0 MV
This extremely high voltage drop is matched by the voltage generated in
the turns of the inductor by the rapidly collapsing magnetic field. Such a
high voltage would cause arcing between the switch contacts and would
probably damage the insulation of the inductor. Therefore, we must design
DC circuits to prevent sudden interruptions of current in an inductor.
Example 17-6
Find the voltage across the switch contacts in Figure 17-14 at the instant
that the switch is opened.
1.0 A
6.0 A
2.0 Ω
+
12 V
−
+
12 Ω
−
4.0 H
+
12 V
−
(a)
−
6.0 A
+
(b)
Figure 17-14 Circuit diagrams for Example 17-6
Solution
With the switch closed, the steady-state current through the inductor is
I=
12 V
E
=
= 6.0 A
R
2.0 Ω
At the same time, a 1.0-A current passes through the 12-Ω resistor, as
shown in Figure 17-14(a). This current does not flow through the inductance of the circuit. When the switch is opened, the inductor momentarily acts as a generator as its magnetic field collapses. The load across
this generator is 12 Ω and 2.0 Ω in series. Since the current through the
inductor cannot change instantly, a current of 6.0 A flows through the
12-Ω resistor, as shown in Figure 17-14(b). The voltage drop across
the 12-Ω resistor is then
v = iR = 6.0 A × 12 Ω = 72 V
17-7 Fall of Current in an Inductive Circuit
503
The 72-V drop across the 12-Ω resistor and the 12-V potential difference of the battery are connected in series aiding across the switch.
Therefore, the total voltage across the switch contacts at the instant we
open the switch is
v = 72 V + 12 V = 84 V
To verify the calculation of the voltage, download Multisim file EX17-6
from the website and follow the detailed instructions in the file.
According to Kirchhoff’s voltage law, the initial induced voltage across
the inductor must equal the voltage drop across the total resistance:
Em = I0RT
(17-9)
where Em is the peak voltage induced by the collapsing magnetic field,
I0 is the current through the inductor at the instant the switch is opened,
and RT is the total resistance of the loop containing the inductor.
In Example 17-6,
Em = 6.0 A × 14 Ω = 84 V
According to Lenz’s law, the induced voltage always opposes any
change in current. When the current rises, the induced voltage cancels
some of the applied voltage. But when the source is removed, the magnetic
field around the inductor collapses. The polarity of the voltage induced by
a collapsing field is opposite to that of the voltage generated by a rising
field. This reversed polarity tends to maintain current flow in the original
direction through the inductor.
The 12-Ω resistor in the circuit of Figure 17-14 acts as a discharge resistor. As
the Example 17-6 shows, this discharge resistor protects both the switch and
the inductor from a high-voltage surge. Some such provision must be made
in all direct-current circuits with appreciable inductance, such as DC motors
and generators. Silicon carbide varistors are excellent for this purpose.
The discharge resistor in Example 17-6 reduces the induced voltage, so
the magnetic field must be collapsing at a much slower rate and, therefore,
it takes longer to discharge the stored energy. Since the magnetic field is
collapsing and the magnetic field strength depends on the current in the
inductor, the current starts to decrease as soon as the switch is opened. We
can define a discharge time constant, τ, as the time the current takes to drop
to zero if it continues to fall at its initial rate of change. If the current does
drop at a constant rate equal to its initial rate (as shown by the dashed line
in Figure 17-15), then
initial
di
I0
=
τ
dt
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Chapter 17 Inductance in DC Circuits
Switch opened
100
80
Current (% of I0)
504
60
i
40
20
0
37% of I0
0
1
2
3
Time (time constants)
4
5
Figure 17-15 Fall of current in an inductive DC circuit
Since initial di/dt = Em/L (Equation 17-4) and Em = I0RT (Equation 17-9),
τ=
L
RT
(17-5)
Therefore, we determine the time constant of an LR circuit in exactly the
same manner for rise and fall of current.
The current through an inductor does not actually fall at a constant
rate. Since the current is decreasing, the voltage drop across the resistance
must also decrease. Kirchhoff’s voltage law requires the induced voltage to
equal the voltage drop across the resistance, so the induced voltage must
also decrease. As the induced voltage decreases, the magnetic field must
collapse more slowly. Thus, the rate of change of current becomes progressively smaller. The graph of falling current in Figure 17-15 has the same
­exponential shape (but inverted) as the curve for rising current.
Example 17-7
The switch in the circuit of Figure 17-14 is closed for 3.0 s and then
opened. Find the inductor current 0.25 s after the switch is opened.
Solution
Step 1
With the switch closed,
Charge τ =
L
4.0 H
=
= 2.0 s
R
2.0 Ω
17-7 Fall of Current in an Inductive Circuit
505
3.0 s
= 1.5τ
2.0 s/T
Therefore,
From the graph for current rise in Figure 17-6, iL = 78% of Im when
t = 1.5τ.
E
12 V
= 4.7 A
Hence,
iL = 0.78 × = 0.78 ×
R
2.0 Ω
Step 2
With the switch open, the time constant becomes
Discharge τ =
L
4.0 H
= 0.286 s
=
RT
12 Ω + 2.0 Ω
Expressing 0.25 s in time constants,
0.25 s
= 0.875τ
0.286 s/T
From the graph for current decay in Figure 17-6, iL = 42% of I0 when
t = 0.88τ.
iL = 0.42 × 4.7 A = 2.0 A
Hence,
To verify the calculation of the current, download Multisim file EX17-7
from the website and follow the instructions in the file.
Figure 17-16 shows another way to prevent sparking at switch contacts
in inductive circuits. The potential difference across the capacitor is zero
while the switch is closed. Since the capacitor voltage cannot rise instantly
when the switch opens, the switch contacts have time to move far enough
apart to prevent arcing. Some of the energy released by the collapsing
magnetic field of the inductor transfers to the capacitor as it charges. The
capacitor then partially discharges, building up a smaller magnetic field
around the inductor opposite to the original magnetic field. Several cycles
of energy transfers take place until the stored energy is all dissipated in the
C
+
E
Inductive
load
−
Figure 17-16 Suppressing a voltage surge with a capacitor
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Chapter 17 Inductance in DC Circuits
resistance of the load. Chapter 25 discusses the transient response of DC
circuits containing both inductance and capacitance in more detail.
See Problems 17-18 to 17-20 and Review Questions 17-37 to 17-41.
17-8 Algebraic Solution
for Discharge Current
When we write the Kirchhoff’s voltage-law equation for the circuit of
Figure 17-14(b), the applied voltage in the circuit is zero. Therefore,
0 = iRT + L
di
dt
As shown in Appendix 2-4, using calculus to solve for the instantaneous
current gives
i = I0e − x
(17-10)
where I0 is the initial current, e = 2.718 . . ., and x = tRT/L = t/τ.
Example 17-7A
The switch in the circuit of Figure 17-14 is closed for 3.0 s and then
opened. Find the inductor current 0.25 s after the switch is opened.
Solution
Step 1
With the switch closed,
x=t×
R
2.0 Ω
= 3.0 s ×
= 1.5τ
L
4.0 H
Therefore,
E
12 V
iL = ( 1 − e − x ) =
× ( 1 − e−1.5 ) = 6.0 A × ( 1 − 0.223 ) = 4.66 A
R
2.0 Ω
Step 2
When the switch has been open for 0.25 s,
and
x=t×
RT
14 Ω
= 0.25 s ×
= 0.875 time constant
L
4.0 H
iL = I0 × e − x = 4.66 A × e − 0.875 = 4.66 × 0.417 = 1.9 A
See Problems 17-21 and 17-22.
17-9 Transient Response
Circuit Check
B
CC 17-3. An inductor with an inductance of 40 H and a resistance of 2.5 kΩ
is connected across a 120-V DC power supply that has negligible
internal resistance. Calculate the maximum energy stored in
the magnetic field of the inductor.
CC 17-4. The switch in Figure 17-17 is set to position 1 for 1.0 s and then
switched to position 2. Calculate:
(a) the charge and discharge time constants
(b) the current flowing after 1.5 s
(c) the time the voltage across the 1.25 Ω resistor takes to fall to
2.5 V on discharge
1.2 H
0.80 H
1
2
+
10 V
−
0.50 Ω
1.25 Ω
Figure 17-17
17-9 Transient Response
As we have seen throughout this chapter, the current in an LR network
does not instantly change to a new steady state when we suddenly change
the source voltage. The interval while the current reaches the final steadystate current is the transient response of the LR network.
So far, we have used one exponential equation and curve for rising
­inductor currents and a different equation and curve for falling inductor
currents. However, we can derive a single equation for any LR transient as
long as we are careful in specifying the initial steady-state current (I0) and
the final steady-state current (IF). For example, in Step 2b of Example 17-3A,
I0 is 1.84 A (at t = 1.0 s) and IF is 10 A. The instantaneous ­current during the
transient period is equal to the initial current plus an i­ ncreasing percentage
of the additional current. In general,
iL = I0 + ( IF − I0 ) ( 1 − e − x )
(17-11)
507
Chapter 17 Inductance in DC Circuits
Simplifying gives the universal equation for inductor current in a DC LR
­network:
iL = I0 + IF − I0 − IFe−x + I0e−x
iL = IF + ( I0 − IF ) e − x
(17-12)
where iL is the instantaneous inductor current, I0 is the initial inductor
­current, IF is the final steady-state inductor current (after at least five
time constants), and x is the elapsed time measured in time constants.
Equation 17-11 expresses instantaneous current as initial current plus
a proportion of the difference between the initial and final currents.
­Equa­tion 17-12 expresses the same instantaneous current in terms of the
final steady-state current and an exponential transient caused by the change
in steady-state current.
Example 17-8
In the circuit of Figure 17-18(a), the switch is closed at t = 0 and opened
again at t = 1.0 s.
(a) Find the inductor current at t = 1.0 s.
(b) Find the voltage across the switch contacts at t = 1.1 s.
(c) Draw a graph of the voltage across the switch from t = 0 to t = 5 s.
2.0 H
1.0 Ω
+
10.0 V
−
4.0 Ω
Switch Voltage, vsw (volts)
508
20.6 V
20
16
12
8
4
0
1
2
3
4
Time, t (seconds)
(a)
(b)
Figure 17-18 Transient response in a DC LR network
Solution
With the switch open, the steady-state current is
IF =
and
τ=
10.0 V
= 2.0 A
4.0 Ω + 1.0 Ω
L
2.0 H
=
= 0.40 s
RT
5.0 Ω
5
17-10 Characteristics of Inductive DC Circuits
509
With the switch closed, the steady-state current is
IF =
and
τ=
10.0 V
= 10 A
1.0 Ω
2.0 H
= 2.0 s
1.0 Ω
(a) When the switch closes at t = 0, the initial current through the coil
is 2.0 A. At t = 1.0 s,
i = IF + (I0 − IF)e−t/τ
= 10 + (2.0 − 10)e−1.0/2.0
= 10 − 8 × 0.6065 = 5.15 A
(b) At t = 1.1 s, the switch has been open for 0.1 s and
i = IF + (I0 − IF)e−t/τ
= 2.0 + (5.15 − 2.0)e−0.1/0.40
= 2.0 + 3.15 × 0.7788 = 4.453 A
Since the switch is in parallel with the 4.0-Ω resistor,
vsw = 4.453 A × 4.0 Ω = 18 V
(c) While the switch is open, the 2.0-A steady-state current through
the 4.0-Ω resistor creates a steady voltage drop of 8.0 V. When the
switch closes, the voltage across it instantly becomes zero. When
the switch opens at t = 1.0 s, the 5.15-A current creates a voltage
drop of 5.15 A × 4.0 Ω = 20.6 V across the 4.0-Ω resistor, creating
the transient shown in Figure 17-18(b). Since 5τ = 2.0 s when the
switch is open, the transient will have disappeared when t = 3 s.
To verify the calculation of the current in part (a) and the voltage in
part (b), download Multisim file EX17-8 from the website and follow
the ­instructions in the file.
See Problems 17-23 to 17-26 and Review Question 17-42.
17-10 Characteristics
of Inductive DC Circuits
The most important characteristic of inductance in a DC circuit is that the
inductance develops an induced voltage that always tends to oppose any
change in current. As a result, the current in an inductive circuit takes time
to either rise or fall. The length of time needed for a given change in current
depends on both the inductance and the resistance of the circuit.
In order to show the
results more clearly,
some of the answers
in this example are
given with more
­significant digits
than are justified by
the data.
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Chapter 17 Inductance in DC Circuits
When we close the switch in an inductive circuit, it takes time for the
current to reach its steady-state level. This delay can be an advantage in
that it reduces the surge in the current drawn from the source. However,
an interruption in the current in an inductive circuit can induce a damaging high-voltage surge unless the circuit is designed so that current decays
gradually. For this reason, switches for DC circuits are designed so that
their contacts move apart as rapidly as possible to prevent arcing.
There are applications for the high voltage produced by the collapsing magnetic field of an inductor. The ignition coil of a car stores energy
gradually as current from the car battery builds up a magnetic field
in the core of the coil. Then the current is suddenly interrupted by the
breaker points, and the rapidly collapsing magnetic field induces a 10-kV
­voltage, which produces a spark between the electrodes at the end of a
spark plug.
Inductance can also be used to reduce, or filter, the variations in load
current that occur when the load is fed from a DC source with a pulsating
terminal voltage, such as a rectifier. As the source voltage in Figure 17-19
rises above its average value, the inductor stores additional energy in its
expanding magnetic field, which opposes the increase in current. When
the source voltage drops, the inductor opposes the corresponding drop in
current and releases the stored energy. The inductor in such applications is
often called a filter choke because it restricts the variations in load current
and power.
Filter choke
Pulsating
DC source
+
e
Load
−
Figure 17-19 Filter choke
17-11 Troubleshooting
Three faults that commonly occur in inductors are
(i) open windings
(ii) shorted windings
(iii) partially shorted windings
Often these faults are the result of overheating, which can be caused by
poor ventilation, a short within the inductor, or a fault in the power supply.
It is good practice to check the power supply voltage when one replaces
an inductor that has overheated.
17-11 Troubleshooting
Open Windings
The heat from excessive current through a winding can cause the wire to
burn out, or a break can occur where the wire had been nicked or scraped.
An ohmmeter connected across the terminals of an open winding will
show infinite resistance. Note that the inductor usually has to be disconected from the circuit in order to measure the resistance accurately.
Shorted Windings
Occasionally the insulation on the winding may melt because of overheating, causing the turns to short to each other. The inductor then has a resistance close to zero ohms. Again, an ohmmeter can detect this fault once the
inductor is removed from the circuit. The resistance of the inductor should
be measured on the lowest scale since the normal resistance is often only a
few ohms.
Partially Shorted Windings
Sometimes overheating or physical damage causes just a few adjacent turns
to short together. Such partial shorts reduce the resistance of the ­inductor
somewhat, but the difference may be difficult to detect with an ohmmeter.
A more accurate test can be done with an LC meter, which measures the
inductance and capacitance of a component directly.
See Review Question 17-43.
511
512
Chapter 17 Inductance in DC Circuits
Summary
• The rate of change of current through an ideal inductor is constant.
• The initial rate of change of current in a series DC circuit containing
­resistance and inductance is determined by the ratio of the applied voltage to the inductance.
• The final value of current in a series DC circuit containing resistance and inductance is determined by the ratio of the applied voltage to the resistance.
• The time constant of a circuit containing resistance and inductance is the
ratio of inductance to the equivalent resistance.
• For practical purposes, the current in a DC circuit containing resistance
and inductance reaches its steady-state value after a time interval of five
time constants.
• Instantaneous currents and voltages in a DC circuit containing resistance
and inductance may be determined by means of either an exponential
graph or the corresponding equation.
• The energy stored in an inductor is dependent on the inductance and the
current flowing through it.
• Inductor faults may be detected with an ohmmeter or an LC meter.
B = beginner
I = intermediate
A = advanced
Problems
B
B
B
B
B
I
Section 17-3
Time Constant
17-1. If a circuit has a resistance of 8.0 Ω and an inductance of 2 H, how
long will the current take to reach its maximum value once the circuit is switched on?
17-2. In a solenoid with an inductance of 40 mH, it takes 25 ms for the current to reach its maximum value. Find the resistance of the solenoid.
17-3. What voltage must be applied to a 600-μH inductor if the initial rate
of change of current must be 8.0 kA/s?
17-4. Calculate the steady-state current for a 100-mH inductor with a resistance of 8.0 Ω, given that the initial rate of rise of current is 600 A/s.
17-5. A choke with an inductance of 15 H and a resistance of 8 Ω is connected through a switch to a 36-V source. Find
(a) the initial current when the switch is closed
(b) initial rate of change of current when the switch is closed
(c) the steady-state current
(d) how long the current takes to reach its maximum value
17-6. An inductor with an inductance of 20 H and a resistance of 500 Ω
is connected in parallel with a 1000-Ω resistor. This combination is
connected across a 120-V DC source with negligible internal resistance.
(a) Calculate the initial current drawn from the source when the
switch is closed.
(b) Calculate the initial rate of change of the current through the
­inductor.
(c) Calculate the steady-state current in the inductor.
(d) Calculate how long the current takes to reach its steady-state value.
Problems
(b) Use Multisim to verify the initial source current in part (a), the
steady-state inductor current in part (b), and the time in part (d).
A
I
I
I
I
I
I
I
Section 17-4
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Graphical Solution for Inductor Current
17-7. (a)In the circuit of Problem 17-1, how long does the current take to
reach 63% of its maximum value after the switch is closed?
(b) How long does the current take to reach 50% of its maximum value?
(c) Find the maximum value of the current if it is 2.6 A when the
switch has been closed for 0.5 s.
17-8. (a) When will the current in the inductor of Problem 17-4 be 6 A?
(b) Find the current through the inductor 10 ms after the circuit is
switched on.
17-9. A 50-mH inductor with a resistance of 8 Ω is connected to a 12-V
source. How long will the inductor current take to reach 500 mA?
Section 17-5
Algebraic Solution for Inductor Current
17-10. (a) When will the current through the choke in Problem 17-5 be 3 A?
(b) Find the choke current 4 s after the switch is closed.
(c) Use Multisim to verify your calculations in part (a) and part (b).
17-11. A 20-Ω resistor is connected in series with the choke of Problem
17-5. This resistor is short-circuited 2.0 s after the switch is closed.
Find the induced voltage across the choke 2.0 s after the 20-Ω resistor is short-circuited.
17-12. (a) A coil with a resistance of 20 Ω is connected to a 60-V DC source.
After 25 ms the current is 1.2 A. Determine the inductance of the coil.
(b) Use Multisim to verify the inductance calculated in part (a).
17-13. One type of time-delay relay is designed to operate 75 ms after the
­circuit is energized, with this delay equaling the time constant of the
circuit. The supply voltage is 24 V and the relay trips when the current is 250 mA. Calculate the resistance and inductance of the relay.
17-14. For the circuit shown in Figure 17-20, find
(a) an equation for the current as a function of time
(b) the current flowing after 0.25 s
(c) the time it takes for current to reach 3 A
(d) the induced voltage after 0.25 s
t=0s
10 Ω
+
100 V
−
I
513
1.0 H
Figure 17-20
17-15. Sketch a graph of current versus time for the circuit in Figure 17-20.
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Chapter 17 Inductance in DC Circuits
B
B
I
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A
A
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Section 17-6
Energy Stored by an Inductor
17-16. (a) H
ow much energy is stored by the choke in Problem 17-5 when
the current reaches its steady-state value?
(b)At what rate is energy being drawn from the source when the
current through the choke reaches its maximum value?
17-17. (a)How much energy is stored in the magnetic field of the inductor in
Problem 17-6 when the current has reached its steady-state value?
(b) At what rate is electrical energy being converted into heat when
the current in the circuit has reached its maximum value?
Section 17-7
Fall of Current in an Inductive Circuit
17-18. The field coils of a DC motor connected across a 120-V source have
an inductance of 8.0 H and a resistance of 40 Ω.
(a) What is the maximum value for a discharge resistor that can be
connected across the field coils to prevent the voltage ­between
the coils from exceeding 200 V?
(b) How long will this maximum resistance take to discharge the
energy stored in the field coils?
(c) How much energy is dissipated by the discharge resistor after
the source is disconnected?
(d) Use Multisim to verify the resistance calculated in part (a).
17-19. (a)Find the voltage across the switch contacts when the switch in
the circuit of Problem 17-6 is opened.
(b) How long does it take the coil to discharge its stored energy?
(c) If the switch is opened 1.0 s after it was first closed, how long
does the current in the coil take to drop 25 mA?
17-20. In the filter network of Figure 17-21, the 80-V source and the 40-mH
inductor both have negligible internal resistance.
(a) Find the output voltage 5 μs after the switch closes.
(b) If the switch remains closed for at least five time constants, how
long after the switch opens will the output voltage become 50 V?
(c) Use Multisim to verify the voltage resistance calculated in part
(a) and the time calculated in part (b).
40 mH
+
80 V
−
5.0 kΩ
5.0 kΩ
Figure 17-21
A
Section 17-8
Algebraic Solution for Discharge Current
17-21. The inductor in Figure 17-22 has negligible resistance.
(a) Find the output voltage 5 μs after the switch closes.
Problems
515
(b) If the switch remains closed for at least five time constants,
how long after the switch opens will the output voltage become
+50 V with respect to ground?
5.0 kΩ
5.0 kΩ
40 mH
−
80 V
+
Figure 17-22
A
17-22. After switch S1 in Figure 17-23 has been closed for 1.0 s, it is opened
and S2 is closed simultaneously.
(a) Calculate the time constant.
(b) Write an equation for the current as a function of time.
(c) Calculate the current flowing 30 ms after S1 opens.
(d) Calculate the time the current takes to reach 3 A.
(e) Calculate the induced voltage 30 ms after S1 opens.
(f) Calculate the total energy stored in the magnetic field.
(g) Sketch and label a graph of the current as a function of time.
(h) Use Multisim to verify the calculation of the current in part (c),
the time in part (d), and the induced voltage in part (e).
walkthrough
0.50 H
S1
+
100 V
−
circuitSIM
S2
20 Ω
Figure 17-23
A
A
A
Section 17-9
Transient Response
17-23. (a)The switch in Figure 17-21 opens 10 μs after it closes. Find the output voltage with respect to ground 10 μs after the switch opens.
(b) If the switch stays for 50 μs in each position alternately, draw
a labelled graph of two complete cycles of the output voltage
waveform.
17-24. (a) Repeat Problem 17-23 for the circuit of Figure 17-22.
(b) Use Multisim to verify the output voltage calculated in part (a).
17-25. The switch in Figure 17-24 is closed for 2.0 s and then opened. Find
the inductor current 1.0 s after the switch is opened.
circuitSIM
walkthrough
516
Chapter 17 Inductance in DC Circuits
1.0 Ω
5.0 H
Discharge
resistor
+
20 V
−
4.0 Ω
1.2 Ω
Figure 17-24
circuitSIM
walkthrough
A
17-26. (a) Solve Problem 17-25 by replacing the source with an equivalent
­constant-current source.
(b) Use Multisim to verify the inductor current calculated in part (a).
Review Questions
Section 17-1
Current in an Ideal Inductor
17-27. Why is the graph in Figure 17-2 a straight line?
17-28. What is the significance of lowercase letter symbols for variables?
17-29. Why are lowercase letter symbols used when one is representing
the rise of current in inductive circuits?
Section 17-2
Rise of Current in a Practical Inductor
17-30. Why must the rise of current follow the exponential curve of
­Fig­ure 17-4?
Section 17-3
Time Constant
17-31. What effect does the resistance of an inductor have on the following?
(a) the initial rate of change of current
(b) the final steady-state current
(c) the time the current takes to reach its steady-state value
17-32. What effect does the inductance of an inductor have on the following?
(a) the initial rate of change of current
(b) the final steady-state current
(c) the time the current takes to reach its steady-state value
17-33. Justify this description of a time constant: “A time constant is the
time it takes the current in an inductive circuit to rise to about 63%
of its final steady-state value.”
17-34. Why does the voltage applied to an inductive circuit have no effect
on the time the current takes to reach a steady-state value?
Section 17-4
Graphical Solution for Inductor Current
17-35. Draw a detailed graph of the actual inductor current in Example 17-3.
Section 17-6
Energy Stored by an Inductor
17-36. Why is there no further increase in the energy stored in an ­inductor when
the current has reached its steady-state value? Where is the e­ nergy from
the source going when a steady current flows through an inductor?
Integrate the Concepts
Section 17-7
Fall of Current in an Inductive Circuit.
17-37. What is the effect of suddenly disconnecting the voltage source in
an inductive DC circuit?
17-38. How does a discharge resistor prevent insulation breakdown in an
inductor?
17-39. In the instant after the circuit is switched off, why is the current
through an inductor the same as the current the instant before the
switch is opened?
17-40. Why are silicon carbide varistors particularly suitable for use as
­discharge resistors in LR circuits?
17-41. What effect does the resistance of the discharge loop of an inductive
circuit have on the following?
(a) the initial current when the switch is opened
(b) the initial rate of fall of current
(c) the time the current takes to fall to zero
Section 17-9
Transient Response
17-42. Derive Equation 17-12 by considering what happens in the circuit
of Figure 17-8 when S2 is opened but S1 remains closed.
Section 17-11
Troubleshooting
17-43. Outline procedures for detecting three common faults in inductors.
Integrate the Concepts
In the inductive DC circuit shown in Figure 17-25, switch S1 is closed at time
t = 0 while switch S2 stays open. After the inductor current reaches a steady
state, S1 is opened at the same time as S2 is closed. Calculate:
(a) the initial rate of change of current
(b) the steady-state current before S2 is closed
(c) the initial time constant
(d) the current after 10 μs
(e) the maximum energy stored in the inductor
(f) the discharge time constant
(g) the current 0.5τ after S2 is closed
S1
4.0 kΩ
2.0 kΩ
S2
+
12 V
−
Figure 17-25
30 mH
517
518
Chapter 17 Inductance in DC Circuits
Practice Quiz
1.
The resistance of an ideal inductor is
(a) zero
(b) low
(c) high
(d) infinite
Complete the statements in questions 2–5.
2.
The current in an inductor is _____________________ proportional to
the voltage across the inductor.
3.
At room temperature a practical inductor has a resistance that ­appears
in _____________________ with the inductor.
4.
The initial rate of change of current in an ideal inductor is
_____________________ the initial rate of change of current in a practical inductor.
5.
The time constant for an LR circuit is _____________________ proportional to inductance.
6.
How long will the current take to reach 50% of the maximum value
when the switch is closed in Figure 17-26?
(a) 69.0 ns
(b) 69.3 ns
(c) 70.0 ns
(d) 70.2 ns
R
10.0 kΩ
12.0 V
L
1.00 mH
Figure 17-26
7.
What will the current be 110 ns after the switch in Figure 17-26 has
been closed?
(a) 1.0 mA
(b) 1.2 mA
(c) 0.95 mA
(d) 0.8 mA
Practice Quiz
8.
How long will the current take to reach the maximum value when the
switch is closed in Figure 17-27?
(a) 6.9 ms
(b) 69 ms
(c) 34 ms
(d) 3.4 ms
R
1.0 kΩ
L1
22 H
24 V
L2
10 H
Figure 17-27
9.
What will the current in the circuit of Figure 17-27 be 7 ms after the
switch has been closed?
(a) 1.8 mA
(b) 15 mA
(c) 1.5 mA
(d) 18 mA
10.
The energy stored by an inductor is
(a) directly proportional to the resistance
(b) inversely proportional to the inductance
(c) directly proportional to the square of the maximum current
flowing through it
(d) inversely proportional to the square of the maximum current
flowing through it
11.
When 15 A of current flows through an 8.0-H inductor, the energy
stored in it is
(a) 60 J
(b) 900 J
(c) 1.8 kJ
(d) 120 J
12.
Which of the following equations describes the current flowing
through an inductor during discharge?
(a) iL =
(b) iL =
E
E tR
+ e− L
R
R
E
E tR
− e− L
R
R
519
520
Chapter 17 Inductance in DC Circuits
(c) iL =
(d) iL =
13.
E tR
eL
R
E −tR
e L
R
The switch in Figure 17-28 has been closed for 15 ms and then opened.
The current through the resistor, R1, 2.0 ms after the switch has been
opened is
(a) 12 mA
(b) 24 mA
(c) 16 mA
(d) 11 mA
R1
1.0 kΩ
24 V
R2
1.0 kΩ
L1
22 H
L2
10 H
Figure 17-28
14.
List three types of faults that can occur in an inductor.
PART
IV
Alternating Current
In Part IV and Part V, we encounter voltage sources with EMFs that change continuously. Part IV describes alternating-current circuits and introduces mathematical techniques for analyzing the properties and behaviour of these circuits.
18
Alternating Current
19
Reactance
20
Phasors
21
Impedance
22
Power in Alternating-Current Circuits
Photo source: © iStock.com/omada
18
Alternating Current
In this chapter, we first examine how an alternating voltage is
generated. We then investigate the characteristics of the sine
wave, the most common form of alternating voltage.
Chapter Outline
18-1
A Simple Generator
18-3
The Sine Wave
18-5
Instantaneous Value of a Sine Wave
18-7
Instantaneous Current in a Resistor
18-9
Periodic Waves
18-2
524
The Nature of the Induced Voltage
526
18-4 Peak Value of a Sine Wave
18-6
18-8
The Radian
529
532
524
529
533
Instantaneous Power in a Resistor
536
18-10 Average Value of a Periodic Wave
540
537
18-11 RMS Value of a Sine Wave
540
Key Terms
slip ring 524
brush 524
alternating voltage 525
alternator 525
cycle 526
period 526
phasor 526
phase angle 526
angular velocity 526
sine wave 528
amplitude 529
peak value 529
frequency 529
hertz 529
radian 532
periodic wave 537
peak-to-peak value
equivalent
DC value 541
effective value 541
root-mean-square
(RMS) value 541
form factor 543
539
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• explain how an alternating voltage is
produced across a loop of wire rotating within
a magnetic field
• graph a sine-wave voltage
• calculate the instantaneous value of a sinewave voltage
• convert frequency to angular velocity
• calculate the instantaneous current through
a resistor in a simple AC circuit
Photo sources: © Lyroky/Alamy; Stock Photo
• calculate the instantaneous power in a
resistor in an AC circuit
• graph different types of periodic waves
• calculate the average value of a half-cycle of
a sine wave
• calculate the RMS value of a sine wave
524
Chapter 18 Alternating Current
18-1 A Simple Generator
Although batteries are convenient voltage sources for portable devices,
generators produce almost all of the electric power used in homes and
­industry. These generators use electromagnetic induction to produce a
­potential difference. Chapter 16 described Faraday’s discovery that a voltage is induced in a conductor that moves perpendicular to a magnetic field.
Figure 18-1 shows a simple arrangement that allows a loop of conductor to
rotate continuously within a stationary magnetic field. Electrical connection
to the rotating loop is maintained through a pair of slip rings and brushes.
Direction
of rotation
N
Conductor formed
into a loop
Brush
Uniform
magnetic
field
Load
S
Slip ring
Figure 18-1
Simple AC generator
As the loop rotates, the two sides of the loop always cut across the magnetic lines of force in opposite directions. As a result, the two induced voltages are in series aiding and the sum of the two voltages appears across
the brushes. If we connect a load resistor to the brushes of this simple
generator, the induced voltage will cause current to flow in the closed circuit c­ onsisting of the loop and the load. This current produces a magnetic
field, which, according to Lenz’s law, opposes the rotation of the loop. To
keep the loop rotating, this magnetic force has to be offset by a mechanical force. The mechanism that produces the mechanical force (a turbine,
for ­example) transfers energy to the loop. Thus the generator converts
­mechanical energy into the electrical energy that flows to the load. When a
conductor cuts across magnetic flux at the rate of 1 Wb/s, an EMF of 1 V is
induced in the conductor.
See Review Questions 18-32 to 18-34 at the end of the chapter.
18-2 The Nature
of the Induced Voltage
As the conductor loop in Figure 18-1 rotates through a complete revolution
(360°), each side of the loop first cuts across the magnetic lines of force in
one direction and then moves back across the same magnetic lines of force
in the opposite direction. Consequently, the induced voltage appearing at
18-2 The Nature of the Induced Voltage
the brushes reverses its polarity with each half-revolution, and the generator develops an alternating voltage. A generator that produces alternating current is called an alternator.
Figure 18-2(a) is a cross-sectional view showing 12 positions along the
path of one side of the conductor in the generator of Figure 18-1. It is customary to consider the normal or positive direction of rotation as being
counterclockwise, starting at the three o’clock position. In other words, the
angle of rotation is measured counterclockwise from the positive x-axis of
a grid with its origin at the centre of rotation. The 12 positions are spaced
evenly at 30° intervals.
N
5
4 3 2
+
0
7
11
8 9 10
EMF
6
1
0
1 2 3 4 5 6 7 8 9 10 11
S
−
(a)
Figure 18-2
Position
(b)
Nature of induced voltage
Faraday’s law states that the voltage induced in a rotating conductor is
directly proportional to the rate at which it cuts across the magnetic lines of
force. Thus, the induced voltage depends on the motion of the conductor
in relation to the direction of the magnetic field. At position 0, the motion
of the conductor is parallel to the magnetic lines of force. At that moment,
the conductor does not cut any lines of force. Hence, the induced voltage is
0 V at the instant the conductor passes through point 0. At position 1, the
conductor has started to cut across the lines of force, and some voltage is
induced into the conductor. At position 2, the conductor is cutting across
the magnetic lines of force at an even steeper angle. Thus, the rate of cutting
and, therefore, the voltage induced into the conductor increases further.
At position 3, the conductor is cutting across the magnetic lines of force at
right angles, giving the greatest possible rate of cutting. Therefore, maximum voltage is induced into the conductor as it passes through point 3.
From positions 3 to 6 in Figure 18-2, the rate of cutting across magnetic
lines of force becomes progressively less, with the induced voltage reaching 0 V again just as the conductor passes through point 6, where it is again
­momentarily moving parallel to the lines of force. From positions 6 to 12,
the motion of the conductor is symmetrical to its motion from position 0
to 6, but the conductor is now cutting across the magnetic lines of force in
525
526
Chapter 18 Alternating Current
the ­opposite direction. As a result, the polarity of the induced voltage is
­reversed during the second half of the revolution. The + sign on the graph
of Figure 18-2(b) indicates the reference polarity of the terminal voltage.
The – sign simply means that the polarity of the voltage is reversed with
­respect to the reference polarity. One complete rotation of the loop constitutes one cycle. The time taken for the loop to complete one cycle is the
­period of the waveform.
See Review Question 18-35.
18-3 The Sine Wave
The international
standard symbol
for phase angle is
the Greek letter ϕ
(phi). Current North
­American standards
permit the use of
­either ϕ or θ (the
Greek letter theta) to
represent the phase
angle between current
and voltage.
Figure 18-2(b) shows the magnitude and polarity of the induced voltage in
a conductor as it rotates in a uniform magnetic field. To understand how
circuits with an alternating applied voltage behave, we must be able to determine the value of the applied voltage at any instant. For such analysis,
we can treat the alternating voltage as a phasor, a quantity that has a magnitude with an associated angle, much as a vector has a magnitude and a
direction. As with vectors, this book uses boldface type to indicate a phasor
and lightface type to indicate the magnitude of the phasor.
In Figure 18-3, the phasor OX represents the position of the rotating
conductor in the generator of Figure 18-1. This phasor pivots around the
origin O. The end that is free to rotate is indicated by an arrowhead. By
mathematical convention, phasor angles are measured counterclockwise
from the reference axis, which is the positive x-axis. Thus, the position of a
phasor is given by the phase angle ϕ through which it has rotated from the
reference axis.
X
O
Rotation
ϕ
Reference axis
Figure 18-3
Phasor representation of the rotating conductor
Since OX pivots around O, the perpendicular phasor XY in Figure 18-4
represents the motion of the conductor at the moment it passes through the
position represented by OX. If the angular velocity (the rate of rotation)
of the conductor is constant, the speed of the conductor is also constant.
Then the magnitude of the motion phasor XY is the same for any position of
the conductor. The motion represented by XY can be resolved into a vertical component XZ and a horizontal component XW. The magnetic field is
­directed downward, so XZ is parallel to the magnetic lines of force and no
induced voltage results from this component of the motion. Since XW cuts
18-3 The Sine Wave
Y
Z
ϕ
W
O
ϕ
X
ϕ
V
Reference axis
Figure 18-4
Phasor diagram of the position and motion of a rotating conductor
across the magnetic lines of force at right angles, the induced voltage is directly proportional to the length of XW. When ϕ = 90°, XW = XY. Therefore,
the ratio of XW to XY represents the ratio between the rate of cutting of the
lines of force at angle ϕ and the maximum rate of cutting.
⦟
⦟
Since WX is parallel to OV, WXO = XOV = ϕ.
⦟
⦟
⦟
Since ⦟XWY = 90°, ⦟XYW = ϕ.
Since ⦟OVX = 90° = ⦟YWX, all the corresponding angles of triangles XOV
Since YXO = 90°, WXY = 90° − ϕ = OXV.
and XYW are equal. Therefore, these two triangles are similar and
XW
XV
=
XY
OX
Hence, the rate of cutting of magnetic lines of force at any instant is directly proportional to XV/OX. If OX remains constant, this ratio depends
only on the phase angle ϕ. In the right triangle OVX, XV is the side opposite the angle ϕ and OX is the hypotenuse, so the ratio XV/OX is the sine of
the phase angle ϕ. Therefore,
The voltage induced into a conductor rotating in a uniform magnetic field is directly proportional to the sine of the phase angle.
The geometrical construction in Figure 18-4 gives the relative magnitude
of an induced voltage for phase angles less than 90°. We can use a similar
procedure for phase angles greater than 90°. In Figure 18-5(a), the phasor
OX has rotated through an angle of 120°. A perpendicular line drawn from
point X to the reference axis forms a right triangle in which side XV is
­opposite an angle of 180° − ϕ, or 60°. Therefore,
sin 120° = sin (180° − 120°) = sin 60°
527
528
Chapter 18 Alternating Current
90°
90°
X
60°
V
O
180°
ϕ = 120°
0°
180°
ϕ = 210°
V
O
0°
X
270°
(a)
Figure 18-5
Determining the sine of angles greater than 90°
Similarly, when ϕ = 210°, the right triangle has an angle opposite XV
of 210° − 180°, or 30°, as shown in Figure 18-5(b). However, XV is below
the reference axis, and sin 210° = −sin 30°. Referring to Figure 18-2, we
see that XV now represents a negative value of induced voltage because
the loop is moving in the opposite direction across the magnetic lines of
force. Whenever XV is below the reference axis, it represents a negative
quantity.
Figure 18-6 shows how sin ϕ varies as ϕ increases. We can plot this sine
curve by using a scientific calculator or a spreadsheet to calculate values of
sin ϕ. An alternating voltage that varies in accordance with the sine curve is
called a sine wave.
+1
sin ϕ
Most scientific
­calculators have a sin
key that makes it
easy to find the sine
of any angle. Spread­
sheets also have
­built-in trigonometric
­functions.
270°
(b)
0
90°
−1
Figure 18-6
180° 270° 360°
Phase angle, ϕ
The sine curve
Figure 18-7 shows another method of producing a sine curve. This
method is a projection of the geometric construction in Figure 18-5. Since
the length of phasor OX is constant, the altitude VX of the tip of the phasor
is proportional to the sine of angle ϕ. Therefore, VX is directly pro­portional
to the induced voltage. When we project VX onto a graph such that point
V1 is on the horizontal axis at a point corresponding to the value of ϕ,
point X1 is at the point that represents sin ϕ. If we repeat this geometric projection at 5° intervals, we have enough points to draw a smooth
sine curve.
18-5 Instantaneous Value of a Sine Wave
529
X1
X
O
ϕ
V
Figure 18-7
ϕ
V1
Drawing a sine curve by geometric projection
See Review Questions 18-36 and 18-37.
18-4 Peak Value of a Sine Wave
As shown in Figure 18-6, the greatest value of the induced voltage occurs
when ϕ = 90°. When ϕ = 270°, the alternating voltage reaches the same
maximum value, but the polarity is reversed. The magnitude of the alternating voltage at these two angles is the amplitude or peak value (Em) of
the AC waveform.
This peak value is measured in volts and does not have a polarity. The
peak value of a given sine wave is a constant that indicates the maximum
values of the sine wave. The peak value itself does not vary with time. In
the simple AC generator, the peak voltage depends on the maximum rate at
which the rotating loop cuts magnetic lines of force. This rate is determined
by the strength of the magnetic field, the angular velocity of the loop, and
the number of turns of wire in the loop.
See Review Questions 18-38 and 18-39.
18-5 Instantaneous
Value of a Sine Wave
Figure 18-6 shows the manner in which the induced voltage changes from
instant to instant as the loop rotates in the generator of Figure 18-1. During
one complete revolution, the voltage varies through the whole range of
possible values. As the loop rotates past 360°, the same waveform is generated again. The time the sine wave takes to go through one complete
cycle of instantaneous values is the period of the sine wave. The number of
­cycles completed in one second is the frequency of the sine wave.
The letter symbol for frequency is f.
The SI unit of frequency is the hertz (symbol Hz). A hertz is equal to
one cycle per second.
The induced voltage changes from instant to instant, so we represent
its instantaneous value with the lowercase letter e. The magnitude of the
­alternating voltage at any instant is directly proportional to sin ϕ. Since the
The hertz is named
in honour of the
­German physicist
­Heinrich ­Rudolf
Hertz (1857−94), who
­discovered radio
waves.
Chapter 18 Alternating Current
instantaneous voltage reaches its peak value at 90° and since sin 90° = 1,
the instantaneous voltage at any instant is
e = Em sin ϕ
(18-1)
In the simple two-pole generator of Figure 18-1, one complete revolution
of the loop generates one complete cycle of the voltage sine wave. But the
four-pole generator of Figure 18-8 generates one complete cycle of voltage as the loop moves through the magnetic flux under a north pole and
then through the flux under an adjacent south pole. Thus, a 360° cycle of
the voltage sine wave occurs while the loop rotates through only 180 mechanical degrees. The relationship between electrical degrees and mechanical
degrees depends on the number of pairs of magnetic poles in the generator.
Electronic circuits can generate sine waves of voltage without any mechanical rotation. To apply Equation 18-1, we must always express the phase
angle ϕ in electrical degrees.
S
360
electrical
degrees
Loop
N
Figure 18-8
N
0°
S
Electrical degrees in a four-pole generator
Source: © Lyroky/Alamy Stock Photo
530
An automotive alternator
18-5 Instantaneous Value of a Sine Wave
Example 18-1
At what speed must the shaft of a six-pole alternator turn in order to
produce a 60-Hz sine wave?
Solution
With three pairs of poles, one mechanical revolution generates three
electrical cycles. Therefore, the shaft speed is
60 Hz
= 20 r/s = 1200 r/min
3 cycles/r
Now we consider how to determine the angle ϕ in electrical degrees.
Often, we know the frequency of the sine wave. Since frequency can be expressed in cycles per second, multiplying the frequency by the time elapsed
since the instantaneous value passed through zero gives the angular position expressed in cycles. Multiplying this result by 360 then gives the
phase angle in electrical degrees. Hence, Equation 18-1 becomes
e = Em sin ( 360º × ft )
(18-2)
where e is the instantaneous source voltage, Em is the peak value of the
voltage sine wave, f is the frequency of the sine wave in hertz, and t is
the elapsed time in seconds.
Example 18-2
Find the instantaneous value of a 60-Hz sine wave 10 ms after the start
of a cycle, given that the peak value is 150 V.
Solution
e = Em sin ( 360º × ft )
= 150 sin ( 360º × 60 × 0.010 )
= 150 sin 216º
= −88 V
See Problems 18-1 to 18-6 and Review Questions 18-40 to 18-42.
531
532
Chapter 18 Alternating Current
18-6 The Radian
The Babylonian system of dividing a circle into 360° is often convenient for
measuring angles in geometry and trigonometry. However, this system is
somewhat awkward for calculations such as finding the linear velocity of
a rotating conductor from its angular velocity. In electrical engineering, the
standard practice is to express the angular distance travelled by a rotating
phasor in radians.
The radian (symbol rad) is the SI unit for angles.
A radian is equal to the angle formed by two radii of a circle when the
arc between them has the same length as each radius.
X
1 rad
O
Figure 18-9
With most engineering calculators,
angles can be entered
in ­either degrees or
­radians.
⦟
A
The radian
In Figure 18-9, XOA is 1 rad when the length of arc AX equals the length
of the phasor OX. Since the circumference of a circle is 2πr, there are 2π
radians in one complete cycle. Hence,
90º =
and
π
rad
2
180º = π rad
270º =
3π
rad
2
360º = 2π rad
1 rad ≈ 57.3˚
(18-3)
Angular velocity can be expressed in radians per second.
The letter symbol for angular velocity is the lowercase Greek letter
ω (omega).
The SI units for angular velocity are radians per second.
Since the phase angle is equal to angular velocity × elapsed time, we can
rewrite Equation 18-2 as
e = Em sin ωt
(18-4)
Since there are 2π radians in a cycle,
ω = 2πf
(18-5)
18-7 Instantaneous Current in a Resistor
533
Substituting for ω in Equation 18-4 gives
e = Em sin 2πft
(18-6)
In Equation 18-4 and Equation 18-6, the angle is expressed in radians.
Example 18-3
Write the general equation for the instantaneous voltage of a 60-Hz
generator with a peak voltage of 170 V.
Solution
Since f = 60 Hz,
Therefore,
2πf = 2 × 3.14 × 60 = 377 rad/s
e = Em sin 2πft = 170 sin 377 t
Since 60 Hz is the standard power-line frequency for North America, we
will often use this conversion:
60 Hz = 377 rad/s
(18-7)
Example 18-2A
Find the instantaneous value of a 60-Hz sine wave 10 ms after the start
of a cycle, given that the peak value is 150 V.
Solution
e = Em sin 2πft = 150 sin ( 2π × 60 × 0.010 rad ) = 150 sin 1.2π = 88 V
See Problems 18-7 to 18-10 and Review Questions 18-43 and 18-44.
18-7 Instantaneous
Current in a Resistor
Now that we are acquainted with the sinusoidal nature of a basic AC voltage source, we can determine the current that a sine-wave voltage will
­produce in the simple circuit of Figure 18-10. At a particular instant, the
voltage developed by the generator has a certain magnitude and polarity.
If we consider only this one particular instant, the AC circuit of Figure 18-10
is just like a DC circuit. However, at the next instant, the magnitude of the
­alternating voltage changes, while the applied voltage in a DC circuit is
­constant. At any given instant, all the equations summarized in Table 6-1
apply to the AC circuit of Figure 18-10.
Since the 60-Hz frequency of the power
grid is normally very
precisely controlled,
this book will treat
60 Hz as an exact
­measurement in most
calculations.
Chapter 18 Alternating Current
e = Em sin ωt
Figure 18-10
R
Simple alternating-current circuit
Therefore, we can apply Ohm’s law to find the instantaneous current
through the resistance in Figure 18-10:
i=
e
R
(18-8)
where i is the instantaneous current through the resistance, e is the
instantaneous voltage applied to the resistance, and R is the resistance
of the c­ ircuit.
Since resistance is determined by such physical factors as the type of
material and its dimensions, the resistance of any circuit is a constant at
a given temperature. Because R is a constant in Equation 18-8, the instantaneous current must stay exactly in step with the instantaneous voltage in order to satisfy Ohm’s law. If the instantaneous voltage is a sine
wave, the instantaneous current must also be a sine wave, as shown in
Figure 18-11. The current must reach its peak value at the same instant
that the voltage reaches its maximum. The current and the voltage must
become zero at the same instant, and the current must reverse its direction at the same instant that the voltage across the resistance reverses
its polarity. Since the current through a resistor and the voltage across it
have exactly the same phase angle at every instant, the current is said to
be in phase with the voltage.
+
Em
Instantaneous values
534
e = Em sin ωt
Im
i = Im sin ωt
0
π
2
π
3π
2
2π
Phase angle, ϕ
−
Figure 18-11
Instantaneous current through a resistor
18-7 Instantaneous Current in a Resistor
Substituting e = Em sin ωt in Equation 18-8 gives
i=
Em
sin ωt
R
(18-9)
Since the peak value of the instantaneous current depends on the peak
value of the applied voltage,
Im =
Em
R
i = Im sin ωt
and
(18-10)
(18-11)
where i is the instantaneous current through a resistor and Im is the peak
current.
See Problems 18-11 to 18-14 and Review Questions 18-45 to 18-49.
Circuit Check
A
A 100-Hz voltage sine wave has an instantaneous value of
− 45 V at 18 ms from the start of a cycle. Find the peak value of
this voltage.
CC 18-2. Find the lowest frequency a 250-V sine wave could have if
the instantaneous voltage is 75 V at 3 ms from the start of a
cycle.
CC 18-3. For the voltage wave in Figure 18-12, find
(a) the frequency
(b) the maximum value
(c) the instantaneous voltage at t = 9 ms
CC 18-1.
30 V
3 ms
Figure 18-12
15 ms
535
Chapter 18 Alternating Current
18-8 Instantaneous
Power in a Resistor
We can apply Equations 6-2, 6-3, and 6-4 to the instantaneous values in an
AC circuit. Therefore,
p = vi = i2R =
v2
R
(18-12)
where p is the instantaneous power in a resistor in watts, v is the instantaneous voltage drop across the resistance in volts, i is the instantaneous
current through the resistance in amperes, and R is the resistance of the
circuit in ohms.
Since R is a constant for a given circuit and since the instantaneous voltage and current in the basic circuit are both sine waves, the instantaneous
power must be a sine-squared wave. Figure 18-13 shows a graph of the
­instantaneous power in a resistor. Note that the instantaneous power in
a resistor is always positive because squaring a negative quantity results
in a positive quantity. We can interpret positive power as electric energy
being converted into some other form of energy (such as heat in a resistor).
Conversely, negative power represents some other form of energy being
converted into electric energy. Since a resistor cannot generate electric
energy, the instantaneous power in a resistor is never negative.
As Figure 18-13 shows, the instantaneous power in the basic AC system
pulsates, swinging from zero to maximum and back twice each cycle. The
instantaneous power pulsates at twice the frequency of the voltage and
current. The power that a current produces in a resistor does not depend
on which way the current flows through it. Therefore, the pulse of energy
transferred to the resistor during the first half of each cycle is the same
as the energy pulse during the second half. This pulsating characteristic of
instantaneous power can cause vibration in some types of small AC motors.
+
Instantaneous values
536
Pm
Im
p = Pm sin2 ωt
0
π
2
π
3π
2
2π
i = Im sin ωt
Phase angle, ϕ
−
Figure 18-13
Instantaneous power in a resistor
18-9 Periodic Waves
We can also show the pulsating nature of the power in an AC system by
deriving an equation for instantaneous power. Substituting I = Im sin ωt in
Equation 18-12, we get
p = ( Im sin ωt ) 2 × R = I2mR × sin2 ωt
Since peak values are maximum instantaneous values, Equation 18-12
gives
Pm = I2mR
(18-13)
Therefore,
p = Pm sin2 ωt
(18-14)
where p is the instantaneous power in a resistance and Pm is the peak
power.
The relationship sin2 θ = 12(1 − cos 2θ) is a trigonometric identity, which
means that the relationship applies for all values of θ. Therefore,
1
sin2 ωt = (1 − cos 2ωt)
2
and
1
1
p = Pm − Pm cos 2ωt
2
2
(18-15)
This equation tells us three things about the instantaneous power waveform that we can check by examining Figure 18-13:
1.
In any right-angled triangle, the cosine of one acute angle is also
the sine of the other acute angle. Therefore, the general shape of the
cosine curve and the sine curve are the same, and cos θ = sin (θ − 90°).
Consequently, the fluctuations in instantaneous power are sinusoidal.
2.Since the angle in the cosine term is 2ωt, the frequency of the sinusoidal
variation in power is twice the frequency of the instantaneous voltage
and current.
3.Since the limits of the value of a cosine are +1 when ϕ = 0° and −1
when ϕ = 180°, the value of the expression 12(1 − cos 2ωt) can vary from
0 to +1. Therefore, the instantaneous power in a resistance is always
positive.
See Review Questions 18-50 to 18-53.
18-9 Periodic Waves
Although the sine wave is by far the most important AC waveform, there
are many other types of periodic waves. In electric circuits, a periodic
wave is any time-varying quantity, such as voltage, current, or power, that
537
Chapter 18 Alternating Current
Time
− 1 period
(a) Sine wave
0
Time
1
period
(b) Sine-squared wave
+
0
+
−
Time
Instantaneous value
0
Instantaneous value
+
+
0
− 1 period
(c) Square wave
− 1 period
(d) Pulse wave
+
+
0
Time
− 1 period
(e) Sawtooth wave
Figure 18-14
Instantaneous value
Instantaneous value
Instantaneous value
c­ ontinually repeats exactly the same sequence of values with each cycle
taking ­exactly the same time. Figure 18-14 shows how the instantaneous
values of six common periodic waves vary as a function of time. The graphs
of the sine wave and sine-squared wave are the same as the two waves
shown in Figure 18-13, but here the horizontal axis represents time rather
than the phase angle. Since ϕ = ωt, the two curves have the same shape in
both ­figures. As shown in Figure 18-14, the period of each waveform is the
length of time it takes the instantaneous voltage or current to complete one
cycle of values. Figure 18-14(b) shows four complete cycles of a periodic
wave. All other graphs in Figure 18-14 show two complete ­cycles.
The instantaneous values of a periodic wave change much too rapidly
to be registered by a conventional electric meter. However, a cathode-ray
tube or a liquid-crystal display (LCD) in an oscilloscope can quite readily
respond to high-frequency instantaneous voltages. If the instantaneous
voltage of a certain periodic wave, such as the sine wave of Figure 18-14(a),
is applied to the vertical deflection circuit of an oscilloscope, the display
will show a dot of light moving up and down exactly in step with the
voltage. For most waves, this motion is still much too rapid for the eye to
Instantaneous value
538
0
Time
Time
− 1 period
(f) Half-wave-rectified wave
Some common periodic waves
18-9 Periodic Waves
Source: © AEMC Instruments, a leader in electrical test and measurement
instruments
follow, so what we see is a vertical line with a length proportional to the
peak-to-peak value of the voltage of the periodic wave. In the sawtooth
periodic wave of Figure 18-14(e), the instantaneous voltage is a linear function of elapsed time. Hence, if the horizontal deflection circuit of the oscilloscope generates a sawtooth wave having a period exactly twice that of
the waveform fed to the vertical deflection circuit, the screen will display
two cycles of that waveform. Multi-trace oscilloscopes can display two or
more waveforms at the same time.
See Review Question 18-54.
Source: © Tihis/Dreamstime/Getstock
A handheld multi-trace LCD oscilloscope
A dual-trace CRT oscilloscope
539
540
Chapter 18 Alternating Current
18-10 Average Value of a Periodic Wave
By definition, an alternating current or voltage is one in which the arithmetic
average, or mean, of the instantaneous values over a cycle is zero. From an
examination of Figure 18-14, it is fairly obvious that the sine, square, and
sawtooth waveforms are alternating because the instantaneous values are
symmetrical about the horizontal axis of the graph. The sine-squared wave,
pulse wave, and half-wave-rectified wave are not alternating currents because their instantaneous values always have the same polarity. For complex periodic waves we can determine the mean of the i­ nstanta­neous values
graphically by examining the area under a graph.
Since the sine wave is symmetrical, each positive value during the
first half-cycle has a matching negative value during the next half-cycle.
Therefore, the mean value of a complete cycle of a sine wave is zero, and
the mean value of the first half-cycle is equal in magnitude but opposite in
polarity to the mean of the second half-cycle. Appendix 2-6 uses integral
calculus to show that the mean value of the first half-cycle of a sine wave
is 2/π, or about 0.637.
Since i = Im sin ωt, the half-cycle average value of a sinusoidal current is
Iav =
2
Im ≈ 0.637Im
π
(18-16)
Similarly, the half-cycle average value of a sinusoidal source voltage is
Eav = 0.637Em =
2
Em
π
(18-17)
As we discovered in Section 18-9, the sine-squared wave of Figure 18-14(b)
has a full-cycle average value of one-half the peak value of the sine-squared
wave. The pulse waveform has an average value that depends on the ratio
of the pulse duration to the period of the wave. The full-cycle a­ verage value
of the half-wave-rectified wave of Figure 18-14(f) is the mean of 0.637Em for
one half-cycle and zero for the next half-cycle, or 0.3183Em.
See Problem 18-15 and Review Question 18-55.
18-11 RMS Value of a Sine Wave
All electric circuits convert electric energy into some other form of energy,
such as heat, light, or mechanical energy. For many of these energy conversions, it does not matter whether the energy source for the circuit produces
direct current or alternating current. Therefore, we find equivalent steadystate values for alternating current and voltage that allow us to use the
same relationships among voltage, current, resistance, power, work, and so
18-11 RMS Value of a Sine Wave
on, that we use for DC circuits. We can determine an equivalent DC value
or effective value of an alternating current experimentally by finding the
direct current that produces heat in a given resistance at the same rate as
when the resistance is connected to a source of alternating voltage. We can
also use algebraic analysis to find such equivalent values.
To determine this equivalent DC value of an alternating current, we first
find the average power in a resistive AC circuit. Since p = i2R (Equation 18-12),
Pav = ( full-cycle average of i2 ) × R
In a DC circuit, P = I2R. Since the effective values of an alternating current
are to represent DC equivalent values, we can use the same letter symbols
for both, that is, uppercase italic letters without subscripts. Therefore, in an
AC circuit,
P = I2R
where P is average power and I is the DC equivalent for the alternating
­current.
√ √
Solving for I gives
I=
P
=
R
( full-cycle average of i2 ) × R
R
= √ full-cycle average of i2
Thus I is equal to the square root of the mean of the squares of the instantaneous current over a full cycle. Root-mean-square (RMS) value, effective
value, and equivalent DC value all mean the same thing. For the rest of the
book, we use the term RMS.
Figure 18-13 shows that the instantaneous power for a sine wave of
alternating current swings alternately and symmetrically between zero
and peak power. Therefore, the average power in a resistor through
which a sine-wave alternating current is flowing is simply one-half the
peak power.
We can also determine this relationship from the general equation for
instantaneous power: p = Pm sin2 ωt (Equation 18-14). As we have already
shown, sin2 ωt = 12(1 − cos 2ωt). If we average a sine or cosine function over
a complete cycle, the average must be zero, since for every positive value
during the first half-cycle, there is a matching equivalent negative value
during the second half-cycle. Hence, when averaged over a complete cycle,
Equation 18-14 becomes
1
P = Pm
(18-18)
2
For a sine-wave alternating current in an AC circuit,
P = I2R and Pm = I2mR
541
542
Chapter 18 Alternating Current
But
Therefore,
1
1
P = Pm = I2mR
2
2
1
I2R = I2mR
2
1
I2 = I2m
2
I=
Im
≈ 0.707Im
√2
(18-19)
The RMS (effective) value of a sine wave of current is 1/√2, or about
0.707, times the peak value.
The RMS value of a sine-wave voltage should be such that the average power
is the product of the RMS voltage across and RMS current through the resistance of the circuit, just as P = VI in the equivalent DC circuit. Since peak
­instantaneous power occurs at the instant when both the voltage across and
the current through a resistance circuit reach their peak values,
Pm = VmIm
Substituting into Equation 18-18 gives
P = VI = V
Therefore,
and
V=
Im
1
1
= Pm = Vm Im
2
2
√2
Vm
≈ 0.707 Vm
√2
(18-20)
Vm = √ 2 V ≈ 1.414V
Since RMS values are used for the majority of AC circuit measurements
and computations, all further AC currents and voltages are given as RMS
values unless otherwise specified.
Example 18-4
Find the peak voltage of a 120-V 60-Hz electric service.
Solution
Em = 1.414E = 1.414 × 120 V = 170 V
18-11 
RMS Value of a Sine Wave
A term we may encounter occasionally when working with AC waveforms is form factor. Form factor is the ratio of the RMS value to the halfcycle average value of an AC wave. For a sine wave,
form factor =
I
0.707Im
≈
= 1.11
Iav 0.637Im
(18-21)
See Problems 18-16 to 18-31 and Review Questions 18-56 to 18-58.
Circuit Check
B
CC 18-4. A voltage wave with the equation e = 100 sin 500t V is applied to a
50-Ω resistor. Calculate the instantaneous current at
t = 3.5 ms.
CC 18-5. A voltage wave with the equation e = 163 sin 377t V is ­applied
to a 100-Ω resistor. Calculate:
(a) the RMS value of voltage
(b) the RMS current in the resistor
(c) the average power and the RMS power delivered
543
544
Chapter 18 Alternating Current
Summary
• A sine wave of alternating voltage is produced across a loop of wire as it
rotates within a uniform magnetic field.
• The peak value of the sine wave of alternating voltage produced by an
AC generator occurs when the loop of wire is parallel to the magnetic
lines of force.
• Zero voltage is induced across the loop of wire when it is perpendicular
to the magnetic lines of force.
• The instantaneous value of the voltage produced by an AC generator is
proportional to the sine of the angle of rotation of the loop of wire.
• The instantaneous current in an AC circuit obeys Ohm’s law.
• The instantaneous power in a resistor in an AC circuit is a sine-squared
wave.
• The RMS value of a sine wave of current in the resistor in an AC circuit equals the DC current that would produce the same amount of
heat.
B = beginner
I = intermediate
A = advanced
Problems
B
Section 18-5
18-1.
B
18-2.
B
18-3.
I
18-4.
B
18-5.
B
18-6.
B
B
I
B
If the peak voltage of a 50-Hz AC generator is 210 V, what is the instantaneous voltage after the loop has rotated through 25 degrees?
If the peak voltage of a 1.0-kHz audio oscillator is 20 V, what is the
instantaneous voltage when ϕ = 210°?
An AM radio transmitter generates a sinusoidal carrier wave with a
frequency of 900 kHz and feeds a peak voltage of 1.3 kV to the antenna.
Determine the instantaneous voltage 0.7 μs after the start of a cycle.
The instantaneous voltage from a 400-Hz aircraft generator is 95 V
when t = 1 ms. Find the instantaneous voltage at t = 2 ms.
Find the instantaneous voltage after the loop of the generator in
Problem 18-1 has rotated for 10 ms.
Find the instantaneous voltage from the audio oscillator in
Problem 18-2 when t = 600 μs.
Section 18-6
18-7.
Instantaneous Value of a Sine Wave
The Radian
Determine the instantaneous voltage when the loop of the generator in Problem 18-1 has rotated through 1.0 rad.
18-8. Find the instantaneous voltage from the audio oscillator in
Problem 18-2 when ϕ = 5 rad.
18-9. Write the general voltage equation for a 400-Hz source with a peak
voltage of 200 V.
18-10. An alternating voltage has an expression e = 75 sin 3000t V. Determine
(a) the frequency of the sine wave
(b) the peak value
(c) the instantaneous value at t = 1.5 ms
Problems
I
I
I
B
I
B
B
B
I
I
I
B
I
I
B
B
Section 18-7
Instantaneous Current in a Resistor
Section 18-8
Instantaneous Power in a Resistor
18-11. A 1-kHz sine wave current has an instantaneous value of 6 A when
t = 0.4 ms. Determine
(a) the peak value
(b) the instantaneous value at t = 1.5 ms
18-12. If the instantaneous current drawn from a 50-Hz source is
−263 mA when t = 0.013 s, what is the instantaneous current when
t = 0.017 s?
18-13. A 50-Ω resistor is connected to a voltage source with an output of
e = 75 sin 1400t V. Determine
(a) the maximum power
(b) the instantaneous power at t = 1 ms
(c) the average power
18-14. What is the frequency of the power wave in Problem 18-13?
Section 18-10
Average Value of a Periodic Wave
18-15. The conductor in the simple generator of Figure 18-1 cuts across a
total magnetic flux of 40 mWb during the first 180° of rotation. The
frequency of the AC output is 60 Hz. Find the half-cycle average
­induced voltage.
Section 18-11
RMS
Value of a Sine Wave
18-16. Calculate the peak voltage in the European 230-V 50-Hz system.
18-17. Determine the peak value of a microwave signal with an RMS value
of 100 μV.
18-18. Find the RMS voltage drop across a 15-Ω resistor when the current
flowing through it is 1.5 sin 377t.
18-19. Write the equation for the instantaneous voltage generated by a
117-V 400-Hz alternator.
18-20. If the symmetrical square wave of current shown in Figure 18-14(c)
has a peak value of 40 mA, find the RMS value.
18-21. If the symmetrical sawtooth voltage waveform in Figure 18-14(e)
has a peak voltage of 30 V, find the RMS value.
18-22. The average power of a 250-Ω soldering iron is 55 W.
(a) Find the RMS current through the iron.
(b) Find the peak voltage across the iron.
18-23. What reading does an AC ammeter show when connected in series
with a 16-Ω load with a peak power of 288 W?
18-24. How long does a 10-Ω heater connected to a 120-V 60-Hz source
take to convert 1.0 kW · h of electric energy into heat?
18-25. A fluorescent lamp has a resistance of 50 Ω. If it's connected to a
110-V AC outlet, determine its wattage rating.
18-26. Find the RMS value of the alternating voltage induced into the loop
in Problem 18-15.
545
Chapter 18 Alternating Current
A
A
B
B
A
18-27. For a voltage wave with the equation v = 220 sin 377t, determine
(a) the peak value
(b) the RMS value
(c) the frequency
(d) the instantaneous value when t = 8 ms
(e) how long the voltage takes to reach half its maximum value
(f) the current the voltage produces in a 50-Ω resistor
18-28. For a 100-Hz voltage wave with an RMS voltage of 500 V,
(a) write the equation of the voltage wave
(b) find the instantaneous value of voltage at t = 6 ms
(c) find the time when the voltage first reaches 300 V
18-29. The peak power in a resistor connected to a 500-V 60-Hz AC source
is 1.0 kW. Calculate the resistance.
18-30. A 500-Ω resistor dissipates a peak power of 600 W when connected
to an AC source. Find the RMS current.
18-31. The voltage waveform shown in Figure 18-15 has an RMS value of
150 V. Find the frequency of the voltage. Write the equation for this
voltage.
+
Voltage
546
1.57 ms
Time
−
Figure 18-15
Review Questions
Section 18-1
A Simple Generator
18-32. What conditions govern the generation of a voltage by electromagnetic induction?
18-33. Why does the angular velocity of the rotating loop in Figure 18-1
tend to decrease when a load resistor is connected to the brushes?
18-34. Why is it not possible to maintain a constant polarity of induced
voltage in the loop in Figure 18-1?
Section 18-2
The Nature of the Induced Voltage
18-35. Why does the voltage across a loop of wire increase as it rotates
from a position perpendicular to magnetic flux to a position parallel to the flux?
Section 18-3
The Sine Wave
18-36. Why do we describe the instantaneous voltage waveform from the
simple generator of Figure 18-1 as a sine wave?
18-37. What is meant by the term angular distance?
Review Questions
Section 18-4
Peak Value of a Sine Wave
18-38. Why does doubling the number of turns in the rotating loop in a
generator double the peak voltage?
18-39. Why is the peak value of an AC voltage never a negative quantity?
Section 18-5
Instantaneous Value of a Sine Wave
18-40. How does the angular velocity of the loop in Figure 18-1 affect
(a) the frequency of the voltage waveform
(b) the peak value of the induced voltage
18-41. Why is the answer to Example 18-2 a negative quantity?
18-42. What is the significance of the term sine wave when describing the
behaviour of an electric circuit?
Section 18-6
The Radian
Section 18-7
Instantaneous Current in a Resistor
18-43. With a diagram similar to Figure 18-5, show that sin 3π/4 = sin π/4.
18-44. Why are radian units preferred for the general equation for an
alternating current?
18-45. What is the significance of a negative value of instantaneous
­current?
18-46. What effect does changing the peak value of an alternating current
have on the waveform?
18-47. Why must the instantaneous current through a resistor have the
same waveform as the instantaneous voltage across it?
18-48. What is meant by the statement that the current through a resistor
is in phase with the voltage across it?
18-49. What factors govern the instantaneous value of the current through
a resistor in an AC circuit?
Section 18-8
Instantaneous Power in a Resistor
18-50. Why can we substitute instantaneous AC values in the various
equations we derived for DC circuits containing a voltage source
and a load resistor?
18-51. What is the significance of a negative value of instantaneous power?
18-52. Why is the instantaneous power input to a resistor never negative?
18-53. Why does the instantaneous power in a resistor pulsate at twice the
frequency of the applied voltage?
Section 18-9
Periodic Waves
18-54. What distinguishes a periodic wave from other types of waves?
Section 18-10
Average Value of a Periodic Wave
18-55. Explain why the mean value for a half-cycle of alternating current
is different from the mean value for a complete cycle.
547
548
Chapter 18 Alternating Current
Section 18-11
RMS
Value of a Sine Wave
18-56. Why is the average power in a resistor connected to a sine-wave
source one-half the peak power?
18-57. Why is it necessary to consider average power when deriving an
RMS value for alternating-current and voltage sine waves?
18-58. The equation for instantaneous voltage in an AC system is sometimes written in the form e = 2 E sin ωt. Suggest a reason for using
this form.
Integrate the Concepts
A 2-kHz sine wave generator with an RMS value of 123 V is connected to a
75-Ω resistor.
(a) Determine the voltage at t = 0.1 ms.
(b) Write an equation for the instantaneous current.
(c) Calculate the peak power supplied to the resistor.
(d) Write an equation for the instantaneous power supplied to the
­resistor.
(e) Calculate the RMS values of the voltage and current.
Practice Quiz
1.
Which of the following statements are true?
(a)The alternating voltage is produced across a loop of wire when it
rotates within an electric field
(b)A phasor represents a time-varying quantity in terms of magnitude and direction.
(c) The length of a phasor represents the direction.
(d) One full cycle of a sine wave has 360 degrees.
Complete the statements in questions 2–4.
2.
The voltage induced into a conductor rotating in a uniform magnetic
field is ______ to the sine of the phase angle.
3.
Rotating a phasor counterclockwise from the reference axis
________________ the angle of the phasor.
4.
The peak value of a sine wave of alternating voltage produced
by an AC generator occurs when the loop of wire is moving
____________________ to the magnetic lines of force.
5.
The peak value of the sine wave shown in Figure 18-16 is
(a) 15 V
(b) 7.5 V
(c) 8.0 V
(d) 16 V
Practice Quiz
8
Amplitude (V)
6
4
2
0
−2
−4
−6
−8
0
1
2
3
4
Time (ms)
5
6
7
8
Figure 18-16
6. A full cycle of a sine wave has
(a)π/2 radians
(b)π radians
(c)2π radians
(d)4π radians
7. At what speed must the shaft of a four-pole alternator turn in order to
produce a 500-Hz sine wave?
(a)750 RPM
(b)500 RPM
(c)125 RPM
(d)250 RPM
8. What is the instantaneous value of the sine wave of Figure 18-16 at t = 1 ms?
(a) 1 V
(b) −2 V
(c) −1 V
(d) −0.8 V
9. What is the instantaneous value of the sine wave of Figure 18-16 at
2.1 ms?
(a) 4 V
(b) 1 V
(c) −1 V
(d) −4 V
10. What is the frequency of the sine wave of Figure 18-16?
(a) 1.1 KHz
(b) 900 Hz
(c) 500 Hz
(d) 513 Hz
549
Chapter 18 Alternating Current
11. What is the frequency of the sine wave of Figure 18-17?
(a) 323 Hz
(b) 333 Hz
(c) 160 Hz
(d) 167 Hz
8
6
4
Amplitude (V)
550
2
0
−2
−4
−6
−8
0
0.5
1
1.5
2
Time (ms)
2.5
3
3.5
Figure 18-17
12. What is the instantaneous value of the waveform of Figure 18-17 at t=
2.2 ms?
(a) 8 V
(b) 7.5 V
(c) 6.5 V
(d) 7 V
13. What is the rms value of the sine wave of Figure 18-17?
(a) 10 V
(b) 6.9 V
(c) 6.4 V
(d) 9.5 V
14. What is the general equation for the instantaneous voltage of a 50-Hz
generator with a peak voltage of 220 V?
(a) 156 sin 314t V
(b) 220 sin 314t V
(c) 110 sin 314t V
(d) 220 sin 377t V
Practice Quiz
15. What is the instantaneous current through the 20-Ω resistor in the circuit of Figure 18-18?
(a) 11 sin 377t A
(b) 1.1 sin 377t A
(c) 11 sin 314t A
(d) 1.1 sin 314t A
V
220 Vpk
60Hz
0°
+
R
20 Ω
−
Figure 18-18
16. What is the expression for the instantaneous power in the circuit
of Figure 18-18?
(a) 11 sin 377t W
(b) 2.42 sin 377t kW
(c) 11 sin2 377t W
(d) 2.42 sin2 377t kW
551
19
Reactance
In Chapter 18 we found that all the laws for DC circuits apply when
an AC source is connected to a resistance. In this chapter we shall
see that these laws also apply when an AC source is connected
to an inductance or a capacitance. In an AC circuit, the current
through a resistance and the voltage across it are in phase.
However, this chapter will show that the AC current and voltage are
not in phase for an ­inductance or capacitance.
Chapter Outline
19-1
Instantaneous Current in an Ideal Inductor 554
19-3
Factors Governing Inductive Reactance 556
19-5
Capacitive Reactance 559
19-2
19-4
19-6
Inductive Reactance 555
Instantaneous Current in a Capacitor 558
Factors Governing Capacitive Reactance 560
19-7Resistance, Inductive Reactance, and Capacitive
Reactance
562
Key Terms
reactance 556
inductive reactance
556
displacement current 560
capacitive reactance 560
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• predict the instantaneous current produced
by a sine wave of voltage across an ideal
inductor
• calculate the inductive reactance of an ideal
inductor given the AC voltage across the
inductor and AC current through it
• describe the effects of inductance and
frequency on the inductive reactance of an
ideal inductor
• calculate the reactance of an ideal inductor
Photo sources: © iStock.com/ Evgeny Rannev
• predict the instantaneous current produced
by a sine wave of voltage across a capacitor
• describe the effects of capacitance and
frequency on the capacitive reactance of a
capacitor
• calculate the reactance of a capacitor
• summarize the differences in the behaviour
of resistance, inductance, and capacitance in
AC circuits
554
Chapter 19 Reactance
19-1 Instantaneous Current
in an Ideal Inductor
L
A
Figure 19-1 Inductance
in an AC circuit
If there is no resistance in the circuit, any voltage drop that appears across
the terminals of an inductor must be due to the voltage induced in the coil
by a changing current through it. From Equation 17-1, the instantaneous
voltage across the inductor in Figure 19-1 must be
vL = L
Therefore,
L
di
dt
di
= Em sin ωt
dt
di
Em
=
sin ωt
dt
L
and
(19-1)
Since Kirchhoff’s voltage law requires the voltage across the inductor to
be exactly equal to the applied voltage at every instant, we can represent
the instantaneous voltage vL across the inductor by the blue sine curve in
­Figure 19-2. According to Faraday’s law of electromagnetic induction, the
magnitude of this inductive voltage at any instant is directly proportional
Maximum
positive voltage
Instantaneous Values
e = Em sin ωt
In the circuit of Figure 19-1, we assume that the inductor has negligible
­resistance. To satisfy Kirchhoff’s voltage law, at every instant the inductive
voltage across the coil in Figure 19-1 must exactly equal the applied voltage.
Hence,
vL = e = Em sin ωt
Maximum negative
rate of change
of current
Zero rate
of change
vL
+
iL
0
π
2
π
3π
2
2π
5π
2
ωt
Zero
voltage
−
Maximum positive
rate of change
of current
Figure 19-2
Maximum
negative voltage
Instantaneous current in an inductor
19-2 Inductive Reactance
to the rate of change of current through the coil at that instant. When the
maximum positive voltage appears across the inductor (at the point where
ϕ = π/2 rad), the current must be changing at the greatest positive rate.
Since ϕ = ω t, the slope of a graph of current versus phase angle is directly
proportional to the rate of change of current with time. In Figure 19-2, the
maximum rate of change is indicated by the steepest slope.
One quarter-cycle later (when ϕ = π rad), the instantaneous voltage is
momentarily zero, and the current must momentarily stop changing.
­Regardless of how great the current may be, as long as it is neither rising
nor falling, there is no change in the magnetic flux linking the turns of the
coil and, hence, no induced voltage. Therefore, the slope of the instantaneous current curve must be horizontal at this instant. Between π/2 rad and
π rad, the voltage across the inductor has to decrease along a sine curve,
and thus the rate of change of current has to decrease in the same manner.
Another quarter-cycle later (when ϕ = 3π/2 rad), the i­ nstantaneous inductive voltage reaches its most negative value, and therefore the instantaneous
current must be changing at its maximum negative rate.
The instantaneous current through the inductor must have the sine-wave
shape shown by the red curve in Figure 19-2 in order to induce the sinewave voltage that appears across the inductor. But the instantaneous current sine wave reaches its positive peak a quarter-cycle (π/2 radians) after
the instantaneous voltage across the inductor reaches its maximum. The
zero and minimum values of the current also occur a quarter-cycle after the
corresponding points in the voltage wave. Therefore,
For a sine-wave voltage drop to appear across an ideal inductor, the
current through it must be a sine wave that lags behind the inductive voltage drop by π/2 radians.
Therefore, the instantaneous current in the circuit of Figure 19-1 is
iL = Im sin ω t −
(
π
2
)
(19-2)
where the phase angle ωt − π/2 is measured in radians.
Appendix 2-7 shows how to derive this equation by using integral c­ alculus.
See Review Questions 19-22 to 19-26 at the end of the chapter.
19-2 Inductive Reactance
Since the current through the inductor in Figure 19-1 is a sine wave, it must
have an RMS value. The ammeter reading shows this constant RMS value.
For a given alternating-voltage source and a given inductance, the ratio
V/I is constant.
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556
Chapter 19 Reactance
This constant V/I ratio represents reactance, an opposition to the flow of
alternating current. However, the current in an ideal inductor lags behind
the voltage across it by π/2 radians, whereas the current in a resistor is in
phase with the voltage drop. Furthermore, an inductor does not convert
electric energy into heat as a resistor does. Thus, the reactance of an inductor differs from resistance.
Inductive reactance is the opposition of inductance to alternating
current.
The letter symbol for inductive reactance is XL.
XL =
VL
IL
(19-3)
Since inductive reactance is a V/I ratio like resistance, the ohm is the unit
of inductive reactance:
An AC circuit has an inductive reactance of one ohm when an alternating current of one ampere RMS through the inductance creates an
inductive voltage drop of one volt RMS across the inductance.
See Problems 19-1 and 19-2 and Review Question 19-27.
19-3 Factors Governing
Inductive Reactance
If the applied voltage in the circuit of Figure 19-1 remains sinusoidal, there are
only three factors that can vary independently: the peak voltage of the sine wave
of applied voltage, the frequency of the applied voltage, and the inductance.
If we double the peak value of the applied voltage while keeping the frequency constant, the maximum rate of change of current must be doubled
in order to develop the required inductive voltage drop. Since the frequency is unchanged, the current reaches its maximum value in the same
length of time, but the slope of the current is twice as steep at every instant.
Therefore, the current rises to twice the original peak value. Consequently,
doubling the amplitude of the applied voltage also doubles the current.
Since XL = VL/IL, there is no change in the inductive reactance. The same
logic applies for any change in the applied voltage.
The magnitude of the applied alternating voltage has no effect on
the inductive reactance of an AC circuit.
If we double the frequency of the applied voltage and leave the peak
value the same, the instantaneous current in the inductance must have
the same maximum rate of change and the same average rate of change
over each quarter-cycle. Because the frequency is doubled, the period (the
19-3 Factors Governing Inductive Reactance
­ uration of a complete cycle) is half the original period. Therefore, the curd
rent rises or falls for half as long in each part of the cycle, and the amplitude
of the current sine wave is reduced by half. Since XL = VL/IL, the inductive
­reactance has been doubled. By considering other ratios of frequencies in
the same way, we find that
Inductive reactance is directly proportional to frequency.
If we double the inductance but leave the source unchanged, Equa­
tion 19-1 shows that the maximum rate of change of current must be cut in
half to develop the same voltage. Therefore, the amplitude of the current
is reduced by half and the inductive reactance is doubled. Since the same
reasoning applies for all other inductances,
Inductive reactance is directly proportional to the inductance.
We can derive a formula for inductive reactance by considering the
a­ verage rate of change of current. The current rises from zero to Im in one
quarter-cycle, and the duration of one cycle is 1/f. Therefore,
average
di
=
dt
1
4
Im
= 4 f Im
× ( 1/f )
From Equation 18-17, we know that Eav = 2Em /π. Substituting into Equation
19-1 gives
4fIm =
from which
Em × 2/π
L
Em
= 2πf L
Im
XL = 2πf L
(19-4)
where XL is inductive reactance in ohms, f is frequency in hertz, and L is
­inductance in henrys.
Since ω = 2πf (Equation 18-5),
XL = ωL
where ω is angular velocity in radians per second.
Appendix 2-7 uses integral calculus to derive this equation.
See Problems 19-3 to 19-10 and Review Question 19-28.
(19-5)
557
558
Chapter 19 Reactance
Circuit Check
A
CC 19-1.
If a voltage of 30 V AC causes a current of 2 A to flow in an
ideal inductor, how much inductive reactance is in the circuit?
CC 19-2. What is the value of the inductor in question CC 19-1 if the
frequency of the source is 60 Hz?
CC 19-3. At what frequency will a 470-nH inductor have a reactance of 100 Ω?
19-4 Instantaneous
Current in a Capacitor
e = Em sin ωt
C
A
Figure 19-3 Capacitance
in an alternating-current
circuit
If we connect a capacitor across a sine-wave voltage source, as in
Figure 19‑3, Kirchhoff’s voltage law requires the voltage across the capacitor to be exactly the same as the applied voltage at every instant. The
voltage across a capacitor can change only if the capacitor charges or discharges. Consequently, the capacitor in Figure 19-3 must charge and discharge in such a manner that the voltage across it is a sine wave equal to
the applied voltage at every instant.
Since q = Cv (Equation 12-6),
dq
dt
By definition, i = dq/dt. Therefore,
=C
i=C
dv
dt
dv
dt
(19-6)
Since capacitance depends on such physical factors as the area of the plates
and the dielectric constant of the material between the plates, the capacitance of a given circuit does not depend on the elapsed time. Since we can
treat C in Equation 19-6 as a constant, this equation shows that the instantaneous current in Figure 19-3 is directly proportional to the rate at which
the voltage across the capacitor is changing.
The blue sine curve in Figure 19-4 represents the instantaneous voltage
across the capacitor. This curve shows that the maximum voltage across
the capacitor occurs π/2 radians after the maximum rate of change of voltage. At the exact moment when the voltage across the capacitor is greatest,
the voltage is neither rising nor falling. Therefore, the instantaneous current must be zero at this instant. The maximum rate of change of voltage
occurs when the voltage sine curve is steepest. At this instant the voltage
is zero, indicating that the capacitor has just finished discharging its stored
charge and is about to start building up an opposite charge. Therefore, the
instantaneous current has its maximum positive value at the instant when
Instantaneous Values
19-5 Capacitive Reactance
Zero
current
+
0
π
2
π
Maximum
positive current
Maximum
positive
voltage
vC
3π
2
2π
5π
2
i C ωt
−
Figure 19-4
Instantaneous current in a capacitor
the voltage across the capacitor changes from a negative polarity to a positive polarity. Similarly, the current reaches its maximum negative value
just as the voltage changes from a positive to a negative polarity.
The instantaneous current must have the sine-wave shape shown by the
red curve in Figure 19-4 in order for the voltage across the inductor to
match the applied voltage at every instant. The instantaneous current is at
its maximum positive value at the instant that the voltage across the capa­
citance is just starting to increase from zero. When the voltage across the
­capacitance has reached its positive peak π/2 rad later, the instantaneous
current has fallen back to zero. Therefore,
For a sine-wave voltage to be developed across a capacitor, the current through it must be a sine wave that leads the instantaneous
voltage by π/2 radians.
Therefore, the instantaneous current in the circuit of Figure 19-3 is
ic = Im sin ωt +
(
π
2
)
(19-7)
where the phase angle ω t + π/2 is measured in radians.
See Review Questions 19-29 and 19-30.
19-5 Capacitive Reactance
Since the current through the capacitor in Figure 19-3 is a sine wave, it
has an RMS value, which can be measured by an AC ammeter connected
as shown in Figure 19-3. This current reading does not mean that current
is flowing through the insulating dielectric of the capacitor. In an AC circuit, instan­taneous current flows in one direction for one half-cycle and
then in the ­opposite direction for the next half-cycle. During each cycle,
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560
Chapter 19 Reactance
the ­capacitor builds up a positive charge, discharges, builds up a negative
charge, and then ­discharges again. The sine-wave alternating current in
the circuit is a displacement current that transfers charge to and from the
capacitance. Thus, capacitors can “pass” alternating current while blocking
direct current.
For a given alternating-voltage source and a given capacitance, both
the sine-wave voltage across and the sine-wave current “through” the
­capacitor have fixed RMS values. Therefore, the ratio V/I is a constant. As
with ­resistance and inductive reactance, this constant V/I ratio represents
­opposition to alternating current.
Capacitive reactance is the opposition of
alternating current.
The letter symbol for capacitive reactance is XC.
XC =
VC
IC
capacitance
to
(19-8)
An AC circuit has a capacitive reactance of one ohm when alternating current of one ampere RMS creates an alternating voltage of one
volt RMS across the capacitance of the circuit.
See Problems 19-11 and 19-12.
19-6 Factors Governing
Capacitive Reactance
By varying the capacitance and the applied voltage in the circuit
of Figure 19-3, we can determine the factors that govern capacitive
­reactance.
If we double the amplitude of the applied voltage while keeping the
frequency constant, the capacitor has to store twice as much charge in the
same time interval. Therefore, the peak charging current must be twice as
great. Since XC = VC/IC, the capacitive reactance remains unchanged when
both VC and IC are doubled. The same logic applies for any change in the
applied voltage.
The magnitude of the applied alternating voltage has no effect on
the capacitive reactance of an AC circuit.
If we double the frequency of the applied voltage and leave the peak value
the same, the capacitor must store the same charge but in half the time.
Therefore, the peak charging current must be twice as great. Consequently,
IC is doubled. Since VC is unchanged and XC = VC/IC, the capacitive reactance is reduced by half. We can generalize this result:
19-6 Factors Governing Capacitive Reactance
Capacitive reactance is inversely proportional to frequency.
Finally, we double the capacitance and leave the source unchanged. Since
Q = CV, the capacitor must store twice as much charge at the peak of the
voltage sine wave. Since the frequency is unchanged, the capacitor must
build up the doubled charge in the same length of time. Therefore, the
peak charging current must be twice as great. Again, IC is doubled and VC
is unchanged, so the capacitive reactance is reduced by half. We can ­extend
this reasoning to any capacitance.
Capacitive reactance is inversely proportional to the capacitance.
We can derive a formula for reactive inductance by considering the average rate of change of current, as we did for inductive reactance. First we
rearrange Equation 19-6.
dv
i
=
dt
C
(19-9)
We can substitute the average rate of voltage rise for dv/dt and an average
value for i. During the first quarter of a cycle the voltage across the capacitor rises from zero to Em. The duration of a complete cycle is 1/f. Therefore,
average
dv
Δv
Em
=
= 1 1 = 4f Em
dt
Δt
4 × f
Equation 18-16 gives the half-cycle average for i. Since the first half of a sine
wave is symmetrical, the half-cycle average is equal to the quarter-cycle
­average, Iav = 2Im/π. Substituting for both sides of Equation 19-9 gives
4f Em =
Im × 2/π
C
Em
1
=
Im
2πfC
and
XC =
1
2πf C
(19-10)
where XC is capacitive reactance in ohms, f is frequency in hertz, and
C is capacitance in farads.
Since ω = 2πf,
XC =
1
ωC
Appendix 2-8 gives a calculus derivation for this equation.
See Problems 19-13 to 19-19 and Review Question 19-31.
(19-11)
561
562
Chapter 19 Reactance
Circuit Check
B
CC 19-4. What is the capacitive reactance of a 470-pF capacitor at a
frequency of 455 kHz?
CC 19-5. What value of capacitance will cause a 25-mA current to flow
when connected to a 22-V 5-kHz source?
19-7 Resistance, Inductive Reactance,
and Capacitive Reactance
Resistance, inductive reactance, and capacitive reactance are all forms of
opposition to alternating current. Each results from one of the basic properties of an electric circuit, and each can be defined in terms of a V/I ratio.
Although the instantaneous currents vary sinusoidally, the opposition to
current in a given circuit at a given frequency is independent of the elapsed
time. Therefore, we represent resistance, inductive reactance, and capacitive reactance with uppercase letters.
There are important differences in the behaviour of resistance, inductance, and capacitance in AC circuits:
•Resistance converts electric energy into heat, but inductance and capacitance alternately store energy during a charging quarter-cycle and return
this energy to the circuit during the next (discharging) quarter-cycle.
•
Resistance is independent of frequency, but inductive reactance is
­directly proportional to frequency and capacitive reactance is inversely
proportional to frequency.
•The current through a resistor is in phase with the voltage across it; the
current through an ideal inductor lags behind the voltage across it by 90°,
or π/2 radians; and the alternating current “through” a capacitor leads
the voltage across it by 90°.
These characteristics are summarized in Table 19-1.
TABLE 19-1 R, L, and C in AC circuits
Resistance
Inductive Reactance
Capacitive Reactance
R=
XL =
XC =
VR
IR
VL
= ωL
IL
VC
1
=
IC
ωC
IR is in phase with VR
IL lags VL by 90º
IC leads VC by 90º
19-7 Resistance, Inductive Reactance, and Capacitive Reactance
It is important to identify clearly the type of opposition when we discuss
circuits. In addition to using a different letter symbol, we sometimes
­indicate the type of opposition after the units, as shown in the following
examples.
AC
Example 19-1
Find the resistance of a 660-W toaster operated from a 110-V 60-Hz
source.
Solution
R=
( 110 V ) 2
E2
=
= 18.3 Ω(resistive)
P
660 W
Example 19-2
What inductive reactance develops a 40.0-V induced voltage when the
current through it has an RMS value of 80.0 mA?
Solution
XL =
VL
40 V
=
= 0.50 kΩ ( inductive )
IL
80 mA
Example 19-3
Calculate the reactance of a 68-pF capacitor when it is used in a
4.5-MHz circuit.
Solution
XC =
1
1
=
= 0.52 kΩ ( capacitive )
2πf C
2 × π × 4.5 MHz × 68 pF
Rather than consider DC and AC circuit analysis techniques separately,
some textbooks treat steady-state direct current as a special zero-frequency
case of alternating current. At zero frequency, ω = 0. Hence, XL = ωL = 0
and XC = 1/ω C = ∞. These values show that the steady-state DC response
of inductance is a short circuit and the steady-state DC response of capacitance is an open circuit.
See Problems 19-20 and 19-21 and Review Questions 19-32 to 19-34.
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564
Chapter 19 Reactance
Summary
• The current through an ideal inductor with a sine wave of voltage across
it is a sine wave that lags behind the voltage by π/2 radians.
• Inductive reactance is the ratio of the AC voltage across an ideal inductor
to the AC current through it.
• Inductive reactance is proportional to both frequency and inductance.
• The current through a capacitor with a sine wave of voltage across it is a
sine wave that leads the voltage by π/2 radians.
• Capacitive reactance is the ratio of the AC voltage across a capacitor to
the AC current through it.
• Capacitive reactance is inversely proportional to both frequency and
­capacitance.
• Resistance, inductance, and capacitance are the three basic properties
that produce opposition to alternating currents.
B = beginner
I = intermediate
A = advanced
Problems
B
B
B
B
B
B
B
B
B
B
B
Section 19-2
19-1.
19-2.
Inductive Reactance
Find the inductive reactance of an ideal inductor connected across
a 220-V AC source when the current flowing through it is 5.0 A.
What value of inductive reactance limits the current through it to
24 mA when the voltage across it is 35 V RMS?
Section 19-3
Factors Governing Inductive Reactance
19-3.
19-4.
19-5.
Calculate the 60-Hz reactance of a 4.0-H choke.
Find the reactance of a 250-mH choke at a frequency of 4.5 MHz.
At what frequency does a 0.5-H inductor have a reactance of
2.0 kΩ?
19-6. At what frequency will a 100-µH inductor have a reactive inductance of 1.5 kΩ?
19-7. What inductance has a reactance of 1400 Ω at 475 kHz?
19-8. What inductance draws a 160-mA RMS current when connected to
a 110-V 25-Hz source?
19-9. The solenoid of a contactor for starting an industrial motor draws
2.0 A from a 220-V 60-Hz source. Assuming negligible resistance,
find the inductance of the solenoid.
19-10. Find the current through a 30-mH inductor connected across a 50-V
5.0-kHz source.
Section 19-5
Capacitive Reactance
19-11. What capacitive reactance draws a 3.0-mA alternating current from
a 50-V 400-Hz source?
Review Questions
B
B
B
B
B
B
B
B
I
I
19-12. What is the current flowing through a capacitor that has a capacitive reactance of 250 Ω when connected to a 25-V RMS source?
Section 19-6
Factors Governing Capacitive Reactance
19-13. Find the reactance of a 0.05-μF coupling capacitor at 400 Hz.
19-14. Find the reactance of a 27-pF capacitor at a frequency of 88 MHz.
19-15. At what frequency does an 8.0-μF capacitor have a reactance
of 160 Ω?
19-16. At what frequency does a 2.0-nF capacitor have a reactance
of 80 Ω?
19-17. What capacitance has a reactance of 47 Ω at 500 Hz?
19-18. The input resistance of a transistor audio amplifier is 47 kΩ. What
value of coupling capacitor should be used if its reactance is to
equal the input resistance at 20 Hz?
19-19. A 9.7-kV 60-Hz transmission line has a capacitance connected across
it drawing 50 A of current. What is the value of the capacitance?
Section 19-7 Resistance, Inductive Reactance, and
Capacitive Reactance
19-20. Find the capacitor that has a reactance at 455 kHz equal to that of a
470-µF inductor.
19-21. At what frequency does the reactance of a 560-pF capacitor equal
that of a 470-µH inductor?
Review Questions
Section 19-1
Instantaneous Current in an Ideal Inductor
Section 19-2
Inductive Reactance
19-22. Why must the current sine wave be exactly π/2 rad out of phase
with the applied voltage waveform across an ideal inductor?
19-23. Referring to the definition of inductance in Chapter 16, explain
why the current must lag behind the voltage by π/2 radians rather
than lead it.
19-24. What is the advantage of expressing the phase difference between
the voltage across and the current through an ideal inductor as
π/2 rad rather than 90°?
19-25. Why is it convenient to label the horizontal axis of the graph in
­Figure 19-2 in radians?
19-26. Why does the ammeter in Figure 19-1 show a constant reading?
19-27. Why is it possible to express inductive reactance in ohms?
Section 19-3
Factors Governing Inductive Reactance
19-28. What is the meaning of the symbol ω in the formula XL = ω L?
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566
Chapter 19 Reactance
Section 19-4
Instantaneous Current in a Capacitor
Section 19-6
Factors Governing Capacitive Reactance
19-29. With reference to the definition of capacitance in Chapter 12, explain why the current leads the voltage across the capacitor by π/2
radians rather than lagging behind it.
19-30. Explain how the polarity of the voltage across a capacitor can
change without any change in the direction of the current charging
the capacitor.
19-31. Explain why an increase in frequency increases the reactance of an
inductor but decreases the reactance of a capacitor.
Section 19-7 Resistance, Inductive Reactance, and
Capacitive Reactance
19-32. Why does the term inductive reactance not apply to DC circuits?
19-33. When we connect the terminals of a capacitor to an ohmmeter,
the steady-state reading shows an infinitely high resistance.
Nevertheless, the ammeter in Figure 19-3 indicates a constant current in the circuit. Explain.
19-34. A “black box” with a pair of terminals marked “1000 Ω ” contains
a single electrical component. Suggest a laboratory procedure for
­determining whether the component is a resistor, a capacitor, or an
inductor.
Integrate the Concepts
An AC voltage source with a 20-V peak-to-peak value at a frequency of
200 Hz is connected across a capacitor with a value of 10 µF.
(a) Determine the value of the capacitive reactance.
(b) Calculate the peak value of the current.
(c) Write an equation for the current.
(d) What is the effect of tripling the frequency of the circuit?
(e)How does reducing the capacitance to 1 µF affect the circuit?
The same voltage source is connected across a 15-mH inductor instead of
the capacitor.
(f) Determine the inductive reactance of the circuit.
(g) Calculate the peak value of the current.
(h) Write an equation for the current.
(i) What effect does doubling the frequency have on the circuit?
(j)If the inductance in the circuit is doubled, what effect it will have on
the circuit?
Practice Quiz
Practice Quiz
1.
For the circuit of Figure 19-5, the instantaneous current of the 10-mH
coil is given by
(a) 11.7 sin 314t kA
(b) 23.4 sin 314 kA
(c) 11.7 sin(314t − 90°) kA
(d) 23.4 sin(314t − 90°) kA
V
117 Vpk
50Hz
0°
+
L
10 mH
−
Figure 19-5
2.
Inductive reactance is
(a) directly proportional to the frequency
(b) inversely proportional to the frequency
(c) independent of the frequency
(d) none of the above
3.
Which of the following equations describes inductive reactance?
1
(a)XL = 2πL
(b)XL = 2πL
(c)XL = 2πf1 L
(d)XL = 2πfL
4.
5.
If the inductor in Figure 19-5 were 30 mH, what would be the inductive reactance?
(a) 3.77 Ω
(b) 5.65 Ω
(c) 9.42 Ω
(d) 4.71 Ω
If the frequency is doubled, the inductive reactance of an ideal inductor will
(a) decrease by half
(b) stay the same
(c)double
(d)quadruple
567
568
Chapter 19 Reactance
6.
The instantaneous current flowing through the 82-nF capacitor in
Figure 19-6 is given by
(a) −3.01 sin 314t mA
(b) 3.01 sin 314t mA
(c) 3.01 sin (314t – 90°) mA
(d) 3.01 sin (314t + 90°) mA
XMM1
+
V
117 Vpk
50Hz
0°
+
−
C
82 nF
−
Figure 19-6
7.
Capacitive reactance is
(a) directly proportional to the frequency
(b) inversely proportional to the frequency
(c) independent of the frequency
(d) none of the above
8.
Which of the following equations describes capacitive reactance?
(a)Xc = 21f c
(b)Xc = 2fC
(c)Xc = 2π1f c
(d)XL = 2πfC
9.
The reactance of the 82-nF capacitor shown in Figure 19-6 is
(a) 32.35 Ω
(b) 38.32 kΩ
(c) 38.32 Ω
(d) 32.35 kΩ
10. If the frequency is doubled, the capacitive reactance of a capacitor will
(a) decrease by half
(b) stay the same
(c)double
(d)quadruple
Practice Quiz
11. Resistance is
(a) directly proportional to the frequency
(b) inversely proportional to the frequency
(c) independent of the frequency
(d) none of the above
12. Current flowing through a resistor is
(a) 90° out of phase with the voltage across it
(b) −90° out of phase with the voltage across it
(c) in phase with the voltage across it
(d) 180° out of phase with the voltage across it
13. Current flowing through a capacitor
(a) leads the voltage across it by 90°
(b) lags the voltage across it by 90°
(c) is in phase with the voltage across it
(d) is 180° out of phase with the voltage across it
14. Current flowing through an ideal inductor
(a) leads the voltage across it by 90°
(b) lags the voltage across it by 90°
(c) is in phase with the voltage across it
(d) is 180° out of phase with the voltage across it
569
20
Phasors
Our next step is to analyze AC circuits containing both resistance
and reactance. Since this analysis involves computations with
phasors, we need a type of algebra that can handle numbers in
two dimensions. This chapter shows how we can use complex
numbers to represent phasors and carry out circuit calculations
involving phasor quantities. However, some aspects of the algebra
of complex numbers are sig­nificantly different from the ordinary
algebra we used to ­analyze DC circuits.
Chapter Outline
20-1
Addition of Sine Waves 572
20-2 Addition of Instantaneous Values 573
20-3Representing a Sine Wave by a Phasor Diagram 575
20-4 Letter Symbols for Phasor Quantities
576
20-5 Phasor Addition by Geometrical Construction 576
20-6 Addition of Perpendicular Phasors 578
20-7 Expressing Phasors with Complex Numbers
581
20-8 Phasor Addition using Rectangular Coordinates 585
20-9 Subtraction of Phasor Quantities 587
20-10Multiplication and Division of Phasor Quantities 589
Key Terms
time domain 572
frequency domain 573
phasor diagram 575
scalar 576
polar coordinates 581
rectangular coordinates 581
real component 581
quadrature component 581
imaginary component 581
complex plane 581
complex number 582
complex conjugate 590
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• graph the resultant of two sine waves having
the same frequency
• represent a sine wave by a phasor in polar
coordinates
• represent a sine wave by a phasor in
rectangular coordinates
• add phasors with the aid of a phasor diagram
Photo sources: iStock.com/wragg
• add phasors by adding their rectangular
coordinates
• subtract phasors with the aid of a phasor diagram
• subtract phasors by subtracting their
rectangular coordinates
• multiply and divide phasors expressed in
polar coordinates
Chapter 20 Phasors
20-1 Addition of Sine Waves
R
VR
Resistor Voltage
Figure 20-1 shows a simple AC series circuit containing resistance and
­inductance. The sine-wave voltage source causes a sine wave of current
to flow in the circuit. Since all the components are connected in series, the
current in the inductance and the current in the resistance must have the
same magnitude at every instant. In fact, the sine-wave currents in the inductance and the resistance are one and the same, so we can represent both
with the letter symbol i.
+
0
Time
−
e
L
VL
Inductor Voltage
572
+
0
Time
−
Figure 20-1
AC circuit with resistance and inductance in series
The sine wave of current through the resistance causes a sine-wave voltage drop across it that is exactly in phase with the current sine wave. As
shown in Chapter 19, the instantaneous current through inductance lags
­behind the instantaneous voltage applied to it by π/2 radians. Consequently,
the sine-wave voltage across the inductance and the sine-wave voltage
across the resistance are π/2 radians out of phase, and the instantaneous
voltage across the inductance reaches its positive peak π/2 radians earlier
than the instantaneous voltage across the resistance.
Kirchhoff’s voltage law states that the sum of the voltage drops in a series
circuit must equal the applied voltage. In DC circuits where the source voltage is independent of time, we can find the sum of the voltage drops by
simple arithmetic. Even when dealing with the charging of a capacitor in
Chapter 13 and the rise of current in an inductor in Chapter 17, we were
able to find the sum of the instantaneous voltage drops by simple arithmetic
addition.
Now we need to find the sum of two out-of-phase sine-wave voltages.
As described in Chapter 18, we can represent a sine-wave voltage either by
its instantaneous value, v = Vm sin ωt, or by its RMS value, V = 0.707 Vm.
Since the instantaneous value varies continuously and depends on the
elapsed time, it is said to belong to the time domain. In the time domain,
20-2 Addition of Instantaneous Values
we express the angular distance travelled by a sine wave in radians. We can
use ordinary algebra for calculations with instantaneous values.
Since the RMS value of a sine wave does not vary with the exact instant in
time, we treat a sine-wave alternating voltage as a phasor quantity having
a magnitude and a phase angle. Such quantities are said to belong to the
frequency domain. Calculations with phasors require a specialized form
of algebra, which is described later in the chapter.
See Review Question 20-23 at the end of the chapter.
20-2 Addition of Instantaneous Values
Since Kirchhoff’s voltage law applies to the circuit in Figure 20-1 at every ­instant,
e = vT = vR + vL
(20-1)
Instantaneous Voltage
Therefore, we can add the instantaneous values of sine waves by simple
arithmetic addition. Although the sum of the instantaneous voltages at one
­instant does not give us the overall picture of the addition of complete sine
waves, we can repeat the calculation at intervals over a complete cycle to determine the nature of the resultant wave. To help with this time-consuming
task, we can draw the sine curves for the two sets of instantaneous values
on the same graph, as in Figure 20-2. Since angular velocity is constant for a
given frequency, we can show the phase angle of the source, ωt, instead of
time on the horizontal axis.
vT
+
Em
vR
vL
π
3π
2
0
π
2
2π
ωt
−
Figure 20-2
Addition of two out-of-phase sine waves
At t = 0, the instantaneous voltage across the resistance is zero; therefore, the instantaneous value of the total voltage is the same as the voltage
across the inductance at that instant. As time elapses, the instantaneous
voltage across the inductance is decreasing and the instantaneous voltage across the resistance is increasing. At approximately 1 rad, their sum
reaches its peak value. At π/2 rad, the instantaneous voltage across the
­inductance is zero, and the instantaneous value of the total voltage equals
573
Chapter 20 Phasors
the ­instantaneous voltage across the resistance. Approximately 1 rad
later, the instantaneous values of the voltages across the resistance and
the inductance are equal in magnitude but opposite in polarity, so vT = 0.
Careful examination of Figure 20-2 shows that the waveform for vT is a
sine wave with the same period as vR and vL. In fact, any sum of sine waves
with the same frequency has this property.
The sum of sine waves of the same frequency is also a sine wave of
the same frequency.
Equation 20-1 can be expanded to
e = VmR sin ωt + VmL sin ( ωt + π/2 )
(20-2)
Since vR and vL do not reach their peak values at the same instant, the peak
value of vT occurs at a time when the instantaneous values of the two component voltages are less than their peak values. Therefore,
Em ≠ VmR + VmL
Since the RMS value of any sine wave equals the peak value divided by √ 2,
E ≠ VR + VL
Therefore, arithmetic addition does not apply to peak and RMS values if the
sine waves are out of phase, as in Figure 20-2. However, if the sine waves
are exactly in phase, as in Figure 20-3, the resultant and component sine
waves do reach their peak values at the same instant, and
Em = Vm1 + Vm2 and E = VT = V1 + V2
Thus, the RMS value of the sum of in-phase sine waves is simply the sum of
the RMS values of the individual sine waves.
+
v1
e
v2
Instantaneous Value
574
e
v1
v2
0
π
2π
ωt
−
Figure 20-3
Addition of instantaneous values of two in-phase sine waves
See Problems 20-1 to 20-3 and Review Questions 20-24 to 20-29.
20-3 Representing a Sine Wave by a Phasor Diagram
20-3 Representing a Sine Wave
by a Phasor Diagram
Once we accept that the addition of sine waves of the same frequency results in a sine wave of the same frequency, the tedious procedure required
to produce Figures 20-2 and 20-3 is not warranted. The original information from which we constructed those graphs was simply their RMS values
and the phase angle between them. RMS value and phase angle will also
serve to identify the resultant.
In Figure 18-3, the length of the motion phasor represents the distance
from the conductor to its centre of rotation, and the position of the phasor
indicates the angular distance the loop has travelled. If we draw phasors
for the two sine-wave voltages in the circuit of Figure 20-1, the rotating
phasor for the instantaneous voltage across the inductance will always stay
­exactly π/2 radians ahead of the rotating phasor for the instantaneous voltage across the resistance. Consequently, these voltage phasors do not move
in relation to each other.
Sine waves of the same frequency can be represented by “stationary”
phasors with the angle between the phasors representing the phase
angle between the sine waves.
When drawing phasors for electrical quantities, we make the length of
the phasors proportional to the RMS value of the sine wave they r­ epresent.
These AC phasors belong to the frequency domain, where angles are ­usually
expressed in degrees.
In the simple series circuit of Figure 20-1, the current is common to
all components. Therefore, in the phasor diagram for a series AC circuit,
such as in Figure 20-4, we draw a phasor representing the RMS value of
the ­current along the reference axis of the diagram (pointing toward three
­o’clock). Since the voltage across the resistance is in phase with the current
through it, the phasor VR points in the same direction as the current phasor.
Since the voltage across the inductance leads the current through it by 90°,
we orient VL 90° counterclockwise from the reference axis.
VL
VR
Figure 20-4
I
Reference
phasor
Phasor diagram for the circuit of Figure 20-1
The phasor diagram of Figure 20-4 is much easier to draw than the instantaneous voltage graphs of Figure 20-2, and it shows the RMS values and
phase angles at a glance.
See Review Questions 20-30 to 20-32.
575
576
Chapter 20 Phasors
20-4 Letter Symbols
for Phasor Quantities
A mathematical
­vector can have
any number of
­dimensions.
Technical journals,
­including IEEE
­publications, usually
represent a phasor
quantity with a
­lightface italic letter
(like a scalar), with
the magnitude
­indicated by vertical
bars, as in V = |V|∠ϕ .
In studying the behaviour of electric and magnetic circuits, we encounter
three types of quantities.
Scalars are quantities that have magnitude but no direction or angle.
­Resistance, inductance, and capacitance are scalars. Voltage and current in
DC circuits are scalars, as are instantaneous values of voltage and current in
AC circuits. Scalars are represented by lightface italic letter symbols (such
as R, L, C, V, I, v, and i). Scalar quantities can be added by simple algebraic
addition.
Vectors are quantities that have both magnitude and direction. Electrostatic
force, electric field intensity, and magnetic flux density are all vector quantities. This book uses boldface italic letter symbols (such as F, E, and B) to
represent vector quantities, and the same letter symbols in lightface italic to
represent magnitudes of vectors. For example, the magnitude of the vector
F is the scalar quantity F. Most electric vector quantities have three dimensions. However, if all the vectors for a particular calculation lie in the same
plane, we can treat them as being two-dimensional. Because vectors are
multidimensional, most operations in vector algebra, such as addition and
multiplication, differ substantially from the simple arithmetic operations
for scalars.
Phasors are quantities that have both magnitude and direction. Phasor
notation is a convenient method for representing sine waves in AC circuits.
For example, a sine-wave voltage can be represented as V = V∠ϕ, where V
is the RMS magnitude and ϕ is the phase angle. We use bold Roman letters
for phasors to distinguish them from vectors, shown by bold italic letters.
As with vectors, the same letter in lightface italic indicates the magnitude
of a phasor.
Using phasor notation we can rewrite Equation 20-2 as
ET = VR + VL
(20-3)
See Review Question 20-33.
20-5 Phasor Addition
by Geometrical Construction
In a phasor diagram, the resultant of two phasor quantities is the diagonal
from the origin to the opposite corner of the parallelogram with the two
phasors as sides.
The resultant can be constructed with ruler and compass, as shown
in Figure 20-5. Using a radius equal to the length of E1, we trace an arc
20-5 Phasor Addition by Geometrical Construction
with the tip of the E2 phasor as the centre. Next, using a radius equal
to the length of E2, we trace an arc with the tip of E1 as the centre. We
then draw the resultant by joining the origin to the intersection of the
two arcs.
E2
ET
ϕ
E1
Figure 20-5
Reference
axis
Geometrical construction of a resultant phasor
Applying this technique to the series circuit of Figure 20-1 shows how
the magnitudes and phase angles of E, I, VR, and VL are related. Figure 20-6
compares the sine-wave graph and the phasor diagram for this series circuit.
Although we can determine RMS values and phase angles from the sinewave graphs in Figure 20-6(a), it is much easier to get this information from
the phasor diagram in Figure 20-6(b).
vT
Instantaneous Voltage
+
vR
vL
π
3π
2
0
π
2
VL
2π
ωt
VT = E
VR
I
Reference
phasor
−
(a)
(b)
Figure 20-6 Sine-wave graph and phasor diagram for the circuit of Figure 20-1
If we draw the phasors carefully, we can measure the RMS voltages and the
phase angle between the applied voltage E and the current I to ­two-figure
accuracy. Although graphical solutions are not accurate enough for many
of the AC circuit problems in later chapters, we can use approximate values
from a phasor diagram to check the calculated answers. Phasor diagrams
also help us to get the correct sign for the sine, cosine, or tangent of angles
larger than 90°.
577
578
Chapter 20 Phasors
Example 20-1
Three AC voltage sources with the same frequency are connected in
­series. Find the total applied voltage given that the source voltages
are
E1 = 80 V ∠0º , E2 = 100 V ∠150º , and E3 = 40 V ∠45º
Solution
Step 1
Construct a phasor for each voltage.
Step 2
Construct a resultant phasor for any pair of phasors. In Figure 20-7 it is
convenient to start with E1 and E3.
Step 3
Construct the resultant for the third source phasor and the resultant of
Step 2.
Step 4
Measure the length and phase angle of the final resultant. These
measurements show that the total applied voltage is approximately 79 V ∠80°.
ET
E2
E3
E1 + E3
E1
Reference
axis
Figure 20-7 Phasor diagram for Example 20-1
See Problems 20-4 to 20-8.
20-6 Addition of Perpendicular Phasors
In Section 20-2, we saw that the RMS values of sine waves that are in phase
can be added arithmetically. Therefore, phasors that have the same phase
angle can be simply added together. Similarly, when the angle between
a pair of phasors is exactly 180°, the phasor sum is simply the difference
of the two phasors. For two phasors with an angle of exactly 90° between
them, we can use trigonometry to find the phasor sum.
20-6 Addition of Perpendicular Phasors
579
In the simple series circuit of Figure 20-1, the voltage drop across the
r­ esistance is exactly in phase with the current and the voltage across
the ­inductance leads the current by exactly 90°. Therefore, the two voltage ­phasors are exactly 90° out of phase. In the geometrical construction for their resultant in Figure 20-8(a), the parallelogram becomes a
rectangle with side AC exactly equal to magnitude of the VL. Therefore,
the triangle in Figure 20-8(b) shows exactly the same data as the conventional phasor diagram of Figure 20-8(a), which has all the phasors
starting from the origin. Such triangular diagrams are useful for analyzing AC circuits that have both resistance and reactance.
B
C
C
e
us
ten
o
p
VT
Hy
ϕ VR
VT
VL
O
ϕ
VR
A
I
O
Adjacent side
(a)
Figure 20-8
VL Opposite
side
A
(b)
Addition of phasors at right angles
Since triangle OAC is a right triangle,
tan ϕ =
opposite side
adjacent side
=
AC
VL
=
OA
VR
Given the RMS voltages across the inductance and the resistance, we can
find the phase angle ϕ using trigonometric tables or a calculator with a
tan–1 function.
ϕ = tan−1
VL
VR
(20-4)
where tan–1 means “the angle having a tangent of.”
We now have a choice of several methods for finding the magnitude of
the total voltage, ET. From trigonometry,
sin ϕ =
Therefore,
opposite side
hypotenuse
ET =
VL
sin ϕ
=
AC
OC
(20-5)
Arctan is another term
for tan–1.
580
Chapter 20 Phasors
cos ϕ =
Similarly,
Therefore,
adjacent side
hypotenuse
VT =
=
OA
OC
VR
cos ϕ
(20-6)
Another method uses the Pythagorean theorem, which states that the
square of the length of the hypotenuse of a right triangle is equal to the
sum of the squares of the lengths of the other two sides. In Figure 20-8(b),
OC2 = OA2 + AC2
VT = √ V2R + V2L
(20-7)
Note that it is customary to state the resistance component before the reactive component in Equation 20-7. Combining Equations 20-7 and 20-4 gives
ET = √V2R + V2L tan−1
VL
VR
(20-8)
Example 20-2
Given that the voltage drop across the resistance in the circuit of
Figure 20-1 is 15 V RMS and the voltage across the inductance is 10 V
RMS, find the magnitude of the total voltage and its phase angle with
respect to the current.
Solution
ϕ = tan− 1
ET =
or
or
ET =
VL
10
= tan− 1
= 33.7º
VR
15
VR
15
=
= 18 V
cos ϕ
0.832
VL
10
=
= 18 V
sin ϕ
0.555
ET = √ V2R + V2L = √ 225 + 100 = 18 V
ET = 18 V∠+ 33.7°
The positive angle shows that the voltage leads the current.
See Review Question 20-34.
20-7 Expressing Phasors with Complex Numbers
581
20-7 Expressing Phasors with
Complex Numbers
Quadrature or
imaginary axis
When we describe a phasor by stating its magnitude and phase angle with
respect to the reference axis, we are using polar coordinates. These coordinates specify the position of the tip of the phasor in terms of a distance along
a radius from the origin and an angle measured from the ­reference axis.
Polar coordinates correspond to the measurements we o
­ btain with instruments such as voltmeters, ammeters, and ­wattmeters.
By applying trigonometric relationships to the phasor diagram of
­Figure 20-8, we determined the polar coordinates of the total voltage from
two perpendicular components, one of which lies along the reference axis.
We can use the same method to find rectangular coordinates (or perpendicular components) for any phasor in the polar form E∠ϕ. As shown in
Figure 20-9, the reference axis component (or real component) is the magnitude of the phasor multiplied by the cosine of its phase. The quadrature component (or imaginary component) is the magnitude of the phasor
multiplied by the sine of its phase angle.
E
E
E sin ϕ
ϕ
E cos ϕ
Figure 20-9
Reference or
real axis
Determining the rectangular coordinates of a phasor
The rectangular coordinates of a phasor indicate distances along
the axes of the complex plane. Figure 20-10 shows how angles in polar
coordinates correspond to the four possible directions for coordinates in the
complex plane.
In the complex plane, multiplying a phasor by the j operator rotates it
90° counterclockwise. To rotate the phasor another 90°, we simply multiply once more by the j operator. Hence, multiplying a phasor by j2 changes its direction by 180°. In polar coordinates, reversing the direction of a
phasor is the same as multiplying by –1. Hence j2 = –1. Since j = √−1 has
no ­solution, numbers preceded by the +j or –j operator are called ­imaginary
numbers. Multiplying a phasor by j three times rotates it a total of 270°.
Since j3 = j2 × j and j2 = −1, the operator –j corresponds to a rotation of 270°.
Mathematicians use
i (from imaginary)
to represent √ −1.
However, j is used in
electrical calculations
to avoid confusion
with i, the symbol for
instantaneous current.
582
Chapter 20 Phasors
+j
90°
E
ϕ
0°
Reference
axis
180°
Reference
axis
−
270°
or −90°
(a)
+
−j
(b)
Figure 20-10 Corresponding directions in (a) polar coordinates
and (b) rectangular coordinates
A phasor can also be
expressed in terms of a
complex exponential:
E = E cos ϕ + jE sin ϕ
= Ee jϕ.
Although this
exponential form is
useful for advanced
AC theory, the
rectangular-coordinate
form is more
convenient for our
calculations.
The rectangular coordinates of a phasor consist of a real number (which
can be positive or negative) and an imaginary number identified by either
+j or –j. The real component is always listed before the imaginary component. The formula for converting a voltage phasor from polar coordinates to
rectangular coordinates is
E∠ϕ = +E cos ϕ + jE sin ϕ
(20-9)
This equation demonstrates how any sine wave quantity can be expressed in
terms of a complex number that has real and imaginary components.
Example 20-3
Express 18 V ∠33.7º in rectangular coordinates.
Solution
Step 1
Sketch a phasor diagram to find the correct operators to use with the
rectangular coordinates (see Figure 20-11).
+j
18 V
18 sin 33.7°
33.7°
18 cos 33.7°
Figure 20-11
+
Phasor diagram for Example 20-3
20-7 Expressing Phasors with Complex Numbers
Step 2
Use trigonometry to find the components of E:
E = +E cos ϕ + jE sin ϕ
= +18 cos 33.7° + j18 sin 33.7°
= 15 + j10 V
Most scientific calculators have a function for converting between
polar and rectangular coordinates. We enter the polar coordinates and
then use the conversion function to display the rectangular coordinates
without having to calculate cos ϕ and sin ϕ.
Example 20-4
State 4.0 A ∠240º in rectangular coordinates.
Solution
Step 1
Sketch a phasor diagram, as in Figure 20-12, to check the signs and
approximate values of the coordinates.
+j
240°
−
+
60°
4.0 A
−j
Figure 20-12
Phasor diagram for Example 20-4
Step 2
Use either the trigonometric or the coordinate-conversion functions of
a calculator. These functions automatically provide the correct signs for
the rectangular coordinates:
I = −2.0 − j3.5 A
583
584
Chapter 20 Phasors
Example 20-5
Convert 48 ∠−45º to rectangular coordinates.
Solution
A negative angle is simply one measured clockwise from the reference
axis. From Figure 20-13,
48 ∠−45º = 48 cos (−45º ) + j48 sin (−45º ) = + 33.9 − j33.9
Again a calculator will handle the negative angle automatically.
−45°
+
48
−j
Figure 20-13
Phasor diagram for Example 20-5
Equation 20-8 indicates a procedure for converting from rectangular
coordinates to polar coordinates. The magnitude of the polar quantity is
simply the square root of the sum of the squares of the rectangular coordinates, leaving out the j operator. We find the phase angle by drawing a
diagram of the perpendicular components. The angle between the phasor
and its reference axis component is
θ = tan−1
quadrature coordinate
reference axis coordinate
We then use the diagram to determine the actual phase angle with respect
to the reference axis.
Example 20-6
State −60 + j30 V in polar form.
Solution
If we use the trigonometry function keys of a calculator, we must pay
careful attention to the component diagram of Figure 20-14. Since the
calculator cannot distinguish between 30/−60 and −30/60, it displays
a value of −26.565° for angle θ. From Figure 20-14, it can be seen that
ϕ = 180° − 26.6° = +153.4°
20-8 Phasor Addition Using Rectangular Coordinates
V = √(− 60 ) 2 + 302 = 67.1 V
or
V=
30
30
=
= 67.1 V
sin 26.6º
0.995
V = 67 V∠+153.4º
Alternatively, we can use the coordinate-conversion function of a
calculator to find the polar coordinates in a single step.
+j
ϕ
+j30
−
θ
−60
Figure 20-14
Reference
+
axis
Phasor diagram for Example 20-6
See Problems 20-9 and 20-10 and Review Questions 20-35 to 20-41.
20-8 Phasor Addition Using
Rectangular Coordinates
Although there are trigonometric procedures for adding phasors having
­angles between them of other than 0° or 90°, it is usually much easier to
­express the phasors in rectangular coordinates and then find the sum of
the components. Since all the real components of the phasors are horizontal, we can simply add them algebraically. Similarly, the imaginary
­components are all vertical, so we can also add them algebraically. These
two sums are the real and imaginary components of the resultant phasor.
All that r­emains is to convert these rectangular coordinates into polar
form. This procedure is the conventional method of phasor addition for
AC circuit problems.
Example 20-7
Find the sum of E1 = 80 V ∠+60º and E2 = 80 V ∠−135º .
Solution
Step 1
Sketch a phasor diagram, as shown in Figure 20-15.
585
586
Chapter 20 Phasors
+j
E1
ET
60°
45°
−
+
−135°
E2
−j
Figure 20-15
Phasor diagram for Example 20-7
Step 2
Convert each voltage into its rectangular coordinates:
E1 = 80 cos 60°
+j80 sin 60°
= +40
+j69.28 V
E2 = 80 cos −135° +j80 sin −135° = −56.57 − j56.57 V
Step 3
Add the components:ET = −16.57 + j12.71 V
Step 4
Convert the resultant into polar coordinates:
12.71
θ = tan−1
= −37.49º
−16.57
From Figure 20-15,
ϕ = 180º − 37.49º = +142.51º
ET = √ 16.572 + 12.712 = 20.88 V
ET = 21 V ∠+143°
As in the preceding examples, we can save several steps by using the
­coordinate-conversion function of a calculator.
See Problems 20-11 to 20-16.
Circuit Check
A
CC 20-1. A voltage sine wave with a peak value of 300 V leads another
voltage sine wave of 200-V peak value by 30º. Calculate the
RMS resultant voltage of these voltage waves.
CC 20-2. Three AC currents in parallel branches of a circuit are given by
i1 = 2 sin 100t, i2 = 3 sin 100t +
(
π
3π
, and i3 = 4 sin 100t −
.
3
4
)
(
)
20-9 Subtraction of Phasor Quantities
(a) Construct a phasor diagram showing the currents.
(b)Use the diagram to determine the RMS value of the total ­current.
(c) Use phasor addition to calculate the RMS total current.
20-9 Subtraction of Phasor Quantities
When two phasors are expressed in rectangular coordinates, one can be subtracted from the other by simple algebraic subtraction of first the real (or reference axis) components and then the imaginary (or quadrature) components.
The resulting rectangular coordinates can then be converted into polar form.
Example 20-8
Subtract 4 − j5 from 3 + j6 and express the result in polar coordinates.
Solution
Step 1
Subtract the components:
3 + j6
4 − j5
−1 + j11
Subtracting gives
We can also subtract by reversing the + and − signs in the subtrahend
and then performing an algebraic addition:
3 + j6
−4 + j5
−1 + j11
Step 2
Sketch a phasor diagram to help determine the phase angle of the polar
form (see Figure 20-16). Then convert to polar coordinates:
−1 + j11 = √ 12 + 112 ∠180º − tan − 1 ( 11/1 )
= 11.05∠+95.2°
+j11
ϕ
−1
Figure 20-16
Reference
axis
Phasor diagram for Example 20-8
587
588
Chapter 20 Phasors
The second procedure in Step 1 of Example 20-8 suggests a
method for subtracting phasors on a phasor diagram. As shown in
Figure 20-17, +4 − j5 and −4 + j5 represent phasors that are exactly
equal in length but 180° apart in angular location. Therefore, we can
use geometrical construction to subtract two phasor quantities by
reversing the direction of the subtrahend phasor and then constructing
the parallelogram for the adding phasors.
+j5
+4
−4
−j5
Figure 20-17
Reversing the signs of rectangular coordinates
Example 20-9
Use geometrical construction to subtract E1 = 40 V ∠30º from E2 = 50 V
∠135º.
Solution
From Figure 20-18, E2 − E1 is about 73 V ∠120°.
90°
E2
E1
E2 − E1
180°
0°
−E1
270°
Figure 20-18
Subtraction of phasors by geometrical construction
See Problems 20-17 and 20-18 and Review Question 20-42.
20-10 Multiplication and Division of Phasor Quantities
20-10 Multiplication and
Division of Phasor Quantities
Multiplication and division of phasor quantities in polar form is the simplest of all the complex algebra processes.
To multiply quantities expressed in polar form, multiply their magnitudes and add their phase angles algebraically.
E1 × E2 = E1 ∠ϕ1 × E2 ∠ϕ2 = E1E2 ∠ϕ1 + ϕ2
(20-10)
To divide phasor quantities expressed in polar form, divide their
magnitudes and subtract their phase angles algebraically.
E1 ÷ E2 = E1∠ϕ1 ÷ E2 ∠ϕ2 =
E1
ϕ1 − ϕ2
E2 ∠
(20-11)
Example 20-10
Calculate 80∠40º × 11∠−15º and 80 ∠40º ÷ 11∠−15º.
Solution
80∠40º × 11∠−15º = 80 × 11∠40 + ( −15 ) = 880∠25°
80∠40º ÷ 11∠−15º =
80
40 − ( −15 ) = 7.27∠55°
11 ∠
Although the procedure for multiplying and dividing phasor quantities expressed in rectangular coordinates is straightforward, it does require
skill in manipulating algebraic terms. It is often safer to do phasor multiplication and division by the simpler polar method. With a calculator that has
coordinate-conversion functions, such conversions are the quickest way to
do phasor multiplication and division. Example 20-11 demonstrates the
procedure for multiplication and division in rectangular form.
589
590
Chapter 20 Phasors
Example 20-11
Calculate (8 + j6) × (2 − j4) and (8 + j6) ÷ (2 − j4).
8 + j6
Solution
Multiply by 2
2 − j4
16 + j12
Multiply by −j4 − j32 − j2 24
Collect like terms
16 − j20 − j2 24
Since the operator j represents a rotation of 90° from the reference
axis, j2 represents a rotation of 180° from the reference axis. Therefore,
j2 = −1, and
16 − j20 − j2 24 = 16 − j20 + 24 = 40 − j20
The procedure for division is more involved. First, we set up the
phasor quantities involved in fraction form. Then we need to eliminate any j term in the denominator. If we multiply both the numerator
and denominator by the same quantity, the value of the fraction does
not change. To eliminate the j term, we multiply the fraction by the
complex conjugate of the denominator. The complex conjugate of a
number has the same real component as the number but the opposite
imaginary component.
( 8 + j6 ) ÷ ( 2 − j4 ) =
=
=
8 + j6
2 − j4
( 8 + j6 ) × ( 2 + j4 )
( 2 − j4 ) × ( 2 + j4 )
−8 + j44
20
= −0.4 + j2.2
Note that the division is straightforward once the j term is cleared
from the denominator.
See Problems 20-19 to 20-22 and Review Question 20-43.
591
Problems
Summary
• The sum of two sine waves having the same frequency is also a sine
wave of the same frequency.
• The peak value of the sum of two sine waves having the same frequency is not the sum of their peak values unless the sine waves are in
phase.
• The polar coordinates of a phasor give its magnitude and phase angle.
• The rectangular coordinates of a phasor give the values of its real and
imaginary components.
• Phasors may be added or subtracted on a phasor diagram.
• Phasors may be added or subtracted when expressed in rectangular
­coordinates.
• Phasors are simpler to multiply or divide when they are expressed in
polar coordinates.
Problems
Solve Problems 20-1 to 20-3 by drawing sine curves on graph paper and adding
­instantaneous values to determine the resultants.
I
Section 20-2
20-1.
I
20-2.
I
20-3.
I
A 45-V sine wave leads a 125-V sine wave by 75°. Determine the
RMS voltage value of the resultant wave.
A series circuit made up of a capacitor with a reactance of a 100 Ω
and an inductor with a reactance of 75 Ω has a 3-A sine-wave current flowing through it. Find the RMS total voltage.
The currents in the two branches of a parallel circuit are i1 = 20 sin
314t and i2 = 50 sin (314t − π/4). Determine the total current.
Section 20-5
20-4.
I
20-5.
I
20-6.
I
20-7.
I
20-8.
Addition of Instantaneous Values
Phasor Addition by Geometrical Construction
Use phasor diagrams to solve Problems 20-1 and 20-2 by geometrical construction.
Three sources of alternating voltage each generate 120 V RMS, but
E2 leads E1 by 120° and E3 lags behind E1 by 120°. Use a phasor
diagram to determine the total voltage when all three sources are
connected in series.
Use a phasor diagram to determine the total voltage in Problem 20-5
when the leads to source E1 are reversed.
Determine the sum of the following phasors in degrees using geometrical construction: 100∠+50º , 35∠+90º , 50∠+p/5 rad, 20∠−4p
An induction motor draws a 10-A current from a 120-V 60-Hz
source. This current lags behind the applied voltage by 30°.
B = beginner
I = intermediate
A = advanced
592
Chapter 20 Phasors
A capacitor in parallel with the motor draws a 5-A current from the
source. Draw a schematic diagram, and use a phasor diagram and geometrical construction to find the total current drawn from the source.
B
B
Section 20-7
Expressing Phasors with Complex Numbers
Convert the following polar coordinates to rectangular:
(a) 120∠−25º
(b) 45∠180º
(c) −35∠90º
(d) 130∠3 rad
(e) 7.5∠π/7 rad
(f) 2∠−45º
20-10. Convert the following to polar coordinates:
(a) 3 + j4
(b) −14 − j10
(c) 12.7 + j0
(d) −0.8 + j1.2
(e) 0 − j18
(f) 0.67 − j0.43
20-9.
Give answers for the following problems in polar coordinates.
B
B
B
B
B
B
B
B
B
B
Section 20-8
20-11.
20-12.
20-13.
20-14.
20-15.
20-16.
Phasor Additon Using Rectangular Coordinates
Add 13 − j7 and 11 + j18.
Add 1.4 + j0.8, −0.9 + j2.1, and 0 − j3.9.
Add 170∠200º and 88∠−75º.
Add 1.8∠125º, 2.7∠−157º, and 1.3∠−66º.
Add 40 + j72 and 60∠100º.
Add 120 + j0, 120∠−120º , and −60 + j104.
Section 20-9
Subtraction of Phasor Quantities
20-17. Subtract 25 − j17 from 35 + j50 and express the result in polar
coordinates.
20-18. Subtract −15 + j10 from 14 − j7 and express the result in polar
­coordinates.
Section 20-10
Multiplication and Division of Phasor Quantities
20-19. Calculate
(a) 4.1∠−64º × 13∠13º
(b) 4.1∠−64º ÷ 13∠13º
20-20. Calculate
(a) (3 + j4) × (5 − j6)
(b) (3 + j4) ÷ (5 − j6)
Review Questions
I
I
20-21. Determine the equivalent impedance of a parallel circuit formed by
Z 1 = 37 Ω ∠45º and Z2 = 45 Ω ∠−37º.
20-22. Calculate
( 47 + j13 ) ( −21 − j32 )
51∠−111º + 19∠+70º
Review Questions
Section 20-1
Addition of Sine Waves
20-23. Why is a knowledge of complex algebra required in solving AC circuit problems?
Section 20-2
Addition of Instantaneous Values
20-24. RMS values in an AC circuit are defined as DC equivalent values.
Why then can we not add RMS values by simple algebra as we can
in a DC circuit?
20-25. Why is it possible to add instantaneous values in an AC circuit by
simple algebra?
20-26. When writing an equation for the instantaneous value of the sum
of two sine waves, how is the phase difference between them
indicated?
20-27. Is it possible to add the instantaneous values of two sine waves if
their frequencies are different? Explain.
20-28. What does the equation IT = I1 + I2 tell us about the current sine waves?
20-29. What is meant by simple algebraic addition?
Section 20-3
Representing a Sine Wave by a Phasor Diagram
20-30. What advantage does a phasor diagram have over a linear graph of
instantaneous values plotted against time?
20-31. What information does a linear graph show that is not shown by a
phasor diagram?
20-32. What condition is necessary for adding RMS quantities in an AC
­circuit?
Section 20-4
Letter Symbols for Phasor Quantities
20-33. What notation is used to distinguish scalar, vector, and phasor ­quantities?
Section 20-6
Addition of Perpendicular Phasors
20-34. When can we use the Pythagorean theorem to add RMS values?
Section 20-7
20-35.
20-36.
20-37.
20-38.
Expressing Phasors with Complex Numbers
Show with a phasor diagram that sin 210° = −sin 30°.
Show with a phasor diagram that cos 135° = −cos 45°.
What is meant by polar coordinates?
What is meant by rectangular coordinates?
593
Chapter 20 Phasors
20-39. Describe how to convert the polar coordinates of a phasor into
­rectangular coordinates.
20-40. What is the significance of the operator j when one is dealing with
phasor quantities?
20-41. What is the advantage of expressing phasor quantities in rectangular coordinates?
Section 20-9
Subtraction of Phasor Quantities
20-42. Explain the procedure for subtracting phasor quantities by geometrical construction on a phasor diagram.
Section 20-10 Multiplication and Division of Phasor Quantities
20-43. Outline two methods for finding the product of phasors that are
­expressed in rectangular coordinates.
Integrate the Concepts
For the circuit of Figure 20-19:
(a) Draw reasonably accurate graphs of e1 and e2.
(b) Draw a reasonably accurate graph of the sum of e1 and e2.
(c) Express e1 and e2 as phasors in polar coordinates.
(d) Add e1 and e2 by using a phasor diagram.
(e) Express e1 and e2 as complex quantities.
(f) Use rectangular coordinates to determine the resultant of e1 + e2.
(g) Determine e1 − e2.
(h) Calculate e1 × e2.
(i) Calculate e1/e2.
V1
100 V peak
60 Hz
30°
Figure 20-19
+
−
−
V2
+
594
175 V peak
60 Hz
150°
230 Ω
Practice Quiz
Practice Quiz
1.
The sum of two sine waves with the same frequency is
(a) a sine wave that has double the frequency
(b) a sine wave that has the same frequency
(c) a sine wave that has half the frequency
(d) not a sine wave
2.
For the circuit of Figure 20-20, the voltage across the resistor can be
expressed as
(a) 153.5 sin (314t − 92.2°)
(b) 180.9 sin (377t + 57.9°)
(c) 180.9 sin (314t − 57.9°)
(d) 153.3 sin (314t − 92.2°)
V2
145 V peak
60 Hz
0°
−
+
+
−
V1
120 V peak
60 Hz
0°
2.2 kΩ
Figure 20-20
3.
4.
5.
The polar form for 10 + j8 is
(a)12.81∠38.66º
(b)12.81∠51.34º
(c)12.81∠−38.66º
(d)12.81∠−51.34º
The rectangular form for 25∠−60º is
(a)12.5 − j21.65
(b)−12.5 − j21.65
(c)12.5 + j21.65
(d)−12.5 + j21.65
The sum of 330∠75º and 15∠−90º is
(a)364.5∠75.6º
(b)335.5∠74.3º
(c)335.5∠−74.3º
(d)364.5∠−75.6º
595
596
Chapter 20 Phasors
6.
The sum of the phasors in Figure 20-21 is
(a)4.24∠45º
(b)3.16∠71.57º
(c)3.16∠18.43º
(d)4.24∠−45º
+j
2
1
−
−3
+
0
−2
1
−1
2
−1
−2
−3
−j
Figure 20-21
7.
8.
9.
The difference between 75∠270º and 90∠−25º is
(a)89.55∠24.4º
(b)139.4∠−54.2º
(c)139.4∠125.8º
(d)89.55∠−155.6º
Subtracting −5 + j10 from 3 − j19 gives
(a)−2 − j9
(b)8 − j29
(c)−2 + j9
(d)−8 + j29
Subtracting −5 + j10 from 10∠45º gives
(a)12.42∠13.64º
(b)17.2∠83.08º
(c)12.42∠−13.64º
(d)17.2∠−83.08º
10. The product of 100∠80º and 50∠90º is
(a)5000∠170º
(b)5∠170º
(c)5000∠−170º
(d)5∠−170º
Practice Quiz
11. The product of 375 + j15 Ω and 525 − j50 Ω in polar coordinates is:
(a)0.72 Ω ∠7.7º
(b)197.924 Ω ∠3.2º
(c)197.924 Ω ∠−3.2º
(d)0.72 Ω + ∠−7.7º
12. Calculate Z1/Z2 for Z1 = 100∠80º and Z2 = 50∠90º.
(a)−2∠170º
(b)2∠10º
(c)2∠− 10º
(d)2∠−170º
13. Determine Z1/Z2 for Z1 = 470 − j50 and Z2 = 325 + j45.
(a)1.4 − j0.35
(b)1.44 − j0.046
(c)150,500 + j37,400
(d)1.4 + j0.046
14. Determine Z1 + Z2 × Z3/(Z2 + Z3) for Z1 = 50 − j25 Ω, Z2 = 10 + j60 Ω
and Z3 = 220 − j45 Ω.
(a)99.83 Ω ∠−56.1º
(b)108.7 Ω ∠−46.5º
(c)74.9 Ω ∠−19.5º
(d)80.2 Ω ∠−21.1º
597
21
Impedance
Circuit analysis for DC circuits containing resistance along
with inductance or capacitance can be done with fairly
straightforward algebra because the calculations involve
only real numbers. In AC circuits containing inductance or
capacitance, however, the calculations involve complex numbers
(phasors). This chapter uses phasors to combine resistance
and reactance into a single quantity that represents the total
opposition to alternating current in a given circuit.
Chapter Outline
21-1
Resistance and Inductance in Series
21-3
Practical Inductors
21-5
Resistance, Inductance, and Capacitance in Series
21-2
21-4
21-6
Impedance
601
604
Resistance and Capacitance in Series
600
607
Resistance, Inductance, and Capacitance in Parallel
21-7
Conductance, Susceptance, and Admittance
21-9
Troubleshooting
21-8
Impedance and Admittance
619
615
613
608
610
Key Terms
impedance 601
impedance diagram
602
total-current method
admittance 614
611
susceptance 614
admittance diagram
615
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• draw the phasor diagram for a series AC ­circuit
containing resistance, inductance,
and capacitance
• calculate resistor, inductor, and capacitor
voltages for a series AC circuit containing
­resistance, inductance, and capacitance
• draw the impedance diagram for a series AC
circuit ­containing resistance, inductance, and
capacitance
• calculate the total impedance of a series
AC circuit
• calculate the total impedance of a practical
inductor
• calculate the net reactive voltage of a series
AC circuit
Photo sources: © David J. Green/Alamy Stock Photo
• calculate the net equivalent reactance of a
series AC circuit
• draw the phasor diagram for a parallel AC
circuit ­containing resistance, inductance, and
capacitance
• calculate resistor, inductor, and capacitor
­currents for a parallel AC circuit containing
resistance, inductance, and capacitance
• calculate the equivalent impedance of a
parallel AC circuit
• draw the admittance diagram for a parallel
AC circuit
• calculate the total admittance of a parallel
AC circuit
• detect faults in resistors, inductors, and
capacitors in AC circuits
600
Chapter 21 Impedance
21-1 Resistance and
Inductance in Series
In Chapter 20 we noted that all rules and laws that apply to DC circuits also
apply to AC circuits, although we keep in mind the sinusoidal nature of AC
voltage sources. A key rule for series circuits is the one by which we define
a series circuit:
The same current must flow in all components in a series circuit.
Because current is the only parameter common to all components in a ­series
circuit, we often use the current as the reference in phasor diagrams for
­series circuits. The phase angles of other quantities are then calculated with
respect to the common current, as shown in Figure 21-1.
VL
VL
E
E
VR
ϕ
VR
(a)
Reference
I
phasor
(b)
Figure 21-1 Phasor diagram of voltage and current relationships in a series
circuit containing resistance and inductance
A second characteristic of series circuits is that the sum of the various
voltage drops must equal the applied voltage. This voltage law applies to
the simple algebraic sum of the instantaneous values in an AC circuit. The
voltage law for series circuits also applies to RMS voltages provided we use
phasor addition:
In a series AC circuit, the phasor sum of the various voltage drops
must equal the applied voltage.
Since the voltage drop across the resistance must be in phase with the
­current, VR lies on the phasor reference axis in Figure 21-1. Similarly, the
voltage across the inductance leads the current by 90°, so VL lies on the +j
axis. Therefore,
E = VR + jVL
(21-1)
21-2 Impedance
From the resultant in Figure 21-1, we can see that the polar form of
Equation 21-2 is
E = √ VR2 + V2L tan − 1
VL
VR
(21-2)
21-2 Impedance
The current in the series circuit of Figure 21-1 must be a sine wave with a
magnitude and phase angle with respect to the applied voltage such that the
phasor sum of the voltage drop across the resistance and the voltage across
the inductance equals the applied voltage. Therefore, for a given AC circuit,
the magnitude of current is constant, and consequently the ratio ­between the
applied voltage and the current is also constant. Just as VR/I represents
the opposition of resistance to alternating current and VL/I represents the
­opposition of inductance to alternating current, the ratio VT/I or E/I represents the total opposition of the circuit to alternating current. In resistors, the
voltage and current are in phase, and in inductors and capacitors the voltage
and current are exactly 90° out of phase. But as shown in Figure 21-1, the
angle ϕ between the applied voltage and the current is between 0° and +90°
for a ­resistor and an inductor connected in series. Therefore, this total opposition is not equivalent to either a resistance or a reactance.
The total opposition to current in an AC circuit is called impedance.
The letter symbol for impedance is Z.
The SI unit for impedance is the ohm.
Z=
E
I
(21-3)
where Z is the impedance of an AC circuit in ohms, E is the applied voltage in volts, and I is the current in amperes.
Because the current is common to all components in a series circuit, we
can divide every term in Equation 21-1 by I:
E
VR
VL
=
+j
I
I
I
Therefore,
Z = R + jXL
(21-4)
We can also divide each term in Equation 21-2 by I:
Z = √ R2 + X2L tan−1
XL
R
(21-5)
601
602
Chapter 21 Impedance
We can derive similar equations for an AC circuit with resistance and
­capacitance in series. Therefore,
The impedance of a series circuit is the phasor sum of the resistance
and reactance.
Like resistance and reactance, impedance is a constant V/I ratio, not a
s­inusoidally varying quantity. But unlike resistance, inductive reactance,
and capacitive reactance, the phase angle is not the same for all values of
impedance. The equation Z = 5 kΩ tells us only the magnitude of the total
opposition to electric current. To obtain the phase angle between the voltage
and the current, we must recognize that impedance is a complex number that
can be represented by a phasor. Hence, we can expand Equation 21-3:
Z=
E∠ϕ
= Z∠ϕ
I∠0º
(21-6)
We could represent impedance with a standard common-origin phasor
diagram as in Figure 21-2(a). However, to preserve a distinction between
sinusoidal phasor quantities (such as voltage and current) and nonsinusoidal phasor quantities, we use a triangular phasor diagram for impedances,
as shown in Figure 21-2(b). In such impedance diagrams, resistance is
always drawn in the reference direction and inductive reactance in the +j
direction. It follows that capacitive reactance is drawn in the −j direction.
Z
+j
Z
Z
XL
ϕ
Reference
direction
R
(a)
XL
ϕ
R
(b)
Figure 21-2 Impedance diagram for a series circuit containing resistance
and inductance
Example 21-1
A solenoid having an inductance of 0.50 H and a resistance of 100 Ω is
connected to a 120-V 60-Hz source. Find the current.
Solution
Step 1
First, calculate the reactance of the solenoid:
XL = ωL = 377 × 0.50 = 188.5 Ω (inductive)
21-2 Impedance
603
Step 2
Draw and label a circuit diagram. Then draw an impedance diagram
for the circuit, as shown in Figure 21-3.
L = 0.50 H
XL
120 V
60 Hz
188.5 Ω
Z
E
R
XL = 188.5 Ω
100 Ω
ϕ
R = 100 Ω
(b)
(a)
Figure 21-3
Impedance diagram for Example 21-1
Step 3
Use the impedance diagram to find the rectangular coordinates of the
­impedance:
Z = R + jXL = 100 + j188.5 Ω
Step 4
Solve for the impedance in polar form.
ϕ = tan− 1
XL
188.5
= tan−1
= 62º
R
100
Z = √ R2 + X2L = √ 1002 + 188.52 = 213 Ω
Z = 213 Ω ∠+62º
Step 5
Solve for the magnitude of the current:
E
120 V
I= =
= 0.563 A
Z
213 Ω
The impedance diagram in Figure 21-3 shows that the current lags
behind the applied voltage by 62°.
Alternatively, we can by rearrange Equation 21-6 and use phasor
division to solve for both the magnitude and phase angle of the current. Since we are calculating this angle with respect to the applied
voltage, the phase angle of the applied voltage is zero.
I=
120 V∠0º
E
=
= 0.563 A ∠−62°
Z
213 Ω ∠62º
To verify the calculation of the magnitude and phase angle of the
current, download Multisim file EX21-1 from the website and follow
the instructions in the file.
See Problems 21-1 to 21-7 and Review Questions 21-60 to 21-66 at the
end of the chapter.
Although calculator shortcuts are
possible for most of
the ­examples in this
­chapter, full solutions
are given to show the
­relationships between
polar and rectangular coordinates.
Students can practice
using the coordinateconversion functions
of a ­calculator to solve
these ­examples and
compare the results to
the solutions given.
circuitSIM
walkthrough
604
Chapter 21 Impedance
21-3 Practical Inductors
Source: © David J. Green/Alamy
In practical circuits, the wire in an inductor’s winding has some resistance.
Therefore, the practical inductor is not a pure reactance with a 90° angle.
Hence the voltage drop across an inductor leads the current through it
by something less than 90°. By expressing the impedance of an inductor
in ­rectangular coordinates, we can determine the resistance and reactance
components.
Example 21-2
The choke in the circuit of Figure 21-4(a) has an impedance of 80 Ω ∠80º.
Find the total impedance when this choke is connected in series with a
75-Ω resistor.
Z coil = 80 Ω 80°
A choke wound on a
ferrite core.
Vcoil
V
Zc
oil
ZT
E
75 Ω
Zcoil sin ϕ
ϕ
VR
R
Zcoil cos ϕ
(a)
Figure 21-4
(b)
Practical inductor and resistor in series
Solution
Draw an impedance diagram as shown in Figure 21-4(b). Then, express
the impedance of the choke in rectangular coordinates:
Zcoil = 80 cos 80° + j80 sin 80° = 14 + j79 Ω
Since the two resistance components are both along the + axis of the
impedance diagram, they can be added by simple arithmetic:
ZT = 75 + 14 + j79 Ω = 89 + j79 Ω
Changing back to polar form gives
ϕ = tan−1
79
= 41.6º
89
ZT = √ 892 + 792 = 119 Ω
ZT = 119 Ω ∠+ 41.6°
21-3 Practical Inductors
We can also use trigonometric ratios to find the magnitude of the impedance:
Z=
R
X
=
cos ϕ
sin ϕ
(21-7)
Example 21-3
A coil and a resistor each have an 80-V drop across them when connected in series to a 120-V 60-Hz source. The current in the circuit is
0.50 A. Find the resistance of the coil.
Solution
Step 1
ZT =
Zcoil =
120 V
E
=
= 240 Ω
I
0.50 A
R=
80 V
Vcoil
=
= 160 Ω
I
0.50 A
80 V
VR
=
= 160 Ω
I
0.50 A
Step 2
Since the coil has both resistance and inductance, the phase angle of its
­impedance must be somewhat less than +90°, as shown in Figure 21-5.
Zcoil = 160 Ω ∠ϕ
In rectangular coordinates, the impedance of the coil is
Zcoil = 160 cos ϕ + j160 sin ϕ
0.50 A
A
E
Zc
ZT
oil
Vcoil = 80 V
VR = 80 V
ϕ
R
Zcoil cos ϕ
(a)
Figure 21-5
(b)
Impedance diagram for Example 21-3
Zcoil sin ϕ
605
606
Chapter 21 Impedance
ZT = 160 + 160 cos ϕ + j160 sin ϕ
Step 3
Therefore, the magnitude of the total impedance is
This example can also
be solved by using
the cosine rule, a
trigonometric formula
that relates the side
lengths of any ­triangle
to the cosine of one of
its angles.
ZT = 240 = √ ( 160 + 160 cos ϕ ) 2 + ( 160 sin ϕ ) 2
Squaring both sides of the equation gives
57 600 = 25 600 + 51 200 cos ϕ + 25 600 cos2 ϕ + 25 600 sin2 ϕ
cos ϕ =
32 000 − 25 600 ( cos2 ϕ + sin2 ϕ )
5 − 4 ( cos2 ϕ + sin2 ϕ )
=
51 200
8
Step 4
Since sin2 ϕ + cos2 ϕ = 1,
cos ϕ =
Step 5
The resistance of the coil is
circuitSIM
walkthrough
5−4
= 0.125
8
Rcoil = Zcoil cos ϕ = 160 × 0.125 = 20 Ω
To verify the calculation of the resistance of the coil, download Multisim
file EX21-3 from the website, and follow the instructions in the file.
See Problems 21-8 to 21-12 and Review Questions 21-66 to 21-68.
Circuit Check
CC 21-1.
A
A 30-mH inductor is connected in series with a 50-Ω resistor
and a 50-V 60-Hz source.
(a) Determine the magnitude and phase angle of the current.
(b) Determine the voltage drop across the inductor and the resistor.
(c) Sketch the phasor diagram.
CC 21-2. A series RL circuit is connected to a 100-V source. A current of
10 A flows in the circuit. Given that the voltage across the resistor is 50 V and the inductance is 30 mH, find the frequency of
the voltage.
CC 21-3. A 4.0-H inductor with a coil resistance of 200 Ω is connected
in series with a 1.0-kΩ resistor and a 200-V 60-Hz source.
Calculate the
(a) phase angle of the inductor
(b) phase angle of the circuit
(c) current flowing in the circuit
(d) voltage drop across the inductor
21-4 Resistance and Capacitance in Series
21-4 Resistance and Capacitance in Series
In Figure 21-6(a), the current again is common to all components. Since the
potential difference across the capacitor lags behind the current through it by
90°, the total-voltage phasor is in the fourth quadrant in the phasor diagram
in Figure 21-6(b). For an AC circuit with resistance and capacitance in series,
E = VR − jVC
VC
VR
E
ϕ
(21-8)
Reference
I
phasor
VR
E
VC
(a)
Figure 21-6
(b)
Series circuit containing resistance and capacitance
or
E = √ VR2 + VC2 −tan − 1
Z = R − jXC
and
or
VC
VR
Z = √R2 + X2C −tan − 1
(21-9)
(21-10)
XC
R
(21-11)
Inductive reactance has a +j direction in impedance diagrams, and capacitive
reactance always has a −j direction. Therefore, an inductive impedance has
a positive angle between 0 and +90° and a capacitive impedance has a
­negative angle between 0 and −90°.
Example 21-4
What capacitance connected in series with a 500-Ω resistor limits the
­current drawn from a 48-mV 465-kHz source to 20 μA?
Solution
Z=
E
48 mV
=
= 2.4 kΩ
I
20 μA
607
608
Chapter 21 Impedance
Applying the Pythagorean theorem to the impedance diagram in Figure
21-7 gives
XC = √ Z2 − R2 = √ 24002 − 5002 = 2350 Ω
XC =
Since
C=
1
2πf C
1
1
=
= 0.15 nF
2πf XC
2 × π × 465 kHz × 2350 Ω
20 μA
R = 500 Ω
A
C
48 mV
465 kHz
Z = 2400 Ω
XC
500 Ω
(a)
Figure 21-7
circuitSIM
(b)
Impedance diagram for Example 21-4
To verify the calculation of the capacitance, download Multisim file
EX21-4 from the website and follow the instructions in the file.
walkthrough
See Problems 21-13 to 21-22.
21-5 Resistance, Inductance,
and Capacitance in Series
In rectangular form the sum of the voltages in the circuit of Figure 21-8 is
E = VR + jVL − jVC
(21-12)
VL
VL
E
VR
VC
(a)
VX
E
ϕ
VR
I
VC
(b)
Figure 21-8 A series circuit containing resistance, inductance, and
capacitance
21-5 Resistance, Inductance, and Capacitance in Series
The voltage across the inductance leads the common current by 90°, and
the voltage across the capacitance lags behind the current by 90°. Since these
two voltages are exactly 180° out of phase, they can be added algebraically.
Therefore, the imaginary terms in Equation 21-12 can be combined:
E = VR + j ( VL − VC )
(21-13)
Unless the two reactive voltages are exactly equal, the circuit has an equivalent reactive voltage, VX, that is either inductive or capacitive depending
on which of the original reactive voltages is greater. If VL is greater than VC,
the circuit in Figure 21-8 appears to contain only inductance and resistance
when measured at the terminals of the source. The overall circuit cannot
be both inductive and capacitive at the same time. The current must either
lead or lag behind the applied voltage.
Converting Equation 21-13 into polar coordinates gives
E = √ VR2 + VX2 tan − 1
VX
VR
where VX is the net reactive voltage, VL − VC.
(21-14)
Dividing Equation 21-13 and Equation 21-14 by the common current
gives the impedance of a series circuit containing resistance, inductance,
and capacitance:
Z = R + jXL − jXC = R + j ( XL − XC )
or
Z = √ R2 + X2eq tan − 1
Xeq
R
where Xeq is the net equivalent reactance, XL − XC.
(21-15)
(21-16)
Example 21-5
Find the total impedance at 20 kHz of a 1.5-mH inductor, a 100-Ω
­resistor, and an 80-nF capacitor connected in series.
Solution
Step 1
First find the reactances of the inductor and the capacitor:
XL = 2πf L = 2 × π × 20 kHz × 1.5 mH = 188.5 Ω (inductive)
and
XC =
1
1
=
= 99.47 Ω (capacitive)
2πf C
2 × π × 20 kHz × 80 nF
609
610
Chapter 21 Impedance
Step 2
Draw a schematic diagram and an impedance diagram, as in Figure 21-9.
XL = 188.5 Ω
XL
188.5 Ω
XR
100 Ω
XC
99.5 Ω
Z
ϕ
R = 100 Ω
XC = 99.5 Ω
(a)
Figure 21-9
Xeq = XL − XC
(b)
Schematic and impedance diagrams for Example 21-5
Step 3
Find the impedance in rectangular coordinates:
Z = R + jXL − jXC = 100 + j188.5 − j99.5 = 100 + j89 Ω
Step 4
Convert the impedance to polar coordinates:
ϕ = tan − 1
Z=
Xeq
R
= tan − 1
89
= +41.7º
100
R
100 Ω
=
= 134 Ω
cos ϕ
cos 41.7º
Z = 134 Ω ∠ + 42°
See Problems 21-23 to 21-32 and Review Questions 21-69 to 21-73.
21-6 Resistance, Inductance,
and Capacitance in Parallel
The characteristic of a parallel circuit is that the same voltage appears across
all parallel branches. We use this common voltage as the reference phasor
in phasor diagrams for any parallel AC circuits. Ohm’s law then gives the
current through each branch of the circuit in Figure 21-10.
21-6 Resistance, Inductance, and Capacitance in Parallel
IR =
For the resistance,
E ∠0º
E
= ∠ 0º
R ∠ 0º
R
Therefore, IR is in phase with the applied voltage.
IT
IC
IC
IR
IT
IL
IX
IL
ϕ
Reference
E
phasor
IR
(a)
(b)
Figure 21-10 A parallel circuit containing resistance, inductance, and capacitance
IL =
For the inductance,
E ∠0º
E
=
−90º
XL ∠+90º
XL ∠
and IL lags behind the reference voltage by 90°.
For the capacitance,
IC =
E ∠0º
E
=
+90º
XC ∠−90º
XC ∠
and IC leads the reference voltage by 90°.
The total current in the parallel circuit is the phasor sum of the branch
­currents:
IT = IR − jIL + jIC = IR + j ( IC − IL )
(21-17)
Converting Equation 21-17 to polar coordinates gives
IT = √ I2R + I2X tan− 1
where IX is the net reactive current IC − IL.
IX
IR
(21-18)
We can apply Equation 21-6 to determine the equivalent impedance of
the parallel circuit:
Zeq =
E
IT
(21-19)
This method of solving for the equivalent impedance of a parallel circuit
is called the total-current method. If the applied voltage is not known,
we can assume any convenient value in order to find the equivalent
­impedance.
611
612
Chapter 21 Impedance
Example 21-6
Find the equivalent impedance at 20 kHz of a 1.5-mH inductance, a
100-Ω resistance, and an 80-nF capacitance all connected in parallel.
Solution
Step 1
Calculate the reactances:
XL = 2πf L = 2 × π × 20 kHz × 1.5 mH = 188.5 Ω (inductive)
XC =
Note that dividing a
quantity having the
same direction as
the reference phasor
by a +j quantity creates a −j quantity,
and dividing by a −j
quantity creates a +j
quantity.
1
1
=
= 99.5 Ω (capacitive)
2πf C
2 × π × 20 kHz × 0.08 mF
Step 2
Assume a convenient value of applied voltage, say E = 200 V ∠0º .
IR =
Then
IL =
IC =
and
E ∠ 0º
200 ∠ 0º
=
= 2.00 A ∠ 0º
R
100 ∠ 0º
E ∠0º
200 ∠0º
=
= 1.06 A ∠−90º
+jXL
188.5 ∠+90º
E ∠0º
200 ∠0º
=
= 2.01 A ∠+90º
−jXC
99.5 ∠−90º
To facilitate the phasor division, we expressed the direction associated with R, XL, and XC in polar form rather than with rectangular
coordinate operators.
Step 3
Draw a circuit diagram and a phasor diagram, as in Figure 21-11.
IC = 2.01 A
IT
200 V
20 kHz
XC
99.5 Ω
R
100 Ω
XL
188.5 Ω
IX = 0.95 A
IL = 1.06 A
(a)
Figure 21-11
ϕ
IR = 2.0 A
(b)
Diagrams for Example 21-6
Step 4
Calculate the total current in rectangular coordinates.
IT = IR − jIL + jIC = 2.00 − j1.06 + j2.01 = 2.00 + j0.95A
E
21-7 Conductance, Susceptance, and Admittance
613
Step 5
Convert the total-current polar coordinates:
ϕ = tan − 1
I=
IX
0.95
= tan− 1
= +25.4º
IR
2.00
IR
2.00 A
=
= 2.214 A
cos ϕ
cos ( +25.4º )
IT = 2.21 A ∠+25.4°
Step 6
Calculate the equivalent impedance:
Zeq =
200 ∠0º
E
=
= 91 Ω ∠−25°
IT
2.21 ∠+25.4º
Note that the equivalent impedance is capacitive in this example
because the capacitive branch draws a greater current from the source
than does the inductive branch. When the same three components were
connected in series in E
­ xample 21-5, the total impedance was inductive
because there was a greater voltage across the inductance than across
the capacitance.
To verify the calculation of the magnitude and phase angle of the
current in Step 5, download Multisim file EX21-6 from the website and
follow the detailed instructions in the file.
See Problems 21-33 to 21-39 and Review Questions 21-74 and 21-75.
21-7 Conductance, Susceptance,
and Admittance
For parallel DC circuits, we found it convenient to think in terms of conductance, the reciprocal of resistance. Conductance is a measure of how easily
a DC circuit passes current. We can use a similar approach for AC circuits.
Since I = E/Z, we can rewrite Equation 21-17 as
Dividing by E gives
E ∠ 0º
E
E
E
E
= = −j
+j
Z
Z
R
XL
XC
1
1
1
1
= −j
+j
Z
R
XL
XC
(21-20)
circuitSIM
walkthrough
614
Chapter 21 Impedance
In this equation, we can replace 1/R with conductance, G. There are similar reciprocals for impedance and reactance.
Admittance is the overall ability of an electric circuit to pass alternating current. The symbol for admittance is Y.
Admittance is the reciprocal of impedance: Y = 1/Z
Susceptance is the ability of inductance or capacitance to pass alternating current. The letter symbol for susceptance is B.
Susceptance is the reciprocal of reactance: B = 1/X
The SI unit for both admittance and susceptance is the siemens, the same
as for conductance.
When we divide 1∠0º by a phasor quantity with a +90° angle, the quotient has a −90° angle. Hence, the reciprocal of +jXL is −jBL. Similarly, the
reciprocal of −jXC is +jBC. Thus, substituting admittance, conductance, and
­susceptance into Equation 21-20 gives
Y = G − jBL + jBC = G + j ( BC − BL )
(21-21)
The polar coordinates of admittance are
ϒ = √G2 + B2eq
tan− 1
Beq
G
(21-22)
where Beq is the net equivalent susceptance, BC − BL.
Since BL = 1/XL ,
BL =
1
1
=
2πfL
ωL
(21-23)
Similarly,
BC = 2πfC = ωC
(21-24)
Example 21-6A
Find the equivalent impedance at 20 kHz of a 1.5-mH inductance, a
100-Ω resistance, and an 80-nF capacitance all connected in parallel.
Solution
Step 1
Calculate the conductance and the susceptances:
1
1
G= =
= 10.0 mS
R
100 Ω
BC = 2πfC = 2 × π × 20 kHz × 80 nF = 10 mS ( capacitive )
BL =
1
1
=
= 5.3 mS ( inductive )
2πfL
2 × π × 20 kHz × 1.5 mH
21-8 Impedance and Admittance
Step 2
Draw an admittance diagram for the circuit, as shown in Figure
21-12. It is impossible to draw an impedance diagram for the circuit of
Figure 21-11(a) because the equivalent impedance of a parallel circuit
is smaller than the r­esistance, but the hypotenuse of a right triangle
cannot be smaller than ­either of the other two sides.
BC = 10 mS
Y
Beq = BC − BL
ϕ
G = 10 mS
BL = 5.3 mS
Figure 21-12
Admittance diagram for Example 21-6A
Step 3
Using rectangular coordinates, calculate the admittance:
Y = G + j(BC − BL) = 10 + j(10.0 − 5.3) mS = 10 + j4.7 mS
Step 4
Convert the admittance to polar form:
ϕ = tan−1
Y=
Beq
G
= tan−1
4.7
= +25.2º
10
G
0.01 s
=
= 11.05 mS
cos ϕ
cos 25.2º
Y = 11.05 mS ∠+25.2º
Step 5
Calculate the equivalent impedance:
Z eq =
1 ∠0º
1
=
= 91 Ω ∠−25°
Y
11.05 mS ∠+25.2º
See Problems 21-40 to 21-51 and Review Questions 21-76 to 21-78.
21-8 Impedance and Admittance
In the simple parallel circuit of Figure 21-10(a), each branch contained only
one type of opposition to alternating current. Suppose, however, that one of
the branches is a practical inductor, which has both resistance and i­ nductive
615
616
Chapter 21 Impedance
reactance. The current through this branch would now be ­inversely proportional to its impedance. The following example illustrates the procedure
for finding the conductance and susceptance of a practical inductor.
Example 21-7
Find the conductance and susceptance of an inductor with a resistance
of 30 Ω and an inductive reactance of 40 Ω (see Figure 21-13).
50 Ω 53.1°
R
0.02 S −53.1°
30 Ω
Z
XL
Y
G
BL
40 Ω
Figure 21-13
Finding the conductance and susceptance of an inductor
Solution
Step 1
Find the impedance of the inductor:
Z = √ R2 + X2 tan−1 R = √ 302 + 402 tan− 1
X
40
= 50 Ω ∠+53.1º
30
Step 2
Calculate the admittance:
Y=
1
1
=
= 20 mS ∠−53.1º
Z
50 Ω ∠+53.1º
Step 3
Split the impedance into its rectangular coordinates:
Therefore,
and
Y = G − jB = Y cos ϕ − jY sin ϕ
G = Y cos ϕ = 20 cos 53.1° mS = 20 × 0.600 = 12 mS
BL = Y sin ϕ = 20 sin 53.1° = 20 × 0.800 = 16 mS
The rectangular coordinates of the impedance are the resistance and
reactance of the equivalent series circuit, while the rectangular coordinates of the admittance are the conductance and susceptance of the
equivalent parallel circuit. These “similar but opposite” characteristics
of series and parallel AC circuits resemble those of series and parallel
resistances in DC circuits (see Section 7-9). Table 21-1 compares the characteristics of series and parallel AC circuits.
21-8 Impedance and Admittance
617
TABLE 21-1 Series and parallel AC circuit characteristics
Characteristic
Series Components
Parallel Components
IC
VL
E
Phasor diagram
VX
ϕ
VR
VC
Sum of phasors
Defining equation
Reference
phasor
ϕ
IT
IR
IL
VT = V1 + V2 + V3 + . . .
Z=
I
IX
Reference
E
phasor
IT = I1 + I2 + I3 + . . .
Y=
VT
I
IT
V
XL
Z
Impedance diagram
Xeq
Not possible
ϕ
R
XC
BC
Admittance diagram
Y
Not possible
Beq
ϕ
G
BL
Rectangular form
Z = R + jXeq
Y = G + jBeq
Resistive component
R (in ohms)
G (in siemens)
Reactive component
1
where XL = ωL and XC =
ωc
where BC = ωC and BL =
Polar form
+j Xeq = +j ( XL − XC )
Z = Z ∠ϕ
Magnitude
Z = √ R2 + X2eq
Phase angle
ϕ = tan−1
Xeq
R
See Problems 21-52 to 21-59 and Review Question 21-79.
+jBeq = +j ( BC − BL )
Y = Y ∠ϕ
Y = √ G2 − B2eq
ϕ = tan−1
Beq
G
1
ωL
618
Chapter 21 Impedance
Circuit Check
B
CC 21-4. Determine the total impedance of the network shown in
Figure 21-14.
L1
10 mH
2.0 V
peak
R1
10 kΩ
+
10 kHz
−
C1
10 μF
Figure 21-14
CC 21-5. For the circuit of Figure 21-15, determine the total admittance
in polar coordinates.
+
3 V peak
50 kHz
−
0°
R
1 kΩ
C
0.68 μF
L
33 μH
Figure 21-15
CC 21-6. Determine the total impedance of the circuit shown in
Figure 21-16.
L2
10 mH
C1
a
L1
10 mH
2.0 V
peak
470 nF
+
10 kHz
−
b
C3
10 nF
Figure 21-16
470 nF
C2
21-9 Troubleshooting
21-9 Troubleshooting
The most common faults in circuits containing resistance, inductance, and
capacitance are shorted and open components. The voltage-current relationships of resistors, inductors, and capacitors are a key to detecting these
faults. These relationships may be stated as follows:
• A current produces a voltage drop across a resistor.
• A changing current (such as alternating current) produces a voltage
across an inductor.
• A changing voltage (such as alternating voltage) produces a current
through a capacitor.
Testing a Series Circuit
If the source voltage in the series AC circuit of Figure 21-9 is 500 V with a
phase angle of 0°, the current is
I=
500 ∠0º
E
= 3.73 A ∠−41.7º
=
Z
134 ∠41.7º
Consequently, the resistor, inductor, and capacitor voltages are
VR = IR = 3.73 A × 100 Ω = 373 V
VL = IXL = 3.73 A × 188.5 Ω = 703 V
VC = IXC = 3.73 A × 99.5 Ω = 371 V
If one of the components is open, an AC ammeter inserted into the circuit
will show that no current is flowing. If the resistor is normal, VR = 0 V since
I = 0 A. If the inductor is normal, VL = 0 V since the current is not changing. If the capacitor is normal, zero current means that VC is not changing,
and therefore the AC voltage across the capacitor is also zero. When connected across the open component, an AC voltmeter will show a voltage
drop equal to the source voltage (500 V). The reading for the other two
components will be 0 V.
However, if one of the components is fully shorted, an AC voltmeter
connected across it will read 0 V, and the sum of voltage drops across the
normal components will equal the source voltage. If one component is partially shorted, its voltage drop will be lower than calculated above, and
the voltage drops across the other two components will be higher than the
values calculated.
Testing a Parallel Circuit
If the parallel AC circuit of Figure 21-11 is operating normally, the branch
currents will be as calculated in Example 21-6. If the resistor, inductor, or
619
620
Chapter 21 Impedance
capacitor is open, an ammeter inserted in the branch with the open component will read zero current, while the currents in the other branches will
be unchanged or will perhaps increase somewhat if the voltage source is
not regulated.
If a component is partially shorted, the current through it will be greater
than normal. The currents through the other branches will either be
unchanged or will decrease, depending on whether the increased load from
the shorted component reduces the source voltage. If a component is fully
shorted, the circuit will draw a very large current from the AC source, probably burning out a fuse or tripping a circuit breaker. We will then remove
each component to test it, starting with any component that showed signs
of overheating or other physical damage. We can use an ohmmeter to test
the resistor and an LC meter to test the inductor and capacitor.
See Review Questions 21-80 and 21-81.
Problems
621
Summary
• In a series AC circuit, the applied voltage equals the phasor sum of all the
voltage drops.
• Impedance is the total opposition to alternating current in an AC circuit.
• The total impedance of a series AC circuit is the phasor sum of the resistance and the reactance.
• A practical inductor consists of the resistance of the coil in series with its
inductance.
• In a series AC circuit, the net reactive voltage is the algebraic sum of the
inductance and capacitance voltages.
• In a series AC circuit, the net equivalent reactance is the algebraic sum of
the inductive and capacitive reactances.
• In a parallel AC circuit, the total current is the phasor sum of all the
branch currents.
• Admittance is the overall ability of an electric circuit to pass alternating
current.
• Susceptance is a measure of the ability of inductance or capacitance to
pass alternating current.
• Admittance is the reciprocal of impedance.
• The rectangular coordinates of impedance are resistance and reactance.
• The rectangular coordinates of admittance are conductance and
susceptance.
• The most common faults in AC circuits are open and shorted resistors,
inductors, and capacitors.
Problems
Draw a circuit diagram for each problem. Also draw either a phasor diagram or an
impedance diagram for each series circuit, and a phasor diagram or an admittance
diagram for each parallel circuit.
I
Section 21-1
21-1.
Resistance and Inductance in Series
Calculate the magnitude of the unknown voltage in the AC circuits
shown in Figure 21-17.
10 mV
10 V
E=?
VR = ?
16 mV
20 V
(a)
Figure 21-17
(b)
B = beginner
I = intermediate
A = advanced
622
Chapter 21 Impedance
I
21-2.
B
21-3.
B
21-4.
B
21-5.
I
21-6.
I
21-7.
circuitSIM
walkthrough
circuitSIM
walkthrough
I
I
A
circuitSIM
walkthrough
I
A
circuitSIM
walkthrough
(a)An inductor connected to a 120-V 60-Hz source has an IR drop
of 72 V across the resistance of its winding. Find the magnitude
and the angle of the voltage across its inductance with respect
to the applied voltage.
(b) Use Multisim to verify your calculations in part (a).
The operating coil of a contactor is connected to a 240-V 60-Hz
source and draws a 2-A current that lags behind the source voltage
by 60°. Determine the impedance of the coil in:
(a) polar coordinates
(b) rectangular coordinates
When a 400-Hz sine-wave voltage of 4.0 V ∠0º is applied to the voice
coil of a loudspeaker, the resulting current is 0.50 A ∠−20º Express
the impedance of the loudspeaker in both polar and rectangular
­coordinates.
Determine the total impedance of a series circuit consisting of a
150-Ω resistor and a 10-mH inductor at a frequency of 750 Hz. Give
your answer in polar coordinates.
Two components in series have an impedance of 15 kΩ ∠45º at a
frequency of 20 kHz. Determine the value of the two components.
Determine the series equivalent circuit of a device that has an
impedance of 5kΩ ∠+ 30º at a frequency of 200 Hz.
Section 21-3
21-8.
Practical Inductors
(a)Find the resistance of a 4.0-H choke with a 750-Ω impedance at
25 Hz. Calculate the phase angle of the impedance.
(b) Use Multisim to verify the resistance of the choke.
21-9. A certain choke draws a current of 1.2 A when connected to a
120-V 60-Hz source. When connected to a 12-V battery, it draws
0.5 A of current. Determine the resistance and inductance of the
choke.
21-10. (a)Another choke draws a 5.0-A current when connected to a
117-V 25‑Hz source. When connected to a 120-V 60-Hz source,
this choke draws a 3.0-A current. Determine the resistance and
inductance of the choke.
(b) Use Multisim to verify the resistance and inductance of the choke.
21-11. A GE 120-V 500-W lamp is operated on a 220-V 60-Hz system by
connecting it in series with a choke with a resistance of 50 Ω. Determine the inductance the choke must have in order to provide the
appropriate voltage to the lamp.
21-12. (a)A 15-W fluorescent lamp operates with a ballast inductance in
series with it. When the total applied voltage is 120 V 60 Hz, the
voltage across the lamp is 56 V RMS and the total voltage across
the ballast inductor is 100 V RMS. Find the resistance and inductance of the ­ballast, given that the lamp has no reactance.
(b) Use Multisim to verify the resistance and inductance of the
ballast.
Problems
B
B
B
B
B
I
I
I
B
I
I
Section 21-4
Resistance and Capacitance in Series
21-13. The voltages across a capacitor and a resistor connected in series in
an AC circuit are measured as 60 V and 80 V RMS respectively. Find
the magnitude and phase angle of the total voltage with respect to
the common current.
21-14. (a)Find the magnitude and phase angle of the current that an
impedance of 80 − j95 Ω draws from a 117-V AC source.
(b) Use Multisim to verify the calculation of the magnitude and
phase angle of the current.
21-15. The input circuit of a transceiver draws a current of 7 μA ∠17º
from a 2.5-MHz signal generator when the source voltage is
350 μV ∠10º. Determine the impedance of the transceiver and
express the result in polar coordinates.
21-16. Calculate the total impedance at 1.0 kHz of a 0.50-μF capacitor and
a 500-Ω resistor in series. Give the answer in polar coordinates.
21-17. Find the impedance in polar coordinates of a 200-pF capacitor in
­series with a 1.2-kΩ resistor when the frequency of the applied voltage is 1.5-MHz.
21-18. At what frequency will a 20-nF capacitor and a 120-kΩ resistor in
­series have an impedance of 200 kΩ?
21-19. Determine the equivalent series circuit for a 4.5-MHz impedance of
500 Ω ∠−15º.
21-20. What value of capacitor must be connected in series with a 560-Ω
­resistor to limit the heat produced in the resistor to 5 W when the
source voltage is 120 V at 60 Hz?
21-21. A 100-V 60-Hz source is connected in series with a 920-Ω resistor
and a 47-uF capacitor. Determine:
(a) the current of the circuit
(b) the voltage across each component
(c) the phase angle of the impedance
21-22. A heating element is rated 2.0 kW at 120 V. What capacitance could
be connected in series with the element to power it from a 220-V
60-Hz supply?
Section 21-5 Resistance, Inductance, and Capacitance in
Series
21-23. Determine the output voltage (magnitude and angle) of the components in the circuits shown in Figure 21-18.
4.5 V
80 mV
+
100 mV
+
Vout = ?
10 V 0°
E
623
−
−
(a)
Figure 21-18
200 mV 0°
E
(b)
Vout = ?
circuitSIM
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624
Chapter 21 Impedance
I
I
I
B
A
circuitSIM
walkthrough
I
A
circuitSIM
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I
A
I
21-24. Find the total 60-Hz impedance in polar form of a 10-μF capacitor
in series with a 1.0-H choke having a resistance of 100 Ω.
21-25. For an RLC series circuit made up of a 820-Ω resistor, a 22-mH
inductor and a 0.33-uF capacitor, determine the total impedance in
polar coordinates when the frequency is
(a) 250 Hz
(b) 3 kHz
(c) 15 kHz
21-26. At a frequency of 400 Hz, what value of capacitor must be connected in series with an impedance of 75 Ω ∠+60º to give a total
impedance of 75 Ω ∠−60º?
21-27. Determine the total impedance of a series circuit formed by Z1 = 350
Ω ∠45º , Z2 = 720 Ω ∠−75º and Z3 = 500 + j120 Ω.
21-28. (a)A relay coil requires a 100-mA current through it in order to
close the contacts of the relay. If it is operated from a dc source,
the required voltage is 24 V. If the relay is operated from a 60-Hz
source, the r­ equired voltage is 160 V. What capacitance in series
with the relay coil will allow its operation from a 120-V 60-Hz
supply? (Give two answers.)
(b) Use Multisim to verify the capacitance.
21-29. At a frequency of 400 Hz, what value of capacitance must be connected in series with an impedance of 75 Ω ∠+60º to form
(a) an inductive impedance of 50 Ω
(b) a capacitive impedance of 50 Ω
21-30. A coil and capacitor are connected in series across a 400-Hz source.
The voltage drops are 48 V across the coil, 52 V across the capacitor,
and 6.0 V across the whole circuit. The current in the circuit is 100 mA.
(a) Determine the inductance and resistance of the coil, and the capacitance of the capacitor.
(b) Determine the total impedance in polar form.
(c) Use Multisim to verify the resistance and inductance of the coil
and the capacitance of the capacitor.
21-31. A 100-Ω resistor is connected in series with a 0.50-H inductor and
a 220-V 60-Hz power supply. What size capacitor must added in
series to give a current of 2.0 A?
21-32. An inductive circuit of 50 Ω resistance and 0.08 H inductance is
connected in series with a capacitor across a 200-V 50-Hz supply.
The current is 3.8 A leading. Find the value of the capacitor.
Section 21-6 Resistance, Inductance, and Capacitance in
Parallel
21-33. Determine the magnitude and phase angle of the current
with respect to the applied voltage in the ac circuits shown in
Figure 21-19.
Problems
IT = ?
IT = 5.0 A
IL = ?
50 mA
+
625
+
750 mA
E
−
E
3.75 A
−
(a)
(b)
Figure 21-19
I
I
I
I
B
B
I
B
B
21-34. (a)A given impedance draws a 5.0-A current from a 120-V 60-Hz
source. The current through the impedance lags behind the
voltage across it by 60°. Find the total current drawn from the
source when a 100-mF capacitor is connected in parallel with
the given impedance.
(b) Use Multisim verify the total current drawn from the source.
21-35. Use the values of problem 21-27 to determine the equivalent total impedance of the circuit when all loads are connected in parallel.
21-36. A 75-W bulb is connected in parallel with a 68-μF capacitor and a
110-V 60-Hz ac source.
(a) Determine the equivalent impedance in polar coordinates.
(b) Determine the total current drawn from the source.
21-37. The bias circuit for one transistor in an audio amplifier has a 2.2-kΩ
resistor and a 0.20-μF capacitor in parallel. Find the rectangular
­coordinates for the equivalent impedance of these components at
1.0 kHz.
21-38. A 200-Ω resistor, 0.50-H inductor, and 10-µF capacitor are connected
in parallel to a 150-V 60-Hz source. Calculate the total current and
the phase angle.
21-39. Find the equivalent impedance for the circuit in Problem 21-38.
Section 21-7
circuitSIM
walkthrough
Conductance, Susceptance, and Admittance
21-40. (a)If the magnitudes of the applied voltages in the circuits shown
in Figure 21-19 are 75 V RMS, calculate the admittance and
impedance of each circuit in polar form.
(b) Use Multisim to verify the admittance and impedance of each
circuit.
21-41. Determine the total admittance and equivalent impedance in polar
coordinates of a 20-mS conductance in parallel with a 75-mS inductive susceptance at a frequency of 50 Hz.
21-42. Calculate polar coordinates for the total admittance and equivalent impedance at 45 MHz of a 300-μS conductance in parallel with a
30-pF capacitor.
circuitSIM
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626
Chapter 21 Impedance
B
B
B
I
B
B
B
B
I
I
I
I
I
I
I
I
A
21-43. Calculate the 60-Hz admittance of a 0.50-H inductance in parallel
with a 12-mS conductance.
21-44. Find the equivalent impedance of a device with an admittance of
0.050 − j0.015 S. Is this device inductive or capacitive?
21-45. What values of conductance and susceptance must be connected in
parallel to form an admittance of 500 mS ∠−30º?
21-46. For an RLC parallel circuit made up of an 820-Ω resistor, a 22-mH
inductor, and a 0.33-μF capacitor, determine the total admittance in
polar form at frequencies of
(a) 250 Hz
(b) 3 kHz
(c) 15 kHz
21-47. Calculate the polar coordinates for the equivalent impedances of
the parallel components in Problem 21-46.
21-48. Find the polar coordinates for the total admittance of a circuit with
36 + j24 mS in parallel with 42 − j20 mS.
21-49. Find the equivalent impedance of an AC circuit consisting of two parallel branches with admittances of 64 μS ∠+15º and 48 μS ∠−45º.
21-50. Find the total admittance of a 1.0-kΩ resistor, 20-mH inductor, and
10-nF capacitor connected in parallel at a frequency of 10 kHz.
21-51. Find the equivalent impedance for the circuit in Problem 21-50
when the frequency is increased to 20 kHz.
Section 21-8
Impedance and Admittance
21-52. What values of resistance and capacitance in parallel have a
775-μS ∠55º admittance when the frequency applied is 15 kHz?
21-53. Determine the parallel equivalent circuit of a device that has an
­impedance of 2.4 kΩ ∠+30º at a frequency of 200 Hz. Compare this
equivalent circuit with the answer to Problem 21-7.
21-54. Determine the parallel equivalent circuit at 4.5 MHz of an impedance of 150 Ω ∠−75º. Compare this equivalent circuit with the
­answer to Problem 21-19.
21-55. Determine the series components that have an equivalent admittance at 1 kHz of 100 + j55 mS.
21-56. Determine the series components that have an equivalent admittance at 25 MHz of 120 + j90 S.
21-57. What value of inductance must be connected in parallel with a 50-Ω
resistor to form a total admittance at 60 Hz of 85 mS?
21-58. What value of resistance must be connected in parallel with an
­inductance of 1.5 H to produce a 400-Ω impedance at 60 Hz?
21-59. (a)Determine values for a set of parallel-connected components
that has an admittance of 100 mS ∠−60º when connected to a
60-Hz source.
(b) Determine values for a set of series-connected components that
are equivalent to the parallel circuit in part (a).
Review Questions
Review Questions
Section 21-2
Impedance
21-60. Why is the additional term impedance required for dealing with the
opposition of a circuit to alternating current?
21-61. Why can we use the Pythagorean theorem to calculate the total
impedance of a series circuit from its resistance and reactance?
21-62. Why is it necessary to state an angle as well as a magnitude when
an impedance is given in polar form?
21-63. Why does a positive angle with an impedance always represent a
circuit with inductive reactance?
21-64. How is the angle of an impedance related to the voltage across and
the current through the impedance?
21-65. Why is it not necessary to state an angle when an impedance is
given in rectangular coordinates?
Section 21-3
Practical Inductors
21-66. Why must the angle of the impedance of a practical inductor be
less than +90°?
21-67. Explain why expressing the impedance of a practical inductor in
­rectangular coordinates gives the actual resistance and reactance of
the inductor.
21-68. Describe how to determine the total impedance of a practical
­inductor and a resistor in series.
Section 21-5 Resistance, Inductance and Capacitance in Series
21-69. What is the significance of a total impedance having a 0° angle?
21-70. Under what circumstances does the total impedance decrease when
a capacitor is added in series with a given impedance?
21-71. Under what circumstances is the magnitude of the RMS current
drawn from a certain source unchanged when an inductance is
connected in series with a given impedance?
21-72. Explain why an AC circuit containing both inductance and capacitance appears to the source as either an inductive circuit or a
capacitive circuit.
21-73. What is meant by equivalent reactance?
Section 21-6
Parallel
Resistance, Inductance, and Capacitance in
21-74. What is meant by the equivalent impedance of a parallel AC circuit?
21-75. Why is quadrant I the capacitive quadrant of a parallel-circuit phasor diagram, whereas quadrant IV is the capacitive quadrant of a
series-circuit phasor diagram?
Section 21-7
Conductance, Susceptance, and Admittance
21-76. Why is it impossible to draw an impedance diagram for a parallel circuit?
627
628
Chapter 21 Impedance
21-77. What is the advantage of thinking in terms of admittance, conductance, and susceptance in dealing with parallel AC circuits?
21-78. Why does BL have a −j operator for phasor addition, whereas XL
has a +j operator?
Section 21-8
Impedance and Admittance
Section 21-9
Troubleshooting
21-79. Explain why Z = 20 + j40 Ω is not the equivalent impedance of
Y = 50 − j25 mS.
21-80. Describe a troubleshooting technique for detecting a shorted component in the circuit of Figure 21-20(a).
IT
R
IR
I
jXL
E
E
IL
IC
jXL
R
−jXC
−jXC
(a)
(b)
Figure 21-20
21-81. Describe a troubleshooting technique for detecting an open component in the circuit of Figure 21-20(b).
Integrate the Concepts
In the circuits of Figure 21-20, the components are a 75-Ω resistor, a 9.1-mH
inductor, and a 0.33-μF capacitor. The applied voltage is 240 V RMS with a
phase angle of 10° and a frequency of 2.0 kHz.
(a) Calculate the total impedance for the circuit in Figure 21-20(a).
(b) Draw the impedance diagram for the circuit in Figure 21-20(a).
(c) Determine the current in Figure 21-20(a).
(d) Determine IR, IL , and IC for the circuit in Figure 21-20(b).
(e) Draw the phasor diagram for the circuit in Figure 21-20(b).
(f) Determine the total current drawn for the circuit of Figure 21-20(b).
Practice Quiz
1.
Which of the following statements is true?
(a)Impedance is the opposition to current in an AC circuit and has
units of ohms.
(b)Admittance is the ability of an electric circuit to pass AC current
and has units of ohms.
Practice Quiz
2.
The phase angle of the impedance of the circuit shown in Figure 21-21 is
(a)21.8°
(b)−11.8°
(c)−21.8°
(d)11.8°
R1
3 kΩ
XL
1.2 kΩ
+
25 VRMS
0°
−
Figure 21-21
3.
4.
5.
6.
For what value of XL will the the phase angle of the impedance in
Figure 21-21 become 57°?
The current flowing through the resistor in Figure 21-21 is
(a) 7.74 mA ∠−21.8º
(b) 7.74 mA ∠11.8º
(c) 7.74 mA ∠−11.8º
(d) 7.74 mA ∠21.8º
The voltage drop across the inductor of Figure 21-21 is
(a) 9.29 V ∠−78.2º
(b) 9.29 V ∠−68.2º
(c) 9.29 V ∠78.2º
(d) 9.29 V ∠68.2º
The total impedance of the circuit of Figure 21-22 is
(a) 10.8 kΩ ∠−63.6º
(b) 10.9 kΩ ∠63.8º
(c) 10.9 kΩ ∠−63.8º
(d) 10.8 kΩ ∠63.6º
R1
2.2 kΩ
L
10 VRMS
0°
+
−
Figure 21-22
R
Coil
10 kΩ
75°
629
630
Chapter 21 Impedance
7.
The phase angle of the circuit shown in Figure 21-23 is
(a)84.3°
(b)−84.3°
(c)5.7°
(d)−5.7°
R1
2.2 kΩ
+ X
C
22 kΩ
+
10 VRMS
45°
−
Figure 21-23
8.
If the value of XC in Figure 21-23 is changed to 2.2 kΩ, the phase angle
of the circuit becomes 45°. True or false?
9.
The total reactance of the circuit shown in Figure 21-24 is
(a) j19.5 kΩ
(b) j19.8 kΩ
(c) −j19.5 kΩ
(d) −j19.8 kΩ
R1
2.2 kΩ
10 VRMS 45°
+
+
XC
22 kΩ
−
XL
2.5 kΩ
Figure 21-24
10. The impedance of the circuit of Figure 21-24 is
(a) 19.6 kΩ ∠−83.6º
(b) 29.4 kΩ ∠−41.6º
(c) 19.6 kΩ ∠83.6º
(d) 29.4 kΩ ∠41.6º
11. The current flowing through the circuit of Figure 21-24 is
(a)340 μA ∠−38.6º
(b)510 μA ∠128.6º
(c)340 μA ∠128.6º
(d)510 μA ∠−30.6º
Practice Quiz
12. The voltage drop across the inductor of Figure 21-24 is
(a) 1.27 V ∠38.6º
(b) 11.2 V ∠38.6º
(c) 11.2 V ∠−141.4º
(d) 1.27 V ∠−141.4º
13. The impedance of the circuit of Figure 21-25 is
(a)98.97 Ω ∠+8.61º
(b)108.77 Ω ∠+7.83º
(c)98.97 Ω ∠−8.61º
(d)108.77 Ω ∠−7.83º
R1
10 Ω
20 VRMS
60°
+
R2
100 Ω
XL
220 Ω
XC
330 Ω
−
Figure 21-25
14. The total current flowing through the circuit of Figure 21-25 is
(a) 183.87 mA ∠−7.83º
(b) 183.87 mA ∠−52.17º
(c) 183.87 mA ∠7.83º
(d) 183.87 mA ∠52.17º
15. The voltage across R2 in the circuit of Figure 21-25 is
(a) 18.18 V ∠43.55º
(b) 18.18 V ∠0.79º
(c) 18.18 V ∠60.79º
(d) 18.18 V ∠−16.45º
16. The current flowing through the inductor of Figure 21-25 is
(a) 82.64 mA ∠−29.21º
(b) 82.64 mA ∠−46.45º
(c) 82.64 mA ∠29.21º
(d) 82.64 mA ∠133.55º
17. The susceptance of the parallel portion in the circuit of Figure 21-25 is
(a) 9.19 mS ∠7.83º
(b) 10.11 mS ∠−8.62º
(c) 10.11 mS ∠8.62º
(d) 9.19 mS ∠−7.83º
631
22
Power in AlternatingCurrent Circuits
Most of the examples in this chapter deal with power in AC circuits
at the standard North American power-line f­ requency of 60 Hz.
Recall that power is the rate of doing work or converting energy
from one form to another. Much of the terminology in this chapter
has come from power-system applications.
Chapter Outline
22-1
22-2
Power in a Resistor
634
Power in an Ideal Inductor 635
22-3 Power in a Capacitor 637
22-4 Power in a Circuit Containing Resistance and
Reactance
639
22-5 The Power Triangle 641
22-6 Power Factor 645
22-7
Power-Factor Correction 648
Key Terms
real power 634
active power 634
reactive power 637
quadrature (or imaginary)
power 637
volt ampere reactive 637
apparent power (S) 640
volt ampere 640
power triangle 642
phasor power 643
power factor 645
power-factor angle 645
lagging power factor 645
leading power factor 645
reactive factor 647
power-factor correction 651
power-factor
improvement 651
Learning Outcomes
At the conclusion of this chapter, you will be able to:
• draw graphs of instantaneous voltage,
­current, and power for a resistor, ideal
inductor, or capacitor ­connected to an AC
circuit
• calculate the real power in a circuit with an
AC source
• explain why the average power in an ideal
inductor or capacitor in an AC circuit is zero
• calculate the reactive power in a circuit
­containing ­resistance, inductive reactance,
and capacitive reactance
• draw graphs of instantaneous voltage,
current, and power in an AC circuit containing
resistance and i­nductive reactance
• calculate the apparent power supplied by an
AC source to a circuit containing resistance,
Photo sources: CP Photo/Graham Hughes
•
•
•
•
•
inductive reactance, and capacitive
reactance
draw the power triangle for an AC circuit
­containing ­resistance, inductive reactance,
and capacitive reactance
calculate the power factor of an AC circuit
­containing ­resistance, inductive reactance,
and capacitive reactance
differentiate between lagging and leading
power factors
explain why increasing the power factor can
improve ­circuit performance
calculate the capacitance needed to
increase the power factor of a circuit to a
specified value
634
Chapter 22 Power in Alternating-Current Circuits
22-1 Power in a Resistor
At any particular instant, all the power equations we derived for DC ­cir­cuits
must apply to AC circuits. Therefore,
p = vi
(22-1)
Since the voltage across a resistor and the current through it are in phase,
v is always positive when i is positive and negative when i is negative.
Therefore the instantaneous power in a resistor is never negative (see
Figure 22-1). As shown in Section 18-8, the instantaneous power in a resistor in an AC circuit is a sine wave that has a mean magnitude of 1/2Pm and
­fluctuates at twice the frequency of the alternating current and voltage.
Section 18-11 used this mean value of power in a resistor as the basis for
the effective, or RMS, values of sine waves of voltage and current. It follows
that the mean power in the resistance of an AC circuit is simply the product
of the RMS value of the voltage drop across the resistance and the RMS value
of the current through it:
Instantaneous Value
P = VRIR (22-2)
p
+
0
π
2
π
i
3π
2 v
2π
ωt
−
Real power is also
called true power
because a resistance
converts electric
energy into heat and
light that dissipates
from the circuit, while
the energy conversions in inductance
and ­capacitance in AC
­circuits do not transfer
any energy from the
circuit.
Figure 22-1
Instantaneous power in a resistor
The average power input to the resistance of an AC circuit is referred to
as real power or active power. The symbol for mean power in an AC ­circuit
is P, the same as for power in a DC circuit.
With RMS values, R = VR / IR for AC circuits, just as for DC circuits. T
­ herefore,
P = VRIR = IR2R =
VR2
R
(22-3)
where VR is the RMS voltage across the resistance and IR is the RMS ­current
through only the resistance portion of the circuit.
22-2 Power in an Ideal Inductor
Example 22-1
Find the resistance that gives off 144 W of heat when the current through
the resistance is 2.0 A RMS.
Solution
R=
P
144 W
= 36 Ω
2 = (
I
2.0 A ) 2
See Problems 22-1 to 22-4 and Review Questions 22-38 to 22-41 at the
end of the chapter.
22-2 Power in an Ideal Inductor
Instantaneous Values
We can determine the nature of the instantaneous power in an ideal inductor
by plotting the product of the instantaneous voltage and current versus
time (see Figure 22-2). As explained in Section 19-2, the instantaneous
current in an ideal inductor lags behind the voltage across it by π/ 2 radians.
p
+
i
0
π
2
π
2π
3π
2 v
ωt
−
Figure 22-2
Instantaneous power in an ideal inductor
When ωt = 0, the instantaneous voltage across the inductance of the circuit is zero, and the rate of change of current is also zero. Between 0 and
π/2 rad, the instantaneous voltage is increasing. To generate this voltage,
the rate of change of current is also increasing as the instantaneous current i­ncreases from its minimum value to zero. During this interval the
instantaneous current is actually flowing against the applied voltage as the
magnetic field around the inductor collapses, thus returning the energy
stored in its magnetic field to the source. Since v is positive and i is negative, the instantaneous power, vi, is negative, indicating that energy is leaving the ­inductance. At π/2 rad, the current momentarily becomes zero, so
the ­instantaneous power is also momentarily zero.
Between π/2 and π rad, the instantaneous voltage across the inductor is
positive but decreasing. The instantaneous current is positive with a positive rate of change that induces the positive voltage in the inductance. The
product vi is now positive, and energy is stored in the magnetic field of
the inductor. At π radians the instantaneous voltage and the instantaneous
power momentarily become zero.
635
636
Chapter 22 Power in Alternating-Current Circuits
The instantaneous power again becomes negative between π and
3π/ 2 rad when the instantaneous current is decreasing, and the collapsing
magnetic field returns energy to the system. Between 3π/ 2 and 2π rad, both
v and i are negative, so vi is positive as energy is again transferred to the
magnetic field of the inductor.
As shown by the green curve in Figure 22-2, the instantaneous power in
the inductance varies sinusoidally at twice the frequency of the instantaneous voltage and current, like the power in the resistance of an AC circuit.
But the instantaneous power in the inductance fluctuates symmetrically
between positive and negative values as the ideal inductor alternately
stores and returns energy. Therefore, the average power in the inductance
of an AC circuit is zero.
From Equation 19-2, iL = Im sin (ωt − π/2). Substituting for v and i in
Equation 22-1 gives
p = v × i = Vm sin ωt × Im sin ωt −
(
Since sin(θ − 90°) = −cos θ and sin θ cos θ = 12 sin 2θ,
π
2
)
p = −Vmlm sin ωt cos ωt
p=
and
1
VmIm sin 2ωt
2
(22-4)
Hence, the instantaneous power in an ideal inductor is a sine wave
that has twice the frequency of the instantaneous voltage and current.
At the beginning of each cycle the power is zero and then decreases to
a negative value. Over a complete cycle, the average of sin 2ωt is zero,
confirming that the average power in an ideal inductor in an AC circuit is
zero.
In Figure 22-3, the meters show the RMS values of the voltage across and
the current through an ideal inductor. For the resistance of an AC circuit,
the average power is VRIR. However, for the inductance, VLIL is the average
amplitude of the power transferred to and from the magnetic field of the
inductance.
A
E
Figure 22-3
V
L
Measuring VL and IL
22-3 Power in a Capacitor
The product of the RMS voltage and current in an ideal inductor is
called the reactive power, quadrature power, or imaginary power of
the inductor. The symbol for reactive power is Q.
Q = VLIL
(22-5)
To help distinguish reactive power from real power, the watt is not used as
the unit for reactive power.
The unit of reactive power is the volt ampere reactive (var).
XL =
Since
VL
IL
Q = VLIL = I2LXL =
(19-3)
VL2
(22-6)
XL
where Q is the reactive power in vars, VL is the RMS voltage across the
inductance in volts, and IL is the RMS current through the inductance in
­amperes.
Example 22-2
Find the reactive power of an ideal 0.50-H inductor that draws 0.50 A
from a 60-Hz source.
Solution
XL = ωL = 377 × 0.50 H = 188.5 Ω (inductive)
Q = I2LXL = ( 0.50 A ) 2 × 188.5 Ω = 47 var
See Problems 22-5 to 22-7 and Review Questions 22-42 and 22-43.
22-3 Power in a Capacitor
The instantaneous current in a capacitor leads the instantaneous voltage
across it by π/2 rad. As with resistance and inductance, the instantaneous
power for capacitance in an AC circuit is a sine wave with twice the frequency of the instantaneous voltage and current (see Figure 22-4). During
the first quarter-cycle (from ωt = 0 to ωt = π/2), the potential difference
across the capacitor is rising and the instantaneous power is positive.
During this interval, the capacitor takes energy from the source and builds
up a charge on its plates. During the next quarter cycle, the voltage across
the capacitor decreases and the capacitor discharges the stored energy
637
Chapter 22 Power in Alternating-Current Circuits
Instantaneous Values
638
p
+
v
i
0
π
2
π
3π
2
2π
ωt
−
Figure 22-4
Instantaneous power in a capacitor
back into the ­system. Therefore, the instantaneous power is now negative.
­Similarly, the power is positive during the third quarter of the cycle and
negative during the last quarter.
Because an ideal capacitor must discharge as much energy in the second
and fourth quarter-cycles as it stores in the first and third quarter-cycles,
the average power in the capacitance of an AC circuit is zero. Substituting
for v and i in Equation 22-1 gives
p=
1
VmIm sin 2ωt 2
(22-7)
As with inductance, the product of the RMS voltage across the capacitance and
the RMS current in the capacitance is not the average power in the capacitance.
The product of the RMS voltage and current in a capacitor is the
­reactive power of the capacitor.
Q = VCIC = I2C XC =
VC2
XC
(22-8)
Example 22-3
Find the capacitance that has a reactive power of 50 var when connected across a 120-V 60-Hz source.
Solution
Since Q = VC2/XC,
XC =
Since XC = 1 /ωC,
C=
( 120 V ) 2
VC2
=
= 288 Ω ( capacitive )
Q
50 var
1
1
=
= 9.2 × 10 − 6 F = 9.2 µF
ωXC
377 rad/s × 288 Ω
See Problems 22-8 and 22-9 and Review Question 22-44.
22-4 Power in a Circuit Containing Resistance and Reactance
Circuit Check
CC 22-1. An AC current signal is given by the equation i = 250 sin 377t
mA and it flows through a 75-Ω resistor. Determine the real
power delivered to the resistor.
CC 22-2. A 10-µF and a 33-µF capacitor are connected in series across
a ­120-V 60-Hz source. Calculate the total reactive power
­delivered to the circuit.
A
22-4 Power in a Circuit Containing
Resistance and Reactance
If the series AC circuit in Figure 22-5 has equal resistance and inductive
reactance, the instantaneous current lags behind the applied instantaneous
voltage by π/4 radians (45°). As shown by the green curve in F
­ igure 22-6,
the instantaneous power is negative from 0 to π/4 rad since the inductance
is returning more energy to the system than the resistance is converting
to heat. For the interval from π/4 rad to π rad, the instantaneous power is
positive. At the peak of the instantaneous-power wave, both the resistance
and the inductance are taking energy from the source.
A
R
E
E
V
L
VL
45°
VR
(a)
Instantaneous Values
Figure 22-5
I
(b)
Ac circuit with resistance and inductance in series
p
+
i
0
π
4
π 5π
4 v
2π
ωt
−
Figure 22-6
inductance
Instantaneous power in an AC circuit with resistance and
The instantaneous-power graph for a circuit containing both resistance
and reactance is again a sine wave with twice the frequency of the voltage
and current. In Figure 22-6, the instantaneous-power graph is neither a­ lways
639
640
Chapter 22 Power in Alternating-Current Circuits
positive (as it would be for just resistance) nor symmetrically positive and
negative (as it would be for just reactance). The average power is p
­ ositive, and
it corresponds to the real power in the resistance of the ­circuit. Over a complete cycle, the average of the reactive power is zero. At any given m
­ oment,
p = VmIm sin ωt sin ( ωt + ϕ ) (22-9)
where ϕ is the phase angle of the current with respect to the applied voltage.
Since sin (a + b) = sin a cos b + cos a sin b,
p = VmIm sin ωt(sin ωt cos ϕ + cos ωt sin ϕ)
= cos ϕ (VmIm sin2 ωt) + sin ϕ (VmIm sin ωt cos ωt)
The expressions in parentheses are the instantaneous real power in the
­resistance and the instantaneous reactive power in the reactance of the circuit, respectively. Therefore,
p = pr cos ϕ + pX sin ϕ
(22-10)
where pr is the instantaneous real power in the resistance of the circuit
and pX is the instantaneous reactive power in the reactance of the circuit.
In the example above, the product of total RMS voltage and total RMS
­current represents neither real power nor reactive power. If we know only
the voltmeter and ammeter readings in Figure 22-5, we cannot determine
the real and reactive components of the power.
The product of the total RMS voltage and the total RMS current in an
AC circuit is called the apparent power of the circuit.
The symbol for apparent power is S.
S = VTIT (22-11)
At this point, we can talk only about the magnitude of the apparent power.
Because apparent power is neither real power in watts nor reactive power
in vars, apparent power is always expressed simply as the product of volts
and amperes.
The volt ampere (VA) is the unit of apparent power.
Since
VT
=Z
IT
S = VTIT = I2TZ =
(21-3)
V2T
Z
(22-12)
22-5 The Power Triangle
Example 22-4
A radio transmitter supplies a current of 8.0 A to an antenna system
with an impedance of 500 Ω ∠+ 20º. Find the apparent power in this
system.
Solution
S = I 2 Z = (8.0 A)2 × 500 Ω = 32 × 103 VA = 32 kVA
See Problems 22-10 to 22-14 and Review Questions 22-45 and 22-46.
22-5 The Power Triangle
The real power in the circuit of Figure 22-5 can be found from the product
of VR and I, and the reactive power from the product of VL and I. Since VL
leads VR by 90°,
VT = √V2R + V2L
(21-2)
Multiplying each side of Equation 21-2 by the common current gives
VTI = √( VRI ) 2 + ( VLI ) 2
Therefore,
S = √ P2 + Q2
(22-13)
Similarly, in the parallel circuit of Figure 22-7, the current in the capacitor branch leads the current in the resistor branch by exactly 90°. For this
­circuit, Equation 21-18 simplifies to
IT = √ IR2 + IC2
Multiplying each side by the common voltage gives
VIT = √ ( VIR ) 2 + ( VIC ) 2
Therefore,
S = √ P2 + Q2 (22-13)
641
642
Chapter 22 Power in Alternating-Current Circuits
IT
E
R
C
IC
ϕ
IR
(b)
(a)
Figure 22-7
E
AC circuit with resistance and capacitance in parallel
Note that Equation 22-13 applies to both series and parallel circuits. Since
power is the product of current (the common phasor of series c­ ircuits) and
voltage (the common phasor of parallel circuits), the same equations for
power apply to series and parallel circuits as well as to series-parallel circuits. Section 22-7 shows how these equations provide a simple method of
solving series-parallel circuits without having to reduce them to equivalent
series or parallel circuits.
Although instantaneous power varies sinusoidally, it does so at twice the
frequency of the voltage and current. Consequently, we cannot draw power
phasors on the same phasor diagram with voltage and current. Therefore,
it is customary to represent the Pythagorean relationship of Equation 22-13
with right triangles, as shown in Figure 22-8.
In a power triangle, the horizontal side represents real power. The vertical side represents reactive power and forms a right angle at the right-hand
end of the side representing real power. Since Equation 22-13 applies to
either a series or a parallel circuit, we can develop a power triangle from
either an impedance diagram or an admittance diagram. If we start with an
impedance diagram, in which inductive reactance is drawn in the +j direction, the power triangle appears as in Figure 22-8(a). But if we start from
an ­admittance diagram, in which inductive susceptance is drawn in the −j
direction, the power triangle has the form shown in F
­ igure 22-8(b). To avoid
confusion between inductive and capacitive r­eactive power in power triangle ­diagrams, we must choose one of these formats for all AC circuits. We
will follow the modern North American convention of showing inductive
­reactive power in the +j direction. With this convention, the source current
is the reference phasor when we are dealing with power in AC circuits.
A)
r (V
t
ren
pa
Ap
e
pow
ϕ
Inductive
reactive
power (vars)
Real power (watts)
ϕ
Ap
par
ent
pow
er
(VA
)
Inductive
reactive
power (vars)
Real power (watts)
(a)
(b)
Figure 22-8 Power triangle with (a) common current as reference axis and
(b) voltage as reference axis
22-5 The Power Triangle
We can now determine the phase angle, ϕ, of the apparent power. With
the power triangle drawn as shown in Figure 22-8(a), S has the same angle
as the impedance of the circuit and as the conjugate of the admittance of the
circuit (in other words, the angle of the admittance is −ϕ). When S is treated
as a complex number, it is called the phasor power of the circuit. Thus,
S = P + jQ (22-14)
We can now see where the terms real power for active power and quadrature
(or imaginary) power for reactive power originated. Since inductive reactive
power has a +j operator, capacitive reactive power has a −j operator.
Since the voltage across the capacitance in a series circuit is 180° out of
phase with the voltage across the inductance, the net reactive voltage is the
difference between the two reactive voltages. Similarly, in a simple parallel
circuit, the net reactive current is the difference between the capacitive and
inductive branch currents. Since Q = VXIX, the net reactive power in both
series and parallel AC circuits is the difference between the inductive reactive
power and the capacitive reactive power.
In Figure 22-2 the instantaneous power in an inductance is positive when
the current is rising and building up a magnetic field around the i­ nductor.
But Figure 22-4 shows that, when the current is rising, the capacitance is
discharging its stored energy back into the system. Therefore, in an AC circuit containing both inductance and capacitance, the capacitance always
returns energy to the circuit when the inductance takes energy from the
circuit, and vice versa. Consequently, some energy is transferred back and
forth between the inductance and the capacitance of the circuit, and the net
reactive power to and from the source is the difference between the inductive reactive power and the capacitive reactive power.
Example 22-5
A circuit has two branches connected in parallel to a 120-V 60-Hz
source. Branch 1 consists of a 75-Ω resistance and a 100-Ω inductive reactance in series, and branch 2 consists of a 200-Ω capacitive reactance.
Determine the apparent power of the circuit.
Solution I
Step 1
Draw a circuit diagram as shown in Figure 22-9. Then calculate the current in branch 1:
Z1 = R + jXL = 75 + j100 = 125 Ω ∠+53.1º
I1 =
E
120 V
=
= 0.960 A
Z1
125 Ω
643
Source: Courtesy of Global
Specialties
644
Chapter 22 Power in Alternating-Current Circuits
Inductive Q
XL
A wattmeter for measuring
power in an AC circuit
ϕ
100 Ω
120 V
60 HZ
XC
R
200 Ω
75 Ω
S
92.2 var
Net Q
P = 69.1 W
(a)
Figure 22-9
72 var
Capacitive Q
(b)
Diagrams for Example 22-5
Step 2
Draw a power triangle and calculate the net reactive power.
Real power of the circuit is
P = I21R = 0.960 2 × 75 = 69.1 W
Reactive power of the inductance is
Q = I21XL = 0.960 2 × 100 = 92.2 var
Reactive power of the capacitance is
Q=
E2
1202
= 72.0 var
=
XC
200
Net reactive power of the circuit is
Q = 92.2 − 72.0 = 20.2 var (inductive)
Step 3
Calculate the apparent power:
S = √P2 + Q2 = √69.12 + 20.22 = 72 VA
Solution II
A somewhat similar procedure involves solving Steps 2 and 3 in terms
of current rather than power.
Step 1
Z1 = R + jXL = 75 + j100 = 125 Ω ∠+ 53.1°
I1 =
120∠0°
E
=
= 0.960 A ∠−53.1°
Z1
125∠+53.1°
I1 = 0.58 − j0.77 A
22-6 Power Factor
Step 2
IC =
645
120 ∠0º
E
=
= 0.60 A ∠+90° = 0 + j0.6 A
−jXC
200 ∠−90°
IT = I1 + IC = ( 0.58 − j0.77 ) + ( 0 + j0.60 ) = 0.58 − j0.17 A
and
Step 3
IT = √0.582 + 0.172 = 0.60 A
S = ETIT = 120 × 0.60 = 72 VA
To verify the apparent power of the circuit, download Multisim file
EX22-5 from the website and follow the instructions in the file.
See Problems 22-15 and 22-16 and Review Questions 22-47 to 22-52.
22-6 Power Factor
Since the hypotenuse of a right triangle must be longer than either of the
other two sides, the apparent power that a generator must supply to a reactive load is always greater than the real power that the load can convert
into some other form of energy. This relationship can be quite important
since most industrial loads have appreciable inductive reactance.
The power factor is the ratio between the real power and the apparent
power of a load in an AC circuit.
The power triangle of Figure 22-8(a) shows that the ratio of real power to
apparent power is the cosine of the power-factor angle between real power
and apparent power. Tracing the construction of the power triangle back
through the impedance diagram to the phasor diagram for current and voltage, we find that the power-factor angle is the same as the phase angle, ϕ,
between the voltage across and the current through the load.
From the definition of power factor,
power factor =
P
= cos ϕ
S
(22-15)
We distinguish between inductive and capacitive loads by stating that inductive loads always have a lagging power factor because their currents lag
behind their voltages, and that capacitive loads always have a leading power
factor because their currents lead their voltages. Since the real power cannot
be greater than the apparent power, the power factor cannot be greater than
1. It can be expressed as either a decimal fraction or a ­percentage.
circuitSIM
walkthrough
646
Chapter 22 Power in Alternating-Current Circuits
Example 22-6
Find the real power and power factor of a load whose impedance is
60 Ω ∠+60° when connected to a 120-V 60-Hz source.
Solution I
Step 1
Convert the impedance to rectangular coordinates to find an equivalent circuit consisting of resistance and reactance in series, as shown in
Figure 22-10.
Z = 60 cos 60° + j60 sin 60°
Therefore,
120 V
60 HZ
Figure 22-10
Step 2
Step 3
R = 60 cos 60° = 60 × 0.500 = 30 Ω
60 Ω 60° Z
120 V
60 HZ
XL = 60 sin 60°
R = 60 cos 60°
Schematic diagram for Example 22-6
I=
120 V
E
=
= 2.0 A
Z
60 Ω
P = I2R = 4.0 × 30 = 120 W
S = I2Z = 4.0 × 60 = 240 VA
cos ϕ =
P
120
=
= 0.50 lagging
S
240
Although Solution I uses the definition of power factor, there is a more
­direct way to solve this example.
Solution II
Step 1
As shown in Figure 22-11,
Step 2
power factor = cos ϕ = cos 60° = 0.50 lagging
S=
E2
1202
=
= 240 VA
Z
60
S=
240
VA
22-6 Power Factor
647
Q
60°
P = 120 W
Figure 22-11
Power triangle for Example 22-6
Step 3
Since cos ϕ = P/S,
P = S cos ϕ = 240 × 0.50 = 0.12 kW
Multisim Solution
To verify the real power and power factor of the load, download Multisim
file EX22-6 from the website and follow the instructions in the file.
As shown in the second solution, we can express real power in terms
of power factor and apparent power. Since S = EI, the real power in an AC
­circuit is
P = EI cos ϕ (22-16)
Once we know the real power and apparent power of an AC load, we can
determine reactive power:
Q = √ S2 − P2 (22-13)
Q = S√ 1 − cos2ϕ (22-17)
Substituting P = S cos ϕ gives
The ratio of reactive power to apparent power can be useful when we wish
to solve directly for the reactive power of the load.
The reactive factor is the ratio of the reactive power to the apparent
power of an AC load.
circuitSIM
walkthrough
648
Chapter 22 Power in Alternating-Current Circuits
From the power triangle in Figure 22-8(a) we can see that
reactive factor =
Q
= sin ϕ S
(22-18)
See Problems 22-17 to 22-29 and Review Questions 22-53 to 22-56.
22-7 Power-Factor Correction
The consequences of a low power factor in an industrial load are illustrated by Figure 22-12. Since the motor in Figure 22-12(a) has a 70% lagging
power factor, the apparent power is
S=
840
P
=
= 1.2 kVA
cos ϕ
0.70
Therefore, the current drain on the source is
I=
S
1.2 kVA
=
= 10 A
V
120 V
10 A
7A
A
A
840 W
M 70% lagging
power factor
120 V
60 HZ
(a)
Figure 22-12
840 W
M 100%
power factor
120 V
60 HZ
(b)
Effect of load power factor
If we replace the original motor by one capable of developing a real power
of 840 W at a 100% power factor, as in Figure 22-12(b), the apparent power
is reduced by 840 VA and the current is only 7.0 A. The motor with a 70%
power factor requires a current of 10 A to do the same work that the motor
with a 100% power factor can do while drawing 7 A from the source.
The size of the wires required for the windings of a generator and for the
conductors connecting the load to the generator depends on the current
the wires have to carry. The energy lost as heat in the wires depends on the
square of the current. Therefore it is more efficient and more economical
to feed power to a load with a 100% power-factor load than to a load with
a feed of a 70% power factor. Electrical utility companies are very much
concerned with maintaining a high overall power factor in their systems.
22-7 Power-Factor Correction
The additional current for the motor with a 70% lagging power alternately transfers energy from the system into the magnetic field of the inductance and then returns the stored energy back to the system. Although
the average power of this process is zero, the reactive current component
still has to flow through the windings and connecting conductors. Thus
the reactive current increases the size of wire needed to deliver a given real
power to the load.
We can determine the reactive current components by expressing the
current of each of the motors in rectangular coordinates. In the motor with
a 70% lagging power factor, the current lags behind the applied voltage by
cos−1 0.70 = 45.6°, and
I = 10 cos 45.6° − j10 sin 45.6° = 7.0 − j 7.14 A
In a motor with a 100% power factor, the current is in phase with the
­applied voltage:
I = 7.0 − j 0 A
The current in each motor has the same reference axis component. This
component carries the real power to the load. The 70% power-factor load
has an additional quadrature component that carries the reactive power.
In an AC circuit containing both inductance and capacitance, the capacitance returns energy to the system while the inductance takes energy from
the system, and vice versa. If a capacitor is connected to the motor with
a 70% lagging power factor, the reactive power can flow back and forth
­between the capacitor and the inductance of the motor, thus reducing or
eliminating the reactive component of the current that travels all the way
from the source to the load and back. This arrangement would allow us to
use lower power-factor loads yet obtain the advantages of a high system
power factor.
Consider the effect of connecting a 308-μF capacitor in series with the
motor that has a 70% lagging power factor, as shown in Figure 22-13.
The impedance of the motor is
ZM = V/ I =
120
= 12 Ω
10
ϕ = cos−1 0.7 = 45.6°
ZM = 12 Ω ∠+45.6° = 12 cos 45.6°+ j12 sin 45.6° = 8.4 + j8.6 Ω
XC =
1
1
=
= 8.6 Ω ( capacitive )
ωC
377 × 308 µF
649
650
Chapter 22 Power in Alternating-Current Circuits
Practical Circuits
Power-Factor Correction
Source: CP Photo/Graham Hughes
The power factor is a measure of how effectively devices operated by a consumer convert
electric current from the local utility to useful
power output, such as heat, light, or mechanical motion. Power-factor correction is the practice of raising the power factor of an inductive
load, such as a motor, by inserting a capacitor
in parallel with it. The lower the power factor,
the greater the current required to deliver a
given amount of energy.
Electricity suppliers usually bill commercial
and industrial customers a power factor surcharge if the power factor of the customer’s
load drops below 90%. The power supplied to
these customers is measured with a meter that
monitors both the usable power and the reactive power, and hence can indicate the power
factor each month.
Power meter
14.3 A
A
120 V
60 HZ
M
70% lagging
power factor
308 μF
Figure 22-13
Effect of a series capacitor on a lagging power-factor load
Because the motor and capacitor are in series,
ZT = 8.4 + j8.6 − j8.6 = 8.4 + j0 Ω = 8.4 Ω ∠0º
22-7 Power-Factor Correction
Since the phase angle of the total impedance is now 0°, the current drawn
from the generator is in phase with the source voltage. However,
I=
V
120 V
= 14.3 A
=
ZT
8.4 Ω
Therefore, connecting a capacitor in series with an inductive load reduces the net reactive power but increases the current drain on the source.
Moreover, since the impedance of the motor itself is still 12 Ω, the voltage
across the motor in Figure 22-13 becomes
VM = IZM = 14.3 A × 12 Ω = 172 V, or 0.17 kV
Although connecting the capacitor in series with the motor raises the system
power factor to 100%, this arrangement aggravates the original problem by
increasing the current drawn from the source and applying excessive voltage to the motor. To be practical, power-factor correction or improvement
must reduce the apparent power drawn from the source without altering
the voltage across the load.
One of the characteristics of a parallel circuit is that a change in one
branch does not affect the current in the other branches. In Figure 22-14,
a capacitor is connected in parallel with the motor that has a 70% lagging
power factor. Since the motor is still connected directly across the 120-V
source, the motor current is still 10 A, lagging behind the applied voltage
by 45.6°. Because of the lagging power factor, the motor has a reactive
power of
Q = VI sin ϕ = 120 × 10 × 0.714 = 860 var
7A
7.14 A
A
A
A 10 A
120 V
60 HZ
Figure 22-14
840 W
M 70% lagging
power factor
Power-factor correction
158 μF
651
652
Chapter 22 Power in Alternating-Current Circuits
If the capacitor in parallel with the motor also has a reactive power of
860 var, all of the reactive power is transferred back and forth between
the capacitor and the motor, and the net reactive power measured at the
source is zero. Since the voltage across and the current through the motor
are unchanged, the real power is still 840 W. Therefore the overall apparent
power that the source must supply is 840 VA, and the current drawn from
the generator is 7.0 A. Note that the motor current is still 10 A. As far as the
source is concerned, the motor and capacitor in parallel are equivalent to a
motor with a power factor of 100%.
We can find the required capacitance by noting that the capacitor
­current must equal the reactive component of the load current or that
the susceptance of the capacitor must equal the susceptance component
of the load admittance. However, the simplest procedure is to work in
terms of reactive power since the power equations apply to both series
and parallel circuits.
To obtain a power factor of 1 when a load has a lagging power factor,
a capacitor with a reactive power equal to the reactive power of the
load is connected in parallel with the load.
We can use a
carefully drawn
phasor diagram to
verify that adding
the capacitance does
reduce the total
­current drawn from
the source without
affecting the load
current.
Example 22-7
Find the capacitance that minimizes the source current when connected in parallel with a motor drawing 10 A at a 70% lagging power
factor from a 120-V 60-Hz source.
Solution
Step 1
The reactive power of the motor is
Q = VI√ 1 − cos2 ϕ = 120 × 10√ 1 − 0.702 = 857 var (inductive)
Therefore, the capacitor must have a reactive power of 857 var when
connected across a 120-V 60-Hz source.
Step 2
Since Q = V 2/XC,
XC =
( 120 V ) 2
V2
=
= 16.8 Ω ( capacitive )
Q
857 var
22-7 Power-Factor Correction
C=
1
1
=
= 1.58 × 10 − 4 F = 158 µF
ωXC
377 × 16.8 Ω
To verify the capacitance, download Multisim file EX22-7 from the
website and follow the instructions in the file.
Example 22-8
A fluorescent lamp and its ballast inductance draw a 1.0-A current
at a 50% lagging power factor from a 120-V 60-Hz source. Find the
overall power factor when a 26.5-μF capacitor is connected across the
fixture.
Solution
Step 1
Calculate S, P, and Q for the lamp and its ballast, and draw the power
­triangle for them:
S = VI = 120 V × 1.0 A = 120 VA
P = S cos ϕ = 120 VA × 0.50 = 60 W
Q = √ S2 − P2 = √ 1202 − 602 = 104 var ( inductive )
Step 2
Calculate the reactance and reactive power for the capacitor:
XC =
and
QC =
1
1
=
= 100 Ω ( capacitive )
ωC
377 × 26.5 µF
V2
1202
=
= 144 var ( capacitive )
XC
100
Step 3
From the power triangle of Figure 22-15,
Final Q = 144 − 104 = 40 var (capacitive)
Step 4
653
Final S = √ P2 + Q 2 = √ 60 2 + 40 2 = 72 VA
overall power factor = cos ϕ =
60
P
=
= 83% leading
S
72
circuitSIM
walkthrough
Chapter 22 Power in Alternating-Current Circuits
0V
A
654
Q of
capacitor
144 var
So
f la
mp
=
12
Q of
ballast
104 var
P = 60 W
Ov
era
ll S
Figure 22-15
circuitSIM
walkthrough
Net Q
40 var
capacitive
Power triangle for Example 22-8
To verify the overall power factor, download Multisim file EX22-8 from
the website and follow the instructions in the file.
Example 22-9
Find the capacitance that raises the overall power factor to 91% lagging
when connected in parallel with a load drawing 1.0 kW at a 70.7% lagging power factor from a 208-V 60-Hz source.
Solution
Step 1
Original S =
P
1000 W
=
= 1.41 kVA
cos ϕ
0.707
Load Q = √ S2 − P2 = √ 1.412 − 1.02 = 1.0 kvar ( inductive )
Step 2
Final S =
P
1000 W
=
= 1.1 kVA
cos ϕ
0.91
Overall Q = √S2 − P2 = √1.12 − 1.02 = 456 var ( inductive )
Step 3
From the power triangle of Figure 22-16,
QC = 1000 − 456 = 544 vars (capacitive)
Therefore,
XC =
V2
2082
=
= 79.5 Ω ( capacitive )
QC
544
22-7 Power-Factor Correction
C=
and
ri
O
al
n
gi
1
1
=
= 3.3 × 10 − 5 F = 33 µF
ωXC
377 × 79.5 Ω
Q of
capacitor
544 var
S
al
Fin
655
Q of load
1.0 kvar
S
Overall Q
456 var
P = 1.0 kW
Figure 22-16
Power triangle for Example 22-9
circuitSIM
To verify the capacitance, download Multisim file EX22-9 from the
website, and follow the instructions in the file.
walkthrough
See Problems 22-30 to 22-35 and Review Questions 22-57 to 22-60.
Circuit Check
B
CC 22-3. For the network shown in Figure 22-17, determine
(a) the total real power
(b) the reactive power
(c) the apparent power
(d) the power factor
(e) the capacitance that will make the power factor unity if
connected in parallel with the network
4 – j2 Ω
10 A 0°
120 Hz
Figure 22-17
6 – j8 Ω
8 – j7 Ω
656
Chapter 22 Power in Alternating-Current Circuits
CC 22-4. For the circuit of Figure 22-18, find the
(a) total real power
(b) reactive power
(c) apparent power
(d) power factor
(e) current in phasor form
I
+
Load 1
0 var
100 W
Load 2
E = 100 V 30°
1500 var
200 W
(ind)
Load 3
700 var
200 W
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Figure 22-17
CC 22-5. The load on a 110-V 60-Hz supply is 5.0 kW (resistive), 10 kvar
(inductive), and 5.0 kvar (capacitive).
(a) Find the total apparent power.
(b) Determine the power factor of the combined loads.
(c) Find the current drawn from the supply.
(d) Calculate the capacitance required to establish a unity
power factor.
(e) Find the current drawn from the supply when the power
­factor is unity.
Problems
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Summary
• The average power in the resistance of an AC circuit is the real or active
power of the circuit.
• The average power in an ideal inductor or capacitor in an AC circuit is zero.
• Reactive power in an ideal inductor or capacitor due to an AC source is
the product of the RMS voltage and current.
• The unit for reactive power is the volt ampere (reactive) or var.
• Apparent power is the power supplied by an AC source to a circuit
­con­taining resistance and reactance.
• The unit of apparent power is the volt ampere (VA).
• A circuit containing resistance and reactance has both real and reactive
power.
• A power triangle shows the relationships among apparent power, real
power, and reactive power in an AC circuit.
• The power factor of a load in an AC circuit is the ratio of real power to
­apparent power.
• The power factor is the cosine of the phase angle between the voltage
across and the current through the load.
• The reactive factor is the ratio of reactive power to apparent power.
• The reactive factor is the sine of the phase angle between the voltage
across and the current through the load.
• A capacitor connected in parallel with an inductive load can increase the
power factor and reduce the total current drawn from the AC source.
Problems
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Section 22-1
Power in a Resistor
22-1. A toaster draws a 6.0-A current from a 110-V 60-Hz source.
(a) Calculate the real power of the toaster.
(b) Calculate the peak value of the instantaneous-power input to
the toaster.
22-2. The voltage drop across the heater of a certain cathode-ray tube is
6.3 V RMS when the current through it is 0.3 A. Find the average rate
of conversion of electric energy into heat.
22-3. Determine the real power dissipated by a 120-Ω resistor when a
sine-wave current of i = 2.7 sin 377t A flows through it.
22-4. What power rating must a 300-Ω dummy antenna resistor possess if
the RMS voltage drop across it is 96 V?
Section 22-2
Power in an Ideal Inductor
22-5. An inductance of 3.0 mH passes a 1.0-kHz sine-wave current of 20 mA.
(a) Find the average power input to the inductor.
(b) Find the reactive power.
(c) Find the peak rate at which the inductor stores energy.
22-6. A 20-mH inductor is connected to a 10-V 1-kHz source. Calculate the
reactive power.
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22-7. At what frequency will a 0.25-H inductor take a reactive power of
25 var from a 40-V source?
Section 22-3
Power in a Capacitor
22-8. A 68-μF capacitor is connected across a 230-V 50-Hz source.
(a) Determine the reactive power of the capacitor.
(b) If a 33-μF capacitor is connected in parallel with it, determine the
total reactive power of the circuit.
22-9. An industrial capacitor is rated at 7.5 kvar when connected to a 208-V
60-Hz circuit. Find the capacitance.
Section 22-4 Power in a Circuit Containing Resistance and
Reactance
22-10. (a) A solenoid having an inductance of 0.50 H and a resistance of
24 Ω is connected to a 120-V 60-Hz source.
(a) Calculate the apparent power input to the solenoid.
(b) Calculate the real power input to the solenoid.
(c) Calculate the reactive power input to the solenoid.
(d) Use Multisim to verify the apparent power in part (a) and the
real power in part (b).
22-11. Find the average power input to the antenna in Example 22-4.
22-12. The reactive power in an impedance of 480 Ω ∠−60º is 300 mvar.
At what rate is energy being dissipated by the impedance?
22-13. A coil with an internal resistance of 8 Ω and a 22-mH inductance is
connected across a 120-V 60-Hz source. Find
(a) the circuit current in polar coordinates
(b) the real, reactive, and apparent power of the circuit
22-14. A 250-V 60-Hz power supply is connected to a circuit consisting of
a 40-Ω resistance in series with a 0.80-H inductance and an 8.0-μF
capacitance. Calculate
(a) the current and phase angle
(b) the apparent power, real power, and reactive power
Section 22-5
The Power Triangle
22-15. A series RLC circuit is formed by a 680-pF capacitor, a 470-mH inductor, and a 500-Ω resistor connected to a 300-V AC source. At
what frequency will the net reactive power be zero?
22-16. (a)A series RC circuit has a current of 1.5 A, an applied voltage of
120 V at 60 Hz, and a power dissipation of 50 W. Find the resistance and capacitance.
(b) Use Multisim to verify the resistance and capacitance.
Section 22-6
Power Factor
22-17. A relay coil with a power factor of 75% lagging draws 1 A from a 12-V
500-Hz source. Find the resistance and inductance of the relay coil.
Problems
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22-18. (a)What series combination of resistance and inductance draws
the same current from a 120-V 60-Hz source as a 500-W load
with a 60% lagging power factor?
(b) Use Multisim to verify the resistance and inductance.
22-19. What parallel combination of resistance and inductance draws the
same current from a 120-V 60-Hz source as a 500-W load with a
60% lagging power factor?
22-20. (a)What current must a 110-V alternator supply to operate a 2.0-kW
load with a 75% lagging power factor?
(b) What real power could the alternator supply for the same magnitude of current in its windings to a load with a power factor
of 1?
22-21. An apparent power input of 25 kVA must be provided to the “work
coil” of a radio-frequency induction heater to generate heat in the
steel work piece at the rate of 5 kW. Find the reactive factor of the
load.
22-22. A crossover network used to feed the woofer (low-frequency driver)
and the tweeter (high-frequency driver) in a speaker system has an
8.0-Ω resistor and a capacitor in parallel. The power input to these
two components at 1.0 kHz is 500 mW with a 20% leading power
factor. Find the capacitance.
22-23. An impedance coil having a 0.2 lagging power factor is connected
in series with a 300-W lamp in order to supply the lamp with 120 V
from a 208-V 60-Hz source. Find the voltage across the terminals of
the impedance coil.
22-24. (a)An induction motor that draws 2.0 A from 120-V 60-Hz source
at a 0.8 lagging power factor is connected in parallel with a
100-W lamp. Calculate the overall power factor.
(b) Use Multisim to verify the overall power factor.
22-25. The ratio of the reactive power to the real power of a coil is known
as its quality factor, Q. Determine the quality factor of the coil in
Problem 22-15 at a frequency of 2.0 kHz.
22-26. (a)A series RLC circuit connected to a 500-V 60-Hz supply draws a
current of 10 A. A power-factor meter indicates a leading power
factor of 0.8. If the voltage across the capacitor is 800 V, find the
value of each component.
(b) Use Multisim to verify the resistance, inductance, and capacitance.
22-27. A parallel RLC circuit connected to a 203-V 50-Hz source draws 3 A
of current. A power-factor meter reads 0.8 leading and the inductive reactive power is 400 var. Determine the resistance, capacitance,
and inductance of the circuit.
22-28. An RL-series circuit draws 1.2 A from a 250-V AC source. A wattmeter in the circuit reads 135 W. What is the power factor and angle
of lag?
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22-29. An inductive circuit draws a current of 30.0 A at a power factor of
0.5 lagging when connected to a 5-kV 30-Hz supply. Find the resistance and inductance of the circuit.
Section 22-7
Power-Factor Correction
22-30. An induction motor draws 6.0 A at a 0.8 lagging power factor from
a 208-V 60-Hz source.
(a) What value of capacitance must be connected in parallel with
the motor to raise the overall power factor to 1?
(b) Calculate the magnitudes of the motor current, capacitor current, and source current with the capacitor in place.
(c) Use Multisim to verify the calculation of the capacitance in part
(a) and the currents in part (b).
22-31. What value of capacitance is necessary to produce an overall power
factor of 0.96 lagging with the motor in Problem 22-30?
22-32. (a)Calculate the overall power factor when a 50-mF capacitor is
connected in parallel with the motor in Problem 22-30.
(b) Use Multisim to verify the overall power factor.
22-33. A synchronous motor capable of operating with a leading power
factor draws 15 kW from a distribution transformer while driving
an air compressor. The remainder of the load on the transformer is
80 kW at a 0.85 lagging power factor.
(a) How much capacitive reactive power must the synchronous
motor produce to raise the overall power factor to 0.96 lagging?
(b) Find the reactive factor of the synchronous motor when it operates in this manner.
22-34. The power factor of a load connected to a 120-V 60-Hz source is
raised from 0.707 lagging to 0.866 lagging by connecting a 53-μF
capacitor across the load. Find the real power of the load.
22-35. The power factor of a load connected to a 120-V 60-Hz source is
raised from 0.866 lagging to 0.966 leading by connecting a 110.5-μF
capacitor in parallel with the load. Determine the RMS load current.
22-36. A 500-V 60-Hz induction motor has a power factor of 0.707 lagging.
The motor draws a current of 10 A when fully loaded. Calculate the
required capacitor to raise the power factor to 0.9 lagging. Find the
new line current.
22-37. A lighting load operates from a 208-V 60-Hz supply and is rated
1.0 kW with a power factor of 0.9 lagging. A 40-µF capacitor is connected in parallel with this load. Calculate the total power factor
and the percent change in line current.
Review Questions
Section 22-1
Power in a Resistor
22-38. Why can we state that p = vi in any AC circuit?
22-39. Why is the instantaneous power in resistance always a positive
quantity?
Review Questions
22-40. Why can we say that the average power in the resistance in a sinewave AC system is one-half the peak power?
22-41. Why is the product VR × IR called the real power?
Section 22-2
Power in an Ideal Inductor
22-42. Describe how you could use an AC voltmeter and an AC ammeter
to determine the peak instantaneous power in an ideal inductor in
an AC circuit.
22-43. Why is the average power of an ideal inductor zero?
Section 22-3
Power in a Capacitor
22-44. What is the meaning of the term reactive power?
Section 22-4 Power in a Circuit Containing Resistance and
Reactance
22-45. What is the meaning of the term apparent power?
22-46. Why is apparent power not expressed in watts?
Section 22-5
The Power Triangle
22-47. What is the apparent power of an ideal resistor?
22-48. What is the apparent power of an ideal capacitor?
22-49. Show that the net reactive power of an AC circuit is the difference
between the inductive reactive power and the capacitive reactive
power.
22-50. Why is the total real power of a network always the sum of the individual real powers regardless of how the components are connected?
22-51. Why is the apparent power of an AC circuit the square root of the
sum of the squares of the real and reactive powers rather than the
simple sum?
22-52. Why is the angle between the real power and apparent power in a
power triangle the same as the angle between the total voltage and
total current of a load?
Section 22-6
Power Factor
22-53. Why do we need to consider power factors when designing industrial wiring?
22-54. What is meant by a leading power factor?
22-55. Of what practical use is the reactive factor?
22-56. Derive an expression for reactive factor in terms of power factor.
Section 22-7
Power-Factor Correction
22-57. Why are AC generators and transformers rated in kilovolt amperes
rather than kilowatts?
22-58. Why is connecting a capacitor in series with an inductive load not
a satisfactory means of power-factor correction?
22-59. What is the purpose of power-factor correction?
22-60. Why does adding capacitance in parallel with an inductive load
change the power factor?
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A series RLC circuit consisting of a 50-Ω