1
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
2
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
3
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
4
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
5
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
6
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
7
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
8
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
9
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Vc(0) = -2V, i(0) = 4A
𝛼 =
𝑅
2𝐿
= 2, 𝜔𝑜2 =
1
𝐿𝐶
−3𝑡
= 3, 𝑠1,2 = −2 ± 1 = −1, −3
a. 𝑖 = 𝐴𝑒 −𝑡 + 𝐵𝑒
∴ 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
−𝑡
0
= 𝑒 − 3 − 5𝑒 −3𝑡 + 5 + 2
∴ −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
Chapter 09: Steady-State Power Analysis
10
Irwin, Engineering Circuit Analysis, 11e ISV
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
Chapter 09: Steady-State Power Analysis
11
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
106
= −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
𝑍𝑐 =
Chapter 09: Steady-State Power Analysis
12
Irwin, Engineering Circuit Analysis, 11e ISV
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.
Chapter 09: Steady-State Power Analysis
13
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
14
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
15
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
16
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
17
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
18
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
19
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
40∠30𝑜
= 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
𝑉 = (10 + 𝑗10)
Chapter 09: Steady-State Power Analysis
20
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
21
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
22
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
1
= 3 + 1 + 𝑗3 = 4 + 𝑗3Ω
0.1 − 𝑗0.3
2 + 𝑗5
5√29
𝐼𝑔𝑛𝑜𝑟𝑖𝑛𝑔 30𝑜 𝑜𝑛 𝑉𝑆 , 𝐼𝑅 = 5
, |𝐼𝑅 | =
6 + 𝑗8
10
𝑍𝑅 = 3 +
1 5√29
× 3 = 10.875𝑊
2 10)2
𝑜 (2+𝑗5)(4+𝑗3)
a. 𝑃3Ω = (
b. 𝑉𝑆 = 5∠0
∴ 𝑃𝑆,𝑔𝑒𝑛 =
6+𝑗8
= 13.463∠51.94𝑜 𝑉
1
× 13.463 × 5𝑐𝑜𝑠51.94 = 20.75𝑊
2
Chapter 09: Steady-State Power Analysis
23
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
24
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The current through the impedance is
𝑰=
𝑽
𝒁
=
120∠0°
30−𝑗70
= 1.576∠66.8°
The average power is
1
1
2
2
𝑃 = 𝑉𝑚 𝐼𝑚 cos(𝜃𝑣 − 𝜃𝑖 ) = (120)(1.576) cos(0 − 66.8°) = 37.24 𝑊
Chapter 09: Steady-State Power Analysis
25
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
26
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
27
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
28
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The current Iis given by
𝑰=
5∠30°
4−𝑗2
= 1.118∠56.57°
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.
Chapter 09: Steady-State Power Analysis
29
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
30
Irwin, Engineering Circuit Analysis, 11e ISV
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
Chapter 09: Steady-State Power Analysis
31
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
32
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
33
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
34
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
35
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
a. Zth = 80||j60 =
𝑗480 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
Chapter 09: Steady-State Power Analysis
36
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
37
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
38
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
39
Irwin, Engineering Circuit Analysis, 11e ISV
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,
𝑽𝑻𝒉 =
8−𝑗6
4+8−𝑗6
(10) = 7.454∠ − 10.3°
The load impedance draws the maximum power from the circuit when
𝒁𝑳 = 𝒁∗𝑻𝒉 = 2.933 − 𝑗4.467 Ω
The maximum average power is
𝑃𝑚𝑎𝑥 =
|𝑽𝑻𝒉 |2
8𝑅𝑇ℎ
= 2.368 𝑉
Chapter 09: Steady-State Power Analysis
40
Irwin, Engineering Circuit Analysis, 11e ISV
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,
𝑽 𝑇ℎ =
𝑗20
𝑗20+40−𝑗30
(150∠30°) = 72.76∠134° 𝑉
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
𝑰=
𝑽𝑇ℎ
𝒁𝑇ℎ +𝑅𝐿
=
72.76∠134°
33.66+𝑗22.35
= 1.8∠100.42°
The maximum average power absorbed by RL is
1
1
2
2
𝑃𝑚𝑎𝑥 = |𝑰|2 𝑅𝐿 = (1.8)2 (24.25) = 39.29 𝑊
Chapter 09: Steady-State Power Analysis
41
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
Chapter 09: Steady-State Power Analysis
42
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
43
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
44
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
45
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
46
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
47
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
48
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
49
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
50
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
51
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
52
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
53
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
54
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
1
𝑡 144
a. √ ∫0
𝑇
1
𝑡 144
b. √ ∫0
𝑇
1
2
𝑡 144
c. √ ∫0
𝑇
1
2
2
𝑡 144
d. √ ∫0
𝑇
2
144
(1 + 𝑐𝑜𝑠2000𝑡)𝑑𝑡 = √
2
144
(1 − 𝑐𝑜𝑠2000𝑡)𝑑𝑡 = √
2
144
(1 + 𝑐𝑜𝑠1000𝑡)𝑑𝑡 = √
2
= 8.485
= 8.485
= 8.485
(1 + 𝑐𝑜𝑠1000𝑡 − 176𝑜 )𝑑𝑡 = √
Chapter 09: Steady-State Power Analysis
144
2
= 8.485
55
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
56
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The period of the waveform is T = 4 Over a period, we can write thecurrent waveform as
𝑖(𝑡) = {
5𝑡,
0<𝑡<2
−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
Chapter 09: Steady-State Power Analysis
57
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
58
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
59
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The period of the voltage waveform is 𝑇 = 2𝜋 and
𝑣(𝑡) = {
10 sin 𝑡 , 0 < 𝑡 < 𝜋
0,
𝜋 < 𝑡 < 2𝜋
The rms value is obtained as
1
𝑇
50
(𝜋 − sin 2𝜋 − 0) = 25,
1
𝜋
2𝜋
2
𝑉𝑟𝑚𝑠
= ∫0 𝑣 2 (𝑡) = [∫0 (10𝑠𝑖𝑛𝑡)2 𝑑𝑡 + ∫𝜋 02 𝑑𝑡
𝑇
2𝜋
=
2𝜋
1
2
𝑉𝑟𝑚𝑠 = 5 𝑉
The average power absorbed is
𝑃=
2
𝑉𝑟𝑚𝑠
𝑅
=
52
10
= 2.5 𝑊
Chapter 09: Steady-State Power Analysis
60
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
61
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
62
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
63
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
64
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
65
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
66
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
67
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
68
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
69
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
70
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(a)
2
𝑉𝑟𝑚𝑠
= 202 +
602
2
= 2200,
𝐼𝑟𝑚𝑠 = √12 +
(b)
0.52
2
=>
𝑉𝑟𝑚𝑠 = 49.6 𝑉
= 1.061 𝐴
𝑝(𝑡) = 𝑣(𝑡)𝑖(𝑡) = 20 + 60 cos 100𝑡 − 10 sin 100𝑡 − 30(sin 100𝑡)(cos 100𝑡);
Clearly the average power = 20W
Chapter 09: Steady-State Power Analysis
71
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
72
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
73
Irwin, Engineering Circuit Analysis, 11e ISV
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,
𝒁=
𝑽
𝑰
=
120∠−20°
4∠10°
= 30∠ − 30° = 25.98 − 𝑗15Ω
The load impedance 𝒁 can be modeled by a 25.98Ωresistor in serieswith a capacitor with
𝑋𝐶 = −15 = −
Or,
1
𝜔𝐶
𝐶 = 212.2 𝜇𝐹
Chapter 09: Steady-State Power Analysis
74
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
75
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
76
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
77
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
78
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
SS = 1600 + j500VA
a. 𝐼𝑠∗ =
1600+𝑗500
400
= 4 + 𝑗1.25 ∴ 𝐼𝑠 = 4 + 𝑗1.25
400
= 𝑗3.33𝐴 𝑟𝑚𝑠 ∴ 𝐼𝐿 = 𝐼𝑠 − 𝐼𝑐 = 4 − 𝑗1.25 − 𝑗3.33
−𝑗120
∴ 𝐼𝐿 = 4 − 𝑗4.583𝐴 𝑟𝑚𝑠
∴ 𝑆𝐿 = 400(4 + 𝑗4.583) = 1600 + 𝑗1833𝑉𝐴
𝐼𝑐 =
𝑡𝑎𝑛−1 1833.3
b. 𝑃𝐹𝐿 = cos (
1600
) = 0.6575𝑙𝑎𝑔
c. 𝑆𝑠 = 1600 + 𝑗500 = 1676∠17.35 𝑉𝐴
∴ 𝑃𝐹𝑠 = 𝑐𝑜𝑠17.35 = 0.9535𝑙𝑎𝑔
Chapter 09: Steady-State Power Analysis
79
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
80
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
81
Irwin, Engineering Circuit Analysis, 11e ISV
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
a. 𝐼𝑒𝑓𝑓 =
𝑆𝑡𝑜𝑡𝑎𝑙
𝑉𝑒𝑓𝑓
=
550.4
115
= 4.786 𝑟𝑚𝑠
b. PF of composite load = cos(24.71) = 0.9084 lagging
Chapter 09: Steady-State Power Analysis
82
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
83
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
84
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
85
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
86
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
Chapter 09: Steady-State Power Analysis
87
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
88
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
a.
b.
c.
d.
e.
f.
𝐼 = 4∠35𝑜 𝐴
= 80 × 10𝑐𝑜𝑠35 = 655.3𝑊
𝑉 = 20𝐼 + 80∠35𝑜 , 𝑉𝑟𝑚𝑠 , 𝑃𝑠,𝑔𝑒𝑛
PR = I2R = 16 x 20 =320W
Pload = 655.3-320 = 335.3W
APs,gen = 80x10 = 800VA
APR = PR = 320VA
let 𝐼 = 10∠0𝑜 − 4∠35𝑜 𝐴 = 7.104∠ − 18.84𝐴 𝑟𝑚𝑠
∴ 𝐴𝑃𝐿 = 80 × 7.104 = 568.3𝑉𝐴
g. 𝑃𝐹𝐿 = 𝑐𝑜𝑠𝜃𝐿 =
𝑃𝐿
𝐴𝑃𝐿
=
335.3
568.3
= 0.599
Since IL lags V, PFL is lagging.
Chapter 09: Steady-State Power Analysis
89
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The total impedance is
𝒁 = 6 + 4||(−𝑗2) = 6 +
−𝑗2×4
4−𝑗2
= 6.18 − 𝑗1.6 = 7∠ − 13.24°
The power factor is
𝑝𝑓 = cos(−13.24) = 0.9734
Since the impedance is capacitive, the rms value of the current is
𝑰𝑟𝑚𝑠 =
𝑽𝑟𝑚𝑠
𝒁
=
30∠0°
7∠−13.24°
= 4.286∠13.24°
The average power supplied by the source is
𝑃 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 𝑝𝑓 = (30)(4.286)0.9734 = 125 𝑊
Chapter 09: Steady-State Power Analysis
90
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
−𝑗2||6 =
6×(−𝑗2)
6−𝑗2
= 0.6 − 𝑗1.8
3 + 𝑗4 + (−𝑗2)||6 = 3.6 + 𝑗2.2
The circuit is reduced to that shown below,
𝑰𝟎 =
3.6+𝑗2.2
8.6+𝑗2.2
= 2(∠300 ) = 0.95∠47.080
𝑽𝟎 = 5𝐼0 = 4.75∠47.080
1
1
2
2
𝑺 = 𝑽𝟎 𝑰∗𝑺 = (4.75∠47.080 )(2∠ − 300 )
𝑺 = 4.75∠17.080 = 4.543 + 𝑗1.396 𝑉𝐴
Chapter 09: Steady-State Power Analysis
91
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
92
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
93
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
94
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
95
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
96
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
a. 𝐼𝑠 =
120
𝑗192
4+12+𝑗16
= 9.214∠ − 26.25 𝐴 𝑟𝑚𝑠
Thus, PFs = cos26.25 = 0.8969lag
b. Ps = 120 x 9.214 x 0.8969 = 991.7W
c. 𝑍𝐿 = 4 +
𝑗48
3+𝑗4
=4+
1
25
(192 + 𝑗144)
∴ 𝑍𝐿 = 11.68 + 𝑗5.76Ω, 𝑌𝐿 =
∴ 𝑗120𝜋𝐶 =
11.68 − 𝑗5.76
11.682 + 5.762
𝑗5.76
, 𝐶 = 90.09µ𝐹
11.682 + 5.762
Chapter 09: Steady-State Power Analysis
97
Irwin, Engineering Circuit Analysis, 11e ISV
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
2
2
𝑃 = 𝑉𝑚 𝐼𝑚 cos(𝜃𝑣 − 𝜃𝑖 ) = 120(10) cos[45° − (−10°)] = 344.2 𝑊
Which is the constant part of 𝑝(𝑡)above
Chapter 09: Steady-State Power Analysis
98
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
99
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
100
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
101
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
102
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
103
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
104
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
105
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
106
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
107
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
108
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
109
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
110
Irwin, Engineering Circuit Analysis, 11e ISV
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.
Chapter 09: Steady-State Power Analysis
111
Irwin, Engineering Circuit Analysis, 11e ISV
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
Chapter 09: Steady-State Power Analysis
112
Irwin, Engineering Circuit Analysis, 11e ISV
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
Chapter 09: Steady-State Power Analysis
113
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
V = 339∠ − 66𝑉, 𝜔 = 100𝜋 rad/s connected to Z = 1000Ω
a. 𝑉𝑒𝑓𝑓 =
b. 𝑃𝑚𝑎𝑥 =
339
= 239.7𝑟𝑚𝑠
√2
3392
1000
= 114.9𝑊
c. 𝑃𝑚𝑖𝑛 = 0
d. Apparent power = 𝑉𝑒𝑓𝑓 𝐼𝑒𝑓𝑓 =
(
339 339
)( )
√2
√2
100
=
2
𝑉𝑒𝑓𝑓
1000
= 57.46𝑉𝐴
e. Since the load is purely resistive, it draws zero resistive power.
f. S = 57.46VA
Chapter 09: Steady-State Power Analysis
114
Irwin, Engineering Circuit Analysis, 11e ISV
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
𝑰∗𝑟𝑚𝑠 =
10.272+𝑗6.204
120∠0°
= 85.6 + 𝑗51.7 𝐴 = 100∠31.13°
Thus, 𝑰𝑟𝑚𝑠 = 100∠ − 31.13°, and the peak current is
𝐼𝑚 = √2𝐼𝑟𝑚𝑠 = √2100 = 141.4 𝐴
(c) The load impedance is
𝒁=
𝑽𝑟𝑚𝑠
𝑰𝑟𝑚𝑠
=
120∠0°
100∠−31.13°
= 1.2∠31.13° Ω
which is an inductive impedance.
Chapter 09: Steady-State Power Analysis
115
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
116
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
117
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
118
Irwin, Engineering Circuit Analysis, 11e ISV
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
Chapter 09: Steady-State Power Analysis
119
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
120
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
121
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
122
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
123
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
124
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
125
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
126
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
127
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
128
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
129
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
Chapter 09: Steady-State Power Analysis
130
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
131
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
132
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
Chapter 09: Steady-State Power Analysis
133
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
134
Irwin, Engineering Circuit Analysis, 11e ISV
.
SOLUTION:
Chapter 09: Steady-State Power Analysis
135
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
136
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 09: Steady-State Power Analysis
137
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 09: Steady-State Power Analysis
138
Irwin, Engineering Circuit Analysis, 11e ISV
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
Adding the results of Eqs. (i) to (iii) gives
Total bill for the month = $8,000 + $4,000 + $7,500 = $19,500
Chapter 09: Steady-State Power Analysis
(iii)
139
Irwin, Engineering Circuit Analysis, 11e ISV
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
Chapter 09: Steady-State Power Analysis