NOTE: We could either:

• Set up the equations, take Laplace of each, then solve simultaneously; or
• Set up the equations, solve simultaneously, then take Laplace.

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):

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).