CH 09

1
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis

2
SOLUTION:

3
SOLUTION:

4
SOLUTION:

5

6
SOLUTION:

7

8
SOLUTION:

9
SOLUTION:
Vc(0) = -2V, i(0) = 4A

𝑅 1
𝛼 = = 2, 𝜔𝑜2 = = 3, 𝑠1,2 = −2 ± 1 = −1, −3
2𝐿 𝐿𝐶
a. 𝑖 = 𝐴𝑒 −𝑡 + 𝐵𝑒 −3𝑡
∴ 𝐴 + 𝐵 = 4; 𝑖(0+ ) = 𝑣𝐿 (0+ ) = (−4 × 4 + 2) = −14
∴ −𝐴 − 38 = −14 ∴ 𝐵 = 5, 𝐴 = 1, 𝑖 = −𝑒 −𝑡 + 5𝑒 −3𝑡 𝐴
𝑡
∴ −𝑣𝐶 = 3 ∫ ( − 𝑒 −𝑡 + 5𝑒 −3𝑡 )𝑑𝑡 − 2 = 3(−𝑒 −𝑡 − 5𝑒 −3𝑡 )𝑡0 − 2
0
= 𝑒 − 3 − 5𝑒 −3𝑡 + 5 + 2
−𝑡
∴ −𝑣𝐶 = 3𝑒 −𝑡 − 5𝑒 −3𝑡
𝑃𝑐 (0+ ) = (3 − 5)(−1 + 5) = −8𝑊
b. 𝑃𝑐 (0.2) = (3𝑒 −0.2 − 5𝑒 −0.6 )(−𝑒 0.2 + 5𝑒 −0.6 ) = −0.5542𝑊
c. 𝑃𝑐 (0.4) = (3𝑒 −0.4 − 5𝑒 −1.2 )(−𝑒1.2 + 5𝑒 −0.4 ) = 0.4220𝑊

10
SOLUTION:
For t>0
𝑅
𝑖(𝑡) = 8𝑒 − 𝐿 𝑡 = 8𝑒 −8𝑡
𝑎. p(0+) = (82) (1)=64W
𝑏. At t = 1s, i = 8e-2 = 1.083A; p(1) = i2R = 1.723W
𝑐. At t = 2s, i = 8e-4 = 146.5mA; p(2) = i2R = 21.47W

11
SOLUTION:
106
𝑍𝑐 = = −𝑗80Ω,
𝑗500 × 25
50(−𝑗80)
= 42.40∠ − 32.01𝑜 Ω
50 − 𝑗80
∴ 𝑉 = 84.80∠32.01𝑜 𝑉, 𝐼𝑅 = 1.696∠−32.01𝑜 𝐴
𝐼𝑅 = 1.0600∠57.99𝑜 𝐴
ps(π/2ms) = 84.80cos(45o-32.01o)2cos45o = 116.85W
pR = 50 x 1.6962cos2(45o-32.01o) = 136.55W
pc = 84.80cos(45o-32.01o) = 1.06cos(45o+57.99o) = -19.69W

12
SOLUTION:
a. P = 276 x 130 = 358.8mW

b. V(t) = 2.76 cos1000t V (given); we need to know the I-V relationship for this non
linear device.

13
SOLUTION:

14
SOLUTION:

15

16
SOLUTION:

17
SOLUTION:

18

19
SOLUTION:
40∠30𝑜
𝑉 = (10 + 𝑗10) = 52.44∠69.18𝑜 𝑉
5∠50𝑜 + 8∠ − 20𝑜
P10,gen = ½ x 10 x 52.44cos69.18o = 93.19W
Pj10,gen = ½ x 10 x 52.44cos(90o- 69.18o)= 245.1W
1 52.44 2
𝑃5∠50𝑎𝑏𝑠 = ( ) cos(50𝑜 ) = 176.8𝑊
2 5
1 52.44 2
𝑃8∠−20𝑎𝑏𝑠 = ( ) cos(−20𝑜 ) = 161.5𝑊
2 8

20
SOLUTION:

21

22
SOLUTION:
1
𝑍𝑅 = 3 + = 3 + 1 + 𝑗3 = 4 + 𝑗3Ω
0.1 − 𝑗0.3
2 + 𝑗5 5√29
𝐼𝑔𝑛𝑜𝑟𝑖𝑛𝑔 30𝑜 𝑜𝑛 𝑉𝑆 , 𝐼𝑅 = 5 , |𝐼𝑅 | =
6 + 𝑗8 10
1 5√29
a. 𝑃3Ω = ( × 3 = 10.875𝑊
2 10)2
𝑜 (2+𝑗5)(4+𝑗3)
b. 𝑉𝑆 = 5∠0 = 13.463∠51.94𝑜 𝑉
6+𝑗8
1
∴ 𝑃𝑆,𝑔𝑒𝑛 = × 13.463 × 5𝑐𝑜𝑠51.94 = 20.75𝑊
2

23
SOLUTION:

24
SOLUTION:
The current through the impedance is

𝑽 120∠0°
𝑰= = = 1.576∠66.8°
𝒁 30−𝑗70
The average power is

1 1
𝑃 = 𝑉𝑚 𝐼𝑚 cos(𝜃𝑣 − 𝜃𝑖 ) = (120)(1.576) cos(0 − 66.8°) = 37.24 𝑊
2 2

25
SOLUTION:

26
SOLUTION:

27

28
SOLUTION:
The current Iis given by

5∠30°
𝑰= = 1.118∠56.57°
4−𝑗2
The average power supplied by the voltage source is

1
𝑃 = (5)(1.118) cos(30° − 56.57°) = 2.5 𝑊
2
The current through the resistor is
𝑰𝑅 = 𝑰 = 1.118∠56.57°
and the voltage across it is
𝑽𝑹 = 4 𝑰𝑅 = 4.472∠56.57°
The average power absorbed by the resistor is

1
𝑃 = (4.472)(1.118) = 2.5 𝑊
2
which is the same as the average power supplied. Zero average power is absorbed by the capacitor.

29
SOLUTION:

30
SOLUTION:
𝑉𝑥 − 20 𝑉𝑥 − 𝑉𝑐
+ = 2𝑉𝐶
2 2
And
𝑉𝑐 𝑉𝑐 − 𝑉𝑥
2= +
−𝑗2 3
Which simplify to
5𝑉𝑥 − 14𝑉𝑐 = 60
And 𝑗𝑉𝑥 + (3 − 𝑗2)𝑉𝑐 = 0
Solving
𝑉𝑥 = 9.233∠ − 83.88𝑉 and 𝑉𝑐 = 5.122∠ − 140.2𝑉
1
𝑃𝑔𝑒𝑛 = × 9.233 × (2 × 5.122) cos(−83.88 + 140.2) = 26.22
2

31
SOLUTION:

32
SOLUTION:

33
SOLUTION:

34
SOLUTION:

35
SOLUTION:
𝑗480 80−𝑗60
a. Zth = 80||j60 =
80+𝑗60 80+𝑗60
= 28.8 + j38.4Ω
∴ 𝑍𝐿𝑚𝑎𝑥 = 28.8 − 𝑗38.4Ω
b. 𝑉𝑡ℎ = 5(28.8 + 𝑗38.4) = 144 + 𝑗192𝑉
144 + 𝑗192
∴ 𝐼𝐿 =
2 × 28.8
1
(1442 + 1922 )
2
𝑎𝑛𝑑 𝑃𝐿,𝑚𝑎𝑥 = × 28.8 = 250𝑊
4 × 28.82

36
SOLUTION:

37

38
SOLUTION:

39
SOLUTION:
First we obtain the Thevenin equivalent at the load terminals. To get𝒁𝑻𝒉 consider the circuit shown in
figure (a) below. We find
𝒁𝑻𝒉 = 𝑗5 + 4||(8 − 𝑗6) = 2.933 + 𝑗4.467 Ω
To find 𝑽𝑻𝒉 consider the circuit shown in fig (b) above. By voltage division,
8−𝑗6
𝑽𝑻𝒉 = (10) = 7.454∠ − 10.3°
4+8−𝑗6
The load impedance draws the maximum power from the circuit when
𝒁𝑳 = 𝒁∗𝑻𝒉 = 2.933 − 𝑗4.467 Ω
The maximum average power is
|𝑽𝑻𝒉 |2
𝑃𝑚𝑎𝑥 = = 2.368 𝑉
8𝑅𝑇ℎ

40
SOLUTION:
We first find the Thevenin equivalent at the terminals of RL.
𝒁 𝑇ℎ = (40 − 𝑗30)||𝑗20 = 9.412 + 𝑗22.35 Ω
By Voltage division,
𝑗20
𝑽 𝑇ℎ = (150∠30°) = 72.76∠134° 𝑉
𝑗20+40−𝑗30
The value of RL that will absorb the maximum average power is
𝑅𝐿 = |𝒁 𝑇ℎ | = √9.4122 + 22.352 = 24.25 Ω
The current through the load is

𝑽𝑇ℎ 72.76∠134°
𝑰= = = 1.8∠100.42°
𝒁𝑇ℎ +𝑅𝐿 33.66+𝑗22.35
The maximum average power absorbed by RL is

1 1
𝑃𝑚𝑎𝑥 = |𝑰|2 𝑅𝐿 = (1.8)2 (24.25) = 39.29 𝑊
2 2

41
SOLUTION:
(See Next Page)

42

43
SOLUTION:

44

45
SOLUTION:

46

47

48
SOLUTION:

49

50
SOLUTION:

51
SOLUTION:

52

53
SOLUTION:

54
SOLUTION:
1 𝑡 144 144
a. √ ∫0 (1 + 𝑐𝑜𝑠2000𝑡)𝑑𝑡 = √ = 8.485
𝑇 2 2
1 𝑡 144 144
b. √ ∫0 (1 − 𝑐𝑜𝑠2000𝑡)𝑑𝑡 = √ = 8.485
𝑇 2 2
1 𝑡 144 144
c. √ ∫0 (1 + 𝑐𝑜𝑠1000𝑡)𝑑𝑡 = √ = 8.485
𝑇 2 2
1 𝑡 144 144
d. √ ∫0 (1 + 𝑐𝑜𝑠1000𝑡 − 176𝑜 )𝑑𝑡 = √ = 8.485
𝑇 2 2

55
SOLUTION:

56
SOLUTION:
The period of the waveform is T = 4 Over a period, we can write thecurrent waveform as
5𝑡, 0<𝑡<2
𝑖(𝑡) = {
−10, 2 < 𝑡 < 4
The rms value is
1 𝑇 1 2 4
𝐼𝑟𝑚𝑠 = √ ∫0 𝑖 2 𝑑𝑡 = √ [∫0 (5𝑡)2 𝑑𝑡 + ∫2 (−10)2 𝑑𝑡] = 8.165 𝐴
𝑇 4
The power absorbed by a 2-Ω resistor is

2
𝑃 = 𝐼𝑟𝑚𝑠 𝑅 = (8.165)2 (2) = 133.3 𝑊

57
SOLUTION:

58

59
SOLUTION:
The period of the voltage waveform is 𝑇 = 2𝜋 and
10 sin 𝑡 , 0 < 𝑡 < 𝜋

𝑣(𝑡) = {
0, 𝜋 < 𝑡 < 2𝜋
The rms value is obtained as
2 1 𝑇 1 𝜋 2𝜋
𝑉𝑟𝑚𝑠 = ∫0 𝑣 2 (𝑡) = [∫0 (10𝑠𝑖𝑛𝑡)2 𝑑𝑡 + ∫𝜋 02 𝑑𝑡
𝑇 2𝜋
50 1
= (𝜋 − sin 2𝜋 − 0) = 25,
2𝜋 2
𝑉𝑟𝑚𝑠 = 5 𝑉
The average power absorbed is

2
𝑉𝑟𝑚𝑠 52
𝑃= = = 2.5 𝑊
𝑅 10

60
SOLUTION:

61
SOLUTION:

62
SOLUTION:

63
SOLUTION:

64
SOLUTION:

65

66
SOLUTION:

67
SOLUTION:

68
SOLUTION:

69
SOLUTION:

70
SOLUTION:
2 602
(a) 𝑉𝑟𝑚𝑠 = 202 + = 2200, => 𝑉𝑟𝑚𝑠 = 49.6 𝑉
2
0.52
𝐼𝑟𝑚𝑠 = √12 + = 1.061 𝐴
2
(b) 𝑝(𝑡) = 𝑣(𝑡)𝑖(𝑡) = 20 + 60 cos 100𝑡 − 10 sin 100𝑡 − 30(sin 100𝑡)(cos 100𝑡);
Clearly the average power = 20W

71
SOLUTION:

72
SOLUTION:

73
SOLUTION:
The power factor is
𝑝𝑓 = cos(𝜃𝑣 − 𝜃𝑖 ) = cos(−20° − 10°) = 0.866
The pf is leading because the current leads the voltage. The load impedance may be obtained as,
𝑽 120∠−20°
𝒁= = = 30∠ − 30° = 25.98 − 𝑗15Ω
𝑰 4∠10°
The load impedance 𝒁 can be modeled by a 25.98Ωresistor in serieswith a capacitor with

1
𝑋𝐶 = −15 = −
𝜔𝐶
Or, 𝐶 = 212.2 𝜇𝐹

74
SOLUTION:

75
SOLUTION:

76
SOLUTION:

77
SOLUTION:

78
SOLUTION:
SS = 1600 + j500VA
1600+𝑗500
a. 𝐼𝑠∗ = = 4 + 𝑗1.25 ∴ 𝐼𝑠 = 4 + 𝑗1.25
400
400
𝐼𝑐 = = 𝑗3.33𝐴 𝑟𝑚𝑠 ∴ 𝐼𝐿 = 𝐼𝑠 − 𝐼𝑐 = 4 − 𝑗1.25 − 𝑗3.33
−𝑗120
∴ 𝐼𝐿 = 4 − 𝑗4.583𝐴 𝑟𝑚𝑠
∴ 𝑆𝐿 = 400(4 + 𝑗4.583) = 1600 + 𝑗1833𝑉𝐴
𝑡𝑎𝑛−1 1833.3
b. 𝑃𝐹𝐿 = cos ( ) = 0.6575𝑙𝑎𝑔
1600
c. 𝑆𝑠 = 1600 + 𝑗500 = 1676∠17.35 𝑉𝐴
∴ 𝑃𝐹𝑠 = 𝑐𝑜𝑠17.35 = 0.9535𝑙𝑎𝑔

79
SOLUTION:

80
SOLUTION:

81
SOLUTION:
𝜃1 = 𝑐𝑜𝑠 −1 (0.92) = 23.07𝑜 , 𝜃2 = 𝑐𝑜𝑠 −1 (0.8) = 36.87𝑜 , 𝜃3 = 0

100∠23.07
𝑆1 = = 100 + 𝑗42.59𝑉𝐴
0.92
250∠36.87
𝑆1 = = 250 + 𝑗187.5𝑉𝐴\
0.8
500∠0
𝑆1 = = 500𝑉𝐴
1
𝑆𝑡𝑜𝑡𝑎𝑙 = 𝑆1 + 𝑆2 + 𝑆3 = 500 + 𝑗230.1𝑉𝐴 = 550∠24.71𝑉𝐴

𝑆𝑡𝑜𝑡𝑎𝑙 550.4
a. 𝐼𝑒𝑓𝑓 = = = 4.786 𝑟𝑚𝑠
𝑉𝑒𝑓𝑓 115
b. PF of composite load = cos(24.71) = 0.9084 lagging

82
SOLUTION:

83

84
SOLUTION:

85

86
SOLUTION:
(See Next Page)

87

88
SOLUTION:
𝐼 = 4∠35𝑜 𝐴
a. 𝑉 = 20𝐼 + 80∠35𝑜 , 𝑉𝑟𝑚𝑠 , 𝑃𝑠,𝑔𝑒𝑛 = 80 × 10𝑐𝑜𝑠35 = 655.3𝑊
b. PR = I2R = 16 x 20 =320W
c. Pload = 655.3-320 = 335.3W
d. APs,gen = 80x10 = 800VA
e. APR = PR = 320VA
f. let 𝐼 = 10∠0𝑜 − 4∠35𝑜 𝐴 = 7.104∠ − 18.84𝐴 𝑟𝑚𝑠
∴ 𝐴𝑃𝐿 = 80 × 7.104 = 568.3𝑉𝐴
𝑃𝐿 335.3
g. 𝑃𝐹𝐿 = 𝑐𝑜𝑠𝜃𝐿 = = = 0.599
𝐴𝑃𝐿 568.3
Since IL lags V, PFL is lagging.

89
SOLUTION:
The total impedance is
−𝑗2×4
𝒁 = 6 + 4||(−𝑗2) = 6 + = 6.18 − 𝑗1.6 = 7∠ − 13.24°
4−𝑗2
The power factor is
𝑝𝑓 = cos(−13.24) = 0.9734
Since the impedance is capacitive, the rms value of the current is

𝑽𝑟𝑚𝑠 30∠0°
𝑰𝑟𝑚𝑠 = = = 4.286∠13.24°
𝒁 7∠−13.24°
The average power supplied by the source is
𝑃 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 𝑝𝑓 = (30)(4.286)0.9734 = 125 𝑊

90
SOLUTION:
6×(−𝑗2)
−𝑗2||6 = = 0.6 − 𝑗1.8
6−𝑗2
3 + 𝑗4 + (−𝑗2)||6 = 3.6 + 𝑗2.2
The circuit is reduced to that shown below,
3.6+𝑗2.2
𝑰𝟎 = = 2(∠300 ) = 0.95∠47.080
8.6+𝑗2.2
𝑽𝟎 = 5𝐼0 = 4.75∠47.080
1 1
𝑺 = 𝑽𝟎 𝑰∗𝑺 = (4.75∠47.080 )(2∠ − 300 )
2 2
𝑺 = 4.75∠17.080 = 4.543 + 𝑗1.396 𝑉𝐴

91
SOLUTION:

92
SOLUTION:

93

94
SOLUTION:

95

96
SOLUTION:
120
a. 𝐼𝑠 = 𝑗192 = 9.214∠ − 26.25 𝐴 𝑟𝑚𝑠
4+12+𝑗16
Thus, PFs = cos26.25 = 0.8969lag

b. Ps = 120 x 9.214 x 0.8969 = 991.7W
𝑗48 1
c. 𝑍𝐿 = 4 + =4+ (192 + 𝑗144)
3+𝑗4 25
11.68 − 𝑗5.76
∴ 𝑍𝐿 = 11.68 + 𝑗5.76Ω, 𝑌𝐿 =
11.682 + 5.762
𝑗5.76
∴ 𝑗120𝜋𝐶 = , 𝐶 = 90.09µ𝐹
11.682 + 5.762

97
SOLUTION:
The instantaneous power is given by
𝑝(𝑡) = 𝑣𝑖 = 1200 cos(377𝑡 + 45°) cos(377𝑡 − 10°)
= 600 cos(754𝑡 + 35°) + cos 55° = 344.2 + 600 cos 754𝑡 + 35° 𝑊
The average power is

1 1
𝑃 = 𝑉𝑚 𝐼𝑚 cos(𝜃𝑣 − 𝜃𝑖 ) = 120(10) cos[45° − (−10°)] = 344.2 𝑊
2 2
Which is the constant part of 𝑝(𝑡)above

98
SOLUTION:

99

100
SOLUTION:

101

102
SOLUTION:

103

104
SOLUTION:

105

106
SOLUTION:

107
SOLUTION:

108

109
SOLUTION:

110
SOLUTION:
𝑍1 = 30∠15Ω, 𝑍1 = 40∠40Ω
a. 𝑍𝑡𝑜𝑡𝑎𝑙 = 40∠40 + 30∠15 = 68.37∠29.31

∴ 𝑃𝐹 = 𝑐𝑜𝑠29.3 = 0.8719𝑙𝑎𝑔
b. 𝑉 = 𝐼𝑍𝑡𝑜𝑡 = 683.8∠29.31Ω so
𝑆 = 𝑉𝐼 ∗ = (683.8∠29.31)(10∠0) = 6838∠28.31𝑉𝐴
Thus the apparent power = S = 6.838kVA
c. The impedance has a positive angle, it therefore has a net inductive character.

111
SOLUTION:
a. 500VA, PF = 0.75 lead

∴ 𝑆 = 500∠ − 𝑐𝑜𝑠 −1 0.75 = 375 − 𝑗330.7𝑉𝐴
b. 500W, PF = 0.75 lead :
500
∴ 𝑆 = 500 − sin(𝑐𝑜𝑠 −1 0.75) = 500 − 𝑗441𝑉𝐴
𝑗0.75
c. -500VAR, PF = 0.75(lead)
∴ 𝜃 = −𝑐𝑜𝑠 −1 0.75 = −41.41
𝑃500
∴ = 566.9𝑊
𝑡𝑎𝑛41.41
𝑆 = 566.9 − 𝑗500𝑉𝐴

112
SOLUTION:
cos-10.8 = 36.87, cos-10.9 = 25.84

a. 𝑆𝑡𝑜𝑡𝑎𝑙 = 1200∠36.87 + 1600∠25.84 + 900
= 960 + 𝑗720 + 1440 + 𝑗697.4 + 900
= 3300 + 𝑗1417.4 = 3592∠23.25𝑉𝐴
3591.5
∴ 𝐼𝑠 = = 15.62𝐴 𝑟𝑚𝑠
230
b. PFs = cos23.245 = 0.9188
c. S = 3300 + j1417 VA

113
SOLUTION:
V = 339∠ − 66𝑉, 𝜔 = 100𝜋 rad/s connected to Z = 1000Ω

339
a. 𝑉𝑒𝑓𝑓 = = 239.7𝑟𝑚𝑠
√2
3392
b. 𝑃𝑚𝑎𝑥 = = 114.9𝑊
1000
c. 𝑃𝑚𝑖𝑛 = 0
339 339 2
( )( ) 𝑉𝑒𝑓𝑓
√2 √2
d. Apparent power = 𝑉𝑒𝑓𝑓 𝐼𝑒𝑓𝑓 = = = 57.46𝑉𝐴
100 1000
e. Since the load is purely resistive, it draws zero resistive power.
f. S = 57.46VA

114
SOLUTION:
Given that𝑝𝑓 = cos 𝜃 = 0.856 we obtain the power angle as= cos −1 0.856 = 31.13° . If the apparent
power isS = 12000 VA,then the average or real power is
𝑃 = 𝑆𝑐𝑜𝑠𝜃 = 12000 × 0.856 = 10.272 𝑘𝑊
While the reactive power is
𝑄 = 𝑆 sin 𝜃 = 12000 × 0.517 = 6.204 𝑘𝑊
(b) The complex power is,
𝑺 = 𝑃 + 𝑗𝑄 = 10.272 + 𝑗6.204 𝑘𝑉𝐴
From 𝑺 = 𝑽𝑟𝑚𝑠 𝑰∗𝑟𝑚𝑠 , we obtain
10.272+𝑗6.204
𝑰∗𝑟𝑚𝑠 = = 85.6 + 𝑗51.7 𝐴 = 100∠31.13°
120∠0°
Thus, 𝑰𝑟𝑚𝑠 = 100∠ − 31.13°, and the peak current is
𝐼𝑚 = √2𝐼𝑟𝑚𝑠 = √2100 = 141.4 𝐴
(c) The load impedance is

𝑽𝑟𝑚𝑠 120∠0°
𝒁= = = 1.2∠31.13° Ω
𝑰𝑟𝑚𝑠 100∠−31.13°
which is an inductive impedance.

115
SOLUTION:

116
SOLUTION:

117

118
SOLUTION:
APL = 10000VA , PFL = 0.8lag, |IL|=40A rms

Let 𝐼𝐿 = 40∠0 A rms; PL = 10000 x 0.8 = 8000W
Let ZL = RL + jXL
8000
∴ 𝑅𝐿 = = 5Ω
402
𝑐𝑜𝑠𝜃𝐿 = 0.8 𝑙𝑎𝑔 ∴ 𝜃𝐿 = 𝑐𝑜𝑠 −1 0.8 = 36.87
∴ 𝑋𝐿 = 5𝑡𝑎𝑛36.87 = 3.75Ω, 𝑍𝐿 = 5 + 𝑗3.75, 𝑍𝑡𝑜𝑡 = 5.2 + 𝑗3.75Ω
1
∴ 𝑉𝑆 = 40(5.2 + 𝑗3.75) = 256.4∠35.80𝑉, 𝑌𝑡𝑜𝑡 =
5.2 + 𝑗3.75
= 0.12651 − 𝑗0.09124𝑆, 𝑌𝑛𝑒𝑤 = 0.12651 + 𝑗(120𝜋𝐶 − 0.09124)
𝑃𝐹𝑛𝑒𝑤 = 0.9𝑙𝑎𝑔, 𝜃𝑛𝑒𝑤 = 25.84 ∴ 𝑡𝑎𝑛25.84 = 0.4853
0.099124 − 120𝜋𝐶
=
0.12651
∴ 𝐶 = 79.48µ𝐹

119
SOLUTION:

120
SOLUTION:

121
SOLUTION:

122
SOLUTION:

123
SOLUTION:

124
SOLUTION:

125

126
SOLUTION:

127

128
SOLUTION:

129
SOLUTION:
(See Next Page)

130

131
SOLUTION:

132
SOLUTION:
(See Next Page)

133

134
SOLUTION:

135

136
SOLUTION:

137

138
SOLUTION:
The demand charge is
$5.00 X 1,600 = $8,000 (i)
The energy charge for the first 50,000 kWh is
$0.08 X 50,000 = $4,000 (ii)
The remaining energy is 200,000 kWh – 50,000 kWh =150,000 kWh, and the corresponding energy
charge is
$0.05 X 150,000 = $7,500 (iii)
Adding the results of Eqs. (i) to (iii) gives
Total bill for the month = $8,000 + $4,000 + $7,500 = $19,500

139
SOLUTION:
The energy consumed is
W = 300 kW X 520 h = 156,000 kWh
The operating power factor pf = 80%= 0.8 is 5 X 0.01 below theprescribed power factor of 0.85. Since
there is 0.1 percent energy chargefor every 0.01, there is a power-factor penalty charge of 0.5
percent.This amounts to an energy charge of
∆W = (156,000 X 5 X 0.01)/100 = 780 kWH
The total energy is
Wt = W + ∆W = 156,000 + 780 = 156,780 kWh
The cost per month is given by
Cost = 6 cents X Wt= $0.06 X 156,780 = $9,406.80

CH 09

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1

Irwin, Engineering Circuit Analysis, 11e ISV

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Vc(0) = -2V, i(0) = 4A

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

a. P = 276 x 130 = 358.8mW

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

The current through the impedance is

The average power is

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

The current Iis given by

The average power supplied by the voltage source is

The current through the resistor is

and the voltage across it is

The average power absorbed by the resistor is

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

Chapter 09: Steady-State Power Analysis

𝒁𝑻𝒉 = 𝑗5 + 4||(8 − 𝑗6) = 2.933 + 𝑗4.467 Ω

𝒁𝑳 = 𝒁∗𝑻𝒉 = 2.933 − 𝑗4.467 Ω

The maximum average power is

Chapter 09: Steady-State Power Analysis

We first find the Thevenin equivalent at the terminals of RL.

𝒁 𝑇ℎ = (40 − 𝑗30)||𝑗20 = 9.412 + 𝑗22.35 Ω

The value of RL that will absorb the maximum average power is

𝑅𝐿 = |𝒁 𝑇ℎ | = √9.4122 + 22.352 = 24.25 Ω

The current through the load is