Quiescent implies i1, i2 and their derivatives are zero for t = 0, ie

i1(0) = i2(0) = i1’(0) = i2’(0) = 0.

For loop 1:

`10i_1+10(i_1-i_2)=4`

`20i_1-10i_2=4`

`10i_1-5i_2=2`

`i_1=1/10(2+5i_2)\ \ \ ...(1)`

For loop 2:

`20i_2+0.1(di_2)/(dt)+10(i_2-i_1)=0`

`-10i_1+30i_2+0.1(di_2)/(dt)=0`

`-100i_1+300i_2+(di_2)/(dt)=0`

Substituting our result from (1) gives:

`-100xx1/10(2+5i_2)+300i_2+(di_2)/(dt)` `=0`

`-10(2+5i_2)+300i_2+(di_2)/(dt)=0`

`250i_2+(di_2)/(dt)=20`

Taking Laplace transform:

`250I_2+(sI_2-i_2(0))=20/s`

`(250+s)I_2=20/s` since `i_2(0)=0`

`I_2=20/(s(s+250))`

Let `20/(s(s+250))=A/s+B/(s+250)`

So `20=A(s+250)+Bs`

`20=250A \ \ →\ \ A=20/250=2/25`

`A+B=0 \ \ →\ \ B=-2/25`

So `20/(s(s+250))=2/25(1/s-1/(s+250))`

Taking Inverse Laplace:

So `i_2=2/25(1-e^(-250t))`

0.010.020.030.040.020.040.060.080.1ti(t)Open image in a new page

Graph of `i_2=2/25(1-e^(-250t))`.

The current builds up to `0.08` A (but never quite reaches it).

Easy to understand math videos:
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