Q. An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}.$ If the voltage is $V$ it draws a power $P.$ Then
A) $P = \left( \frac{V_{0}}{V} \right)^{2}P_{0}$
B) $P = \left( \frac{V}{V_{0}} \right)^{2}P_{0}$
C) $P = \left( \frac{V}{V_{0}} \right)P_{0}$
D) $P = \left( \frac{V_{0}}{V} \right)P_{0}$
Explanation
(b) $P \propto V^{2} \Rightarrow \frac{P}{P_{0}} = \left( \frac{V}{V_{0}} \right)^{2} \Rightarrow P = \left( \frac{V}{V_{0}} \right)^{2}P_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.