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

I_{L} = I_{P}

**At half load, I = I**_{L} / 2 = 10.5 A

Total short circuit loss = 3 I^{2}R

= 3 × 10.5^{2} × 5

= **1653.7 W**