Given the circuit below and the associated set of voltages and currents, the reactive Power developed by dependent voltage source is 10 j2 N + 10 j3 N I 12 Ωξ 39 I, -j16 N. V, = 150 /0° V V1 = (78 – j104) V %3D I = (-26 – j52) A I, = (-2 + j6) A I, = (-24 – j58) A %3! V2 = (72 + j104) V %3D V3 = (150 – j130) V %3D Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall Select one: a. None of these b. 7050 VAR
Given the circuit below and the associated set of voltages and currents, the reactive Power developed by dependent voltage source is 10 j2 N + 10 j3 N I 12 Ωξ 39 I, -j16 N. V, = 150 /0° V V1 = (78 – j104) V %3D I = (-26 – j52) A I, = (-2 + j6) A I, = (-24 – j58) A %3! V2 = (72 + j104) V %3D V3 = (150 – j130) V %3D Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall Select one: a. None of these b. 7050 VAR
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
Related questions
Question
![N
Given the circuit below and the associated set of
voltages and currents, the reactive Power
developed by dependent voltage source is
1Ω
j2 N +
1Ω
j3 N
12 Ωξ
+
39 I,
-j16 N;
V, = 150 /0° V
V1 = (78 – j104) V
V2 = (72 + j104) V
I = (-26 – j52) A
I, = (-2 + j6) A
V3 = (150 – j130) V
I = (-24 – j58) A
Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall
Select one:
a. None of these
b. 7050 VAR
c. 705 VAR
d. 5070 VAR](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50404f28-931e-4669-82f7-50357b91aea4%2F0f54306e-c4dc-42e2-a44c-2cd0fd7620eb%2F3lu0ia6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:N
Given the circuit below and the associated set of
voltages and currents, the reactive Power
developed by dependent voltage source is
1Ω
j2 N +
1Ω
j3 N
12 Ωξ
+
39 I,
-j16 N;
V, = 150 /0° V
V1 = (78 – j104) V
V2 = (72 + j104) V
I = (-26 – j52) A
I, = (-2 + j6) A
V3 = (150 – j130) V
I = (-24 – j58) A
Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall
Select one:
a. None of these
b. 7050 VAR
c. 705 VAR
d. 5070 VAR
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