Given the circuit below and the associated set of voltages and currents, the real Power developed by the dependent voltage source is V3 1Ω j2 2 + 1Ω j3Ω I 12 Ωξ V, 39 I, -j16 N = 150 /0° V V1 = (78 – j104) V V = %3D I = (-26 – j52) A I, = (-2 + j6) A %3D - V2 = (72 + j104) V %3D V3 = (150 – j130) V I = (-24 – j58) A %3D Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall (+ I
Given the circuit below and the associated set of voltages and currents, the real Power developed by the dependent voltage source is V3 1Ω j2 2 + 1Ω j3Ω I 12 Ωξ V, 39 I, -j16 N = 150 /0° V V1 = (78 – j104) V V = %3D I = (-26 – j52) A I, = (-2 + j6) A %3D - V2 = (72 + j104) V %3D V3 = (150 – j130) V I = (-24 – j58) A %3D Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall (+ I
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
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.2P: Convert the following instantaneous currents to phasors, using cos(t) as the reference. Give your...
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Please solve this question quickly in electric
![Given the circuit below and the associated set of voltages and currents, the real Power
developed by the dependent voltage source is
V3
1Ω
j2 N +
1Ω
j3Ω
I
12 Ωξ
Vs
39 I,
-j16 NF
Vs
= 150 /0° V
I = (-26 – j52) A
I, = (-2 + j6) A
V1 = (78 – j104) V
V2 = (72 + j104) V
V3 = (150 – j130) V
I = (-24 – j58) A
%3D
Copyright © 2011 Pearson Education, Inc, publishing as Prentice Hall
Select one:
a. None of these
b. 2500 W
c. 5850 W
d. 8000 W](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88a3afeb-9633-4d29-8cad-20e68ae2a130%2F638a93e1-df12-4f46-a785-b34abbcb1d81%2Fz5utca_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given the circuit below and the associated set of voltages and currents, the real Power
developed by the dependent voltage source is
V3
1Ω
j2 N +
1Ω
j3Ω
I
12 Ωξ
Vs
39 I,
-j16 NF
Vs
= 150 /0° V
I = (-26 – j52) A
I, = (-2 + j6) A
V1 = (78 – j104) V
V2 = (72 + j104) V
V3 = (150 – j130) V
I = (-24 – j58) A
%3D
Copyright © 2011 Pearson Education, Inc, publishing as Prentice Hall
Select one:
a. None of these
b. 2500 W
c. 5850 W
d. 8000 W
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