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

<p>To find the current in the secondary coil of the transformer, we first need to calculate the output power, taking into account the efficiency. The efficiency ($\eta$) of the transformer is given by the ratio of the output power ($P_{\text{out}}$) to the input power ($P_{\text{in}}$) times 100%.</p> <p>The given efficiency is $\eta = 80\%$ or $\eta = 0.8$ in decimal form. The input power is also given as $P_{\text{in}} = 4 \text{kW}$ or $P_{\text{in}} = 4000 \text{W}$.</p> <p>Let&#39;s calculate $P_{\text{out}}$ using the efficiency formula:</p> <p>$$ P_{\text{out}} = \eta \times P_{\text{in}} = 0.8 \times 4000 \text{W} = 3200 \text{W} $$</p> <p>Now that we have $P_{\text{out}}$, we can calculate the secondary current ($I_{\text{secondary}}$) using the formula:</p> <p>$P_{\text{out}} = V_{\text{secondary}} \times I_{\text{secondary}}$</p> <p>We are given $V_{\text{secondary}} = 240 \text{V}$.</p> <p>Isolating $I_{\text{secondary}}$ gives us:</p> <p>$I_{\text{secondary}} = \frac{P_{\text{out}}}{V_{\text{secondary}}}$</p> <p>Substituting the known values:</p> <p>$I_{\text{secondary}} = \frac{3200 \text{W}}{240 \text{V}}$</p> <p>$I_{\text{secondary}} = \frac{3200}{240}$</p> <p>$I_{\text{secondary}} = 13.33 \text{A}$</p> <p>Therefore, the current in the secondary coil is $13.33 \text{A}$, which corresponds to Option B.</p>