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A 400 kVA, 11 KV star connected alternator has resistance of 5 ohm per phase. At half load, total short circuit load loss will be:
1. 2582.6 W
2. 1365.8 W
3. 1652.9 W
4. None of the abvoe

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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

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