L3 5 4 Thevenin Example 2

Example determining Thevenin Equivalent Circuit

All right, here's another example of
finding the Thevenin equivalent of

this circuit.

Again, we have two values
that we need to find.

We need to determine
the open circuit voltage,

which is the voltage,
Across there with no load connected.

And then we also need to determine
the Thevenin equivalent resistance,

which in this one we'll do
a couple of different ways.

All right, first of all, Vth open circuit.

That's the voltage at this node,
noting that the current

leaving this node going in that direction,
call it I, is = 0.

So let's call this node here V sub A,
we got V open circuit there and

let's write the node
equations of these two nodes.

Noting first of all, before we get
ahead of ourselves, that this dependent

current source is pushing current into the
node going this direction with an amount

of 1.2 times I0, where I0 is the current
flowing through the 3-ohm resistor.

So before we get started, let's just note

that I0 = 15-V sub A divided by 3.

We're gonna need that
here in just a second.

All right, some of the current's leaving
V sub A going in that direction.

We have V sub A -15 divided by 3 +,
the current coming down to

the 6-ohm resistor is
gonna be VA divided by 6.

The current leaving the sub A node
going through the 4-ohm resistor,

is going to be VA, the voltage on this
side minus the voltage on that side,

which is VOC, the open circuit voltage,
divided by 4.

Now we have coming into the node,
it's gonna be -1.2 times I0.

But I0 we have already expressed
as 15 -VA divided by 3.

So there's one, two, three, four branches,
one, two, three, four terms,

the sum of those four
things needs to be =0.

Right in the node equation at the other
node there which we're calling VOC,

we have two nonzero currents.

We've got the current leading
this node going to the 4 ohm

resistor in this direction is
Voc-V sub A divided by 4 plus

the current leaving it's not going
that way which is 1.2 time I0.

So +1.2 times I0 which
we've already express as 15

minus V sub A divided by 3,
as we've all ready noted,

the current leaving going
this direction is 0.

So, the sum of those two
currents must equal 0.

We're gonna skip the grimy algebra,

but after you've simplified this down,

you come up with VA (1.15) +

VOC(-0.25) = 11.

The top equation simplifies to that,

and the second equation simplifies

to VA(- 0.65 )+ Voc

(0.25) = -6.

And when you plug that in
your matrix solver and

however you do it, you can in
that V sub A equals 10 volts and

VOC which is our Thevenin
voltage equals two volts.

Now we have three different methods
of finding the Thevenin resistance.

The first method would be to
short out the terminals AB and

determine the short circuit current, and

then our Thevenin equals V open
circuit divided by I short circuit.

We're gonna do that here in just a minute.

The second method only works when you
have independent sources only, or

when you only have independent sources.

The fact that we have this
dependent source here makes it so

that method 2 doesn't work.

Method 3 is also gonna work,
we'll do that here in just a second.

Method 3 involves shorting
out this voltage or

deactivating the voltage source and
then applying a test voltage.

So let's start with
the short circuit method.

Let's short this out,
call that I short circuit and

now determine the current
in that short circuit.

We need to note a couple of things.

First of all, as we short this out,

we now know that
the voltage there equals 0.

So we only have one node where
we have an unknown voltage.

We're gonna calculate V sub A and
once we know V sub A,

we'll then write a node equation
at this node noting that V0 = 0.

But the equation will then be
in terms of I short circuit,

and we'll be able to
calculate I short circuit.

Let's go ahead and do it.

It's probably easier to show it
than it is to talk about it.

So let's just work on over here.

Write in the note equation at the A node,
we have

V sub A minus 15 divided by 3,

plus V sub A divided by 6,
plus V sub A divided by four,

minus 1.2 times,

now notice that I zero, even though we've

shorted this out I zero is still
15 minus V sub A divided by three.

So it would be minus one
point two times (15-Va/3) and

the sum of those terms has to equal zero.

Now when you go through and
simplify that and

solve for V sub A you get
that V sub A = 9.565 Volts.

Now that we know V sub A with
this output short circuited,

we can go ahead and
write a node equation at this node.

We'll do it in terms of V sub A,
but A sub A is gonna equal 9.565.

Let's just do it and
show you what we get here.

So we're going to have the current leaving
this node going in that direction.

It's going to be 0, oops.

Yeah 0, the voltage there is 0.

0- v sub a, which is 9.565 divided by 4.

Plus the current going in this

direction, which is 1.2 times I0,
I0 is 15,

minus V sub A, so that's going

to be 15 minus 9.565, divided by 3.

Then we have I short-circuit leaving.

So it would be plus I
short-circuit equals 0.

Well let's just bring that short circuit
on the other side as a equals minus I,

short-circuit, running out of room here.

All right,
let's continue on now and solve for

I short-circuit, and when you do that,

you get I short circuit
is equal to 0.2173 Amps.

We now know V open circuit,
V open circuit is 2.

I short circuit is that
R Thevenin is equal

to the ratio of V open
circuit I short circuit,

which equals 2 divided by 0.2173 and

that equals 9.2 ohms.

And we can now draw our
Thevenin equivalent circuit,

it consists of a two volt source,

with a 9.2 ohm resistor in series, and

there are our a and b terminals.

Thus, this nice concise little circuit.

If all we're interested in is
the terminal characteristics,

this models this more complex circuit.

Now lets just show that we get the same
R Thevenin using method three.

Method three requires us to
deactivate the independent sources.

So this is a voltage source return it to
zero volts, that is effectively shorting

it out and we then apply a test
voltage called a V test.

Which then drives a current
I test into the circuit.

We have two nodes we got the node here,
a node here we will call this one again

V sub A noting that this V sub A is
different than the V sub A we did before.

The V sub A that we did before had a 15
volt source connected at this point.

This circuit has that 15
volts source shorted out.

So we gonna use the same name V sub A and

over here this is no
longer 0 this is now v10.

Let's go ahead and write the node

equation for this A node.

So the current leaving, noting also
we're gonna need this I0 in this case,

is the current going in that
direction that's going to

equal 0-V sub A divided by 3, or
just negative V sub A over 3.

All right,
now let's write the node equation there.

The current leaving this
node going in this direction

is going to be Va- 0
because that's now shorted

out divided by 3 + Va
divided by 6 + Va minus.

We're now talking about the current
leaving this node going

through the four ohm resistor.

It's gonna be V sub a minus Vtest

over 4 minus the current entering this
node through the dependent source.

It's going to be minus cuz it's
entering 1.2 times I zero and

I zero is negative v sub a over three.

One, two, three, four terms, the sum
of those four terms must equal zero.

Now, let's write the node equation
over here at the test note.

We have three branches, three currents,

the current leaving the node going in this
direction is going to be V test minus V

sub a divided by 4 plus....

Now this current here is leaving
the node so it would be plus 1.2 I

zero, but I zero we've determined
to be negative V sub a over 3e,

ngative V sub a over 3.

Now, IT is coming in, so
that would be negative I test, so

let's just take it to the other
side as a positive I test and

say at the end of those two terms
will equal positive I test.

Now you notice we have three unknowns,
V sub a,

v test, and I test, but
have only two equations.

But that's all right,
cuz what we're gonna try and

do is through algebraic manipulation, come
up with an expression here in this upper

equation that gives us v
sub a in terms of v-test.

We'll then plug in down here in
the second equation for v sub a.

All the terms then will have either
factors of v-test or i-test.

We will factor out the v-test to be able
to make the ratio of v-test over i-test

and again that is what our Thevenin
is the ratio of v-test over i-test.

So this upper equation cleans

up to give us 1.15 V sub

A Is equal

to 0.25 VT.

So V sub A is equal to 0.2174v test.

Let's take those values in then,
and plug them in down here.

In fact, let's just clean this
one up a little bit down here.

Let's rewrite this bottom one as
if we were gonna have v-test times

one-fourth, that's the only v-test term.

Then we're gonna have plus v sub a times,

we've got a negative one and

a fourth, and we have a negative

1.2 over 3 equals I-test.

Now we'll take this expression for

V sub A in terms of V test, and
plug it down here for V sub A.

And we have 1/4Vt + .2174 v test,

times,a negative one fourth,

minus 1.2 over 3,

that equals, i test.

Now you go through and clean this up, and

we get 0.10869V

test equals I test.

Now, for me the ratio of V test
over I test gives us our Thevenin.

So R Thevenin equals V test
over I test which equals one

over 0.10869,

which once again equals 9.2 Ohms.

Thus we get the same R Thevenin
from method three as we got for

method one in the previous page,

method two doesn't work because
there is a depended source in there.