2 Thevenin

Okay now we want to find Thevenin
using the external voltage method, so

our Thevenin is going
to be equal to Vex/Iex.

We do is we go back to our circuit, and

we add in an arbitrary
voltage source we can Vex.

Since it's arbitrary,
we can say it's equal to 1 volt, and

that makes our equation very
simple Then we will have an IEX.

We don't know what that is yet.

We solve the problem in much the same
way as we did before, so watch for

the differences.

Our first equation is going to
be VC minus 68 volts divide by

6 ohms plus Vc minus 4 IX

divided by 2 omes plus vc minus va,

divided by 6 omes and that's equal to
0.No change in our first equation.

The second equation is going to
be va A minus by VC divided by

six ohms plus VA minus zero
divided by four ohms plus

alright if I had done BA minus BEX I would
of had to divide by zero because there's

zero resistance in this arm instead what
I'm going to do is I'm going to add in

the outgoing current the current
coming were going to subtract it out.

Iex =0.

Our third equation is going to be the same
we had before in order to find our Ix

unknown, which will be Vc-Va
divided by 6 ohmes=Ix.

Now let's take stock of our knowns and
unknowns.

We have an unknown voltage Va and
Va and unknown current Ix and Iex.

So I need one more equation.

Well since this is a current I'm looking
for let's just put a note right here and

I can add up my ingoing and
outgoing currents.

So Ix is comming in, Iex is comming in and

then what's going out is this current
right here, let's call it I4.

And that is going to be.

Sorry, that's going to be equal
to Va divided by four ohms.

These are the incoming currents,
that's the outgoing current.

Then we put this together
into a matrix equation,

it's going to be a four
by four matrix equation.

So equation one, two, three,
and four for unknowns Va.

Vc, Ix, and Iex.

My unknowns will be Va, Vc, Ix, and

Iex and And
that will have a set of four constants.

So my first equation,
what's multiplied by Va?

It's minus one-sixth.

What's multiplied by Vc?

It's one-sixth plus
one-half plus one-sixth.

What's multiplied by Ix?

Minus four halves.

And what's multiplied by Iex0?

What's my constant?

Take that over there.

It'll be 68/ 6.

Ok second equation,
lets multiply Va 1/6 Vc

1/6, I left out a 1/4 here.

Then minus one times I ex and
nothing times no constants

Third equation minus one sixth,
Vc is one sixth and

Ix is -1, the zero basically brought
the Ix over, zero right there.

Fourth the equation 1 times Ix,
1 times IEx, let's bring this over here,

it will be minus one fourth and
nothing times Vc and again no constant.

When I solve this What I will find
is that Iex which will come out my

vector is going to be equal
to be equal to 0.196.

Then, when I take our Thevenin which is
equal to Vex which was 1 divided by 0.196,

I will find our Thevenin Is equal to,
let me just invert that-

five wholes.

So that's how we find the Rth Thevenin
using the external voltage method.

Again we could measure,
we could simulate or

we could calculate as we did in this case.