JEE Physics 2026 Question with Solution

Given: Power delivered is $1000\,\text{W}$, voltage is $250\,\text{V}$, and total resistance of the line is $2\,\Omega$.

Find: The percentage efficiency of the transmission line.

Step 1: Calculate the current in the transmission line.

$$I = \frac{P}{V} = \frac{1000}{250} = 4\,\text{A}$$

Step 2: Find the power loss in the transmission line.

$$P_{\text{loss}} = I^2R = 4^2 \times 2 = 32\,\text{W}$$

Step 3: Calculate the efficiency of transmission.

$$\eta = \frac{P_{\text{output}}}{P_{\text{input}}}\times 100 = \frac{1000}{1000 + 32}\times 100 \approx 96.9\%$$

Therefore, the percentage efficiency of the transmission line is $96.9\%$. The correct option is B.

Given: $P = 1000\,\text{W}$, $V = 250\,\text{V}$, $R = 2\,\Omega$.

Find: Transmission efficiency.

First find the line current using

$$I = \frac{P}{V} = 4\,\text{A}$$

Then compute line loss:

$$P_{\text{loss}} = I^2R = 32\,\text{W}$$

Now input power is output power plus line loss, so efficiency becomes

$$\eta = \frac{1000}{1032} \times 100 \approx 96.9\%$$

This works because efficiency depends on how much of the supplied power is lost as $I^2R$ heat in the line. Therefore, the correct option is B.