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(C) The power consumed by a bulb with resistance $R$ connected to a potential difference $V$ is given by $P = \frac{V^2}{R}$.For the first bulb,$P_1 =

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$\frac{P_1 P_2}{P_1+P_2}$

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(C) The power consumed by a bulb with resistance $R$ connected to a potential difference $V$ is given by $P = \frac{V^2}{R}$.For the first bulb,$P_1 = \frac{V^2}{R_1}$,which implies $R_1 = \frac{V^2}{P_1}$.For the second bulb,$P_2 = \frac{V^2}{R_2}$,which implies $R_2 = \frac{V^2}{P_2}$.When the bulbs are connected in series,the total resistance is $R_{eq} = R_1 + R_2$.The total power consumed in series is $P_{series} = \frac{V^2}{R_{eq}} = \frac{V^2}{R_1 + R_2}$.Substituting the values of $R_1$ and $R_2$,we get $P_{series} = \frac{V^2}{\frac{V^2}{P_1} + \frac{V^2}{P_2}} = \frac{1}{\frac{1}{P_1} + \frac{1}{P_2}}$.Simplifying this,we get $P_{series} = \frac{P_1 P_2}{P_1 + P_2}$.

Two electric bulbs are connected one by one across a potential difference $V$. At that time,the power consumed in them is $P_1$ and $P_2$ respectively. Now,if the potential difference $V$ is applied across the series combination of these bulbs,what will be the total power consumed?

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