L3 5 3 Thevenin Exampe 1

Example determining Thevenin Equivalent Circuit

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.