A transformer has $N_p=1200$ turns on the primary and $N_s=300$ turns on the secondary. The primary is connected to a 240 V rms AC supply at 50 Hz and a resistive load of 600 Ω is connected across the secondary. Assuming copper loss is 2 % and core loss is 1 % of the primary power, calculate (i) the secondary rms voltage, (ii) the secondary rms current, (iii) the transformer efficiency.

Step-by-step solution

1. Voltage ratio: $V_s=V_p\frac{N_s}{N_p}=240\times \frac{300}{1200}=60$ V.
2. Secondary current: $I_s=V_s/R_{\text{load}}=\frac{60}{600}=0.1$ A.
3. Ideal primary current: $I_{p,\text{ideal}}=I_s\frac{N_s}{N_p}=0.1\times \frac{300}{1200}=0.025$ A.
4. Ideal primary power: $P_{\text{ideal}}=V_p{I}_{p,\text{ideal}}=240\times 0.025=6$ W.
5. Let $P_{\text{in}}$ be actual primary input. Losses: $P_{\text{cu}}=0.02P_{\text{in}}$, $P_{\text{core}}=0.01P_{\text{in}}$.\ Thus $P_{\text{in}}=P_{\text{out}}+P_{\text{cu}}+P_{\text{core}}=6+0.03P_{\text{in}}$. Solve: $0.97P_{\text{in}}=6;\Rightarrow ;P_{\text{in}}\approx 6.20$ W.
6. Output power: $P_{\text{out}}=V_s{I}_s=60\times 0.1=6$ W (same as ideal because losses are small).
7. Efficiency: $\eta =\frac{P_{\text{out}}}{P_{\text{in}}}=\frac{6}{6.20}\approx 0.9695=96.95$ (≈ 97 %).

Concept tested: Transformer efficiency with losses

Published 16 September 2026 · How this solution was prepared and checked