1. Given data

   • Efficiency of transformer: $\eta = 80\% = 0.8$
   • Input voltage: $V_p = 10\,\text{V}$
   • Input power: $P_{in} = 4\,\text{kW} = 4000\,\text{W}$
   • Secondary voltage: $V_s = 240\,\text{V}$

2. Use efficiency relation Efficiency of a transformer is $\eta = \frac{P_{out}}{P_{in}}$ Therefore, $P_{out} = \eta P_{in} = 0.8 \times 4000 = 3200\,\text{W}$

3. Relate output power to secondary voltage and current $P_{out} = V_s I_s$ So, $I_s = \frac{P_{out}}{V_s} = \frac{3200}{240} = 13.33\,\text{A}$

4. Match with options The secondary current is $I_s = 13.33\,\text{A}$ which corresponds to Option B.

5. Verification of options

   • A: $1.33\,\text{A}$ → too small
   • B: $13.33\,\text{A}$ → correct
   • C: $1.59\,\text{A}$ → incorrect
   • D: $15.1\,\text{A}$ → incorrect

Therefore, the correct answer is B.