In this video, we're going to connect those circuit elements together to make
an entire circuit. Right now, all we know about the circuit
elements are what their individual VI characteristics are.
How the current and voltage are related to each other because of the element's
construction. When you connect them together, we have
to not only connect them physically, but we have to have the laws that govern the
voltages and currents when elements are connected together.
And these are called the Kirchhoff's Laws, and as you might expect, there's a
current law and a voltage law. And what we're going to find is that once
we have the VR relations KCL, Kirchhoff's Current Law, and KVL, Kirchhoff's Voltage
Law, we can solve any circuit. So let's start with a very simple
circuit. And here we have three elements connected
together. I have a voltage source and two resistors
never connecting together one after another, in what's called a series
configuration. And what series means is that, if I was
electron, this is the way I like to think about it.
An electron, if I try to go out of one element, and I have no choice, but to go
through the next one, that means you're in series.
So we have an entire series, configuration of a voltage source and two
resistors. All right.
Now, I would point out that we have three circuit elements and we have a voltaging
incurrent for each element, which means, we have six unknowns and in order to
define six unknowns, you're going to need six equations.
And all we have right now are the Vi relations, so I'm calling the voltage
across R2 our output forward just because I wanted to and here's the Vi relation
for the source and here it is for the 2 resistors and of course anything could be
a function of time as indicated. Alright, but 3 equations isn't 6, we need
3 more. And that's where the interconnection laws
come in. Alright, so the first one we're going to
talk about is Kirchhoff's Current Law, KCL, which basically says, the sum of
currents entering a node, or leaving a node, equals zero.
So, what's a node? So we look back at our circuit.
A node is a connection point. That's where 2 or more elements are
connected together and they're really attached.
So, in this circuit, I see a node there, I see a node here where R1 and R2 are
connected and I see a big node here where essentially R2 is connected to the other
end of the voltage source. And this wire going between them as we
know is just a short circuit, an ideal wire.
And it doesn't contribute much to anything.
There's no voltage drop, and the current just goes through it as it would
normally. So I'm going to consider this one being
So we have three nodes. Now what I'm going to do is I'm going to
write KCL for each node and I'm going to do it in terms of currents entering the
node. So let's consider them one at a time.
So if we're the blue node, the current entering from the south,
the current going in that way is -i. And the current entering from the east is
-i1. Right? You just re-flip the sign to
reverse the direction of the current arrows.
And so, in terms of currents entering a node, -i-i1, is 0.
Okay. Go on to the next node, and the red node,
the current entering from the west is the is i1, the current entering from the
south is -i out, so i1 - iout 0. And then finally, for the, green node.
currents entering, we have I entering here, and I entering there.
So, those, we have 3 nodes we had 3 KCL equations.
Well, I want you to notice something about these KCL equations.
If I were to add these two equations together,
take a minus sign and negate it, I would get the other equation.
And that's not good. Remember we're trying to find six
equations and six unknowns. And if the equations are not what's
called linearly independent, they don't do us much good.
There's some sense one too many equations here.
So, what's going on, and what's going on is a detail about KCL and construction of
every circuit. So, if you think about this circuit in 2
pieces. So we have the big yellowish kind of area
here, and our green node at the bottom. Because KCL is satisfied at each of the
individual nodes here. The sum of currents leaving the yellow
surface has to be is, i + iout. Well those currents have to leave and go
into the green node by definition. So that gives us the other KCL equation
of currents entering. So, it turns out that in any circuit, if
you have n nodes, you have n-1KCL equations that are
linearly independent and the only nature writing for any set of n-1 nodes, you can
pick anything more. As you're going to as we go through the
course, we typically write circuits such that there's always sort of I note at the
bottom where lots of things are connected together, and that tends to be the one we
don't write KCL for. And so I'm going to throw out that
equation. I don't need it.
It can be derived from the other two. So let's move on.
we're going to need another equation. We've got three VI relations, two KCL
equations that are independent of each other.
we need that last equation so we can solve our circuit, and of course that's
going to come from the voltage law. So Kirchhoff's voltage law is that the
sum of voltages around any closed loop in a circuit must add up to zero.
So here's our circuit, and we have three voltages here, that are defined.
And what we're going to do is go around a little loop, and what, the way you do
that is if we define we're going around the loop clockwise, that's what the
graphic says, we're going to take The voltage signs according if it goes plus
to minus. So, I'm going to bring in the one with
the plus sign, the out, with a plus sign. But over here, d is defined minus the
plus, so I have to bring it in with a minus.
And that's my, KVL equation. So, from one point of view I'm very
happy, in terms of solving my circuit. I've got 3Vi relations, 2 KCL equations,
and 1 KVL equation, so that's 6 equations; we should be able to solve our
circuit. But the similar issues arises with KVL
that does in KCL. How many K, KVL equations are there
actually for any circuit? Here there's only one loop, but how about this
circuit? How many loops are there? Well, I get one loop, involving On the voltage
source; r1 and r2. And, I get another loop out here, in
around it, that's surrounded by r2 and r3.
However, I point out, there's another loop.
And that is, a green loop, that's also there too.
Well, it turns out, the number of KVL equations you need Is determined by how
many open areas there are in the circuit. If you thought about this as a a window,
and these were pieces of wood holding together the pieces of glass, the number
of KVL equations is the number of pieces of glass.
In this case, two. So, the number of KVL equations I would
write would be two, and I'd write it for each pane of glass, if you will.
So, that's how you find the number of KVL equations that any, that need, that you
need to solve for any circuit. All right, so let's solve that circuit.
So we have 3Vi relations, we had 2 KCL equations and we have 1 KVL equation.
Aren't we happy? All we understand is finding, what the output voltage is for
the given source voltage and we have 6 equations and 6 unknowns to plow through.
So we want to eliminate everything. We will eliminate all the currents, we
will eliminate the one, and we want a just relationship between Vi and V.
And with a little bit of going through this and thinking about it, is actually
pretty easy to do, and I won't go through all that agony.
I'll just say that after some manipulation.
you should be a little worried about that.
It's easily shown, that what you get for an output, is that it is the output
voltages proportional to the input voltage, and constant proportionality is
given by the ratio of R2 to the sum, R1 + R2.
It turns out, we have derived this in general, right? We've, have, don't have
specific resistors. I usually derive this in general.
So anytime you see a voltage source attached to two resistors in series like
this, you'll always find that the voltage across a resistor of phases rule.
And in fact, I'm going to ask you to tell me, what is the voltage v1? And I want
you to do that without doing any calculations.
I want you, you should be able to just tell me what that answer is.
All right. Well that relationship that we, for,
voltage across a series connection of resistors is called voltage divider, and
the resistors the voltage, divide it. Is where the terminology comes from.
So the bigger R2 is the closer the ratio here is to 1.
The smaller r2 is relative to r1 the smaller this proportionality constant
gets. So, what does our circuit do? So,
remember what we're trying to do is build systems and we're trying to build them at
this point out of circuits. So in terms of black diagram models, we
have a source that's producing some voltage VN, and there's the source, it's
literally the voltage sources. The system is the entire circuit it's
attached to, and we've taken as an output the voltage across R2 And what our
system, the input output relationship, we have shown is given by that expression.
So what does our system do? Have we encountered a system like that already?
And the answer is we certainly have. It serves as a kind of amplifier.
The gain of our amplifier is simply R2 divided by R1 + R2 and because our gain
has to be less than 1, resistors are always positive valued, it turns out we
have an attenuator. And so what our little system, our
circuit does In terms of system theory, is that it serves as an attenuator that
makes voltages smaller. So, let's do a little power calculation.
I'm very interested in what is consuming power.
And what is the source of the power that's being consumed.
So as we know power for each element is the voltage that we have * it's current
and I want to consider that for each element.
So we know for resistor that the voltage, the power rather, is the voltage squared
divided by the resistance, that's v times i for a resistor, and since we solved for
the output voltage, we might as well use this version of the formula.
So what we get is that the voltage, output,
the output voltage, just to remind you was R2 over R1 + R2 - VN, so you square
that and divide by R2 you get that, and similarly the hour
dissipated across resistor one is if you had a similar expression where R2 R2 is
replaced by R1, makes things pretty easy to see.
Because these are positive valued powers, right? Resistors are always positive
value and we're the [UNKNOWN] the voltage coming in.
The terms of that means these 2 resistors dissipate power, they consume it.
Well it has to come from somewhere, so let's see what's going on at the source.
Yeah. The output current is what we need.
We need to find We have the voltage v that's just given by the source and we
won't need to find that current i is going in.
Well, for a series combination, I think it's pretty clear that i1 = iout and that
i = -i1. So, if I just find the iout which is
given by the voltage that we found divided by R2.
We know that that is iout, and I will be equal to -iout and we get for a power
produced by the source, is the negative of VN squared divided by R1+R2, and
negative means it produces power. It is the one that is producing power in
a circuit. If you add these two up, if you add
everything up, you get zero, and that's the way it should be.
It has to be that power is conserved. There are going to be parts of a circuit
that consume power, and there's going to be parts that produce it and they have to
be in balance or something is wrong with, with the calculations.
And so, no net power is produced or consumed by a simple circuit.
That's the way the laws of physics are. So, and this is the general formula that
if you, for a given circuit, if you sum over all the circuit elements the sum of
the b k, i k's have to be zero. We'll prove that later.
So, along with the v i relations, KVL and KCL always give you exactly the number of
equations you need, to solve any circuit. So you have to go through them all, and
of course you don't need all the set of equations.
What I mean by that, if our, n-node circuit, you need n-1 KCL equations.
And for, a number of openings in a circuit are only, the number of loops you
need for which you write KVR. Now, I do not want to spend my life,
solving lots and lots of sets of equations just to find an input, output
relationship. So we really need methods to streamline
our calculations, and we're going to talk about those soon.
And it turns out, they're very easy, to use, and, not only simplifying our life,
but give us insight, into how circuits behave.
Now, a little bit later I'm going to prove, that only KVL and KCL are needed,
to show that every circuit Conserves power.
Some of the BK, IKs is indeed zero. That's really interesting and we'll see
of that in a succeeding video.