0:00:01.430,0:00:04.560
All right, here's another example of[br]finding the Thevenin equivalent of

0:00:04.560,0:00:06.390
this circuit.

0:00:06.390,0:00:09.150
Again, we have two values[br]that we need to find.

0:00:09.150,0:00:11.753
We need to determine[br]the open circuit voltage,

0:00:11.753,0:00:17.580
which is the voltage,[br]Across there with no load connected.

0:00:17.580,0:00:21.660
And then we also need to determine[br]the Thevenin equivalent resistance,

0:00:21.660,0:00:24.340
which in this one we'll do[br]a couple of different ways.

0:00:24.340,0:00:27.400
All right, first of all, Vth open circuit.

0:00:27.400,0:00:32.579
That's the voltage at this node,[br]noting that the current

0:00:32.579,0:00:38.300
leaving this node going in that direction,[br]call it I, is = 0.

0:00:40.000,0:00:44.685
So let's call this node here V sub A,[br]we got V open circuit there and

0:00:44.685,0:00:48.380
let's write the node[br]equations of these two nodes.

0:00:49.680,0:00:54.769
Noting first of all, before we get[br]ahead of ourselves, that this dependent

0:00:54.769,0:01:00.336
current source is pushing current into the[br]node going this direction with an amount

0:01:00.336,0:01:06.079
of 1.2 times I0, where I0 is the current[br]flowing through the 3-ohm resistor.

0:01:06.079,0:01:11.191
So before we get started, let's just note

0:01:11.191,0:01:15.885
that I0 = 15-V sub A divided by 3.

0:01:15.885,0:01:17.988
We're gonna need that[br]here in just a second.

0:01:17.988,0:01:21.112
All right, some of the current's leaving[br]V sub A going in that direction.

0:01:21.112,0:01:27.596
We have V sub A -15 divided by 3 +,[br]the current coming down to

0:01:27.596,0:01:32.648
the 6-ohm resistor is[br]gonna be VA divided by 6.

0:01:32.648,0:01:37.590
The current leaving the sub A node[br]going through the 4-ohm resistor,

0:01:37.590,0:01:42.690
is going to be VA, the voltage on this[br]side minus the voltage on that side,

0:01:42.690,0:01:46.662
which is VOC, the open circuit voltage,[br]divided by 4.

0:01:46.662,0:01:53.618
Now we have coming into the node,[br]it's gonna be -1.2 times I0.

0:01:53.618,0:02:01.120
But I0 we have already expressed[br]as 15 -VA divided by 3.

0:02:01.120,0:02:05.990
So there's one, two, three, four branches,[br]one, two, three, four terms,

0:02:05.990,0:02:08.683
the sum of those four[br]things needs to be =0.

0:02:08.683,0:02:14.271
Right in the node equation at the other[br]node there which we're calling VOC,

0:02:14.271,0:02:16.551
we have two nonzero currents.

0:02:16.551,0:02:21.798
We've got the current leading[br]this node going to the 4 ohm

0:02:21.798,0:02:27.152
resistor in this direction is[br]Voc-V sub A divided by 4 plus

0:02:27.152,0:02:33.500
the current leaving it's not going[br]that way which is 1.2 time I0.

0:02:33.500,0:02:39.135
So +1.2 times I0 which[br]we've already express as 15

0:02:39.135,0:02:44.425
minus V sub A divided by 3,[br]as we've all ready noted,

0:02:44.425,0:02:49.150
the current leaving going[br]this direction is 0.

0:02:49.150,0:02:53.534
So, the sum of those two[br]currents must equal 0.

0:02:53.534,0:02:59.040
We're gonna skip the grimy algebra,

0:02:59.040,0:03:05.062
but after you've simplified this down,

0:03:05.062,0:03:10.052
you come up with VA (1.15) +

0:03:10.052,0:03:14.367
VOC(-0.25) = 11.

0:03:14.367,0:03:21.351
The top equation simplifies to that,

0:03:21.351,0:03:27.900
and the second equation simplifies

0:03:27.900,0:03:32.919
to VA(- 0.65 )+ Voc

0:03:32.919,0:03:37.296
(0.25) = -6.

0:03:37.296,0:03:41.225
And when you plug that in[br]your matrix solver and

0:03:41.225,0:03:46.360
however you do it, you can in[br]that V sub A equals 10 volts and

0:03:46.360,0:03:50.820
VOC which is our Thevenin[br]voltage equals two volts.

0:03:54.690,0:03:57.620
Now we have three different methods[br]of finding the Thevenin resistance.

0:03:58.625,0:04:01.785
The first method would be to[br]short out the terminals AB and

0:04:01.785,0:04:03.965
determine the short circuit current, and

0:04:03.965,0:04:07.165
then our Thevenin equals V open[br]circuit divided by I short circuit.

0:04:07.165,0:04:08.965
We're gonna do that here in just a minute.

0:04:08.965,0:04:13.295
The second method only works when you[br]have independent sources only, or

0:04:13.295,0:04:15.135
when you only have independent sources.

0:04:15.135,0:04:18.745
The fact that we have this[br]dependent source here makes it so

0:04:18.745,0:04:20.870
that method 2 doesn't work.

0:04:20.870,0:04:24.140
Method 3 is also gonna work,[br]we'll do that here in just a second.

0:04:24.140,0:04:27.312
Method 3 involves shorting[br]out this voltage or

0:04:27.312,0:04:32.460
deactivating the voltage source and[br]then applying a test voltage.

0:04:32.460,0:04:36.550
So let's start with[br]the short circuit method.

0:04:36.550,0:04:42.700
Let's short this out,[br]call that I short circuit and

0:04:42.700,0:04:48.490
now determine the current[br]in that short circuit.

0:04:48.490,0:04:50.170
We need to note a couple of things.

0:04:50.170,0:04:54.280
First of all, as we short this out,

0:04:54.280,0:04:59.489
we now know that[br]the voltage there equals 0.

0:04:59.489,0:05:03.546
So we only have one node where[br]we have an unknown voltage.

0:05:03.546,0:05:08.725
We're gonna calculate V sub A and[br]once we know V sub A,

0:05:08.725,0:05:15.300
we'll then write a node equation[br]at this node noting that V0 = 0.

0:05:15.300,0:05:17.541
But the equation will then be[br]in terms of I short circuit,

0:05:17.541,0:05:19.750
and we'll be able to[br]calculate I short circuit.

0:05:19.750,0:05:20.410
Let's go ahead and do it.

0:05:20.410,0:05:24.980
It's probably easier to show it[br]than it is to talk about it.

0:05:24.980,0:05:28.200
So let's just work on over here.

0:05:28.200,0:05:33.240
Write in the note equation at the A node,[br]we have

0:05:33.240,0:05:37.896
V sub A minus 15 divided by 3,

0:05:37.896,0:05:43.436
plus V sub A divided by 6,[br]plus V sub A divided by four,

0:05:43.436,0:05:49.400
minus 1.2 times,

0:05:49.400,0:05:54.730
now notice that I zero, even though we've

0:05:54.730,0:05:59.980
shorted this out I zero is still[br]15 minus V sub A divided by three.

0:05:59.980,0:06:07.581
So it would be minus one[br]point two times (15-Va/3) and

0:06:07.581,0:06:12.660
the sum of those terms has to equal zero.

0:06:12.660,0:06:17.879
Now when you go through and[br]simplify that and

0:06:17.879,0:06:24.522
solve for V sub A you get[br]that V sub A = 9.565 Volts.

0:06:27.915,0:06:32.339
Now that we know V sub A with[br]this output short circuited,

0:06:32.339,0:06:36.500
we can go ahead and[br]write a node equation at this node.

0:06:37.900,0:06:42.676
We'll do it in terms of V sub A,[br]but A sub A is gonna equal 9.565.

0:06:42.676,0:06:44.510
Let's just do it and[br]show you what we get here.

0:06:45.860,0:06:49.580
So we're going to have the current leaving[br]this node going in that direction.

0:06:49.580,0:06:54.988
It's going to be 0, oops.

0:06:54.988,0:06:58.350
Yeah 0, the voltage there is 0.

0:06:58.350,0:07:05.572
0- v sub a, which is 9.565 divided by 4.

0:07:07.020,0:07:11.570
Plus the current going in this

0:07:12.650,0:07:17.710
direction, which is 1.2 times I0,[br]I0 is 15,

0:07:17.710,0:07:23.250
minus V sub A, so that's going

0:07:23.250,0:07:30.150
to be 15 minus 9.565, divided by 3.

0:07:30.150,0:07:32.820
Then we have I short-circuit leaving.

0:07:32.820,0:07:35.721
So it would be plus I[br]short-circuit equals 0.

0:07:35.721,0:07:40.163
Well let's just bring that short circuit[br]on the other side as a equals minus I,

0:07:40.163,0:07:42.920
short-circuit, running out of room here.

0:07:42.920,0:07:47.910
All right,[br]let's continue on now and solve for

0:07:47.910,0:07:52.275
I short-circuit, and when you do that,

0:07:52.275,0:07:58.040
you get I short circuit[br]is equal to 0.2173 Amps.

0:07:58.040,0:08:01.714
We now know V open circuit,[br]V open circuit is 2.

0:08:01.714,0:08:07.333
I short circuit is that[br]R Thevenin is equal

0:08:07.333,0:08:13.594
to the ratio of V open[br]circuit I short circuit,

0:08:13.594,0:08:19.537
which equals 2 divided by 0.2173 and

0:08:19.537,0:08:23.084
that equals 9.2 ohms.

0:08:23.084,0:08:29.179
And we can now draw our[br]Thevenin equivalent circuit,

0:08:29.179,0:08:32.972
it consists of a two volt source,

0:08:32.972,0:08:37.712
with a 9.2 ohm resistor in series, and

0:08:37.712,0:08:42.340
there are our a and b terminals.

0:08:42.340,0:08:45.460
Thus, this nice concise little circuit.

0:08:45.460,0:08:49.020
If all we're interested in is[br]the terminal characteristics,

0:08:49.020,0:08:52.740
this models this more complex circuit.

0:08:54.810,0:08:58.970
Now lets just show that we get the same[br]R Thevenin using method three.

0:08:58.970,0:09:03.070
Method three requires us to[br]deactivate the independent sources.

0:09:03.070,0:09:07.750
So this is a voltage source return it to[br]zero volts, that is effectively shorting

0:09:07.750,0:09:13.410
it out and we then apply a test[br]voltage called a V test.

0:09:16.430,0:09:24.630
Which then drives a current[br]I test into the circuit.

0:09:24.630,0:09:29.763
We have two nodes we got the node here,[br]a node here we will call this one again

0:09:29.763,0:09:34.676
V sub A noting that this V sub A is[br]different than the V sub A we did before.

0:09:34.676,0:09:39.200
The V sub A that we did before had a 15[br]volt source connected at this point.

0:09:39.200,0:09:42.350
This circuit has that 15[br]volts source shorted out.

0:09:42.350,0:09:44.427
So we gonna use the same name V sub A and

0:09:44.427,0:09:47.110
over here this is no[br]longer 0 this is now v10.

0:09:47.110,0:09:52.383
Let's go ahead and write the node

0:09:52.383,0:09:57.420
equation for this A node.

0:09:57.420,0:10:04.674
So the current leaving, noting also[br]we're gonna need this I0 in this case,

0:10:04.674,0:10:09.809
is the current going in that[br]direction that's going to

0:10:09.809,0:10:15.859
equal 0-V sub A divided by 3, or[br]just negative V sub A over 3.

0:10:15.859,0:10:18.410
All right,[br]now let's write the node equation there.

0:10:18.410,0:10:24.206
The current leaving this[br]node going in this direction

0:10:24.206,0:10:29.484
is going to be Va- 0[br]because that's now shorted

0:10:29.484,0:10:34.920
out divided by 3 + Va[br]divided by 6 + Va minus.

0:10:34.920,0:10:37.640
We're now talking about the current[br]leaving this node going

0:10:37.640,0:10:38.850
through the four ohm resistor.

0:10:38.850,0:10:42.890
It's gonna be V sub a minus Vtest

0:10:45.900,0:10:52.960
over 4 minus the current entering this[br]node through the dependent source.

0:10:52.960,0:10:58.860
It's going to be minus cuz it's[br]entering 1.2 times I zero and

0:10:58.860,0:11:03.020
I zero is negative v sub a over three.

0:11:04.540,0:11:09.730
One, two, three, four terms, the sum[br]of those four terms must equal zero.

0:11:11.430,0:11:15.290
Now, let's write the node equation[br]over here at the test note.

0:11:16.340,0:11:18.750
We have three branches, three currents,

0:11:18.750,0:11:24.430
the current leaving the node going in this[br]direction is going to be V test minus V

0:11:24.430,0:11:29.330
sub a divided by 4 plus....

0:11:29.330,0:11:34.620
Now this current here is leaving[br]the node so it would be plus 1.2 I

0:11:34.620,0:11:39.459
zero, but I zero we've determined[br]to be negative V sub a over 3e,

0:11:39.459,0:11:44.390
ngative V sub a over 3.

0:11:44.390,0:11:47.020
Now, IT is coming in, so[br]that would be negative I test, so

0:11:47.020,0:11:49.740
let's just take it to the other[br]side as a positive I test and

0:11:49.740,0:11:54.110
say at the end of those two terms[br]will equal positive I test.

0:11:54.110,0:11:57.740
Now you notice we have three unknowns,[br]V sub a,

0:11:57.740,0:12:03.270
v test, and I test, but[br]have only two equations.

0:12:03.270,0:12:04.660
But that's all right,[br]cuz what we're gonna try and

0:12:04.660,0:12:09.700
do is through algebraic manipulation, come[br]up with an expression here in this upper

0:12:09.700,0:12:14.910
equation that gives us v[br]sub a in terms of v-test.

0:12:14.910,0:12:18.070
We'll then plug in down here in[br]the second equation for v sub a.

0:12:19.200,0:12:24.890
All the terms then will have either[br]factors of v-test or i-test.

0:12:24.890,0:12:29.520
We will factor out the v-test to be able[br]to make the ratio of v-test over i-test

0:12:29.520,0:12:36.250
and again that is what our Thevenin[br]is the ratio of v-test over i-test.

0:12:36.250,0:12:41.025
So this upper equation cleans

0:12:41.025,0:12:45.227
up to give us 1.15 V sub

0:12:45.227,0:12:51.026
A Is equal

0:12:51.026,0:12:55.820
to 0.25 VT.

0:12:55.820,0:13:02.636
So V sub A is equal to 0.2174v test.

0:13:02.636,0:13:08.220
Let's take those values in then,[br]and plug them in down here.

0:13:08.220,0:13:10.900
In fact, let's just clean this[br]one up a little bit down here.

0:13:10.900,0:13:18.271
Let's rewrite this bottom one as[br]if we were gonna have v-test times

0:13:18.271,0:13:23.456
one-fourth, that's the only v-test term.

0:13:23.456,0:13:30.701
Then we're gonna have plus v sub a times,

0:13:30.701,0:13:35.670
we've got a negative one and

0:13:35.670,0:13:41.260
a fourth, and we have a negative

0:13:41.260,0:13:46.870
1.2 over 3 equals I-test.

0:13:46.870,0:13:50.014
Now we'll take this expression for

0:13:50.014,0:13:55.055
V sub A in terms of V test, and[br]plug it down here for V sub A.

0:13:55.055,0:14:03.126
And we have 1/4Vt + .2174 v test,

0:14:03.126,0:14:09.179
times,a negative one fourth,

0:14:09.179,0:14:13.438
minus 1.2 over 3,

0:14:13.438,0:14:17.720
that equals, i test.

0:14:17.720,0:14:20.940
Now you go through and clean this up, and

0:14:20.940,0:14:30.050
we get 0.10869V

0:14:30.050,0:14:35.060
test equals I test.

0:14:35.060,0:14:41.020
Now, for me the ratio of V test[br]over I test gives us our Thevenin.

0:14:41.020,0:14:46.190
So R Thevenin equals V test[br]over I test which equals one

0:14:46.190,0:14:50.670
over 0.10869,

0:14:50.670,0:14:54.340
which once again equals 9.2 Ohms.

0:14:55.913,0:14:59.743
Thus we get the same R Thevenin[br]from method three as we got for

0:14:59.743,0:15:01.643
method one in the previous page,

0:15:01.643,0:15:04.173
method two doesn't work because[br]there is a depended source in there.