`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)=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)`