>> In these next few videos,
we're going to demonstrate the application of Thevenin's equivalent circuits
in this phasor domain using complex impedances.
You'll recall from our previous discussions on Thevenin Equivalency,
that the idea was that if we had some circuit,
some relatively complicated circuit,
and we were only interested in the terminal characteristics of that circuit,
we could replace that circuit with its Thevenin equivalent
which consisted of a single voltage source
and back then when we are talking about
a single resistance called the Thevenin resistance,
but now since we are talking about impedances in this complex domain,
the Thevenin equivalent circuit involves
a voltage source - a phasor voltage -
and a complex impedance.
By equivalent, we mean that if you attach a load to the actual circuit,
a certain current draws,
and as you start drawing current, the terminal voltage starts to drop.
This Thevenin equivalent circuit will model that current flowing
and the voltage drop that will happen as current starts to flow
and be equivalent to or be the same current and voltage drop
that you would experience in the more complicated circuit.
Again, it's simply a method of modeling
a more complicated circuit by a less complicated circuit
that has the same terminal characteristics.
You'll recall that to do so,
we had to determine what the open circuit voltage was,
the open-circuit voltage being simply the voltage across the terminals AB
when no load is connected.
That's the Thevenin voltage.
The Thevenin impedance, we can find in one of three different ways.
We can determine the equivalent impedance seen looking
back in to the circuit with all independent sources deactivated;
for one method.
Another method involves putting a short circuit,
actually grounding or shorting out the AB terminals,
and calculating that short-circuit current,
and then this Thevenin impedance is equal to
the ratio of the open-circuit voltage to the short circuit current.
We'll do an example using that.
And the third method involves deactivating the source,
any independent sources,
applying an external source to the circuit terminals,
and calculating the ratio of the voltage, that external voltage,
to the current that then flows.
Okay, let's go ahead and do this.
Here we've got a time domain circuit.
Relatively complicated, not terribly but a good one for our example.
The first thing we need to do is convert it
to its phasor domain representation.
So, this 10cosine(500t) has been converted
to a complex impedance voltage of 10e to the j0
and similarly each of the impedances are represented over here,
and our task then is to determine the Thevenin equivalent circuit
for this circuit.
To do so, we need the open-circuit voltage.
The open circuit voltage is the voltage that we would
measure across the terminals with no load connected.
Because there is no load connected,
there is no current flowing through that inductor
and any current that is produced by this voltage source
will be going through a series combination of the capacitor and the resistor.
So the open-circuit voltage is simply going to be
the voltage across this 20-Ohm resistor.
Again, because there's no current flowing through that inductor,
there'll be no voltage drop across that inductor
and the open-circuit voltage would just be
the voltage across that 20-Ohm resistor.
Easy way to get that, it appears to me,
would be to use a voltage divider,
and say then that V open circuit is equal to
10 times 20 divided by 20 minus j25,
and when you do that, you get that the open-circuit voltage
is equal to 6.25e to the j51.34.
Because we are going to be writing so many of these phasors,
rather than writing in the exponential form,
I'm going to start writing them as a magnitude and an angle of 51.34.
So this form and this form are both polar forms.
This just represents the magnitude and the phase
and here we've got the complex exponential
that again demonstrates the mathematics behind it,
what is actually happening
but whatever form we choose,
we now have our Thevenin equivalent voltage
because the Thevenin equivalent voltage is
simply equal to the open circuit voltage
which in this case is the 6.25 angle 51.34 degrees.
In the next video, we'll demonstrate the three different methods
of determining the Thevenin equivalent impedance