Full-load and half-load efficiency of 100 kVA transformer

$\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu,FL}}$, with $S = 100$ kVA, $P_i = 960$ W (constant), $P_{cu,FL} = 1200$ W.

(i) Full load, unity pf ($x = 1$)

$$\begin{aligned} \text{Output} &= 1 \times 100 \times 1 = 100\ \text{kW} = 100000\ \text{W} \\ \text{Losses} &= 960 + 1^2 \times 1200 = 2160\ \text{W} \\ \eta &= \frac{100000}{100000 + 2160}\times 100 = 97.89\% \end{aligned}$$

(ii) Half load, 0.8 pf ($x = 0.5$)

$$\begin{aligned} \text{Output} &= 0.5 \times 100 \times 0.8 = 40\ \text{kW} = 40000\ \text{W} \\ P_{cu} &= 0.5^2 \times 1200 = 300\ \text{W} \\ \text{Losses} &= 960 + 300 = 1260\ \text{W} \\ \eta &= \frac{40000}{40000 + 1260}\times 100 = 96.95\% \end{aligned}$$

Answer: (i) 97.89%, (ii) 96.95%.

Note: maximum efficiency occurs at $x = \sqrt{960/1200} = 0.894$, i.e. at about 89.4% of full load, where copper loss equals 960 W.