Question

A current of 5 ampere is flowing through a wire of for 193 seconds calculate the number of electrons flowing through the cross section of wire for 193 seconds

Answer 1

Ans:- i=dq/dt
i=5A, t=193 sec

Putting the value
So, 5=dq/193
dq= 5*193 = 965

1electron =1.6*10^-19 charge

965 cahrge = 965/(1.6*10^-19) electron
=6.03*10^19 ans

Answer 2

Explanation:

5 Ampere =

$\frac{5c}{sec}$

it means 5 coulumb of charges passes from the crosses section in 1 seconds .

we just have calculate the total number of charges passes from the cross section in 193 seconds

:- JUST USE UNITARY METHOD

5C= 1 SEC

193*5C = 193*1SEC

965 C =193 Sec

IT MEANS TOTAL 965 COULOMB CHARGES PASSES FROM THE CROSS SECTION IN 193 SECONDS.

NOW WE CAN EASILY CALCULATE THE TO TOTAL NUMBERS OF ELECTRONS ,

WE KNEW THAT 1 COULOMB CHARGES CONTAINS

$6.25 \times 10 {}^{18 \: } electrons$

then 965 COULOMB charges will contain

$965 \times 6.25 \times 10 {}^{18} electrons = 6.03125 \times 10 {}^{21} electrons$