Question

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

• A. P/3
• B. 3P
• C. 9P
• D. P/9

Answer 1

Answer: (C)

By Joule's law, the power dissipated through a resistor $R$,
having a potential difference $V$ is
$P=\frac{V^{2}}{R}$
When bulbs are connected in series
$R'=R+R+R=3 R$
Power dissipated is
$P=\frac{V^{2}}{3 R}$ ...(i)
When they are connected in parallel
$\frac{1}{R''}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}$
$\Rightarrow R''=\frac{R}{3}$
Power dissipated, $P' =\frac{V^{2}}{R / 3}$ ...(ii)
From Eqs. (i) and (ii), we have
$P'=9 P$