INSTRUCTOR'S
SOLUTIONS MANUAL
Power System Analysis and Design 7th Edition
By J. Duncan Glover, Thomas Overbye
TU
TO
R
G
U
R
U
, Table of Content
Chapter 1. Introduction
Chapter 2. Fundamentals
Chapter 3. Power Transformers
Chapter 4. Transmission Line Parameters
Chapter 5. Transmission Lines: Steady-State Operation
Chapter 6. Power Flows
TU
Chapter 7. Symmetrical Faults
Chapter 8. Symmetrical Components
Chapter 9. Unsymmetrical Faults
Chapter 10. System Protection
TO
Chapter 11. Transient Stability
Chapter 12. Power System Controls
Chapter 13. Transmission Lines: Transient Operation
R
Chapter 14. Power Distribution
G
U
R
U
,Chapter 2
Fundamentals
TU
ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS
2.1 b 2.19 a
2.2 a 2.20 A. c
2.3 c B. a
TO
2.4 a C. b
2.5 b 2.21 a
2.6 c 2.22 a
2.7 a 2.23 b
2.8 c 2.24 a
2.9 a 2.25 a
R
2.10 c 2.26 b
2.11 a 2.27 a
2.12 b 2.28 b
G
2.13 b 2.29 a
2.14 c 2.30 (i) c
2.15 a (ii) b
U
2.16 b (iii) a
2.17 A. a (iv) d
B. b 2.31 a
R
C. a 2.32 a
2.18 c
U
1
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 2.1 (a) A1 = 5∠30° = 5 [ cos30° + j sin 30°] = 4.33 + j 2.5
4
(b) A2 = −3 + j 4 = 9 + 16 ∠ tan −1 = 5 ∠126.87° = 5e j126.87°
−3
(c) A3 = ( 4.33 + j 2.5 ) + ( −3 + j 4 ) = 1.33 + j 6.5 = 6.635∠78.44°
(d) A4 = ( 5∠30° )( 5 ∠126.87° ) = 25 ∠156.87° = −22.99 + j 9.821
(e) A5 = ( 5∠30° ) / ( 5∠ − 126.87° ) = 1∠156.87° = 1 e j156.87°
(a) I = 400∠ − 30° = 346.4 − j 200
TU
2.2
(b) i(t ) = 5sin (ω t + 15° ) = 5cos (ω t + 15° − 90° ) = 5cos (ω t − 75° )
( 2 ) ∠ − 75° = 3.536∠ − 75° = 0.9151− j3.415
I = 5
(c) I = ( 4 2 ) ∠ − 30° + 5∠ − 75° = ( 2.449 − j1.414 ) + (1.294 − j 4.83 )
= 3.743 − j 6.244 = 7.28∠ − 59.06°
TO
2.3 (a) Vmax = 359.3V; I max = 100 A
(b) V = 359.3 2 = 254.1V; I = 100 2 = 70.71A
(c) V = 254.1∠15° V; I = 70.71 ∠ − 85° A
R
− j6 6∠ − 90°
2.4 (a) I1 = 10∠0° = 10 = 7.5∠ − 90° A
8 + j6 − j6 8
I 2 = I − I1 = 10∠0° − 7.3∠ − 90° = 10 + j 7.5 = 12.5∠36.87° A
G
V = I 2 ( − j 6 ) = (12.5∠36.87° ) ( 6∠ − 90° ) = 75∠ − 53.13° V
(b)
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2.5 (a) υ (t ) = 277 2 cos (ω t + 30° ) = 391.7cos (ω t + 30° ) V
(b) I = V / 20 = 13.85∠30° A
i(t ) = 19.58cos (ω t + 30° ) A
2
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SOLUTIONS MANUAL
Power System Analysis and Design 7th Edition
By J. Duncan Glover, Thomas Overbye
TU
TO
R
G
U
R
U
, Table of Content
Chapter 1. Introduction
Chapter 2. Fundamentals
Chapter 3. Power Transformers
Chapter 4. Transmission Line Parameters
Chapter 5. Transmission Lines: Steady-State Operation
Chapter 6. Power Flows
TU
Chapter 7. Symmetrical Faults
Chapter 8. Symmetrical Components
Chapter 9. Unsymmetrical Faults
Chapter 10. System Protection
TO
Chapter 11. Transient Stability
Chapter 12. Power System Controls
Chapter 13. Transmission Lines: Transient Operation
R
Chapter 14. Power Distribution
G
U
R
U
,Chapter 2
Fundamentals
TU
ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS
2.1 b 2.19 a
2.2 a 2.20 A. c
2.3 c B. a
TO
2.4 a C. b
2.5 b 2.21 a
2.6 c 2.22 a
2.7 a 2.23 b
2.8 c 2.24 a
2.9 a 2.25 a
R
2.10 c 2.26 b
2.11 a 2.27 a
2.12 b 2.28 b
G
2.13 b 2.29 a
2.14 c 2.30 (i) c
2.15 a (ii) b
U
2.16 b (iii) a
2.17 A. a (iv) d
B. b 2.31 a
R
C. a 2.32 a
2.18 c
U
1
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 2.1 (a) A1 = 5∠30° = 5 [ cos30° + j sin 30°] = 4.33 + j 2.5
4
(b) A2 = −3 + j 4 = 9 + 16 ∠ tan −1 = 5 ∠126.87° = 5e j126.87°
−3
(c) A3 = ( 4.33 + j 2.5 ) + ( −3 + j 4 ) = 1.33 + j 6.5 = 6.635∠78.44°
(d) A4 = ( 5∠30° )( 5 ∠126.87° ) = 25 ∠156.87° = −22.99 + j 9.821
(e) A5 = ( 5∠30° ) / ( 5∠ − 126.87° ) = 1∠156.87° = 1 e j156.87°
(a) I = 400∠ − 30° = 346.4 − j 200
TU
2.2
(b) i(t ) = 5sin (ω t + 15° ) = 5cos (ω t + 15° − 90° ) = 5cos (ω t − 75° )
( 2 ) ∠ − 75° = 3.536∠ − 75° = 0.9151− j3.415
I = 5
(c) I = ( 4 2 ) ∠ − 30° + 5∠ − 75° = ( 2.449 − j1.414 ) + (1.294 − j 4.83 )
= 3.743 − j 6.244 = 7.28∠ − 59.06°
TO
2.3 (a) Vmax = 359.3V; I max = 100 A
(b) V = 359.3 2 = 254.1V; I = 100 2 = 70.71A
(c) V = 254.1∠15° V; I = 70.71 ∠ − 85° A
R
− j6 6∠ − 90°
2.4 (a) I1 = 10∠0° = 10 = 7.5∠ − 90° A
8 + j6 − j6 8
I 2 = I − I1 = 10∠0° − 7.3∠ − 90° = 10 + j 7.5 = 12.5∠36.87° A
G
V = I 2 ( − j 6 ) = (12.5∠36.87° ) ( 6∠ − 90° ) = 75∠ − 53.13° V
(b)
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2.5 (a) υ (t ) = 277 2 cos (ω t + 30° ) = 391.7cos (ω t + 30° ) V
(b) I = V / 20 = 13.85∠30° A
i(t ) = 19.58cos (ω t + 30° ) A
2
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
