Q. If voltage across a bulb rated 220 Volt-100 Watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
A) 20%
B) 2.5%
C) 5%
D) 10%
Explanation
(c) Resistance of bulb is constant
$P = \frac{v^{2}}{R} \Rightarrow \frac{\mathrm{\Delta}P}{P} = \frac{2\mathrm{\Delta}V}{V} + \frac{\mathrm{\Delta}R}{R}$
$\frac{\mathrm{\Delta}P}{P} = 2 \times 2.5 + 0 = 5\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.