GATE Papers >> EEE >> 2018 >> Question No 31

Question No. 31 EEE | GATE 2018

The positive, negative and zero sequence impedances of a 125 MVA, three-phase, 15.5 kV, star-grounded, 50 Hz generator are j0.1 pu, j0.05 pu and j0.01 pu respectively on the machine rating base. The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the generator is j0.01 pu, then the magnitude of fault current for a b-phase to ground fault (in kA) is __________ (up to 2 decimal places).


Answer : 73.0 to 74.0


Solution of Question No 31 of GATE 2018 EEE Paper

I(pu) = $\frac3{{\mathrm Z}_1+{\mathrm Z}_2+{\mathrm Z}_0+3{\mathrm Z}_\mathrm n}=\frac3{0.01+0.05+0.01+3\times0.01}=\frac3{0.19}$

If = $\frac3{0.19}\times$ Ibace = $\frac3{0.19}\times\frac{125}{\sqrt3{\displaystyle\times}{\displaystyle15}{\displaystyle.}{\displaystyle5}}$ = 73.5166 kA

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