We have not seen how to solve "2 mesh" networks before. We consider the total voltage of the inner loop and the total voltage of the outer loop. We then solve the resulting two equations simultaneously.

We use the basic formula: `Ri+L(di)/(dt)=V`

Parallel RL circuit diagram

Considering the inner loop:

`10(i_1+i_2)+5i_1+0.01(di_1)/(dt)=` `150 sin 1000t`

`15\ i_1+10\ i_2+0.01(di_1)/(dt)=` `150 sin 1000t`

`3i_1+2i_2+0.002(di_1)/(dt)=` `30 sin 1000t\ \ \ ...(1)`

Now, considering the outer loop:

`10(i_1+i_2)+5i_2=150 sin 1000t`

`10i_1+15i_2=150 sin 1000t`

`2i_1+3i_2=30 sin 1000t\ \ \ ...(2)`

We now solve (1) and (2) simultaneously:

(1) × 3 − (2) × 2 gives:

`5i_1+0.006(di_1)/(dt)=30 sin 1000t`

Solving this using SNB with the boundary condition i1(0) = 0 gives:

`i_1(t)=-2.95 cos 1000t+` `2.46 sin 1000t+` `2.95e^(-833t)`

The graph of our answer is:

0.010.0224-2-4tiOpen image in a new page

Graph of current `i_1` at time `t`. It's in steady state by around `t=0.007`.

Now, from equation (2), we have:

`i_2=1/3(30 sin 1000t-2i_1)`

`=1/3(30 sin 1000t-` `2[-2.95 cos 1000t+` `2.46 sin 1000t+` `{:{:2.95e^(-833t)])`

`=8.36 sin 1000t+` `1.97 cos 1000t-` `1.97e^(-833t)`

The graph of i2 is:

0.010.022468-2-4-6-8tiOpen image in a new page

Graph of current `i_2` at time `t`. It's also in steady state by around `t=0.007`.