Question

$n$ identical light bulbs, each designed to draw $P$ power from a certain voltage supply, are joined in series across that supply. The total power which they will draw is

• A. $nP$
• B. $P$
• C. $P/n$
• D. $P/n^2$

Answer 1

The correct option is C $P/n$

Given: each bulb consumes $P$ power

We know that ,

$\text{Power}=\frac{{\text{(Voltage)}}^2}{\text{Resistance}}.......\left(1\right)$

Let potential drop across each bulb be $V$.

Using eq.$\left(1\right)$, resistance offered by a bulb is

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

When $n$ bulbs connected in series across the same voltage source, total resistance offered by $n$ bulbs is,

$R_{eff}=nR=\frac{n{V}^2}{P}$

$\therefore Power=\frac{V^2}{R_{eff}}=\frac{V^{2}P}{n{V}^2}=\frac{P}{n}$

Hence, option $\left(c\right)$ is correct.

Key concept: If the same resistances are connected in parallel with the same voltage source $V$, $P_p=\frac{V^2}{\left(R/n\right)}=\frac{n{V}^2}{R}=nP$