Question

Three identical bulbs are connected in series and these together dissipate a power $P$. If now these bulbs are connected in parallel, then the power dissipated will be

• A. $\frac{P}{3}$
• B. $3P$
• C. $9P$
• D. $\frac{P}{9}$

Answer 1

The correct option is C $9P$

Let the resistance of each bulb be $R$.

When all three bulbs are in series,

${\left(R_{eff}\right)}_s=3R$

$\therefore$ Power dissipation in series will be

$P=\frac{V^2}{{\left(R_{eff}\right)}_s}=\frac{V^2}{3R}........\left(1\right)$

When all three bulbs are in parallel,

${\left(R_{eff}\right)}_p=\frac{R}{3}$

$\therefore$ Power dissipation in parallel,

$P{}^{\prime }=\frac{V^2}{{\left(R_{eff}\right)}_p}=\frac{3V^2}{R}.......\left(2\right)$

From eq (1) and (2), we get,

$P^{{}^{\prime }}=3\times 3P=9P$

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