WEBVTT

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>> In these next few videos,

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we're going to demonstrate the application
of Thevenin's equivalent circuits

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in this phasor domain
using complex impedances.

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You'll recall from our previous
discussions on Thevenin Equivalency,

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that the idea was that
if we had some circuit,

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some relatively complicated circuit,

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and we were only interested in the
terminal characteristics of that circuit,

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we could replace that circuit
with its Thevenin equivalent

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which consisted of a single voltage source

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and back then when we are talking about

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a single resistance called
the Thevenin resistance,

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but now since we are talking about
impedances in this complex domain,

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the Thevenin equivalent circuit involves

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a voltage source - a phasor voltage -

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and a complex impedance.

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By equivalent, we mean that if you
attach a load to the actual circuit,

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a certain current draws,

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and as you start drawing current,
the terminal voltage starts to drop.

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This Thevenin equivalent circuit
will model that current flowing

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and the voltage drop that will
happen as current starts to flow

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and be equivalent to or be
the same current and voltage drop

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that you would experience in
the more complicated circuit.

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Again, it's simply a method of modeling

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a more complicated circuit by
a less complicated circuit

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that has the same terminal characteristics.

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You'll recall that to do so,

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we had to determine what
the open circuit voltage was,

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the open-circuit voltage being simply
the voltage across the terminals AB

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when no load is connected.

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That's the Thevenin voltage.

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The Thevenin impedance, we can find
in one of three different ways.

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We can determine the equivalent
impedance seen looking

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back in to the circuit with
all independent sources deactivated;

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for one method.

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Another method involves
putting a short circuit,

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actually grounding or shorting
out the AB terminals,

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and calculating that short-circuit current,

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and then this Thevenin
impedance is equal to

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the ratio of the open-circuit voltage
to the short circuit current.

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We'll do an example using that.

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And the third method involves
deactivating the source,

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any independent sources,

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applying an external source
to the circuit terminals,

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and calculating the ratio of
the voltage, that external voltage,

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to the current that then flows.

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Okay, let's go ahead and do this.

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Here we've got a time domain circuit.

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Relatively complicated, not terribly
but a good one for our example.

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The first thing we need to do is convert it

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to its phasor domain representation.

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So, this 10cosine(500t) has been converted

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to a complex impedance voltage
of 10e to the j0

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and similarly each of the impedances
are represented over here,

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and our task then is to determine
the Thevenin equivalent circuit

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for this circuit.

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To do so, we need the open-circuit voltage.

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The open circuit voltage is
the voltage that we would

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measure across the terminals
with no load connected.

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Because there is no load connected,

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there is no current flowing
through that inductor

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and any current that is
produced by this voltage source

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will be going through a series combination
of the capacitor and the resistor.

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So the open-circuit voltage
is simply going to be

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the voltage across this 20-Ohm resistor.

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Again, because there's no current
flowing through that inductor,

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there'll be no voltage drop
across that inductor

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and the open-circuit voltage would just be

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the voltage across that 20-Ohm resistor.

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Easy way to get that, it appears to me,

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would be to use a voltage divider,

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and say then that V
open circuit is equal to

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10 times 20 divided by 20 minus j25,

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and when you do that, you get
that the open-circuit voltage

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is equal to 6.25e to the j51.34.

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Because we are going to be
writing so many of these phasors,

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rather than writing in
the exponential form,

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I'm going to start writing them as
a magnitude and an angle of 51.34.

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So this form and this form
are both polar forms.

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This just represents
the magnitude and the phase

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and here we've got the complex exponential

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that again demonstrates
the mathematics behind it,

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what is actually happening

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but whatever form we choose,

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we now have our Thevenin equivalent voltage

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because the Thevenin equivalent voltage is

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simply equal to the open circuit voltage

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which in this case is the
6.25 angle 51.34 degrees.

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In the next video, we'll demonstrate
the three different methods

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of determining the Thevenin
equivalent impedance