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) Resistance of bulb,
$R = \frac{V^{2}}{P} = \frac{220 \times 220}{300} = \frac{484}{3}\Omega$
New power, $\ P^{'} = \frac{(110)^{2}}{R} = \frac{110 \times 110}{484/3} = 75$ watt reduction
Of power $= \frac{300 - 75}{300} \times 100 = 75\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.