For the circuit shown below, Given that V 10020°V,I = 30245 Arms'then the values of P, and Q3 areandrmsgiven asIs600 VARCapacitive1200 W800 WVs1400 VARInductiveP, =0.6 kWQ3 = 2.3 kVAr capacitiveP, = 0.6 kWQa = 0.7 kVAr inductiveP2 = 0.12 kWQ2 = 2.92 kVAr capacitiveActivate WincowP, = 1 kW

Given circuit shown below Vs=100∠0° VrmsIs=30∠45° Arms

For the circuit shown below, Given that V 100Z0°Vm and I = 30245°Ams then the values of P, and Q3 are %3D given as Is VAR 600 Capacitive 1200 W 1400 VAR 800 W Vs Inductive ○ Pッ %3D0.6kw Q3 = 2.3 kVAr capacitive P, = 0.6 kW Q3 = 0.7 kVAr inductive P = 0.12 kW Q = 2.92 kVAr capacitive

For the circuit shown below, Given that V 100Z0°Vm and I = 30245°Ams then the values of P, and Q3 are %3D given as Is VAR 600 Capacitive 1200 W 1400 VAR 800 W Vs Inductive ○ Pッ %3D0.6kw Q3 = 2.3 kVAr capacitive P, = 0.6 kW Q3 = 0.7 kVAr inductive P = 0.12 kW Q = 2.92 kVAr capacitive

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

For the circuit shown below, Given that V 10020°V,
I = 30245 Arms'
then the values of P, and Q3 are
and
rms
given as
Is
600 VAR
Capacitive
1200 W
800 W
Vs
1400 VAR
Inductive
P, =0.6 kW
Q3 = 2.3 kVAr capacitive
P, = 0.6 kW
Qa = 0.7 kVAr inductive
P2 = 0.12 kW
Q2 = 2.92 kVAr capacitive
Activate Wincow
P, = 1 kW

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