Q. A bulb of 220 V and 300 W is connected across 110 V circuit, the percentage reduction in power is

A) 100%
B) 25%
C) 70%
D) 75%
Explanation

(d) Power $P = \frac{V^{2}}{R}\ \Rightarrow 300 = \frac{220\ \times 220}{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ R = \frac{22\ \times 22}{3}$
Again $P = \frac{110\ \times 110\ \times 3}{22\ \times 22}$
$\ \ \ \ \ \ \ \ \ \ \ \ P = 75$
$\ \ \ \ \ \ \ \ P\% = \frac{75\ \times 100}{300} = 25\%$
The percentage reduction in power
$\ \ \ \ \ \ \ \ \ \ \ \ P = 100 - 25 = 75\%$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.