Discussion
A transformer with 400 W of iron loss and 1,600 W of full-load copper loss reaches maximum efficiency at
Answer: D
Correct answer: half of full load
Maximum efficiency occurs where the copper loss equals the iron loss: \[ x^2 P_{cu,fl} = P_i \quad\Rightarrow\quad x = \sqrt{\frac{400}{1600}} = \sqrt{0.25} = 0.5. \] So the peak is at half load. A design like this suits a transformer that spends most of its life lightly loaded, since the peak is placed where it actually operates.
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