If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the circuit?

Respuesta :

Answer:

I = 0.96 A

Explanation:

No of electrons, [tex]n=1.8\times 10^{16}[/tex]

Time, t = 3 ms = [tex]3\times 10^{-3}\ s[/tex]

We need to find the electric current. We know that electric charge per unit time is equal to the electric current.

[tex]I=\dfrac{q}{t}[/tex]

q = ne (Quantization of electric charge)

[tex]I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A[/tex]

So, the electric current is 0.96 A.