Solve Integration Problem: Charge Flowed After 5 Seconds

[happyjoe]

Hey everyone. This has been really bugging me for 4 hours now, and I cannot figure out where I am going wrong. If anyone can help me that would be greatly appreciated.

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 = \int I dt


5
\int 2e^-(0.2t) dt
0

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

= -10e^-(0.2t)

5
\int -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 .

 

[montoyas7940]

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


= -10e^-1 - 0

Your mistake is: e^0 does not equal 0, e^0 equals 1.

 

[happyjoe]

Wow, I can't believe I made a mistake that simple, lol

Thanks for your help! You really helped me out there!