Respuesta :
Given:
At t = 0 s and [tex]\tau[/tex] = 0.50s
[tex]I_{o}[\tex] = 7.75 A
I = [tex]I_{o}e^-{\frac{t}{\tau}}[/tex]
Solution:
Now, to calculate the charge flow in the interval t = 0 to t = 2[tex]\tau[/tex]
We know that, electric current is the rate of flow of electric charge and is given by:
[tex]I = \frac{\Delta Q}{\Delta T}[/tex]
[tex]\Delta Q = I\Delta T = I_{o}e^-{\farc{t}{\tau}}[/tex] (1)
Integrating the above eqn in the interval t = 0 to t = 2[tex]\tau[/tex]
[tex]\int_{0}^{Q}\Delta Q = \int_{0}^{2\tau}I_{o}e^-{\frac{t}{\tau}}\Delta T[/tex]
Q = [tex][-\tau (I_{o}e^-\frac{t}{\tau})]_{0}^{2\tau}[/tex]
Q = [tex]I_{o}[-\tau e^{-2} + \tau][/tex]
Q = [tex]7.75[-0.50 e^{-2} + 0.50][/tex]
Q = 3.35 C
Therefore, Q = 3.35 C of charge flows through the conductor in the given interval.
