Efficiency of 5 kVA transformer at 2 kW load

Given (secondary side): $R_{02} = 0.04\ \Omega$, $X_{02} = 0.24\ \Omega$. Maximum efficiency occurs at $I_2 = 40$ A.

Iron loss

At maximum efficiency, copper loss = iron loss:

$$P_i = I_2^2R_{02} = 40^2 \times 0.04 = 64\ \text{W}$$

Load current for 2 kW at 0.85 pf (at 110 V)

$$I_2 = \frac{P}{V_2\cos\phi} = \frac{2000}{110 \times 0.85} = 21.39\ \text{A}$$

(Rated current is $5000/110 = 45.45$ A, so this is 47% load.)

Copper loss at this load

$$P_{cu} = I_2^2R_{02} = 21.39^2 \times 0.04 = 18.30\ \text{W}$$

Efficiency

$$\begin{aligned} \eta &= \frac{2000}{2000 + 64 + 18.30}\times 100 \\ &= \frac{2000}{2082.30}\times 100 = 96.05\% \end{aligned}$$

The reactance $X_{02}$ is not needed because it causes no power loss.

Answer: Efficiency at 2 kW, 0.85 pf lagging = 96.05%.