- #1

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The current flowing through a capacitor is:

I = Ioe^-(kt)

Io = 2 amps

k = 0.2s

T = time

Find the charge flowed after the first 5 seconds

Q = [tex]\int[/tex] I dt

5

[tex]\int[/tex] 2e^-(0.2t) dt

0

= 1/-0.2e^-(0.2t)

= -10e^-(0.2t)

5

[tex]\int[/tex] -10e^-(0.2t)

0

[ -10e^-(0.2t)] = {-10e^-(0.2x5)} - {-10e^-(0.2x0)}

= -10e^-1 - 0

Then I end up with a negative number for Q . If anyone could point me in the right direction i'd be thankful .