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All righty,
let's work through some examples of

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finding Thevenin equivalent circuits.

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Take a look at this circuit right here.

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We want to determine
the Thevenin equivalent circuit

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between the terminals a and b.

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That means we need to find
the open circuit voltage and

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we need to determine the Thevenin
equivalent resistance.

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Now, you'll notice the open circuit
voltage is the voltage across here, but

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because it's open circuit, that means that
there's no current flowing through here.

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I equals zero, so if there's no
current flowing through that resister,

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there's no voltage drop
across that resister.

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And the voltage at the terminals,
the open circuit voltage then,

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is, in fact,
the voltage across this 30 ohm resistor.

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Again, because there's
no voltage drop here,

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the voltage that hit this point is
the same as the voltage at this point.

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And again, because there's no current
going here, the current source here,

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0.5 amps flowing is driving these
two in series at this point.

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And we have then that V open
circuit is simply equal

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to 0.5 times, 0.5 is the current,

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times the 30 ohms is equal to 15 volts.

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We now need to determine
the Thevenin equivalent resistance,

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which we're gonna do in
each of the three methods.

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Or at least consider whether each of those
three different methods will work for

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this circuit.

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Method one works when we have at least one
independent source, which we have here.

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Method one involves finding
the short circuit current.

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And then, forming the ratio, V,
open circuit divided by I, short circuit.

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So if we short this out,
short the terminals,

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the current flowing through
that short is I, short circuit.

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Now, how do we find I,
short circuit in this circuit?

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Well, by shorting that, out that brings
this, the voltage here, down to ground.

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We have 0.5 amps flowing into now
a node which consists of two resistors,

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this 30 ohm resistor and
this 5 ohm resistor.

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Some of the current's gonna through the 5
ohm resistor, that's our short circuit

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current, the rest of the current is
gonna come through this 30 ohm resistor.

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We have a current divider here.

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So using current divider,
we can calculate short circuit directly.

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I short circuit is equal to 0.5 times,
and remember from the current divider,

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the term that goes in the numerator is
the resistance in the other branch.

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So that would be 30 ohms divided
by the sum of those two,

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30 plus 5 is 35 ohms.

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And we get then the short circuit current

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is equal to 0.4286, 0.4286 amps.

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And then RTh is equal to Voc

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divided by Isc, which is 15

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divided by 0.4286.

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Which, when we get the calculator out on
that, you'll find that equals 35 ohms.

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So, method 1.

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Method 2,
let's clear this up just a little bit.

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All righty,

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now let's use method 2 to determine
the Thevenin equivalent resistance.

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In method 2, which works if you have,

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only, independent sources.

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Method two, you deactivate the sources.

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In this case,
we have an independent current source.

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We turn the independent
current source to zero.

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And, that is effectively,
open circuiting this branch,

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which means that there will be no
current flowing through this branch.

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Such that as we look
back in to the circuit,

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this is just hanging unconnected and
does not enter into this.

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Effectively, we have,
because this is open circuit here,

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this is effectively
an infinite resistance here.

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And we see coming back into this then,
the 5 ohms in series with the 35.

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So method 2,

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We get that R Thevenin is simply
the 30 + 5 ohms = 35 ohms,

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same value we got for method 1.

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Now, let's take a look at method 3.

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Method 3 works regardless of
the types of sources that we have,

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but in method 3,
we do deactivate the independent sources.

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So once again,
we have this independent current source.

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By deactivating it, we open the circuit.

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And then, method 3 says that
we apply a test voltage,

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which then forces a test
current into the AB terminals,

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and we then algebraically work to
make the ratio of V test over I test.

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So method 3 then Is form the ratio V

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test over I test to give us R Thevenin.

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All righty, with that open circuit,
this is a simple situation.

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I, going in here,
we can write a KVL loop going

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around this loop in the direction of IT.

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So KVL, we have then, minus v test
going in the direction of current flow.

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We have then a voltage drop of plus five
I sub t coming on down here plus 30.

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That's a plus sign, 30 I sub t,
the sum of those terms equals zero.

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Let's bring in V test over to
the other side of the positive V test,

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factor out the I test.

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We had I test x 5 + 30 is 35 = V test.

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Divide both sides by I test,
that gives us our R Thevenin,

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which is equal to the ratio of
V test over I test = 35 ohms.

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So we get the same results in
any one of the three methods,

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and we then, can or now can,

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Draw our Thevenin equivalent circuit,
where V Thevenin

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is equal to 15 volts, and
R Thevenin is equal to 35 ohms.

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Notice that the original
circuit had a current source.

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But the Thevenin Equivalent Circuit
replaces that source and

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all of the resistance, the effects of
the resistance within this circuit,

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are represented by a single voltage source
and the Thevenin Equivalent resistance.