Q. Two electric bulbs, each designed to operate with a power of 500 W in 220 V line are connected in series in a 110 V line. The power generated by each bulb will be
A) 31.25 W
B) 40 W
C) 60 W
D) 3.125 W
Explanation
(a) Resistance of each bulb $R = V^{2}$/$P$.
When connected in series total resistance of bulbs $= 2R$
Current in each bulb , $I = V^{'}/2R$;
Power generated by each bulb $= I^{2}\ R$
$= \left( \ \frac{V^{'}}{2R} \right)^{2} \times R = \frac{V'}{4R} = \frac{V'2}{(V^{2}/P)}$
$= \frac{(110)^{2} \times 500}{4 \times (220)^{2}} =$31.25 W
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.