`1/Cinti\ dt+Ri=V`

`1/10^(-6) inti\ dt+10^(3)i=5`

Multiplying throughout by 10−6 gives:

`inti\ dt+10^(-3)i=5xx10^(-6)`

Taking Laplace transform:

`(I/s+1/s[inti\ dt]_(t=0))+10^(-3)I` `=(5xx10^(-6))/s`

Now in this example, we are told `Vc(0)=1.0`.

So

`Vc(0)=1/C[intidt]_(t=0)` `=1/C[intidt]_(t=0)=1`

That is: `1/10^(-6)[inti\ dt]_(t=0)=1` Therefore: `[inti\ dt]_(t=0)=10^(-6)`

NOTE: `[intidt]_(t=0)=q_0`.

`(I/s+10^(-6)/s)+10^(-3)I=(5xx10^(-6))/s`

Collecting I terms and subtracting `10^(-6)/s` from both sides:

`(1/s+10^(-3))I` `=(5xx10^(-6))/s-10^(-6)/s` `=(4xx10^(-6))/s`

Multiply throughout by `s`:

` (1+10^(-3)s)I=4xx10^(-6)`

Solve for `I`:

` I=(4xx10^(-6))/(1+10^(-3)s)` `=(4xx10^(-3)) 1/(1000+s)`

Finding the inverse Laplace transform gives us the current at time `t`:

`i=4xx10^(-3)e^(-1000t)`

0.0050.01-0.0050.0010.0020.0030.0040.005ti(t)Open image in a new page

Graph of `i=4xx10^(-3)e^(-1000t)`.