# Circuits 1 help with integration

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1. Oct 4, 2016

### sammyqw

1. The problem statement, all variables and given/known data

http://imgur.com/a/qlQ5z

2. Relevant equations

i=i(o)+1/L integration(v0) dt formula is in the attemp at a solution.
3. The attempt at a solution
http://imgur.com/a/HVjl1
For the interval 2<t<infinite . I understand that voltage at 2ms is 0 , but where did they get 2x10^-3? is that the 2ms to seconds? and how did they 50ma after the integration? I only get 10ma not 50. Thanks in advance.

2. Oct 4, 2016

### sammyqw

Nevermind I think I got it. After getting 25t we do i(0.002seconds)=25(0.002)= 0.05a then for the interval 2<t<infinity i= i (0) +1/L... substituting i(o) is 0.05a so i=0.05 + integration of 0 which is 0= so the final answer is 0.05a

3. Oct 9, 2016

### CWatters

Looks correct to me. I checked it this way..

V = Ldi/dt
so
di/dt = V/L
= 5*10-3/200*10-6
=25 amps per second (or 25t if you prefer)

Calculate I after 2mS..
25*2*10-3 = 0.05A
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