Question

A 5.00-a current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. copper has 8.5 * 1028 free electrons per cubic meter. (a) how many electrons pass through the light bulb each second

Answer 1

Answer:

$\frac{dN}{dt} = 3.125 \times 10^{19}$ electrons per second

Explanation:

As we know that electric current is defined as rate of flow of electric charge

so here we will have

$i = \frac{dq}{dt}$

here we know that

$q = Ne$

now from above equation

$i = e\frac{dN}{dt}$

here we know

$\frac{dN}{dt}$ = number of electrons passing per second

e = charge of an electron

i = 5.00 Ampere

now from above equation we have

$\frac{dN}{dt} = \frac{i}{e}$

$\frac{dN}{dt} = \frac{5}{1.6 \times 10^{-19}}$

$\frac{dN}{dt} = 3.125 \times 10^{19}$

so above is the total number of electrons passing through wire per second

Answer 2

The correct answer is: $3.125*10^{19}$ electrons/second

Explanation:

5A current is passing through the copper wire and the light bulb; it means that 5 Coulombs of charge per second is passing through the wire (as current = coulombs/second). To find the electrons per second, the following formula is used:

Electrons per second = $n_e=\frac{5}{e}=\frac{5}{ 1.60\cdot 10^{-19}}=3.125*10^{19}$