We solve this 2 ways:

1. Setting up the equations and getting SNB to help solve them.

2. Directly using SNB to solve the 2 equations simultaneously.

Solution 1

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

Parallel circuit diagram

Considering the left-hand loop, the flow of current through the 8 Ω resistor is opposite for `i_1` and `i_2`. We regard `i_1` as having positive direction:

`0.2(di_1)/(dt)+8(i_1-i_2)=` `30 sin 100t\ \ \ ...(1)`

Now, we consider the right-hand loop and regard the direction of `i_2` as positive:

`8(i_2-i_1)+4i_2=0`

`12i_2-8i_1=0`

`i_2=2/3i_1\ \ \ ...(2)`

We now solve (1) and (2) simultaneously by substituting `i_2=2/3i_1` into (1) so that we get a DE in `i_1` only:

`0.2(di_1)/(dt)+8(i_1-2/3i_1)=` `30 sin 100t`

`0.2(di_1)/(dt)+8/3i_1=30 sin 100t`

Solving using Scientific Notebook gives:

`i_1(t)` `=-1.474 cos 100t+` `0.197 sin 100t+1.474e^(-13.3t)`

The graph of our solution is:

0.10.20.30.40.512-1tiOpen image in a new page

Graph of current `i_1` at time `t`. It's in steady state by around `t=0.25`.

Now, from equation (2), we have:

`i_2=2/3i_1`

`=2/3(-1.474 cos 100t+` `0.197 sin 100t+` `{:1.474e^(-13.3t))`

`=-0.983 cos 100t+` `0.131 sin 100t+` `0.983e^(-13.3t)`

This is of course the same graph, only it's `2/3` of the amplitude:

0.10.20.30.40.512-1tiOpen image in a new page

Graph of current `i_2` at time `t`. It's also in steady state by around `t=0.25`.

Solution 2 - Using SNB directly

If we try to solve it using Scientific Notebook as follows, it fails because it can only solve 2 differential equations simultaneously (the second line is not a differential equation):

`0.2(di_1)/(dt)+8(i_1-i_2)=30 sin 100t`

` i_2=2/3i_1`

`i_1(0)=0`

` i_2(0)=0`

But if we differentiate the second line as follows (making it into a differential equation so we have 2 DEs in 2 unknowns), SNB will happily solve it using Compute → Solve ODE... → Exact:

`0.2(di_1)/(dt)+8(i_1-i_2)=30 sin 100t`

`(di_2)/(dt)=2/3(di_1)/(dt)`

`i_1(0)=0`

`i_2(0)=0`

Exact solution is:

`i_1(t)=-4.0xx10^-9` `+1.4738 e^(-13.333t)` `-1.4738 cos 100.0t` `+0.19651 sin 100.0t`

` i_2(t)=0.98253 e^(-13.333t)` `-3.0xx10^-9` `-0.98253 cos 100.0t` `+0.131 sin 100.0t`

Note the curious extra (small) constant terms `-4.0xx10^-9` and `-3.0xx10^-9`.

Easy to understand math videos:
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