Primary current and efficiency of 11 kV/380 V Δ/Y transformer

Primary is delta (phase voltage = 11 kV, phase current = line current/$\sqrt3$); secondary is star (phase current = line current). Winding resistances are per phase.

Load apparent power:

$$S = \frac{40}{0.8} = 50\ \text{kVA}$$

Primary line current

Neglecting no-load current, input VA = output VA:

$$\begin{aligned} I_{1L} &= \frac{S}{\sqrt3\,V_{1L}} = \frac{50000}{\sqrt3 \times 11000} = 2.624\ \text{A} \\ I_{1ph} &= \frac{I_{1L}}{\sqrt3} = 1.515\ \text{A} \end{aligned}$$

Secondary current

$$I_{2L} = I_{2ph} = \frac{50000}{\sqrt3 \times 380} = 75.97\ \text{A}$$

Check by turns ratio: $K = \frac{380/\sqrt3}{11000} = 0.01995$, $I_{1ph} = 0.01995 \times 75.97 = 1.515$ A.

Copper losses

$$\begin{aligned} P_{cu1} &= 3I_{1ph}^2R_1 = 3 \times 1.515^2 \times 25 = 172.18\ \text{W} \\ P_{cu2} &= 3I_{2ph}^2R_2 = 3 \times 75.97^2 \times 0.01 = 173.13\ \text{W} \\ P_{cu} &= 172.18 + 173.13 = 345.31\ \text{W} \end{aligned}$$

(Leakage reactances cause no power loss.)

Efficiency

$$\begin{aligned} \text{Losses} &= P_i + P_{cu} = 1000 + 345.31 = 1345.31\ \text{W} \\ \eta &= \frac{40000}{40000 + 1345.31}\times 100 = 96.75\% \end{aligned}$$

Answer: Primary line current = 2.624 A; approximate efficiency = 96.75%.