A transformer has an efficiency of 80% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is:

Find the secondary current of a transformer with 80% efficiency, 10 V primary, 4 kW input power, and 240 V secondary voltage.

Recall that efficiency $$\eta = \frac{P_{\text{output}}}{P_{\text{input}}}$$.

With an input power of 4000 W and an efficiency of 0.80, the output power is $$P_{\text{output}} = \eta \times P_{\text{input}} = 0.80 \times 4000 = 3200\text{ W}$$.

Applying $$P = VI$$ to the secondary winding gives $$I_2 = \frac{P_{\text{output}}}{V_2} = \frac{3200}{240} = 13.33\text{ A}$$.

The correct answer is Option B: 13.33 A.