All righty, let's work through some examples of
finding Thevenin equivalent circuits.
Take a look at this circuit right here.
We want to determine the Thevenin equivalent circuit
between the terminals a and b.
That means we need to find the open circuit voltage and
we need to determine the Thevenin equivalent resistance.
Now, you'll notice the open circuit voltage is the voltage across here, but
because it's open circuit, that means that there's no current flowing through here.
I equals zero, so if there's no current flowing through that resister,
there's no voltage drop across that resister.
And the voltage at the terminals, the open circuit voltage then,
is, in fact, the voltage across this 30 ohm resistor.
Again, because there's no voltage drop here,
the voltage that hit this point is the same as the voltage at this point.
And again, because there's no current going here, the current source here,
0.5 amps flowing is driving these two in series at this point.
And we have then that V open circuit is simply equal
to 0.5 times, 0.5 is the current,
times the 30 ohms is equal to 15 volts.
We now need to determine the Thevenin equivalent resistance,
which we're gonna do in each of the three methods.
Or at least consider whether each of those three different methods will work for
this circuit.
Method one works when we have at least one independent source, which we have here.
Method one involves finding the short circuit current.
And then, forming the ratio, V, open circuit divided by I, short circuit.
So if we short this out, short the terminals,
the current flowing through that short is I, short circuit.
Now, how do we find I, short circuit in this circuit?
Well, by shorting that, out that brings this, the voltage here, down to ground.
We have 0.5 amps flowing into now a node which consists of two resistors,
this 30 ohm resistor and this 5 ohm resistor.
Some of the current's gonna through the 5 ohm resistor, that's our short circuit
current, the rest of the current is gonna come through this 30 ohm resistor.
We have a current divider here.
So using current divider, we can calculate short circuit directly.
I short circuit is equal to 0.5 times, and remember from the current divider,
the term that goes in the numerator is the resistance in the other branch.
So that would be 30 ohms divided by the sum of those two,
30 plus 5 is 35 ohms.
And we get then the short circuit current
is equal to 0.4286, 0.4286 amps.
And then RTh is equal to Voc
divided by Isc, which is 15
divided by 0.4286.
Which, when we get the calculator out on that, you'll find that equals 35 ohms.
So, method 1.
Method 2, let's clear this up just a little bit.
All righty,
now let's use method 2 to determine the Thevenin equivalent resistance.
In method 2, which works if you have,
only, independent sources.
Method two, you deactivate the sources.
In this case, we have an independent current source.
We turn the independent current source to zero.
And, that is effectively, open circuiting this branch,
which means that there will be no current flowing through this branch.
Such that as we look back in to the circuit,
this is just hanging unconnected and does not enter into this.
Effectively, we have, because this is open circuit here,
this is effectively an infinite resistance here.
And we see coming back into this then, the 5 ohms in series with the 35.
So method 2,
We get that R Thevenin is simply the 30 + 5 ohms = 35 ohms,
same value we got for method 1.
Now, let's take a look at method 3.
Method 3 works regardless of the types of sources that we have,
but in method 3, we do deactivate the independent sources.
So once again, we have this independent current source.
By deactivating it, we open the circuit.
And then, method 3 says that we apply a test voltage,
which then forces a test current into the AB terminals,
and we then algebraically work to make the ratio of V test over I test.
So method 3 then Is form the ratio V
test over I test to give us R Thevenin.
All righty, with that open circuit, this is a simple situation.
I, going in here, we can write a KVL loop going
around this loop in the direction of IT.
So KVL, we have then, minus v test going in the direction of current flow.
We have then a voltage drop of plus five I sub t coming on down here plus 30.
That's a plus sign, 30 I sub t, the sum of those terms equals zero.
Let's bring in V test over to the other side of the positive V test,
factor out the I test.
We had I test x 5 + 30 is 35 = V test.
Divide both sides by I test, that gives us our R Thevenin,
which is equal to the ratio of V test over I test = 35 ohms.
So we get the same results in any one of the three methods,
and we then, can or now can,
Draw our Thevenin equivalent circuit, where V Thevenin
is equal to 15 volts, and R Thevenin is equal to 35 ohms.
Notice that the original circuit had a current source.
But the Thevenin Equivalent Circuit replaces that source and
all of the resistance, the effects of the resistance within this circuit,
are represented by a single voltage source and the Thevenin Equivalent resistance.