Question

40 Coulombs of electricity was applied over a 793 second time period. What was the current expressed in mA?

Answer 1

Current `I` , charge `Q` and time `t` are related as,

`I=Q/t`

Given, `Q=40C,t=793s`

So, `I=40/793=50.43 mA` (1 `A=10^3 mA`)