E= 16 x = j0.2 X2 = j0 22 e= j0.15 Line X, = x, = j0.11 Xe = j0 33 - j0.16 X2 = j0.17 Xo j0.06 X, -X2 = j0.10 X =X2 = Xg = 0.10 ele

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Calculate the fault current in each phase for a L-G fault occurs at primary of transformer-2 for the system shown in below fig.All the reactaces are given in p.u on 100 MVA base
E= 16
x = j0.2
2=10 22
- J0.15
Line
- j0.16
Xp = j033
* = 0.17 A/Y
Xo - j0.06
१=x; = j0. 10
Transcribed Image Text:E= 16 x = j0.2 2=10 22 - J0.15 Line - j0.16 Xp = j033 * = 0.17 A/Y Xo - j0.06 १=x; = j0. 10
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