dc answers
TRANSCRIPT
Chapter 1
Chapter 1 Quantities and Units
Section 1-2 Scientific Notation
1. (a) 3000 = 3 103 (b) 75,000 = 7.5 104 (c) 2,000,000 = 2 106
2. (a) 500
1 = 0.002 = 2 103
(b) 2000
1 = 0.0005 = 5 104
(c) 000,000,5
1 = 0.0000002 = 2 107
3. (a) 8400 = 8.4 103 (b) 99,000 = 9.9 104 (c) 0.2 106 = 2 105 4. (a) 0.0002 = 2 104 (b) 0.6 = 6 101 (c) 7.8 102 (already in scientific notation) 5. (a) 32 103 = 3.2 104 (b) 6800 106 = 6.8 103 (c) 870 108 = 8.7 1010 6. (a) 2 105 = 200,000 (b) 5.4 109 = 0.0000000054 (c) 1.0 101 = 10 7. (a) 2.5 106 = 0.0000025 (b) 5.0 102 = 500 (c) 3.9 101 = 0.39 8. (a) 4.5 106 = 0.0000045 (b) 8 109 = 0.000000008 (c) 4.0 1012 = 0.0000000000040
2
Chapter 1
3
9. (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 10. (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 11. (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
12. (a) 2
3
105.2
100.1
= 0.4 103 2 = 0.4 101 = 4
(b) 8
6
1050
105.2
= 0.05 106 (8) = 0.05 102 = 5
(c) 5
8
102
102.4
= 2.1 108 (5) = 2.1 1013
13.
4 2 3 2 3 2
7 5 12 12 12 12
2 4 4
(a) 8 10 4 2 10 80 10 4 2 10 84 10 2 10 4.20 10
(b) 3 10 5 10 9 10 15 10 9 10 6 10
(c) 2.2 10 1.1 5.5 10 11 10
2
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
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. (a) 3102.2
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
22. (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
23. (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 27. (a) 7.5 pA = 7.5 1012 A (b) 3.3 GHz = 3.3 109 Hz (c) 280 nW = 2.8 107 W 28. (a) 5 A = 5 106 A (b) 43 mV = 43 103 V (c) 275 k = 275 103 (d) 10 MW = 10 106 W
Section 1-4 Metric Unit Conversions
29. (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
30. (a) A101
A101
A1
mA16
3
= 1 103 = 1000
(b) V101
V1005.0
mV1
kV05.03
3
= 0.05 106 = 50,000
(c)
6
3
101
1002.0
M 1
k02.0 = 0.02 103 = 2 105
(d) W101
W10155
kW1
mW1553
3
= 155 106 = 1.55 104
5
Chapter 1
31. (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
32. (a)
k 2.12
k10
k 10k 2.2
k 10 = 0.8197
(b) 6
3
1050
10250
V50
mV250
= 5000
(c) 3
6
102
101
kW 2
MW 1
= 500
Section 1-5 Measured Numbers 33. 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 34. (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) C 1
J 10
Q
WV = 10 V (b)
C 2
J 5
Q
WV = 2.5 V
(c) C 25
J 100
Q
WV = 4 V
6. C 100
J 500
Q
WV = 5 V
7. C 40
J 800
Q
WV = 20 V
8. W = VQ = (12 V)(2.5 C) = 30 J
9. I = t
Q
Q = It = (2 A)(15 s) = 30 C
V = C 30
J 1000
Q
W = 33.3 V
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.
7
Chapter 2
Section 2-4 Current 13. A current source provides a constant current of 100 mA regardless of the load value.
14. (a) s 1
C 75
t
QI = 75 A
(b) s0.5
C 10
t
QI = 20 A
(c) s 2
C 5
t
QI = 2.5 A
15. s 3
C 0.6
t
QI = 0.2 A
16. t
QI
t = A 5
C 10
I
Q = 2 s
17. Q = It = (1.5 A)(0.1 s) = 0.15 C
18. I = t
Q
Q = Celectrons/ 1025.6
electrons 1057418
15
= 9.18 102 C
I = s 10250
C1018.93
2
= 367 mA
Section 2-5 Resistance
19. (a) G =
5
11
R = 0.2 S = 200 mS
(b) G =
25
11
R = 0.04 S = 40 mS
(c) G =
100
11
R = 0.01 S = 10 mS
20. (a) R = S 1.0
11
G = 10
(b) R = S 5.0
11
G = 2
(c) R = S 02.0
11
G = 50
8
Chapter 2
21. (a) Red, violet, orange, gold: 27 k 5% (b) 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. (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% 26. 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. (a) 0.47 : yellow, violet, silver, gold (b) 270 k: red, violet, yellow, gold (c) 5.1 M: green, brown, green, gold 28. (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% 29. (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
9
Chapter 2
30. 500 , There is equal resistance on each side of the contact.
31. 4K7 = 4.7 k 32. (a) 4R7J = 4.7 5% (b) 5602M = 56 k 20% (c) 1501F = 1500 1%
Section 2-6 The Electric Circuit
Figure 2-1
33. See 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.
10Figure 2-3
Chapter 2
38. See Figure 2-4.
Figure 2-4
Section 2-7 Basic Circuit Measurements 39. See Figure 2-5.
Figure 2-5
Figure 2-6
40. See Figure 2-6. 41. Position 1: V1 = 0 V, V2 = VS Position 2: V1 = VS, V2= 0 V 42. See Figure 2-7.
Figure 2-7
11
Chapter 2
43. See Figure 2-8.
Figure 2-8
44. See Figure 2-8. 45. On the 600 V scale (middle AC/DC scale): 250 V 46. R = 10 10 = 100 47. (a) 2 10 = 20 (b) 15 100 k = 1.50 M (c) 45 100 = 4.5 k 48. 0.9999 + 0.0001 = 1.0000 Resolution = 0.00001 V
12
Chapter 2
13
See Figure 2-9. 49.
Figure 2-9
Chapter 3 Ohm’s Law Note: Solutions show conventional current direction.
Section 3-1 The Relationship of Current, Voltage, and Resistance 1. (a) When voltage triples, current triples. (b) When voltage is reduced 75%, current is reduced 75%. (c) When resistance is doubled, current is halved. (d) When resistance is reduced 35%, current increases 54%. (e) When voltage is doubled and resistance is halved, current quadruples. (f) When voltage and resistance are both doubled, current is unchanged.
2. I = R
V
3. V = IR
4. R = I
V
5. See Figure 3-1.
I = 100
V 0 = 0 A
I = 100
V 10 = 100 mA
I = 100
V 20 = 200 mA
I = 100
V 30 = 300 mA
I = 100
V 40 = 400 mA
I = 100
V 50 = 500 mA
I = 100
V 60 = 600 mA
I = 100
V 70 = 700 mA
I = 100
V 80 = 800 mA
I = 100
V 90 = 900 mA
Figure 3-1
The graph is a straight line indicating a linear relationship between V and I.
I = 100
V 100 = 1 A
14
Chapter 3
6. R = mA 15
V 1 = 200
(a) I = 200
V 1.5 = 7.5 mA
(b) I = 200
V 2 = 10 mA
(c) I = 200
V 3 = 15 mA
(d) I = 200
V 4 = 20 mA
(e) I = 200
V 10 = 50 mA
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.
R1 = A 2
V 1
I
V = 500 m
R2 = A 1
V 1
I
V = 1
R3 = A 0.5
V 1
I
V = 2
8. See Figure 3-2.
Figure 3-2
15
Chapter 3
I = 2 V
8.2 k = 0.244 mA
I = 4 V
8.2 k = 0.488 mA
I = 6 V
8.2 k = 0.732 mA
I = 8 V
8.2 k = 0.976 mA
I = 10 V
8.2 k = 1.22 mA
9. See Figure 3-3.
Figure 3-3
I = 2 V
1.58 k = 1.27 mA
I = 4 V
1.58 k = 2.53 mA
I = 6 V
1.58 k = 3.80 mA
I = 8 V
1.58 k = 5.06 mA
I = 10 V
1.58 k = 6.33 mA
16
Chapter 3
10. (a) I = k 3.3
V 50 = 15.2 mA
(b) I = k 3.9
V 75 = 19.2 mA
(c) I = k 4.7
V 100 = 21.3 mA
Circuit (c) has the most current and circuit (a) has the least current.
11. R = mA 50
V 10
mA 30S
V = 0.2 k = 200
VS = (200 )(30 mA) = 6 V (new value) The battery voltage decreased by 4 V (from 10 V to 6 V). 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.
(a) I = 1V 10
= 10 A (b) I = 5
V 10 = 2 A (c) I =
20
V 10 = 0.5 A
I = 1
V 20 = 20 A I =
5
V 20 = 4 A I =
20
V 20 = 1 A
I = 1
V 30 = 30 A I =
5
V 30 = 6 A I =
20
V 30 = 1.5 A
I = 1V 40
= 40 A I = 5
V 40 = 8 A I =
20
V 40 = 2 A
I = 1V 50
= 50 A I = 5
V 50 = 10 A I =
20
V 50 = 2.5 A
I = 1V 60
= 60 A I = 5
V 60 = 12 A I =
20
V 60 = 3 A
I = 1V 70
= 70 A I = 5
V 70 = 14 A I =
20
V 70 = 3.5 A
I = 1
V 80 = 80 A I =
5
V 80 = 16 A I =
20
V 80 = 4 A
I = 1V 90
= 90 A I = 5
V 90 = 18 A I =
20
V 90 = 4.5 A
I = 1
V 100 = 100 A I =
5
V 100 = 20 A I =
20
V 100 = 5 A
17
Chapter 3
(d) I = 100
V 10 = 0.1 A
Figure 3-4
I = 100
V 20 = 0.2 A
I = 100
V 30 = 0.3 A
I = 100
V 40 = 0.4 A
I = 100
V 50 = 0.5 A
I = 100
V 60 = 0.6 A
I = 100
V 70 = 0.7 A
I = 100
V 80 = 0.8 A
I = 100
V 90 = 0.9 A
I = 100
V 100 = 1 A
14. Yes, the lines on the IV graph are straight lines.
Section 3-2 Current Calculations
15. (a) I =
1
V5
R
V = 5 A
(b) I =
10
V15
R
V = 1.5 A
(c) I =
100
V50
R
V = 500 mA
(d) I =
k15
V30
R
V = 2 mA
(e) I =
M6.5
V250
R
V = 44.6 A
18
Chapter 3
16. (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 =
k2.2
kV1
R
V = 455 mA
(e) I =
M10
kV66
R
V = 6.6 mA
17. I =
10
V12
R
V = 1.2 A
18. R = 3300 5% Rmax = 3300 + (0.5)(3300 ) = 3465 Rmin = 3300 (0.5)(3300 ) = 3135
Imax =
3135
V 12
min
s
R
V = 3.83 mA
Imin =
3465
V 12
max
s
R
V = 3.46 mA
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
Imin =
k 51.7
V 25
maxR
V = 484 A
Imax =
k 42.3
V 25
minR
V = 591 A
Inom =
k 47
V 25
R
V = 532 A
20. R = 37.4
I = 12 V
37.4
V
R
= 0.321 A
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
Imax =
8
V 20)(
min
maxR
R
V = 2.5 A
A fuse with a rating of less than 2.5 A must be used. A 2-A fuse is suggested.
19
Chapter 3
23. 3 V
9.1 mA330
VI
R
Section 3-3 Voltage Calculations
24. 180 A 27 k 4.86 VV IR 25. (a) V = IR = (2 A)(18 ) = 36 V
(b) V = IR = (5 A)(56 ) = 280 V (c) V = IR = (2.5 A)(680 ) = 1.7 kV
(d) V = IR = (0.6 A)(47 ) = 28.2 V (e) V = IR = (0.1 A)(560 ) = 56 V 26. (a) V = IR = (1 mA)(10 ) = 10 mV
(b) V = IR = (50 mA)(33 ) = 1.65 V (c) V = IR = (3 A)(5.6 k) = 16.8 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 27. VS = IR = (3 A)(27 ) = 81 V 28. (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
29. Wire resistance = RW = CM1624.3
ft) ft)(24/ CM 4.10( = 0.154
(a) I =
100.154
V 6
WRR
V = 59.9 mA
(b) VR = (59.9 mA)(100 ) = 5.99 V
(c) VRW =
2WR
I = (59.9 mA)(0.154 /2) = 4.61 mV
Section 3-4 Resistance Calculations
30. (a) R = A 2
V 10
I
V = 5
(b) R = A 45
V 90
I
V = 2
(c) R = A 5
V 50
I
V = 10
20
Chapter 3
(d) R = A 10
V 5.5
I
V = 550 m
(e) R = A 0.5
V 150
I
V = 300
31. (a) R = A 5
kV 10
I
V = 2 k
(b) R = mA 2
V 7
I
V = 3.5 k
(c) R = mA 250
V 500
I
V = 2 k
(d) R = A 500
V 50
I
V = 100 k
(e) R = mA 1
kV 1
I
V = 1 M
32. R = mA 2
V 6
I
V = 3 k
33. (a) RFIL: = A 0.8
V 120
I
V = 150
34. Measure the current with an ammeter connected as shown in Figure 3-5, then calculate the
unknown resistance as R = 12 V/I.
Figure 3-5
35. R = mA 750
V 100
I
V = 133
R = A 1
V 100
I
V = 100
The source can be shorted if the rheostat is set to 0 .
36. Rmin + 15 = A 2
V 120 = 60 . Thus Rmin = 60 15 = 45
The rheostat must actually be set to slightly greater than 45 so that the current is limited to slightly less than 2 A.
37. Rmin + 15 = 110 V
1 A = 110
Rmin = 110 15 = 95
21
Chapter 3
22
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.
23
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 = h 5
J 7500
t
W
s 18000
J 7500
s/h) h)(3600 (5
J 7500 = 417 mW
5. P = ms 50
J 1000 = 20 kW
6. (a) 1000 W = 1 103 W = 1 kW (b) 3750 W = 3.75 103 W = 3.75 kW (c) 160 W = 0.160 103 W = 0.160 kW (d) 50,000 W = 50 103 W = 50 kW
7. (a) 1,000,000 W = 1 106 W = 1 MW (b) 3 106 W = 3 MW (c) 15 107 W = 150 106 = 150 MW (d) 8700 kW = 8700 103 W = 8.7 106 W = 8.7 MW
8. (a) 1 W = 1000 103 W = 1000 mW (b) 0.4 W = 400 103 W = 400 mW (c) 0.002 W = 2 103 = 2 mW (d) 0.0125 W = 12.5 103 W = 12.5 mW
9. (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
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 11. Energy = W = Pt = (100 mW)(24 h)(3600 s/h) = 8.64 103 J
Chapter 4
24
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. s/h) 0 W/kW)(360(1000
Ws6700 = 0.00186 kWh
16. W = Pt P = I2R = (5 A)2(47 ) = 1175 W
t = W1175
J 25
P
W = 0.0213 s = 21.3 ms
Section 4-2 Power in an Electric Circuit
17. RL = A 2
V 75
I
V = 37.5
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 =
680
V) 60( 22
R
V= 5.29 W
23. P =
56
V) 5.1( 22
R
V = 40.2 mW
24. P = I2R
R = 22 A) (2
W100
I
P = 25
25. (a) P =
10
V) 12( 22
R
V = 14.4 W
W = Pt = (14.4 W)(2 min)(1/60 h/min) = 0.48 Wh
(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.
Chapter 4
25
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:
221
11
222
22
223
33
3.25 V0.59 W Choose next highest standard value of 1 W.
18
6.5 V1.1W Choose next highest standard value of 2 W.
39
10 V1.5 W Choose next highest stand
68
RR
RR
RR
VP
R
VP
R
VP
R
ard value of 2 W.
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.
Section 4-5 Power Supplies and Batteries
31. VOUT = ) W)(501( LLRP = 7.07 V
32. PAVG = 22 (1.25) V
10
V
R
= 156 mW
Figure 4-1
Chapter 4
26
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 = h 10
Ah 80 = 8 A
36. I = h 48
mAh 650 = 13.5 mA
37. PLost = PIN POUT = 500 mW 400 mW = 100 mW
% efficiency = OUT
IN
400 mW100% 100%
500 mW
P
P
= 80%
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
% efficiency = OUT
IN
100%P
P
PIN = OUT 2 W100% 100%
% efficiency 60%
P
= 3.33 W
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
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).
Figure 5-2
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.
27
Chapter 5
Section 5-2 Total Series Resistance 6. RT = 1 + 2.2 + 5.6 + 12 + 22 = 42.8 7. (a) RT = 560 + 1000 = 1560
(b) RT = 47 + 56 = 103
(c) RT = 1.5 k + 2.2 k + 10 k = 13.7 k
(d) RT = 1 M + 470 k + 1 k + 2.2 M = 3.671 M 8. (a) RT = 1 k + 5.6 k + 2.2 k = 8.8 k (b) RT = 4.7 + 10 + 12 + 1 = 27.7 (c) 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
15. I = T
12 V
120
V
R
= 100 mA
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) RT = R1 + R2 + R3 = 2.2 k + 5.6 k + 1 k = 8.8 k
I = T
5.5 V
8.8 k
V
R
= 625 A
(b) RT = R1 + R2 + R3 = 1 M + 2.2 M + 560 k = 3.76 M
I = T
16 V
3.76 M
V
R
= 4.26 A
20. (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. RT = 3(470 ) = 1.41 k
(a) I = T
48 V
1.41 k
V
R
= 34 mA
(b) VR = 48 V
3 = 16 V
(c) P = (34 mA)2(470 ) = 0.543 W
22. RT = mA 2.23
V 5
I
V = 2.24 k
Reach = T 2.24 k
4 4
R = 560
23. R1 = mA 65.8
V 21.71 I
V = 330 R2 =
mA65.8
V 14.52 I
V = 220
R1 = mA 65.8
V 6.583 I
V = 100 R4 =
mA65.8
V 30.94 I
V = 470
24. V1 = IR1 = (12.3 mA)(82 ) = 1.01 V
R2 = 2 12 V 2.21 V 1.01 V
12.3 mA
V
I
= 714
R3 = 3 2.21 V
12.3 mA
V
I = 180
25. (a) RT = R1 + R2 + R3 + R4
R4 = 12 V
7.84 mA (R1 + R2 + R3) =
12 V
7.84 mA 1200 = 1531 1200 = 331
(b) Position B: I = 2 3 4
12 V 12 V
1311 R R R
= 9.15 mA
Position C: I = 3 4
12 V 12 V
841 R R
= 14.3 mA
Position D: I = 4
12 V 12 V
331 R
= 36.3 mA
(c) No
26. Position A: RT = R1 = 1 k
I = T
9 V
1 k
V
R
= 9 mA
Position B: RT = R1 + R2 + R5 = 1 k + 33 k + 22 k = 56 k
I = T
9 V
56 k
V
R
= 161 A
Position C:
30
Chapter 5
RT = R1 + R2 + R3 + R4 + R5 = 1 k + 33 k + 68 k + 27 k + 22 k = 151 k
I = T
9 V
151 k
V
R
= 59.6 A
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) 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 (b) 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
33. I = 56
V 11.2 = 200 mA
R4 = mA 200
V 4.4 = 22
34. R1 = mA 10
V 5.61 I
V = 560
R2 = 22
2
mA) (10
mW 22
I
P = 220
RT = mA 10
V 9 = 900
R3 = RT R1 R2 = 900 560 200 = 120 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
31
Chapter 5
Position B:
1.8 k + 1 k + 820 = 3.62 k
5 V
= 1.8 k + 1 k = 2.8 k
.6 V
k s R :
6. A)(1.8 k) = 6.03 V
)(1.8 k) = 6.71 V
: )(1.8 k) = 8.1 V
1.8 k) = 10.8 V
ection 5-7 Voltage Dividers
RT = R1 + R2 + R3 = Voltage drop across R1 through R3: V = IRT = (3.73 mA)(3.62 k) = 13. V5 = 18 V 13.5 V = 4.5 V Position C: RT = R1 + R2 Voltage drop across R1 and R2: V = IRT = (4.5 mA)(2.8 k) = 12 V5 = 18 V 12.6 V = 5.4 V Position D: RT = R1 = 1.8 Voltage drop acros 1
V = IRT = (6 mA)(1.8 k) = 10.8 V V5 = 18 V 10.8 V = 7.2 V 3 Position A: V1 = (3.35 m 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 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 V2 = (4.5 mA)(1 k) = 4.5 V V5 = 5.4 V Position D: V1 = (6 mA)( V5 = 7.2 V
S
27
T
27100
560
V
V
37. = 4.82%
VAB = V 12156
56
38. (a) = 4.31 V
(b) VAB = V 8k5.6
k5.5
= 6.77 V
9. VA = VS = 15 V 3
32
Chapter 5
2 3S
1 2 3
13.3 k
18.9 k
R R V
R R R
VB =
15 V = 10.6 V
VC =
3S
1 2 3
3.3 k15 V
18.9 k
RV
R R R
= 2.62 V
VOUT(min) = 3S
1 2 3
680 12 V
2150
RV
R R R
40. = 3.80 V
VOUT(max) = 2 3S
1 2 3
1680 12 V
2150
R RV
R R R
= 9.38 V
1. RT = 15R
R
4
V 9015
R
V = R
= 6 V
V2R = V 9015
2
R
R = 12 V
V3R = V 9015
3
R
R = 18 V
V4R = V 9015
4
R
R = 24 V
V5R = V 9015
5
R
R = 30 V
2. VAF = 100 V 4
V 100k 6.108
k 6.86
BFR
VBF =
AFAF
VR
= 79.7 V
VCF = V 100k 6.108
k 6.76
AF
AF
CF VR
R = 70.5 V
VDF = V 100k 6.108
k 6.20
AF
AF
DF VR
R = 19.0 V
VEF = V 100k 6.108
k 6.5
AF
AF
EF VR
R = 5.16 V
3. I =
k 5.6
V 10
1
1
R
V4 = 1.79 mA
2 IR2 = (1.79 m 1.79 V
V = A)(1 k) = V3 = IR3 = (1.79 mA)(560 ) = 1.0 V
V4 = IR4 = (1.79 mA)(10 k) = 17.9 V
33
Chapter 5
44. See Figure 5-5 for one possible solution: 00 k
T
Figure 5-5
RT = 18 k + 33 k + 22 k + 27 k = 1
I = k 100
V 30 = 300 A
V 30k 100
k 82
VA = = 24.6 V
VB = V 30k 100
k 49
= 14.7 V
VC = V 30k 100
k 27
= 8.1 V
P1 = = (300 A)2 27 k = 2.43 mW
5. See Figure 5-6 for one possible solution.
OUT(max)
2T 1I R
P2 = = (300 A)2 22 k = 1.98 mW 2T 2I R
P3 = = (300 A)2 33 k = 2.97 mW 2T 3I R
P4 = = (300 A)2 18 k = 1.62 mW 2T 4I R
All resistors can be 1/8 W. 4 RT = 12.1 k
V 120k 12.1
k 10
V = .1
= 100.2 V
VOUT(min) =
V 120k 12.1
k 1
= 9.92 V
These values are within 1% of the specified values.
IMAX = T
120 V 120 V = 9.9 mA Figure 5-6 12.1 kR
ection 5-8 Power in Series Circuits
S 46. PT = 5(50 mW) = 250 mW
7. VT = V1 + V2 + V3 + V4 = 10 V + 1.79 V + 1 V + 17.9 V = 30.69 V
8. Since P = I R and since each resistor has the same current, the 5.6 k resistor is the limiting
Imax =
4 PT = VTI = (30.69 V)(1.79 mA) = 54. 9 mW
24element in terms of power dissipation.
k 5.6
W0.25
k 5.6maxP
= 6.68 mA
5.6 k ) = 37.4 V
6.1 V = 86.2 V
V = (6.68 mA)(5.6 k 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 + 2
34
Chapter 5
M 5.6
V 12
1
1
R
V49. I = = 2.14 A
R2 = A 2.14
V 4.82
I
V
2
= 2.2 M
P3 = I R
3
3
3P R =
22 A)142
W 5.21
I = 4.7 M
RT = R1 + R2 + R3 = 5.6 M + 2.2 M + 4.7 M = 12.5 M
0. (a) P = I2R
5
R = 2I
P
R + 1 2 3R + R = 2400
222
W2
1 W
1W
1
48
I I I
= 2400
2
W8
7
I
= 2400
I 2 =
2400
W8
7
= 0.0003646 A2
I = 2A 0003646.0 = 19.1 mA
(b) VT = IRT = (19.1 mA)(2400 ) = 45.8 V
R1 =
221
mA) (19.1
W0.125
I
P (c) = 343
R2 = 22
2
mA) (19.1
W0.25
I
P = 686
R3 = 22
3
mA) (19.1
W5.0
I
P = 1.37 k
ection 5-9 Voltage Measurements S 51. VAG = 100 V (voltage from point A to ground)
k Resistance between A and C: RAC = 5.6 k + 5.6 k = 11.2 Resistance between C and ground: RCG = 1 k + 1 k = 2 k
35
Chapter 5
VCG = V 100k 2.13
k 2
= 15.2 V
VDG = V 15.2k 2
k 1
k 2
k 1
CGV = 7.58 V
VAC = V 100k 2.13
k 2.11
= 84.9 V
VBC = V 84.9k 2.11
k 6.5
k 2.11
k 6.5
ACV = 42.5 V
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
I = T
T
6 V 1.816 M
V
R
= 3.3 A
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) Zero current indicates an open. R4 is open since all the voltage is dropped across it.
(b) S
1 2 3
10 V
300
V
R R R
= 33.3 mA
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
37
58. R2 = 0 RT = R1 + R3 + R4 + R5 = 400
IT = S
T
10 V
400
V
R
= 25 mA
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) R11 has burned out because it has the highest resistance value (P = I2R). (b) Replace R11 (10 k). (c) RT = 47.73 k
Imax =
k 10
W5.0
11
11
R
P = 7.07 mA
Vmax = ImaxRT = (7.07 mA)(10 k) = 70.7 V
Multisim Troubleshooting and Analysis 63. 7.481 k 66. 6 V 64. R2 is open. 67. R1 is shorted 65. R3 = 22
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. V1 = V2 = V3 = V4 = 12 V
IT = T
T
12 V
550
V
R
= 21.8 mA
The total current divides equally among the four equal parallel resistors.
I1 = I2 = I3 = I4 = 4
mA 21.8 = 5.45 mA
5. 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. Position A: IT = 15V
730 = 20.6 mA
Position B: IT = 15V
688 = 21.8 mA
Position C: IT = 15V
643 = 23.3 mA
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
R2 = S
2
47 mV
2.14 mA
V
I = 22
R3 = S
3
47 mV
0.47 mA
V
I = 100
I4 = IT (I1 + I2 + I3) = 5.03 mA 3.61 mA = 1.42 mA
R4 = S
4
47 mV
1.42 mA
V
I = 33
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
I220 = T 2.5 V
220 220
V
= 11.4 mA
14. IT = 4IRUN + 2ITAIL = 4(0.5A) + 2(1.2 A) = 4.4 A
39
Chapter 6
15. (a) IT = 4IRUN + 2ITA + 2IBRAKE = 4(0.5A) + 2(1.2 A) + 2(1 A) = 6.4 A IL
(b) IGND = IT = 6.4 A Section 6-4 Total Parallel Resistance
16. RT =
M22
1
M12
1
M 6.5
1
M 2.2
1
M 1
11
= 568 k
17. (a) RT =
k 1560
)k 1)(560( = 359
(b) RT = 5647
) 56)(47( = 25.6
(c) RT =
k10
1
k 2.2
1
k 1.5
11
= 819
(d) RT =
M7.2
1
k 1
1
k 470
1
M 1
11
= 997
18. (a) RT = 220560
) 220)(560( = 158
(b) RT =
k 56k27
)k 56)(k27( = 18.2 k
(c) RT =
k 2.2k5.1
)k 2.2)(k5.1( = 892
19. RT = 12
k 6.8 = 0.567 k = 567
20. Five 470 resistors in parallel:
R1 = 470
5
= 94
Ten 1000 resistors in parallel:
R2 = 1000
10
= 100
Two 100 resistors in parallel:
R3 = 100
2
= 50
21. RT = 1
1 1 1
94 100 50
= 24.6
40
Chapter 6
22. RT = 21
21
RR
RR
RT(R1 + R2) = R1R2 RTR1 + RTR2 = R1R2 RTR1 = R1R2 RTR2 RTR1 = R2(R1 RT)
R2 = T 1
1 T
(389.2 )(680 )
680 389.2
R R
R R
= 910
23. (a) RT = R1 = 510 k
(b) RT = R1 R2 =
k470
1
k 510
11
= 245 k
(c) RT = R1 = 510 k
RT = R1 R2 R3 =
k910
1
k470
1
k510
11
= 193 k
Section 6-5 Application of Ohm’s Law
24. (a) RT =
27
1
33
1
33
11
= 10.2
IT = T
10 V
10.2
V
R
= 980 mA
(b) RT =
560
1
k 4.7
1
k 1
11
= 334
IT = T
25 V
334
V
R
= 74.9 mA
25. RT = 240
3 3
R = 80
IT = 120 V
80 = 1.5 A
26. RT = S
T
5 V
1.11 mA
V
I = 4.5 k
Reach = 4RT = 4(4.5 k) = 18 k
27. I = S
filament
110 V
2.2 k
V
R
= 50 mA
When one bulb burns out, the others remain on.
41
Chapter 6
28. (a) I2 = IT I1 = 150 mA 100 mA = 50 mA
R1 = mA 100
V 10 = 100
R2 = mA 50
V 10 = 200
(b) I3 = k 1
V 100 = 100 mA
I2 = 680
V 100 = 147 mA
I1 = IT I2 I3 = 500 mA 247 mA = 253 mA
R1 = mA 253
V 100 = 395
29. Imax = 0.5 A
RT(min) = max
15 V 15 V
0.5 AI = 30
x
x
R
R
68
)68( = RT(min)
(68 )Rx = (30 )(68 + Rx) 68Rx = 2040 + 30Rx 68Rx 30Rx = 2040 38Rx = 2040 Rx = 53.7 30. Position A:
I1 = k 560
V 24 = 42.9 A
I2 = k 220
V 24 = 109 A
I3 = k 270
V 24 = 88.9 A
IT = 42.9 A + 109 A + 88.9 A = 241 A Position B: I1 = 42.9 A I2 = 109 A I3 = 88.9 A
I4 = M 1
V 24 = 24 A
I5 = k 820
V 24 = 29.3 A
I6 = M 2.2
V 24 = 10.9 A
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. I3 = k 1.2
V 100 = 83.3 mA
I2 = 250 mA 83.3 mA = 166.7 mA IT = 250 mA + 50 mA = 300 mA
R1 = mA 50
V 100 = 2 k
R2 = mA 7.166
V 100 = 600
Section 6-6 Current Sources in Parallel 32. (a) IL = 1 mA + 2 mA = 3 mA (b) IL = 50 A 40 A = 10 A (c) 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. I1 = 2T
1 2
2.7 k3 A
3.7 k
RI
R R
= 2.19 A
I2 = 1T
1 2
1 k3 A
3.7 k
RI
R R
= 0.811 A
35. (a) I1 = 2T
1 2
2.2 M10 A
3.2 M
RI
R R
= 6.88 A
I2 = IT I1 = 10 A 6.88 A = 3.12 A
(b) Ix = TT
x
RI
R
RT = 525
I1 = mA 101000
525
= 5.25 mA
I2 = mA 10k2.2
525
= 2.39 mA
43
Chapter 6
I3 = mA 10k3.3
525
= 1.59 mA
I4 = mA 10k8.6
525
= 0.772 mA
36. RT =
RRRR 4
1
3
1
2
111
=
4
1
3
1
2
11/R = 0.48R
IR = T 0 4810 mA 10 mA
R . R
R R
= 4.8 mA; I2R = T 0 48
10 mA 10 mA2 2
R . R
R R
= 2.4 mA;
I3R = T 0 4810 mA 10 mA
3 3
R . R
R R
= 1.59 mA; I4R = T 0 48
10 mA 10 mA4 4
R . R
R R
= 1.2 mA
37. RT = 773 I3 = IT I1 I2 I3 = 15.53 mA 3.64 mA 6.67 mA 3.08 mA = 2.14 mA
I1 = TT
1
RI
R
R1 = TT
1
77315.53 mA
3.64 mA
RI
I
= 3.3 k
R2 = TT
2
77315.53 mA
6.67 mA
RI
I
= 1.8 k
R3 = TT
3
77315.53 mA
2.14 mA
RI
I
= 5.6 k
R4 = TT
4
77315.53 mA
3.08 mA
RI
I
= 3.9 k
38. (a) IT = 10 mA, IM = 1 mA VM = IMRM = (1 mA)(50 ) = 50 mV
ISH1 = 9 mA
RSH1 = M
SH1
50 mV
9 mA
V
I = 5.56
(b) IT = 100 mA, IM = 1 mA
VM = IMRM = (1 mA)(50 ) = 50 mV ISH2 = 99 mA
RSH2 = M
SH2
50 mV
99 mA
V
I = 0.505
44
Chapter 6
39. (a) RSH = 50 mV
50 A = 1 m
(b) ISH = 50 mV
1 m = 50 A
Imeter = 50 mV
10 k = 5 A
Section 6-8 Power in Parallel Circuits 40. PT = 5(250 mW) = 1.25 W
41. (a) RT =
M 2.2M 1
)M 2.2)(M (1 = 687.5 k
PT = I 2RT = (10 A)2(687.5 k) = 68.8 W
(b) RT =
k 8.6
1
k 3.3
1
k 2.2
1
k 1
11
11111
4321 RRRR
= 525
PT = I 2RT = (10 mA)2(525 ) = 52.5 mW 42. P = VI
Ieach = 75 W
120 V
P
V = 625 mA
IT = 6(625 mA) = 3.75 A 43. P1 = PT P2 = 2 W 0.75 W = 1.25 W
VS = T
T
2 W
200 mA
P
I = 10 V
I2 = 2
S
0.75 W
10 V
P
V = 75 mA
R2 = S
2
10 V
75 mA
V
I = 133
I1 = IT I2 = 200 mA 75 mA = 125 mA
R1 = S
1
10 V
125 mA
V
I = 80
44. (a) PT = = (50 mA)2 1 k = 2.5 W 2T TI R
Number of resistors = n = T
each
2.5 W
0.25 W
P
P = 10
(b) RT = n
R
R = nRT = 10(1 k) = 10 k
45
Chapter 6
(c) I = T 50 mA
10
I
n = 5 mA
(d) VS = ITRT = (50 mA)(1 k) = 50 V
Section 6-10 Troubleshooting
45. Ieach = 75 W
120 V
P
V = 625 mA
IT = 5(625 mA) = 3.13 A
46. RT =
270
1
560
1
k1
1
100
1
220
11
= 47.5
IT = 47.5
V 10 = 210.5 mA
The measured current is 200.4 mA, which is 10.1 mA less than it should be. Therefore, one of the resistors is open.
R? = mA 10.1
V 10
I
V = 990 1 k
The 1 k resistor (R3) is open.
47. RT =
k2.8
1
k10
1
k7.4
11
= 2.3 k
IT = k 2.3
V 25 = 10.87 mA
The meter indicates 7.82 mA. Therefore, a resistor must be open.
I3 = k 8.2
V 25 = 3.05 mA
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.
48. I1 = k 4.7
V 25 = 5.32 mA
I2 = k 10
V 25 = 2.5 mA
I3 = k 8.2
V 25 = 3.05 mA
R1 is open producing a total current of IT = I2 + I3 = 2.5 mA + 3.05 mA = 5.55 mA
46
Chapter 6
47
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
(d) R1-4 = R1 R2 R3 R4 R11 R12 = 10 k 2.2 k 2.2 k 3.3 k 18 k 1 k = 422
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.
Chapter 7 Series-Parallel Circuits Note: Solutions show conventional current direction.
Section 7-1 Identifying Series-Parallel Relationships
Figure 7-1
1. See Figure 7-1.
2. See Figure 7-2.
Figure 7-2
3. (a) R1 and R4 are in series with the parallel combination of R2 and R3. (b) R1 is in series with the parallel combination of R2, R3, and R4.
(c) 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. (b) All of the resistors are in parallel. (c) 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.
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. RT = 21
21
RR
RR
R2 = 1 T
1 T
(1 k )(667 )
1 k 667
R R
R R
= 2.0 k
9. (a) RT = R1 + R4 + 2
2R = 56 + 27 +
2
100 = 133
(b) RT =
432
1 1111
RRR
R
= 680 +
180
1
330
1
680
11
= 680 + 99.4 = 779
(c) RT = R1 (R2 R3 + R4 R5) = R1 (2.154 k + 3.59 k) = 852 10. (a) RT = R1 (R2 + R3 R4) = 1 k (1 k + 2.2 k 3.3 k) = 699
(b) RT =
M6.2
1
M 3.3
1
M 1
1
M 1
11
= 406 k
(c) RA = R1 + R2 + 43
43
RR
RR
= 1 k + 1 k +
k7.4k10
)k 7.4)(k (10 = 5.2 k
RB = R5 + R8 + 76
76
RR
RR
= 3.3 k + 1.8 k +
2
k 8.6 = 8.5 k
RT =
k 8.5
1
k 2.5
11
111
BA RR
= 3.23 k
11. (a) IT = 133
V 1.5 = 11.3 mA
I1 = I4 = 11.3 mA
I2 = I3 = 2
mA 11.3 = 5.64 mA
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
(b) IT = 779
V 3 = 3.85 mA
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
I2 =
680
mV 383
2
2
R
V = 563 A
I3 =
330
mV 383
3
3
R
V = 1.16 mA
I4 =
180
mV 383
4
4
R
V = 2.13 mA
(c) I1 = k 1
V 5 = 5 mA Iright =
k5.74
V 5 = 871 A
I2 = A871k 9.5
k 3.3
= 303 A
I3 = A871k 9.5
k 6.2
= 568 A
I4 = A871k 15.6
k 5.6
= 313 A
I5 = A871k 15.6
k 10
= 558 A
V1 = VS = 5 V V2 = V3 = (303 A)(6.2 k) = 1.88 V V4 = V5 = (313 A)(10 k) = 3.13 V
12. (a) IT = 699
V 1 = 1.43 mA
I1 = mA 43.1k 3.32
k 2.32
= 1 mA
V1 = (1 mA)(1 k) = 1 V
I2 = mA43.1k 3.32
k 1
= 431 A
V2 = (431 A)(1 k) = 431 mV
I3 = A431k 5.5
k 3.3
= 259 A
V3 = (259 A)(2.2 k) = 570 mV V4 = V3 = 570 mV
I4 = k 3.3
mV 570 = 173 A
51
Chapter 7
(b) V1 = V2 = V3 = V4 = 2 V
I1 = M 1
V 2 = 2 A
I2 = M 3.3
V 2 = 606 nA
I3 = M 6.2
V 2 = 323 nA
I4 = M 1
V 2 = 2 A
(c) IT = k 23.3
V 5 = 1.55 mA
I5= mA 55.1k 13.7
k 5.2
= 588 A
V5 = (588 A)(3.3 k) = 1.94 V
I6 = I7 =
A588
25I
= 294 A
V6 = V7 = (294 A)(6.8 k) = 2 V I8 = I5 = 588 A V8 = (588 A)(1.8 k) = 1.06 V
I1 = I2 = mA55.1k 13.7
k 8.5
= 962 A
V1 = V2 = (962 A)(1 k) = 962 mV
I3 = A962k 7.14
k 4.7
= 308 A
V3 = V4 = (308 A)(10 k) = 3.08 V
I4 = A962k 7.14
k 10
= 654 A
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. VAG = 100 V RAC = (4.7 k + 5.6 k) 10 k = 5.07 k RCG = 2 k 1.8 k = 947
VAC = V 100k 02.6
k 07.5
= 84.2 V
VCG = V 100k 02.6
947
= 15.7 V
VDG = V 15.7k 2
k 1
k 2
k 1
CGV = 7.87 V
VBC = V 84.2k 3.10
k 6.5
k 3.10
k 6.5
ACV = 45.8 V
VBG = VCG + VBC = 15.7 V + 45.8 V = 61.5 V
16. VA = V 50k 716
k 56
= 3.91 V VB = V 50k 716
k 616
= 43.0 V
VC = 50 V VD = V 50M 1.1
k 100
= 4.55 V
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
IT = k 1.64
V 100 = 60.9 mA
Current in the right branch:
IR = T
103060.9 mA
2740L
L R
RI
R R
= 22.9 mA
Current in the left branch:
IL = T
171060.9 mA
2740R
L R
RI
R R
= 38.0 mA
53
Chapter 7
54
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) I2 = 1T
1 2
RI
R R
1 mA = T2
47 k
47 kI
R
47 k + R2 = (47 k)IT Also,
IT = 2T
2
220(47 k )
33 k(47 k )
VRRR
Substituting the expression for IT into 47 k + R2 = (47 k)IT.
47 k + R2 = 2
2
22047 k
(47 k )33 k
47 k
R
R
(47 k + R2) 2
2
(47 k )33 k
47 k
R
R
= 47 k(220)
(80 k)R2 = 47 k(220) (47 k)(33 k)
R2 = 47 k (220 33 k )
80 k
= 109.9 k 110 k
(b) P2 = = (1 mA)2 110 k = 0.11 W = 110 mW 2
2 2I R 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
Chapter 7
24. V2 = V5 V6 = 5 V 1 V = 4 V
I2 = I6 = V 4
W2 = 0.5 A
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
V1 = A 2
W20 = 10 V
V4 = V3 V1 = 10 V V8 = V4 V5 = 5 V
R1 = A2
V 10 = 5
R2 = A0.5
V 4 = 8
R3 = A2
V 20 = 10
R4 = A1
V 10 = 10
R5 = A0.5
V 5 = 10
R6 = A0.5
V 1 = 2
R7 = A 4
V 20 = 5
R8 = A 1
V 5 = 5
Section 7-3 Voltage Dividers with Resistive Loads
25. VOUT(unloaded) = V 15k 112
k 56
= 7.5 V
56 k in parallel with a 1 M load is
Req =
M 1k 56
)M 1)(k (56 = 53 k
VOUT(loaded) = V 15k 109
k 56
= 7.29 V
26. See Figure 7-6.
VA = V 12k 9.9
k 6.6
= 8 V
VB = V 12k 9.9
k 3.3
= 4 V
With a 10 k resistor connected from tap A to ground:
RAB =
k 10k 6.6
)k 10)(k 6.6( = 3.98 k
VA(loaded) = V 12 28.7
k 98.3
= 6.56 V
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
OUT(NL) V = 2 3R R S
1 2 3
VR R R
=
8.3 k 22 V
18.3 k
= 9.98 V
With a 100 k load:
T 1
R = R + 2 3( )
2 3
L
L
R R R
R R = 10 k +
(8.3 k )(100 k )
108.3 k
R
= 17.7 k
VOUT = 7.7 k
22 V17.7 k
= 9.57 V
9. RAB = 2(8.3 k )(33 k )
8.3 k 33 k
= 6.63 k
VAB = 6.63 k
22 V10 k 6.63 k
= 8.77 V
0. RT = 10 k + 5.6 k + 2.7 k = 18.3 k 3
I = 22 V
= 1.2 mA 18.3 k
RT = 10 k + (8.3 k )(33 k )
8.3 k 33 k
= 16.6 k
I = 22 V
16.6 k
= 1.33 mA
1. See Figure 7-7. R1 + 2R2 = 2 k 3
RT = mA 5
V 10 = 2 2R2 + 2R2 =k 2 k
2 + R3
0 00
With a 1 k load on the lower tap:
00 = 333
Figure 7-7
R1 = R 4R2 = 2 k R2 = R3 R2 = R3 = 50 R1 = 2R2 R1 = R2 + R3 = 10 1 k 5
IT = 333500k 1
V 10 = 5.46 mA
V 1.82 V lower tap = (333 )(5.46 mA) = 4.55 V
With a 1 k load on the upper tap:
Vupper tap = (500 + 333 )(5.46 mA) =
IT = 2/k1k1
V 10
= 6.67 mA
Vuppe 6.67 mA) = 3r tap = (500 )( .33 V
Vlower tap = 2
V 33.3 = 1.67 V
56
Chapter 7
32. Position 1: R = 10 k + 30 k 68 k = 10 k + 20.82 k = 30.8 k T
V1 = V 120k 8.30
= 81.0 V k 20.8
V 81k 30
k 20
V2 = = 54.0 V
V 81k 30
k 10
= 27.0 V V3 =
Position 2: RT = 20 k + 20 k 68 k = 20 k + 15.5 k = 35.5 k
V1 = V 120k 5.35
= 86.2 V k 10 k 5.15
V 81k 5.35
k 15.5
= 52.4 V V2 =
V 52.4k 20
k 10
= 26.2 V V3 =
Posit R = 30 k 68 k = 30 k
ion 3: T + 10 k + 8.72 k = 38.7 k
V1 = V 120k 7.38
= 89.0 V k 20 k 72.8
V 81k 7.38
k 72.8k 10
58.0 V V2 = =
V 81k 7.38
k 8.72
V3 = = 27.0
3. (a) VG =
V
2DD
1 2
270 k16 V
2.47 M
RV
R R
3 = 1.75 V
G + 1.5 V = 1.75 V + 1.5 V = 3.2 VS = V 5 V
(b) I1 = DD G 16 V 1.75 VV V = 6.48 A
1 2.2 MR
1 = G
2
1.75 V
270 k
V
R
= 6.48 A I2 = I
IS = SV 17 mA
S
3.25 V
1.5 kR = 2.
IS = 2.17 mA
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
ID = (c) VD = VDD IDRD = VDG = VD VG = 5.8 V 1.75 V = 4.05 V
57
Chapter 7
34. Imax = 100 mA
RT = mA 100
=V 24
240
2R 6 V
T
24 VR
=
24R = 6R 2 T
R2 = 24
)240(6 = 60
40 6 load:
.6
R1 = 2 0 = 180 With R2 RL = 60 1000 = 56
VOUT = V 246.56180
= 5.6.56
74 V
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:
on the 1000 V range
6. (b) Rinternal = (20,000 /V)(1 V) = 20 k
7.
Rinternal = (20,000 /V)(1 V) = 20 k on the 1 V range Rinternal = (20,000 /V)(1000 V) = 20 M 3 (a) Rinternal = (20,000 /V)(0.5 V) = 10 k (c) Rinternal = (20,000 /V)(5 V) = 100 k (d) Rinternal = (20,000 /V)(50 V) = 1 M (e) Rinternal = (20,000 /V)(100 V) = 2 M (f) Rinternal = (20,000 /V)(1000 V) = 20 M
3 V 1.5133
274
R
V 1.54321
4
RRRR
VR = 0.305 V actual
(a) Use the 0.5 V range to measure 0.305 V
27 10 k = 26.93
.
(b) Rinternal = (20,000 /V)(0.5 V) = 10 k
V 1.593.1324
RV 93.26
0.304 V wit mh eter connected
.305 V 0.304 V = 0. 0 001 V less with meter
58
Chapter 7
38. V 3779.4
99.4V 3
1432
432
4
RRRR
RRRVR = 0.383 V actual
(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
V 34.778
4.984
RV = 0.379 V with meter connected
0.383 V 0.379 V = 0.004 V less with meter 39. RMETER = 10 V(10,000/V) = 100 k
22
2
2R2
1 2
(100 k )(100 k )50 k
200 k
50 k0.333
100 k 50 k
METERMETER
METER
METERS S
METER
R RR R
R R
R RV V V
R R R SV
40.
22
2
2R2
1 2
(100 k )(10 M )99 k
10.1M
99 k0.498
100 k 99 k
METERMETER
METER
METERS S
METER
R RR R
R R
R RV V V
R R RV
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 = 271
V 60 = 221 mA
(c) I2 = mA 2215.524
271
= 114 mA
59
Chapter 7
I910 = mA 114910
5.468
= 58.7 mA
(d) The voltage across the 437.5 parallel combination of the 560 and the two series
1 k resistors is determined as follows:
I4 = mA 1145.965
5.468
= 55 mA
V437.5 = I4(437.5 ) = (55 mA)(437.5 ) = 24.06 V
VAB = V 06.24k 2
k 1
= 12 V
Figure 7-8
42. The total resistance is determined in the steps shown in Figure 7-9. RT = 6.66 k
VA = V 18k 66.6
k 06.1
= 2.86 V
VB = V 86.2k 05.2
k 05.1
= 1.47 V
VC = V 47.1k 2
k 1
= 735 mV
60Figure 7-9
Chapter 7
61
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):
I2 = mA 16.1820
8.420
= 8.27 mA
From Figure 7-10(b):
I3 = I8 = mA 16.15.864
8.420
= 7.84 mA
From Figure 7-10(a):
I4 = mA 7.84820
5.424
= 4.06 mA
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.
V5 = I5R5 = (3.78 mA)(100 ) = 0.378 V V1 = ITR1 = (16.1 mA)(100 ) = 1.61 V V2 = I2R2 = (8.27 mA)(820 ) = 6.78 V V6 = I6R6 = (3.78 mA)(680 ) = 2.57 V V3 = I3R3 = (7.84 mA)(220 ) = 1.73 V V7 = I7R7 = (3.78 mA)(100 ) = 0.378 V V4 = I4R4 = (4.06 mA)(820 ) = 3.33 V V8 = I8R8 = (7.84 mA)(220 ) = 1.73 V
V9 = I9R9 = (16.1 mA)(100 ) = 1.61 V
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
RT(both) = T1 61.7
2 2
R = 30.9
IT = 30.9
V 30 = 971 mA
46. (a) VOUT = 8
V 12
8
V = 1.5 V
(b) VOUT = 16
V 12
16
V = 0.75 V
47. (a) VOUT = 2
V 12
4
V 12
24
VV = 3 V + 6 V = 9 V
(b) VOUT = 16
V 12
4
V 12
164
VV = 3 V + 0.75 V = 3.75 V
(c) VOUT = 16
V 12
8
V 12
4
V 12
2
V 12
16842
VVVV
= 6 V + 3 V + 1.5 V + 0.75 V = 11.25 V
Section 7-6 The Wheatstone Bridge
48. Rx =
4
2
R
RRV = (18 k)(0.02) = 360
49. VLEFT = S
SG3 119.94 12 V
SG1 + SG3 120.06 119.94 V
= 5.997 V
VRIGHT = S
SG4 120.06 12 V
SG2 + SG4 119.94 120.06 V
= 6.003 V
VOUT = VRIGHT VLEFT = 6.003 V 5.997 V = 6 mV (Right side positive with respect to left side) 50. At 60 C, RTHERM = 5 k
VLEFT = 3S
1 3
27 k9 V
32 k
RV
R R
= 7.59 V
VRIGHT = 4S
2 4
27 k9 V
54 k
RV
R R
= 4.50 V
VOUT = VLEFT VRIGHT = 7.59 V 4.50 V = 3.09 V
62
Chapter 7
Section 7-7 Troubleshooting
51. Req =
k 7.4680
)k 4.7)(680( = 594
RT = 560 + 470 + 594 = 1624 The voltmeter reading should be
V? = V 121624
594
= 4.39 V
The voltmeter reading of 6.2 V is incorrect. 52. The circuit is redrawn in figure 7-11 and points are labeled.
RBG =
k 100k 47k 10
)k 100)(k 47k 10( = 36.3 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
VAG = T
69.3 k18 V 18 V
96.3 kAGR
R
= 12.95 V
VCG = V 6.79k 57
k 47
k 57
k 47
BGV = 5.60 V
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
64
54. The circuit is redrawn in Figure 7-12.
VA = V 150k 16
k 6V 150
k 10k 12k 12
k 12k 12
= 56.25 V
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.
VA = V 150k 22
k 12
= 81.8 V
VB = V 150k 8.7
k 2.2
= 42.3 V
The meter is correct.
Figure 7-12
55. V3.3 k = V) 10(k 2.62
k 62.1
= 6.18 V
The 7.62 V reading is incorrect.
V2.2 = V) 18.6(k 3.2
k 2.2
= 4.25
The 5.24 V reading is incorrect. The 3.3 k resistor must be open. If it is, then
V3.3 k = V) 10(k 4.2
k 2.3
= 7.62 V
V2.2 k = V) 62.7(k 3.2
k 2.2
= 5.24 V
56. 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
Chapter 8 Circuit Theorems and Conversions Note: Solutions show conventional current direction.
Section 8-3 Source Conversions
1. IS = S
S
300 V
50
V
R
= 6 A
Figure 8-1
RS = 50 See Figure 8-1.
2. (a) IS = 100
kV 5 = 50 A
(b) IS = 2.2
V 12 = 5.45 A
Figure 8-2
3. RS = 1.6 V
8.0 A = 0.2
4. See Figure 8-2. 5. VS = ISRS = (600 mA)(1.2 k) = 720 V
Figure 8-3
RS = 1.2 k See Figure 8-3. 6. (a) VS = (10 mA)(4.7 k) = 47 V (b) VS = (0.01 A)(2.7 k) = 27 V
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
IT = k 1.955
V 2 = 1.02 mA
I3 = mA 02.1k 89.3
k 2.2
= 577 A
65
Chapter 8
I5 = A 77.5k 2.3
k 1
= 180 A
Next, zero the 2 V source by replacing it with a short as in Figure 8-4(b). RT = 1.955 k
IT = k 1.955
V 3 = 1.53 mA
I5 = mA 53.1k 89.3
k 1.69
= 655 A
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):
I2 = mA 1.02k 3.89
k 1.69
= 443 A (downward)
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):
ILeft = mA 1.53k 3.89
k 2.2
= 865 A
I2 = A 865k 3.2
k 1
= 270 A (downward)
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):
I1 = 680
100 mA852.6
= 79.8 mA
I3 = 220
79.8 mA1020
= 17.2 mA
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
587.6 = 34.0 mA downward
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):
I1 = mA 1006.852
680
= 79.8 mA
I3 = mA 79.81020
220
= 17.2 mA
Next, zero the current source by replacing it with an open as shown in Figure 8-6(b): RT = 587.6
IT = 587.6
V 20 = 34.0 mA
I3 = mA 0.341480
680
= 15.6 mA
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):
IL = A 1k 6.1
k 2.2
= 361 mA (down)
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):
VL = V 60k 06.2
k 1.5
= 43.7 V
IL =
k 1.5
V 7.43
L
L
R
V = 29.1 mA
Total current through RL: IL = 0 A + 0 A + 29.1 mA = 29.1 mA
68
Chapter 8
Figure 8-8
13. 2 3Ref(max)
1 2 3
7.8 k30 V 15 V 30 V 15 V
12.5 k
R RV
R R R
= 3.72 V
3Ref(min)
1 2 3
6.8 k30 V 15 V 30 V 15 V
12.5 k
RV
R R R
= 1.32 V
14. 2 3Ref(max)
1 2 3
16.8 k30 V 15 V 30 V 15 V
21.5 k
R RV
R R R
= 8.44 V
3Ref(min)
1 2 3
6.8 k30 V 15 V 30 V 15 V
21.5 k
RV
R R R
= 5.51 V
15. 75 V source. See Figure 8-9(a): Req = R2 R3 (R4 + R5) = 17.2 k
VA = eq
eq 1
17.2 k75 V 75 V
99.2 k
R
R R
= 13 V
VB = V 13k 101
k 91
54
5
= 11.7 V
AV
RR
R
50 V source. See Figure 8-9(b): Req = R1 R2 (R4 + R5) = 25 k
VA = eq
eq 3
25 k50 V 50 V
58 k
R
R R
= 21.6 V
69
Chapter 8
VB = 5
4 5
91 k( 21.6 V)
101 kA
RV
R R
= 19.5 V
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
IT = k 117.6
V 100 = 850 A
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):
IL =
k 23.6
V 12
k 18k 5.6
V 12 = 508 A
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
IT = k 11.2
V 12 = 1.07 mA
IL = mA 1.07k 2.26
k 8.2
= 335 A
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
IT = k 12.47
V 6 = 481 A
IL = A 481k 6.23
k 5.6
= 114 A
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
IT = k 9.43
V 12 = 1.27 mA
IL = 2 3T
3.83 k1.27 mA
18 kL
L
R R RI
R
= 270 A
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
IT = k 11.35
V 6 = 529 A
IL = 1 3T
3.15 k529 A
18 kL
L
R R RI
R
= 93 A
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
IT = k 14.85
V 9 = 608 A
IL = 1 2T
2.81 k608 A
18 kL
L
R R RI
R
= 95 A
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
IT(S1) = k 14.0
V 32 = 2.28 mA
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) = k 11.1
V 15 = 1.35 mA
72
Chapter 8
Section 8-5 Thevenin’s Theorem 18. (a) RTH = 27 + 75 147 = 76.7 (b) RTH = 100 270 = 73
VTH = V 25222
75
= 8.45 V VTH = V 3370
100
= 811 mV
(c) RTH = 56 k 100 k = 35.9 k (b) RTH = 2.2 k (1 k + 2.2 k = 1.3 k
VTH = V) 10V (15k156
k56
= 1.79 V IAB = A 0.1k 4.5
k 2.2
= 40.7 mV
VTH = IAB(2.2 k) = (40.7 mA)(2.2 k) = 89.5 V
19. First, convert the circuit to its Thevenin equivalent as shown in the steps of Figure 8-11. RTH = 13.97 k
VA = V 32k 12.14
k 4.12
= 9.34 V
VTH = V 9.34k 6.15
k 5.6
k 6.15
k 5.6
AV = 3.35 V
IL = TH
TH
3.35 V
28.97 kL
V
R R
= 116 A
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
VTH = V 50k 8.25
k 2.65V 50
k 65.2k 6.5
k 65.2
= 16.1 V
Determine V4 due to the 50 V source using the Thevenin circuit in Figure 8-12(b).
V4 = 4TH
TH 4
10 k16.1 V
11.8 k
RV
R R
= 13.6 V
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
I3 = mA 10k 25.8
k 2.65
= 3.2 mA
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).
V4 = 4TH
TH 4
10 k17.9 V
11.8 k
RV
R R
= 15.2 V
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
75
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)):
VA = V 1876
776
= 886 mV
5 V source (Figure 8-13(b)):
VA = V 52292
3.92
= 200 mV
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
IT = k 13.8
V 48 = 3.48 mA
I4 = 2T
2 3 4
6.8 k3.48 mA
21.5 k
RI
R R R
= 1.1 mA
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
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
IA = TH
TH
42.8 V
9.72 kL
V
R R
= 4.41 mA
When R8 = 5 k:
RL = 12 k (8.2 k + 5 k) = 6.29 k
IA = TH
TH
42.8 V
10.8 kL
V
R R
= 3.97 mA
When R8 = 10 k: RL = 12 k (8.2 k + 10 k) = 7.23 k
IA = TH
TH
42.8 V
11.7 kL
V
R R
= 3.66 mA
Figure 8-14
76
Chapter 8
23. See Figure 8-15.
VTH = VA VB = V 12k 02.2
k 1.2V 12
k 2.3
k 2.2
= 8.25 V 7.13 V = 1.12 V
RTH = 1 k 2.2 k + 820 1.2 k = 688 + 487 = 1175
IL = TH
TH
1.12 V
11,175 L
V
R R
= 100 A
Figure 8-15
24. See Figure 8-16. VR3 = (0.2 mA)(15 k) = 3 V
R4 = S 3
4
10 V 3 V
0.2 mARV V
I
= 35 k
VA = 2S
1 2
12 k10 V
22 k
RV
R R
= 5.46 V
VB = 4S
3 4
35 k10 V
50 k
RV
R R
= 7V 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). (b) See Figure 8-17(b). RN = 76.7 RN = 73
RT = 166.9 IN = 270
V 3 = 11.1 mA
IT = 166.9
V 25 = 150 mA
IN = T
75 75150 mA
102 102I
= 110 mA
(c) See Figure 8-17(c). (d) See Figure 8-17(d).
RN =
k 156
)k 100)(k 56( = 35.9 k RN =
k4.5
)k 2.2)(k 2.3( = 1.3 k
IN = k 100
V 5 = 50 A IN = A 0.1
k 2.3
k 2.2
= 68.8 mA
Figure 8-17
78
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
IT = k 14.0
V 32 = 2.29 mA
I1 = mA 2.29k 2.19
k 6.5
= 668 A
IN = A 668k 6.15
k 6.5
= 240 A
Finally, the current through RL is determined by connecting RL to the Norton equivalent circuit
as shown in Figure 8-18(c).
IL = A 240k 0.29
k 0.14
= 116 A
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
IT = k 6.51
V 50 = 7.68 mA
IN = mA 7.68k 11
k 10
41
4
TI
RR
R= 6.98 mA
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):
IR5 = NN
N 5
1.92 k6.98 mA
5.22 k
RI
R R
= 2.57 mA (from B to A)
The 10 mA source acting alone. Short AB to get IN. See Figure 8-19(d):
IN = mA 10k 59.4
k 3.59mA 10
k 10k 6.5k 1
k 10k 6.5mA 10
431
43
RRR
RR = 7.82 mA
RN = 1.92 k See Figure 8-19(e):
IR5 = mA 7.82k 22.5
k 1.9
= 2.85 mA (from B to A)
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
IT = k 3.89
V 48 = 12.3 mA
I2 = T
9.11 k 9.11 k12.3 mA
6.8 k 9.11 k 6.8 k 9.11 kI
= 7.07 mA
I4 = mA 12.3k 9.15
k 6.8
= 5.27 mA
I3 = mA 5.27k 9.15
k 7.89
= 2.62 mA
IN = I2 + I3 = 7.07 mA + 2.62 mA = 9.69 mA See Figure 8-20(c):
I1 = mA 9.69k 6.13
k 4.46
= 3.18 mA
Figure 8-20
81
Chapter 8
29. Using the results of Problem 23:
Figure 8-21
IN = TH
TH
1.12 V
1175
V
R
= 953 A
RN = RTH = 1175 See 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
IT = k 27.3
V 12 = 440 A
IN1 = A 440k 3.23
k 2.8
= 156 A down
See Figure 8-22(c):
IN2 = mA 10k 97.20
k 15mA 10
k 8.2k 22k 15
k 15
= 7.15 mA down
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).
IN1 = 330
V 3 = 9.1 mA (down)
Find IN2 due to the 8 V source, as shown in Figure 8-23(b).
IN2 =
100
V 8 = 80 mA (up)
Find IN3 due to the 5 V source, as shown in Figure 8-23(c).
IN1 = 220
V 5 = 22.7 mA (down)
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 32. (a) RL = RS = 12 (b) RL = RS = 8.2 k (c) RL = RS = 4.7 + 1 2 = 6.37 (d) RL = RS = 47 + 680 = 727 33. See Figure 8-24.
Figure 8-24
As seen by RL: RS = 8.2 + 2.94 = 11.1 For maximum power transfer: RL = RS = 11.1 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:
VTH = V 1.5 49.12
7.79V 1.5
16.4 15 4.7
16.4 15
= 936 mV
IL = TH
TH
936 mV
23.4L
V
R R
= 40 mA
IL due to the 1 mA source:
I15 = mA 1 65.18
3.65mA 1
16.4 7.4 15
16.4 7.4
= 196 A
VTH = I15(15 ) = (196 A)(15 ) = 2.94 mV
IL = TH
TH
2.94 mV
23.4 L
V
R R
= 126 mA
IL(total) = 40 mA + 126 A = 40.126 mA
PL = = (40.126 mA)212.21 = 19.7 mW LLRI 2
84
Chapter 8
35. For maximum power transfer, RTH = RLADDER The voltage across RTH = 24 V (one half of VTH)
RTH = A 0.5
V 24 = 48
RLADDER = 48 RLADDER = ((R4 (R5 + R6) + R3) R2) + R1
10
69
6947
471069
69
4
4
4
4
R
RR
R
= 26
57
69
69
47
2610
69
69
4
4
4
4
R
R
R
R
53.21105747
26
47
261
69
69
4
4
R
R
69R4 = 69(48.17) + 48.17R4 R4(69 48.17) = 69(48.17)
R4 = 17.4869
)17.48(69
= 160
Section 8-8 Delta-Wye (Y) and Wye-Delta (Y-) Conversions
36. (a) R1 =
M 06.3
)M 1)(k 560(
CBA
CA
RRR
RR = 183 k
R2 =
M 06.3
)M 1)(M 5.1(
CBA
CB
RRR
RR = 490 k
R3 =
M 06.3
)M 5.1)(k 560(
CBA
BA
RRR
RR = 275 k
(b) R1 =
9.5
) 2.2)( 1(
CBA
CA
RRR
RR = 373 m
R2 =
9.5
) 7.2)( 2.2(
CBA
CB
RRR
RR = 1.01
R3 =
9.5
) 7.2)( 1(
CBA
BA
RRR
RR = 4.58 m
85
Chapter 8
37. (a) RA =22
876
22
)18)(22()18)(12()22)(12(
2
323121
R
RRRRRR= 39.8
RB = 12
876
12
)18)(22()18)(12()22)(12(
1
323121
R
RRRRRR= 73
RC = 18
876
18
)18)(22()18)(12()22)(12(
3
323121
R
RRRRRR= 48.7
(b) RA =
k3.3
)k7.4)(k3.3()k7.4)(k8.6()k3.3)(k8.6(
2
323121
R
RRRRRR= 21.2 k
RB =
k8.6
)k7.4)(k3.3()k7.4)(k8.6()k3.3)(k8.6(
1
323121
R
RRRRRR= 10.3 k
RC =
k7.4
)k7.4)(k3.3()k7.4)(k8.6()k3.3)(k8.6(
3
323121
R
RRRRRR= 14.9 k
38. Convert the delta formed by R3, R4, and R5 to a Wye configuration. See Figure 8-25:
RY1 =
k 1.43
)k 12)(k 22(
543
43
RRR
RR = 6.13 k
RY2 =
k 1.43
)k 1.9)(k 22(
543
53
RRR
RR = 4.65 k
RY3 =
k 1.43
)k 1.9)(k 12(
543
54
RRR
RR = 2.53 k
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
IT = T
136 V 136 V
14.3 kR
= 9.5 mA
IR1 = IRY1 = mA 5.9k 78.59
k 65.43
2211
22
T
YY
Y IRRRR
RR= 6.94 mA
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:
Figure 8-25
IR4 =
k 12
V 6.66
4R
VB = 5.55 mA
IR5 =
k 9.1
V 16.36
5R
VC = 3.97 mA
IR3 =
k 22
V 36.16V 6.66
3R
VV CB = 1.38 mA
86
Chapter 8
87
Multisim Troubleshooting and Analysis 39. R1 is leaky.
0. VTH = 17.478 V; RTH = 247.279
1. IN = 0.383 mA; RN = 9.674 k
2. R3 is shorted.
3. IAB = 1.206 mA; VAB = 3.432 V
4 4 4 4
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
I1 = 100
5030 2I
100
503075 2I
+ 90I2 = 15
22.5 37.5I2 + 90I2 = 15 52.5I2 = 7.5 I2 = 143 mA 100I1 + 50(0.143) = 30 I1 = 371 mA
2. (a) 32
64 = 12 12 = 0 (b)
50
19 = 45 0 = 45
(c) 12
1512
= 12 (30) = 18 (d)
2030
50100
= 2000 1500 = 3500
3. (a) I1 = 143
1212
37
21
36
24
= 0 A (b) I2 = 143
286
37
21
67
41
= 2 A
4. (a)
10
4
0
2
5
1
0102
145
201
= (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)
3.0
2.1
1
1.0
1.0
5.0
53.01.0
5.12.11.0
8.015.0
= (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) 25 0 20 25 0
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
0 2.12 0.98 0 2.1
1 3.49 1.05 1 3.49
2
= (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
0
5.0
3
75.0
2
12.03
5.1075.0
05.02
= (0 + 2.25 + 0) (0 + 0.6 0.375) = 2.25 0.225 = 2.025
I3 = 35.2
025.2 = 862 mA
7. The characteristic determinant is:
2 6 10 2
3 7 8 3 7
10 5 12 10 5
6
= (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
3 7 8 3 7
0 5 12 0 5
374
6
= (9)(7)( 12) ( 6)( 8)(0) (10)(3)(5) [(0)(7)(10) (5)( 8)(9) ( 12)(3)( 6)]
374
= 606 144 462
374 374
= 1.24 A
89
Chapter 9
I2 =
2 9 10 2 9
3 3 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
374 374
= 2.05 A
I3 =
2 6 9 2
3 7 3 3 7
10 5 0 10 5
374
6
= (2)(7)(0) ( 6)(3)(10) (9)(3)(5) [(10)(7)(9) (5)(3)(2) (0)(3)( 6)]
374
= 45 660 705
374 374
= 1.89 A
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
I2 = 2.18
2.812 3I
2.18
2.81210 3I
+ 5.6I3 = 6
2.18
82120 3I + 5.6I3 = 6
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
10047
12 AA VV
= 0.1
100(12 VA) 47VA = 470 1200 100VA 47VA = 470 147VA = 730 VA = 4.97
I1 =
47
V 7.03
47
V 4.97 V 12 = 150 mA
I2 = 100
V 4.97 = 49.7 mA
Figure 9-1
I3 = 100 mA (current source) 15. Current source zeroed (open). See Figure 9-1(a).
VAB = V2 = 2S
1 2
10012 V
147
RV
R R
= 8.16 V
Voltage source zeroed (shorted). See Figure 9-1(b). VAB = V3 = I3R3 = (100 mA)(68 ) = 6.8 V
I2 = 1S
1 2
47100 mA
147
RI
R R
= 31.97 mA
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
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
I1 = 200,839,1
9400
600,313800,152,2
11208280
1380560
5601560
13802
5606
= 5.11 mA
I2 = 200,839,1
33603180
200,839,1
2560
61560
= 3.52 mA
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
43,595.8I1 =
1.5 10 0 1.5 10
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)]
I1 = 1302.735 624.135 678.6
43,595.8 43,595.8
= 15.6 mA
43,595.8I2 =
57 1.5 0 57 1.5
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)]
I2 = 3368.7 697.35 2671.35
43,595.8 43,595.8
= 61.3 mA
Substituting into the third equation to get I3: 19.7I3 = 1.5 + 4.7I2
I3 = 1.5 4.7( 0.0613 A)
19.7
= 61.5 mA
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
I1 = 49.273
152
61.471.321
152
)9.6)(9.6()19)(9.16(
)19)(8(
199.6
9.69.16
190
9.68
= 555 mA
I2 = 49.273
2.55
61.471.321
2.55
)9.6)(9.6()19)(9.16(
)9.6)(8(
199.6
9.69.16
09.6
89.16
= 202 mA
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.
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 The characteristic determinant is:
16.9 4.7 2.2 16.9 4.7
4.7 22.9 10 4.7 22.9
2.2 10 16.1 2.2 10
= (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
3867.576I2 =
16.9 8 2.2 16.9 8
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
3867.576 3867.576
= 202 mA
3867.576I2 =
16.9 4.7 8 16.9 4.7
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
3867.576 3867.576
= 201 mA
IBA = I2 I3 = 202 mA 201 mA = 1 mA
95
Chapter 9
25. See Figure 9-4.
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. 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:
14768
40
82
30 AAA VVV
= 0
014760
40
688282
30 AAA VVV
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
Figure 9-5
VAB = VA = 626,27
040,782 = 28.3 V
96
Chapter 9
29. Use VAB = 28.3 V from Problem 28.
I1 =
82
V 3.28V 30
82
V 30 ABV = 20.6 mA
I2 =
68
V 40 V 28.3
68
V 40ABV = 172 mA
I3 =
147
V 28.3
147ABV
= 193 mA
Figure 9-6
30. See Figure 9-6. I1 I2 I3 = 0 I3 + I4 I5 = 0
Substituting into the first equation and simplifying:
271047
5.1 BAAA VVVV
= 0
2727104747
5.1 BAAA VVVV = 0
47
5.1
279.126
479.12627
BAAA VVVV
47
5.1
279.126
9.200 BA VV
1.58VA 0.037VB = 0.0319 Substituting into the second equation and simplifying:
15
5.1
7.4
3
27
BBBA VVVV = 0
5
5.1
157.47.4
3
2727 BBBA VVVV
= 0
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
Figure 9-7
Node B: I3 I4 I5 = 0
I1 = 1
V 9
R
V A
I2 = 2R
VA
I3 = 3R
VV BA
I4 = 4
V 4.5
R
VB
I5 = 5
V 1.5
R
VB
Node A: 912756
9 BAAA VVVV
= 0
9191275656
9 BAAA VVVV = 0
056
9
91592,137
151250962457
BAAA VVVV
56
9
91
1
592,137
9065
BA VV = 0
0.0659VA + 0.0109VB = 0.1607
Node B: 82
15
33
5.4
91
BBBA VVVV = 0
082
15
8233
5.4
339191 BBBA VVVV
2706
)15)(33()5.4)(32(
246,246
300374622706
91
AAAA VVVV
= 0
2706
864
246,246
171,131
91 BA VV
= 0
0.0109VA 0.0535VB = 0.3193
The characteristic determinant is:
0535.00109.0
0109.00659.0
= 0.0035 0.0001 = 0.0034
0.0034VA = 0535.03193.0
0109.01607.0
= 0.0086 0.0035 = 0.0051
VA = 0034.0
0051.0 = 1.5 V
98
Chapter 9
0.0034VB = 3193.00109.0
1607.00659.0
= 0.0210 0.0018 = 0.0192
VB = 0034.0
0192.0 = 5.65 V
32. See Figure 9-8.
Figure 9-8
Node A: I1 I2 + I3 + I4 = 0 Node B: I2 + I5 I6 = 0 Node C: I3 + I7 + I8 = 0
I1 =
k 1
V 24 AV I5 =
k 1
V 24 BV
I2 = k 1
BA VV I6 =
k 1
V 18BV
I3 = k 1
AC VV I7 =
k 1
V 10 CV
I4 = k 1AV
I8 =
k 1
V 18 CV
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:
301
031
114
= (4)(3)(3) (1)(3)(1) (1)(1)(3) = 36 + 3 + 3 = 30
30VA =
3028
0342
1124
= (24)(3)(3) (28)(3)(1) (1)(42)(3) = 2166 84 126
= 426
VA = 30
426
= 14.2 V
99
Chapter 9
30VB =
3281
0421
1244
= (4)(42)(3) + (1)(28)(1) (1)(42)(1) (42)(1)(3)
= 504 28 + 42 72 = 562
VB = 30
562
= 18.7 V
30VC =
2801
4231
2414
= (4)(3)(28) + (1)(42)(1) (1)(3)(24) (1)(1)(28)
= 336 42 72 28 = 422
VC = 30
422
= 14.1 V
33. See Figure 9-9.
I7 = k 2
V 4.32 = 2.16 mA
VC = +4.32 V 20 V = 15.7 V
I6 =
k 20
V 10.43
k 20
V) 7.15( V 5.25= 522 A
I4 = k 16
V 5.25 = 328 A
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
I2 = k 10
V 3.70 = 370 A
I5 = I7 I4 I2 = 2.16 mA 328 A 370 A = 1.46 mA VB = (1.46 mA)(4 k) = 5.85 V
I3 =
k12
V 2.14
k 12
V) 5.85( V 70.3
k 12BA VV
= 179 A
I8 = I3 + I5 = 179 A + 1.46 mA = 1.64 mA
100 Figure 9-9
Chapter 9
101
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.
Chapter 10 Magnetism and Electromagnetism Note: Solutions show conventional current direction.
Section 10-1 The Magnetic Field
1. Since B = A
, when A increases, B (flux density) decreases.
2. B = 2m 0.5
Wb1500
A
= 3000 Wb/m2 = 3000 T
3. B = A
There are 100 cm per meter:
2
2
cm 000,10
m 1
cm 100
m 1
Converting 150 cm 2 to m2:
A =
2
22
cm 000,10
m 1cm 150 = 0.015 m2
= BA = (2.5 103 T)(0.015 m2) = 37.5 Wb 4. 1 T = 104 gauss
B = 4
1 T(0.6 gauss)
10 gauss
= 60 T
5. B = 410 gauss
(100,000 T)1 T
= 1000 gauss
Section 10-2 Electromagnetism 6. The compass needle turns 180.
7. r = 0
0 = 4 107 Wb/At m
r = mWb/At104
mWb/At107507-
6
= 597
8. Reluctance = )m m)(0.08Wb/At10(150
m 28.0
A
127-
= 233,333 At/Wb
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) The electromagnetic field causes the plunger to move when the solenoid is activated. (b) The spring force returns the plunger to its inactive position. 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.
Section 10-4 Magnetic Hysteresis 14. Fm = 150 At
H = m 0.2
At 150
l
Fm = 750 At/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) H = (500 t)(0.25A)
0.3 mmF NI
l l = 417 At/m
(b) = Al
NI
reluctance
Fm
/
= ro = (250)(4 107) = 3142 107 Wb/At m A = (2 cm)(2 cm) = (0.02 m)(0.02 m) = 4 104 m2
=
6
7 4 2
(500 t)(0.25 A) 125 At
2.39 10 At/Wb0.3 m
3142 10 Wb/At m 4 10 m
= 5.23 Wb
(c) B = 4 2
5.23 Wb
4 10 mA
= 0.13 T
17. 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
104
20. Vind =
dt
dN
= 50(3500 103 Wb/s = 175 V
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. A L F 12 A + 1 A = 13 AI I I 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%
Chapter 11 Introduction to Alternating Current and Voltage
Section 11-1 The Sinusoidal Waveform
1. (a) f = s 1
11
T = 1 Hz (d) f =
ms1
11
T = 1 kHz
(b) f = ms 0.2
11
T = 5 Hz (e) f =
s 500
11
T = 2 kHz
(c) f = ms 50
11
T = 20 Hz (f) f =
s 10
11
T = 100 kHz
2. (a) T = Hz 1
11
f = 1 s (d) T =
kHz 1
11
f = 1 ms
(b) T = Hz 60
11
f = 16.7 ms (e) T =
kHz 200
11
f = 5 s
(c) T = Hz 500
11
f = 2 ms (f) T =
MHz 5
11
f = 200 ns
3. T = cycles 5
s10 = 2 s
4. T = kHz 50
11
f = 20 s
ms 0.02
ms 10 = 500 cycles
5. 1 1
0.1ms10 kHz
Tf
Time for 100 cycles = 100(0.1 ms) = 10 ms
Section 11-2 Sinusoidal Voltage and Current Values 6. (a) Vrms = 0.707Vp = 0.707(12 V) = 8.48 V
(b) Vpp = 2Vp = 2(12 V) = 24 V
(c) Vavg = 0 V over a full cycle. Vavg = 0.637(12 V) = 7.64 over a half cycle.
105
Chapter 11
7. (a) Ip = 1.414Irms = 1.414(5 mA) = 7.07 mA
(b) Iavg = 0 A over a full cycle
Iavg = 0.637Ip = 0.637(7.07 mA) = 4.5 mA over a half cycle
(c) Ipp = 2Ip = 2(7.07 mA) = 14.14 mA
8. 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) 6
30180
rad rad (d)
4
3
135180
rad rad
(b) 4
45180
rad rad (e)
9
10
200180
rad rad
(c) 90
39
78180
rad rad (f)
3
5
300180
rad rad
10. (a)
rad
3.57rad
8
= 22.5 (d)
rad
3.57rad
5
3 = 108
(b)
rad
3.57rad
3
= 60 (e)
rad
3.57rad
5
6 = 216
(c)
= 90 (f) rad
3.57rad
2
rad
57.3rad) 8.1( = 324
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
10
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) i = Ipsin = (100 mA)sin 35 = 57.4 mA (b) i = Ipsin = (100 mA)sin 95 = 99.6 mA (c) i = Ipsin = (100 mA)sin 190 = 17.4 mA (d) i = Ipsin = (100 mA)sin 215 = 57.4 mA (e) i = Ipsin = (100 mA)sin 275 = 99.6 mA (f) i = Ipsin = (100 mA)sin 360 = 0 mA 16. Vp = 1.414Vrms = 1.414(6.37 V) = 9 V (e) = 180
(a) 8
= 22.5 v = (9 V)sin 180 = 0 V
(f) 2
3 = 270
v = (9 V)sin 22.5 = 3.44 V
(b) 4
= 45 v = (9 V)sin 270 = 9 V
(g) 2 = 360 v = (9 V)sin 360 = 0 V v = (9 V)sin 45 = 6.36 V
(c) 2
= 90
v = (9 V)sin 90 = 9 V
(d) 4
3 = 135
v = (9 V)sin 135 = 6.36 V 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) vB = (15 V)sin ( 30) = (15 V)sin(30 30) = (15 V)sin(0) = 0 V (b) vB = (15 V)sin ( 30) = (15 V)sin(45 30) = (15 V)sin(15) = 3.88 V (c) vB = (15 V)sin ( 30) = (15 V)sin(90 30) = (15 V)sin(60) = 13.0 V (d) vB = (15 V)sin ( 30) = (15 V)sin(180 30) = (15 V)sin(150) = 7.5 V (e) vB = (15 V)sin ( 30) = (15 V)sin(200 30) = (15 V)sin(170) = 2.60 V (f) vB = (15 V)sin ( 30) = (15 V)sin(300 30) = (15 V)sin(270) = 15 V
7
Chapter 11
19. T = kHz 2.2
11
f = 4.55 s
At t = 0.12 ms = 120 s:
= 120 s
360455 s
= 94.9
Vp = 0.707
V 25 = 35.4 V
v = (35.4 V)sin 94.9 = 35.3 V At t = 0.2 ms = 200 s:
= 200 s
360455 s
= 158
v = (35.4 V)sin 158 = 13.3 V v = 35.4 V 13.3 V = 22.1 V
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) f =
2
60
2
= 9.55 Hz
(b) f =
2
360
2
= 57.3 Hz
(c) f =
2
2
2
= 0.318 Hz
108
Chapter 11
(d) f =
2
1256
2
= 200 Hz
23. (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
Section 11-6 Analysis of AC Circuits
24. (a) Irms =
k1
V 10707.0707.0
R
Vp = 7.07 mA
(b) Iavg = mA 102
= 6.37 mA
(c) Ip = k1
V10 = 10 mA
(d) Ipp = 2(10 mA) = 20 mA (e) 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 =
k1
V 16V 16
1R = 16 mA
Irms = 0.707
2
mA 16707.0
2ppI
= 5.66 mA
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
P = Pac + PDC =
100
V) 200(
100
V) 1.106( 2222
L
S
L
rms
R
V
R
V = 112.6 W + 400 W = 513 W
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)
rps = s/m 60
rpm 3600 = 60 rps
f = (2 pole pairs)(60 rps) = 120 Hz
33. rps = 2
Hz400
pairs pole
f = 200 rps
34. f = (number of pole pairs)rps
400 Hz 400 Hz#pole pairs 8
3000 rpm 50 rps60 s/m
f
rps
# 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. T = kHz 2
1 = 0.5 ms = 500 s
% duty cycle = %100s 500
s 1%100
T
tW = 0.2%
39. Vavg = baseline + (duty cycle)(amplitude)
duty cycle = s6
s1
T
tW = 0.167
Vavg = 5 V + (0.167)(5 V) = 5.84 V
40. (a) % duty cycle = %100s 4
s 1%100
T
tW = 25%
(b) % duty cycle = %100ms 30
ms 20%100
T
tW = 66.7%
41. (a) 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
42. (a) f = s 4
11
T = 250 kHz
(b) f = ms 30
11
T = 33.3 Hz
43. (a) f = s 20
11
T = 50 kHz
(b) f = ms 100
11
T = 10 Hz
44. Average value = ms 7
ms) V)(1 6 V 5 V 4 V 3 V 2 V 1 V (0
period
curveunder area
= ms 7
msV21 = 3 V
45. f = s 40
11
T = 25 kHz (fundamental)
3rd harmonic = 75 kHz 5th harmonic = 125 kHz 7th harmonic = 175 kHz 9th harmonic = 225 kHz 11th harmonic = 275 kHz 13th harmonic = 325 kHz
111
Chapter 11
112
46. f = s 40
11
T= 25 kHz
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
fin = s 200
1
= 5 kHz
Rtot = 560 + (470 (560 + 470 )) = 560 + 323 = 883
Vp(out) = V 5883
323
1030
470
883
323
560470
470)(
inpV = 835 mV
fout = fin = 5 kHz 49. Vp(out) = (3 div)(0.2 V/div) = 0.6 V Tout = (10 div)(50 ms/div) = 500 ms
fout = ms 500
1 = 2 Hz
Rtot = 1 k + 1 k 3.2 k = 1762
Vp(in) = V) V)(0.6 3.2(762
1762V) 2.3(
762
1762)(
outpV = 4.44 V
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. 53. No fault. 54. R1 is open. 55. VMIN = 2.000 Vp; VMAX = 22.000 Vp
56. VMIN = 4.000 Vp; VMAX = 16.000 Vp
Chapter 12 Capacitors
Section 12-1 The Basic Capacitor
1. (a) C = V 10
C 50
V
Q = 5 F
(b) Q = CV = (0.001 F)(1 kV) = 1 C
(c) V = F 200
mC 2
C
Q = 10 V
2. (a) (0.1 F)(106 pF/F) = 100,000 pF (b) (0.0025 F)(106 pF/F) = 2500 pF (c) (4.7 F)(106 pF/F) = 4,700,000 pF 3. (a) (1000 pF)(106 F/pF) = 0.001 F (b) (3500 pF)(106 F/pF) = 0.0035 F (c) (250 pF)(106 F/pF) = 0.00025 F 4. (a) (0.0000001 F)(106 F/F) = 0.1 F (b) (0.0022 F)(106 F/F) = 2200 F (c) (0.0000000015 F)(106 F/F) = 0.0015 F
5. W = 21 1(1 1000 F)(500 V)
2 2CV = 125 J
6. W = 2
2
1CV
C = 22 V) (100
mJ) 2(102
V
W = 2 F
7. (a) Air: = r0 = 1(8.85 1012 F/m) = 8.85 1012 F/m (b) Oil: = r0 = 4.0(8.85 1012 F/m) = 35.4 1012 F/m (c) Glass: = r0 = 7.5(8.85 1012 F/m) = 66.4 1012 F/m (d) Teflon: = r0 = 2.0(8.85 1012 F/m) = 17.7 1012 F/m
8. C = 12 3 12
5
(8.85 10 F/m) (1.44 10 )(5)(8.85 10 F/m)
6.35 10 mrA
d
= 0.001 F
9. C = d
A r )F/m1085.8( 12
= 2 1
4
(0.05 m )(1.0)(8.85 10 F/m)
4.5 10 m
2
= 983 pF
113
Chapter 12
10. C = 128.85 10rA
d
A = 5
12 12
(1)(8 10 )
8.85 10 (2.5)(8.85 10 )r
Cd
= 3.6 106 m2
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!
11. C = 12 12
5
8.85 10 (0.09)(2.5)(8.85 10 )
(8.0 10 )rA
d
= 24.9 nF = 0.0249 F
12. T = 50 C (200 ppm/C)50 C = 10,000 ppm
C = ppm) 1010(101
101 36
3
= 10 pF
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
C = ppm) 105.12(101
1000 36
= +12.5 pF
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).
Figure 12-1
16. See Figure 12-1. 17. Aluminum, tantalum; electrolytics are polarized, others are not. 18. (a) Encapsulation (b) Dielectric (ceramic disk) (c) Plate (metal disk) (d) Conductive leads 19. (a) 0.022 F (b) 0.047 F (c) 0.001 F (d) 220 pF
114
Chapter 12
Section 12-3 Series Capacitors
20. CT = 5
pF 1000= 200 pF
21. (a) CT =
F 2.2
1
F 1
11
= 0.688 F
(b) CT =
pF 390
1
pF 560
1
pF 100
11
= 69.7 pF
(c) CT =
F 22
1
F 47
1
F 4.7
1
F 10
11
= 2.64 F
22. (a) CT = 0.688 F
V1F = T 0.688 F10 V 10 V
1 F 1 F
C
= 6.88 V
V2.2F = T 0.688 F10 V 10 V
2.2 F 2.2 F
C
= 3.13 V
(b) CT = 69.7 pF
V100pF = T 69.7 pF100 V 100 V
100 pF 100 pF
C
= 69.7 V
V560pF = T 69.7 pF100 V 100 V
560 pF 560 pF
C
= 12.4 V
V390pF = T 69.7 pF100 V 100 V
390 pF 390 pF
C
= 17.9 V
(c) CT = 2.64 F
V10F = T 2.64 F30 V 30 V
10 F 10 F
C
= 7.92 V
V4.7F = T 2.64 F30 V 30 V
4.7 F 4.7 F
C
= 16.9 V
V47F = T 2.64 F30 V 30 V
47 F 47 F
C
= 1.69 V
V22F = T 2.64 F30 V 30 V
22 F 22 F
C
= 3.60 V
115
Chapter 12
23. Vx = TS
x
CV
C
CT = S
(1 F)(8 V)
12 Vx xC V
V
= 0.667 F
Cx = TS
0.667 F12 V
4 V
CV
V
= 2 F
24. QT = Q1 = Q2 = Q3 = Q4 = 10 C
V1 = F 4.7
C 10
1
1
C
Q = 2.13 V
V2 = F 1
C 10
2
2
C
Q = 10 V
V3 = F 2.2
C 10
3
3
C
Q = 4.55 V
V4 = F 10
C 10
4
4
C
Q = 1 V
Section 12-4 Parallel Capacitors 25. (a) CT = 47 pF + 10 pF + 1000 pF = 1057 pF (b) 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
27. (a) CT =
F .33F 2.2
1
F 10
1
F 10
11
= 2.62 F
(b) CT =
pF 470
1
F 0.001
11
pF 470
1
pF 1000
11
pF 100
1
pF 100
11
= 50 pF + 319.7 pF + 319.7 pF = 689 pF
116
Chapter 12
(c) CT = F1F1
F1
F 1
1
F 1
11
11
= 1.6 F
470 pF
Figure 12-2
28. (a) CT = 2.62 F
VAB = V 5F5
F62.2
= 2.62 V
(b) See Figure 12-2(a): For this part of the circuit:
CT =
Fp470
1
F001.0
11
= 319.7 pF
VAB = V 10F001.0
pF 319.7
= 3.20 V
(c) See Figure 12-2(b).
CAB = 1.5 F For this part of the circuit:
CT =
F 1.5
1
F 1
11
= 0.6 F
VAB = V 10F 1.5
F 0.6
= 4 V
29. (a) CT = C1,2 + C3,4 = 0.00872 F + 0.0256 F = 0.0343 F QT = CTVT = (0.0343 F)(12 V) = 0.411 C
(b) V1 = 2T
1 2
0.068 F12 V
0.01 F 0.068 F
CV
C C
= 10.47 V
V2 = 1T
1 2
0.01 F12 V
0.01 F 0.068 F
CV
C C
= 1.54 V
V3 = 4T
3 4
0.056 F12 V
0.047 F 0.056 F
CV
C C
= 6.52 V
V4 = 3T
3 4
0.047 F12 V
0.047 F 0.056 F
CV
C C
= 5.48 V
117
Chapter 12
Section 12-5 Capacitors in DC Circuits
30. (a) = RC = (100 )(1 F) = 100 s (b) = RC = (10 M)(47 pF) = 470 s (c) = RC = (4.7 k)(0.0047 F) = 22.0 s (d) = RC = (1.5 M)(0.01 F) = 15 ms
31. (a) 5 = 5RC = 5(56 )(47 F) = 13.2 ms (b) 5 = 5RC = 5(3300 )(0.015 F) = 247.5 s (c) 5 = 5RC = 5(22 k)(100 pF) = 11 s (d) 5 = 5RC = 5(5.6 M)(10 pF) = 280 s
32. = RC = (10 k)(0.001 F) = 10 s (a) vC = VF(1 et/RC) = 15 V(1 e10s/10s) = 15 V(1 e1) = 9.48 V (b) vC = VF(1 et/RC) = 15 V(1 e20s/10s) = 15 V(1 e2) = 13.0 V (c) vC = VF(1 et/RC) = 15 V(1 e30s/10s) = 15 V(1 e3) = 14.3 V (d) vC = VF(1 et/RC) = 15 V(1 e40s/10s) = 15 V(1 e4) = 14.7 V (e) vC = VF(1 et/RC) = 15 V(1 e50s/10s) = 15 V(1 e5) = 14.9 V 33. = RC = (1 k)(1.5 F) = 1.5 ms (a) vC = Vie
t/RC = (25 V)e1.5ms/1.5ms = (25 V)e1 = 9.20 V (b) vC = Vie
t/RC = (25 V)e4.5ms/1.5ms = (25 V)e3 = 1.24 V (c) vC = Vie
t/RC = (25 V)e6ms/1.5ms = (25 V)e4 = 0.458 V (d) vC = Vie
t/RC = (25 V)e7.5ms/1.5ms = (25 V)e5 = 0.168 V 34. (a) vC = VF(1 et/RC) = 15 V(1 e2s/10s) = 15 V(1 e0.2) = 2.72 V (b) vC = VF(1 et/RC) = 15 V(1 e5s/10s) = 15 V(1 e0.5) = 5.90 V (c) vC = VF(1 et/RC) = 15 V(1 e15s/10s) = 15 V(1 e1.5) = 11.7 V 35. (a) vC = Vie
t/RC = (25 V)e0.5ms/1.5ms = (25 V)e0.333 = 17.9 V (b) vC = Vie
t/RC = (25 V)e1ms/1.5ms = (25 V)e0.667 = 12.8 V (c) vC = Vie
t/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
et/RC = F
CF
V
vV
ln et/RC =
F
CF
V
vVln
RC
t =
F
CF
V
vVln
t = RC
F
C
V
v1ln
t = (2.2 k)(0.01 F) ln
V 12
V 61 = 15.2 s
118
Chapter 12
37. t = RC
F
C
V
v1ln = (10 k)(0.001 F)
V 15
V 81ln = 7.62 s
38. t = RC
i
C
V
vln = (1 k)(1.5 F)
V 25
V 3ln = 3.18 ms
39. 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
40. t = RC
F
C
V
v1ln
R =
10
2.71pF)ln (1000
s10
1lnF
C
V
vC
t = 7.86 k
Figure 12-3
41. 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
119
Chapter 12
Section 12-6 Capacitors in AC Circuits
42. (a) XC = )F kHz)(0.047 1(2
1
2
1
fC = 3.39 k
(b) CT = 10 F + 15 F = 25 F
XC = T
1 1
2 2 (1 Hz)(25 F)fC
= 6.37 k
(c) CT =
F 1
1
F 1
11
= 0.5 F
XC = T
1 1
2 2 (60 Hz)(0.5 F)fC
= 5.31 k
43. CT for each circuit was found in Problem 27.
(a) XC = T
1 1
2 2 (2 kHz)(2.62 F)fC
= 30.4
(b) XC = T
1 1
2 2 (2 kHz)(689 pF)fC
= 116 k
(c) XC = T
1 1
2 2 (2 kHz)(1.6 F)fC
= 49.7
44. (a) For XC = 100 :
f = )F )(0.047100(2
1
2
1
CXC
= 33.9 kHz
For XC = 1 k:
f = )F )(0.047k1(2
1
2
1
CXC
= 3.39 kHz
(b) For XC = 100 :
f = )F )(25100(2
1
2
1
CXC
= 63.7 Hz
For XC = 1 k:
f = )F )(25k1(2
1
2
1
CXC
= 6.37 Hz
(c) For XC = 100 :
f = )F )(0.5100(2
1
2
1
CXC
= 3.18 kHz
For XC = 1 k:
f = )F )(0.5k1(2
1
2
1
CXC
= 318 Hz
120
Chapter 12
45. XC = mA 100
V 20
rms
rms I
V = 200
46. Vrms = IrmsXC
XC = )F 7kHz)(0.004 10(2
1
= 3.39 k
Vrms = (1 mA)(3.39 k) = 3.39 V
47. XC = fC2
1 = 3.39 k
Ptrue = 0 W
Pr = =(1 mA)2(3.39 k) = 3.39 mVAR CXI 2rms
48. 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
IC1 =
k 66.3
V 10
CT
s
X
V = 151 A
VC1 = V 10F 0.01
F 0.008
1
s
T VC
C= 8.00 V
VC2 = Vs VC1 = 10 V 8.00 V = 2.00 V XC2 = 24.1 k
IC2 =
k 24.1
V 2.00
2
2
C
C
X
V = 83.0 A
VC3 = V 2.00F 0.015
F 0.0122
3
6543
CV
C
C = 1.6 V
XC3 = 35.4 k
IC3 =
k 35.4
V 1.6
3
3
C
C
X
V = 45.2 A
VC4 = VC2 VC3 = 2.00 V 1.6 V = 400 mV XC4 = 11.3 k
IC4 =
k 11.3
mV 400
4
4
C
C
X
V = 35.4 A
VC5 = mV 400F 0.01
F 0.0064
5
65
CV
C
C= 240 mV
XC5 = 53.1 k
IC5 = IC6 =
k 53.1
mV 240
5
5
C
C
X
V = 4.52 A
VC6 = VC4 VC5 = 400 mV 240 mV = 160 mV
121
Chapter 12
49. VC2 = VC3 = (4 mA)XC3 = (4 mA)(750 ) = 3 V
f = )F0015.0)(750(2
1
2
1
33
CXC
= 141.5 kHz
XC2 = )F0022.0(kHz) 141.5(2
1
2
1
2
fC = 511.3
IC2 =
3.511
V 3
2
2
C
C
X
V = 5.87 mA
IC1 = ICT = IC2 + IC3 = 5.87 mA + 4 mA = 9.87 mA VC1 = 5 V 3 V = 2 V
XC1 = mA87.9
V 2
1
1 C
C
I
V = 203
C1 = )203(kHz) 141.5(2
1
2
1
1
CfX = 0.00541 F
50. CT(3,5,6)
F015.0
1
F01.0
1
F015.0
11
1111
653
CCC
= 0.0043 F
CT(2,3,5,6) = 0.022 F + 0.0043 F = 0.0263 F
CT =
F0263.0
1
F01.0
11
111
6,5,3,21
CC
= 0.00725 F
VC1 = T
1
0.00725 F10 V 10 V
0.01 F
C
C
= 7.25 V
VC2 = T
2
0.00725 F10 V 10 V
0.022 F
C
C
= 3.30 V
VC3 = T(3,5,6)2
3
0.0043 F3.30 V
0.015 FC
CV
C
= 945 mV
VC5 = T(3,5,6)2
5
0.0043 F3.30 V
0.01 FC
CV
C
= 1.42 V
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
123
Section 12-8 Switched-Capacitor Circuits
53. R = 10 s
2200 pF
T
C
= 4.55 k
54. T = 1 1
8 kHzf = 125 s
R = 125 s
100 pF
T
C
= 1.25 M
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.
Chapter 13 Inductors
Section 13-1 The Basic Inductor 1. (a) 1 H 1000 mH/H = 1000 mH (b) 250 H 0.001 mH/H = 0.25 mH (c) 10 H 0.001 mH/H = 0.01 mH (d) 0.0005 H 1000 mH/H = 0.5 mH 2. (a) (300 mH)(103) = 300,000 H (b) (0.08 H)(106) = 80,000 H (c) (5 mH)(103) = 5000 H (d) (0.00045)(103) = 0.45 H
3. vind =
s
mA 10H5
dt
diL = 50 mV
4. v =
dt
diL
mH 25
mV 50
L
v
dt
di = 2000 A/s
5. vind =
s 1
mA 200mH 100
dt
diL = 20 mV
6. L = 2N A
l
N = )m1010)(102.1(
m) mH)(0.05 (30256
A
Ll
= 3536 turns
7. W = 2 21 1(4.7 mH)(20 mA)
2 2LI = 0.94 J
8. L = 2N A
l
; Inductor 2 has 4 times the inductance of inductor 1.
124
Chapter 13
9. L = 2N A
l
1 = r0 = 200 0 2 = r0 = 150 0
2
1
= 150 3
200 4
Therefore, coil 2 has 3/4 the inductance of coil 1.
L2 = 1
3
4L
10. A = r2 = (0.0035 m)2 = 38.5 106 m2
L = 2N A
l
=
2 6 6100 (4 10 H/m)(38.5 10 m )
0.035 m
2
= 138 H
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 =
H15
1
H25
1
H50
1
H75
11
= 7.14 H
16. 8 mH = mH 12
)mH 21(
1
1
L
L
( 8 mH)L1 + (8 mH)(12 mH) = (12 mH)L1 (4 mH)L1 = 96 mH2
L1 = mH 4
mH 96 2
= 24 mH
125
Chapter 13
17. (a) LT = 1 H + H5 H10
H) H)(5 (10
= 4.33 H
(b) LT = 2
mH 100 = 50 mH
(c) LT = 1
1 1 1
100 H 200 H 400 H
= 57.1 H
18. (a) LT = mH100
mH) mH)(40 (60
mH 150
mH) mH)(50 (100 = 33.3 mH + 24 mH = 57.3 mH
(b) LT = mH18
mH) mH)(6 (12 = 4 mH
(c) LT = 4 mH + mH 6
mH) mH)(4 (2= 5.33 mH
Section 13-4 Inductors in DC Circuits
19. (a) =
100
H100
R
L = 1 s
(b) =
k7.4
mH10
R
L = 2.13 s
(c) =
M5.1
H3
R
L = 2 s
20. (a) 5 =
56
H5055
R
L = 4.46 s
(b) 5 =
3300
Hm1555
R
L = 22.7 s
(c) 5 =
k22
Hm10055
R
L = 22.7 s
21. =
k0.1
mH10
R
L = 10 s
(a) vL = (15 V)e10s/10s = (15 V)e1 = 5.52 V (b) vL = (15 V)e20s/10s = (15 V)e2 = 2.03 V (c) vL = (15 V)e30s/10s = (15 V)e3 = 747 mV (d) vL = (15 V)e40s/10s = (15 V)e4 = 275 mV (e) vL = (15 V)e50s/10s = (15 V)e5 = 101 mV
126
Chapter 13
22. IF = 15 V
1.0 k
V
R
= 15 mA
(a) iL = 15 mA(1 - e10s/10s ) = 15 mA(1 - e1 )= 9.48 mA (b) iL = 15 mA(1 - e20s/10s )= 15 mA(1 - e2 )= 13.0 mA (c) iL = 15 mA(1 - e30s/10s )= 15 mA(1 - e3 )= 14.3 mA (d) iL = 15 mA(1 - e40s/10s )= 15 mA(1 - e4 )= 14.7 mA (e) iL = 15 mA(1 - e50s/10s )= 15 mA(1 - e5 )= 14.9 mA
23. 75 mH
9.15 s8.2 kΩ
L
R
24. The time constant is 75 mH
8.2 k
L
R
= 9.15 s. For the increasing exponential, the final
current is IF = S 10 V
8.2 k
V
R
= 1.22 mA
(a) At 10 s, i = 1.22 mA(1 e10s/9.15s) = 0.81 mA (b) At 20 s, i = 1.22 mA(1 e20s/9.15s) = 1.08 mA (c) 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) At 65 s, i = 1.22 mA(e15s/9.15s) = 0.237 mA (b) At 75 s, i = 1.22 mA(e25s/9.15s) = 0.079 mA (c) At 85 s, i = 1.22 mA(e35s/9.15s) = 0.027 mA 27. VL = (15 V)et/10 s
et/10 s = V 15
V 5
t = (10 s)
15
5ln
t = 11.0 s 28. (a) The polarity is positive at the top of the inductor.
(b) IF = 24 V 24 V
8.2 kWR
= 1.22 mA
29. The time constant is found by first thevenizing the bridge. Figure 13-1 shows the Thevenin circuit.
127
Chapter 13
Figure 13-1
= 3.3 mH
4.57 k
L
R
= 0.722 s
30. (a) The current at 1.0 s is I = 0.569 mA(1 e1.0s/0.722s) = 0.426 mA (b) The current after 5 is 0.569 mA.
Figure 13-2
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
= 3.3 mH
4.72 k
L
R
= 0.699 s
i = 0.569 mA(e1.0s/0.699s) = 136 A
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. LT = L1 + H 60
H) H)(40 (20H50
32
32
LL
LL = 63.3 mH
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
IT = rms 10 V
995 mLT
V
X
= 10.1 A
IL2 = 3
T2 3
628 m10.1 A
314 m 628 mL
L L
XI
X X
= 6.73 A
IL3 = 2
T2 3
314 m10.1 A
314 m 628 mL
L L
XI
X X
= 3.37 A
35. (a) LT = 57.3 mH
mA 500
V 10
I
V XL = = 20
128
Chapter 13
129
fL XL = 2 T
f = T2 L
20
2 (57.3 mH)L
= 55.5 Hz
4 mH L
X
(b) LT =X = 20
f = T2 L
20
2 (4 mH)
= 796 Hz
5.33 mH L
LX
(c) LT =X = 20
)mH33.5(2
20 f =
2 TL
= 597 Hz
LT m
Pr = = (10.1 mA)2(995 m) = 101 VAR
7. 2(3 kXL3 = 2(3 kHz)(3 mH) = 56.5
.83 V
LX
36. X = 995
LTXI 2rms
3 XL1 = Hz)(5 mH) = 94.2 VL3 = IL3XL3 = (50 mA)(56.5 ) = 2 VL1 = 10 V 2.83 V = 7.17 V
I 1 = L1 7.17 VLV = 76.1 mA
1 94.2LX IL2 = I mA 50 mA
L1 IL3 = 76.1 = 26.1 mA
Multisim Troubleshooting and Analysis 38. VL1 = 1.158 V; VL2 = 3.579 V; VL3 = 5.263 V
V L5
L(5kHz) 0.032 mA
is o
.
39. VL1 = 12.953 V; VL2 = 11.047 V; L3 = 5.948 V.099 V; V = 5.099 V VL4 = 5
0. .855 mA; I =4 IL(10kHz) = 0.016 A; IL(20kHz) = 7
41. L3 pen. 42. L2 is shorted
Chapter 14 Transformers
Section 14-1 Mutual Inductance
1. LM = H)4)(H1(75.0 spLLk = 1.5 H
2. LM = 21LLk
k = )H2)(H8(
H1
21
LL
LM = 0.25
Section 14-2 The Basic Transformer
3. n = 250
1000
p
s
N
N = 4
n = 400
100
p
s
N
N = 0.25
4. Ns = 2Np = 2(25) = 50 turns 5. See Figure 14-1.
(a) Vs = pp
s VN
N
= 10(10 V) = 100 V rms (b) Vs = p
p
s VN
N
= 2(50 V) = 100 V rms
(c) Vs = pp
s VN
N
= 0.2(100 V) = 20 V rms
Figure 14-1
130
Chapter 14
Section 14-3 Step-Up and Step-Down Transformers
6. n = V 240
V 720
p
s
N
N = 3
7. Vs = pp
s VN
N
= 5(120 V) = 600 V
8. p
s
p
s
V
V
N
N
Vp = V 6010
1
s
s
p VN
N= 6 V
9. V 120
V 30
p
s
p
s
N
N
V
V = 0.25
10. Vs = (0.2)(1200 V) = 240 V
11. p
s
p
s
V
V
N
N
Vp =
1
10s
s
p VN
N6 V = 60 V
12. (a) VRL = V 12020
1
p
p
s VN
N = 6 V
(b) VRL = 0 V (transformers do not couple dc)
(c) VRL = pp
s VN
N
= 4(10 V) = 40 V
13. (a) VL = (0.1)Vs = (0.1)(100 V) = 10 V (b) Vp = 20VL = 20(12 V) = 240 V
Section 14-4 Loading the Secondary
14. s
p
p
s
N
N
I
I
Is = mA 1003
1
3
1
pp
s
p IIN
N= 33.3 mA
VL = 3(20 V) = 60 V
RL = mA 33.3
V 60
s
L
I
V = 1.8 k
131
Chapter 14
15. (a) p
s
N
N = 0.5
Rreflect =
300
5.0
122
Ls
p RN
N = 1200
Ip = 1200
V 30 = 25 mA
(b) Is = ps
p IN
N
= 2(25 mA) = 50 mA
(c) Vs = pp
s VN
N
= 0.5(30 V) = 15 V
(d) PL = VsIs = (15 V)(50 mA) = 750 mW
Section 14-5 Reflected Load
16. Rp =
680
5
122
Ls
p RN
N = 27.2
17. Rp = 300 , RL = 1 k
300
k 1
p
L
p
s
R
R
N
N = 1.83
Section 14-6 Impedance Matching
18. Rp = Ls
p RN
N2
L
p
s
p
R
R
N
N
2
44
16
L
p
s
p
R
R
N
N = 2
n = 2
1
p
s
N
N = 0.5
132
Chapter 14
19. Rp = Ls
p RN
N2
= 16
n2 =
16
42
p
L
p
s
R
R
N
N
n = 25.016
4
p
s
N
N = 0.5
Ip = 1616
V 25 = 781 mA
Is = mA 7815.0
11
pIn
= 1562 mA
Pspeaker = = (1562 mA)24 = 9.76 W Ls RI 2
20. Position 1: RL = 560 + 220 + 1 k = 1780
10
1780
p
L
p
s
R
R
N
N = 13.34
Ns = Ns1 + Ns2 + Ns3 = 13.34Np = 13.34(1000) = 13,340 turns (total secondary turns) Position 2: RL = 220 + 1 k = 1220
10
1220
p
L
p
s
R
R
N
N = 11.05
Ns2 + Ns3 = 11.05Np = 11.05(1000) = 11,050 turns Position 3: RL = 1 k
10
1000
p
L
p
s
R
R
N
N = 10 Thus, Ns2 = 11,050 10,000 = 1,050 turns
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. % efficiency = %100 W100
W94.5%100
in
out
P
P= 94.5 %
23. Coefficient of coupling = 1 0.02 = 0.98
133
Chapter 14
24. (a) IL(max) = V 600
kVA1
s
a
V
P = 1.67 A
(b) RL(max) = A 1.67
V 600
(max)
L
s
I
V = 359
(c) XC = L
s
I
V = 359
Cmax = )359)(Hz 60(2
1
= 7.39 F
25. kVA = (2.5 kV)(10 A) = 25 kVA 26. (a) Vp = 2400 V
V 2400
V 120
p
s
p
s
V
V
N
N = 0.05
(b) Is = V 120
kVA 5
s
a
V
P = 41.7 A
(c) Ip = V 2400
kVa 5 = 2.08 A
Section 14-8 Tapped and Multiple-Winding Transformers
27. V1 = 50
120 V500
= 12.0 V
V2 = 100
120 V500
= 24.0 V
V3 = 100
120 V500
= 24.0 V
V4 = V2 + V3 = 48.0 V
28. V 12
V 241 p
s
p
s
V
V
N
N = 2
V 12
V 62 p
s
N
N = 0.5
V 12
V 33 p
s
N
N = 0.25
29. Vs = V 120500
200
p
p
s VN
N = 48 V
V 550
250
p
p
s VN
N = 25 V Vs =
134
Chapter 14
135
30. (a) See Figure 14-2.
100 T: Vs =
Figure 14-2
V 2402000
100
= (b)
12 V
200 T: Vs = V 2402000
200
= 24 V
500 T: Vs = V 2402000
500
= 60 V
1000 T: Vs = V 2402000
1000
= 120 V
31. (a) Ns = 400 T + 300 T = 700 turns
VRL = 700
60 V1200
sp
p
NV
N
= 35 V
IRL = 35 V
12RL
L
V
R
= 2.92 A
VC = 300
60 V1200
= 15 V
IC = 15 V
10C
C
V
X
= 1.5 A
(b) 2 2
700 300
1 1 1 1 1
(2.94)(12 ) (16)(10 )p p pL CL
R N NR X
N N
1 1
35.3 160
= = 28.3 mS + 6.25 mS = 29.0 mS
Rp = 1
29.0 mS = 34.5
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.
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) 3, j5 (b) +7, j1 (c) +10, +j10 5. (a) 5, +j3 and +5, j3 (b) 1, j7 and 1, +j7 (c) 10, +j10 and +10, j10
6. (a) 3 + j5 (b) 2 + j1.5 (c) 10 j14
7. C = 22 1510 = 18.0
8. (a) 40 j40 =
40
40tan4040 122 = 56.645
(b) 50 j200 =
50
200tan20050 122 = 20676
(c) 35 j20 =
35
20tan2035 122 = 40.329.7
(d) 98 + j45 =
98
45tan4598 122 = 10824.7
9. (a) 100050 = 643 j766 (b) 15160 = 14.1 + j5.13 (c) 25135 = 17.7 j17.7 (d) 3180 = 3 + j0 10. (a) 10120 = 10240 (b) 3285 = 32275 (c) 5310 = 550 11. (a) 40 j40 is in the fourth quadrant. (b) 50 j200 is in the fourth quadrant. (c) 35 j20 is in the fourth quadrant. (d) 98 + j45 is in the first quadrant. 12. (a) 10120 is in the second quadrant. (b) 3285 is in the first quadrant. (c) 5310 is in the fourth quadrant. 13. (a) 12(180 65) = 12115 (b) 20(180 + 50) = 20230 (c) 100(360 170) = 100190 (d) 50(360 200) = 50160
137
Chapter 15
14. (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 15. (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 16. (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
17. (a)
395.2
508 = 3.2(50 39) = 3.211
(b)
109
9163 = 7(91 10) = 7101
(c)
6.404.18
3028
j1214
3028= 1.52(30 (40.6)) = 1.5270.6
(d)
6.269.17
9.3650
8j16
30j40 = 2.79(36.9 26.6) = 2.7963.5
138
Chapter 15
18. (a)
372.1
238.1655.2 =
371.2
j0.70)66.1(j2.27)06.1(
=
372.1
4.10103.3
371.2
j2.976.0 = 2.5364.4
(b)
0.615.51
)46.604.172)(15100(
j4525
j150)85)(15100( = 335106
(c) )50j90)(35(125
j100)50)(7517590250(
= )5035)(75.3504.154(
)43.638.111)(04.169j45.29j250(
=)75.8505.5391(
)43.638.111)(83.8348.421(
= 8.7465.4
(d)
2
4j
2
4j77.7
2
4j
4
8j
1.1
)8.3()5.1( 2
= 7.77 + 2 + j2 = 9.77 + j2 = 9.9711.6
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) Z = R jXC = 270 j100 = 28820.3 (b) Z = R jXC = 680 j1000 = 1.2155.8 k 22. (a) RT = R1 + R2 = 100 k + 47 k = 147 k
CT =
F022.0
1
F01.0
11
111
21
CC
= 0.00688 F
XCT = )F 0.00688)(Hz 001(2
1
2
1
TfC = 231 k
Z = RT jXCT = 147 k j231 k = 27357.5 k Z = 273 , = 57.5
(b) CT = C1 + C2 = 470 pF + 470 pF = 940 pF
XCT = )pF 940)(kHz 20(2
1
2
1
TfC = 8.47 k
Z = R jXCT = 10 k j8.47 k = 13.140.3 k Z = 13.1 k, = 40.3
139
Chapter 15
(c) RT = R1 + R2 R3 = 680 + 720 = 1400
CT =
F0032.0
1
pF 1000
1111
321
CCC
= 762 pF
XCT = )pF 762)(kHz 001(2
1
2
1
TfC = 2089
Z = RT jXCT = 1400 j2089 = 2.5256.2 k Z = 2.52 k, = 56.2
23. (a) XC = )F 0.0022)(Hz 100(2
1
2
1
fC = 723 k
Z = 56 k j723 k
(b) XC = )F 0.0022)(Hz 005(2
1
2
1
fC = 145 k
Z = 56 k j145 k
(c) XC = )F 0.0022)(kHz 1(2
1
2
1
fC = 72.3 k
Z = 56 k j72.3 k
(d) XC = )F 0.0022)(kHz 5.2(2
1
2
1
fC = 28.9 k
Z = 56 k j28.9 k
24. (a) XC = )F 0.0047)(Hz 100(2
1
2
1
fC = 339 k
Z = 56 k j339 k
(b) XC = )F 0.0047)(Hz 500(2
1
2
1
fC = 67.7 k
Z = 56 k j67.7 k
(c) XC = )F 0.0047)(kHz 1(2
1
2
1
fC = 33.9 k
Z = 56 k j33.9 k
(d) XC = )F 0.0047)(kHz 5.2(2
1
2
1
fC = 13.5 k
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
I =
3.20288
V 010 = 34.720.3 mA
(b) From Problem 21(b): Z = 1.2155.8 k
I =
k8.5521.1
V 05 = 4.1355.8 mA
27. Start with the current in polar form from Problem 26:
(a) I = 10 0 V
288 20.3
= 34.720.3 mA = (34.7 mA)cos 20.3o + j(34.7 mA)sin 20.3o
= 32.5 mA + j12.0 mA
(b) I = 5 0 V
1.21 55.8 k
= 4.1355.8 mA = (4.13 mA)cos 55.8o + j(4.13 mA)sin 55.8o
= 2.32 mA + j3.42mA 28. (a) From Problem 22(a): Z = 147 k j231 k = 27357.5
IT =
5.57273
V 050 = 18357.5 A
(b) From Problem 22(b): Z = 10 k j8.47 k = 13.140.3 k
IT =
k3.401.13
V 08 = 61140.3 A
(c) From Problem 22(c): Z = 1400 j2089 = 2.5256.2 k
IT =
k2.5652.2
V 205 = 1.9876.2 mA
141
Chapter 15
29. Start with the current in polar form from Problem 28
(a) IT = 50 0 V
273 57.5
= 18357.5 A = (183 A)cos 57.5o + j(183 A)sin 57.5o
= 98.3 A + j154 A
(b) IT = 8 0 V
13.1 40.3 k
= 61140.3 A = (611 A)cos 40.3o + j(611 A)sin 40.3o
= 466 A + j395 A
(c) IT = 5 20 V
2.52 56.2 k
= 1.9876.2 mA = (1.98 mA)cos 76.2o + j(1.98 mA)sin 76.2o
= 0.472 mA + j1.92 mA 30. Using the results of Problem 22:
(a) =
k 147
k 231tantan 11
R
XC = 57.5
(b) =
k 10
k 47.8tantan 11
R
XC = 40.3
(c) =
1400
2089tantan 11
R
XC = 56.2
31. XC = fC2
1 14.5 k
=
k 56
k 5.14tantan 11
R
XC = 14.5
32. See Figure 15-3.
CT =
F 0.22
1
F 0.1
11
= 0.069 F
XC = )F 0.069)(kHz 15(2
1
= 154
ZT = 50 j154 = 16272.0
IT =
0.72162
V 02
T
s
Z
V = 12.372.0 mA
XC1 = )F 0.1)(kHz 15(2
1
= 106
XC2 = )F 0.22)(kHz 15(2
1
= 48.2
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
142
Chapter 15
Figure 15-3
33. (a) XC = )F 100)(Hz 20(2
1
= 79.6
Z = 56 j79.6 = 97.354.9
(b) IT =
9.543.97
010 = 10354.9 mA
(c) VR = V 0109.543.97
056
= 5.7654.9 V
(d) VC = V 0109.543.97
906.79
= 8.1835.1 V
34. Z = mA 10
V 10
I
Vs = 1 k
XC = )F 0.027)(kHz 10(2
1
= 589.5
22CXR = 1 k
R2 + (589.5 )2 = (1 k)2
R = 22 )5.589()k1( = 808
= tan1
808
5.589 = 36.1
35. ZT = A 5
V 10020
Ptrue = I2RT
RT = 22 A) (5
W400
I
Ptrue = 16
RX = RT R1 = 16 4 = 12
143
Chapter 15
222CTT XRZ
XC = 144)16()20( 2222 TT RZ = 12
C = ) 12)(kHz 1(2
1
= 13.3 F
36. (a) XC = )F039.0)(Hz 1(2
1
= 4.08 M
= 90 + tan1
k 3.9
M 4.08tan90 1
R
XC = 0.055
(b) XC = )F039.0)(Hz 100(2
1
= 40.8 k
= 90 + tan1
k 3.9
k 40.8tan90 1
R
XC = 5.46
(c) XC = )F039.0)(kHz 1(2
1
= 4.08 k
= 90 + tan1
k 3.9
k 4.08tan90 1
R
XC = 43.7
(d) XC = )F039.0)(kHz 10(2
1
= 408
= 90 + tan1
k 3.9
408tan90 1
R
XC = 84.0
37. Use the formula, Vout = V. 1
T
C
Z
X See Figure 15-4.
Figure 15-4
Frequency (kHz)
XC
ZT
Vout
0 1 V 1 4.08 k 5.64 k 723 mV 2 2.04 k 4.40 k 464 mV 3 1.36 k 4.13 k 329 mV 4 1.02 k 4.03 k 253 mV 5 816 3.98 k 205 mV 6 680 3.96 k 172 mV 7 583 3.94 k 148 mV 8 510 3.93 k 130 mV 9 453 3.93 k 115 mV 10 408 3.92 k 104 mV
144
Chapter 15
38. (a) XC = )F10)(Hz 1(2
1
2
1
fC = 15.9 k
= tan1
10
k 15.9tan 1
R
XC = 90.0
(b) XC = )F10)(Hz 100(2
1
= 159
= tan1
10
159tan 1
R
XC = 86.4
(c) XC = )F10)(kHz 1(2
1
= 15.9
= tan1
10
15.9tan 1
R
XC = 57.9
(d) XC = )F10)(kHz 10(2
1
= 1.59
= tan1
10
1.59tan 1
R
XC = 9.04
39. Use the formula, Vout = V. 1
TZ
R See Figure 15-5.
Figure 15-5
Frequency (kHz)
XC ZT Vout
0 0 V 1 15.9 18.8 5.32 V 2 7.96 12.8 7.82 V 3 5.31 11.3 8.83 V 4 3.98 10.8 9.29 V 5 3.18 10.5 9.53 V 6 2.65 10.4 9.66 V 7 2.27 10.3 9.76 V 8 1.99 10.2 9.80 V 9 1.77 10.2 9.84 V
10 1.59 10.1 9.87 V 40. For Figure 15-91 in the text (See Figure 15-6):
XC = )F039.0)(kHz 5(2
1
= 816
Z = 3.9 k j816 = 398411.8
I =
8.113984
V 01
Z
Vs = 25111.8 A
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)
C X = 1
= 15.9 2 (1 kHz)(10 F)
Z = 10 j15.9 = 18.857.8
I = 10 0 V
sV = 53257.8
18.8 57.8 Z mA
VR = IR = (53257.8 mA)(100 ) = 5.3257.8 V .2 V
ection 15-5 Impedance and Admittance of Parallel RC Circuits
VC = IXC = (53257.8 mA)(15.990 ) = 8.4632 Figure 15-7
Part 2: Parallel Circuits
S
k 59 33.2
)k 902)(k 02.1(
k 2j k 2.1
)k 902)(k 02.1(
j
)90)(0(
C
C
XR
XR42. Z =
= 1.0331 k
3. BC = 2fC = 2f(C1 + C2) = 2(2 kHz)(0.32 F) = 4.02 mS 4
146
Chapter 15
G =
1480
11
321 RRR = 0.676 mS
Y = G + jBC = 0.676 mS + j4.02 mS = 4.0880.5 mS
Z = mS 80.54.08
11
Y = 24580.5
Z = 245 , = 80.5
44. (a) XC = )F 0.32)(kHz 5.1(2
1
= 332
Z =
4.773246.121517
)90332)(01480(
j3321480
)90332)(01480(
Z = 324 , = 77.4
(b) XC = )F 0.32)(kHz 3(2
1
= 166
Z =
6.8316540.61489
)90166)(01480(
j1661480
)90166)(01480(
Z = 165 , = 83.6
(c) XC = )F 0.32)(kHz 5(2
1
= 99.5
Z =
2.863.9985.31483
)905.99)(01480(
j99.51480
)905.99)(01480(
Z = 99.3 , = 86.2
(d) XC = )F 0.32)(kHz 10(2
1
= 49.7
Z =
1.887.4992.11481
)907.49)(01480(
j49.71480
)907.49)(01480(
Z = 49.7 , = 88.1
Section 15-6 Analysis of Parallel RC Circuits
45. ZT =
j9068
)9090)(068( = 54.337.1
VC = VR = Vs = 10 0 V
IT =
1.373.54
V 010 = 18437.1 mA
IR =
068
V 010 = 1470 mA
147
Chapter 15
IC =
9090
V 010 = 11190 mA
46. XC1 = )F 0.047)(kHz 50(2
1
= 67.7
XC2 = )F 0.022)(kHz 50(2
1
= 145
IC1 =
907.67
V 08
C1
s
X
V = 11890 mA
IC2 =
90145
V 08
C2
s
X
V = 55.290 mA
IR1 =
0220
V 08
1
s
R
V = 36.40 mA
IR2 =
0180
V 08
2
s
R
V = 44.40 mA
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 47. (a) XC = XC1 XC2 = 21 15 = 8.75
BC =
75.8
11
CX = 114 mS
G =
10
11
R = 100 mS
YT = 100 mS + j114 mS = 15248.8 mS
ZT = mS 8.48152
1
TY
1 = 6.5948.8
(b) IR =
010
mV 0100 = 100 mA
(c) ICT =
9075.8
mV 0100 = 11.490 mA
(d) IT =
8.4859.6
mV 0100 = 15.248.8 mA
(e) = 48.8
148
Chapter 15
48. (a) CT = C1 + C2 = 0.047 F + 0.022 F = 0.069 F
XCT = )F 0.069)(Hz 500(2
1
= 4613
Z =
5.397255
)904613)(k06.5( = 3.5650.5 k
(b) IR =
k06.5
mV 0100
R
Vs = 17.90 A
(c) ICT =
904613
mV 0100
CT
s
X
V = 21.790 A
(d) IT =
5.503560
mV 0100
Z
Vs = 28.150.5 A
(e) = 50.5 49. RT = 22 k, CT = 32 pF
XCT = )pF 32)(kHz 001(2
1
= 49.7 k
Z =
k1.664.54
)k907.49)(k022( = 20.123.9 k = 18.4 k j8.14 k
Req = 18.4 k, XCeq = 8.14 k
Ceq = )k kHz)(8.14 100(2
1
= 196 pF
50. XC = )F 0.01)(kHz 1(2
1
= 15.9 k
=
T
C
R
X1tan
T
C
R
X = tan
RT =
30tan
k 9.15
tanCX
= 27.6 k
RT = 21
21
RR
RR
RT(R1 + R2) = R1R2 R1RT + R2RT = R1R2 R1(RT R2) = R2RT
R1 =
k 4.19
)k 6.27)(k 47(
2
2
T
T
RR
RR = 66.7 k
149
Chapter 15
Part 3: Series-Parallel Circuits Section 15-7 Analysis of Series-Parallel RC Circuits 51. See Figure 15-8.
XC1 = )F1.0)(kHz 15(2
1
2
1
1
fC = 106
XC2 = )F047.0)(kHz 15(2
1
2
1
2
fC = 226
XC3 = )F22.0)(kHz 15(2
1
2
1
3
fC = 48.2
ZC2R1 = R1 jXC2 = 470 j226 = 52225.7 ZC3 = jXC3 = 48.290 ZR2R3 = R2 + R3 = 330 + 180 = 5100 Combining the three parallel branches:
ZA =
0510
1
902.48
1
7.25522
11
111
R2R3C3C2R1 ZZZ
1
=mS 1.96 mS j20.7 mS j0.833 mS 73.1
1
mS096.1mS907.20mS7.2592.1
1
= mS3.808.21
1
mS j21.5 mS 69.3
1
= 45.980.3 = 7.73 j45.2
ZT = XC1 + ZA = j106 + 7.73 j45.2 = 7.73 j151 = 15187.1
VC1 = V 012 1.87151
90106 V 012
T
C1
Z
X= 8.422.9 V
VZA = V 012 1.87151
3.809.45 V 012
T
A
Z
Z= 3.656.8 V
VC2 = V 8.665.3 7.25522
90226
V
Z
XZA
C2R1
C2 = 1.5857.5 V
VR1 = V 8.665.3 7.25522
0470
V
Z
RZA
C2R1
1 = 3.2932.5 V
VC3 = VZA = 3.656.8 V
VR2 = V 8.665.3 0510
0330
V
Z
RZA
R2R3
2 = 2.366.8 V
V 8.665.3 0510
0180
V
Z
RZA
R2R3
3VR3 = = 1.296.8 V
150
Chapter 15
2. From Problem 51: ZT = 7.73 j151
3.
Figure 15-8
2.9 57.5
5 The j term is larger, therefore th
e circuit is predominantly capacitive.
5 See Figure 15-9. Using the results of Problem 51:
IT = 1.87151 TZ
V 012 V 012 = 79.587.1 mA
IC2R1 =
= 6.9932.5 mA
7.25522
V 8.665.3
Z
V
C2R1
ZA
902.48
V 8.665.3 IC3 =
Z.796.8 mA
V
C3
ZA = 75
IR2R3 =
= 7.166.8 mA
0510
V 8.665.3
Z
V
R2R3
ZA
Figure 15-9
151
Chapter 15
54. RT = R1 + R2 R3 = 47 + 42.9 = 89.9
XC = )F47.0)(kHz 1(2
1
= 339
ZT = 89.9 j339 = 35175.1
(a) IT =
1.75351
V 015
T
s
Z
V = 42.775.1 mA
(b) = 75.1
(c) VR1 = V 0151.75351
V 047
s
T
1 VZ
R = 2.0175.1 V
(d) VR2 = V 0151.75351
V 042.9
s
T
32 VZ
RR = 1.8375.1 V
(e) VR3 = VR2 = 1.8375.1 V
(f) VC = V 0151.75351
09339
s
T
C VZ
X = 14.514.9 V
55. For I = 0 A, VA = VB and VR1 = VR2
XC1 = )F047.0)(kHz 1(2
1
= 3.39 k
VR1 = VR2
s
C
s VX
V
2
2222 )k 1(
k 1
)k 39.3()k 2.2(
k 2.2
Cancelling the Vs terms and solving for XC2:
2
2222 )k 1(
k 1
)k 39.3()k 2.2(
k 2.2
CX
k2.2
)k 39.3()k 2.2()k 1()k 1(
2222
22
CX
Squaring both sides to eliminate the radicals and solving for XC2:
2
2222
22
)k 2.2(
)k 39.3()k 2.2()k 1()k 1(
CX
XC2 = 2
2
222
)k 1()k 2.2(
)k 39.3()k 2.2()k 1(
= 1541
C2 = )1541)(kHz 1(2
1
= 0.103 F
152
Chapter 15
56. XC3 = )F022.0)(kHz 5.2(2
1
= 2.89 k
ZC3R6 = R6 jXC3 = 820 j2.89 k = 374.2 k
XC2 = )F047.0)(kHz 5.2(2
1
= 1.35 k
ZC = ZC2R4 =
k 5663.1
)0910)(k 9035.1(
j 24 CXR4C2RX
= 75434 = 625 j422
ZR5C2R4 = R5 + ZC = 1 k + 625 j422 = 1625 j422 = 1.6814.6 k ZB = (R5 + ZC) ZC3R6 = 1.6814.6 k 374.2 k
=
k 9.554
)k2.743)(k 6.1468.1( = 1.2632.9 k = 1.06 k j684
ZR3ZB = R3 + ZB = 680 + 1.06 k j684 = 1.74 k j684 = 1.8721.5 k ZA = R2 ZR3ZB = 2200 1.8721.5 k
=
k 2.1908.2
)k5.2187.1)( 0220( = 1982.3 = 197.8 j7.95
XC1 = )F015.0)(kHz 5.2(2
1
= 4.24 k
ZT = R1 jXC1 + ZA = (1 k j4.24 k) + (197.8 j7.95 ) = 1197.8 j4248 = 4.4174.3 k
VA = V 010 k 3.7441.4
3.2198
V
Z
Zs
T
A = 44972.0 mV
VB = mV 0.72449 k 5.2187.1
k 9.3226.1
V
Z
ZA
R3ZB
B = 30360.6 mV
VC = mV 6.60303 k 6.1468.1
34754
V
Z
ZB
R5C2R4
C = 13641.2 mV
VD = mV 6.60303 k 2.743
0820
V
Z
RB
C3R6
6 = 83135 mV
57. Use the voltages found in Problem 56:
IC3 = IR6 =
0820
mV 13583
6
D
R
V = 101135 A = 71.4 A + j71.4 A
IR4 =
0910
mV 2.41136
4
C
R
V = 14941.2 A = 112 A + j98 A
IC2 =
k 035.1
mV 2.41136
C2
C
X
V = 101131 A = 66 A + j76 A
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
IR2 =
0220
mV 0.72449
2
A
R
V = 2.0472.0 mA = 0.630 mA + j1.94 mA
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
59. Pa = 2222true )VAR 3.5() W2( rPP = 4.03 VA
60. From Problem 33:
IT = 103 mA, XC = 79.6
Ptrue = = (103 mA)2(56 ) = 594 mW RIT2
Pr = = (103 mA)2(79.6 ) = 845 mVAR CT XI 2
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 = = (42.7 mA)2(89.9 ) = 164 mW TT RI 2
Pr = = (42.7 mA)2(339 ) = 618 mVAR CT XI 2
Pa = = (42.7 mA)2(351 ) = 640 mVA TT ZI 2
PF = cos(75.1) = 0.257
63. (a) ILA = 50
V 240 = 4.8 A
ILB = 72
V 240 = 3.33 A
(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
PrA = = (4.8 A)226.3 = 606 VAR CALA XI 2
PrB = = (3.33 A)222.5 = 250 VAR CBLB XI 2
(c) RA = (50 )cos(31.8) = 42.5 RB = (72 )cos(18.2) = 68.4
PtrueA = = (4.8 A)242.5 = 979 W ALARI 2
PtrueB = = (3.33 A)268.4 = 759 W BLBRI 2
155
Chapter 15
(d) PaA = 22 VAR) (606 W)979( = 1151 VA
PaB = 22 VAR) (250 W)759( = 799 VA
(e) Load A
Section 15-9 Basic Applications
64. fr = 1
2 6RC =
1
2 6(10 k )(0.0022 F) = 9278 Hz
65. 22
1
1
Cin
out
XR
R
V
V
= 0.707
R = 22707.0 CXR
707.0
22 RXR C = 1.414R
= (1.414)2R2 22CXR
= 2R2 R2 = R2(2 1) = R2 2CX
XC = R
fC2
1 = R
C = )k Hz)(100 20(2
1
2
1
fR = 0.0796 F
66. XC = F)kHz)(0.047 3(2
1
= 1.13 k
Vin(B) =
22)(
22
)(
)(
)k 13.1()k 10
k 10Aout
CBin
Bin VXR
R50 mV = 49.7 mV
Signal loss = 50 mV 49.7 mV = 300 V
Section 15-10 Troubleshooting 67. After removing C, the circuit is reduced to Thevenin’s equivalent:
Rth =
k7.9
)k5)(k7.4( = 2.42 k
Vth = V 10k7.9
k5
= 5.15 V
Assuming no leakage in the capacitor:
XC = )F10)(Hz 10(2
1
= 1592
156
Chapter 15
Vout = V 0107.184962
901592V 010
C
C
jXR
X= 3.2171.3 V
With the leakage resistance considered:
Vout = V 05.153.332897
901592
th
Cth
C VjXR
X= 2.8356.7 V
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
Vth = V 01067.11
k067.1
inleak21
leak2 VRRR
RR = 1430 mV
XC = )F7.4)(Hz 10(2
1
= 3.38 k
Vout = mV 0143k1.6767.3
k9038.3
th
Cth
C VjXR
X = 13222.9 mV
(b) XC = )Fp470)(Hz 100(2
1
= 3.39 M
Req = R1 (R2 + R3) = 2.2 k 2 k = 1.05 k ZT = Req + XC Rleak
= 1.050 k +
M j3.39k 2
)k 02)(M 9039.3(
= 1.050 k +
M9039.3
)k 02)(M 9039.3(
= 1.050 k + 20 k = 3.050 k
VR1 = V 05k 005.3
k 005.1
in
T
eq VZ
R = 1.720 V
Vout = V 01.72k02
k01
R1
32
3 VRR
R = 8600 mV
69. (a) Vout = 0 V (less than normal)
(b) XC = )F7.4)(Hz 10(2
1
= 3.39 k
Vout = V 01k7.186.10
k9039.3
in
C
C VjXR
X= 320-71.3 mV (greater than normal)
(c) Vout = V 01k020
k010
in
21
2 VRR
R= 5000 mV (greater than normal)
(d) Vout = 0 V (less than normal output)
157
Chapter 15
158
70. (a) Vout = 0 V (less than normal)
(b) Vout = V 05k02
k01
in
32
3 VRR
R = 2.50 V (greater than normal)
(c) XC = 3.39 M
Vout = V 05M9039.3
k01
in
C32
3 VjXRR
R= 1.4790 mV
(greater than normal) (d) Vout = 0 V (less than normal)
(e) Vout = V 05M9039.3
k02.2
in
C1
1 VjXR
R= 3.2490 mV
(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
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 3. (a) Z = R + jXL = 100 + j50 = 11226.6 (b) Z = R + jXL = 1.5 k + j1 k = 1.8033.7 k 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. (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 6. (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 7. 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. RT = 806 XLT = 2fLT = 2(10 kHz)(4.11 mH) = 258
VRT = 2 2 2 2
806 5 V
(806 ) (258 )
Ts
T LT
RV
R X
= 4.76 V
9. VLT = 2 2 2 2
258 5 V
(806 ) (258 )
LTs
T LT
XV
R X
= 1.52 V
XL3 = 2fL3 = 2(10 kHz)(1000 H) = 62.8
VL3 = 3 62.8 1.52 V
258 L
LTLT
XV
X
= 0.370 V
160
Chapter 16
10. (a) From Problem 3(a): Z = 11226.6
I =
6.26112
V 010 = 89.426.6 mA
(b) From Problem 3(b): Z = 1.8033.7 k
I =
k 7.3380.1
V 05 = 2.7833.7 mA
11. (a) From Problem 4(a): Z = 11555.0
IT =
0.55115
V 05 = 43.555.0 mA
(b) From Problem 4(b): Z = 68134.6 k
IT =
k 6.34681
V 08 = 11.834.6 mA
12. XL = 2fL = 2(60 Hz)(0.1 H) = 37.7
=
47
7.37tantan 11
R
X L = 38.7
13. = 38.7 from Problem 12. Double L: XL = 2fL = 2(60 Hz)(0.2 H) = 75.4
=
47
4.75tantan 11
R
X L = 58.1
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
VL = V 0107.383.60
907.37V 010
j37.747
907.37
s
L VX
LjXR = 6.2553.1 V
VR = V 0107.383.60
047
sVR
LjXR = 7.7938.7 V
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
VR =
sVZ
RV 05
1.147.154
0150
= 4.8514.1 V
VL =
sL V
Z
XV 05
1.147.154
907.37
= 1.2275.9 V
(b) f = 200 Hz XL = 2(200 Hz)(100 mH) = 125.7 Z = R + jXL = 150 + j125.7 = 195.740.0
VR =
sVZ
RV 05
0.407.195
0150
= 3.8340.0 V
VL =
sL V
Z
XV 05
0.407.195
907.125
= 3.2150.0 V
(c) f = 500 Hz XL = 2(200 Hz)(100 mH) = 314 Z = R + jXL = 150 + j314 = 34864.5
VR =
sVZ
RV 05
5.64348
0150
= 2.1664.5 V
VL =
sL V
Z
XV 05
5.64348
90314
= 4.5125.5 V
(d) f = 1 kHz XL = 2(1 kHz)(100 mH) = 628 Z = R + jXL = 150 + j628 = 645.776.6
VR =
sVZ
RV 05
6.767.645
0150
= 1.1676.6 V
VL =
sL V
Z
XV 05
6.767.645
90628
= 4.8613.4 V
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. (a) XL = 2(1 Hz)(10 mH) = 62.8 m
= 1 1 62.8 mtan tan
39 LX
R
= 0.0923
(b) XL = 2(100 Hz)(10 mH) = 628 m
= 1 1 628 mtan tan
39 LX
R
= 9.15
(c) XL = 2(1 kHz)(10 mH) = 6.28
= 1 1 6.28tan tan
39 LX
R
= 58.2
(d) XL = 2(10 kHz)(10 mH) = 62.8
= 1 1 62.8tan tan
39 LX
R
= 86.4
= tan1 1 39 tan
62.8 mL
R
X
= 89.9 18. (a)
(b) = tan1 1 39 tan
628 mL
R
X
= 80.9
(c) = tan1 1 39 tan
6.28L
R
X
= 31.8
(d) = tan1 1 39 tan
62.8L
R
X
= 3.60
19. (a) XL = 2(1 Hz)(10 mH) = 62.8 m
Z = 39 + j62.8 m = o39 0
oLo
62.8 90 m50 0 mV = 80.5 90 V
39 0
oo
out inV V
X
Z
(b) XL = 2(100 Hz)(10 mH) = 628 m
` Z = 39 + j628 m = o39 0
oLo
628 90 m50 0 mV 805 90 V
39 0
oo
out inV V
X
Z
163
Chapter 16
(c) XL = 2(1 kHz)(10 mH) = 6.28
Z = 39 + j6.28 = o39.5 9.14
o oLo
6.28 9050 0 mV = 7.95 80.9 mV
39.5 9.14
o
out inV V
X
Z
(d) XL = 2(10 kHz)(10 mH) = 62.8
Z = 39 + j62.8 = o73.9 58.2
o oLo
62.8 9050 0 mV = 42.5 31.8 mV
73.9 58.2
o
out inV V
X
Z
Part 2: Parallel Circuits Section 16-4 Impedance and Admittance of Parallel RL Circuits 20. XL = 2fL = 2(2 kHz)(800 H) = 10.1
YT = G jBL =
10.1
1j
12
1 = 83.3 mS j99.0 mS = 12949.9 mS
Z = mS 9.49129
11
TY = 7.7549.9
21. From Problem 20:
Z = 1 1
129 49.9 mS
TY = 7.7549.9 = (7.75 cos 49.9o + j(7.75 sin49.9o
= 4.99 + j5.93 22. (a) f = 1.5 kHz XL = 2fL = 2(1.5 kHz)(800 H) = 7.54
YT = G jBL =
7.54
1j
12
1 = 83.3 mS j133 mS = 15758.0 mS
Z = mS 0.58157
11
TY = 6.3758.0
(b) f = 3 kHz
XL = 2fL = 2(3 kHz)(800 H) = 15.1
YT = G jBL =
15.1
1j
12
1 = 83.3 mS j66.2 mS = 10638.5 mS
Z = mS 5.38106
11
TY = 9.4338.5
(c) f = 5 kHz
XL = 2fL = 2(5 kHz)(800 H) = 25.1
164
Chapter 16
YT = G jBL =
25.1
1j
12
1 = 83.3 mS j39.8 mS = 92.325.5 mS
Z = mS 5.253.92
11
TY = 10.825.5
(d) f = 10 kHz
XL = 2fL = 2(10 kHz)(800 H) = 50.3
YT = G jBL =
50.3
1j
12
1 = 83.3 mS j19.9 mS = 85.613.4 mS
Z = mS 4.136.85
11
TY = 11.713.4
23. XL = 2fL
f = H)(8002
12
2
L
X L = 2.39 kHz
Section 16-5 Analysis of Parallel RL Circuits
24. IR =
k 02.2
V 010 = 4.550 mA
IL =
k 905.3
V 010 = 2.8690.0 mA
IT = IR + IL = 4.55 mA j2.86 mA = 5.3732.2 mA 25. (a) XL = 2fL = 2(2 kHz)(25 mH) = 314
Z = 2222 )314(560(
)314)(560(
L
L
XR
RX = 274
= 90
R
X L1tan = 90
560
314tan 1 = 60.7
Z = 27460.7
(b) IR =
0560
mV 050 = 89.30 mA
(c) IL =
90314
mV 050 = 15990 mA
(d) IT =
7.60274
mV 050 = 18260.7 mA
(e) = 60.7 (from part a)
165
Chapter 16
26. (a) XL = 2fL = 2(2 kHz)(330 H) = 4.15
Z =
4.15
1j
56
1
1
1j
1
1
LXR
= mS8.85242
1
mS j241 mS 17.9
1
= 4.1385.8
(b) IR =
056
mV 050 = 8930 mA
(c) IL =
9015.4
mV 050 = 12.090 mA
(d) IT = IR + IL = 893 mA j12 A = 12.085.8 A
(e) = 85.8
27. ZT = 2222
21
21
)k 5()k 5.11(
)k 5)(k 5.11(
)(
)(
L
L
XRR
XRR = 4.59 k
= 90
k 5.11
k 5tan90tan 11
R
X L = 66.5
ZT = 4.5966.5 = 1.83 k + j4.21 k 1.83 k resistance in series with 4.21 k inductive reactance. 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
VR1 = V 02589.33.64
056
s
T
1 VZ
R= 21.83.89 V
VR2 = V 02589.33.64
5.137.18
s
T
R3L1L22 VZ
ZR= 7.279.61 V
VR3 = V 61.927.73.629.70
033
R2
R3L1L2
3 VZ
R= 3.3853.3 V
166
Chapter 16
VL1 = VL2 = V 61.927.73.629.70
908.62
R2
R3L1L2
L2L1 VZ
XX= 6.4437.3 V
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:
ZB =
2.527.89
3.621560
9.70j3322
)3.629.70)(022( 2
= 17.410.1 = 17.1 + j3.05
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:
IR1 = IT =
056
V89.38.21
1
R1
R
V = 3893.89 mA
IR2 =
022
V61.927.7
2
R2
R
V = 3309.61 mA
IR3 =
033
V3.5338.3
3
R3
R
V = 102-53.3 mA
IL1 = IL2 =
09125.6
V3.3744.6
L1
L1
X
V = 51.3-52.7 mA
52.7 IR1 = IT
3.89
51.3 mAFigure 16-4
32. 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:
Z2 =
k 2.2476.9
k 5.358.22
k j4k9.8
)k5.3588.6)(k03.3( = 2.3411.3 k = 2.29 k + j459
Combining XL1 in series with Z2: Z3 = 2.29 k + j5.46 k = 5.9167.5 k Combining R1 in parallel with Z3:
ZT =
k 6.5746.6
k 5.6709.7
k j5.46k46.3
)k5.6791.5)(k02.1( = 1.109.90 k
(a) IT =
k09.91.10
V018
T
S
Z
V = 16.49.90 mA
(b) = 9.90 (IT = lags Vs) (c) VR1 = Vs = 180 V
(d) VR2 = V 018k 5.6791.5
k 3.1134.2
R1
3
2 VZ
Z = 7.1356.2 V
(e) VR3 = V 2.5613.7k 5.3588.6
k 06.5
R2
1
3 VZ
R = 5.8091.7 V
168
Chapter 16
(f) VL1 = V 018k 5.6791.5
k 905
S
3
L1 VZ
X = 15.222.5 V
(g) VL2 = V 2.5613.7k 5.3588.6
k 904
R2
1
L2 VZ
X = 4.151.70 V
33. 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 ) =
j1525.96
) 90120)( 1.18102(
=
6.57180
) 90120)( 1.18102( = 68.050.5
(a) IT =
5.500.68
V040
T
s
Z
V= 58850.5 mA
(b) VL1 = V 0401.18102
2.342.56
s2L11
2L1 VRXR
RX= 22.016.1 V
(c) VA = V 04090120
9045
s
L3L2
L3 VXX
X= 150 V
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.
IL1 =
90100
V1.160.22
L1
L1
X
V = 22073.9 mA
IR2 =
068
V1.160.22
2
B
R
V = 32416.1 mA
IL2 = IL3 =
90120
V 040
L3L2
s
XX
V = 33390 mA
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
VL2 = V 04090120
9075V 040
L3L2
L2
XX
X = 250 V
VL3 = V 04090120
9045V 040
L3L2
L3
XX
X = 150 V
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
=
k 41.2
k 14.3tantan 11
R
X L = 52.5 Vout lags Vin
VR1 = V 1)k 14.3()k 41.2(
k 41.22222
in
LT
T VXR
R= 609 mV
VR3 = mV 609k 6.5k 27.3
k 27.3
)(
)(1
5432
543
RVRRRR
RRR = 225 mV
Vout = VR5 = mV 225k 8.6k 9.3
k 8.63
54
5
RV
RR
R = 143 mV
V 1
mV 143
in
out
V
V = 0.143
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 Z = XL2 + R2 Z3 2
= 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
VR1 = V 01k6.596.4
k5.1947.2
= 53740.1 mV
VR2 = mV 1.40537k7.5938.6
k1.1432.3
= 27985.7 mV
Vout = mV 7.85279k7.4226.9
k08.6
= 205128 mV
Phase shift = 128, Attenuation = V 1
mV 205
in
out
V
V= 0.205
37. R1 = A 1
V 12 = 12
R2 = A 1
kV 2.5 = 2.5 k
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.
Figure 16-7
Part 4: Special Topics Section 16-7 Power in RL Circuits
38. Pa = 2222 )mVAR 340()mW 100( rtrue PP = 354 mVA
39. XL = 2(60 Hz)(0.1 H) = 37.7 Z = R + jXL = 47 + j37.7 = 60.338.7
IT =
7.383.60
V 010
Z
Vs = 165.838.7 mA
Ptrue = = (165.8 mA)2(47 ) = 1.29 W RIT2
Pr = = (165.8 mA)2(37.7 ) = 1.04 VAR LT XI 2
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 = = (16.5 mA)2(1.08 k) = 290 mW RIR2
Pr = = (16.4 mA)2(189 ) = 50.8 mVAR LT XI 2
Pa = = (16.4 mA)2(1.10 k) = 296 mVA TT ZI 2
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 = = (588 mA)2(43.3 ) = 15.0 W RIT2
Section 16-8 Basic Applications
43. Use the formula, Vout = .inT
VZ
R
See Figure 16-9.
Figure 16-10
Figure 16-9
Frequency (kHz)
XL Ztot Vout
0 0 39.0 1 V 1 62.8 73.9 528 mV 2 126 132 296 mV 3 189 193 203 mV 4 251 254 153 mV 5 314 317 123 mV
44. Use the formula, Vout = .inT
L VZ
X
See Figure 16-10.
Frequency
(kHz) XL ZT Vout
0 0 39.0 0 V 1 62.8 73.9 42.5 mV 2 126 132 47.8 mV 3 189 193 49.0 mV 4 251 254 49.4 mV 5 314 317 49.6 mV
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
VR = 39 0
1 0 V504.16 85.6
in
RV
Z = 77.485.6 mV
VL = 502.65 90
1 0 V504.16 85.6
= 9974.44 mV
L
in
XV
Z For Figure 16-62 in the text (See Figure 16-11(b)):
VR = 39 0
50 0 mV504.16 85.6
= 3.8785.6 mV
in
RV
Z
VL = 502.65 90
50 0 mV504.16 85.6
= 49.94.44 mV
L
in
XV
Z
(a) (b)
Figure 16-11
Section 16-9 Troubleshooting 46. VR1 = VL1 = 18 V VR2 = VR3 = VL2 = 0 V 47. (a) Vout = 0 V (b) Vout = 0 V (c) 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
I =
8.2526.173
V 05 = 28.925.8 mA
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
I =
5.645.83
V 05 = 59.964.5 mA
Vout = I(R1 R3) = (59.964.5 mA)35.90 = 2.1564.5 V
173
Chapter 16
174
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
Chapter 17 RLC Circuits and Resonance
Part 1: Series Circuits Section 17-1 Impedance of Series RLC Circuits
1. XC = )F047.0)(kHz 5(2
1
2
1
fC= 677
XL = 2fL = 2(5 kHz)(5 mH) = 157 Z = R + jXL jXC = 10 + j157 j677 = 10 520 = 52088.9 Net reactance = jXL jXC = j520 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.
XT = 2XL 5.1422
35)80(2
22
X
ZT = 47 j142.5 = 15071.7
4. Z = 22 )( CL XXR = 100
R2 + (XL XC)2 = 1002 (XL XC)2 = 1002 R2
XL XC = 2222 )47()100(100 R = 88.3
Section 17-2 Analysis of Series RLC Circuits 5. ZT = R + jXL jXC = 47 + j80 j35 = 47 + j45 = 65.143.8
IT =
8.431.65
V 04
T
s
Z
V = 61.443.8 mA
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
Figure 17-1
6. Use the results of Problem 5. See 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
(a) IT =
1.65335
V 012
T
s
Z
V = 35.865.1 mA
(b) Ptrue = = (35.8 mA)2(141 ) = 181 mW TT RI 2
(c) Pr = = (35.8 mA)2(304 ) = 390 mVAR TT XI 2
(d) Pa = 22true )()( rPP = 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 = pF) mH)(47 (12
1
2
1
LC= 734 kHz
XL = 2frL = 2(734 kHz)(1 mH) = 4.61 k XC = XL = 4.61 k Ztot = R = 22
I =
22
V 12
tot
s
Z
V = 545 mA
11. VC = VL = 100 V at resonance
Z = R = mA 50
V 10
max
s
I
V = 200
XL = XC = mA 50
V 100
max
L
I
V= 2 k
176
Chapter 17
12. fr = F)mH)(0.015(0.0082
1
2
1
LC= 459 kHz
XL = 2(459 kHz)(0.008 mH) = 23.1
Q =
10
1.23
R
X L = 2.31
BW = 31.2
kHz 459
Q
fr = 199 kHz
f1 = fr 2
BW = 459 kHz
2
kHz 199 = 359.5 kHz
f2 = fr + 2
BW = 459 kHz +
2
kHz 199 = 558.5 kHz
13. Imax =
10
V 7.07
R
Vs = 707 mA at resonance
Ihalf-power = 0.707Imax = 0.707(707 mA) = 500 mA 14. At f1:
XC = F) kHz)(0.015 53592
1
= 29.5
XL = 2(359.5 kHz)(0.008 mH) = 18.1 XC XL = 29.5 18.1 = 11.4
=
10
4.11tan 1 = 48.7 current leading
At f2:
XC = F) kHz)(0.015 55582
1
= 19.0
XL = 2(588.5 kHz)(0.008 mH) = 28.1 XL XC = 28.1 19.0 = 9.1
=
10
1.9tan 1 = 42.3 current lagging
Figure 17-2
r = 0 15. Refer to Figure 17-2.
fr = LC2
1 Choose C = 0.001 F
(a) fr = 500 kHz
LC
fr 22
4
1
L = F)001.0(kHz) 500(4
1
4
12222
Cfr
= 101 H
(b) fr = 1000 kHz
L = F)001.0(kHz) 1000(4
1
4
12222
Cfr
= 25.3 H
177
Chapter 17
(c) fr = 1500 kHz
L = F)001.0(kHz) 1500(4
1
4
12222
Cfr
= 11.3 H
(d) fr = 2000 kHz
L = F)001.0(kHz) 2000(4
1
4
12222
Cfr
= 6.33 H
Part 2: Parallel Circuits Section 17-4 Impedance of Parallel RLC Circuits 16. XL = 2fL = 2(12 kHz)(15 mH) = 1131
XC = )F022.0)(kHz 12(2
1
2
1
fC= 603
Z =
90603
1
901131
1
0100
11
= mS j1.66 mS j0.884mS 10
1
= 99.74.43
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
2fL < fC2
1
f2 < LC24
1
f < LC24
1
f < LC2
1
f <F) mH)(0.022 2
1
f < 8.76 kHz
178
Chapter 17
Section 17-5 Analysis of Parallel RLC Circuits
19. IT =
43.47.99
V 05
Z
Vs = 50.24.43 mA
IR =
0100
V 05
R
Vs = 500 mA
IL =
901131
V 05
L
s
X
V = 4.4290 mA
IC =
90603
V 05
C
s
X
V = 8.2990 mA
VR = VL = VC = 50 V 20. XL = j9.42 k; XC = j72.3 k ZT = 100 j9.42 k j72.3 k = 58.953.9
21. IR =
0100
V 05 = 500 mA; IL =
k9042.9
V 05 = 53190 A
IC =
k 903.72
V 05 = 69.190 A; IT =
9.539.58
V 05 = 84.953.9 mA
Section 17-6 Parallel Resonance 22. ZT = (infinitely high)
23. fr = pF) mH)(47 (502
1
2
1
2
12
LCLCL
CRW
= 104 kHz
XL = 2frL = 2(104 kHz)(50 mH) = 32.7 k
Q =
20
k7.32
W
L
R
X = 1635
Zr = RW(Q2 + 1) = 20 (16352 + 1) = 53.5 M 24. Zr = 53.5 M and fr = 104 kHz from Problem 23.
Itot = M5.53
V 6.3 = 11.8 A
XL = 2frL = 2(104 kHz)(50 mH) = 32.7 k
IC = IL = 22 )k7.32()20(
V 6.3
= 164 mA
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. (a) ZT = jXL + C
C
XR
XR
j
)j(
= j100 +
3.34266
33000j100j
150j220
)150j)(220( = j100 + 12455.7
= j100 + 69.9 j102.4 = 69.9 j2.4 = 69.91.97
(b) ZT = jXL + C
C
XR
XR
j
)j(
= j8 k + j120 k
15.6 39.8
= j8 k + 7.6950.2 k
= j8 k + 4.92 k j5.91 k = 4.92 k + j2.09 k = 5.3523.0 k 27. From Problem 26: (a) = 1.97 (capacitive) (b) = 23.0 (inductive) 28. 1.5 H = 1500 mH XL = 2(2 kHz)(1500 mH) = 18.9 k
XC = )F0047.0)(k 2(2
1
= 16.9 k
ZLR2C = jXL (R2 jXC) = LC
CL
XXR
XRX
jj
)j(j
2
2
=
k 19.51.22
)k 5.377.27)(k 909.18( = 23.747.3 k = 16.1 k + j17.4 k
ZT = R1 + ZLR2C = 33 k + 16.1 k + j17.4 k = 49.1 k + j17.4 k = 52.1 19.5 k
VR1 = V 012k5.191.52
k033
s
T
1 VZ
R= 7.6019.5 V
VL = V 012k5.191.52
k3.477.23
s
T
LR2C VZ
Z= 5.4627.8 V
VR2 = V 8.2746.5k5.377.27
k022
j2
L
2 VR
CXR= 4.3463.3 V
VC = V 8.2746.5k5.377.27
k909.16
j2
L
C VX
CXR= 3.3324.7 V
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
Z2 = XL Z1 =
k j2k22
)k 909.18)(k 5.377.27( = 23.747.3 k = 16.1 k + j17.4 k
ZT = R1 + Z2 = 33 k + 16.1 k + j17.4 k = 49.1 k + j17.4 k = Req + jXeq Xeq = 2fL
1.38 H49.1 k
Figure 17-3
L = kHz) 2(2
k 4.17
2
f
X eq = 1.38 H
30. XC = )F47)(Hz 60(2
1
= 56.4
XL = 2(60 Hz)(390 mH) = 147 ZA = R1 + jXL = 100 j147 = 17855.8
ZB = R2 ZA =
j147 200
) 8.58178)( 0100( = 71.819.5 = 67.7 + j24.0
ZT = XC + ZB = j56.4 + 67.7 + j24.0 = 67.7 j32.4 = 74.825.7
VR2 = V 01157.258.74
5.198.71
s
T
B VZ
Z = 11045.2 V
I2 =
0100
V2.45110
2
R2
R
V = 1.1045.2 A
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 =
k 10
1j
k 5
1j
k 10
11
= mS451414.0
1
mS j0.1 mS 0.1
1
= 7.0745 k = 5 k + j5 k
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
Chapter 17
33. From Problem 32, ZT = 8.3 k + j4 k = 9.2125.7 k
IT =
k7.2521.9
V010
T
s
Z
V = 1.0925.7 mA
ZA = 7.0745 k from Problem 32.
VZA = V 010k7.2521.9
k4507.7
s
T
A VZ
Z = 7.6719.3 V
IR1 = IC1 = IT = 1.0925.7 mA
IR2 =
k010
V3.1967.7
2
ZA
R
V = 76719.3 A
IC2 =
k9010
V3.1967.7
C2
ZA
X
V = 767109.3 A
IL =
k905
V3.1967.7
L
ZA
X
V = 1.53-70.7 mA
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
Va = VL1 = V 012j226180
90226
= 9.3838.5 V
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
XC = )F22.0)(kHz 3(2
1
= 241
XL1 = 2(3 kHz)(12 mH) = 226 XL2 = 2(3 kHz)(8 mH) = 151
182
Chapter 17
Za + Zb + Zc = 100 j226 + j151 = 100 + j377 = 39075
Z1 =
75390
) 0100)( 90226(
cba
ca
ZZZ
ZZ = 57.915
Z2 =
75390
) 0100)( 90151(
cba
cb
ZZZ
ZZ = 38.515
Z3 =
75390
) 90151)( 90226(
cba
ba
ZZZ
ZZ = 86.9105
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
VR1Z1 = VCZ2 = V 0129.8103
9.38159
= 18.530.0 V
VR1 = V 0.305.1864.3236
0180
R1Z1
R1Z1
1 VZ
R = 14.133.6 V
VC = V 0.305.189.80234
90241
CZ2
CZ2
C VZ
X = 19.139.1 V
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
I100 =
0100
V 12622.5
0100abV
= 52.2126 mA
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:
fr = )F15.0)(mH 10(2
1
2
1
1
CL= 4.11 kHz
XL1 = 2(4.11 kHz)(10 mH) = 258 XC = 258
XL2 = 2(4.11 kHz)(25 mH) = 646
Zr = RW1 + jXL1 +CLW
LWC
XXR
XRX
jj
)j(j
22
22
= 2 + j258 + 258j646j4
)646j4(58j2
= 47590
Vout = V 010904750860
90475
s
r
r VZR
Z
= 4.8361.0 V
183
Chapter 17
For parallel resonance:
fr = )F15.0)(mH 25(2
1
2
1
2
12
LCLCL
CRW
= 2.60 kHz
XL = 408
Q =
4
408
2
2
W
L
R
X = 102
Zr = RW )1102(41 22 Q = 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.
fr LC2
1 2 1
4rf
LC
C Lfr
24
1
For fr = 8 MHz, 9 MHz, 10 MHz, and 11 MHz
C1 = H)10(MHz) 8(4
12
= 39.6 pF
Figure 17-5
C2 = H)10(MHz) 9(4
12
= 31.3 pF
C3 = H)10(MHz) 10(4
12 = 25.3 pF
C4 = H)10(MHz) 11(4
12 = 20.9 pF
Part 4: Special Topics Section 17-8 Bandwidth of Resonant Circuits
39. Q =
25
k 2
R
X L = 80
BW = 80
kHz 5
Q
fr = 62.5 Hz
40. BW = f2 f1 = 2800 Hz 2400 Hz = 400 Hz
fr = 2
Hz 2800 Hz 2400
221
ff = 2600 Hz
184
Chapter 17
185
41. Pf1 = (0.5)Pr = (0.5)(2.75 W) = 1.38 W
42. Q = Hz 800
kHz 8
BW
fr = 10
XL(res) = QRW = 10(10 ) = 100
L = kHz) 82
100
2
r
L
f
X = 1.99 mH
XC = XL at resonance
C = ) kHz)(100 82
1
2
1
Cr Xf = 0.2 F
43. BW = Q
fr
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
Chapter 18 Passive Filters
Section 18-1 Low-Pass Filters
1. Vout =
j500k 2.2
90500100 V = 0.491 V j2.16 V = 2.2277.2 V rms
2. (a) 100 Hz is passed (b) 1 kHz is passed (c) 2 kHz is passed (d) 3 kHz is borderline (e) 5 kHz is rejected
3. (a) XC = F) Hz)(10 60(2
1
= 265
Vout = V 010j265 100
90265
= 9.3620.7 V
(b) XC = F) Hz)(8.2 400(2
1
= 48.5
Vout = V 010j48.5 47
905.48
= 7.1844.1 V
(c) XL = 2(1 kHz)(5 mH) = 31.4
Vout = V 010j31.4 330
0330
= 9.965.44 V
(d) XL = 2(2 kHz)(80 H) = 1
Vout = V 010j1 10
010
= 9.955.74 V
4. (a) fc = F) )(10 100(2
1
= 159 Hz
XC = F) Hz)(10 159(2
1
= 100
Vout = V 05j100 100
90100
= 3.5445 V
(b) fc = F) )(8.2 47(2
1
= 413 Hz
XC = F) Hz)(8.2 413(2
1
= 47.0
Vout = V 05j47 47
9047
= 3.5445 V
186
Chapter 18
18
(c) fc = ) mH/330 5(2
1
= 10.5 kHz
XL = 2(10.5 kHz)(5 mH) = 330
Vout = V 05j330 330
0330
= 3.5445 V
(d) fc = )10/H 08(2
1
= 20.0 k
XL = 2(20.0 kHz)(80 H) = 10
Vout = V 05j10 10
010
= 3.5445 V
5. fc = RC2
1
C = cRf2
1
(a) C = Hz) 60)(220(2
1
= 12.1 F
(b) C = Hz) 500)(220(2
1
= 1.45 F
(c) C = kHz) 1)(220(2
1
= 0.723 F
(d) C = kHz) 5)(220(2
1
= 0.144 F
6. Position 1: Position 3:
CT = 1000 pF fc =
TRC2
1
fc = pF) 1000)(k10(2
1
= 15.9 kHz
CT =
F047.0F022.0
1
F01.0
11
Position 4:
CT =
pF 1000
1
F001.0
11
= 500 pF = 0.00873 F
fc = F) 00873.0)(k10(2
1
= 1.82 kHz
fc = pF) 500)(k10(2
1
= 31.8 kHz
Position 2: CT = 0.022 F + 0.047 F = 0.069 F
fc = F) 069.0)(k10(2
1
= 231 Hz
7
Chapter 18
7. See Figure 18-1.
Figure 18-1
8. (a)
V 1
V 1log20log20
in
out
V
V = 0 dB
(b)
V 5
V 3log20log20
in
out
V
V = 4.44 dB
(c)
V 10
V 7.07log20log20
in
out
V
V = 3.01 dB
(d)
V 25
V 5log20log20
in
out
V
V = 14.0 dB
9. dB =
in
out
V
Vlog20
20
dBlog 1
in
out
V
V
Vout =
20
dBlog 1
inV
(a) Vout =
20
1logV) 8( 1 = 7.13 V
(b) Vout =
20
3logV) 8( 1 = 5.67 V
(c) Vout =
20
6logV) 8( 1 = 4.01 V
(d) Vout =
20
20logV) 8( 1 = 0.800 V
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. 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
13. Vout =
j500k 2.2
k 02.2100 V = 9.7512.8 V
14. (a) 1 Hz is rejected. (b) 20 Hz is rejected. (c) 50 Hz is borderline. (d) 60 Hz is passed. (e) 30 kHz is passed.
15. (a) XC = F) Hz)(10 60(2
1
= 265
Vout = V 010j265 100
0100
= 3.5369.3 V
(b) XC = F) Hz)(4.7 400(2
1
= 84.7
Vout = V 010j84.7 47
047
= 4.8561.0 V
(c) XL = 2(1 kHz)(5 mH) = 31.4
Vout = V 010j31.4 330
904.31
= 94784.6 mV
(d) XL = 2(2 kHz)(80 H) = 1
Vout = V 010j1 10
901
= 99584.3 mV
16. fc = RC2
1, fc =
)/(2
1
RL
(a) fc = F) )(10 100(2
1
= 159 Hz; Vout = 7.07 V
189
Chapter 18
190
(b) fc = F) )(4.7 47(2
1
= 720 Hz; Vout = 7.07 V
(c) fc = ) mH/330 5(2
1
= 10.5 kHz; Vout = 7.07 V
(d) fc = ) H/10 80(2
1
= 19.9 kHz; Vout = 7.07 V
Figure 18-2
720 Hz
17. See Figure 18-2. 18. Position 1: Position 3:
RT = 1 k + 3.3 k + 1 k = 5.3 k RT = 860 + 1 k = 1.86 k
fc = F) 015.0)(k3.5(2
1
= 2.00 kHz fc =
F) 015.0)(k86.1(2
1
= 5.70 kHz
Position 2: Position 4: RT = 3.3 k + 1 k = 4.3 k RT = 2.2 k + 3.3 k + 1 k = 6.5 k
fc = F) 015.0)(k5.6(2
1
= 1.63 kHz CT =
F 01.0
1
F015.0
11
= 0.006 F
fc = F) 006.0)(k3.4(2
1
= 6.17 kHz
Chapter 18
Section 18-3 Band-Pass Filters
19. (a) f0 = F) mH)(0.01 12(2
1
2
1
LC = 14.5 kHz
(b) f0 = F) mH)(0.022 2(2
1
2
1
LC = 24.0 kHz
20. (a) RT = 10 + 75 = 85
f0 = F) mH)(0.01 12(2
1
2
1
LC = 14.5 kHz
XL = 2(14.5 kHz)(12 mH) = 1.10 k
Q =
85
k 1.1
T
L
R
X = 13
BW = 13
kHz 5.140 Q
f = 1.12 kHz
(b) RT = 10 + 22 = 32
f0 = F) mH)(0.022 2(2
1
2
1
LC = 24.0 kHz
XL = 2(24.0 kHz)(2 mH) = 302
Q =
32
302
T
L
R
X = 9.44
BW = 9.44
kHz 0.240 Q
f = 2.54 kHz
21. Using the results of Problems 19 and 20:
(a) f2 = f0 + 2
kHz 1.12kHz 5.14
2
BW = 14.5 kHz + 560 kHz = 15.06 kHz
f1 = f0 2
kHz 1.12kHz 5.14
2
BW = 14.5 kHz 560 Hz = 13.94 kHz
(b) f2 = f0 + 2
kHz 2.54kHz 0.24
2
BW = 24.0 kHz + 1.27 kHz = 25.3 kHz
f1 = f0 2
kHz 2.24kHz 0.24
2
BW = 24.0 kHz 1.27 kHz = 22.7 kHz
22. Center frequency = f0 = LCL
CRW
2
12
Since RW is assumed to be zero, f0 = LC2
1.
191
Chapter 18
(a) f0 = F) H)(10 1(2
1
= 50.3 Hz
(b) f0 = pF) H)(25 5.2(2
1
= 20.1 MHz
23. (a) f0 = F) H)(10 1(2
H 1
F)10()4(1
2
122
LCL
CRW
= 50.3 Hz
XL = 2(50.3 Hz)(1 H) = 316
Q =
4
316
W
L
R
X = 79
Ztank = RW(Q2 + 1) = 4 (792 + 1) = 24,968
Vout = V 120680968,24
968,24
= 117 V
(b) f0 = pF) H)(25 2(2
1
2
1
LC = 10.1 MHz
XL = 2(20.1 MHz)(2.5 H) = 316
Q =
4
316
W
L
R
X = 79
Ztank = RW(Q2 + 1) = 4 (792 + 1) = 24,968
Vout = V 1201000968,24
968,24
= 115 V
24. Position 1:
f0 = pF) H)(1000 50(2
1
2
1
LC = 712 kHz
Position 2:
f0 = F) H)(0.01 100(2
1
2
1
LC = 159 kHz
Position 3:
f0 = F) H)(0.001 270(2
1
2
1
LC = 306 kHz
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
XC = mA 20
V 5.2 = 125
C = CXf02
1
= 0.064 F
Q = W
L
R
X = 40
RW = 40
LX = 0.025XL = 0.025(2f0L)
f0 = LCL
RW
2
12
LC
LCf
LCL
CLf
LCL
R
f
W
2
20
2
2
20
2
2
20
4
))2(025.0(1
4
))2)(025.0(1
4
1
Note: in the above derivation, (0.025(2))2 = 0.025
= LCf 220 4 LCf 2
025.01
= 1 )025.04( 220 LCf
L = )025.04(
122
0 Cf = 989 H
Section 18-4 Band-Stop Filters
26. (a) f0 = F) H)(0.0022 100(2
1
2
1
LC = 339 kHz
(b) f0 = F) mH)(0.047 5(2
1
2
1
LC = 10.4 kHz
27. (a) f0 F) H)(6.8 5.0(2
1
2
1
LC = 86.3 Hz
(b) f0 pF) H)(47 10(2
1
2
1
LC = 7.34 MHz
28. (a) f0 = LCL
CRW
2
12
= 86.3 Hz
XL = 2(86.3 Hz)(0.5 H) = 271
Q =
8
271
W
L
R
X = 33.9
Ztank = RW(Q2 + 1) = 8 ((33.9)2 +1)= 9.20 k
V 50k2.10
k 1
= 4.90 V Vout =
193
Chapter 18
194
(b) f0 = pF) H)(47 10(2
H 10
pF)1
47()8( 2
= 7.34 MHz
(7.34 MHz)(10
Q =
XL = 2 H) = 461
8WR2 2
461LX = 57.6
Z = R (Q + 1) = 8 (57.6 + 1) = 26.6 k
Vout =
tank W
V 50k8.28
k2.2
= 3.82 V
and, f0 = 1200
f0 =
29. For the pass b kHz:
CL12
1
CL
f 20
124
1
L1 = F)22.0(kHz) 1200(4
1
4
1222
02
Cf
= 0.08 H
For the stop band, f0 = 456 kHz:
0 f = CL22
1
F)22.0(kHz) 456(4
1
4
1222
02
L2 = Cf
= 0.554 H
Multisim Troubleshooting and Analysis 30. C1 is open. 31. C is leaky.
2. R is open.
3. C1 is shorted.
4. L is open.
5. No fault.
6. f = 107.637 kHz
7. BW 88.93 MHz
2
3 3
3 3 2
3 3 r
3
Chapter 19 Circuit Theorems in AC Analysis
Section 19-1 The Superposition Theorem 1. Z1 = R2 R3 = 6880 Z2 = R1 + Z1 = 16880
Z3 = XL Z2 =
8.492617
)k902)(01688( = 1.2940.2 k = 985 + j833
ZT1 = XC + Z3 = j1 k + 985 + 833 = 985 j167 = 9999.6
IT1 =
9.6999
V 02
T1
1
Z
V = 29.6 mA
IR1 = mA 6.92k8.4962.2
k02
T1
2L
L IZX
X= 1.5349.8 mA
IR3(V1) = mA 8.4953.1k02.3
k01
R1
32
2 IRR
R= 46949.8 A = 303 A + j358 A
With V1 reduced to zero (shorted):
Z1 = XC XL =
k901
)k902)(k901( = 290 k
Z2 = R1 + Z1 = 1 k j2 k = 2.2463.4 k
Z3 = R3 Z2 =
k j2k2.3
)k4.6324.2)(k02.2(
=
k3277.3
k4.6393.4 = 1.3131.4 k = 1.12 k j0.68 k
ZT2 = R2 + Z3 = 2.12 k j0.68 k = 2.2317.8 k
IT2 =
8.172.23
V 303
T2
2
Z
V = 1.3547.8 mA
IR3(V2) = mA 8.4735.1k3277.3
k4.6324.2
= 80216.4 A = 769 A + j226 A
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
IR2(V1) = mA 8.4953.1k 2.3
k 2.2
R1
32
3 IRR
R = 1.0549.8 mA = 678 A + j802 A
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
Z1 = R1 (R2 + XL) =
k j1.9k5.6
)k 9.1jk 7.4)(k08.1(
=
k3.1677.6
)k221.5)(k08.1( = 1.365.7 k = 1.35 k j0.135 k
IZ1 = mA 0100k j2.28k35.1
k9041.2
S
1C
C IZX
X
= mA 0100k4.5965.2
k9041.2
= 90.930.6 mA
IR1(I) = Z1C21
C2 IXRR
XR
= mA 6.309.90k3.2093.6
k1.2728.5
= 69.323.8 mA = 63.4 mA j28.0 mA
With Is reduced to zero (opened):
ZT = R1 + (XC (R2 + XL)) = 1.80 k +
k j0.51k7.4
)k221.5)(k9041.2(
= 1.8 k + 2.6059.2 k = 3.8535.5 k
IR1(V) =
k 35.53.85
V 075
T
s
Z
V = 19.535.5 k = 15.9 mA + j11.3 mA
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:
IL(2) = A 01k1.231.5
k902 mA 01
k j2k7.4
k902
s2
CL
C IXR
X
= 39266.7 mA IL = IL(1) + IL(2) = 0 A + 39266.9 A = 39266.9 mA
(b) XC1 = XC2 = XC3 = pF) kHz)(100 5.2(2
1
= 637 k
With V2 zeroed (shorted), the impedance “seen” by V1 is developed as follows: ZA = RL jXC3 = 5 M j637 k = 5.047.26 M
ZB = R2 ZA =
M06.603.6
)M26.704.5)(M01( = 8351.2 k = 835 k j17.5 k
ZC = XC2 + ZB = j637 k + 835 k j17.5 k = 835 k j654 k = 1.0638 M
ZD = R1 ZC =
M6.1995.1
)M3806.1)(M01( = 54518.5 k = 517 k j172 k
ZT(1) = XC1 + ZD = j637 k + 517 k j172 k = 517 k j809 k = 96057.4 k
IT(1) =
k 4.57960
V 6040
T(1)
1
Z
V = 41.7117.4 A
IC2(1) = A 4.1177.41M8.1994.1
M01
T(1)
C1
1 IZR
R = 21.4137 A
IL(1) = A 1374.21M06.603.6
M01
C2(1)
A2
2 IZR
R = 3.55143 A
With V1 zeroed (shorted), the impedance “seen” by V2 is developed as follows:
ZA = R1 XC1 =
M5.3219.1
)k90637)(M01( = 53557.5 k = 287 k j450 k
ZB = XC2 + ZA = 287 k + j1.09 M = 1.1375.2 M
ZD = (RL + XC3) ZB =
M j1.73M29.5
)M2.7513.1)(k j637M5(
=
M1.1856.5
)M2.7513.1)(M7.774.4( = 1.0264.4 M = 442 k j921 k
ZT(2) = R2 + ZC = 1.44 M j921 k = 1.7132.6M
IT(2) =
M.6321.71
V 3020
T2
2
Z
V = 11.762.6 A
IL(2) = A 6.627.11M1.1856.5
M2.7513.1
T(2)
C3LB
B IXRZ
Z = 2.375.46 A
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). . R + R 5 = 1 k + 3.9 k + 10 k + 5.1 k = 20 k
=
RT = + R4 + R2 3
ITk 20
= 1 mA V 20
I R = 16.1 V (1 mA)(1 k) = 15.1 V
1(b).
T = R (R 1.2 k + 2.36 k = 3.56 k
peak) =
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) T 2
VA(dc) = 0 V See Figure 19- VA(peak) = 9 V R R1 + + R4 R6) =3 2
IT(k 3.56
= 2. A V 9
53 m
VB(peak) = V I R = 9 V (2.53 mA)(1.2 k) = 9 V 3.04 V = 5.96 V
IR2(peak) =
A(peak) T(peak) 1
mA 2.53k 9.9)
6423
RR
k 3.9(
3
peakTIRR
R = 1 mA
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)):
ZT = 1890 +
7.7851
)6.716.31)(9020(
= 1890 + 12.482.9 = j18 + 1.53 + j12.3 = 1.53 j5.69 = 5.9074.9
IT = IC(Vs) =
9.7490.5
V3012 = 2.04105 A = 522 mA + j1.97 A
With the voltage source zeroed (see Figure 19-2(b)): Impedance of the L1, L2, C branch:
Z = j30 +
j2
)9018)(9020( = j30 +
902
0360 = j30 j180 = j150
IL2 = mA 1205002.863.150
010 mA 120500
j15010
010
= 33.3206 mA
IC(Is) = mA 2063.33902
9020
= 333206 mA = 298 mA j147 mA
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
IR(VS) = 1 2T(VS)
1
j ( ) 12.4 97.7I 2.04 105 1.26 112.7 A
20 90L L
L
X R jX
jX
With the voltage source zeroed (shorted): Impedance of LC branch is ZLC = 150-90
LC
LC
1500 9010 0
150 90
R Ζ
ΖZ
IR(IS) = TS
10 00.5 120 A = 0.5 120 A
10 0
Z
IR
IRT = IR(VS) – IR(IS) = 1.26 -0.24 A + j0.727 A 112.7 A 0.5 120 A = 766 71.7 A
Section 19-2 Thevenin’s Theorem 8. From Problem 5, V = V = 4.96 V D(peak) C(peak)
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) Vth = V 025j75100
9075
= 1553.1 V s
C1
C VjXR
X
Zth = R2 +
9.36125
)9075)(0100( 027
C1
C1
jXR
XR = 63 j48
= 79.237.3
(b) Vth = V 0390980
90400
= 1.220 V s
L2L1
L1 VXX
X
Zth =
90980
)90580)(90400(
L2L1
L2L1
XX
XX = 23790 = j237 = 23790
(c) VT = V1 + V2 = 15 V + 8.66 V + j5 V = 24.211.9 V
200
Chapter 19
Vth = V 9.112.24k0200
k0100
T
21
2 VRR
R= 12.111.9 V
Zth = XC + R1 R2 = 50 k j20 k = 53.921.8 k
10. XC1 = XC2 = F) Hz)(0.047 100(2
1
= 33.86 k
Find Zth looking from the open terminals after removing RL:
ZA = R1 XC1 =
k0.574.40
)k9086.33)(k022( = 18.4433.0 k = 15.5 k j10 k
ZB = R2 + ZA = 22 k + 15.5 k j10 k = 37.5 k j10 k = 38.814.9 k
ZC = XC2 ZB =
k5.497.57
)k9.148.38)(k9086.33( = 22.7755.4 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
Find Vth looking from the source after removing RL: ZA = R2 jXC2 = 22 k j33.86 k = 40.457 k
ZB = XC1 ZA =
k722.71
)k574.40)(k9086.33( = 19.275 k = 4.97 k j18.6 k
ZT = R1 + ZB = 26.97 k j18.6 k = 32.734.5 k
IT =
k5.347.32
V032
T
s
Z
V = 0.9834.5 mA
IR2 = mA 5.3498.0k j67.7k22
k9086.33
T
AC1
C1 IZX
X = 0.4716.5 mA
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 =
k9.72.136
V5.739.15
thL
th
ZR
V = 11765.6 A
Figure 19-3
201
Chapter 19
11. The circuit is redrawn in Figure 19-4(a) for easier analysis. Combining R1, R2, and XL:
ZA = R1 + R2 XL = 10 k +
k j3 k3.3
)k903)(k03.3( = 2.49 k + j1.64 k
= 2.9833.4 k Combining R3 and ZA:
ZB = R3 ZA =
k5.76.12
)k4.3398.2)(k010( = 2.3725.9 k
Figure 19-4
Combining XC and ZB:
Zth = XC ZB =
k3.6252.4
)k9.2537.2)(k905( = 2.621.8 k 2.62 k j0.082 k
Vth = V 050k7.615.4
k9.2537.2
= 26.387.6 V s
BC
B VZX
Z
The Thevenin circuit with R4 connected is shown in Figure 19-4(b).
VR4 = V 6.873.26k64.032.7
k07.4
= 16.988.2 V th
th4
4 VZR
R
12. Refer to Figure 19-5 (note that R3 has been removed).
ZA = R1 XL =
425.134
)9090)(0100( = 66.948 = 44.8 + j49.7
ZB = R2 + ZA = 194.8 + j49.7 = 20114.3
Zth = XC ZB =
8.19207
)3.14201)(90120( = 11755.9
Looking from Vs:
ZT = XL + R2 (R1 + XC) = j90 +
6.25277
)7.38192)(0100(
= j90 + 69.313.1 = j90 + 67.5 j15.7 = 67.5 + j74.3 = 10047.7
VL = V 0757.47100
9090
= 67.542.3 V
sT
L VZ
X
202
Chapter 19
VR2 = V) 3.425.67V 075(120j150
0150
)V(VXR
RLs
C2
2
=
7.38192
0150 (75 V 49.9 V j45.4 V) = (0.78138.7)(51.961.1 V)
= 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. Using Zth and Vth from Problem 9 in each part:
(a) In =
3.372.79
V1.5315
th
th
Z
V = 18915.8 mA
Zn = Zth = 79.237.3
(b) In =
90237
V022.1
th
th
Z
V = 5.1590 mA
Zn = Zth = 23790
(c) In =
k8.219.53
V9.111.12
th
th
Z
V = 22433.7 A
Zn = Zth = 53.921.8 k 14. 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.
IT =
k2.286.39
V032
T
s
Z
V = 80828.2 A
203
Chapter 19
IR2 = A 2.28808k5.497.57
909.33
TC232C1
C1 IXRRX
X = 47512.3 A
In = IR3 = A 3.12475k574.40
909.33
R2
C23
C2 IXR
X = 39945.3 A
The Norton equivalent circuit with RL connected is shown in Figure 19-6.
IRL = A 3.45399k9.72.136
k2.286.39
n
nL
n IZR
Z = 11665.7 A
39.628.2 k
39945.3 A
Figure 19-6
15. First remove R4 and determine Zn looking in at the resulting open terminals.
ZA = R3 XC =
k6.262.11
)k905)(k010( = 4.4663.4 = 2 k j4 k
ZB = R1 + R2 XL = 10 k +
k3.4246.4
)k903)(k03.3(
= 10 k + 2.2247.7 k = 2.49 k + j1.64 k = 2.9833.4 k
Zn = ZA ZB =
k7.271.5
)k4.3398.2)(k4.6346.4(= 2.612.30 k =2.61 k j0.105 k
Looking from the source with R4 shorted: ZT = XC = 590 k
In =
k905
V050
T
s
Z
V= 1090 mA
The Norton equivalent circuit with R4 connected is shown in Figure 19-7.
IR4 = mA 9010k823.031.7
k30.261.2
n
n4
n IZR
Z= 3.5788.5 mA
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) XC = )F0047.0)(kHz 3(2
1
2
1
fC= 11.3 k
ZL = RL + jXL = 6.8 k + j11.3 k
L = kHz) 3(2
k 3.11
2
f
X L = 599 mH
(b) ZL = 8.2 k + j5 k (c) XL = 75.4 , XC = 60.3
Zth = R + XC XL = 500 +
901.15
)903.60)(904.75( = 50 j301
ZL = 50 + j301
L = Hz) 120(2
301
= 0.4 H
17. For maximum load power, ZL equals the complex conjugate of Zth. ZA = R1 jXC1 = 8.2 j10 = 12.950.6
ZB = R2 XC2 =
5.124.18
)904)(018( = 3.9177.5 = 0.846 j3.82
ZC = ZB ZA =
7.565.16
)5.7791.3)(6.509.12( = 3.0671.4 = 0.976 j2.90
Zth = R3 + ZC = 9.18 j2.90 ZL = 9.18 + j2.90
18. First convert the delta to a wye:
X1 = X2 = X3 =
121212
)12)(12( = 4
The circuit is redrawn in Figure 19-8(a). Remove ZL and Thevenize:
Zth = j4 + (6.8 + j4 ) j4 = j4 +
6.495.10
)904)(5.309.7( j4
j88.6
) j4)(j48.6(
= j4 + 370.9 = j4 + 0.98 + j2.8 = 0.98 + j6.8
Vth = V 0106.495.10
5.309.7V 010
j88.6
j48.6
= 7.519.1 V
For maximum power to ZL: ZL = 0.98 j6.8
IL =
096.1
V1.195.7
Lth
th
ZZ
V = 3.8319.1 A
PL(true) = = (3.83 A)2(0.98 ) = 14.4 W LLRI 2
205
Chapter 19
20
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.
Zth =
0j12220
)90120)(0220(
90j100
)9090)(0100( = 6748 + 105.361.4
= 44.8 + j49.8 + 50.4 j92.5 = 95.2 j42.7 ZL = 95.2 + j42.7
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
6
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) 5RC = 5(56 )(47 F) = 13.2 ms (b) 5RC = 5(3300 k)(0.015 F) = 248 s (c) 5RC = 5(22 k)(100 pF) = 11 s (d) 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) v 0.865(20 V) = 17.3 V (b) v 0.950(20 V) = 19.0 V (c) v 0.982(20 V) = 19.6 V (d) v 0.993(20 V) = 19.9 V (considered full charge of 20 V) 5. See Figure 20-1.
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) Looking from the capacitor, the Thevenin resistance is R1 R2 = 5 k. = (5 k)(4.7 F) = 23.5 ms
Figure 20-3
(b) Vout(max) = V 20k 20
k 10
= 10 V
See 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.
Figure 20-5 23.5 ms
6.32 V
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
Figure 20-6
5 = 5(47 ms)= 235 ms See Figure 20-6. 12. See Figure 20-7.
Figure 20-7
13. T = kHz 10
11
f = 100 s
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.
Figure 20-8
209
Chapter 20
14. The steady-state output equals the average value of the square wave input which is
2
V 30
2inV
= 15 V (with a small ripple voltage)
Section 20-4 Response of RC Differentiators to a Single Pulse 15. See Figure 20-9.
Figure 20-9
Figure 20-10
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. 17. (a) Looking from the source and capacitor:
Figure 20-11
RT =
k 2.4
)k 1k 1)(k 2.2( = 1.05 k
= RTC = (1.05 k)(470 pF) = 493.5 ns 5 = 5(493.5 ns) = 2.467 s
(b) Vout(max) = V 10k 2
k 1
= 5 V
See Figure 20-11.
Section 20-5 Response of RC Differentiators to Repetitive Pulses 18. = (1 k)(1 F) = 1 ms See Figure 20-12.
210Figure 20-12
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 10 = 1 ms
Figure 20-13
5 = 5 ms Vout(max) = 0.637(8 V) = 5.06 V See Figure 20-13.
Figure 20-14
21. = 1
mH 50 = 50 ms
5 = 250 ms See Figure 20-14. 22. LT = 8 H + 4 H = 12 H
RT =
256
)156)(100( = 60.9
=
60.9
H12
T
T
R
L = 197 ns
This circuit is an integrator.
Section 20-7 Response of RL Differentiators to Pulse Inputs
Figure 20-15
23. (a) =
22
H100 = 4.55 s
(b) See Figure 20-15.
211
Chapter 20
212
24. (a) =
22
H100 = 4.55 s
Figure 20-16
(b) See Figure 20-16.
Section 20-8 Relationship of Time Response to Frequency Response
25. fh = rt
35.0, 5 = 50 s
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
fh = s0.22
35.0
= 15.9 kHz
26. fh = ns 42
35.035.0
ft = 8.33 MHz
Section 20-9 Troubleshooting 27. (b) Vout = Vin: C is open or R could be shorted. (c) C is leaky or C is greater than 0.22 F or R is greater than 3.3 k. (d) Resistor open or capacitor shorted. 28. (a) No problem since 5<tW. (b) C is leaky. (c) C is open or R is shorted.
Multisim Troubleshooting and Analysis 29. C1 open or R1 shorted. 30. No fault. 31. R1 or R2 open. 32. L1 or L2 open.
Chapter 21 Three-Phase Systems in Power Applications
Section 21-1 Generators in Power Applications
1. IL = 100 V
265.8
V
Z
= 376 mA
2. = tan1 175
200
= 41.2
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
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 = A) 905(3 = 8.6690 A
ILc = A) 1505(3 = 8.66150 A 6. 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:
Ia = IZa = k 321
V 0500 = 50032 mA
Ib = IZb =
k 321
V 120500 = 50088 mA
Ic = IZc =
k 321
V 120500 = 500152 mA
(c) Line currents: (d) Load currents: (e) Load voltages: ILa = 50032 mA IZa = 50032 mA VZa = Va = 5000 V ILb = 50088 mA IZb = 50088 mA VZb = Vb = 500120 V ILc = 500152 mA IZc = 500152 mA VZc = Vc = 500120 V
214
Chapter 21
8. (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:
Ia =
45135
V 0100 = 74145 mA
Ib = 60100
V 120100 = 160 A
Ic =
20200
V 120100 = 500140 mA
(c) Line currents: (d) Load currents: (e) Load voltages: ILa = Ia = 74145 mA IZa = Ia = 74145 mA VZa = Va = 1000 V ILb = Ib = 160 A IZb = Ib = 160 A VZb = Vb = 100120 V ILc = Ic = 500140 mA IZc = Ic = 500140 mA VZc = Vc = 100120 V
(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 9. (a) Line voltages:
VL(ab) = 3 Va(30) = 3 (50(030)) V = 86.630 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 (b) Phase currents: First find the load currents:
IZa =
70600
V 1506.86)(
a
caL
Z
V = 144220 mA = 110 mA + j92.6 mA
IZb =
70600
V 906.86)(
b
bcL
Z
V = 14420 mA = 135 mA + j49.3 mA
IZc =
70600
V 306.86)(
c
abL
Z
V = 144100 mA = 25.0 mA + j142 mA
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
(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
10. (a) Line voltages: (b) Phase currents:
VL(ab) = 120120 V Ia = IZa =
5010
V0120 = 1250 A
VL(ca) = 1200 V Ib = Ib =
5010
V120120 = 12170 A
VL(bc) = 120120 V Ic = Ic =
5010
V120120 = 1270 A
(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: (e) Load voltages: IZa = 1250 A VZa = 1200 V IZb = 12170 A VZb = 120120 V IZc = 1270 A VZc = 120120 V 11. (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:
VZa = V )30120(3
aV = 19190 V
VZb = V )300(3
bV = 19130 V
VZc = V )30120(3
cV = 191150 V
IZa =
605
V 90191
a
Za
Z
V = 38.2150 A
216
Chapter 21
IZb =
605
V 30191
b
Zb
Z
V = 38.230 A
IZc =
605
V 150191
c
Zc
Z
V = 38.290 A
Section 21-4 Three-Phase Power 12. PT = 3(1200 W) = 3.6 kW
13. Figure 21-34 in text:
IZ = k 1
V 500 = 500 mA
PL = 3VZIZcos = 2(500 V)(500 mA)cos 32 = 636 W Figure 21-35 in text:
IZa = 135
V 100 = 741 mA
PZa = VZaIZacos = (100 V)(741 mA)cos 45 = 52.4 W
IZb = 200
V 100 = 500 mA
PZb = VZbIZbcos = (100 V)(500 mA)cos 20 = 47.0 W
IZc = 100
V 100 = 1 A
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 = V) 50(3 = 86.6 V
IZ = 600
V 86.6 = 144 mA
PL = 3VZIZcos = 3(86.65 V)(144 mA)cos 70 = 12.8 W Figure 21-37 in text:
IZ = 10
V 120 = 12 A
PL = 3VZIZcos = 3(120 V)(12 A)cos 50 = 2.78 kW Figure 21-38 in text:
VZ = 3
V 330 = 191 V
IZ = 5
V 191 = 38.2 A
PL = 3VZIZcos = 3(191 V)(38.2 A)cos 60 = 10.9 kW
217
Chapter 21
218
14. VZ = 3
V 120 = 69.3 V
IZ =
4.141
V 69.3
)100()100(
V 3.6922Z
VZ = 490 mA
= 45 PL = 3VZIZcos = 3(69.3 V)(490 mA)cos 45 = 72 W 15. ZL = 141.445 IL = IZ
VZ = 3
V 120
3LV
= 69.8 V
IZ =
141.4
V 8.69
L
Z
Z
V = 494 mA
PT = 3 VLILcos
Peach = 3
mA) V)(494 (120cos
3
LLIVcos 45 = 24.2 W
16. P1 = VLILcos( + 30) P2 = VLILcos( 30)
=
100
100tan 1 = 45
P1 = (120 V)(494 mA)cos(45 + 30) = 15.3 W P2 = (120 V)(494 mA)cos(45 30) = 56.8 W