Q. A $500\ W$ heating unit is designed to operate from a $115\ volt$ line. If the line voltage drops to 110 volt, the percentage drop in heat output will be
A) $10.20\%$
B) $8.1\%$
C) $8.6\%$
D) $7.6\%$
Explanation
(c) $P_{consumed} = \left( \frac{V_{A}}{V_{R}} \right)^{2} \times P_{R} = \left( \frac{110}{115} \right)^{2} \times 500 = 457.46\ W$
So, percentage drop in power output
$= \frac{(500 - 457.46)}{500} \times 100 = 8.6\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.