Circuits 1 help with integration

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SUMMARY

The discussion focuses on solving a circuit integration problem involving the equation i = i(o) + (1/L) ∫(v0) dt. The user clarifies the conversion of 2 milliseconds to seconds as 2x10^-3 and confirms the calculation of current after integration, arriving at a final current of 0.05A. The user also derives the rate of change of current as 25 amps per second using the formula V = L(di/dt), with V set at 5mV and L at 200μH.

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Homework Statement



http://imgur.com/a/qlQ5z
jJjYb53.png


Homework Equations



i=i(o)+1/L integration(v0) dt formula is in the attemp at a solution.

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.
 
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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 :o
 
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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