Hi everyone,

Rather simple question:

The current entering the positive terminal of a device is i(t) = 3e^(-2t) A and the voltage across the device is v(t) = 5di/dt V.

Find the charge delivered to the device between t = 0 and t = 2 s.

since i = dq/dt, q = integral of i with respect to t

I'll have my integral sign be [

So q = [ i dt = [ 3e^(-2t) dt = (-3/2)e^(-2t) evaluated from 0 to 2:

-3/2e^(-4) - (-3/2) = 1.47 C

Answer in the back of the book = 1.297 C

Who's wrong?

Thanks in advance.

 

I think your answer is correct. If you differentiate the result of the integration; do you get back the original integrand? Check! There is no problem with the numerical evaluation.

FWIW the book's answer just happens to be what you get if you compute

-3/2 * [ e^(-2) - 1] = 1.297 Coulombs

This would also be correct if t was equal to 1 in the original problem.
Have a bit more confidence next time. I think you've earned it.

 

ihaveaquestion,

I get the same answer as you do.

Ratch