L7 5 2 3 Thevenin Impedance Short Circuit Current Method

Determining Thevenin impedance using the short circuit current

>> 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.