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.