WEBVTT

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>> Similar to the concept that
current division applies in

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exactly the same way or in an analogous
way as we've ever to put it in

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an analogous way to what we saw when
we had two resistances in parallel.

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So we have this current
I sub S flowing into a node.

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Part of that current is going to
come down here in flow through Z_1,

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and the other part of the current will
flow down into Z_2 and of course,

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I_1 plus I_2 must equal I sub S. So we

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have I_1 is equal to I sub S times.

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So we're talking about the current
coming down here and just as it

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was when we had resistances in series,

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this current here was proportional
to this impedance now.

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So the larger this impedance,

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the greater the current flowing down here.

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You can think of it as the more restricted
this path is for a current to flow,

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the more current you'll have over here.

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So I_1 is proportional to Z_2,

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and it is then I_1 equals I sub S
times Z_2 over Z_1 plus Z_2.

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Similarly, I_2, the current flowing
through here is going to be proportional

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to Z_1 or times I sub S
times Z_1 over Z_1 plus Z_2,

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and again, it's a pretty simple
exercise to show that I_1

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plus I_2 must equal I sub S. That comes
just from Kirchhoff's Current Law.

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Again let's just do an example
using I sub S equals that Z_1,

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Z_2, the same Z_1 and Z_2
we were using before.

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So in this case I_1 is going to
equal I sub S which is 2e to

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the j25 times Z_2 which is

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5 minus j divided by the sum

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Z_1 plus Z_2 which is
three plus j2 plus 5 minus j.

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When you do the calculations on that,

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you get then that I_1 is equal to

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1.26e to the j6.57 degrees.

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That's I_1 and we can do I_2 right here,

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I_2 then is equal to I sub S
to e to the j25 times Z_1,

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which is 3 plus j2 over Z_1 plus Z_2,

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or 3 plus j2 plus 5 minus j.

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You did the calculations on that,

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and you get that I_2 is equal to

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0.894e to

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the j51.57.

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Again, I'll leave it to you to show
that I_1 plus I_2 is equal to I sub S,

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which was 2e to the j25.