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(B) Let the resistance of each bulb be $R$ and the voltage of the source be $V$.The power rating of each bulb is given by $P = V^2 / R$,which implies

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$P/n$

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(B) Let the resistance of each bulb be $R$ and the voltage of the source be $V$.The power rating of each bulb is given by $P = V^2 / R$,which implies the resistance of each bulb is $R = V^2 / P$.When $n$ such bulbs are connected in series,the total resistance of the circuit becomes $R_t = nR$.The current flowing through the series circuit is $I = V / R_t = V / (nR)$.The total power drawn by the series combination is $P_t = I^2 R_t$.Substituting the values of $I$ and $R_t$,we get $P_t = (V / nR)^2 	imes (nR) = (V^2 / n^2 R^2) 	imes (nR) = V^2 / (nR)$.Since $V^2 / R = P$,we can write $P_t = (V^2 / R) / n = P / n$.

$n$ identical bulbs,each designed to draw a power $P$ from a certain voltage supply,are joined in series across that supply. The total power which they will draw is

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