Answer:
I =62.5 μ A
Explanation:
Given that
Charge ,q = 2.5 x 10⁻³ C
time ,t= 40 s
We know that current I given as
[tex]I=\dfrac{q}{t}[/tex]
Now by putting the values in the above equation we get
[tex]I=\dfrac{2.5\times 10^{-3}}{40}\ C/s[/tex]
We know that
1 A = 1 C/s
Therefore the current I will be
I=0.0000625 A
I = 0.0625 m A
The current in μ A will be
I =62.5 μ A
