JEE Physics 2025 Question with Solution

Given: There are $n$ identical electric bulbs. Each bulb, when connected independently across the mains supply, draws power $p$.

Find: The total power drawn when all $n$ bulbs are connected in series across the same mains supply.

For identical bulbs in series, the total resistance is

$$R_{\text{total}} = R_1 + R_2 + \dots + R_n = nR$$

where $R$ is the resistance of each bulb.

The total power drawn from a mains supply of voltage $V$ is

$$P_{\text{total}} = \frac{V^2}{R_{\text{total}}}$$

Substituting $R_{\text{total}} = nR$, we get

$$P_{\text{total}} = \frac{V^2}{nR}$$

Now for one bulb connected independently across the mains supply,

$$p = \frac{V^2}{R}$$

Using this relation,

$$P_{\text{total}} = \frac{p}{n}$$

Therefore, the total power drawn by the combination is $\frac{p}{n}$. The correct option is C.

Given: The bulbs are identical and connected in series.

Find: Total power drawn by the series combination.

When bulbs are connected in series, the same current flows through each bulb and the equivalent resistance increases to $nR$. Since the supply voltage remains the mains voltage, increasing resistance reduces the current drawn from the source.

Hence the total power becomes

$$P_{\text{total}} = \frac{V^2}{nR}$$

But for a single bulb,

$$p = \frac{V^2}{R}$$

So,

$$P_{\text{total}} = \frac{p}{n}$$

Thus, the total power is reduced by a factor of $n$, and the correct option is C.