In this video, we're going to extend our 
knowledge of Equivalent Circuits to the 
most general case. 
And last time we asked, what does the 
source see? And we figured out what we 
mean by sources seeing things, but in 
this video we're going to talk about what 
the circuit elements see. 
This is going to open a door to, the 
generalization of what we've already 
talked about namely Thevenin and the 
Mayer-Norton equivalent circuits. 
These are very important to understand 
and appreciate the knowledge they give 
us. 
So let's Talk about a little bit of an 
extension of what we've already talked 
about. 
You recall this was our original circuit. 
We had a source, and our system was a 
series resistance and we now, we know 
That, for that case V-out was R2/1+R2 * 
VN, we use voltage divider to find it. 
Well, if you're going to actually use 
that voltage for something it has to be 
attached to the sync. 
Remember our whole system thinking that 
we have to send our message to the sink. 
And the input to most circuits that we 
care about look like a simple resistor. 
And so the question is, if I attach my 
circuit to. 
The receiving circuit which is going to 
look like a resistor, does that change 
the output voltage V out. 
You may say well the voltage v out is 
across both the RL and across our 
original R2. 
But in reality it's really across the 
parallel combination. 
Well that changes the resistance, so you 
should expect there might be at least the 
possibility of some change in V out, so I 
want to explore that. 
By the way, RL, this is standard 
terminology, this L here stands for load, 
and that's a term that comes from the 
power in the circuit. 
Well, how do we think about this? And 
here's the way I want to do this. 
I want to try to figure out what does the 
load resistor see? What is the circuit 
that it sees looking back in that way? 
That's going to be a key for this, so 
that's what I meant by what are the 
circuit elements C. 
Well, it's not just a simple resistor of 
course, there's a voltage source in here, 
so the series and parallel rules don't 
quite apply. 
However, we can use the Principle of 
Superposition to figure this out. 
If you think about this for a little bit 
If I drew a box around here, this just 
looks like a regular old generic circuit 
with a current define going in the 
positive side, of the voltage defined for 
it. 
And I could figure out the, voltage if 
you just gave me the current. 
But because of the presence of the source 
There's going to be an additional term, 
which is the part of v that's due to the 
source. 
So the principal of superposition says 
that voltage V will be equal to the sum 
of the voltage you get when you set the 
voltage source that's inside, V in to 0. 
Plus the part you get when you set the 
input current to 0. 
And that will give you the, the part of 
V, this is to the V in. 
Okay? So, let's do the first one. 
When we set the voltage source to 0, 
that's the same as a short circuit. 
So remember our voltage source supplies 
some voltage for any current. 
Well, if you set that source to zero now 
the voltage is zero for any current. 
That's the definition of an ideal wire, 
or a short circuit. 
So now, what do we see looking into The 
circuit, we see R1 in parallel with R2. 
And so we know that at least that 
component of the output voltage is given 
by a simple resistive like term. 
And now, if you go on to setting I to 0. 
So I'm going to turn the source back on. 
Our source is alive and well. 
In a setting i=0, means there's nothing 
attached to the terminals, and what we 
have is the same circuit that we had 
before that we already analyzed. 
You simply use voltage divider to figure 
out what vout is and this is what we got. 
Last time. 
So, v, is going to be the sum of 2 terms. 
If you look, think about, that, as a 
system, and you want to know what the v i 
relation is, you can use this super 
position principle. 
And we see that, at least for this case, 
that's what we get. 
We get that v is equal to some resistance 
linked term times the current plus a 
source linked term. 
Well, that turns out, opens the door to a 
really interesting result. 
So, let's define something. 
I'm going to call that r e q, equivalent 
resistance. 
I'm going to call the second term v e q, 
equivalent voltage. 
And I want to point out, that something, 
that of this form. 
Which can be written like that, that 
looks, that is the same via our 
relationship that applies to this 
circuit. 
And of course the subscripts EQ here mean 
equivalent. 
So these two circuits cannot, you cannot 
tell them apart Just from the terminals. 
They look identical because their VI 
relationships are exactly the same. 
So I can express in particular as a very 
simple circuit, a very complicated 
circuit, containing voltage sources. 
And resistors. 
We're going to show you in a second that 
you can in addition to alter sources, you 
can have current sources too. 
So this is called the Thevenin equivalent 
circuit, and it's named for a late 19th 
century enigneer Who developed the 
equivalent circuit idea, while he was 
learning to teach a course at the eco 
polytectic in Paris. 
So, that's quite interesting. 
But there's even more. 
So, here's our simple little circuit that 
we've been talking about. 
And here's the, Thevenin equivalent. 
relationship for it. 
I'm going to turn that equation around. 
I'm going to write this as i equals 
something, and I'm going to call that 
last term, IEQ. 
And I want to point out that that 
equisite of that equation Describes 
behavior of this circuit. 
Where there is a current source, in 
parallel with REQ, it's the same REQ that 
we had, when we talked about the Thevenin 
equivalent, there's no difference. 
This is called the Mayer-Norton. 
Equivalent circuit. 
Mayer-Norton is named for two engineers, 
Edward Norton who is from the US and Hans 
Ferdinand Mayer who is from Germany, they 
developed this idea, at the, almost 
exactly same time, but independently. 
So. 
We have two, different ways, of thinking 
about, a circuit. 
So. 
We think about these, terminals, this 
pair of terminals out here, as a place 
where you're about to attach something, 
some other circuit. 
And you want to know. 
What does that circuit look like? In 
order to simplify your calculations, you 
can describe it either as a voltage 
source in series with a resistor, or you 
can describe it as a current source in 
parallel with the same resistor. 
Whichever seems to be the easiest 
approach for your application. 
They both apply. 
They, you can do either one. 
So, this really simplifies things. 
In fact, this is a profound result. 
This says that no matter how complicated 
A circuit is in side consisting of 
sources, current sources, voltage sources 
and resistors. 
It all can be summarized as a simple 
voltage source and series with equivalent 
resistance. 
This equivalent resistance is sometimes 
cal, called the output resistance. 
And we'll see that this is an important 
quantity in just a second. 
Now, and these, this equation ties these 
two equivalent circuits together, so that 
VEQ is REQ times Ieq so if you find one. 
You, you've got all you need to find the 
other, no problem. 
So, one issue with finding equivalent 
circuits is, can you really, do it. 
Find the, equivalent voltage, equivalent 
resistance. 
Without looking inside the box can you 
make measurements just at the terminals 
to figure out what those quantities are? 
And the answer is yes. 
The first thing we do is to find what we 
call the open-circuit voltage and that 
means we set this current to zero. 
And we look at the voltage we get. 
Well, if you look at this equivalent 
circuit. 
If you set that current to 0, that means 
there's no voltage drop across that 
resistor, so v is just equal to v(eq). 
So if you do that for a real circuit You 
just open-circuit the terminals, don't 
attach anything, and you just measure the 
voltage. 
That voltage that you measure is the 
equivalent, the Thevenin equivalent 
voltage. 
Similarly, you can short-circuit the 
current, so what that means Is that you 
short circuit these terminals, you put a 
short across there. 
That sets the voltage to, zero of course 
and then you measure the current. 
And, when you do that, you look at this 
equivalent, circuit, that short circuit 
current, is just going to be minus VEQ, 
over REQ, and that's the negative. 
Of the Myer-Norton equivalent current 
source so, I'm making measurements at the 
terminal, the open circuit voltage and 
the short circuit current. 
You do not have to look inside the box, 
to figure out what the equivalent 
circuits are, so let's 
go on, and to a simple example. 
And so, to show these ideas, I want to 
find the equivalent circuits for this 
example circuit. 
And I could find, explicitly, the. 
Thevenin equivalent. 
I could find the minor Norton equivalent, 
but I'm just going to find, the 
equivalent, quantities, the Veq, Req, and 
Ieq. 
so, the first thing we can do is set all 
the sources to zero. 
If you set the source to zero in the 
Thevenin equivalent, that's your short 
circuit, and what you see from the 
terminals is just that resistor. 
If you set the current source to zero, 
that's the same as opening it up, and 
then again, you also see Req. 
So, here if you look inside the box we 
see we have a current source, a zero 
valued current source. 
It's the same as an open circuit, no 
current flows through any voltage. 
That's the definition of an open circuit. 
And what I see when I look this way. 
Into there, I see R3 is in parallel with 
this series combination. 
So we can use our series and parallel 
formulas and we get that for Req. 
So, that's the source resistance of the 
Rs, of our circuit. 
Now, suppose we want to find VEQ. 
I'm just trying to use our terminal only 
results, I'm going to open circuit 
voltage so there's nothing across here, 
this is 0. 
And now we measure the voltage. 
Well this voltage. 
Is, going to be given by some dependence 
on this current source, and since it's a 
current source, it's a whole lot easier 
to find the current through here, and 
then I'll multiply that by R3 to get the 
voltage. 
Well. 
Because we have this in parallel with 
this now, from the point of view of the 
volt of the current source. 
I'm going to use current divider which is 
going to be the other resistor divided by 
the sum of that one plus this series 
combination. 
And when you all get said and done, 
that's what you get. 
So, here's the, current divider part, and 
then I just multiply by R3 to get our 
voltage. 
So that is the Thevinin equivalent, 
voltage, for this circuit. 
It's a little more complicated than you 
might think, but it's pretty easy to 
find. 
And then finally I'm going to find IEQ, 
which I'm going to do by finding the 
short circuit current. 
And I just put a wire across here, and, 
so what that does by setting the voltage 
to zero, that shorts out this, so that 
resistor isn't there, doesn't matter if 
it's there or not. 
And so, again, I'm going to use current 
divider, because I have a current source 
here. 
Well, the current i is going to be, i is 
going to be the negative of that current, 
and I used current divider, so current 
divider says that current is R1 divided 
by the sum of the resistances times I in. 
And I stick a negative sign in front 
because of the directionality of the 
current and I get this for isc, the short 
circuit current which, I know, is the 
negative of Ieq. 
So, we could have divided The equivalent 
voltage, by the equivalent, current, 
defined req. 
We u-, in finding req up here, what we 
did was look inside the box. 
And we don't really have to do that. 
But it's nice to know we have a way of 
checking your answers. 
So if the source is to zero to find Req. 
Find the open circuit voltage to find an 
evident equivalent voltage. 
find the short circuit current. 
Find the Meyer-Norton equivalent current 
source, and then your ratios should work 
out to be the other. 
Everything should check. 
Okay. 
So, what did we discover? Well, we 
discovered that all circuits consisting 
of sources and resistors are equivalent 
to one source, be it a voltage source or 
a current source, that's in series or in 
parallel with a single resistor. 
No matter how complicated that circuit 
is. 
From the point of view of the connection 
points, the terminals. 
It is a very, very simple circuit. 
So, that means that, we can use either of 
these equivalent, circuits. 
The Thévenin or Mayer-Norton form. 
The voltage source or the current source 
form. 
Whichever seems to be appropriate. 
It makes solving the problem easy. 
What I mean by that. 
Is you're now going to attach something 
to the terminals, and, and des, determine 
things. 
And, depending on what you're interested 
in finding, how complicated the circuit 
is you attach, you would use one or the 
other. 
Now. 
Remember how we got started on this. 
Remember we were attaching our, Sync to 
our original circuit That we've been 
talking about. 
And the issue was what's called Loading. 
And, Loading is a term that means, that 
if I attach to the wrong kind of load 
resistor for the given application. 
This V out will change. 
That's called loading down the original 
circuit. 
Well, I want to figure out what values of 
RL will not contribute to loading. 
What kind of resistor values can the sink 
have so the The alk doesn't change from 
what it is if nothing is attached. 
Well, the, we now know that our system 
has, looks like this. 
This, which is, I'm going to use the 
Thevenin equivalent. 
Because I have voltage that's the output. 
And the voltage I want is across RL. 
Well, what is v out for this really 
simple circuit? Well, it's the voltage 
divider. 
It's very easy to find. 
And, what values of RL. 
We'll allow V out to be = VEQ. 
Because when RL is infinity, that's the 
same as attaching nothing. 
and that's what we started. 
So I know that the output that I want is 
VEQ, so I think it's pretty clear that 
what we want is RL to be big. 
But big relative to what? 
And the answer is it has to be big 
relative to Req, which is R1 in parallel 
with r 2. 
So as long as the load is big compared to 
the Whatever this parallel, resistance 
turns out to be, then the, you can attach 
the sync to the circuit and it won't 
change V out. 
However, if RL is smaller than that 
quantity or comparable to it, that 
outputs going to change and that may not 
be what you want It depends on the 
application. 
This little, simple example points out 
the power of using equivalent circuits.