WEBVTT

00:00:01.430 --> 00:00:04.560
All right, here's another example of
finding the Thevenin equivalent of

00:00:04.560 --> 00:00:06.390
this circuit.

00:00:06.390 --> 00:00:09.150
Again, we have two values
that we need to find.

00:00:09.150 --> 00:00:11.753
We need to determine
the open circuit voltage,

00:00:11.753 --> 00:00:17.580
which is the voltage,
Across there with no load connected.

00:00:17.580 --> 00:00:21.660
And then we also need to determine
the Thevenin equivalent resistance,

00:00:21.660 --> 00:00:24.340
which in this one we'll do
a couple of different ways.

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

00:00:27.400 --> 00:00:32.579
That's the voltage at this node,
noting that the current

00:00:32.579 --> 00:00:38.300
leaving this node going in that direction,
call it I, is = 0.

00:00:40.000 --> 00:00:44.685
So let's call this node here V sub A,
we got V open circuit there and

00:00:44.685 --> 00:00:48.380
let's write the node
equations of these two nodes.

00:00:49.680 --> 00:00:54.769
Noting first of all, before we get
ahead of ourselves, that this dependent

00:00:54.769 --> 00:01:00.336
current source is pushing current into the
node going this direction with an amount

00:01:00.336 --> 00:01:06.079
of 1.2 times I0, where I0 is the current
flowing through the 3-ohm resistor.

00:01:06.079 --> 00:01:11.191
So before we get started, let's just note

00:01:11.191 --> 00:01:15.885
that I0 = 15-V sub A divided by 3.

00:01:15.885 --> 00:01:17.988
We're gonna need that
here in just a second.

00:01:17.988 --> 00:01:21.112
All right, some of the current's leaving
V sub A going in that direction.

00:01:21.112 --> 00:01:27.596
We have V sub A -15 divided by 3 +,
the current coming down to

00:01:27.596 --> 00:01:32.648
the 6-ohm resistor is
gonna be VA divided by 6.

00:01:32.648 --> 00:01:37.590
The current leaving the sub A node
going through the 4-ohm resistor,

00:01:37.590 --> 00:01:42.690
is going to be VA, the voltage on this
side minus the voltage on that side,

00:01:42.690 --> 00:01:46.662
which is VOC, the open circuit voltage,
divided by 4.

00:01:46.662 --> 00:01:53.618
Now we have coming into the node,
it's gonna be -1.2 times I0.

00:01:53.618 --> 00:02:01.120
But I0 we have already expressed
as 15 -VA divided by 3.

00:02:01.120 --> 00:02:05.990
So there's one, two, three, four branches,
one, two, three, four terms,

00:02:05.990 --> 00:02:08.683
the sum of those four
things needs to be =0.

00:02:08.683 --> 00:02:14.271
Right in the node equation at the other
node there which we're calling VOC,

00:02:14.271 --> 00:02:16.551
we have two nonzero currents.

00:02:16.551 --> 00:02:21.798
We've got the current leading
this node going to the 4 ohm

00:02:21.798 --> 00:02:27.152
resistor in this direction is
Voc-V sub A divided by 4 plus

00:02:27.152 --> 00:02:33.500
the current leaving it's not going
that way which is 1.2 time I0.

00:02:33.500 --> 00:02:39.135
So +1.2 times I0 which
we've already express as 15

00:02:39.135 --> 00:02:44.425
minus V sub A divided by 3,
as we've all ready noted,

00:02:44.425 --> 00:02:49.150
the current leaving going
this direction is 0.

00:02:49.150 --> 00:02:53.534
So, the sum of those two
currents must equal 0.

00:02:53.534 --> 00:02:59.040
We're gonna skip the grimy algebra,

00:02:59.040 --> 00:03:05.062
but after you've simplified this down,

00:03:05.062 --> 00:03:10.052
you come up with VA (1.15) +

00:03:10.052 --> 00:03:14.367
VOC(-0.25) = 11.

00:03:14.367 --> 00:03:21.351
The top equation simplifies to that,

00:03:21.351 --> 00:03:27.900
and the second equation simplifies

00:03:27.900 --> 00:03:32.919
to VA(- 0.65 )+ Voc

00:03:32.919 --> 00:03:37.296
(0.25) = -6.

00:03:37.296 --> 00:03:41.225
And when you plug that in
your matrix solver and

00:03:41.225 --> 00:03:46.360
however you do it, you can in
that V sub A equals 10 volts and

00:03:46.360 --> 00:03:50.820
VOC which is our Thevenin
voltage equals two volts.

00:03:54.690 --> 00:03:57.620
Now we have three different methods
of finding the Thevenin resistance.

00:03:58.625 --> 00:04:01.785
The first method would be to
short out the terminals AB and

00:04:01.785 --> 00:04:03.965
determine the short circuit current, and

00:04:03.965 --> 00:04:07.165
then our Thevenin equals V open
circuit divided by I short circuit.

00:04:07.165 --> 00:04:08.965
We're gonna do that here in just a minute.

00:04:08.965 --> 00:04:13.295
The second method only works when you
have independent sources only, or

00:04:13.295 --> 00:04:15.135
when you only have independent sources.

00:04:15.135 --> 00:04:18.745
The fact that we have this
dependent source here makes it so

00:04:18.745 --> 00:04:20.870
that method 2 doesn't work.

00:04:20.870 --> 00:04:24.140
Method 3 is also gonna work,
we'll do that here in just a second.

00:04:24.140 --> 00:04:27.312
Method 3 involves shorting
out this voltage or

00:04:27.312 --> 00:04:32.460
deactivating the voltage source and
then applying a test voltage.

00:04:32.460 --> 00:04:36.550
So let's start with
the short circuit method.

00:04:36.550 --> 00:04:42.700
Let's short this out,
call that I short circuit and

00:04:42.700 --> 00:04:48.490
now determine the current
in that short circuit.

00:04:48.490 --> 00:04:50.170
We need to note a couple of things.

00:04:50.170 --> 00:04:54.280
First of all, as we short this out,

00:04:54.280 --> 00:04:59.489
we now know that
the voltage there equals 0.

00:04:59.489 --> 00:05:03.546
So we only have one node where
we have an unknown voltage.

00:05:03.546 --> 00:05:08.725
We're gonna calculate V sub A and
once we know V sub A,

00:05:08.725 --> 00:05:15.300
we'll then write a node equation
at this node noting that V0 = 0.

00:05:15.300 --> 00:05:17.541
But the equation will then be
in terms of I short circuit,

00:05:17.541 --> 00:05:19.750
and we'll be able to
calculate I short circuit.

00:05:19.750 --> 00:05:20.410
Let's go ahead and do it.

00:05:20.410 --> 00:05:24.980
It's probably easier to show it
than it is to talk about it.

00:05:24.980 --> 00:05:28.200
So let's just work on over here.

00:05:28.200 --> 00:05:33.240
Write in the note equation at the A node,
we have

00:05:33.240 --> 00:05:37.896
V sub A minus 15 divided by 3,

00:05:37.896 --> 00:05:43.436
plus V sub A divided by 6,
plus V sub A divided by four,

00:05:43.436 --> 00:05:49.400
minus 1.2 times,

00:05:49.400 --> 00:05:54.730
now notice that I zero, even though we've

00:05:54.730 --> 00:05:59.980
shorted this out I zero is still
15 minus V sub A divided by three.

00:05:59.980 --> 00:06:07.581
So it would be minus one
point two times (15-Va/3) and

00:06:07.581 --> 00:06:12.660
the sum of those terms has to equal zero.

00:06:12.660 --> 00:06:17.879
Now when you go through and
simplify that and

00:06:17.879 --> 00:06:24.522
solve for V sub A you get
that V sub A = 9.565 Volts.

00:06:27.915 --> 00:06:32.339
Now that we know V sub A with
this output short circuited,

00:06:32.339 --> 00:06:36.500
we can go ahead and
write a node equation at this node.

00:06:37.900 --> 00:06:42.676
We'll do it in terms of V sub A,
but A sub A is gonna equal 9.565.

00:06:42.676 --> 00:06:44.510
Let's just do it and
show you what we get here.

00:06:45.860 --> 00:06:49.580
So we're going to have the current leaving
this node going in that direction.

00:06:49.580 --> 00:06:54.988
It's going to be 0, oops.

00:06:54.988 --> 00:06:58.350
Yeah 0, the voltage there is 0.

00:06:58.350 --> 00:07:05.572
0- v sub a, which is 9.565 divided by 4.

00:07:07.020 --> 00:07:11.570
Plus the current going in this

00:07:12.650 --> 00:07:17.710
direction, which is 1.2 times I0,
I0 is 15,

00:07:17.710 --> 00:07:23.250
minus V sub A, so that's going

00:07:23.250 --> 00:07:30.150
to be 15 minus 9.565, divided by 3.

00:07:30.150 --> 00:07:32.820
Then we have I short-circuit leaving.

00:07:32.820 --> 00:07:35.721
So it would be plus I
short-circuit equals 0.

00:07:35.721 --> 00:07:40.163
Well let's just bring that short circuit
on the other side as a equals minus I,

00:07:40.163 --> 00:07:42.920
short-circuit, running out of room here.

00:07:42.920 --> 00:07:47.910
All right,
let's continue on now and solve for

00:07:47.910 --> 00:07:52.275
I short-circuit, and when you do that,

00:07:52.275 --> 00:07:58.040
you get I short circuit
is equal to 0.2173 Amps.

00:07:58.040 --> 00:08:01.714
We now know V open circuit,
V open circuit is 2.

00:08:01.714 --> 00:08:07.333
I short circuit is that
R Thevenin is equal

00:08:07.333 --> 00:08:13.594
to the ratio of V open
circuit I short circuit,

00:08:13.594 --> 00:08:19.537
which equals 2 divided by 0.2173 and

00:08:19.537 --> 00:08:23.084
that equals 9.2 ohms.

00:08:23.084 --> 00:08:29.179
And we can now draw our
Thevenin equivalent circuit,

00:08:29.179 --> 00:08:32.972
it consists of a two volt source,

00:08:32.972 --> 00:08:37.712
with a 9.2 ohm resistor in series, and

00:08:37.712 --> 00:08:42.340
there are our a and b terminals.

00:08:42.340 --> 00:08:45.460
Thus, this nice concise little circuit.

00:08:45.460 --> 00:08:49.020
If all we're interested in is
the terminal characteristics,

00:08:49.020 --> 00:08:52.740
this models this more complex circuit.

00:08:54.810 --> 00:08:58.970
Now lets just show that we get the same
R Thevenin using method three.

00:08:58.970 --> 00:09:03.070
Method three requires us to
deactivate the independent sources.

00:09:03.070 --> 00:09:07.750
So this is a voltage source return it to
zero volts, that is effectively shorting

00:09:07.750 --> 00:09:13.410
it out and we then apply a test
voltage called a V test.

00:09:16.430 --> 00:09:24.630
Which then drives a current
I test into the circuit.

00:09:24.630 --> 00:09:29.763
We have two nodes we got the node here,
a node here we will call this one again

00:09:29.763 --> 00:09:34.676
V sub A noting that this V sub A is
different than the V sub A we did before.

00:09:34.676 --> 00:09:39.200
The V sub A that we did before had a 15
volt source connected at this point.

00:09:39.200 --> 00:09:42.350
This circuit has that 15
volts source shorted out.

00:09:42.350 --> 00:09:44.427
So we gonna use the same name V sub A and

00:09:44.427 --> 00:09:47.110
over here this is no
longer 0 this is now v10.

00:09:47.110 --> 00:09:52.383
Let's go ahead and write the node

00:09:52.383 --> 00:09:57.420
equation for this A node.

00:09:57.420 --> 00:10:04.674
So the current leaving, noting also
we're gonna need this I0 in this case,

00:10:04.674 --> 00:10:09.809
is the current going in that
direction that's going to

00:10:09.809 --> 00:10:15.859
equal 0-V sub A divided by 3, or
just negative V sub A over 3.

00:10:15.859 --> 00:10:18.410
All right,
now let's write the node equation there.

00:10:18.410 --> 00:10:24.206
The current leaving this
node going in this direction

00:10:24.206 --> 00:10:29.484
is going to be Va- 0
because that's now shorted

00:10:29.484 --> 00:10:34.920
out divided by 3 + Va
divided by 6 + Va minus.

00:10:34.920 --> 00:10:37.640
We're now talking about the current
leaving this node going

00:10:37.640 --> 00:10:38.850
through the four ohm resistor.

00:10:38.850 --> 00:10:42.890
It's gonna be V sub a minus Vtest

00:10:45.900 --> 00:10:52.960
over 4 minus the current entering this
node through the dependent source.

00:10:52.960 --> 00:10:58.860
It's going to be minus cuz it's
entering 1.2 times I zero and

00:10:58.860 --> 00:11:03.020
I zero is negative v sub a over three.

00:11:04.540 --> 00:11:09.730
One, two, three, four terms, the sum
of those four terms must equal zero.

00:11:11.430 --> 00:11:15.290
Now, let's write the node equation
over here at the test note.

00:11:16.340 --> 00:11:18.750
We have three branches, three currents,

00:11:18.750 --> 00:11:24.430
the current leaving the node going in this
direction is going to be V test minus V

00:11:24.430 --> 00:11:29.330
sub a divided by 4 plus....

00:11:29.330 --> 00:11:34.620
Now this current here is leaving
the node so it would be plus 1.2 I

00:11:34.620 --> 00:11:39.459
zero, but I zero we've determined
to be negative V sub a over 3e,

00:11:39.459 --> 00:11:44.390
ngative V sub a over 3.

00:11:44.390 --> 00:11:47.020
Now, IT is coming in, so
that would be negative I test, so

00:11:47.020 --> 00:11:49.740
let's just take it to the other
side as a positive I test and

00:11:49.740 --> 00:11:54.110
say at the end of those two terms
will equal positive I test.

00:11:54.110 --> 00:11:57.740
Now you notice we have three unknowns,
V sub a,

00:11:57.740 --> 00:12:03.270
v test, and I test, but
have only two equations.

00:12:03.270 --> 00:12:04.660
But that's all right,
cuz what we're gonna try and

00:12:04.660 --> 00:12:09.700
do is through algebraic manipulation, come
up with an expression here in this upper

00:12:09.700 --> 00:12:14.910
equation that gives us v
sub a in terms of v-test.

00:12:14.910 --> 00:12:18.070
We'll then plug in down here in
the second equation for v sub a.

00:12:19.200 --> 00:12:24.890
All the terms then will have either
factors of v-test or i-test.

00:12:24.890 --> 00:12:29.520
We will factor out the v-test to be able
to make the ratio of v-test over i-test

00:12:29.520 --> 00:12:36.250
and again that is what our Thevenin
is the ratio of v-test over i-test.

00:12:36.250 --> 00:12:41.025
So this upper equation cleans

00:12:41.025 --> 00:12:45.227
up to give us 1.15 V sub

00:12:45.227 --> 00:12:51.026
A Is equal

00:12:51.026 --> 00:12:55.820
to 0.25 VT.

00:12:55.820 --> 00:13:02.636
So V sub A is equal to 0.2174v test.

00:13:02.636 --> 00:13:08.220
Let's take those values in then,
and plug them in down here.

00:13:08.220 --> 00:13:10.900
In fact, let's just clean this
one up a little bit down here.

00:13:10.900 --> 00:13:18.271
Let's rewrite this bottom one as
if we were gonna have v-test times

00:13:18.271 --> 00:13:23.456
one-fourth, that's the only v-test term.

00:13:23.456 --> 00:13:30.701
Then we're gonna have plus v sub a times,

00:13:30.701 --> 00:13:35.670
we've got a negative one and

00:13:35.670 --> 00:13:41.260
a fourth, and we have a negative

00:13:41.260 --> 00:13:46.870
1.2 over 3 equals I-test.

00:13:46.870 --> 00:13:50.014
Now we'll take this expression for

00:13:50.014 --> 00:13:55.055
V sub A in terms of V test, and
plug it down here for V sub A.

00:13:55.055 --> 00:14:03.126
And we have 1/4Vt + .2174 v test,

00:14:03.126 --> 00:14:09.179
times,a negative one fourth,

00:14:09.179 --> 00:14:13.438
minus 1.2 over 3,

00:14:13.438 --> 00:14:17.720
that equals, i test.

00:14:17.720 --> 00:14:20.940
Now you go through and clean this up, and

00:14:20.940 --> 00:14:30.050
we get 0.10869V

00:14:30.050 --> 00:14:35.060
test equals I test.

00:14:35.060 --> 00:14:41.020
Now, for me the ratio of V test
over I test gives us our Thevenin.

00:14:41.020 --> 00:14:46.190
So R Thevenin equals V test
over I test which equals one

00:14:46.190 --> 00:14:50.670
over 0.10869,

00:14:50.670 --> 00:14:54.340
which once again equals 9.2 Ohms.

00:14:55.913 --> 00:14:59.743
Thus we get the same R Thevenin
from method three as we got for

00:14:59.743 --> 00:15:01.643
method one in the previous page,

00:15:01.643 --> 00:15:04.173
method two doesn't work because
there is a depended source in there.