The number of electrons passing in the circuit is [tex]2.96\cdot 10^{27}[/tex]
Explanation:
The current intensity is defined as
[tex]I=\frac{q}{t}[/tex]
where
I is the current
q is the amount of charge passing through a given point in a circuit in a time of t
In this problem, we know that
I = 15 A
[tex]t = 1 y \cdot (365)(24)(60)(60)=3.15\cdot 10^7 s[/tex] is the time
Solving for q, we find the amount of charge that passes in the circuit:
[tex]q=It=(15)(3.15\cdot 10^7)=4.73\cdot 10^8 C[/tex]
Now we know that the charge of one electron is
[tex]e=1.6\cdot 10^{-19}C[/tex]
So, the number of electrons passing through the circuit is
[tex]N=\frac{q}{e}=\frac{4.73\cdot 10^8 C}{1.6\cdot 10^{-19}C}=2.96\cdot 10^{27}[/tex]
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