Discussion
A transformer with 2 kW of iron loss and 8 kW of copper loss at full load reaches maximum efficiency at
Answer: B
Correct answer: 50 per cent of full load
Maximum efficiency occurs where the variable loss equals the constant loss. With copper loss varying as the square of the loading \( x \), \[ 8x^2 = 2 \quad\Rightarrow\quad x^2 = 0.25 \quad\Rightarrow\quad x = 0.5. \] So this transformer is most efficient at half load, and in general \( x = \sqrt{P_{iron}/P_{copper}} \) - which is why distribution transformers are commonly sized to run well below their nameplate.
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