Question

A steedy curent of 0.25 A flows through a conductor how many electron flows through the section of the conductor is 1 sec

Answer 1

q=I×t
q=0.25×1=0.25c
1c=1.6×10^-16
0.25c=0.4×10^-16

Answer 2

Hii friend,

Given,

I => 0.25 A

e = 1.6 ×10^-19

T =1 secs

n =>?

I =q/t

I => ne/t

n = It /e

n =0.25 ×1/1.6×10^-19

n =0.15625 ×10^19 electrons.

Hope this helps you a little!!