In this video we're gonna demonstrate the technique for determining the Thevenin
equivalent circuit by looking at a relatively simple circuit and
going through the different steps you will be using to do this.
The first thing we need to determine is the open circuit voltage.
The Thevenin voltage or the voltage in our Thevenin model.
Consists of a Thevenin voltage which we pointed out is just the open
circuit voltage.
So in this case we need to know what the voltage between a and
b is with nothing connecting a and b.
As you look at this, you can see then,
that the open circuit voltage is the voltage across R2.
Now, we can use any method we want to determine the voltage across R2.
We could calculate the current running through these.
Which, because this open circuit,
we should point out there is no current- [SOUND] Going through that branch.
That means that these two resistors are in series.
And by definition the current is [INAUDIBLE] the same through both of them.
So we could calculate the current through those two resistors and
we multiply R2 by that current and
that would give us the Open circuit volts through the volts across R2,
we're gonna also use our voltage divider which was developed, we derive a voltage
formula by taking advantage of the fact that those two cells are in series.
So let's just do that.
V open circuit, is equal to then, V sub s, times.
R2 over R1 + R2.
For this circuit, that is our V Thevenin.
So to get the open circuit voltage using any of the voltage,
any of the certain analysis techniques that you have to determine the voltage
across the open terminals.
Next, we need to determine the thevenin resistance.
And at that point, we will have our thevenin equivalent circuit.
Now we're gonna see there are at least three different methods for
determining the thevenin resistance.
And we're going to Demonstrate each of those models on the simple circuit.
The first method takes advantage of the definition of R Thevenin,
you'll recall from our previous video we said that R Thevenin was defined
as the ratio of the open circuit voltage to the short circuit current.
Now in the previous slide we determine that the open circuit voltage
which is the voltage across our to in this case which was vs so
times R two over R one plus R two.Now we need
to determine what the short circuit current is by short circuit currently mean
if we short The wire between the two terminals A and
B and then determine what that current is that flows through that short circuit.
That's the short circuit current.
Well, in this case the short circuit pulls the voltage here to zero.
So, there would be no current going through R2.
And the short circuit current then is simply the current that would be flowing
through R 1 or I short circuit is equal to V sub S divided by R 1.
So you've been using this first method method 1 we say
then that open the R thevenin is equal to the ratio of the open circuit voltage
short circuit current, we form that And say that R 7 is equal to V
open circuit divided by I short circuit which is equal to V sub
S times R 2 over R 1 plus R 2.
That's our open circuit voltage
divided by the short circuit current which is V sub S.
Over R one.
Well, the v of s is cancelled.
We take this one over our one in the denominator, invert and multiply, and
we get then that our Is equal
to R 1 R 2 over R 1 + R 2 At least it works
if you've got an independent source in the circuit that you're attempting to model.
Method one then is determine the open circuit voltage and
the short circuit current using any circuit analysis method you'd like.
And then forming the ratio of the open circuit voltage to the short
circuit current.
The second method works when all you have or
the only types of sources you have are independent sources.
It works only with independent
Sources and in this method, or to apply this method,
we deactivate any of the sources present.
And by deactivate,
we mean the same thing that we did back when we were talking about super position.
When you deactivate a voltage source, you turn the voltage source to zero.
Or in this case, or
a voltage source turned to 0 is effectively a short circuit.
So having deactivated the independent sources present, we then calculate
the resistance seen looking back into the circuit.
Well, looking back into it with this point pulled to ground by deactivating
the voltage source, we see then that R1 and R2 are in parallel with each other.
And the resistance that we see going back into the ab terminals is
R1R2 over R1 + R2.
Which is the same result we got on the previews.
Using the previews method.
The third method, involves applying a test voltage at the terminals, and
pushing the current back into the circuit, pushing the current back into the circuit.
After deactivating the sources.
This is somewhat like going back to our power training analogy, or
an automobile analogy.
It'd be like sort of turning off the engine, and
then blowing back into the exhaust pipe, and measuring
the resistance to the movement of air going back into the Into the exhaust pipe.
So, Method 3, deactivate the independent sources.
If you have dependent sources present, don't deactivate them.
Deactivate just the independent sources, and then
apply a test voltage We'll call it V Test, VT for test.
That's not the Thevenin volt, it's just V Test which will
force a test current to flow back into the circuit.
Again, after we have deactivated the source.
The resistance that we feel going back into it is simply the ratio
of the test voltage divided by the test current.
So our approach now requires us to
come up with an equation or more than one equation.
It allows us then to by just algebraic manipulation
form the ratio of Vtest over Itest.
Let's just do it.
It's easier to show that it is to tell about it.
When we apply this,
after deactivating the voltage source, we see that we have only one node here.
And that node is in fact, at least in this circuit, The voltage,
our test voltage being Test.
So, let's write a node equation here at this node in terms of the node voltage
V Test.
The current leaving this node going through R1 is going to be V Test divided
by R1, is the current leaving going this way.
The current coming down here through R2 Is gonna be added to it
plus V test divided by R2.
Now you'll notice that I test is directed into the nodes, so
that would be minus I test equals 0.
Now let's factor out the common V test here Times 1 over R1
plus 1 over R2 and take I test to the other side as a positive I test.
Getting a common denominator we have then V test time
R1 plus R2 over R1 times R2.
Is equal to I test.
Now that we've got a factor of the test on this side,
I test on that side divide both sides by I test, divide both sides by this term
here and we get that then v test over I test
Which is our R thevenin is = R1,
R2 over R1 + R2.
Which is the same resistance we determined through the other two methods.
So, again, this approach drives home the idea that what this Thevenin resistance
represents is the restriction or the resistance seen
looking back into the terminals of our circuit.
Let's review then, determining we find the open circuit voltage
using any method of circuit analysis we want
we then use one of three methods to determine our thevenin.
The first method works When you have at least one independent source present.
Method one then is too short between the output terminals and
determine the short circuit current and
then form the ratio V open circuit divided by high short circuit.
That's method one.
Method two works only if you have independent sources
and with only independent sources present, And deactivate the independent sources.
In this case, we had a voltage source that we shorted out.
Had it been a current source, we would have opened circuited that current source,
and then determined the resistance,
the equivalent resistance seen looking back in from a to b.
Method two works only if independent sources are present.
Method three works when you have both independent and dependent sources.
If you have dependent sources, if you have independent sources you deactivate Any
of the independent sources by replacing voltage sources with short circuits, and
current sources with open circuits, and
then apply a test voltage to the terminals.
And then using algebraic manipulation techniques,
derive the ratio of V test over I test.
Three different options.
Some options work under all circumstances,
other options work only at given certain circumstances.
And in the next two or three videos, we'll go through and give examples demonstrating
each of these three techniques in somewhat more complicated circuits.