We're continuing on with the lesson on Systematic Solution Methods, so this is
part two of the two part lesson. Just to remind you where we are in this
module, we've covered all the basic material and now we're doing the
Systematic Solution Methods. The lesson objective, is to demonstrate
Thevenin equivalent and Norton equivalent circuits.
And also source transformations, which are a way of applying the Thevenin and
Norton equivalent circuits. Looking back at this overview of the
different methods that we have, recall that mesh analysis, and node analysis,
are primarily algebraic methods. Thevenin and Norton equivalents are
primarily graphical methods where we find simple equivalent circuits.
And we do source transformations to reduce our, our schematic to a simpler
form. We've talked about when to apply mesh and
node Analysis in the first part of this lesson.
Now I wanted to mention when we would apply Thevenin and Norton equivalent.
And that's when we don't care about intermediate values, and maybe only the
output voltage or current is important. Let's look at the Thevenin equivalent
fundamental concepts here. I've got a circuit, and this circuit
could contain a lot of elements in it, it could be very messy.
It has to contain linear elements, so just no diodes or transistors, but linear
elements. It can be very messy, and I want to
simplify it and say if I'm only interested in what's happening at these
terminals, then I can reduce this entire circuit.
To something much easier, an equivalent voltage source and resistance.
And we call this Thevenin voltage, and the Thevenin resistance.
And then notice the current going, flowing through this resistance is i sub
sc. So then if I look at applying or putting
a resistor or some other load at this terminal.
The voltage and resistance, the voltage and current through this point right
here, these points a and b, is going to be the same as when I went back to the
original circuit. So, the process is based on this formula,
right here. The Thevenin Equivalent voltage is equal
to the Thevenin Equivalent resistance times Is of sc.
So I've gotta solve for these quantities. Since I've got an equation, I really only
have to solve for two of the three. [NOISE] So if I wanted to solve for the
Thevenin voltage I open circuit across the terminals ab so these ab terminals.
I open circuit across it, and then I find that open circuit voltage.
The voltage v sub a minus v sub b. And that's my Thevenin voltage.
Now to find the i sub sc, I take those same terminals, and short circuit across
them, and I find the current through them.
R Thevenin, I can find by zeroing out all my sources.
I zero out all my independent sources, so voltage sources if I zero them out.
Becomes zero, they're shorted out, there's no voltage, so that means a short
circuit. Current sources, when there's no current,
that's equivalent to an open circuit. I can only apply, I can only find
Thevenin equivalent resistance in this method when there are no dependent
sources present in my circuit. So I only have to use two of the three of
these to find these variables, and then use the equation to find the third one.
Let's look like an example of Thevenin's equivalent method to solve a circuit
problem. In this particular case, I'm going to
select part of the circuit to reduce down.
Now let's suppose I'm interested in this output.
I'm going to call them a and b, the two nodes that will determine, that will
separate my circuit into the part that I want to reduce, and the part that I
want to keep. So let's say everything to the left of
this line, I want to reduce down. And maybe the reason that I don't want to
reduce is, part of the circuit down is because maybe why, I want to try
different values in here and see how it works.
So, that way I can reduce down everything else in the circuit, and then just plug
in these values. And it's easier to solve them
individually, rather than having to re-solve the entire circuit each time.
In this case, I don't have any dependent sources, so it's usually easier to solve,
Thevenin's equivalent circuit by finding the R resistance, the Thevinin's
resistance. By zeroing out all of the sources.
I'm going to redraw this schematic, zeroing out the sources.
And that means, that I'm going to short circuit my voltages, so along to the left
there. I shorted out that voltage source.
And in here I've got a current source, and if I want zero current, that means I
open circuit it. Again, I short circuit across that
voltage, and then I leave off that R sub 0, because that's the part of the circuit
that I'm not worried about. I'm trying to reduce everything else.
So now I've got a simple circuit. Simple schematic with resistors and
series in a parallel. So, the resistors in parallel, I'm going
to replace over here with. Resistors in parallel will be, 2 times 4
or 8 over 2 plus 4 or 6. And that's going to be in series with a
ten ohm resistor. And that is equivalent to this circuit
here, which is 11 and 1 3rds. So R Thevenin is equal to 11 and 1 3rds
Ohms. Now, I'm going to find the Thevenin
voltage. And here I have a choice, I could have
found the I sub SC or I could find the Thevenin voltage, either one is fine.
So in this particular method to find V Thevenin.
I open circuit across the terminals a and b, and find that voltage.
So, let me redraw the schematic. And I leave the sources in.
But I open circuit here. So this circuit is fairly easy to analyze
because of this open circuit, all the current that flows through this leg here
has to continue on and flow through here. So the current in here exactly matches
the current there. So looking at Kirchoff's Current Law,
that means that all the current goes, that flows into this leg is exact current
that flows through that. So I've got this loop where I know the
voltage and I know the resistance, so I can use Ohm's law to find this current.
And this current, because I've got one volt and a total of six Ohms the current
has got to be 1 over 6. And that would be in Amps.
Now, I still need to find the voltage drop between a and b.
And that'll be the Thevenin voltage. So, I know this current is 1 6th, and I
know this current has got to be 0.2 going in that direction.
So if I try a voltage, a Kirchoff's voltage law around this.
Oh, I need to add in this, I have a, I have a voltage source here that I should
not have gotten rid of. It's still there.
So, let's go ahead and do a Kirchoff's voltage law around here.
Let me start from this direction. I'm going in this direction.
I hit the plus first, so I add a plus 2 and then going in this direction, I'm
going against the current flow that I've defined, so that's minus 4 times 1 6th.
And if I'm following along this way, again, I'm going against the current
flow, so it's a minus 10 times 0.2 equals 0 plus, plus v sub s equals 0.
Plus v sub Thevenin equals 0. If I solve for vThevenin I get 4 6th.
So the final picture is, I'm going to erase this here, because I've run out of
room, and I'll just show the final picture.
And it's based on this Thevenin. Voltage in that Thevenin resistance.
So this circuit, looking from the left this way, is equivalent to this circuit.
And that's 2 3rds volts. 11 and 1 3rd ohm.
Terminals a and b. And then I can then attach, if I want,
whatever resistances, that's the load I want to.
So in this particular case. We found the resistors.
The, the R Thevenin and V Thevenin. I could have also found the i sub sc.
And I applied Thevenins that way. There's a, there's actually another way
to, to do this. There's actually another way to do this.
And that's the use of the Norton equivalent circuit.
I have that basic relationship, the V and R and i relationship there.
And I can apply it either as a resistor with in series with the voltage source,
or as resistor in series with a current source.
And these two, these two. Circuits are equivalent.
Thévenin, the Thévenin equivalent circuit is equivalent to Norton equivalent
circuit. And what I mean by equivalent is if I
look at, if I look at the voltage drop across here, and the current going
through here. They're the same in the Norton equivalent
circuit as they are in the Thévenin equivalent circuit.
They both satisfy this relationship. So, the interesting thing is that I can
substitute these two configurations in a schematic, at will.
So, the Source Tranformation Method. Is to take these two configurations, and
substitute them, interchange them in a circuit.
Whenever I see the, the voltage in series with the resistor, I change it out with a
current source in series with the resistor, whichever is to my benefit.
And by doing that, I can do, I can come up with a graphical.
Way of simplifying my schematic. Now this method differs from the Mesh
node analysis, because that was some, those, those methods were algebraic in
nature. This is one where it's just graphical.
Let's show you an example of that. Looking at this circuit.
I can see this two, this right here a voltage in series with a, with a resistor
and a current. Source in series with the resistor.
This right now is a Norton equivalent circuit and this is a Thevenin equivalent
circuit. What I ultimately want to do is reduce
this down to something everything in series.
So I'd like to reduce to everything in series.
So if I look at this, I can reduce this to something in series.
So let's start with that one. Actually, let's do both of them together.
I've got 1 volt and 2 ohms, the 2 ohms stays the same.
And then the 1 volt is divided by the 2 ohms to give me current of one half.
So I've replaced this Thevenin equivalent with a Norton equivalent.
And that is in series with the 4 ohm resistor.
The reason I did this is because I wanted to combine these resistors, and I know
how to combine resistors in series, in parallel with one another.
Now this top configuration, the Norton configuration, I'm going to replace with
the voltage source in series with the resistor.
And notice my polarity here. The head of the arrow, it's easy maybe to
remember it this way, the head of the arrow is going to be the same direction
and the same polarity as the plus sign. So looking back here, the head of the
arrow, is in the same, polarity or direction as the plus sign.
[BLANK_AUDIO]. And again, this is my a terminal, and my
b node. Now, let me combine these two resistors
in parallel with one another, and that would be equivalent to a resistance.
Let me go ahead and draw that, that would be 8 over 6.
And this one right here, I didn't fill in the value there, the value of this
voltage source would be this current source times that resistance.
So that would be 2 volts. Ok, now I can go back the Thevenin over
here so I've swapped. Here from a, a Thevenin to a Norton, now
I'm going back to a Thevenin. And the whole point is to reduce this
down, step by step, so I get something that's just in series and it's easier to
analyze. The Thevenin here, would be a half this
current times that resistance, which would be 4 over 6, or 2 3rds.
Again, notice that my polarity that I chose on that matches the arrow
direction. The resistance is, recopied there.
So, now I've got. Two resistances in series so, they add.
And then I've got three voltage sources in series and they add.
But notice that two of them are in the same direction, the same polarity, and
one is the opposite. So, this one is going to subtract off.
And this reduces to, I've got a, a plus 2 plus 2 3rds minus 2.
So that ends up to be 2 3rds and then this is 10 plus 8 6ths is going to be 11
and 1 3rd ohms. And that's my simple circuit, my, the
Thevenin equivalent circuit. And it was done by source transformations
where I interchanged the Thevenin and Norton Equivalent circuits.
Those configurations, at will, to reduce down the circuit to something that is
simpler to analyze, like a series circuit.
This was, the final result was a series circuit.
So, in summary, between these two lessons.
We've covered mesh and node analysis. And it's systematic ways to find
independent simultaneous equations. It, it's an algebraic method.
Then, we've applied Thevenin and Norton methods to replace most of the circuit
with a simple equivalent circuit. And we use source transformations to do
this more or less graphically. As I've mentioned before.
As far as this course is concerned, I don't care which of these methods that
you use. They're all systematic methods, and some
students just prefer one method versus another.
There are extra worked problems given on each of these methods, but as you do the
homework, and as you do the quizzes, all I ultimately care about is that you're
able to solve the problems. So, our next lesson, we will cover
maximum power transfer, and I do want to mention that we are going to use Thevenin
equivalent circuits to find the load to maximize the power delivered.