Question

$N$ identical light bulbs, each designed to draw power $P$ 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. $\frac{P}{N}$
• D. $\frac{P}{N^2}$

Answer 1

The correct option is C $\frac{P}{N}$

Let $V$ be the voltage of the source and $R$ be the resistance of each bulb. Then, we have
$R=\frac{V^2}{P}\Rightarrow P=\frac{V^2}{R}$
when $N$ bulbs are joined in series across $V$ current in each bulb is given by $I=\frac{V}{NR}$. Power drawn by each bulb is
$I^{2}R=\left(\frac{V^2}{N^2{R}^2}\right)R=\frac{V^2}{N^{2}R}=\frac{P}{N^2}$
Hence. total power drawn is
$N_P=N\left(\frac{P}{N^2}\right)=\frac{P}{N}$
Hence, the correct answer is (C).