The correct answer is: [tex]3.125*10^{19}[/tex] 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 = [tex]n_e=\frac{5}{e}=\frac{5}{ 1.60\cdot 10^{-19}}=3.125*10^{19}[/tex]
Answer:
[tex]\frac{dN}{dt} = 3.125 \times 10^{19}[/tex] electrons per second
Explanation:
As we know that electric current is defined as rate of flow of electric charge
so here we will have
[tex]i = \frac{dq}{dt}[/tex]
here we know that
[tex]q = Ne[/tex]
now from above equation
[tex]i = e\frac{dN}{dt}[/tex]
here we know
[tex]\frac{dN}{dt}[/tex] = number of electrons passing per second
e = charge of an electron
i = 5.00 Ampere
now from above equation we have
[tex]\frac{dN}{dt} = \frac{i}{e} [/tex]
[tex]\frac{dN}{dt} = \frac{5}{1.6 \times 10^{-19}}[/tex]
[tex]\frac{dN}{dt} = 3.125 \times 10^{19}[/tex]
so above is the total number of electrons passing through wire per second
