Discussion
If Pi be the iron loss and Pcu be the copper loss on full load. The following condition has to be satisfied to obtain maximum efficiency at \(\frac{3}{4}\) full load.
Answer: C
The condition is \(P_{cu} = \frac{16}{9} P_{i}\).
A transformer is most efficient at the load where the variable copper loss equals the constant iron loss. Copper loss varies as the square of the load fraction x, so at a fraction x of full load it is \(x^{2}P_{cu}\).
\[ \begin{aligned} x^{2} P_{cu} &= P_{i} \\ \left(\frac{3}{4}\right)^{2} P_{cu} &= P_{i} \\ P_{cu} &= \frac{16}{9} P_{i} \end{aligned} \]
9/16 is the same relation turned upside down, and 3/4 and 4/3 forget that copper loss goes with the square of the load, not the load itself.
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