All right, here's another example of finding the Thevenin equivalent of
this circuit.
Again, we have two values that we need to find.
We need to determine the open circuit voltage,
which is the voltage, Across there with no load connected.
And then we also need to determine the Thevenin equivalent resistance,
which in this one we'll do a couple of different ways.
All right, first of all, Vth open circuit.
That's the voltage at this node, noting that the current
leaving this node going in that direction, call it I, is = 0.
So let's call this node here V sub A, we got V open circuit there and
let's write the node equations of these two nodes.
Noting first of all, before we get ahead of ourselves, that this dependent
current source is pushing current into the node going this direction with an amount
of 1.2 times I0, where I0 is the current flowing through the 3-ohm resistor.
So before we get started, let's just note
that I0 = 15-V sub A divided by 3.
We're gonna need that here in just a second.
All right, some of the current's leaving V sub A going in that direction.
We have V sub A -15 divided by 3 +, the current coming down to
the 6-ohm resistor is gonna be VA divided by 6.
The current leaving the sub A node going through the 4-ohm resistor,
is going to be VA, the voltage on this side minus the voltage on that side,
which is VOC, the open circuit voltage, divided by 4.
Now we have coming into the node, it's gonna be -1.2 times I0.
But I0 we have already expressed as 15 -VA divided by 3.
So there's one, two, three, four branches, one, two, three, four terms,
the sum of those four things needs to be =0.
Right in the node equation at the other node there which we're calling VOC,
we have two nonzero currents.
We've got the current leading this node going to the 4 ohm
resistor in this direction is Voc-V sub A divided by 4 plus
the current leaving it's not going that way which is 1.2 time I0.
So +1.2 times I0 which we've already express as 15
minus V sub A divided by 3, as we've all ready noted,
the current leaving going this direction is 0.
So, the sum of those two currents must equal 0.
We're gonna skip the grimy algebra,
but after you've simplified this down,
you come up with VA (1.15) +
VOC(-0.25) = 11.
The top equation simplifies to that,
and the second equation simplifies
to VA(- 0.65 )+ Voc
(0.25) = -6.
And when you plug that in your matrix solver and
however you do it, you can in that V sub A equals 10 volts and
VOC which is our Thevenin voltage equals two volts.
Now we have three different methods of finding the Thevenin resistance.
The first method would be to short out the terminals AB and
determine the short circuit current, and
then our Thevenin equals V open circuit divided by I short circuit.
We're gonna do that here in just a minute.
The second method only works when you have independent sources only, or
when you only have independent sources.
The fact that we have this dependent source here makes it so
that method 2 doesn't work.
Method 3 is also gonna work, we'll do that here in just a second.
Method 3 involves shorting out this voltage or
deactivating the voltage source and then applying a test voltage.
So let's start with the short circuit method.
Let's short this out, call that I short circuit and
now determine the current in that short circuit.
We need to note a couple of things.
First of all, as we short this out,
we now know that the voltage there equals 0.
So we only have one node where we have an unknown voltage.
We're gonna calculate V sub A and once we know V sub A,
we'll then write a node equation at this node noting that V0 = 0.
But the equation will then be in terms of I short circuit,
and we'll be able to calculate I short circuit.
Let's go ahead and do it.
It's probably easier to show it than it is to talk about it.
So let's just work on over here.
Write in the note equation at the A node, we have
V sub A minus 15 divided by 3,
plus V sub A divided by 6, plus V sub A divided by four,
minus 1.2 times,
now notice that I zero, even though we've
shorted this out I zero is still 15 minus V sub A divided by three.
So it would be minus one point two times (15-Va/3) and
the sum of those terms has to equal zero.
Now when you go through and simplify that and
solve for V sub A you get that V sub A = 9.565 Volts.
Now that we know V sub A with this output short circuited,
we can go ahead and write a node equation at this node.
We'll do it in terms of V sub A, but A sub A is gonna equal 9.565.
Let's just do it and show you what we get here.
So we're going to have the current leaving this node going in that direction.
It's going to be 0, oops.
Yeah 0, the voltage there is 0.
0- v sub a, which is 9.565 divided by 4.
Plus the current going in this
direction, which is 1.2 times I0, I0 is 15,
minus V sub A, so that's going
to be 15 minus 9.565, divided by 3.
Then we have I short-circuit leaving.
So it would be plus I short-circuit equals 0.
Well let's just bring that short circuit on the other side as a equals minus I,
short-circuit, running out of room here.
All right, let's continue on now and solve for
I short-circuit, and when you do that,
you get I short circuit is equal to 0.2173 Amps.
We now know V open circuit, V open circuit is 2.
I short circuit is that R Thevenin is equal
to the ratio of V open circuit I short circuit,
which equals 2 divided by 0.2173 and
that equals 9.2 ohms.
And we can now draw our Thevenin equivalent circuit,
it consists of a two volt source,
with a 9.2 ohm resistor in series, and
there are our a and b terminals.
Thus, this nice concise little circuit.
If all we're interested in is the terminal characteristics,
this models this more complex circuit.
Now lets just show that we get the same R Thevenin using method three.
Method three requires us to deactivate the independent sources.
So this is a voltage source return it to zero volts, that is effectively shorting
it out and we then apply a test voltage called a V test.
Which then drives a current I test into the circuit.
We have two nodes we got the node here, a node here we will call this one again
V sub A noting that this V sub A is different than the V sub A we did before.
The V sub A that we did before had a 15 volt source connected at this point.
This circuit has that 15 volts source shorted out.
So we gonna use the same name V sub A and
over here this is no longer 0 this is now v10.
Let's go ahead and write the node
equation for this A node.
So the current leaving, noting also we're gonna need this I0 in this case,
is the current going in that direction that's going to
equal 0-V sub A divided by 3, or just negative V sub A over 3.
All right, now let's write the node equation there.
The current leaving this node going in this direction
is going to be Va- 0 because that's now shorted
out divided by 3 + Va divided by 6 + Va minus.
We're now talking about the current leaving this node going
through the four ohm resistor.
It's gonna be V sub a minus Vtest
over 4 minus the current entering this node through the dependent source.
It's going to be minus cuz it's entering 1.2 times I zero and
I zero is negative v sub a over three.
One, two, three, four terms, the sum of those four terms must equal zero.
Now, let's write the node equation over here at the test note.
We have three branches, three currents,
the current leaving the node going in this direction is going to be V test minus V
sub a divided by 4 plus....
Now this current here is leaving the node so it would be plus 1.2 I
zero, but I zero we've determined to be negative V sub a over 3e,
ngative V sub a over 3.
Now, IT is coming in, so that would be negative I test, so
let's just take it to the other side as a positive I test and
say at the end of those two terms will equal positive I test.
Now you notice we have three unknowns, V sub a,
v test, and I test, but have only two equations.
But that's all right, cuz what we're gonna try and
do is through algebraic manipulation, come up with an expression here in this upper
equation that gives us v sub a in terms of v-test.
We'll then plug in down here in the second equation for v sub a.
All the terms then will have either factors of v-test or i-test.
We will factor out the v-test to be able to make the ratio of v-test over i-test
and again that is what our Thevenin is the ratio of v-test over i-test.
So this upper equation cleans
up to give us 1.15 V sub
A Is equal
to 0.25 VT.
So V sub A is equal to 0.2174v test.
Let's take those values in then, and plug them in down here.
In fact, let's just clean this one up a little bit down here.
Let's rewrite this bottom one as if we were gonna have v-test times
one-fourth, that's the only v-test term.
Then we're gonna have plus v sub a times,
we've got a negative one and
a fourth, and we have a negative
1.2 over 3 equals I-test.
Now we'll take this expression for
V sub A in terms of V test, and plug it down here for V sub A.
And we have 1/4Vt + .2174 v test,
times,a negative one fourth,
minus 1.2 over 3,
that equals, i test.
Now you go through and clean this up, and
we get 0.10869V
test equals I test.
Now, for me the ratio of V test over I test gives us our Thevenin.
So R Thevenin equals V test over I test which equals one
over 0.10869,
which once again equals 9.2 Ohms.
Thus we get the same R Thevenin from method three as we got for
method one in the previous page,
method two doesn't work because there is a depended source in there.