A current of `7.5 A` is maintained in wire of `45 s`. In this time
(a) how much charge and
(b) how many electrons flow through the wire?

Explanation

`a. q = it = (7.5A) (45 s) = 337.5C`
b. The number of electron N is given by
`N = (q)(e) = (337.5C)/(1.6xx10^(-19)C) =2.1xx^(21)`
where `e = 1.6xx10^(-19)C` is the charge of an electron.