NOTE: We could either:

It is easier in this example to do the second method. In many examples, it is easier to do the first method.

For the first loop, we have:




`2i_1+2i_2+inti_1=1\ \ \ ...(1)`

For the second loop, we have:




`i_2=1/10(1-2i_1)\ \ \ ...(2)`

Substituting `(2)` into `(1)` gives:




Multiply throughout by `5`:


Next we take the Laplace Transform of both sides.




In this example, `q_0=0`. So `Lap{inti_1dt}=I_1/s`



` (8s+5)I_1=4`

` I_1=4/(8s+5)` `=1/2 1/(s+5/8)`

Now taking Inverse Laplace:


And using result (2) from above, we have:




For charge on the capacitor, we first need voltage across the capacitor:


`=5-10(1/2e^(-5/8t)+` `{:1/10(1-e^(-5/8t)))`


So, since `V_C=q/C`, we have:



`=0.8(1-e^(-5/8t))` C

Graph of q(t):

123456780. image in a new page

Graph of `q(t)=0.8(1-e^(-5/8t))`.

The charge builds up to `0.8` C (but never quite reaches it).