Q. An electric bulb is rated $60W,\ 220V.$ The resistance of its filament is
A) $708\ \Omega$
B) $870\ \Omega$
C) $807\ \Omega$
D) $780\ \Omega$
Explanation
(c) $R = \frac{V^{2}}{P} = \frac{(220)^{2}}{60} = 807\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.