Question
Show that one ampere is equivalent to a flow of 6.25 × 1018 elementary charges per second.

Answer

$I =1 A , t =1 s , e =1.6 \times 10^{-19} C$
$q = ne $
$I =\frac{q}{t}=\frac{n e}{t}$
Number of electrons, $n=\frac{ I t}{e}=\frac{1 \times 1}{1.6 \times 10^{-19}}=6.25 \times 10^{18}$.

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