I323+RIlok0.JVIR 1:"123IKmillianpsmilliampsmilliSmacross RiR2R2Ik1231R2 =|R3Voltage DropVoltage DropACROSSVoltage Drop acrossPOWER dissipated RiPow En dissipated R₂Power dissipated R3TOTAL PORNER dissipated.-CIDEVITON MEكمل√₁₂milliampsmilliampsmilliampsvoltsVoltsVoltsmilliants

Answered: Image /qna-images/answer/96273662-763e-463b-aed9-235739ecadd9.jpg

Answered: I +1 V₁ 0.JV IR 1 = 182 RI lok R2 IK…

Engineering

Electrical Engineering

I +1 V₁ 0.JV IR 1 = 182 RI lok R2 IK 123 Ik milli amps √₁₂

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

Chapter7: Symmetrical Faults

Section: Chapter Questions

Problem 7.21MCQ

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Question

I
323
+
RI
lok
0.JV
IR 1:
"
123
IK
milli
anps
milliamps
milli
Sm
across Ri
R2
R2
Ik
123
1R2 =
|R3
Voltage Drop
Voltage Drop
ACROSS
Voltage Drop across
POWER dissipated Ri
Pow En dissipated R₂
Power dissipated R3
TOTAL PORNER dissipated.
-CIDEVIT
ON ME
كمل
√₁₂
milliamps
milliamps
milliamps
volts
Volts
Volts
milliants

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