A transformer has 1500 turns on the primary and 300 turns on the secondary. The primary is connected to a 240 V${}_{\text{rms}}$ AC supply at 50 Hz and a resistive load of 800 Ω is connected across the secondary. Copper loss is 3 % and core loss is 2 % of the primary power. Find (i) the secondary rms voltage, (ii) the secondary rms current, and (iii) the efficiency of the transformer.

Step-by-step solution

1. Use turns ratio to find secondary voltage: $V_s=V_p\frac{N_s}{N_p}=240\frac{300}{1500}=48$V.
2. Secondary current: $I_s=\frac{V_s}{R}=\frac{48}{800}=0.06$A.
3. Primary current using inverse turns ratio: $I_p=I_s\frac{N_s}{N_p}=0.06\frac{300}{1500}=0.012$A.
4. Primary power input: $P_{in}=V_p{I}_p=240\times 0.012=2.88$W.
5. Copper loss: $P_c=0.03P_{in}=0.0864$W. Core loss: $P_{core}=0.02P_{in}=0.0576$W. Total loss $=0.144$W.
6. Output power: $P_{out}=V_s{I}_s=48\times 0.06=2.88$W.
7. Efficiency: $\eta =\frac{P_{out}}{P_{out}+\text{losses}}\times 100\%=\frac{2.88}{2.88+0.144}\times 100\%\approx 95.3\%$.

Concept tested: Transformer operation with losses

Published 17 September 2026 · How this solution was prepared and checked