The one-line diagram of a power system is shown in the figure below. A single-line-to- ground (SLG) fault occurs at bus 2. Assume the fault impedance Zj=j0.08 pu. Calculate the fault current (in per unit). 2 M Line X, = X2 = 40 N 1 Xo = 1200 100 MVA 13.8 kV 100 MVA 100 MVA X, = 0.15 X2 = 0.17 X, = 0.05 per unit 138-kV Y/13.8-kV A X = 0.10 per unit 13.8-kVA/138-kVY 100 MVA X = 0.10 per unit 13.8 kV X, = 0.20 X2 = 0.21 0 10

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The one-line diagram of a power system is shown in the figure below. A single-line-to-
ground (SLG) fault occurs at bus 2. Assume the fault impedance Z,= j0.08 pu. Calculate
the fault current (in per unit).
T,
2
M
Line
X, = X2 = 40 N
Xo =1200
%3D
100 MVA
13.8 kV
100 MVA
100 MVA
X, = 0.15 13.8-kVA/138-kVY
X = 0.10 per unit
138-kV Y/13.8-kV A
100 MVA
X2 = 0.17
X = 0.10 per unit
13.8 kV
X, = 0.20
X2 = 0.21
Xo = 0.10
X, = 0.05 per unit
Xo = 0.05 per unit
Transcribed Image Text:The one-line diagram of a power system is shown in the figure below. A single-line-to- ground (SLG) fault occurs at bus 2. Assume the fault impedance Z,= j0.08 pu. Calculate the fault current (in per unit). T, 2 M Line X, = X2 = 40 N Xo =1200 %3D 100 MVA 13.8 kV 100 MVA 100 MVA X, = 0.15 13.8-kVA/138-kVY X = 0.10 per unit 138-kV Y/13.8-kV A 100 MVA X2 = 0.17 X = 0.10 per unit 13.8 kV X, = 0.20 X2 = 0.21 Xo = 0.10 X, = 0.05 per unit Xo = 0.05 per unit
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