A transformer has $N_p=500$ turns on the primary and $N_s=125$ turns on the secondary. The primary is connected to a $240{\text{V}}_{\text{rms}}$ source at $50\text{Hz}$. The secondary supplies a resistive load of $R_L=400\Omega$. Copper loss in the windings is $4\%$ of the input power and core loss is $3\%$ of the input power. Determine the efficiency of the transformer (in %).

Step-by-step solution

1. Find the turns ratio: $a=\frac{N_p}{N_s}=\frac{500}{125}=4$.
2. Compute the secondary rms voltage using the ideal transformer relation: $V_s=\frac{V_p}{a}=\frac{240}{4}=60\text{V}$.
3. Calculate the secondary current: $I_s=\frac{V_s}{R_L}=\frac{60}{400}=0.15\text{A}$.
4. Determine the output (load) power: $P_{\text{out}}=V_s{I}_s=60\times 0.15=9\text{W}$ (resistive load, unity PF).
5. Total loss percentage = copper loss + core loss = $4\%+3\%=7\%$.
6. Relate input power to output power: $P_{\text{in}}=\frac{P_{\text{out}}}{1-0.07}=\frac{9}{0.93}=9.68\text{W}$.
7. Compute efficiency: $\eta =\frac{P_{\text{out}}}{P_{\text{in}}}\times 100=\frac{9}{9.68}\times 100\approx 93\%$.

Concept tested: Transformer efficiency with copper and core losses

Published 20 September 2026 · How this solution was prepared and checked