>> Similar to the concept that current division applies in
exactly the same way or in an analogous way as we've ever to put it in
an analogous way to what we saw when we had two resistances in parallel.
So we have this current I sub S flowing into a node.
Part of that current is going to come down here in flow through Z_1,
and the other part of the current will flow down into Z_2 and of course,
I_1 plus I_2 must equal I sub S. So we
have I_1 is equal to I sub S times.
So we're talking about the current coming down here and just as it
was when we had resistances in series,
this current here was proportional to this impedance now.
So the larger this impedance,
the greater the current flowing down here.
You can think of it as the more restricted this path is for a current to flow,
the more current you'll have over here.
So I_1 is proportional to Z_2,
and it is then I_1 equals I sub S times Z_2 over Z_1 plus Z_2.
Similarly, I_2, the current flowing through here is going to be proportional
to Z_1 or times I sub S times Z_1 over Z_1 plus Z_2,
and again, it's a pretty simple exercise to show that I_1
plus I_2 must equal I sub S. That comes just from Kirchhoff's Current Law.
Again let's just do an example using I sub S equals that Z_1,
Z_2, the same Z_1 and Z_2 we were using before.
So in this case I_1 is going to equal I sub S which is 2e to
the j25 times Z_2 which is
5 minus j divided by the sum
Z_1 plus Z_2 which is three plus j2 plus 5 minus j.
When you do the calculations on that,
you get then that I_1 is equal to
1.26e to the j6.57 degrees.
That's I_1 and we can do I_2 right here,
I_2 then is equal to I sub S to e to the j25 times Z_1,
which is 3 plus j2 over Z_1 plus Z_2,
or 3 plus j2 plus 5 minus j.
You did the calculations on that,
and you get that I_2 is equal to
0.894e to
the j51.57.
Again, I'll leave it to you to show that I_1 plus I_2 is equal to I sub S,
which was 2e to the j25.