Q41GATE 2001MCQ

A 50 Hz alternator is rated 500 MVA, 20 kV, with $X_{d}$ = 1.0 per unit and ${X_{d}}''$ per unit. It supplies a purely resistive load of 400 MW at 20 kV. The load is connected directly across the generator terminals when a symmetrical fault occurs at the load terminals. The initial rms current in the generator in per unit is

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📖 Explanation

\begin{aligned} P_{3-\phi (p.u.)}&=V_{p.u.}I_{p.u.} \cos \theta \\ P_{p.u.}&=\frac{400}{500}=0.8p.u.\\ (\because \; \text{Base MVA }&\text{=500 MVA})\\ P_{actual}&=400MW\\ \text{since, }P_{p.u.}&=1\angle 0^{\circ} p.u.\\ \therefore \; I_{p.u.}&=0.8\\ E_g''&=1\angle 0^{\circ}+j0.2 (0.8\angle 0^{\circ})\\ &=1.01\angle 9.09^{\circ}p.u.\\ I_f''&=\frac{1.01}{0.2}=5.05p.u. \end{aligned}