Question

The current enters to the positive terminal of a device is i(t)=20e−4t amps, and the voltage across the device is v(t)=3di(t)dt volt. The amount of the charge delivered (in C) to the device between t=0 and t=2 seconds.

Answer 1

Relation between charge flow and current is given by

$i = \frac{dq}{dt}$

here we can rearrange it as

$q = \int idt$

now here we know that current is given as function of time

$i = 20e^{-4t}$

now to find the charge from from t = 2 to t = 2 s we need to put the equation of current in the above equation.

$q = \int 20e^{-4t} dt$

here in above equation time limits are t = 0 to t = 2 s

$q = \frac{20}{-4} (e^{-8} - e^0)$

$q = 5(1 - e^{-8})C$

q = 4.998 C

so the total charge flow from t = 0 to t = 2 s will be q = 4.998 C