Solution Manual For Electronics Fundamentals: A Systems Approach, 1st Edition
Solution Manual For Electronics Fundamentals: A Systems Approach, 1st Edition simplifies tough problems, making them easier to understand and solve.
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Online Instructor’s Manual
for
Electronics Fundamentals
A Systems Approach
Thomas L. Floyd
David M. Buchla
for
Electronics Fundamentals
A Systems Approach
Thomas L. Floyd
David M. Buchla
1
PART ONE
Solutions to End-of-Chapter Problems
PART ONE
Solutions to End-of-Chapter Problems
2
1. The circuit is first tested with a computer design and simulation program, which can simulate
the performance and look for potential problems. When the simulation is satisfactory, a
prototype circuit is constructed, tested, and modified as needed before putting it into
production.
2. Semiconductor and component manufacturers as well as printed circuit board manufacturers.
3. Electronic assemblies have become more complex but also more reliable, so there is less need
for repair. It is generally cheaper for manufacturers to replace a board than troubleshoot it to
the component level. Skills needed by technicians tend to be broader skills than in the past.
4. Electrical systems deal primarily with power; electronic systems involve signals and a logical
sequence of processes.
5. Advantages are that the digital signal can be processed and stored easily; it is also less subject
to noise.
6. A block diagram shows signal flow in a system; a flowchart shows a logical process.
7. (a) An electronic oscillator generates a repetitive electronic signal
(b) An oscillator does not have a signal input.
8. (a) High Voltage Direct Current
CHAPTER 1
SYSTEMS, QUANTITIES AND UNITS
SECTION 1-1 The Electronics Industry
SECTION 1-2 Introduction to Electronic Systems
SECTION 1-3 Types of Circuits
1. The circuit is first tested with a computer design and simulation program, which can simulate
the performance and look for potential problems. When the simulation is satisfactory, a
prototype circuit is constructed, tested, and modified as needed before putting it into
production.
2. Semiconductor and component manufacturers as well as printed circuit board manufacturers.
3. Electronic assemblies have become more complex but also more reliable, so there is less need
for repair. It is generally cheaper for manufacturers to replace a board than troubleshoot it to
the component level. Skills needed by technicians tend to be broader skills than in the past.
4. Electrical systems deal primarily with power; electronic systems involve signals and a logical
sequence of processes.
5. Advantages are that the digital signal can be processed and stored easily; it is also less subject
to noise.
6. A block diagram shows signal flow in a system; a flowchart shows a logical process.
7. (a) An electronic oscillator generates a repetitive electronic signal
(b) An oscillator does not have a signal input.
8. (a) High Voltage Direct Current
CHAPTER 1
SYSTEMS, QUANTITIES AND UNITS
SECTION 1-1 The Electronics Industry
SECTION 1-2 Introduction to Electronic Systems
SECTION 1-3 Types of Circuits
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(b) HVDC is used for long distance and underwater power transmission.
9. A carrier is a high frequency radio wave that can be modulated (changed) by a lower
frequency signal.
10. A stair-step output; each step represents a different digital value.
11. (a) 3000 = 3 103 (b) 75,000 = 7.5 104 (c) 2,000,000 = 2 106
12. (a)500
1 = 0.002 = 2 103
(b)2000
1 = 0.0005 = 5 104
(c)000,000,5
1 = 0.0000002 = 2 107
13. (a) 8400 = 8.4 103 (b) 99,000 = 9.9 104 (c) 0.2 106 = 2 105
14. (a) 0.0002 = 2 104 (b) 0.6 = 6 101
(c) 7.8 102 (already in scientific notation)
15. (a) 2.5 106 = 0.0000025 (b) 5.0 102 = 500 (c) 3.9 101 = 0.39
16. (a) 4.5 106 = 0.0000045
(b) 8 109 = 0.000000008
(c) 4.0 1012 = 0.0000000000040
17. (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
18. (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
19. (a) (5 103)(4 105) = 5 4 103 + 5 = 20 108 = 2 109
SECTION 1-4 Scientific and Engineering Notation
(b) HVDC is used for long distance and underwater power transmission.
9. A carrier is a high frequency radio wave that can be modulated (changed) by a lower
frequency signal.
10. A stair-step output; each step represents a different digital value.
11. (a) 3000 = 3 103 (b) 75,000 = 7.5 104 (c) 2,000,000 = 2 106
12. (a)500
1 = 0.002 = 2 103
(b)2000
1 = 0.0005 = 5 104
(c)000,000,5
1 = 0.0000002 = 2 107
13. (a) 8400 = 8.4 103 (b) 99,000 = 9.9 104 (c) 0.2 106 = 2 105
14. (a) 0.0002 = 2 104 (b) 0.6 = 6 101
(c) 7.8 102 (already in scientific notation)
15. (a) 2.5 106 = 0.0000025 (b) 5.0 102 = 500 (c) 3.9 101 = 0.39
16. (a) 4.5 106 = 0.0000045
(b) 8 109 = 0.000000008
(c) 4.0 1012 = 0.0000000000040
17. (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
18. (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
19. (a) (5 103)(4 105) = 5 4 103 + 5 = 20 108 = 2 109
SECTION 1-4 Scientific and Engineering Notation
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(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
20. (a)2
3
105.2
100.1
= 0.4 103 2 = 0.4 101 = 4
(b)6
8
2.5 10
5.0 10
= 0.5 106 (8) = 0.5 102 = 50
(c)5
8
102
102.4
= 2.1 108 (5) = 2.1 1013
21. (a) 89,000 = 89 103
(b) 450,000 = 450 103
(c) 12,040,000,000,000 = 12.04 1012
22. (a) 2.35 105 = 235 103
(b) 7.32 107 = 73.2 106
(c) 1.333 109 (already in engineering notation)
23. (a) 0.000345 = 345 106
(b) 0.025 = 25 103
(c) 0.00000000129 = 1.29 109
24. (a) 9.81 103 = 9.81 103
(b) 4.82 104 = 482 106
(c) 4.38 107 = 438 109
25. (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
26. (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
(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
20. (a)2
3
105.2
100.1
= 0.4 103 2 = 0.4 101 = 4
(b)6
8
2.5 10
5.0 10
= 0.5 106 (8) = 0.5 102 = 50
(c)5
8
102
102.4
= 2.1 108 (5) = 2.1 1013
21. (a) 89,000 = 89 103
(b) 450,000 = 450 103
(c) 12,040,000,000,000 = 12.04 1012
22. (a) 2.35 105 = 235 103
(b) 7.32 107 = 73.2 106
(c) 1.333 109 (already in engineering notation)
23. (a) 0.000345 = 345 106
(b) 0.025 = 25 103
(c) 0.00000000129 = 1.29 109
24. (a) 9.81 103 = 9.81 103
(b) 4.82 104 = 482 106
(c) 4.38 107 = 438 109
25. (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
26. (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
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(c) (100)(55 103) = 5500 103 = 5.5
27. (a)3
102.2
50
= 22.7 103
(b)6
3
1025
105
= 0.2 10(3 (6)) = 0.2 109 = 200 106
(c)3
3
10660
10560
= 0.848 10(3 3) = 0.848 100 = 848 103
28. (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
29. (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
30. (a) 31 103 A = 31 mA (b) 5.5 103 V = 5.5 kV (c) 20 1012 F = 20 pF
31. (a) 3 106 F = 3 F (b) 3.3 106 = 3.3 M (c) 350 109 A = 350 nA
32. (a) 5 A = 5 106 A (b) 43 mV = 43 103 V
(c) 275 k = 275 103 (d) 10 MW = 10 106 W
33. (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
SECTION 1-6 Metric Unit Conversions
SECTION 1-5 Units and Metric Prefixes
(c) (100)(55 103) = 5500 103 = 5.5
27. (a)3
102.2
50
= 22.7 103
(b)6
3
1025
105
= 0.2 10(3 (6)) = 0.2 109 = 200 106
(c)3
3
10660
10560
= 0.848 10(3 3) = 0.848 100 = 848 103
28. (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
29. (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
30. (a) 31 103 A = 31 mA (b) 5.5 103 V = 5.5 kV (c) 20 1012 F = 20 pF
31. (a) 3 106 F = 3 F (b) 3.3 106 = 3.3 M (c) 350 109 A = 350 nA
32. (a) 5 A = 5 106 A (b) 43 mV = 43 103 V
(c) 275 k = 275 103 (d) 10 MW = 10 106 W
33. (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
SECTION 1-6 Metric Unit Conversions
SECTION 1-5 Units and Metric Prefixes
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34. (a)A101
A101
A1
mA1 6
3
= 1 103 = 1000
(b)V101
V1005.0
mV1
kV05.0 3
3
= 0.05 106 = 50,000
(c)
6
3
101
1002.0
M1
k02.0 = 0.02 103 = 2 105
(d)W101
W10155
kW1
mW155 3
3
= 155 106 = 1.55 104
35. (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
36. (a)
k2.12
k10
k10k2.2
k10 = 0.8197
(b)6
3
1050
10250
V50
mV250
= 5000
(c)3
6
102
101
kW2
MW1
= 500
37. (a) 1.00 103 has 3 significant digits. (b) 0.0057 has 2 significant digits.
(c) 1502.0 has 5 significant digits. (d) 0.000036 has 2 significant digits.
(e) 0.105 has 3 significant digits. (f) 2.6 102 has 2 significant digits.
38. (a) 50,505 50.5 103 (b) 220.45 220
(c) 4646 4.65 103 (d) 10.99 11.0
(e) 1.005 1.00
SECTION 1-7 Measured Numbers
34. (a)A101
A101
A1
mA1 6
3
= 1 103 = 1000
(b)V101
V1005.0
mV1
kV05.0 3
3
= 0.05 106 = 50,000
(c)
6
3
101
1002.0
M1
k02.0 = 0.02 103 = 2 105
(d)W101
W10155
kW1
mW155 3
3
= 155 106 = 1.55 104
35. (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
36. (a)
k2.12
k10
k10k2.2
k10 = 0.8197
(b)6
3
1050
10250
V50
mV250
= 5000
(c)3
6
102
101
kW2
MW1
= 500
37. (a) 1.00 103 has 3 significant digits. (b) 0.0057 has 2 significant digits.
(c) 1502.0 has 5 significant digits. (d) 0.000036 has 2 significant digits.
(e) 0.105 has 3 significant digits. (f) 2.6 102 has 2 significant digits.
38. (a) 50,505 50.5 103 (b) 220.45 220
(c) 4646 4.65 103 (d) 10.99 11.0
(e) 1.005 1.00
SECTION 1-7 Measured Numbers
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BASIC PROBLEMS
1. Q = (charge per electron)(number of electrons) = (1.6 1019 C/e)(50 1031e) = 80 1012 C
2. (6.25 1018 e/C)(80 106 C) = 5 1014 e
3. The magnitude of the charge on a proton (p) is equal to the magnitude of the charge on the
electron (e). Therefore, (1.6 10-19 C/p)(29 p) = 4.64 10-18 C
4. (1.6 10-19 C/p)(17 p) = 2.72 10-18 C
5. (a)C1
J10
Q
W
V = 10 V (b)C2
J5
Q
W
V = 2.5 V (c)C25
J100
Q
W
V = 4 V
6.C100
J500
Q
W
V = 5 V
7.C40
J800
Q
W
V = 20 V
8. W = VQ = (12 V)(2.5 C) = 30 J
9.2.5 J
0.2 C
W
V Q
= = = 12.5 V
10.0.2 C
10 s
Q
I t
= = = 20 mA
11. (a)s1
C75
t
Q
I = 75 A (b)s0.5
C10
t
Q
I = 20 A (c)s2
C5
t
Q
I = 2.5 A
12.s3
C0.6
t
Q
I = 0.2 A
CHAPTER 2
VOLTAGE, CURRENT, AND RESISTANCE
SECTION 2-2 Electrical Charge
SECTION 2-3 Voltage
SECTION 2-4 Current
BASIC PROBLEMS
1. Q = (charge per electron)(number of electrons) = (1.6 1019 C/e)(50 1031e) = 80 1012 C
2. (6.25 1018 e/C)(80 106 C) = 5 1014 e
3. The magnitude of the charge on a proton (p) is equal to the magnitude of the charge on the
electron (e). Therefore, (1.6 10-19 C/p)(29 p) = 4.64 10-18 C
4. (1.6 10-19 C/p)(17 p) = 2.72 10-18 C
5. (a)C1
J10
Q
W
V = 10 V (b)C2
J5
Q
W
V = 2.5 V (c)C25
J100
Q
W
V = 4 V
6.C100
J500
Q
W
V = 5 V
7.C40
J800
Q
W
V = 20 V
8. W = VQ = (12 V)(2.5 C) = 30 J
9.2.5 J
0.2 C
W
V Q
= = = 12.5 V
10.0.2 C
10 s
Q
I t
= = = 20 mA
11. (a)s1
C75
t
Q
I = 75 A (b)s0.5
C10
t
Q
I = 20 A (c)s2
C5
t
Q
I = 2.5 A
12.s3
C0.6
t
Q
I = 0.2 A
CHAPTER 2
VOLTAGE, CURRENT, AND RESISTANCE
SECTION 2-2 Electrical Charge
SECTION 2-3 Voltage
SECTION 2-4 Current
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13.;
t
Q
I t =A5
C10
I
Q = 2 s
14. Q = I t = (1.5 A)(0.1 s) = 0.15 C
15. A: Blue, gray, red, silver: 6800 10%
B: Orange, orange, black, silver: 33 10%
C: Yellow, violet, orange, gold: 47,000 5%
16. A: Rmin = 6800 0.1(6800 ) = 6800 680 = 6120
Rmax = 6800 + 680 = 7480
B: Rmin = 33 0.1(33 ) = 33 3.3 = 29.7
Rmax = 33 + 3.3 = 36.3
C: Rmin = 47,000 (0.05)(47,000 ) = 47,000 2350 = 44,650
Rmax = 47,000 + 2350 = 49,350
17. (a) 1st band = red, 2nd band = violet, 3rd band = brown, 4th band = gold
(b) 330 ; orange, orange, brown, (B)
2.2 k: red, red, red (D)
39 k: orange, white, orange (A)
56 k: green, blue, orange (L)
100 k: brown, black, yellow (F)
18. (a) 36.5 2%
(b) 2.74 k 0.25%
(c) 82.5 k 1%
19. (a) Brown, black, black, gold: 10 ± 5%
(b) Green, brown, green, silver: 5,100,000 ± 10% = 5.1 M ± 10%
(c) Blue, gray, black, gold: 68 ± 5%
20. (a) 0.47 ± 5%: yellow, violet, silver, gold
(b) 270 k ± 5%: red, violet, yellow, gold
(c) 5.1 M ± 5%: green, brown, green, gold
21. (a) Red, gray, violet, red, brown: 28,700 ± 1% = 28.7 k ± 1%
(b) Blue, black, yellow, gold, brown: 60.4 ± 1%
(c) White, orange, brown, brown, brown: 9310 ± 1% = 9.31 k ± 1%
22. (a) 14.7 k ± 1%: brown, yellow, violet, red, brown
(b) 39.2 ± 1%: orange, white, red, gold, brown
(c) 9.76 k ± 1%: white, violet, blue, brown, brown
23. (a) 220 = 22 (b) 472 = 4.7 k
(c) 823 = 82 k (d) 3K3 = 3.3 k
(e) 560 = 56 (f) 10M = 10 M
24. 500 , equal resistance on each side of the contact.
SECTION 2-5 Resistance
13.;
t
Q
I t =A5
C10
I
Q = 2 s
14. Q = I t = (1.5 A)(0.1 s) = 0.15 C
15. A: Blue, gray, red, silver: 6800 10%
B: Orange, orange, black, silver: 33 10%
C: Yellow, violet, orange, gold: 47,000 5%
16. A: Rmin = 6800 0.1(6800 ) = 6800 680 = 6120
Rmax = 6800 + 680 = 7480
B: Rmin = 33 0.1(33 ) = 33 3.3 = 29.7
Rmax = 33 + 3.3 = 36.3
C: Rmin = 47,000 (0.05)(47,000 ) = 47,000 2350 = 44,650
Rmax = 47,000 + 2350 = 49,350
17. (a) 1st band = red, 2nd band = violet, 3rd band = brown, 4th band = gold
(b) 330 ; orange, orange, brown, (B)
2.2 k: red, red, red (D)
39 k: orange, white, orange (A)
56 k: green, blue, orange (L)
100 k: brown, black, yellow (F)
18. (a) 36.5 2%
(b) 2.74 k 0.25%
(c) 82.5 k 1%
19. (a) Brown, black, black, gold: 10 ± 5%
(b) Green, brown, green, silver: 5,100,000 ± 10% = 5.1 M ± 10%
(c) Blue, gray, black, gold: 68 ± 5%
20. (a) 0.47 ± 5%: yellow, violet, silver, gold
(b) 270 k ± 5%: red, violet, yellow, gold
(c) 5.1 M ± 5%: green, brown, green, gold
21. (a) Red, gray, violet, red, brown: 28,700 ± 1% = 28.7 k ± 1%
(b) Blue, black, yellow, gold, brown: 60.4 ± 1%
(c) White, orange, brown, brown, brown: 9310 ± 1% = 9.31 k ± 1%
22. (a) 14.7 k ± 1%: brown, yellow, violet, red, brown
(b) 39.2 ± 1%: orange, white, red, gold, brown
(c) 9.76 k ± 1%: white, violet, blue, brown, brown
23. (a) 220 = 22 (b) 472 = 4.7 k
(c) 823 = 82 k (d) 3K3 = 3.3 k
(e) 560 = 56 (f) 10M = 10 M
24. 500 , equal resistance on each side of the contact.
SECTION 2-5 Resistance
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25. There is current through Lamp 2.
26. See Figure 2-1.
Figure 2-1
27. See Figure 2-2(a).
Figure 2-2 Figure 2-3
28. See Figure 2-2(b).
29. Position 1: V1 = 0 V, V2 = VS
Position 2: V1 = VS, V2 = 0 V
30. See Figure 2-3.
31. On the 600 V DC scale: 250 V
32. R = (10)(10 ) = 100
33. (a) 2(100 ) = 200
(b) 15(10 M) = 150 M
(c) 45(100 ) = 4500
34. See Figure 2-4.
SECTION 2-6 The Electric Circuit
SECTION 2-7 Basic Circuit Measurements
Figure 2-4
25. There is current through Lamp 2.
26. See Figure 2-1.
Figure 2-1
27. See Figure 2-2(a).
Figure 2-2 Figure 2-3
28. See Figure 2-2(b).
29. Position 1: V1 = 0 V, V2 = VS
Position 2: V1 = VS, V2 = 0 V
30. See Figure 2-3.
31. On the 600 V DC scale: 250 V
32. R = (10)(10 ) = 100
33. (a) 2(100 ) = 200
(b) 15(10 M) = 150 M
(c) 45(100 ) = 4500
34. See Figure 2-4.
SECTION 2-6 The Electric Circuit
SECTION 2-7 Basic Circuit Measurements
Figure 2-4
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ADVANCED PROBLEMS
35. I =t
Q
Q = I t = (2 A)(15 s) = 30 C
V =C30
J1000
Q
W = 33.3 V
36. I =t
Q
Q = (number of electrons) / (number of electrons/coulomb)
Q =e/C1025.6
e10574 18
15
= 9.184 102 C I =s10250
C10184.9 3
2
t
Q = 0.367 A
37. Total wire length = 100 ft
Resistance per 1000 ft = (1000 ft)(6 /100 ft) = 60
Smallest wire size is AWG 27 which has 51.47 /1000 ft
38. (a) 4R7J = 4.7 5%
(b) 560KF = 560 k 1%
(c) 1M5G = 1.5 M 2%
39. The circuit in (b) can have both lamps on at the same time.
40. There is always current through R5.
41. See Figure 2-5.
Figure 2-5
42. See Figure 2-5.
ADVANCED PROBLEMS
35. I =t
Q
Q = I t = (2 A)(15 s) = 30 C
V =C30
J1000
Q
W = 33.3 V
36. I =t
Q
Q = (number of electrons) / (number of electrons/coulomb)
Q =e/C1025.6
e10574 18
15
= 9.184 102 C I =s10250
C10184.9 3
2
t
Q = 0.367 A
37. Total wire length = 100 ft
Resistance per 1000 ft = (1000 ft)(6 /100 ft) = 60
Smallest wire size is AWG 27 which has 51.47 /1000 ft
38. (a) 4R7J = 4.7 5%
(b) 560KF = 560 k 1%
(c) 1M5G = 1.5 M 2%
39. The circuit in (b) can have both lamps on at the same time.
40. There is always current through R5.
41. See Figure 2-5.
Figure 2-5
42. See Figure 2-5.
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11
43. See Figure 2-6.
Figure 2-6
44. See Figure 2-7.
43. See Figure 2-6.
Figure 2-6
44. See Figure 2-7.
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BASIC PROBLEMS
1. I is directly proportional to V and will change the same percentage as V.
(a) I = 3(1 A) = 3 A
(b) I = 1 A (0.8)(1 A) = 1 A 0.8 A = 0.2 A
(c) I = 1 A + (0.5)(1 A) = 1 A + 0.5 A = 1.5 A
2. (a) When the resistance doubles, the current is halved from 100 mA to 50 mA.
(b) When the resistance is reduced by 30%, the current increases from 100 mA to
I = V/0.7R = 1.429(V/R) = (1.429)(100 mA) 143 mA
(c) When the resistance is quadrupled, the current decreases from 100 mA to 25 mA.
3. Tripling the voltage triples the current from 10 mA to 30 mA, but doubling the resistance
halves the current to 15 mA.
4. (a) I =
1
V5
R
V = 5 A (b) I =
10
V15
R
V = 1.5 A
(c) I =50 V
100
V
R = 0.5 A (d) I =
k15
V30
R
V = 2 mA
(e) I =
M7.4
V250
R
V = 53.2 A
5. (a) I =
k7.2
V9
R
V = 3.33 mA (b) I =
k10
V5.5
R
V = 550 A
(c) I =
k68
V40
R
V = 588 A (d) I =
k2
kV1
R
V = 500 mA
(e) I =
M10
kV66
R
V = 6.60 mA
6. I =
10
V12
R
V = 1.2 A
CHAPTER 3
OHM’S LAW, ENERGY, AND POWER
SECTION 3-1 Ohm’s Law
SECTION 3-2 Application of Ohm’s Law
BASIC PROBLEMS
1. I is directly proportional to V and will change the same percentage as V.
(a) I = 3(1 A) = 3 A
(b) I = 1 A (0.8)(1 A) = 1 A 0.8 A = 0.2 A
(c) I = 1 A + (0.5)(1 A) = 1 A + 0.5 A = 1.5 A
2. (a) When the resistance doubles, the current is halved from 100 mA to 50 mA.
(b) When the resistance is reduced by 30%, the current increases from 100 mA to
I = V/0.7R = 1.429(V/R) = (1.429)(100 mA) 143 mA
(c) When the resistance is quadrupled, the current decreases from 100 mA to 25 mA.
3. Tripling the voltage triples the current from 10 mA to 30 mA, but doubling the resistance
halves the current to 15 mA.
4. (a) I =
1
V5
R
V = 5 A (b) I =
10
V15
R
V = 1.5 A
(c) I =50 V
100
V
R = 0.5 A (d) I =
k15
V30
R
V = 2 mA
(e) I =
M7.4
V250
R
V = 53.2 A
5. (a) I =
k7.2
V9
R
V = 3.33 mA (b) I =
k10
V5.5
R
V = 550 A
(c) I =
k68
V40
R
V = 588 A (d) I =
k2
kV1
R
V = 500 mA
(e) I =
M10
kV66
R
V = 6.60 mA
6. I =
10
V12
R
V = 1.2 A
CHAPTER 3
OHM’S LAW, ENERGY, AND POWER
SECTION 3-1 Ohm’s Law
SECTION 3-2 Application of Ohm’s Law
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13
7. (a) I =
k10
V25
R
V = 2.50 mA (b) I =
M2.2
V5
R
V = 2.27 A
(c) I =
k8.1
V15
R
V = 8.33 mA
8. Orange, violet, yellow, gold, brown 37.4 ± 1%
I =S 12 V
= 37.4 Ω
V
R = 0.321 A
9. I =24 V
37.4 Ω = 0.642 A
0.642 A is greater than 0.5 A, so the fuse will blow.
10. (a) V = IR = (2 A)(18 ) = 36 V (b) V = IR = (5 A)(47 ) = 235 V
(c) V = IR = (2.5 A)(620 ) = 1550 V (d) V = IR = (0.6 A)(47 ) = 28.2 V
(e) V = IR = (0.1 A)(470 ) = 47 V
11. (a) V = IR = (1 mA)(10 ) = 10 mV (b) V = IR = (50 mA)(33 ) = 1.65 V
(c) V = IR = (3 A)(4.7 k) = 14.1 kV (d) V = IR = (1.6 mA)(2.2 k) = 3.52 V
(e) V = IR = (250 A)(1 k) = 250 mV (f) V = IR = (500 mA)(1.5 M) = 750 kV
(g) V = IR = (850 A)(10 M) = 8.5 kV (h) V = IR = (75 A)(47 ) = 3.53 mV
12. V = IR = (3 A)(20 m) = 60 mV
13. (a) V = IR = (3 mA)(27 k) = 81 V (b) V = IR = (5 A)(100 M) = 500 V
(c) V = IR = (2.5 A)(47 ) = 117.5 V
14. (a) R =A2
V10
I
V = 5 (b) R =A45
V90
I
V = 2
(c) R =A5
V50
I
V = 10 (d) R =A10
V5.5
I
V = 0.55
(e) R =A0.5
V150
I
V = 300
15. (a) R =A5
kV10
I
V = 2 k (b) R =mA2
V7
I
V = 3.5 k
(c) R =mA250
V500
I
V = 2 k (d) R =A500
V50
I
V = 100 k
(e) R =mA1
kV1
I
V = 1 M
7. (a) I =
k10
V25
R
V = 2.50 mA (b) I =
M2.2
V5
R
V = 2.27 A
(c) I =
k8.1
V15
R
V = 8.33 mA
8. Orange, violet, yellow, gold, brown 37.4 ± 1%
I =S 12 V
= 37.4 Ω
V
R = 0.321 A
9. I =24 V
37.4 Ω = 0.642 A
0.642 A is greater than 0.5 A, so the fuse will blow.
10. (a) V = IR = (2 A)(18 ) = 36 V (b) V = IR = (5 A)(47 ) = 235 V
(c) V = IR = (2.5 A)(620 ) = 1550 V (d) V = IR = (0.6 A)(47 ) = 28.2 V
(e) V = IR = (0.1 A)(470 ) = 47 V
11. (a) V = IR = (1 mA)(10 ) = 10 mV (b) V = IR = (50 mA)(33 ) = 1.65 V
(c) V = IR = (3 A)(4.7 k) = 14.1 kV (d) V = IR = (1.6 mA)(2.2 k) = 3.52 V
(e) V = IR = (250 A)(1 k) = 250 mV (f) V = IR = (500 mA)(1.5 M) = 750 kV
(g) V = IR = (850 A)(10 M) = 8.5 kV (h) V = IR = (75 A)(47 ) = 3.53 mV
12. V = IR = (3 A)(20 m) = 60 mV
13. (a) V = IR = (3 mA)(27 k) = 81 V (b) V = IR = (5 A)(100 M) = 500 V
(c) V = IR = (2.5 A)(47 ) = 117.5 V
14. (a) R =A2
V10
I
V = 5 (b) R =A45
V90
I
V = 2
(c) R =A5
V50
I
V = 10 (d) R =A10
V5.5
I
V = 0.55
(e) R =A0.5
V150
I
V = 300
15. (a) R =A5
kV10
I
V = 2 k (b) R =mA2
V7
I
V = 3.5 k
(c) R =mA250
V500
I
V = 2 k (d) R =A500
V50
I
V = 100 k
(e) R =mA1
kV1
I
V = 1 M
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14
16. R =mA2
V6
I
V = 3 k
17. (a) R =A2
V8
I
V = 4 (b) R =mA4
V12
I
V = 3 k
(c) R =A150
V30
I
V = 0.2 M = 200 k
18. I =3.2 V
3.9
V
R
= = 0.82 A
19. P =26 J
10 s
W
t = = 2.6 W
20. Since 1 watt = 1 joule, P = 350 J/s = 350 W
21. P =h5
J7500
t
Ws18,000
J7500
s3600
h1
h5
J7500
= 0.417 J/s = 417 mW
22. (a) 1000 W = 1 103 W = 1 kW (b) 3750 W = 3.750 103 W = 3.75 kW
(c) 160 W = 0.160 103 W = 0.160 kW (d) 50,000 W = 50 103 W = 50 kW
23. (a) 1,000,000 W = 1 106 W = 1 MW (b) 3 106 W = 3 MW
(c) 15 107 W = 150 106 W = 150 MW (d) 8700 kW = 8.7 106 W = 8.7 MW
24. (a) 1 W = 1000 103 W = 1000 mW (b) 0.4 W = 400 103 W = 400 mW
(c) 0.002 W = 2 103 W = 2 mW (d) 0.0125 W = 12.5 103 W = 12.5 mW
25. (a) 2 W = 2,000,000 W (b) 0.0005 W = 500 W
(c) 0.25 mW = 250 W (d) 0.00667 mW = 6.67 W
26. (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
SECTION 3-3 Energy and Power
16. R =mA2
V6
I
V = 3 k
17. (a) R =A2
V8
I
V = 4 (b) R =mA4
V12
I
V = 3 k
(c) R =A150
V30
I
V = 0.2 M = 200 k
18. I =3.2 V
3.9
V
R
= = 0.82 A
19. P =26 J
10 s
W
t = = 2.6 W
20. Since 1 watt = 1 joule, P = 350 J/s = 350 W
21. P =h5
J7500
t
Ws18,000
J7500
s3600
h1
h5
J7500
= 0.417 J/s = 417 mW
22. (a) 1000 W = 1 103 W = 1 kW (b) 3750 W = 3.750 103 W = 3.75 kW
(c) 160 W = 0.160 103 W = 0.160 kW (d) 50,000 W = 50 103 W = 50 kW
23. (a) 1,000,000 W = 1 106 W = 1 MW (b) 3 106 W = 3 MW
(c) 15 107 W = 150 106 W = 150 MW (d) 8700 kW = 8.7 106 W = 8.7 MW
24. (a) 1 W = 1000 103 W = 1000 mW (b) 0.4 W = 400 103 W = 400 mW
(c) 0.002 W = 2 103 W = 2 mW (d) 0.0125 W = 12.5 103 W = 12.5 mW
25. (a) 2 W = 2,000,000 W (b) 0.0005 W = 500 W
(c) 0.25 mW = 250 W (d) 0.00667 mW = 6.67 W
26. (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
SECTION 3-3 Energy and Power
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15
27.W
P t
in wattsW
V Q
Q
I t
W
P VI t
So, (1 V)(1 A) = 1 W
28.1 J
1 s
W
P t
= 1 W
1 kW = 1000 W =1000 J
1 s
1 kW-second = 1000 J
1 kWh = 3600 1000 J
1 kWh = 3.6 106 J
29. P = VI = (5.5 V)(3 mA) = 16.5 mW
30. P = VI = (115 V)(3 A) = 345 W
31. P = I2R = (500 mA)2(4.7 k) = 1.18 kW
32. P = I2R = (5.0 A)2(20 10-3 ) = 500 mW
33. P =
620
)V60( 22
R
V = 5.81 W
34. P =
56
)V5.1( 22
R
V = 0.0402 W = 40.2 mW
35. P = I2R
R =22 A)(2
W100
I
P = 25
SECTION 3-4 Power in an Electric Circuit
27.W
P t
in wattsW
V Q
Q
I t
W
P VI t
So, (1 V)(1 A) = 1 W
28.1 J
1 s
W
P t
= 1 W
1 kW = 1000 W =1000 J
1 s
1 kW-second = 1000 J
1 kWh = 3600 1000 J
1 kWh = 3.6 106 J
29. P = VI = (5.5 V)(3 mA) = 16.5 mW
30. P = VI = (115 V)(3 A) = 345 W
31. P = I2R = (500 mA)2(4.7 k) = 1.18 kW
32. P = I2R = (5.0 A)2(20 10-3 ) = 500 mW
33. P =
620
)V60( 22
R
V = 5.81 W
34. P =
56
)V5.1( 22
R
V = 0.0402 W = 40.2 mW
35. P = I2R
R =22 A)(2
W100
I
P = 25
SECTION 3-4 Power in an Electric Circuit
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16
36. 5 106 watts for 1 minute = 5 103 kWminhrmin/160
kWmin105 3
= 83.3 kWh
37.6700 W/s
(1000 W/kW)(3600 s/h) = 0.00186 kWh
38. (50 W)(12 h) = 600 Wh
50 W = 0.05 kW
(0.05 kW)(12 h) = 0.6 kWh
39.1.25 V
10L
V
I R
= 0.125 A
P = VI = (1.25 V)(0.125 A) = 0.156 W = 156 mW
40. P =W
t
156 mW =156 mJ
1 stot (156 mJ/s)(90 h)(3600 s/h)W
= 50,544 J
41. P = I2R = (10 mA)2(6.8 k) = 0.68 W
Use the next highest standard power rating of 1 W.
42. If the 8 W resistor is used, it will be operating in a marginal condition.
To allow for a safety margin of 20%, use a 12 W resistor.
43. (a) + at top, at bottom of resistor (b) + at bottom, at top of resistor
(c) + on right, on left of resistor
44. VOUT =)50)(W1( LLRP = 7.07 V
45. Ampere-hour rating = (1.5 A)(24 h) = 36 Ah
SECTION 3-5 The Power Rating of Resistors
SECTION 3-6 Energy Conversion and Voltage Drop in a Resistance
SECTION 3-7 Power Supplies and Batteries
36. 5 106 watts for 1 minute = 5 103 kWminhrmin/160
kWmin105 3
= 83.3 kWh
37.6700 W/s
(1000 W/kW)(3600 s/h) = 0.00186 kWh
38. (50 W)(12 h) = 600 Wh
50 W = 0.05 kW
(0.05 kW)(12 h) = 0.6 kWh
39.1.25 V
10L
V
I R
= 0.125 A
P = VI = (1.25 V)(0.125 A) = 0.156 W = 156 mW
40. P =W
t
156 mW =156 mJ
1 stot (156 mJ/s)(90 h)(3600 s/h)W
= 50,544 J
41. P = I2R = (10 mA)2(6.8 k) = 0.68 W
Use the next highest standard power rating of 1 W.
42. If the 8 W resistor is used, it will be operating in a marginal condition.
To allow for a safety margin of 20%, use a 12 W resistor.
43. (a) + at top, at bottom of resistor (b) + at bottom, at top of resistor
(c) + on right, on left of resistor
44. VOUT =)50)(W1( LLRP = 7.07 V
45. Ampere-hour rating = (1.5 A)(24 h) = 36 Ah
SECTION 3-5 The Power Rating of Resistors
SECTION 3-6 Energy Conversion and Voltage Drop in a Resistance
SECTION 3-7 Power Supplies and Batteries
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17
46. I =h10
Ah80 = 8 A
47. I =h48
mAh650 = 13.5 mA
48. PLOST = PIN POUT = 500 mW 400 mW = 100 mW
% efficiency =%100
mW500
mW400
%100
IN
OUT
P
P = 80%
49. POUT = (efficiency)PIN = (0.85)(5 W) = 4.25 W
50. The 4th bulb from the left is open.
51. If should take five (maximum) resistance measurements.
ADVANCED PROBLEMS
52. Assume that the total consumption of the power supply is the input power plus the power lost.
POUT = 2 W
% efficiency =%100
IN
OUT
P
P
PIN =%100
%60
W2
%100
efficiency%
OUT
P = 3.33 W
The power supply itself uses
PIN POUT = 3.33 W 2 W = 1.33 W
Energy = W = Pt = (1.33 W)(24 h) = 31.9 Wh 0.032 kWh
53. Rf =A0.8
V120
I
V = 150
SECTION 3-8 Introduction to Troubleshooting
Troubleshootinhg
46. I =h10
Ah80 = 8 A
47. I =h48
mAh650 = 13.5 mA
48. PLOST = PIN POUT = 500 mW 400 mW = 100 mW
% efficiency =%100
mW500
mW400
%100
IN
OUT
P
P = 80%
49. POUT = (efficiency)PIN = (0.85)(5 W) = 4.25 W
50. The 4th bulb from the left is open.
51. If should take five (maximum) resistance measurements.
ADVANCED PROBLEMS
52. Assume that the total consumption of the power supply is the input power plus the power lost.
POUT = 2 W
% efficiency =%100
IN
OUT
P
P
PIN =%100
%60
W2
%100
efficiency%
OUT
P = 3.33 W
The power supply itself uses
PIN POUT = 3.33 W 2 W = 1.33 W
Energy = W = Pt = (1.33 W)(24 h) = 31.9 Wh 0.032 kWh
53. Rf =A0.8
V120
I
V = 150
SECTION 3-8 Introduction to Troubleshooting
Troubleshootinhg
Loading page 19...
18
54. Measure the current with an ammeter connected as shown in Figure 3-1. Then calculate the
unknown resistance with the formula, R = 12 V/I.
Figure 3-1
55. Calculate I for each value of V:
I1 =100
V0 = 0 A I2 =100
V10 = 100 mA
I3 =100
V20 = 200 mA I4 =100
V30 = 300 mA
I5 =100
V40 = 400 mA I6 =100
V50 = 500 mA
I7 =100
V60 = 600 mA I8 =100
V70 = 700 mA
I9 =100
V80 = 800 mA I10 =100
V90 = 900 mA
I11 =100
V100 = 1 A
Figure 3-2
The graph is a straight line as shown in
Figure 3-2. This indicates a linear
relationship between I and V.
54. Measure the current with an ammeter connected as shown in Figure 3-1. Then calculate the
unknown resistance with the formula, R = 12 V/I.
Figure 3-1
55. Calculate I for each value of V:
I1 =100
V0 = 0 A I2 =100
V10 = 100 mA
I3 =100
V20 = 200 mA I4 =100
V30 = 300 mA
I5 =100
V40 = 400 mA I6 =100
V50 = 500 mA
I7 =100
V60 = 600 mA I8 =100
V70 = 700 mA
I9 =100
V80 = 800 mA I10 =100
V90 = 900 mA
I11 =100
V100 = 1 A
Figure 3-2
The graph is a straight line as shown in
Figure 3-2. This indicates a linear
relationship between I and V.
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19
56. R =mA5
V1S
I
V = 200
(a) I =
200
V1.5S
R
V = 7.5 mA (b) I =
200
V2S
R
V = 10 mA
(c) I =
200
V3S
R
V = 15 mA (d) I =
200
V4S
R
V = 20 mA
(e) I =
200
V10S
R
V = 50 mA
57. R1 =A2
V1
I
V = 0.5 R2 =A1
V1
I
V = 1 R3 =A0.5
V1
I
V = 2
58.mA50
V10
mA30
2
V
V2 =mA50
mA)V)(30(10 = 6 V new value
The voltage decreased by 4 V, from 10 V to 6 V.
59. The current increase is 50%, so the voltage increase must be the same; that is, the voltage must
be increased by (0.5)(20 V) = 10 V.
The new value of voltage is V2 = 20 V + (0.5)(20 V) = 20 V + 10 V = 30 V
60. Wire resistance: RW =CM1624.3
ft)/ft)(24CM(10.4 = 0.154
(a)
154.100
V6
WRR
V
I = 59.9 mA
(b) VR = (59.9 mA)(100 ) = 5.99 V
(c)WRV = 6 V 5.99 V = 0.01 V
For one length of wire, V =2
V01.0 = 0.005 V
61. 300 W = 0.3 kW
30 days = (30 days)(24 h/day) = 720 h
Energy = (0.3 kW)(720 h) = 216 kWh
62.days31
kWh1500 = 48.39 kWh/day
P =h/day24
kWh/day48.39 = 2.02 kW
56. R =mA5
V1S
I
V = 200
(a) I =
200
V1.5S
R
V = 7.5 mA (b) I =
200
V2S
R
V = 10 mA
(c) I =
200
V3S
R
V = 15 mA (d) I =
200
V4S
R
V = 20 mA
(e) I =
200
V10S
R
V = 50 mA
57. R1 =A2
V1
I
V = 0.5 R2 =A1
V1
I
V = 1 R3 =A0.5
V1
I
V = 2
58.mA50
V10
mA30
2
V
V2 =mA50
mA)V)(30(10 = 6 V new value
The voltage decreased by 4 V, from 10 V to 6 V.
59. The current increase is 50%, so the voltage increase must be the same; that is, the voltage must
be increased by (0.5)(20 V) = 10 V.
The new value of voltage is V2 = 20 V + (0.5)(20 V) = 20 V + 10 V = 30 V
60. Wire resistance: RW =CM1624.3
ft)/ft)(24CM(10.4 = 0.154
(a)
154.100
V6
WRR
V
I = 59.9 mA
(b) VR = (59.9 mA)(100 ) = 5.99 V
(c)WRV = 6 V 5.99 V = 0.01 V
For one length of wire, V =2
V01.0 = 0.005 V
61. 300 W = 0.3 kW
30 days = (30 days)(24 h/day) = 720 h
Energy = (0.3 kW)(720 h) = 216 kWh
62.days31
kWh1500 = 48.39 kWh/day
P =h/day24
kWh/day48.39 = 2.02 kW
Loading page 21...
20
63. The minimum power rating you should use is 12 W so that the power dissipation does not
exceed the rating.
64. (a) P =
10
V)(12 22
R
V = 14.4 W
(b) W = Pt = (14.4 W)(2 min)(1/60 h/min) = 0.48 Wh
(c) Neither, the power is the same because it is not time dependent.
65. VR(max) = 120 V 100 V = 20 V
Imax =
8
V20
min
(max)
R
VR = 2.5 A
A fuse with a rating of less than 2.5 A must be used. A 2 A fuse is recommended.
66.0.5 W
0.030
P
I R
4.08 A
67. Power will increase by four times.
66. The materials required for the Load Test Box are as follows:
Item Component Qty
1 Resistor: 5.0 , 10 W 1
2 Resistor: 16 , 5 W 1
3 Resistor: 100 , 2.0 W 1
4 Resistor: 150 , 3.0 W 1
5 1 pole, 4 position rotary switch 1
6 Knob 1
7 Enclosure (4” x 4” 2” Al) 1
8 Banana plug terminals 2
9 Fuse (1.5 A) and fuse holder 1
10 PC board (etched with pattern) 1
11 Screws, washers, nuts 4
12 Standoffs 4
63. The minimum power rating you should use is 12 W so that the power dissipation does not
exceed the rating.
64. (a) P =
10
V)(12 22
R
V = 14.4 W
(b) W = Pt = (14.4 W)(2 min)(1/60 h/min) = 0.48 Wh
(c) Neither, the power is the same because it is not time dependent.
65. VR(max) = 120 V 100 V = 20 V
Imax =
8
V20
min
(max)
R
VR = 2.5 A
A fuse with a rating of less than 2.5 A must be used. A 2 A fuse is recommended.
66.0.5 W
0.030
P
I R
4.08 A
67. Power will increase by four times.
66. The materials required for the Load Test Box are as follows:
Item Component Qty
1 Resistor: 5.0 , 10 W 1
2 Resistor: 16 , 5 W 1
3 Resistor: 100 , 2.0 W 1
4 Resistor: 150 , 3.0 W 1
5 1 pole, 4 position rotary switch 1
6 Knob 1
7 Enclosure (4” x 4” 2” Al) 1
8 Banana plug terminals 2
9 Fuse (1.5 A) and fuse holder 1
10 PC board (etched with pattern) 1
11 Screws, washers, nuts 4
12 Standoffs 4
Loading page 22...
21
69. See Figure 3-3.
Figure 3-3
70. R is open.
71. No fault
72. R1 is shorted.
73. Lamp 4 is shorted.
74. Lamp 6 is open.
Multisim Troubleshooting Problems
Troubleshootinhg
69. See Figure 3-3.
Figure 3-3
70. R is open.
71. No fault
72. R1 is shorted.
73. Lamp 4 is shorted.
74. Lamp 6 is open.
Multisim Troubleshooting Problems
Troubleshootinhg
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22
BASIC PROBLEMS
1. See Figure 4-1.
Figure 4-1
2. The groups of series resistors are
R1, R2, R3, R9 R4; R13, R7, R14, R16; R6, R8, R12; R10, R11, R15, R5
See Figure 4-2.
Figure 4-2
3.1 8 13 7 14 16R R R R R
= 68 k + 33 k + 47 k + 22 k
= 170 k
4.2 3 12 8 6 10 Ω+ 18 Ω + 22 ΩR R R R = 50
CHAPTER 4
SERIES CIRCUITS
SECTION 4-1 Resistors in Series
BASIC PROBLEMS
1. See Figure 4-1.
Figure 4-1
2. The groups of series resistors are
R1, R2, R3, R9 R4; R13, R7, R14, R16; R6, R8, R12; R10, R11, R15, R5
See Figure 4-2.
Figure 4-2
3.1 8 13 7 14 16R R R R R
= 68 k + 33 k + 47 k + 22 k
= 170 k
4.2 3 12 8 6 10 Ω+ 18 Ω + 22 ΩR R R R = 50
CHAPTER 4
SERIES CIRCUITS
SECTION 4-1 Resistors in Series
Loading page 24...
23
5. RT = 82 + 56 = 138
6. (a) RT = 560 + 1.0 k = 1560
(b) RT = 47 + 33 = 80
(c) RT = 1.5 k + 2.2 k + 10 k = 13.7 k
(d) RT = 1.0 k + 1.8 k + 100 k + 1.0 M = 1,102,800 (round to 1.10 M)
7. (a) RT = 1.0 k + 4.7 k + 2.2 k = 7.9 k
(b) RT = 10 + 10 + 12 + 1.0 = 33
(c) RT = 1.0 M + 560 k + 1.0 M + 680 k + 10 M = 13.24 M
See Figure 4-3.
Figure 4-3
8. RT = 12(5.6 k) = 67.2 k
9. RT = 6(47 ) + 8(100 ) + 2(22 ) = 282 + 800 + 44 = 1126
10. RT = R1 + R2 + R3 + R4 + R5
R5 = RT (R1 + R2 + R3 + R4)
= 20 k (4.7 k + 1.0 k + 2.2 k + 3.9 k)
= 20 k 11.8 k
= 8.2 k
11. (a) R18 = R13 + R7 + R14 + R16
= 68 k + 33 k + 47 k + 22 k = 170 k
(b) R23 = R12 + R8 + R6
= 10 + 18 + 22 = 50
(c) R47 = R10 + R11 + R15 + R5
= 2.2 k + 8.2 k + 1.0 k + 1.0 k = 12.4 k
(d) R56 = R1 + R2 + R3 + R9 + R4
= 220 + 330 + 390 + 470 + 560 = 1.97 k
12. RT = R18 + R23 + R47 + R56
= 170 k + 50 + 12.4 k + 1.97 k = 184.42 k
SECTION 4-2 Total Series Resistance
5. RT = 82 + 56 = 138
6. (a) RT = 560 + 1.0 k = 1560
(b) RT = 47 + 33 = 80
(c) RT = 1.5 k + 2.2 k + 10 k = 13.7 k
(d) RT = 1.0 k + 1.8 k + 100 k + 1.0 M = 1,102,800 (round to 1.10 M)
7. (a) RT = 1.0 k + 4.7 k + 2.2 k = 7.9 k
(b) RT = 10 + 10 + 12 + 1.0 = 33
(c) RT = 1.0 M + 560 k + 1.0 M + 680 k + 10 M = 13.24 M
See Figure 4-3.
Figure 4-3
8. RT = 12(5.6 k) = 67.2 k
9. RT = 6(47 ) + 8(100 ) + 2(22 ) = 282 + 800 + 44 = 1126
10. RT = R1 + R2 + R3 + R4 + R5
R5 = RT (R1 + R2 + R3 + R4)
= 20 k (4.7 k + 1.0 k + 2.2 k + 3.9 k)
= 20 k 11.8 k
= 8.2 k
11. (a) R18 = R13 + R7 + R14 + R16
= 68 k + 33 k + 47 k + 22 k = 170 k
(b) R23 = R12 + R8 + R6
= 10 + 18 + 22 = 50
(c) R47 = R10 + R11 + R15 + R5
= 2.2 k + 8.2 k + 1.0 k + 1.0 k = 12.4 k
(d) R56 = R1 + R2 + R3 + R9 + R4
= 220 + 330 + 390 + 470 + 560 = 1.97 k
12. RT = R18 + R23 + R47 + R56
= 170 k + 50 + 12.4 k + 1.97 k = 184.42 k
SECTION 4-2 Total Series Resistance
Loading page 25...
24
13. I =
120
V12
T
S
R
V = 0.1 A
14. I = 5 mA at all points in the circuit.
15. (a) RT = 2.2 k + 5.6 k + 1.0 k = 8.8 k
I =
k8.8
V5.5
TR
V = 625 A
(b) RT = 1.0 M + 2.2 M + 560 k = 3.76 M
I =M76.3
V16 = 4.26 A
The ammeters are connected in series. See Figure 4-4.
Figure 4-4
16. (a) V1 =V5.5
k8.8
k.22
S
T
1
V
R
R = 1.375 V
V2 =V5.5
k8.8
k6.5
S
T
2
V
R
R = 3.5 V
V3 =V5.5
k8.8
k.01
S
T
3
V
R
R = 625 mV
(b) V1 =V16
M76.3
M0.1
S
T
1
V
R
R = 4.26 V
SECTION 4-4 Application of Ohm’s Law
SECTION 4-3 Current in a Series Circuit
13. I =
120
V12
T
S
R
V = 0.1 A
14. I = 5 mA at all points in the circuit.
15. (a) RT = 2.2 k + 5.6 k + 1.0 k = 8.8 k
I =
k8.8
V5.5
TR
V = 625 A
(b) RT = 1.0 M + 2.2 M + 560 k = 3.76 M
I =M76.3
V16 = 4.26 A
The ammeters are connected in series. See Figure 4-4.
Figure 4-4
16. (a) V1 =V5.5
k8.8
k.22
S
T
1
V
R
R = 1.375 V
V2 =V5.5
k8.8
k6.5
S
T
2
V
R
R = 3.5 V
V3 =V5.5
k8.8
k.01
S
T
3
V
R
R = 625 mV
(b) V1 =V16
M76.3
M0.1
S
T
1
V
R
R = 4.26 V
SECTION 4-4 Application of Ohm’s Law
SECTION 4-3 Current in a Series Circuit
Loading page 26...
25
V2 =V16
M76.3
M2.2
S
T
2
V
R
R = 9.36 V
V3 =V16
M3.76
k056
S
T
3
V
R
R = 2.38 V
17. (a) RT = 3(470 ) = 1410
I =48 V
1410 = 34.0 mA
(b)(34.0 mA)(470 Ω)RV IR = 16 V
(c)2 2
min (34.0 mA) 470 ΩP I R = 0.543 W
18. RT =mA1
V5
T
S
I
V = 5 k
Reach =4
k5 = 1.25 k
19. See Figure 4-5.
Figure 4-5
20. The total voltage is 6 V + 6 V + 6 V 6 V = 12 V
21. VS = 5.5 V + 8.2 V + 12.3 V = 26 V
22. VS = V1 + V2 + V3 + V4 + V5
V5 = VS (V1 + V2 + V3 + V4) = 20 V (1.5 V + 5.5 V + 3 V + 6 V) = 20 V 16 V = 4 V
23. (a) By Kirchhoff’s voltage law:
15 V = 2 V + V2 + 3.2 V + 1 V + 1.5 V + 0.5 V
SECTION 4-5 Voltage Sources in Series
SECTION 4-6 Kirchhoff’s Voltage Law
V2 =V16
M76.3
M2.2
S
T
2
V
R
R = 9.36 V
V3 =V16
M3.76
k056
S
T
3
V
R
R = 2.38 V
17. (a) RT = 3(470 ) = 1410
I =48 V
1410 = 34.0 mA
(b)(34.0 mA)(470 Ω)RV IR = 16 V
(c)2 2
min (34.0 mA) 470 ΩP I R = 0.543 W
18. RT =mA1
V5
T
S
I
V = 5 k
Reach =4
k5 = 1.25 k
19. See Figure 4-5.
Figure 4-5
20. The total voltage is 6 V + 6 V + 6 V 6 V = 12 V
21. VS = 5.5 V + 8.2 V + 12.3 V = 26 V
22. VS = V1 + V2 + V3 + V4 + V5
V5 = VS (V1 + V2 + V3 + V4) = 20 V (1.5 V + 5.5 V + 3 V + 6 V) = 20 V 16 V = 4 V
23. (a) By Kirchhoff’s voltage law:
15 V = 2 V + V2 + 3.2 V + 1 V + 1.5 V + 0.5 V
SECTION 4-5 Voltage Sources in Series
SECTION 4-6 Kirchhoff’s Voltage Law
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26
V2 = 15 V (2 V + 3.2 V + 1 V + 1.5 V + 0.5 V) = 15 V 8.2 V = 6.8 V
See Figure 4-6(a).
(b) VR = 8 V; V2R = 16 V; V3R = 24 V; V4R = 32 V
See Figure 4-6(b).
24.100
500
22
= 4.4%
25. (a) VAB =V12
147
47
= 3.84 V
(b) VAB =V8
k6.5
k5.5
V8
k3.3k2.2k0.1
k3.3k2.2
= 6.77 V
26.SAV V = 15 V2 3 S
1 2 3
13.3 kΩ 15 V
18.9 k
B
R R
V V
R R R
= 10.6 V3 S
1 2 3
3.3 kΩ 15 V
18.9 k
C
R
V V
R R R
= 2.62 V
27.3
min S
1 2 3
680 Ω 12 V
2150
R
V V
R R R
= 3.80 V2 3
max S
1 2 3
1680 Ω 12 V
2150
R R
V V
R R R
= 9.38 V
SECTION 4-7 Voltage Dividers
Figure 4-6
V2 = 15 V (2 V + 3.2 V + 1 V + 1.5 V + 0.5 V) = 15 V 8.2 V = 6.8 V
See Figure 4-6(a).
(b) VR = 8 V; V2R = 16 V; V3R = 24 V; V4R = 32 V
See Figure 4-6(b).
24.100
500
22
= 4.4%
25. (a) VAB =V12
147
47
= 3.84 V
(b) VAB =V8
k6.5
k5.5
V8
k3.3k2.2k0.1
k3.3k2.2
= 6.77 V
26.SAV V = 15 V2 3 S
1 2 3
13.3 kΩ 15 V
18.9 k
B
R R
V V
R R R
= 10.6 V3 S
1 2 3
3.3 kΩ 15 V
18.9 k
C
R
V V
R R R
= 2.62 V
27.3
min S
1 2 3
680 Ω 12 V
2150
R
V V
R R R
= 3.80 V2 3
max S
1 2 3
1680 Ω 12 V
2150
R R
V V
R R R
= 9.38 V
SECTION 4-7 Voltage Dividers
Figure 4-6
Loading page 28...
27
28. RT = R + 2R + 3R + 4R + 5R = 15R
VR =V9
15
R
R = 0.6 V VR =V9
15
2
R
R = 1.2 V VR =V9
15
3
R
R = 1.8 V
VR =V9
15
4
R
R = 2.4 V VR =V9
15
5
R
R = 3.0 V
29. V5.6k = 10 V (by measurement); I =k6.5
V10 = 1.79 mA;
V1k = (1.79 mA)(1 k) = 1.79 V; V560 (1.79 mA)(560 ) = 1 V;
V10k = (1.79 mA)(10 k) = 17.9 V
30. PT = 5(50 mW) = 250 mW
31. RT = 5.6 k + 1 k + 560 + 10 k = 17.16 k
P = I2RT = (1.79 mA)2(17.16 k) = 0.055 W = 55 mW
32. Voltage from point A to ground (G): VAG = 10 V
Resistance between A and G: RAG = 5.6 k + 5.6 k + 1.0 k + 1.0 k = 13.2 k
Resistance between B and G: RBG = 5.6 k + 1.0 k + 1.0 k = 7.6 k
Resistance between C and G: RCG = 1.0 k + 1.0 k = 2 k
VBG =V10
k2.13
k6.7
V10
AG
BG
R
R = 5.76 V
VCG =V10
k2.13
k2
V10
AG
CG
R
R = 1.52 V
VDG =V10
k2.13
k0.1
V10
AG
DG
R
R = 0.758 V
33. Measure the voltage at point A with respect to ground and the voltage at point B with respect to
ground. The difference of these two voltages is VR2.
VR2 = VA
VB
SECTION 4-8 Power in Series Circuits
SECTION 4-9 Voltage Measurements
28. RT = R + 2R + 3R + 4R + 5R = 15R
VR =V9
15
R
R = 0.6 V VR =V9
15
2
R
R = 1.2 V VR =V9
15
3
R
R = 1.8 V
VR =V9
15
4
R
R = 2.4 V VR =V9
15
5
R
R = 3.0 V
29. V5.6k = 10 V (by measurement); I =k6.5
V10 = 1.79 mA;
V1k = (1.79 mA)(1 k) = 1.79 V; V560 (1.79 mA)(560 ) = 1 V;
V10k = (1.79 mA)(10 k) = 17.9 V
30. PT = 5(50 mW) = 250 mW
31. RT = 5.6 k + 1 k + 560 + 10 k = 17.16 k
P = I2RT = (1.79 mA)2(17.16 k) = 0.055 W = 55 mW
32. Voltage from point A to ground (G): VAG = 10 V
Resistance between A and G: RAG = 5.6 k + 5.6 k + 1.0 k + 1.0 k = 13.2 k
Resistance between B and G: RBG = 5.6 k + 1.0 k + 1.0 k = 7.6 k
Resistance between C and G: RCG = 1.0 k + 1.0 k = 2 k
VBG =V10
k2.13
k6.7
V10
AG
BG
R
R = 5.76 V
VCG =V10
k2.13
k2
V10
AG
CG
R
R = 1.52 V
VDG =V10
k2.13
k0.1
V10
AG
DG
R
R = 0.758 V
33. Measure the voltage at point A with respect to ground and the voltage at point B with respect to
ground. The difference of these two voltages is VR2.
VR2 = VA
VB
SECTION 4-8 Power in Series Circuits
SECTION 4-9 Voltage Measurements
Loading page 29...
28
34. RT = R1 + R2 + R3 + R4 + R5
= 560 k + 560 k + 100 k + 1.0 M + 100 k = 2.32 M
VT = 15 V
VA =V15
M32.2
M76.1
T
T
V
R
RAG = 11.4 V
VB =V15
M32.2
M2.1
T
T
V
R
RBG = 7.76 V
VC =V15
M32.2
M1.1
T
T
V
R
RCG = 7.11 V
VD =V15
M32.2
k100
T
T
V
R
RDG = 647 mV
35.11.38 V 7.11 VAC A CV V V= - = - = 4.27 V
36.7.11 V 11.38 VCA C AV V V= - = - = 4.27 V
37. (a) Zero current indicates an open. R4 is open since all the voltage is dropped across it.
(b)
300
V10
321
S
RRR
V = 33.3 mA
R4 and R5 have no effect on the current. There is a short from A to B.
38. RT = 10 k + 8.2 k + 12 k + 2.2 k + 5.6 k = 38 k
The meter reads about 28 k. It should read 38 k. The 10 k resistor is shorted.
ADVANCED PROBLEMS
39. V1 = IR1 = (10 mA)(680 ) = 6.8 V
V2 = IR2 = (10 mA)(1.0 k) = 10 V
V4 = IR4 = (10 mA)(270 ) = 2.7 V
V5 = IR5 = (10 mA)(270 ) = 2.7 V
V3 = VS (V1 + V2 + V4 + V5)
V3 = 30 V (6.8 V + 10 V + 2.7 V + 2.7 V) = 30 V 22.2 V = 7.8 V
R3 =mA10
V7.83
I
V = 0.78 k = 780
40. RT = 3(5.6 k) + 1.0 k + 2(100 ) = 18 k
Three 5.6 k resistors, one 1 k resistor, and two 100 resistors
41. VA = 10 V, RT = 22 k + 10 k + 47 k + 12 k + 5.6 k = 96.6 k
SECTION 4-10 Troubleshooting
34. RT = R1 + R2 + R3 + R4 + R5
= 560 k + 560 k + 100 k + 1.0 M + 100 k = 2.32 M
VT = 15 V
VA =V15
M32.2
M76.1
T
T
V
R
RAG = 11.4 V
VB =V15
M32.2
M2.1
T
T
V
R
RBG = 7.76 V
VC =V15
M32.2
M1.1
T
T
V
R
RCG = 7.11 V
VD =V15
M32.2
k100
T
T
V
R
RDG = 647 mV
35.11.38 V 7.11 VAC A CV V V= - = - = 4.27 V
36.7.11 V 11.38 VCA C AV V V= - = - = 4.27 V
37. (a) Zero current indicates an open. R4 is open since all the voltage is dropped across it.
(b)
300
V10
321
S
RRR
V = 33.3 mA
R4 and R5 have no effect on the current. There is a short from A to B.
38. RT = 10 k + 8.2 k + 12 k + 2.2 k + 5.6 k = 38 k
The meter reads about 28 k. It should read 38 k. The 10 k resistor is shorted.
ADVANCED PROBLEMS
39. V1 = IR1 = (10 mA)(680 ) = 6.8 V
V2 = IR2 = (10 mA)(1.0 k) = 10 V
V4 = IR4 = (10 mA)(270 ) = 2.7 V
V5 = IR5 = (10 mA)(270 ) = 2.7 V
V3 = VS (V1 + V2 + V4 + V5)
V3 = 30 V (6.8 V + 10 V + 2.7 V + 2.7 V) = 30 V 22.2 V = 7.8 V
R3 =mA10
V7.83
I
V = 0.78 k = 780
40. RT = 3(5.6 k) + 1.0 k + 2(100 ) = 18 k
Three 5.6 k resistors, one 1 k resistor, and two 100 resistors
41. VA = 10 V, RT = 22 k + 10 k + 47 k + 12 k + 5.6 k = 96.6 k
SECTION 4-10 Troubleshooting
Loading page 30...
29
VB = VA V22k = 10 V V10
k6.96
k22
= 10 V 2.28 V = 7.72 V
VC = VB V10k = 7.72 V V10
k6.96
k10
= 7.72 V 1.04 V = 6.68 V
VD = VC V47k = 6.68 V V10
k6.96
k47
= 6.68 V 4.87 V = 1.81 V
VE = VD V12k = 1.81 V V10
k6.96
k12
= 1.81 V 1.24 V = 0.57 V
VF = 0 V
42. V2 = IR2 = (20 mA)(100 ) = 2 V
R5 =mA20
V6.6S
I
V = 330
R6 =22
6
mA)(20
mW112
I
P = 280
V6 = IR6 = (20 mA)(280 ) = 5.6 V
V1 = VS (20 V + V6) = 30 V (20 V + 5.6 V) = 4.4 V
R1 =mA20
V4.41
I
V = 220
V3 + V4 = 20 V V2 V5 = 20 V 2 V 6.6 V = 11.4 V
V3 = V4 =2
V11.4 = 5.7 V R3 = R4 =mA20
V5.73
I
V = 285
43. VS = IRT = (250 mA)(1.5 k) = 375 V
Inew = 250 mA 0.25(250 mA) = 250 mA 62.5 mA = 188 mA
Rnew =mA188
V375
new
S
I
V 2000
500 must be added to the existing 1500 to reduce I by 25%.
44. P = I2R
Imax =
120
W5.0
R
P = 0.0645 A = 64.5 mA
Since all resistors in series have the same current, use the largest R to determine the maximum
current allowable because the largest R has the greatest power.
Thus, the 120 resistor burns out first.
45. (a) PT =W
2
1
W
4
1
W
8
1 = 0.125 W + 0.25 W + 0.5 W = 0.875 W
I =
2400
W875.0
T
T
R
P = 19.1 mA
(b) VS = ITRT = (19.1 mA)(2400 ) = 45.8 V
(c) R =2
I
P
VB = VA V22k = 10 V V10
k6.96
k22
= 10 V 2.28 V = 7.72 V
VC = VB V10k = 7.72 V V10
k6.96
k10
= 7.72 V 1.04 V = 6.68 V
VD = VC V47k = 6.68 V V10
k6.96
k47
= 6.68 V 4.87 V = 1.81 V
VE = VD V12k = 1.81 V V10
k6.96
k12
= 1.81 V 1.24 V = 0.57 V
VF = 0 V
42. V2 = IR2 = (20 mA)(100 ) = 2 V
R5 =mA20
V6.6S
I
V = 330
R6 =22
6
mA)(20
mW112
I
P = 280
V6 = IR6 = (20 mA)(280 ) = 5.6 V
V1 = VS (20 V + V6) = 30 V (20 V + 5.6 V) = 4.4 V
R1 =mA20
V4.41
I
V = 220
V3 + V4 = 20 V V2 V5 = 20 V 2 V 6.6 V = 11.4 V
V3 = V4 =2
V11.4 = 5.7 V R3 = R4 =mA20
V5.73
I
V = 285
43. VS = IRT = (250 mA)(1.5 k) = 375 V
Inew = 250 mA 0.25(250 mA) = 250 mA 62.5 mA = 188 mA
Rnew =mA188
V375
new
S
I
V 2000
500 must be added to the existing 1500 to reduce I by 25%.
44. P = I2R
Imax =
120
W5.0
R
P = 0.0645 A = 64.5 mA
Since all resistors in series have the same current, use the largest R to determine the maximum
current allowable because the largest R has the greatest power.
Thus, the 120 resistor burns out first.
45. (a) PT =W
2
1
W
4
1
W
8
1 = 0.125 W + 0.25 W + 0.5 W = 0.875 W
I =
2400
W875.0
T
T
R
P = 19.1 mA
(b) VS = ITRT = (19.1 mA)(2400 ) = 45.8 V
(c) R =2
I
P
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Subject
Electrical Engineering & Electronics