1
00:00:00,000 --> 00:00:03,240
>> Similar to the concept that
current division applies in

2
00:00:03,240 --> 00:00:06,360
exactly the same way or in an analogous
way as we've ever to put it in

3
00:00:06,360 --> 00:00:12,625
an analogous way to what we saw when
we had two resistances in parallel.

4
00:00:12,625 --> 00:00:19,450
So we have this current
I sub S flowing into a node.

5
00:00:19,550 --> 00:00:26,050
Part of that current is going to
come down here in flow through Z_1,

6
00:00:26,050 --> 00:00:30,770
and the other part of the current will
flow down into Z_2 and of course,

7
00:00:30,770 --> 00:00:35,780
I_1 plus I_2 must equal I sub S. So we

8
00:00:35,780 --> 00:00:43,035
have I_1 is equal to I sub S times.

9
00:00:43,035 --> 00:00:45,350
So we're talking about the current
coming down here and just as it

10
00:00:45,350 --> 00:00:49,655
was when we had resistances in series,

11
00:00:49,655 --> 00:00:55,160
this current here was proportional
to this impedance now.

12
00:00:55,160 --> 00:00:57,820
So the larger this impedance,

13
00:00:57,820 --> 00:00:59,730
the greater the current flowing down here.

14
00:00:59,730 --> 00:01:05,135
You can think of it as the more restricted
this path is for a current to flow,

15
00:01:05,135 --> 00:01:07,415
the more current you'll have over here.

16
00:01:07,415 --> 00:01:10,845
So I_1 is proportional to Z_2,

17
00:01:10,845 --> 00:01:19,570
and it is then I_1 equals I sub S
times Z_2 over Z_1 plus Z_2.

18
00:01:19,670 --> 00:01:26,000
Similarly, I_2, the current flowing
through here is going to be proportional

19
00:01:26,000 --> 00:01:34,150
to Z_1 or times I sub S
times Z_1 over Z_1 plus Z_2,

20
00:01:34,150 --> 00:01:37,685
and again, it's a pretty simple
exercise to show that I_1

21
00:01:37,685 --> 00:01:44,690
plus I_2 must equal I sub S. That comes
just from Kirchhoff's Current Law.

22
00:01:44,690 --> 00:01:49,440
Again let's just do an example
using I sub S equals that Z_1,

23
00:01:49,440 --> 00:01:51,945
Z_2, the same Z_1 and Z_2
we were using before.

24
00:01:51,945 --> 00:01:59,060
So in this case I_1 is going to
equal I sub S which is 2e to

25
00:01:59,060 --> 00:02:07,685
the j25 times Z_2 which is

26
00:02:07,685 --> 00:02:16,350
5 minus j divided by the sum

27
00:02:16,350 --> 00:02:25,620
Z_1 plus Z_2 which is
three plus j2 plus 5 minus j.

28
00:02:25,620 --> 00:02:28,060
When you do the calculations on that,

29
00:02:28,060 --> 00:02:32,160
you get then that I_1 is equal to

30
00:02:32,160 --> 00:02:41,580
1.26e to the j6.57 degrees.

31
00:02:41,580 --> 00:02:46,215
That's I_1 and we can do I_2 right here,

32
00:02:46,215 --> 00:02:52,620
I_2 then is equal to I sub S
to e to the j25 times Z_1,

33
00:02:52,620 --> 00:02:58,470
which is 3 plus j2 over Z_1 plus Z_2,

34
00:02:58,470 --> 00:03:04,275
or 3 plus j2 plus 5 minus j.

35
00:03:04,275 --> 00:03:06,675
You did the calculations on that,

36
00:03:06,675 --> 00:03:09,030
and you get that I_2 is equal to

37
00:03:09,030 --> 00:03:14,510
0.894e to

38
00:03:14,510 --> 00:03:20,220
the j51.57.

39
00:03:20,220 --> 00:03:26,280
Again, I'll leave it to you to show
that I_1 plus I_2 is equal to I sub S,

40
00:03:26,280 --> 00:03:31,930
which was 2e to the j25.