JEE Main 2024 — Electromagnetic Induction Question with Solution

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.