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(C) Let the voltage of the mains supply be $V$. The resistance of each bulb is $R = V^2/P$.When $n$ identical bulbs are connected in series,the total

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$P/n$

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(C) Let the voltage of the mains supply be $V$. The resistance of each bulb is $R = V^2/P$.When $n$ identical bulbs are connected in series,the total resistance of the combination is $R_{eq} = nR$.The total power $P_s$ drawn by the series combination is given by $P_s = V^2 / R_{eq}$.Substituting $R_{eq} = nR$,we get $P_s = V^2 / (nR)$.Since $R = V^2/P$,we substitute this into the equation: $P_s = V^2 / (n(V^2/P))$.Simplifying this,we get $P_s = P/n$.

There are $n$ number of identical electric bulbs,each designed to draw a power $P$ independently from the mains supply. If they are now joined in series across the same mains supply,what is the total power drawn by the combination?

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