EEE GATE 2018 Question No 31 Power Systems - Symmetrical and Unsymmetrical Fault Analysis

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

GATE 2018 EEE Power Systems Symmetrical and Unsymmetrical Fault Analysis

Solution of Question No 31 of GATE 2018 EEE Paper

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

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

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

Answer : 73.0 to 74.0

Questions of Symmetrical and Unsymmetrical Fault Analysis

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