A current of 4.8,A is flowing through a condu | vedclass

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(A) The current $I$ is defined as the rate of flow of charge,given by $I = \frac{q}{t}$.Since the total charge $q$ is the product of the number of ele

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$3 	imes 10^{19}$

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(A) The current $I$ is defined as the rate of flow of charge,given by $I = \frac{q}{t}$.Since the total charge $q$ is the product of the number of electrons $n$ and the elementary charge $e$ $(q = ne)$,we can write $I = \frac{ne}{t}$.To find the number of electrons per second,we rearrange the formula to solve for $\frac{n}{t}$:$\frac{n}{t} = \frac{I}{e}$.Given $I = 4.8\,A$ and $e = 1.6 	imes 10^{-19}\,C$,we substitute these values:$\frac{n}{t} = \frac{4.8}{1.6 	imes 10^{-19}} = 3 	imes 10^{19}$.Therefore,the number of electrons flowing per second is $3 	imes 10^{19}$.

$A$ current of $4.8\,A$ is flowing through a conductor. The number of electrons flowing per second is:

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