Full-load efficiency of 500 kVA transformer

Efficiency $\eta = \frac{\text{output}}{\text{output} + P_i + P_{cu}}$; copper loss varies with the square of load.

Step 1: Losses at half load (upf)

$$\begin{aligned} \text{Output}_{1/2} &= 0.5 \times 500 \times 1 = 250\ \text{kW} \\ \text{Input}_{1/2} &= \frac{250}{0.90} = 277.78\ \text{kW} \\ \text{Total loss} &= 277.78 - 250 = 27.78\ \text{kW} \end{aligned}$$

Step 2: Copper loss at half and full load

$$\begin{aligned} P_{cu,1/2} &= 27778 - 2000 = 25778\ \text{W} \\ P_{cu,FL} &= \frac{P_{cu,1/2}}{(0.5)^2} = 4 \times 25778 = 103111\ \text{W} = 103.11\ \text{kW} \end{aligned}$$

Step 3: Efficiency at full load (upf)

$$\begin{aligned} \text{Output}_{FL} &= 500\ \text{kW} \\ \text{Losses} &= 2 + 103.11 = 105.11\ \text{kW} \\ \eta_{FL} &= \frac{500}{500 + 105.11}\times 100 = 82.63\% \end{aligned}$$

Answer: Full-load efficiency at unity pf = 82.63%.

(The efficiency falls at full load because copper loss, which grows as the square of load, is much larger than iron loss here. The data are as given in the question; a real 500 kVA transformer would have far higher efficiency.)