>> All right. In our previous video,
we found that the Thevenin equivalent voltage, V_Th,
was equal to 6.25 angle 51.34 degrees.
Once again, I'm going to be using this shorthand notation
of representing the polar coordinates of the impedances of the phasors
as opposed to writing 6.25e to the j51.34.
All right. With the Thevenin voltage,
we need now simply to determine
the Thevenin equivalent impedance
and we're going to do that each of three different ways.
This first method is what we'll refer to as the Equivalent Impedance method.
It works when we have only independent sources or no sources.
This equivalent impedance method that we're going to show first
does not work if the circuit you're analyzing has dependent sources.
So, this approach requires us to deactivate any independent sources present.
In this case, there's an independent voltage source here
which we deactivate by turning it to zero
and thus replacing it with a short circuit.
Then, looking back into the circuit from the terminals AB
and determining what the equivalent impedance is that we would see
looking back in that way.
Well, hopefully you can see pretty easily that with the source deactivated-
replaced with short-circuit,
it brings this capacitor and this resistor in parallel,
-and that parallel combination is in series with the j50-Ohm resistor
such that the impedance seen looking back into these terminals,
Z Thevenin, will equal Z parallel plus the j50-Ohm inductor.
So, Z parallel is equal to
20 times minus j25 divided by 20 minus j25,
which turns out to equal 12.2 minus j9.8 Ohms.
So finally, Z Thevenin then is equal to Zp;
12.2 minus j9.8 plus the j50,
and that then equals 12.2 plus j40.2 Ohms.
Z Thevenin, bringing it up here,
then rewriting it just to show what we've got to show our incomplete model.
Our complete model involves the Z Thevenin equaling 12.2 plus j40.2 ohms
and our claim then is that this Thevenin equivalent circuit
has the same terminal characteristics at the AB terminals
as our original more complicated circuit.