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>> Our third and final method for finding

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the Thevenin equivalent
impedance of the circuit,

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involves applying an external source
to the terminals of our circuit,

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after deactivating
the any independent sources.

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So this circuit that we're
finding the Thevenin circuit for,

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involved or has a single
independent voltage source.

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We're going to deactivate
that independent source

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again by replacing it with a short-circuit.

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This method then involves applying
an external voltage source just V,

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call it V_ex for external,

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which would then cause a current to flow.

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We'll call it I_ex,

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for the external current of the current,

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is due to this external voltage source,

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and Z Thevenin then is going to equal,

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the ratio of V_ex to I_ex.

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This method works under any circumstances,

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whether you have dependent
or independent sources.

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This third method will work.

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Sometimes it gets to be
algebraically a little bit

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cumbersome involving multiple equations
and multiple unknowns,

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but it will always work.

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This method is equivalent to,

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and can maybe something of a visual,

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of a gasoline engine and an exhaust system.

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If you wanted to measure or to model,

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the back pressure due to
the resistance of the exhaust system.

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This would be like turning off the engine,

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and then putting your mouth over
the exhaust pipe and blowing

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into it and measuring
the current that flows,

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then that resistance of
the exhaust system would be

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equal to the pressure that
you're pushing against it,

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divided by the amount of
air that flowed into it.

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So our Thevenin impedance will be,

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we're going to deactivate
the source, the independent sources.

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If there were dependent sources in here,

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we would leave them active,

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and having deactivated
the independent sources,

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pushing against this or applying a voltage,

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and taking the ratio of
the voltage to the current.

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So here's our circuit then.

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We've deactivated the source,

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we're applying the external voltage source,

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and we need to do some algebra and write
some equations that will allow us to

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come up with a ratio of the external
voltage to the external current.

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With this circuit here,

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and with the source deactivated,

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we see now that this 20 ohm and

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the negative j25 ohm capacitor
are in parallel with each other.

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So let's call that Z parallel,

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and by reducing those to
a parallel combination,

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we'll have these two,

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the parallel equivalent
in series with that,

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and we'll then be able to write
the expression for the external current,

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will just be equal to this Thevenin
or this external voltage,

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divided by the series
combination of those two,

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of this plus the parallel equivalent.

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So Z parallel is equal to,

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negative j25 times 20,

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divided by 20 minus j25,

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and that again works out
to be 12.2 minus j9.76.

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So we have then,

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just redrawing to make obvious
what we're doing here.

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You've got this inductor j50,

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and we have this equivalent impedance,

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of 12.2 minus j9.76.

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This is our external voltage,

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and there'll be
an external current flowing.

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We can now write an expression for I_ex.

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The external current is just
equal to the external voltage,

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divided by the sum of those,

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which is j50 plus 12.2 minus j9.76,

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and the ratio V_ex over I_ex then,

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can be gotten by multiplying
both sides of the equation by

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this denominator and dividing
both sides by I_ex, and we get then,

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that V_ex over I_ex is equal

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to 12.2 plus j40.2,

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and that of course is
our Thevenin equivalent voltage.

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So we've seen now,

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three different methods of
calculating the Thevenin impedance,

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and which one works best becomes
a matter of art and experience.

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You'll have some opportunities to do enough

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of these and you'll start to get a feel for

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the circumstances that make each of
these different methods most applicable.