Question

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?

Answer 1

Answer:

I = 0.96 A

Explanation:

No of electrons, $n=1.8\times 10^{16}$

Time, t = 3 ms = $3\times 10^{-3}\ s$

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

$I=\dfrac{q}{t}$

q = ne (Quantization of electric charge)

$I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A$

So, the electric current is 0.96 A.