We use the following equation from before:

`Ri+L(di)/(dt)+1/Cinti\ dt=E`

and obtain:

`20i+0.05(di)/(dt)+1/10^(-4)inti\ dt` `=100 cos 200t`

After multiplying throughout by 20, we have:

`400i+(di)/(dt)+20xx10^4intidt` `=2000 cos 200t`

Taking Laplace transform and using the fact that i(0) = 0:

`400I+sI-i(0)+200000I/s` `=2000 s/(s^2+200^2)`

`400sI+s^2I+200000I` `=2000 s^2/(s^2+200^2)`

`(s^2+400s+200000)I` `=2000 s^2/(s^2+200^2)`

`I=` ` (2000 s^2)/((s^2+400s+200000)(s^2+200^2))`

Using Scientific Notebook to find the partial fractions:

`(2000 s^2)/((s^2+400s+200000)(s^2+200^2))`

`=(2000-s)/(s^2+400s+200000)+(-400+s)/(s^2+40000)`

So

`I=-(s-2000)/(s^2+400s+200000)` `+(s-400)/(s^2+200^2)`

`=-(s-2000)/(s^2+400s+40000+160000)` `+(s-400)/(s^2+200^2)`

`=-(s-2000)/((s+200)^2+400^2)` `+(s-400)/(s^2+200^2)`

`=-(s+200)/((s+200)^2+400^2)` `+2200/((s+200)^2+400^2)` `+s/(s^2+200^2)-400/(s^2+200^2)`

`=-(s+200)/((s+200)^2+400^2)` `+11/2 400/((s+200)^2+400^2)` `+s/(s^2+200^2)-2 200/(s^2+200^2)`

So

`i=-e^(-200t)cos 400t` `+11/2e^(-200t)sin 400t` `+cos 200t` `- 2 sin 200t`


NOTE: Scientific Notebook can do all this for us very easily. In one step, we have:

`Lap^{:-1:}` `{(2000 s^2)/((s^2+400s+200000)(s^2+200^2))}`

`=-e^(-200t)cos 400t+` `11/2e^(-200t)sin 400t+cos 200t-` ` 2 sin 200t`

0.020.040.060.080.10.120.14123-1-2ti(t)Open image in a new page

Graph of `i=` `-e^(-200t)cos 400t+11/2e^(-200t)sin 400t+` `cos 200t-` ` 2 sin 200t`.

Transient part: `-e^(-200t)cos 400t+11/2e^(-200t)sin 400t` (using a different scale on the horizontal axis)

0.020.04123-1ti(t)Open image in a new page

Graph of `i=-e^(-200t)cos(400t)+11/2e^(-200t)sin(400t)`.

Steady state part: `cos 200t-2 sin 200t`

0.020.040.060.080.10.120.14123-1-2ti(t)Open image in a new page

Graph of `i=cos 200t-2 sin 200t`.