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We use Thevenin Equivalent Circuits when we want to simplify part of a circuit.

For example, we look over here.

I could have a circuit that's really messy, lots of resistors,

lots of sources in there.

It could be very messy.

And I got these two terminals coming out, A and B.

I want to replace this entire circuit with one that is much simpler.

It's just got a voltage source.

One voltage source and one resistor.

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And the behavior of the circuit is the same

across those two terminals as the original one is.

So we use Thevenin Equivalent Circuits again when we want to simplify

part of a circuit to simplify our analysis.

The other reason we use Thevenin Equivalent Circuits is to find the load

that will maximize power to the load.

So in other words if I wanted to put a load resistor across right here or

to the original circuit right there then I would want to find

the Thevenin Equivalent Circuit and

that would tell me what load will maximize the power.

How do we build our Thevenin Equivalent Circuit?

Well it's shown over here, we're going need to define a few terms here.

We define V Thevenin, which is the voltage source here.

The R Thevenin is the resistance rate here.

Those are the elements of our Thevenin Equivalent Circuit and then I short

circuit or i sub sc is the current that relates these two through Ohm's law.

Now how do we find these from an actual circuit?

So if I took an actual circuit and I had my terminals coming out and I measured

the open circuit across here, and the open circuit voltage, that would be V Thevenin.

The other way I can find the other thing I need to find is the, I short circuit,

which is shorting out or just putting a wire across this and

measuring the current across those terminals.

That's i sub short circuit, and those two are both necessary in finding here.

And if I found those two through measurements, then I just can come back

through this relationship Ohm's law and find R Thevenin.

A lot of times in actual circuits it's easier to find

R Thevenin maybe analytically by doing a simple method here.

What I have to do is go to my original circuit,

this might be my original circuit.

I zero outsources

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And this method only holds when no dependent sources are present.

So if there are no dependent sources, I zero out those sources.

And by zeroing out sources, the voltage source has to be 0,

that means it's shorted.

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And I open circuit the current source,

now I want to find the equivalent resistance.

In this case I've got 2 in parallel with 4 in series with 10 and

that is 11.33 ohmes.

Now to find V Thevenin I actually draw the same circuit here because what I'm

trying to do is find the open circuit voltage across here so let me draw that.

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So we've drawn it and now I want to find this voltage right here.

And that's V Thevenin.

In this particular case, all the current will flow through this,

since this is an open circuit.

All the current flows through here, that means I know this voltage.

If I know this voltage and these current matches right here,

that means all this current goes in this direction and I know this voltage.

So I know this voltage, this voltage, this voltage, I can do a KVL around here and

solve for V thevenin.

It's going to be equal to two thirds volts.

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And similarly I can draw the circuit for i sub sc.

In this particular case, I draw the same circuit, but

in this case I draw a line across here and I short it out and

then I saw through this current right there I sub S C.

So if I did that in this case I would get 0.0589 amps.

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And I only really have to solve two of these three circuit problems because I

can use Ohm's law for the other ones.

So I can relate V Thevenin to R Thevenin and

i sc through Ohm's law right there.

Now once I solve this I can draw the V Thevenin and

that is 2/3 for V Thevenin,

R Thevenin is 11.33 omz, and

that's the Thevenin equivalent circuit.

So, this circuit here, is the same as the original circuit.

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In this particular case I short out this resistor,

so that R Thevenin is equal to 2 ohms.

I want to show a couple of examples of finding

V Thevenin in this particular case,

we want to find V Thevenin which is the voltage across here.

Since this is an open circuit there is no voltage drop across here.

That means that the voltage from here to here that's a plus and

that's a minus is V Thevenin, that means I just have to voltage devider lock R2

over R1 plus R2 times VS is equal to V thevatin.

And then I solve for R thevanin the same way we've been solving for it.

Let me show an other example very similar, finding V thevenin.

Again, I want to find this voltage right here.

But it's kind of messy to do this, so sometimes it's easier to find

i short circuit and then use Ohm's law to find V Thévenin.

And we'll show that here.

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So that's why it makes it easier.

So all I have left with is this circuit right here.

R1, Vs And this current right there.

So, I can solve for i sub sc from this simplified circuit, and

once I solve for i sub sc and I solve for R Thevenin,

then I say V Thevenin is equal to R Thevenin, and Times i sub sc.

So in this case, it just turned out easier to solve our i sub sc first and

then use Ohm's Law to find V Thevenin.

So one of the other uses of the Thevenin Equivalent Circuit that

I mentioned all ready was to find the maximum power transfer.

So if I've taken a circuit that I've already reduced down to its Thevenin

equivalent, I want to say, well what sort of resistor can I put across here

when I call it a load resistor, that would maximize the power?

Well the answer is the Thevenin resistance.

If I've already solved for that Thevenin resistance, then this particular value of

resistor, the load resistor, maximizes power to this.

Now let's go ahead and show the derivation of that.

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It's based on a power analysis.

Power is equal to voltage across here, so we're trying to maximize this power.

So it's a voltage across this resistor times the current in here.

And I can go through a little bit of analysis, substitute in for i.

From Ohm's Law and then right here, substitute in for

V sub ab and that comes from the Voltage Divider Law.

And I have something now, the power is equal to this.

From this equation I can find the optimum R sub L maximize the power.

Now where is this useful?

Sometimes it's useful in matched impedances.

Sometimes you've got cables, and you want to maximize power at the load

end of the cable, and we call things like matched impedance or matched resistance.

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So the key concepts that we've covered so far, is the most important thing being,

this is what the Thevenin equivalent circuit looks like.

And we showed how to find the V thevanin and

the I short circuit when we short circuit this across here.

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The usages we remember that the uses when one part of our circuit is fixed and

we want to take something fairly complicated like this and

replace it with something very simple.

So then if we wanted to add something else like a resistor here

then we don't have to reanalyze the entire circuit.

And the other use is when we want to maximize power to the load.