Q. A bulb rated at $(100W - 200V)$ is used on a $100V$ line. The current in the bulb is
A) $\frac{1}{4}amp$
B) $4\ amp$
C) $\frac{1}{2}amp$
D) $2\ amp$
Explanation
(a) $P = \frac{V^{2}}{R} \Rightarrow 100 = \frac{(200)^{2}}{R} \Rightarrow R = \frac{4 \times 10^{4}}{10^{2}} = 400\ \Omega$
Now,$\ i = \frac{V}{R} = \frac{100}{400} = \frac{1}{4}amp$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.