A power transmission line feeds input power at 2400 V to a step-down transformer with its primary windings having 5000 turns. The output power is delivered at 240 V by the transformer. If the current in the primary coil of the transformer is 5 A and its efficiency is 95% then the current in the secondary will be

47.5 A

The correct answer is Option (2) → 47.5 A

Given:

Primary voltage, $V_p = 2400\ \text{V}$

Secondary voltage, $V_s = 240\ \text{V}$

Primary current, $I_p = 5\ \text{A}$

Efficiency, $\eta = 95\% = 0.95$

Step-down transformer relation (ideal):

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$, $\frac{I_s}{I_p} = \frac{N_p}{N_s}$

But with efficiency:

$P_{\text{out}} = \eta P_{\text{in}} = \eta V_p I_p$

Output power: $P_{\text{out}} = V_s I_s$

So, $I_s = \frac{P_{\text{out}}}{V_s} = \frac{\eta V_p I_p}{V_s}$

Substitute values:

$I_s = \frac{0.95 \cdot 2400 \cdot 5}{240} = \frac{11400}{240} = 47.5\ \text{A}$

Current in secondary coil = 47.5 A