Answer to Question 6

 

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The transformer in the worked example is operating at half power.  The power factor remains the same.  What is the efficiency now?

Useful power = 150 kVA 0.85 = 127.5 kW

 

Iron loss remains at 2.0 kW

Copper loss = 0.25 2.5 kW = 0.625 kW

Total loss = 2.0 kW + 0.625 kW = 2.625 kW

 

Efficiency = 1 - (2.625 kW 127.5 kW) = 0.979 = 97.9 %