Q. An electric bulb of $100\ watt$ is connected to a supply of electricity of $220\ V.$ Resistance of the filament is
A) $484\ \Omega$
B) $100\ \Omega$
C) $22000\ \Omega$
D) $242\ \Omega$
Explanation
(a) $\frac{V^{2}}{R} = P \Rightarrow R = \frac{V^{2}}{P} = \frac{220 \times 220}{100} = 484\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.