12 ΩξV,39 I,-j16 N=V = 150 /0° V%3DV = (78 – j104) VV, = (72 + j104) VV, = (150 – j130) VI, = (-26 – j52) AI, = (-2 + j6) A1 = (-24 – j58) ACopyright 6 2011 Pearson Education, Inc publishing as Prentice HallSelect one:a. 1960 Wb. -1690 WC. 1690 Wd. None of these Given the circuit below and the associated set ofvoltages and currents, the real Power dissipatedin the 1 Ohms resistor of the left side mesh is2Ω10j3 N12 0V39 I,-j16 N=V, = 150/0° VV = (78 – j104) VV2 = (72 + j104) VI, = (-26 – j52) AI, = (-2 + j6) A1 = (-24 – j58) AV3 = (150 – j130) VCopyright 2011 Pearson Education, Inc. publishing as Prentice HallSelect one:a. 1960 W

Answered: Image /qna-images/answer/27b5f24f-ea34-4d69-a5f7-00026a0a9fb2.jpg

V, 39 I, -j16 N V = 150 /0° V V = (78 – j104) V V, = (72 + j104) V I, = (-26 – |52) A I, = (-2 + j6) A I = (-24 – j58) A V, = (150 – j130) V Copyright 2011 Pearson Education, Inc publishing as Prentice Hal Select one: a. 1960 W b. -1690 W C. 1690 W d. None of these

V, 39 I, -j16 N V = 150 /0° V V = (78 – j104) V V, = (72 + j104) V I, = (-26 – |52) A I, = (-2 + j6) A I = (-24 – j58) A V, = (150 – j130) V Copyright 2011 Pearson Education, Inc publishing as Prentice Hal Select one: a. 1960 W b. -1690 W C. 1690 W d. None of these

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 2.17MCQ: Consider the load convention that is used for the RLC elements shown in Figure 2.2 of the text. A....

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Question

12 Ωξ
V,
39 I,
-j16 N=
V = 150 /0° V
%3D
V = (78 – j104) V
V, = (72 + j104) V
V, = (150 – j130) V
I, = (-26 – j52) A
I, = (-2 + j6) A
1 = (-24 – j58) A
Copyright 6 2011 Pearson Education, Inc publishing as Prentice Hall
Select one:
a. 1960 W
b. -1690 W
C. 1690 W
d. None of these

Given the circuit below and the associated set of
voltages and currents, the real Power dissipated
in the 1 Ohms resistor of the left side mesh is
2Ω
10
j3 N
12 0
V
39 I,
-j16 N=
V, = 150/0° V
V = (78 – j104) V
V2 = (72 + j104) V
I, = (-26 – j52) A
I, = (-2 + j6) A
1 = (-24 – j58) A
V3 = (150 – j130) V
Copyright 2011 Pearson Education, Inc. publishing as Prentice Hall
Select one:
a. 1960 W

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