Q. A $100\ watt$ bulb working on $200\ volt$ and a $200\ watt$ bulb working on $100\ volt$ have
A) Resistances in the ratio of 4 :1
B) Maximum current ratings in the ratio of 1 :4
C) Resistances in the ratio of 2 :1
D) Maximum current ratings in the ratio of 1 :2
Explanation
(b) $R = \frac{V^{2}}{P} \Rightarrow R_{1} = \frac{200 \times 200}{100} = 400\ \Omega\ and$
$R_{2} = \frac{100 \times 100}{200} = 50\ \Omega.$ Maximum current rating$\ i = \frac{P}{V}$
So$\ i_{1} = \frac{100}{200}$ and $i_{2} = \frac{200}{100} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.