>> Our third and final method for finding
the Thevenin equivalent impedance of the circuit,
involves applying an external source to the terminals of our circuit,
after deactivating the any independent sources.
So this circuit that we're finding the Thevenin circuit for,
involved or has a single independent voltage source.
We're going to deactivate that independent source
again by replacing it with a short-circuit.
This method then involves applying an external voltage source just V,
call it V_ex for external,
which would then cause a current to flow.
We'll call it I_ex,
for the external current of the current,
is due to this external voltage source,
and Z Thevenin then is going to equal,
the ratio of V_ex to I_ex.
This method works under any circumstances,
whether you have dependent or independent sources.
This third method will work.
Sometimes it gets to be algebraically a little bit
cumbersome involving multiple equations and multiple unknowns,
but it will always work.
This method is equivalent to,
and can maybe something of a visual,
of a gasoline engine and an exhaust system.
If you wanted to measure or to model,
the back pressure due to the resistance of the exhaust system.
This would be like turning off the engine,
and then putting your mouth over the exhaust pipe and blowing
into it and measuring the current that flows,
then that resistance of the exhaust system would be
equal to the pressure that you're pushing against it,
divided by the amount of air that flowed into it.
So our Thevenin impedance will be,
we're going to deactivate the source, the independent sources.
If there were dependent sources in here,
we would leave them active,
and having deactivated the independent sources,
pushing against this or applying a voltage,
and taking the ratio of the voltage to the current.
So here's our circuit then.
We've deactivated the source,
we're applying the external voltage source,
and we need to do some algebra and write some equations that will allow us to
come up with a ratio of the external voltage to the external current.
With this circuit here,
and with the source deactivated,
we see now that this 20 ohm and
the negative j25 ohm capacitor are in parallel with each other.
So let's call that Z parallel,
and by reducing those to a parallel combination,
we'll have these two,
the parallel equivalent in series with that,
and we'll then be able to write the expression for the external current,
will just be equal to this Thevenin or this external voltage,
divided by the series combination of those two,
of this plus the parallel equivalent.
So Z parallel is equal to,
negative j25 times 20,
divided by 20 minus j25,
and that again works out to be 12.2 minus j9.76.
So we have then,
just redrawing to make obvious what we're doing here.
You've got this inductor j50,
and we have this equivalent impedance,
of 12.2 minus j9.76.
This is our external voltage,
and there'll be an external current flowing.
We can now write an expression for I_ex.
The external current is just equal to the external voltage,
divided by the sum of those,
which is j50 plus 12.2 minus j9.76,
and the ratio V_ex over I_ex then,
can be gotten by multiplying both sides of the equation by
this denominator and dividing both sides by I_ex, and we get then,
that V_ex over I_ex is equal
to 12.2 plus j40.2,
and that of course is our Thevenin equivalent voltage.
So we've seen now,
three different methods of calculating the Thevenin impedance,
and which one works best becomes a matter of art and experience.
You'll have some opportunities to do enough
of these and you'll start to get a feel for
the circumstances that make each of these different methods most applicable.