In this video, we're going to talk about using what are called formal circuit
methods that are needed to solve circuits that you can't use a series in parallel
rules on. all the circuits we've seen to date, using voltage divider and current
divider are all you need to really solve circuits and that's applies in a vast
majority of the cases but there are a few that you can't. They had a bit different
structure and I'll show you an example of one in just a second. So instead of going
back to the v-i relations, KVL, KCL and solving a load of equations, we're going
to develop something called the node method, which very efficiently uses the
v-i relations, KVL and KCL. It turns out it's so efficient that you may want to use
it to check your answers that you get by other techniques. It's very nice, and very
general. One of the surprises is that you can use the node method to prove that
circuits conserve power. the assumptions underlying that proof are pretty
interesting, and we'll see how that works out but first, a circuit that's a little
harder than it looks. So here's the circuit, and let's look at its structure.
See if it fits into the mold of series and parallel circuits. Well, let's start over
here. This resistor looks like it could be in parallel with this series combination,
but these two aren't in series. Because there's this, this branch going off into
that 2 ohm resistor. So, these aren't in series and because of the, that branch,
this is not in parallel with those two either. And these two sort of look to be
in series but they're not because of this side branch et cetera because if you go
through it nothing is in series with anything, nothing is in parallel with
anything, it's kind of interesting. We still want to solve circuits like this and
we can do it, we know that given the v-i relations, KVL and KCL we can solve any
circuit. It's just that is very tedious to solve it that way. because, lets see, this
circuit has 1, 2, 3, 4, 5, 6 elements? That means 12 unknowns. And I don't r
eally want to write all those equations, all 12 equations and 12 unknowns. So lets
talk about the node method. Node method is very easy and very simple. So here's our
circuit and what you do is you start by defining what are called node voltages.
You pick a node as a reference and usually it's the one that you don't write KCL for
if you use the old techniques. There's usually this big node at the bottom, and
I'm going to pick that as being the reference. And that symbol is used for
reference and I have defined a nide voltage is almost all the other nodes. So,
here's a node here, a node here, and a node there. And what this, these voltages,
e1 means, it's the voltage between that node and the reference. E2 means the
voltage between this node and the reference. And since the voltage source is
attached to this node in the reference, we know that this node voltage is vn, so I
don't need to define the node voltage there, it's not necessary. However,
something a little bit more interesting, that when we define node voltages It also
means it doesn't matter which path we took to go between this node and the reference.
That is the voltage there. So that saying that this voltage is equal to that
voltage, plus that Voltage, so you can go around this way, or you can go around that
way, you get to that reference and that voltage is e1. Well that is, that implies
KVL, so we're using the KVL relations implicitly when we define node voltages as
just being the voltage between some point in the circuit, a node and the reference.
And this voltage applies no matter how what path you take to get between the two.
So the n minus 1 here refers to the fact that we don't need to write a define a
node voltage for the reference, it's the reference. And it turns out in this case,
we don't need to write three node voltages. This has, this is a four node
circuit because we already know one of the node voltage. So we only need to define
two, E1 and E2. That is going to turn out to be our only set of unknowns. We do n't
need anything else. So that means we only need two equations to write those two
unknowns. We've used KVL, so the next thing we're going to do is use KCL at each
node and we're going to write the case seal equations using the node voltages and
the v-i relations all at once. So that uses up all the things, tools that we
need. We know that the, defining node voltages using up KVL, we're going to
write KCL use up the v-i relations while we do it. So how do we do that? So here's
the set of KCL for this circuit. So let's start with e1, the node here, node 1,
there's a current going that way, current going that way and a current going that
way and I'm going to write the KCL equation, and all those currents sum to
zero. But as I write the currents I'm going to go ahead and use the v-i
relations. Now the current going that way means plus to minus that way, and so that
is e1 minus v-n is that voltage divided by 1, which is the resistance. So that's what
I mean by using the v-i relations at the same time. So the west going current is
e1-, vn over 1. The software in current is simply e1 over 1. And the east going
current is e1 minus e2 divided by 1 again. And those have to be 0. That's KCL at that
node. So we can write KCL at the other node. , where e2 is so it's e2 minus e1
divided by 1, e1 divided by 1, and e2 minus v in divided by 2 is the last KCL
equations. So, we've exhaust, we've used up all the
circuit laws. There are KVL, KCL, NDR relations. Turns out they're all
implicitly or explicitly used. We have wound up with two equations and two
unknowns. And all I have to do is solve them, and you, there's no tricks to
solving them. You just solve this set of linear equations. Now I want to point out
something that also makes the node method almost spill proof. And that is, that you
look at the equations for node 1, since we did some of currency leaving, the node
voltage for that node always appears with the plus sign. The node voltage at the
other nodes always appears with a minus sign. Always, always, always, always,
always, if you obey this convention. So, and see that happens here, too. e2 is
always with a plus sign, and the other node voltages are always there with a
minus sign. And, so at elast you get the signs right for the voltages, in your KCL
equations you gotta be careful and use the right value for the resistence, other than
that, it's foolproof, and hopefully you can solve the equations without error. So,
2 equations, 2 unknowns and all we have to do is just solve them for the node
voltages and I did this. You should check to make sure I got the right answers, and
that e1 is going to be six-thirteenths of v in. It just turns out to be
six-thirteenths. No good reason why it should be, but it is and e2 is
five-thirteenths. So just solve, 2 equations, 2 unknowns and you get the
result in terms of Vin. And now what you do, is you express the output in terms of
the node voltage. And for this circuit example, I choose that current as being my
output. And I think it's pretty easy to see that that's just e2 divided by 1 and
so the answer is five-thirteenths v in. A little note about units here this current
should be in amperes, and it looks like I have a voltage here, and I do but it turns
iyt the five-thirteenths, if you look back up at this answer, it turns out to be at
no dimensions, that 1 has a dimension in ohms. When you get numeric answers like
this, it tends to hide the fact that the units are correct, it may look a little
strange but if you did everything correctly the units should all work out
and your answer would be correct. Okay, let's use the node method to prove that
all circuits conserve power, very powerful statement. So, here is a section of a
generic circuit, not going to tell you, and I don't even know, what these elements
are, they're just elements. And it's a piece of a circuit, because i7 is going
somewhere, 5, 6, and 4 They're going off. There's some elements that have those
numbers, and, I don't need to define exactly what each circuit is, because I
want a very general answer. I'm using this core of the circuit to illustrate the core
of the circuit to illustrate that I want to make. Now, what I want to, what I'm
headed for is to show. That the VK iK is equal to zero. So what I'm going to do is
express the voltages across each element in terms of node voltages. The direction
for that voltage is going to be defined by the direction I chose for the currents.
So. I3 is going that way. So I'm going to
define d3 to be that. So that I get a consistent answer that goes according to
the rules by which we define the I relations. So, voltage across here goes
like that. And voltage across there is going to be like that, okay? So here's
what we're going to do, we're going to write VKIK for each element using the node
voltages. Okay, so let's look at element 1, let's see i1 goes that way, that means
it's plus to minus that way. So that means it's eb minus ea. And that's right up
there. And v 2, let's see plus or minus that way, node c minus node a. Right, okay
times i 2 and ec minus eb times 93 that looks right. And, let's see for circuit
element five which is somewhere out here is eb minus whatever's at the end of it,
times i5 and same thing for i6. I didn't write i4 and i7. we won't need them. Okay,
so the next thing I want to do is because I'm considering the sum of all these
terms, which are now expressed in terms of node voltages, I want to group them
according to node. What I'm going to do is group these equations, according to eb.
I'm going to find all the terms involved in eb and write them down, collect them
together. So the eb terms we have e b times, these currents. So let's check our
answers here we have an i1 from this equation. There's no eb in this equation.
here's a minus i3, okay. eb times an i5, okay, and an eb times an i6, okay. Yeah,
we got them all. Okay. Notice that this equation is the KCL
equation for node b. So, i1, i is leaving, i3 is entering. So it gets a minus sign.
i5 is leaving, i6 is leavin g. So that is KCL and what's going to happen when you
group the sum of the VKs, IKs. You group them according to node voltage what you're
going to get is that KCL because KCL applies at every node, you're going to get
a KCL equation. So, what, since all these terms are zero, for every node voltage,
the sum of the VKIK's is zero. Amazing. So, we have proven what me wanted. That.
But this is summed of all elements in a circuit including sources. That the vk,
ik's multiplying together, add 'em up you get zero. Okay, so that proves that all
circuits conserve power. But what did we use to prove this? What we used were the,
the node voltages and then, when we grouped terms, we used KCL. Using node
voltages implies KVL. So turns out, the only thing that is required through, that
circuits convert, conserve power is KVL and KCL. But that means that v-i relations
are immaterial. So this proves that for resis-, circuits consisting of resistors,
inductors, capacitors. Non linear elements, like diodes. All kinds of
things. It really doesn't matter. It's just that KVL and KCL are sufficient to
show that circuits conserve power. That's really interesting, I think. It really is
a very interesting and important result that all circuits conserve power now
matter what, how you build them and no matter what elements are used to construct
them. Okay, so it turns out you can use the node method for circuits involving
impedances. So and that's because complex amplitudes obey KVL and KCL. And, so you
can use the node method for an RLC circuit. So I took the simple circuit we
used in our example, and replaced an elements by some inductors and capacitors.
So, we begin using the convention that capital letters denote complex amplitudes.
The node voltages are now complex amplitudes. That's a capital V although it
may not look like it. You define a reference and you do exactly what we did
before. Here are the two, KCL equations for nodes 1 and node 2. And now you,
instead of resistance, you're dividing by the impedance. So, let's see e1 minus Vin,
well that's this voltage. And the impedance of the inductor is 1 times S.
I'm using the S instead of j 2 pi f again. because it really cleans up the notation
in the equations. And I'll substitute that back in there when we're done. E 1 is
sitting on top of the capacitor. So is E 1 divided by the impedance of the one farad
capacitor. And E 1 minus E 2. And divided by the resistance of the one resistor, et
cetera. You should check the equation to make sure that I didn't make any errors,
and of course, the final step is to express the output which is the current in
terms of no voltages in mean and this case, it's e2 divided by the impedance of
the inductor, which is 1s. Okay. So, everything follows. So you can solve
any circuit using a node method. Really, just resistors, or at whether RLC
circuits. It really doesn't matter. you can solve all of them. And similarly,
because it works, the proof that we just used to prove conservation of power
applies also to complex power. , the complex power. And this is the, the
average power for the Kth element. That also applies.
The reason is, is that if you have the KCL equations and if you just take the
conjugates of them, that's what can be, the conjugates also, adding the
conjugates, conjugates of complex amplitudes of current also equal zero. So
this shows that power in all senses that we've talked about, complex amplitudes or
power in the time domain all have to add up to be 0. All circuits from any
viewpoint satisfy conservation of All right, so the node method we can solve any
RL circuit using it and it's especially used when voltage and current dividers
show currents can't be applied as we start talking about electronics, you'll see
there's a really good reason why we need to talk about the node method. Like I
said, it can be used to check your answers. it's so easy to use. It's so
efficient. It's hard to write down the wrong node. I'm sorry, the wrong TCL
equations. So it's a pretty good way of c hecking answers. And its because of the
generality of the node method that applies to all circuits, it's the underlying
reason we could use it to prove that circuits conserve power.