Instructor’s Resource Manual to accompany
Principles of Electric Circuits 9th Edition Tom Floyd
Upper Saddle River, New Jersey Columbus, Ohio
______________________________________________________________________________ Copyright © 2010 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice Hall™ is a trademark of Pearson Education, Inc. Pearson® is a registered trademark of Pearson plc Prentice Hall® is a registered trademark of Pearson Education, Inc.
Instructors of classes using Floyd, Principles of Electric Circuitss, ninth edition, may reproduce material from the Instructor’s Manual for classroom use.
10 9 8 7 6 5 4 3 2 1
ISBN-13: ISBN-10:
978-0-13-507330-8 0-13-507330-8
Part 1 Solutions for End-of-Chapter Problems 1 Part 2 Solutions for Application Activities
219
Part 3 Summary of Multisim Circuit Results
257
Part 4 Test Item File
281
Part 5 Lab Solutions for Experiments in Basic Circuits
461
Part 6 Lab Solutions for Experiments in Electric Circuits
522
PART 1 Solutions for End-of-Chapter Problems
Chapter 1
Chapter 1 Quantities and Units Section 1-2 Scientific Notation 1.
(a)
3000 = 3 103
2.
(a)
1 = 0.002 = 2 103 500
(b)
1 = 0.0005 = 5 104 2000
(c)
1 = 0.0000002 = 2 107 5,000,000
3.
(a)
8400 = 8.4 103
(b)
99,000 = 9.9 104
4.
(a)
0.0002 = 2 104
(b)
0.6 = 6 101
(c)
7.8 102 (already in scientific notation)
(a)
32 103 = 3.2 104
(b)
6800 106 = 6.8 103
(c)
870 108 = 8.7 1010
(a)
2 105 = 200,000
(b)
5.4 109 = 0.0000000054
(c)
1.0 101 = 10
7.
(a)
2.5 106 = 0.0000025
8.
(a)
4.5 106 = 0.0000045
(b)
8 109 = 0.000000008
(c)
4.0 1012 = 0.0000000000040
5.
6.
(b)
75,000 = 7.5 104
(b)
5.0 102 = 500
2
(c)
2,000,000 = 2 106
(c)
0.2 106 = 2 105
(c)
3.9 101 = 0.39
Chapter 1 9.
10.
11.
12.
(a)
9.2 106 + 3.4 107 = 9.2 106 + 34 106 = 4.32 107
(b)
5 103 + 8.5 101 = 5 103 + 0.00085 103 = 5.00085 103
(c)
5.6 108 + 4.6 109 = 56 109 + 4.6 109 = 6.06 108
(a)
3.2 1012 1.1 1012 = 2.1 1012
(b)
2.6 108 1.3 107 = 26 107 1.3 107 = 24.7 107
(c)
1.5 1012 8 1013 = 15 1013 8 1013 = 7 1013
(a)
(5 103)(4 105) = 5 4 103 + 5 = 20 108 = 2.0 109
(b)
(1.2 1012)(3 102) = 1.2 3 1012 + 2 = 3.6 1014
(c)
(2.2 109)(7 106) = 2.2 7 10 9 6 = 15.4 1015 = 1.54 1014
(a)
1.0 103 = 0.4 103 2 = 0.4 101 = 4 2.5 102
(b)
2.5 106 = 0.05 106 (8) = 0.05 102 = 5 50 108
(c)
4.2 108 = 2.1 108 (5) = 2.1 1013 2 105
13.
(a) (b) (c)
8 10 4 2 10 80 10 4 2 10 84 10 3 10 5 10 9 10 15 10 9 10 6 10 2.2 10 1.1 5.5 10 1110 4
2
3
7
5
12
12
2
4
2
12
12
4
Section 1-3 Engineering Notation and Metric Prefixes 14.
The powers of ten used in engineering notation are multiples of 3: 10-12, 10-9, 10-6, 10-3, 103, 106, 109, 1012
15.
(a) 89000 = 89 103 (b) 450,000 = 450 103 (c) 12,040,000,000,000 = 12.04 1012
3
3
2 102 4.20 102
Chapter 1 16.
(a) 2.35 105 = 235 103 (b) 7.32 107 = 73.2 106 (c) 1.333 109 (already in engineering notation)
17.
(a) 0.000345 = 345 106 (b) 0.025 = 25 103 (c) 0.00000000129 = 1.29 109
18.
(a) 9.81 103 = 9.81 103 (b) 4.82 104 = 482 106 (c) 4.38 107 = 438 109
19.
(a) 2.5 103 + 4.6 103 = (2.5 + 4.6) 103 = 7.1 103 (b) 68 106 + 33 106 = (68 + 33) 106 = 101 106 (c) 1.25 106 + 250 103 = 1.25 106 + 0.25 106 = (1.25 + 0.25) 106 = 1.50 106
20.
(a) (32 103)(56 103) = 1792 10(3 + 3) = 1792 100 = 1.792 103 (b) (1.2 106)(1.2 106) = 1.44 10(6 6) = 1.44 1012 (c) (100)(55 103) = 5500 103 = 5.5
21.
22.
(a)
50 = 22.7 103 2.2 103
(b)
5 103 = 0.2 10(3 (6)) = 0.2 109 = 200 106 25 10 6
(c)
560 103 = 0.848 10(3 3) = 0.848 100 = 848 103 660 103
(a)
89,000 = 89 103 = 89 k
(b)
450,000 = 450 103 = 450 k
(c)
12,040,000,000,000 = 12.04 1012 = 12.04 T
4
Chapter 1 (a)
0.000345 A = 345 106 A = 345 A
(b)
0.025 A = 25 103 A = 25 mA
(c)
0.00000000129 A = 1.29 109 A = 1.29 nA
24.
(a)
31 103 A = 31 mA
(b)
5.5 103 V = 5.5 kV
(c)
20 1012 F = 20 pF
25.
(a)
3 106 F = 3 F
(b)
3.3 106 = 3.3 M
(c)
350 109 A = 350 nA
26.
(a)
2.5 1012 A = 2.5 pA
(b)
8 109 Hz = 8 GHz
(c)
4.7 103 = 4.7 k
(a)
7.5 pA = 7.5 1012 A
(b)
3.3 GHz = 3.3 109 Hz
(c)
280 nW = 2.8 107 W
(a)
5 A = 5 106 A
(b)
43 mV = 43 103 V
(c)
275 k = 275 103
(d)
10 MW = 10 106 W
23.
27.
28.
Section 1-4 Metric Unit Conversions 29.
30.
(a)
(5 mA) (1 103 A/mA) = 5 103 A = 5000 A
(b)
(3200 W)(1 103 W/W) = 3.2 mW
(c)
(5000 kV)(1 103) MV/kV = 5 MV
(d)
(10 MW)(1 103 kW/MW) = 10 103 kW = 10,000 kW
(a)
1 mA 1 103 A = 1 103 = 1000 1 A 1 10 6 A
(b)
0.05 kV 0.05 103 V = 0.05 106 = 50,000 3 1 mV 1 10 V
(c)
0.02 k 0.02 103 = 0.02 103 = 2 105 1 M 1 106
(d)
155 mW 155 103 W = 155 106 = 1.55 104 3 1 kW 1 10 W
5
Chapter 1 31.
32.
(a)
50 mA + 680 A = 50 mA + 0.68 mA = 50.68 mA
(b)
120 k + 2.2 M = 0.12 M + 2.2 M = 2.32 M
(c)
0.02 F + 3300 pF = 0.02 F + 0.0033 F = 0.0233 F
(a)
10 k 10 k = 0.8197 2.2 k 10 k 12.2 k
(b)
250 mV 250 103 = 5000 50 V 50 10 6
(c)
1 MW 1 106 = 500 2 kW 2 103
Section 1-5 Measured Numbers 33.
34.
The significant digits are shown in bold face. (a) Three: 1.00 x 103
(b) Two: 0.0057
(c) Five: 1502.0
(d) Two: 0.000036
(e) Three: 0.105
(f) Two: 2.6 x 102
(a) 50,505 rounds to 5.05 x 104
(b) 220.45 rounds to 220
(c) 4646 rounds to 4.65 x 103
(d) 10.99 rounds to 11.0
(e) 1.005 rounds to 1.00
6
Chapter 2 Voltage, Current, and Resistance Note: Solutions show conventional current direction.
Section 2-2 Electrical Charge 1.
29 e 1.6 1019 C/e = 4.64 1018 C
2.
17 e 1.6 1019 C/e = 2.72 1018 C
3.
Q = (charge per electron)(number of electrons) = (1.6 1019 C/e)(50 1031e) = 80 1012 C
4.
(6.25 1018 e/C)(80 106 C) = 5 1014 electrons
Section 2-3 Voltage 5.
(a)
V
W 10 J = 10 V Q 1C
(c)
V
W 100 J =4V Q 25 C
(b)
V
W 5J = 2.5 V Q 2C
6.
V
W 500 J =5V Q 100 C
7.
V
W 800 J = 20 V Q 40 C
8.
W = VQ = (12 V)(2.5 C) = 30 J
9.
I=
10.
Four common sources of voltage are dc power supply, solar cell, generator, and battery.
11.
The operation of electrical generators is based on the principle of electromagnetic induction.
12.
A power supply converts electricity in one form (ac) to another form (dc). The other sources convert other forms of energy into electrical energy.
Q t Q = It = (2 A)(15 s) = 30 C W 1000 J V= = 33.3 V Q 30 C
7
Chapter 2
Section 2-4 Current 13. 14.
15.
16.
A current source provides a constant current of 100 mA regardless of the load value. Q 75 C = 75 A t 1s Q 10 C (b) I = 20 A t 0.5 s Q 5C (c) I = 2.5 A t 2s Q 0.6 C = 0.2 A I t 3s I
(a)
Q t Q 10 C =2s t= I 5A I
17.
Q = It = (1.5 A)(0.1 s) = 0.15 C
18.
I=
Q t
574 1015 electrons = 9.18 102 C 6.25 1018 electrons/C 9.18 102 C = 367 mA I= 250 103 s Q=
Section 2-5 Resistance 19.
(a) (b) (c)
20.
(a) (b) (c)
1 1 = 0.2 S = 200 mS R 5 1 1 G= = 0.04 S = 40 mS R 25 1 1 G= = 0.01 S = 10 mS R 100
G=
1 1 = 10 G 0.1 S 1 1 R= =2 G 0.5 S 1 1 R= = 50 G 0.02 S R=
8
Chapter 2
21.
(a) (b)
Red, violet, orange, gold: 27 k 5% Brown, gray, red, silver: 1.8 k 10%
22.
(a)
Rmin = 27 k 0.05(27 k) = 27 k 1350 = 25.65 k Rmax = 27 k + 0.05(27 k) = 27 k + 1350 = 28.35 k
(b)
Rmin = 1.8 k 0.1(1.8 k) = 1.8 k 180 = 1.62 k Rmax = 1.8 k + 0.1(1.8 k) = 1.8 k + 180 = 1.98 k
23.
330 : orange, orange, brown. gold 2.2 k: red, red, red, gold 56 k: green, blue, orange, gold 100 k: brown, black, yellow, gold 39 k: orange, white, orange, gold
24.
(a)
brown, black, black, gold: 10 5%
(b)
green, brown, green, silver: 5.1 M 10%
(c)
blue, gray, black, gold: 68 5%
25.
26.
(a) red, violet, orange, silver : 27 k + 10%
(b) brown, black, brown, silver: 100 + 10%
(c) green, blue, green , gold: 5.6 M + 5%
(d) blue, gray, red, silver: 6.8 k + 10%
(e) orange, orange, black, silver: 33 + 10%
(f) yellow, violet, orange, gold: 47 k + 5%
330 : (b) orange, orange, brown; 2..2 k: (d) red, red, red; 56 k: (l) green, blue, orange; 100 k: (f) brown, black, yellow; 39 k: (a) orange, white, orange
27.
28.
29.
(a)
0.47 : yellow, violet, silver, gold
(b)
270 k: red, violet, yellow, gold
(c)
5.1 M: green, brown, green, gold
(a)
red, gray, violet, red, brown: 28.7 k 1%
(b)
blue, black, yellow, gold, brown: 60.4 1%
(c)
white, orange, brown, brown, brown: 9.31 k 1%
(a) (b) (c)
14.7 k 1%: brown, yellow, violet, red, brown 39.2 1%: orange, white, red, gold, brown 9.76 k 1%: white, violet, blue, brown, brown
9
Chapter 2
30.
500 , There is equal resistance on each side of the contact.
31.
4K7 = 4.7 k
32.
(a) (b) (c)
4R7J = 4.7 5% 5602M = 56 k 20% 1501F = 1500 1%
Section 2-6 The Electric Circuit 33.
See Figure 2-1.
Figure 2-1 34.
See Figure 2-2.
Figure 2-2
35.
Circuit (b) in Figure 2-68 can have both lamps on at the same time.
36.
There is always current through R5.
37.
See Figure 2-3.
Figure 2-3
10
Chapter 2
38.
See Figure 2-4.
Figure 2-4
Section 2-7 Basic Circuit Measurements 39.
See Figure 2-5.
Figure 2-5
40.
See Figure 2-6.
41.
Position 1: V1 = 0 V, V2 = VS Position 2: V1 = VS, V2= 0 V
42.
Figure 2-6
See Figure 2-7.
Figure 2-7
11
Chapter 2
43.
See Figure 2-8.
44.
See Figure 2-8.
Figure 2-8
45.
On the 600 V scale (middle AC/DC scale): 250 V
46.
R = 10 10 = 100
47.
(a) (b) (c)
48.
0.9999 + 0.0001 = 1.0000 Resolution = 0.00001 V
2 10 = 20 15 100 k = 1.50 M 45 100 = 4.5 k
12
Chapter 2 49.
See Figure 2-9.
Figure 2-9
13
Chapter 3 Ohm’s Law Note: Solutions show conventional current direction.
Section 3-1 The Relationship of Current, Voltage, and Resistance 1.
(a) (b) (c) (d) (e) (f)
2.
I=
3.
V = IR
4.
R=
5.
See Figure 3-1. I= I= I= I= I= I= I= I= I= I= I=
When voltage triples, current triples. When voltage is reduced 75%, current is reduced 75%. When resistance is doubled, current is halved. When resistance is reduced 35%, current increases 54%. When voltage is doubled and resistance is halved, current quadruples. When voltage and resistance are both doubled, current is unchanged. V R
V I
0V 100 10 V 100 20 V 100 30 V 100 40 V 100 50 V 100 60 V 100 70 V 100 80 V 100 90 V 100 100 V 100
=0A
= 100 mA = 200 mA = 300 mA = 400 mA = 500 mA = 600 mA = 700 mA Figure 3-1
= 800 mA = 900 mA
The graph is a straight line indicating a linear relationship between V and I.
=1A
14
Chapter 3 6.
R=
(a) (b) (c) (d) (e)
1V = 200 15 mA 1.5 V I= = 7.5 mA 200 2V I= = 10 mA 200 3V I= = 15 mA 200 4V I= = 20 mA 200 10 V I= = 50 mA 200
7.
Pick a voltage value and find the corresponding value of current by projecting a line up from the voltage value on the horizontal axis to the resistance line and then across to the vertical axis. V 1V R1 = = 500 m I 2A V 1V R2 = =1 I 1A V 1V R3 = =2 I 0.5 A
8.
See Figure 3-2.
Figure 3-2
15
Chapter 3 I= I= I= I= I= 9.
2V 8.2 k 4V 8.2 k 6V 8.2 k 8V 8.2 k 10 V 8.2 k
= 0.244 mA = 0.488 mA = 0.732 mA = 0.976 mA = 1.22 mA
See Figure 3-3.
Figure 3-3 I= I= I= I= I=
2V 1.58 k 4V 1.58 k 6V 1.58 k 8V 1.58 k 10 V 1.58 k
= 1.27 mA = 2.53 mA = 3.80 mA = 5.06 mA = 6.33 mA
16
Chapter 3 10.
(a) (b) (c)
50 V = 15.2 mA 3.3 k 75 V = 19.2 mA I= 3.9 k 100 V I= = 21.3 mA 4.7 k I=
Circuit (c) has the most current and circuit (a) has the least current. 11.
VS 10 V = 0.2 k = 200 30 mA 50 mA VS = (200 )(30 mA) = 6 V (new value) The battery voltage decreased by 4 V (from 10 V to 6 V). R=
12.
The current increase is 50%, so the voltage increase must also be 50%. VINC = (0.5)(20 V) = 10 V V2 = 20 V + VINC = 20 V + 10 V = 30 V (new value)
13.
See Figure 3-4. 10 V (a) I= = 10 A 1 20 V = 20 A I= 1 30 V = 30 A I= 1 40 V = 40 A I= 1 50 V I= = 50 A 1 60 V = 60 A I= 1 70 V = 70 A I= 1 80 V I= = 80 A 1 90 V = 90 A I= 1 100 V = 100 A I= 1
(b)
10 V =2A 5 20 V I= =4A 5 30 V =6A I= 5 40 V =8A I= 5 50 V I= = 10 A 5 60 V = 12 A I= 5 70 V = 14 A I= 5 80 V I= = 16 A 5 90 V = 18 A I= 5 100 V = 20 A I= 5
I=
17
(c)
10 V = 0.5 A 20 20 V I= =1A 20 30 V I= = 1.5 A 20 40 V =2A I= 20 50 V I= = 2.5 A 20 60 V =3A I= 20 70 V = 3.5 A I= 20 80 V I= =4A 20 90 V = 4.5 A I= 20 100 V =5A I= 20
I=
Chapter 3 (d)
I= I= I= I= I= I= I= I= I= I=
14.
10 V 100 20 V 100 30 V 100 40 V 100 50 V 100 60 V 100 70 V 100 80 V 100 90 V 100 100 V 100
= 0.1 A = 0.2 A = 0.3 A = 0.4 A = 0.5 A = 0.6 A = 0.7 A = 0.8 A = 0.9 A =1A Figure 3-4
Yes, the lines on the IV graph are straight lines.
Section 3-2 Current Calculations 15.
(a)
I=
(b)
I=
(c)
I=
(d)
I=
(e)
I=
V R V R V R V R V R
5V =5A 1 15 V = 1.5 A 10 50 V = 500 mA 100 30 V = 2 mA 15 k 250 V = 44.6 A 5 . 6 M
18
Chapter 3 16.
(a)
I=
(b)
I=
(c)
I=
(d)
I=
(e)
I=
V R V R V R V R V R
9V = 3.33 mA 2.7 k 5.5 V = 550 A 10 k 40 V = 588 A 68 k 1 kV = 455 mA 2 .2 k 66 kV = 6.6 mA 10 M
V 12 V = 1.2 A R 10
17.
I=
18.
R = 3300 5% Rmax = 3300 + (0.5)(3300 ) = 3465 Rmin = 3300 (0.5)(3300 ) = 3135 V 12 V Imax = s = 3.83 mA Rmin 3135 V 12 V Imin = s = 3.46 mA Rmax 3465
19.
R = 47 k 10% Rmin = 47 k 0.1(4.7 k) = 42.3 k Rmax = 47 k + 0.1(4.7 k) = 51.7 k V 25 V Imin = = 484 A Rmax 51.7 k V 25 V Imax = = 591 A Rmin 42.3 k V 25 V Inom = = 532 A R 47 k
20.
R = 37.4 V 12 V I= = 0.321 A R 37.4
21.
I = 0.642 A Yes, the current exceeds the 0.5 A rating of the fuse.
22.
VR(max) = 120 V 100 V = 20 V VR ( max ) 20 V Imax = = 2.5 A Rmin 8 A fuse with a rating of less than 2.5 A must be used. A 2-A fuse is suggested.
19
Chapter 3 23.
I
V 3V 9.1mA R 330
Section 3-3 Voltage Calculations 24.
V IR 180A 27k 4.86V
25.
(a) (b) (c) (d) (e)
V = IR = (2 A)(18 ) = 36 V V = IR = (5 A)(56 ) = 280 V V = IR = (2.5 A)(680 ) = 1.7 kV V = IR = (0.6 A)(47 ) = 28.2 V V = IR = (0.1 A)(560 ) = 56 V
26.
(a) (b) (c) (d) (e) (f) (g) (h)
V = IR = (1 mA)(10 ) = 10 mV V = IR = (50 mA)(33 ) = 1.65 V V = IR = (3 A)(5.6 k) = 16.8 kV V = IR = (1.6 mA)(2.2 k) = 3.52 V V = IR = (250 A)(1 k) = 250 mV V = IR = (500 mA)(1.5 M) = 750 kV V = IR = (850 A)(10 M) = 8.5 kV V = IR = (75 A)(47 ) = 3.53 mV
27.
VS = IR = (3 A)(27 ) = 81 V
28.
(a) (b) (c)
29.
Wire resistance = RW = (a) (b) (c)
V = IR = (3 mA)(27 k) = 81 V V = IR = (5 A)(100 M) = 500 V V = IR = (2.5 A)(47 ) = 117.5 V (10.4 CM / ft)(24 ft) = 0.154 1624.3 CM V 6V I= = 59.9 mA R RW 100.154 VR = (59.9 mA)(100 ) = 5.99 V R VRW = I W = (59.9 mA)(0.154 /2) = 4.61 mV 2
Section 3-4 Resistance Calculations 30.
(a) (b) (c)
V 10 V =5 I 2A V 90 V R= =2 I 45 A V 50 V R= = 10 I 5A
R=
20
Chapter 3 (d) (e)
31.
V 5.5 V = 550 m I 10 A V 150 V R= = 300 I 0.5 A R=
(a)
R=
(b)
R=
(c)
R=
(d)
R=
(e)
R=
V I V I V I V I V I
10 kV = 2 k 5A 7V = 3.5 k 2 mA 500 V = 2 k 250 mA 50 V = 100 k 500 A 1 kV = 1 M 1 mA
6V V = 3 k I 2 mA
32.
R=
33.
(a)
34.
Measure the current with an ammeter connected as shown in Figure 3-5, then calculate the unknown resistance as R = 12 V/I.
35.
36.
RFIL: =
V 120 V = 150 I 0.8 A
100 V V = 133 I 750 mA Figure 3-5 V 100 V = 100 R= I 1A The source can be shorted if the rheostat is set to 0 .
R=
Rmin + 15 =
120 V = 60 . Thus Rmin = 60 15 = 45 2A
The rheostat must actually be set to slightly greater than 45 so that the current is limited to slightly less than 2 A. 37.
110 V = 110 1A Rmin = 110 15 = 95
Rmin + 15 =
21
Chapter 3 Section 3-5 Introduction to Troubleshooting 38.
The 4th bulb from the left is open.
39.
It should take five (maximum) resistance measurements.
Multisim Troubleshooting and Analysis 40.
RB is open.
41.
RA = 560 k, RB = 2.2 M, RC = 1.8 k, RD = 33
42.
No fault. I = 1.915 mA, V = 9.00 V
43.
V = 18 V, I = 5.455 mA, R = 3.3 k
44.
R is leaky.
22
Chapter 4 Energy and Power Section 4-1 Energy and Power 1.
volt = joule/coulomb ampere = coulomb/s VI = (joule/coulomb)(coulomb/s) = joule/s
2.
1 kWh = (1000 joules/s)(3600 s) = 3.6 106 joules
3.
1 watt = 1 joule/s P = 350 J/s = 350 W
4.
P=
5.
P=
6.
(a) (b) (c) (d)
1000 W = 1 103 W = 1 kW 3750 W = 3.75 103 W = 3.75 kW 160 W = 0.160 103 W = 0.160 kW 50,000 W = 50 103 W = 50 kW
7.
(a) (b) (c) (d)
1,000,000 W = 1 106 W = 1 MW 3 106 W = 3 MW 15 107 W = 150 106 = 150 MW 8700 kW = 8700 103 W = 8.7 106 W = 8.7 MW
8.
(a) (b) (c) (d)
1 W = 1000 103 W = 1000 mW 0.4 W = 400 103 W = 400 mW 0.002 W = 2 103 = 2 mW 0.0125 W = 12.5 103 W = 12.5 mW
9.
(a) (b) (c) (d)
2 W = 2,000,000 W 0.0005 W = 500 W 0.25 mW = 250 W 0.00667 mW = 6.67 W
10.
(a) 1.5 kW = 1.5 103 W = 1500 W (b) 0.5 MW = 0.5 106 W = 500,000 W (c) 350 mW = 350 103 W = 0.350 W (d) 9000 W = 9000 106 W = 0.009 W Energy = W = Pt = (100 mW)(24 h)(3600 s/h) = 8.64 103 J
11.
W 7500 J t 5h 7500 J 7500 J = 417 mW (5 h)(3600 s/h) 18000 s
1000 J = 20 kW 50 ms
23
Chapter 4 12.
300 W = 0.3 kW (30 days)(24 h/day) = 720 h (0.3 kW)(720 h) = 216 kWh
13.
1500 kWh/31 days = 48.39 kWh/day (48.39 kWh/day)/24 h) = 2.02 kW/day
14.
5 106 watt-minutes = 5 103 kWminutes (5 103 kWmin)(1 h/60 min) = 83.3 kWh
15.
6700 Ws = 0.00186 kWh (1000 W/kW)(3600 s/h)
16.
W = Pt P = I2R = (5 A)2(47 ) = 1175 W W 25 J t= = 0.0213 s = 21.3 ms P 1175 W
Section 4-2 Power in an Electric Circuit 17.
RL =
V 75 V = 37.5 I 2A
18.
P = VI = (5.5 V)(3 mA) = 16.5 mW
19.
P = VI = (120 V)(3 A) = 360 W
20.
P = I2R = (500 mA)2(4.7 k) = 1.175 kW
21.
P = I2R = (100 A)2(10 k) = 100 W
22.
P=
V 2 (60 V)2 = 5.29 W R 680
23.
P=
V 2 (1.5 V)2 = 40.2 mW R 56
24.
P = I2R P 100 W = 25 R= 2 I (2 A)2
25.
(a)
P=
(b)
If the resistor is disconnected after 1 minute, the power during the first minute is equal to the power during the two minute interval. Only energy changes with time.
V 2 (12 V) 2 = 14.4 W R 10 W = Pt = (14.4 W)(2 min)(1/60 h/min) = 0.48 Wh
24
Chapter 4 Section 4-3 Resistor Power Ratings 26.
From Activity 1:
VR1 = 3.25 V and R1 = 18 VR2 = 6.5 V and R2 = 39 VR3 = 10 V and R3 = 68
The power rating for each resistor is determined as follows:
V 2 3.25V 0.59WChoose next highest standard value of 1 W. PR1 R1 18 R1 2
V 2 6.5V 1.1WChoose next highest standard value of 2 W. PR 2 R 2 39 R2 2
V 2 10V 1.5WChoose next highest standard value of 2 W. PR 3 R 3 68 R3 2
27.
A 2 W resistor should be used to provide a margin of safety. A resistor rating greater than the actual maximum power should always be used.
28.
P = I2R = (10 mA)2(6.8 k) = 0.68 W Use at least the next highest standard rating of 1 W.
29.
Use the 12 W resistor to allow a minimum safety margin of greater than 20%. If the 8 W resistor is used, it will be operating in a marginal condition and its useful life will be reduced.
Section 4-4 Energy Conversion and Voltage Drop in Resistance 30.
See Figure 4-1.
Figure 4-1
Section 4-5 Power Supplies and Batteries 31.
VOUT =
32.
PAVG =
PL RL (1 W)(50 ) = 7.07 V V 2 (1.25) 2 V = 156 mW R 10
25
Chapter 4
33.
W = Pt = (0.156 W)(90 h) = (0.156 W)(324,000 s) = 50,544 J
34.
Ampere-hour rating = (1.5 A)(24 h) = 36 Ah
35.
I=
80 Ah =8A 10 h
36.
I=
650 mAh = 13.5 mA 48 h
37.
PLost = PIN POUT = 500 mW 400 mW = 100 mW P 400 mW % efficiency = OUT 100% 100% = 80% 500 mW PIN
38.
POUT = (efficiency)PIN = (0.85)(5 W) = 4.25 W
39.
Assume that the total consumption of the power supply is the input power plus the power lost. POUT = 2 W P % efficiency = OUT 100% PIN POUT 2W PIN = 100% 100% = 3.33 W 60% % efficiency Energy = W = Pt = (3.33 W)(24 h) = 79.9 Wh 0.08 kWh
Multisim Troubleshooting and Analysis 40.
V = 24 V, I = 0.035 A, R = 680
41.
V = 5 V, I = 5 mA, R = 1 k
42.
I = 833.3 mA
26
Chapter 5 Series Circuits Note: Solutions show conventional current direction.
Section 5-1 Resistors in Series 1.
See Figure 5-1.
Figure 5-1
2.
R1, R2, R3, R4, and R9 are in series (pin 5 to 6). R7, R13, R14 and R16 are in series (pin 1 to 8). R6, R8, and R12 are in series (pin 2 to 3). R5, R10, R11, and R15 are in series (pin 4 to 7). See Figure 5-2.
3.
R1-8 = R13 + R7 + R14 + R16 = 68 k + 33 k + 47 k + 22 k = 170 k
4.
R2-3 = R12 + R8 + R6 = 10 + 18 + 22 = 50
5.
R1, R7, R8, and R10 are in series. R2, R4, R6, and R11 are in series. R3, R5, R9, and R12 are in series.
Figure 5-2
27
Chapter 5 Section 5-2 Total Series Resistance 6.
RT = 1 + 2.2 + 5.6 + 12 + 22 = 42.8
7.
(a) (b) (c) (d)
RT = 560 + 1000 = 1560 RT = 47 + 56 = 103 RT = 1.5 k + 2.2 k + 10 k = 13.7 k RT = 1 M + 470 k + 1 k + 2.2 M = 3.671 M
8.
(a) (b) (c)
RT = 1 k + 5.6 k + 2.2 k = 8.8 k RT = 4.7 + 10 + 12 + 1 = 27.7 RT = 1 M + 560 k + 5.6 M + 680 k + 10 M = 17.84 M
9.
RT = 12(5.6 k) = 67.2 k
10.
RT = 6(56 ) + 8(100 ) + 2(22 ) = 336 + 800 + 44 = 1180
11.
RT = R1 + R2 + R3 + R4 + R5 R5 = RT (R1 + R2 + R3 + R4) = 17.4 k (5.6 k + 1 k + 2.2 k + 4.7 k) = 17.4 k 13.5 k = 3.9 k
12.
RT = 3(5.6 k) + 1 k + 2(100 ) = 16.8 k + 1 k + 200 = 18 k Three 5.6 k resistors, one 1 k resistor, and two 100 resistors. Other combinations are possible.
13.
RT = 1 k + 5.6 k + 2.2 k + 4.7 + 10 + 12 + 1 + 1 M + 560 k + 5.6 M + 680 k + 10 M = 17.848827.7 M 17.8 M
14.
Position 1: RT = R1 + R3 + R5 = 510 + 820 + 680 = 2.01 k Position 2: RT = R1 + R2 + R3 + R4 + R5 = 510 + 910 + 820 + 750 + 680 = 3.67 k
Section 5-3 Current in a Series Circuit V 12 V = 100 mA RT 120
15.
I=
16.
I = 5 mA at all points in the series circuit.
17.
See Figure 5-3. The current through R2, R3, R4, and R9 is also measured by this set-up.
28
Chapter 5
Figure 5-3 18.
See Figure 5-4.
Figure 5-4
Section 5-4 Application of Ohm’s Law 19.
(a)
(b)
20.
RT = R1 + R2 + R3 = 2.2 k + 5.6 k + 1 k = 8.8 k V 5.5 V = 625 A I= RT 8.8 k RT = R1 + R2 + R3 = 1 M + 2.2 M + 560 k = 3.76 M V 16 V I= = 4.26 A RT 3.76 M
(a)
I = 625 A V1 = IR1 = (625 A)(2.2 k) = 1.375 V V2 = IR2 = (625 A)(5.6 k) = 3.5 V V3 = IR3 = (625 A)(1 k) = 0.625 V
(b)
I = 4.26 A V1 = IR1 = (4.26 A)(1 M) = 4.26 V V2 = IR2 = (4.26 A)(2.2 M) = 9.36 V V3 = IR3 = (4.26 A)(560 k) = 2.38 V
29
Chapter 5 21.
22.
23.
RT = 3(470 ) = 1.41 k V 48 V (a) I= = 34 mA RT 1.41 k 48 V = 16 V 3
(b)
VR =
(c)
P = (34 mA)2(470 ) = 0.543 W
V 5V = 2.24 k I 2.23 mA R 2.24 k Reach = T = 560 4 4 V 21.7 V R1 = 1 = 330 I 65.8 mA V 6.58 V R1 = 3 = 100 I 65.8 mA
RT =
V2 14.5 V = 220 I 65.8 mA V 30.9 V R4 = 4 = 470 I 65.8 mA
R2 =
24.
V1 = IR1 = (12.3 mA)(82 ) = 1.01 V V 12 V 2.21 V 1.01 V = 714 R2 = 2 I 12.3 mA V 2.21 V R3 = 3 = 180 I 12.3 mA
25.
(a)
(b)
(c) 26.
R T = R 1 + R 2 + R3 + R 4 12 V 12 V (R1 + R2 + R3) = 1200 = 1531 1200 = 331 R4 = 7.84 mA 7.84 mA 12 V 12 V = 9.15 mA Position B: I = R2 R3 R4 1311 12 V 12 V Position C: I = = 14.3 mA R3 R4 841 12 V 12 V = 36.3 mA Position D: I = R4 331 No
Position A: RT = R1 = 1 k V 9V I= = 9 mA RT 1 k Position B: RT = R1 + R2 + R5 = 1 k + 33 k + 22 k = 56 k V 9V I= = 161 A RT 56 k Position C:
30
Chapter 5 RT = R1 + R2 + R3 + R4 + R5 = 1 k + 33 k + 68 k + 27 k + 22 k = 151 k V 9V I= = 59.6 A RT 151 k
Section 5-5 Voltage Sources in Series 27.
VT = 5 V + 9 V = 14 V
28.
VT = 12 V 3 V = 9 V
29.
(a)
VT = 10 V + 8 V + 5 V = 23 V
(b)
VT = 50 V + 10 V + 25 V = 85 V
Section 5-6 Kirchhoff’s Voltage Law 30.
VS = 5.5 V + 8.2 V + 12.3 V = 26 V
31.
VS = V1 + V2 + V3 + V4 + V5 20 V = 1.5 V + 5.5 V + 3 V + 6 V + V5 V5 = 20 V (1.5 V + 5.5 V + 3 V + 6 V) = 20 V 16 V = 4 V
32.
(a)
(b)
By Kirchhoff’s voltage law: 15 V = 2 V + V2 + 3.2 V + 1 V + 1.5 V + 0.5 V V2 = 15 V (2 V + 3.2 V + 1 V + 1.5 V + 0.5 V) = 15 V 8.2 V = 6.8 V VR = 8 V, V2R = 2(8 V) = 16 V, V3R = 3(8 V) = 24 V, V4R = 4(8 V) = 32 V VS = VR + VR + V2R + V3R + V4R = 11(VR) = 88 V
11.2 V = 200 mA 56 4.4 V R4 = = 22 200 mA
33.
I=
34.
R1 =
35.
Position A: RT = R1 + R2 + R3 + R4 = 1.8 k + 1 k + 820 + 560 = 4.18 k Voltage drop across R1 through R4: V = IRT = (3.35 mA)(4.18 k) = 14 V V5 = 18 V 14 V = 4 V
V1 5.6 V = 560 I 10 mA 22 mW P R2 = 22 = 220 (10 mA)2 I 9V RT = = 900 10 mA R3 = RT R1 R2 = 900 560 200 = 120
31
Chapter 5
36.
Position B: RT = R1 + R2 + R3 = 1.8 k + 1 k + 820 = 3.62 k Voltage drop across R1 through R3: V = IRT = (3.73 mA)(3.62 k) = 13.5 V V5 = 18 V 13.5 V = 4.5 V Position C: RT = R1 + R2 = 1.8 k + 1 k = 2.8 k Voltage drop across R1 and R2: V = IRT = (4.5 mA)(2.8 k) = 12.6 V V5 = 18 V 12.6 V = 5.4 V Position D: RT = R1 = 1.8 k Voltage drop across R1: V = IRT = (6 mA)(1.8 k) = 10.8 V V5 = 18 V 10.8 V = 7.2 V Position A: V1 = (3.35 mA)(1.8 k) = 6.03 V V2 = (3.35 mA)(1 k) = 3.35 V V3 = (3.35 mA)(820 ) = 2.75 V V4 = (3.35 mA)(560 ) = 1.88 V V5 = 4.0 V Position B: V1 = (3.73 mA)(1.8 k) = 6.71 V V2 = (3.73 mA)(1 k) = 3.73 V V3 = (3.73 mA)(820 ) = 3.06 V V5 = 4.5 V Position C: V1 = (4.5 mA)(1.8 k) = 8.1 V V2 = (4.5 mA)(1 k) = 4.5 V V5 = 5.4 V Position D: V1 = (6 mA)(1.8 k) = 10.8 V V5 = 7.2 V
Section 5-7 Voltage Dividers 37.
V27 27 100 = 4.82% VT 560
38.
(a) (b)
39.
56 12 V = 4.31 V VAB = 156 5.5 k 8 V = 6.77 V VAB = 6.5 k
VA = VS = 15 V
32
Chapter 5 R2 R3 13.3 k VB = 15 V = 10.6 V VS 18.9 k R1 R2 R3 R3 3.3 k VC = 15 V = 2.62 V VS R R R 18.9 k 2 3 1 40.
R3 680 VOUT(min) = 12 V = 3.80 V VS 2150 R1 R2 R3 R2 R3 1680 VOUT(max) = 12 V = 9.38 V VS 2150 R1 R2 R3
41.
RT = 15R R VR = 90 V = 6 V 15 R 2R V2R = 90 V = 12 V 15 R 3R V3R = 90 V = 18 V 15 R 4R V4R = 90 V = 24 V 15 R 5R V5R = 90 V = 30 V 15 R
42.
VAF = 100 V R 86.6 k VBF = BF VAF 100 V = 79.7 V 108.6 k RAF R 76.6 k VCF = CF VAF 100 V = 70.5 V 108.6 k RAF R 20.6 k VDF = DF VAF 100 V = 19.0 V 108.6 k RAF
R VEF = EF RAF 43.
5.6 k VAF 100 V = 5.16 V 108.6 k
V1 10 V = 1.79 mA R1 5.6 k V2 = IR2 = (1.79 mA)(1 k) = 1.79 V V3 = IR3 = (1.79 mA)(560 ) = 1.0 V V4 = IR4 = (1.79 mA)(10 k) = 17.9 V
I=
33
Chapter 5 44.
See Figure 5-5 for one possible solution: RT = 18 k + 33 k + 22 k + 27 k = 100 k 30 V IT = = 300 A 100 k 82 k VA = 30 V = 24.6 V 100 k 49 k VB = 30 V = 14.7 V 100 k 27 k VC = 30 V = 8.1 V 100 k
Figure 5-5
P1 = I T2 R1 = (300 A)2 27 k = 2.43 mW
P2 = I T2 R2 = (300 A)2 22 k = 1.98 mW P3 = I T2 R3 = (300 A)2 33 k = 2.97 mW P4 = I T2 R4 = (300 A)2 18 k = 1.62 mW All resistors can be 1/8 W. 45.
See Figure 5-6 for one possible solution. RT = 12.1 k 10.1 k VOUT(max) = 120 V = 100.2 V 12.1 k 1 k VOUT(min) = 120 V = 9.92 V 12.1 k These values are within 1% of the specified values. 120 V 120 V IMAX = = 9.9 mA RT 12.1 k
Figure 5-6
Section 5-8 Power in Series Circuits 46.
PT = 5(50 mW) = 250 mW
47.
VT = V1 + V2 + V3 + V4 = 10 V + 1.79 V + 1 V + 17.9 V = 30.69 V PT = VTI = (30.69 V)(1.79 mA) = 54. 9 mW
48.
Since P = I2R and since each resistor has the same current, the 5.6 k resistor is the limiting element in terms of power dissipation. 0.25 W Pmax = 6.68 mA 5.6 k 5.6 k V5.6 k = (6.68 mA)(5.6 k) = 37.4 V V1.2 k = (6.68 mA)(1.2 k) = 8.02 V V2.2 k = (6.68 mA)(2.2 k) = 14.7 V V3.9 k = (6.68 mA)(3.9 k) = 26.1 V VT(max) = 37.4 V + 8.02 V + 14.7 V + 26.1 V = 86.2 V
Imax =
34
Chapter 5
49.
50.
12 V V1 = 2.14 A R1 5.6 M V 4.8 V R2 = 2 = 2.2 M I 2.14 A P3 = I2R3 P 21.5 W = 4.7 M R3 = 32 I 214 A)2 RT = R1 + R2 + R3 = 5.6 M + 2.2 M + 4.7 M = 12.5 M
I=
(a)
P = I2R P R= 2 I R1 + R2 + R3 = 2400 1 1 1 W W W 8 4 2 = 2400 I2 I2 I2 7 W 8 = 2400 I2 7 W 8 2 = 0.0003646 A2 I = 2400 I=
0.0003646 A 2 = 19.1 mA
(b)
VT = IRT = (19.1 mA)(2400 ) = 45.8 V
(c)
R1 =
0.125 W P1 = 343 2 I (19.1 mA)2 0.25 W P R2 = 22 = 686 (19.1 mA)2 I 0.5 W P R3 = 32 = 1.37 k I (19.1 mA)2
Section 5-9 Voltage Measurements 51.
VAG = 100 V (voltage from point A to ground) Resistance between A and C: RAC = 5.6 k + 5.6 k = 11.2 k Resistance between C and ground: RCG = 1 k + 1 k = 2 k
35
Chapter 5 2 k VCG = 100 V = 15.2 V 13.2 k 1 k 1 k VDG = VCG 15.2 V = 7.58 V 2 k 2 k 11.2 k VAC = 100 V = 84.9 V 13.2 k 5.6 k 5.6 k VBC = VAC 84.9 V = 42.5 V 11.2 k 11.2 k VBG = VCG + VBC = 15.2 V + 42.5 V = 57.7 V
52.
Measure the voltage at point A with respect to ground and the voltage at point B with respect to ground. The difference is VR2. VR2 = VB VA
53.
RT = R1 + R2 + R3 + R4 + R5 = 56 k + 560 k + 100 k + 1 M + 100 k = 1.816 M VT = 15 V 9 V = 6 V V 6V I= T = 3.3 A RT 1.816 M V1 = IR1 = (3.3 A)(56 k) = 185 mV VA = 15 V V1 = 15 V 185 mV = 14.82 V V2 = IR2 = (3.3 A)(560 k) = 1.85 V VB = VA V2 = 14.82 V 1.85 V = 12.97 V V3 = IR3 = (3.3 A)(100 k) = 330 mV VC = VB V3 = 12.97 V 330 mV = 12.64 V V4 = IR4 = (3.3 A)(1 M) = 3.3 V VD = VC V4 = 12.64 V 3.3 V = 9.34 V
54.
VAC = VA – VC = 14.82 V – 12.64 V = 2.18 V
55.
VCA = VC – VA = 12.64 V – 14.82 V = -2.18 V
Section 5-10 Troubleshooting 56.
There is no current through the resistors which have zero volts across them; thus, there is an open in the circuit. Since R2 has voltage across it, it is the open resistor. 12 V will be measured across R2.
57.
(a) (b)
Zero current indicates an open. R4 is open since all the voltage is dropped across it. VS 10 V = 33.3 mA R1 R2 R3 300 R4 and R5 have no effect on the current. There is a short from A to B, shorting out R4 and R5.
36
Chapter 5 58.
R2 = 0 RT = R1 + R3 + R4 + R5 = 400 V 10 V IT = S = 25 mA RT 400
59.
The results in Table 5-1 are correct.
60.
If 15 k is measured between pins 5 and 6, R3 and R5 are shorted as indicated in Figure 5-7.
61.
In this case, there is a short between the points indicated in Figure 5-7.
Figure 5-7
62.
(a) (b) (c)
R11 has burned out because it has the highest resistance value (P = I2R). Replace R11 (10 k). RT = 47.73 k 0.5 W P11 Imax = = 7.07 mA R11 10 k Vmax = ImaxRT = (7.07 mA)(10 k) = 70.7 V
Multisim Troubleshooting and Analysis 63.
7.481 k
66.
6V
64.
R2 is open.
67.
R1 is shorted
65.
R3 = 22
37
Chapter 6 Parallel Circuits Note: Solutions show conventional current direction.
Section 6-1 Resistors in Parallel 1.
See Figure 6-1.
Figure 6-1
2.
R1, R2 and R5 are not individually in parallel with the other resistors. The series combination of R1, R2, and R5 is in parallel with the other resistors.
3.
R1, R2, R5, R9, R10 and R12 are in parallel. R4, R6, R7, and R8 are in parallel. R3 and R11 are in parallel.
Section 6-2 Voltage in a Parallel Circuit 4.
5.
V1 = V2 = V3 = V4 = 12 V V 12 V IT = T = 21.8 mA RT 550 The total current divides equally among the four equal parallel resistors. 21.8 mA = 5.45 mA I1 = I2 = I3 = I4 = 4 The resistors are all in parallel across the source. The voltmeters each measure the voltage across a resistor, so each meter indicates 100 V.
6.
Position A: RT = R1 R4 = (1.0 k) (2.7 k) = 730 Position B: RT = R1 R3 = (1.0 k) (2.2 k) = 688 Position C: RT = R1 R2 = (1.0 k) (1.8 k) = 643
7.
Position A: V1 = 15 V, V2 = 0 V, V3 = 0 V, V4 = 15 V Position B: V1 = 15 V, V2 = 0 V, V3 = 15 V, V4 = 0 V Position C: V1 = 15 V, V2 = 15 V, V3 = 0 V, V4 = 0 V
38
Chapter 6 8.
15V = 20.6 mA 730 15V Position B: IT = = 21.8 mA 688 15V = 23.3 mA Position C: IT = 643 Position A: IT =
Section 6-3 Kirchhoff’s Current Law 9.
IT = 250 mA + 300 mA + 800 mA = 1350 mA = 1.35 A
10.
IT = I1 + I2 + I3 + I4 + I5 I5 = IT (I1 + I2 + I3 + I4) = 500 mA (50 mA + 150 mA + 25 mA + 100 mA) = 500 mA 325 mA = 175 mA
11.
VS = I1R1 = (1 mA)(47 ) = 47 mV V 47 mV R2 = S = 22 I 2 2.14 mA V 47 mV = 100 R3 = S I 3 0.47 mA I4 = IT (I1 + I2 + I3) = 5.03 mA 3.61 mA = 1.42 mA V 47 mV R4 = S = 33 I 4 1.42 mA
12.
IT = 1.25 A + 0.833 A + 0.833 A + 10 A = 12.92 A I4 = 15 A 12.92 A = 2.08 A See Figure 6-2.
Figure 6-2
13.
VT = ITRT = (100 mA)(25 ) = 2500 mV = 2.5 V VT 2.5 V I220 = = 11.4 mA 220 220
14.
IT = 4IRUN + 2ITAIL
= 4(0.5A) + 2(1.2 A) = 4.4 A
39
Chapter 6 15.
(a) IT = 4IRUN + 2ITAIL + 2IBRAKE = 4(0.5A) + 2(1.2 A) + 2(1 A) = 6.4 A (b) IGND = IT = 6.4 A
Section 6-4 Total Parallel Resistance 16.
RT =
17.
(a) (b) (c)
(d)
18.
(a) (b) (c)
1 = 568 k 1 1 1 1 1 1 M 2.2 M 5.6 M 12 M 22 M (560 )(1 k) = 359 560 1 k (47 )(56 ) RT = = 25.6 47 56 1 RT = = 819 1 1 1 1.5 k 2.2 k 10 k 1 RT = = 997 1 1 1 1 1 M 470 k 1 k 2.7 M
RT =
(560 )(220 ) = 158 560 220 (27 k)(56 k) RT = = 18.2 k 27 k 56 k (1.5 k)(2.2 k) = 892 RT = 1.5 k 2.2 k
RT =
6.8 k = 0.567 k = 567 12
19.
RT =
20.
Five 470 resistors in parallel: 470 R1 = = 94 5 Ten 1000 resistors in parallel: 1000 = 100 R2 = 10 Two 100 resistors in parallel: 100 R3 = = 50 2
21.
RT =
1 = 24.6 1 1 1 94 100 50
40
Chapter 6 R1R2 R1 R2 RT(R1 + R2) = R1R2 RTR1 + RTR2 = R1R2 RTR1 = R1R2 RTR2 RTR1 = R2(R1 RT) RT R1 (389.2 )(680 ) R2 = = 910 R1 RT 680 389.2
22.
RT =
23.
(a)
RT = R1 = 510 k
(b)
RT = R1 R2 =
(c)
1 = 245 k 1 1 510 k 470 k RT = R1 = 510 k 1 RT = R1 R2 R3 = = 193 k 1 1 1 510 k 470 k 910 k
Section 6-5 Application of Ohm’s Law 24.
(a)
(b)
R 240 = 80 3 3 120 V IT = = 1.5 A 80
25.
RT =
26.
RT =
27.
1 = 10.2 1 1 1 33 33 27 V 10 V IT = = 980 mA RT 10.2 1 = 334 RT = 1 1 1 1 k 4.7 k 560 V 25 V IT = = 74.9 mA RT 334 RT =
VS 5V = 4.5 k I T 1.11 mA Reach = 4RT = 4(4.5 k) = 18 k VS
110 V = 50 mA Rfilament 2.2 k When one bulb burns out, the others remain on.
I=
41
Chapter 6 28.
(a)
I2 = IT I1 = 150 mA 100 mA = 50 mA 10 V = 100 R1 = 100 mA 10 V R2 = = 200 50 mA
(b)
I3 =
100 V = 100 mA 1 k 100 V = 147 mA I2 = 680 I1 = IT I2 I3 = 500 mA 247 mA = 253 mA 100 V = 395 R1 = 253 mA
29.
Imax = 0.5 A 15 V 15 V RT(min) = = 30 I max 0.5 A (68 ) Rx = RT(min) 68 Rx (68 )Rx = (30 )(68 + Rx) 68Rx = 2040 + 30Rx 68Rx 30Rx = 2040 38Rx = 2040 Rx = 53.7
30.
Position A: 24 V = 42.9 A I1 = 560 k 24 V = 109 A I2 = 220 k 24 V I3 = = 88.9 A 270 k IT = 42.9 A + 109 A + 88.9 A = 241 A Position B: I1 = 42.9 A I2 = 109 A I3 = 88.9 A 24 V = 24 A I4 = 1 M 24 V = 29.3 A I5 = 820 k 24 V I6 = = 10.9 A 2.2 M IT = 42.9 A + 109 A + 88.9 A + 24 A + 29.3 A + 10.9 A = 305 A
42
Chapter 6 Position C: I4 = 24 A I5 = 29.3 A I6 = 10.9 A IT = 24 A + 29.3 A + 10.9 A = 64.2 A 31.
100 V = 83.3 mA 1.2 k I2 = 250 mA 83.3 mA = 166.7 mA IT = 250 mA + 50 mA = 300 mA 100 V R1 = = 2 k 50 mA 100 V = 600 R2 = 166.7 mA I3 =
Section 6-6 Current Sources in Parallel 32.
(a) (b) (c)
IL = 1 mA + 2 mA = 3 mA IL = 50 A 40 A = 10 A IL = 1 A 2.5 A + 2 A = 0.5 A
33.
Position A: IR = 2.25 mA Position B: IR = 4.75 mA Position C: IR = 4.75 mA + 2.25 mA = 7 mA
Section 6-7 Current Dividers 34.
35.
R2 2.7 k I1 = IT 3 A = 2.19 A 3.7 k R1 R2 R1 1 k I2 = IT 3 A = 0.811 A 3.7 k R1 R2
(a)
(b)
R2 2.2 M I1 = IT 10 A = 6.88 A 3.2 M R1 R2 I2 = IT I1 = 10 A 6.88 A = 3.12 A R Ix = T I T Rx RT = 525 525 10 mA = 5.25 mA I1 = 1000 525 10 mA = 2.39 mA I2 = 2.2 k
43
Chapter 6 525 10 mA = 1.59 mA I3 = 3 .3 k 525 10 mA = 0.772 mA I4 = 6.8 k 36.
37.
1 1 1 = R / 1 = 0.48R 1 1 1 1 2 3 4 R 2 R 3R 4 R RT RT 0.48 R 0.48 R IR = 10 mA 10 mA = 4.8 mA; I2R = 10 mA = 2.4 mA; 10 mA R 2R R 2R R RT 0.48R 0.48 R I3R = T 10 mA 10 mA = 1.59 mA; I4R = 10 mA = 1.2 mA 10 mA 3R 4R 3R 4R 1
RT =
RT = 773 I3 = IT I1 I2 I3 = 15.53 mA 3.64 mA 6.67 mA 3.08 mA = 2.14 mA R I1 = T I T R1 R R1 = T I1 R R2 = T I2
R R3 = T I3 R R4 = T I4 38.
IT IT IT IT
773 15.53 mA = 3.3 k 3.64 mA
773 15.53 mA = 1.8 k 6.67 mA 773 15.53 mA = 5.6 k 2.14 mA 773 15.53 mA = 3.9 k 3.08 mA
(a)
IT = 10 mA, IM = 1 mA VM = IMRM = (1 mA)(50 ) = 50 mV ISH1 = 9 mA V 50 mV RSH1 = M = 5.56 I SH1 9 mA
(b)
IT = 100 mA, IM = 1 mA VM = IMRM = (1 mA)(50 ) = 50 mV ISH2 = 99 mA V 50 mV RSH2 = M = 0.505 I SH2 99 mA
44
Chapter 6 39.
(a) (b)
50 mV = 1 m 50 A 50 mV ISH = = 50 A 1 m 50 mV Imeter = = 5 A 10 k RSH =
Section 6-8 Power in Parallel Circuits 40.
PT = 5(250 mW) = 1.25 W
41.
(a)
RT =
(b)
RT =
42.
(1 M)(2.2 M) = 687.5 k 1 M 2.2 M PT = I 2RT = (10 A)2(687.5 k) = 68.8 W
1 1 = 525 1 1 1 1 1 1 1 1 R1 R2 R3 R4 1 k 2.2 k 3.3 k 6.8 k PT = I 2RT = (10 mA)2(525 ) = 52.5 mW
P = VI Ieach =
P 75 W = 625 mA V 120 V
IT = 6(625 mA) = 3.75 A 43.
P1 = PT P2 = 2 W 0.75 W = 1.25 W P 2W VS = T = 10 V I T 200 mA P 0.75 W I2 = 2 = 75 mA 10 V VS V 10 V R2 = S = 133 I 2 75 mA I1 = IT I2 = 200 mA 75 mA = 125 mA V 10 V R1 = S = 80 I1 125 mA
44.
(a)
(b)
PT = I T2 RT = (50 mA)2 1 k = 2.5 W P 2.5 W Number of resistors = n = T = 10 Peach 0.25 W R RT = n R = nRT = 10(1 k) = 10 k
45
Chapter 6 I T 50 mA = 5 mA 10 n
(c)
I=
(d)
VS = ITRT = (50 mA)(1 k) = 50 V
Section 6-10 Troubleshooting 45.
Ieach =
P 75 W = 625 mA V 120 V
IT = 5(625 mA) = 3.13 A 46.
1 = 47.5 1 1 1 1 1 220 100 1 k 560 270 10 V = 210.5 mA IT = 47.5 The measured current is 200.4 mA, which is 10.1 mA less than it should be. Therefore, one of the resistors is open. RT =
V 10 V = 990 1 k I 10.1 mA The 1 k resistor (R3) is open. 1 RT = = 2.3 k 1 1 1 4.7 k 10 k 8.2 k 25 V IT = = 10.87 mA 2.3 k The meter indicates 7.82 mA. Therefore, a resistor must be open. 25 V I3 = = 3.05 mA 8.2 k I = IT IM = 10.87 mA 7.82 = 3.05 mA This shows that I3 is missing from the total current as read on the meter. Therefore, R3 (8.2 k) is open.
R? =
47.
48.
25 V = 5.32 mA 4.7 k 25 V I2 = = 2.5 mA 10 k 25 V I3 = = 3.05 mA 8.2 k R1 is open producing a total current of IT = I2 + I3 = 2.5 mA + 3.05 mA = 5.55 mA I1 =
46
Chapter 6 49.
Connect ohmmeter between the following pins: Pins 1-2 Correct reading: R = 1 k 3.3 k = 767 R1 open: R = 3.3 k R2 open: R = 1 k Pins 3-4 Correct reading: R = 270 390 = 159.5 R3 open: R = 390 R4 open: R = 270 Pins 5-6 Correct reading: R = 1 M 1.8 M 680 k 510 k = 201 k R5 open: R = 1.8 M 680 k 510 k = 251 k R6 open: R = 1 M 680 k 510 k = 226 k R7 open: R = 1 M 1.8 M 510 k = 284 k R8 open: R = 1 M 1.8 M 680 k = 330 k
50.
Short between pins 2 and 4: (a) R1-2 = R1 R2 R3 R4 R11 R12 + R5 R6 R7 R8 R9 R10 = 10 k 2.2 k 2.2 k 3.3 k 18 k 1 k + 4.7 k 4.7 k 6.8 k
5.6 k 1 k 5.6 k = 940 (b)
R2-3 = R5 R6 R7 R8 R9 R10 = 4.7 k 4.7 k 6.8 k 5.6 k 1 k 5.6 k = 518
(c)
R3-4 = R5 R6 R7 R8 R9 R10 = 4.7 k 4.7 k 6.8 k 5.6 k 1 k 5.6 k = 518 R1-4 = R1 R2 R3 R4 R11 R12 = 10 k 2.2 k 2.2 k 3.3 k 18 k 1 k = 422
(d)
51.
Short between pins 3 and 4: (a) R1-2 = (R1 R2 R3 R4 R11 R12) + (R5 R6 R7 R8 R9 R10) = 940 (b) R2-3 = R5 R6 R7 R8 R9 R10 = 518
(c)
R2-4 = R5 R6 R7 R8 R9 R10 = 518
(d)
R1-4 = R1 R2 R3 R4 R11 R12 = 422
Multisim Troubleshooting and Analysis 52.
RT = 547.97
53.
R2 is open.
54.
R1 = 890
55.
VS = 3.3 V
56.
R1 is open.
47
Chapter 7 Series-Parallel Circuits Note: Solutions show conventional current direction.
Section 7-1 Identifying Series-Parallel Relationships 1.
See Figure 7-1.
Figure 7-1
2.
See Figure 7-2.
Figure 7-2 3.
(a) (b) (c)
R1 and R4 are in series with the parallel combination of R2 and R3. R1 is in series with the parallel combination of R2, R3, and R4. The parallel combination of R2 and R3 is in series with the parallel combination of R4 and R5. This is all in parallel with R1.
4.
(a)
R2 is in series with the parallel combination of R3 and R4. This series-parallel combination is in parallel with R1. All of the resistors are in parallel. R1 and R2 are in series with the parallel combination of R3 and R4. R5 and R8 are in series with the parallel combination of R6 and R7. These two series-parallel combinations are in parallel with each other.
(b) (c)
48
Chapter 7 5.
See Figure 7-3.
Figure 7-3
6.
See Figure 7-4.
Figure 7-4
7.
See Figure 7-5.
Figure 7-5
49
Chapter 7 Section 7-2 Analysis of Series-Parallel Resistive Circuits 8.
9.
R1R2 R1 R2 R1 RT (1 k)(667 ) R2 = = 2.0 k R1 RT 1 k 667 RT =
(c)
R2 100 = 133 = 56 + 27 + 2 2 1 1 = 680 + 99.4 = 779 RT = R1 = 680 + 1 1 1 1 1 1 680 330 180 R2 R3 R4 RT = R1 (R2 R3 + R4 R5) = R1 (2.154 k + 3.59 k) = 852
(a)
RT = R1 (R2 + R3 R4) = 1 k (1 k + 2.2 k 3.3 k) = 699
(b)
RT =
(c)
RA = R1 + R2 +
(a)
IT =
(a)
(b)
10.
11.
RT = R1 + R4 +
1 = 406 k 1 1 1 1 1 M 1 M 3.3 M 6.2 M
R3 R4 (10 k)(4.7 k) = 1 k + 1 k + = 5.2 k R3 R4 10 k 4.7 k R6 R7 6 .8 k RB = R5 + R8 + = 8.5 k = 3.3 k + 1.8 k + R6 R7 2 1 1 RT = = 3.23 k 1 1 1 1 RA RB 5.2 k 8.5 k 1.5 V = 11.3 mA 133 I1 = I4 = 11.3 mA 11.3 mA I2 = I3 = = 5.64 mA 2 V1 = (11.3 mA)(56 ) = 633 mV V4 = (11.3 mA)(27 ) = 305 mV V2 = V3 = (5.64 mA)(100 ) = 564 mV
50
Chapter 7
12.
3V = 3.85 mA 779 V1 = (3.85 mA)(680 ) = 2.62 V V2 = V3 = V4 = VS ITR1 = 3 V (3.85 mA)(680 ) = 383 mV I1 = IT = 3.85 mA V 383 mV I2 = 2 = 563 A R2 680 V 383 mV I3 = 3 = 1.16 mA R3 330 V 383 mV I4 = 4 = 2.13 mA R4 180
(b)
IT =
(c)
I1 =
(a)
IT =
5V 5V Iright = = 5 mA = 871 A 1 k 5.74 k 3.3 k I2 = 871 A = 303 A 9.5 k 6.2 k I3 = 871 A = 568 A 9.5 k 5.6 k I4 = 871 A = 313 A 15.6 k 10 k I5 = 871 A = 558 A 15.6 k V1 = VS = 5 V V2 = V3 = (303 A)(6.2 k) = 1.88 V V4 = V5 = (313 A)(10 k) = 3.13 V 1V = 1.43 mA 699 2.32 k I1 = 1.43 mA = 1 mA 3.32 k V1 = (1 mA)(1 k) = 1 V 1 k I2 = 1.43 mA = 431 A 3.32 k V2 = (431 A)(1 k) = 431 mV 3.3 k I3 = 431 A = 259 A 5.5 k V3 = (259 A)(2.2 k) = 570 mV V4 = V3 = 570 mV 570 mV I4 = = 173 A 3.3 k
51
Chapter 7 (b)
V1 = V2 = V3 = V4 = 2 V 2V I1 = = 2 A 1 M 2V I2 = = 606 nA 3.3 M 2V I3 = = 323 nA 6.2 M 2V I4 = = 2 A 1 M
(c)
IT =
5V = 1.55 mA 3.23 k 5.2 k I5= 1.55 mA = 588 A 13.7 k V5 = (588 A)(3.3 k) = 1.94 V I 588 A I6 = I7 = 5 = 294 A 2 V6 = V7 = (294 A)(6.8 k) = 2 V I8 = I5 = 588 A V8 = (588 A)(1.8 k) = 1.06 V 8.5 k I1 = I2 = 1.55 mA = 962 A 13.7 k V1 = V2 = (962 A)(1 k) = 962 mV 4.7 k I3 = 962 A = 308 A 14.7 k V3 = V4 = (308 A)(10 k) = 3.08 V 10 k I4 = 962 A = 654 A 14.7 k
13.
SW1 closed, SW2 open: RT = R2 = 220 SW1 closed, SW2 closed: RT = R2 R3 = 220 2.2 k = 200 SW1 open, SW2 open: RT = R1 + R2 = 100 + 220 = 320 SW1 open, SW2 closed: RT = R1 + R2 R3 = 100 + 200 = 300
14.
RAB = (10 k + 5.6 k) 4.7 k = 15.6 k 4.7 k = 3.61 k The 1.8 k and the two 1 ks are shorted).
52
Chapter 7 15.
16.
VAG = 100 V RAC = (4.7 k + 5.6 k) 10 k = 5.07 k RCG = 2 k 1.8 k = 947 5.07 k VAC = 100 V = 84.2 V 6.02 k 947 VCG = 100 V = 15.7 V 6.02 k 1 k 1 k VDG = VCG 15.7 V = 7.87 V 2 k 2 k 5.6 k 5 .6 k VBC = VAC 84.2 V = 45.8 V 10.3 k 10.3 k VBG = VCG + VBC = 15.7 V + 45.8 V = 61.5 V
56 k VA = 50 V = 3.91 V 716 k VC = 50 V
616 k VB = 50 V = 43.0 V 716 k 100 k VD = 50 V = 4.55 V 1.1 M
17.
Measure the voltage at point A with respect to ground and the voltage at point B with respect to ground. The difference is VR2. VR2 = VB VA
18.
RT = (10 k (4.7 k + 5.6 k)) + (1.8 k (1 k + 1 k)) = 10 k 10.3 k + 1.8 k 2 k = 5.07 k + 947 k = 6.02 k
19.
RT = (R1 + R2 + R3) R4 (R5 + R6) = (100 k + 560 k + 56 k) 1.0 M (1.0 M + 100 k) = 716 k 1.0 M 1.1 M = 303 k
20.
Resistance of the right branch: RR = R2 + R5 R6 + R7 + R8 = 330 + 600 + 680 + 100 = 1710 Resistance of the left branch: RL = R3 + R4 = 470 + 560 = 1030 Total resistance: RT = R1 + RL RR = 1 k + 643 = 1.64 k 100 V IT = = 60.9 mA 1.64 k Current in the right branch: RL 1030 IR = IT 60.9 mA = 22.9 mA 2740 RL RR Current in the left branch: RR 1710 IL = IT 60.9 mA = 38.0 mA 2740 RL RR
53
Chapter 7 With respect to the negative source terminal: VA = ILR4 = (38.0 mA)(560 ) = 21.3 V VB = IR(R7 + R8) = (22.9 mA)(780 ) = 17.9 V VAB = VA VB = 21.3 V 17.9 V = 3.4 V 21.
(a)
R1 I2 = IT R1 R2 47 k 1 mA = IT 47 k R2 47 k + R2 = (47 k)IT Also, V 220 IT = RT 33 k (47 k) R2 (47 k) R2
Substituting the expression for IT into 47 k + R2 = (47 k)IT. 220 47 k + R2 = 47 k 33 k (47 k) R2 47 k R2 (47 k) R2 (47 k + R2) 33 k = 47 k(220) 47 k R2 (80 k)R2 = 47 k(220) (47 k)(33 k) 47 k(220 33 k) R2 = = 109.9 k 110 k 80 k (b)
P2 = I 22 R2 = (1 mA)2 110 k = 0.11 W = 110 mW
22.
RAB = R1 (R2 + R7 + R8) = 1 k (2.2 k + 3.3 k + 4.7 k) = 1 k 10.2 k = 911 RAG = R8 (R1 + R2 + R7) = 4.7 k (1 k + 2.2 k + 3.3 k) = 4.7 k 6.5 k = 2.73 k RAC = (R1 + R2) (R7 + R8) = (1 k + 2.2 k) (3.3 k + 4.7 k) = 3.2 k 8 k = 2.29 k RAD = RAC + R3 (R4 + R5 + R6) = 2.29 k + 1 k 10.2 k = 3.20 k RAE = RAC + (R3 + R4) (R5 + R6) = 2.29 k + 3.2 k 8 k = 4.58 k RAF = RAC + R6 (R3 + R4 + R5) = 2.29 k + 4.7 k 6.5 k = 5.02 k
23.
RAB = (R1 + R2) R4 R3 = 6.6 k 3.3 k 3.3 k = 1.32 k Note: R5 and R6 is shorted out (ACD) and is not a factor in the total resistance. RBC = R4 (R1 + R2) R3 = 1.32 k RCD = 0
54
Chapter 7 24.
V2 = V5 V6 = 5 V 1 V = 4 V 2W = 0.5 A I2 = I6 = 4V I5 = I8 I6 = 1 A 0.5 A = 0.5 A I1 = I2 + I5 + I4 = 0.5 A + 0.5 A + 1 A = 2 A I3 = IT I1 = 4 A 2 A = 2 A V7 = VS V3 = 40 V 20 V = 20 V 20 W = 10 V V1 = 2A V4 = V3 V1 = 10 V V8 = V4 V5 = 5 V
R1 = R2 = R3 = R4 = R5 = R6 =
10 V =5 2A 4V =8 0.5 A 20 V = 10 2A 10 V = 10 1A 5V = 10 0.5 A 1V =2 0.5 A
20 V =5 4A 5V R8 = =5 1A R7 =
Section 7-3 Voltage Dividers with Resistive Loads 25.
26.
56 k VOUT(unloaded) = 15 V = 7.5 V 112 k 56 k in parallel with a 1 M load is (56 k)(1 M) = 53 k Req = 56 k 1 M 56 k VOUT(loaded) = 15 V = 7.29 V 109 k
See Figure 7-6. 6.6 k VA = 12 V = 8 V 9.9 k 3.3 k VB = 12 V = 4 V 9.9 k With a 10 k resistor connected from tap A to ground: (6.6 k)(10 k) RAB = = 3.98 k 6.6 k 10 k 3.98 k VA(loaded) = 12 V = 6.56 V 7.28
Figure 7-6
27.
The 47 k will result in a smaller decrease in output voltage because it has less effect on the circuit resistance than does the smaller resistance.
55
Chapter 7 28.
RT = 10 k + 5.6 k + 2.7 k = 18.3 k R2 R3 8.3 k VOUT(NL) = 22 V = 9.98 V VS = 18.3 k R1 R2 R3
With a 100 k load: ( R2 R3 ) RL (8.3 k)(100 k) RT = R1 + = 17.7 k = 10 k + 108.3 k R2 R3 RL 7.7 k VOUT = 22 V = 9.57 V 17.7 k (8.3 k)(33 k) = 6.63 k 8.3 k 33 k 6.63 k VAB = 22 V = 8.77 V 10 k 6.63 k
29.
RAB =
30.
RT = 10 k + 5.6 k + 2.7 k = 18.3 k 22 V = 1.2 mA I= 18.3 k (8.3 k)(33 k) RT = 10 k + = 16.6 k 8.3 k 33 k 22 V = 1.33 mA I= 16.6 k
31.
See Figure 7-7. 10 V = 2 k RT = 5 mA R1 = R2 + R3 R2 = R3 R1 = 2R2
R1 + 2R2 = 2 k
2R2 + 2R2 = 2 k 4R2 = 2 k R2 = R3 = 500 R1 = R2 + R3 = 1000
Figure 7-7
With a 1 k load on the lower tap:
1 k 500 = 333 10 V = 5.46 mA IT = 1 k 500 333 Vlower tap = (333 )(5.46 mA) = 1.82 V Vupper tap = (500 + 333 )(5.46 mA) = 4.55 V
56
With a 1 k load on the upper tap: 10 V IT = = 6.67 mA 1 k 1 k / 2 Vupper tap = (500 )(6.67 mA) = 3.33 V 3.33 V = 1.67 V Vlower tap = 2
Chapter 7 32.
Position 1: RT = 10 k + 30 k 68 k = 10 k + 20.82 k = 30.8 k 20.8 k V1 = 120 V = 81.0 V 30.8 k 20 k V2 = 81 V = 54.0 V 30 k 10 k V3 = 81 V = 27.0 V 30 k Position 2: RT = 20 k + 20 k 68 k = 20 k + 15.5 k = 35.5 k 10 k 15.5 k V1 = 120 V = 86.2 V 35.5 k 15.5 k V2 = 81 V = 52.4 V 35 . 5 k 10 k V3 = 52.4 V = 26.2 V 20 k Position 3: RT = 30 k + 10 k 68 k = 30 k + 8.72 k = 38.7 k 20 k 8.72 k V1 = 120 V = 89.0 V 38.7 k 10 k 8.72 k V2 = 81 V = 58.0 V 38.7 k 8.72 k V3 = 81 V = 27.0 V 38.7 k
33.
(a)
R2 270 k VG = VDD 16 V = 1.75 V 2.47 M R1 R2 VS = VG + 1.5 V = 1.75 V + 1.5 V = 3.25 V VDD VG 16 V 1.75 V = 6.48 A R1 2.2 M V 1.75 V I2 = I1 = G = 6.48 A R2 270 k V 3.25 V IS = S = 2.17 mA RS 1.5 k ID = IS = 2.17 mA
(b)
I1 =
(c)
VD = VDD IDRD = 16 V (2.17 mA)(4.7 k) = 16 V 10.2 V = 5.8 V VDS = VD VS = 5.8 V 3.25 V = 2.55 V VDG = VD VG = 5.8 V 1.75 V = 4.05 V
57
Chapter 7 34.
Imax = 100 mA 24 V = 240 RT = 100 mA R2 24 V = 6 V RT
24R2 = 6RT 6(240 ) R2 = = 60 24 R1 = 240 60 = 180 With load: R2 RL = 60 1000 = 56.6 56.6 24 V = 5.74 V VOUT = 180 56.6
Section 7-4 Loading Effect of a Voltmeter 35.
The voltmeter presents the least load when set on the 1000 V range. For example, assuming 20,000 /V: Rinternal = (20,000 /V)(1 V) = 20 k on the 1 V range Rinternal = (20,000 /V)(1000 V) = 20 M on the 1000 V range
36.
(a) (b) (c) (d) (e) (f)
37.
Rinternal = (20,000 /V)(0.5 V) = 10 k Rinternal = (20,000 /V)(1 V) = 20 k Rinternal = (20,000 /V)(5 V) = 100 k Rinternal = (20,000 /V)(50 V) = 1 M Rinternal = (20,000 /V)(100 V) = 2 M Rinternal = (20,000 /V)(1000 V) = 20 M
R4 1.5 V 27 1.5 V = 0.305 V actual VR4 133 R R R R 2 3 4 1 (a) Use the 0.5 V range to measure 0.305 V. (b)
Rinternal = (20,000 /V)(0.5 V) = 10 k 27 10 k = 26.93 26.93 1.5 V 0.304 V with meter connected VR4 132.93 0.305 V 0.304 V = 0.001 V less with meter
58
Chapter 7 38.
R2 R3 R4 3 V 99.4 3 V = 0.383 V actual VR4 779.4 R R R R 1 2 3 4 (a)
Use the 0.5 V range to measure 0.383 V.
(b)
Rinternal = (20,000 /V)(0.5 V) = 10 k 99.4 10 k = 98.4 98.4 3 V = 0.379 V with meter connected VR4 778.4 0.383 V 0.379 V = 0.004 V less with meter
39.
RMETER = 10 V(10,000/V) = 100 k
R2
RMETER
R2 RMETER (100k)(100k) 50k R2 RMETER 200k
R2 RMETER VR2 R1 R2 RMETER 40.
R2
RMETER
50k VS VS 0.333VS 100 k 50 k
R2 RMETER (100k)(10M) 99k R2 RMETER 10.1M
R2 RMETER VR2 R1 R2 RMETER
99k VS VS 0.498V 100k 99k
Section 7-5 Ladder Networks 41.
The circuit in Figure 7-76 in the text is redrawn here in Figure 7-8 to make the analysis simpler. (a)
RT = 560 524.5 = 271
(b)
IT =
(c)
271 221 mA = 114 mA I2 = 524.5
60 V = 221 mA 271
59
Chapter 7 468.5 114 mA = 58.7 mA I910 = 910 (d)
The voltage across the 437.5 parallel combination of the 560 and the two series 1 k resistors is determined as follows: 468.5 114 mA = 55 mA I4 = 965.5 V437.5 = I4(437.5 ) = (55 mA)(437.5 ) = 24.06 V 1 k VAB = 24.06 V = 12 V 2 k
Figure 7-8
42.
The total resistance is determined in the steps shown in Figure 7-9. RT = 6.66 k 1.06 k VA = 18 V = 2.86 V 6.66 k 1.05 k VB = 2.86 V = 1.47 V 2.05 k 1 k VC = 1.47 V = 735 mV 2 k
Figure 7-9
60
Chapter 7
43.
The circuit is simplified in Figure 7-10 to determine RT. RT = 621 From Figure 7-10(e): IT = I9 = IT = 16.1 mA From Figure 7-10(c): 420.8 16.1 mA = 8.27 mA I2 = 820 From Figure 7-10(b): 420.8 16.1 mA = 7.84 mA I3 = I8 = 864.5 From Figure 7-10(a): 424.5 7.84 mA = 4.06 mA I4 = 820 From the original circuit: I5 = I6 = I7 = I3 I4 = 7.84 mA 4.06 mA = 3.78 mA
Figure 7-10
44.
The currents were found in Problem 43. V1 = ITR1 = (16.1 mA)(100 ) = 1.61 V V2 = I2R2 = (8.27 mA)(820 ) = 6.78 V V3 = I3R3 = (7.84 mA)(220 ) = 1.73 V V4 = I4R4 = (4.06 mA)(820 ) = 3.33 V
V5 = I5R5 = (3.78 mA)(100 ) = 0.378 V V6 = I6R6 = (3.78 mA)(680 ) = 2.57 V V7 = I7R7 = (3.78 mA)(100 ) = 0.378 V V8 = I8R8 = (7.84 mA)(220 ) = 1.73 V V9 = I9R9 = (16.1 mA)(100 ) = 1.61 V
61
Chapter 7 45.
The two parallel ladder networks are identical; so, the voltage to ground from each output terminal is the same; thus, VOUT = 0 V. Working from the right end, RT and then IT are determined as follows: (12 + 12 ) 18 = 10.3 (22 + 10.3 ) 27 = 14.7 RT1 = 47 + 14.7 = 61.7 R RT(both) = T1 61.7 = 30.9 2 2 30 V = 971 mA IT = 30.9
46.
(a) (b)
47.
(a) (b) (c)
V 12 V = 1.5 V 8 8 V 12 V VOUT = = 0.75 V 16 16
VOUT =
V V 12 V 12 V =3V+6V=9V 4 2 4 2 V V 12 V 12 V VOUT = = 3 V + 0.75 V = 3.75 V 4 16 4 16 V V V V 12 V 12 V 12 V 12 V VOUT = 2 4 8 16 2 4 8 16 = 6 V + 3 V + 1.5 V + 0.75 V = 11.25 V VOUT =
Section 7-6 The Wheatstone Bridge 48.
49.
50.
R Rx = RV 2 = (18 k)(0.02) = 360 R4 SG3 119.94 VLEFT = VS 12 V = 5.997 V SG1 + SG3 120.06 119.94 SG4 120.06 VRIGHT = VS 12 V = 6.003 V SG2 + SG4 119.94 120.06 VOUT = VRIGHT VLEFT = 6.003 V 5.997 V = 6 mV (Right side positive with respect to left side)
At 60 C, RTHERM = 5 k R3 27 k VLEFT = VS 9 V = 7.59 V 32 k R1 R3 R4 27 k VRIGHT = VS 9 V = 4.50 V 54 k R2 R4 VOUT = VLEFT VRIGHT = 7.59 V 4.50 V = 3.09 V
62
Chapter 7 Section 7-7 Troubleshooting (680 )(4.7 k) = 594 680 4.7 k RT = 560 + 470 + 594 = 1624 The voltmeter reading should be 594 12 V = 4.39 V V? = 1624 The voltmeter reading of 6.2 V is incorrect.
51.
Req =
52.
The circuit is redrawn in figure 7-11 and points are labeled. (10 k 47 k)(100 k) RBG = = 36.3 k 10 k 47 k 100 k RAG = 33 k + RBG = 33 k + 36.3 k = 69.3 k RT = 27 k + RAG = 27 k + 69.3 k = 96.3 k R 69.3 k VAG = AG 18 V 18 V = 12.95 V 96.3 k RT 47 k 47 k VCG = VBG 6.79 V = 5.60 V 57 k 57 k VAC = VAG VCG = 12.95 V 5.60 V = 7.35 V Both meters are correct.
Figure 7-11
53.
The 2.5 V reading indicated on one of the meters shows that the series-parallel branch containing the other meter is open. The 0 V reading on the other meter shows that there is no current in that branch. Therefore, if only one resistor is open, it must be the 2.2 k.
63
Chapter 7 54.
The circuit is redrawn in Figure 7-12. 12 k 12 k 150 V 6 k 150 V = 56.25 V VA = 12 k 12 k 10 k 16 k The meter reading of 81.8 V is incorrect. The most likely fault is an open 12 k resistor. This will cause the voltage at point A to be higher than it should be. To verify, calculate VA assuming an open 12 k resistor. 12 k VA = 150 V = 81.8 V 22 k 2.2 k VB = 150 V = 42.3 V 7 .8 k The meter is correct.
Figure 7-12
55.
56.
1.62 k V3.3 k = (10 V) = 6.18 V 2.62 k The 7.62 V reading is incorrect. 2.2 k V2.2 = (6.18 V) = 4.25 3.2 k The 5.24 V reading is incorrect. The 3.3 k resistor must be open. If it is, then 3.2 k V3.3 k = ( 10 V) = 7.62 V 4.2 k 2.2 k V2.2 k = (7.62 V) = 5.24 V 3.2 k If R2 opens, VA = 15 V, VB = 0 V, and VC = 0 V
Multisim Troubleshooting and Analysis 57.
RT = 296.744
58.
R4 is open.
59.
R3 = 560 k
60.
No fault.
61.
R5 is shorted.
62.
RX = 550
64
Chapter 8 Circuit Theorems and Conversions Note: Solutions show conventional current direction.
Section 8-3 Source Conversions VS 300 V =6A RS 50 RS = 50 See Figure 8-1.
1.
IS =
2.
(a) (b)
5 kV = 50 A 100 12 V IS = = 5.45 A 2.2
Figure 8-1
IS =
1.6 V = 0.2 8.0 A
3.
RS =
4.
See Figure 8-2.
Figure 8-2 5.
VS = ISRS = (600 mA)(1.2 k) = 720 V RS = 1.2 k See Figure 8-3.
6.
(a) (b)
VS = (10 mA)(4.7 k) = 47 V VS = (0.01 A)(2.7 k) = 27 V Figure 8-3
Section 8-4 The Superposition Theorem 7.
First, zero the 3 V source by replacing it with a short as in Figure 8-4(a). RT = 1.955 k 2V IT = = 1.02 mA 1.955 k 2.2 k I3 = 1.02 mA = 577 A 3.89 k
65
Chapter 8 1 k I5 = 5.77 A = 180 A 3.2 k Next, zero the 2 V source by replacing it with a short as in Figure 8-4(b). RT = 1.955 k 3V IT = = 1.53 mA 1.955 k 1.69 k I5 = 1.53 mA = 655 A 3.89 k Since both components of I5 are in the same direction, the total I5 is I5(total) = 180 A + 665 A = 845 A
Figure 8-4
8.
From Problem 7: RT = 1.955 k and IT = 1.02 mA Current in R2 due to the 2 V source acting alone. See Figure 8-5(a): 1.69 k I2 = 1.02 mA = 443 A (downward) 3.89 k From Problem 7: RT = 1.955 k and IT = 1.53 mA Current in R2 due to the 3 V source acting alone. See Figure 8-5(b): 2.2 k ILeft = 1.53 mA = 865 A 3.89 k 1 k I2 = 865 A = 270 A (downward) 3.2 k The total current through R2 is I2 = 443 A + 270 A = 713 A
Figure 8-5
66
Chapter 8 9.
From Problem 7: From the 2 V source: I4 = I3 – I5 = 577 A – 180 A = 397 A downward through R4 From the 3 V source: I4 = IT = 1.53 mA upward through R4
I4(TOT) = 1.53 mA – 397 A = 1.13 mA upward 10.
First, zero the voltage source by replacing it with a short as shown in Figure 8-6(a):
680 100 mA = 79.8 mA 852.6
I1 =
220 79.8 mA = 17.2 mA 1020
I3 =
I2 = I1 –I3 = 79.9 mA – 17.2 mA = 62.7 mA downward Next, zero the current source by replacing it with an open as shown in Figure 8-6(b): RT = 587.6
I2 = IT =
20 V = 34.0 mA downward 587.6
I2(TOT) = 62.7 mA + 34.0 mA = 96.7 mA 11.
First, zero the voltage source by replacing it with a short as shown in Figure 8-6(a): 680 100 mA = 79.8 mA I1 = 852.6 220 79.8 mA = 17.2 mA I3 = 1020 Next, zero the current source by replacing it with an open as shown in Figure 8-6(b): RT = 587.6 20 V IT = = 34.0 mA 587.6
680 34.0 mA = 15.6 mA I3 = 1480 The total I3 is the difference of the two component currents found in the above steps because they are in opposite directions. I3(total) = 17.2 mA 15.6 mA = 1.6 mA
67
Chapter 8
Figure 8-6
12.
(a)
Current through RL due to the 1 A source. See Figure 8-7(a): 2.2 k IL = 1 A = 361 mA (down) 6.1 k Current through RL due to the 2 A source is zero because of infinite resistance (open) of the 1 A source. See Figure 8-7(b): IL = 0 A Total current through RL: IL(total) = 361 mA + 0 A = 361 mA
Figure 8-7
(b)
Current through RL due to the 40 V source is zero because of zero resistance (short) of the 60 V source. See Figure 8-8(a): IL = 0 A Current through RL due to the 0.5 A source is zero because of zero resistance of the 60 V source. See Figure 8-8(b): IL = 0 A Current through RL due to the 60 V source. See Figure 8-8(c): 1.5 k VL = 60 V = 43.7 V 2.06 k 43.7 V V IL = L = 29.1 mA RL 1.5 k Total current through RL: IL = 0 A + 0 A + 29.1 mA = 29.1 mA
68
Chapter 8
Figure 8-8
13.
R2 R3 7.8 k VRef(max) 30 V 15 V = 3.72 V 30 V 15 V 12.5 k R1 R2 R3 R3 6.8 k VRef(min) 30 V 15 V = 1.32 V 30 V 15 V 12.5 k R1 R2 R3
14.
R2 R3 16.8 k VRef(max) 30 V 15 V = 8.44 V 30 V 15 V 21.5 k R1 R2 R3 R3 6.8 k VRef(min) 30 V 15 V = 5.51 V 30 V 15 V 21.5 k R1 R2 R3
15.
75 V source. See Figure 8-9(a): Req = R2 R3 (R4 + R5) = 17.2 k Req 17.2 k VA = 75 V 75 V = 13 V R R 99.2 k 1 eq R5 91 k VA VB = 13 V = 11.7 V 101 k R4 R5
50 V source. See Figure 8-9(b): Req = R1 R2 (R4 + R5) = 25 k Req 25 k VA = 50 V 50 V = 21.6 V R R 58 k 3 eq
69
Chapter 8 R5 91 k VB = VA ( 21.6 V) = 19.5 V 101 k R4 R5 100 V source. See Figure 8-9(c): Req = R1 R2 R3 = 16.6 k RT = 10 k + 91 k + 16.6 k = 117.6 k 100 V IT = = 850 A 117.6 k VA = (850 A)(16.6 k) = 14.1 V VB = (850 A)(91 k) = 77.4 V Superimposing voltages at each point: VA = 13 V 21.6 V + 14.1 V = 5.5 V VB = 11.7 V 19.5 V 77.4 V = 85.2 V VAB = 5.5 V (85.2 V) = 90.7 V
Figure 8-9
70
Chapter 8 16.
SW1 closed. See Figure 8-10(a): 12 V 12 V IL = = 508 A 5.6 k 18 k 23.6 k SW1 and SW2 closed. See Figure 8-10(b): Current from the 12 V source (6 V source zeroed) RT = R1 + R2 RL = 5.6 k + 8.2 k 18 k = 11.2 k 12 V = 1.07 mA IT = 11.2 k 8.2 k IL = 1.07 mA = 335 A 26.2 k Current from the 6 V source (12 V source zeroed): RT = R2 + R1 RL = 8.2 k + 5.6 k 18 k = 12.47 k 6V IT = = 481 A 12.47 k 5.6 k IL = 481 A = 114 A 23.6 k IL(total) = 335 A + 114 A = 449 A SW1, SW2, and SW3 closed. See Figure 8-10(c). Current from the 12 V source (6 V and 9 V sources zeroed): RT = R1 + R2 R3 RL = 5.6 k + 8.2 k 12 k 18 k = 9.43 k 12 V IT = = 1.27 mA 9.43 k R R R 3.83 k IL = 2 3 L I T 1.27 mA = 270 A RL 18 k Current from the 6 V source (9 V and 12 V sources zeroed): RT = R2 + R1 R3 RL = 8.2 k + 5.6 k 12 k 18 k = 11.35 k 6V IT = = 529 A 11.35 k R R R 3.15 k IL = 1 3 L I T 529 A = 93 A RL 18 k Current from the 9 V source (6 V and 12 V sources zeroed): RT = R3 + R1 R2 RL = 12 k + 5.6 k 8.2 k 18 k = 14.8 k 9V IT = = 608 A 14.85 k R R R 2.81 k IL = 1 2 L I T 608 A = 95 A RL 18 k IL(total) = 270 A + 93 A + 95 A = 458 A
71
Chapter 8
Figure 8-10
17.
VS1 “sees” a total resistance of RT = 10 k + (5.6 k (10 k + (5.6 k ((10 k + 5.6 k) + (10 k (5.6 k + (10 k 5.6 k))))))) = 10 k + (5.6 k (10 k + (5.6 k (15.6 k + (10 k (5.6 k + 3.59 k)))))) = 10 k + (5.6 k (10 k + (5.6 k (15.6 k + (10 k 9.19 k))))) = 10 k + (5.6 k (10 k + (5.6 k (15.6 k + 4.79 k)))) = 10 k + (5.6 k (10 k + (5.6 k 20.4 k))) = 10 k + (5.6 k (10 k + 4.39 k)) = 10 k + (5.6 k (14.4 k) = 10 k + 4.03 k = 14.0 k 32 V IT(S1) = = 2.28 mA 14.0 k VS2 “sees” a total resistance of RT = 5.6 k + (10 k (5.6 k + (10 k ((10 k + 5.6 k) + (5.6 k (10 k + (5.6 k 10 k))))))) = 5.6 k + (10 k (5.6 k + (10 k (15.6 k + (5.6 k (10 k + 3.59 k)))))) = 5.6 k + (10 k (5.6 k + (10 k (15.6 k + (5.6 k 13.6 k))))) = 5.6 k + (10 k (5.6 k + (10 k (15.6 k + (3.97 k)))) = 5.6 k + (10 k (5.6 k + (10 k (19.6 k))) = 5.6 k + (10 k (5.6 k + 6.62 k)) = 5.6 k + (10 k 12.2 k) = 5.6 k + 550 k = 11.1 k IT(S2) =
15 V = 1.35 mA 11.1 k
72
Chapter 8 Section 8-5 Thevenin’s Theorem 18.
19.
(a)
RTH = 27 + 75 147 = 76.7 75 25 V = 8.45 V VTH = 222
(b)
RTH = 100 270 = 73 100 3 V = 811 mV VTH = 370
(c)
RTH = 56 k 100 k = 35.9 k 56 k (15 V 10 V) = 1.79 V VTH = 156 k
(b)
RTH = 2.2 k (1 k + 2.2 k = 1.3 k 2.2 k IAB = 0.1 A = 40.7 mV 5.4 k VTH = IAB(2.2 k) = (40.7 mA)(2.2 k) = 89.5 V
First, convert the circuit to its Thevenin equivalent as shown in the steps of Figure 8-11. RTH = 13.97 k 4.12 k VA = 32 V = 9.34 V 14.12 k 5.6 k 5.6 k VTH = VA 9.34 V = 3.35 V 15.6 k 15.6 k VTH 3.35 V IL = = 116 A RTH RL 28.97 k
Figure 8-11
73
Chapter 8 20.
First, zero (open) the current source, remove R4, and redraw the circuit as shown in Figure 8-12(a). RTH = R3 (R1 + R2 R5) = 5.6 k (1 k + 1.65 k) = 5.6 k 2.65 k = 1.8 k 2.65 k 2.65 k VTH = 50 V 50 V = 16.1 V 5.6 k 2.65 k 8.25 k Determine V4 due to the 50 V source using the Thevenin circuit in Figure 8-12(b). R4 10 k V4 = VTH 16.1 V = 13.6 V 11.8 k RTH R4 Next, zero (short) the voltage source, remove R4, and redraw the circuit as shown in Figure 8-12(c). RTH = R3 (R1 + R2 R5) = 5.6 k (1 k + 1.65 k) = 5.6 k 2.65 k = 1.8 k 2.65 k I3 = 10 mA = 3.2 mA 8.25 k VTH = V3 = I3R3 = (3.2 mA)(5.6 k) = 17.9 V Determine V4 due to the current source using the Thevenin circuit in Figure 8-12(d). R4 10 k V4 = VTH 17.9 V = 15.2 V 11.8 k RTH R4 Use superposition to combine the V4 voltages to get the total voltage across R4: V4 = 13.6 V + 15.2 V = 28.8 V
Figure 8-12
74
Chapter 8 21.
Looking back from the amplifier input: RTH = R1 R2 R3 = 100 2.2 k 1.2 k = 88.6 1 V source (Figure 8-13(a)): 776 1 V = 886 mV VA = 876 5 V source (Figure 8-13(b)): 92.3 VA = = 200 mV 2292 5 V VTH = 886 mV + 200 mV = 1.09 V
Figure 8-13
22.
Consider R6 (R7 + R8) to be the load. Thevenize to the left of point A as shown in Figure 8-14(a). RTH = R5 + R4 (R3 + (R1 R2)) = 1 k 4.7 k (10 k + 6.8 k 9.1 k) = 1 k + 4.7 k 13.89 k = 4.51 k See Figure 8-14(b) to determine VTH: RT = (R3 + R4) R2 + R1 = (10 k + 4.7 k) 6.8 k + 9.1 k) = 4.65 k + 9.1 k = 13.8 k 48 V IT = = 3.48 mA 13.8 k R2 6.8 k I4 = IT 3.48 mA = 1.1 mA R R R 21.5 k 3 4 2 V4 = I4R4 = (1.1 mA)(4.7 k) = 5.17 V VX = 48 V V4 = 48 V 5.17 V = 42.8 V VTH = VA = VX = 42.8 V
75
Chapter 8 The Thevenin circuit is shown in Figure 8-14(c). The current into point A is determined for each value of R8. When R8 = 1 k: RL = 12 k (8.2 k + 1 k) = 5.21 k VTH 42.8 V IA = = 4.41 mA RTH RL 9.72 k
When R8 = 5 k: RL = 12 k (8.2 k + 5 k) = 6.29 k VTH 42.8 V IA = = 3.97 mA RTH RL 10.8 k When R8 = 10 k: RL = 12 k (8.2 k + 10 k) = 7.23 k VTH 42.8 V IA = = 3.66 mA RTH RL 11.7 k
Figure 8-14
76
Chapter 8 23.
See Figure 8-15. 1.2 k 2.2 k VTH = VA VB = 12 V = 8.25 V 7.13 V = 1.12 V 12 V 2.02 k 3.2 k RTH = 1 k 2.2 k + 820 1.2 k = 688 + 487 = 1175 VTH 1.12 V IL = = 100 A RTH RL 11,175
Figure 8-15
24.
See Figure 8-16. VR3 = (0.2 mA)(15 k) = 3 V V VR 3 10 V 3 V = 35 k R4 = S I4 0.2 mA R2 12 k VA = VS 10 V = 5.46 V 22 k R1 R2 R4 35 k VB = VS 10 V = 7V 50 k R3 R4
VTH = VBA = VB VA = 7 V 5.46 V = 1.54 V RTH = R1 R2 + R3 R4 = 5.46 k + 10.5 k = 15.96 k
Figure 8-16
77
Chapter 8 Section 8-6 Norton’s Theorem 25.
(a)
See Figure 8-17(a). RN = 76.7
(b)
RT = 166.9
See Figure 8-17(b). RN = 73 3V IN = = 11.1 mA 270
25 V = 150 mA 166.9 75 75 IN = IT 150 mA = 110 mA 102 102
IT =
(c)
See Figure 8-17(c). (56 k)(100 k) RN = = 35.9 k 156 k 5V IN = = 50 A 100 k
(d)
Figure 8-17
78
See Figure 8-17(d). (3.2 k)(2.2 k) RN = = 1.3 k 5.4 k 2.2 k IN = 0.1 A = 68.8 mA 3.2 k
Chapter 8 26.
First, RN is found by circuit simplification as shown in Figure 8-18(a). RN = 14.0 k The current IN through the shorted AB terminals is found as shown in Figure 8-18 (b). RT = 14.0 k as viewed from the source 32 V IT = = 2.29 mA 14.0 k 5.6 k I1 = 2.29 mA = 668 A 19.2 k 5.6 k IN = 668 A = 240 A 15.6 k Finally, the current through RL is determined by connecting RL to the Norton equivalent circuit as shown in Figure 8-18(c). 14.0 k IL = 240 A = 116 A 29.0 k
Figure 8-18
79
Chapter 8 27.
The 50 V source acting alone. Short AB to get IN. See Figure 8-19(a): RT = R3 + R1 R4 = 5.6 k + 1 k 10 k = 6.51 k 50 V IT = = 7.68 mA 6.51 k R4 10 k IT IN = 7.68 mA = 6.98 mA R R 11 k 4 1 See Figure 8-19(b): RN = R2 (R1 + R3 R4) = 3.3 k (1 k + 5.6 k 10 k) = 3.3 k 4.59 k = 1.92 k See Figure 8-19(c): RN 1.92 k IR5 = IN 6.98 mA = 2.57 mA (from B to A) 5.22 k RN R5 The 10 mA source acting alone. Short AB to get IN. See Figure 8-19(d): R3 R4 5.6 k 10 k 10 mA 10 mA 3.59 k 10 mA = 7.82 mA IN = R R R 1 k 5.6 k 10 k 4.59 k 3 4 1 RN = 1.92 k See Figure 8-19(e): 1.9 k IR5 = 7.82 mA = 2.85 mA (from B to A) 5.22 k V5 = I5R5 = (5.42 mA)(3.3 k) = 17.9 V
Figure 8-19
80
Chapter 8 28.
See Figure 8-20(a): RN = R2 (R3 + R4 (R5 + R6 (R7 + R8 ))) = 6.8 k (10 k + 4.7 k (1 k + 6.89 k)) = 6.8 k (10 k + 2.95 k) = 4.46 k See Figure 8-20(b): RT = R2 (R4 + R3 (R5 + R6 (R7 + R8))) = 6.8 k (4.7 k + 10 k (1 k + 6.89 k)) = 6.8 k (4.7 k + 4.41 k) = 3.89 k 48 V = 12.3 mA IT = 3.89 k 9.11 k 9.11 k I2 = IT 12.3 mA = 7.07 mA 6.8 k 9.11 k 6.8 k 9.11 k 6.8 k I4 = 12.3 mA = 5.27 mA 15.9 k 7.89 k I3 = 5.27 mA = 2.62 mA 15.9 k IN = I2 + I3 = 7.07 mA + 2.62 mA = 9.69 mA See Figure 8-20(c): 4.46 k I1 = 9.69 mA = 3.18 mA 13.6 k
Figure 8-20
81
Chapter 8 29.
Using the results of Problem 23: V 1.12 V = 953 A IN = TH RTH 1175 RN = RTH = 1175 See Figure 8-21. Figure 8-21
30.
See Figure 8-22(a): RN = 10 k (15 k + 8.2 k 22 k) = 6.77 k See Figure 8-22(b): RT = 8.2 k 15 k + 22 k = 27.3 k 12 V = 440 A IT = 27.3 k 8 .2 k IN1 = 440 A = 156 A down 23.3 k See Figure 8-22(c): 15 k 10 mA 15 k 10 mA = 7.15 mA down IN2 = 15 k 22 k 8.2 k 20.97 k See Figure 8-22(d): IN = IN1 + IN2 = 156 A + 7.15 mA = 7.31 mA
Figure 8-22
82
Chapter 8 31.
RN = 220 100 330 = 56.9 Find IN1 due to the 3 V source, as shown in Figure 8-23(a). 3V = 9.1 mA (down) IN1 = 330 Find IN2 due to the 8 V source, as shown in Figure 8-23(b). 8V = 80 mA (up) IN2 = 100 Find IN3 due to the 5 V source, as shown in Figure 8-23(c). 5V IN1 = = 22.7 mA (down) 220 The Norton equivalent is shown in Figure 8-23(d). IN(tot) = IN1 + IN2 + IN3 = 9.1 mA 80 mA + 22.7 mA = 48.2 mA
56.9
Figure 8-23
83
Chapter 8 Section 8-7 Maximum Power Transfer Theorem RL = RS = 12 RL = RS = 8.2 k RL = RS = 4.7 + 1 2 = 6.37 RL = RS = 47 + 680 = 727
32.
(a) (b) (c) (d)
33.
See Figure 8-24. As seen by RL: RS = 8.2 + 2.94 = 11.1 For maximum power transfer: RL = RS = 11.1
Figure 8-24
34.
Refer to Problem 33 and Figure 8-24. RL+ = RL + 0.1RL = 11.1 + 1.11 = 12.21 RTH = RS = 11.1 IL due to the 1.5 V source: 15 16.4 1.5 V 7.79 1.5 V = 936 mV VTH = 4.7 15 16.4 12.49 VTH 936 mV IL = = 40 mA 23.4 RTH RL IL due to the 1 mA source: 4.7 16.4 1 mA 3.65 1 mA = 196 A I15 = 15 4.7 16.4 18.65 VTH = I15(15 ) = (196 A)(15 ) = 2.94 mV VTH 2.94 mV IL = = 126 mA 23.4 RTH RL IL(total) = 40 mA + 126 A = 40.126 mA PL = I L2 RL = (40.126 mA)212.21 = 19.7 mW
84
Chapter 8 35.
For maximum power transfer, RTH = RLADDER The voltage across RTH = 24 V (one half of VTH) 24 V = 48 RTH = 0.5 A RLADDER = 48 RLADDER = ((R4 (R5 + R6) + R3) R2) + R1 69 R4 10 47 69 R4 = 26 69 R4 47 10 69 R4 69 R4 26 69 R4 10 57 69 R4 47 69 R4 69 R4 26 26 1 57 10 21.53 69 R4 47 47 69R4 = 69(48.17) + 48.17R4 R4(69 48.17) = 69(48.17) 69(48.17) = 160 R4 = 69 48.17
Section 8-8 Delta-Wye (Y) and Wye-Delta (Y-) Conversions 36.
RA RC (560 k)(1 M) = 183 k RA RB RC 3.06 M RB RC (1.5 M)(1 M) R2 = = 490 k RA RB RC 3.06 M RA RB (560 k)(1.5 M) R3 = = 275 k RA RB RC 3.06 M
(a)
R1 =
(b)
R1 =
RA RC (1 )(2.2 ) = 373 m RA RB RC 5.9 RB RC (2.2 )(2.7 ) R2 = = 1.01 RA RB RC 5 .9 RA RB (1 )(2.7 ) R3 = = 4.58 m RA RB RC 5 .9
85
Chapter 8 37.
38.
R1R2 R1R3 R2 R3 (12 )(22 ) (12 )(18 ) (22 )(18 ) 876 = 39.8 R2 22 22 R R R1R3 R2 R3 (12 )(22 ) (12 )(18 ) (22 )(18 ) 876 RB = 1 2 = 73 R1 12 12 R R R1R3 R2 R3 (12 )(22 ) (12 )(18 ) (22 )(18 ) 876 RC = 1 2 = 48.7 R3 18 18
(a)
RA =
(b)
RA =
R1R2 R1R3 R2 R3 (6.8 k)(3.3 k) (6.8 k)(4.7 k) (3.3 k)(4.7 k) = 21.2 k R2 3.3 k R R R1R3 R2 R3 (6.8 k)(3.3 k) (6.8 k)(4.7 k) (3.3 k)(4.7 k) RB = 1 2 = 10.3 k R1 6.8 k R R R1R3 R2 R3 (6.8 k)(3.3 k) (6.8 k)(4.7 k) (3.3 k)(4.7 k) RC = 1 2 = 14.9 k R3 4 .7 k
Convert the delta formed by R3, R4, and R5 to a Wye configuration. See Figure 8-25: R3 R4 (22 k)(12 k) RY1 = = 6.13 k 43.1 k R3 R4 R5 (22 k)(9.1 k) R3 R5 RY2 = = 4.65 k 43.1 k R3 R4 R5 R4 R5 (12 k)(9.1 k) RY3 = = 2.53 k 43.1 k R3 R4 R5 RT = (R1 + RY1) (R2 + RY2) + RY3 = (10 k + 6.13 k) (39 k + 4.65 k) + 2.53 k = 11.78 k + 2.53 k = 14.3 k 136 V 136 V IT = = 9.5 mA 14.3 k RT R2 RY 2 43.65 k IT IR1 = IRY1 = 9.5 mA = 6.94 mA 59.78 k R1 RY 1 R2 RY 2 IR2 = IRY2 = IT IR1 = 9.5 mA 6.94 mA = 2.56 mA VB = VA IR1R1 = 136 V (6.94 mA)(10 k) = 66.6 V VC = VA IR2R2 = 136 V (2.56 mA)(39 k) = 36.16 V In the original circuit: V 66.6 V IR4 = B = 5.55 mA R4 12 k V 36.16 V IR5 = C = 3.97 mA R5 9.1 k V V 66.6 V 36.16 V IR3 = B C = 1.38 mA R3 22 k
Figure 8-25
86
Chapter 8 Multisim Troubleshooting and Analysis 39.
R1 is leaky.
40.
VTH = 17.478 V; RTH = 247.279
41.
IN = 0.383 mA; RN = 9.674 k
42.
R3 is shorted.
43.
IAB = 1.206 mA; VAB = 3.432 V
87
Chapter 9 Branch, Loop, and Node Analysis Note: Solutions show conventional current direction.
Section 9-1 Simultaneous Equations in Circuit Analysis 1.
100I1 + 50I2 = 30 75I1 + 90I2 = 15 30 50 I 2 I1 = 100 30 50 I 2 75 + 90I2 = 15 100 22.5 37.5I2 + 90I2 = 15 52.5I2 = 7.5 I2 = 143 mA 100I1 + 50(0.143) = 30 I1 = 371 mA
2.
(a)
4 6 = 12 12 = 0 2 3
(b)
9 1 = 45 0 = 45 0 5
(c)
12 2
(d)
100 50 = 2000 1500 = 3500 30 20
15 = 12 (30) = 18 1
1 4
4 2 3.
(a)
4.
(a)
I1 =
6 3 12 12 =0A 1 2 3 14 7 3
(b)
I2 =
7 6 6 28 =2A 1 2 3 14 7 3
1 0 2 1 0 5 4 15 4 2 10 0 2 10 = (1)(4)(0) + (0)(1)(2) + (2)(5)(10) [(2)(4)(2) + (10)(1)(1) + (0)(5)(0)] = (0 + 0 100) (16 + 10 + 0) = 100 + 6 = 94
(b)
0 .5 1 0.8 0.5 1 0.1 1.2 1.5 0.1 1 .2 0.1 0.3 5 0 .1 0 .3 = (0.5)(1.2)(5) + (1)(1.5)(0.1) + (0.8)(0.1)(0.3) [(0.8)(1.2)(0.1) + (0.3)(1.5)(0.5) + (5)(0.1)(1)] = (3 0.15 + 0.024) (0.096 0.255 + 0.5) = 2.874 0.371 = 2.50
88
Chapter 9 5.
(a)
0 20 25 0
25
10 12 5 10 12 8 30 16 8 30 = 25(12)(16) + (0)(5)(8) + (20)(10)(30) [(8)(12)(20) + (30)(5)(25) + (16)(10)(0)] = 10800 5670 = 16,470 (b)
1.08 1.75
0.55 1.08 1.75
2.12 0.98 0 3.49 1.05 1
0 1
2.12 3.49
= (1.08)(2.12)(1.05) + (1.75)(0.98)(1) + (0.55)(0)(3.49) [(1)(2.12)(0.55) + (3.49)(0.98)(1.08) + (1.05)(0)(1.75)] = 4.119 + 2.528 = 1.591 6.
The characteristic determinant was evaluated as 2.35 in Example 9-4. The determinant for I3 is as follows: 2 0.5
0
2 0.5
0.75 0 1.5 0.75 0 = (0 + 2.25 + 0) (0 + 0.6 0.375) = 2.25 0.225 = 2.025 3 0.2 1 3 0 .2 I3 = 7.
2.025 = 862 mA 2.35
The characteristic determinant is: 2 6 10 2 6 3 7 8 3 7 10 5 12 10 5 = (2)(7)(12) + (6)(8)(10) + (10)(3)(5) [(10)(7)(10) + (5)(8)(2) + (12)(3)(6)] = 462 836 = 374
I1 =
9 6 10 9 6 3 7 8 3 7 0 5 12 0 5
374 (9)(7)(12) (6)(8)(0) (10)(3)(5) [(0)(7)(10) (5)(8)(9) (12)(3)(6)] = 374 606 144 462 = 1.24 A = 374 374
89
Chapter 9 2 9 3 3 I2 =
10 2 9 8 3 3
10 0 12 10 0
374 (2)(3)(12) (9)( 8)(10) (10)(3)(0) [(10)(3)(10) (0)( 8)(2) ( 12)(3)(9)] = 374 792 24 768 = = 2.05 A 374 374 2 6 9 2 6 3 7 3 3 7
I3 =
10
5 0 10
5
374 (2)(7)(0) (6)(3)(10) (9)(3)(5) [(10)(7)(9) (5)(3)(2) (0)(3)(6)] = 374 45 660 705 = = 1.89 A 374 374
8.
The calculator results are:
V1 = 1.61301369863 V2 = 1.69092465753 V3 = 2.52397260274 V4 = 4.69691780822 9.
X1 = .371428571429 (I1 = 371 mA) X2 = .142857142857 (I2 = 143 mA)
10.
X1 = 1.23529411765 (I1 = 1.24 A) X2 = 2.05347593583 (I2 = 2.05 A) X3 = 1.88502673797 (I3 = 1.89 A)
Section 9-2 Branch Current Method 11.
The sum of the currents at the node is zero. Currents into the node are assumed positive and currents out of the node are assumed negative.
I1 I2 I3 = 0 12.
I1 I2 I3 = 0 8.2I1 + 10I2 =12 10I2 + 5.6I3 = 6 Solving by substitution: I1 = I2 + I3 8.2(I2 + I3) + 10I2 = 12 8.2I2 + 8.2I3 = 10I2 = 12
90
Chapter 9 18.2I2 + 8.2I3 = 12 12 8.2 I 3 I2 = 18.2 12 8.2 I 3 10 + 5.6I3 = 6 18.2 120 82 I 3 + 5.6I3 = 6 18.2 10.11I3 = 0.59 I3 = 58.4 mA 10I2 + 5.6(0.058) = 6 10I2 + 0.325 = 6 I2 = 633 mA I1 = I2 + I3 = 633 mA + 58.4 mA = 691 mA 13.
The branch currents were found in Problem 12. I1 = 691 mA I2 = 633 mA I3 = 58.4 mA V1 = I1R1 = (691 mA)(8.2 ) = 5.66 V (+ on left) V2 = I2R2 = (633 mA)(10 ) = 6.33 V (+ at top) V3 =I3R3 = (58.4 mA)(5.6 ) = 325 mV (+ on left)
14.
I1 I2 = 100 mA 12 VA VA = 0.1 47 100 100(12 VA) 47VA = 470 1200 100VA 47VA = 470 147VA = 730 VA = 4.97 12 V 4.97 V 7.03 V I1 = = 150 mA 47 47 4.97 V I2 = = 49.7 mA 100 I3 = 100 mA (current source)
15.
Current source zeroed (open). See Figure 9-1(a). R2 100 VAB = V2 = VS 12 V = 8.16 V 147 R1 R2
Voltage source zeroed (shorted). See Figure 9-1(b). VAB = V3 = I3R3 = (100 mA)(68 ) = 6.8 V R1 47 I2 = IS 100 mA = 31.97 mA 147 R1 R2 VAG = V2 = (31.97 mA)(100 ) = 3.197 V VAB = VAG VBG = 3.197 6.8 V = 9.997 V Superimposing: VAB = 8.16 V + (9.997 V) = 1.84 V
91
Figure 9-1
Chapter 9 Section 9-3 Loop Current Method 16.
The characteristic determinant is: 0.045 0.130 0.066 0.045 0.130 0.177 0.042 0.109 0.177 0.042 0.078 0.196 0.290 0.078 0.196 = (0.045)(0.042)(0.290) + (0.130)(0.109)(0.078) + (0.066)(0.177)(0.196) [(0.078(0.042)(0.066) + (0.196)(0.109)(0.045) + (0.290)(0.177)(0.130)] = 0.00394 0.00785 = 0.00391
17.
1560I1 560I2 = 6 560I1 + 1380I2 = 2 6 560 I1 =
2 1380 8280 1120 9400 = 5.11 mA 1560 560 2,152,800 313,600 1,839,200 560 1380
1560 6 560 2 3180 3360 I2 = = 3.52 mA 1,839,200 1,839,200 18.
Using the loop currents from Problem 17: I1 k = I1 = 5.11 mA I820 = I2 = 352 mA I560 = I1 I2 = 5.11 mA + 3.52 mA = 1.59 mA
19.
Using the branch currents from Problem 18: V1 k = I1 k(1 k) = (5.11 mA)(1 k) = 5.11 V (+ on right) V560 = I560 (560 ) = (1.59 mA)(560 ) = 890 mV (+ on bottom) V820 = I820 (820 ) = (3.52 mA)(820 ) = 2.89 V (+ on right)
20.
57I1 10I2 = 1.5 10I1 + 41.7I2 4.7I3 = 3 4.7I2 + 19.7I3 = 1.5
92
Chapter 9 21.
The equations were developed in Problem 20. The characteristic determinant is as follows with the k units omitted for simplicity: 57 10 0 57 10 10 41.7 4.7 10 41.7 0 4.7 19.7 0 4.7 = (57)(41.7)(19.7) + (10)(4.7)(0) + (0)(10)(4.7) [(0)(41.7)(0) + (4.7)(4.7)(57) + (19.7)(10)(10)] = 46,824.93 3,229.13 = 43,595.8 1.5 10 0 1.5 10 43,595.8I1 = 3 41.7 4.7 3 41.7 1.5 4.7 19.7 1.5 4.7 = (1.5)(41.7)(19.7) + (10)(4.7)(1.5) + (0)(3)(4.7) [(1.5)(41.7)(0) + (4.7)(4.7)(1.5) + (19.7)(3)(10)] 1302.735 624.135 678.6 = 15.6 mA I1 = 43,595.8 43,595.8
57 1.5 0 57 1.5 43,595.8I2 = 10 3 4.7 10 3 0 1.5 19.7 0 1.5 = (57)(3)(19.7) + (1.5)(4.7)(0) + (0)(10)(1.5) [(0)(3)(0) + (1.5)(4.7)(57) + (19.7)(10)(1.5)] 3368.7 697.35 2671.35 = 61.3 mA I2 = 43,595.8 43,595.8
Substituting into the third equation to get I3: 19.7I3 = 1.5 + 4.7I2 1.5 4.7(0.0613 A) = 61.5 mA I3 = 19.7
93
Chapter 9 22.
Use the loop currents from Problem 21: I47 = I1 = 15.6 mA I27 = I2 = 61.3 mA I15 = I3 = 61.5 mA I10 = I1 I2 = 15.6 mA (61.3 mA) = 76.9 mA I4.7 = I2 I3 = 61.3 mA 61.5 mA = 123 mA
23.
See Figure 9-2. The loop equations are: (10 + 4.7 + 2.2)I1 (4.7 + 2.2)I2 = 8 V (2.2 + 4.7 + 8.2 + 3.9)I2 (2.2 + 4.7)I1 = 0 V 16.9I1 6.9I2 = 8 6.9I1 + 19I2 = 0 8 6.9 0 19 (8)(19) 152 152 = 555 mA I1 = 16.9 6.9 (16.9)(19) (6.9)(6.9) 321.1 47.61 273.49 6 .9 19 16.9 8 I2 =
6 .9 0 (8)(6.9) 55.2 55.2 = 202 mA 16.9 6.9 (16.9)(19) (6.9)(6.9) 321.1 47.61 273.49 6 .9 19
VA = (I1 I2)2.2 = (555 mA 202 mA) 2.2 = (353 mA)2.2 = 776.6 mV VB = I2(3.9 ) = (202 mA)(3.9 ) = 787.8 mV VAB = VA VB = 776.6 mV 787.8 mV = 11.2 mV
Figure 9-2
94
Chapter 9 24.
See Figure 9-3. The loop equations are: (10 + 4.7 + 2.2)I1 4.7I2 2.2I3 = 8 V (4.7 + 8.2 + 10)I2 4.7I1 10I3 = 0 (2.2 + 10 + 3.9)I3 2.2I1 10I2 = 0 16.9I1 4.7I2 2.2I3 = 8 V 4.7I1 + 22.9I2 10I3 = 0 2.2I1 10I2 + 16.1I3 = 0
Figure 9-3
The characteristic determinant is: 4.7 2.2 16.9
4.7
4.7 22.9 10 4.7 2.2 10 16.1 2.2
22.9 10
16.9
= (16.9)(22.9)(16.1) + (4.7)(10)(2.2) + (2.2)(4.7)(10) [(2.2)(22.9)(2.2) + (10)(10)(16.9) + (16.1)(4.7)(4.7)] = 6024.061 2156.485 = 3867.576 16.9 8 2.2 16.9 8 3867.576I2 = 4.7 0 10 4.7 0 2.2 0 16.1 2.2 0 = (16.9)(0)(16.1) + (8)(10)(2.2) + (2.2)(4.7)(0) [(2.2)(0)(2.2) + (0)(10)(16.9) + (16.1)(4.7)(8)] I2 =
176 605.36 781.36 = 202 mA 3867.576 3867.576
16.9 4.7 8 16.9 4.7 3867.576I2 = 4.7 22.9 0 4.7 22.9 2.2 10 0 2.2 10 = (16.9)(22.9)(0) + (4.7)(0)(2.2) + (8)(4.7)(10) [(2.2)(22.9)(8) + (10)(0)(16.9) + (0)(4.7)(4.7)] I3 =
376 403.04 779.04 = 201 mA 3867.576 3867.576
IBA = I2 I3 = 202 mA 201 mA = 1 mA
95
Chapter 9 25.
See Figure 9-4. (R1 + R2 + R3)IA R2IB R3IC = 0 R2IA + (R2 + R4)IB R4IC = VS R3IA R4IB + (R3 + R4 + RL)IC = 0 5.48IA 3.3IB 1.5IC = 0 3.3IA + 4.12IB 0.82IC = 15 1.5IA 0.82IB + 4.52IC = 0 Coefficients are in k. Figure 9-4
26. Using a calculator to solve for the loop currents:
IA = 7.63 mA, IB = 10.6 mA, IC = 4.46 mA IRL = IC = 4.46 mA 27. IR3 = IA – IC = 7.63 mA – 4.46 mA = 3.17 mA
VR3 = IR3R3 = (3.17 mA)(1.5 k) = 4.76 V
Section 9-4 Node Voltage Method 28.
See Figure 9-5. The current equation at node A is: I1 I2 I3 = 0 Using Ohm’s law substitutions for the currents: 30 VA VA 40 VA =0 82 68 147 30 VA VA 40 VA 0 82 82 68 60 147 Multiply each term in the last equation by (82)(68)(147) = 819,672 to eliminate the denominators. 9996(30) 9996VA 12,054VA + 12,054 5576VA = 0 782,040 27,626VA = 0 782,040 = 28.3 V VAB = VA = 27,626
Figure 9-5
96
Chapter 9 29.
Use VAB = 28.3 V from Problem 28. 30 V VAB 30 V 28.3 V I1 = = 20.6 mA 82 82 V 40 V 28.3 V 40 V I2 = AB = 172 mA 68 68 28.3 V VAB I3 = = 193 mA 147 147
30.
See Figure 9-6. I1 I2 I3 = 0 I3 + I4 I5 = 0
Figure 9-6
Substituting into the first equation and simplifying: 1.5 VA VA VA VB =0 47 10 27 1.5 VA VA VA VB =0 47 47 10 27 27 27VA 126.9VA 47VA VB 1.5 126.9 27 47 200.9VA VB 1.5 126.9 27 47 1.58VA 0.037VB = 0.0319 Substituting into the second equation and simplifying: VA VB 3 VB VB 1.5 =0 27 4 .7 15 3 VA VB V V 1.5 B B =0 27 27 4.7 4.7 15 5 0.037VA 0.037VB 0.213VB 0.067VB + 0.738 0.037VA 0.317VA = 0.738
97
Chapter 9 31.
See Figure 9-7. Node A: I1 I2 I3 = 0 Node B: I3 I4 I5 = 0 I1 =
9 V VA R1
VA R2 V VB I3 = A R3 V 4.5 V I4 = B Figure 9-7 R4 V 1.5 V I5 = B R5 9 VA VA VA VB Node A: =0 56 27 91 9 VA VA VA VB =0 56 56 27 91 91 2457VA 5096VA 1512VA VB 9 0 137,592 91 56 I2 =
9 9065 1 =0 VA VB 56 91 137,592 0.0659VA + 0.0109VB = 0.1607 Node B:
VA VB VB 4.5 VB 15 =0 91 33 82 VA VB VB 4.5 VB 15 0 91 91 33 33 82 82 VA 2706VA 7462VA 3003VA (32)(4.5) (33)(15) =0 91 246,246 2706 VA 131,171VB 864 =0 91 246,246 2706 0.0109VA 0.0535VB = 0.3193
The characteristic determinant is: 0.0659 0.0109 = 0.0035 0.0001 = 0.0034 0.0109 0.0535
0.0034VA =
VA =
0.1607 0.0109 = 0.0086 0.0035 = 0.0051 0.3193 0.0535
0.0051 = 1.5 V 0.0034
98
Chapter 9 0.0034VB =
VB = 32.
0.0659
0.1607
0.0109 0.3193
= 0.0210 0.0018 = 0.0192
0.0192 = 5.65 V 0.0034
See Figure 9-8. Node A: I1 I2 + I3 + I4 = 0 Node B: I2 + I5 I6 = 0 Node C: I3 + I7 + I8 = 0 24 V VA 1 k VA VB I2 = 1 k V VA I3 = C 1 k VA I4 = 1 k
I1 =
24 V VB 1 k VB 18 V I6 = 1 k 10 V VC I7 = 1 k 18 V VC I8 = 1 k
I5 =
Figure 9-8
The k and V units are omitted for simplicity and the denominators are all 1. Node A: (24 VA) (VA VB) + (VC VA) VA = 0 4VA + VB + VC = 24 Node B: (VA VB) + (24 VB) +(VB 18) = 0 VA 3VB = 42 Node C: (VC VA) + (10 VC) + (18 VC) = 0 VA 3VC = 28
The characteristic determinant is: 4 1 1 1 3 0 = (4)(3)(3) (1)(3)(1) (1)(1)(3) = 36 + 3 + 3 = 30 1 0 3 24 1 1 30VA = 42 3 0 = (24)(3)(3) (28)(3)(1) (1)(42)(3) = 2166 84 126 28 0 3 = 426
VA =
426 = 14.2 V 30
99
Chapter 9 4 24 1 1 42 0 = (4)(42)(3) + (1)(28)(1) (1)(42)(1) (42)(1)(3) 30VB = 1 28 3 = 504 28 + 42 72 = 562
VB =
562 = 18.7 V 30 4
30VC =
1 24 1 3 42 = (4)(3)(28) + (1)(42)(1) (1)(3)(24) (1)(1)(28) 1 0 28 = 336 42 72 28 = 422
VC = 33.
422 = 14.1 V 30
See Figure 9-9. 4.32 V = 2.16 mA I7 = 2 k VC = +4.32 V 20 V = 15.7 V 5.25 V ( 15.7 V) 10.43 V I6 = = 522 A 20 k 20 k 5.25 V = 328 A I4 = 16 k I1 = I6 I4 = 522 A 328 A = 193 A VA = 5.25 V + (193 A)(8 k) = 5.25 V + 1.55 V = 3.70 V 3.70 V I2 = = 370 A 10 k I5 = I7 I4 I2 = 2.16 mA 328 A 370 A = 1.46 mA VB = (1.46 mA)(4 k) = 5.85 V V VB 3.70 V (5.85 V) 2.14 V I3 = A = 179 A 12 k 12 k 12 k I8 = I3 + I5 = 179 A + 1.46 mA = 1.64 mA
100
Figure 9-9
Chapter 9 Multisim Troubleshooting and Analysis 34.
No fault.
35.
No fault.
36.
VA = 0.928 V; VB = 5.190 V
37.
R4 is open.
38.
V1 = 4.939 V; V2 = 2.878 V
39.
Lower fuse is open.
40.
R3 is open.
41.
R4 is open.
101
Chapter 10 Magnetism and Electromagnetism Note: Solutions show conventional current direction.
Section 10-1 The Magnetic Field 1.
Since B =
2.
B=
3.
B=
4.
A
A
, when A increases, B (flux density) decreases.
1500 Wb = 3000 Wb/m2 = 3000 T 0.5 m 2
A There are 100 cm per meter: 1m 1 m2 100 cm 10,000 cm 2 Converting 150 cm 2 to m2: 1 m2 = 0.015 m2 A = 150 cm 2 2 10 , 000 cm 3 = BA = (2.5 10 T)(0.015 m2) = 37.5 Wb
1 T = 104 gauss
1T B = (0.6 gauss) 4 = 60 T 10 gauss 5.
104 gauss B = (100,000 T) = 1000 gauss 1T
Section 10-2 Electromagnetism 6.
The compass needle turns 180.
7.
r =
8.
Reluctance =
0 0 = 4 107 Wb/At m 750 106 Wb/At m = 597 r = 4 10-7 Wb/At m 1 0.28 m = 233,333 At/Wb -7 A (150 10 Wb/At m)(0.08 m 2 )
102
9.
Fm = NI = (50 t)(3 A) = 150 At
Section 10-3 Electromagnetic Devices 10.
The plunger is retracted when the solenoid is activated.
11.
(a) (b)
12.
When SW1 is closed, there is current through the relay coil, moving the armature from contact 1 to contact 2. This action causes current through lamp 1 to stop and current to begin through lamp 2.
13.
When there is current through the coil of a d’Arsonval meter movement, it creates a magnetic field around the coil. This reinforces the permanent field on one side of the coil and weakens it on the other, causing the coil to move because of the differential field strength.
The electromagnetic field causes the plunger to move when the solenoid is activated. The spring force returns the plunger to its inactive position.
Section 10-4 Magnetic Hysteresis 14.
Fm = 150 At F 150 At H= m = 750 At/m l 0.2 m
15.
The flux density can be changed without altering the core characteristics by changing the current or changing the number of turns.
16.
(a) (b)
(c) 17.
Fm NI (500 t)(0.25A) = 417 At/m l l 0.3 m Fm NI = reluctance l / A = ro = (250)(4 107) = 3142 107 Wb/At m A = (2 cm)(2 cm) = (0.02 m)(0.02 m) = 4 104 m2 (500 t)(0.25 A) 125 At = = 5.23 Wb 6 2.39 10 At/Wb 0.3 m 3142 107 Wb/At m 4 104 m 2 5.23 Wb = 0.13 T B= A 4 104 m 2 H=
Material A has the most retentivity.
Section 10-5 Electromagnetic Induction 18.
The induced voltage doubles when the rate of change of magnetic flux doubles.
19.
The strength of the magnetic field, the length of the conductor exposed to the field, and the velocity of the conductor relative to the field.
103
Chapter 10 20.
d Vind = N = 50(3500 103 Wb/s = 175 V dt
21.
Lenz’s law defines the polarity of the induced voltage.
22.
The magnetic field is not changing, therefore, there is no induced voltage.
Section 10-6 The DC Generator 23.
The commutator and brush assembly electrically connect the loop to the external circuit.
24.
60 rps 2 peaks/rev = 120 peaks/s
25.
See Figure 10-1.
Figure 10-1
26.
I A I L I F 12 A + 1 A = 13 A
27.
(a) PL = IL VL = (12 A)(14 V) = 168 W (b) PF = IF VL = (1 A)(14 V) = 14 W
Section 10-7 The DC Motor 28.
(a) P = 0.105Ts = (0.105)(3.0 N-m)(1200 rpm) = 378 W (b) 378 W/746 W/hp = 0.51 hp
29.
PT = Pint + PL = 12 W + 50 W = 62 W PL = 50 W Efficiency = PL/PT = 50 W/62 W = 81%
104
Chapter 11 Introduction to Alternating Current and Voltage Section 11-1 The Sinusoidal Waveform 1.
(a)
(b)
(c)
2.
(a) (b) (c)
3.
T=
4.
T=
5.
T
1 1 = 1 Hz T 1s 1 1 f= = 5 Hz T 0.2 ms 1 1 f= = 20 Hz T 50 ms
(d)
f=
(e)
(f)
1 1 = 1s f 1 Hz 1 1 T= = 16.7 ms f 60 Hz 1 1 = 2 ms T= f 500 Hz
T=
(d) (e) (f)
1 1 = 1 kHz T 1 ms 1 1 = 2 kHz f= T 500 s 1 1 f= = 100 kHz T 10 s f=
1 1 = 1 ms f 1 kHz 1 1 T= = 5 s f 200 kHz 1 1 T= = 200 ns f 5 MHz
T=
10 s = 2 s 5 cycles
1 1 = 20 s f 50 kHz 10 ms = 500 cycles 0.02 ms
1 1 0.1ms f 10kHz
Time for 100 cycles = 100(0.1 ms) = 10 ms
Section 11-2 Sinusoidal Voltage and Current Values 6.
(a) (b) (c)
Vrms = 0.707Vp = 0.707(12 V) = 8.48 V Vpp = 2Vp = 2(12 V) = 24 V Vavg = 0 V over a full cycle. Vavg = 0.637(12 V) = 7.64 over a half cycle.
105
Chapter 11 7.
(a) (b) (c)
8.
Ip = 1.414Irms = 1.414(5 mA) = 7.07 mA Iavg = 0 A over a full cycle Iavg = 0.637Ip = 0.637(7.07 mA) = 4.5 mA over a half cycle Ipp = 2Ip = 2(7.07 mA) = 14.14 mA
Vp = 25 V Vpp = 2Vp = 50 V Vrms = 0.707Vp = 17.7 V Vavg = 0.637Vp = 15.9 V
Section 11-3 Angular Measurement of a Sine Wave 9.
(a) (b) (c)
10.
(a) (b) (c)
rad 30 180 rad 45 180 rad 78 180
rad 6 rad 4 39 rad 90
57.3 rad = 22.5 8 rad 57.3 rad = 60 3 rad 57.3 rad = 90 2 rad
(d) (e) (f)
(d) (e) (f)
3 rad rad 135 4 180 10 rad rad 200 9 180 5 rad rad 300 3 180 3 57.3 rad = 108 5 rad 6 57.3 rad = 216 5 rad 57.3 (1.8 rad) = 324 rad
11.
= 45 30 = 15 A leading B
12.
With respect to 0: sine wave with a peak at 75 is shifted 15 to left. Sine wave with a peak at 100 is shifted 10 to right. Phase difference = = 100 75 = 25
13.
See Figure 11-1.
Figure 11-1
106
Chapter 11 Section 11-4 The Sine Wave Formula 14.
Vp = 1.414(20 V) = 28.28 V (a) v = Vpsin = (28.28 V)sin15 = 7.32 V (b) v = Vpsin = (28.28 V)sin 33 = 15.4 V (c) v = Vpsin = (28.28 V)sin 50 = 21.7 V (d) v = Vpsin = (28.28 V)sin 110 = 26.6 V (e) v = Vpsin = (28.28 V)sin 70 = 26.6 V (f) v = Vpsin = (28.28 V)sin 145 = 16.2 V (g) v = Vpsin = (28.28 V)sin 250 = 26.6 V (h) v = Vpsin = (28.28 V)sin 325 = 16.2 V
15.
(a) (b) (c) (d) (e) (f)
16.
Vp = 1.414Vrms = 1.414(6.37 V) = 9 V (a) = 22.5 8 v = (9 V)sin 22.5 = 3.44 V (b) = 45 4 v = (9 V)sin 45 = 6.36 V = 90 (c) 2 v = (9 V)sin 90 = 9 V 3 = 135 (d) 4 v = (9 V)sin 135 = 6.36 V
i = Ipsin = (100 mA)sin 35 = 57.4 mA i = Ipsin = (100 mA)sin 95 = 99.6 mA i = Ipsin = (100 mA)sin 190 = 17.4 mA i = Ipsin = (100 mA)sin 215 = 57.4 mA i = Ipsin = (100 mA)sin 275 = 99.6 mA i = Ipsin = (100 mA)sin 360 = 0 mA
17.
vB = (15 V)sin (30 + 30) = 13.0 V vB = (15 V)sin (30 + 45) = 14.5 V vB = (15 V)sin (30 + 90) = 13.0 V vB = (15 V)sin (30 + 180) = 7.5 V vB = (15 V)sin (30 + 200) = 11.5 V vB = (15 V)sin (30 + 300) = 7.5 V
18.
(a) (b) (c) (d) (e) (f)
(e) (f) (g)
= 180 v = (9 V)sin 180 = 0 V 3 = 270 2 v = (9 V)sin 270 = 9 V 2 = 360 v = (9 V)sin 360 = 0 V
vB = (15 V)sin ( 30) = (15 V)sin(30 30) = (15 V)sin(0) = 0 V vB = (15 V)sin ( 30) = (15 V)sin(45 30) = (15 V)sin(15) = 3.88 V vB = (15 V)sin ( 30) = (15 V)sin(90 30) = (15 V)sin(60) = 13.0 V vB = (15 V)sin ( 30) = (15 V)sin(180 30) = (15 V)sin(150) = 7.5 V vB = (15 V)sin ( 30) = (15 V)sin(200 30) = (15 V)sin(170) = 2.60 V vB = (15 V)sin ( 30) = (15 V)sin(300 30) = (15 V)sin(270) = 15 V
107
Chapter 11 19.
1 1 = 4.55 s f 2.2 kHz At t = 0.12 ms = 120 s: 120 s = 360 = 94.9 455 s 25 V Vp = = 35.4 V 0.707 v = (35.4 V)sin 94.9 = 35.3 V At t = 0.2 ms = 200 s: 200 s = 360 = 158 455 s v = (35.4 V)sin 158 = 13.3 V v = 35.4 V 13.3 V = 22.1 V
T=
Section 11-5 Introduction to Phasors 20.
See Figure 11-2.
Figure 11-2 21.
See Figure 11-3.
Figure 11-3
22.
= 2f (a) (b) (c)
60 = 9.55 Hz 2 2 360 f= = 57.3 Hz 2 2 2 f= = 0.318 Hz 2 2 f=
108
Chapter 11
23.
1256 = 200 Hz 2
(d)
f=
(a)
v = Vpsin(t /4) = (1 V)sin[2(5 kHz)(30 s) /4] = (1 V)sin(0.3 0.25) = (1 V)sin(0.05) = (1 V)(0.156) = 156 mV
(b)
v = Vpsin(t /4) = (1 V)sin[2(5 kHz)(75 s) /4] = (1 V)sin(0.75 0.25) = (1 V)sin(0.5) = (1 V)(1) = 1 V
(c)
v = Vpsin(t /4) = (1 V)sin[2(5 kHz)(125 s) /4] = (1 V)sin(1.25 0.25) = (1 V)sin() = (1 V)(0) = 0 V
2
Section 11-6 Analysis of AC Circuits 24.
(a) (b) (c) (d) (e)
Vp 10 V = 7.07 mA Irms = 0.707 0.707 R 1 k 2 Iavg = 10 mA = 6.37 mA 10 V Ip = = 10 mA 1 k Ipp = 2(10 mA) = 20 mA i = Ip = 10 mA
25.
V2(rms) = V4 V3 = 65 V 30 V = 35 V V2(p) = 1.414(35 V) = 49.5 V V2(AVG) = 0.637(49.5 V) = 31.5 V V1(rms) = Vs V4 = 120 V 65 V = 55 V V1(p) = 1.414(55 V) = 77.8 V V1(AVG) = 0.637(77.8 V) = 49.6 V
26.
Ipp =
16 V 16 V = 16 mA R1 1 k
I pp 16 mA 0.707 Irms = 0.707 = 5.66 mA 2 2 VR4 = IrmsR4 = (5.66 mA)(560 ) = 3.17 V rms Applying Kirchhoff’s voltage law: VR1 + VR2 + VR3 + VR4 = Vs 0.707(8 V) + 5 V + VR3 + 3.17 V = 0.707(30 V) VR3 = 21.21 V 5.66 V 5 V 3.17 V = 7.38 V 27.
Vp = (1.414)(10.6 V) = 15 V Vmax = 24 V + Vp = 39 V Vmin = 24 V Vp = 9 V
109
Chapter 11 28.
Vp = (1.414)(3 V) = 4.242 V VDC = Vp = 4.24 V
29.
Vmin = VDC Vp = 5 V 6 V = 1 V
30.
Vrms = 0.707Vp = 0.707(150 V) = 106.1 V V2 V 2 (106.1 V) 2 (200 V) 2 = 112.6 W + 400 W = 513 W P = Pac + PDC = rms S RL RL 100 100
Section 11-7 The Alternator (AC Generator) 31.
f = (number of pole pairs)(rps) = (1)(250 rps) = 250 Hz
32.
f = (number of pole pairs)(rps) 3600 rpm rps = = 60 rps 60 s/m f = (2 pole pairs)(60 rps) = 120 Hz f 400 Hz rps = = 200 rps pole pairs 2
33. 34.
f = (number of pole pairs)rps
#pole pairs
f 400Hz 400Hz 8 rps 3000rpm 50rps 60s/m
# poles = 2(# pole pairs) = 2 X 8 = 16
Section 11-8 The AC Motor 35.
A one-phase motor requires a starting winding or other means to produce torque for starting the motor, whereas a three-phase motor is self-starting.
36.
The field is set up by current in the stator windings. As the current reaches a peak in one winding, the other windings have less current and hence less effect on the field. The result is a rotating field.
Section 11-9 Nonsinusoidal Waveforms 37.
The approximate values determined from the graph are: tr 3.5 ms 0.5 ms = 3.0 ms tf 16 ms 13 ms = 3.0 ms tW 14.5 ms 2.5 ms = 12.0 ms Amplitude = 5 V
110
Chapter 11 38.
1 = 0.5 ms = 500 s 2 kHz 1 s t 100% = 0.2% % duty cycle = W 100% T 500 s T=
39.
Vavg = baseline + (duty cycle)(amplitude) t 1 s = 0.167 duty cycle = W T 6 s Vavg = 5 V + (0.167)(5 V) = 5.84 V
40.
(a) (b)
41.
42.
Vavg = baseline + (duty cycle)(amplitude) Vavg = 1 V + (0.25)(2.5 V) = 0.375 V
(b)
Vavg = baseline + (duty cycle)(amplitude) Vavg = 1 V + (0.67)(3 V) = 3.01 V
(a)
(a) (b)
44.
45.
1 s 100% = 25% 100% 4 s 20 ms 100% 100% = 66.7% 30 ms
(a)
(b)
43.
t % duty cycle = W T t % duty cycle = W T
1 1 = 250 kHz T 4 s 1 1 f= = 33.3 Hz T 30 ms f=
1 1 = 50 kHz T 20 s 1 1 f= = 10 Hz T 100 ms f=
area under curve (0 V 1 V 2 V 3 V 4 V 5 V 6 V)(1 ms) period 7 ms 21V ms =3V = 7 ms
Average value =
1 1 = 25 kHz (fundamental) T 40 s 3rd harmonic = 75 kHz 5th harmonic = 125 kHz 7th harmonic = 175 kHz 9th harmonic = 225 kHz 11th harmonic = 275 kHz 13th harmonic = 325 kHz f=
111
Chapter 11 46.
f=
1 1 = 25 kHz T 40 s
Section 11-10 The Oscilloscope 47.
Vp = (3 div)(0.2 V/div) = 600 mV T = (10 div)(50 ms/div) = 500 ms
48.
Vp(in) = (1 div)(5 V/div) = 5 V Tin = (2 div)(0.1 ms/div) = 200 s 1 fin = = 5 kHz 200 s Rtot = 560 + (470 (560 + 470 )) = 560 + 323 = 883 470 323 323 470 5 V = 835 mV V p (in ) Vp(out) = 1030 883 470 560 883 fout = fin = 5 kHz
49.
Vp(out) = (3 div)(0.2 V/div) = 0.6 V Tout = (10 div)(50 ms/div) = 500 ms 1 fout = = 2 Hz 500 ms Rtot = 1 k + 1 k 3.2 k = 1762 1762 1762 (3.2 V)(0.6 V) = 4.44 V (3.2 V)V p ( out ) Vp(in) = 762 762 fout = fin = 2 Hz
Multisim Troubleshooting and Analysis 50.
VR1 = 199.411 Vpp = 70.509 Vrms; VR2 = 111.685 Vpp = 39.487 Vrms
51.
VR1 = 16.717 Vpp = 5.911 Vrms; VR2 = 36.766 Vpp = 13.005 Vrms; VR3 = 14.378 Vpp = 5.084 Vrms
52.
R2 open.
55.
VMIN = 2.000 Vp; VMAX = 22.000 Vp
56.
VMIN = 4.000 Vp; VMAX = 16.000 Vp
53.
No fault.
54.
112
R1 is open.
Chapter 12 Capacitors Section 12-1 The Basic Capacitor 1.
(a)
(b) (c)
Q 50 C = 5 F V 10 V Q = CV = (0.001 F)(1 kV) = 1 C Q 2 mC V= = 10 V C 200 F C=
2.
(a) (b) (c)
(0.1 F)(106 pF/F) = 100,000 pF (0.0025 F)(106 pF/F) = 2500 pF (4.7 F)(106 pF/F) = 4,700,000 pF
3.
(a) (b) (c)
(1000 pF)(106 F/pF) = 0.001 F (3500 pF)(106 F/pF) = 0.0035 F (250 pF)(106 F/pF) = 0.00025 F
4.
(a) (b) (c)
(0.0000001 F)(106 F/F) = 0.1 F (0.0022 F)(106 F/F) = 2200 F (0.0000000015 F)(106 F/F) = 0.0015 F
5.
W=
6.
1 1 CV 2 (1 1000 F)(500 V) = 125 J 2 2
1 W = CV 2 2 2W 2(10 mJ) C= 2 = 2 F (100 V) 2 V
7.
(a) (b) (c) (d)
8.
C=
9.
C=
Air: = r0 = 1(8.85 1012 F/m) = 8.85 1012 F/m Oil: = r0 = 4.0(8.85 1012 F/m) = 35.4 1012 F/m Glass: = r0 = 7.5(8.85 1012 F/m) = 66.4 1012 F/m Teflon: = r0 = 2.0(8.85 1012 F/m) = 17.7 1012 F/m
A r (8.85 1012 F/m) (1.44 103 )(5)(8.85 1012 F/m) = 0.001 F d 6.35 105 m
A r (8.85 1012 F/m) d 2 (0.05 m )(1.0)(8.85 1012 F/m) = = 983 pF 4.5 104 m
113
Chapter 12 10.
A r 8.85 1012 d Cd (1)(8 105 ) = 3.6 106 m2 A= r 8.85 1012 (2.5)(8.85 1012 ) C=
l = A = 1.9 103 m (almost 1.2 miles on a side!) The capacitor is too large to be practical and will not fit in the Astrodome!
A r 8.85 1012 (0.09)(2.5)(8.85 1012 ) = 24.9 nF = 0.0249 F d (8.0 105 )
11.
C=
12.
T = 50 C (200 ppm/C)50 C = 10,000 ppm 1 103 C = (10 103 ppm) = 10 pF 6 1 10 C75 = 1000 pF 10 pF = 990 pF
13.
T = 25 C (500 ppm/C)25 C = 12,500 ppm (1 106 pF/F)(0.001 F) = 1000 pF 1000 (12.5 103 ppm) = +12.5 pF C = 6 1 10
Section 12-2 Types of Capacitors 14.
The plate area is increased by increasing the number of layers of plate material and dielectric.
15.
Ceramic has the highest dielectric constant (r = 1200).
16.
See Figure 12-1.
Figure 12-1 17.
Aluminum, tantalum; electrolytics are polarized, others are not.
18.
(a) (b) (c) (d)
Encapsulation Dielectric (ceramic disk) Plate (metal disk) Conductive leads
19.
(a) (b) (c) (d)
0.022 F 0.047 F 0.001 F 220 pF
114
Chapter 12 Section 12-3 Series Capacitors 20.
CT =
21.
(a)
(b)
(c)
22.
(a)
(b)
(c)
1000 pF = 200 pF 5 1
= 0.688 F 1 1 1 F 2.2 F 1 = 69.7 pF CT = 1 1 1 100 pF 560 pF 390 pF 1 = 2.64 F CT = 1 1 1 1 10 F 4.7 F 47 F 22 F
CT =
CT = 0.688 F C 0.688 F V1F = T 10 V 10 V = 6.88 V 1 F 1 F CT 0.688 F V2.2F = 10 V 10 V = 3.13 V 2.2 F 2.2 F CT = 69.7 pF CT 69.7 pF V100pF = 100 V 100 V = 69.7 V 100 pF 100 pF
CT 69.7 pF V560pF = 100 V 100 V = 12.4 V 560 pF 560 pF CT 69.7 pF V390pF = 100 V 100 V = 17.9 V 390 pF 390 pF CT = 2.64 F C 2.64 F V10F = T 30 V 30 V = 7.92 V 10 F 10 F CT 2.64 F V4.7F = 30 V 30 V = 16.9 V 4.7 F 4.7 F C 2.64 F V47F = T 30 V 30 V = 1.69 V 47 F 47 F C 2.64 F V22F = T 30 V 30 V = 3.60 V 22 F 22 F
115
Chapter 12 23.
C Vx = T VS Cx CV (1 F)(8 V) = 0.667 F CT = x x 12 V VS C Cx = T V
24.
0.667 F 12 V = 2 F VS 4V
QT = Q1 = Q2 = Q3 = Q4 = 10 C 10 C Q = 2.13 V V1 = 1 C1 4.7 F Q 10 C = 10 V V2 = 2 C2 1 F 10 C Q V3 = 3 = 4.55 V C3 2.2 F Q 10 C =1V V4 = 4 C4 10 F
Section 12-4 Parallel Capacitors 25.
(a) (b)
CT = 47 pF + 10 pF + 1000 pF = 1057 pF CT = 0.1 F + 0.01 F + 0.001 F + 0.01 F = 0.121 F
26.
(a)
Q = CV Q47pF = (47 pF)(10 V) = 470 1012 C = 470 pC Q10pF = (10 pF)(10 V) = 100 1012 C = 100 pC Q1000pF = (1000 pF)(10 V) = 10,000 1012 C = 0.01 C
(b)
Q = CV Q0.1F = (0.1 F)(5 V) = 0.5 C Q0.01F = (0.01 F)(5 V) = 0.05 C Q0.001F = (0.001 F)(5 V) = 0.005 C Q10000pF = (10000 pF)(5 V) = 0.05 C
(a)
= 2.62 F 1 1 1 10 F 10 F 2.2 F 3.3 F 1 1 1 CT = 1 1 1 1 1 1 100 pF 100 pF 1000 pF 470 pF 0.001 F 470 pF = 50 pF + 319.7 pF + 319.7 pF = 689 pF
27.
(b)
CT =
1
116
Chapter 12 (c)
1
CT =
1 1 1 1 1 F 1 F
28.
(a)
(b)
1 F = 1.6 F 1 F
1 F
CT = 2.62 F 2.62 F 5 V = 2.62 V VAB = 5 F
470 pF
See Figure 12-2(a): For this part of the circuit: 1 CT = = 319.7 pF 1 1 0.001 F 470 pF 319.7 pF 10 V = 3.20 V VAB = 0.001 F
Figure 12-2 (c)
See Figure 12-2(b). CAB = 1.5 F For this part of the circuit: 1 CT = = 0.6 F 1 1 1 F 1.5 F 0.6 F 10 V = 4 V VAB = 1.5 F
29.
(a)
(b)
CT = C1,2 + C3,4 = 0.00872 F + 0.0256 F = 0.0343 F QT = CTVT = (0.0343 F)(12 V) = 0.411 C C2 0.068 F V1 = VT 12 V = 10.47 V 0.01 F 0.068 F C1 C2 C1 0.01 F V2 = VT 12 V = 1.54 V 0.01 F 0.068 F C1 C2 C4 0.056 F V3 = VT 12 V = 6.52 V 0.047 F 0.056 F C3 C4 C3 0.047 F V4 = VT 12 V = 5.48 V 0.047 F 0.056 F C3 C4
117
Chapter 12 Section 12-5 Capacitors in DC Circuits 30.
(a) (b) (c) (d)
31.
(a) (b) (c) (d)
5 5 5 5
32.
= RC = (10 k)(0.001 F) = 10 s vC = VF(1 et/RC) = 15 V(1 e10s/10s) = 15 V(1 e1) = 9.48 V (a) (b) vC = VF(1 et/RC) = 15 V(1 e20s/10s) = 15 V(1 e2) = 13.0 V vC = VF(1 et/RC) = 15 V(1 e30s/10s) = 15 V(1 e3) = 14.3 V (c) (d) vC = VF(1 et/RC) = 15 V(1 e40s/10s) = 15 V(1 e4) = 14.7 V vC = VF(1 et/RC) = 15 V(1 e50s/10s) = 15 V(1 e5) = 14.9 V (e)
33.
= RC = (1 k)(1.5 F) = 1.5 ms vC = Viet/RC = (25 V)e1.5ms/1.5ms = (25 V)e1 = 9.20 V (a) (b) vC = Viet/RC = (25 V)e4.5ms/1.5ms = (25 V)e3 = 1.24 V (c) vC = Viet/RC = (25 V)e6ms/1.5ms = (25 V)e4 = 0.458 V (d) vC = Viet/RC = (25 V)e7.5ms/1.5ms = (25 V)e5 = 0.168 V
34.
(a) (b) (c)
vC = VF(1 et/RC) = 15 V(1 e2s/10s) = 15 V(1 e0.2) = 2.72 V vC = VF(1 et/RC) = 15 V(1 e5s/10s) = 15 V(1 e0.5) = 5.90 V vC = VF(1 et/RC) = 15 V(1 e15s/10s) = 15 V(1 e1.5) = 11.7 V
35.
(a) (b) (c)
vC = Viet/RC = (25 V)e0.5ms/1.5ms = (25 V)e0.333 = 17.9 V vC = Viet/RC = (25 V)e1ms/1.5ms = (25 V)e0.667 = 12.8 V vC = Viet/RC = (25 V)e2ms/1.5ms = (25 V)e1.333 = 6.59 V
36.
vC = VF(1 et/RC) = VF VFet/RC VFet/RC = VF vC V vC et/RC = F VF
= RC = (100 )(1 F) = 100 s = RC = (10 M)(47 pF) = 470 s = RC = (4.7 k)(0.0047 F) = 22.0 s = RC = (1.5 M)(0.01 F) = 15 ms = 5RC = 5(56 )(47 F) = 13.2 ms = 5RC = 5(3300 )(0.015 F) = 247.5 s = 5RC = 5(22 k)(100 pF) = 11 s = 5RC = 5(5.6 M)(10 pF) = 280 s
V vC ln et/RC = ln F VF V vC t = ln F RC VF v t = RC ln1 C VF
6V t = (2.2 k)(0.01 F) ln 1 = 15.2 s 12 V
118
Chapter 12 37.
v t = RC ln1 C VF
38.
v t = RC ln C Vi
39.
40.
41.
8V = (10 k)(0.001 F) ln1 = 7.62 s 15 V
3V = (1 k)(1.5 F) ln = 3.18 ms 25 V
Looking from the capacitor, the Thevenin resistance is: RTH = R3 + R1 R2 R4 = 1 k + 1 k 2.2 k 1.5 k = 1.47 k = RTHC = (1.47 k)0.0022 F) = 3.00 s v t = RC ln1 C VF t 10 s R= = 7.86 k vC 7 .2 (1000 pF)ln1 C ln1 10 VF See Figure 12-3(a).
1 = (R1 + R2)C = (57 k)(1 F) = 57 ms 2 = (R2 + R3)C = (43 k)(1 F) = 43 ms
vC = 20 V(1 e10 ms/57 ms) = 3.22 V See Figure 12-3(b). vC = (3.22 V)e5 ms/43 ms = 2.85 V
Figure 12-3
119
Chapter 12 Section 12-6 Capacitors in AC Circuits 42.
(a) (b)
(c)
1 1 = 3.39 k 2fC 2(1 kHz)(0.047 F) CT = 10 F + 15 F = 25 F 1 1 XC = = 6.37 k 2fCT 2(1 Hz)(25 F) 1 CT = = 0.5 F 1 1 1 F 1 F 1 1 XC = = 5.31 k 2fCT 2(60 Hz)(0.5 F) XC =
43.
CT for each circuit was found in Problem 27. 1 1 = 30.4 (a) XC = 2fCT 2(2 kHz)(2.62 F) 1 1 (b) XC = = 116 k 2fCT 2(2 kHz)(689 pF) 1 1 = 49.7 (c) XC = 2fCT 2(2 kHz)(1.6 F)
44.
(a)
For XC = 100 : 1 1 = 33.9 kHz f= 2X C C 2(100 )(0.047 F) For XC = 1 k: 1 1 = 3.39 kHz f= 2X C C 2(1 k)(0.047 F)
(b)
For XC = 100 : 1 1 = 63.7 Hz f= 2X C C 2(100 )(25 F) For XC = 1 k: 1 1 = 6.37 Hz f= 2X C C 2(1 k)(25 F)
(c)
For XC = 100 : 1 1 = 3.18 kHz f= 2X C C 2(100 )(0.5 F) For XC = 1 k: 1 1 f= = 318 Hz 2X C C 2(1 k)(0.5 F)
120
Chapter 12 Vrms 20 V = 200 I rms 100 mA
45.
XC =
46.
Vrms = IrmsXC 1 = 3.39 k 2(10 kHz)(0.0047 F) Vrms = (1 mA)(3.39 k) = 3.39 V
XC =
47.
48.
1 = 3.39 k 2fC Ptrue = 0 W 2 Pr = I rms X C =(1 mA)2(3.39 k) = 3.39 mVAR XC =
C5-6 = 0.006 F, C4-5-6 = 0.053 F, C3-4-5-6 = 0.012 F, C2-3-4-5-6 = 0.034 F CT = 0.008 F, XCT = 66.3 k V 10 V IC1 = s = 151 A X CT 66.3 k C 0.008 F 10 V = 8.00 V VC1 = T Vs 0.01 F C1 VC2 = Vs VC1 = 10 V 8.00 V = 2.00 V XC2 = 24.1 k V 2.00 V IC2 = C 2 = 83.0 A X C 2 24.1 k
C 0.012 F 2.00 V = 1.6 V VC3 = 3 4 5 6 VC 2 0.015 F C3 XC3 = 35.4 k V 1.6 V IC3 = C 3 = 45.2 A X C 3 35.4 k VC4 = VC2 VC3 = 2.00 V 1.6 V = 400 mV XC4 = 11.3 k V 400 mV IC4 = C 4 = 35.4 A X C 4 11.3 k C 0.006 F 400 mV = 240 mV VC5 = 5 6 VC 4 0.01 F C5 XC5 = 53.1 k V 240 mV IC5 = IC6 = C 5 = 4.52 A X C 5 53.1 k VC6 = VC4 VC5 = 400 mV 240 mV = 160 mV
121
Chapter 12 49.
VC2 = VC3 = (4 mA)XC3 = (4 mA)(750 ) = 3 V 1 1 = 141.5 kHz f= 2X C 3C3 2(750 )(0.0015 F) 1 1 XC2 = = 511.3 2fC2 2(141.5 kHz)(0.0022 F) V 3V = 5.87 mA IC2 = C 2 X C 2 511.3 IC1 = ICT = IC2 + IC3 = 5.87 mA + 4 mA = 9.87 mA VC1 = 5 V 3 V = 2 V V 2V XC1 = C1 = 203 I C1 9.87 mA 1 1 = 0.00541 F C1 = 2fX C1 2(141.5 kHz)(203 )
50.
CT(3,5,6)
1 1 = 0.0043 F 1 1 1 1 1 1 C3 C5 C6 0.015 F 0.01 F 0.015 F CT(2,3,5,6) = 0.022 F + 0.0043 F = 0.0263 F 1 1 CT = = 0.00725 F 1 1 1 1 C1 C2,3,5,6 0.01 F 0.0263 F
C VC1 = T C1 C VC2 = T C2
0.00725 F 10 V 10 V = 7.25 V 0.01 F 0.00725 F 10 V 10 V = 3.30 V 0.022 F
CT(3,5,6) VC3 = C3 CT(3,5,6) VC5 = C5
0.0043 F VC 2 3.30 V = 945 mV 0.015 F 0.0043 F VC 2 3.30 V = 1.42 V 0.01 F
VC6 = 945 mV
Section 12-7 Capacitor Applications 51.
The ripple voltage is reduced when the capacitance is increased.
52.
The reactance of the bypass capacitor should ideally be 0 in order to provide a short to ground for ac.
122
Chapter 12 Section 12-8 Switched-Capacitor Circuits 53. 54.
T 10 s = 4.55 k C 2200 pF 1 1 T= = 125 s f 8 kHz T 125 s = 1.25 M R= C 100 pF R=
Multisim Troubleshooting and Analysis 55.
VC = 3.103 V; VC2 = 6.828 V; VC3 = 2.069 V
56.
VC1 = 48.837 V; VC2 = 51.163 V; VC3 = 51.163 V; VC4 = 51.163 V
57.
IC(1kHz) = 1.383 mA; IC(500Hz) = 0.691 mA; IC(2kHz) = 2.768 mA
58.
C4 is open.
59.
C4 is shorted.
123
Chapter 13 Inductors Section 13-1 The Basic Inductor 1.
(a) (b) (c) (d)
1 H 1000 mH/H = 1000 mH 250 H 0.001 mH/H = 0.25 mH 10 H 0.001 mH/H = 0.01 mH 0.0005 H 1000 mH/H = 0.5 mH
2.
(a) (b) (c) (d)
(300 mH)(103) = 300,000 H (0.08 H)(106) = 80,000 H (5 mH)(103) = 5000 H (0.00045)(103) = 0.45 H
3.
4.
10 mA di = 50 mV vind = L 5 H dt s di v = L dt di v 50 mV = 2000 A/s dt L 25 mH
6.
di 200 mA vind = L 100 mH = 20 mV dt 1s N 2A L= l (30 mH)(0.05 m) Ll N= = 3536 turns A (1.2 10 6 )(10 10 5 m 2 )
7.
W=
1 2 1 LI (4.7 mH)(20 mA) 2 = 0.94 J 2 2
8.
L=
N 2 A ; Inductor 2 has 4 times the inductance of inductor 1. l
5.
124
Chapter 13 9.
N 2A l 1 = r0 = 200 0 2 = r0 = 150 0 L=
2 150 3 = 200 4 1 Therefore, coil 2 has 3/4 the inductance of coil 1. 3 L2 = L1 4 10.
A = r2 = (0.0035 m)2 = 38.5 106 m2 N 2A 1002 (4 106 H/m)(38.5 106 m 2 ) L= = = 138 H l 0.035 m
Section 13-3 Series and Parallel Inductors 11.
LT = 5 H + 10 H + 20 H + 40 H + 80 H = 155 H
12.
Lx = 50 mH 10 mH 22 mH = 18 mH
13.
LT = L1 + L2 + L3 = 50 mH + 500 H + 0.01 mH = 50.5 mH
14.
Position 1: LT = 330 H + 680 H = 1010 H Position 2: LT = 680 H + 800 H = 1480 H Position 3: LT = 800 H Position 4: LT = 1.5 mH + 800 H = 2300 H
15.
LT =
16.
1 = 7.14 H 1 1 1 1 75 H 50 H 25 H 15 H
L1 (12 mH) L1 12 mH ( 8 mH)L1 + (8 mH)(12 mH) = (12 mH)L1 (4 mH)L1 = 96 mH2 96 mH 2 L1 = = 24 mH 4 mH
8 mH =
125
Chapter 13 17.
(a) (b) (c)
18.
(a) (b) (c)
(10 H)(5 H) = 4.33 H 10 H 5 H 100 mH LT = = 50 mH 2 1 LT = = 57.1 H 1 1 1 100 H 200 H 400 H
LT = 1 H +
(100 mH)(50 mH) (60 mH)(40 mH) = 33.3 mH + 24 mH = 57.3 mH 150 mH 100 mH (12 mH)(6 mH) LT = = 4 mH 18 mH (2 mH)(4 mH) LT = 4 mH + = 5.33 mH 6 mH LT =
Section 13-4 Inductors in DC Circuits 19.
(a) (b) (c)
20.
21.
L 100 H = 1 s R 100 L 10 mH = = 2.13 s R 4.7 k 3H L = = 2 s R 1.5 M
=
(a)
50 H L = 4.46 s 5 = 5 5 R 56
(b)
15 mH L = 22.7 s 5 = 5 5 R 3300
(c)
100 mH L = 22.7 s 5 = 5 5 R 22 k
= (a) (b) (c) (d) (e)
L 10 mH = 10 s R 1 .0 k vL = (15 V)e10s/10s = (15 V)e1 = 5.52 V vL = (15 V)e20s/10s = (15 V)e2 = 2.03 V vL = (15 V)e30s/10s = (15 V)e3 = 747 mV vL = (15 V)e40s/10s = (15 V)e4 = 275 mV vL = (15 V)e50s/10s = (15 V)e5 = 101 mV
126
Chapter 13 22.
IF = (a) (b) (c) (d) (e)
V 15V = 15 mA R 1.0 k iL = 15 mA(1 - e10s/10s ) = 15 mA(1 - e1 )= 9.48 mA iL = 15 mA(1 - e20s/10s )= 15 mA(1 - e2 )= 13.0 mA iL = 15 mA(1 - e30s/10s )= 15 mA(1 - e3 )= 14.3 mA iL = 15 mA(1 - e40s/10s )= 15 mA(1 - e4 )= 14.7 mA iL = 15 mA(1 - e50s/10s )= 15 mA(1 - e5 )= 14.9 mA
L 75mH 9.15s R 8.2kΩ
23.
24.
The time constant is
L 75 mH = 9.15 s. For the increasing exponential, the final R 8.2 k V 10 V = 1.22 mA current is IF = S R 8.2 k
(a) (b) (c)
At 10 s, i = 1.22 mA(1 e10s/9.15s) = 0.81 mA At 20 s, i = 1.22 mA(1 e20s/9.15s) = 1.08 mA At 30 s, i = 1.22 mA(1 e30s/9.15s) = 1.17 mA
25.
vL = (15 V)et/(L/R) = (15 V)e2s/10s = (15 V)e0.2 = 12.3 V vL = (15 V)et/(L/R) = (15 V)e5s/10s = (15 V)e0.5 = 9.10 V vL = (15 V)et/(L/R) = (15 V)e15s/10s = (15 V)e1.5 = 3.35 V
26.
For the decreasing exponential, the initial current is 1.22 mA and the final current is 0. The current is solved by subtracting 50 s from the given times to account for the time when the falling square wave occurs. (a) (b) (c)
At 65 s, i = 1.22 mA(e15s/9.15s) = 0.237 mA At 75 s, i = 1.22 mA(e25s/9.15s) = 0.079 mA At 85 s, i = 1.22 mA(e35s/9.15s) = 0.027 mA
27.
VL = (15 V)et/10 s 5V et/10 s = 15 V 5 t = (10 s) ln 15 t = 11.0 s
28.
(a) (b)
29.
The polarity is positive at the top of the inductor. 24 V 24 V = 1.22 mA IF = RW 8.2 k
The time constant is found by first thevenizing the bridge. Figure 13-1 shows the Thevenin circuit.
127
Chapter 13 =
L 3.3 mH = 0.722 s R 4.57 k Figure 13-1 The current at 1.0 s is I = 0.569 mA(1 e1.0s/0.722s) = 0.426 mA The current after 5 is 0.569 mA.
30.
(a) (b)
31.
When the switch is open, the circuit appears as in Figure 13-2. RT = (R1 + R3) (R2 + R4) = 8 k 11.5 k = 4.72 k L 3.3 mH = 0.699 s R 4.72 k i = 0.569 mA(e1.0s/0.699s) = 136 A
=
Figure 13-2
Section 13-5 Inductors in AC Circuits 32.
The total inductance for each circuit was found in Problem 17. (a) XL = 2fLT = 2(5 kHz)(4.33 H) = 136 k (b) XL = 2fLT = 2(5 kHz)(50 mH) = 1.57 k (c) XL = 2fLT = 2(5 kHz)(57.1 H) = 1.79
33.
The total inductance for each circuit was found in Problem 18. (a) XL = 2fLT = 2(400 Hz)(57.3 mH) = 144 (b) XL = 2fLT = 2(400 Hz)(4 mH) = 10.1 (c) XL = 2fLT = 2(400 Hz)(5.33 mH) = 13.4
34.
L2 L3 (20 H)(40 H) 50 H = 63.3 mH L2 L3 60 H XL(T) = 2fLT = 2(2.5 kHz)(63.3 H) = 995 m XL2 = 2fL2 = 2(2.5 kHz)(20 H) = 314 m XL3 = 2fL2 = 2(2.5 kHz)(40 H) = 628 m V 10 V = 10.1 A IT = rms X LT 995 m
LT = L 1 +
X L3 628 m IL2 = IT 10.1 A = 6.73 A 314 m 628 m X L 2 X L3 X L2 314 m IL3 = IT 10.1 A = 3.37 A 314 m 628 m X L 2 X L3
35.
(a)
LT = 57.3 mH 10 V V = 20 XL = I 500 mA
128
Chapter 13
(b)
(c)
XL = 2fLT XL 20 = 55.5 Hz f= 2LT 2(57.3 mH) LT = 4 mH XL = 20 XL 20 = 796 Hz f= 2LT 2(4 mH) LT = 5.33 mH XL = 20 20 XL f= = 597 Hz 2LT 2(5.33 mH)
36.
XLT = 995 m 2 Pr = I rms X LT = (10.1 mA)2(995 m) = 101 VAR
37.
XL1 = 2(3 kHz)(5 mH) = 94.2 XL3 = 2(3 kHz)(3 mH) = 56.5 VL3 = IL3XL3 = (50 mA)(56.5 ) = 2.83 V VL1 = 10 V 2.83 V = 7.17 V V 7.17 V IL1 = L1 = 76.1 mA X L1 94.2 IL2 = IL1 IL3 = 76.1 mA 50 mA = 26.1 mA
Multisim Troubleshooting and Analysis 38.
VL1 = 1.158 V; VL2 = 3.579 V; VL3 = 5.263 V
39.
VL1 = 12.953 V; VL2 = 11.047 V; VL3 = 5.948 V VL4 = 5.099 V; VL5 = 5.099 V
40.
IL(10kHz) = 0.016 A; IL(20kHz) = 7.855 mA; IL(5kHz) = 0.032 mA
41.
L3 is open.
42.
L2 is shorted.
129
Chapter 14 Transformers Section 14-1 Mutual Inductance 1.
LM = k L p Ls 0.75 (1 H )(4 H) = 1.5 H
2.
LM = k L1L2
k=
LM L1L2
1 H = 0.25 (8 H)(2 H)
Section 14-2 The Basic Transformer 3.
n=
N s 1000 =4 Np 250
n=
N s 100 = 0.25 N p 400
4.
Ns = 2Np = 2(25) = 50 turns
5.
See Figure 14-1. N (a) Vs = s V p = 10(10 V) = 100 V rms Np N (c) Vs = s V p = 0.2(100 V) = 20 V rms Np
(b)
Figure 14-1
130
N Vs = s V p = 2(50 V) = 100 V rms Np
Chapter 14 Section 14-3 Step-Up and Step-Down Transformers N s 720 V =3 N p 240 V
6.
n=
7.
N Vs = s V p = 5(120 V) = 600 V Np
8.
N s Vs N p Vp Np 1 Vs 60 V = 6 V Vp = 10 Ns
9.
Vs N 30 V s = 0.25 V p N p 120 V
10.
Vs = (0.2)(1200 V) = 240 V
11.
N s Vs N p Vp Np 10 Vs 6 V = 60 V Vp = 1 Ns
12.
(a) (b) (c)
13.
(a) (b)
N 1 VRL = s V p 120 V = 6 V Np 20 VRL = 0 V (transformers do not couple dc) N VRL = s V p = 4(10 V) = 40 V Np VL = (0.1)Vs = (0.1)(100 V) = 10 V Vp = 20VL = 20(12 V) = 240 V
Section 14-4 Loading the Secondary 14.
Is N p I p Ns Np 1 1 I p I p 100 mA = 33.3 mA Is = 3 3 Ns VL = 3(20 V) = 60 V V 60 V RL = L = 1.8 k I s 33.3 mA
131
Chapter 14 15.
Ns = 0.5 Np
(a)
2
Np 1 RL Rreflect = 300 = 1200 0 .5 Ns 30 V Ip = = 25 mA 1200 2
Np I p = 2(25 mA) = 50 mA Is = Ns N Vs = s V p = 0.5(30 V) = 15 V Np PL = VsIs = (15 V)(50 mA) = 750 mW
(b) (c) (d)
Section 14-5 Reflected Load 16.
17.
Np Rp = Ns
2
1 RL 680 = 27.2 5 2
Rp = 300 , RL = 1 k Ns RL 1 k = 1.83 Np Rp 300
Section 14-6 Impedance Matching 2
18.
Np RL Rp = Ns 2
Rp Np N R L s Np Ns n=
Rp RL
16 4 =2 4
Ns 1 = 0.5 Np 2
132
Chapter 14 2
19.
Np RL = 16 Rp = Ns 2
N R 4 n = s L Np R p 16 4 N 0.25 = 0.5 n= s 16 Np 2
Ip =
25 V = 781 mA 16 16
1 1 Is = I p 781 mA = 1562 mA 0 .5 n Pspeaker = I s2 RL = (1562 mA)24 = 9.76 W 20.
Position 1: RL = 560 + 220 + 1 k = 1780 Ns RL 1780 = 13.34 Np Rp 10
Ns = Ns1 + Ns2 + Ns3 = 13.34Np = 13.34(1000) = 13,340 turns (total secondary turns) Position 2: RL = 220 + 1 k = 1220 Ns RL 1220 = 11.05 Np Rp 10 Ns2 + Ns3 = 11.05Np = 11.05(1000) = 11,050 turns Position 3: RL = 1 k Ns RL 1000 = 10 Thus, Ns2 = 11,050 10,000 = 1,050 turns Np Rp 10 Ns3 = 10Np = 10(1000) = 10,000 turns
Ns1 = 13,340 11,050 = 2,290 turns
Section 14-7 Transformer Ratings and Characteristics 21.
PL = Pp Plost = 100 W 5.5 W = 94.5 W
22.
P % efficiency = out Pin
23.
Coefficient of coupling = 1 0.02 = 0.98
94.5 W 100% 100% = 94.5 % 100 W
133
Chapter 14 24.
(a) (b)
(c)
Pa 1 kVA = 1.67 A Vs 600 V Vs 600 V RL(max) = = 359 I L (max) 1.67 A
IL(max) =
Vs = 359 IL 1 Cmax = = 7.39 F 2(60 Hz)(359 )
XC =
25.
kVA = (2.5 kV)(10 A) = 25 kVA
26.
(a)
(b) (c)
Vp = 2400 V N s Vs 120 V = 0.05 N p V p 2400 V Pa 5 kVA = 41.7 A Vs 120 V 5 kVa Ip = = 2.08 A 2400 V Is =
Section 14-8 Tapped and Multiple-Winding Transformers 27.
50 120 V = 12.0 V 500 100 V2 = 120 V = 24.0 V 500 100 V3 = 120 V = 24.0 V 500
V1 =
V4 = V2 + V3 = 48.0 V 28.
N s1 Vs 24 V =2 N p V p 12 V Ns2 6V = 0.5 N p 12 V N s3 3V = 0.25 N p 12 V
29.
N 200 Vs = s V p 120 V = 48 V Np 500 N 250 Vs = s V p 5 V = 25 V Np 50
134
Chapter 14 30.
(a) (b)
31.
(a)
See Figure 14-2. 100 100 T: Vs = 240 V = 12 V 2000 200 200 T: Vs = 240 V = 24 V 2000 500 500 T: Vs = 240 V = 60 V 2000 1000 1000 T: Vs = 240 V = 120 V 2000
Figure 14-2
Ns = 400 T + 300 T = 700 turns N 700 VRL = s V p 60 V = 35 V N 1200 p V 35 V = 2.92 A IRL = RL RL 12 300 VC = 60 V = 15 V 1200 V 15 V = 1.5 A IC = C X C 10
(b)
1 1 1 1 1 2 2 (2.94)(12 ) (16)(10 ) Rp N p Np RL X CL N 700 N 300 1 1 = = 28.3 mS + 6.25 mS = 29.0 mS 35.3 160 1 = 34.5 Rp = 29.0 mS
Section 14-9 Troubleshooting 32.
Open primary winding. Replace the transformer.
33.
If the primary shorts, excessive current is drawn which potentially can burn out the source and/or the transformer unless the primary is fused.
34.
Some, but not all, of the secondary windings are shorted.
Multisim Troubleshooting and Analysis 35.
Turns ratio = 0.5
36.
Secondary winding is open.
37.
R2 is open.
135
Chapter 15 RC Circuits Section 15-1 The Complex Number System 1.
A complex number indicates both magnitude and angle of quantity.
2.
See Figure 15-1.
Figure 15-1 3.
See Figure 15-2.
Figure 15-2
136
Chapter 15 4.
(a) (b) (c)
3, j5 +7, j1 +10, +j10
5.
(a) (b) (c)
5, +j3 and +5, j3 1, j7 and 1, +j7 10, +j10 and +10, j10
6.
(a) (b) (c)
3 + j5 2 + j1.5 10 j14
7.
C = 102 152 = 18.0
8.
(a) (b) (c) (d)
40 402 402 tan 1 = 56.645 40 200 50 j200 = 502 2002 tan 1 = 20676 50 20 35 j20 = 352 202 tan 1 = 40.329.7 35 45 98 + j45 = 982 452 tan 1 = 10824.7 98
40 j40 =
9.
(a) (b) (c) (d)
100050 = 643 j766 15160 = 14.1 + j5.13 25135 = 17.7 j17.7 3180 = 3 + j0
10.
(a) (b) (c)
10120 = 10240 3285 = 32275 5310 = 550
11.
(a) (b) (c) (d)
40 j40 is in the fourth quadrant. 50 j200 is in the fourth quadrant. 35 j20 is in the fourth quadrant. 98 + j45 is in the first quadrant.
12.
(a) (b) (c)
10120 is in the second quadrant. 3285 is in the first quadrant. 5310 is in the fourth quadrant.
13.
(a) (b) (c) (d)
12(180 65) = 12115 20(180 + 50) = 20230 100(360 170) = 100190 50(360 200) = 50160
137
Chapter 15 14.
15.
16.
17.
(a)
(9 + j3) + (5 + j8) = 14 + j11
(b)
(3.5 j4) + (2.2 + j6) = 5.7 + j2
(c)
(18 + j23) + (30 j15) = 12 + j8
(d)
1245 = 8.49 + j8.49 2032 = 17.0 + j10.6 (8.49 + j8.49) + 17.0 + j10.6) = 25.5 + j19.1
(e)
3.875 = 0.984 + j3.67 (0.984 + j3.67) + (1 + j1.8) = 1.98 + j5.47
(f)
6030 = 52 j30 (52 j30) + (50 j39) = 102 j69
(a)
(2.5 + j1.2) (1.4 + j0.5) = 1.1 + j0.7
(b)
(45 j23) (36 + j12) = 81 j35
(c)
(8 j4) 325 = (8 j4) (2.72 + j1.27) = 5.28 j5.27
(d)
48135 3360 = (33.9 + j33.9) (16.5 j28.6) = 50.4 + j62.5
(a)
(4.548)(3.290) = 14.4138
(b)
(120220)(95200) = 11,40020
(c)
4 j3 = 536.9 (3150)(536.9) = 15113
(d)
67 + j84 = 107.551.4 (107.551.4)(10240) = 10,96591.4
(e)
15 j10 = 1833.7 25 j30 = 39.1129.8 (1833.7)(39.1129.8) = 704164
(f)
0.8 + j0.5 = 0.9432 1.2 j1.5 = 1.9251.3 (0.9432)(1.9251.3) = 1.8119.3
(a) (b) (c) (d)
850 = 3.2(50 39) = 3.211 2.539 63 91 = 7(91 10) = 7101 910 2830 2830 = 1.52(30 (40.6)) = 1.5270.6 14 j12 18.4 40.6 40 j30 50 36.9 = 2.79(36.9 26.6) = 2.7963.5 16 j8 17.926.6
138
Chapter 15 18.
(a)
(b)
(c)
(d)
2.565 1.8 23 (1.06 j2.27) (1.66 j0.70) = 1.237 1.237 0.6 j2.97 3.03101.4 = 2.5364.4 = 1.237 1.237 (10015)(85 j150) (10015)(172.460.46) = 335106 25 j45 51.561.0 (25090 17575)(50 j100) ( j250 45.29 j169.04)(111.8 63.43) = (125 j90)(3550) (154.0435.75)(3550) (421.4883.83)(111.8 63.43) = = 8.7465.4 (5391.0585.75)
(1.5) 2 (3.8) 8 4 4 4 j j 7.77 j j = 7.77 + 2 + j2 = 9.77 + j2 = 9.9711.6 1.1 4 2 2 2
Part 1: Series Circuits Section 15-2 Sinusoidal Response of Series RC Circuits 19.
fVR = 8 kHz, fVC = 8 kHz
20.
The current is sinusoidal because the voltage is sinusoidal.
Section 15-3 Impedance of Series RC Circuits 21.
(a) (b)
Z = R jXC = 270 j100 = 28820.3 Z = R jXC = 680 j1000 = 1.2155.8 k
22.
(a)
RT = R1 + R2 = 100 k + 47 k = 147 k 1 1 CT = = 0.00688 F 1 1 1 1 0.01 F 0.022 F C1 C2 1 1 = 231 k XCT = 2fCT 2(100 Hz)(0.00688 F) Z = RT jXCT = 147 k j231 k = 27357.5 k Z = 273 , = 57.5
(b)
CT = C1 + C2 = 470 pF + 470 pF = 940 pF 1 1 XCT = = 8.47 k 2fCT 2(20 kHz)(940 pF) Z = R jXCT = 10 k j8.47 k = 13.140.3 k Z = 13.1 k, = 40.3
139
Chapter 15 (c)
23.
(a)
(b)
(c)
(d)
24.
(a)
RT = R1 + R2 R3 = 680 + 720 = 1400 1 1 1 = 762 pF CT = 1 1 C1 C2 C3 1000 pF 0.0032 F 1 1 = 2089 XCT = 2fCT 2(100 kHz)(762 pF) Z = RT jXCT = 1400 j2089 = 2.5256.2 k Z = 2.52 k, = 56.2 1 1 = 723 k 2fC 2(100 Hz)(0.0022 F) Z = 56 k j723 k
XC =
1 1 = 145 k 2fC 2(500 Hz)(0.0022 F) Z = 56 k j145 k
XC =
1 1 = 72.3 k 2fC 2(1 kHz)(0.0022 F) Z = 56 k j72.3 k
XC =
1 1 = 28.9 k 2fC 2(2.5 kHz)(0.0022 F) Z = 56 k j28.9 k
XC =
1 1 = 339 k 2fC 2(100 Hz)(0.0047 F) Z = 56 k j339 k XC =
1 1 = 67.7 k 2fC 2(500 Hz)(0.0047 F) Z = 56 k j67.7 k
(b)
XC =
(c)
XC =
(d)
XC =
1 1 = 33.9 k 2fC 2(1 kHz)(0.0047 F) Z = 56 k j33.9 k 1 1 = 13.5 k 2fC 2(2.5 kHz)(0.0047 F) Z = 56 k j13.5 k
140
Chapter 15 25.
(a)
R = 33 , XC = 50
(b)
Z = 30025 = 272 j127 R = 272 , XC = 127
(c)
Z = 1.867.2 k = 698 j1.66 k R = 698 , XC = 1.66 k
(d)
Z = 78945 = 558 j558 R = 558 , XC = 558
Section 15-4 Analysis of Series RC Circuits 26.
(a)
From Problem 21(a): Z = 28820.3 100 V = 34.720.3 mA I= 288 20.3
From Problem 21(b): Z = 1.2155.8 k 50 V I= = 4.1355.8 mA 1.21 55.8 k Start with the current in polar form from Problem 26:
(b)
27.
(a)
I=
100 V = 34.720.3 mA = (34.7 mA)cos 20.3o + j(34.7 mA)sin 20.3o 288 20.3
= 32.5 mA + j12.0 mA
(b)
I=
50 V = 4.1355.8 mA = (4.13 mA)cos 55.8o + j(4.13 mA)sin 55.8o 1.21 55.8 k
= 2.32 mA + j3.42mA 28.
(a)
From Problem 22(a): Z = 147 k j231 k = 27357.5 500 V IT = = 18357.5 A 273 57.5
(b)
From Problem 22(b): Z = 10 k j8.47 k = 13.140.3 k 80 V IT = = 61140.3 A 13.1 40.3 k
(c)
From Problem 22(c): Z = 1400 j2089 = 2.5256.2 k 520 V IT = = 1.9876.2 mA 2.52 56.2 k
141
Chapter 15 29.
Start with the current in polar form from Problem 28 (a)
IT =
500 V = 18357.5 A = (183 A)cos 57.5o + j(183 A)sin 57.5o 273 57.5
= 98.3 A + j154 A
(b)
IT =
80 V = 61140.3 A = (611 A)cos 40.3o + j(611 A)sin 40.3o 13.1 40.3 k
= 466 A + j395 A
(c)
IT =
520 V = 1.9876.2 mA = (1.98 mA)cos 76.2o + j(1.98 mA)sin 76.2o 2.52 56.2 k
= 0.472 mA + j1.92 mA
30.
31.
32.
Using the results of Problem 22: X 231 k = tan 1 C tan 1 (a) = 57.5 R 147 k X 8.47 k (b) = tan 1 C tan 1 = 40.3 R 10 k X 2089 (c) = tan 1 C tan 1 = 56.2 R 1400 1 14.5 k 2fC X 14.5 k = tan 1 C tan 1 = 14.5 R 56 k See Figure 15-3. 1 = 0.069 F CT = 1 1 0.1 F 0.22 F 1 XC = = 154 2(15 kHz)(0.069 F) ZT = 50 j154 = 16272.0 20 V V IT = s = 12.372.0 mA Z T 162 72.0 1 = 106 XC1 = 2(15 kHz)(0.1 F) 1 XC2 = = 48.2 2(15 kHz)(0.22 F) VC1 = ITXC1 = (12.372.0 mA)(10690 ) = 1.3018.0 V VC2= ITXC2= (12.372.0 mA)(48.290 ) = 59318.0 mV VR1 = VR2 = ITRT = (12.372.0 mA)(500 ) = 61572.0 mV XC =
142
Chapter 15
Figure 15-3 33.
34.
(a)
1 = 79.6 2(20 Hz)(100 F) Z = 56 j79.6 = 97.354.9 XC =
100 = 10354.9 mA 97.3 54.9
(b)
IT =
(c)
560 100 V = 5.7654.9 V VR = 97.3 54.9
(d)
79.6 90 100 V = 8.1835.1 V VC = 97.3 54.9
Vs 10 V = 1 k I 10 mA 1 XC = = 589.5 2(10 kHz)(0.027 F)
Z=
R 2 X C2 = 1 k R2 + (589.5 )2 = (1 k)2 R=
(1 k) 2 (589.5 ) 2 = 808 589.5 = 36.1 808
= tan1
35.
100 V 20 5A Ptrue = I2RT P 400 W RT = true = 16 2 I (5 A)2 RX = RT R1 = 16 4 = 12
ZT =
143
Chapter 15 ZT2 RT2 X C2 XC = C=
36.
(a)
ZT2 RT2 (20 ) 2 (16 ) 2 144 = 12 1 = 13.3 F 2(1 kHz)(12 )
XC =
1 = 4.08 M 2(1 Hz)(0.039 F)
XC 1 4.08 M 90 tan = 0.055 R 3.9 k
= 90 + tan1
(b)
XC =
1 = 40.8 k 2(100 Hz)(0.039 F)
XC 1 40.8 k 90 tan = 5.46 R 3.9 k
= 90 + tan1
(c)
XC =
1 = 4.08 k 2(1 kHz)(0.039 F)
XC R
= 90 + tan1
(d)
XC =
1 4.08 k 90 tan = 43.7 3.9 k
1 = 408 2(10 kHz)(0.039 F) XC 1 408 = 84.0 90 tan 3.9 k R
= 90 + tan1
37.
X Use the formula, Vout = C ZT Frequency (kHz) XC 0 1 4.08 k 2 2.04 k 3 1.36 k 4 1.02 k 5 816 6 680 7 583 8 510 9 453 10 408
1 V. See Figure 15-4. ZT 5.64 k 4.40 k 4.13 k 4.03 k 3.98 k 3.96 k 3.94 k 3.93 k 3.93 k 3.92 k
Vout 1V 723 mV 464 mV 329 mV 253 mV 205 mV 172 mV 148 mV 130 mV 115 mV 104 mV
144
Figure 15-4
Chapter 15 38.
(a)
XC =
1 1 = 15.9 k 2fC 2(1 Hz)(10 F)
XC 1 15.9 k = 90.0 tan 10 R 1 XC = = 159 2(100 Hz)(10 F)
= tan1 (b)
XC 1 159 = 86.4 tan 10 R 1 XC = = 15.9 2(1 kHz)(10 F)
= tan1 (c)
XC 1 15.9 = 57.9 tan 10 R 1 XC = = 1.59 2(10 kHz)(10 F)
= tan1
(d)
XC 1 1.59 = 9.04 tan 10 R
= tan1
39.
40.
R Use the formula, Vout = ZT Frequency XC (kHz) 0 1 15.9 2 7.96 3 5.31 4 3.98 5 3.18 6 2.65 7 2.27 8 1.99 9 1.77 10 1.59
1 V. See Figure 15-5. ZT Vout 18.8 12.8 11.3 10.8 10.5 10.4 10.3 10.2 10.2 10.1
0V 5.32 V 7.82 V 8.83 V 9.29 V 9.53 V 9.66 V 9.76 V 9.80 V 9.84 V 9.87 V
Figure 15-5
For Figure 15-91 in the text (See Figure 15-6): 1 XC = = 816 2(5 kHz)(0.039 F) Z = 3.9 k j816 = 398411.8 10 V V I= s = 25111.8 A Z 3984 11.8 VR = IR = (25111.8 A)(3.90 k) = 97911.8 mV VC = IXC = (25111.8 A)(81690 ) = 20578.2 mV
145
Chapter 15
Figure 15-6 41.
For Figure 15-92 in the text (See Figure 15-7) 1 = 15.9 XC = 2(1 kHz)(10 F) Z = 10 j15.9 = 18.857.8 V 100 V I= s = 53257.8 mA Z 18.8 57.8 VR = IR = (53257.8 mA)(100 ) = 5.3257.8 V VC = IXC = (53257.8 mA)(15.990 ) = 8.4632.2 V
Figure 15-7
Part 2: Parallel Circuits Section 15-5 Impedance and Admittance of Parallel RC Circuits ( R0)( X C 90) (1.20 k)(2 90 k) (1.20 k)(2 90 k) 1.2 k j2 k 2.33 59 k R jX C = 1.0331 k
42.
Z=
43.
BC = 2fC = 2f(C1 + C2) = 2(2 kHz)(0.32 F) = 4.02 mS
146
Chapter 15 1 1 = 0.676 mS R1 R2 R3 1480 Y = G + jBC = 0.676 mS + j4.02 mS = 4.0880.5 mS 1 1 Z= = 24580.5 Y 4.0880.5 mS Z = 245 , = 80.5
G=
44.
(a)
1 = 332 2(1.5 kHz)(0.32 F) (14800 )(332 90 ) (14800 )(332 90 ) Z= 324 77.4 1480 j332 1517 12.6 Z = 324 , = 77.4 XC =
1 = 166 2(3 kHz)(0.32 F) (14800 )(166 90 ) (14800 )(166 90 ) Z= 165 83.6 1480 j166 1489 6.40 Z = 165 , = 83.6
(b)
XC =
(c)
XC =
(d)
XC =
1 = 99.5 2(5 kHz)(0.32 F) (14800 )(99.5 90 ) (14800 )(99.5 90 ) 99.3 86.2 Z= 1480 j99.5 1483 3.85 Z = 99.3 , = 86.2 1 = 49.7 2(10 kHz)(0.32 F) (14800 )(49.7 90 ) (14800 )(49.7 90 ) 49.7 88.1 Z= 1480 j49.7 1481 1.92 Z = 49.7 , = 88.1
Section 15-6 Analysis of Parallel RC Circuits 45.
(680)(90 90 ) = 54.337.1 68 j90 VC = VR = Vs = 10 0 V
ZT =
100 V = 18437.1 mA 54.3 37.1 100 V IR = = 1470 mA 680 IT =
147
Chapter 15 IC =
46.
47.
100 V = 11190 mA 90 90
1 = 67.7 2(50 kHz)(0.047 F) 1 XC2 = = 145 2(50 kHz)(0.022 F) V 80 V IC1 = s = 11890 mA XC1 67.7 90 80 V V IC2 = s = 55.290 mA XC2 145 90 V 80 V IR1 = s = 36.40 mA R 1 2200 80 V V IR2 = s = 44.40 mA R 2 1800 IT = IR1 + IR2 + IC1 + IC2 = 36.4 mA + 44.4 mA + j118 mA + j55.2 mA = 80.8 mA + j173.2 mA = 19165.0 mA IT = 191 mA, = 65.0
XC1 =
(a)
XC = XC1 XC2 = 21 15 = 8.75 1 1 BC = = 114 mS X C 8.75 1 1 G= = 100 mS R 10 YT = 100 mS + j114 mS = 15248.8 mS 1 1 ZT = = 6.5948.8 YT 15248.8 mS
(b)
IR =
(c)
ICT =
(d)
IT =
(e)
= 48.8
1000 mV = 100 mA 100 1000 mV = 11.490 mA 8.75 90 1000 mV = 15.248.8 mA 6.59 48.8
148
Chapter 15 48.
(a)
CT = C1 + C2 = 0.047 F + 0.022 F = 0.069 F 1 XCT = = 4613 2(500 Hz)(0.069 F) (5.60 k)(4613 90 ) = 3.5650.5 k Z= 7255 39.5
(b)
IR =
(c)
ICT =
1000 mV Vs = 21.790 A XCT 4613 90
(d)
IT =
1000 mV Vs = 28.150.5 A Z 3560 50.5
(e)
= 50.5
Vs 1000 mV = 17.90 A R 5.60 k
49.
RT = 22 k, CT = 32 pF 1 XCT = = 49.7 k 2(100 kHz)(32 pF) (220 k)(49.7 90 k) = 20.123.9 k = 18.4 k j8.14 k Z= 54.4 66.1 k Req = 18.4 k, XCeq = 8.14 k 1 Ceq = = 196 pF 2 (100 kHz)(8.14 k)
50.
XC =
1 = 15.9 k 2(1 kHz)(0.01 F) X
= tan 1 C RT XC = tan RT 15.9 k XC = 27.6 k RT = tan tan 30 R1R2 RT = R1 R2 RT(R1 + R2) = R1R2 R1RT + R2RT = R1R2 R1(RT R2) = R2RT (47 k)(27.6 k) R2 RT R1 = = 66.7 k R2 RT 19.4 k
149
Chapter 15
Part 3: Series-Parallel Circuits Section 15-7 Analysis of Series-Parallel RC Circuits 51.
See Figure 15-8. 1 1 XC1 = = 106 2fC1 2(15 kHz)(0.1 F) 1 1 XC2 = = 226 2fC2 2(15 kHz)(0.047 F) 1 1 XC3 = = 48.2 2fC3 2(15 kHz)(0.22 F) ZC2R1 = R1 jXC2 = 470 j226 = 52225.7 ZC3 = jXC3 = 48.290 ZR2R3 = R2 + R3 = 330 + 180 = 5100
Combining the three parallel branches:
1 1 1 1 1 1 1 ZC2R1 ZC3 Z R2R3 522 25.7 48.2 90 5100 1 1 = 1.9225.7 mS 20.790 mS 1.960 mS 1.73 mS j0.833 mS j20.7 mS 1.96 mS 1 1 = 45.980.3 = 7.73 j45.2 = 3.69 mS j21.5 mS 21.880.3 mS
ZA =
1
ZT = XC1 + ZA = j106 + 7.73 j45.2 = 7.73 j151 = 15187.1 X 106 90 VC1 = C1 120 V 120 V = 8.422.9 V 151 87.1 ZT Z 45.9 80.3 VZA = A 120 V 120 V = 3.656.8 V 151 87.1 ZT X 226 90 VC2 = C2 VZA 3.656.8 V = 1.5857.5 V 522 25.7 ZC2R1 R1 4700 VZA VR1 = 3.656.8 V = 3.2932.5 V 522 25.7 ZC2R1 VC3 = VZA = 3.656.8 V R2 3300 VZA VR2 = 3.656.8 V = 2.366.8 V Z 5100 R2R3 R3 1800 VZA VR3 = 5100 Z R2R3
3.656.8 V = 1.296.8 V
150
Chapter 15
57.5
Figure 15-8
52.
From Problem 51: ZT = 7.73 j151 The j term is larger, therefore the circuit is predominantly capacitive.
53.
See Figure 15-9. Using the results of Problem 51: 120 V 120 V IT = = 79.587.1 mA ZT 151 87.1
VZA 3.656.8 V = 6.9932.5 mA Z C2R1 522 25.7 V 3.656.8 V IC3 = ZA = 75.796.8 mA ZC3 48.2 90 IC2R1 =
IR2R3 =
VZA 3.656.8 V = 7.166.8 mA Z R2R3 5100
Figure 15-9
151
2.9
Chapter 15 54.
RT = R1 + R2 R3 = 47 + 42.9 = 89.9 1 XC = = 339 2(1 kHz)(0.47 F) ZT = 89.9 j339 = 35175.1 150 V V (a) IT = s = 42.775.1 mA Z T 351 75.1 (b) = 75.1 R 470 V 150 V = 2.0175.1 V VR1 = 1 Vs (c) 351 75.1 ZT
(d) (e) (f)
55.
R R 42.90 V 150 V = 1.8375.1 V VR2 = 2 3 Vs 351 75 . 1 Z T VR3 = VR2 = 1.8375.1 V X 339 90 150 V = 14.514.9 V VC = C Vs 351 75.1 ZT
For I = 0 A, VA = VB and VR1 = VR2 1 XC1 = = 3.39 k 2(1 kHz)(0.047 F) VR1 = VR2 2.2 k 1 k V (2.2 k) 2 (3.39 k) 2 s (1 k) 2 X 2 C2 Cancelling the Vs terms and solving for XC2: 2.2 k 1 k (2.2 k) 2 (3.39 k) 2 2 2 (1 k) X C 2
V s
(1 k) 2 (2.2 k) 2 (3.39 k) 2 2.2 k Squaring both sides to eliminate the radicals and solving for XC2: (1 k) 2 (2.2 k) 2 (3.39 k) 2 (1 k) 2 X C2 2 ( 2 .2 k ) 2 (1 k) 2 X C2 2
XC2 = C2 =
(1 k) 2 (2.2 k) 2 (3.39 k) 2 (1 k) 2 = 1541 2 (2.2 k) 1 = 0.103 F 2(1 kHz)(1541 )
152
Chapter 15 56.
1 = 2.89 k 2(2.5 kHz)(0.022 F) ZC3R6 = R6 jXC3 = 820 j2.89 k = 374.2 k 1 XC2 = = 1.35 k 2(2.5 kHz)(0.047 F) XC2 R 4 (1.35 90 k)(9100 ) ZC = ZC2R4 = = 75434 = 625 j422 R4 jX C 2 1.63 56 k ZR5C2R4 = R5 + ZC = 1 k + 625 j422 = 1625 j422 = 1.6814.6 k ZB = (R5 + ZC) ZC3R6 = 1.6814.6 k 374.2 k (1.68 14.6 k)(3 74.2 k) = = 1.2632.9 k = 1.06 k j684 4 55.9 k ZR3ZB = R3 + ZB = 680 + 1.06 k j684 = 1.74 k j684 = 1.8721.5 k ZA = R2 ZR3ZB = 2200 1.8721.5 k (2200 )(1.87 21.5 k) = 1982.3 = 197.8 j7.95 = 2.08 19.2 k 1 XC1 = = 4.24 k 2(2.5 kHz)(0.015 F) ZT = R1 jXC1 + ZA = (1 k j4.24 k) + (197.8 j7.95 ) = 1197.8 j4248 = 4.4174.3 k
XC3 =
Z 198 2.3 VA = A Vs 100 V = 44972.0 mV 4.41 74.3 k ZT ZB 1.26 32.9 k VA VB = 44972.0 mV = 30360.6 mV 1.87 21.5 k Z R3ZB ZC 754 34 VB VC = 30360.6 mV = 13641.2 mV 1.68 14.6 k Z R5C2R4 R6 820 0 VB VD = 30360.6 mV = 83135 mV 3 74.2 k ZC3R6 57.
Use the voltages found in Problem 56: V 83135 mV IC3 = IR6 = D = 101135 A = 71.4 A + j71.4 A R6 8200 V 13641.2 mV IR4 = C = 14941.2 A = 112 A + j98 A R4 9100 V 13641.2 mV IC2 = C = 101131 A = 66 A + j76 A XC2 1.350 k IR5 = IR4 + IC2 = (112 A + j98 A) + (66 A + j76 A) = 46 A + j174 A = 18075.1 A IR3 = IR5 + IC3 = (46 mA + j174 A) + (71.4 A + j71.4 A)
153
Chapter 15 = 25.4 A + j245 A = 24684.3 A V 44972.0 mV IR2 = A = 2.0472.0 mA = 0.630 mA + j1.94 mA R2 2200 IR1 = IC1 = IR2 + IR3 = (0.630 mA + j194 mA) + (25.4 A + j245 A) = 605 A + j2185 A = 2.2774.5 mA 58.
See Figure 15-10.
Figure 15-10
154
Chapter 15
Part 4: Special Topics Section 15-8 Power in RC Circuits 2 Ptrue Pr2 (2 W ) 2 (3.5 VAR ) 2 = 4.03 VA
59.
Pa =
60.
From Problem 33: IT = 103 mA, XC = 79.6 Ptrue = IT2 R = (103 mA)2(56 ) = 594 mW
Pr = IT2 X C = (103 mA)2(79.6 ) = 845 mVAR 61.
Using the results from Problem 49: Req = 18.4 k XCeq = 8.14 k Zeq = Req jXCeq = 18.4 k j8.14 k = 20.123.9 k = 23.9 PF = cos = cos (23.9) = 0.914
62.
From Problem 54: IT = 42.7 mA, RT = 89.9 , XC = 339 , ZT = 35175.1 Ptrue = IT2 RT = (42.7 mA)2(89.9 ) = 164 mW
Pr = IT2 X C = (42.7 mA)2(339 ) = 618 mVAR Pa = IT2 ZT = (42.7 mA)2(351 ) = 640 mVA PF = cos(75.1) = 0.257 63.
240 V = 4.8 A 50 240 V ILB = = 3.33 A 72
(a)
ILA =
(b)
PFA = cos = 0.85; = 31.8 PFB = cos = 0.95; = 18.2 XCA = (50 )sin(31.8) = 26.3 XCB = (72 )sin(18.2) = 22.5 2 PrA = I LA X CA = (4.8 A)226.3 = 606 VAR 2 PrB = I LB X CB = (3.33 A)222.5 = 250 VAR
(c)
RA = (50 )cos(31.8) = 42.5 RB = (72 )cos(18.2) = 68.4 2 PtrueA = I LA RA = (4.8 A)242.5 = 979 W 2 PtrueB = I LB RB = (3.33 A)268.4 = 759 W
155
Chapter 15 (d)
(e)
PaA =
(979 W)2 (606 VAR)2 = 1151 VA
PaB =
(759 W) 2 (250 VAR)2 = 799 VA
Load A
Section 15-9 Basic Applications 1 1 = = 9278 Hz 2 6(10 k)(0.0022 F) 2 6RC
64.
fr =
65.
Vout1 Vin1
R R X C2 2
= 0.707
R = 0.707 R 2 X C2 R = 1.414R 0.707 R 2 X C2 = (1.414)2R2 R 2 X C2
X C2 = 2R2 R2 = R2(2 1) = R2 XC = R 1 =R 2fC 1 1 C= = 0.0796 F 2fR 2(20 Hz)(100 k) 66.
XC =
1 = 1.13 k 2(3 kHz)(0.047 F)
Rin ( B ) 10 k V Vin(B) = ( ) out A 2 2 2 Rin ( B ) X C 2 10 k) (1.13 k) Signal loss = 50 mV 49.7 mV = 300 V
50 mV = 49.7 mV
Section 15-10 Troubleshooting 67.
After removing C, the circuit is reduced to Thevenin’s equivalent: (4.7 k)(5 k) = 2.42 k Rth = 9.7 k 5 k 10 V = 5.15 V Vth = 9 .7 k Assuming no leakage in the capacitor: 1 XC = = 1592 2(10 Hz)(10 F)
156
Chapter 15 XC 1592 90 100 V 100 V = 3.2171.3 V Vout = 4962 18.7 R jXC With the leakage resistance considered: 1592 90 XC Vth 5.150 V = 2.8356.7 V Vout = 2897 33.3 R th jXC 68.
(a)
The leakage resistance effectively appears in parallel with R2. Thevenizing from the capacitor: Rth = R1 R2 Rleak = 10 k 10 k 2 k = 1.43 k R 2 R leak Vin 1.670 k 10 V = 1430 mV Vth = 11.670 R R R 2 leak 1 1 = 3.38 k XC = 2(10 Hz)(4.7 F) 3.38 90 k XC Vth 1430 mV = 13222.9 mV Vout = 3.67 67.1 k R th jXC
(b)
1 = 3.39 M 2(100 Hz)(470 pF) Req = R1 (R2 + R3) = 2.2 k 2 k = 1.05 k ZT = Req + XC Rleak (3.39 90 M)(20 k) = 1.050 k + 2 k j3.39 M (3.39 90 M)(20 k) = 1.050 k + 3.39 90 M = 1.050 k + 20 k = 3.050 k R eq 1.050 k Vin VR1 = 50 V = 1.720 V 3.050 k ZT
XC =
R3 10 k VR1 1.720 V = 8600 mV Vout = 20 k R2 R3 69.
(a) (b)
(c)
(d)
Vout = 0 V (less than normal) 1 XC = = 3.39 k 2(10 Hz)(4.7 F) XC 3.39 90 k Vin 10 V = 320-71.3 mV (greater than normal) Vout = 10.6 18.7 k R jXC R2 100 k Vin 10 V = 5000 mV (greater than normal) Vout = 200 k R1 R 2
Vout = 0 V (less than normal output)
157
Chapter 15 70.
(a) (b) (c)
(d) (e)
Vout = 0 V (less than normal) R3 10 k Vin 50 V = 2.50 V (greater than normal) Vout = 20 k R2 R3 XC = 3.39 M 10 k R3 Vin 50 V = 1.4790 mV Vout = 3.39 90 M R 2 R 3 jXC (greater than normal) Vout = 0 V (less than normal) 2.20 k R1 Vin 50 V = 3.2490 mV Vout = 3 . 39 90 M R jX 1 C (greater than normal)
Multisim Troubleshooting and Analysis 71.
No fault.
72.
C1 is leaky.
73.
R1 is open.
74.
No fault.
75.
No fault.
76.
C2 is open.
77.
fc = 48.41 Hz
78.
fc = 3.422 kHz
158
Chapter 16 RL Circuits Part 1: Series Circuits Section 16-1 Sinusoidal Response of RL Circuits 1.
fVR = 15 kHz, fVL = 15 kHz
2.
The current is sinusoidal because the voltage is sinusoidal.
Section 16-2 Impedance of Series RL Circuits Z = R + jXL = 100 + j50 = 11226.6 Z = R + jXL = 1.5 k + j1 k = 1.8033.7 k
3.
(a) (b)
4.
See Figure 16-1. (a) RT = 56 + 10 = 66 LT = 50 mH + 100 mH = 150 mH XL = 2fLT = 2(100 Hz)(150 mH) = 94.2 Z = RT + jXL = 66 + j94.2 = 11555.0 Z = 115 , = 55.0 (b)
LT = 5 mH 8 mH = 3.08 mH XL = 2fLT = 2(20 kHz)(3.08 mH) = 387 Z = RT + jXL = 560 + j387 = 68134.6 k Z = 681 , = 34.6
Figure 16-1
159
Chapter 16 5.
6.
7.
(a)
XL = 2fL = 2(100 Hz)(0.02 H) = 12.6 Z = 12 + j12.6 = 17.446.4
(b)
XL = 2fL = 2(500 Hz)(0.02 H) = 62.8 Z = 12 + j62.8 = 64.079.2
(c)
XL = 2fL = 2(1 kHz)(0.02 H) = 126 Z = 12 + j126 = 12784.6
(d)
XL = 2fL = 2(2 kHz)(0.02 H) = 251 Z = 12 + j251 = 25187.3
(a)
Z = 20 + j45 = R + jXL R = 20 , XL = 45
(b)
Z = 50035 = 410 + j287 = R + jXL R = 410 , XL = 287
(c)
Z = 2.572.5 k = 752 + j2.38 k = R + jXL R = 4752 , XL = 2.38 k
(d)
Z = 99845 = 706 + j706 = R + jXL R = 706 , XL = 706
L1 L2 = 3.11 mH, R1 R2 = 476 RT = R1 + R1 R2 = 330 + 476 = 806 LT = L3 + L1 L2 = 1000 H + 3.11 mH = 4.11 mH
Section 16-3 Analysis of Series RL Circuits 8.
9.
RT = 806 XLT = 2fLT = 2(10 kHz)(4.11 mH) = 258
RT VRT = 2 R X2 LT T
806 Vs 2 2 (806 ) (258 )
5 V = 4.76 V
X LT VLT = R2 X 2 LT T
258 Vs (806 ) 2 (258 ) 2
5 V = 1.52 V
XL3 = 2fL3 = 2(10 kHz)(1000 H) = 62.8 X 62.8 VL3 = L 3 VLT 1.52 V = 0.370 V 258 X LT
160
Chapter 16 10.
11.
(a)
From Problem 3(a): Z = 11226.6 100 V I= = 89.426.6 mA 11226.6
(b)
From Problem 3(b): Z = 1.8033.7 k 50 V I= = 2.7833.7 mA 1.8033.7 k
(a)
From Problem 4(a): Z = 11555.0 50 V IT = = 43.555.0 mA 11555.0
(b)
From Problem 4(b): Z = 68134.6 k 80 V IT = = 11.834.6 mA 68134.6 k
12.
XL = 2fL = 2(60 Hz)(0.1 H) = 37.7 37.7 X = 38.7 = tan 1 L tan 1 R 47
13.
= 38.7 from Problem 12. Double L: XL = 2fL = 2(60 Hz)(0.2 H) = 75.4 75.4 X = 58.1 = tan 1 L tan 1 R 47
increases by 19.4 from 38.7 to 58.1 14.
See Figure 16-2. The circuit phase angle was determined to be 38.7 in Problem 12. This is the phase angle by which the source voltage leads the current; it is the same as the angle between the resistor voltage and the source voltage. The inductor voltage leads the resistor voltage by 90. Assume that 10 V is the rms value of Vs.
XL = 37.7 XL 37.790 37.790 Vs 100 V 100 V = 6.2553.1 V VL = 47 j37.7 60.338.7 R jX L R 470 Vs 100 V = 7.7938.7 V VR = 60.338.7 R jX L
161
Chapter 16
Figure 16-2 15.
(a)
f = 60 Hz XL = 2(60 Hz)(100 mH) = 37.7 Z = R + jXL = 150 + j37.7 = 154.714.1 1500 R 50 V = 4.8514.1 V VR = Vs Z 154.714.1 37.790 X 50 V = 1.2275.9 V VL = L Vs Z 154.714.1
(b)
f = 200 Hz XL = 2(200 Hz)(100 mH) = 125.7 Z = R + jXL = 150 + j125.7 = 195.740.0 1500 R 50 V = 3.8340.0 V VR = Vs Z 195.740.0 125.790 X 50 V = 3.2150.0 V VL = L Vs Z 195.740.0
(c)
f = 500 Hz XL = 2(200 Hz)(100 mH) = 314 Z = R + jXL = 150 + j314 = 34864.5 1500 R 50 V = 2.1664.5 V VR = Vs Z 34864.5 31490 X 50 V = 4.5125.5 V VL = L Vs Z 34864.5
(d)
f = 1 kHz XL = 2(1 kHz)(100 mH) = 628 Z = R + jXL = 150 + j628 = 645.776.6 1500 R 50 V = 1.1676.6 V VR = Vs Z 645.776.6 62890 X 50 V = 4.8613.4 V VL = L Vs Z 645.776.6
162
Chapter 16 16.
Vs = VL1 + VL2 + VR1 + VR2 = 1590 V + 8.590 V + 6.90 V + 20 V = 8.90 V + 23.590 V = 8.9 V + j23.5 V = 25.169.3 V
Vs = 25.1 V, = 69.3 17.
18.
(a)
XL = 2(1 Hz)(10 mH) = 62.8 m X 62.8 m = tan 1 L tan 1 = 0.0923 39 R
(b)
XL = 2(100 Hz)(10 mH) = 628 m X 628 m = tan 1 L tan 1 = 9.15 39 R
(c)
XL = 2(1 kHz)(10 mH) = 6.28 X 6.28 = tan 1 L tan 1 = 58.2 39 R
(d)
XL = 2(10 kHz)(10 mH) = 62.8 X 62.8 = tan 1 L tan 1 = 86.4 39 R
(a)
= tan1
(b) (c)
19.
R XL
1 39 tan = 89.9 62.8 m R 1 39 = tan1 tan = 80.9 628 m XL R 1 39 = tan1 tan = 31.8 6.28 XL R XL
= tan1
(a)
XL = 2(1 Hz)(10 mH) = 62.8 m Z = 39 + j62.8 m = 390o
X Vout L Z (b) `
1 39 tan = 3.60 62.8
(d)
62.890o m o o V 500 mV = 80.590 V in o 390
XL = 2(100 Hz)(10 mH) = 628 m Z = 39 + j628 m = 390o
X Vout L Z
62890o m o o V 500 mV 80590 V in o 390
163
Chapter 16 (c)
XL = 2(1 kHz)(10 mH) = 6.28 Z = 39 + j6.28 = 39.59.14o
X Vout L Z (d)
6.2890o o o V 500 mV = 7.9580.9 mV in o 39.59.14
XL = 2(10 kHz)(10 mH) = 62.8 Z = 39 + j62.8 = 73.958.2o
X Vout L Z
62.890o o o V 500 mV = 42.531.8 mV in o 73.958.2
Part 2: Parallel Circuits Section 16-4 Impedance and Admittance of Parallel RL Circuits 20.
XL = 2fL = 2(2 kHz)(800 H) = 10.1 1 1 YT = G jBL = = 83.3 mS j99.0 mS = 12949.9 mS j 12 10.1 1 1 = 7.7549.9 Z= YT 129 49.9 mS
21.
From Problem 20: Z=
1 1 = 7.7549.9 = (7.75 cos 49.9o + j(7.75 sin49.9o YT 129 49.9 mS
= 4.99 + j5.93 22.
(a)
f = 1.5 kHz XL = 2fL = 2(1.5 kHz)(800 H) = 7.54 1 1 YT = G jBL = = 83.3 mS j133 mS = 15758.0 mS j 12 7.54 1 1 = 6.3758.0 Z= YT 157 58.0 mS
(b)
f = 3 kHz XL = 2fL = 2(3 kHz)(800 H) = 15.1 1 1 YT = G jBL = = 83.3 mS j66.2 mS = 10638.5 mS j 12 15.1 1 1 = 9.4338.5 Z= YT 106 38.5 mS
(c)
f = 5 kHz XL = 2fL = 2(5 kHz)(800 H) = 25.1
164
Chapter 16 1 1 YT = G jBL = j = 83.3 mS j39.8 mS = 92.325.5 mS 12 25.1 1 1 Z= = 10.825.5 YT 92.3 25.5 mS
(d)
23.
f = 10 kHz XL = 2fL = 2(10 kHz)(800 H) = 50.3 1 1 YT = G jBL = = 83.3 mS j19.9 mS = 85.613.4 mS j 12 50.3 1 1 = 11.713.4 Z= YT 85.6 13.4 mS
XL = 2fL X 12 f= L = 2.39 kHz 2L 2(800 H)
Section 16-5 Analysis of Parallel RL Circuits 100 V = 4.550 mA 2.20 k 100 V = 2.8690.0 mA IL = 3.590 k IT = IR + IL = 4.55 mA j2.86 mA = 5.3732.2 mA
24.
IR =
25.
(a)
XL = 2fL = 2(2 kHz)(25 mH) = 314 RX L (560 )(314 ) = 274 Z= 2 2 R XL (560 2 (314 ) 2 XL 1 314 = 60.7 = 90 tan R 560 Z = 27460.7
= 90 tan 1
(b)
IR =
500 mV = 89.30 mA 5600
(c)
IL =
500 mV = 15990 mA 31490
(d)
IT =
500 mV = 18260.7 mA 27460.7
(e)
= 60.7 (from part a)
165
Chapter 16 26.
(a)
XL = 2fL = 2(2 kHz)(330 H) = 4.15 1 1 Z= 1 1 1 1 j j R X L 56 4.15 =
27.
1 1 = 4.1385.8 17.9 mS j241 mS 242 85.8 mS
(b)
IR =
500 mV = 8930 mA 560
(c)
IL =
500 mV = 12.090 mA 4.1590
(d)
IT = IR + IL = 893 mA j12 A = 12.085.8 A
(e)
= 85.8
ZT =
( R1 R2 ) X L ( R1 R2 )
2
X L2
(11.5 k)(5 k) (11.5 k) 2 (5 k) 2
= 4.59 k
XL 1 5 k 90 tan = 66.5 R 11.5 k ZT = 4.5966.5 = 1.83 k + j4.21 k 1.83 k resistance in series with 4.21 k inductive reactance.
= 90 tan 1
28.
IT = IR1 + IR2R3 + IL1 = 50 mA + 30 mA + 8.390 mA = 80 mA + 8.390 mA = 8 mA j8.3 mA = 11.546.1 mA IT = 11.5 mA, = 46.1
Part 3: Series-Parallel Circuits Section 16-6 Analysis of Series-Parallel RL Circuits 29.
See Figure 16-3. XL1 = XL2 = 2(400 Hz)(50 mH) = 125.6 ZR3L1L2 = R3 + XL1 XL2 = 33 + j62.8 = 70.962.3 R2 ZR3L1L2 = 220 70.962.3 = 18.713.5 = 18.2 + j4.37 ZT = R1 + R2 ZR3L1L2 = 56 + 18.2 + j4.37 = 64.33.89 R 560 250 V = 21.83.89 V VR1 = 1 Vs Z 64 . 3 3 . 89 T R Z R3L1L2 18.713.5 Vs VR2 = 2 64.33.89 250 V = 7.279.61 V Z T R3 330 VR2 7.279.61 V = 3.3853.3 V VR3 = 70 . 9 62 . 3 Z R3L1L2
166
Chapter 16 X XL2 62.890 VR2 VL1 = VL2 = L1 70.962.3 7.279.61 V = 6.4437.3 V Z R3L1L2
Figure 16-3
30.
LT = L1 L2 = 25 mH XLT = 2(400 Hz)(25 mH) = 62.8 Combining R3, L1, and L2: ZA = 33 + j62.8 = 70.962.3 Combining ZA with R2 in parallel: (220 )(70.962.3 ) 156062.3 2 ZB = = 17.410.1 = 17.1 + j3.05 22 33 j70.9 89.752.2 ZT = R1 + ZB = 56 + 17.1 + j3.05 = 73.1 + j3.05 The circuit is predominantly resistive because the resistance is greater than the reactance in the expression for ZT.
167
Chapter 16 31.
See Figure 16-4. Using the results of Problem 29: 21.8 3.89 V V IR1 = IT = R1 = 3893.89 mA R1 560 V 7.279.61 V IR2 = R2 = 3309.61 mA R2 220 3.38 53.3 V V IR3 = R3 = 102-53.3 mA R3 330 V 6.4437.3 V IL1 = IL2 = L1 = 51.3-52.7 mA XL1 125.690
52.7
Figure 16-4
32.
3.89 IR1 = IT
51.3 mA
XL1 = 2(80 kHz)(10 mH) = 5 k XL2 = 2(80 kHz)(8 mH) = 4 k Z1 = 5.6 k + j4 k = 6.8835.5 k Combining R2 in parallel with Z1: (3.30 k)(6.8835.5 k) 22.835.5 k Z2 = = 2.3411.3 k = 2.29 k + j459 8.9 k j4 k 9.7624.2 k Combining XL1 in series with Z2: Z3 = 2.29 k + j5.46 k = 5.9167.5 k Combining R1 in parallel with Z3: (1.20 k)(5.9167.5 k) 7.0967.5 k ZT = = 1.109.90 k 3.46 k j5.46 k 6.4657.6 k
(a) (b) (c) (d) (e)
VS 180 V = 16.49.90 mA Z T 1.109.90 k = 9.90 (IT = lags Vs) VR1 = Vs = 180 V Z 2.3411.3 k VR2 = 2 VR1 180 V = 7.1356.2 V 5.9167.5 k Z3
IT =
R 5.60 k VR3 = 3 VR2 7.13 56.2 V = 5.8091.7 V 6.8835.5 k Z1
168
Chapter 16 (f) (g)
33.
X 590 k VL1 = L1 VS 180 V = 15.222.5 V 5.9167.5 k Z3 X 490 k VL2 = L2 VR2 7.13 56.2 V = 4.151.70 V 6.8835.5 k Z1
The circuit is rearranged in Figure 16-5 for easier analysis. ZT = (R1 + XL1 R2) (XL2 + XL3) = (50 0 + 56.234.2 ) 12090 = (50 + 46.5 + j31.6 ) (12090) (10218.1 )(12090 ) = (10218.1 ) (12090 ) = 96.5 j152 (10218.1 )(12090 ) = 68.050.5 = 18057.6 (a) (b) (c)
IT =
400 V Vs = 58850.5 mA Z T 68.050.5
XL1 R 2 Vs 56.234.2 400 V = 22.016.1 V VL1 = R X 10218.1 L1 R 2 1 XL3 4590 Vs 400 V = 150 V VA = 12090 XL2 XL3
VB = VL1 = 22.016.1 V VAB = VA VB = 150 V 22.016.1 V = 15 V 21.1 V j6.10 V = 6.10 j6.10 = 8.63135 V
Figure 16-5
34.
See Figure 16-6. 22.016.1 V V IL1 = L1 = 22073.9 mA XL1 10090 V 22.016.1 V IR2 = B = 32416.1 mA R2 680 400 V Vs IL2 = IL3 = = 33390 mA X L2 X L3 12090 IR1 = IR2 + IL1 = 32416.1 mA + 22073.9 mA = (311 mA + j89.8 mA) + (61.0 mA j211 mA) = 372 mA j121 mA = 39118.0 mA
169
Chapter 16 VR1 = IR1R1 = (39118.0 mA)(500 ) = 19.618.0 V VR2 = VL1 = 22.016.1 V XL2 7590 400 V 400 V = 250 V VL2 = 12090 XL2 XL3
XL3 4590 400 V 400 V = 150 V VL3 = 12090 XL2 XL3
16.1
Figure 16-6
35.
R4 + R5 = 3.9 k + 6.8 k = 10.7 k R2 (R4 + R5) = 4.7 k 10.7 k = 3.27 k R2 + R3 (R4 + R5) = 5.6 k + 3.27 k = 8.87 k RT = R1 (R2 + R3 (R4 + R5)) = 3.3 k 8.87 k = 2.41 k XL = 2(10 kHz)(50 mH) = 3.14 k X 3.14 k = tan 1 L tan 1 = 52.5 Vout lags Vin R 2.41 k 2.41 k RT V 1 V = 609 mV VR1 = R 2 X 2 in (2.41 k) 2 (3.14 k) 2 L T
R3 ( R4 R5 ) 3.27 k V VR3 = 609 mV = 225 mV R R ( R R ) R1 3.27 k 5.6 k 2 3 4 5 R5 6 .8 k VR 3 Vout = VR5 = 225 mV = 143 mV 3.9 k 6.8 k R4 R5 Vout 143 mV = 0.143 Vin 1V
36.
XL1 = 3.14 k, XL2 = 4.7 k, XL3 = 6.38 k Z3 = R3 + jXL3 = 6.8 k + j6.28 k = 9.2642.7 k Z2 = XL2 + R2 Z3 = 4.790 k + 4.70 k 9.2642.7 k = 4.790 k + 3.32 14.06 k = 6.3859.7 k Z1 = R1 Z2 = 3.30 k 6.3859.7 k = 2.4719.5 k ZT = XL1 + Z1 = 3.1490 k + 2.4719.5 k = 4.659.6 k
170
Chapter 16 2.4719.5 k 10 V = 53740.1 mV VR1 = 4.659.6 k 3.3214.1 k 537 40.1 mV = 27985.7 mV VR2 = 6.3859.7 k 6.80 k 279 85.7 mV = 205128 mV Vout = 9.2642.7 k V 205 mV Phase shift = 128, Attenuation = out = 0.205 Vin 1V
37.
12 V = 12 1A 2.5 kV = 2.5 k R2 = 1A When the switch is thrown from position 1 to position 2, the inductance will attempt to keep 1 A flowing through R2 for a short time. This design neglects the arcing of the switch, assuming instantaneous closure from position 1 to position 2. The value of L is arbitrary since no time constant requirements are imposed. See Figure 16-7.
R1 =
Figure 16-7
Part 4: Special Topics Section 16-7 Power in RL Circuits 2 Ptrue Pr2 (100 mW ) 2 (340 mVAR) 2 = 354 mVA
38.
Pa =
39.
XL = 2(60 Hz)(0.1 H) = 37.7 Z = R + jXL = 47 + j37.7 = 60.338.7 100 V V IT = s = 165.838.7 mA Z 60.338.7 Ptrue = IT2 R = (165.8 mA)2(47 ) = 1.29 W Pr = IT2 X L = (165.8 mA)2(37.7 ) = 1.04 VAR
40.
= 32.2 from Problem 22. PF = cos = cos (32.2) = 0.846
171
Chapter 16 41.
See Figure 16-8. From Problem 32: ZT = 1.109.90 k = 1.08 k + j189 IT = 16.49.90 mA Ptrue = I R2 R = (16.5 mA)2(1.08 k) = 290 mW Pr = IT2 X L = (16.4 mA)2(189 ) = 50.8 mVAR Pa = IT2 ZT = (16.4 mA)2(1.10 k) = 296 mVA PF = cos(9.90) = 0.985
Figure 16-8
42.
From Problem 33: ZT = 68.050.5 = 43.3 + j52.5 , IT = 58850.5mA R = 43.3 . Ptrue = IT2 R = (588 mA)2(43.3 ) = 15.0 W
Section 16-8 Basic Applications 43.
R Use the formula, Vout = ZT Frequency XL (kHz) 0 0 1 62.8 2 126 3 189 4 251 5 314
Vin . See Figure 16-9. Ztot Vout 39.0 73.9 132 193 254 317
1V 528 mV 296 mV 203 mV 153 mV 123 mV Figure 16-9
44.
X Use the formula, Vout = L Vin . See Figure 16-10. ZT Frequency (kHz) 0 1 2 3 4 5
XL
ZT
Vout
0 62.8 126 189 251 314
39.0 73.9 132 193 254 317
0V 42.5 mV 47.8 mV 49.0 mV 49.4 mV 49.6 mV Figure 16-10
172
Chapter 16 45.
For Figure 16-61 in the text (See Figure 16-11(a)): XL = 2(8 kHz)(10 mH) = 502.65 Z = 39 + j502.65 = 504.1685.6 390 R VR = Vin 10 V = 77.485.6 mV Z 504.16 85.6 X 502.65 90 VL = L Vin 10 V = 9974.44 mV 504.16 85.6 Z For Figure 16-62 in the text (See Figure 16-11(b)): 390 R VR = Vin 500 mV = 3.8785.6 mV 504.16 85.6 Z X 502.6590 VL = L Vin 500 mV = 49.94.44 mV Z 504.1685.6
(a)
(b)
Figure 16-11
Section 16-9 Troubleshooting 46.
VR1 = VL1 = 18 V VR2 = VR3 = VL2 = 0 V
47.
(a) (b) (c)
Vout = 0 V Vout = 0 V XL1 = 2(1 MHz)(8 H) = 50.26 XL2 = 2(1 MHz)(4 H) = 25.13 XLT = 50.26 + 25.13 = 75.39 RT = R2 + R3 = 156 Z = RT + jXLT = 156 + j75.39 = 173.2625.8 50 V = 28.925.8 mA I= 173.2625.8 Vout = IR3 = (28.925.8 mA)560 = 1.6225.8 V
(d)
R1 R3 = 100 56 = 35.9 Z = 35.9 + j75.39 = 83.564.5 50 V = 59.964.5 mA I= 83.564.5 Vout = I(R1 R3) = (59.964.5 mA)35.90 = 2.1564.5 V
173
Chapter 16 Multisim Troubleshooting and Analysis 48.
No fault.
49.
L1 is leaky.
50.
No fault.
51.
L1 is open.
52.
R2 is open.
53.
No fault.
54.
fc = 16.05 MHz
55.
fc = 53.214 kHz
174
Chapter 17 RLC Circuits and Resonance Part 1: Series Circuits Section 17-1 Impedance of Series RLC Circuits 1.
1 1 = 677 2fC 2(5 kHz)(0.047 F) XL = 2fL = 2(5 kHz)(5 mH) = 157 Z = R + jXL jXC = 10 + j157 j677 = 10 520 = 52088.9 Net reactance = jXL jXC = j520 XC =
2.
Z = R + j(XL XC) = 47 + j45 = 65.143.8
3.
Doubling f doubles XL and halves XC, thus increasing the net reactance and, therefore, the impedance magnitude increases. X2 35 2(80 ) 142.5 2 2 ZT = 47 j142.5 = 15071.7
XT = 2XL
4.
Z=
R 2 ( X L X C ) 2 = 100
R2 + (XL XC)2 = 1002 (XL XC)2 = 1002 R2 XL XC = 1002 R 2 (100 ) 2 (47 ) 2 = 88.3
Section 17-2 Analysis of Series RLC Circuits 5.
ZT = R + jXL jXC = 47 + j80 j35 = 47 + j45 = 65.143.8 V 40 V IT = s = 61.443.8 mA Z T 65.143.8 VR = ITR = (61.443.8 mA)(470 ) = 2.8943.8 V VL = ITXL = (61.443.8 mA)(8090 ) = 4.9146.2 V VC = ITXC = (61.443.8 mA)(3590 ) = 2.15134 V
175
Chapter 17 6.
Use the results of Problem 5. See Figure 17-1.
Figure 17-1 7.
RT = R1 R2 = 220 390 = 141 LT = L1 + L2 = 0.5 mH + 1.0 mH = 1.5 mH CT = C1 + C2 = 0.01 F + 1800 pF = 0.0118 F XLT = 236 , XCT = 540 Ztot = RT + j(XLT XCT) = 141 j304 = 33565.1 V 120 V (a) IT = s = 35.865.1 mA ZT 335 65.1 (b)
Ptrue = IT2 RT = (35.8 mA)2(141 ) = 181 mW
(c)
Pr = IT2 X T = (35.8 mA)2(304 ) = 390 mVAR
(d)
Pa =
( Ptrue ) 2 ( Pr ) 2 = 430 mVA
Section 17-3 Series Resonance 8.
At the resonant frequency, XL = XC. In text Figure 1759, XL = 80 and XC = 35 . For resonance to occur, XL must decrease and XC must increase. Therefore, the resonant frequency is lower than the frequency, producing the indicated values.
9.
VR = Vs = 12 V
10.
fr =
= 734 kHz 2 LC 2 (1 mH)(47 pF) XL = 2frL = 2(734 kHz)(1 mH) = 4.61 k XC = XL = 4.61 k Ztot = R = 22 V 12 V = 545 mA I= s Z tot 22
11.
VC = VL = 100 V at resonance V 10 V = 200 Z=R= s I max 50 mA V 100 V = 2 k XL = XC = L I max 50 mA
1
1
176
Chapter 17 12.
fr =
1 2 LC
1 2 (0.008 mH)(0.015 F)
= 459 kHz
XL = 2(459 kHz)(0.008 mH) = 23.1 X 23.1 Q= L = 2.31 R 10 f 459 kHz BW = r = 199 kHz Q 2.31 BW 199 kHz f 1 = fr = 459 kHz = 359.5 kHz 2 2 BW 199 kHz = 459 kHz + = 558.5 kHz f2 = fr + 2 2 13.
14.
Vs 7.07 V = 707 mA at resonance R 10 Ihalf-power = 0.707Imax = 0.707(707 mA) = 500 mA Imax =
At f1: 1 = 29.5 23595 kHz)(0.015 F) XL = 2(359.5 kHz)(0.008 mH) = 18.1 XC XL = 29.5 18.1 = 11.4 11.4 = 48.7 current leading = tan 1 10 At f2: 1 XC = = 19.0 25585 kHz)(0.015 F) XL = 2(588.5 kHz)(0.008 mH) = 28.1 XL XC = 28.1 19.0 = 9.1 9 .1 = 42.3 current lagging = tan 1 10 XC =
r = 0 15.
Refer to Figure 17-2. 1 fr = Choose C = 0.001 F 2 LC (a) fr = 500 kHz 1 f r2 2 4 LC 1 1 L= 2 = 101 H 2 2 4 f r C 4 (500 kHz)2 (0.001 F)
Figure 17-2 (b)
fr = 1000 kHz 1 1 L= = 25.3 H 2 2 2 4 f r C 4 (1000 kHz)2 (0.001 F)
177
Chapter 17 (c)
fr = 1500 kHz 1 1 2 = 11.3 H L= 2 2 4 f r C 4 (1500 kHz)2 (0.001 F)
(d)
fr = 2000 kHz 1 1 2 = 6.33 H L= 2 2 4 f r C 4 ( 2000 kHz)2 (0.001 F)
Part 2: Parallel Circuits Section 17-4 Impedance of Parallel RLC Circuits 16.
XL = 2fL = 2(12 kHz)(15 mH) = 1131 1 1 XC = = 603 2fC 2(12 kHz)(0.022 F) 1 Z= 1 1 1 1000 113190 603 90 1 = 99.74.43 = 10 mS j0.884 mS j1.66 mS
17.
From Problem 16, Z = 99.74.43 The small negative phase angle indicates a slightly capacitive circuit.
18.
The circuit was found to be capacitive in Problem 17. A decrease in frequency to a point where XL is slightly less than XC will result in an inductive circuit. XL < XC
1 2fC 1 f2 < 2 4 LC 1 f< 42 LC 1 f< 2 LC
2fL <
f<
1
2 mH)(0.022 F) f < 8.76 kHz
178
Chapter 17 Section 17-5 Analysis of Parallel RLC Circuits 50 V Vs = 50.24.43 mA Z 99.7 4.43 V 50 V IR = s = 500 mA R 1000 50 V V IL = s = 4.4290 mA XL 113190 V 50 V IC = s = 8.2990 mA XC 603 90 VR = VL = VC = 50 V
19.
IT =
20.
XL = j9.42 k; XC = j72.3 k ZT = 100 j9.42 k j72.3 k = 58.953.9
21.
IR =
50 V 50 V = 500 mA; IL = = 53190 A 1000 9.4290 k 50 V 50 V = 69.190 A; IT = = 84.953.9 mA IC = 72.3 90 k 58.9 53.9
Section 17-6 Parallel Resonance 22.
ZT = (infinitely high) RW2 C 1 1 L fr = = 104 kHz 2 LC 2 LC 2 (50 mH)(47 pF) XL = 2frL = 2(104 kHz)(50 mH) = 32.7 k 32.7 k X Q= L = 1635 RW 20 Zr = RW(Q2 + 1) = 20 (16352 + 1) = 53.5 M
1
23.
24.
Zr = 53.5 M and fr = 104 kHz from Problem 23. 6.3 V Itot = = 11.8 A 53.5 M XL = 2frL = 2(104 kHz)(50 mH) = 32.7 k 6.3 V = 164 mA IC = IL = 2 (20 ) (32.7 k) 2
25.
Ptrue = (164 mA)2 20 = 538 mW; Pr = 0 VAR Pa = (11.8 A)2 53.5 M = 7.45 mVA
179
Chapter 17
Part 3: Series Parallel RLC Circuits Section 17-7 Analysis of Series-Parallel RLC Circuits 26.
27.
28.
R ( jX C ) R jX C (220 )( j150 ) j33000 = j100 + j100 = j100 + 12455.7 266 34.3 220 j150 = j100 + 69.9 j102.4 = 69.9 j2.4 = 69.91.97
(a)
ZT = jXL +
(b)
ZT = jXL +
R ( jX C ) R jX C j120 k = j8 k + 7.6950.2 k = j8 k + 15.6 39.8 = j8 k + 4.92 k j5.91 k = 4.92 k + j2.09 k = 5.3523.0 k
From Problem 26: = 1.97 (capacitive) (a)
(b)
= 23.0 (inductive)
1.5 H = 1500 mH XL = 2(2 kHz)(1500 mH) = 18.9 k 1 XC = = 16.9 k 2(2 k)(0.0047 F) jX L ( R2 jX C ) ZLR2C = jXL (R2 jXC) = R2 jX C jX L (18.990 k)(27.7 37.5 k) = 23.747.3 k = 16.1 k + j17.4 k = 22.15.19 k ZT = R1 + ZLR2C = 33 k + 16.1 k + j17.4 k = 49.1 k + j17.4 k = 52.1 19.5 k R 330 k 120 V = 7.6019.5 V VR1 = 1 Vs 52.119.5 k ZT Z 23.747.3 k 120 V = 5.4627.8 V VL = LR2C Vs 52.119.5 k ZT R2 220 k VL 5.4627.8 V = 4.3463.3 V VR2 = 27.7 37.5 k R2 jX C XC VC = R2 jX C
16.9 90 k VL 5.4627.8 V = 3.3324.7 V 27.7 37.5 k
180
Chapter 17 29.
See Figure 17-3. XC = 16.9 k, XL = 18.9 k Z1 = R2 jXC = 22 k j16.9 k = 27.737.5 k (27.7 37.5 k)(18.990 k) Z2 = XL Z1 = = 23.747.3 k = 16.1 k + j17.4 k 22 k j2 k ZT = R1 + Z2 = 33 k + 16.1 k + j17.4 k = 49.1 k + j17.4 k = Req + jXeq Xeq = 2fL X eq 17.4 k L= = 1.38 H 2f 2( 2 kHz) 49.1 k
Figure 17-3 30.
1 = 56.4 2(60 Hz)(47 F) XL = 2(60 Hz)(390 mH) = 147 ZA = R1 + jXL = 100 j147 = 17855.8 (1000 )(17858.8 ) ZB = R2 ZA = = 71.819.5 = 67.7 + j24.0 200 j147 ZT = XC + ZB = j56.4 + 67.7 + j24.0 = 67.7 j32.4 = 74.825.7 Z 71.819.5 1150 V = 11045.2 V VR2 = B Vs 74.8 25.7 ZT
XC =
I2 =
VR2 11045.2 V = 1.1045.2 A R2 1000
31.
From Problem 30: I2 = 1.1045.2 A The phase angle between I2 and the source voltage is 45.2 with I2 leading.
32.
ZA = R2 XL XC2 =
1 1 1 1 j j 10 k 5 k 10 k 1 1 = = 7.0745 k = 5 k + j5 k 0.1 mS j0.1 mS 0.1414 45 mS
ZB = R1 jXC1 = 3.3 k j1 k ZT = ZA + ZB = 8.3 k + j4 k = RT + jXT RT = 8.3 k, XT = 4 k (inductive)
181
1.38 H
Chapter 17 33.
From Problem 32, ZT = 8.3 k + j4 k = 9.2125.7 k V 100 V IT = s = 1.0925.7 mA Z T 9.2125.7 k ZA = 7.0745 k from Problem 32. Z 7.0745 k 100 V = 7.6719.3 V VZA = A Vs 9.2125.7 k ZT IR1 = IC1 = IT = 1.0925.7 mA V 7.6719.3 V IR2 = ZA = 76719.3 A R2 100 k 7.6719.3 V V IC2 = ZA = 767109.3 A XC2 10 90 k V 7.6719.3 V IL = ZA = 1.53-70.7 mA XL 590 k VR1 = ITR1 = (1.0925.7 mA)(3.30 k) = 3.6025.7 V VR2 = VL = VC2 = VZA = 7.6719.3 V VC1 = ITXC1 = (1.0925.7 mA)(190 k) = 1.09116 V
34.
For Vab = 0 V, Va must equal Vb. XL1 = 226 , XL2 = 151 22690 120 V = 9.3838.5 V Va = VL1 = 180 j226 It is not possible for Vab to be 0 V because the LC branch has no resistance; thus, the voltage a to b can only have a phase angle of 0, 90, or 90 (the branch is either resonant, purely inductive, or purely capacitive depending on the value of XL). Therefore, it is not possible for Va to equal Vb in both magnitude and phase, which are necessary conditions.
35.
See Figure 17-4.
Figure 17-4 1 = 241 2(3 kHz)(0.22 F) XL1 = 2(3 kHz)(12 mH) = 226 XL2 = 2(3 kHz)(8 mH) = 151
XC =
182
Chapter 17 Za + Zb + Zc = 100 j226 + j151 = 100 + j377 = 39075 (22690 )(1000 ) Za Zc Z1 = = 57.915 39075 Za Zb Zc (15190 )(1000 ) Zb Zc Z2 = = 38.515 Za Zb Zc 39075 (22690 )(15190 ) Za Zb Z3 = = 86.9105 39075 Za Zb Zc
Combining R1 + Z1 in parallel with XC + Z2: (1800 + 57.915) (24190 + 38.515 ) = (180 + 55.9 + j14.98 ) (j241 + 37.2 + j9.96 ) = (236 + j15.0 ) (37.2 j231 ) = (2363.64 ) (23480.9 ) = 15938.9 ZT = 15938.9 + 86.9105 = 124 j99.8 22.5 + j83.9 = 101.5 j15.9 = 1038.9 159 38.9 120 V = 18.530.0 V VR1Z1 = VCZ2 = 103 8.9 R1 1800 VR1Z1 18.5 30.0 V = 14.133.6 V VR1 = 2363.64 Z R1Z1 X 241 90 18.5 30.0 V = 19.139.1 V VC = C VCZ2 234 80.9 ZCZ2 Vab = VR1 VC = 14.133.6 V 19.139.1 V = (11.7 V j7.80 V) (14.8 V j12.0 V) = 3.10 V + j4.20 V = 5.22126 V 5.22126 V Vab I100 = = 52.2126 mA 1000 1000 36.
There are two resonant frequencies. One is associated with the parallel circuit containing C and L2. The other is associated with the series circuit consisting of C and L1.
37.
For series resonance: 1 1 fr = = 4.11 kHz 2 L1C 2 (10 mH)(0.15 F)
XL1 = 2(4.11 kHz)(10 mH) = 258 XC = 258 XL2 = 2(4.11 kHz)(25 mH) = 646 jX C ( RW 2 jX L 2 ) j258 (4 j646 ) Zr = RW1 + jXL1 + = 2 + j258 + RW 2 jX L 2 jX C 4 j646 j258 = 47590 Zr 475 90 Vs Vout = 100 V 8600 475 90 R Zr = 4.8361.0 V
183
Chapter 17 For parallel resonance: RW2 C 1 1 L fr = = 2.60 kHz 2 LC 2 LC 2 (25 mH)(0.15 F) 1
XL = 408 X 408 = 102 Q = L2 RW 2 4
Zr = RW Q 2 1 4 (1022 1) = 41.6 k XL1 = 2FrL1 = 2(2.6 kHz)(10 mH) = 163 Since Zr is much greater than R, RW1, or XL1 and is resistive, the output voltage is approximately: Vout 100 V
38.
See Figure 17-5. The winding resistance is neglected because it contributes negligibly to the outcome of the calculations. 1 1 f r2 fr 4 LC 2 LC 1 C 2 4 f r L For fr = 8 MHz, 9 MHz, 10 MHz, and 11 MHz 1 C1 = = 39.6 pF 4 (8 MHz)2 (10 H) 1 = 31.3 pF C2 = 4 (9 MHz) 2 (10 H) 1 C3 = = 25.3 pF 4 (10 MHz) 2 (10 H) 1 = 20.9 pF C4 = 4 (11 MHz) 2 (10 H) Figure 17-5
Part 4: Special Topics Section 17-8 Bandwidth of Resonant Circuits X L 2 k = 80 R 25 f 5 kHz BW = r = 62.5 Hz Q 80
39.
Q=
40.
BW = f2 f1 = 2800 Hz 2400 Hz = 400 Hz f f 2 2400 Hz 2800 Hz fr = 1 = 2600 Hz 2 2
184
Chapter 17 41.
Pf1 = (0.5)Pr = (0.5)(2.75 W) = 1.38 W
42.
Q=
43.
BW =
fr 8 kHz = 10 BW 800 Hz XL(res) = QRW = 10(10 ) = 100 XL 100 = 1.99 mH L= 2f r 28 kHz) XC = XL at resonance 1 1 = 0.2 F C= 2f r X C 28 kHz)(100 ) fr Q If Q is doubled, the bandwidth is halved to 200 Hz.
Multisim Troubleshooting and Analysis 44.
No fault.
45.
C1 is leaky.
46.
R1 is open.
47.
C1 is leaky.
48.
L1 is open.
49.
No fault.
50.
fc = 504.89 kHz
51.
fc = 338.698 kHz
185
Chapter 18 Passive Filters Section 18-1 Low-Pass Filters 1.
500 90 100 V = 0.491 V j2.16 V = 2.2277.2 V rms Vout = 2.2 k j500
2.
(a) (d)
100 Hz is passed 3 kHz is borderline
3.
(a)
XC =
(b)
(c)
4.
(b) (e)
1 kHz is passed 5 kHz is rejected
1 = 265 2(60 Hz)(10 F)
265 90 100 V = 9.3620.7 V Vout = 100 j265 1 XC = = 48.5 2(400 Hz)(8.2 F) 48.5 90 100 V = 7.1844.1 V Vout = 47 j48.5 XL = 2(1 kHz)(5 mH) = 31.4 3300 100 V = 9.965.44 V Vout = 330 j31.4
(d)
XL = 2(2 kHz)(80 H) = 1 100 100 V = 9.955.74 V Vout = 10 j1
(a)
fc =
(b)
100 90 50 V = 3.5445 V Vout = 100 j100 1 fc = = 413 Hz 2(47 )(8.2 F) 1 XC = = 47.0 2(413 Hz)(8.2 F)
1 = 159 Hz 2(100 )(10 F) 1 XC = = 100 2(159 Hz)(10 F)
47 90 50 V = 3.5445 V Vout = 47 j47
186
(c)
2 kHz is passed
Chapter 18
5.
fc =
(d)
fc =
1 = 20.0 k 2(80 H / 10 ) XL = 2(20.0 kHz)(80 H) = 10 100 50 V = 3.5445 V Vout = 10 j10
1 2RC 1 C= 2Rf c fc =
(a) (b) (c) (d) 6.
1 = 10.5 kHz 2(5 mH/330 ) XL = 2(10.5 kHz)(5 mH) = 330 3300 50 V = 3.5445 V Vout = 330 j330
(c)
1 = 12.1 F 2(220 )(60 Hz) 1 C= = 1.45 F 2(220 )(500 Hz) 1 C= = 0.723 F 2(220 )(1 kHz) 1 = 0.144 F C= 2(220 )(5 kHz)
C=
Position 1: 1 fc = 2RCT
CT =
Position 3: CT = 1000 pF 1 = 15.9 kHz 2(10 k)(1000 pF) Position 4: 1 = 500 pF CT = 1 1 0.001 F 1000 pF 1 fc = = 31.8 kHz 2(10 k)(500 pF)
fc =
1
1 1 0.01 F 0.022 F 0.047 F = 0.00873 F 1 = 1.82 kHz fc = 2(10 k)(0.00873 F) Position 2: CT = 0.022 F + 0.047 F = 0.069 F 1 = 231 Hz fc = 2(10 k)(0.069 F)
187
Chapter 18 7.
See Figure 18-1.
Figure 18-1
8.
(a) (b)
(c) (d)
9.
V 20 log out Vin V 20 log out Vin
1V 20 log = 0 dB 1V 3V 20 log = 4.44 dB 5V
V 20 log out Vin V 20 log out Vin
7.07 V 20 log = 3.01 dB 10 V 5V 20 log = 14.0 dB 25 V
V dB = 20 log out Vin Vout dB log 1 Vin 20 dB Vout = Vin log 1 20
(a) (b) (c) (d)
1 Vout = (8 V) log 1 = 7.13 V 20 3 Vout = (8 V) log 1 = 5.67 V 20 6 Vout = (8 V) log 1 = 4.01 V 20 20 Vout = (8 V) log 1 = 0.800 V 20
188
Chapter 18 10.
The output decreases at the rate of 20 dB/decade (a) 10 kHz is 2 decades above fc: Vout = 20 dB (b) 100 kHz is 2 decades above fc: Vout = 40 dB (c) 1 MHz is 3 decades above fc: Vout = 60 dB
11.
The output decreases at the rate of 20 dB/decade (a) 10 kHz is in the pass bandc: Vout = 0 dB (b) 100 kHz is the cutoff frequency fc: Vout = 3 dB (ideally 0 dB) (c) 1 MHz is 1 decade above fc: Vout = 20 dB
Section 18-2 High-Pass Filters 12.
13.
The output increases at the rate of 20 dB/decade (a) 10 kHz is 1 decade below fc: Vout = -20 dB (b) 100 kHz is the cutoff frequency fc: Vout = 3 dB (ideally 0 dB) (c) 1 MHz is in the pass band: Vout = 0 dB 2.20 k 100 V = 9.7512.8 V Vout = 2.2 k j500
14.
(a) (d)
1 Hz is rejected. 60 Hz is passed.
15.
(a)
XC =
(b)
(c)
(d)
(b) (e)
20 Hz is rejected. 30 kHz is passed.
1 = 265 2(60 Hz)(10 F)
1000 100 V = 3.5369.3 V Vout = 100 j265 1 XC = = 84.7 2(400 Hz)(4.7 F) 470 100 V = 4.8561.0 V Vout = 47 j84.7 XL = 2(1 kHz)(5 mH) = 31.4 31.490 100 V = 94784.6 mV Vout = 330 j31.4 XL = 2(2 kHz)(80 H) = 1 190 100 V = 99584.3 mV Vout = 10 j1 1 1 , fc = 2RC 2 ( L / R ) 1 fc = = 159 Hz; 2(100 )(10 F) fc =
16.
(a)
Vout = 7.07 V
189
(c)
50 Hz is borderline.
Chapter 18 (b) (c) (d) 17.
1 = 720 Hz; 2(47 )(4.7 F) 1 = 10.5 kHz; fc = 2(5 mH/330 ) 1 fc = = 19.9 kHz; 2(80 H/10 )
fc =
Vout = 7.07 V Vout = 7.07 V Vout = 7.07 V
See Figure 18-2.
720 Hz
Figure 18-2
18.
Position 1: RT = 1 k + 3.3 k + 1 k = 5.3 k 1 = 2.00 kHz fc = 2(5.3 k)(0.015 F) Position 2: RT = 3.3 k + 1 k = 4.3 k 1 = 0.006 F CT = 1 1 0.015 F 0.01 F 1 = 6.17 kHz fc = 2(4.3 k)(0.006 F)
Position 3: RT = 860 + 1 k = 1.86 k 1 = 5.70 kHz fc = 2(1.86 k)(0.015 F) Position 4: RT = 2.2 k + 3.3 k + 1 k = 6.5 k 1 fc = = 1.63 kHz 2(6.5 k)(0.015 F)
190
Chapter 18 Section 18-3 Band-Pass Filters 19.
20.
21.
1
1
(a)
f0 =
(b)
f0 =
(a)
RT = 10 + 75 = 85 1 1 f0 = = 14.5 kHz 2 LC 2 (12 mH)(0.01 F) XL = 2(14.5 kHz)(12 mH) = 1.10 k X 1.1 k Q= L = 13 RT 85 f 14.5 kHz BW = 0 = 1.12 kHz Q 13
(b)
RT = 10 + 22 = 32 1 1 f0 = = 24.0 kHz 2 LC 2 (2 mH)(0.022 F) XL = 2(24.0 kHz)(2 mH) = 302 X 302 Q= L = 9.44 RT 32 f 24.0 kHz BW = 0 = 2.54 kHz Q 9.44
2 LC 1 2 LC
2 (12 mH)(0.01 F) 1 2 (2 mH)(0.022 F)
= 14.5 kHz = 24.0 kHz
Using the results of Problems 19 and 20: BW 1.12 kHz 14.5 kHz (a) f2 = f0 + = 14.5 kHz + 560 kHz = 15.06 kHz 2 2 BW 1.12 kHz 14.5 kHz f1 = f0 = 14.5 kHz 560 Hz = 13.94 kHz 2 2 (b)
BW 2.54 kHz 24.0 kHz = 24.0 kHz + 1.27 kHz = 25.3 kHz 2 2 BW 2.24 kHz f1 = f0 = 24.0 kHz 1.27 kHz = 22.7 kHz 24.0 kHz 2 2 f2 = f0 +
RW2 C L Center frequency = f0 = 2 LC 1
22.
Since RW is assumed to be zero, f0 =
1 2 LC
.
191
Chapter 18 (a)
f0 =
(b)
f0 =
1 2 (1 H)(10 F) 1
= 50.3 Hz
2 (2.5 H)(25 pF)
= 20.1 MHz
RW2 C (4 ) 2 (10 F) 1 L 1H f0 = = 50.3 Hz 2 (1 H)(10 F) 2 LC XL = 2(50.3 Hz)(1 H) = 316 X 316 Q= L = 79 RW 4 Ztank = RW(Q2 + 1) = 4 (792 + 1) = 24,968 24,968 120 V = 117 V Vout = 24,968 680 1
23.
(a)
(b)
24.
1
1
= 10.1 MHz 2 LC 2 (2 H)(25 pF) XL = 2(20.1 MHz)(2.5 H) = 316 X 316 Q= L = 79 RW 4 Ztank = RW(Q2 + 1) = 4 (792 + 1) = 24,968 24,968 120 V = 115 V Vout = 24,968 1000 f0 =
Position 1: 1 1 f0 = = 712 kHz 2 LC 2 (50 H)(1000 pF) Position 2: 1 1 = 159 kHz f0 = 2 LC 2 (100 H)(0.01 F) Position 3: 1 1 = 306 kHz f0 = 2 LC 2 (270 H)(0.001 F) f12 = 712 kHz 159 kHz = 553 kHz f23 = 306 kHz 159 kHz = 147 kHz f13 = 712 kHz 159 kHz = 405 kHz
Responses do not overlap.
192
Chapter 18 25.
f0 = (BW) Q = (500 Hz)40 = 20 kHz 2.5 V XC = = 125 20 mA 1 = 0.064 F C= 2f 0 X C X Q = L = 40 RW X RW = L = 0.025XL = 0.025(2f0L) 40 RW2 L f0 = 2 LC 1
RW2 (0.025)(2f 0 L)) 2 C 1 2 1 (0.025(2 ) 2 f 0 ) LC 2 L L f0 2 4 LC 42 LC 4 2 LC 2 Note: in the above derivation, (0.025(2)) = 0.025 f 02 4 2 LC = 1 0.25 f 02 LC 1
f 02 LC ( 4 2 0.025) = 1 1 L= 2 = 989 H 2 f 0 C (4 0.025)
Section 18-4 Band-Stop Filters 26.
(a) (b)
27.
1
1
= 339 kHz 2 (100 H)(0.0022 F) 1 = 10.4 kHz f0 = 2 LC 2 (5 mH)(0.047 F) f0 =
(a)
f0
(b)
f0
2 LC 1
1 2 LC 1 2 LC
1 2 (0.5 H)(6.8 F) 1 2 (10 H)(47 pF)
= 86.3 Hz = 7.34 MHz
RW2 C L = 86.3 Hz f0 = 2 LC XL = 2(86.3 Hz)(0.5 H) = 271 X 271 Q= L = 33.9 RW 8 Ztank = RW(Q2 + 1) = 8 ((33.9)2 +1)= 9.20 k 1 k 50 V = 4.90 V Vout = 10.2 k 1
28.
(a)
193
Chapter 18 1
(b)
f0 =
(8 ) 2 (47 pF) 10 H
2 (10 H)(47 pF)
= 7.34 MHz
XL = 2(7.34 MHz)(10 H) = 461 461 X Q= L = 57.6 RW 8 Ztank = RW(Q2 + 1) = 8 (57.62 + 1) = 26.6 k 2 .2 k 50 V = 3.82 V Vout = 28.8 k 29.
For the pass band, f0 = 1200 kHz: 1 f0 = 2 L1C
1 4 L1C 1 1 2 = 0.08 H L1 = 2 2 4 f 0 C 4 (1200 kHz)2 (0.22 F) For the stop band, f0 = 456 kHz: 1 f0 = 2 L2C f 02
L2 =
2
1 1 2 = 0.554 H 2 4 f 0 C 4 (456 kHz) 2 (0.22 F) 2
Multisim Troubleshooting and Analysis 30.
C1 is open.
31.
C2 is leaky.
32.
R3 is open.
33.
C1 is shorted.
34.
L2 is open.
35.
No fault.
36.
fr = 107.637 kHz
37.
BW 88.93 MHz
194
Chapter 19 Circuit Theorems in AC Analysis Section 19-1 The Superposition Theorem 1.
Z1 = R2 R3 = 6880 Z2 = R1 + Z1 = 16880 (16880 )(290 k) Z3 = XL Z2 = = 1.2940.2 k = 985 + j833 261749.8 ZT1 = XC + Z3 = j1 k + 985 + 833 = 985 j167 = 9999.6 V 20 V IT1 = 1 = 29.6 mA Z T1 999 9.6
XL 20 k I T1 29.6 mA = 1.5349.8 mA IR1 = 2.6249.8 k XL Z 2 R2 10 k I R1 1.5349.8 mA = 46949.8 A = 303 A + j358 A IR3(V1) = 3.20 k R2 R3
With V1 reduced to zero (shorted): (1 90 k)(290 k) Z1 = XC XL = = 290 k 190 k Z2 = R1 + Z1 = 1 k j2 k = 2.2463.4 k (2.20 k)(2.2463.4 k) Z3 = R3 Z2 = 3.2 k j2 k 4.93 63.4 k = = 1.3131.4 k = 1.12 k j0.68 k 3.77 32 k ZT2 = R2 + Z3 = 2.12 k j0.68 k = 2.2317.8 k V 330 V = 1.3547.8 mA IT2 = 2 Z T2 2.23 17.8 2.24 63.4 k 1.3547.8 mA = 80216.4 A = 769 A + j226 A IR3(V2) = 3.77 32 k IR3(tot) = IR3(V1) + IR3(V2) = 1.07 mA + j584 A = 1.22 28.6 mA
195
Chapter 19 2.
Use the results of Problem 1: With V2 reduced to zero (shorted): IR1 = 1.53 49.8 mA R3 2 .2 k I R1 IR2(V1) = 1.5349.8 mA = 1.0549.8 mA = 678 A + j802 A 3 .2 k R2 R3 With V1 reduced to zero (shorted): IR2(V2) = IT2 = 1.3547.8 mA = 907 A = j1 mA The total current through R2 is: IR2 = IR2(V2) + IR2(V2) = 1.59 mA + j1.80 mA = 2.448.5 mA VR2 = IR2R2 = (2.448.5 mA)(10k) = 2.448.5 V The total voltage across the R2 branch is: VT = V2 + VR2 = 330 V + 2.448.5 V = (2.6 V + j1.5 V) + (1.59 V + 1.8 V) = 4.19 V + j3.3 V = 5.3338.2 V
3.
With Vs reduced to zero (shorted): XL = 1.9 k, XC = 2.41 k (1.80 k)(4.7 k j1.9 k) Z1 = R1 (R2 + XL) = 6.5 k j1.9 k =
(1.80 k)(5.122 k) = 1.365.7 k = 1.35 k j0.135 k 6.7716.3 k
XC 2.41 90 k I S 1000 mA IZ1 = 1.35 k j2.28 k X C Z1 2.41 90 k 1000 mA = 90.930.6 mA = 2.65 59.4 k R 2 XC I Z1 IR1(I) = R 1 R 2 XC 5.2827.1 k 90.9 30.6 mA = 69.323.8 mA = 63.4 mA j28.0 mA = 6.9320.3 k With Is reduced to zero (opened): ( 2.41 90 k)(5.122 k) ZT = R1 + (XC (R2 + XL)) = 1.80 k + 4.7 k j0.51 k = 1.8 k + 2.6059.2 k = 3.8535.5 k V 750 V IR1(V) = s = 19.535.5 k = 15.9 mA + j11.3 mA Z T 3.85 35.5 k The total current through R1 is: IR1(tot) = IR1(I) + IR1(V) = 79.3 mA j16.7 mA = 81.011.9 mA
196
Chapter 19 4.
(a)
With Is2 zeroed (open), there is no current through RL due to Is1, so IL(1) = 0 A. With Is1 zeroed (open), the current through RL due to Is2 is: XC 2 90 k 2 90 k I s2 10 mA 10 A IL(2) = 4.7 k j2 k 5.1 23.1 k R L XC = 39266.7 mA IL = IL(1) + IL(2) = 0 A + 39266.9 A = 39266.9 mA
(b)
1 = 637 k 2(2.5 kHz)(100 pF) With V2 zeroed (shorted), the impedance “seen” by V1 is developed as follows: ZA = RL jXC3 = 5 M j637 k = 5.047.26 M (10 M)(5.04 7.26 M) ZB = R2 ZA = = 8351.2 k = 835 k j17.5 k 6.03 6.06 M ZC = XC2 + ZB = j637 k + 835 k j17.5 k = 835 k j654 k = 1.0638 M (10 M)(1.06 38 M) = 54518.5 k = 517 k j172 k ZD = R1 ZC = 1.95 19.6 M ZT(1) = XC1 + ZD = j637 k + 517 k j172 k = 517 k j809 k = 96057.4 k V1 4060 V IT(1) = = 41.7117.4 A Z T(1) 960 57.4 k XC1 = XC2 = XC3 =
R1 10 M I T(1) 41.7117.4 A = 21.4137 A IC2(1) = 1.94 19.8 M R1 ZC R2 10 M I C2(1) 21.4137 A = 3.55143 A IL(1) = 6.03 6.06 M R 2 ZA With V1 zeroed (shorted), the impedance “seen” by V2 is developed as follows: (10 M)(637 90 k) ZA = R1 XC1 = = 53557.5 k = 287 k j450 k 1.19 32.5 M ZB = XC2 + ZA = 287 k + j1.09 M = 1.1375.2 M (5 M j637 k)(1.13 75.2 M) ZD = (RL + XC3) ZB = 5.29 M j1.73 M (4.74 7.7 M)(1.13 75.2 M) = = 1.0264.4 M = 442 k j921 k 5.56 18.1 M ZT(2) = R2 + ZC = 1.44 M j921 k = 1.7132.6M IT(2) =
V2 2030 V = 11.762.6 A Z T2 1.71 32.6 M
1.13 75.2 M ZB I T(2) 11.762.6 A = 2.375.46 A IL(2) = 5.56 18.1 M Z B R L XC3 IL(tot) = IL(1) + IL(2) = 3.55143 A + 2.375.46 A = (2.88 A + j2.13 A) + (2.36 A j0.225 A) = 0.478 A + j2.35 A = 2.40101.5 A
197
Chapter 19 5.
See Figure 19-1(a). RT = R2 + R3 + R4 + R5 = 1 k + 3.9 k + 10 k + 5.1 k = 20 k 20 V IT = = 1 mA 20 k IR3 = ITR3 = (1 mA)(3.9 k) = 3.9 V VB(dc) = 20 V 3.9 V = 16.1 V VD(dc) = 0 V VC(dc) = VB(dc) ITR2 = 16.1 V (1 mA)(1 k) = 15.1 V VA(dc) = 0 V See Figure 19-1(b). VA(peak) = 9 V RT = R1 + R3 (R2 + R4 R6) = 1.2 k + 2.36 k = 3.56 k 9V IT(peak) = = 2.53 mA 3.56 k VB(peak) = VA(peak) IT(peak)R1 = 9 V (2.53 mA)(1.2 k) = 9 V 3.04 V = 5.96 V R3 3.9 k I IR2(peak) = 2.53 mA = 1 mA R R R R T ( peak ) 9.9 k 3 2 4 6 VC(peak) = VD(peak) = VB(peak) IR2(peak)R2 = 5.96 V (1 mA)(1 k) = 4.96 V
198
Chapter 19
Figure 19-1
6.
With the current source zeroed (see Figure 19-2(a)): (2090 )(31.671.6 ) ZT = 1890 + 5178.7 = 1890 + 12.482.9 = j18 + 1.53 + j12.3 = 1.53 j5.69 = 5.9074.9 1230 V = 2.04105 A = 522 mA + j1.97 A IT = IC(Vs) = 5.90 74.9 With the voltage source zeroed (see Figure 19-2(b)): Impedance of the L1, L2, C branch: 3600 (2090 )(18 90 ) = j30 j180 = j150 = j30 + Z = j30 + 290 j2 100 100 500120 mA 500120 mA = 33.3206 mA IL2 = 10 j150 150.3 86.2 2090 33.3206 mA = 333206 mA = 298 mA j147 mA IC(Is) = 290 The total capacitor current is: IC(tot) = IC(Vs) + IC(Is) = 821 mA j1.82 A = 2.00114 A
199
Chapter 19
Figure 19-2
7.
From Problem 6 With current source zeroed (open): IT(VS) = 2.04105 A
jX L1 ( R jX L 2 ) 12.497.7 IT(VS) 2.04105 1.26112.7A jX L1 2090
IR(VS) =
With the voltage source zeroed (shorted): Impedance of LC branch is ZLC = 150-90
Ζ
R Ζ LC 1500 90 100 150 90 Z LC
100 IS 0.5120A = 0.5120A 100 IRT = IR(VS) – IR(IS) = 1.26112.7A 0.5120A -0.24 A + j0.727 A = 766 71.7A ZT R
IR(IS) =
Section 19-2 Thevenin’s Theorem 8.
From Problem 5, VD(peak) = VC(peak) = 4.96 V Vth = VD(rms) = 0.707(4.96 V) = 3.51 V Rth = R4 (R2 + R1 R3) = 10 k (1.0 k + 1.2 k 3.9 k) = 10 k 2.14 k = 1.76 k
9.
(a)
(b)
(c)
XC 75 90 Vs 250 V = 1553.1 V Vth = 100 j75 R1 jXC (1000 )(75 90 ) R 1 XC 270 Zth = R2 + = 63 j48 125 36.9 R 1 jXC = 79.237.3 XL1 40090 Vs 30 V = 1.220 V Vth = 98090 XL1 XL2 (40090 )(58090 ) XL1 XL2 Zth = = 23790 = j237 = 23790 98090 XL1 XL2
VT = V1 + V2 = 15 V + 8.66 V + j5 V = 24.211.9 V
200
Chapter 19 R2 1000 k VT 24.211.9 V = 12.111.9 V Vth = 2000 k R1 R 2 Zth = XC + R1 R2 = 50 k j20 k = 53.921.8 k
10.
1 = 33.86 k 2(100 Hz)(0.047 F) Find Zth looking from the open terminals after removing RL: (220 k)(33.86 90 k) = 18.4433.0 k = 15.5 k j10 k ZA = R1 XC1 = 40.4 57.0 k ZB = R2 + ZA = 22 k + 15.5 k j10 k = 37.5 k j10 k = 38.814.9 k (33.86 90 k)(38.8 14.9 k) ZC = XC2 ZB = = 22.7755.4 k 57.7 49.5 k = 12.9 k j18.7 k ZD = R3 + ZC = 34.9 k j18.7 k = 39.628.2 k Zth = ZD = 39.628.2 k = 34.9 k j18.7 k
XC1 = XC2 =
Find Vth looking from the source after removing RL: ZA = R2 jXC2 = 22 k j33.86 k = 40.457 k (33.86 90 k)(40.4 57 k) = 19.275 k = 4.97 k j18.6 k ZB = XC1 ZA = 71.2 72 k ZT = R1 + ZB = 26.97 k j18.6 k = 32.734.5 k 320 V V IT = s = 0.9834.5 mA Z T 32.7 34.5 k XC1 33.86 90 k I T 0.9834.5 mA = 0.4716.5 mA IR2 = 22 k j67.7 k XC1 Z A IC2 = IR2 = 0.4716.5 mA Vth = VC2 = IC2XC2 = (0.4716.5)(33.8690 k) = 15.973.5 V The Thevenin equivalent circuit with RL connected is shown in Figure 19-3. The current through RL is:
IL =
15.9 73.5 V Vth = 11765.6 A R L Z th 136.2 7.9 k
Figure 19-3
201
Chapter 19 11.
The circuit is redrawn in Figure 19-4(a) for easier analysis. Combining R1, R2, and XL: (3.30 k)(390 k) = 2.49 k + j1.64 k ZA = R1 + R2 XL = 10 k + 3.3 k j3 k = 2.9833.4 k Combining R3 and ZA: (100 k)(2.9833.4 k) ZB = R3 ZA = = 2.3725.9 k 12.67.5 k
Figure 19-4
Combining XC and ZB: (5 90 k)(2.3725.9 k) = 2.621.8 k 2.62 k j0.082 k Zth = XC ZB = 4.52 62.3 k ZB 2.3725.9 k Vs 500 V = 26.387.6 V Vth = 4.5 61.7 k XC Z B The Thevenin circuit with R4 connected is shown in Figure 19-4(b). R4 4.70 k Vth 26.387.6 V = 16.988.2 V VR4 = 7.32 0.64 k R 4 Z th
12.
Refer to Figure 19-5 (note that R3 has been removed). (1000 )(9090 ) = 66.948 = 44.8 + j49.7 ZA = R1 XL = 134.542 ZB = R2 + ZA = 194.8 + j49.7 = 20114.3 (120 90 )(20114.3 ) = 11755.9 Zth = XC ZB = 207 19.8 Looking from Vs: (1000 )(192 38.7 ) ZT = XL + R2 (R1 + XC) = j90 + 277 25.6 = j90 + 69.313.1 = j90 + 67.5 j15.7 = 67.5 + j74.3 = 10047.7 X VL = L ZT
9090 Vs 750 V = 67.542.3 V 10047.7
202
Chapter 19 R2 1500 (Vs VL ) (750 V 67.542.3 V) VR2 = 150 j120 R 2 XC 1500 = (75 V 49.9 V j45.4 V) = (0.78138.7)(51.961.1 V) 192 38.7 = 40.522.4 V Vth = Vab = VR2 + VL = (37.4 V 15.4 V) + (49.9 V + j45.4 V) = 87.2 V + j30 V = 92.219 V
Figure 19-5
Section 19-3 Norton’s Theorem 13.
14.
Using Zth and Vth from Problem 9 in each part: 15 53.1 V V (a) In = th = 18915.8 mA Z th 79.2 37.3 Zn = Zth = 79.237.3 Vth 1.220 V = 5.1590 mA Z th 23790 Zn = Zth = 23790
(b)
In =
(c)
In =
12.111.9 V Vth = 22433.7 A Z th 53.9 21.8 k Zn = Zth = 53.921.8 k
From Problem 10, Zn = Zth = 39.628.2 k The total impedance seen by the source with the terminals shorted is the same (in this case) as Zn. 320 V V IT = s = 80828.2 A Z T 39.6 28.2 k
203
Chapter 19 XC1 I 33.9 90 80828.2 A = 47512.3 A IR2 = X R R X T 57.7 49.5 k 2 3 C2 C1 XC2 33.9 90 I R2 475 12.3 A = 39945.3 A In = IR3 = 40.4 57 k R 3 XC2 The Norton equivalent circuit with RL connected is shown in Figure 19-6. Zn 39.6 28.2 k I n 399 45.3 A = 11665.7 A IRL = 136.2 7.9 k R L Zn
39.628.2 k 39945.3 A
Figure 19-6
15.
First remove R4 and determine Zn looking in at the resulting open terminals. (100 k)(5 90 k) ZA = R3 XC = = 4.4663.4 = 2 k j4 k 11.2 26.6 k (3.30 k)(390 k) ZB = R1 + R2 XL = 10 k + 4.4642.3 k = 10 k + 2.2247.7 k = 2.49 k + j1.64 k = 2.9833.4 k (4.46 63.4 k)(2.9833.4 k) Zn = ZA ZB = = 2.612.30 k =2.61 k j0.105 k 5.1 27.7 k Looking from the source with R4 shorted: ZT = XC = 590 k 500 V V In = s = 1090 mA Z T 5 90 k The Norton equivalent circuit with R4 connected is shown in Figure 19-7. Zn 2.61 2.30 k I n 1090 mA = 3.5788.5 mA IR4 = 7.31 0.823 k R 4 Zn VR4 = IR4R4 = (3.5788.5 mA)(4.70 k) = 16.888.5 V
Figure 19-7
204
Chapter 19
Section 19-4 Maximum Power Transfer Theorem 16.
(a)
1 1 = 11.3 k 2fC 2(3 kHz)(0.0047 F) ZL = RL + jXL = 6.8 k + j11.3 k X 11.3 k L= L = 599 mH 2f 2(3 kHz) XC =
(b)
ZL = 8.2 k + j5 k
(c)
XL = 75.4 , XC = 60.3
Zth = R + XC XL = 500 +
(75.490 )(60.3 90 ) = 50 j301 15.190
ZL = 50 + j301 301 L= = 0.4 H 2(120 Hz) 17.
For maximum load power, ZL equals the complex conjugate of Zth. ZA = R1 jXC1 = 8.2 j10 = 12.950.6 (180 )(4 90 ) = 3.9177.5 = 0.846 j3.82 ZB = R2 XC2 = 18.4 12.5 (12.9 50.6 )(3.91 77.5 ) = 3.0671.4 = 0.976 j2.90 ZC = ZB ZA = 16.5 56.7 Zth = R3 + ZC = 9.18 j2.90 ZL = 9.18 + j2.90
18.
First convert the delta to a wye: (12 )(12 ) X1 = X2 = X3 = =4 12 12 12 The circuit is redrawn in Figure 19-8(a). Remove ZL and Thevenize: (6.8 j4 )( j4 ) (7.930.5 )(490 ) j4 6.8 j8 10.549.6 = j4 + 370.9 = j4 + 0.98 + j2.8 = 0.98 + j6.8 7.930.5 6.8 j4 100 V = 7.519.1 V 100 V Vth = 10.549.6 6.8 j8 For maximum power to ZL: ZL = 0.98 j6.8 7.5 19.1 V Vth = 3.8319.1 A IL = Z th Z L 1.960
Zth = j4 + (6.8 + j4 ) j4 = j4 +
PL(true) = I L2 RL = (3.83 A)2(0.98 ) = 14.4 W
205
Chapter 19
Figure 19-8 19.
The circuit is redrawn in Figure 19-9 to determine Zth. The load impedance (real part) must equal the Thevenin impedance (real part) and the reactive parts must be equal in magnitude but opposite in sign. That is, the impedances must be complex conjugates. (1000 )(9090 ) (2200 )(120 90 ) = 6748 + 105.361.4 100 j90 220 j120 = 44.8 + j49.8 + 50.4 j92.5 = 95.2 j42.7 ZL = 95.2 + j42.7 Zth =
Figure 19-9
Multisim Troubleshooting and Analysis 20.
R2 is open.
21.
C2 is leaky.
22.
C1 is open.
23.
No fault.
24.
VTH = 750.281.40 mV ZTH = 11.970 k
25.
IN = 30.142113.1 A ZN = 30.364.28 k
206
Chapter 20 Time Response of Reactive Circuits Section 20-1 The RC Integrator 1.
= RC = (2.2 k)(0.047 F) = 103 s
2.
(a) (b) (c) (d)
5RC = 5(56 )(47 F) = 13.2 ms 5RC = 5(3300 k)(0.015 F) = 248 s 5RC = 5(22 k)(100 pF) = 11 s 5RC = 5(5.6 M)(10 pF) = 280 s
Section 20-2 Response of an Integrator to a Single Pulse 3.
VC 0.632(20 V) = 12.6 V
4.
(a) (b) (c) (d)
5.
See Figure 20-1.
v v v v
0.865(20 V) = 17.3 V 0.950(20 V) = 19.0 V 0.982(20 V) = 19.6 V 0.993(20 V) = 19.9 V (considered full charge of 20 V)
Figure 20-1 6.
= RC = (1 k)(1 F) = 1 ms vout = 0.632(8 V) = 5.06 V See Figure 20-2 for output waveform. The time to reach steady-state with repetitive pulses is 5 ms.
Figure 20-2
207
Chapter 20 7.
(a)
(b)
Looking from the capacitor, the Thevenin resistance is R1 R2 = 5 k. = (5 k)(4.7 F) = 23.5 ms 10 k Vout(max) = 20 V = 10 V 20 k See Figure 20-3.
Figure 20-3
8.
See Figure 20-4.
Figure 20-4 9.
From Problem 7 23.5 ms The input pulse width equals one time constant, therefore Vout = 0.632(10V) = 6.32 V See Figure 20-5. 6.32 V
23.5 ms
Figure 20-5
Section 20-3 Response of RC Integrators to Repetitive Pulses 10.
Transient time = 5RC = 5(4.7 k)(10 F) = 235 ms
208
Chapter 20
11.
= (4.7 k)(10 F) = 47 ms 5 = 5(47 ms)= 235 ms See Figure 20-6.
Figure 20-6
12.
See Figure 20-7.
Figure 20-7
13.
1 1 = 100 s f 10 kHz tW = 0.25(100 s) = 25 s 1st pulse: 0.632(1 V) 632 mV Between 1st and 2nd pulses: 0.05(0.632 V) = 31.6 mV 2nd pulse: 0.632(1 V 0.0316 V) + 0.0316 V = 644 mV Between 2nd and 3rd pulses: 0.05(0.644 V) = 32.2 mV 3rd pulse: 0.632(1 V 0.0322 V) + 0.0322 V = 644 mV See Figure 20-8.
T=
Figure 20-8
209
Chapter 20 14.
The steady-state output equals the average value of the square wave input which is Vin 30 V = 15 V (with a small ripple voltage) 2 2
Section 20-4 Response of RC Differentiators to a Single Pulse 15.
See Figure 20-9.
Figure 20-9
16.
= (1 k)(1 F) = 1 ms Steady-state is reached in 5 = 5 ms. At 1 ms, V (0.368)(8 V) = 2.94 V See Figure 20-10.
Figure 20-10 17.
(a)
(b)
Looking from the source and capacitor: ( 2.2 k)(1 k 1 k) = 1.05 k RT = 4 .2 k = RTC = (1.05 k)(470 pF) = 493.5 ns 5 = 5(493.5 ns) = 2.467 s 1 k Vout(max) = 10 V = 5 V 2 k See Figure 20-11.
Figure 20-11
Section 20-5 Response of RC Differentiators to Repetitive Pulses 18.
= (1 k)(1 F) = 1 ms See Figure 20-12.
Figure 20-12 210
Chapter 20
19.
Since 5>> tW, the output shape is an approximate reproduction of the input but with a zero average value.
Section 20-6 Response of RL Integrators to Pulse Inputs 20.
10 mH = 1 ms 10 5 = 5 ms Vout(max) = 0.637(8 V) = 5.06 V See Figure 20-13. =
Figure 20-13
50 mH = 50 ms 1 5 = 250 ms See Figure 20-14.
21.
=
22.
LT = 8 H + 4 H = 12 H (100 )(156 ) RT = = 60.9 256 L 12 H = 197 ns = T RT 60.9 This circuit is an integrator.
Figure 20-14
Section 20-7 Response of RL Differentiators to Pulse Inputs 23.
100 H = 4.55 s 22
(a)
=
(b)
See Figure 20-15.
Figure 20-15
211
Chapter 20 24.
100 H = 4.55 s 22
(a)
=
(b)
See Figure 20-16.
Figure 20-16
Section 20-8 Relationship of Time Response to Frequency Response 0.35 , 5 = 50 s tr 0.9 = 1(1 et/RC) t2 = RCln(0.1) = (10 s)ln(0.1) = 23 s t1 = RCln(0.9) = (10 s)ln(0.9) = 1.05 s tr = 23 s 1.05 s = 22.0 s 0.35 fh = = 15.9 kHz 22.0 s
25.
fh =
26.
fh =
0.35 0.35 = 8.33 MHz tf 42 ns
Section 20-9 Troubleshooting 27.
(b) (c) (d)
Vout = Vin: C is open or R could be shorted. C is leaky or C is greater than 0.22 F or R is greater than 3.3 k. Resistor open or capacitor shorted.
28.
(a) (b) (c)
No problem since 5
Multisim Troubleshooting and Analysis 29.
C1 open or R1 shorted.
30.
No fault.
31.
R1 or R2 open.
32.
L1 or L2 open.
212
Chapter 21 Three-Phase Systems in Power Applications Section 21-1 Generators in Power Applications V 100 V = 376 mA Z 265.8
1.
IL =
2.
= tan1
3.
220 A = 1.88 A + j0.684 A 3140 A = 2.3 A + j1.93 A 1.5100 A = 0.26 A j1.48 A In = (1.88 A + j0.684 A) + (2.3 A + j1.93 A) + (0.26 A j1.48 A) = (1.88 A 2.3 A 0.26 A) + j(0.684 A + 1.93 A 1.48 A) = 0.68 A + j1.134 A = 1.32121 A
175 = 41.2 200
Section 21-2 Types of Three-Phase Generators 4.
VL(ba) = 600120 V 6000 V = 300 V + j520 V 600 V = 900 V + j520 V = 1.04150 kV VL(ca) = 600120 V 6000 V = 300 V j520 V 600 V = 900 V j520 V = 1.04150 kV VL(cb) = 600120 V 600120 V = 300 V j520 V + 300 V j520 V = j1.04 kV = 104 90 kV
5
ILa = Ia Ib = 50 A 5120 A = 5 A (2.5 A + j4.33 A) = 7.5 A j4.33 A = 8.6630 A ILb = 3 (590 A) = 8.6690 A ILc =
6.
3 (5 150 A) = 8.66150 A
See Figure 21-1. IL1 = Ia Ib = 50 A 5120 A = 5 A + 2.5 A j4.33 A = 7.5 A j4.33 A = 8.6630 A IL2 = Ib Ic = 5120 A 5120 A = 2.5 A + j4.33 A + 2.5 A j4.33 A = j8.66 A = 8.6690 A IL3 = Ic Ia = 5120 A 50 A = 2.5 A j4.33 A 5 A = 7.5 A j4.33 A = 8.66150 A
213
Chapter 21
Figure 21-1
Section 21-3 Three-Phase Source/Load Analysis 7.
(a)
Line voltages: VL(ab) = 3 Va(30) = 3 (500(030)) V = 86630 V VL(ca) = 3 Vc(30) = 3 (500(12030)) V = 866150 V VL(bc) = 3 Vb(30) = 3 (500(12030)) V = 86690 V
(b)
Phase currents: 5000 V = 50032 mA 132 k 500120 V = 50088 mA Ib = IZb = 132 k 500 120 V Ic = IZc = = 500152 mA 132 k
Ia = IZa =
(c)
Line currents: ILa = 50032 mA ILb = 50088 mA ILc = 500152 mA
(d)
Load currents: IZa = 50032 mA IZb = 50088 mA IZc = 500152 mA
214
(e)
Load voltages: VZa = Va = 5000 V VZb = Vb = 500120 V VZc = Vc = 500120 V
Chapter 21 8.
9.
(a)
Line voltages: VL(ab) = 3 Va(30) = 3 (100(030)) V = 17330 V VL(ca) =
3 Vc(30) = 3 (100(12030)) V = 173150 V
VL(bc) =
3 Vb(30) = 3 (100(12030)) V = 17390 V
(b)
Phase currents: 1000 V = 74145 mA Ia = 13545 100120 V = 160 A Ib = 10060 100 120 V Ic = = 500140 mA 20020
(c)
Line currents: (d) ILa = Ia = 74145 mA ILb = Ib = 160 A ILc = Ic = 500140 mA
(f)
Neutral current: In = IZa + IZb + IZc = 74145mA + 160 A + 500140 mA = (524 mA j524 mA) + (383 mA j321 mA) + (500 mA + j866 mA) = 641 mA j20.9 mA = 6411.86 mA
(a)
Line voltages: VL(ab) = 3 Va(30) = 3 (50(030)) V = 86.630 V
(b)
Load currents: (e) IZa = Ia = 74145 mA IZb = Ib = 160 A IZc = Ic = 500140 mA
Load voltages: VZa = Va = 1000 V VZb = Vb = 100120 V VZc = Vc = 100120 V
VL(ca) =
3 Vc(30) = 3 (50(12030)) V = 86.6150 V
VL(bc) =
3 Vb(30) = 3 (50(12030)) V = 86.690 V
Phase currents: First find the load currents: VL ( ca ) 86.6 150 V IZa = = 144220 mA = 110 mA + j92.6 mA Za 60070 VL (bc ) 86.690 V IZb = = 14420 mA = 135 mA + j49.3 mA Zb 60070 VL ( ab ) 86.6 30 V IZc = = 144100 mA = 25.0 mA + j142 mA Zc 60070
Ia = IZa IZc = (110 mA + j92.6 mA) (25.0 mA j142 mA) = 85 mA + j235 mA = 250110 mA Ib = IZc IZb = (25.0 mA j142 mA) (135 mA + j49.3 mA) = 160 mA + j191.3 mA = 250130 mA Ic = IZb IZa = (135 mA + j49.3 mA) (110 mA + j92.6 mA) = 245 mA j43.3 mA = 25010 mA
215
Chapter 21
10.
(c)
Line currents: ILa = Ia = 250110 mA ILb = Ib = 250130 mA ILc = Ic = 25010 mA
(d)
Load currents were found in part (b).
(e)
Load voltages: VZa = VL(ca) = 86.6150 V VZb = VL(bc) = 86.690 V VZc = VL(ab) = 86.630 V
(a)
Line voltages:
(b)
VL(ab) = 120120 V VL(ca) = 1200 V VL(bc) = 120120 V
11.
Phase currents: 1200 V Ia = IZa = = 1250 A 1050 120 120 V = 12170 A Ib = Ib = 1050 120120 V Ic = Ic = = 1270 A 1050
(c)
Line currents: IL1 = 120 A 12120 A = 12 A (6 A + j10.4 A) = 18 A j10.4 A = 20.830 A IL2 = 12120 A 120 A = (6 A j10.4 A) 12 A = 18 A j10.4 A = 20.8150 A IL3 = 12120 A 12120 A = (6 A + j10.4 A) (6 A j10.4 A) = 20.890 A
(d)
Line currents: IZa = 1250 A IZb = 12170 A IZc = 1270 A
(a)
Line voltages: VL(ab) = Va = 330120 V VL(ca) = Vc = 330120 V VL(bc) = Vb = 3300 V
(b)
Load currents: First find the load voltages: V VZa = a (120 30) V = 19190 V 3 V VZb = b (0 30) V = 19130 V 3 V VZc = c (120 30) V = 191150 V 3 V 191 90 V IZa = Za = 38.2150 A Za 560
(e)
Load voltages: VZa = 1200 V VZb = 120120 V VZc = 120120 V
216
Chapter 21 VZb 19130 V = 38.230 A Zb 560 191150 V V IZc = Zc = 38.290 A Zc 560
IZb =
Section 21-4 Three-Phase Power 12.
PT = 3(1200 W) = 3.6 kW
13.
Figure 21-34 in text: 500 V IZ = = 500 mA 1 k PL = 3VZIZcos = 2(500 V)(500 mA)cos 32 = 636 W Figure 21-35 in text: 100 V IZa = = 741 mA 135 PZa = VZaIZacos = (100 V)(741 mA)cos 45 = 52.4 W 100 V IZb = = 500 mA 200 PZb = VZbIZbcos = (100 V)(500 mA)cos 20 = 47.0 W 100 V =1A IZc = 100 PZc = VZcIZccos = (100 V)(1 A)cos 60 = 50.0 W PL = PZa + PZb + PZc = 52.4 W + 47.0 W + 50.0 W = 149 W Figure 21-36 in text: VZ = 3 (50 V) = 86.6 V 86.6 V IZ = = 144 mA 600 PL = 3VZIZcos = 3(86.65 V)(144 mA)cos 70 = 12.8 W Figure 21-37 in text: 120 V IZ = = 12 A 10 PL = 3VZIZcos = 3(120 V)(12 A)cos 50 = 2.78 kW Figure 21-38 in text: 330 V VZ = = 191 V 3 191 V IZ = = 38.2 A 5 PL = 3VZIZcos = 3(191 V)(38.2 A)cos 60 = 10.9 kW
217
Chapter 21 14.
VZ = IZ =
120 V 3 VZ Z
= 69.3 V 69.3 V (100 ) (100 ) 2
2
69.3 V = 490 mA 141.4
= 45 PL = 3VZIZcos = 3(69.3 V)(490 mA)cos 45 = 72 W 15.
ZL = 141.445 IL = IZ 120 V V VZ = L = 69.8 V 3 3 69.8 V V IZ = Z = 494 mA Z L 141.4 PT =
3 VLILcos
(120 V)(494 mA) V I Peach = L L cos cos 45 = 24.2 W 3 3 16.
P1 = VLILcos( + 30) P2 = VLILcos( 30) 100 = 45 = tan 1 100 P1 = (120 V)(494 mA)cos(45 + 30) = 15.3 W P2 = (120 V)(494 mA)cos(45 30) = 56.8 W
218
PART 2 Solutions for Application Activities
Chapter22 Tech Application Activity Chapter TIP The Test Bench 1. The protoboard components corresponding to the schematic are shown in Figure 2-1. The connections are correct as shown. 2.
The power supply provides power, the fuse protects the circuit from excessive current, the switch turns the lamp on or off, the rheostat adjusts the brightness of the lamp by varying the current.
Switch (SW)
Rheostat (R)
Fuse
Lamp
Figure 2-1 Measuring Current with the Multimeter 3. The meter is placed in series between the switch and the rheostat. 4. The lamp is brightest for measurement B, which indicates the highest current. 5. The current will increase from A to B if the source voltage is increased or if the resistance of the rheostat is decreased. 6. A blown fuse, open switch, or open lamp will cause the current to be 0 mA. Measuring Voltage with the Multimeter 7. Voltage is measured across the lamp. 8. The voltmeter is connected across the lamp. 9. The lamp is brighter for measurement A because more voltage across the lamp means more current. 10. A decrease in the source voltage or an increase in resistance of the rheostat will cause the voltage to decrease. Measuring Resistance with the Multimeter 11. The resistance of the rheostat is measured. 12. The lamp is brighter for measurement B because the lower resistance results in more current.
220
Review 13. Less voltage causes less light because the current is reduced. 14. A lower resistance will result in more light.
221
TIP Chapter 3 Application Activity Resistance Calculations 1. Using the graphs in Figure 3-19 in the text: At 600 rpm: I1 = 0.185 A; VM = 8.75 V VR1 = 12 V – VM = 12 V – 8.75 V = 3.25 V
R1
VR1 3.25V = 17.6 Standard value = 18 I1 185mA
At 400 rpm: I2 = 0.168 A; VM = 5.5 V VR2 = 12 V – VM = 12 V – 5.5 V = 6.5 V
R2
VR 2 6.5V = 38.7 Standard value = 39 I 2 168mA
At 200 rpm: I3 = 0.15 A; VM = 2 V VR3 = 12 V – VM = 12 V – 2 V = 10 V
R3
VR 3 10V = 66.7 Standard value = 68 I 3 150mA
The Schematic 2. See Figure 3-1. +12 V R1
18
R2
39
R3
68
OUT
Figure 3-1
3.
See Figure 3-1.
Test Procedure 4.
1. Apply +12 V. 2. Check for +12 V at OUT terminal for each switch position.
5.
A 12 V power supply and a DMM.
Troubleshooting 6. R1 is open. 7.
An open connection at the 12 V terminal or the OUT terminal
8.
The ohmmeter is not calibrated properly or all resistors are at their maximum tolerance value.
222
Review 9. Ohm’s law was used to find the resistance values, given the current and voltage. 10.
Based on the graph in text Figure 3-19(a), there is a linear relationship between the current and the motor speed.
223
Chapter44 Application Activity Chapter Power Ratings 1. PR1 =
(5 V) 2 = 2.5 W 10
(Use 5 W)
PR2 =
(5 V) 2 = 1.14 W 22
(Use 2 W)
PR3 =
(5 V) 2 = 532 mW 47
(Use 1 W)
PR4 =
(5 V) 2 = 250 mW 100
(Use ½ W)
PR5 =
(5 V) 2 = 114 mW 220
(Use ¼ W)
PR6 =
(5 V) 2 = 53 mW 470
(Use ¼ W)
PR7 =
(5 V) 2 = 25 mW 1.0 k
(Use ¼ W)
PR8 =
(5 V) 2 = 11.4 mW 2.2 k
(Use ¼ W)
PR9 =
(5 V) 2 = 5.32 mW 4.7 k
(Use ¼ W)
Materials List and Estimate of the Total Cost of the Project 2. Item Quantity Unit Cost Total Cost Resistor, 10 , 5W 1 $0.33 $0.33 Resistor, 22 , 2W 1 $0.10 $0.10 Resistor, 47 , 1W 1 $0.09 $0.09 Resistor, 100 , ½W 1 $0.09 $0.09 Resistor, 220 , ¼W 1 $0.08 $0.08 Resistor, 470 , ¼W 1 $0.08 $0.08 Resistor, 1.0 k, ¼W 1 $0.08 $0.08 Resistor, 2.2 k, ¼W 1 $0.08 $0.08 Resistor, 4.7 k, ¼W 1 $0.08 $0.08 Switch, rotary 1 $10.30 $10.3 Knob, switch 1 $3.30 $3.30 Terminal, dual 1 $0.20 $0.20 Post, binding 2 $1.78 $3.56 Standoff 4 $0.50 $2.00 Board, pc 1 $1.78 $1.78 Enclosure 1 $8.46 $8.46 3. Total cost = $30.53 + tax.
224
The Schematic 4. See Figure 4-1.
R1 10 5W
R1 22 2W
R1 47 1W
R1 100
R1 220
R1 470
R1 1.0 k
R1 2.2 k
R1 4.7 k
½ W
¼W
¼W
¼W
¼W
¼W
1
2
Figure 4-1
5. See Figure 4-1. Test Procedure 6. 1. Measure the resistance between terminals 1 and 2 for each of the switch positions. 2. Compare measured values to specified values. 7.
Ohmmeter
Troubleshooting 8. R1 open or bad connection 9.
Connection to switch wiper open or connection to terminal 1 or 2 open
10.
Ohmmeter not properly calibrated
Review 11. Watt’s law was used to calculate the power in each resistor, given voltage and resistance. 12.
The power ratings of R1 through R5 are insufficient for 7 V because the actual power is too close to or exceeds the rated value. The power ratings of R6 through R9 are adequate because the actual power is sufficiently less than the rated value.
225
Chapter 5 Application Activity
Chapter 5
The Schematic of the Circuit 1.
See Figure 5-1.
Figure 5-1 The Voltages 2. The total resistance. RTOT = R1 + R2 + R3 + R4 + R5 + R6 = 3.9 k + 4.7 k + 1.5 k + 1 k + 3.3 k + 2.2 k = 16.6 k Calculations of output voltages: R 3.9 k V2 = 1 VS 12 V = 2.82 V 16.6 k RTOT
R R2 8.6 k V7 = 1 VS 12 V = 6.22 V 16.6 k RTOT R R2 R3 10.1 k V6 = 1 VS 12 V = 7.30 V RTOT 16.6 k R R2 R3 R6 11.1 k V5 = 1 VS 12 V = 8.02 V RTOT 16.6 k R R2 R3 R6 R5 14.4 k V4 = 1 VS 12 V = 10.4 V RTOT 16.6 k 3. All voltages are within 5% of specified values.
226
Since the output voltages of the existing circuit meet the specifications, no resistor value changes are required. The maximum power occurs in R2: 12 V 12 V = 723 A I= RTOT 16.6 k Pmax = I2R2 = (723 A)24.7 k = 2.46 mW The 1/4 W rating of each resistor is more than adequate. The Battery 4. (723 A)(h) = 6.5 Ah 6.5 Ah = 8990 h h= 723 A 8990 h = 374.6 days Number of days = 24 h/day A Test Procedure 5. Equipment: A 12 V dc source and a voltmeter. 1. Connect the positive terminal of the 12 V source to pin 3 of the circuit board. Connect the negative terminal of the source to pin 1 of the circuit board. 2. Measure the voltage at each of the pins as follows: Pin 1: 0 V Pin 2: 2.8 5% Pin 3: 12 V Pin 4: 10.4 V 5% Pin 5: 8.0 V 5% Pin 6: 7.3 V 5% Pin 7: 6.2 V 5% Troubleshooting 6. 1. No voltage at any of the pins on the circuit board: The battery is dead or the connection from the battery to pin 3 is open. 2. 12 V at pins 3 and 4. All other pins have 0 V: Resistor R5 is open. 3. 12 V at all pins except 0 V at pin 1: Resistor R1 is open. 4. 12 V at pin 6 and 0 V at pin 7: Resistor R3 is open. 5. 3.3 V at pin 2: The battery voltage is too high (14.3 V instead of 12 V) or resistors R1 and R2 are reversed (R1 = 4.7 k and R2 = 3.9 k). This is shown by the following calculation: 4.7 k V2 = 12 V = 3.4 V 16.6 k Review
(12V) 2 8.67W 16.6k
7.
PT
8.
The voltage at each pin is one-half the voltage calculated in Activity 2. Pin 4: 5.2 V; Pin 5: 4.01 V; Pin 6: 3.65 V; Pin 7: 3.11 V; Pin 2: 1.41 V Pin 3 connects to ground.
9.
227
Chapter 6 Application Activity
Chapter 6
Another Approach 1, Determine the maximum power dissipated by RSH in Figure 6-56 for each range setting.
For the 25 mA range, no power is dissipated because there is no current in RSH. For the 250 mA range, the maximum current in RSH is 225 mA and the voltage is 150 mV. P = IV = (225 mA)(150 mV) = 33.8 mW
For the 2.5 A range, the maximum current in RSH is 2.475 A and the voltage across it is 150 mV. P = IV = (2.475 mA)(150 mV) = 371 mW 2.
How much voltage is there from A to B in Figure 6-56 when the switch is set to the 2.5 A range and the current is 1 A?
In the 2.5 A range the fraction of the total current in the meter is 25 mA/2.5 A = 1%. The meter current is thus 1% of 1 A = 10 mA, which is in both R1 and the meter. The meter resistance is 6 and the resistance of R1 is 60.4 . The voltage from A to B is VAB = I(R1 + RM) = (10 mA)(66.4 ) = 66.4 mV 3.
The meter indicates 250 mA. How much does the voltage across the meter circuit from A to B change when the switch is moved from the 250 mA position to the 2.5 A position?
In the 250 mA position, the voltage between A and B is 150 mV, as given previously. In the 2.5 A position, the actual current in the meter is 1% of the input current. Thus, when the input current is 250 mA, the meter current is 2.5 mA. In this case, VAB = I(RM) = (2.5 mA)(6 ) = 15 mV
The change is 150 mV 15 mV = 135 mV. 4.
Assume the meter movement has a resistance of 4 instead of 6 . Specify any changes necessary in the circuit of Figure 6-56.
The meter will drop 100 mV at maximum current for any switch position. 100 mV = 444 m (2/3 of its former value) 225 mA VAB = (2.475)(444 m) = 1.1 V
RSH –
The voltage across R1 is
VR1 = 1.1 V 100 mV = 1.0 V 1.0 V = 40 R1 = 25 mA
228
Review 5. RSH has the most current. 6.
25 mA range: RAB = RM = 6 250 mA range: RAB = RM RSH = 6 670 m = 603 m 2.5 A range: RAB = (R1 + RM) RSH = (60.4 + 6 ) 670 m = 663 m
The circuit in text Figure 6-56 negates the effect of the switch contact resistance.
8.
The current indicated is 150 mA.
9.
25 mA range: 7.5 mA 250 mA range: 75 mA 2.5 A range: 750 mA
229
Chapter 7 Application Activity
er 6
The Control Circuit 1. Using the circuit board as a guide, complete the schematic in Figure 7-58. The op-amp inputs are connected to a Wheatstone bridge. Show the values for all resistors. See Figure 7-1.
Figure 7-1
The Thermistor 2. Calculate the resistance of the thermistor at a temperature of 40C using the exponential equation and confirm that your calculation is correct by comparing your result with Figure 7-59. Remember that temperatures in the equation are in K. (K = C + 273). T0 T T0T
RT = R0 e 3.
(25 k)e
298 K 313 K 4616K (298 K)(313 K)
= 11.9 k
Calculate the resistance setting of R3 to balance the bridge at 25C.
At balance,
R3 + R4 = RT = 25 k Therefore,
R3 = 25 k 20 k = 5.0 k 4.
Calculate the output voltage of the bridge (input to the op-amp) when the temperature of the thermistor is 40C. Assume the bridge was balanced at 25C and that the only change is the resistance of the thermistor.
Thermistor resistance at 40C = 11.9 k. Using the voltage divider equation,
230
11.9 k VB = 15 V = 4.84 V 11.9 k 25 k VAB = VA VB = 7.5 V 4.84 V = 2.66 V
5.
If you needed to set the reference temperature to 0C, what simple change would you make to the circuit? Show with a calculation that your change will work and draw the revised schematic.
At 0, the resistance of the thermistor is 103.4 k. One option is to change R4 to 100 k, so that the bridge can be balanced at this lower temperature. 103.4 k 27 k VAB = VA VB = 15 V 103.4 k 100 k 3.4 k 15 V = 0 V 54 k The schematic is the same as Figure 7-1 except values. Review 6.
7.5V P 25k
2
2.25mW
7.
The thermistor resistance decreases as temperature increases as a result of the negative temperature coefficient.
8.
Maximum power for R1 and R2:
V 2 7.5V mW P1 P2 R 27k 2
Maximum power for R4, assuming R3 and RT are 0.
V 2 15V mW P4 R 20k 2
Yes, all the fixed resistors dissipate much less than 1/8 watt 9.
The output of the op-amp is either at its maximum positive level or its maximum negative level depending on the condition of the bridge. At the positive level, the red LED is on. At the negative level, the green LED is on.
231
Chapter 8 Application Activity
er 6
Temperature Measuring Circuit 1. Substituting into the equation for RT, calculate the resistance of the thermistor at a temperature of 50C (full-scale deflection of the meter). T0 T T0T
RT = R0 e
(25 k)e
298 K 323 K 4616K (298 K)(323 K)
= 7.54 k
At 50C, the resistance of the thermistor is 7.54 k. 2.
Thevenize the bridge between terminals A and B by keeping the ground reference and forming two facing Thevenin circuits. Assume the thermistor temperature is 50C and its resistance is the value calculated previously. Draw the Thevenin circuit for this temperature but do not show a load.
Figure 8-1 shows the Thevenin circuit with no load.
Figure 8-1 3.
Show the load resistance for the Thevenin circuit you drew in the last step. The load is a resistor in series with the ammeter, which will have a current of 50 A at full scale (50C). You can find the value of the required load resistance by applying the superposition theorem to the two sources and calculating the total resistance from Ohm’s law (using the full-scale deflection as the current). Subtract the Thevenin resistance of each arm from the total resistance to obtain the required load resistance. Neglect the meter resistance. Show the value calculated on the Thevenin circuit.
Use 4.71 V/50 A to determine RTOT and subtract Thevenin arm resistances to obtain RL. See Figure 8-2.
Figure 8-2
232
4.
Calculate the thermistor resistance for the lower and upper limits of temperature (30C) and 40C). Draw Thevenin circuits for each temperature and calculate the current through the load resistor.
At 30C, the thermistor resistance is 19.35 k. Figure 8-3 shows the Thevenin circuit.
Figure 8-3
At 40C, the thermistor resistance is 11.89 k. Figure 8-4 shows the Thevenin circuit.
Figure 8-4
Although not asked for specifically, the current as a function of temperature is nearly linear for this circuit because the nonlinearity of the Wheatstone bridge is nearly compensated for by the nonlinearity of the thermistor. The plot of meter current versus temperature is shown in Figure 8-5 by the solid line. The dashed line is a linear fit for comparison.
Figure 8-5
233
The Meter Scale 5. Indicate how you would mark the meter to have a quick visual indication of the temperature in the tank.
Answers will vary. One way is shown in Figure 8-6 when out of range is indicated as red bands.
Figure 8-6
Review T0 T T0T
298 K 308 K 4616K (298 K)(308 K)
(25 k)e = 15.1 k 7.5V 4.71V IM = 28.3 A 16.5k 71.6k 10.4k
6.
RT = R0 e
7.
RTOT
4.71V 47.1k 100A
RL = 47.1 k – (16.5 k + 6.14 k) = 24.5 k Reduce RL to 24.5 k
234
er 6 Chapter 9 Application Activity Analysis 1. Substitute the values that were given in Figure 9-25 into the standard form equations. Solve the equations to find VIN and VL (Resistance can be entered in k.) 1 1 1 1 1 VA VB 0.010 Node A: 10 300 1000 1000 10 = (0.10433)VA + (0.0010)VB = 0.0010
1 1 1 100,000 1 Node B: VA VB = 0 1000 0.010 1 0.010 1000 = +(1.0 107)VA + (101.001)VB = 0
VA = VIN = 10.11 V VB = VL = 1.001 V 2.
Calculate the input current, IIN, and the current in the feedback resistor, IF.
IIN = 1.00 A V VB 10.11 V (1.001 V) 1.00A IF = A 1 M RF Review 3. The output voltage is unaffected. 4.
The output voltage is doubled.
235
Chapter 10 Application Activity
hapter 6
System Interconnections 1. The system is connected according to the following point-to-point wiring list with reference to the components in text Figure 10-48. From Battery-I Relay Bd.-C Relay Bd.-E Mag. Sw.-M Mag. Sw.-O Battery-J Tog. Sw.-U Siren-T Relay Bd.-A Relay Bd.-F Mag. Sw.-N Mag. Sw.-P Relay Bd.-B Relay Bd.- G Relay Bd.-H
To Relay Bd.-C Relay Bd.-E Mag. Sw.-M Mag. Sw.-O Mag. Sw.-Q Tog. Sw.-U Siren-S Relay Bd.-D Relay Bd.-F Mag. Sw.-N Mag. Sw.-P Mag. Sw - R Tog. Sw.-V Wall Sw.-K Wall Sw.-L
A Test Procedure 2. A test procedure is as follows: 1. Turn the system ON/OFF switch to the ON position. 2. Turn the wall switch off. 3. Close the first magnetic switch to simulate the opening of a window or door. 4. Verify that the alarm and the house lights are on. 5. Turn the system ON/OFF switch to the OFF position. 6. Repeat procedures 1 through 5 for each remaining magnetic switches. Review 3. The detection switches, when closed, indicate an intrusion through a window or door. 4.
Contact B latches the relay and keeps it energized when intrusion is detected.
236
Chapter 11 Application Activity
hapter 6
Oscilloscope Measurements 1. 1 T = (10 div)(0.1 s/div) = 1 s 1 1 f= = 1 MHz T 1 s Vp = (1.5 div)(2 mV/div) = 3.0 mV Vrms = (0.707)(3.0 mV) = 2.12 mV 2
T = (6.8 div)(0.1 s/div) = 680 ns 1 1 f= = 1.47 MHz T 680 ns Vp = (1.6 div)(5 mV/div) = 8.0 mV Vrms = (0.707)(8.0 mV) = 5.66 mV
4
T = (10 div)(10 s/div) = 100 s 1 1 f= = 10 kHz T 100 s Vp = (1.2 div)(5 mV/div) = 6.0 mV Vrms = (0.707)(6.0 mV) = 4.24 mV
5
T = (5 div)(20 s/div) = 100 s 1 1 = 10 kHz f= T 100 s Vp = (1.8 div)(0.2 V/div) = 360 mV Vrms = (0.707)(360 mV) = 255 mV
6
T = (5 div)(20 s/div) = 100 s 1 1 f= = 10 kHz T 100 s Vp = (1.0 div)(0.5 V/div) = 500 mV Vrms = (0.707)(500 mV) = 354 mV
Amplifier Analysis 2.
From the oscilloscope measurements at points 4 and 5: The input voltage to the preamplifier is Vin = 4.24 mV rms The output voltage from the preamplifier is Vout = 255 mV rms The voltage gain is 255 mV V 60 Av = out 4.24 mV Vin
3.
The volume of the speaker is adjusted by applying a greater or lesser voltage to the audio power amplifier input. This is accomplished by the potentiometer acting as an adjustable voltage divider. The rms amplitude at the speaker is 354 mV.
237
Review 4. RF stands for radio frequency. 5.
IF stands for intermediate frequency.
6.
The carrier frequency is always higher than the audio frequency.
7.
The amplitude varies in an AM signal.
238
Chapter 12 Application Activity The Printed Circuit Board on the Schematic 1. The circuit board and schematic in Figure 12-59 agree. Testing Board 1 2. Oscilloscope measurements (Figure 12-60): Vdc = 5.1 DIV 1 V/DIV = 5.1 V Vpp = 2.8 DIV 1 V/DIV = 2.8 V
Vrms = 0.707Vp = 0.707(1.4 V) 1 V T = 2 DIV 0.1 ms/DIV = 0.2 ms 1 1 f= = 5 kHz T 0.2 ms The ac and dc voltages at the base should be: At 1 V rms signal is applied to the circuit board and coupled through C1 to the base. 27 k VB = 24 V = 5.1 V 127 k The frequency should be 5 kHz. The signal displayed on the scope is correct. Testing Board 2 3. Oscilloscope measurements (Figure 12-61): Vdc = 5.1 DIV 1 V/DIV = 5.1 V Vrms = 0 V The ac and dc voltages at the base should be: At 1 V rms signal is applied to the circuit board which should be coupled through C1 to the base. 27 k VB = 24 V = 5.1 V 127 k The frequency should be 5 kHz. The dc voltage at the base is correct but the signal voltage is missing. This indicates that the coupling capacitor C1 is open or there is no signal at the input terminal of the printed circuit board. Testing Board 3 4. Oscilloscope measurements (Figure 12-62): Vdc = 0 V Vp = 1.4 DIV 1 V/DIV = 1.4 V Vrms = 0.707Vp = 0.707(1.4 V) 1 V T = 2 DIV 2 s/DIV = 4 s 1 1 f= = 250 kHz T 4 s The ac and dc voltages at the base should be: At 1 V rms signal is applied to the circuit board and coupled through C1 to the base. 27 k VB = 24 V = 5.1 V 127 k
239
The ac voltage is correct but there is no dc level. This indicates that R1 is open or there is no dc supply voltage. Review 5. The coupling capacitor prevents the source impedance from affecting the dc voltage but passes the input signal. 6.
An ac voltage riding on a dc voltage is at point C. An ac voltage only is at the output.
240
Chapter 13 Application Activity
hapter 6
The Winding Resistance 1. These are given as RW = 85 and R = 10 k. V 10 V 10 V I= = 992 A RT 10 k 85 10.085 Inductance of Coil 1 2. Oscilloscope measurement (Figure 13-41) 5 = 7 DIV 0.5 s/DIV = 3.5 s L 3.5 s = = 0.7 s R 5 3. L (10 k)(0.7 s) = 7 mH Inductance of Coil 2 4. Oscilloscope measurement (Figure 13-42) 5 = 7 DIV 20 s/DIV = 140 s L 140 s = = 28 s R 5 5. L (10 k)(28 s) = 280 mH 6.
Measurement of the time constant from the scope display can lead to some inaccuracy.
7.
Using a sinusoidal input: 1. Apply a sinusoidal signal of known amplitude and frequency to the RL circuit. 2. Measure the voltage across the resistor. 3. Calculate the current: I = VR/R 4. Calculate the impedance magnitude: Z = VS/I 5. 6.
Calculate the inductive reactance: XL = Calculate the inductance: L = XL = 2f
Z 2 R2
Review 8. 5 = 2 s 0.5Tmin = 2 s Tmin = 4 s
f max 9.
5 = 80 s 0.5Tmin = 80 s Tmin = 160 s
f max 10.
1 1 250kHz Tmin 4s
1 1 6.25kHz Tmin 160s
If f > fmax, the inductor current would not reach the final value because 0.5T < 5.
241
Chapter 14 Application Activity
hapter 6
Measuring Voltages on Power Supply Board 1 1. The voltages on the meters in Figure 14-40 are correct. The power supply is working properly. Measuring Voltages on Power Supply Boards 2, 3, and 4. 2. Board 2: The 110 V ac is applied to the circuit but there is no voltage on the primary. The fuse is open Board 3: There are 10 V ac across the secondary but no dc voltage from the regulator. The circuit board is faulty (cannot be isolated to a component with these measurements). Board 4: There is no secondary voltage. The transformer has an open winding. Review 3. Use and ohmmeter to check for open windings. Shorted windings, which are rare, are indicated by an incorrect secondary voltage. 4.
A short will cause the fuse to blow.
242
Chapter 15 Application Activity
hapter 6
The Amplifier Input Circuit 1.
Rin = R1 R2 = 10 k 47 k =
1 = 8.2 k 1 1 10 k 47 k
The Response at Frequency f1 2. The channel 2 scope probe is connected to the amplifier input (point B in Figure 15-79(a)). The frequency is: T = 5 DIV 0.2 ms/DIV = 1 ms 1 1 f1 = = 1 kHz T 1 ms The scope is ac coupled so the dc voltage at point B is not displayed. The peak-to-peak voltage at point B that should be displayed on channel 2 is: 1 1 XC = = 1.59 k 2f1C 2 kHz)(0.1 F)
Rin V VB(pp) = in 2 2 Rin X C 8.2 k = (8.2 k) 2 (1.59 k) 2
1 V = 0.98 V 1 V
The Response at Frequency f2 3. The frequency is: T = 5 DIV 2 ms/DIV = 10 ms 1 1 f2 = = 100 Hz T 10 ms The scope is ac coupled so the dc voltage at point B is not displayed. The peak-to-peak voltage at point B that should be displayed on channel 2 is: 1 1 XC = = 15.9 k 2f 2C 2 Hz)(0.1 F)
Rin V VB(pp) = in 2 2 Rin X C 8.2 k = (8.2 k) 2 (15.9 k) 2 4.
1 V = 0.46 V
The ac voltage at point B is less at 100 Hz than at 1 kHz because the reactance has increased.
243
The Response at Frequency f3 5. The frequency is: T = 4 DIV 5 ms/DIV = 20 ms 1 1 f3 = = 50 Hz T 20 ms The scope is ac coupled so the dc voltage at point B is not displayed. The peak-to-peak voltage at point B that should be displayed on channel 2 is: 1 1 XC = = 31.8 k 2f 3C 2 Hz)(0.1 F)
Rin V VB(pp) = in 2 2 R X in C 8.2 k = (8.2 k) 2 (31.8 k) 2 6.
1 V = 0.25 V
The ac voltage at point B is less at 50 Hz than at 100 Hz because the reactance has increased.
Response Curve for the Amplifier Input Circuit 7. At the cutoff (critical) frequency, XC = Rin = 8.2 k 1 = 8.2 k 2f cC 1 1 fc = = 194 Hz 2X C C 2 k)(0.1 F) VB(pp) = 0.707(1 V) = 0.707 V 8.
The response curve is shown in Figure 15-1.
Figure 15-1 9.
The high pass characteristic of the input circuit is indicated by the increase in voltage as the frequency increases. The cutoff frequency can be decreased by increasing the value of the coupling capacitor.
244
10. Increase the value of C1. Review 11. A lower value of coupling capacitor will increase the frequency at which a significant drop in voltage occurs. 12.
R2 10k VB 18V= 18V 3.16Vdc 57k R1 R2
13.
VB = 10 V rms
With R1 open, there is no dc voltage at point B.
245
Chapter 16 Application Activity
hapter 6
Resistance Measurements of Module 1 1. The resistance measurements are: 1.610 k from input to ground 1.500 k from output to ground The larger resistance measurement has to be the series combination of the resistor and the winding resistance of the coil. The smaller resistance measurement has to be the resistor value. Therefore, the values are as follows: R = 1.50 k RW = 1.610 k 1.50 k = 110 See Figure 16-1.
RW = 110
Figure 16-1
AC Measurements of Module 1 2. The frequency of the input and ouput signals is T = 2.5 DIV 5 s/DIV = 12.5 s 1 1 f= = 80 kHz T 12.5 s The peak-to-peak voltages are Vin(pp) = 1 V (channel 2 on the scope) Vout(pp) 0.707 V (channel 1 on the scope) Since Vout is 70.7% of Vin the measurements are at the cutoff frequency. Therefore, XL = R = 1.5 k 2FcL = 1.5 k 1.5 k L= = 3.0 mH 2(80 kHz)
246
Resistance Measurements of Module 2 3. The resistance measurements are: 100.0 from output to ground 22.10 k from input to ground The smaller resistance measurement has to be the winding resistance of the coil. The larger resistance measurement has to be the resistor value plus the winding resistance: R = 22 k RW = 100 See Figure 16-2.
100
Figure 16-2
AC Measurements of Module 2 4. The frequency of the input and ouput signals is T = 3.2 DIV 0.1 ms/DIV = 0.32 ms 1 1 f= = 3.13 kHz T 0.32 ms The peak-to-peak voltages are Vin(pp) = 1 V (channel 2 on the scope) Vout(pp) 0.707 V (channel 1 on the scope) Since Vout is 70.7% of Vin the measurements are at the cutoff frequency. Therefore, XL = R = 22 k 2FcL = 22 k 22 k L= = 1.12 H 2(3.13 kHz) Review 5. You will measure 0 V on the output. 6.
You will measure the input voltage on the output.
247
Chapter 17 Application Activity
hapter 6
Capacitance in the Resonant Circuit 1. CTOT = CVaractor + C2 For f = 535 kHz: 1 CTOT = 2 = 88.5 pF 4 (535 kHz) 2 (1 mH) For f = 1605 kHz: 1 CTOT = = 9.83 pF 2 4 (1605 kHz)2 (1 mH) Since the varactor goes down to 5 pF and up to 200 pF, which ideally more than covers the required frequency range, capacitor C2 is only for “tweaking” at 1605 kHz. C2 CTOT CVaractor = 9.83 pF 5 pF = 4.83 pF. Therefore, set the nominal value of C2 to pF. 2.
At f = 535 kHz: CVaractor = CTOT C2 = 88.5 pF 3 pF = 85.5 pF At f = 1605 kHz: CVaractor = CTOT C2 = 9.83 pF 3 pF = 6.83 pF
Testing the Resonant Circuit 3.
General procedure: Tune the circuit by varying VED from the power supply. Vary the frequency of the function generator and monitor scope for maximum output to verify frequency set by tuned circuit. Point-to-point connections: Function generator A to 2 (ground clip to 3) Power supply B to Circuit board 3 Power supply C to Circuit board 1 Power supply D to Circuit board 3 Power supply E to Circuit board 4 Oscilloscope F to Circuit board 5 (ground clip to 3)
4.
Setting 1: V=2V CVaractor 150 pF CTOT = 150 pF + 3 pF = 153 pF 1 1 = 407 kHz fr = 2 LCTOT 2 mH)(153 pF) Setting 2: V=4V CVaractor 85 pF CTOT = 85 pF + 3 pF = 88 pF 1 1 fr = = 537 kHz 2 LCTOT 2 mH)(88 pF)
248
Setting 3: V=5V CVaractor 62 pF CTOT = 62 pF + 3 pF = 65 pF 1 1 fr = = 624 kHz 2 LCTOT 2 mH)(65 pF) Setting 4: V=6V CVaractor 42 pF CTOT = 42 pF + 3 pF = 45 pF 1 1 fr = = 750 kHz 2 LCTOT 2 mH)(45 pF) Setting 5: V=8V CVaractor 15 pF CTOT = 15 pF + 3 pF = 18 pF 1 1 fr = = 1190 kHz 2 LCTOT 2 mH)(18 pF)
Review 5. The AM frequency range is 535 kHz to 1605 kHz. 6.
The rf amplifier rejects all signals but the one from the desired station. It then amplifies the selected signal.
7.
A particular AM frequency is selected by varying the varactor capacitance with a dc voltage.
249
Chapter 18 Application Activity
hapter 6
Filter Measurement and Analysis 1.
Top left screen: 1 1 f1 = = 100 Hz (2 ms/Div)(5 Div) 10 ms V1 = (0.5 V/Div)(4 Div) = 2 V p-p Top center screen: 1 1 = 1 kHz f2 = (0.5 ms/Div)(2 Div) 1 ms V2 = (0.5 V/Div)(4 Div) = 2 V p-p Top right screen: 1 1 = 5 kHz f3 = (50 s/Div)(4 Div) 200 s V3 = (0.2 V/Div)(7 Div) = 1.4 V p-p Bottom screen: 1 1 f4 = = 50 kHz (10 s/Div)(2 Div) 20 s V4 = (20 mV/Div)(7.2 Div) = 144 mV p-p This is a low-pass filter and the idealized Bode plot is in Figure 18-1. 2 V peak-to-peak is the maximum output voltage and is the 0 dB reference level. The other dB levels are: 2V 1.4 V At f3: 20 log At f1: 20 log = 0 dB = 3.1 dB 2V 2V 2V 144 mV At f4: 20 log At f2: 20 log = 0 dB = 22.9 dB 2V 2V
Figure 18-1
2.
Top left screen: 1 1 f1 = = 100 Hz (2 ms/Div)(5 Div) 10 ms
250
V1 = (0.5 mV/Div)(5 Div) = 25 mV p-p Top center screen: 1 1 f2 = = 800 Hz (0.5 ms/Div)(2 Div) 1.25 ms V2 = (20 mV/Div)(7 Div) = 140 mV p-p Top right screen: 1 1 f3 = = 8 kHz (50 s/Div)(2.5 Div) 125 s Figure 18-1 V3 = (0.2 V/Div)(7 Div) = 1.4 V p-p Middle left screen: 1 1 f4 = = 10 kHz (20 s/Div)(5 Div) 100 s V4 = (0.5 V/Div)(4 Div) = 2 V p-p Middle right screen: 1 1 = 11.6 kHz f5 = (20 s/Div)(4.3 Div) 86 s V5 = (0.2 V/Div)(7 Div) = 1.4 V p-p Bottom screen: 1 1 = 116 kHz f6 = (2 s/Div)(4.3 Div) 8.6 s V6 = (20 mV/Div)(7 Div) = 140 mV p-p This is a band-pass filter and the idealized Bode plot is in Figure 18-2. 2 V peak-to-peak is the maximum output voltage and is the 0 dB reference level. The other dB levels are: 25 mV At f1: 20 log = 38 dB 2V 140 mV At f2: 20 log = 23.1 dB 2V 1.4 V At f3: 20 log = 3.1 dB 2V 2V At f4: 20 log = 0 dB 2V 1.4 V At f5: 20 log = 3.1 dB 2V 140 mV Figure 18-2 At f6: 20 log = 23.1 dB 2V
Review 3. The waveforms indicate that the output amplitude decreases with an increase in frequency as in a low-pass filter. 4.
The waveforms indicate that the output amplitude is maximum at 10 kHz and drops off above and below as in a band-pass filter.
251
Chapter 19 Application Activity
hapter 6
Filter Measurement and Analysis 1 1 1. f= = 20 kHz (10 s/Div)(5 Div) 50 s Circuit A: 1 XC1 = = 79.6 2(20 kHz)(0.1 F) 1 = 7.96 k XC2 = 2(20 kHz)(1000 pF) ZT = 100 + (j79.6 ) (12 k j7.96 k) 100 j79.6 = 12838.5 79.6 90 10 V = 6.2251.5 V p-p VC1 128 38.5 12 k 120 k 6.22 51.5 V = 5.18 V p-p VC1 Vout = 12 k j7.96 k 14.4 33.6 k Circuit B: 1 XC1 = = 36.2 2(20 kHz)(0.22 F) 1 XC2 = = 5.31 k 2(20 kHz)(1500 pF) ZT = 1 k + (j36.2 ) (12 k j5.31 k) 1 k j36.2 = 10 k 36.2 90 10 V = 36290 mV p-p VC1 1 0 k
12 k 120 k 362 90 mV = 36166.1 mV p-p VC1 Vout = 12 k j5.31 k 13.1 23.9 k The output voltage measured on the scope is Vout (1 V/div)(5 Div) = 5 V p-p This is approximately the circuit A output. 2.
Circuit A 1 = 15.9 kHz 2(100 )(0.1 F) 1 fc2 = = 13.3 kHz 2(12 k)(1000 pF) 15.9 kHz 13.3 kHz fcenter = 14.6 kHz 2 At 20 kHz, the circuit is operating well out of its pass band.
fc1 =
252
3.
At the center frequency of 14.6 kHz: XC1 = 109 , XC2 = 10.9 k Zth = 12 k [(j10.9 k) + (100 ) (j109 )] = 5.46 k j5.95 k The power is maximum for a load that is the complex conjugate of Zth: ZL = 5.46 k + j5.95 k
Review 4. From #1 Circuit B: Vout = 36166.1 mV p-p. 5.
1 1.59kHz 2(1 k)(0.1 F) 1 fc 2 8.84kHz 2(12 k)(1500 pF) f c1
fo
1.59 kHz 8.84 kHz 5.2kHz 2
253
Chapter 20 Application Activity Capacitor Values 1.
vthresh = 5 V 1 e t d / RC
3.5 V = 5 V (5 V)e t d / RC 1.5 V = (5 V)e t d / RC
1 .5 t / RC ln ln e d 5 t 1.2 = d RC t RC = d 1 .2 td C= (1.2)(47 k) Switch position A (specified delay time = 10 ms) 10 ms C1 = = 0.18 F (1.2)(47 k) td = 1.2(47 k)(0.18 F) = 10 ms Switch position B (specified delay time = 25 ms) 25 ms C2 = = 0.44 F (use 0.47 F) (1.2)(47 k) td = 1.2(47 k)(0.47 F) = 26.5 ms Switch position C (specified delay time = 40 ms) 40 ms C3 = = 0.71 F (use 0.68 F) (1.2)(47 k) td = 1.2(47 k)(0.68 F) = 38.4 ms Switch position D (specified delay time = 65 ms) 65 ms C4 = = 1.15 F (use 1.2 F) (1.2)(47 k) td = 1.2(47 k)(1.2 F) = 67.7 ms Switch position E (specified delay time = 85 ms) 85 ms C5 = = 1.51 F (use 1.5 F) (1.2)(47 k) td = 1.2(47 k)(1.5 F) = 84.6 ms
254
Circuit Connections 2 & 3. From 1 3 4 6 7 8 15 16 17 18 19 20 21 22
To 5 9 8 14 10 19 2 13 15 12 17 11 1 14
Test Procedure 4.& 5. For each switch position, the pulse waveform must be set so the pulse is equal to the delay time for that position and the frequency is set to a value to allow sufficient time between pulses for the capacitor to completely discharge. During the pulse, the output voltage should increase to 3.5 V for each switch setting and then decrease to 0 V before the next pulse. See Figure 20-1.
Figure 20-1
1. 2.
Set the function generator to produce a pulse waveform output. Adjust the frequency of the function generator to a value equal to or less than required for the maximum time delay and capacitor discharge time determined as follows:
maximum delay time = 85 ms time constant used to produce the maximum delay time: = RC = 47 k 1.5 F = 70.5 ms Tmax = td(max) + 5RC = 85 ms + 5(70.5 ms) = 437.5 ms 1 fmin = = 2.28 Hz 437.5 ms
255
3.
Adjust the duty cycle of the pulse waveform for each switch position according to the following table and verify on the scope that the output voltage reaches 3.5 V at the end of the pulse in each case.
Switch Position A B C D E
Delay Time 10 ms 25 ms 40 ms 65 ms 85 ms
Pulse Width 10 ms 25 ms 40 ms 65 ms 85 ms
Duty Cycle 2.28 % 5.69 % 9.11 % 14.8 % 19.4 %
Scope Sec/Div 1 ms 5 ms 5 ms 10 ms 10 ms
Review 6. A capacitor must be added and the switch changed to one with six positions. 7.
C6 =
100 ms = 1.77 F (use 1.8 F) (1.2)(47 k)
256
PART 3 Summary of Multisim Circuit Results Prepared by Gary Snyder
257
Multisim Results for Principles of Electric Circuits, 9th Ed. by Tom Floyd CHAPTER 3 Circuit File
Results V = 30 V:
E03-01
IT = 6.384 mA
V = 40 V:
IT = 8.512 mA R = 22 kΩ: E03-03
IT = 4.545 mA
R = 33 Ω:
IT = 3.03 mA R = 5.6 kΩ: E03-06
IT = 5.357 mA
R = 2.2 kΩ:
IT = 13.639 mA IT = 5.000 mA: E03-12
VR = 280 mV
IT = 1.500 mA:
VR = 49.5 mV
IT = 9.997 mA: E03-14
V = 33 V
IT = 25 mA:
V = 82.5 V
V = 150 V, I = 4.519 mA: E03-18
R = 33 kΩ V = 75 V, I = 1.108 mA: R = 68 kΩ
258
Circuit File
Results IT = 12.001 mA:
P03-40
No fault
IT = 888.178 nA:
RB is open
IT = 45.461 mA:
No fault RA = 560 kΩ RB = 2.2 MΩ RC = 1.8 kΩ RD = 33 Ω V=9V I = 1.915 mA
P03-41 P03-42 P03-43
V = 18 V, I = 5.453 mA: R = 3.3 kΩ V = 12 V R = 680 Ω I = 137.646 mA R is leaky
P03-44
CHAPTER 4 Circuit File P04-40 P04-41 P04-42
Results VR = 24 V, IT = 35.293 mA, R = 680 Ω: P = 846 mW VR = 5 V, IT = 5 mA, R = 1 kΩ: P = 25 mW P = 10 W V = 12 V I = 833.332 mA Current agrees CHAPTER 5
Circuit File
Results R4 = 100 Ω:
E05-07
R4 = 200 Ω:
IT = 19.28 mA IT = 17.987 mA
IT = 1.000 mA: E05-08
IT = 1.000 mA:
VS = 9.5 V VS = 7.8 V
259
Circuit File
Results IT = 1 mA:
E05-09A IT = 5 mA: E05-09B IT = 1 mA: E05-09C
E05-11 E05-12 E05-13A E05-13B IT = 10.004 mA: E05-14
IT = 9.999 mA:
VS = 9 V VR1 = 1 V VR2 = 3.3 V VR3 = 4.7 V VS = 45 V VR1 = 5 V VR2 = 16.5 V VR3 = 23.5 V VS = 6.5 V VR1 = 1 V VR2 = 3.3 V VR3 = 2.2 V VT(1) = 18 V VT(2) = 6 V VA = -12 V VB = +12 V VR1 = 5 V VR2 = 10 V VS = 15 V VS = 30 V VR1 = 10 V VR2 = 20 V R4 = 3.43 KΩ R4 = 930 Ω
R2 = 56 Ω: E05-16
E05-17
R2 = 180 Ω:
VR1 = 6.410 V VR2 = 3.590 V
VR1 = 3.571 V VR2 = 6.429 V R1 = 100 Ω, R2 = 220 Ω, R3 = 6.8 Ω: VR1 = 1.000 V VR2 = 2.200 V VR3 = 6.800 V R1 = 680 Ω, R2 = 680 Ω, R3 = 680 Ω: VR1 = 3.333 V VR2 = 3.333 V VR3 = 3.333 V
260
Circuit File
Results VS = 25 V:
E05-18
VS = 50 V:
VAB = 2.003 V VAC = 18.399 V VBC = 16.395 V VBD = 22.997 V VCD = 6.601 V VAB = 4.006 V VAC = 36.797 V VBC = 32.791 V VBD = 45.994 V VCD = 13.203 V
Ground = Point E:
Ground = Point D:
E05-22
Ground = Point C:
Ground = Point A:
P05-63 P05-64 P05-65 P05-66
VA = 100.000 V VB = 75.000 V VC = 50.000 V VD = 25.000 V VE = 0 V VA = 75.000 V VB = 50.000 V VC = 25.000 V VD = 0 V VE = -25.000 V VA = 50.000 V VB = 25.000 V VC = 0 V VD = -25.000 V VE = -50.000 V
VA = 0 V VB = -25.000 V VC = -50.000 V VD = -75.000 V VE = -100.000 V RT = 7.481 kΩ VR1 = 2.699 mV, VR2 = 17.990 V, VR3 = 5.935 mV, VR4 = 1.547 mV R2 is open. R3 = 22 Ω VS1 = 6 V
261
Circuit File
P05-67
Results VR1 = 2.927 pV VR2 = 7.902 V VR3 = 3.512 V VR4 = 9.659 V VR5 = 2.927 V R1 is shorted.
CHAPTER 6 Circuit File Results E06-03 R4 in circuit: VR1 = VR2 = VR3 = VR4 = VR5 = 25.000 V R4 removed from circuit: VR1 = VR2 = VR3 = VR5 = 25.000 V E06-12 R2 = 56 kΩ: IT = 2.786 mA R2 = 120 kΩ: IT = 1.833 mA IR1 = 1 mA E06-13 Original circuit: IR1 = 20 mA IR2 = 9.091 mA IR3 = 36 mA 910 Ω added to circuit: IR1 = 20 mA IR2 = 9.091 mA IR3 = 36 mA IR4 = 22 mA E06-14 R3 = 1 kΩ: IT = 10.000 mA, VS = 1.364 V R3 = 680 Ω: IT = 10 mA, VS = 1.282 V E06-18 IR1 = 0.032 A IR2 = 0.068 A IT = 0.100 A P06-52 RT = 547.97 Ω P06-53 IR1 = 2.398 mA, IR2 = 3.553 µA, IR3 = 1.087 mA, IR4 = 1.599 mA R2 is open. P06-54 R2 = 890 Ω P06-55 VS = 3.3 V P06-56 IR1 = 0.000 A IR2 = 6.667 µA IR3 = 15 µA R1 is open. 262
CHAPTER 7 Circuit File
Results Original circuit:
E07-08 4.7 kΩ added to circuit:
IR2 = 5.690 mA IR3 = 3.793 mA IR4 = 2.377 mA
R1 = 150 Ω: E07-10
R1 = 220 Ω:
VA = 52.093 V, VR1 = 27.907 V VA = 44.800 V, VR1 = 35.200 V VS = 2.035 V VR1 = VR2 = 0.687 V VR3 = 1.348 V VR4 = 0.981 V VR5 = 0.366 V
E07-11 R2 in circuit:
E07-12
E07-13
E07-14
E07-15
VR1 = VR2 = 4.694 V VR3 = 2.179 V VR4 = 1.127 V VR5 = 0.716 V VR6 = 0.411 R2 removed from circuit: VAB = 5.481 V VBC = 1.661 V VCD = 0.858 V No load: VOUT = 3.401 V RL = 10 kΩ: VOUT = 2.577 V RL = 100 kΩ: VOUT = 3.296 V RL = 1 MΩ: VOUT = 3.391 V IR3 = 0.886 mA IRL1 = 0.114 mA IRL2 = 0.055 mA R2 = 100 Ω: VOUT = 5.357 V R2 = 100 kΩ: VOUT = 5.323 V R2 = 1 MΩ: VOUT = 5.034 V
263
Circuit File
Results R1 = 1 kΩ:
E07-16
R1 = 2.2 kΩ:
IR1 = 8.846 mA IR2 = 4.409 mA IR3 = 4.437 mA IR4 = 2.151 mA IR5 = 2.286 mA IR6 = 2.287 mA VA = 36.152 V VB = 21.510 V VC = 10.751 V IR1 = 7.155 mA IR2 = 3.567 mA IR3 = 3.588 mA IR4 = 1.740 mA IR5 = 1.851 mA IR6 = 1.852 mA VA = 29.247 V VB = 17.399 V VC = 8.702 V
RV = 1.2 kΩ: E07-17
RX = 1.8 kΩ RV = 2.2 kΩ: RX = 3.3 kΩ VR3 = 9.6 V VR2 = VS = 24 V R2 is open. VR7 = 3.025 pV R7 is shorted. RT = 296.744 Ω IR4 = -1.554 pA R4 is open. R3 = 560 kΩ
E07-19 E07-20 E07-21 P07-57 P07-58 P07-59 No load:
P07-60
RL = 47 kΩ:
VR1 = 0.578 V VR2 = 3.931 V VR3 = 2.254 V VR4 = 3.237 V VR1 = 0.599 V VR2 = 4.071 V VR3 = 2.335 V VR4 = 2.996 V
264
Circuit File P07-61 P07-62
Results VR5 = 8.165 nV IR5 = 8.165 µA R5 is shorted. VOUT = -0.034 V for RV = 550 Ω
265
CHAPTER 8 Circuit File
E08-01
E08-06
Results RS = 10 Ω, RL = 100 Ω: VOUT = 90.909 V RS = 10 Ω, RL = 560 Ω: VOUT = 98.245 V RS = 10 Ω, RL = 1 kΩ: VOUT = 99.010 V RS = 50 Ω, RL = 10 kΩ: VOUT = 99.502 V Original circuit: VS1 currents: IR1 = 0.067 A IR2 = 0.033 A IR3 = 0.033 A VS2 currents: IR1 = 0.017 A IR2 = 0.017 A IR3 = -0.033 A IR2 total: 0.050 A VS2 polarity reversed: VS1 currents: IR1 = 0.067 A IR2 = 0.033 A IR3 = 0.033 A VS2 currents: IR1 = -0.017 A IR2 = -0.017 A IR3 = 0.033 A IR2 total: 0.017 A
266
Circuit File
Results Original circuit: VS1 current: VS2 current: Total:
E08-09
VS1 = -12 V: VS1 currents VS2 currents: Total:
IR3 = 3.704 mA IR3 = -2.778 mA IR3 = 925.926 µA IR3 = -2.222 mA IR3 = -2.778 mA IR3 = -5 mA
Original circuit: E08-10
560 Ω parallel resistor:
VTH = 4.083 V RTH = 1.408 kΩ VTH = 2.361 V RTH = 1.236 kΩ
Original circuit: E08-12A
E08-12B
E08-14
VRL = 3.479 V IRL = 3.479 mA R1 = 2.2 kΩ, R2 = 3.3 kΩ, R3 = 3.9 kΩ, R4 = 2.7 kΩ: VRL = 1.170 V IRL = 1.170 mA R1 = 2.2 kΩ, R2 = 3.3 kΩ, R3 = 3.9 kΩ, R4 = 2.7 kΩ: VRL = 1.170 V IRL = 1.170 mA Original circuit: IT = 0.063 A IN = 0.020 A R2 doubled: IT = 0.052 A IN = 0.025 A
267
Circuit File
Results RL = 0 Ω: RL = 25 Ω: RL = 50 Ω:
E08-18
RL = 75 Ω: RL = 100 Ω: RL = 125 Ω:
IT = 0.133 A IT = 0.100 A IT = 0.080 A IT = 0.067 A IT = 0.057 A IT = 0.050 A
Original circuit: E08-21
P08-39 P08-40 P08-41 P08-42 P08-43
VRC = -0.924 V IRC = -0.051 mA RA = 27 kΩ, RB = 33 kΩ, RC = 100 kΩ, RD = 39 kΩ, RE = 47 kΩ: VRC = -0.030 V IRC = -0.302 µA IR1 = 0.267 A IR2 = 0.101 A IR3 = 0.167 A Value of R2 shifted to 10 Ω VTH = 17.478 V RTH = 247.3 Ω IN = 383 µA RN = 9.674 kΩ VR2 = 3.180 pV R2 is shorted VRL = 3.432 V, IRL = 1.206 mA for RL = 2.847 kΩ CHAPTER 9
Circuit File
Results Original circuit:
E09-07
E09-08
IR1 = 13.898 mA IR2 = 15.765 mA IR3 = 1.867 mA Circuit with VS2 polarity reversed: IR1 = 17.185 mA IR2 = 8.742 mA IR3 = 8.443 mA IR1 = 13.898 mA IR2 = 15.765 mA IR3 = -1.867 mA 268
Circuit File
Results IT = 35.067 mA IR1 = 18.861 mA IR2 = 19.252 mA IR3 = 16.206 mA IR4 = 15.812 mA IRL = -391.867 µA IR1 = 511.591 µA IR2 = -373.035 µA IR3 = 79.492 µA IR4 = 455.177 µA IR5 = 432.554 µA VR1 = 7.675 V VR2 = 8.269 VR3 = -593.771 mV VR4 = 3.731 V VR5 = 4.325 V
E09-09
E09-10A
E09-10B Original circuit:
E09-11
E09-12
E09-13A E09-13B P09-34
P09-35
IR1 = 13.898 Ma IR2 = 15.765 mA IR3 = 1.867 mA Circuit with VS2 polarity reversed: IR1 = 17.185 mA IR2 = 8.742 mA IR3 = 8.443 mA VB = 5.776 V VC = 6.167 V IR1 = 18.86 mA IR2 = 19.252 mA IR3 = 16.204 mA IR4 = 15.812 Ma IRL = -391.687 µA Original circuit: VRL = 4.325 V VB = 3.731 V RL = 15 kΩ VRL = 5.02 V VB = 4.039 V IR1 = 137.967 mA IR2 = 159.797 mA IR3 = -21.831 mA IR1 = 297.873 mA IR2 = 264.889 mA IR3 = 247.648 mA IR4 = 17.241 mA
269
Circuit File
Results VA = -928.066 mV VB = -5.19 V IR4 = 0 A R4 is open. V1 = 4.939 V V2 = -2.878 VU2 = 14.743 V U2 is open. VR3 = 14.743 V R3 is open. IR4 = 888.178 nA R4 is open.
P09-36 P09-37 P09-38 P09-39 P09-40 P09-41
CHAPTER 10 No circuit files. CHAPTER 11 Circuit File
Results Original circuit: IT = 76.938 mARMS VR1 = 76.923 VRMS VR2 = 43.077 VRMS PT = 9.23 W
E11-13A VS = 10 VPEAK: E11-13B
E11-15A E11-15B P11-50
P11-51
IT = 4.534 mARMS VR1 = 4.533 VRMS VR2 = 2.538 VRMS PT = 32.1 mW Maximum = 19.998 V Minimum = 2.002 V Maximum = 15.998 V Minimum = -3.998 V VR1 = 76.923 VRMS VR2 = 43.077 VRMS VR1 = 108.786 VPEAK VR2 = 60.920 VPEAK VR1 = 5.911 VRMS VR2 = 13.005 VRMS VR3 = 5.084 VRMS VR1 = 8.360 VPEAK VR2 = 18.932 VPEAK VR3 = 7.189 VPEAK
270
Circuit File P11-52 P11-53 P11-54 P11-55 P11-56
Results VR2 = 24.0 VRMS R2 is open. IR1 = 12.5 mARMS IR2 = 4.545 mARMS IR1 = 0 A R1 is open. VR1 (maximum) = 19.998 V, VR1 (minimum) = 2.002 V AC = 9.998 VPEAK DC = 12.000 VDC VR1 (maximum) = 15.998 V, VR1 (minimum) = -3.998 V AC = 9.998 VPEAK DC = 6.000 VDC CHAPTER 12
Circuit File E12-12 E12-14 E12-17 E12-19 P12-55
P12-56
P12-57 P12-58 P12-59
Results VC (50 µs) = 22.894 V VC (15 µs) = 8.379 V VC = 75 V at t = 1.386 ms VC = 86.467 V at t = 2.000 ms VC = 50 V at t = 693 µs VC = 95.021 V at t = 3.000 ms XC = 33.9 kΩ f = 3.386 kHz for XC = 10 kΩ IT = 1.763 mARMS at 10 kHz IT = 4.455 mARMS at 25 kHz VC1 = 3.103 VRMS VC2 = 6.828 VRMS VC3 = 2.069 VRMS VC1 = 48.837 VRMS VC2 = 51.163 VRMS VC3 = 51.163 VRMS VC4 = 51.163 VRMS IT = 13.828 mARMS at 1 kHz IT = 6.913 mARMS at 100 Hz IT = 27.681 mARMS at 2 kHz IC4 = 39.635 nARMS C4 is open. VC1 = 35.951 nVRMS C1 is shorted.
271
CHAPTER 13 Circuit File
Results Original circuit:
E13-07
E13-08 E13-09
E13-11
E13-14
P13-38
P13-39
P13-40
IL = 6.256 mARMS at 8.3333 µs IL = 8.655 mARMS at 16.667 µs IL = 9.518 mARMS at 25.000 µs IL = 9.828 mARMS at 33.333 µs IL = 9.938 mARMS at 41.667 µs R = 680 Ω, L = 100 µH: IL = 6.237 mARMS at 147.059 ns IL = 8.668 mARMS at 294.118 ns IL = 9.531 mARMS at 441.177 ns IL = 9.828 mARMS at 588.235 ns IL = 9.938 mARMS at 735.294 ns IT = 158 µARMS at 0.1 ms f(max) = 220 kHz VR1(max) = 9.987 VRMS IL = 10.290 mARMS at 1 µs IL = 13.434 mARMS at 2 µs IL = 14.335 mARMS at 3 µs IL = 14.599 mARMS at 4 µs IL = 14.675 mARMS at 5 µs IL = 4.375 mARMS at 6 µs IL = 1.280 mARMS at 7 µs IL = 0.367 mARMS at 8 µs IL = 0.107 mARMS at 9 µs IL = 0.039 mARMS at 10 µs IL = 6.538 mARMS at 0.5 µs Original circuit: IT = 785.512 µARMS VS = 12 VRMS at 4.9 kHz, L1 = 680 mH: IT = 571.417 µARMS VL1 = 1.158 VRMS VL2 = 3.579 VRMS VL3 = 5.263 VRMS VL1 = 12.593 VRMS VL2 = 11.047 VRMS VL3 = 5.948 VRMS VL4 = 5.099 VRMS VL5 = 5.099VRMS IT = 15.864 mARMS at 10 kHz IT = 7.855 mARMS at 20 kHz IT = 31.805 mARMS at 5 kHz Inductive reactance is proportional to frequency
272
Circuit File
Results IL5 = 4.479 nARMS, VL5 = 4.475 VRMS L5 is open. IL2 = 0.437 ARMS, VL3 = 10.000 VRMS L2 is shorted.
P13-41 P13-42
CHAPTER 14 Circuit File E14-04 Original circuit:
Results VSEC = 359.964 VRMS
NSEC/NPRI = 4: E14-05
VSEC = 479.953 VRMS
Original circuit: NSEC/NPRI = 2.08333:
VSEC = 23.998 VRMS
VSEC = 57.594 VRMS VSEC = 7.499 VRMS NSEC/NPRI = 0.5 VPRI = 24 VRMS, IPRI = 12.731 mARMS VSEC = 9.504 nVRMS, ISEC = 4.998 pARMS Secondary winding is open. VR2 = 29.997 VRMS, IR2 = 30.023 nARMS R2 is open.
P14-35 P14-36 P14-37
CHAPTER 15 Circuit File
Results Original circuit:
E15-10 E15-12A E15-12B E15-12C E15-13A E15-13B E15-14A E15-14B E15-14C E15-15A
f = 5 kHz:
IT = 1.89 mA 64.8°
IT = 3.77 mA 33.1° Phase = -71.6° Phase = -23.1° Phase = -65.1° Phase lag increases as frequency increases. VOUT = 9.19 VRMS -23.1° VOUT = 4.21 VRMS -65.1° Output voltage amplitude decreases as frequency increases Phase = 34.1° Phase = 55.4° Phase = 16.1° Phase lead decreases as frequency increases. VOUT = 5.685 VRMS 55.4°
273
Circuit File E15-15B E15-18A E15-18B E15-21A E15-21B E15-21C E15-22A E15-22B E15-22C E15-22D E15-22E E15-22F E15-22G P15-71 P15-72 P15-73 P15-74
P15-75
P15-76 P15-77 P15-78
Results VOUT = 1.369 VRMS 82.1° Output voltage amplitude decreases as frequency decreases. IT = 4.994 mA 24.5° IT = 6.146 mA 42.4° IT = 6.71 mAPEAK 26.2° VZ1 = 7.037 VPEAK 8.67° VZ2 = 3.188 VPEAK -19.2° IT = 23.1 mAPEAK 67.2° IT = 33.8 mAPEAK 37.3° IT = 54.1 mAPEAK 48.2° VC1 = 1.815 VPEAK -22.8° VR1 = 762 mVPEAK 67.2° VC2 = 1.206 VPEAK -52.7° VR2 = 1.587 VPEAK 37.3° IT = 1.81 mARMS No fault. VR1 = 14.912 VRMS VC1 = 88.3651 mVRMS C1 is leaky. IC1 = 3.183 mARMS IR1 = 5.63 pARMS R1 is open. IC1 = 318.252 µARMS IR1 = 499.981 µARMS No fault. IT = 1.419 mARMS VZ1 = 3.835 VRMS VZ2 = 1.218 VRMS No fault. IT = 1.244 mARMS IC2 = 2.079 pARMS IR2 = 1.244 mARMS C2 is open. Low-pass filter, fC = 48.114 Hz High-pass filter, fC = 3.394 kHz CHAPTER 16
Circuit File E16-02A E16-02B E16-04A E16-04B
Results VS = 2.36 VPEAK 32.4° 420 µAPEAK -32.4° Phase = 71.6° Phase = 65.2°
274
Circuit File E16-04C E16-05A E16-05B E16-06A E16-06B E16-06C E16-07A E16-07B E16-10A E16-10B E16-12A E16-12B E16-13A E16-13B E16-13C E16-13D E16-13E P16-48 P16-49 P16-50 P16-51
P16-52 P16-53 P16-54 P16-55
Results Phase = 65.6° VOUT = 2.10 VRMS 65.2° VOUT = 3.78 VRMS 40.8° Phase = -18.4° Phase = -32.1° Phase = -32.0° VOUT = 8.46 VRMS -32.2° VOUT = 12.32 VRMS -52.1° IT = 14.04 mAPEAK -69.1° IT = 8.42 mARMS -56.8° IT = 14.04 mAPEAK -70.9° VOUT = 1.98 VRMS -10.5° VR1 = 4.65 VRMS -62.3° VC1 = 8.85 VRMS 27.7° VC1 = 6.23 VRMS -51.5° VC1 = 7.82 VRMS 38.5° IT = 20.2 mARMS -58.9° IT = 2.914 mARMS No fault. VR1 = 14.968 VRMS VL1 = 31.848 mVRMS L1 is leaky. IR1 = 999.984 µARMS IL1 = 1.047 mARMS No fault. IR1 = 12.82 mARMS IL1 = 5.571 pARMS L1 is open. IT = 13.737 mARMS IL2 = 13.737 mARMS IR2 = 15.959 pARMS R2 is open. IT = 115.047 µARMS No fault. Low-pass filter, fC = 15.878 MHz High-pass filter, fC = 52.649 kHz CHAPTER 17
Circuit File E17-05A E17-05B
Results f0 = 331.131 kHz f0 = 22.44 kHz
275
Circuit File E17-06A E17-06B E17-11A E17-11B E17-12 E17-18A E17-18B E17-18C P17-44 P17-45
P17-46
P17-47
P17-48 P17-49 P17-50 P17-51
Results IT = 3.333 mARMS VL1 = 3.376 VRMS VC1 = 3.29 VRMS VR1 = 4.999 VRMS Phase = -45.6° f0 = 2.323 kHz Q = 14.6 No measurable change in f0 with smaller winding resistance. f0 decreases with larger winding resistance. VC1 = 1.857 VPEAK -68.4° C1 = 47 pF: BW = 8 kHz BW = 1.589 kHz C1 = 1000 pF: BW = 7.95 kHz IT = 1.316 mARMS No fault. VC1 = 1.034 VRMS VL1 = 3.363 VRMS VR1 = 10.566 VRMS C1 is leaky. IR1 = 5.591 pARMS IC1 = 3.183 mARMS IL1 = 785.511 µARMS R1 is open. IL1 = 15.916 mARMS IR1 = 30.302 mARMS IC1 = 20.963 mARMS C1 is leaky. VL1 = 3 VRMS IL1 = 3.006 nARMS L1 is open. IT = 7.71 mARMS No fault. f0 = 503.535 kHz f0 = 339.625 kHz CHAPTER 18
Circuit File E18-04A E18-04B E18-05
Results Low-pass filter with fC = 74.321 kHz, -20 dB/decade roll-off Low-pass filter with fC = 349.31 kHz, -20 dB/decade roll-off High-pass filter with fC = 10.297 kHz, -20 dB/decade roll-off
276
Circuit File E18-06A E18-06B E18-08A E18-08B E18-09A E18-09B E18-11A E18-11B E18-12A E18-12B P18-30
P18-31
P18-32
P18-33
P18-34 P18-35 P18-36 P18-37
Results C1 = 234 nF VOUT = 4.892 VPEAK 5.58° @ 10 kHz C1 = 732 nF VOUT = 4.892 VPEAK 5.58° @ 10 kHz VOUT @ f0 = 9.09 VRMS BW = 17.468 kHz BW = 18.746 kHz f0 = 5.033 MHz f0 = 1.592 MHz VOUT @ f0 = 4.69 mVRMS (should be 3.45 mV) BW = 107 Hz (should be 92 Hz) VOUT @ f0 = 15.4 mVRMS BW = 104 Hz f0 = 5.77 MHz VOUT = 1.185 V minimum VOUT = 9.859 V maximum VOUT @ f0 = 1.94 VRMS IT = 13.868 mARMS IC1 = 11.05 pARMS IC2 = 13.868 mARMS C1 is open. IT = 0.086 ARMS IC1 = 6.365 mARMS IR1 = 0.086 ARMS IC2 = 0.075 ARMS IR2 = IC3 = 0.022 ARMS C2 is leaky. IT = 6.41 mARMS IR3 = 0.114 pARMS IC1 = 6.41 mARMS R3 is open. IT = 7.574 mARMS VC1 = 7.575 pVRMS VR1 = VR2 = 10 VRMS C1 is shorted. IR2= IL2 = 5.374 NARMS VL1 = 5.369 VRMS L2 is open. IT = 7.997 mARMS No fault. f0 = 107.4 kHz BW = 87.9 MHz
277
CHAPTER 19 Circuit File E19-01A E19-01B E19-03A E19-03B E19-04A E19-04B E19-05A E19-05B E19-06A E19-06B E19-16A
E19-16B P19-20
P19-21
P19-22
P19-23
P19-24 P19-25
Results IR1 = 7.626 mAPEAK 26.1° IR1 = 2.686 mAPEAK 26.4° IRL = 4.999 mADC IRL = 2.396 mAPEAK 47.3° IRL = 2.999 mADC IRL = 2.396 mAPEAK 47.3° VTH = 11.2 VPEAK 63.4° VTH = 18.2 VPEAK 43.2° VTH = 4.743 VPEAK -18.4° VTH = 4.031 VPEAK -36.3° VTH = 3.536 VPEAK 45.0° VTH = 4.550 VPEAK 24.4° f = 10 kHz: 439.4 µW f = 30 kHz: 8.949 mW f = 50 kHz: 2.379 W f = 80 kHz: 10.37 mW f = 100 kHz: 4.424 mW Z* = 47 Ω + j 72.3 Ω L1 = 115 µH VC2 = VR2 = 4.764 VRMS IR2 = 0 A R2 is open. IT = 51.819 mARMS IC1 = 51.54 mARMS IC2 = IR2 = 282.836 µARMS C1 is leaky. IT = 185.169 µARMS IC1 = 5.683 pARMS IR2 = 185.169 µARMS C1 is open. VC1 = 8.696 VRMS VR1 = 3.363 VRMS VR2 = 8.407 VRMS VL1 = 4.354 VRMS IR3 = 9.967 mARMS No fault. VTH = 750.5 mVPEAK 82.0° ZTH = 11.97 kΩ 4.31° IN = 171.6 µAPEAK -10.6° ZN = 71.9 kΩ 18.7°
278
CHAPTER 20 Circuit File E20-01A E20-01B E20-03 E20-04A E20-04B E20-05A E20-05B E20-07A E20-07B E20-08A E20-08B E20-10A E20-10B P20-29 P20-30 P20-31 P20-32
Results VOUT = 6.324 V maximum VOUT = 42.447 mV @ 600 µs VOUT = 8.649 V maximum VOUT = 399.518 µV @ 1.2 ms Pulse 1: Begin = 0 V, End = 906 mV Pulse 2: Begin = 672 mV, End = 1.456 V Pulse 3: Begin = 1.079 V +5V input: +5V inductive pulse decaying to 0 V 0 V input: -5V inductive pulse decaying to 0 V +5V input: +5V inductive pulse decaying to 0 V 0 V input: -5V inductive pulse decaying to 0 V Pulse: Begin = +25.0 V, End = +2.052 V Decay: Begin = -22.95 V VOUT @ 15 ms = -155.2 mV Pulse: Begin = +25.0 V, End = +892 mV Decay: Begin = -24.11 V VOUT @ 15 ms = -30.68 mV +10V input: RL energizing curve from 0 V to 10 V 0 V input: RL de-energizing curve from 10 V to 0 V VR1 = 19.988 V minimum VOUT = 6.094 V maximum VOUT = 55.4 mV @ 6 ms R1 = 2.5 kΩ Pulse: Begin = +25.0 V, End = +1.821 V Decay: Begin = -22.07 V VOUT @ 10 µs = -131.6 mV R1 = 20 kΩ VOUT = VIN C1 is open. VOUT is expected RC charge/discharge waveform. No fault. Time constant ≈ 1 ms IR1 = 4.296 pARMS R1 is open. Time constant ≈ 200 µs IL2 = 9.11 pARMS L2 is open. CHAPTER 21 No circuit files.
279
PART 4 Test Item File
Principles of Electric Circuits Chapter 1: Quantities and Units 1)
The unit of measurement for conductance is the coulomb. (True/False)
2)
Kilo equals 1,000 times the base unit. (True/False)
3)
Inductors store energy in an electrostatic field. (True/False)
4)
An electronic device which stores an electric charge is known as an inductor. (True/False)
5)
The symbol μ is an abbreviation for 10-6 or micro. (True/False)
6)
Which of the following are common applications of electronics? A) Computers B) Communications systems C) Automation D) Consumer products E) All of the above
7)
The symbol for current is: A) C B) I C) A D) V
8)
The unit of measurement for current is the: A) Volt B) Ampere C) Watt D) Ohm
9)
The symbol for voltage is: A) C B) R C) A D) V
10)
The unit of measurement for voltage is the: A) Volt B) Ampere C) Watt D) Ohm
11)
The symbol for a resistor is: A) C B) R C) A D) V
282
12)
The shortcut symbol for ohms is: A) α B) Ω C) δ D) .
13)
The unit of measurement for resistance is the: A) Volt B) Ampere C) Watt D) Ohm
14)
Which of the following is NOT a typical electrical component? A) Resistor B) Capacitor C) Megatron D) Inductor E) Transformer
15)
An ohmmeter is used for measuring: A) current B) resistance C) voltage D) power
16)
Which of the following metric prefixes is NOT commonly used in electronics work? A) kilo B) milli C) tera D) micro E) pico
17)
Express the number 10,000 in proper scientific notation. A) 10.0 × 103 B) 100.0 × 102 C) 1.0 × 104 D) 1.0 × 103
18)
Convert 4.7 mA to amperes. A) 0.00047 A B) 0.0047 A C) 4,700 A D) 47,000 A
19)
Convert 120 mW to W. A) 120,000 W B) 1,200 W C) 0.00012 W D) 0.12 W
283
20)
Convert 10,000 ohms to kΩ. A) 1 kΩ B) 10 kΩ C) 100 kΩ D) 1000 kΩ
21)
Convert 75 μV to mV. A) 7500 mV B) 75,000 mV C) 0.075 mV D) 0.000075 mV
22)
Convert 5.7 mW to μW. A) 0.057 μW B) 0.00057 μW C) 57,000 μW D) 5,700 μW
23)
Convert 6.8 × 10-5 W to the closest standard metric prefix. A) 68 μW B) 0.68 μW C) 680 μW D) 6.8 μW
24)
Convert 3.95 × 10-4 A to the closest standard metric prefix. A) 3.95 mA B) 39.5 mA C) 0.395 mA D) 395 mA
Convert the following: 25) 2 × 10-3 Amp = ________ A) 2 microamps B) 2 amps C) 2 milliamps D) 0.5 milliamps 26)
4.7 kΩ = ________ A) 4.7 × 10-3 Ω B) 4.7 × 103 Ω C) 47 × 10-3 Ω D) 4.7 × 10-4 Ω
27)
3.9 kΩ = ________ A) 39 × 10-3 Ω B) 3.9 × 105 Ω C) 3.9 × 103 Ω D) 3.9 × 10-4 Ω
284
28)
980 microvolts = ________ A) 9.80 millivolts B) 98 × 103 V C) 980 × 10-3 V D) both A and C E) none of the above
29)
2.2 kV = ________ A) 2,200 Volts B) 22 × 103 V C) 2.2 × 10-3 V D) 2.2 × 10-4 V
30)
Siemens is a unit for: A) resistance B) conductance C) power D) voltage
31)
The ________ is a device that magnetically couples ac voltages from one point in a circuit to another. A) dc power supply B) capacitor C) transformer D) inductor
32)
A resistor is a device that ________. A) opposes the flow of electrons B) stores power C) is a type of semiconductor D) stores electrostatic charge
33)
The shorthand method that uses a base number between 1 and 10 is called: A) decimal B) scientific notation C) prefix D) engineering notation
34)
The symbol for power is: A) Z B) W C) Q D) P
35)
An oscilloscope is used primarily for measuring ________. A) current B) ac voltage C) power D) conductance
285
36)
Express 0.004730 = ________, ________. A) 4.73 × 10-3, 4.73 milli B) 4.73 × 10-6, 4.73 micro C) 473 × 10-3, 4.73 milli D) M473 × 103, .473 milli
37)
Express 5.6 × 10-2 in milli, basic units, and micro. A) 5.6 milli, 0.056, 56000 micro B) 56 milli, 0.056, 56000 micro C) 560 milli, 5.600, 5600 micro D) 5600 milli, 56, 560 pico
38)
Multiply (99.2 × 10-6)(48 × 101) = ________, ________. A) 476 × 10-3, 47.6 micro B) 4.76 × 10-4, 47.6 milli C) 4.76 × 10-2, 47.6 milli D) 4.76 × 10-2, 47.6 nano
39)
Add (430 × 106) + (9.75 × 108) = ________, ________. A) 1.4 × 109, 1.4 Giga B) 14 × 109, 1.4 Giga C) 14 × 109, 1.4 Giga D) 1.4 × 109, 1.4 Mega
40)
Subtract (3462 × 100) - (2.22 × 102) = ________, ________. A) 3.24 × 103, 3.24 milli B) 3.24 × 103, 3.24 kilo C) 3.24 × 102, 3.24 kilo D) 3.24 × 104, 3.24 kilo
41)
Divide A) B) C) D)
42)
65 103 = 2.3 102 2.83 × 10-3, 283 micro 2.83 × 10-5, 283 micro 2.83 × 10-4, 283 micro 2.83 × 10-2, 283 micro
Convert 4,600,000 Ω to Mega Ω. A) 4.6 Mega Ω B) 460 Mega Ω C) 4600 Mega Ω D) 46 Mega Ω
286
43)
2 μF = ________ A) 2000 nF B) 200 pF C) 2 × 10-6 F D) both A and C
44)
Express the number 51,000,000,000 in proper scientific notation. A) 5.1 × 1010 B) 51 × 108 C) 5.1 × 109 D) 5.1 × 1011
45)
The SI system is A) based on a system of fundamental units B) used for engineering work C) used for scientific work D) an international system E) all of the above
287
1)
FALSE
24)
C
2)
TRUE
25)
C
3)
FALSE
26)
B
4)
FALSE
27)
C
5)
TRUE
28)
E
6)
E
29)
A
7)
B
30)
B
8)
B
31)
C
9)
D
32)
A
10)
A
33)
B
11)
B
34)
D
12)
B
35)
B
13)
D
36)
A
14)
C
37)
B
15)
B
38)
C
16)
C
39)
A
17)
C
40)
B
18)
B
41)
C
19)
D
42)
A
20)
B
43)
D
21)
C
44)
A
22)
D
45)
E
23)
A
288
Principles of Electric Circuits Chapter 2: Voltage, Current, and Resistance 1)
The smallest particle of an element that still retains the characteristics of the element is called an atom. (True/False)
2)
Electrons are negatively charged particles, and are contained in the nucleus of the atom. (True/False)
3)
In the neutral state all atoms contain the same number of protons and electrons. (True/False)
4)
The outermost shell of an atom is called the valence shell. (True/False)
5)
The copper atom contains 29 free electrons. (True/False)
6)
The nucleus contains protons and electrons, each with an opposite charge. (True/False)
7)
The process of gaining a valence electron is known as ionization. (True/False)
8)
Valence electrons are tightly bound in insulators. (True/False)
9)
Current value depends on the amount of charge moving past a point in a unit of time. (True/False)
10)
The movement of electrons between two points is called voltage. (True/False)
11)
The best conductor in the following list is: A) Silver B) Copper C) Silicon D) Germanium
12)
The materials with the fewest free electrons are classified as: A) conductors B) semiconductors C) insulators D) silly putty
13)
The unit of electrical charge is measured in: A) amperes B) free electrons C) positive or negative charges D) Coulombs
14)
How is a negative ion produced? A) the atom loses an electron from the valence shell B) the atom gains a proton in the nucleus C) the atom gains an electron in the nucleus D) the atom gains an electron in the valence shell
289
15)
A voltage source contains which of the following? A) potential energy B) potential difference C) electromotive force D) all of the above
16)
Which of the following is NOT a voltage source? A) battery B) semiconductor C) solar cell D) generator
17)
The rate of flow of free electrons in a conductive material is called: A) ampere B) current C) voltage D) coulomb
18)
Five coulombs of charge flow past a point in 0.5 seconds. How many amperes of current are flowing? A) 5A B) 10 A C) 15 A D) 20 A
19)
The opposition to current is called: A) resistance B) conductance C) insulators D) ohms
20)
The typical resistor to use for higher power applications would be the: A) carbon-composition B) carbon film C) metal oxide D) metal film E) wire wound
21)
A resistor coded with Red-Violet-Black-Gold bands respectively, would be of what value and tolerance? A) 270 Ω +/- 5% B) 27 Ω +/- 5% C) 2.7 Ω +/- 5% D) 0.27 Ω +/- 5%
22)
A potentiometer is a: A) variable resistor B) voltage-control device C) 3-terminal device D) all of the above
290
23)
Which of the following is NOT needed to form an electric circuit? A) voltage source B) load C) switch D) conductive path for current
24)
To measure current with an ammeter, connect the ammeter: A) across the voltage source B) across the resistance C) in the current path D) across the load
25)
DMM's are the most widely used type of electronic measuring instrument because of: A) better accuracy B) greater ease of reading C) greater reliability D) more functions provided E) all of the above
26)
Determine the tolerance and resistance of a resistor labeled 470KJ. A) 47 Ω +/- 5% B) 470 kΩ +/- 5% C) 47 kΩ +/- 5% D) 4700 Ω +/- 5%
27)
A rheostat is a: A) variable resistor B) voltage-control device C) 3-terminal device D) all of the above
28)
What is the maximum resolution of a 3 1/2-digit DMM when measuring volts? A) 0.1 V B) 0.01 V C) 0.001 V D) 0.0001 V
29)
The adjustable contact of a 1 k linear potentiometer is set at 3/4 of full rotation from the lowerend terminal. What is the resistance between the adjustable contact and the upper-end terminal? A) 1 kΩ B) 750 Ω C) 250 Ω D) 500 Ω
30)
If you need a 5.1 Ω resistor with +/- 5% tolerance, what color code should you look for? A) BLU-BRN-BL-GOLD B) GRN-BRN-BLK-GOLD C) GRN-BRN-GOLD-GOLD D) GRN-BLK-GOLD-GOLD
291
31)
The negative and positive charge symbols are assigned (in that order) to the: A) proton and electron B) electron and proton C) atom and nucleus D) electron and element
32)
Valence electrons of an atom determine the: A) electrical stability only B) atomic number C) chemical and electrical stability D) chemical stability only
33)
What is the most commonly used conductor in electronics? A) aluminum B) copper C) gold D) silver
34)
A conductor has 3.12 x 1020 electrons pass through it. How much charge is this? A) 3.12 coulombs B) 20 coulombs C) 50 coulombs D) 62.4 coulombs
35)
The capacity of a battery cell is measured by the amount of ________ that can be supplied over time. A) current B) coulomb C) voltage D) joules
36)
If a fluid system is compared to an electrical system, the fluid pump will correspond to a: A) generator B) conductor C) battery D) both A and C
37)
The force that causes conductor electrons to move is: A) conductance B) voltage C) current D) resistance
38)
Current equals: coulombs A) time B) coulombs × time voltage C) time D) voltage × time
292
39)
Electrical current is defined as the movement of which subatomic particles? A) neutrons B) electrons C) protons D) molecules
40)
Siemens is a unit for: A) conductance B) current C) voltage D) resistance
41)
__________ would NOT increase the resistance of the wire. A) A lower AWG size wire B) A smaller diameter wire C) Winding pattern D) Use in a hotter environment
42)
Which statement is NOT true? A) A potentiometer can be used as a rheostat. B) A rheostat can be used as a potentiometer. C) A potentiometer is a three terminal device. D) A rheostat is a two terminal device.
43)
The fifth band in a precision resistor represents the: A) failure rate B) reliability C) tolerance D) multiplier
44)
A green fifth-band on a precision resistor indicates a tolerance value of ±: A) 0.5% B) 0.25% C) 1% D) 2%
45)
What is the tolerance of a resistor manufactured to a minimum value of 5445 ohms and a maximum value of 5555 ohms? A) 1% B) 2% C) 10% D) 20%
46)
What is the resistance value and tolerance with color bands of gray, red, black, and gold? A) 82 ohms, ± 5% B) 82 ohms, ± 10% C) 820 ohms, ± 5% D) 820 ohms, ± 10%
293
47) The switch shown is a: A) DPST B) SPST C) SPDT D) DPDT 48)
Ground is the reference point in electric circuits and has a potential of ________ with respect to other points in the circuit. A) 0 volts B) source voltage C) an electron D) none of the above
49)
In Bohr’s model, the ________ level is called the ground state and represents the ________ atom with a single electron in the first shell. A) lowest, least stable B) highest, most stable C) lowest, most stable D) lowest, least stable
50)
Which is NOT one of the three categories of materials used in electronics? A) conductors B) crystals C) semiconductors D) insulators
51)
Which of the following is a true statement? A) The more collisions, the more easier the flow of electrons. B) Collisions do not cause electrons to lose some of their energy. C) The property of a material to oppose current flow is called work. D) The more collisions, the more restrictive the flow of electrons.
52)
Which of the following is a true statement? A) Work is the product of voltage (V) and charge (Q). B) Voltage is charge (Q) divided by energy (W). C) Voltage is not determined by charge. D) Voltage is not determined by work.
53)
Voltage can be produced by means of : A) chemical energy B) light energy C) magnetic energy combined with mechanical motion D) all of the above
54)
Which of the following is NOT a true statement? A) Batteries are usually classified according to their chemical makeup. B) Flat, round batteries are often called button or coin batteries. C) AAA batteries are physically larger than AA batteries. D) A lead-acid battery is a secondary (rechargeable) battery.
294
55)
Transistors used as switches A) Can open and close a circuit path. B) Are often called semiconductor switches. C) Can be controlled by voltage D) all of the above E) none of the above
56)
Resistance of the human body averages between A) 50K ohms and 100K ohms B) 10K ohms and 50K ohms C) zero ohms and 100K ohms D) 90K ohms and 100K ohms
295
1)
TRUE
20)
E
39)
B
2)
FALSE
21)
B
40)
A
3)
TRUE
22)
D
41)
A
4)
TRUE
23)
C
42)
B
5)
FALSE
24)
C
43)
C
6)
FALSE
25)
E
44)
A
7)
FALSE
26)
B
45)
A
8)
TRUE
27)
A
46)
A
9)
TRUE
28)
C
47)
C
10)
FALSE
29)
C
48)
A
11)
A
30)
C
49)
C
12)
C
31)
B
50)
B
13)
D
32)
C
51)
D
14)
D
33)
B
52)
A
15)
D
34)
C
53)
D
16)
B
35)
A
54)
C
17)
B
36)
D
55)
D
18)
B
37)
B
56)
B
19)
A
38)
A
296
Principles of Electric Circuits Chapter 3: Ohm's Law 1)
The voltage across a circuit resistance is inversely related to the current through that resistance. (True/False)
2)
Increasing voltage while maintaining the same resistance will cause current to decrease. (True/False)
3)
A change in voltage will result in a linearly proportional change in current. (True/False)
4)
A mathematical relationship between current, voltage, and resistance was determined by George Ohm? (True/False)
5)
A short could best be described as a unintended path for voltage. (True/False)
Figure 3-1 6)
Given the circuit in Figure 3-1, how many amps of current flow through the circuit? A) 1.0 A B) 1.22 A C) 0.82 A D) 8.2 A
Figure 3-2 7)
Given the circuit in Figure 3-2, how much current is flowing through the circuit? A) 3.84 A B) 0.26 mA C) 3.84 mA D) 3.84 μA
297
Figure 3-3 8)
Given the circuit in Figure 3-3, how much current flows through the circuit? A) 3.84 A B) 0.26 mA C) 3.84 mA D) 3.84mA
Figure 3-4 9)
Given the circuit in Figure 3-4, how much current flows through the circuit? A) 1.5 A B) 1.5 mA C) 0.67 mA D) 0.69 A
10)
To obtain voltage in volts, you must express the value of I in ________ and the value of R in ________. A) Milliamperes, Ohms B) Amperes, Ohms C) Amperes, Kilohms D) Milliamperes, Megohms
Figure 3-5 11)
Given the circuit in Figure 3-5, What voltage is required to produce a current of 2.0 amps? A) 50 V B) 12.5 V C) 5.0 V D) 1.25 V
298
Figure 3-6 12)
Given the circuit in Figure 3-6, What is the voltage across the resistor? A) 50 V B) 2.0 mV C) 50 mV D) 2.0 V
Figure 3-7 13)
Given the circuit in Figure 3-7, what is the battery voltage? A) 110 V B) 110 mV C) 22.7 V D) 110 μV
Figure 3-8 14)
Given the circuit in Figure 3-8, what is the voltage across the resistor? A) 1.42 V B) 15.5 V C) 15.5 mV D) 51.2 V
15)
To obtain the resistance in ohms, you must express the value of I in ________ and the value of V in ________. A) Amperes, Millivolts B) Milliamperes, Volts C) Amperes, Volts D) Microamperes, Millivolts
299
Figure 3-9 16)
Given the circuit in Figure 3-9, what resistance is required to draw the specified current? A) 5.7 Ω B) 25.2 Ω C) 5.7 kΩ D) 11.4 Ω
Figure 3-10 17)
Given the circuit in Figure 3-10, what is the value of the resistor? A) 121 Ω B) 2.69 Ω C) 4.78 kΩ D) 2.69 kΩ
Figure 3-11 18)
Given the circuit in Figure 3-11, what is the value of the resistor? A) 1Ω B) 1 kΩ C) 100 kΩ D) 1 MΩ
300
Figure 3-12 19)
Given the circuit in Figure 3-12, what is the value of the resistor? A) 125 Ω B) 125 kΩ C) 800 kΩ D) 8 MΩ
20)
If the voltage across a resistor has tripled, the current will: A) stay the same B) double C) triple D) decrease
Figure 3-13 21)
Given the circuit in Figure 3-13 what resistance is required to draw the specified current? A) 120 Ω B) 12 kΩ C) 120 kΩ D) 11.2 kΩ
Figure 3-14 22)
Given the circuit in Figure 3-14 what is the value of the resistor? A) 12 Ω B) 120 Ω C) 12 kΩ D) 120 kΩ
301
23)
Given the circuit in Figure 3-15, what is the circuit current? A) 0.0001 A B) 0.001 A C) 0.01 A D) 0.00001 A
Figure 3-16 24)
Given the circuit in Figure 3-16, what is the circuit current? A) 10 μA B) 100 μA C) 1 μA D) 1 mA
25)
If the voltage in a circuit has doubled, the current will: A) stay the same B) double C) triple D) decrease
26)
How is Ohm's law expressed in words? A) The current is linearly proportional to the resistance and inversely proportional to the voltage. B) The voltage is linearly proportional to the resistance and inversely proportional to the current. C) The current is linearly proportional to the voltage and inversely proportional to the resistance. D) The voltage is linearly proportional to the current and inversely proportional to the resistance.
27)
A battery applies 45 volts to a circuit, while an ammeter reads 10 mA. Later, the ammeter reading drops to 7 mA. Assuming the resistance has not changed, the voltage must have: A) increased to 58.5 volts B) decreased to 31.5 volts C) decreased 30% from its old value D) both B and C are correct E) stayed the same
302
28)
If a variable voltage source is set to 12 volts. To what voltage must it be adjusted to decrease the circuit current by 25%? A) Increase it to 16 volts. B) Increase it to 20 volts. C) Decrease it to 9 volts. D) Decrease it to 6 volts.
29)
A certain resistor has a color code of (Red/Red/Red/Gold). If this resistor were connected across the terminals of a 12 volt car battery, what current would flow through the resistor? A) 5.19 mA B) 5.45 mA C) 5.74 mA D) 6.0 mA
30)
You found a real bargain on a dashboard fan at a flea market. When you got home, you discovered it was a 6 volt fan that would draw 500 mA at 6 V. Your car has a 12 V system. You need a resistor that will drop 6 V at 500 mA. What value resistor do you need? A) 6.2 Ω B) 12 Ω C) 6.0 Ω D) 18 Ω
31)
Which of the following is a form of Ohm's Law? A)
R =V
B)
I
C)
I=V×R
D)
V=
I R = V
R I
32)
How much current flows when a 15 ohm resistor is connected across a 5 V source? A) 0.33 A B) 0.5 A C) 3.3 A D) 3.0 A
33)
What electromotive force would cause 20 A of current to flow through a 500 ohm resistor? A) 25 V B) 2.5 V C) 1 kV D) 10 kV
34)
If V = 50 V and R = 100 kΩ the current equals ________. A) 0.5 mA B) 50 mA C) 500 μA D) both A and C
303
35)
The current through a 10 kΩ resistor is 14 μA. What is the voltage drop across the resistor? A) 140 mV B) 14 mV C) 14 V D) 1.4 V
36)
How many ohms of resistance allows a current of 720 μA to flow, when 3.6 kV is applied? A) 200 nΩ B) 5 kΩ C) 200 kΩ D) 5 MΩ
37)
How many amperes will flow in a circuit that contains a 110 V source and 20 kΩ of resistance? A) 5.5 mA B) 5.5 A C) M55 mA D) 55 mA
38)
How much voltage is needed to produce a current of 5 A when total opposition is 50 Ω? A) 25 V B) 250 V C) 2.5 V D) 2.5 kV
39)
If an ammeter in a circuit is reading 20 mA and the applied voltage is 25 V, what would the value of the load resistor be? A) 12.5 kΩ B) 12.5 Ω C) 1.25 kΩ D) 125 Ω
40)
When circuit voltage increases and resistance decreases at the same rate, current will: A) remain the same B) double C) decrease by half D) not enough information
41)
Voltage and current are: A) inversely proportional B) directly proportional C) ideally the same value D) quantities that subtract
42)
If an electric circuit requires 1 A at 50 V, how much current will it require if the voltage is increased to 75 V? A) 1.5 A B) 15 mA C) 15 μA D) M15 V
304
43)
If the resistors voltage is 3 V, with 100 mA flowing through it, what is the resistors value? A) 0.03 Ω B) 0.3 Ω C) 3.0 Ω D) 30 Ω
44)
What EMF would cause 20 A of current to flow through a 500 Ω resistor? A) 0.04 V B) 2.5 V C) 1 kV D) 10 kV
305
1)
FALSE
23)
A
2)
FALSE
24)
D
3)
TRUE
25)
B
4)
TRUE
26)
C
5)
FALSE
27)
B
6)
B
28)
C
7)
C
29)
B
8)
D
30)
B
9)
D
31)
A
10)
B
32)
A
11)
A
33)
D
12)
C
34)
D
13)
A
35)
A
14)
B
36)
D
15)
C
37)
A
16)
A
38)
B
17)
D
39)
C
18)
D
40)
A
19)
B
41)
B
20)
C
42)
A
21)
B
43)
D
22)
D
44)
D
306
Principles of Electric Circuits Chapter 4: Energy and Power 1)
Energy is the ability to do work. (True/False)
2)
Power is the rate at which energy is used. (True/False)
3)
Voltage forces electrical energy through a circuit. (True/False)
4)
If 50 Joules are used in ten seconds, the power will be 500 Watts. (True/False)
5)
One watt equals one joule per second. (True/False)
6)
The power dissipated by a resistor doubles if the applied voltage doubles. (True/False)
7)
A 1 kW load that operates for 1 hour consumes the same amount of energy as a 100 W load that operates for 10 hours. (True/False)
8)
Efficiency is determined by dividing the power output by the power input. (True/False)
9)
The rate at which work is performed is called power. (True/False)
10)
The symbol for power is: A) P B) W C) I D) J
11)
Power is measured in: A) Joules B) Watts C) Energy D) Amperes
12)
How many watts are used when 6000 joules of energy are consumed in 5 hours? A) 1200 W B) 200 W C) 0.33 W D) 6000 W
13)
Convert 0.072 W to mW. A) 7.2 mW B) 72 mW C) 720 mW D) 0.000072 mW
307
14)
Express 0.15 mW as μW. A) 0.00015 μW B) 1.5 μW C) 15 μW D) 150 μW
15)
Power companies charge for energy used on the basis of: A) kW B) kWh C) MWh D) Wh
16)
The collision of electrons when passing through resistance results in energy loss given off as: A) light B) voltage C) watts D) heat
17)
Which of the following is NOT a correct formula for calculating power? A) P = IV B) P = I2R C) P = VR D) P = V2/R
18)
The amount of current in a circuit is 10 mA, while the resistance is 1 kΩ. Determine the power. A) 0.1 W B) 10 W C) 100 W D) 0.0001 W
19)
The amount of voltage in a circuit is 50 V, while the current is 0.2 mA. Determine the power. A) 2W B) 10 W C) 0.01 W D) 250 W
20)
Voltage in a circuit is 100 V, the resistance is 10 kΩ. Determine the power. A) 10 W B) 1W C) 0.01 W D) 0.1 W
21)
The maximum amount of power a resistor can safely handle without damage is called: A) surface area B) heat C) power rating D) watts
308
22)
If one needed a very high power rating, an appropriate choice for the resistor would be: A) carbon-composition B) carbon-film C) metal-film D) wire wound
23)
An ohmmeter scale typically decreases from left to right and is called a ________. A) backwards scale B) reverse scale C) back-off scale D) linear scale
24)
Which of the following is NOT a necessary procedure for using an ohmmeter to check a resistor: A) turn off the voltage source B) isolate the resistor to be checked C) select the proper range switch setting D) observe polarity when connecting the meter leads
25)
The change in energy from one end of a resistor to another creates: A) voltage loss B) voltage drop C) voltage rise D) voltage gain
26)
A 12 V battery is connected to a 400 Ω load. If it supplied current for 1000 hours, what was its Ah rating? A) 3 Ah B) 30 Ah C) 60 Ah D) 120 Ah
27)
A resistor which has a large physical size means: A) it contains more resistance in ohms B) it contains more carbon C) it can dissipate more heat D) it can handle less power
28)
A 5.1 kΩ resistor has burned out in a circuit. You are to replace it with one of like ohmic value. If the resistor should carry 15 mA, which of the following standard resistors should one use? A) 5.1 kΩ, 1/2 W B) 5.1 kΩ, 1 W C) 5.1 kΩ, 2 W D) 5.1 kΩ, 5 W
309
29)
A power supply provides a continuous 200 W to a load, with an efficiency of 80%. In a 24hour period, how many KWh of energy does it consume? A) 6 KWh B) 250 KWh C) 600 KWh D) 4.8 KWh
Figure 4-1 30)
Given the circuit in Figure 4-1, the power consumed by the lamp is: A) 100 W B) 220 W C) 144 W D) 14 W
Figure 4-2 31)
Given the circuit in Figure 4-2, the power delivered to the lamp is: A) 1.0 W B) 12 W C) 1.2 W D) 120 W
32)
Given the circuit in Figure 4-2, what is the resistance of the lamp? A) 1.2 Ω B) 120 Ω C) 12 Ω D) Not enough data to calculate the resistance
310
Figure 4-3 33)
Given the circuit in Figure 4-3, which of the following resistor power values would you use in the circuit? A) 1W B) 15 W C) 20 W D) 5W
34)
If you installed a 33 Ω, 1 watt resistor in the circuit in Figure 4-3, the resistor will probably: A) produce a little heat B) work fine C) burn up D) short
35)
If a 12 Ω draws 1 A of current, what should be its wattage rating? A) 1/4 W B) 3/4 W C) 1.2 W D) 12 W
36)
A system with an input power of 50 W and an output of 20 W has an efficiency of: A) 20% B) 25% C) 40% D) 50%
37)
One kilowatt-hour is equivalent to: A) 1.0 × 103 joules B) 60 × 104 joules C) 3.6 × 105 joules D) 3.6 × 106 joules
38)
How many joules of energy will a 10 W lamp dissipate in one minute? A) 10 joules B) 60 joules C) 600 joules D) 3600 joules
39)
What is the total dissipated power from two series-connected 100 W lamps? A) 50 W B) 100 W C) 200 W D) 10 kW
311
40)
How much power does a 50 V amplifier use when 5 A of current flows through it? A) 100 mW B) 10 W C) 250 mW D) 250 W
41)
How long would it take for a 60 W lamp to consume 30.24 kilowatt-hours of power? A) 0.505 hours B) 3 hours C) 21 hours D) 3 weeks
42)
50 mA of current flows through a 10 kΩ resistor. How much power is dissipated? A) 0.25 μW B) 5 μW C) 25 W D) 25 mW
43)
The power rating of a resistor is ________ related to ________. A) directly, its surface area B) indirectly, the type of material used C) directly, the type of cooling used
44)
Which of the following is a correct formula for calculating power? A) P = IR2 V2 R
B)
P=
C) D)
P = EV P = WI
45)
How much energy does a 6 V battery have when it stores 12 coulombs of charge? A) 5 joules B) 2 joules C) 72 joules D) 432 joules
46)
If a lamp transforms 120 joules of energy into light and heat every 2 seconds, how much power does it use? A) 60 W B) 120 W C) 240 W D) 480 W
47)
What value of power is dissipated by a 5 kΩ resistor, when 30 nA flows through it? A) 6 nW B) 4.5 W C) 6.67 mW D) 150 W
312
48)
What is the kilowatt-hour consumption of a 40 W lamp if it remains "on" for 1750 hours? A) 70.00 B) 70,000 C) 43.75 D) 43,750
49)
For a circuit that contains a 50 V power supply and a 100 Ω resistor, what power rating should be assigned to the resistor? A) 125W B) 25 W C) 2.5 W D) 250 W
50)
A voltage drop is A) the loss of energy by electrons as they flow through a resistance B) called a difference of potential C) the comparison of energy lost from one end of a resistance to the other D) all of the above
51)
The length of time that a battery can deliver a certain amount of average current to a load at the rated voltage is called the A) voltage rating B) ampere-hour rating C) current rating D) minimum power rating
313
1)
TRUE
27)
C
2)
TRUE
28)
C
3)
FALSE
29)
A
4)
FALSE
30)
C
5)
TRUE
31)
B
6)
FALSE
32)
C
7)
TRUE
33)
C
8)
FALSE
34)
C
9)
TRUE
35)
D
10)
A
36)
C
11)
B
37)
D
12)
C
38)
C
13)
B
39)
C
14)
D
40)
D
15)
B
41)
D
16)
D
42)
C
17)
C
43)
A
18)
A
44)
B
19)
C
45)
C
20)
B
46)
A
21)
C
47)
B
22)
D
48)
A
23)
C
49)
B
24)
D
50)
D
25)
B
51)
B
26)
B
314
Principles of Electric Circuits Chapter 5: Series Circuits 1)
A series connection provides two or more paths for current to flow. (True/False)
2)
Assembly diagrams show how to connect components electrically. (True/False)
3)
The current entering a point in a circuit is equal to the current leaving the point. (True/False)
4)
A voltage divider is a series arrangement of resistors. (True/False)
5)
Ground is a common reference point in a circuit. (True/False)
6)
In a series circuit, current is not the same when measured at different points in the circuit. (True/False)
7)
A series circuit has the same voltage across every component. (True/False)
8)
The total resistance of a series circuit is equal to the average of all the resistance values. (True/False)
9)
As you add resistors to a series circuit, total current will decrease. (True/False)
10)
The product of all the voltage drops in a series circuit will equal source voltage. (True/False)
Figure 5-1
Figure 5-2
Figure 5-3 11)
Which of the circuits (Figure 5-1, 5-2 or 5-3) is a circuit with resistors in series? A) Figure 5-1 B) Figure 5-2 C) Figure 5-3
315
Figure 5-4 12)
Given the circuit in Figure 5-4, what is the current through resistor R2? A) 1A B) 100 mA C) 0.2 A D) 1.2 A
13)
The total resistance in the circuit in Figure 5-4 is: A) 10 Ω B) 90 Ω C) 90 Ω / 10 Ω D) 90 Ω + 10 Ω
Figure 5-5 14)
The total resistance in the circuit in Figure 5-5 is: A) 709.2 kΩ B) 927 Ω C) 2907 Ω D) 23,416.2 Ω
Figure 5-6 15)
The total resistance in the series circuit in Figure 5-6 is: A) 498 Ω B) 534 Ω C) 578 Ω D) 590 Ω
316
16)
What is the current flowing in the circuit in Figure 5-6? A) 0.05 mA B) 25 mA C) 2.5 mA D) 0.5 mA
Figure 5-7 17)
Given the circuit in Figure 5-7, the oltage drop across R1 is 18 V. What is the voltage drop across R2? A) 36 V B) 20 V C) 54 V D) 45 V
Figure 5-8 18)
Given the circuit in Figure 5-8, the voltage drop across R1 is 60 V. The drop across R2 is 45 V. What is the resistance value of R2? A) 400 Ω B) 225 Ω C) 345 Ω D) 450 Ω
317
Figure 5-9 19)
The total current flowing in the circuit in Figure 5-9 is: A) 1.06 A B) 0.53 A C) 0.35 A D) 2.82 A
Figure 5-10 20)
Figure 5-10 shows 5 D-size batteries connected in a series circuit. What is the total voltage measured between points A and B? A) 7.5 V B) 4.5 V C) 3.0 V D) 1.5 V
Figure 5-11 21)
Given the circuit in Figure 5-11, the voltage drop across R1 is 75 V. What is the voltage drop across R2? A) 75 V B) 20 V C) 100 V D) 25 V
22)
Given the circuit in Figure 5-11, what is the resistance value of R2? A) 75 Ω B) 20 Ω C) 100 Ω D) 25 Ω
318
Figure 5-12 23)
Given the circuit in Figure 5-12, the circuit current is: A) 0.022 A B) 2.22 mA C) 220 mA D) 22.2 mA
24)
Given the circuit in Figure 5-12, the voltage drop across R1 is: A) 33.3 V B) 1.22 V C) 3.33 V D) 2.22 V
25)
Given the circuit in Figure 5-12, the voltage drop across R2 is: A) 4.44 V B) 22.2 V C) 3.33 V D) 2.22 V
26)
Given the circuit in Figure 5-12, the voltage drop across R3 is: A) 4.44 V B) 0.333 V C) 3.33 V D) 2.22 V
Figure 5-13
319
27)
Given the special series circuit, called a voltage divider, in Figure 5-13, the output voltage (Vout) is: A) 2.5 V B) 10 V C) 5.0 V D) 3.75 V
28)
Given the voltage divider circuit, in Figure 5-13. Suppose resistors R1 and R2 are changed to 10Ω. What is the new output voltage (Vout)? A) 1.0 V B) 2.25 V C) 5.0 V D) 3.75 V
Figure 5-14 29)
Given the voltage divider, in Figure 5-14, the output voltage (Vout) is: A) 0.91 V B) 1.09 V C) 6.0 V D) 2.75 V
30)
Given the voltage divider circuit, in Figure 5-14, what is the circuit current? A) 1.09 mA B) 10.9 mA C) 6.0 mA D) 2.75 mA
31)
What is the effect on total current if the voltage supplying a circuit is halved, and the circuit resistance stays the same? A) It goes to zero. B) It decreases by half. C) It doubles. D) It remains the same.
32)
Kirchhoff's voltage law states that: A) the algebraic sum of the potential rises and drops around a closed loop is zero 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
320
33)
Given a series circuit containing resistors of different values, which statement is not true? A) The current through each resistor is the same. B) The sum of the voltage drops across each resistive element will be equal to the voltage source. C) The total resistance is the sum of the value of the resistors. D) The voltage drop across each resistor is the same.
34)
The total resistance of a series circuit is equal to: A) the average of all the resistance values B) the sum of all the resistance values C) the smallest resistance value D) the largest resistance value
35)
What is the supply voltage when a 5 kΩ and a 4 kΩ resistor are in series and 6 V is measured across the smaller resistor? A) 6V B) 12 V C) 13.5 V D) 16 V
36)
A voltage divider is always a: A) series-parallel circuit B) bridge circuit C) series circuit D) parallel circuit
37)
The voltage drop across a series resistor is proportional to what other value? A) Total resistance B) Wattage rating C) The amount of time the circuit is on D) Its own resistance
38)
If the total voltage applied in a series circuit is 120 V and the voltage drop across R1 is 20 V and R2 drops 30 V, what is the voltage drop for R3? A) 40 V B) 50 V C) 70 V D) 120 V
39)
Which measuring device requires no circuit power to obtain a reading? A) Ammeter B) Ohmmeter C) Voltmeter D) Wattmeter
40)
A series circuit has: A) only one current path B) two current paths C) more than two current paths D) different currents at different points
321
41)
In a series circuit, the voltage measured across an open will be: A) zero B) infinite voltage C) the normal voltage drop D) source voltage
42)
If the voltage of a lamp is to be measured by a voltmeter, then: A) open the circuit and connect the leads across the lamp B) connect the leads across the lamp without opening the circuit C) connect the voltmeter in the same current path as the lamp D) remove the lamp and put the voltmeter in the same current path as the lamp was in
43)
The voltage drop across a series resistor is proportional to what other value? A) The direction of current flow B) The wattage rating C) The amount of time the circuit is on D) Its own resistance
44)
What determines whether connected resistors are in series, parallel, or series-parallel? A) Power source B) Current flow C) Resistance D) Voltage source
45)
What is the total current of a series circuit with a 9 V battery, a 4.7 kΩ, a 6.1 kΩ resistor, and a 450 Ω resistor? A) 800 μA B) 1.9 mA C) 20 mA D) 23.4 mA
46)
Which of the following is a true statement concerning potentiometers? A) Potentiometers have only two terminals. B) Potentiometers do not have a schematic symbol. C) Potentiometers can be used as voltage dividers. D) Potentiometers can only be classified as linear
322
1)
FALSE
25)
A
2)
FALSE
26)
C
3)
TRUE
27)
C
4)
TRUE
28)
C
5)
TRUE
29)
B
6)
FALSE
30)
A
7)
FALSE
31)
B
8)
FALSE
32)
A
9)
TRUE
33)
D
10)
FALSE
34)
B
11)
A
35)
C
12)
B
36)
C
13)
D
37)
D
14)
C
38)
C
15)
D
39)
B
16)
B
40)
A
17)
D
41)
D
18)
B
42)
B
19)
C
43)
D
20)
D
44)
B
21)
D
45)
A
22)
D
46)
C
23)
B
24)
D
323
Principles of Electric Circuits Chapter 6: Parallel Circuits 1)
A parallel connection provides two or more paths for current to flow. (True/False)
2)
The voltage drop across each resistance in a parallel circuit is equal to the supply voltage. (True/False)
3)
The total (equivalent) resistance in a parallel circuit is always less than the lowest value resistor in the circuit. (True/False)
4)
Each resistor in a parallel circuit demands a current (I = V/R), and is independent of all other resistors in parallel with it. (True/False)
5)
The total power taken from the power supply, and dissipated by the resistors in a parallel circuit, is the sum of the powers dissipated by each resistor in the circuit. (True/False)
6)
A parallel circuit is also used as a current divider. (True/False)
7)
If one of the resistors opens in a parallel circuit, total resistance of the circuit will increase. (True/False)
8)
When parallel resistors are of three different values, the smallest resistor will dissipate the greatest amount of power. (True/False)
9)
When two or more components are connected across the same voltage source, they are in parallel. (True/False)
10)
A junction is any point in a circuit where two or more circuit paths come together. (True/False)
Figure 6-1 Figure 6-2
Figure 6-3 11)
Which of the circuits (Figure 6-1, 6-2, or 6-3) is a circuit with resistors in parallel? A) Figure 6-1 B) Figure 6-2 C) Figure 6-3
324
Figure 6-4 12)
Given the circuit in Figure 6-4, find the current through R1. A) 0.187 A B) 0.024 A C) 0.25 A D) 24.8 mA
13)
Given the circuit in Figure 6-4, find the current through R2. A) 0.187 A B) 0.125 A C) 0.25 A D) 24.8 mA
14)
Given the circuit in Figure 6-4, find the total current (It). A) 0.187 A B) 0.024 A C) 375 mA D) 24.8 mA
15)
Given the circuit in Figure 6-4, what is the voltage drop across R2? A) 10 V B) 3.3 V C) 0.25 A D) 25.0 V
Figure 6-5 16)
Given the circuit in Figure 6-5, what is the current flowing through R2? A) 0.5 mA B) 1.0 mA C) 500 mA D) 1.5 mA
325
17)
Given the circuit in Figure 6-5, what is the resistance value of R2? A) 1 kΩ B) 20 kΩ C) 220 Ω D) 1.5 kΩ
18)
Given the circuit in Figure 6-6, what is the total current (It)? A) 200 mA B) 3.1 V C) 31 mA D) 11.0 mA
19)
Given the circuit in Figure 6-6, what is the total resistance of the three parallel resistors? A) 1.05 kΩ B) 322.5 Ω C) 105.5 Ω D) 16 kΩ
20)
Given the circuit in Figure 6-6, what is the voltage drop across R2? A) 12 V B) 3.31 V C) 0.22 V D) 10.0 V
Figure 6-7 21)
Given the circuit in Figure 6-7, find the total current. A) 0.02 A B) 0.3 A C) 0.2 A D) 0.3 mA
326
Figure 6-8 22)
Given the circuit in Figure 6-8, find the total power. A) 187 mW B) 0.125 W C) 0.035 W D) 310 mW
23)
Given the circuit in Figure 6-8, select a minimum wattage rating for R2. A) 1/4 W B) 1/2 W C) 1W D) 5W
Figure 6-9 24)
Given the circuit in Figure 6-9, what is the total (equivalent) resistance of the circuit? A) 100 Ω B) 100 Ω × 3 C) 300 Ω/3 D) 600 Ω E) A and C are correct.
25)
Given the circuit in Figure 6-9, if R3 is changed to 100 Ω, the total circuit current will: A) double B) remain the same C) increase D) decrease
327
Figure 6-10 26)
Given the circuit in Figure 6-10, what is the output current (Iout)? A) 0.01 A B) 1.25 mA C) Iin/3 D) 2.75 mA
27)
Given the circuit in Figure 6-10, what is the total (equivalent) resistance? A) 520 Ω B) 5.2 kΩ C) 6.4 kΩ D) 4.7 kΩ
28)
Given the circuit in Figure 6-10, what voltage must be applied between points A and B to get the circuit current of 2.75 mA? A) 4.7 V B) 14.3 V C) 2.75 V D) 1.43 V
29)
In a parallel resistive circuit: A) there is more than one current path between two points. B) the voltage applied divides between the branches. C) the total branch power exceeds the source power. D) the total circuit conductance is less than the smallest branch conductance.
30)
The voltage across any branch of a parallel circuit: A) varies as the total current varies. B) is inversely proportional to total circuit resistance. C) is equally applied to all branch conductances. D) is dropped in proportion to each branch resistance.
31)
Kirchhoff's law for parallel circuits states that the: A) sum of all branch voltages equal zero. B) total circuit resistance is less than the smallest branch resistor. C) sum of currents into a junction is equal to the difference of all the branch currents. D) sum of the total currents flowing out of a junction equals the sum of the total currents flowing into that junction.
328
32)
The total resistance of a parallel circuit: A) equals the sum of all the branch resistances. B) is the inverse of the total circuit conductance. C) is always greater than the largest branch resistance. D) equals the sum of the total circuit conductance.
33)
The total resistance of two resistors in parallel is equal to the: A) sum divided by the product of the two resistors. B) source voltage divided by any of the two branch resistors. C) product of their R values divided by the sum of the two resistors. D) value of one branch resistor divided by the sum of their R values.
34)
The general current divider equation used to find a branch current is equal to the: A) resistance total divided by that branch resistor, multiplied by the total current. B) branch voltage divided by the total resistance. C) total resistance divided by the product of that branch resistance and the total current. D) branch resistance divided by the total resistance, multiplied by the total current.
35)
Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum drive output voltage of fifteen volts. How much audio power must the amplifier be able to deliver to the speakers when outputting this maximum voltage? A) 28 W B) 56 W C) 112.5 W D) 225 W
36)
What is the total resistance of a 3 kΩ, 4 kΩ, and 12 kΩ resistors that are in parallel with each other? A) 1.5 kΩ B) 2 kΩ C) 6.3 kΩ D) 19 kΩ
37)
To determine the value of a resistor to place in shunt with a known resistor to obtain a certain value of the total parallel combination: A) subtract the known from the desired total. B) multiply the desired by the known, then divide this product by their sum. C) multiply the desired value by the known value, then divide this product by their difference. D) add the desired to the known.
38)
To find one branch current of two resistors in shunt, the: A) total current is divided by the ratio of the other branch resistor divided by their sum. B) product of both is divided by their sum. C) product of both is divided by their difference. D) total current is multiplied by the ratio of the other branch resistor, divided by the sum of the branch resistances.
329
39)
If two lamps, L1 and L2, are connected in parallel and L1 opens, what would happen? A) L1 would be off and L2 would be on. B) L1 would be on and L2 would be off. C) Both lamps would be on. D) Both lamps would be off.
40)
If 10 A of current entered a two branch junction, and 2 A flowed into Branch "A", how much current will flow in branch "B"? A) 10A B) 5A C) 8A D) 12 A
41)
Circuit analysis should reveal that the largest resistance in a parallel circuit has the: A) least current B) most current C) least voltage D) most voltage
42)
What procedure should be followed when troubleshooting with an ammeter or voltmeter? A) short the leads and adjust B) check the meters external power supply C) start with the highest scale and adjust down to a lower scale D) start with the lowest scale and adjust down to a higher scale
43)
While checking a circuit for current flow, it is noted that R1 current has increased; since it is in parallel with R2, the problem is: A) R2 is open B) R1 is shorted C) R1 is open D) R2 is shorted
44)
What devices can be used to find shorts in a circuit? A) pulser B) potentiometer C) current tracer D) both A and C E) none of the above
45)
Shunt resistors: A) are connected in series for higher current range settings on a meter. B) are connected in parallel for higher current range settings on a meter. C) are not used in ammeters. D) are resistors connected in series.
330
46)
The regulator in a dc power supply maintains a ________ dc output voltage with a ________ input voltage. A) changing, constant B) zero, maximum C) constant, changing D) changing, changing
331
1)
TRUE
24)
E
2)
TRUE
25)
C
3)
TRUE
26)
D
4)
TRUE
27)
A
5)
TRUE
28)
D
6)
TRUE
29)
A
7)
FALSE
30)
C
8)
TRUE
31)
D
9)
TRUE
32)
B
10)
TRUE
33)
C
11)
C
34)
A
12)
C
35)
C
13)
B
36)
A
14)
C
37)
C
15)
D
38)
D
16)
A
39)
A
17)
B
40)
C
18)
C
41)
A
19)
B
42)
C
20)
D
43)
B
21)
B
44)
D
22)
D
45)
B
23)
A
46)
C
332
Principles of Electric Circuits Chapter 7: Series-Parallel Circuits 1)
A series-parallel circuit is a combination of both series and parallel paths. (True/False)
2)
When a load resistor is connected across a voltage divider output, the output voltage increases. (True/False)
3)
Troubleshooting is the process of identifying and locating a fault in a circuit. (True/False)
4)
Open circuits and short circuits are typical circuit faults. (True/False)
5)
Resistors normally short when they burn out. (True/False)
6)
Series components in a series-parallel circuit may be in series with other components, or with other combinations of components. (True/False)
7)
Parallel components in a series-parallel circuit may be in parallel with other individual components, or with other combinations of components. (True/False)
8)
Observed voltages not being shared identify components in parallel in a series-parallel circuit. (True/False)
9)
A loaded voltage divider is a common application of a series-parallel circuit. (True/False)
10)
When analyzing series-parallel circuit currents, you should start with the current in the branch farthest from the source. (True/False)
11)
A series-parallel circuit is a: A) combination of series and parallel paths. B) combination of series paths. C) combination of parallel paths. D) None of the above are correct.
12)
When finding the total resistance of a series-parallel combination, the most important step is: A) to know the value of the source voltage. B) to know how to use Ohm's law. C) to know the value of the total current. D) to define the series and parallel relationships.
13)
Which of the following is needed to solve most resistive circuit analysis problems? A) Ohm's law B) the voltage divider formula C) the current divider formula D) all of the above
333
14)
When analyzing a series-parallel circuit, the circuit should be: A) reduced down to one series circuit. B) reduced down to one parallel circuit. C) reduced down to find the total current first. D) reduced down to find the total voltage first.
Figure 7-1 15)
Given the circuit in Figure 7-1, which resistors are in parallel? A) R1 and R4 B) R2 and R3 C) R2 and R4 D) R1 and R3
Figure 7-2 16)
Given the circuit in Figure 7-2, which resistors are in series? A) R3 and R4 B) R3, R4 and R5 C) R1, R2 and R3 D) R1, R2 and R5
334
Figure 7-3 17)
Given the circuit in Figure 7-3, calculate the total resistance. A) 122 Ω B) 152 Ω C) 162 Ω D) 242 Ω
Figure 7-4 18)
Given the circuit in Figure 7-4, calculate the current at Point A. A) 2A B) 1.2 A C) 0.67 A D) 0.8 A
19)
Given the circuit in Figure 7-4, calculate the current through the 40 Ω resistor. A) 2A B) 1.2 A C) 0.67 A D) 0.8 A
335
Figure 7-5 20)
Given the circuit in Figure 7-5, calculate the total resistance. A) 3 kΩ B) 7 kΩ C) 12 kΩ D) 34 kΩ
21)
Given the circuit in Figure 7-5, calculate the current through the 10 kΩ resistor. A) 48 mA B) 47.2 mA C) 0.6 mA D) 4.8 mA
22)
Given the circuit in Figure 7-5, calculate the voltage drop across the 15 kΩ resistor. A) 12 V B) 6V C) 2.4 V D) 7.2 V
23)
Voltage dividers are used to: A) change several voltages into one voltage. B) change several voltages into other values. C) convert several voltages into a common supply voltage. D) obtain two or more, lower, voltages from a common supply voltage.
24)
An unloaded voltage divider consists of: A) all series connections. B) any parallel circuit. C) all series-parallel connections. D) a single parallel circuit only.
336
Figure 7-6 25)
Given the circuit in Figure 7-6, calculate the output voltage. A) 10 V B) 10.7 V C) 8.6 V D) 6.4 V
Figure 7-7 26)
A voltmeter with an ohms per volt rating of 200 Ω/V is connected as shown in Figure 7-7. It is set on the 150 volt range. What voltage will the meter indicate? A) 45 V B) 60 V C) 105 V D) 30 V
27)
A Wheatstone bridge can be used as a: A) sensitive voltmeter B) precision ohmmeter C) precision ammeter D) voltage divider
337
Figure 7-8 28)
Given the circuit in Figure 7-8, what is the current through R3? A) 0.35 A B) 208 mA C) 35 mA D) 0.35 mA
29)
Given the circuit in Figure 7-8, what current would flow if R2 were open? A) 68 mA B) 175 mA C) 208 mA D) 140 mA
Figure 7-9 30)
Determine what is wrong in the circuit in Figure 7-9. A) R2 open B) R1 shorted C) R4 open D) R3 open E) none of the above
338
31)
Components or combinations of components with common currents, in a series-parallel circuit, are in: A) parallel with each other. B) series with each other. C) either series or parallel with each other. D) none of the above
32)
The first goal to accomplish in analyzing a complex series-parallel circuit is to: A) equate all parallel components. B) equate all series components. C) solve for all the voltage drops. D) solve for the total current and resistance.
33)
When analyzing circuit current in a series-parallel circuit, you start: A) anywhere in the circuit. B) at the source with total current. C) at the farthest resistor from the source. D) in any parallel branch.
34)
In a series-parallel circuit, individual component power dissipations are calculated using: A) a percent of the total power depending on resistor ratios. B) total current squared multiplied by the resistor values. C) a percent of the voltage division ratio squared. D) individual component parameters.
35)
Where can the voltage divider formula be useful in a series-parallel circuit? A) with a group of series connected resistors B) with a group of parallel connected resistors C) with a group of series-parallel connected resistors D) with series-opposing power supplies
36)
________ is often used to analyze multiple-source circuits. A) Ohm's Law B) Thevenin's Theorem C) Superposition D) Kirchhoff's Law
37)
A Thevenin equivalent circuit contains ________. A) a voltage source in series with a resistance B) a current source in series with a resistance C) resistors in a series-parallel configuration D) many voltage sources
339
38)
The loading effect of a voltmeter can be ignored if ________. A) the meter's resistance is at least 10 times greater than the resistance across which it is connected B) the meter's resistance is at least 10 times less than the resistance across which it is connected C) the measured voltage is very high D) the measured voltage is very low
39)
What is the voltage of R1 when the circuit with 24 V applied, has series resistors R1 (1 kΩ) and R2 (2 kΩ) which are in parallel with R3 (3 kΩ)? A) 8V B) 12 V C) 16 V D) 24 V
40)
When R1 = 7 kΩ, R2 = 20 kΩ, R3 = 36 kΩ, and R4 = 45 kΩ. What is the total circuit resistance if R1 is in series with a parallel combination of R2, R3, and R4? A) 4 kΩ B) 17 kΩ C) 41 kΩ D) 108 kΩ
41)
The current that flows through an unloaded voltage divider is called the: A) resistor current B) voltage current C) bleeder current D) load current
42)
When a Wheatstone bridge is in a balanced condition, the center voltmeter in the bridge will read: A) zero volts B) half the source voltage C) the same as source voltage D) twice the source voltage
43)
An R-2R ladder circuit is used for: A) analog-to-analog conversion B) analog-to-digital conversion C) digital-to-analog conversion D) digital-to-digital conversion
44)
The total resistance of an R-2R ladder circuit equals: A) R B) 2R C) R + 2R D) R × 2R
340
45)
When a Wheatstone bridge is balanced, the ________ of resistance on one side of the bridge is equal to the ________ of resistance on the other side of the bridge. A) average, product B) ratio, ratio C) sum, difference D) product, sum
46)
A strain gauge A) is a transducer. B) can be extremely delicate. C) can be used in Wheatstone bridge circuits. D) all of the above
47)
A Wheatstone bridge A) does not have much sensitivity. B) is widely used in measurement applications. C) is a null instrument. D) B and C are correct E) none of the above
341
1)
TRUE
25)
C
2)
FALSE
26)
D
3)
TRUE
27)
B
4)
TRUE
28)
C
5)
FALSE
29)
B
6)
TRUE
30)
E
7)
TRUE
31)
C
8)
FALSE
32)
D
9)
TRUE
33)
B
10)
FALSE
34)
D
11)
A
35)
A
12)
D
36)
C
13)
D
37)
B
14)
A
38)
A
15)
B
39)
A
16)
D
40)
B
17)
B
41)
C
18)
D
42)
A
19)
B
43)
C
20)
C
44)
A
21)
C
45)
B
22)
B
46)
D
23)
D
47)
C
24)
A
342
Principles of Electric Circuits Chapter 8: Circuit Theorems and Conversions 1)
An ideal voltage source has infinite internal resistance. (True/False)
2)
A practical voltage source has infinite internal resistance. (True/False)
3)
An ideal current source has infinite internal resistance. (True/False)
4)
A practical current source has infinite internal resistance. (True/False)
5)
Maximum power transfer occurs when the load resistance equals the source resistance. (True/False)
6)
The superposition theorem is used to simplify circuit analysis where two or more sources are present. (True/False)
7)
Thevenin's theorem is used to simplify complex networks to a simple voltage source with its source resistance. (True/False)
8)
The first step in Nortonizing a circuit is to short out the load. (True/False)
9)
Open circuit voltage refers to: A) the voltage taken across a high value load resistor. B) the voltage taken across a low value load resistor. C) zero internal resistance. D) the output voltage with no load.
10)
The internal resistance of an ideal voltage source is: A) zero B) low C) high D) infinite
11)
Calculate the voltage output of a source, when the source voltage equals 50 V, the source internal resistance is 10 Ω and the load resistance is 50 Ω. A) 8.33 V B) 41.67 V C) 10 V D) 40 V
12)
An ideal current source produces a(n) ________ value of current through a load, regardless of the value of the load. A) proportional B) increasing C) constant D) decreasing
343
13)
An ideal current source has ________ internal resistance. A) zero B) low C) high D) infinite
Figure 8-1 14)
Given the circuit in Figure 8-1, find the total current through the load. A) 12 mA B) 1.2 mA C) 20 mA D) 1.0 mA
15)
What is the equivalent current source when VS = 100 V, RS = 40 Ω? A) IS = 2.5 A, RS = infinite B) IS = 2.0 A, RS = 40 Ω C) IS = 2.5 A, RS = 40 Ω D) IS = 2.5 A, RS = zero Ω
16)
The superposition method for calculating voltages and currents requires: A) opening the load resistance. B) replacing the load resistance with a short circuit. C) taking one source at a time, with other voltage sources replaced by an open circuit. D) taking one source at a time, and replacing other sources with their internal resistances.
344
Figure 8-2 17)
Given the circuit in Figure 8-2, Calculate the total current through R2, using the superposition method. A) 16 mA B) 67 mA C) 83 mA D) 133 mA
18)
If two currents are in opposing directions through a branch of a circuit, in what direction will the net current flow? A) In the direction of the larger current B) In the direction of the smaller current C) Opposing currents will always cancel each other out. D) The net current always flows down through a resistor.
19)
Maximum power is delivered to the load under what conditions? A) when the load resistance equals zero B) when the load resistance is open C) when the load resistance is greater than the source resistance D) when the load resistance equals the source resistance
Figure 8-3 20)
Given the circuit in Figure 8-3, Calculate VTH and RTH. A) VTH = 10 V, RTH = 150 Ω B) VTH = 10 V, RTH = 50 Ω C) VTH = 6.7 V, RTH = 150 Ω D) VTH = 5.0 V, RTH = 50 Ω
345
Figure 8-4 21)
Given the circuit in Figure 8-4, Thevenize the bridge circuit between points "A" and "B". Find the current flow through the load (RL). A) 3.2 mA B) 1.92 mA C) 0.45 mA D) 0.69 mA
Figure 8-5 22)
Given the circuit in Figure 8-5, Calculate the current (IN,) and RN, with RL disconnected. A) IN = 296 mA, RN = 59 Ω B) IN = 27.1 mA, RN = 59 Ω C) IN = 27.1 mA, RN = 14.6 Ω D) IN = 296 mA, RN = 174 Ω
346
Figure 8-6 23)
Convert the Delta network in Figure 8-6 into a Wye network. A) R1 = 25.7 kΩ, R2 = 174.9 kΩ, R3 = 64.8 kΩ B) R1 = 25.7 kΩ, R2 = 143.8 kΩ, R3 = 78.3 kΩ C) R1 = 42.1 kΩ, R2 = 174.9 kΩ, R3 = 78.3 kΩ D) R1 = 37.9 kΩ, R2 = 143.8 kΩ, R3 = 64.8 kΩ
24)
The first steps to Nortonizing a circuit are: A) short RL, determine IL, make IL = IN B) open RL, determine IL, make IL = IN C)
short RL, determine RN, make
= IN
D)
open RL, determine RN, make
= IN
25)
The concept that states the equivalency of two voltage sources means that for any given load resistance connected to the two sources, the same load voltage and load current are produced by both sources is called: A) load equivalency B) terminal equivalency C) junction equivalency D) loop equivalency
26)
Which is the correct formula for converting a current source to a voltage source? A) IS = VSRS B) RS = ISVS C) VS = ISRS D)
27)
VS =
To use the superposition theorem: A) note the directions of current flow. B) insert an equivalent current source. C) analyze one source at a time. D) both A and C
347
28)
To convert a Norton equivalent circuit to a Thevenin equivalent circuit: A) let the Norton resistance equal the Thevenin resistance. B) the product of Norton current and Norton resistance equals the Thevenin voltage. C) place the Norton and Thevenin voltage sources in series. D) both A and B
29)
The first steps to Thevenizing a circuit are: A) open RL, determine VL, make VL = Vth B) short RL, determine IL, make IL = Ith Vs Rth
= Ith
C)
open RL, determine Rth, make
D)
open RL, determine IL, make RL × IL = Vth
30)
What is the symbol usually used inside a circle to designate a current source? A) a sine wave B) an arrow C) a circle D) a current wave
31)
Which of these statements is true of two or more current sources in parallel? A) They violate Kirchhoff's current law. B) They may be replaced by one current source. C) A series resistor must be placed in each branch. D) The magnitude of the combined current is always less than the smallest individual current.
32)
Thevenin's theorem states that the Thevenin voltage is equal to: A) open circuit voltage at the networks terminals B) short circuit voltage at the networks terminals C) open circuit current at the networks terminals D) short circuit current at the networks terminals
33)
Power effects in a dc network cannot be determined using superposition because: A) open sources and shorted neither consume nor produce power. B) power computations require a voltage and a current source in each circuit. C) power is proportional to the square of the current or voltage. D) all voltage and current sources are ideal devices that consume no power.
34)
When using the superposition theorem on a two source network, if the current produced by one source is in one direction, while that produced by the other source is in the opposite direction through the same resistor: A) all voltage sources were not properly converted to current sources. B) the absolute values of the two currents add algebraically, and the direction is the same as the direction of the larger current. C) a mistake in the sign of the result occurred. D) the resulting current is the difference of the two and has the direction of the larger current.
348
35)
Under maximum power transfer conditions, which one is true? A) The algebraic sum of all resistances in the source equals the algebraic sum of all the resistances in the load. B) The Thevenin resistance of the source equals the equivalent resistance of the load. C) The equivalent load resistance is very large compared to the equivalent of the source. D) The equivalent load resistance is very small compared to the equivalent of the source.
36)
Norton's theorem states that you can replace a dc network with an equivalent circuit consisting of: A) a voltage source and a series resistor B) a current source and a series resistor C) a current source and a parallel resistor D) a voltage source and a parallel resistor
37)
Delta to wye configurations are used in: A) bridge circuits B) polyphase circuits C) resonance circuits D) non-sinusoidal networks
38)
When converting a wye configuration with all equal resistors to a delta configuration, the delta resistor will always be ________ times its related wye resistor. A) 6 B) 5 C) 4 D) 3
39)
When a thermistor is used in a basic Wheatstone bridge measuring circuit, the bridge becomes ________ as the thermistor’s resistance changes with ________. A) unbalanced, changes in temperature B) balanced, changes in current C) balanced, changes in light D) balanced, changes in pressure
349
1)
FALSE
21)
C
2)
FALSE
22)
B
3)
TRUE
23)
A
4)
FALSE
24)
A
5)
TRUE
25)
B
6)
TRUE
26)
C
7)
TRUE
27)
D
8)
TRUE
28)
D
9)
D
29)
A
10)
A
30)
B
11)
B
31)
B
12)
C
32)
A
13)
D
33)
C
14)
B
34)
D
15)
C
35)
B
16)
D
36)
C
17)
C
37)
A
18)
A
38)
D
19)
D
39)
A
20)
A
350
Principles of Electric Circuits Chapter 9: Branch, Loop, and Node Analyses 1)
A loop is a closed current path in a circuit. (True/False)
2)
A node is a junction of two or more current paths. (True/False)
3)
The branch method is based on loop currents. (True/False)
4)
Simultaneous equations can be solved by substitution or by using determinants. (True/False)
5)
The loop current is not necessarily the actual current in a branch. (True/False)
6)
Mesh currents flow in a counter-clockwise direction through each loop. (True/False)
7)
Loop/Mesh analysis uses simultaneous equations to calculate the current through each component. (True/False)
8)
A node is where the voltage is connected to the circuit. (True/False)
9)
To convert from a wye to an equivalent delta, superimpose the two and take the sum of the products of the wye, then divide by the opposite wye resistor from the delta resistor you are trying to find. (True/False)
10)
Delta to wye configurations are used in bridge circuits. (True/False)
11)
Which of the following are useful for analyzing circuits with two or more voltage or current sources? A) branch current method B) mesh current method C) node-voltage method D) all of the above
12)
The branch current method involves using: A) loops and junctions B) loops and nodes C) junctions and nodes D) voltages and nodes
13)
What is the significance of a negative sign in the answer for current, when solving network problems? A) It means you must subtract the negative value from the other value. B) It means there is a negative current flow. C) It means that the negative current flows in the opposite direction from the assigned current. D) It means you must chose the correct direction for the actual, net, current flow.
351
14)
The rule, when solving for several unknown quantities is: A) Use Ohm's law. B) You must have the same number of equations as there are unknown values. C) You must use determinants. D) You must use substitution.
15)
In the equation; 5I1 - 8I2 = 10, which is the coefficient for I1? A) 5 B) 10 C) 8 D) -8
16)
The cofactor method for determinants is best suited for: A) more than 3 unknowns B) less than 3 unknowns
17)
What is a loop current? A) Loop currents are actual currents. B) Loop currents are assigned currents.
18)
The loop current direction must be drawn: A) clockwise B) counter-clockwise C) It doesn't matter, as long as you are consistent.
19)
An advantage of mesh currents over branch currents is: A) There are no negative answers for current. B) The mesh current method can be applied to circuits with any number of loops. C) Given a particular circuit, the mesh method requires only two equations, while the branch method requires three equations. D) There are no advantages.
20)
What circuit law is used in the mesh current method? A) Ohm's law B) the power law C) Kirchhoff's current law D) Kirchhoff's voltage law
21)
What circuit law is used in the node voltage method? A) Ohm's law B) the power law C) Kirchhoff's current law D) Kirchhoff's voltage law
352
22)
23)
What is the characteristic determinant from the following equations? 5I1 - 8I2 = 10 l0I1 + 5I2 = 20 A) 10 B) -90 C) 55 D) 105 Solve for I1 and I2 in the following equation: 100I1 + 30I2 = 30 75I1 - 45I2 = 20 A) I1 = 0.289 A, I2 = -0.037 A B) I1 = 0.037 A, I2 = 0.368 A C) I1 = 0.289 A, I2 = 0.037 A D) I1 = 28.9 mA, I2 = 3.7 mA
Figure 9-1 24)
Given the circuit in Figure 9-1, find the approximate current flow through the 500 Ω resistor. A) 5.16 mA B) 1.69 mA C) 3.47 mA D) 8.63 mA
353
Figure 9-2 25)
Given the circuit in Figure 9-2, find the approximate voltage drop across the 68 Ω resistor. A) 1.36 V B) 11.69 V C) 12.92 V D) 14.28 V
26)
A branch is a path that connects: A) two loops B) two nodes C) two components D) two multiple loop circuits
27)
In nodal analysis, how many equations will need to be solved if the circuit contains five nodes (including the reference node)? A) 3 B) 4 C) 5 D) 6
28)
Third order determinants can be evaluated by: A) expansion method alone B) cofactor method alone C) either the expansion method or the cofactor method D) characteristic determinant method alone
29)
How many general steps are used in applying the branch current method? A) 1 B) 3 C) 4 D) 5
354
30)
Simultaneous equations can be solved by which two methods? A) substitution and coefficient B) substitution and determinant C) determinant and mesh D) substitution and loop
31)
The cofactor method can be used to evaluate determinants with orders higher than: A) 3 B) 4 C) 5 D) 6
32)
A branch current is an ________ current through a branch, whereas: loop currents are ________. A) assumed, mathematical quantities B) actual, circuit dependent C) actual, mathematical D) both A and B
Figure 9-3 33)
In Figure 9-3 what value of R will result in a balanced bridge? A) 3Ω B) 300 Ω C) 30 Ω D) 60 Ω
34)
See Figure 9-3. If R = 100 Ω, compute the current through the 50 Ω resistor. Use the delta-wye conversion method. A) 4.56 Ω B) 45.6 Ω C) .456 Ω D) 456 Ω
35)
See Figure 9-3. Using R = 100 Ω, compute the current through the 50 Ω resistor. A) 21.47 mA B) 21.47 A C) 2.14 mA D) 21.47 μA
355
Figure 9-4 36)
See Figure 9-4. Which of the following terms describes the voltage across the 7 Ω resistor when using mesh analysis? A) (7 Ω) I1 B) (7 Ω) I2 C) (7 Ω) (I1 - I2) D) (7 Ω) (I1 + I2)
37)
See Figure 9-4. Which statement is true if the loop current I2 is found to be a negative number? A) The nodal analysis approach should have been used, not the mesh analysis approach. B) The determinant used to compute the current should have been third order, not second order. C) The 10 V battery and 10 Ω resistor should have been converted to a current source. D) The original direction assumed for I2 is wrong.
38)
See Figure 9-4. If nodal analysis were to be used to solve for unknown voltages in this circuit, how many nodes would be needed (including the reference node)? A) 1 B) 2 C) 3 D) 4
Figure 9-5 39)
See Figure 9-5. Which equation describes node X? A) I3 = I2 - I1 B) I3 = I2 + I1 C) I1 + I2 + I3 = 0 D) -I1 - I2 - I3 = 0
356
40)
See Figure 9-5. If the branch current method is used, which equation describes loop 1? A) 0 = +5 V - (3 Ω)I1 + (7 Ω)I2 B) 0 = +5 V + (3 Ω)I1 - (7 Ω)I2 C) 0 = -5 V - (3 Ω)I1 + (7 Ω)I2 D) 0 = -5 V - (3 Ω)I1 - (7 Ω)I2
41)
See Figure 9-5. The equation obtained from performing mesh analysis on mesh #1 is: A) 5 V + 3I2 + 7(I2 - I1) = 0 B) 5 V + 3I2 + 7(I1 - I2) = 0 C) 5 V + 3I2 - 7(I2 - I1) = 0 D) 5 V - 3I2 + 7(I2 - I1) = 0
42)
Give the two equations, 2I1=8-5I2 and 0=4I2-5I1+6, in standard form. A) 3I1+5I2=6 2I1+4I2=8 B) 2I1+5I2=8 3I1+4I2=6 C) 2I1+3I2=6 4I1=2I2=8
43)
The node voltage method for the Bridged-T circuit: A) results in two equations. B) results in two unknowns. C) creates a source node and a reference node. D) all of the above E) none of the above
357
1)
TRUE
23)
C
2)
TRUE
24)
B
3)
FALSE
25)
B
4)
TRUE
26)
B
5)
TRUE
27)
B
6)
FALSE
28)
C
7)
TRUE
29)
D
8)
FALSE
30)
B
9)
TRUE
31)
A
10)
TRUE
32)
A
11)
D
33)
C
12)
B
34)
B
13)
C
35)
A
14)
B
36)
C
15)
A
37)
D
16)
A
38)
B
17)
B
39)
B
18)
C
40)
D
19)
C
41)
B
20)
D
42)
B
21)
C
43)
D
22)
D
358
Principles of Electric Circuits Chapter 10: Magnetism and Electromagnetism 1)
Unlike magnetic poles attract each other, and like magnetic poles repel each other. (True/False)
2)
Materials that can be magnetized are called ferromagnetic. (True/False)
3)
A permanent magnet is basically a coil of wire wound around a magnetic core. (True/False)
4)
Induced voltage is directly proportional to the relative motion between a conductor and a magnetic field. (True/False)
5)
The amount of magnetomotive force is inversely proportional to the amount of current flow. (True/False)
6)
Magnetic fields can transfer forces without objects physically touching. (True/False)
7)
Magnetic flux lines are never continuous outside of the magnet. (True/False)
8)
The right hand rule is used for current-carrying conductors using electron flow. (True/False)
9)
If the current flow was visualized going into this page, the magnetic field would be drawn clockwise around the conductor. (True/False)
10)
A tesla is equal to a meter per weber. (True/False)
11)
A group of magnetic force lines going from the north pole of a magnet to the south pole is called: A) magnetic flux B) electromagnetism C) force energy D) permeability
12)
The ________ is the unit of magnetic flux. A) Tesla B) Weber C) Maxwell D) reluctance
13)
The symbol for flux density is: A) B B) T C) A D) O
14)
Permanent magnets are characterized by the fact that: A) their magnetic domains are randomly oriented. B) their magnetic domains are aligned north-to-south. C) they produce an electrostatic field. D) they lose their magnetism very quickly.
359
15)
The ease with which domains align and a magnetic field is established, in a given material, is called the material's: A) permeability B) reluctance C) retentivity D) magnetomotive force
16)
An example of an electromagnetic device is: A) the solenoid B) the loudspeaker C) a tape recording head D) the relay E) all of the above
17)
Magnetomotive force is: A) directly proportional to the number of turns in the coil, and inversely proportional to the current through the coil. B) directly proportional to the number of turns in the coil, and directly proportional to the current through the coil. C) inversely proportional to the number of turns in the coil, and directly proportional to the current through the coil. D) inversely proportional to the number of turns in the coil, and inversely proportional to the current through the coil.
18)
There is 1.5 A of current through a coil consisting of 8 turns, what is the magnetomotive force (mmF)? A) 5.33 ampere-turns (At) B) 6.5 ampere-turns (At) C) 9.5 ampere-turns (At) D) 12 ampere-turns (At)
19)
What is the amount of flux in an 8-turn coil with 1.5 A of current if the reluctance is .04 x 106 At/Wb? A) 300 μWb B) 0.48 μWb C) 150 μWb D) 1.24 μWb
20)
What constitutes an electromagnet? A) a permanent magnet with a coil around it B) a permanent magnet C) a core material that can be easily magnetized with a coil of wire around it D) a core material with high reluctance with a coil of wire around it
21)
An example of an electromagnetic device that also uses a permanent magnet is: A) the solenoid B) the loudspeaker C) a tape recording head D) the relay E) all of the above
360
22)
The most common movement for analog meters is the: A) D'arsonval movement B) iron-vane movement C) electrodynamometer D) hot wire movement
23)
The magnetizing force (H) is not dependent on which of following? A) the number of turns B) the amount of current flow C) the length of the material D) the type of material
24)
The ability of a material to maintain a magnetized state without the presence of a magnetizing force is called: A) permeability B) reluctance C) residual magnetism D) retentivity
25)
Another name for a hysteresis loop is: A) field intensity B) flux density C) the B-H curve D) magnetomotive force
26)
According to Faraday's law, if the rate of flux change quadruples, what happens to the induced voltage? A) It doubles. B) It quadruples. C) It halves. D) It increases by a factor of eight.
27)
Which of the following statements about flux lines is not correct? A) They are continuous. B) They flow in the shortest and/or easiest path. C) They never cross each other. D) Magnetic flux flows from the north pole to the south pole within the magnet.
28)
Magnetic flux lines travel from: A) south to north outside the magnet. B) north to south outside the magnet. C) south to north inside the magnet. D) north to south inside the magnet.
29)
Magnetizing force "H," is measured in: A) ampere-turns per meter. B) Webers per square meter. C) number of turns per square meter. D) ampere-turns per Weber.
361
30)
The opposition to magnetic flux line flow is called: A) permeability. B) retention. C) reluctance. D) Webers per meter.
31)
The field strength of an electromagnet is influenced by the: A) number of coil turns. B) current flowing. C) Hysteresis. D) length of the coil.
32)
The force between current carrying parallel conductors: A) is one of attraction for the same direction of current flow. B) is one of repulsion for the same direction of current flow. C) is cancelled no matter what the direction of current flow. D) is neutral because the force fields from each conductor cancel each other out.
33)
The Greek symbol ________ is used to indicate the ________ of a material in relation to flux lines. A) 1, reluctance B) μ, permeability C) β, Webers D) δ, Gauss
34)
Ferrites are: A) paramagnetic materials. B) diamagnetic materials. C) low electrical resistance, high permeability materials. D) high electrical resistance, high permeability materials.
35)
Faraday's law states that the: A) direction of induced voltage produces an opposition. B) direction of an induced current produces an aiding effect. C) emf depends on the rate of cutting flux. D) emf is related to the direction of current.
36)
The B-H curve plots: A) flux density versus magnetizing force. B) flux intensity versus flux density. C) retentivity versus saturation. D) permeability versus reluctance.
37)
Reluctance varies: A) inversely with its length. B) directly with its permeability. C) inversely with its permeability. D) directly with its area.
362
38)
Relative permeability of a magnetic material is a ratio of its: A) area to its field density. B) flux lines to its ampere-turns. C) field intensity to its field density. D) area to its length.
39)
Calculate the induced voltage across a coil with 150 turns that is being cut by a magnetic field that is changing at a rate of 7 Wb/s. A) 1050 V B) 10.5 V C) 1.05 V D) 105 V
40)
Mmf in an electromagnet is increased by: A) increasing coil turns B) increasing coil length C) decreasing permeability D) decreasing the voltage
41)
What is the mmf of a material whose magnetic flux is equal to 670 microwebers with a reluctance of 44.8 x 103 At/Wb? A) 10 At B) 20 At C) 30 At D) 40 At
42)
Calculate the reluctance of a toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer radius of 2.25 cm. The permeability of the steel is 2 x 10-4 Wb/At-m. A) 1.96x10-5 At/Wb B) 31.9x106 At/Wb C) 2.29x106 At/Wb D) 4.03x105 At/Wb
363
1)
TRUE
23)
D
2)
TRUE
24)
D
3)
FALSE
25)
C
4)
TRUE
26)
B
5)
FALSE
27)
D
6)
TRUE
28)
B
7)
FALSE
29)
A
8)
FALSE
30)
C
9)
FALSE
31)
D
10)
FALSE
32)
A
11)
A
33)
B
12)
B
34)
D
13)
A
35)
C
14)
B
36)
A
15)
A
37)
C
16)
E
38)
D
17)
B
39)
A
18)
D
40)
A
19)
A
41)
C
20)
C
42)
B
21)
B
22)
A
364
Principles of Electric Circuits Chapter 11: Introduction to Alternating Current and Voltage 1)
A sine wave is the only waveform that contains no harmonics. All other waveforms are composed of a fundamental sine wave and harmonically related sine waves. (True/False)
2)
A harmonic frequency is always an integral (odd or even) multiple of the fundamental frequency. (True/False)
3)
Frequency is the reciprocal of the period. (True/False)
4)
There are 360 degrees in a complete cycle of a sine wave. (True/False)
5)
A given rms value of a sine wave is equal to the same value of DC voltage with respect to the work it can do. (True/False)
6)
Increasing positive angles of a phasor move clockwise from the reference point. (True/False)
7)
When the phasor moves in the clockwise direction, the various angle positions are labeled as negative values. (True/False)
8)
The rate of change of a voltage sine wave is due to the change in flux density. (True/False)
9)
The average value of a sine wave is zero. (True/False)
10)
A sawtooth voltage waveform has a linear ramp changing voltage level with respect to time. (True/False)
Figure 11-1 11)
Given the sine wave in Figure 11-1, identify the peak-to-peak voltage: A) value A B) value B C) value C D) value D E) value E
365
12)
Given the sine wave in Figure 11-1, identify the negative peak voltage: A) value A B) value B C) value C D) value D E) value E
13)
Given the sine wave in Figure 11-1, identify the 1/2 cycle: A) value A B) value B C) value C D) value D E) value E
14)
Given the sine wave in Figure 11-1, identify the positive peak voltage: A) value A B) value B C) value C D) value D E) value E
15)
Given the sine wave in Figure 11-1, identify the zero volts: A) value A B) value B C) value C D) value D E) value E
Figure 11-2
366
16)
You count 5 divisions on the horizontal axis on the scope. The Time-per-division knob is set to 1.0 ms/Div. Given the drawing in Figure 11-2, what is the frequency of Waveform B? A) 1 kHz B) 5 kHz C) 0.2 kHz D) 0.5 kHz
17)
Given the drawing in Figure 11-2, and the Time-per-division knob is set to 1.0 ms/Div, what is the frequency of Waveform A? A) 100 Hz B) 2 kHz C) 400 Hz D) 5 kHz
18)
For Waveform A, you count 4 divisions on the vertical axis on the scope. The Volts-perdivision knob is set to 50 mV/Div. Given the drawing in Figure 11-2, what is the Peak-to-peak amplitude of Waveform A? A) 50 mV B) 500 mV C) 0.2 V D) 250 mV
Figure 11-3 19)
Given the drawing in Figure 11-3, the phase shift between Waveform A and Waveform B is: A) 45° B) 60° C) 90° D) 180°
20)
Refer to Figure 11-3. Waveforms A and B have the same frequency, but appear to have different amplitudes. Is this possible? A) Yes B) No
21)
Refer to Figure 11-3. Which of the following statements is correct? A) Waveform A lags Waveform B. B) Waveform A leads Waveform B.
367
22)
Refer to Figure 11-3. The phase shift between Waveform A and Waveform B expressed in radians is: A) π/2 rad B) π/4 rad C) 2π rad D) 2π/6 rad
23)
One radian equals: A) 45° B) 90° C) 57.3° D) 114.6°
24)
A sine wave reaches its rms value of 70.7% of the peak voltage at an angle of how many radians? A) 2π rad B) π/4 rad C) π/2 rad D) π rad
25)
A complete sine wave contains: A) π radians B) 2π radians C) 3π/2 radians D) π/2 radians
Figure 11-4 26)
The period of the waveform in Figure 11-4 is: A) 1 kHz B) 100 Hz C) 1.0 mS D) 0.5 mS
27)
If the rms value of the waveform in Figure 11-4 is 10 V rms, what is the peak voltage? A) 28.28 V B) 14.14 V C) 20 V D) 7.07 V
368
28)
If the rms value of the waveform in Figure 11-4 is 100 V rms, what is the peak voltage? A) 288.28 V B) 141.4 V C) 200 V D) 70.07 V
29)
If the rms value of the waveform in Figure 11-4 is 12 V rms, what is the peak voltage? A) 16.97 V B) 33.94 V C) 8.48 V D) 15.3 V
30)
The frequency of the waveform in Figure 11-4 is: A) 1 Hz B) 1 kHz C) 1000 cycles per second D) Both B and C are correct.
31)
How much time has elapsed when the waveform in Figure 11-4 reaches its rms value? A) 0.25 mS B) 0.5 mS C) 0.707 mS D) 0.125 mS
32)
The equation for finding the frequency, when the period is known is: A) T = 0.707f B) T = π/2(f) C) T = 1/f D) f = 1/T
33)
The equation for finding the period, when the frequency is known is: A) T = 0.707f B) T = π/2(f) C) T = 1/f D) f = 1/T
34)
The coordinate system consists of: A) two perpendicular lines. B) two axes, x and y. C) four quadrants. D) all of the above.
35)
A sine wave reaches half of its peak value at: A) 45° B) 30° C) 90° D) 60°
369
36)
The phase difference between sine waves of different frequencies is: A) equal to their frequency differences. B) the same throughout time. C) constantly changing. D) the difference in their fixed time displacement.
37)
The negative alternation of a sine wave of current indicates that: A) the current is the opposite direction and polarity across components changes. B) it is different in amplitude from its positive alternation. C) power in the circuit will be less during this time. D) the current has changed direction.
38)
The instantaneous value of a sine wave is: A) equal to 1.414 of the rms value. B) constantly changing. C) 0.707 of its peak value. D) equivalent to a like DC value.
39)
What is the period of a 50 kHz sine wave? A) 5 μs B) 20 μs C) 50 μs D) 5 × 104 s
40)
What angle in degrees is equivalent to π/3 radians? A) 30° B) 60° C) 90° D) 120°
41)
If the effective voltage of an ac receptacle is 120 V, what is the peak-to-peak voltage? A) 84.8 V B) 169.7 V C) 240 V D) 339.4 V
42)
Find the frequency of a periodic wave that has a period of one hour. A) 278 mHz B) 2.78 mHz C) 0.278 mHz D) 27.8 mHz
43)
If a sine wave signal measured 100 mV peak-to-peak, how many volts would be read by a voltmeter? A) 14.14 mV B) 35.4 mV C) 63.7 mV D) 70.7 mV
370
44)
The ________ is defined as the time the output is active divided by the total period it could possibly be active. A) pulse width B) on time C) active ratio D) duty cycle
45)
On an oscilloscope display: A) voltage is on the vertical axis and time is on the horizontal axis. B) a straight diagonal trace means voltage is changing at a steady rate. C) a flat horizontal trace means voltage is constant. D) all the above
46)
Compensating a probe is necessary to: A) balance the electrical properties of the oscilloscope probe with the oscilloscope. B) to provide a reference point for making measurements. C) prevent damaging the circuit being tested. D) both A and C
47)
The time base control of the oscilloscope does which of the following? A) adjusts the vertical scale B) shows you the current time of day C) sets the amount of time represented by the horizontal width of the screen D) sends a clock pulse to the probe
48)
In a circuit that contains a dc voltage source of 15 V and an ac voltage source of 50 Vp and a resistive load, what would be the maximum and minimum voltage measured across the resistive load? A) VMAX = 65 V, VMIN = -35 V B) VMAX = 65 V, VMIN = zero V C) VMAX = 35 V, VMIN = 65 V D) VMAX = zero V, VMIN = 35 V
49)
A four-pole generator has a rotation speed of 400 rps. Determine the frequency of the output voltage. A) 200 Hz B) 400 Hz C) 800 Hz D) 1600 Hz
50)
A phasor represents a time-varying quantity in terms of both magnitude and direction. (True/False)
51)
The angular position of a phasor represents the amplitude of a sine wave. (True/False)
52)
The length of a phasor represents the amplitude. (True/False)
53)
Phasor algebra for sinusoidal quantities is applicable only for waveforms which have different frequencies. (True/False)
371
54)
The length of a phasor is called the magnitude. (True/False)
55)
The length of a phasor is represented by the formula v = Vpsin θ. (True/False)
56)
The opposite side of a right triangle is equal to the hypotenuse times the sine of the angle θ.
57)
Phasors are a convenient graphic way of representing sine wave voltages in terms of their: A) magnitude and amplitude B) magnitude and frequency C) magnitude and direction D) magnitude and period
58)
The length of the phasor "arrow" represents the ________ of a quantity. A) direction B) amplitude C) angle D) frequency
59)
The instantaneous value of a sine wave voltage, at any point, is equal to the vertical distance from the tip of the phasor to the: A) vertical axis B) horizontal axis C) positive peak D) negative peak
60)
The equivalent negative angle for a positive angle of 30 degrees is: A) 330 degrees B) -60 degrees C) -330 degrees D) -150 degrees
61)
Determine the instantaneous value of a sine wave whose rms is 20 V, and phase angle is - 297 degrees. A) -17.8 V B) +17.8 V C) +25.2 V D) -25.2 V
62)
The velocity of rotation is called: A) the frequency B) the wavelength C) the angular velocity D) the speed
63)
Which of the following expresses angular velocity? A) π/T B) f/T C) 2π/T D) 2π × T
372
64)
What is the angular velocity of a phasor representing a sine wave with a frequency of 1.4 kHz? A) 8796 rad/s B) 4398 rad/s C) 880 rad/s D) 440 rad/s
373
1)
TRUE
27)
B
45)
D
2)
TRUE
28)
B
46)
D
3)
TRUE
29)
A
47)
C
4)
TRUE
30)
D
48)
A
5)
TRUE
31)
D
49)
C
6)
FALSE
32)
D
50)
TRUE
7)
TRUE
33)
C
51)
FALSE
8)
FALSE
34)
D
52)
TRUE
9)
TRUE
35)
B
53)
FALSE
10)
TRUE
36)
C
54)
TRUE
11)
B
37)
A
55)
TRUE
12)
D
38)
D
56)
TRUE
13)
E
39)
B
57)
C
14)
C
40)
B
58)
B
15)
A
41)
D
59)
B
16)
C
42)
C
60)
C
17)
C
43)
B
61)
C
18)
C
44)
D
62)
C
19)
C
63)
C
20)
A
64)
A
21)
B
22)
A
23)
C
24)
B
25)
B
26)
C
374
Principles of Electric Circuits Chapter 12: Capacitors 1)
Capacitance is the measure of a capacitor's ability to store an electrical charge. (True/False)
2)
Energy is stored in a capacitor in a magnetic field, concentrated in the dielectric. (True/False)
3)
Five time constants are required to fully charge or discharge a capacitor. (True/False)
4)
Current lags the voltage by 90 degrees in a capacitor. (True/False)
5)
The capacitive reactance of a capacitor is inversely proportional to the frequency, and directly proportional to the capacitance. (True/False)
6)
Increasing the plate area of a capacitor increases the capacitance. (True/False)
7)
Increasing the distance between the plates of a capacitor increases the capacitance. (True/False)
8)
A material with a high dielectric constant increases the capacitance. (True/False)
9)
Exceeding the dielectric strength of a capacitor means you have applied too high a voltage, and probably destroyed the capacitor. (True/False)
Figure 12-1 10)
The largest capacitor in Figure 12-1 has the highest charge voltage. (True/False)
11)
The charge voltage across each capacitor in Figure 12-1 is inversely proportional to the size of the capacitor in microfarads. (True/False)
12)
A capacitor stores voltage in its electromagnetic field. (True/False)
13)
The material used in between the plates of a capacitor is called the dielectric. (True/False)
14)
A discharged capacitor has a positive charge on its plates. (True/False)
375
15)
The current through a capacitor is always maximum at the start of the first time constant. (True/False)
16)
A charged capacitor could be considered a voltage source. (True/False)
17)
All dielectrics are made from conducting material. (True/False)
18)
Determine the capacitance when Q = 75 μC and V = 20 V. A) 3.75 μF B) 150 μF C) 0.27 μF D) 9.8 μF
19)
Which of the following capacitors is polarized and will be destroyed if connected in reverse polarity? A) electrolytic B) ceramic C) mylar D) paper
Figure 12-1 20)
Which capacitor in Figure 12-1 is charged to the highest voltage? A) C1 B) C2 C) C3
21)
The three capacitors in series in Figure 12-1 have an equivalent value: A) equal to the sum of the three values B) equal to the product of the three values C) less than the value of the smallest capacitor
22)
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1. A) 0.8 μF B) 0.58 μF C) 0.060 μF D) 0.01 μF
376
23)
A capacitor has a capacitance of 0.1 μF with air as the dielectric. If the dielectric material is changed from air to a ceramic material with a dielectric constant of 600, what is the new capacitance? (Note: neither air nor ceramic capacitors this large are available). A) 0.006 μF B) 60 μF C) 6.28 μF D) 600 μF
24)
The main disadvantage of an electrolytic capacitor is: A) low capacitance values B) low voltage values C) low leakage current D) high leakage current
25)
A capacitor rated at 10,000 μF would typically have what kind of dielectric? A) tantalum oxide B) aluminum oxide C) ceramic D) Both A and B are correct
Figure 12-2 26)
Calculate the equivalent value of the parallel capacitors in Figure 12-2. A) 4.7 μF B) 15510 μF C) 90 μF D) 227.24 μF
27)
What is the maximum DC voltage that can safely be applied to the circuit in Figure 12-2? A) 100 V B) 35 V C) 50 V D) 185 V
28)
Assuming the capacitors in Figure 12-2 are non-polarized electrolytic capacitors, what is the maximum AC rms voltage that can safely be applied to the circuit? A) 35 V B) 24 V C) 49.5 V D) 100 V
377
Figure 12-3 29)
If the capacitor in Figure 12-3 is 0.1 μF and it is replaced by a 0.47 μF capacitor, which value capacitor will have the higher reactance? A) the 0.1 μF capacitor B) the 0.47 μF capacitor
30)
If the capacitor in Figure 12-3 is 0.1 μF and it is replaced by a 0.47 μF capacitor, which value capacitor will cause the higher circuit current? A) the 0.1 μF capacitor B) the 0.47 μF capacitor
31)
Calculate the reactance of the capacitor if it has a value 1.0 μF and the frequency is 10 kHz. A) 159.0 Ω B) 198.75 Ω C) 15.9 Ω D) 227.87 Ω
32)
A capacitor produces a capacitive reactance of 18.086 kΩ with an applied voltage having a frequency of 400 Hz. What is the capacitor value? A) 22 μF B) 0.47 μF C) 0.022 μF D) 0.22 μF
33)
Three 2 μF capacitors are connected in parallel. What is the total capacitance? A) 0.66 μF B) 6.0 μF C) 66.0 μF D) 2.0 μF
378
Figure 12-4 34)
Given the circuit in Figure 12-4, what is the approximate voltage across the capacitor at the end of 1 second? A) 10 V B) 63 V C) 6.3 V D) 0.707 V
35)
The charge of a capacitor is: A) on the top plate only. B) on the bottom plate only. C) in the dielectric. D) on both plates.
36)
If the distance between the plates of a capacitor is increased, the capacitance ________. A) increases B) decreases C) remains the same D) can't tell with information given
37)
What would we call a 0.000000000001 F capacitor? A) 1F B) 0.001 F C) 1 μF D) 1 pF
38)
A discharged capacitor has ________ charges on its plates. A) no B) unequal C) equal D) negative
39)
Select the equation below that represents the relationship between charge, capacitance, and voltage for a capacitor. A) C = QV B) Q = CV 1 2 fC
C)
XC =
D)
V = IR
379
40)
A capacitor has a plate area of 0.01 square meters. A second capacitor has a plate area of 0.02 square meters. If the first capacitor is 0.1 μF, what is the value of the second capacitor? A) 0.05 μF B) 0.1 μF C) 0.2 μF D) 0.4 μF
41)
One time constant can be solved by: A) multiplying capacitance by resistance B) dividing capacitance by resistance C) subtracting capacitance from reactance D) adding capacitance to resistance
42)
How long does it take to fully charge a 5 μF capacitor in series with a 5 MΩ resistor? A) 25 seconds B) 125 seconds C) 125 milliseconds D) 125 microseconds
43)
Using the formula for capacitive reactance, as frequency decreases: A) capacitive reactance decreases B) capacitive reactance increases C) capacitive reactance remains the same D) capacitive reactance changes by 90 degrees
44)
If three capacitors with values of 33 μF, 40 μF and 88 μF were connected in parallel, what is the total capacitance? A) 15 μF B) 53.7 μF C) 102.1 μF D) 161 μF
45)
A capacitor with a 20 volt charge is being discharged, how many volts are across the capacitor after the third time constant? A) 1V B) 5V C) 12.64 V D) 19 V
46)
Phase relationship of voltage and current in a reactive circuit is: A) an in-phase relationship B) dependant upon the magnitude of the applied voltage C) dependent upon circuit current D) an out-of-phase relationship
47)
Why are capacitors used as coupling components in many circuits? A) because the dielectric has a high dc resistance B) because they pass ac and block dc C) because they pass dc and block ac D) both A and B
380
48)
If the applied frequency is 1 kHz and the capacitance is 10 μF, calculate the capacitive reactance XC. A) 15.9 Ω B) 15.9 kΩ C) 15.9 MΩ D) 0.063 Ω
49)
Power that is stored from the source and then returned to the source is called: A) true power B) reactive power C) apparent power D) instantaneous power
50)
If an open capacitor is checked with an analog ohmmeter, the meter should show: A) zero B) infinity C) movement from zero to infinity D) movement from infinity to zero
Complete the following tables: Frequency = 60 Hz 51)
Capacitance 1.0 μF
Reactance __________
52)
Capacitance 0.10 μF
Reactance __________
53)
Capacitance 100 μF
Reactance __________
54)
Capacitance 1000 μF
Reactance __________
55)
Capacitance 4000 μF
Reactance __________
Complete the following tables: Frequency = 400 Hz 56)
Capacitance 0.01 μF
Reactance __________
57)
Capacitance 0.47 μF
Reactance __________
58)
Capacitance 1.0 μF
Reactance __________
381
59)
Capacitance 10 μF
Reactance __________
60)
Capacitance 100 μF
Reactance __________
Complete the following tables: Frequency = 1.0 kHz 61) Capacitance 0.001 μF
Reactance __________
62)
Capacitance 0.01 μF
Reactance __________
63)
Capacitance 0.022 μF
Reactance __________
64)
Capacitance 1.0 μF
Reactance __________
65)
Capacitance 10.0 μF
Reactance __________
Complete the following tables: Frequency = 10 kHz 66)
Capacitance 0.0001 μF
Reactance __________
67)
Capacitance 0.001 μF
Reactance __________
68)
Capacitance 0.022 μF
Reactance __________
69)
Capacitance 0.01 μF
Reactance __________
70)
Capacitance 1.0 μF
Reactance __________
Complete the following tables: Frequency = 100 kHz 71)
Capacitance 100 pF
Reactance __________
382
72)
Capacitance 1000 pF
Reactance __________
73)
Capacitance 0.002 μF
Reactance __________
74)
Capacitance 0.01 μF
Reactance __________
75)
Capacitance 1.0 μF
Reactance __________
Complete the following tables: Frequency = 1.0 MHz 76)
Capacitance 10 pF
Reactance __________
77)
Capacitance 100 pF
Reactance __________
78)
Capacitance 1000 pF
Reactance __________
79)
Capacitance 100 μF
Reactance __________
80)
Capacitance 1000 μF
Reactance __________
81) An amplifier with a switched-capacitor circuit has an input resistor value of 20K ohms. If you want the switched-capacitor to emulate a 10K ohm resistor, at what frequency must the switch be operated? The value of the capacitor is 1000pF. A) 100KHz B) 50KHz C) 25KHz D) 20KHz 82)
Switched-capacitor circuits A) can be used as radio filters. B) can be used in programmable circuits. C) can be readily changed by reprogramming. D) all of the above E) none of the above
383
1)
TRUE
22)
C
43)
B
64)
159.23 Ω
2)
FALSE
23)
B
44)
D
65)
15.9 Ω
3)
TRUE
24)
D
45)
A
66)
159.23K
4)
FALSE
25)
B
46)
D
67)
15.9 kΩ
5)
FALSE
26)
C
47)
D
68)
723.4 Ω
6)
TRUE
27)
B
48)
A
69)
1.59 kΩ
7)
FALSE
28)
B
49)
B
70)
15.9 Ω
8)
TRUE
29)
A
50)
B
71)
15.9 kΩ
9)
TRUE
30)
B
51)
2.65kΩ
72)
1.59 kΩ
10)
FALSE
31)
C
52)
26.5 kΩ
73)
796.18 Ω
11)
TRUE
32)
C
53)
26.5 Ω
74)
159.2 Ω
12)
FALSE
33)
B
54)
2.65 Ω
75)
1.59 Ω
13)
TRUE
34)
A, C
55)
0.66 Ω
76)
15.9 kΩ
14)
FALSE
35)
D
56)
39.8 Hz
77)
1.59 kΩ
15)
TRUE
36)
B
57)
847 Ω
78)
159.15 Ω
16)
TRUE
37)
D
58)
398 Ω
79)
0.0016 Ω
17)
FALSE
38)
C
59)
39.8 Ω
80)
0.00016 Ω
18)
A
39)
B
60)
3.98 Ω
81)
B
19)
A
40)
C
61)
159.2 kΩ
82)
D
20)
A
41)
A
62)
15.9 kΩ
21)
C
42)
B
63)
7.2 kΩ
384
Principles of Electric Circuits Chapter 13: Inductors 1)
Any conductor with a current flowing through it is an inductor. (True/False)
2)
Energy is stored in an inductor in its electrostatic field. (True/False)
3)
Voltage leads the current in an inductor. (True/False)
4)
An inductor opposes any change in current through it. (True/False)
5)
The Unit of inductance is the Henry. (True/False)
6)
Inductive reactance opposes a sine wave current, thus reducing the current. (True/False)
7)
Inductive reactance is measured in henrys. (True/False)
8)
Inductance is measured in henrys. (True/False)
9)
Inductance is the property that opposes current flow. (True/False)
10)
Self-inductance is another term for cemf. (True/False)
11)
The rate of change of magnetic flux and its proportional induced emf is a form of Lenz's law. (True/False)
12)
The inductor always opposes a change in current, even a transient DC current. (True/False)
13)
Inductors in parallel increase the total cemf in the circuit. (True/False)
14)
The time constant of an inductive circuit is dependent on the circuit's resistance. (True/False)
15)
A length of wire is wound into a coil to: A) concentrate the magnetic fields B) increase the inductance C) increase the reactance D) all of the above
16)
When a voltage is induced into a conductor (or coil) as a result of a change in current through the conductor, the phenomenon is called: A) the Henry B) self inductance C) parasitic inductance D) divine inductance
385
17)
Calculate the induced voltage across a 1 H inductor when the current is changing at a rate of 2 mA per microsecond. A) 2V B) 20 V C) 200 V D) 2000 V
Figure 13-1 18)
What is the total inductance in the circuit in Figure 13-1? A) 602 mH B) 800 mH C) 2100 mH D) 2600 μH
19)
If the frequency in the circuit in Figure 13-1 is increased to 100 kHz, what is the total inductance? A) 602 mH B) 1600 mH C) 5200 mH D) 2600 μH
20)
In addition to pure inductance, what other physical parameters does an inductor contain? A) reactance B) winding resistance C) capacitance D) both B and C
Figure 13-2
386
21)
What is the total inductance in the circuit in Figure 13-2? A) 4.0 H B) 8.0 H C) 1.0 H D) 2.6 H
22)
What is the total reactance in Figure 13-2? A) 3,760.8 Ω B) 376.8 Ω C) 753.6 Ω D) 37.68 Ω
23)
In an inductor: A) the voltage leads the current by 90 degrees. B) the voltage leads the current by 45 degrees. C) the current leads the voltage by 90 degrees. D) there is no phase shift between voltage and current.
Figure 13-3 24) In the circuit in Figure 13-3, one time constant is equal to: A) 0.2 mS B) 20 μS C) 200 μS D) 2.0 μS 25)
In the circuit in Figure 13-3, how much time will elapse, after the switch is closed, before the current reaches its maximum value? A) 1.0 mS B) 100 μS C) 1000 μS D) 10 μS
26)
At the instant the switch is closed in Figure 13-3, the inductor acts as: A) a black hole. B) an open circuit. C) a short circuit. D) a simple resistance.
27)
The direction of the cemf produced by an inductor is explained by: A) Lenz's law. B) Faraday's law. C) Ohm's law. D) Henry's law.
387
28)
When a DC voltage is first switched across an LR circuit: A) all the voltage is dropped across the resistor. B) all the current will flow through the resistor. C) the maximum current flow is dependent on the inductor. D) all the voltage is dropped across the inductor.
29)
Before an inductor can reach a steady state after a current transient: A) the voltage drop across it must remain constant. B) five time constants must have elapsed. C) the circuit must be interrupted. D) none of the above.
30)
The notation A) B) C) D)
di dt
in an inductor refers to:
the rate of change of current with respect to time. the ratio of current to number of turns. the magnetizing force applied per turn. the coil permeability as a function of temperature.
31)
Physical properties that affect inductance are: A) number of turns, voltage applied, and core material. B) core material, number of turns, length of the core, and cross-sectional area of the core. C) area of the coil, permeability, and amount of current flow. D) length of coil, resistance of the coil's wire, and number of turns.
32)
When current flow tries to decrease in an inductor, the magnetic field ________ and the voltage polarity across the inductor ________. A) collapses, remains the same B) expands, reverses C) collapses, reverses D) expands, remains the same
33)
After two time constants on the de-energizing transient, the voltage across the inductor has changed by ________%, leaving ________% remaining. A) 86.5, 13.5 B) 13.5, 86.5 C) 63.2, 36.8 D) 76.7, 23.3
34)
Opposition to current flow without the dissipation of energy in an inductor is called: A) resistance B) inductive reactance C) counter emf D) impedance
35)
If an air core inductor has an inductance of 2 μH, what will the inductance become if an iron core is inserted? Assume that the iron core has a relative permeability (μ) of 1000. A) 63 μH B) 2000 μH C) 2nH D) 2 μH
388
36)
Three parallel inductors of 84 mH, 96 mH, and 160 mH will provide a total inductance of: A) 35 mH B) 113 mH C) 205 mH D) 340 mH
37)
The formula used to determine total inductance when inductors are connected in parallel is similar to: A) series resistors B) series capacitors C) parallel capacitors D) series inductors
38)
When measured, the voltage drop across an open series-connected inductor is: A) very low B) varies with other components C) minus what is dropped across the meter D) source voltage
39)
In an inductive circuit, reactance will: A) increase with frequency increase B) decrease with frequency increase C) be independent from frequency D) depend upon the value of XC
40)
A change of one ampere per second is an inductor that induces a voltage of one volt is considered as which unit value? A) a Lenz B) an Ohm C) a Henry D) a Farad
41)
As the resistance of an LR circuit is increased: A) the circuit time constant is decreased. B) the circuit time constant is increased. C) self-inductance is partially reduced. D) counter emf is greatly increased.
42)
A series RL circuit, with R=33K and L=20mH, is driven by a square wave. What is the highest frequency that can be used and still observe the complete waveform? A) 16.5KHz B) 60KHz C) 606KHz D) 165KHz
389
Complete the following tables. Frequency = 60 Hz 43)
Inductance 10 mH
Reactance __________
44)
Inductance 33 mH
Reactance __________
45)
Inductance 100 mH
Reactance __________
46)
Inductance 2H
Reactance __________
47)
Inductance 10 H
Reactance __________
Complete the following tables. Frequency = 400 Hz 48)
Inductance 0.33 mH
Reactance __________
49)
Inductance 100 mH
Reactance __________
50)
Inductance 1000 mH
Reactance __________
51)
Inductance 1.0 H
Reactance __________
52)
Inductance 2H
Reactance __________
Complete the following tables. Frequency = 1.0 kHz 53)
Inductance 0.6 mH
Reactance __________
54)
Inductance 250 mH
Reactance __________
55)
Inductance 1000 mH
Reactance __________
390
56)
Inductance 1000 μH
Reactance __________
57)
Inductance 200 μH
Reactance __________
Complete the following tables. Frequency = 10 kHz 58)
Inductance 100 mH
Reactance __________
59)
Inductance 25 mH
Reactance __________
60)
Inductance 1000 μH
Reactance __________
61)
Inductance 0.1 H
Reactance __________
62)
Inductance 1.0 H
Reactance __________
Complete the following tables. Frequency = 100 kHz 63)
Inductance 1000 μH
Reactance __________
64)
Inductance 100 μH
Reactance __________
65)
Inductance 10 μH
Reactance __________
66)
Inductance 1 μH
Reactance __________
67)
Inductance 0.1 H
Reactance __________
391
Complete the following tables. Frequency = 1.0 MHz 68)
Inductance 1 mH
Reactance __________
69)
Inductance 10 mH
Reactance __________
70)
Inductance 100 μH
Reactance __________
71)
Inductance 10 μH
Reactance __________
72)
Inductance 1 μH
Reactance __________
392
1)
FALSE
25)
B
49)
251.3 Ω
2)
FALSE
26)
A
50)
2.51 kΩ
3)
TRUE
27)
A
51)
2.51 kΩ
4)
TRUE
28)
D
52)
5.03 kΩ
5)
TRUE
29)
B
53)
3.77 Ω
6)
TRUE
30)
A
54)
1.57 kΩ
7)
FALSE
31)
B
55)
6.28 kΩ
8)
TRUE
32)
C
56)
6.28 Ω
9)
FALSE
33)
A
57)
1.26 Ω
10)
TRUE
34)
B
58)
6.28 kΩ
11)
FALSE
35)
B
59)
1.57 kΩ
12)
TRUE
36)
A
60)
2.83 Ω
13)
FALSE
37)
B
61)
6.28 kΩ
14)
TRUE
38)
D
62)
62.83 kΩ
15)
D
39)
A
63)
628.32 kΩ
16)
B
40)
C
64)
62.83 Ω
17)
D
41)
A
65)
6.28 Ω
18)
D
42)
D
66)
0.628 Ω
19)
D
43)
3.77 Ω
67)
62.8 Ω
20)
D
44)
12.44 Ω
68)
6.28 kΩ
21)
C
45)
37.69 Ω
69)
62.83 kΩ
22)
B
46)
753.98 Ω
70)
628.3 Ω
23)
A
47)
3.77 kΩ
71)
62.83 Ω
24)
B
48)
0.829 Ω
72)
6.28 Ω
393
Principles of Electric Circuits Chapter 14: Transformers 1)
Mutual inductance is defined as the inductance between two coils that are coupled by a magnetic field. (True/False)
2)
A transformer consists of two or more coils, sharing a common core, or placed close together, and are magnetically coupled. (True/False)
3)
A transformer can be used to step-up (increase) voltage. (True/False)
4)
A transformer can be used to step-down (decrease) voltage. (True/False)
5)
A transformer can be used to step-up (increase) power. (True/False)
6)
The product of input voltage and input current in a transformer is equal to the output voltage times the output current, minus any power losses. (True/False)
7)
A transformer can pass a steady state DC from the primary to the secondary. (True/False)
8)
A transformer can pass an AC voltage from the primary into the secondary winding. (True/False)
9)
When mutual inductance is present, current changing in one circuit causes an induced voltage in the other circuit. (True/False)
10)
In a step-up transformer, the secondary has more turns of wire than does the primary. (True/False)
11)
Voltage and current in the primary of a step-down transformer equals the voltage and current in the secondary. (True/False)
12)
Even though two coils do not share a common core, they can have some amount of mutual inductance. (True/False)
13)
Secondary current in a transformer depends on secondary voltage and load resistance values. (True/False)
14)
In a given transformer, 87% of the flux produced in the primary is coupled into the secondary winding, what is the coefficient of coupling? A) 13 B) 0.13 C) 87 D) 0.87
15)
Laminated iron-core (or silicon steel) transformers are normally used for: A) power line (60 Hz) applications. B) audio frequency applications (20 Hz to 20 kHz). C) radio frequency applications, above 100 kHz. D) Both A and B are correct.
394
16)
Air core or ferrite core transformers are normally used for: A) power line (60 Hz) applications. B) audio frequency applications (20 Hz to 20 kHz). C) radio frequency applications, above 100 kHz. D) Both A and B are correct.
Figure 14-1 17)
If the input voltage (Vin) in Figure 14-1 is 120 V rms and the output voltage (Vout) is 24 V rms, what is the turns ratio? A) 120:1 B) 5:1 C) 24:1 D) 1.2:1
18)
If the primary (L1) in Figure 14-1 has 40 turns and the secondary (L2) has 10 turns, what is the reflected resistance into the primary if the load resistance is 120 Ω? A) 1920 Ω B) 7.5 Ω C) 480 Ω D) 30 Ω
19)
The input voltage (Vin) in Figure 14-1 measures 120 V rms and the output voltage (Vout) measures 13.6 V rms. The manufacturer specifies an output voltage of 12.6 V at 1 A, with an input voltage of 120 V. The load resistance is 1kΩ. Why is the measured output voltage higher than the manufacturer specifies? A) The output voltage will be higher if the load current is lower than specified. B) The output voltage will be higher if the load current is higher than specified. C) There is a voltage drop across the secondary winding resistance. D) Both A and C are correct.
20)
The two round dots in Figure 14-1 indicate: A) the two connections are in phase. B) both marked connections are positive, while the unmarked connections are negative. C) both marked connections are negative, while the unmarked connections are positive. D) none of the above.
21)
Which of the following are common applications for a transformer? A) to step-up or step-down voltage B) impedance matching C) isolation from the AC power line D) all of the above
395
Figure 14-2 22)
The transformer shown in Figure 14-2 is called: A) an Isolation transformer B) a step-up transformer C) an Autotransformer D) an impedance transformer
23)
Eddy currents, magnetic flux leakage, winding resistance and hysteresis in a transformer, all contribute to: A) mutual inductance B) power transfer from primary to secondary C) power loss and heat D) voltage step-up or step-down
24)
Which transformer loss is greatly reduced by laminating the iron core? A) winding resistance losses B) Barkhausen losses C) radio frequency losses D) Eddy current losses E) Both A and B are correct
Figure 14-3
396
25)
Given the transformer in Figure 14-3, what is the maximum theoretical current rating of the Secondary? A) 2.08 A B) 41.67 A C) 25.4 A D) 12.78 A
26)
Given the transformer in Figure 14-3, what is the maximum theoretical current rating of the Primary? A) 2.08 A B) 41.67 A C) 25.4 A D) 12.78 A
27)
What is the maximum secondary power of the transformer in Figure 14-3, if the transformer is 85% efficient? A) 4.25 kVA B) 5.85 kVA C) 5 kVA D) 12.78 kVA
28)
What is the mutual inductance between two coils in a transformer when: 1) Primary inductance = 1.0 μH 2) Secondary inductance = 4.0 μH 3) K = 0.8? A) 1.6 μH B) 3.2 μH C) 0.8 μH D) 4.8 μH
29)
A transformer with an air core, ferrite core, or powdered iron core is used for: A) power line frequency applications. B) audio frequency applications. C) radio frequency applications. D) all of the above.
30)
Increasing the number of turns of wire on the secondary of a transformer will: A) decrease the output voltage. B) increase the output voltage. C) have no effect on the output voltage. D) increase the primary input voltage.
31)
With 100 V of primary voltage and a 1:3 turns ratio, the output voltage will be: A) 100 V B) 33 V C) 300 V D) Not enough information is given.
32)
With a primary current of 3 A and a 1:3 turns ratio, the output current will be: A) 3A B) 333 A C) 1A D) 3 mA
397
33)
A transformer has 430 turns in the primary and 200 turns in the secondary. What is the P-S turns ratio? A) 230:1 B) 2.15:1 C) 1:2.15 D) 1:30
34)
A transformer has a 1:6 P-S turns ratio and has a primary current of 12 A. What is the secondary current? A) 72 A B) 12 A C) 6A D) 2A
35)
The ability to transfer energy from one coil to another is dependent upon ________. A) mutual inductance B) coupling C) self-inductance D) impedance matching
36)
A transformer that has only one winding is called a(n) ________. A) tapped transformer B) multiple-winding transformer C) autotransformer D) nonideal transformer
37)
In a transformer, the primary and secondary coils are magnetically linked, but they are: A) always in phase B) electrically isolated C) current connected D) mechanically connected
38)
A voltage is induced in a transformer secondary winding by the action of the: A) primary magnetic field B) secondary magnetic field C) primary turns ratio D) secondary cemf
39)
In a transformer, how will the primary cemf be affected, if the secondary current decreases? A) It will remain the same. B) It will be canceled. C) It will increase. D) It will decrease.
40)
The ratio of the amount of flux linking the secondary coil versus the primary flux is called the: A) coefficient of accuracy B) coefficient of coupling C) a Weber D) turns ratio
398
41)
A transformer would never be selected to: A) step-up or step-down voltage B) isolate stages C) match impedances D) step-up or step-down power
42)
Which coil would be used as a step-up transformer primary, if coil number 1 has 100 more turns than coil number 2? A) Coil windings must be the same. B) coil 1 C) Coil ratio is too small. D) coil 2
43)
What is the number of turns in a primary winding, when the voltage applied to a transformer input is 24 Vac with a 225-turn secondary output of 360 V? A) 15 B) 3325 C) 5400 D) 9.38
44)
When a transformer secondary lead has been connected to a middle loop that is not an end connector, it is called a: A) multiple-tapped secondary B) center-tapped secondary C) multiple winding secondary D) single winding secondary
45)
What transformer-turns relationship will be needed to match an output impedance of 5 kΩ to an 8 Ω speaker? A) 5:1 B) 8:1 C) 25:1 D) 625:1
399
1)
TRUE
24)
D
2)
TRUE
25)
B
3)
TRUE
26)
A
4)
TRUE
27)
A
5)
FALSE
28)
A
6)
TRUE
29)
C
7)
FALSE
30)
B
8)
TRUE
31)
C
9)
TRUE
32)
C
10)
TRUE
33)
B
11)
FALSE
34)
D
12)
TRUE
35)
A
13)
TRUE
36)
C
14)
D
37)
B
15)
D
38)
A
16)
C
39)
C
17)
B
40)
B
18)
A
41)
D
19)
D
42)
D
20)
A
43)
A
21)
D
44)
B
22)
C
45)
C
23)
C
400
Principles of Electric Circuits Chapter 15: RC Circuits 1)
The current leads the voltage in a series resistor capacitor circuit. (True/False)
2)
The term impedance, applied to a circuit containing capacitance and resistance, is the phasor result of the resistance and capacitive reactance. (True/False)
3)
Impedance is defined as the total opposition to current in an AC circuit. (True/False)
4)
The current always leads the voltage in a series resistor capacitor circuit by 45°. (True/False)
5)
A High-pass filter rejects higher frequencies and passes lower frequencies. (True/False)
Figure 15-1 6)
The true power delivered to the load in Figure 15-1 is the power that does the work. (True/False)
7)
The reactive power in Figure 15-1 creates a loss in the form of heat. (True/False)
8)
If a capacitor generates heat, it is probably defective. (True/False)
9)
The susceptance of a given capacitor decreases as frequency increases. (True/False)
10)
The amount of total voltage applied to a capacitor affects its capacitive reactance. (True/False)
11)
A 10 mF capacitor will have more capacitive reactance than a 20 mF capacitor in a 60 Hz circuit. (True/False)
12)
In a series capacitive circuit, the smallest capacitor has the largest voltage drop. (True/False)
13)
In a parallel capacitive circuit, all capacitors have the same amount of charge on them. (True/False)
401
Figure 15-2 14)
The current in the circuit in Figure 15-2 is: A) 49.76 mA. B) 1.4 mA. C) 14 mA. D) 1.0 mA.
15)
What is the phase angle in the circuit in Figure 15-2? A) 45° B) 90° C) 180° D) 60°
16)
What is the sine of the phase angle in the circuit in Figure 15-2? A) 1.0 B) 0.707 C) 0.57 D) 0.333
Figure 15-3 17)
The total current in the circuit in Figure 15-3 is: A) 49.3 mA B) 4.47m C) 1.33 mA D) 13.3 mA
18)
The phase angle in the circuit in Figure 15-3 is: A) -63.4° B) -22.6° C) 11.94° D) -11.94°
402
19)
The total current and phase angle in the circuit in Figure 15-3 expressed in polar form is: A) 4.47, 63.4° mA B) 4.93, -22.6° mA C) 1.33, 22.6° mA D) 1.33, -22.6° mA
20)
The voltage across the capacitor in Figure 15-3 is: A) π/XC B) 8.94 V rms C) 7.07 V rms D) 11.7 V rms The voltage across the resistor in Figure 15-3 is: A) π/XC B) 4.47 V rms C) 7.07 V rms D) 11.7 V rms
21)
Figure 15-4 22)
The total circuit impedance in Figure 15-4 is: A) 22.4 kΩ B) 7.07 kΩ C) 4.46 kΩ D) 0.144 kΩ
23)
The total circuit current in Figure 15-4 is: A) 44.6 mA. B) 0.707 mA. C) 1.4 mA. D) 10.04 mA.
24)
What is the phase angle in the circuit in Figure 15-4? A) 45° B) -45° C) 94.4° D) -60°
25)
The True power delivered to the load in Figure 15-4 is: A) 9.8 mW. B) I2R. C) 4.46 mW. D) Both A and B are correct.
403
26)
The Reactive power in the capacitor in Figure 15-4 is: A) 9.8 mW. 2 B) XC. C) 39.48 mVAR. D) Both B and C are correct.
27)
The Apparent power in the entire circuit in Figure 15-4 is: A) 19.9 mVAR B) 7.06 mVAR C) 39.48 mVA D) 4.46 mVAR
Figure 15-5 28)
The phase angle between the source voltage and the current in Figure 15-5 is: A) approximately zero degrees. B) 90 degrees. C) 22.5 degrees. D) approximately 60 degrees.
29)
Which of the following expresses the impedance (Z) correctly in Figure 15-5 in both rectangular and polar form? A) 1.88-j.78, 2.03 -22.5° B) 1.88-j.78, 7.07 45° C) 1.88-j.78, 2.03 -22.5°kΩ D) 1.88+j.78, 2.03 -22.5°kΩ
30)
What is the impedance of the circuit in Figure 15-5? A) 25 kΩ B) 2.03 kΩ C) 10 kΩ D) 5.897 kΩ
31)
Impedance equals the vector sum of: A) R and C. B) R and I. C) R and V. D) R and XC.
32)
In a series RC circuit, impedance is: A) always smaller than R. B) the same as R. C) larger than R. D) none of the above.
404
33)
Which of the following is the reference vector for parallel RC circuits? A) R B) V C) I D) C
34)
In a two-branch parallel RC circuit, impedance is: A) more than the resistive branch value. B) more than the capacitive reactance branch value. C) less than either branch's opposition to current. D) equal to the value of R.
35)
Express the impedance of a 10 μF capacitor at 60 Hz in rectangular form. A) 0 Ω - j265.3 Ω B) 265.3 Ω - j0 Ω C) 0 Ω - j0.00377 Ω D) 0 Ω + j265.3 Ω
36)
What is the susceptance of a 100 μF capacitor at 1000 Hz? A) 10-4 S B) 104 S C) 0.63 S D) 1.59 S
37)
A series circuit has an applied voltage of 100 V40° and a current of 1 A10°. Determine (ZT) total impedance and express in rectangular form. A) 64.3 Ω + j76.6 Ω B) 86.6 Ω + j50 Ω C) 100 Ω + j30 Ω D) 100 Ω + j50 Ω
38)
Which one of the following is true of ac circuits with reactive elements? A) The magnitude of the voltage across any one element can never exceed the applied voltage. B) The impedance of any one element can never exceed the total network impedance. C) The smaller the resistive element of a circuit, the closer the power factor is to unity. D) Depending on the frequency applied, the circuit can look either inductive or capacitive.
39)
What is the impedance of a RC circuit consisting of a 56 kΩ resistor in series with a 0.033 μF capacitor at a frequency of 450 Hz? A) 10,730 Ω B) 57,019 Ω C) 66,730 Ω D) 45,270 Ω
405
40)
What is the total current in a parallel circuit consisting of a 1.27 μF capacitor, and a 50 Ω resistor, with a 2.5 kHz, 10 Vac power supply? A) 141 mA B) 200 mA C) 282 mA D) 400 mA
41)
The frequency at which the capacitive reactance equals the resistance in a low-pass or highpass filter is called ________. A) cutoff frequency B) unity-gain frequency C) generator frequency D) gain bandwidth product
Figure 15-6 42)
See Figure 15-6. Which one of these statements is true? A) IS = I3 B) I3 = I4 + I6 C) I3 = I2 + I5 D) I4 = I6
43)
See Figure 15-6. Which equation describes the total impedance ZT? A) ZT = Z1 + ZT ' B) ZT = Z 1 + Z 3 + Z 5 + Z 6 ' C) ZT = Z1 Z2 D) ZT = Z1 + (Z2 ZT')
44)
See Figure 15-6. If E = 10 V 30° and IS = 5 A 10°, what is the total impedance ZT? A) 0.47 Ω - j0.17 Ω B) 1.88 Ω + j0.68 Ω C) 38.3 Ω + j32.1 Ω D) 32.1 Ω + j38.3 Ω
406
Figure 15-7 45)
See Figure 15-7. What is the total impedance ZT? A) 17.1 Ω -33.3° B) 39.2 Ω -30.7° C) 50.2 Ω -10.7° D) 49.3 Ω -9.4°
46)
See Figure 15-7. What is the total current IS? A) 0.2 A -10.7° B) 502 A 10.7° C) 5.0 A -10.7° D) 5.0 A -9.4°
47)
In an RC circuit, true power equals: A) VT × RT B) I2 R C) VS × I P D) P
48)
In an RC circuit, the power factor is referred to as: A) leading power factor B) lagging power factor C) true power factor D) apparent power factor
49)
Convert the following complex number from rectangular form to polar form: 8 + j6 A) 10 53.1° B) 10 36.9° C) 10 - 53.1° D) 10 - 36.9°
50)
Convert the following complex number from polar form to rectangular form: 200 - 45° A) 141 + jl41 B) 141 - jl41 C) 100 + j100 D) 100 - j100
407
51)
Add the following complex numbers: 40 + j10 and 25 - j7 A) 65 + jl7 B) 15 + j3 C) 65 + j3 D) 65 - j3
52)
Subtract the following complex numbers: (15 + j15) - (10 + j8) A) 5 + j7 B) 25 - j7 C) 5 - j7 D) 25 + j23
53)
Multiply the following complex numbers: (5 + j3) (2 - j4) A) 22 - j14 B) 22 + j14 C) 10 - j14 D) 22 + j26
54)
Divide the following complex numbers: (10 + j5) / (2 + j4) A) 40 - j30 B) 2 - j1.5 C) 2 + j1.5 D) 40 + j30
55)
Multiply the following complex numbers: (25 - 36.9°) (14 67°) A) 11 30.1° B) 350 - 30.1° C) 350 30.1° D) 350 103.9°
56)
Divide the following complex numbers: (100 27°) / (33 - 14.3°) A) 3 12.7° B) 3 41.3° C) 0.33 12.7° D) 0.33 41.3°
57)
Which one of the following polar values is equivalent to 30 + j40? A) 70 36.9° B) 70 53.1° C) 50 36.9° D) 50 53.1°
58)
Which one of the following rectangular values is equivalent to the polar form 20 55°? A) 11.17 - j16.38 B) 16.38 - j11.47 C) 11.47 + j16.38 D) 16.38 + j11.47
408
59)
Which one of the following values is equivalent to (5 + j3) (4 - j6)? A) 2 - j18 B) 38 - j18 C) 2 + j18 D) 38 + j18
60)
The time required for a phasor to go through 2π radians is the __________ of the sine wave. A) period B) frequency C) domain D) angle
61)
If the voltage v = 50 sin(500t - 75°) is applied across a 25 Ω resistor, which equation describes the resistor current? A) 2 cos(500t - 75°) B) 2 sin(20t - 3°) C) 2 sin(500t - 75°) D) 1250 sin(500t - 75°)
62)
Convert the following complex number from rectangular form to polar form: 40 + j30V A) 50 V -36.87° B) 50 V 36.87° C) 40 V 45° D) 40 V -53.1°
63)
Convert the following complex number from polar form to rectangular form: 40 V 120° A) -20 + j34.6V B) 20 + j34.6V C) 80 + j34.6V D) -80 - j34.6V
64)
Divide the following complex numbers: 500 40° ÷ 10 15° A) 250 -15° B) 250 15° C) 50 25° D) 50 -25°
65)
Add the following complex numbers: 75 + j18 and 50 - j9 A) 25 + j9 B) 125 + j27 C) 25 - j9 D) 125 - j27
66)
Subtract the following complex numbers: 20 -30° - 10 30° A) 17.3 -60° B) 17.3 60° C) 10 0° D) 10 360°
409
67)
Multiply the following complex numbers: 1000 -30° × 200 -20° A) 1200 50° B) 800 -50° C) 200,000 50° D) 200,000 -50°
68)
Which one of the following phasor domain expressions is equivalent to the time domain 50 sin(Y + 15°)? A) 35.35 15° B) 50 15° C) 70.7 15° D) 70.7 -15°
69)
In the complex plane, the horizontal is called the ________ and the vertical is called the ________. A) angular axis, quadrant axis B) imaginary axis, real axis C) real axis, imaginary axis D) phasor axis, wavelength
70)
The j operator is considered a: A) rotational vector B) rotational operator C) complex operator D) polar vector
71)
The phase shift oscillator uses ________ equal component RC networks that produce a required ________ degree phase shift. A) three, 180 B) four, 90 C) three, 90 D) four, 45
410
1)
TRUE
25)
D
49)
B
2)
TRUE
26)
D
50)
B
3)
TRUE
27)
B
51)
C
4)
FALSE
28)
C
52)
A
5)
FALSE
29)
C
53)
A
6)
TRUE
30)
B
54)
B
7)
FALSE
31)
D
55)
C
8)
TRUE
32)
C
56)
B
9)
FALSE
33)
B
57)
D
10)
FALSE
34)
C
58)
C
11)
TRUE
35)
A
59)
B
12)
TRUE
36)
C
60)
A
13)
FALSE
37)
B
61)
C
14)
B
38)
D
62)
C
15)
A
39)
B
63)
A
16)
B
40)
C
64)
C
17)
B
41)
A
65)
A
18)
A
42)
B
66)
A
19)
A
43)
D
67)
D
20)
B
44)
B
68)
A
21)
B
45)
C
69)
C
22)
B
46)
A
70)
B
23)
C
47)
B
71)
A
24)
B
48)
A
411
Principles of Electric Circuits Chapter 16: RL Circuits
Figure 16-1
1)
In the circuit in Figure 16-1, The current lags the voltage. (True/False)
2)
In the circuit in Figure 16-1, assume that XL, = R. The phase angle is 45°. (True/False)
3)
In the circuit in Figure 16-1, increasing the frequency will cause an increase in circuit current. (True/False)
4)
In the circuit in Figure 16-1, increasing the frequency will cause an increase in circuit impedance. (True/False)
5)
In the circuit in Figure 16-1, increasing the inductance value will cause an increase in circuit current. (True/False)
6)
Phase angle is the difference in phase between the applied voltage and the circuit current. (True/False)
7)
For series AC circuits containing both resistance and inductive reactance, total opposition to current flow cannot be found by simply adding ohmic values or oppositions to current. (True/False)
8)
For voltage-current vector diagrams relating to series RL circuits, current is the reference vector because it is the same throughout the circuit. (True/False)
9)
In a series RL circuit, the resultant voltage vector represents the total applied voltage. (True/False)
10)
For parallel AC circuits containing resistance and inductive reactance, total current can only be found using right-triangle methods. (True/False)
Figure 16-2
412
11)
In the circuit in Figure 16-2, the current through the inductor equals the current through the resistor. (True/False)
12)
In the circuit in Figure 16-2, assume that XL = R. The phase angle is 45°. (True/False)
13)
In the circuit in Figure 16-2, increasing the frequency will cause an increase in circuit current. (True/False)
14)
In the circuit in Figure 16-2, increasing the frequency will cause an increase in circuit impedence. (True/False)
15)
In the circuit in Figure 16-2, increasing the inductance value will cause an increase in circuit current. (True/False)
Figure 16-3
16)
Given the circuit in Figure 16-3, what is the total circuit impedence (Z)? A) 7 kΩ B) 2.65 kΩ C) 5.0 kΩ D) 500 Ω
17)
Given the circuit in Figure 16-3, what is the total circuit current? A) 100 mA B) 7.07 mA C) 2.0 mA D) 20 mA
18)
Given the circuit in Figure 16-3, what is the voltage (rms) across the resistor? A) 8V B) 7.07 V C) 4V D) 3V
19)
Given the circuit in Figure 16-3, what is the voltage (rms) across the inductor? A) 10 V B) 7.07 V C) 4V D) 6V
413
Figure 16-4
20)
What is the total circuit impedance in Figure 16-4? A) 141.5 Ω B) 14.15 Ω C) 195 Ω D) 120 Ω
21)
What is the phase angle of the circuit in Figure 16-4? A) 58° B) -58° C) 32° D) -32°
22)
What is the Impedance and phase angle of the circuit in Figure 16-4, expressed in rectangular form? A) 75 Ω + j120 Ω B) 120 Ω + j75 Ω C) j75 Ω + j120 Ω D) 141 Ω + j120 Ω
23)
What is the Impedance and phase angle of the circuit in Figure 16-4, expressed in polar form? A) 141.5 58° Ω B) 141.5 -58° Ω C) 120 32° Ω D) 120 -58° Ω
24)
In a series resistor/inductor circuit: A) the resistor voltage is in phase with the current. B) the same current flows through both the resistor and the inductor. C) the inductor voltage leads the current by 90 degrees. D) All of the above are correct.
25)
The impedance of a series resistor/inductor circuit is defined as: 56 Ω + j75 Ω. What is the resistance value? A) 56 Ω B) 75 Ω C) 93.6 Ω D) 19 Ω
414
Figure 16-5
26)
Given the circuit in Figure 16-5, what is the current through the resistor? A) 0.707 A B) 1.0 A C) 2.0 A D) 3.0 A
27)
Given the circuit in Figure 16-5, what is the current through the inductor? A) 0.707 A B) 1.0 A C) 2.0 A D) 3.0 A
28)
Given the circuit in Figure 16-5, what is the total circuit current? A) 3A B) 5A C) 2.24 A D) 4.69 mA
29)
Given the circuit in Figure 16-5, what is the total circuit impedance? A) 569 Ω B) 22.6 Ω C) 26.8 Ω D) 268 Ω
Figure 16-6
30)
Given the circuit in Figure 16-6, what is the current through the resistor? A) 45.3 mA B) 1100 mA C) 207.5 mA D) 21.8 mA
415
31)
Given the circuit in Figure 16-6, what is the current through the inductor? A) 218 mA B) 533.3 mA C) 21.8 mA D) 300 mA
32)
Given the circuit in Figure 16-6, what is the total circuit current? A) 555.1 mA B) 57.7 mA C) 533.7 mA D) 4.69 mA
33)
Given the circuit in Figure 16-6, what is the total circuit impedance? A) 44.96 Ω B) 222.6 Ω C) 449.6 Ω D) 268.3 Ω
34)
Given the circuit in Figure 16-6, what is the phase angle? A) 43.2° B) 87.6° C) 23.4° D) 45°
35)
Given the circuit in Figure 16-6, does the total lead or lag the total voltage? A) The total current (IT) leads the total voltage (VS). B) The total current (IT) lags the total voltage (VS).
36)
If XL = 100 ohms and R = 100 ohms in a series RL circuit, then impedance will be: A) 200 Ω B) 100 Ω C) 141.1 Ω D) 14.14 Ω
37)
In a series RL circuit, if resistance increases, impedance will: A) increase. B) decrease. C) remain the same. D) drop to zero.
38)
A 12 mH inductor is used in a circuit with a 10 kHz source. What is the inductive reactance? A) 1.33 k ohms B) 753 ohms C) 500 ohms D) 127 ohms
416
39)
A circuit has an apparent power of 240 VA, and a power factor of 0.943. What is the true power dissipated by the circuit? A) 240 W B) 226 W C) 13.6 W D) 84.5 W
40)
In a series RL circuit where R = 100 Ω and XL = 50 Ω, which of the following combinations is equivalent to the given series circuit? A) R = 125 Ω in parallel with XL = 250 Ω B) R = 250 Ω in parallel with XL = 125 Ω C) R = 20 Ω in parallel with XL = 40 Ω D) R = 40 Ω in parallel with XL = 20 Ω
41)
In a series RL circuit where R = 100 Ω and XL = 50 Ω. What is the total impedance of this circuit? A) 111.8 Ω -26.6° B) 111.8 Ω -63.4° C) 111.8 Ω 26.6° D) 111.8 Ω 63.4°
42)
The total opposition that a series or parallel RL circuit offers to current flow is called ________. A) impedance B) susceptance C) inductive impedance D) reactance
43)
Resistance and inductive reactance must be added ________ in a series RL circuit. A) using vectors, or Pythagorean Theorem B) using Kirchhoff's Voltage Law C) inductively D) both A and C
Figure 16-7
417
44)
In Figure 16-7, the plot of the sine wave could represent the ________ and the cosine wave could represent the ________ relationship in a purely inductive circuit A) voltage, current B) current, voltage C) voltage, impedance D) current, impedance
Figure 16-8
45)
In Figure 16-8, for the voltage across the resistor to equal the voltage across the inductor: A) XL must be larger than R. B) XL must be smaller than R. C) XL must equal R. D) impedance must be zero.
46)
The inverse of inductive reactance is: A) impedance B) resistance C) capacitive susceptance D) inductive susceptance
47)
Switching power supplies: A) are efficient at converting ac to dc. B) can change unregulated dc to high frequency pulses C) contain pulse-width modulators D) all of the above
418
1)
TRUE
2)
TRUE
3)
FALSE
4)
TRUE
5)
FALSE
6)
TRUE
7)
TRUE
8)
TRUE
9)
TRUE
10)
TRUE
11)
FALSE
12)
TRUE
13)
FALSE
14)
TRUE
15)
FALSE
16)
C
17)
C
18)
A
19)
D
20)
A
21)
A
22)
A
23)
B
24)
D
419
25)
A
26)
B
27)
C
28)
C
29)
D
30)
D
31)
B
32)
C
33)
A
34)
B
35)
B
36)
C
37)
A
38)
B
39)
B
40)
A
41)
C
42)
A
43)
A
44)
A
45)
C
46)
D
47)
D
Principles of Electric Circuits Chapter 17: RLC Circuits and Resonance
Figure 17-1
1)
If the series circuit in Figure 17-1 is resonant, the inductive and capacitive reactance must be equal. (True/False)
2)
If the series circuit in Figure 17-1 is resonant, the circuit is purely resistive and the phase shift is zero degrees. (True/False)
3)
If the series circuit in Figure 17-1 is resonant, the impedance, as seen by the generator will be very high. (True/False)
4)
If the series circuit in Figure 17-1 is NOT resonant, the impedance will be lower than it is at the resonant frequency. (True/False)
5)
If the series circuit in Figure 17-1 is resonant, increasing the Q will produce a wider bandwidth. (True/False)
6)
In a series resonant circuit, current is maximum and impedance is minimum at resonance. (True/False)
7)
Resonant frequency of a circuit occurs when the inductive reactance is equal to the capacitive reactance. (True/False)
8)
In a series RLC circuit, at above resonance the circuit is more capacitive than it is inductive. (True/False)
9)
By increasing the resistance of a coil you can increase the Q of the coil at resonance. (True/False)
10)
A parallel tuned circuit can be used to couple energy from one circuit to another. (True/False)
Figure 17-2
420
11)
If the parallel circuit in Figure 17-2 is resonant, the inductive and capacitive reactance must be equal. (True/False)
12)
If the parallel circuit in Figure 17-2 is resonant, the circuit is purely resistive and the phase shift is zero degrees. (True/False)
13)
If the parallel circuit in Figure 17-2 is resonant, the impedance, as seen by the generator will be very high. (True/False)
14)
If the parallel circuit in Figure 17-2 is NOT resonant, the impedance will be lower than it is at the resonant frequency. (True/False)
15)
If the parallel circuit in Figure 17-2 is resonant, increasing the Q will produce a wider bandwidth. (True/False)
16)
A series resonant circuit has a low impedance at the resonant frequency. (True/False)
17)
A parallel resonant circuit has a low impedance at the resonant frequency. (True/False)
Figure 17-3
18)
Given the circuit in Figure 17-3, the circuit current is: A) 198 μA B) 100 mA C) 1A D) 0.87 mA
19)
Given the circuit in Figure 17-3, the circuit impedance is: A) 318 Ω B) 345 Ω C) 690 Ω D) 100 Ω
421
Figure 17-4
20)
Given the circuit in Figure 17-4, the circuit impedance is: A) 6.4 Ω B) 8.1 Ω C) 5.88 Ω D) 14.46 Ω
21)
Given the circuit in Figure 17-4, is the circuit mostly inductive or capacitive? A) inductive B) capacitive
Figure 17-5
22)
Given the circuit in Figure 17-5, the circuit impedance is: A) 27.1 kΩ B) 3.3 kΩ C) 6.94 kΩ D) 7.07 kΩ
23)
Given the circuit in Figure 17-5, the circuit phase angle is: A) -1.2° B) 34.5° C) 90° D) -5.4°
24)
Given the circuit in Figure 17-5, the circuit current is: A) 36.9 mA B) 3.03 mA C) 7.64 mA D) 0.1 A
422
25)
Is the circuit in Figure 17-5, at or very close to resonance? A) Yes B) No
Figure 17-6
26)
Given the circuit in Figure 17-6, calculate the resonant frequency: A) 127.3 kHz B) 29.58 kHz C) 447.64 kHz D) 290.6 kHz
27)
If the generator frequency, in Figure 17-6, is higher than the resonant frequency, the circuit acts as: A) an inductive circuit in which the current lags the generator voltage. B) an inductive circuit in which the current leads the generator voltage. C) a capacitive circuit in which the current lags the generator voltage. D) a capacitive circuit in which the current leads the generator voltage.
28)
In a parallel resonant circuit (at the resonant frequency): A) The current demanded from the generator is much larger at the resonant frequency than at frequencies lower or higher than the resonant frequency. B) The current demanded from the generator is much smaller at the resonant frequency than at frequencies lower or higher than the resonant frequency.
29)
In a parallel resonant circuit: A) There is a small current circulating between the capacitor and inductor, and a large current flowing from the generator. B) There is a large current circulating between the capacitor and inductor, and a small current flowing from the generator.
30)
A parallel LRC circuit has: A) low impedance at the resonant frequency. B) a high impedance at the resonant frequency. C) a very high Q at frequencies on either side of resonance. D) The impedance is independent of the frequency.
31)
In a parallel resonant circuit (at the resonant frequency) which of the following is not true? A) The impedance is at its minimum value. B) The total current is at its minimum value. C) The phase angle is zero degrees. D) The impedance is at its maximum value.
423
Figure 17-7
32)
The capacitive reactance in Figure 17-7 is: A) 26.53 Ω B) 265.3 Ω C) 159.7 Ω D) 2.7 kΩ
33)
The inductive reactance in Figure 17-7 is: A) 2.6 Ω B) 750 Ω C) 7.5 Ω D) 2.7 kΩ
34)
Is the circuit in Figure 17-7 at or near resonance? A) Yes B) No
35)
The current through the capacitor in Figure 17-7 is: A) 430 mA B) 4.3 A C) 159.7 mA D) 279.65 mA
36)
The current through the resistor in Figure 17-7 is: A) 350 mA B) 3.5 A C) 15.9 A D) 459.65 mA
37)
The current through the inductor in Figure 17-7 is: A) 15.3 A B) 153.2 mA C) 1.5 A D) 27.9 A
38)
The total circuit current in Figure 17-7 is: A) 3.5 A B) 4.3 A C) 14.87 A D) 15.3 A
424
39)
The phase shift between the source voltage (VS) and total circuit current in Figure 17-7 is: A) 23.4° B) -23.4° C) -76.7° D) -46.45°
40)
The circuit impedance in Figure 17-7 is: A) 70.5 Ω B) 7.5 Ω C) 33 Ω D) 265 Ω
41)
The True power in Figure 17-7 is: A) 43.29 W B) 404.3 W C) 7.5 W D) 200.7 W
42)
The Apparent power in Figure 17-7 is: A) 170.6 VA B) 403.3 VA C) 7.5 VA D) 1760 VA
43)
The Power Factor in Figure 17-7 is: A) 0.23 B) 2.3 C) 0.89 D) 0.45
44)
In a series RLC circuit the current can be found using: VR R VT Z
A)
IT =
B)
IT =
C) D)
both A and B none of the above
45)
Apparent power in an RLC circuit is equal to total voltage times total current when: A) the circuit is at resonance. B) the circuit is not at resonance. C) it is a series or parallel circuit. D) both A and C
46)
At resonance the power factor is: A) zero B) 1 C) negative D) M5
425
47)
The lower and upper end of the band width of a series RLC circuit is where the current has fallen to ________ of the maximum. A) 50% B) 63.6% C) 70.7% D) 29.3%
48)
When a parallel circuit resonates it is said to: A) flywheel B) oscillate C) both A and B D) neither A nor B
49)
At frequencies well above and below the resonant frequency, the series RLC circuit looks ________ above resonance, and the parallel RLC circuit looks ________ below resonance. A) inductive, inductive B) inductive, capacitive C) like an open, like a short D) like a short, like an open
50)
In a series LC circuit, L = 100 μH, C = 0.047 μF, and f = 150 kHz. The value of ZT is: A) 72 Ω -90° B) 72 Ω 90° C) 72 Ω -90° D) 36 Ω 45°
51)
In a series LC circuit, VL = 8.3 V and VC = 10.6 V. VS = ________. A) -2.3 V -90° B) -2.3 V 90° C) 2.3 V -90° D) 8.3 V -90°
52)
In a series RLC circuit, R = 1.1 kΩ, XL = 1.6 kΩ, and XC = 2.9 kΩ. ZT = ________. A) 1.7 kΩ -50° B) 1.7 kΩ 50° C) 4.6 kΩ 50° D) 4.6 kΩ -50°
Figure 17-8
426
53)
In Figure 18-8, calculate ZT. A) 58 Ω -39° B) 58 Ω 39° C) 20 Ω -60° D) 20 Ω 60°
54)
In Figure 18-8, calculate IT. A) 0.75 A -60° B) 0.75 A 60° C) 1.25 A -39° D) 1.25 A 39°
55)
In Figure 18-8, calculate VR. A) 9.5 V 39° B) 9.5 V -39° C) 7.5 V -60° D) 7.5 V 60°
56)
In Figure 18-8, calculate VL. A) 15 V -30° B) 15 V 30° C) 9.5 V -39° D) 9.5 V 39°
57)
In Figure 18-8, calculate VC. A) 4 V 60° B) 2 V 150° C) 4 V -60° D) 2 V -150°
58)
The center frequency of a band-pass filter is always equal to the ________. A) bandwidth B) 3-dB frequency C) bandwidth divided by Q D) geometric mean average of the cutoff frequencies
59)
If the bandwidth of a filter increases, ________. A) Q decreases B) the roll-off rate increases C) ripples appear in the stopband D) the center frequency decreases
60)
Too narrow of a tuned circuit bandwidth may result in A) reception of all unwanted stations. B) low selectivity of the tuned circuit. C) loss of fidelity. D) none of the above
427
61)
Half-power frequencies A) determine selectivity. B) determine bandwidth. C) determine the pass band. D) all of the above E) none of the above
428
1)
TRUE
22)
B
43)
A
2)
TRUE
23)
A
44)
C
3)
FALSE
24)
B
45)
D
4)
FALSE
25)
A
46)
B
5)
FALSE
26)
D
47)
C
6)
TRUE
27)
A
48)
C
7)
TRUE
28)
B
49)
A
8)
FALSE
29)
B
50)
B
9)
FALSE
30)
B
51)
C
10)
TRUE
31)
A
52)
A
11)
TRUE
32)
B
53)
C
12)
TRUE
33)
C
54)
A
13)
TRUE
34)
B
55)
C
14)
TRUE
35)
A
56)
B
15)
FALSE
36)
B
57)
D
16)
TRUE
37)
A
58)
D
17)
FALSE
38)
D
59)
A
18)
B
39)
C
60)
C
19)
D
40)
B
61)
D
20)
A
41)
B
21)
A
42)
D
429
Principles of Electric Circuits Chapter 18: Passive Filters 1)
The effectiveness of a filter in rejecting signals beyond cutoff frequency is a function of the filter roll-off rating. (True/False)
2)
Frequency response is a graph of voltage gain versus frequency. (True/False)
3)
At the cutoff frequency, the voltage gain is 63.6% of the maximum. (True/False)
4)
The range of low frequencies passed by a low-pass filter within a specified limit is called the stopband. (True/False)
Figure 18-1
5)
The circuit in Figure 18-1 is: A) a high-pass filter B) a band-pass filter C) a low-pass filter D) a band-stop filter
6)
The filter in Figure 18-1 has a roll-off of: A) -40 dB per decade B) -20 dB per decade C) -3 dB per decade D) -6 dB per decade
7)
The circuit in Figure 18-1 has a critical frequency of: A) 318.3 kHz B) 31.83 kHz C) 3.18 MHz D) 3.18 kHz
8)
The phase shift of the circuit in Figure 18-1, at the critical frequency is: A) -90° B) 60.254° C) -45° D) 0°
430
9)
What is the output voltage, in Figure 18-1 at a frequency of 50 kHz? A) 4.22 V B) 3.45 V C) 3.17 V D) 1.76 V
10)
The output voltage, in Figure 18-1 is how many dB down from the input voltage, at the critical frequency? A) 0 dB B) -3 dB C) -20 dB D) -0.707 dB
11)
What is the output voltage, in Figure 18-1 at a frequency of 10 kHz? A) 1.50 V B) 4.23 V C) 4.77 V D) 5V
Figure 18-1
Figure 18-2
431
12)
Which of the two response curves in Figure 18-2 (A or B) represents the response of the circuit in Figure 18-1? A) Curve A B) Curve B
Figure 18-1
13)
The capacitive reactance in Figure 18-1 is equal to the resistance at the critical frequency. (True/False)
14)
Given a frequency of 440 Hz, one decade higher would be a frequency of: A) 4,400 Hz B) 220 Hz C) 880 Hz D) 4 kHz
Figure 18-3
15)
The capacitive reactance in Figure 18-3 is equal to the resistance at the critical frequency. (True/False)
16)
The circuit in Figure 18-3 is: A) a high-pass filter B) a band-pass filter C) a low-pass filter D) a band-stop filter
17)
The filter in Figure 18-3 has a roll-off of: A) -40 dB per decade B) -20 dB per decade C) -3 dB per decade D) -6 dB per decade
432
18)
The circuit in Figure 18-3 has a critical frequency of: A) 318.3 kHz B) 5.0 kHz C) 50.0 kHz D) 13.18 kHz
19)
The phase shift of the circuit in Figure 18-3, at the critical frequency is: A) -90° B) 60.254° C) -45° D) 0°
20)
What is the output voltage, in Figure 18-3 at a frequency of 1.0 kHz? A) 0.269 V B) 3.45 V C) 4.92 V D) 1.76 V
21)
The output voltage, in Figure 18-3 is how many dB down from the input voltage, at the critical frequency? A) 0 dB B) -3 dB C) -20 dB D) -0.707 dB
22)
What is the output voltage, in Figure 18-3 at a frequency of 20 kHz? A) 5V B) 1.21 V C) 4.77 V D) 0.5 V
Figure 18-3
433
Figure 18-4
23)
Which of the two response curves in Figure 18-4 (A or B) represents the response of the circuit in Figure 18-3? A) Curve A B) Curve B
24)
The frequency at which the output voltage of a filter is 70.7% of the input voltage is called the: A) critical frequency B) break frequency C) half-power point D) all of the above
Figure 18-5
25)
The circuit in Figure 18-5 is: A) a high-pass filter B) a band-pass filter C) a low-pass filter D) a notch filter
434
26)
The filter in Figure 18-5 has a roll-off of: A) -40 dB per decade B) -20 dB per decade C) -3 dB per decade D) -6 dB per decade
27)
The critical frequency of the circuit in Figure 18-5 is: A) 1.59 kHz B) 159 Hz C) 60 Hz D) 135 Hz
28)
The output voltage of the circuit in Figure 18-5, at the cut-off (critical) frequency is: A) 7.07 V B) 4.9 V C) 0.707 V D) 1.59 V
29)
Which of the following equations is correct for calculating the critical or cut-off frequency for the circuit in Figure 18-5? A) Fc = 1/[2π (L/R)] B) Fc = 1/2π (LR) C) Fc = 1/[2π (LR)] D) Fc = 2π(L/R)
30)
The term "attenuation" refers to a: A) loss or reduction of signal strength B) signal gain C) phase shift D) unity gain
31)
The ________ is defined as the difference between the high and low cut-off frequencies. A) Q frequency range B) bandwidth C) filter width D) roll-off
32)
The higher the Q, the ________ the filter. A) lower the center frequency of B) greater the bandwidth of C) less selective D) more selective
33)
A band-stop filter can be designed by using ________ and a high-pass filter. A) a band-pass filter B) a summer circuit C) an amplifier circuit D) a low-pass filter
435
34)
A band-stop filter filters out ________. A) odd harmonics B) even harmonics C) a band of frequencies D) signals with large voltage swings
35)
Attenuation is usually expressed in ________. A) volts B) MHz C) Decibels D) a value of Q
Figure 18-6
36)
Identify the filter circuit in Figure 18-6 above. A) Band Pass B) Band Stop C) Low Pass D) High Pass
Figure 18-7
37)
Identify the filter circuit in Figure 18-7 above. A) Band Pass B) Band Stop C) Low Pass D) High Pass
Figure 18-8
38)
Identify the filter circuit in Figure 18-8 above. A) Band Pass B) Band Stop C) Low Pass D) High Pass
436
Figure 18-9
39)
Identify the filter circuit in Figure 18-9 above. A) Band Pass B) Band Stop C) Low Pass D) High Pass
Figure 18-10
40)
Identify the filter circuit in Figure 18-10 above. A) Band Pass B) Band Stop C) Low Pass D) High Pass
41)
Decibels are used to provide a ________ between voltage levels. A) value B) comparison C) common level D) reference
42)
In a low-pass RC filter, having a cutoff frequency of 2 kHz, C = 80nF. Determine R. A) 22.7 Ω B) 44.7 Ω C) 98.0 Ω D) 995 Ω
43)
A high-pass RC filter can be changed to a low-pass RC filter by ________. A) adding both another resistor and capacitor B) adding another capacitor C) adding another resistor D) reversing the position of the resistor and capacitor
437
44)
The bandwidth of a resonant circuit is determined by two points on the curve known as ________. A) the half power points B) fcutoff C) Q Point D) both A and B
45)
A low-pass filter and a high-pass filter in parallel with each other, would form a ________ filter. A) band-stop B) band-pass C) band-reject D) both A and B E) none of the above
46)
A low-pass filter and a high-pass filter in series with each other, would form a ________ filter. A) band-stop B) band-pass C) band-reject D) both A and B E) none of the above
47)
Critical frequencies are also called: A) half-voltage frequencies B) half-current points C) -3dB frequencies D) none of the above
438
1)
TRUE
25)
A
2)
TRUE
26)
B
3)
FALSE
27)
B
4)
FALSE
28)
A
5)
A
29)
A
6)
B
30)
A
7)
B
31)
B
8)
C
32)
D
9)
A
33)
D
10)
B
34)
C
11)
A
35)
C
12)
B
36)
B
13)
TRUE
37)
D
14)
A
38)
C
15)
TRUE
39)
A
16)
C
40)
B
17)
B
41)
B
18)
B
42)
D
19)
C
43)
D
20)
C
44)
D
21)
B
45)
D
22)
B
46)
B
23)
A
47)
C
24)
D
439
Principles of Electric Circuits Chapter 19: Circuit Theorems in AC Analysis 1)
The superposition theorem is useful for the analysis of multiple source circuits. (True/False)
2)
Thevenin's theorem provides a method for the reduction of any ac circuit to an equivalent voltage source in parallel with an equivalent impedance. (True/False)
3)
Norton's theorem provides a method for the reduction of any ac circuit to an equivalent form consisting of an equivalent current source in parallel with an equivalent impedance. (True/False)
4)
Maximum power is transferred to a load when the load impedance is a complex conjugate of the output impedance of the driving circuit. (True/False)
5)
In order to get maximum power transfer from a capacitive source, the load must have an impedance that is the complex conjugate of the source impedance. (True/False)
6)
The superposition theorem is useful for the analysis of single source-circuits. (True/False)
7)
The superposition theorem is useful for the analysis only in ac circuits. (True/False)
8)
An equivalent circuit is one that produces the same voltage and current to a given load as the original circuit it replaces. (True/False)
9)
Thevenin's equivalent circuit is the transformation of Norton's equivalent circuit. (True/False)
10)
If two equal currents are in opposing directions at any instant in time, in a given branch, what is the net current at that instant? A) the complex conjugate of the two B) their sum C) zero D) 1A
Figure 19-1
440
11)
Given the circuit in Figure 19-1, find the total circuit current. A) 12 mA B) 9 mA C) 6 mA D) 18.2 mA
Figure 19-2
12)
Given the circuit in Figure 19-2, find the total current in the load (RL). A) 1.69 47.3° mA B) 4 mA C) 5.69 47° mA D) 1.69 47° mA on a dc level of 4 mA
13)
Why is the superposition theorem useful in the analysis of multiple-source circuits? A) because only Ohm's law is required B) the circuit can be analyzed one-source at a time C) because all sources are shorted out D) because all sources are openend
14)
An equivalent voltage source in series with an equivalent impedance means you should solve the problem using: A) the superposition theorem B) Thevenin's theorem C) Norton's theorem D) Millman's theorem
Figure 19-3
441
15)
Given Figure 19-3, find the Thevenin voltage (Vth) for the circuit external to RL. A) 11.2 63.4° V B) 15 53.1° V C) 8.3 63.4° V D) 13.2 53.1° V
16)
Given Figure 19-3, find the Thevenin impedance (Zth). A) 44.7 63.4° Ω B) 54.3 47.2° Ω C) 60 53.1° Ω D) 74.1 46.9° Ω
17)
Assume you need to find the total impedance between two specific terminals, looking into the open terminals, with all sources zeroed. Which theorem is suggested by these conditions? A) Norton's theorem B) Thevenin's theorem C) total impedance D) both A and B
18)
The Norton's equivalent circuit is characterized by the: A) equivalent current source in series with an equivalent impedance B) equivalent voltage source in parallel with an equivalent impedance C) equivalent current source in parallel with an equivalent impedance D) equivalent voltage source in series with an equivalent impedance
19)
Maximum power transfer occurs when the ac impedances are: A) equal B) smaller C) larger D) complex conjugates
20)
What is the complex conjugate of 75 Ω + j45 Ω? A) -75 Ω - j45 Ω B) -75 Ω + j45 Ω C) 75 Ω - j45 Ω D) 45 Ω - j75 Ω
21)
If the output impedance of a certain circuit is 100 Ω - j20 Ω, what value of load impedance will result in maximum power transfer to the load? A) 100 Ω - j20 Ω B) 20 Ω + j100 Ω C) 20 Ω - j100 Ω D) 100 Ω + j20 Ω
22)
For the circuit with output impedance of 100 Ω - j20 Ω, how much power is delivered to a matched load if the voltage (VS) = 25 V rms? A) 1.56 W B) 3.12 W C) 6.25 W D) 12.5 W
442
Figure 19-4
23)
Given the circuit in Figure 19-4, what load value should be used to replace R2 to obtain the maximum power transfer to the load? Determine the type of load, and express the value in rectangular form. A) 168 Ω + j76.8 Ω B) 168 Ω - j76.8 Ω C) 88.1 Ω - j39.4 Ω D) 64.1 Ω - j73.6 Ω
24)
When using the superposition theorem in ac network analysis, which one of the following statements is not true? A) Voltage sources are replaced by open circuits. B) Voltage sources are replaced by short circuits. C) Current sources are replaced by short circuits. D) All impedances are replaced by their complex conjugates.
25)
The only difference in applying network theorems to ac circuits rather than dc circuits is that we will be working with ________ and ________ instead of just resistors and real numbers. A) impedances, phasors B) impedances, vectors C) voltages, loads D) currents, shorts
26)
When applying Thevenin's theorem to a circuit with ________, start by removing the portion of the network across which the Thevenin equivalent circuit is to be found. A) a dependant source B) more than one dependant source C) an independent source D) a conjugate source
27)
Dependant sources in which the controlling variable is not determined by the network for which the Norton equivalent is to be found, ________ the procedure for determining the equivalent circuit. A) alter B) do not alter C) totally change D) transform
443
28)
When the load is properly chosen for maximum power transfer, the load will appear ________. A) totally resistive B) complex C) imaginary D) reactive
29)
The Norton equivalent circuit and Thevenin equivalent circuit can be found from each other by using ________. A) Ohm's law B) the voltage divider rule C) Kirchhoff's law D) a source transformation
30)
The Norton current is found by calculating the ________ between the marked terminals. A) short circuit current B) open circuit voltage C) source current D) load current
31)
To find Vth, determine the ________ between the two specified terminals. A) short circuit current B) short circuit voltage C) open circuit voltage D) open circuit current
32)
The Norton and Thevenin equivalent circuit is applicable at ________ frequency/frequencies. A) several B) only one C) many D) all
33)
To calculate Zth, first ________ all voltage and current sources. A) open B) short C) combine D) remove
Figure 19-5
444
34)
See Figure 19-5. Using the superposition theorem, what is the portion of the current through the capacitor caused by the 5 V 30° voltage source? A) 0.616 A 129.5° B) 0.741 A 55.6° C) 0.822 A 9.5° D) 0.89 A 93.4°
35)
See Figure 19-5. Using the superposition theorem, what is the portion of the current through the capacitor caused by the 10 V 0° voltage source? A) 0.616 A 129.5° B) 0.741 A 55.6° C) 0.822 A 9.5° D) 0.5 A 90°
Figure 19-6
36)
See Figure 19-6. What is the Norton equivalent impedance Zn external to ZL? A) 0.47 Ω -26.6° B) 5.0 Ω 90° C) 10.0 Ω 0° D) 11.18 Ω 26.6°
37)
See Figure 19-6. What is the Norton equivalent current In external to ZL? A) 0 A 0° B) 0.89 A 3.4° C) 0.67 A 30° D) 0.67 A 26.6°
38)
What mathematical relationship allows calculation of the Thevenin impedance Zth? Vopen circuit A) I open circuit B) C) D)
Vshort circuit I open circuit
Vopen circuit I short circuit Vshort circuit I short circuit
445
1)
TRUE
20)
C
2)
FALSE
21)
D
3)
TRUE
22)
A
4)
TRUE
23)
A
5)
TRUE
24)
B
6)
FALSE
25)
A
7)
FALSE
26)
C
8)
TRUE
27)
B
9)
TRUE
28)
A
10)
C
29)
D
11)
A
30)
A
12)
D
31)
C
13)
B
32)
B
14)
B
33)
D
15)
B
34)
A
16)
C
35)
C
17)
D
36)
D
18)
C
37)
B
19)
D
38)
C
Principles of Electric Circuits Chapter 20: Time Response of Reactive Circuits 1)
In an R-C integrating circuit, the output voltage is taken across the capacitor. (True/False)
2)
In an R-L differentiating circuit, the output voltage is taken across the resistor. (True/False)
3)
The transient time of an R-C or R-L circuit equals five time constants. (True/False)
4)
The rising and falling edges of a pulse waveform contain the higher frequency components. (True/False)
5)
The flat portion of a pulse waveform contains the average value. (True/False)
6)
The shape of a repetitive-pulse affects the determination of the pulse repetition frequency. (True/False)
7)
The rise time of an ideal pulse is zero. (True/False)
8)
The time for a capacitor to fully charge or discharge is called the transient ramp. (True/False)
9)
The voltage across a capacitor in an RC integrator circuit cannot change exponentially, it can change only instantaneously. (True/False)
10)
If a periodic pulse waveform is applied to an RC integrator, the output waveshape depends on the relationship of the circuit time constant and the duty cycle of the input pulses. (True/False)
11)
Integration is: A) an averaging process B) a process for establishing a rate of change C) a summing operation D) used for mixing different input signals
12)
When there is a sudden change in voltage, a capacitor, for an instant, appears as: A) an open circuit B) a short circuit C) a fully charged capacitor D) a resistive circuit
13)
The rate of charging and discharging of a capacitor is controlled by: A) the peak input voltage B) the frequency of the input C) the period of the input D) the RC time constant
447
14)
What is the capacitor voltage in an RC integrator when the pulse width is equal to or greater than five time constants? A) 37% of maximum B) 63% of maximum C) fully charged D) 86% of maximum
15)
The output of an RC integrator approaches the shape of the input pulse when: A) the pulse is at maximum B) the pulse width is increased C) the pulse width is decreased D) the pulse is at minimum
16)
When the pulse width is less than five time constants, the capacitor voltage in an RC integrator: A) follows the input voltage B) charges to maximum C) discharges to minimum D) only partially charges
17)
If R = 2.2 kΩ, C = 1 μF, the pulse width = 5 ms, and the peak voltage of the pulse = 50 V, to what voltage will the capacitor charge in an RC integrator? A) 44.85 V B) 22.43 V C) 37.41 V D) 18.57 V
18)
With repetitive pulses, the amount of capacitor charge or discharge in an RC integrator depends upon: A) the circuit time constant B) the input frequency C) the input peak voltage D) both the circuit time constant and the input frequency
19)
The RC differentiator is: A) a basic low-pass filter B) a basic band-pass filter C) a basic high-pass filter D) a basic band-stop filter
20)
If the resistor voltage in a differentiating circuit is down to + 2.4 V at the end of a 14 V input pulse, to what negative value will the resistor voltage go in response to the falling edge of the input? A) -14 V B) -16.4 V C) -11.6 V D) -2.4 V
448
21)
What does the average value of the differentiator output voltage equal during steady state? A) 0V B) peak value C) 0.637 × Peak D) 0.707 × Peak
22)
The fast rising and falling edges of a pulse waveform represent: A) the rise times B) the fall times C) the lower frequency components D) the higher frequency components
23)
The top or flat portion of a pulse represents: A) the average value B) the peak value C) the lower frequency components D) the higher frequency components
24)
The average value of a pulse is: A) its DC component B) zero C) 0.637 × Peak D) the peak value
25)
The RC differentiator is characterized by: A) introduces tilt to the flat portion of a pulse B) reduction of lower frequency components C) eliminates the dc component of the input D) produces a zero average-value output voltage E) all of the above
26)
What is the highest frequency contained in a pulse that has rise and fall times of 17 ns? A) 20.59 MHz B) 205.9 MHz C) 10.3 MHz D) 168 MHz
Figure 20-1
27)
What is the steady state output of the differentiator circuit in Figure 20-1? A) 0V B) 20 V C) 12.74 V D) 6.37 V
449
28)
A certain pulse waveform has a rise-time 6 nS and a fall-time of 9 nS. What is the highest frequency component in the waveform? A) 58.33 MHz B) 38.9 MHz C) 23.33 MHz D) 116.67 MHz
29)
A mathematical operation that determines the rate of change of a curve is called ________. A) differentiation B) integrator C) curve averaging D) mixing
30)
A mathematical operation for finding the area under the curve of a graph is called ________. A) differentiation B) integration C) curve averaging D) mixing
31)
In a repetitive-pulse RC integrator circuit, what would the steady-state voltage equal at the end of the fifth pulse? Assume a VIN of 20 V. A) 1.46 V B) 14.62 V C) 20 V D) 0V
32)
A series of pulses is called: A) periodic B) pulse width C) duty cycle D) rise time
33)
A(n) ________ will decrease the time constant in an RC integrator or differentiator. A) open resistor B) leaky capacitor C) shorted capacitor D) shorted resistor
34)
For a capacitor to completely charge in an RC integrator, the pulse width must be ________ five time constants. A) less than B) much greater than C) greater than or equal to D) much less than
35)
The average value of a waveform is its: A) tw B) ac component C) rising edge minus falling edge D) dc component
450
36)
An RL integrator can act as a ________ filter and an RC differentiator can act as a ________ filter. A) low-pass, low-pass B) high-pass, high-pass C) low-pass, high-pass D) high-pass, low-pass
37)
If the capacitor in an RC integrator shorts, the output: A) is at ground B) would measure the same as the input C) would measure zero volts D) none of the above
Figure 20-2
38)
See Figure 20-2. Assume that the capacitor is initially charged to 3 V before the switch is closed. What is the peak current IC that flows after the switch is closed? A) 0.3 mA B) 0.7 mA C) 1.0 mA D) infinity
39)
See Figure 20-2. Assume that the capacitor is initially charged to 3 V before the switch is closed. What is the current that flows after an elapsed time of 20 ms? A) 0.20 mA B) 0.23 mA C) 0.47 mA D) 0.67 mA
Figure 20-3
40)
See Figure 20-3. What is VC at t = 1 ms? Assume that the capacitor is initially discharged. A) VC ≈ 0 V B) VC ≈ 5 V C) VC ≈ 8.33 V D) VC ≈ 10 V
451
41)
See Figure 20-3. What is the peak current that flows through the capacitor? A) 0 mA B) 1 mA C) 5 mA D) 10 mA
42)
See Figure 20-3. If the square wave pulse width is decreased to 1.0 μs, how will the VC waveform change? A) The corners will be less rounded. B) The peak voltage reached will increase. C) The waveform will become flattened. D) The average value will drop to zero.
452
1)
TRUE
22)
D
2)
FALSE
23)
C
3)
TRUE
24)
A
4)
TRUE
25)
E
5)
FALSE
26)
A
6)
FALSE
27)
A
7)
TRUE
28)
A
8)
FALSE
29)
A
9)
FALSE
30)
B
10)
FALSE
31)
B
11)
A
32)
A
12)
B
33)
B
13)
D
34)
B
14)
C
35)
B
15)
B
36)
C
16)
D
37)
A
17)
A
38)
B
18)
D
39)
C
19)
C
40)
D
20)
C
41)
D
21)
A
42)
C
453
Principles of Electric Circuits Chapter 21: Three-Phase Systems in Power Applications 1)
A simple two-phase generator consists of two conductive loops, separated by 180°, rotating in a magnetic field. (True/False)
2)
A simple three-phase generator consists of three conductive loops, separated by 120°, rotating in a magnetic field. (True/False)
3)
In a (Y) wye-connected generator, there is a 30° difference between each line voltage and the nearest phase voltage. (True/False)
4)
A balanced load is one in which all impedances are equal. (True/False)
5)
For a balanced circuit, the neutral current is non-zero. (True/False)
6)
In general, the three-phase system is more economical for transmitting power at a fixed power loss than is the single-phase system. (True/False)
7)
In a Δ-connected generator, there is a 120° difference between each line current and the nearest phase current. (True/False)
8)
The stator is the rotating assembly in a generator or motor. (True/False)
9)
Polyphase is characterized by two or more sinusoidal voltages, each having a different phase angle. (True/False)
10)
The term squirrel-cage applies to a type of three-phase ac generator. (True/False)
11)
The output of an ac generator has a maximum value of 240 V. At what angle is the instantaneous voltage equal to 85 V? A) 20.7° B) 69.3° C) 19.5° D) 90°
12)
Determine the instantaneous voltage of an ac generator, where the maximum voltage is 208 V, and the angle is 40 degrees. A) 208 V B) 159.3 V C) 133.7 V D) 147 V
13)
A four-pole, single phase generator rotates at 500 revolutions per second. What frequency is produced? A) 500 Hz B) 1000 Hz C) 2000 Hz D) 1500 Hz
454
14)
Which of the following is NOT an advantage of polyphase systems over single-phase systems? A) less copper cross section B) constant power C) constant rotating magnetic field D) higher output power
15)
Which advantage of polyphase systems is most important in mechanical-to-electrical energy conversion? A) less copper cross section B) constant power C) constant rotating magnetic field D) higher output power
16)
Which advantage of polyphase systems is most important in electrical-to-mechanical energy conversion? A) less copper cross section B) constant power C) constant rotating magnetic field D) higher output power
17)
In a certain three-wire, Y-connected generator, the phase voltages are 2 kV. Determine the magnitude of the line voltage. A) 3464 V B) 2000 V C) 1414 V D) 1154 V
18)
In a Δ-connected generator, the phase voltages are 208 V. What are the line voltages? A) 132 V B) 208 V C) 360 V D) 120 V
19)
Which of the following are source/load configurations? A) Y-Y system B) Y-Δ system C) Δ-Δ system D) Δ-Y system E) all of the above
20)
The power in each phase of a balanced three-phase system is 1500 W. What is the total power? A) 500 W B) 1500 W C) 4500 W D) 2278 W
21)
Power is measured in three-phase systems using: A) voltmeters B) ammeters C) wattmeters D) AC voltmeters
455
22)
The secondary of a power transformer feeding an electrical transmission line acts as a ________ to the transmission line. A) source B) load
23)
In a certain Δ-connected balanced load, the line voltages are 230 V and the impedances are 60 35° Ω Determine the magnitude of the phase currents. A) 3.83 A B) 2.78 A C) 4.15 A D) 1.63 A
24)
In a certain Δ-connected balanced load, with line voltages of 230 V and the impedances are 60 35° Ω, determine the magnitude of the load currents. A) 6.63 A B) 4.82 A C) 7.19 A D) 2.82 A
25)
In a certain Δ-connected balanced load, with line voltages of 230 V and impedances of 60 35° Ω, determine the total power. A) 2163.5 W B) 1572.9 W C) 2346.3 W D) 920.2 W
26)
Polyphase generators produce simultaneous multiple sinusoidal voltages that are separated by: A) certain constant phase angles B) certain constant currents C) certain constant frequencies D) certain constant voltages
27)
For a three-phase, four-wire, Y-connected load that is unbalanced, the neutral current is: A) zero. B) equal to the sum of the phase of the phase currents. C) equal to the largest phase current minus the other two phase currents. D) in phase with the smallest phase current.
28)
The magnitude of the phase voltage in a Y-connected generator is 100 V. What is the magnitude of the line voltage? A) 57.5 V B) 86.6 V C) 100 V D) 173.2 V
29)
In a three-phase motor, if two phase voltages are interchanged: A) two of the phase voltages will cancel and the motor will operate as a single phase unit. B) the direction of rotation will reverse. C) the neutral current will increase significantly. D) the motor will stall, possibly burning out one or more phase windings.
456
30)
A single-phase sinusoidal voltage of 120 V is connected to a 70 Ω load. Current in the circuit is: A) 17.1 mA B) 174 mA C) 1.71 A D) 17.1 A
31)
Referring to the previous problem, power consumption is: A) 205 W B) 205 mW C) 20.5 W D) 67.5 W
32)
A two-phase generator is connected to two 70 Ω load. Each coil generates 120 V ac. A common neutral line exists. How much current is flowing through the common neutral line? A) 2.42 A B) 1.85 A C) 3.42 A D) 1.71 A
33)
Referring to the previous problem, the total copper cross section must handle: A) 1.71 A B) 2.41 A C) 3.42 A D) 5.83 A
34)
A three-phase generator is connected to three 70 Ω loads. Each coil generates 120 V ac. A common neutral line exists. How much current is flowing through the common neutral line? A) 5.13 A B) 0A C) 2.26 A D) 3.42 A
35)
Compare the total copper cross sections in terms of current-carrying capacity for a single-phase system and a three-phase 120 V system with an effective load of 50 Ω. A) Single-phase 4.8 A; three-phase 7.2 A B) Single-phase 7.2 A; three-phase 4.8 A C) Single-phase 9.6 A; three-phase 14.4 A D) Single-phase 4.8 A; three-phase 0 A
36)
In a Y-Y source/load configuration, the: A) line current and the load current are in phase, and both are out of phase with the phase current. B) phase current, the line current, and the load current are all equal in each phase. C) phase current and the line current are in phase, and both are 120° out of phase with the load current. D) phase current, the line current, and the load current are 120° out of phase.
457
37)
In a Y-connected source feeding a Δ-connected load: A) each phase of the load has the full line voltage across it. B) each phase of the load has two-thirds of the full line voltage across it. C) each phase of the load has one-third of the full line voltage across it. D) each phase of the load has a voltage across it equal to 3 .
38)
In a Δ-connected source feeding a Y-connected load: A) each phase voltage equals the corresponding load voltage. B) each phase voltage is 60° out of phase with the corresponding load voltage. C) each phase voltage equals the difference between the corresponding load voltages. D) each phase voltage is one-third the corresponding load voltage.
39)
In a Δ-connected source feeding a Δ-connected load, the: A) load voltage and line voltage are one-third the source voltage for a given phase. B) load voltage and line voltage are two-thirds the source voltage for a given phase. C) load voltage and line voltage are all equal for a given phase. D) load voltage and line voltage cancel for a given phase.
40)
In a balanced three-phase load, each phase has: A) a power consumption equal to 3VL I L . B) two-thirds of total power. C) one-third of total power. D) an equal amount of power.
458
1)
FALSE
21)
C
2)
TRUE
22)
A
3)
TRUE
23)
A
4)
TRUE
24)
A
5)
FALSE
25)
A
6)
TRUE
26)
A
7)
FALSE
27)
B
8)
TRUE
28)
B
9)
TRUE
29)
B
10)
FALSE
30)
C
11)
A
31)
B
12)
C
32)
A
13)
B
33)
D
14)
D
34)
B
15)
B
35)
A
16)
C
36)
B
17)
A
37)
A
18)
B
38)
C
19)
E
39)
C
20)
C
40)
D
459
460
PART 5 Laboratory Solutions for Experiments in Basic Circuits Prepared by David Buchla
461
Instructor’s Manual for:
Experiments in Basic Circuits: Theory and Application
- 9th edition
Note: All experiments have been laboratory tested; most are shown with measured values. Student values will vary depending on the components and equipment used.
Contents Multisim Troubleshooting ……. 463 Experiment 1 ……………………. 464 Experiment 2 ……………………. 467 Experiment 3 ……………………. 468 Experiment 4 ……………………. 469 Experiment 5 ……………………. 471 Experiment 6 ……………………. 473 Experiment 7 ……………………. 474 Experiment 8 ……………………. 476 Experiment 9 ……………………. 478 Experiment 10 …………………… 479 Experiment 11 …………………… 481 Experiment 12 …………………… 482 Experiment 13 …………………… 484 Experiment 14 …………………… 486 Experiment 15 …………………… 488 Experiment 16 …………………… 490 Experiment 17 …………………… 492 Experiment 18 …………………… 494 Experiment 19 …………………… 496 Experiment 20 …………………… 497 Experiment 21 …………………… 498 Experiment 22 …………………… 500 Experiment 23 …………………… 501 Experiment 24 …………………… 503 Experiment 25 …………………… 505 Experiment 26 …………………… 507 Experiment 27 …………………… 509 Experiment 28 …………………… 510 Experiment 29 …………………… 512 Experiment 30 …………………… 513 Experiment 31 …………………… 515 Experiment 32 …………………… 517 Experiment 33 …………………… 519 Motor Project (Appendix A) …… 521
462
Summary of Multisim Troubleshooting Solutions The following are the file names and simulated troubles for the Multisim circuits. These solutions are also given with each experiment. File names are based on figure numbers in the lab manual. For example, EXP7-3nf is a no fault file that is shown as Figure 7-3. Files with faults are given with f1, f2, or f3 suffixes, and have hidden faults. The password for the circuit restrictions option to reveal faults is “current”. Circuits are in folders labeled Multisim 9 and Multisim 10 for the version of Multisim you are using. The following is the contents of each Multisim folder: Folder Name: EXP-07 File Name EXP7-3-nf EXP7-3-f1 EXP7-3-f2 EXP7-3-f3
Fault none Potentiometer has open output (pin 2) R3 is 82 (done by adding 100 of “leakage”) VS is open
Folder Name: EXP-11 File Name EXP11-1-nf EXP11-1-f1 EXP11-1-f2 EXP11-1-f3
Fault none R3 is open VS2 is open VS2 is reversed (can be found by inspection)
Folder Name: EXP-13 File Name EXP13-3-nf EXP13-3-f1 EXP13-3-f2 EXP13-3-f3
Fault none - bridge is balanced if R4 is 1.83 k (18%) R2 is shorted Potentiometer is shorted between pins 2 and 3. R3 is 330 (done by adding 500 of “leakage”).
Folder Name: EXP-15 File Name EXP 15-3nf EXP15-3f1 EXP15-3f2 EXP15-3f3
Fault none R3 is open. R4 is 500 (done by adding 1 k of “leakage”). R5 is open.
Folder Name: EXP-25 File Name EXP25-2nf EXP25-2f1: EXP25-2f2: EXP25-2f3:
Fault none C1 is open. VS is open. R1 is 3.9 k (by adding “leakage”).
Folder Name: EXP-26 File Name EXP26-2-nf EXP26-2-f1 EXP26-2-f2 EXP26-2-f3
Fault none RS1 is open C1 is open R1 is shorted
Folder Name: EXP-30 File Name EXP30-3nf EXP30-3f1 EXP30-3f2 EXP30-3f3
Fault none RS is open VS is open C1 has 5 k of “leakage” (dropping Q significantly).
Folder Name: EXP-31 File Name EXP31-3nf EXP31-3f1 EXP31-3f2 EXP31-3f3
Fault none RL is open (note affect on response) L1 open VS is open
463
Experiment 1: Metric Prefixes and Scientific Notation. Data: Examples of metric units used on instrument controls: Table 1-3 Instrument
Control
Metric unit
Meaning
Oscilloscope
SEC/DIV
ms
10-3 seconds
Oscilloscope
SEC/DIV
ms
10-3 seconds
DMM
Range
mV
10-3 volt
DMM
Range
106 ohm
Function Gen
Range
M kHz
103 Hertz
Table 1-4 Dimension A
Length in millimeters 6.4 mm
Length in meters 6.4 x 10-3 m
B
14.8 mm
1.48 x 10-3 m
C
8.3 mm
8.3 x 10-3 m
D
31.8 mm
31.8 x 10-3 m
E
13.5 mm
13.5 x 10-3 m
F
5.2 mm
5.2 x 10-3 m
G
10.4 mm
10.4 x 10-3 m
Table 1-5 Number 0.0829 V
Scientific Notation 8.29 x 10-2 V
Engineering Notation 82.9 x 10-3 V
Metric Value 82.9 mV
48,000 Hz
4.8 x 104 Hz
48 x 103 Hz
48 kHz
2,200,000 0.000 015 A
2.2 x 10 1.5 x 10-5 A
2.2 x 10 15 x 10-6 A
2.2 M
6
6
7,500 W
7.5 x 10 W
7.5 x 10 W
15 A 7.5 kW
0.000 000 033 F
3.3 x 10-8 F
33 x 10-9 F
33 nF
270 000 0.000 010 H
2.7 x 10 1.0 x 10-5 H
270 x 10 10 x 10-6 H
270 k
3
5
3
3
464
10 H
Table 1-6 Metric Value 100 pF
Engineering Notation 100 x 10-9 F
12 kV
12 x 103 V
Table 1-7 Number
-6
Number
1.472
Rule Number 1
4.09 x 106
3
60.90
0.008
85 A 50 GHz
50 x 10 Hz
0.00150
4
84 x 10
33 k 250 mV
33 x 10 250 x 10-3 V
0.00842
2
0.105
85 x 10 A 9
3
5.0
-9
7.8 ns
7.8 x 10 s
2.0 M
2.0 x 106
70 = 400 turns = 300 turns = 200 turns = 100 turns
60
Inductance(mH)
50 40 30
10 0 0
2.0
4.0 6.0 8.0 Length (cm)
10
12
Plot 1-1 Further Investigation: Table 1-9 Metric Unit in Operand
Mathematical Operation
milli kilo nano milli micro micro pico milli
multiplied by multiplied by multiplied by multiplied by divided by divided by divided by divided by
Metric Unit in Operand milli micro kilo Mega nano pico pico Mega
Metric Unit in Result = = = = = = = =
micro milli micro kilo kilo Mega (unit) nano
465
4
Rule Number 2 3,4
3
1.00 x 10
1 3
-9
4
Evaluation and Review Questions: 1. (a) kilowatt = kW (b) milliampere = mA (c) picofarad = pF (d) nanosecond = ns (e) megohm = M (f) microhenry = µH
3. (a) 3.17 x 101 (b) -1.08 x 10-6 (c) 1.88 x 10-2 (d) 5.04 x 10-2 4. (a) -6.29 x 105 (b) 7.56 x 106 (c) 3.08 x 109 (d) -2.2 x 10-5
2. (a) MW = megawatt (b) nA = nanoamp (c) µJ = microjoule (d) mS = millisiemens (e) k = kilohm (f) GHz = gigahertz
5. (a) -0.629 M (b) 7.56 M (c) 3.08 G (d) -22 µ
6. The number 10.0 has three significant figures; the number 10 has only two. Application Problem: A representative physical layout is shown below:
466
Experiment 2: Laboratory Meters and Power Supply Data: Readings: Figure 2-7 (a)__22.5_mA_
Figure 2-7 (b)__15.5 mA___
For the meter in Figure 2-8, assume the OHMS function is selected and the range selected is X 1K ohms. What does the meter indicate for a resistance? _25 k__ What is the meter reading if the range selected is the 12 V DC range? ___1.95 V__ What is the meter reading if the range selected is the 30 V AC range? __4.8 V__ Step 3 and 4: Answers depend on lab equipment. Step Number
Power Supply Meter Reading
Lab Station Meter Reading
6
+5.0 V
+4.98 V
7
+12 V
+12.0 V
8
0V
0.10 V
Further Investigation Answers vary. The calibration curve should be approximately a 45° line, passing through the origin. Unlike other curves, points are connected “dot-to-dot” on a calibration curve in order to allow for interpolation between the nearest measured points. Evaluation and Review Questions: 1. The precision of a 3 1/2 digit DMM is one part in 2000. The precision of a power supply meter depends on the scale but the best is about 1 part in 100. This question can lead to a post-lab discussion of the difference between resolution and accuracy. 2. An autoranging meter automatically switches to the appropriate range to display the measured quantity. 3. A multiple scale is one with more than one range. A complex scale is used for more than one function. 4. A linear scale has equally spaced divisions and a nonlinear scale does not. 5. Each secondary mark has a value of 0.20. The meter reading is 3.20. 6. The ammeter must be in series with the load; start with the highest range and decrease the range as needed.
467
Experiment 3: Measurement of Resistance Data: Answers for Table 3-2 depend on resistors used. Values for Table 3-3 depend on potentiometer but should have a consistent sum in all steps equal to the total resistance between terminals 1 and 3. Further Investigation: Color of Band Resistor 1st
2nd
3rd
4th
Color-Code Value
Minimum Value
Maximum Value
5th
0
brown
red
violet
brown
red
1.27 k+2%
1.24 k
1.30 k
1
green
orange
blue
black
brown
536 +1%
531
541
2
orange
orange
black
yellow
red
3.3 M+2%
3.23 M
3.37 M
3
white
violet
blue
red
brown
97.6 k+1%
96.6 k
98.6 k
4
violet
green
black
gold
red
75.0 +2%
73.5
76.5
5
brown
green
black
black
brown
150 +1%
148.5
151.5
Evaluation and Review Questions: 1. (a) Answers vary. (b) Check the measurements with another meter or measure a known standard resistor with the meter in question. 2. Answer depends on potentiometer used. 3. (a) brown - red - black - silver (b) blue - gray - red - silver (c) white - brown - brown - silver (d) yellow - violet - green - silver (e) brown - black - gold - silver 4. (a) 22 (5%) (c) 510 (5%) (e) 820 k (10%)
(b) 750 (10%) (d) 9.1 (5%)
5. (a) largest value = 28,350
(b) smallest value = 25,650
6. Measured values of resistance should be used to avoid adding the tolerance error to computations.
468
Experiment 4: Ohm's Law Data: Table 4-2 (R1) VS =
2.0 V
4.0 V
6.0 V
8.0 V
10.0 V
I=
2.0 mA
4.0 mA
6.0 mA
8.0 mA
10.0 mA
Table 4-3 (R2) VS =
2.0 V
4.0 V
6.0 V
8.0 V
10.0 V
I=
1.3 mA
2.7 mA
4.0 mA
5.3 mA
6.7 mA
Table 4-4 (R3) VS =
2.0 V
4.0 V
6.0 V
8.0 V
10.0 V
I=
0.9 mA
1.8 mA
2.7 mA
3.6 mA
4.5 mA
R1 0.996
= R1 = R1 = R3
8 6
10 Current (mA)
Current (mA)
10
R2 1.52 k R3 2.20 k
4 2 0
0
2
4 6 8 Volts (V)
8 6 4 2 0
10
Plot 4-1
0
1
2 3 4 Volts (V)
Plot 4-2
Further Investigation: Table 4-5 (Zener diode) VS =
2.0 V
4.0 V
6.0 V
8.0 V
VZ = VZ =
2.0 V
3.99 V
4.97 V
4.98 V
5.0 V
0 mA
0 mA
1.0 mA
3.0 mA
5.0 mA
469
10.0 V
5
Application Problem: S2
S1 -
VS
+ A 0-10 mA V
R1
R2
R3
Evaluation and Review Questions: 1. The slope represents the conductance of each resistor. For R1, the slope is 1.0 mS; for R2 the slope is 0.67 mS, for R3 the slope is 0.45 mS. 2. The slope is lower for larger resistors. 3. (a) The current is doubled. (b) The current is doubled. 4. 2.0 k 5. 0.5 A 6. The current through the bulb is higher when the filament is not hot.
470
Experiment 5: Power in DC Circuits Data: Table 5-1 Component R1
Listed Value 2.7 k
Table 5-3
Measured Value
Variable Resistance Setting, R2 0.5 k 1.0 k 2.0 k 3.0 k 4.0 k 5.0 k 7.5 k 10.0 k
2.70 k
Table 5-2 Measured Value of Resistance
Voltage
Current
2.7 k
11.94 V
4.4 mA
Computed Power P= IV
P= I2R
52.5 mW
52.2 mW
P=
V R
2
52.8 mW
V1 (measured)
V2 (measured)
10.12 V 8.76 V 6.89 V 5.68 V 4.84 V 4.21 V 3.18 V 2.55 V
1.88 V 3.24 V 5.11 V 6.32 V 7.16 V 7.79 V 8.82 V 9.45 V
Power in R2 2
V2 R2 7.0 mW 10.5 mW 13.0 mW 13.3 mW 12.8 mW 12.1 mW 10.4 mW 8.9 mW P2 =
20 Power in R2 Power (mW)
16 12 8 4 00
6 8 2 4 Resistance (k)
10
Plot 5-1 Application Problem:
Further Investigation: Voltage Current
10 8 Measured data shown
8 6
2
4
1
2
0
2 4 6 8 Resistance (k)
10
Power (mW)
4
10 Voltage (V)
Current (mA)
5
6 4 2 00
Plot 5-2 (The product of IV is maximum when R2 is set to 3 k, the closest value to a matching load of 2.7 k
2
4 6 Voltage(V)
8
10
Plot 5-3 (Find I by measuring the voltage across R1. compute PLED from PLED = IVLED). 471
Evaluation and Review Questions: 1. The actual resistance of R2 at the peak power is 2.7 k (matching load), however, the data should support the answer that it is approximately 3 k. 2. (a) 213 mW (b) Yes 3. The total power decreases as R2 increases because there is less current. 4. (a) 15 mA. (b) 337.5 mW (c) A 1/4 watt resistor is rated for a maximum of 250 mW and should not be used. 5. Size. 6. 200
472
Experiment 6: Series Circuits Data: Table 6-1 Component
Table 6-2 Measured Value
R1
Listed Value 1.0 k
996
R2
1.5 k
R3
2.2 k
R4
330 RT (computed)
Computed Value
Measured Value
IT
2.98 mA
3.0 mA
1.52 k
VAB
2.98 V
2.96 V
2.20 k
VBC
4.47 V
4.46 V
332
VCD
6.58 V
6.54 V
5.05 k
VDE
0.98 V
0.99 V
Table 6-3 Step Number 7 8 9
Kirchhoff’s Voltage Law (Measured Values) –15.0 V + 2.96 V + 4.46 V + 6.54 V + 0.99 V = –0.05 V
Answers vary but should be equal to the result from step 7. –15.0 V + 0 V + 15.0 V + 0 V + 0 V = 0 V
(Voltage across open) Further Investigation: A resistance of 1.5 k will limit the current to 8.7 mA, meeting the requirement with a standard resistor. (I = 13 V/1.5 k = 8.7 mA). Application Problem: The computed value of RA is 9 k; RV is 149 k. Evaluation and Review Questions: 1. The numbers used in the summation are the same regardless of the starting point. The commutative property of mathematics applies to Kirchhoff's voltage law. 2. There was no current in the circuit of step 9 yet Kirchhoff's voltage law was found to be valid. (Note that Kirchhoff's voltage law applies to any closed path regardless of whether it is a conductive path or not). 3. The fuse would have 120 V across it. 4. VX = 3 V. 5. (a) 4 V
(b) 0.4 A
6. (a) 320
(b) 32 mW
(c) 20
473
Experiment 7: The Voltage Divider Data: Table 7-1 Resistor
VX (measured)
Measured Value
R1
Listed Value 333
333
RX RT 1.34 V
R2
470
473
1.90 V
1.90 V
R3
680
683
2.75 V
2.75 V
R4
1000
998
4.01 V
4.01 V
Total
2478
2487
10.0 V
10.0 V
VX = VS
1.34 V
+10 V +10 V
R1
+10 V
R1
R2
R1
R3 R3 6.8 V R4
5.0 V
R4 Circuit for step 5
Circuit for step 6
R4
7.5 V
Circuit for step 8
Table 7-2 Computed
Measured
VMIN
2.61 V
2.63 V
VMAX
8.61 V
8.20 V
Further Investigation: (a) Data: Table 7-3 Switch Position A
Measured Voltages RL = 1.0 kRL = 10 kRL = 100 k 8.44 V 8.65 V 6.76 V
B
4.40 V
6.43 V
6.73 V
C
2.53 V
3.80 V
4.01 V
(b) Results: The loading effect is greater with smaller resistors. Application Problem: The computed value of R1 is 1.67 k; R2 is 3.33 k. Nearest standard values (5%) are 1.6 k and 3.3 k.
474
Multisim Application: File EXP7-3f1: fault is: Potentiometer has open output (pin 2) File EXP7-3f2: fault is: R3 is 82 ohms (done by inserting 100 of leakage) File EXP7-3f3: fault is: VS is open Evaluation and Review Questions: (b) Power dissipated is a factor of ten less. 1. (a) Vout is unchanged. 2. (a) Vout = 0 V
(b) Vout = 10.0 V
3. Range increases. Vmin = 0.44 V Vmax = 9.7 V. 4. VA = 10 V
VB = 1.0 V
VC = 0.10 V
5. Vmin = 0 V
Vmax = 6.67 V.
VD = 0.010 V
6. (a) P1 = 640 mW P2 = 320 mW (b) As long as no current is supplied to the output, the power dissipated in the divider is dependent only on the divider resistance and the source voltage.
475
Experiment 8: Circuit Ground Data: Table 8-1 Component
Table 8-2 Measured Value
R1
Listed Value 330
333
VS
+10.0 V
R2
680
683
VA
+1.64 V
1.0 k
998
B
R3
Measured Value
VBC
+3.38 V
VC
+4.98 V
D
Table 8-3 Measured Voltage VA
+10.0 V
VB
+8.36 V
VC
+4.98 V
VD
0.0 V (ref)
Table 8-4 Measured Voltage
Voltage Difference Calculation VAB = VA VB = +1.64 V VBC = VB VC = +3.38 V VCD = VC VD = +4.98 V
VA
+5.03 V
VB
+3.38 V
VC
0.0 V (ref)
VD
-4.98 V
Table 8-5 Measured Voltage VA
+1.64 V
VB
0.0 V (ref)
VC
-3.38 V
VD
-8.36 V
Voltage Difference Calculation VAB = VA VB = +1.64 V VBC = VB VC = +3.38 V VCD = VC VD = +4.98 V
Table 8-6 Measured Voltage
Voltage Difference Calculation VAB = VA VB = +1.64 V VBC = VB VC = +3.38 V VCD = VC VD = +4.98 V
VA
0.0 V (ref)
VB
-1.64 V
VC
-5.02 V
VD
-10.0 V
Voltage Difference Calculation VAB = VA VB = +1.64 V VBC = VB VC = +3.38 V VCD = VC VD = +4.98 V
Further Investigation: Observed voltage differences are unchanged in this investigation. Point C in the circuit will have nearly (but not exactly) the same potential as ground since half the circuit’s resistance is above this point and half is below. Voltages referenced to ground will be nearly the same as in Table 8-4. Application Problem: The approximate voltages shown can be obtained if RA = RC = 1.0 k; RB = 680 ; and RD = 330 . The source voltage is set to 3.475 V.
476
Evaluation and Review Questions: 1. The voltage difference calculations indicate that the voltage difference is independent of the circuit ground. Voltage is usually defined with respect to a reference point called ground, but voltage difference is measured between the points named by the subscripts. 2. Reference ground is the point in a circuit defined as 0 V. All other voltages in a circuit are referenced to this point. 3. 12 V. 4. 70 V. 5. +8.3 V. 6. (a) VGS = 2 V
(b) VDS = +10 V
477
Experiment 9: Parallel Circuits Data: Table 9-1 Component
Measured Value
R1
Listed Value 3.3 k
R2
4.7 k
4.71 k
R3
4.7 k
6.82 k
R4
10 k
9.97 k
3.30 k
Table 9-2 RT (measured) IT (measured)
R1
R1 || R2
R1 || R2 || R3
R1 || R2 || R3 || R4
3.30 k
1.93 k
1.51 k
1.31 k 9.15 mA
Table 9-3 VS R1 3.64 mA I 1=
I (computed)
VS R2 2.55 mA I2=
V I 3 = RS 3 1.76 mA
VS R4 1.20 mA I4=
Table 9-4 Total current with R1 open =
5.5 mA
Evaluation and Review Questions: 1. Subtract the observed current from the original total. The difference is the "missing" current due to an open branch. Apply Ohm's law to find the open resistance. 2. The short causes the current to go very high. If the power supply does not have short circuit current limiting, a fuse will blow or damage will result. 3. (a) Current should be 167 mA.
(b) The 820 resistor is open.
4. I4 is 25 mA entering the junction. 5. The current divider rule applied to the resistors in this experiment is: Ix =
RT I by substitution, I 1 = 3.63 mA Rx x
I 2 = 2.54 mA
I 3 = 1.76 mA
I 4 = 1.20 mA
6. (a) The shunt current is 225 mA. (b) The resistance is 6.7 . It dissipates 338 mW; specify a 1/2 watt resistor.
478
Experiment 10: Series-Parallel Combination Circuits Data: Table 10-1 Component
Measured Value
R1
Listed Value 2.2 k
R2
4.7 k
4.69 k
R3
5.6 k
5.63 k
R4
10 k
9.96 k
2.22 k
Step 3 Equivalent circuit: Equivalent resistance of R2 || R3
R1 VS = +12.0 V
Table 10-2 Computed Voltage Ohm’s Divider Law RT
2.22 k
R2,3 2.56 k
R4 9.96 k
1.82 V
1.81 V
V2,3
2.09 V
2.10 V
2.09 V
V4
8.13 V
8.16 V
8.13 V
I2
0.45 mA
I3
0.37 mA
R1,2 6.91 k
12.0 V
12.0 V
Table 10-3
Equivalent resistance of R3 in series with R4 R3,4 15.59 k
479
14.7 k
1.81 V
Step 10 Equivalent circuit:
VS = +12.0 V
14.7 k 0.82 mA
V1
VT
Equivalent resistance of R1 in series with R2
14.7 k
IT
Measured
Computed
Measured
R1,2
6.91 k
6.90 k
R3,4
15.59 k
15.59 k
RT
4.79 k
4.78 k
IT
2.51 mA
I1,2
1.74 mA
I3,4
0.77 mA
V1
3.86 V
3.84 V
V2
8.16 V
8.16 V
V3
4.34 V
4.33 V
V4
7.67 V
7.67 V
12.0 V
Further Investigation: Student should summarize a procedure for solving the problem. The total resistance seen by the voltage source is 1.74 k and the total current is 6.90 mA. Application Problem: For the L-pad, A = 10:1. By substitution into the equations given, the values for the pad resistors are: A 1 10 1 R1 Rs 600 540 A 10 1 1 600 R2 Rs 150 Rs 10 6 A RT Looking into the pad from the source, the equivalent resistance is 600 (matching load). The voltage divider theorem can be used to show the attenuation is 10. Evaluation and Review Questions: 1. a) The voltage divider rule was applied to an equivalent series circuit. b) The voltage divider rule can be applied to any set of series resistors for which the total voltage across the resistors is known. The voltage divider rule can be applied to each series branch in Figure 10-3 independently to find the voltage drops across each resistor. 2. Answers vary. One possible path (around outside loop) is: 12 V + 4.31 V + 7.69 V = 0 V 3. Currents entering and leaving the junction are equal. This can be shown as: 2.51 mA = 1.74 mA + 0.77 mA 4. (a) A resistor or connection in the R1 - R2 branch is open. (b) Check the resistors to see if a +12 V drop is across one or the other. If not, check for a drop across the connection points. 5. (a) V1 = +16 V (b) VS = +24 V. 6. Series resistors are connected such that there is one path; the identical current is in both. Parallel resistors are connected so that both ends of each resistor are joined; the identical voltage is across both.
480
Experiment 11: The Superposition Theorem Data: Table 11-1 Measured Value
R1
Listed Value 4.7 k
R2
6.8 k
6.80 k
R3
10 k
9.90 k
Table 11-2
4.69 k
Quantity
Computed
Measured
Step 4
RT (VS1 operating alone)
8.74 k
8.73 k
Step 7
RT (VS2 operating alone)
9.98 k
10.1 k
Table 11-3 Computed Current* I1 I2 I3
Computed Voltage* V1 V2 V3
+0.57 +0.34 +0.23 Step 5 +2.67 +2.31 Step 6 -0.68 -1.00 +0.32 Step 8 -3.19 -6.80 Step 9 -0.11 -0.66 +0.55 -0.52 -4.49 Step 10 (totals) * (Note: all currents are shown in mA, all voltages in V).
V1
Measured Voltage* V2 V3
+2.28
+2.69
+2.32
+2.32
+3.18 +5.46
-3.18 -0.50
-6.80 -4.49
+3.18 +5.49
Further Investigation: The results indicate that the superposition theorem does not apply to power. Multisim Application: File EXP11-1f1: fault is: R3 is open File EXP11-1f2: fault is: VS2 is open. File EXP11-1f3: fault is: VS2 is reversed (can be found by inspection) Application Problem: RA = 6.8 k; RB = 10 k potentiometer; RC = 4.7 k. Evaluation and Review Questions: 1. (a) -5.0 V + (-0.51 V) + (-4.49 V) + 10 V = 0 V (b) -0.109 mA = - 0.66 mA + 0.55 mA 2. The actual direction of current is the opposite of the assumed direction. 3. The sign of all results would be reversed but there is no effect on the circuit itself. 4. (a) Replace all sources except one with their internal resistance. (b) Compute the current or voltage due to the one source acting alone. (c) Repeat steps (a) and (b) for all sources. (d) Algebraically sum the results. 5. Current due to VS1 = 54.5 mA. = -145.5 mA. Current due to VS2 Net current due to both sources = -91.0 mA. 6. Current sources have high internal resistance; an ideal one has infinite resistance.
481
Experiment 12: Thevenin's Theorem Data: Table 12-1 Component
Measured Value
R1
Listed Value 270
R2
560
556
R3
680
680
RL1
150
151
RL2
470
471
RL3
820
810
Load Voltage Calculation: (Step 2) RL1 + R2 = 151 + 556 =707
274
VL1
Measured
1.20 V
1.20 V
VL2
2.76 V
2.76 V
VL3
3.72 V
3.70 V
VTH
7.13 V
7.12 V
RTH
751
751
347 347 + 274
VL1 = 5.62 V
151 707
= 5.62 V
= 1.20 V
Thevenin Circuit (step 8)
Table 12-2 and Table 12-3 Computed
VL1,2,3 = 10 V
RTH = 751 VTH = 7.12 V
Note: Tables 12-2 and 12-3 should have identical results.
Further Investigation: When the voltage across the load is one-half the unloaded voltage, the internal Thevenin resistance has same voltage drop as the load resistor. Therefore, the two resistances must be equal. Application Problem: Setting up simultaneous equations, the computed value of RA is 559 and RB is 805 . The resistors in this experiment (RA = 560 and RB = 820 ) give the required results. Evaluation and Review Questions: 1. The original circuit and the Thevenin circuit are equivalent as seen by the load resistor. 2. The load current in a short is 9.5 mA for both circuits. 3. Calculations are simplified. 4. The voltage and resistance measured at the output are not affected by R1, therefore it is not part of the Thevenin circuit.
482
5.
Table 12-4 RL
Power in RL
0.7 k
8.7 mW
1.7 k
12.6 mW
2.7 k
13.3 mW
3.7 k
13.0 mW
4.7 k
12.4 mW
6.(a) VTH = 10 V
RTH = 500
The circuit is the same (except for resistance settings) as the one in Experiment 5 (Figure 5-2), so the calculated points here should fit on Plot 5-1. This emphasizes that the peak power is delivered to a load when the load resistance is equal to the fixed Thevenin resistance.
(b)
RTH = 500 VTH = 25.9 V
483
Experiment 13: The Wheatstone Bridge Data: Table 13-1 Component
Measured Value
R1
Listed Value 1.2 k
R2
2.2 k
2.22 k
1.19 k
R3
1.0 k
1.00 k
RL
470
475
R4
10 k pot.
9.65 k
The unbalanced Wheatstone Bridge: Table 13-2 Computed
Measured
RTH = 1.67 k A VTH = -2.99 V
VTH
-3.05 V*
-2.99 V
RTH VL
1.68 k
1.67 k
0.662 V
0.662 V
RL = 475
B
*(Referenced from A to B)
The balanced Wheatstone Bridge: Table 13-3
RTH = 1.42 k A
Measured VTH
0V
RTH VL
1.42 k
VTH = 0 V B
0V
Further Investigation: In a 150 foot length, the fault can be located within about 2 feet (depending on bridge). Application Problem: Table 13-4 Temperature (oC) 40o
Meter Current (mA)
50o
1.374 k
0.6 mA
60
o
0.993 k
1.30 mA
70
o
0.731 k
2.07 mA
80o
0.547 k
2.87 mA
o
0.416 k
3.61 mA
0.322 k
4.33 mA
90
100o
Current (mA)
10
Thermister Resistance (k) 1.943 k
0 mA
8 6 4 2 040
50 60 70 80 Temperature ( oC)
Plot 13-1
484
90 100
Multisim Application: File EXP13-3f1: fault is: R2 is shorted. File EXP13-3f2: fault is: Potentiometer is shorted between pins 2 and 3. File EXP13-3f3: fault is: R3 is 330 ohms (done by adding 500 of “leakage”). Evaluation and Review Questions: 1. The Thevenin resistance is 297 ; RT = 397 . The Thevenin voltage is 1.2 V. The load current is 3.02 mA. 2. Doubling the load resistance does not double the total resistance; therefore the load current is not halved. 3. There was no voltage across the load. 4. (a) Yes. Current in the load must flow through the bridge arms. (b) No. In a balanced bridge, there is no current in the load. 5. (a) The current in the load is reduced but not by half. This is because doubling the bridge resistors does not double the total resistance of the circuit. (b) Doubling the resistance has no effect on load current since it is zero. The balanced bridge will remain in balance with no load current. 6. (a) All currents would be doubled including the load current. (b) Theoretically, there is no effect on the load current since it is zero in a balanced bridge, however, doubling the source voltage makes detection of the balance point easier.
485
Experiment 14: Norton's Theorem Data: Table 14-1 Resistor
Listed Value
Measured Resistance
R=0
5.0 4.0 I (mA) 3.0 2.0 1.0 0
Measured Current 5.0 mA
RL1
470
463
5.0 mA
RL2
1.0 k
999
5.0 mA
RL3
1.5 k
1.49 k
4.9 mA
RL4
2.2 k
2.23 k
4.8 mA
0
1.0 1.5 R (k)
Plot 14-1
Table 14-2 Resistor
Measured Resistance
R1
Listed Value 1.8 k
R2
560
558
R3
1.2 k
1.19 k
1.81 k
Norton Circuit:
Thevenin Circuit: A
IN = 4.71 mA
RN = 1.275 k
RTH = 1.275 k A VTH = -6.0 V
B
B
Further Investigation: The computed Norton current source is 10.24 mA and the computed Norton resistance is 868 . Application Problem: Solving the simultaneous equations gives RA = 141 and RB = 70.7 . Evaluation and Review Questions: 1. (a) The internal resistance of an ideal current source is infinite. (b) The internal resistance of an ideal voltage source is zero. 2. (a) As long as the load resistance is less than about 2.5 k, the source is nearly ideal. (b) The current would no longer be constant. 3. Thevenin circuit:
Norton circuit:
RTH = 1 VTH = 12 V
IN = 12 A
RN = 1
486
4. Thevenin circuit:
Norton circuit:
RTH = 600 VTH = 10 V
IN = 16.7 mA
RN = 600
5. (a) IN = 1.0 mA
RN = 100 k
(b) More than 99% of the Norton current will go in any load that is equal or less than 1 k. 6. The Norton circuit for the unbalanced Wheatstone bridge is:
IN = 1.86 mA
RN = 1.69 k
The current in a 470 ohm load is 1.45 mA. This is the same circuit that was investigated in Experiment 13 and is equivalent to the circuit calculated in steps 4 and 5 of that experiment.
487
Experiment 15: Circuit Analysis Methods Data: Table 15-1 Resistor
Measured Resistance
R1
Listed Value 10 k
R2
3.6 k
3.60 k
R3
2.0 k
1.97 k
R4
1.0 k
0.976 k
R5
4.7 k
4.72 k
10.1 k
Loop Equations: (Measured values are shown. Currents are in mA.) 10.1 k I A I C 1.97 k I A I C 10 V 0
1.97 k I B I A 3.60 k I B I C 0.976 kI B 0 10.1 k I C I A 4.72 k I C 3.60 k I C I B 0
Standard form: 12.1 k I A 1.97 k I B 10.1 k I C 10 V
1.97 k I A 6.55 k I B 3.60 k I C 0 V
10.1 k I A 3.60 k I B 18.4 k I C 0 V Note: |D| = 420.15 Table 15-2
Table 15-3
Computed Current IA
2.56 mA
IB
1.73 mA
IC
1.74 mA
I1
0.82 mA
I2 I3
0.01 mA
I4
1.73 mA
I5
1.74 mA
0.85 mA
Computed
Measured
V1
8.28 V
8.35 V
V2
0.04 V
0.05 V
V3
1.67 V
1.66 V
V4
1.69 V
1.71 V
V5
8.21 V
8.29 V
Further Investigation: Assigning the left windowpane as IA and the right windowpane as IB, the clockwise loop equations are: -5.0 V + ( IA ) 4.7 k + (IA-IB) 10.0 k = 0. (IB-IA) 10.0 k + (IB) 6.8 k + 10 V = 0. Solving gives IA = -0.109 mA and IB = -0.66 mA. Note that IA = I1 ; IB = I2 ; and IA-IB = I3.
488
Application Problem: The proof is: V V V in – out V out – out 2 2 + =IL R R V V V in – out + V out – out = I L R 2 2 V I L = in R Multisim Application: File EXP15-3f1: fault is: R3 is open. File EXP15-3f2: fault is: R4 is 500 ohms (done by adding 1 k of “leakage”). File EXP15-3f3: fault is: R5 is open. Evaluation and Review Questions: 1. The Thevenin resistance is 2.48 k and the Thevenin voltage is 6.06 V. Adding R4 to the Thevenin circuit, the current, I4 is computed to be 1.74 mA. 2. Setting up loop A in the bottom left windowpane, loop B in the top windowpane, and loop C in the right windowpane, the loop equations are: (IA ) 1 k + (IA-IB) 1 k + (IA-IC) 1 k = 0. (I B ) 1 k + (IB-IC) 1 k + (IB-IA) 1 k = 0. (IC-IA) 1 k + (IC-IB) 1 k + 1 V = 0. Solving the simultaneous equations yields IC = 1 mA. Therefore, the resistance looking back into the bridged tee is 1 k; this implies that the Thevenin resistance is 1 k. 3. No effect on the actual current; all answers will have the opposite sign which cancels the effect of the assumption. 4. Solution of simultaneous equations gives IB = Vin / 5R. The transfer function is 0.2. 5. The bottom of the bridge is the reference. For the node equations, the direction of current in the load resistor is assumed to be from A to B. The equations are: For node A: I 1 – I 2 – IL = 0 V S – V A V A V A – VB – – =0 R1 R2 RL
For node B: I3–I4+IL =0 VS – VB VB VA –VB – + =0 R3 R4 RL
Normally, the unknowns are VA and VB; therefore two equations are sufficient. 6. (a) Loop equations
(b) Node equations
489
Experiment 16: Magnetic Devices Data: Step 1. Answers depend on particular relay used. Data shown is for a small relay rated for 9 V on the coil.
Table 16-1 Pull-in Voltage
Release Voltage
Trial 1
7.4 V
3.4 V
Trial 2
7.4 V
3.2 V
Trial 3
7.5 V
2.9 V
Average
7.43 V
3.17 V
Step 7. Red light turns on when S1 is closed; it stays on when S1 is open. Step 8. Relay "buzzes" and both red and green light are rapidly switched on and off. Further Investigation: The CdS cell can be placed in series with the control voltage and the relay coil. The relay will be energized when light strikes the CdS cell. Application Problem: Answers to questions: 1. Contacts 1-2 are latching contacts that keep power applied after pushbutton is released. 2. The door opening is stopped by the upper limit switch. 3. The STOP pushbutton removes power from CR2, which releases holding contacts 2-2. The door cannot resume motion because the holding contacts and CLOSE pushbutton are both open. 4. This prevents the line voltage from being applied to both motor coils at the same time. 5. The CR2 coil pulls in contacts 2-3 to energize the motor. Schematic: Student schematic should show the second OPEN and CLOSE pushbuttons in parallel with the first pushbuttons and the second STOP pushbutton in series with the original pushbutton.
490
Evaluation and Review Questions: 1. Answers depend on the particular relay tested. The tested relay had a coil resistance of 59 . The average pull-in current was 126 mA. 2. The tested relay had an average release current of 53.7 mA. 3. The hysteresis of the tested relay was 7.43 V - 3.16 V = 4.27 V. 4. (a) SPDT means there is one switch with two contacts. (b) DPST means there are two switches, each with one contact. 5. (a) Relay coil could be open, control voltage could be off or too low, switch S1 is not making contact, relay is stuck in one position. (b) Measure the voltage across the relay coil, if it is correct, relay is inoperative. If the coil voltage is not sufficient to pull the relay in, check S1 and the source voltage. 6. The magnetic core concentrates and intensifies the magnetic flux.
491
Experiment 17: The Oscilloscope Data: Computed values are shown. Student results will depend on the oscilloscope and function generator used. Table 17-1 Power Number of Oscilloscop DMM Supply VOLTS/DIV Divisions of e (measured (measured Setting Setting Deflection voltage) voltage) 1.0 V 0.2 V/div 5.0 div 1.0 V 1.0 V 2.5 V 0.5 V/div 5.0 div 2.5 V 2.5 V 4.5 V 1.0 V/div 4.5 div 4.5 V 4.5 V 8.3 V 2.0 V/div 4.15 div 8.3 V 8.3 V Signal Generator Amplitude 1.0 Vrms 2.2 Vrms 3.7 Vrms 4.8 Vrms
Table 17-2 Number of VOLTS/DIV Divisions (peak-to-peak) Setting 0.5 V/div 1.0 V/div 2.0 V/div 2.0 V/div
Oscilloscop e (measured)
DMM (measured voltage)
2.8 Vpp 6.2 Vpp 10.5 Vpp 13.6 Vpp
1.0 Vrms 2.5 Vrms 3.7 Vrms 4.8 Vrms
(peak-to-peak)
5.6 div 6.2 div 5.25 div 6.8 div
Further Investigation: Most oscilloscopes will have a 1 kHz square wave output at a small connector labeled PROBE COMP. The operator’s manual typically will show how to adjust the probe and representative waveforms of a properly compensated probe. Forgetting to check probe compensation and not initializing the control setup of an oscilloscope are the two most common operator errors. Application Problem:
Waveform Sine Square Triangle Sawtooth *
Table 17-4 Oscilloscope Readings Vp (Measured) Vavg(Computed) Vavg x FF = 1.0 Vp 1.0 Vp 1.0 Vp 1.0 Vp
0.637 Vavg 1.0 Vavg 0.5 Vavg 0.5 Vavg
measured
492
0.707 V 1.0 V 0.575 V 0.575 V
DMM Reading* 0.687 V 1.042 V 0.550 V 0.562 V
Evaluation and Review Questions: 1. (a) Answers vary. (b) The answer depends on the instruments used; however, in general, the DMM is more accurate for dc measurements for typical laboratory equipment. A 3 1/2 digit DMM has a resolution of 1 part in 2000 and an accuracy specification of <0.1%. An analog oscilloscope linearity is typically 3% (with less resolution than the DMM). A digital oscilloscope typically has a resolution of at least 1 part in 256 (8-bit resolution) and a dc accuracy of 3% X reading + 0.1 div + 1mV. 2. Vertical controls: control the vertical axis of the oscilloscope and coupling of the input signal. Trigger controls: determines when the horizontal sweep occurs and the source of triggers. Horizontal controls: control the horizontal axis (typically the time axis) of the oscilloscope. Display controls: control the CRT. 3. Trigger controls. 4. (a) 17.0 Vpp (b) 6.01 Vrms 5. The signal is 56.6 Vpp. Use 10.0 volts/div to spread the signal over 5.66 divisions. 6. For most analog oscilloscope measurements, the resolution of the display is limited by the beam size and ability to read the graticule. By using a larger portion of the display, these errors are minimized. For digital scopes, the waveform should be large to maximize resolution, but not set beyond the display area.
493
Experiment 18: Sine Wave Measurements Data: Signal Generator Dial Frequency 1.25 kHz 1.90 kHz 24.5 kHz 83.0 kHz 600.0 kHz
measured compute d
Table 18-1 Computed Oscilloscop Period e SEC/DIV 0.8ms 0.1 ms/div 0.526ms 0.1 ms/div 40.8 s 5.0 s/div 2.0s/div 12.0 s 0.2 s/div 1.67 s
Table 18-2 Function Gen. Voltage Voltage across R1 1.0 Vpp 0.28 Vpp 1.0 Vpp 0.28 Vpp
Number of Divisions
Measured Period
8.0 div 5.26 div 8.16 div 6.0 div 8.35 div
0.8 ms 0.53 ms 40.8 s 12.0 s 1.67 s
Voltage across R2 0.72 Vpp 0.72 Vpp
Application Problem: The ratio of points touching on the x-axis to points touching on the y-axis of a Lissajous figure is inversely proportional to the frequencies applied to the x- and y-axis. Frequencies which are exact multiples will stop the figure from moving, a difficult task with ordinary signal generators. Evaluation and Review Questions: 1. (a) Answers vary. (b) Signal generator calibration, oscilloscope time-base error, reading error. 2. -1.0 V + 0.28 V + 0.72 V = 0 V 3. (a) 126 ms (b) 7.9 Hz 4. 10 µs/div 5. (a) Connect CH 1 to one side of the component and CH 2 to the other side of the component. (b) Calibrate both channels (verniers in detent position) and set volts/div to the same for each channel. (c) Use difference function to measure voltage across the ungrounded component. (For many oscilloscopes, this is done by using the ADD mode and INVERT CH 2).
494
Note: You may want to emphasize the danger inherent in attempting to measure voltages in a circuit which are not referenced to ground. If the scope ground clip is connected to a point in the circuit which is not ground, the circuit under test can be destroyed. Another danger awaits the unwary in measuring "transformer less" ac circuits such as TV sets and the like. These appliances connect one side of the ac line directly to the chassis. If a two-wire plug is connected to the outlet in reverse, or a technician attempts to "hot-wire" the input ac voltage, the entire chassis can become "hot" with respect to ground. Connecting a scope ground to the chassis can lead to a direct short across the ac power or be the source of a shock. The best solution is to connect the test circuit through an isolation transformer. 6. CHOP/ALTERNATE is a choice on an analog scope for selecting how one beam can be shared between two signals. CHOP is selected when viewing low frequency signals (below about 1 kHz); ALTERNATE is selected for high frequency signals.
495
Experiment 19: Pulse Measurements Data:
Data shown were taken with an H-P 3311A Function Generator and a Tektronix 2246 oscilloscope.
Table 19-1 Oscilloscope BW =
100 MHz
t(r) =
3.5 ns
Table 19-2 Function Generator (square wave output) 64 ns Rise time, t(r) Fall time, t(f)
67 ns
Period, T
10.0 s
Pulse width, tw
5.0 s
% duty cycle
50%
Table 19-3 Function Generator (with 1000 pF capacitor across output) Rise time, t(r) 1.40 s Fall time, t(f)
1.27 s
Table 19-4 Function Generator
(pulse output) 64 ns Rise time, t(r) Fall time, t(f)
67 ns
Period, T
10.0 s
Pulse width, tw
5.0 s
% duty cycle
50%
Application Problem: The sine wave 70.7% point was measured at 630 kHz. The rise time of the amplifier was measured as 0.58 µs; from this measurement, the frequency response is computed to be 603 kHz. (Note that the lower cutoff frequency is about 1 Hz and can be ignored for calculating the bandwidth). Evaluation and Review Questions: 1. Answer depends on the oscilloscope. If scope’s rise time is at least 4 times faster than the measured rise time, the result is not bandwidth limited, otherwise it is (3% error limits). 2. The oscilloscope should have a rise time of 2.5 ns - implying a bandwidth of 140 MHz. 3. When the X10 magnifier is used, the sweep is moving faster to increase the horizontal gain by a factor of ten. The faster sweep means that the time per division across the screen is reduced by a factor of ten. 4. Figure 19-4 indicates 4.0 horizontal divisions between the 10% and 90% levels. The rise time is 8.0 ms. 5. The rise time would be 0.8 ms. 6. Pulses contain low frequency components; ac coupling can cause distortion, especially to low frequency pulses.
496
Experiment 20: Capacitors Data:
(Answers depend on meters used; typical answers shown.)
Capacitor C1 C2 C3 C4
Listed Value 100 F 47 F 1.0 F 0.1 F
C5
0.01 F
Ste p 4 5 6 7 8 9 10
V1
V2
Table 20-1 Ohmmeter Test Pass/Fail
Voltmeter Test Pass/Fail
pass
very slow- fail
pass
very slow- fail
pass
pass
difficult to see pass cannot see - fail
pass
Q1
pass
Table 20-2 Q2
Observations Closing S1 causes series LED to flash. Closing S2 causes parallel LED to flash.
3.4 V 10.1 V
7.0 V 340 C
333 C
The parallel LED flashes for a shorter time.
1010 C
475 C
The parallel LED flashes for a longer time.
10.1 V Series LED flashes; parallel LED is steady.
For Further Investigation: The ungrounded LED shows a square wave; the grounded LED waveform approaches dc. Application Problem: The LEDs alternately blink as one capacitor is charged on the positive cycle and the other on the negative cycle. A 16.5 V dc level (measured) was across the load. When the frequency is increased to 50 Hz, a sawtooth waveform of about 30 mVpp can be observed riding on the dc level. (Scope should be ac coupled to observe this). Evaluation and Review Questions: 1. The total stored charge was less for the series capacitors. 2. The charge would be smaller and the LED would flash for an even shorter time. 3. (a) The total capacitance in parallel is 3.0 µF. (b) The total capacitance in series is 0.67 µF. (c) The larger voltage is across the 1.0 µF capacitor. 4. The charge of 300 µC is distributed over 13.0 µF. The voltage is therefore 23 V. 5. 683 = 68,000 pF = 0.068 F; 102 = 1000 pF = 0.001 F; 224 = 220,000 pF = 0.22 F 6. 47 pF = 47 (under 100 pF – no multiplier digit shown). 0.033 F = 333
497
10,000 pF = 103;
Experiment 21: Capacitive Reactance Data: Table 21-2 Capacitor C1 Voltage across R1, (VR) 0.532 V Total current, IT 0.532 mA Voltage across C, (VC) 0.846 V Capacitive reactance, XC 1.59 k Computed capacitance, C 0.10 F
Capacitor C2 0.283 V 0.283 mA 0.959 V 3.39 k 0.047 F
Table 21-3 Step 7(a) 7(b) 7(c) 7(d)
Voltage across R1, (VR) Total current, IT Voltage across C, (VC) Capacitive reactance, XC Computed capacitance, C
Series Capacitors 0.197 V 0.197 mA 0.98 V 1.59 k 0.10 F
Parallel Capacitors 0.68 V 0.68 mA 0.73 V 3.39 k 0.047 F
For Further Investigation: The output is an ac signal riding on a 10.2 V dc level. At a frequency of 1.0 kHz, the capacitive reactance is 159.2 , which drops approximately 50 mV across the capacitor. At 100 Hz, the reactance is 1.59 k and the voltage across the capacitor is approximately 0.45 V. At 10 kHz, the reactance is 15.9 and the capacitor appears to be nearly a short to the ac signal. Application Problem: The required capacitive reactance is 10 k (ignoring the small difference between Ztotal and R ). At 100 Hz, a reactance of 10 k is produced by a capacitance of 0.16 µF. Any capacitor larger than this satisfies the requirement of dropping at least 90% of the voltage across the resistor at 100 Hz. Evaluation and Review Questions: 1. The total series reactance is the sum of the capacitive reactances from Table 21-3 whereas the parallel reactance is the product-over-sum of the values in Table 21-3. 2. The capacitance of the series capacitors is the product-over-sum of the capacitances listed in Table 21-3 whereas the parallel reactance is the sum of the capacitances listed in Table 21-3. 3. The smaller capacitance would have a higher capacitive reactance.
498
4. Measure the voltage across the resistor and determine the current. Measure the voltage across the capacitor. Apply Ohm's law to these measured values to find the reactance of the capacitor. Find the capacitance by applying the formula: C =
1 2 f X C
5. The higher frequency will result in a lower voltage across the capacitor; hence the computed capacitive reactance will be smaller. The higher frequency just compensates for the reduction in XC in the equation in step 5. Thus the capacitance can be correctly computed. 6. The capacitive reactance is 796 .
499
Experiment 22: Inductors Data: Observations from step 1: Neon bulb fires when switch is opened. Observations from step 2: The bulb fires with as little as 1 V from the power supply. VS =
Table 22-1 Computed Time Time constant, 0.21 ms
VL = VR =
Measured Time 0.22 ms
Plot 22-1 Observations from step 5: In parallel, the time constant is shorter. Consequently, the rise of the resistor voltage and fall of the inductor voltage is faster. In series, the time constant is longer, and the waveforms change at a slower rate. Further Investigation: The measurement of the time constant, as performed in this experiment, can be used with a known resistor to determine the inductance of an unknown inductor. Application Problem: The output voltage is approximately 2.5 V dc with ripple "spikes" of 16 mVpp. The spikes are slightly larger when the load is decreased to 16 k. The voltage across the inductor is a 5 Vpp square wave that is centered about zero volts. Evaluation and Review Questions: 1. The opening of the switch interrupts the current and causes a very large change in current in the circuit. This large change in current induces a voltage transient across the inductor according to Lenz's law. 2. The arc is caused by the induced voltage that appears across the coil in response to the large change in current when the switch is opened. 3. (a) The inductance of two 100 mH inductors connected in series is 200 mH. (b) The inductance of two 100 mH inductors connected in parallel is 50 mH. 4. The time constant would be 10 times larger. 5. A higher frequency does not change the time constant. At higher frequencies, there is less time for the current to change so the voltage across the resistor is approaches a dc level and the inductor voltage approaches a square wave. 6. The polarity of the voltage induced across the inductor is such to keep current in the direction it was going prior to a change.
500
Experiment 23: Inductive Reactance Data: Computed values are shown; measured values were within 2% of these values. Table 23-2 Inductor L1 Voltage across R1, (VR) 0.303 V Total current, IT 0.303 mA Voltage across L, (VL) 0.953 V Inductive reactance, XL 3.14 k Computed inductance, L 100 mH
Inductor L2 0.303 V 0.303 mA 0.953 V 3.14 k 100 mH
Table 23-3 Step 7(a) 7(b) 7(c) 7(d)
Voltage across R1, (VR) Total current, IT Voltage across inductors, (VL) Inductive reactance, XL Computed inductance, L
Series Inductors 0.157 V
Parallel Inductors 0.537 V
0.157 mA 0.987 V 6.28 k 200 mH
0.537 mA 0.843 V 1.57 k 50 mH
For Further Investigation: A small power transformer was tested using a 330 series resistor at a frequency of 20 kHz to increase the voltage across the windings. Results indicated an inductive reactance from center-tap to either side of 220 but an inductive reactance of both sides taken together of 1000 . The computed inductance of each side alone was 1.7 mH but together was 8 mH. Student results will vary from this depending on the transformer tested. Application Problem: The errors for the Maxwell bridge include the accuracy of all components in the bridge, the frequency of the generator, and the ability to detect a null. Evaluation and Review Questions: 1. (a) The computed sum is 6.28 k. (b) The computed product-over-sum is 1.57 k. (c) The inductive reactance of series inductors is additive. The inductive reactance of parallel inductors uses the product-over-sum rule. 2. The inductance of series inductors is additive. The inductance of parallel inductors uses the product-over-sum rule.
501
3. Inductive reactance is proportional to frequency. If the actual frequency is higher than the measured frequency, the inductive reactance will appear higher than its actual value and hence cause the measured inductance to appear high and vice versa. 4. Measure the reactance at a known frequency and compute the inductance. 5. 157 . 6. As long as consistent measurements were made, there will be no difference between rms and peak-to-peak values.
502
Experiment 24: Transformers Data: Measured results for a small transformer (120 V primary to 12.6 V center-tapped secondary) are shown on the tables: Table 24-1 Primary winding resistance, RP Secondary winding resistance, RS Turns ratio, n, (computed) Turns ratio, n, (measured) %difference
18 1 0.11 0.126 13%
Observations from step 5: Primary and secondary voltages are in phase. Vp = 14 Vpp; Vs = 1.75 Vpp Observations from step 6: Secondary voltages are 180° out of phase with each other. Vp = 14 Vpp; Vs (measured from each side of the line to the center-tap) = 0.83 Vpp Step 7 8
Table 24-2 Condition Output directly from generator With impedance-matching transformer
Vspeaker 100 mV 355 mV*
*Triad TY30X impedance matching transformer
Further Investigation: The magnetizing current will decrease as the frequency is raised. Data for a Triad TY30X impedance-matching transformer indicate a current of 1.3 mA at 1 kHz and 0.53 mA at 10 kHz. Application Problem: Answers depend on the particular IF transformer. The J.W. Miller 8812 transformer is designed for 455 kHz and has a primary reactance of 20 k and a secondary reactance of 500 at the design frequency. Based on these figures, the turns ratio is 40:1. Evaluation and Review Questions: 1. (a) Answers will depend on transformer. For the transformer shown in the experiment, the ratio is 1:18. (b) The wire size could be different and measurement error is present, especially for the very small secondary winding resistance. 2. The computed turns ratio assumes an ideal transformer which implies that all of the flux generated in the primary winding passes through the secondary winding. The Summary of Theory cites several factors which contribute to the difference between the ideal and the actual transformer.
503
3. The voltage across the speaker increased because the impedance-matching transformer causes the load to appear larger to the source. This means that more power can be delivered to the load. 4. The power delivered to an ideal transformer is equal to the power delivered to the load. 5. (a) The power delivered to the primary is 24 watts. This is the power delivered by the ideal transformer to the load. (b) The secondary current is 1.0 A. (c) The turns ratio is 0.20. 6. For the same power, voltage and current are inversely proportional. Higher voltages imply lower current for the same power and hence lower I2R loss in the transmission line.
504
Experiment 25: Series RC Circuits Data:
Computed values are shown. Measured values were within 7% of these values.
Frequency 500 Hz 1000 Hz 1500 Hz 2000 Hz 4000 Hz 8000 Hz
Table 25-2 I* VC
VR 0.63V 1.18 V 1.62 V 1.95 V 2.59 V 2.88 V
2.93 V 2.75 V 2.52 V 2.28 V 1.51 V 0.84 V
0.092 0.173 0.238 0.287 0.381 0.423
XC*
Z*
31.8 15.9 10.6 7.79 3.98 1.99
32.5 17.3 12.6 10.5 7.88 7.02
* I in mA, XC and Z in k
Step 8 (Graph of Voltage versus Frequency):
Voltage (V)
3.0
VR
2.0 1.0 0
VC
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
Frequency (kHz)
Plot 25-1 Step 9 (Impedance and Voltage Phasors for 2000 Hz: 2
R(k) 4 6 8 R = 6.8 k
2 4
Z = 17.3 k
00 VC(V)
XC(k)
0 0
2 3
6 8
1
4
XC = 7.96 k
VR(V) 1 2 3 4 VR = 1.95 V VS = 3.0 V
VC = 2.28 V (Voltages are peak-to peak)
Plot 25-2 Multisim Application: File EXP25-2f1: fault is: C1 is open. File EXP25-2f2: fault is: VS is open. File EXP25-2f3: fault is: R1 is 3.9 k (by adding “leakage”).
505
Further Investigation:
Note: Voltages and currents shown in Table 25-2 have been converted to rms values to compute the power phasors shown below.
Ptrue(W) 20 40 60 80 Ptrue=25.6 W
0 Preactive(VAR)
Preactive(VAR)
0
0
20 40
Papp = 64.9 W
60 80
Preactive = 59.6 W
0
Ptrue(W) 50 150 200 Ptrue=123 W
50 Papp = 143 W 150 Preactive = 71 W 200
f = 1000 Hz
f = 4000 Hz
Plot 25-3 Calculation for 1000 Hz:: VR = 0.417 Vrms and VC = 0.972 Vrms 2 (0.417 V) V2 = 25.6 W Ptrue = R = 6.8 k R (0.972 V)2 V2 = 59.4 W Preactive = C = 15.9 k XC (1.06 V)2 VS2 = 64.9 W = Papp = 17.3 k Z
Calculation for 4000 Hz:: VR = 0.915 Vrms and VC = 0.533 Vrms 2 (0.915 V) V2 = 123 W Ptrue = R = 6.8 k R (0.533 V)2 V2 = 71.3 W Preactive = C = 3.98 k XC (1.06 V)2 VS2 = 143 W = Papp = 7.88 k Z
Application Problem: The computed capacitance is 0.94 µF. A 1 µF capacitor should be selected. Evaluation and Review Questions: 2 2 1. Z = 6.8 k + 7.96 k = 10.5 k
VS =
2
2
1.95 V + 2.28 V = 3.0 V
2. Connect the output across the resistor because the maximum voltage at high frequencies is developed across the resistor. 3. (a) The total impedance decreases as the frequency increases. (b) The capacitive reactance decreases but the resistance remains constant; therefore the total impedance becomes more resistive. 4. VC will be smaller at any given frequency and VR will be larger. 5. The open component will indicate VS across it; the other will show no voltage drop. 6. The frequency at which the phase angle is 45° is approximately 2.3 kHz (cutoff).
506
Experiment 26: Parallel RC Circuits Data
(computed values shown)
R1 RS1 RS2 C1
100 k 1.0 k 1.0 k 1000 pF
100 k 1.0 k 1.0 k 1000 pF
2.96 V 35.2 mV 18.7 mV
29.6 A 35.2 A 18.7 A
40 IC(A)
Listed Value
Table 26-1 (f = 1.0 kHz) Measured Voltage Computed Value Drop Current
30 20
IC = 18.7 A
10 00
IT = 35.2 A
IR = 29.6 A 10 20 30 40 IR(A)
Plot 26-1
XC1 = 159 k ZT = 84.6 k IT = 35.2 A (same as the current in RS1)
R1 RS1 RS2 C1
100 k 1.0 k 1.0 k 1000 pF
100 k 1.0 k 1.0 k 1000 pF
2.97 V 47.7mV 37.7 mV
29.7 A 47.7 A 37.7 A
40 IC(A)
Listed Value
Table 26-2 (f = 2.0 kHz) Measured Voltage Computed Value Drop Current
IC = 37.7 A
30
IT = 47.7 A
20 10 00
IR = 29.7 A 10 20 30 40 IR(A)
Plot 26-2 XC1 = 79.6 k ZT = 62.3 k IT = 47.7 A (same as the current in RS1) Multisim Application: File EXP26-2f1.: fault is: RS1 is open. File EXP26-2f2.: fault is: C1 is open. File EXP26-2f3: fault is: R1 is shorted. Further Investigation:
BC(S)
12 9 6
BC = 6.3 A
3 00
Y = 11.7 A G = 9.9 S 3 6 9 12 G (S)
Plot 26-3
507
Application Problem: RA and CA, connected as a low pass filter, control the high frequency response. The parallel combination of RB and CB control the low frequency response because the reactance of CB is large for low frequencies. Evaluation and Review Questions: 1. (a) The total impedance is lowered. (b) The phase angle between the generator voltage and current increases. 2. (a) Current in R1 is still approximately 30 µA. (b) The capacitor current is approximately 5 times larger than in step 4 or 93.5 µA. (c) The total current is approximately 98.2 µA. 3. A smaller capacitor has greater reactance. The capacitive current would be smaller which in turn causes the total current to be less. The current in R1 is unaffected. 4. (a) The cutoff frequency is 5.9 Hz. (b) The branch currents are equal. (c) Above this frequency, the capacitor has more current than the resistor. 5. If the capacitance is larger the cutoff frequency will be lower. 6.
45°
508
Experiment 27: Series RL Circuits Data: VR 1.61 V
Table 27-2 (f = 25 kHz) I VL XL 2.53 V 0.16 mA 15.7 k
ZT 18.6 k
20
20
16 XL = 15.7 k
16 VL = 2.53 V
Z = 18.6 k VL(V)
XL(k)
12 8 4 0
VS = 3.0 V
12 8 4
0
R = 10 k 4 8 12 16 20 R(k)
0
0
4
(a)
VR = 1.61 V 8 12 16 20 VR (V)
(b)
Plot 27-1 Computed Phase Angle 57.5 o
Table 27-3 Measured Time Period Difference T t 40.0 s 6.2 s
Phase Angle Method 1 Method 2 57 o 56 o
Application Problem: The current when the phase angle is 60° is 150 µA (at a frequency of 27.6 kHz). Evaluation and Review Questions: 1. (a) The impedance increases. (b) The impedance increases. 2. (a) The phase angle (seen by the generator) will increase. (b) The phase angle (seen by the generator) will increase. 3. Answers vary. Error for method-1 was 2.6% from the computed value.
4. In rectangular form, the impedance is: Z T = 10 k +j15.7 k. In polar form, the impedance is: Z T = 18.6 57 5. The critical frequency is 15.9 kHz. At this frequency, the phase angle is 45°. 6. (a) The current in the inductor is the same as in the resistor or 30 mA. (b) The inductive reactance is 377 . (c) The voltage across the inductor is 11.3 V. (d) The source voltage is 11.7 V. (e) The phase angle is 75°.
509
Experiment 28: Parallel RL Circuits Data: Listed Value 3.3 k R1 47 RS1 47 RS2 L1 100 mH L1 resistance
0
Table 28-1 Measured Voltage Value Drop 3.3 k 47 47 100 mH
5.88 Vpp 123 mVpp 87 mVpp
Computed Current 1.78 mApp 2.62 mApp 1.85 mApp 1.85 mApp
155
IR(mApp) 0 0.4 0.8 1.2 1.6 2.0
IL(mApp)
0.4 0.8 1.2 1.6 2.0
Plot 28-1 Further Investigation: The loading effects depend on the Thevenin impedance of the generator. The total impedance of the circuit is 4.6 k; a 600 generator represents about 13% of this impedance. Application Problem: The circuit tends to load a 600 generator. The basic idea of charging current can be observed with a 600 ohm generator, however a 50 generator is better. The generator changes the voltage across the inductor by approximately +0.5 V. In accordance with Lenz's law, this causes the current in the 100 mH coil to change linearly by about 0.5 mA during the period of 100 µs. The waveforms are sketched below. Notice the overshoot on the generator (VS) due to the loading. VS 0V
VR 0V
50 s/div
510
Evaluation and Review Questions: 1. (a) IT 1.85 mA 2 1.78 mA 2 2.47 mA (b) The inductor resistance and measurement error contribute to the differences read. 2. Coil resistance reduces the phase angle. 3. Sense resistors should be very small compared to the impedance of the branch they are used in. 4. (a) The total current would be the same as the resistor current. (b) The phase angle would be 0°. (c) The generator voltage would rise due to the reduced drop across its Thevenin resistance. 5. (a) The total current would decrease. (b) The phase angle would decrease. (c) The generator voltage would rise due to the reduced drop across its Thevenin resistance. 6. (a) The phase angle would increase. (b) The phase angle would decrease.
511
Experiment 29: Series Resonance Data: Table 29-1
Table 29-2
L1
Listed Value 100 mH
Measured Resistance 101 mH
C1
0.01 F
0.01 F
R1
100
101
RS1
47 L1 resistance
Computed
47 292
RT
425 W
fr Q
5003 Hz
Measured 5072 Hz
7.48
VRS1
124 mV
f2
5409 Hz
f1 BW
4645 Hz 669 Hz
764 Hz
For Further Investigation: Data: I(mA) 0.74 1.00 1.34 2.04 2.46 2.06 1.49 1.15 0.91
3.0
Current (mA)
f (Hz) 4000 4250 4500 4750 5000 5250 5500 5750 6000
2.0
1.0
0 4.0
4.5
fr Frequency (kHz)
5.5
6.0
Plot 29-1 Application Problem: At 50 kHz, a 100 pF capacitor is required. The load resistor is the Thevenin resistance plus the inductor’s internal resistance. A sharp peak occurs at resonance across the load. (Test circuit peak voltage was 23 Vpp). Power in the load is at its maximum at resonance. Evaluation and Review Questions: 1. (a) Answers vary. (b) Measurement of components, voltage, frequency, and non-ideal components. 2. (a) The impedance of the LC branch at resonance is the resistance of that branch. (b) At resonance, the circuit appears resistive, and the phase shift is zero. 3. (a) The voltages are out of phase and their sum is not greater than the source voltage; Kirchhoff's voltage law is still satisfied at each instant in time. (b) Yes 4. (a) Resonant frequency is the same.
(b) The Q is doubled so bandwidth is halved.
5. (a) 712 kHz.
(b) Q = 22.4 and BW = 31.8 kHz.
6. Below resonance, capacitive reactance dominates the impedance equation. The source current leads the source voltage; the circuit "looks" capacitive.
512
Experiment 30: Parallel Resonance Data: Table 30-1
Table 30-2
L1
Listed Value 100 mH
Measured Resistance 101 mH
C1
0.047 F
0.0505 F
RS1
1.0 k L1 resistance
1.03 k
Computed
Measured
fr Q
2228 Hz
2170 Hz
BW
460 Hz
fi
292
BW 4
4.84 506 Hz
115 Hz
Table 30-3 Computed Frequency fr - 5 fi = 1652 Hz fr - 4 fi = 1768 Hz fr - 3 fi = 1883 Hz fr - 2 fi = 1998 Hz fr - 1 fi = 2113 Hz fr = 2228 Hz fr + 1 fi = 2343 Hz fr + 2 fi = 2458 Hz fr + 3 fi = 2573 Hz fr + 4 fi = 2688 Hz fr + 5 fi = 2803 Hz
VRS1 350 mV 285 mV 230 mV 175 mV 148 mV 134 mV 144 mV 185 mV 235 mV 275 mV 325 mV
I 340 A 275 A 222 A 169 A 143 A 130 A 139 A 179 A 227 A 269 A 314 A
Z 2.94 k 3.64 k 4.50 k 5.92 k 6.99 k 7.69 k 7.19 k 5.59 k 4.40 k 3.72 k 3.18 k
8
Impedance (k)
7 6 5
BW = 506 Hz
4 3 2 fr - 5fi
fr
fr + 5fi
Plot 30-1 Multisim Application: File EXP30-3f1: fault is: RS is open. File EXP30-3f2: fault is: VS is open File EXP30-3f3: fault is: C1 has 5k of leakage resistance (dropping the Q significantly.)
513
Further Investigation: Signal across L1||C1
Generator signal
CA = 0.01 F. Try lowering f while observing signal.
Signal across L1||C1
Generator signal
CA = 1000 pF. Amplitude is reduced but attenuation rate is less.
Application Problem: One way to calibrate the frequency is to note the amplitude of the frequency response at two points, then take the scope out of x-y mode and change the sweep generator to a constant output frequency. Tune the generator until the same amplitude is found and the frequency is measured. Evaluation and Review Questions: 1. (a) In a series resonant circuit, the impedance is minimum at the resonant frequency; in a parallel resonant circuit, the impedance is maximum at the resonant frequency.
(b) In a series resonant circuit, the current is maximum at the resonant frequency; in a parallel resonant circuit, the current is minimum at the resonant frequency. 2. At resonance, current and voltage are in phase. 3. Ignoring the inductor's resistance, the current in the inductor is the source voltage divided by the inductive reactance; current in the capacitor is the source voltage divided by the capacitive reactance. 4. The Q depends on XL and RW for the coil. The circuit Q will be different than that of the inductor if there is additional resistance in the circuit. 5. (a) fr = 5033 Hz. (b) XL = 3.16 k (c) Q = 26.4 (d) BW = 191 Hz. 6. Below resonance, inductive reactance is lower and the source current lags the source voltage; therefore, the circuit "looks" inductive.
514
Experiment 31: Passive Filters Data: Table 31-1 Measured Resistance 101 mH
C1
0.1 F
0.100 F
C2
0.1 F
0.099 F
C3
0.033 F
0.033 F
RL1
680
681
RL2
1.6 k
1.61 k
Table 31-2 Frequency VRL1 4.12 V 500 Hz 3.88 V 1000 Hz 3.30 V 1500 Hz 2.39 V 2000 Hz 1.49 V 3000 Hz 0.88 V 4000 Hz 0.24 V 8000 Hz
6.0 Voltage (VPP)
L1
Listed Value 100 mH
4.0 2.0 0
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
5.0
6.0
7.0
8.0
5.0
6.0
7.0
8.0
Frequency (kHz)
Plot 31-1 Voltage (VPP)
Table 31-3 Frequency VRL1 0.14 V 500 Hz 0.79 V 1000 Hz 2.02 V 1500 Hz 3.40 V 2000 Hz 5.01 V 3000 Hz 5.54 V 4000 Hz 5.85 V 8000 Hz
6.0 4.0 2.0 0
0
1.0
2.0
3.0
4.0
Frequency (kHz)
Plot 31-2 Voltage (VPP)
Table 31-4 Frequency VRL1 1.38 V 500 Hz 3.00 V 1000 Hz 4.12 V 1500 Hz 3.61 V 2000 Hz 2.36 V 3000 Hz 1.70 V 4000 Hz 0.81 V 8000 Hz
6.0 4.0 2.0 0
0
1.0
2.0
3.0
4.0
Frequency (kHz)
Plot 31-3
515
Multisim Application: File EXP31-3f1: fault is: RL is open. File EXP31-3f2: fault is: L1 is shorted. File EXP31-3f3: fault is: VS is open. Further Investigation: A 100 mH inductor and a 0.1 µF capacitor were wired in parallel with a series 680 resistor. The signal generator was kept at a constant 3.0 V. Data across the resistor is tabulated below: Frequency 500 Hz 1000 Hz 1500 Hz 1640 (fr) 2000 Hz
Frequency 3000 Hz 4000 Hz 6000 Hz 8000 Hz 10 kHz
Vresistor 2.68 V 2.04 V 0.51 V 0.40 V 0.44 V
Vresistor 0.64 V 1.34 V 1.86 V 2.14 V 2.31 V
Application Problem: The computed notch frequency is 15.9 kHz based on the listed components. The measured notch frequency of the test circuit was 16.1 kHz with the peak-to-peak voltage at the notch of 40 mVpp. Sample data is listed below (voltages listed are peak-to-peak): Frequency 5.0 kHz 10.0 kHz 13.0 kHz 16.1 kHz 19.0 kHz
Frequency 22.0 kHz 30.0 kHz 40.0 kHz 80.0 kHz
Vresistor 1.75 V 0.70 V 0.34 V 0.04 V 0.20 V
Vresistor 0.40 V 0.80 V 1.20 V 1.85 V
Evaluation and Review Questions: 1. Pi filter cutoff frequency = 1800 Hz. (approximate) T filter cutoff frequency = 2400 Hz. (approximate) 2. The frequency response of both the pi and T filters is steeper than a simple RC filter, indicating better filtering. See the response in Experiment 25 for comparison. 3. (a) Plot 1 (pi filter):____low pass___ (b) Plot 2 (T filter):_____high pass__ (c) Plot 3 (resonant filter):___band pass___ 4. The voltage across the two components (inductor and capacitor) is the difference between the source voltage and the voltage measured across the load resistor. Thus the response curve is that of a notch filter instead of a band-pass filter. 5. The circuit would become a high-pass filter. 6. See text Figure 18-24 for a parallel resonant band-pass filter and Figure 18-32 for a resonant notch (band-stop) filter.
516
Experiment 32: Circuit Theorems for AC Circuits Data: In step 4, the Thevenin impedance, at 17 kHz is 4.3628° (using listed values).
Table 32-1
Listed Value
Measured Resistance 0.105 F 1040 pF
R1
9.1 k
9.09 k
R2
10 k
9.98 k
C1
In step 5, the Thevenin voltage is 1.35 28° V.
Load
C2
Listed Value 0.1 F 1000 pF
Table 32-2 Original Circuit Thevenin Circuit Measured Measured Measured Measured* Voltage Phase Angle Voltage Phase Angle
0.27 V +79 o 0.265 V +47 o 1.46 V 1.46 V 0o 24 o o 0.69 V +14 0.69 V 4.7 k 14 o *The measured phase angle in the Thevenin circuit is referenced to the "Thevenin generator". Because the Thevenin source has a phase shift of 28° with respect to the original source, the measured phase angles for the Thevenin circuit include this difference.
CL LL RL
0.01 F 100 mH
Further Investigation: At 17 kHz, the impedance of C2 in parallel with R2 is 6.83 47° k (using listed values). Writing Kirchhoff's voltage law and ignoring C1 (at 17 kHz) in Figure 32-2 gives: VS
+
VR1
+
VR 2,C 2
0
3.00 V + I 9.10 k + I 6.83 47 k 0 I 3.0 V
14.7 20 k 0.2020 mA
VR1 IR1 1.8 20 Vpp
Application Problem:
Writing the IB loop through
R and C (upper path): 2
R IB XC 2 0.025 j 0.025 0.025 j 0.075 5 k 0.025 j 0.025 j10 k
Vout I B I C
0.025 j 0.025 j 0.025 0.025 0
517
Evaluation and Review Questions: 1. (a) Assuming a 600 generator, VTH = 1.19 38.9° V. C2 can be ignored at 100 Hz. (b) Assuming a 600 generator, VTH = 0.47 72.1° V. C1 can be ignored at 100 kHz. 2. At 17 kHz, the Thevenin impedance consists of 3.86 k of resistance and 2.03 k of capacitive reactance. (a) The resistive part of the load must be equal to 3.86 k. (b) The reactive part of the load must be equal to 2.03 k, which is a 19 mH inductor. 3. Capacitive reactance is a function of frequency; therefore the Thevenin circuit changes. Note - It is informative to plot the effect of the Thevenin voltage and impedance as a function of frequency. As frequency increases, the Thevenin voltage increases, then decreases; the Thevenin impedance starts at 10 k (dc) and continues to decrease. This result gives some insight to why the circuit is a band-pass filter:
f 10 Hz
VTH
ZTH
0.1983° V 9.953° 100 Hz 0.1738° V 7.3419° 1.0 kHz 1.491° V 5.14.5° 10 kHz 1.4317° V 4.817° 100 kHz 0.4672°V 1.572° 1.0 MHz 0.0688° V 0.1688° These values were calculated using 10 k for (R1 + RTH(gen)) and listed values for components.
4. At 17 kHz, the Norton circuit is a 0.33 0° mA current source in parallel with a Z of 4.36 28° k (equal to a 3.86 k resistor 2.03 k of capacitive reactance). 5. Thevenin's theorem can be applied to ac circuits at a specific frequency if the circuit is analyzed using complex numbers. Instead of a Thevenin resistance, a Thevenin impedance is used in the ac circuit and Thevenin voltage source is written as a phasor voltage. 6. It must be the conjugate or 600 + j100 .
518
Experiment 33: Integrating and Differentiating Circuits Data: Table 33-1 Listed Value 100 mH
Measured Resistance 101 mH
C2
0.01 F 1000 pF
0.010 F 1040 pF
R1
10 k
9.98 k
L1 C1
Table 33-2 RC time constant
0.10 ms
0V -1 V +1 V
VR
0V -1 V +1 V
VC
0.106 ms*
+1 V
VS
0V -1 V +1 V
VR
Measured
*includes RTH of 600 for generator.
+1 V
VS
Computed
+0.68 V +0.38 V
0V -1 V +1 V
VC
0V -1 V
0V -1 V
0
1 2 time (ms)
3
0
Plot 33-1
1 2 time (ms)
3
Plot 33-2
Step 6 – (Triangle Waveform Results): Capacitor waveform looks sinusoidal, centered about 0.5 V (170 mVpp).
Step 7 – (Square Wave with 1000 pF Capacitor): Resistor waveform is similar to VR in Plot 33-1 except frequency is 10X higher.
Step 8 – (Square Wave with 100 mH inductor): +1 V
VS
0V -1 V +1 V
VR
Amplitude = 0.6 V
0V -1 V +1 V
VL
0V -1 V
0
0.1 0.2 time (ms)
Some ringing observed on signal. If scope sweep speed is increased, lower rise time 0.3 is observed (approx. 0.4 s)
Plot 33-3 Further Investigation: R TH = 5.0 k. VTH = 1.0 Vpp. VC charges toward 1.0 V with a time constant of 50 µs.
519
Application Problem: The output is a triangular waveform with slight bending (due to the charging curve) that goes from 0.8 V to 1.6 V. When the resistor is set for maximum, the measured rise time is 0.5 µs and the fall time is 7.9 µs. When the resistor is set for minimum, the waveform has a rise time of about 0.5 µs and a fall time of 0.5 µs. Evaluation and Review Questions: 1. (a) The total resistance in the charging circuit includes the Thevenin resistance. By setting the output square wave with the generator disconnected, the Thevenin resistance affects only the RC time constant, not the final charging voltage. (b) Measure the resistance and the time constant for the circuit; solve for C. 2. (a) Answers will vary. (b) Component measurements, oscilloscope calibration, instrument reading error. 3. The capacitor does not have time to fully charge and discharge. 4. (a) An RC integrating circuit:
An RC differentiating circuit:
R
VS
C
C
output
(b) An RL integrating circuit:
VS
output
An RL differentiating circuit:
L
VS
R
R
R
output
VS
L
output
5. The signal path could be open, causing a capacitively coupled path to the scope. At low frequencies, the input signal is differentiated. Note: This can be demonstrated by setting up a signal generator for a large amplitude 100 Hz square wave. Attach a test lead from the generator and use an alligator clip to connect the generator to the insulated tip of an oscilloscope probe. Alternatively, connect the generator to a protoboard and probe on an adjacent row.
6. An integrator.
520
Appendix A Project Constructing a Reed-Switch Motor If students have difficulty with the motor, make sure the rotor spins freely and is in balance. A modified motor that worked very well is shown here. The modification (shown in the power point slides that go with this series) is to use 400 turns on the coil and put two series green LED across the coil. (The LEDs blink as the motor turns, producing an interesting effect). By collapsing the magnetic field faster with the LEDs, the motor spun even faster. The schematic for the modified motor is given here.
Magnetic reed switch
Electromagnet
S2
on/off
S1 LEDs (green)
VS + 3.0 V
Questions 1. The motor works in repulsion. When a magnet passes by the reed switch, it closes causing the coil to produce a like coil near the rotor’s magnet, pushing it away. 2. The repulsion force needs to occur only when the magnet is next to the reed switch. This will only occur if the motor turns counterclockwise (of course it can be changed by reversing the position of the coil and reed switch). 3. Answers vary. One solution is given above.
521
PART 6 Laboratory Solutions for Experiments in Electric Circuits By Brian Stanley
Instructors please note the following The answers given in this manual to questions 5 and 6 in each experiment are generalized responses that one might use as a guide for grading. Similar responses making the same general point or some other pertinent point could be accepted at the instructor's discretion.
523
Experiment 1: 1. (d) Question 5:
2. (b)
3. (d)
4. (c)
Body resistance changed as a result of reduced contact resistance through both increased grip and added moisture.
Question 6:
The average of 7.1 kΩ and 7.9 kΩ equals 7.5 kΩ. Referring to standard value chart in Appendix B, a resistor of value 7.5 kΩ ± 5 % would do the job (Rmax. would be 7.88 kΩ and Rmin. about 7.13 kΩ). ______________________________________ Experiment 2: 1. (d) Question 5:
2. (a)
3. (d)
4. (a)
As shown in the sketch, the wiper needs to be connected to one end of the potentiometer to form a single variable resistance.
Question 6:
A themistor having a negative temperature coefficient is one whose resistance decreases with an increase in temperature whereas a positive temperature coefficient thermistor is one whose resistance increases with an increase in temperature. A resistor with a zero temperature coefficient is one whose resistance does not change with temperature. ______________________________________
524
Experiment 3: 1. (a) Question 5:
2. (b)
3. (b)
4. (b)
A single pole single throw switch would be the one to use for this application. A sketch is shown below.
Question 6:
The switch used in conjunction with an electric door bell would be a good example of an appropriate use of a momentary push button switch. ______________________________________ Experiment 4: 1. (a)
2. (b)
3. (a)
4. (a)
Question 5:
The resolution of a DMM is defined to be the smallest change in voltage that the meter can detect and display. It is typically the last, (or least significant) digit that the meter displays.
Question 6:
As shown in the sketch below, the ammeter might be inserted at the top end of the resistor.
______________________________________
525
Experiment 5: 1. (c) Question 5:
2. (a)
3. (b)
4. (b)
A vertical line drawn on the I versus V graph represents a line of constant voltage, in this case 5 Volts. This line intersects each of the resistor's characteristics at only one point. The intersection that the line makes with the 1 kΩ line, represents the current that flows through that resistor when the voltage across it is 5 Volts. Likewise, the intersection that the line makes with the 2 kΩ line represents the current that flows through that resistor when the voltage across it is 5 Volts.
Question 6:
A horizontal line drawn on the I versus V graph represents a line of constant current, in this case 1 mA. This line intersects each of the resistor's characteristics at one point. The intersection that the line makes with the 1 kΩ line, represents the voltage across that resistor when the current is 1 mA. Likewise, the intersection that the line makes with the 2 kΩ line represents the voltage across that resistor when the current through it 1 mA. ______________________________________ Experiment 6: 1. (b)
2. (b)
3. (b)
4. (c)
Question 5:
For a given current, the 2 kΩ resistor dissipates the most power. This can be verified by drawing a vertical line at some current on the P versus I graph, and noting that it crosses the 2 kΩ characteristic at a greater power level than the 1 kΩ characteristic.
Question 6:
For a given voltage, the 1 kΩ resistor dissipates the most power. This can be verified by drawing a vertical line at some voltage on the P versus V graph, and noting
526
that it crosses the 1 kΩ graph at a greater power level than the 2 kΩ graph. ______________________________________ Experiment 7: 1. (c) Question 5:
2. (b)
3. (d)
4. (b)
The 1/4 W resistor is smaller in size than the 1/2 W resistor. It therefore has a smaller surface area, cannot get rid of its energy as fast, and therefore for a given power dissipation will have a greater surface temperature.
Question 6:
The power rating of a resistor assumes some standard ambient temperature (normally (20o C). If the temperature is actually lower than this, the resistor can actually dissipate a greater amount of power without its own surface temperature rising above a given level. ______________________________________ Experiment 8: 1. (c)
2. (a)
3. (a)
4. (c)
Question 5:
Reducing the value of any resistor in a series circuit will reduce the total resistance, (Total Resistance Law) and therefore increase the total current (by Ohm's Law)
Question 6:
If a resistor's value increases by 10 %, then its value will be 10% higher than nominal. When a resistor's value decreases by 10 % its value will be 10% lower than nominal. Unless the actual number of Ohms that each resistor changes by is the same, there will be a net change in the total resistance of the sum of the two, and so the current will change. If the resistors were originally the same nominal value, then the +10% and the -10% changes will exactly cancel, leaving no change
527
in the total resistance and therefore no change in the current. ______________________________________ Experiment 9: 1. (e)
2. (b)
3. (a)
4. (c)
Question 5:
Shown in the sketch is a series connection of eight cells that make up a total battery voltage of 12 V.
Question 6:
Shown in the sketch are the modifications necessary to achieve a total voltage of 0 V in Figure 9-1f.
______________________________________
528
Experiment 10: 1. (a) Question 5:
2. (b)
3. (a)
4. (d)
If a single resistor is smaller than its nominal value, then the total resistance will be smaller than its nominal value (Total Resistance Law). The result will be an increase in the current flowing through the series connection (Ohm's Law). The remaining resistors will have greater than nominal voltages across them (Ohm's Law) while the resistor that is below nominal in value will have a less
than nominal voltage across it (Kirchhoff's Voltage Law) Question 6:
When a resistor is not exactly equal to its nominal (color-coded) value, this will affect the total resistance, total current, and therefore the voltages across every other resistor in the circuit. A resistor that has a smaller than nominal value will have a smaller than nominal voltage across it, and the rest will have larger than nominal voltages across them. On the other hand if the resistor has a larger than nominal value, then its voltage will be larger than nominal while the remainder will be smaller than nominal. ______________________________________ Experiment 11: 1. (b)
2. (b)
3. (d)
4. (b)
Question 5:
The potentiometer method provides a continuous variation of voltage as the position of the wiper is moved, whereas the fixed resistor voltage divider only gives certain discrete values of voltage.
Question 6:
The voltage VEF would have been about 8.4 Volts, since by Kirchhoff's Voltage Law, these voltages must sum to the source voltage which was 10 V.
______________________________________
529
Experiment 12: 1. (d) Question 5:
2. (d)
3. (d)
4. (a)
If VB is 12 V, then we can assume that voltage is getting to the upper end of R1. Since the potential at B is exactly half of that at C, we must assume that there is one half of the total circuit resistance above and below this point. This could mean that either R2 is short circuit or R3 is short circuit.
Question 6:
When a component in a series circuit becomes a short circuit, the total circuit resistance will decrease, causing an increase in circuit's total current (Ohm's Law) . The voltages across the remaining resistors will be greater than nominal (Ohm's Law), and the power dissipated by each of these resistors will also be larger than it is supposed to be. If the resistors are not of a sufficiently large power rating that they can safely dissipate this additional power, then they might overheat and ignite causing a severe risk of fire. ______________________________________ Experiment 13: 1. (b) Question 5:
2. (a)
3. (c)
4. (b)
The current from the power supply is the largest current in this circuit because it is the sum of all of the parallel branch currents. (Kirchhoff's Current Law)
Question 6:
Since the voltage across each branch of a parallel circuit is equal to that of the power supply, and Ohm's Law says that current is inversely proportional to resistance, then the branch with the smallest resistance will have the largest current while that with the largest resistance will have the smallest current. ______________________________________
530
Experiment 14: 1. (b) Question 5:
2. (b)
3. (a)
4. (a)
The measured resistance of a resistor will be affected if there is another path in which instrument current can flow besides in the resistor itself. If the fingers are placed across the ends of the resistor, then they are effectively in parallel, and therefore provide an additional path for current to flow. Since the resistance of two resistors in parallel is always smaller than either one, then the measured resistance value will be smaller than the true value.
Using RX = R1. RT/(R-RT) with RT the value from Table 14-5 labeled "with body" and R the value from the table labeled "without body" the effective body resistance can be calculated. ______________________________________ Question 6:
Experiment 15: 1. (b) Question 5:
2. (d)
3. (a)
4. (b)
Since by Ohm's Law, I = V/R, and V is the same for each resistor in a parallel connection, the smallest resistance will take the largest current, and the largest resistance the smallest current.
Question 6:
For a constant source current (which would have to be maintained by adjusting the source voltage), if any one of the resistor values was changed, this would change the total resistance, (resistors in parallel law) and therefore every single branch current in the circuit (Current Divider Theorem) ______________________________________
531
Experiment 16: 1. (c) Question 5:
2. (d)
3. (a)
4. (a)
When a resistor in a parallel circuit goes short circuit, the current through it can increase to a very large value limited only by the capacity of the power supply. In practice, the resistance through which this current is flowing, is not zero, but a very small value. By P = I2R, and since I is large, the power could be quite large, and the heat generated could present a fire risk. There is not normally any danger to the remaining parallel resistors since they conduct little or no current, though they would be in danger if they caught fire from the resistor that is shorted.
Question 6:
The 120 Ω resistor must be of sufficient power rating that it is able to dissipate the power in it when any of the resistors are shorted. Since the current is about 100 mA, and the voltage about 12 V, then the resistor will have to dissipate approximately 1.2 W of power. ______________________________________ Experiment 17: 1. (b) Question 5:
2. (a)
3. (d)
4. (a)
If any resistor is reduced in value in a series-parallel circuit, since it will be in either a parallel or series subsection of the circuit, the total resistance of that part of the circuit will decrease, and therefore the total circuit resistance will also decrease.
Question 6:
There are three different combinations of total resistance obtainable with the three resistors given in this experiment and using the circuit given in the figure. These are: i) 10 kΩ, (2 kΩ + 12kΩ in parallel with 24 kΩ) ii) 25.71 kΩ (24 kΩ + 12kΩ in parallel with 2 kΩ) and iii) 13.85 kΩ (12 kΩ + 24 kΩ in parallel with 2 kΩ) ______________________________________
532
Experiment 18: 1. (a) Question 5:
2. (b)
3. (d)
4. (d)
When a load is connected to the output of a voltage divider, such as that in the figure, from the point of view of the source, the load appears in parallel with one of the divider resistors, and therefore reduces the total circuit resistance. As a result, the total circuit current increases. This is inevitable since the load demands current that was previously not required, and all current must ultimately come from the source.
Question 6:
A resistor placed in parallel with another resistor has a greater effect on the equivalent resistance if the resistances are closer in value than if they are far apart in value. So when the resistors in the voltage divider are relatively small, the load resistors (which are much larger) do not affect the total resistance as seen by the source, very much. The current supplied by the source does not change very much, and neither do the voltages across the divider resistors. On the other hand, when the divider resistors are large to begin with, a large-valued load resistor that is comparable in size will affect the overall resistance seen by the source, to a much greater degree, and thus the current will change considerably. The voltages across the divider resistors will now be heavily dependent upon the load values. ______________________________________ Experiment 19: 1. (a) Question 5:
2. (d)
3. (a)
4. (c)
When a DMM/VOM is positioned across either resistor, the resistor and meter are in parallel, and the source the resistor sees this as a decrease in the total resistance. The current drawn by the circuit now increases (by Ohm's Law) and therefore the voltage across the resistor
533
that does not have the meter connected increases (by Ohm's Law). By Kirchhoff's Voltage Law, the voltage across the resistor across which the meter is connected decreases. Question 6:
Since the resistance of the VOM is 20 kΩ/VOLT, in order for it to have a resistance of 10 MΩ, the range would have to be 10 MΩ/(20 kΩ/Volt) = 500 V range ______________________________________ Experiment 20: 1. (d)
2. (c)
3. (b)
4. (a)
Question 5
To the left of point C the circuits are identical. In the first circuit the resistance from point C to ground is 10 k Ω and this is also the same for the second circuit. The resistance from point C to ground is given by the series combination of the two 10 k Ω resistors together with the 20 k Ω in parallel giving 10 k Ω. Therefore the total resistance seen by the source is not changed.
Question 6:
A circuit is shown.
______________________________________
534
Experiment 21: 1. (b) Question 5
2. (b)
3. (a)
4. (a)
When the resistors in the right side of the bridge of Figure 21-1 are reversed, an approximate balance is obtained. The reason for this is that each resistance has a 5 % tolerance and therefore the voltages at points C and D will vary with this tolerance. The potential difference between C and D will also be a function of this tolerance. Unless the resistors in both arms of the bridge are exactly the same ratio, the bridge will show some unbalance.
Question 6:
When the potentiometer resistance is made larger, the voltage VD increases, (becomes more positive) the voltage VC stays the same, and so the voltage at C with respect to D (VCD) decreases. (moves closer toward zero) ______________________________________ Experiment 22: 1. (d) Question 5:
2. (a)
3. (d)
4. (b)
The highest voltage to which point C can rise will occur when the current is a maximum, (Ohm's Law) and this will occur when a single fault condition produces a minimum total circuit resistance. Clearly this will occur for some shorted resistor. By trial and error we can see that the current is maximized by letting R1 go short circuit. (R4 is not allowed to go short circuit since this would produce 0 V at point C) By the voltage divider theorem, the voltage at point C would rise to 9 V
Question 6:
If resistor R4 goes short circuit, it will produce the greatest percentage increases in supply current. This is because it is the largest resistor in the circuit, The current will increase from 1 mA to 2 mA an increase of 100 %. ______________________________________
535
Experiment 23: 1. (a)
2. (a)
3. (b)
4. (b)
Question 5:
A good constant voltage source is one who's terminal or load voltage stays fairly constant over a range of load conditions. In order for this to be the case, the source resistance should be small, especially in comparison with the smallest load resistance that will be connected to the source.
Question 6:
From the graphs or by calculation, the value of load that will cause the terminal voltage to drop to 8 V is 800 Ω. By calculation, Vt = Vs - RsIL so,
IL = (Vs - Vt)/ Rs = 10 mA
and,
RL = Vt/RL = 8 V/10 mA = 800 Ω
______________________________________
536
Experiment 24: 1. (a)
2. (a)
3. (a)
4. (a)
Question 5:
A voltage source in series with a resistor will have a terminal voltage that varies by a small amount provided that the load resistance is always a lot larger than the series resistance. A voltage source in series with a resistor will have a load current that varies by a small amount provided that the load resistance is a lot smaller than the series resistance.
Question 6:
The voltage source-resistor combination will behave as an approximate current source provided the load resistance is less than about 2 kΩ. The model for this source is shown below.
______________________________________ Experiment 25: 1. (d) Question 5:
2. (b)
3. (a)
4. (d)
Superposition will not work for power because power depends on the square of voltage and current. For example, if we take the current down the center resistor of Figure 24-1, I2, and calculate the current components due to each of the sources Vs1 and Vs2, we get 3 mA for I21 and 2 mA for I22. Using I2R for the power due to each source we get 9 mW and 4 mW. Adding, these gives 13 mW. However the total current flowing down the center resistor is 3 mA + 2 mA = 5 mA and when we apply the power formula to this total current we get P = 25 mW. The latter of these two answers is the correct one for power. The 537
reason that the first calculation is in error is because it is the total current that most first be squared and multiplied by R to get the power. Squaring each current separately does not work because I212R + I222R ≠ (I21 + I22)2R. Question 6:
For the current I1 to be equal to zero, the components I11 and I12 must cancel out ! (they must be equal in value and opposite in polarity) Since I12 = -1 mA with Vs2 = 11 V, and I11 is 4 mA with Vs1 = 11 V, then to get I11 equal to only 1 mA (instead of 4 mA) Vs1 needs to be reduced by a factor of four (Ohm's Law) to 2.5 V. Thus if Vs1 is reduced to 2.5 V, the current I1 will be equal to zero. ______________________________________ Experiment 26: 1. (d)
2. (c)
3. (c)
4. (b)
Question 5:
A resistor placed directly in parallel with the ideal source voltage will have no effect on the Thevenin Resistance as seen from the terminals A-B of Figure 25-1. This is because the source is replaced with a short circuit (0 Ω) when the Thevenin Resistance is calculated. Any resistor in parallel with this short circuit will not enter into the calculation.
Question 6:
In order to reduce the effective Thevenin Resistance from 3 kΩ to 2 kΩ a resistor can be placed across the terminals A-B as shown. Using the formula for two parallel resistors, and solving for the unknown, we arrive at 6 kΩ for Ru. As a check, 6 kΩ in parallel with 3 kΩ equals 2 kΩ. Calling the new Thevenin resistance Rth' and the old one Rs, we have, R'th = Rs. Ru/(Rs+Ru)
538
so that,
Ru = R'th. Rs/(Rs- R'th) = 2kΩ.3kΩ/(3kΩ+2kΩ) = 6 kΩ
The new Thevenin equivalent voltage will be 4 V. To get this I use the old 6 V Thevenin voltage together with a series voltage divider consisting of the old 3 kΩ Thevenin resistance, and the new 6 kΩ resistor that is attached across terminals A and B, viz: V = 6 V (6/(3+6)) = 4 V. ______________________________________ Experiment 27: 1. (c) Question 5:
2. (a)
3. (d)
4. (a)
To achieve high levels of power efficiency in this circuit, the source resistance must be much smaller than the load resistance. Here's why! Using P = I2R we have for the efficiency, PL/PT x100% = I2RL/I2RT x100% = RL/RT x 100%. This can only become close to 100% if RL ≈ RT which can only happen if Rs << RL
Question 6:
When the circuit is supplying maximum power to the load, Rs = RL and RT = 2 x Rs Since efficiency = RL/RT then this will be equal to Rs/2Rs x 100% = 50%. ______________________________________ Experiment 28: 1. (b) Question 5:
2. (b)
3. (b)
The modified mesh equations are (3)I1 + (-1) I2 = 11 (-1)I1 + (4) I2 = -22
539
4. (c)
The solution is: I1 = 2 mA and I2 = - 5 mA The resistor currents are: IR1 = I1 = 2 mA IR2 = I1 - I2 = 7 mA, IR3 = - I2 = 5 mA The resistor voltages are: VR1 = 4 V, VR2 = 7 V, VR3 = 15 V Question 6:
The node equation at Node A is now 11-VA - VA + 22 - VA = 0 2 1 3 from which the solution is VA = 7 V
The circuit currents and voltages are the same as for question 5. ______________________________________ Experiment 29: 1. (c) Question 5:
2. (b)
3. (c)
4. (c)
With the input coupling switch set to AC, there will be a capacitor in series with the input channel. This will block any dc current and therefore dc voltage from reaching the vertical input circuitry of the oscilloscope. The oscilloscope will display only a 0 V trace as though there were no dc voltage present at the sampling ends of the probe.
Question 6:
Since the oscilloscope is an earth-grounded instrument, if any parts of a circuit under test are also earth grounded, then the earth side of the oscilloscope test lead must be connected to that point. Failure to do this will result in erroneous readings and can produce an electrically hazardous condition. ______________________________________
540
Experiment 30: 1. (c) Question 5:
2. (d)
3. (c)
4. (a)
If these vernier controls are not in their calibrated positions, then the vertical and horizontal amplifiers are not calibrated, (VOLT/DIV and SEC/DIV markings will not be true) and amplitude and time information taken directly from the oscilloscope screen will be erroneous.
Question 6:
If the earth-grounded side of the oscilloscope lead were connected to the non-grounded side of the function generator (who's other terminal is earth-grounded) then a short circuit across the generator would occur. Besides giving no reading for voltage, this might cause some damage to the function generator output. ______________________________________ Experiment 31: 1. (a) Question 5:
2. (b)
3. (a)
4. (c)
The amplitude control usually controls the peak-to-peak value of the ac portion of the wave form. The smooth frequency control usually allows the user to continuously vary the frequency of the generator. The dc offset control allows the user to introduce a dc voltage into the ac signal thus raising or lowering its baseline value and changing the dc component. Switches are often provided so that the frequency can be changed in discrete steps, usually by factors of ten. These switches are called decade frequency switches.
Question 6:
In order to be able to see whether a signal has a dc component present, the oscilloscope must be set to dc coupling. This will bypass the capacitor that is present in the AC mode. ______________________________________
541
Experiment 32: 1. (b)
2. (a)
3. (b)
4. (b)
Question 5:
The only sure way to ensure that the load voltage remains constant as the load varies is to monitor it with a meter or an oscilloscope, and make appropriate corrections at the amplitude control manually.
Question 6:
Ohmmeters should never be connected to circuits that are powered. Not only do they give erroneous readings under these conditions, but damage to the meter may result. Furthermore, even if the function generator is switched off, one must remember that the source resistance of a generator is only a Thevenin model that allows us to predict the terminal characteristics of the generator. It does not mean that there is a physical resistor equal to the Thevenin Resistance inside the generator
_____________________________________ Experiment 33: 1. (c) Question 5:
2. (a)
3. (a)
4. (c)
The signals being observed in this activity were symmetrical about zero volts and therefore had no dc offset voltage present in them. Changing the oscilloscope setting from AC to DC would therefore have no effect on the trace.
Question 6:
The DMM/VOM will read out rms values and is therefore more convenient when this quantity is of primary interest. The oscilloscope allows us to view peak and peak-to-peak values directly and is the more direct way of determining such values. Of course, when the relationship between the peak and rms values of a wave form is known, either one can be calculated from a knowledge of the other. ______________________________________
542
Experiment 34: 1. (c) Question 5:
2. (c)
3. (b)
4. (a)
An uncalibrated time base is convenient in measuring phase difference because only distances as fractions of a period are being measured. It is much easier to accomplish this if a full period of the wave form can be fitted into a whole number of divisions across the screen. Since phase difference is a relative measure, the actual time it represents is of no concern.
Question 6:
Using ratio and proportion, the phase difference angle Ø/360 = (∆t)/T = (∆t).f. Solving this for ∆t we get ∆t = Ø/(360.f) ______________________________________ Experiment 35: 1. (e)
2. (e)
3. (c)
4. (e)
Question 5:
The following sketches show two square waves as asked for in the question.
Question 6:
The following show two square waves as asked for in the question.
______________________________________
543
Experiment 36: 1. (a)
2. (b)
3. (b)
4. (b)
Question 5:
The lamp was on for a very short period of time. This period of time depends on both the capacitance and the resistance of the lamp.
Question 6:
The capacitor discharges slowly through the meter resistance.
______________________________________ Experiment 37: 1. (b)
2. (b)
3. (b)
4. (e)
Question 5:
The meter indicates a negative voltage during the capacitor discharge because the current is flowing in the opposite direction to that which produces a positive voltage across the meter.
Question 6:
Since the meter appears across the capacitor, Thevenin's theorem can be applied to find the effective resistance that appears in the time constant formula. Since the DMM resistance will appear in parallel with the circuit resistance, its effect would have been to reduce the time constant. At the same time, the capacitor would not have charged up to the full supply voltage but instead would have charged up to a voltage given by the voltage divider formed by the circuit resistance and the DMM resistance.
______________________________________
544
Experiment 38: 1. (b) Question 5:
2. (a)
3. (a)
4. (d)
When S1 is opened, the current that was flowing through the coil is forced to continue flowing through the circuit formed by the coil and the neon lamp. The Energy needed to keep this current flowing comes from the collapsing magnetic field around the inductor.
Question 6:
It was important not to exceed Imax for the inductor because this represents the maximum safe current that the coil can handle without overheating. ______________________________________ Experiment 39: 1. (b)
2. (c)
3. (d)
4. (b)
Question 5:
The primary inductance was greater than the secondary inductance primarily because of the many more turns that are on the primary side of the transformer. Since inductance is proportional to the square of the number of turns on the coil, this accounts for the larger primary inductance.
Question 6:
The primary resistance was greater than the secondary resistance because of the larger number of turns on that coil. In addition, it is likely that the wire thickness on the primary side was thinner because this side carries a smaller current than the secondary side.
______________________________________
545
Experiment 40: 1. (a)
2. (a)
3. (a)
4. (d)
Question 5:
Under matched conditions for maximum power transfer, one half of the source power is dissipated in the source resistance, while the other half is transferred to the load through the transformer. When the impedance is matched, the efficiency is 50%.
Question 6:
The relationship between the turns ratio, n, the load resistance, RL and the source resistance Rs for a matched system, is Rs = RL/n2 so a transformer that
matches a 10 kΩ source to a 16 Ω load will match other source-load combinations provided that their ratio RL/Rs is equal to n2, and provided that other design parameters such as frequency limitations and power ratings are not exceeded. ______________________________________ Experiment 41: 1. (a) Question 5:
2. (b)
3. (a)
4. (b)
The diagram below is valid for the circuit of Figure 41-1 at a frequency of 10 kHz. At this frequency, the relationship that exists between the capacitor, resistor, and generator voltage is given by the right angled 45o triangle, since this is approximately the critical frequency of the system.
546
Question 6:
Frequency response data assumes that the circuit is driven with an ideal constant amplitude voltage source, so that all variations in current and voltage are due to the changes in impedance with frequency. If the generator voltage were not kept constant, the source resistance would have to be taken into account and this would change the total resistance of the circuit ______________________________________ Experiment 42: 1. (c) Question 5:
2. (a)
3. (c)
4. (d)
The diagram below is valid for the circuit of Figure 42-1 at a frequency of 10 kHz. At this frequency, the relationship that exists between the capacitor, resistor, and generator currents is given by the right angled 45o triangle, since this is approximately the critical frequency of the system.
Question 6:
The current monitoring resistors are made small so that each does not have much effect on the impedance of the branch into which they are inserted. If they are made too small, then the voltages across them become vanishingly small and difficult to measure. ______________________________________
547
Experiment 43: 1. (b)
2. (b)
3. (a)
4. (a)
Question 5:
Vcoil may have been larger because the calculations ignored the resistance of the coil, which has a voltage that adds to the reactive drop in the inductive part of the coil.
Question 6:
The phasor diagram illustrates the relationships between all voltages and currents in the circuit.
______________________________________ Experiment 44: 1. (d)
2. (b)
3. (c)
4. (b)
Question 5:
The coil resistance will add to the circuit resistance R and make the base of the impedance triangle larger. This will force the impedance of the circuit to be larger at any given frequency.
Question 6:
The sketches below show how a series RLC circuit can look inductive or capacitive overall depending upon the relative magnitudes of XL and XC.
______________________________________
548
Experiment 45: 1. (d)
2. (b)
3. (b)
4. (c)
Question 5:
A constant input voltage is necessary so that all variations in element voltages and currents can be attributed to variations in the impedances of R, L and C with frequency.
Question 6:
The coil resistance adds to the total circuit resistance, and at resonance, adds to the resistive impedance that is seen by the source. This means that the current might be somewhat smaller than predicted by ignoring coil resistance. In addition, there will be a resistive voltage drop across the resistive component of the coil at resonance.
______________________________________ Experiment 46: 1. (d) Question 5:
2. (b)
3. (b)
4. (c)
If the coil resistance is not small compared with the reactance of the coil at frequencies close to resonance, i.e. if the coil Q factor is not high, then the resistance will have an effect on the resonant frequency of the circuit, lowering it slightly relative to its value if the coil resistance were zero.
Question 6:
The answer given for this question will depend upon the particular coil and resistor combination used. However at resonance, the circuit acts like a resistive voltage divider with the tank circuit acting as a resistor of value L/CRcoil. This resistance can be calculated using ratio and proportion from which with a knowledge of L and C, one can calculate Rcoil. ______________________________________
549
Experiment 47: 1. (c) Question 5:
2. (a)
3. (a)
4. (b)
If log frequency axes were not used, such a wide range of frequency response data would not be able to be effectively plotted. Log frequency scales allow data at very low and very high frequencies to be plotted in a systematic and readable way.
Question 6:
A good application for a low pass filter is in the tone controls of an audio system. Actually, a filter that attenuates clicks and pops, and other high frequency unwanted noise in such a system is called a low pass filter. ______________________________________ Experiment 48: 1. (c) Question 5:
2. (b)
3. (b)
4. (b)
A ratio of 20 dB is a factor of ten in voltage. A filter having a roll-off rate of 20dB/decade is one in which the voltage gain (or output voltage) falls by a factor of ten for each factor of ten change in frequency. In the case of a high pass filter, this means that for frequencies less than fc, the filter's gain drops by a factor of ten for each factor of ten decrease in frequency
Question 6:
A good application for a high pass filter is in the tone controls of an audio system. Actually, a filter that attenuates rumble, turntable noise, and other low frequency unwanted noise in such a system is called a high pass filter. ______________________________________
550
Experiment 49: 1. (c)
2. (b)
3. (d)
4. (d)
Question 5:
The formulas for bandwidth and resonant frequency in this circuit are ∆f = R/2πL and fr = 1/2π√LC respectively. From this we can see that bandwidth does not depend on the capacitance and that the resonant frequency does not depend upon the resistance. This means that the resistance can be used to control the bandwidth and the capacitance can be used to control the resonant frequency.
Question 6:
Because the bandwidth depends on the total resistance of the circuit including the ac resistance of the coil, and because the ac resistance of the coil usually increases with frequency in a rather unpredictable way, this will lower the circuit Q and therefore increase the circuit bandwidth.
______________________________________ Experiment 50: 1. (c) Question 5:
2. (c)
3. (a)
4. (b)
The resistance R controls the bandwidth of this circuit. By changing the value of the resistance, the bandwidth and Q factor will be altered. Specifically, a larger value of R will increase the bandwidth and lower the Q.
Question 6:
If the output voltage were taken from the tank circuit instead of the resistor, the filter response would be a maximum at resonance and so the filter would behave as a bandpass type. ______________________________________
551
Experiment 51: 1. (a)
2. (e)
3. (e)
4. (a)
Question 5:
The term "AC coupled" refers to the fact that the capacitor has a low impedance to the frequencies in the circuit, and therefore allows current to pass to the resistor enabling an ac voltage to be developed across the resistor. To dc however, the capacitor is an open circuit and so any dc voltage in the source will be blocked from getting to the resistor.
Question 6:
The capacitor in Figure 51-2 has a low impedance to ac voltages; much lower than that of the resistor R2. Therefore, ac currents will pass around R2 and no ac voltage can be developed across R2. To dc however, the capacitor acts like an open circuit and so dc voltage can be developed across R2 (no dc current can pass through C) This means that almost all of the source ac voltage will be developed across R1.
______________________________________
552
Experiment 52: 1. (d) Question 5:
2. (a)
3. (c)
4. (a)
The peak-to-peak resistor voltage will be about twice that of the input square wave for the following reason. On the leading (positive going) edge of the input pulse, the resistor voltage jumps up to the peak input voltage and gradually decays toward zero volts. On the trailing (negative going) edge of the input pulse, the voltage again jumps down (negative) from zero by an amount equal to the input transition. The voltage across the capacitor cannot change instantly. Since this jump is downward, the negative and positive peaks will each be the same magnitude as the p-p amplitude of the square wave. Refer to the sketches below.
Question 6:
The charge/discharge time for an RC circuit depends on the time constant which depends only on the values of (R) and (C) in the circuit. The actual amplitude of the input voltage is of no consequence. ______________________________________
553
Experiment 53: 1. (a)
2. (b)
3. (d)
4. (c)
Question 5:
In a balanced Y system, there is no neutral current. If there is no current, there is no need for a conductor to carry this current.
Question 6:
The savings in copper comes from using only three or four wires compared with the six that would be necessary in a multiple single phase system.
______________________________________
554
555
