There are 1.35 × [tex]10^{-19}[/tex] many electrons flow through a point in a wire in 4.50 s if there is a constant current of i = 4.82 a.
The voltmeter detects the significant drop, which would be actually the reduction in potential energy even as electricity passes through the filament, as well as the ammeter analyzes the rate at which electrons flow through globes.
Given data :
T = 4.50 s
i = 4.82 a
n ?
Number of electrons can be calculated as by using the formula:
q = it...(i)
q = ne...(ii)
where, q = charge , n = number of electrons, e = electrons
Comparing equation (i) and (ii), put the values of given data in above equation.
n × 16 × [tex]10^{-19}[/tex] = 4.82 a × 4.50 s
n = 1.35 × [tex]10^{-19}[/tex] .
Therefore, the number of electrons flows 1.35 × [tex]10^{-19}[/tex] .
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