WEBVTT

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>> The second method for determining
the Thevenin equivalent impedance,

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involves calculating the
short circuit current

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that would flow or the
current that would flow,

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if you were to apply a short circuit
between the terminals,

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and then Z Thevenin,

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is equal to the open circuit voltage
divided by the short-circuit current.

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We've already determined
the open-circuit voltage.

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In our previous video,

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we found that the open-circuit voltage
was equal to 6.25, angle 51.34.

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So now, we simply need to calculate
the short-circuit current and

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then find the Thevenin impedance
by taking the ratio of those two.

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This method works anytime you have
at least one independent source,

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either an independent voltage source
or an independent current source,

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and whether you have
dependent sources or not,

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the short-circuit current method will work.

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You'll recall that the equivalent
impedance method that we showed in

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the previous example only worked if you
had independent sources or no sources.

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At previous method did not work
if you had dependent sources.

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This method will work
with dependent sources as

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long as you have at least
one independent source also.

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All right. What will our approach be?

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Well, we need to find the
current flowing through here.

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In this circuit, the short circuit
current is going to be the current

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flowing through this j50 ohm inductor.

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Now the current flowing
through that j50 ohm inductor,

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will be the voltage at
this point divided by the j50.

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So let's call that voltage there.

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Let's call that V_1 and
we can then say that

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I short circuit is equal
to V_1 divided by j50.

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So our problem now becomes
determining what V_1 is.

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Well, again in this circuit,

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this inductor has been brought into
parallel with this 20 ohm resistor,

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and if we combine those two in parallel,

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we will have, in fact,

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I should do that first.

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Z parallel is equal to 20 times
j50 divided by 20 plus j50,

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and that turns out being 17.24 plus j6.9.

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So this circuit then reduces down to,

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our source here of 10 angle zero,

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then minus j25 capacitor.

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Then, we've got this
equivalent parallel impedance,

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Z_p and V_1.

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The voltage that we're looking
for is that voltage there.

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So these two are in series with each other,

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so we can get V_1 using a voltage divider.

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V_1 that is equal to 10 times Z_p,

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which is 17.24 plus j6.9 divided
by the sum of those two,

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17.24 plus j6.9 minus j25.

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V_1 it turns out is then equal

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to 7.43 angle 68.2

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All right. That's V_1.

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We need I short circuit.

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I short circuit is V_1 divided by
j50 or I short circuit is equal to

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7.43 angle 68.2 divided by j50,

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and that equals

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0.149 angle negative 21.8.

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Finally, we've got
the short-circuit current,

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so we can find Z Thevenin by taking
the open-circuit voltage of 6.25 angle

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51.34 and dividing by the short-circuit
current of 0.149 angle negative 21.8,

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and that gives us the same impedance
that we found in the previous method of

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12.2 plus j40.2 ohms.