Using `L(di)/(dt)+Ri+1/Cq=E` from Section 8, we can write the DE in i and q as follows:

`(di)/(dt)+10i+50q=100\ \ \ (1)`

Differentiating gives a 2nd order DE in i:

`(d^2i)/(dt^2)+10(di)/(dt)+50i=0`

Auxiliary equation: `m^2+10m+50=0`

Solution is: `m_1=-5+5j,` `\ m_2=-5-5j`

So

`i=e^(-5t)(A\ cos 5t+B\ sin 5t)`

Now

`[i]_(t=0)=A=0`

(This means at `t = 0`, `i = A = 0` in this case.)

So

`i=e^(-5t)B\ sin 5t`

We need to find the value of B.

Differentiating gives:

`(di)/(dt)=e^(-5t)(5 B\ cos 5t)+` `(B sin 5t)(-5e^(-5t))`

`=5Be^(-5t)(cos 5t-sin 5t)`

At `t = 0`, `(di)/(dt)=5B`

Returning to equation (1): `(di)/(dt)+10i+50q=100`

Now, at time `t = 0`,

`[(di)/(dt)]_(t=0)+10(0)+50(0.1)=100`

So `[(di)/(dt)]_(t=0)=95=5B`

Therefore `B = 19`.

So we have our solution for the current `i`:

`i=19e^(-5t)sin 5t`

Here is the graph of the current at time `t`.

1234123456tiOpen image in a new page

Graph of `i(t)=19e^(-5t)sin 5t`.

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