Answer to Question 6

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 %
