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(B) The current $I$ flowing through the filament of the bulb is given by the formula $I = \frac{P}{V}$.Given $P = 60 \ W$ and $V = 120 \ V$,we have:$I

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$3.125 	imes 10^{18}$

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(B) The current $I$ flowing through the filament of the bulb is given by the formula $I = \frac{P}{V}$.Given $P = 60 \ W$ and $V = 120 \ V$,we have:$I = \frac{60}{120} = 0.5 \ A$.Since current is the rate of flow of charge,$I = \frac{Q}{t} = \frac{ne}{t}$,where $n$ is the number of electrons,$e$ is the elementary charge $(1.6 	imes 10^{-19} \ C)$,and $t$ is time.For $t = 1 \ s$,the number of electrons $n$ is:$n = \frac{I 	imes t}{e} = \frac{0.5 	imes 1}{1.6 	imes 10^{-19}}$.$n = 0.3125 	imes 10^{19} = 3.125 	imes 10^{18}$.Thus,the number of electrons flowing per second is $3.125 	imes 10^{18}$.

How many electrons per second flow through a filament of a $120 \ V$,$60 \ W$ electric bulb?

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