1. Power rating of one bulb on mains

Let the mains voltage be $V$.

Each bulb is designed to draw power $p$ when connected individually across the mains. So for one bulb,

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

where $R$ is the resistance of one bulb.

Thus,

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

2. Equivalent resistance of $n$ identical bulbs in series

In series, resistances add:

$R_{\text{eq}}=nR$

3. Total power drawn from the same mains voltage

The combination is connected across the same mains voltage $V$, so total power is

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

Substitute $R_{\text{eq}}=nR$:

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

Now use $R=\frac{V^2}{p}$:

$P_{\text{total}}=\frac{V^2}{n\left(\frac{V^2}{p}\right)}=\frac{p}{n}$

4. Check options

• A: $np$ ❌
• B: $p$ ❌
• C: $\frac{p}{n^2}$ ❌
• D: $\frac{p}{n}$ ✅

So the correct answer is:

$\boxed{\frac{p}{n}}$