A transformer has $800$ turns on the primary and $200$ turns on the secondary. The primary is connected to a $240{\text{V}}_{\text{rms}}$ supply at $50\text{Hz}$, and the secondary supplies a resistive load of $600\Omega$. Copper loss in the windings is $3$ of the input power and core loss is $2$ of the input power. Find (i) the secondary rms voltage, (ii) the secondary rms current, and (iii) the efficiency of the transformer (in %).

Step-by-step solution

1. Compute the turns ratio $a=\frac{N_s}{N_p}=\frac{200}{800}=0.25$.
2. Use $V_s=V_{p}a=240\times 0.25=60{\text{V}}_{\text{rms}}$ for the secondary voltage.
3. Compute secondary current $I_s=V_s/R=60/600=0.10\text{A}$.
4. Primary current is $I_p=I_{s}a=0.10\times 0.25=0.025\text{A}$ (for an ideal transformer).
5. Input power $P_{\text{in}}=V_p{I}_p=240\times 0.025=6\text{W}$.
6. Total losses are $5$ of input, so $P_{\text{loss}}=0.05\times 6=0.30\text{W}$ and output power $P_{\text{out}}=P_{\text{in}}-P_{\text{loss}}=5.70\text{W}$.
7. Efficiency $\eta =\frac{P_{\text{out}}}{P_{\text{in}}}\times 100\approx 95$.

Concept tested: Transformer voltage, current, efficiency

Published 4 October 2026 · How this solution was prepared and checked