Correct Answer - Option 3 : 1652.9 W
Concept:
\(P = \sqrt 3 \times VI\)
P = power through an alternator
V = voltage across it
I = current through it.
Calculation:
Star connected alternator
400 kVA, 11 kV
resistance of 5 ohms per phase
\(400 × {10^3} = \sqrt 3 × 11 × {10^3} × {I_L}\)
\({I_L} = \frac{{400}}{{11\sqrt 3 }} = 21\;A\)
For star connection
IL = IP
At half load, I = IL / 2 = 10.5 A
Total short circuit loss = 3 I2R
= 3 × 10.52 × 5
= 1653.7 W