Answer:
Q=1.42 C
Explanation:
Given that
[tex]I=I_oe^{\dfrac{-t}{\tau}}[/tex]
When t= 0 ,Io = 3 A
τ = Time constant = 0.5 s
[tex]I=3e^{-\dfrac{t}{\tau}}[/tex]
We know that
[tex]I=\dfrac{dQ}{dt}[/tex]
Q=Charge ,I =Current
Q = ∫I.dt
Given that
t= 0 to t= 3τ= 1.5 s
The charge Q
[tex]Q=\int_{0}^{1.5}3e^{-\dfrac{t}{0.5}}dt[/tex]
Q=1.42 C
Therefore charge flow conductor will be 1.42 C.
