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 equation of current is given as

$i(t)=20e^{−4t}$

now from above equation we can write

$Q = \int i dt$

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

our time limits are t=0 to t=2 s

$Q = \frac{20}{-4}(e^{-8} - 1)$

$Q = 5*(1 - e^{-8})$

$Q = 4.998 C$

SO total charge flow will be Q = 4.998 C

Answer 2

Answer:

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