Q. 3 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) $\frac{P}{3}$
B) $3P$
C) $9P$
D) $\frac{P}{9}$
Explanation
(c) When bulbs are connected in series, $P = \frac{V^{2}}{R^{'}} = \frac{V^{2}}{3R}$
When bulbs are connected in parallel
$P^{'} = \frac{V^{2}}{R"} = \frac{V^{2} \times 3}{R} = 3 \times 3P = 9P$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.