POWER SYSTEM ANALYSIS AND DESIGN
SEVENTH EDITION
J. DUNCAN GLOVER
THOMAS J. OVERBYE
MULUKUTLA S. SARMA
Adam B. Birchfield
, Contents
Chapter 2............................................................................................................................. 1
Chapter 3........................................................................................................................... 24
Chapter 4........................................................................................................................... 63
Chapter 5........................................................................................................................... 85
Chapter 6......................................................................................................................... 125
Chapter 7......................................................................................................................... 160
Chapter 8......................................................................................................................... 166
Chapter 9.........................................................................................................................185
Chapter 10....................................................................................................................... 219
Chapter 11....................................................................................................................... 293
Chapter 12....................................................................................................................... 314
Chapter 13....................................................................................................................... 333
Chapter 14....................................................................................................................... 346
Chapter 15....................................................................................................................... 369
,Chapter 2
Fundamentals
2.1 (a) 𝐴1̅ = 9∠30° = 9[𝑐𝑜𝑠 30° + 𝑗 𝑠𝑖𝑛 30°] = 7.8 + 𝑗4.5
5
(b) 𝐴̄2 = 4 + 𝑗5 = √16 + 25 ∠ 𝑡𝑎𝑛−1 4 = 6.40∠128.66° = 6.40𝑒 𝑗128.66°
(c) 𝐴̅3 = (7.8 + 𝑗4.5) + (−4 + 𝑗5) = (11.8 + 𝑗9.5) = 15.15∠38.8°
(d) ̅̅̅
𝐴4 = (9∠30°)(6.4∠51.34°) = 57.6∠81.34° = 8.673 + 𝑗56.9
9∠30°
(e) ̅̅̅
𝐴5 = = 1.41∠81.34° = 1.41𝑒 𝑗81.34°
6.4∠−52.34°
2.2 (a) 𝐼̄ = 500∠ − 30° = 433.01 − 𝑗250
(b) 𝑖(𝑡) = 4𝑠𝑖𝑛(𝜔𝑡 + 15°) = 4𝑐𝑜𝑠(𝜔𝑡 + 15° − 90°) = 4co s(𝜔𝑡 − 75°)
4
𝐼̅ = ∠ − 75° = 2.83∠ − 75° = 0.73 − 𝑗2.73
√2
(c) 𝐼̄ = (5/√2)∠ − 15° + 4∠ − 60° = (3.42 − 𝑗0.92) + (2 − 𝑗3.46)
= 5.42 − 𝑗4.38 = 6.964∠ − 38.94°
13
2.3 𝑉̅2 = 2 ∠(−(125 × 10−6 )(2𝜋60)) kV = 9.19∠ − 2.7° kV
√
− j6 6 − 90
2.4 (a) I1 = 100 = 10 = 7.5 − 90 A
8 + j6 − j6 8
I 2 = I − I1 = 100 − 7.3 − 90 = 10 + j 7.5 = 12.536.87 A
V = I 2 ( − j 6 ) = (12.536.87 ) ( 6 − 90 ) = 75 − 53.13 V
(b)
1
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, 2.5 (a) (t ) = 277 2 cos ( t + 30) = 391.7cos ( t + 30 ) V
̅
𝑉
(b) 𝐼 ̅ = 45 = 6.155∠30° A
𝑖(𝑡) = 8.7 cos (𝜔𝑡 + 30°) A
(c) Z = j L = j ( 2 60 ) (10 10 −3 ) = 3.77190
I = V Z = ( 277 30 ) ( 3.77190 ) = 73.46 − 60 A
i(t ) = 73.46 2 cos (t − 60 ) =103.9cos (t − 60 ) A
(d) Z = − j 25
I = V Z = ( 27730 ) ( 25 − 90 ) = 11.08120 A
i(t ) = 11.08 2 cos ( t + 120 ) = 15.67cos ( t + 120 ) A
75
2.6 (a) 𝑉̄ = ( 2) ∠ − 15° = 53.03∠ − 15°; does not appear in the answer.
√
(b) 𝜐(𝑡) = 50√2 𝑐𝑜𝑠(𝜔𝑡 + 10°); with = 377,
𝜐(𝑡) = 70.71 𝑐𝑜𝑠(377𝑡 + 10°)
(c) 𝐴̄ = 𝐴∠𝛼; 𝐵̄ = 𝐵∠𝛽; 𝐶̄ = 𝐴̄ + 𝐵̄
c(t ) = a(t ) + b(t ) = 2 Re Ce jt
The resultant has the same frequency .
2.7 (a) The circuit diagram is shown below:
(b) 𝑍̄ ̶ = 3 + 𝑗8 − 𝑗4 = 3 + 𝑗4 = 5∠53.1° 𝛺
(c) I = (1000 ) ( 553.1 ) = 20 − 53.1 A
The current lags the source voltage by 53.1
Power Factor = cos53.1 = 0.6 Lagging
2
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