Q. Three electric bulbs of 200 W, 200 W and 400 W are shown in figure. The resultant power of the combination is
A) 800 W
B) 400 W
C) 200 W
D) 600 W
Explanation
(c) Bulbs $A\ and\ B$ are in parallel.
There, effective power is
$P^{'} = P_{A} + P_{B} = 200 + 200$
$\ \ \ \ \ = 400\ W$
Now $P'$ and bulb $C$ are in serried. So, the resultant power of the combination is
$P^{''} = \frac{400 \times 400}{400 + 400} = 200\ W$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.