Efficiency of 25 kVA transformer at half load and overload

Transformation ratio $K = \frac{220}{2200} = 0.1$. Power factor is not given, so unity power factor is assumed.

Equivalent resistance and full-load copper loss

$$\begin{aligned} R_{01} &= R_1 + \frac{R_2}{K^2} = 1 + \frac{0.01}{0.01} = 2\ \Omega \\ I_{1,FL} &= \frac{25000}{2200} = 11.364\ \text{A} \\ P_{cu,FL} &= I_{1,FL}^2R_{01} = 11.364^2 \times 2 = 258.26\ \text{W} \end{aligned}$$

Iron loss $P_i = 206$ W (constant). Leakage reactances cause no power loss and are not needed.

(i) Half load ($x = 0.5$)

$$\begin{aligned} \text{Output} &= 0.5 \times 25000 \times 1 = 12500\ \text{W} \\ P_{cu} &= 0.5^2 \times 258.26 = 64.57\ \text{W} \\ \eta &= \frac{12500}{12500 + 206 + 64.57}\times 100 = 97.88\% \end{aligned}$$

(ii) 50% overload ($x = 1.5$)

$$\begin{aligned} \text{Output} &= 1.5 \times 25000 \times 1 = 37500\ \text{W} \\ P_{cu} &= 1.5^2 \times 258.26 = 581.10\ \text{W} \\ \eta &= \frac{37500}{37500 + 206 + 581.10}\times 100 = 97.94\% \end{aligned}$$

Answer (upf): (i) 97.88%, (ii) 97.94%.

Maximum efficiency would occur at $x = \sqrt{206/258.26} = 0.893$ of full load, which is why the two values on either side are close. At another pf (say 0.8), multiply the outputs by 0.8 and recompute.