0:00:01.090,0:00:05.120
Okay now we want to find Thevenin[br]using the external voltage method, so

0:00:05.120,0:00:10.250
our Thevenin is going[br]to be equal to Vex/Iex.

0:00:10.250,0:00:12.340
We do is we go back to our circuit, and

0:00:12.340,0:00:16.250
we add in an arbitrary[br]voltage source we can Vex.

0:00:16.250,0:00:19.080
Since it's arbitrary,[br]we can say it's equal to 1 volt, and

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that makes our equation very[br]simple Then we will have an IEX.

0:00:23.530,0:00:25.400
We don't know what that is yet.

0:00:25.400,0:00:28.500
We solve the problem in much the same[br]way as we did before, so watch for

0:00:28.500,0:00:29.880
the differences.

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Our first equation is going to[br]be VC minus 68 volts divide by

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6 ohms plus Vc minus 4 IX

0:00:40.040,0:00:44.630
divided by 2 omes plus vc minus va,

0:00:44.630,0:00:49.850
divided by 6 omes and that's equal to[br]0.No change in our first equation.

0:00:49.850,0:00:54.465
The second equation is going to[br]be va A minus by VC divided by

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six ohms plus VA minus zero[br]divided by four ohms plus

0:00:59.735,0:01:04.694
alright if I had done BA minus BEX I would[br]of had to divide by zero because there's

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zero resistance in this arm instead what[br]I'm going to do is I'm going to add in

0:01:09.375,0:01:12.608
the outgoing current the current[br]coming were going to subtract it out.

0:01:12.608,0:01:14.925
Iex =0.

0:01:14.925,0:01:19.890
Our third equation is going to be the same[br]we had before in order to find our Ix

0:01:19.890,0:01:25.305
unknown, which will be Vc-Va[br]divided by 6 ohmes=Ix.

0:01:25.305,0:01:30.540
Now let's take stock of our knowns and[br]unknowns.

0:01:30.540,0:01:36.050
We have an unknown voltage Va and[br]Va and unknown current Ix and Iex.

0:01:36.050,0:01:38.100
So I need one more equation.

0:01:38.100,0:01:41.860
Well since this is a current I'm looking[br]for let's just put a note right here and

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I can add up my ingoing and[br]outgoing currents.

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So Ix is comming in, Iex is comming in and

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then what's going out is this current[br]right here, let's call it I4.

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And that is going to be.

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Sorry, that's going to be equal[br]to Va divided by four ohms.

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These are the incoming currents,[br]that's the outgoing current.

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Then we put this together[br]into a matrix equation,

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it's going to be a four[br]by four matrix equation.

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So equation one, two, three,[br]and four for unknowns Va.

0:02:17.466,0:02:23.230
Vc, Ix, and Iex.

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My unknowns will be Va, Vc, Ix, and

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Iex and And[br]that will have a set of four constants.

0:02:32.375,0:02:35.335
So my first equation,[br]what's multiplied by Va?

0:02:35.335,0:02:37.285
It's minus one-sixth.

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What's multiplied by Vc?

0:02:38.975,0:02:43.105
It's one-sixth plus[br]one-half plus one-sixth.

0:02:43.105,0:02:45.035
What's multiplied by Ix?

0:02:45.035,0:02:46.405
Minus four halves.

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And what's multiplied by Iex0?

0:02:48.305,0:02:49.955
What's my constant?

0:02:49.955,0:02:50.918
Take that over there.

0:02:50.918,0:02:53.630
It'll be 68/ 6.

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Ok second equation,[br]lets multiply Va 1/6 Vc

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1/6, I left out a 1/4 here.

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Then minus one times I ex and[br]nothing times no constants

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Third equation minus one sixth,[br]Vc is one sixth and

0:03:14.580,0:03:19.940
Ix is -1, the zero basically brought[br]the Ix over, zero right there.

0:03:19.940,0:03:24.730
Fourth the equation 1 times Ix,[br]1 times IEx, let's bring this over here,

0:03:24.730,0:03:30.370
it will be minus one fourth and[br]nothing times Vc and again no constant.

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When I solve this What I will find[br]is that Iex which will come out my

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vector is going to be equal[br]to be equal to 0.196.

0:03:39.880,0:03:47.219
Then, when I take our Thevenin which is[br]equal to Vex which was 1 divided by 0.196,

0:03:47.219,0:03:52.394
I will find our Thevenin Is equal to,[br]let me just invert that-

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five wholes.

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So that's how we find the Rth Thevenin[br]using the external voltage method.

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Again we could measure,[br]we could simulate or

0:04:07.180,0:04:09.050
we could calculate as we did in this case.