QJEEM24Ph7f72: JEE Main 2024 Physics - Class 12 JEEM24Ph7f72 Solution: A transformer has an efficiency of and works at and . If the…

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%.

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 $P_{\text{in}} = 4000 \text{W}$ .

Let's calculate $P_{\text{out}}$ using the efficiency formula:

$P_{\text{out}} = \eta \times P_{\text{in}} = 0.8 \times 4000 \text{W} = 3200 \text{W}$

Now that we have $P_{\text{out}}$ , we can calculate the secondary current ( $I_{\text{secondary}}$ ) using the formula:

$P_{\text{out}} = V_{\text{secondary}} \times I_{\text{secondary}}$

We are given $V_{\text{secondary}} = 240 \text{V}$ .

Isolating $I_{\text{secondary}}$ gives us:

$I_{\text{secondary}} = \frac{P_{\text{out}}}{V_{\text{secondary}}}$

Substituting the known values:

$I_{\text{secondary}} = \frac{3200 \text{W}}{240 \text{V}}$

$I_{\text{secondary}} = \frac{3200}{240}$

$I_{\text{secondary}} = 13.33 \text{A}$

Therefore, the current in the secondary coil is $13.33 \text{A}$ , which corresponds to Option B.