So in this video, we're going to figure out how to solve circuits in a much more
quicker, much more efficient way, than we have by writing lots of equations and
lots of anomalies. And, the way to see how that could be is
by asking a rather interesting question, what does a source see? And of course
multi-sources and current sources don't have eyes,
but I'll show you what that means. When you go down the path, they're asking
what the source sees, that leads to the concept of equivalent circuits, one of
the most powerful concepts and circuit theory. And we'll explore series in
parallel combinations and learn how to think about circuits, in terms of these
structural components. Let's go back to our example that we did
last time. the we have a series combination of
resistors and a voltage source. And what we calculated last time was the
output voltage was proportional to the input voltage with a constant
proportionality, that's the resistor R2 / R1 + R2.
And this as I pointed out last time is called voltage divider, and that applies
any time, you have a series combination of resistors and you're applying a
voltage across them. A voltage across one of them, is that
resistance divided by the sum of the resistance as a rule that you can use to
simplify your life. I asked you last time, what was the
voltage across R1, and I think it's pretty clear, but if you plot a voltage
divider, it's the resistance of the resistor.
I'm talking about divided by sum times v in.
very easy to see, and of course, by KVL, this and this have to add up to be v in,
so everything checks out. Well, what does the source see? And what
do I mean by that? The voltage source, in this case, certainly notice what it's
voltage is. The question is, when it gets to see
through the current. So, the idea is, I'm going to put the
resistors in a box, so that the voltage source can't cheat.
The only thing you can do is figure out it's current, the current going into the
box by one and based on what that, how that's related to the voltage source,
figure out what's inside the box. We know the V out is equal to this,
and we want to know the current i1, but KCL says that i1 = i out,
that's the way it always works out for a series combination,
the current goes straight through them, and so it's easy to see if the currents
were all the same. Well, I can figure out i out from v out
by simply dividing by R2, and, that's what I get.
So, what does the voltage source see? Put in a box,
for all the world, the relationship between the source voltage and the
current, that it's supplying to the circuit looks like that of a resistor.
A single resistor having a value equal to the sum and that's the idea of what's
called an equivalent circuit. So, what does the source see? It looks in
here, through its current and it sees something that behaves just like a single
resistor, whose value is the sum of the resistances.
Of course, it doesn't know that there is a sum. It might see a resistor that looks
like 1 kilo ohm. It doesn't know how many resistors or if
there is only one resistor, that can equally be,
well be the case. But the equivalent resistance of a series
combination is the sum of the resistances.
You can really use that to simplify your calculations for circuits.
And by the way, this result all comes from the vi relations KVL and KCL.
Putting them in a much more concise, understandable form when we think about
equal answers. So, like I said, it's a single resistor.
Now, let's try another simple circuit and this one has what's called a parallel
structure. We have three circuit elements in
parallel beside each other, connected together with wires,
and that's called a parallel circuit. So we have two resistors in parallel and
a current source has been placed in parallel with that.
And what I want to do is I want to find out how the output current, the currents
of R2 is related to the input current. Well, how many nodes, how many loops are
there for writing KVL? And you should see, there are two nodes,
there's a big node here, and a big node down here.
We are only going to write KCL at that node,
and here are two loops, so we get exactly the number of equations
we need. We have two vi relations and the third
one is the, where the current source. We have the KCL equation, one KCL
equation, we have two KVL equations. One thing to point out is that when you
have resistors in parallel, it's pretty clear that the voltage across the them,
each of them is the same. And since the current source is in
parallel with that, all the voltage is here, all the voltages are equal to each
other, so that's not really a question. The question is what is this current and
how it's related to in? Well, all we have to do is put it all together.
it's not hard to show that this is the result and the way you do that is by
looking at this set of equations here, the vi relations, because these two
voltages are equal, these two quantities are equal.
And I can substitute for i1 in this equation by solving this set for i1.
What I get is i1 = R2 / R1 times i out. And if I substitute that in there and
solve, I will get this result, and this is known as current divider as you might
expect, so in a parallel circuit, if you have a current going in, it splits
between the two resistors. So, that the, the part through this resistor is the
other, the value of the other resistor divided by the sum,
and so, it works in a similar way, the voltage divider, except it's not the
resistance of the one that we're talking about, it's the other resistor. Okay.
So what does this look like? Well, again, because we get that for the input-output
relationship, well, the only way the, the current
source can see what's going on is to look at its voltage.
Well, that voltage is just that. Well, that's again the vi relationship
for a resistor. It looks like a single resistor whose
resistance is the product over the sum and I use this shorthand notation for
that result. So R1 parallel R2 means R1 R2 divided by
the sum of R1 and R2. So, again, what the source sees is a
single resistor it does not know, there are two resistors in there. All it
can tell it looks like one resistor. So, if the series in parallel combination
structures are very important to recognize to help you simplify and see
what's going on. So let's go through this little exercise
here. I'm not going to tell you if I'm, what
kind of a source I'm going to attach out here, It doesn't really matter,
the question is what does the source see? Well, we have to figure out the structure
first. It's clear that R1 is in parallel with
all of this, or what's all of this. I see R2 in parallel with R3 at there,
and then that's in series with R4, and that's going to allow us to figure out
what the equivalent resistance is. So, let's start the parallel combination
here, that looks like a single resistor whose
value is R2 parallel R3. This set now looks like a single resistor
which, whose value is R2 parallel R3 plus R4, because this parallel combination is
in series with that. And then, this single resistor, is what
that looks like there, is in parallel with R1,
and so our final result is it's a little complicated, but very simple to write.
And that is that R1 is in parallel with the combination R2 R2 and R3, and then,
which is in series with R4. Well,
I'm going to use those results to solve a circuit the easy way and this is actually
a pretty practical circuit. So what I want to point out here is that
this is the system we've already been talking about.
'Kay? And what usually happens is when you try to pass that signal on to the
sync, that it has an input resistance. In some sense, its equivalent resistance
is something which we're going to call R sub L and L means load. And the question
is, does the relationship between v in and v out change, because there is a load
resistance out there? You might say, well, geez, this voltage
here is the same as that voltage there, so it probably doesn't change.
Well, that's not quite true, and that's because these are in parallel
and these voltages are actually across the parallel combination.
There's something to figure out there. I've essentially given away how you
analyze it. What we discover is that the output
voltage is using our series of parallel combination levels.
This looks like a single resistor, which is R2 in parallel with RL, and then
that's in series with R1, so we simply use voltage divider to
figure it out. And is the concise expression, and when
you simplify everything, you get this for a final answer.
Well, that doesn't much look like the result that we had before, which was R2 /
R1 + R2 * v in. So, when is, is there a situation as we
change RL, that this result looks like that? If you think about that for a
second, there are sort of two extremes to look at,
RL going to zero, RL going to v in. That's the easy cases to look at,
which is pretty clear if you had set RL to zero, the output is zero voltage,
right? Because there's an RL [INAUDIBLE] More interesting is RL going to infinity.
When RL goes to infinity this term down here essentially disappears, because
these terms are getting big, and then the RLs cancel, and sure enough,
you get this result. So, if you want to do develop the system
such that when you attach it to the sync, it has input resistance RL.
It better, it better be that RL is big and it's going to take until the
succeeding video we define what being in comparison what is.
That's in the next video. Alright.
So, in summarize, the resistor equivalence.
So, if you have resistors in series, the equivalent resistance looks like the sum.
And voltage divider, what it looks like, the voltage across any one of them is
that resistance divided by the sum of all of the resistances.
If you have a lot of resistors in parallel, the equivalent conductance is
the sum of the conductances and the current through any one of them is that
conductance divided by the sum of the conductances.
So, it turns out writing the current divider rule, which is what this is over
here. In terms of conductances, it makes it
look a lot like the voltage divider up here,
is when you write in terms of resistances, the expressions get
complicated. It's a whole lot easier to write it in
terms of the conductances in this case. By the way, there's a, a minor error on
this sign. It should be a voltage v here and an i
there. Okay.
[SOUND] So, that's what we've learned about solving circuits.
The equivalent resistance idea comes up because you asked the question, what does
the source see? And it turns out as you can see from all these examples, when you
use the idea of an equivalent resistance along with the ideas of voltage divider
and current divider, you can solve circuits rather quickly.
And notice, we solved that circuit having a load resistor on it without ever
writing KVL, KCL, and V averages. We just wrote down the answer.
That's really very, very nice. So, the whole idea about being able to
solve circuits easily means seeing the structure,
what's in series with what, what's in parallel with what, and finding the
equivalent resistances accordingly.