If R_{1} and R_{2} are the resistances of fil | vedclass

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(A) The power $P$ of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since both

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$R_{2} = 2R_{1}$

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(A) The power $P$ of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since both lamps are designed to operate on the same voltage $V$,we have $R = \frac{V^2}{P}$.For the first lamp,$P_{1} = 400 \ W$,so $R_{1} = \frac{V^2}{400}$.For the second lamp,$P_{2} = 200 \ W$,so $R_{2} = \frac{V^2}{200}$.Dividing $R_{2}$ by $R_{1}$,we get $\frac{R_{2}}{R_{1}} = \frac{V^2 / 200}{V^2 / 400} = \frac{400}{200} = 2$.Therefore,$R_{2} = 2R_{1}$.

If $R_{1}$ and $R_{2}$ are the resistances of filaments of a $400 \ W$ and a $200 \ W$ lamp,designed to operate on the same voltage,then

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