L7 5 2 1 Thevenin Equiv Circ Open Circuit Voltage

Explains Thevenin equivalent circuits. Determining the Thevenin equivalent voltage or Open Circuit Voltage

>> In these next few videos,

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

in this phasor domain
using complex impedances.

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

that the idea was that
if we had some circuit,

some relatively complicated circuit,

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

we could replace that circuit
with its Thevenin equivalent

which consisted of a single voltage source

and back then when we are talking about

a single resistance called
the Thevenin resistance,

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

the Thevenin equivalent circuit involves

a voltage source - a phasor voltage -

and a complex impedance.

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

a certain current draws,

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

This Thevenin equivalent circuit
will model that current flowing

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

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

that you would experience in
the more complicated circuit.

Again, it's simply a method of modeling

a more complicated circuit by
a less complicated circuit

that has the same terminal characteristics.

You'll recall that to do so,

we had to determine what
the open circuit voltage was,

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

when no load is connected.

That's the Thevenin voltage.

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

We can determine the equivalent
impedance seen looking

back in to the circuit with
all independent sources deactivated;

for one method.

Another method involves
putting a short circuit,

actually grounding or shorting
out the AB terminals,

and calculating that short-circuit current,

and then this Thevenin
impedance is equal to

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

We'll do an example using that.

And the third method involves
deactivating the source,

any independent sources,

applying an external source
to the circuit terminals,

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

to the current that then flows.

Okay, let's go ahead and do this.

Here we've got a time domain circuit.

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

The first thing we need to do is convert it

to its phasor domain representation.

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

to a complex impedance voltage
of 10e to the j0

and similarly each of the impedances
are represented over here,

and our task then is to determine
the Thevenin equivalent circuit

for this circuit.

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

The open circuit voltage is
the voltage that we would

measure across the terminals
with no load connected.

Because there is no load connected,

there is no current flowing
through that inductor

and any current that is
produced by this voltage source

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

So the open-circuit voltage
is simply going to be

the voltage across this 20-Ohm resistor.

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

there'll be no voltage drop
across that inductor

and the open-circuit voltage would just be

the voltage across that 20-Ohm resistor.

Easy way to get that, it appears to me,

would be to use a voltage divider,

and say then that V
open circuit is equal to

10 times 20 divided by 20 minus j25,

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

is equal to 6.25e to the j51.34.

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

rather than writing in
the exponential form,

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

So this form and this form
are both polar forms.

This just represents
the magnitude and the phase

and here we've got the complex exponential

that again demonstrates
the mathematics behind it,

what is actually happening

but whatever form we choose,

we now have our Thevenin equivalent voltage

because the Thevenin equivalent voltage is

simply equal to the open circuit voltage

which in this case is the
6.25 angle 51.34 degrees.

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

of determining the Thevenin
equivalent impedance