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`

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 *i*_{1}(0) = 0 gives:

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

The graph of our answer is:

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 *i*_{2} is:

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