Q. An electric bulb rated for 500 W at 100V is used in a circuit having a 200 V supply. The resistance $R$ that must be put in series with the bulb, so that the bulb drawn 500 W is
A) 18$\Omega$
B) 20$\Omega$
C) 40$\Omega$
D) 700$\Omega$
Explanation
(b) $P = VI,\ I = \frac{P}{V}$
or $\ I = \frac{500\ W}{100\ V} = 5A$
Now, $\ 5\ R = 100$
or $\ \ R = 20\Omega$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.