Position 1, after a 'long time': `i_0=10/5=2\ "A"`

Position 2: (`t>=0`)

We apply `sum"emf"` (that is, sum of the electromotive force), and consider the sum of the potential difference across elements.

In position 2, there is no emf.

`5i+0.1(di)/(dt)+1/(200xx10^(-6))inti\ dt` `=0`

`5i+0.1(di)/(dt)+5000inti\ dt=0`

Finding Laplace Transform:

`5I+0.1(sI-i_0)+5000(I/s+q_0/s)` `=0`

`5I+0.1(sI-2)+` `5000(I/s+(10^(-3))/s)` `=0`

`5I+0.1sI-0.2+5000 I/s+5/s=0`

Multiplying by `10s`:



Solving for I and completing the square on the denominator gives us:





`=2((s+25)/((s+25)^2+222.2^2)` `{:-50/((s+25)^25+222.2^2))`

`=2((s+25)/((s+25)^2+222.2^2)` `{:-50/222.2 222.2/((s+25)^2+222.2^2))`

So the transient current is:

`i=2(e^(-25t)cos 222.2t` `{:-50/222.2e^(-25t)sin 222.2t)`

`=e^(-25t)(2 cos 222.2t-` `{:0.45 sin 222.2t)`

We could transform the trigonometric part of this to a single expression.

First, re-arrange the equation by taking the negative out front:

`i=-e^(-25t)(0.45 sin 222.2t -` `{: 2 cos 222.2t)`

Using the "Cosine Minus Case" identity given in the link above which says,

`a sin θ − b cos θ = −R cos (θ + α)`, we aim to solve:

`0.45 sin 222.2t - 2 cos 222.2t` ` = -R cos(222.2t + α) `

In this case we have `a=0.45` and `b=2,` giving us:

`R=sqrt(a^2+b^2)` `=sqrt(0.45^2 + 2^2)=2.05`, and

`alpha="arctan"a/b = "arctan"0.45/2` `=0.22131`

So we can write `0.45 sin 222.2t - 2 cos 222.2t` ` = - 2.05 cos(222.2t+0.22)`

This gives us (on multiplying both sides by `-1`):

`i=2.05 e^(-25t)cos(222.2t+0.22)` A

Here's the graph:

0.511.52-0.5-1-1.5ti(t)Open image in a new page

Graph of `i(t)=2.05 e^(-25t)cos(222.2t+0.22)`.

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