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:

`R(i_1+i_2)+1/Cinti_1=E`

`10(i_1+i_2)+1/0.2inti_1=5`

`10(i_1+i_2)+5inti_1=5`

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


For the second loop, we have:

`10(i_1+i_2)+40i_2=5`

`10i_1+50i_2=5`

`2i_1+10i_2=1`

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


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

`2i_1+1/5(1-2i_1)+inti_1dt=1`

Simplifying:

`8/5i_1+inti_1dt=4/5`

Multiply throughout by `5`:

`8i_1+5inti_1dt=4`

Next we take the Laplace Transform of both sides.

Note:

`Lap{inti_1dt}=I_1/s+1/s[inti_1(t)dt]_(t=0)`

`=I_1/s+q_0/s`

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

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

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

` (8s+5)I_1=4`

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

Now taking Inverse Laplace:

`i_1=1/2e^(-5/8t)`

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

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

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

`=1/10(1-e^(-5/8t))`

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

`V_C=5-10(i_1+i_2)`

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

`=4-4e^(-5/8t)`

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

`q=CxxV_C`

`=0.2xx(4-4e^(-5/8t))`

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

Graph of q(t):

123456780.20.40.60.81tq(t)Open 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).