>> The second method for determining the Thevenin equivalent impedance,
involves calculating the short circuit current
that would flow or the current that would flow,
if you were to apply a short circuit between the terminals,
and then Z Thevenin,
is equal to the open circuit voltage divided by the short-circuit current.
We've already determined the open-circuit voltage.
In our previous video,
we found that the open-circuit voltage was equal to 6.25, angle 51.34.
So now, we simply need to calculate the short-circuit current and
then find the Thevenin impedance by taking the ratio of those two.
This method works anytime you have at least one independent source,
either an independent voltage source or an independent current source,
and whether you have dependent sources or not,
the short-circuit current method will work.
You'll recall that the equivalent impedance method that we showed in
the previous example only worked if you had independent sources or no sources.
At previous method did not work if you had dependent sources.
This method will work with dependent sources as
long as you have at least one independent source also.
All right. What will our approach be?
Well, we need to find the current flowing through here.
In this circuit, the short circuit current is going to be the current
flowing through this j50 ohm inductor.
Now the current flowing through that j50 ohm inductor,
will be the voltage at this point divided by the j50.
So let's call that voltage there.
Let's call that V_1 and we can then say that
I short circuit is equal to V_1 divided by j50.
So our problem now becomes determining what V_1 is.
Well, again in this circuit,
this inductor has been brought into parallel with this 20 ohm resistor,
and if we combine those two in parallel,
we will have, in fact,
I should do that first.
Z parallel is equal to 20 times j50 divided by 20 plus j50,
and that turns out being 17.24 plus j6.9.
So this circuit then reduces down to,
our source here of 10 angle zero,
then minus j25 capacitor.
Then, we've got this equivalent parallel impedance,
Z_p and V_1.
The voltage that we're looking for is that voltage there.
So these two are in series with each other,
so we can get V_1 using a voltage divider.
V_1 that is equal to 10 times Z_p,
which is 17.24 plus j6.9 divided by the sum of those two,
17.24 plus j6.9 minus j25.
V_1 it turns out is then equal
to 7.43 angle 68.2
All right. That's V_1.
We need I short circuit.
I short circuit is V_1 divided by j50 or I short circuit is equal to
7.43 angle 68.2 divided by j50,
and that equals
0.149 angle negative 21.8.
Finally, we've got the short-circuit current,
so we can find Z Thevenin by taking the open-circuit voltage of 6.25 angle
51.34 and dividing by the short-circuit current of 0.149 angle negative 21.8,
and that gives us the same impedance that we found in the previous method of
12.2 plus j40.2 ohms.