If $R_{1}\ and\ R_{2}$ be the resistances of the filaments of 200 W and 100 W electric bulbs operation at 220 V, then $\left( \frac{R_{1}}{R_{2}} \right)$ is

Question

Dr. Abubacker · Accepted Answer

(c) Power of bulb, $P = \frac{V^{2}}{R}$
$	herefore\ \frac{P_{2}}{P_{1}} = \frac{R_{1}}{R_{2}}$
Or $\frac{R_{1}}{R_{2}} = \frac{100}{200} = 0.5$

Q. If $R_{1}\ and\ R_{2}$ be the resistances of the filaments of 200 W and 100 W electric bulbs operation at 220 V, then $\left( \frac{R_{1}}{R_{2}} \right)$ is

Explanation

Reviewed by Dr. Abubacker

Want more practice?