Q. A light bulb is rated $100\ W$ for a $220\ V$ supply. The resistance of the bulb and the peak voltage of the source respectively are
A) $242\ \Omega$ and 311 $V$
B) $484\ \Omega$ and 311 $V$
C) $484\ \Omega$ and 440 $V$
D) $242\ \Omega$ and 440 $V$
Explanation
(b) Resistance of a bulb $= \frac{(Rated\ voltage)^{2}}{Rated\ power}$
$= \frac{(220)^{2}}{100} = 484\ \Omega$
Peak voltage of the source, $V_{0} = 220\sqrt{2}\ V = 311\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.