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.