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In a transformer, if the iron losses and copper losses are 40.5 kW and 50 kW respectively, then at what fraction of load will the efficiency be maximum?
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- 0.8
- 0.57
- 0.70
- 0.9
Correct Option: D
Efficiency is maximum, when
x²Pc = Pi
where, x = fraction of loading
Pc = full load copper loss
Pi = normal voltage iron loss.
It gives x = 0.9.
