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Good morning,
OK.

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Let's get started.
We have one handout today.

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That's your lecture notes.
There's some copies still

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outside for those who haven't
picked one up.

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In general, what I do is,
in the lecture notes,

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I leave out large amounts of
material.

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So, this will enable you to
keep your hands busy while I'm

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lecturing and take down some
notes and so on.

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So, don't assume that
everything that I talk about is

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on here.
Please follow along.

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OK, so as is my usual practice,
let me start with a quick

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review of what we covered so
far.

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So what we did primarily was
looked at this discipline that

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we call the lump matter
discipline, which was very

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similar, very reminiscent of the
point mass simplification in

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physics.
And this discipline,

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this set of constraints we
imposed on ourselves,

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allowed us to move from
Maxwell's equations to a very,

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very simple form of algebraic
equations.

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And specifically,
the discipline took two forms.

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One is, we said that we will
deal with elements for whom the

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rate of change of magnetic flux
is zero outside of the elements,

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and for whom the rate of change
of charge I want to charge

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inside the element was zero.
So, if I took any element,

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any element that I called a
lump circuit element,

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like a resistor or a voltage
source, and I put a black box

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around it, then what I'm saying
is that the net charge inside

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that is going to be zero.
And this is not true in

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general.
We will see examples where,

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if you choose some piece of an
element for example,

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there might be charge buildup,
but net inside the,

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if I put a box around the
entire element,

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I am going to assume that the
rate of change of charge is

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going to be zero.
So, what this did was it

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enabled us to create the lump
circuit abstraction,

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where I could take elements,
some element of the sort,

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this could be a resistor,
a voltage source,

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or whatever,
and I could now ascribe a

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voltage, some voltage across an
element, and also some current,

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"i," that was going into the
element.

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And as I go forward,
when I label the voltages and

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currents across and through
elements, I'm going to be

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following a convention.
OK, the convention is that I'm

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going to label,
if I label V in the following

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manner, then I'm going to label
"i" for that element as a

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current flowing into the
positive terminal.

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It's just a convention.
By doing this,

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it turns out that the power
consumed by the element is "vi"

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is positive.
OK, so by choosing I going in

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this way into the positive
terminal, the power consumed by

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the element is going to be
positive.

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OK, so in general of even
simply following this

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convention, when I label
voltages and currents,

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I'll be labeling the current
into an element entering in

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through the plus terminal.
Remember, of course,

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if the current is going this
way, let's have one amp of

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current flowing this way,
then when I compute the

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current, "i" will come out to be
negative.

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OK, so by making these
assumptions, the assumptions of

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the lumped matter discipline,
I said I was able to simplify

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my life tremendously.
And, in particular what it did

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was it allowed me to take
Maxwell's equations,

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OK, and simplify them into a
very simple algebraic form,

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which has both a voltage law
and a current law that I call

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Kirchhoff's voltage law,
and Kirchhoff's current law.

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KVL simply states that if I
have some circuit,

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and if I measured the voltages
in any loop in the circuit,

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so if I look at the voltages in
any loop, then the voltages in

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the loop would sum to zero.
OK, so I measure voltages in

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the loop, and they will sum to
zero.

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Similarly, for the current,
if I take a node of a circuit,

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if I build the circuit,
a node is a point in the

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circuit where multiple edges
connect.

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If I take a node,
then the current coming into

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that node, the net current
coming into a node is going to

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be zero.
OK, so if I take any node of

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the circuit and sum up all the
currents going into that node,

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they will all net sum to zero.
So, notice what I've done is by

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this discipline,
by this constraint I imposed on

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myself, I was able to make this
incredible leap from Maxwell's

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equations to these really,
really simple algebraic

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equations, KVL and KCL.
And I promise you,

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going forward to the rest of
6.002, if this is all you know,

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you can pretty much solve any
circuit using these two very

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simple relations.
It's actually really,

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really simple.
It's all very simple algebra,

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OK?
So, just to show you an

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example, let me do a little
demonstration.

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Let me build let me build a
small circuit and measure some

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voltages for you,
and show you that the voltages,

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indeed, add up to zero.
So, here's my little circuit.

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So, I'm going to show you a
simple circuit that looks like

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this, and let's go ahead and
measure some voltages and

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currents.
In terms of terminology to

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remember, this is called a loop.
So if I start from the point C

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and I travel through the voltage
source, come to the node A down

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through R1 and all the way down
through R2 back to C,

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that's a loop.
Similarly, this point A is a

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node where resistor R1 the
voltage source V0,

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and R4 are connected.
OK, just make sure your

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terminology is correct.
So, what I'll do is I'll make

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some quick measurements for you,
and show you that these KVL and

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KCL are indeed true.
So, the circuits up there,

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could I have a volunteer?
Any volunteer?

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All you have to do is write
things on the board.

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Come on over.
OK, so let me take some

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measurements,
and why don't you write down

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what I measure on the board?
What I'll do is,

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let me borrow another piece of
chalk here.

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What I'll do is focus on this
loop here, and focus on this

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node and make some measurements.

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All right, so you see the
circuit up there.

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OK, so I get 3 volts for the
voltage from C to A.

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so why don't you write down 3
volts?

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OK, so the next one is -1.6.
And so that will be,

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I'm doing AB,
V_AB.

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OK, and then let me do the last
one.

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It is -1.37.
The measurements,

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I guess, have been this way.
So, what's written is V_AC.

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But it's OK for now.
Don't worry about it.

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So, well, thank you.
I appreciate your help here.

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OK, so within the bonds of
experimental error,

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noticed that if I add up these
three voltages,

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they nicely sum up to zero.
OK, next let me focus on this

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node here.
And at this node,

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let me go ahead and measure
some currents.

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What I'll do now is change to
an AC voltage so that I can go

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ahead and measure the current
without breaking my circuit.

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OK, this time around,
you'll get to see the

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measurements that I'm taking as
well.

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So, what I have here,
I guess you can see it this

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way.
What I have here is three wires

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that I have pulled out from D.
And this is the node D,

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OK?
So, I have three wires coming

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into the node D just to make it
a little bit easier for me to

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measure stuff.
OK, so everybody keep your

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fingers crossed so I don't look
like a fool here.

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I hope this works out.
So, you roughly get,

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what's that,
10 mV.

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OK, so it's about 10 mV peak to
peak out there,

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and let's say that if the
waveform raises on the left-hand

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side, it's positive.
So, it's positive 10 mV.

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And another positive 10 mV,
so that's 20 mV.

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And this time,
it's a negative,

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roughly 20, I guess,
-20.

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So, I'm getting,
in terms of currents,

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I have a -10,
-10, I'm sorry,

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positive 10,
positive 10,

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and a -20 that adds up to zero.
But more interestingly,

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I can show you the same thing
by holding this current

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measuring probe directly across
the node.

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And, notice that the net
current that is entering into

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this node here is zero.
OK, so that should just show

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you that KCL does indeed hold in
practice, and it is not just a

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figment of our imaginations.
So, before I go on,

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I wanted to point one other
thing out.

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Notice that I've written down
two assumptions of the lumped

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matter discipline,
OK?

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There is a total assumption of
the lump matter discipline,

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and that assumption is,
in spirit, at least,

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shared by the point mass
simplification in physics as

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well.
Can someone tell me what that

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assumption is?
A total assumption,

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which I did not mention,
which you can read in your

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notes in section 8.2 in the
appendix, what's a total

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assumption that is shared in
spirit with the point mass

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simplification?
Anybody?

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A total assumption to be made
here is that in all the signals

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that we will study in this
course, we've made the

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assumption that the signal
speeds of interest,

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transition speeds,
and so on, are much slower than

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the speed of light.
OK, that my signal transition

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speeds of interest are much
slower than the speed of light.

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Remember, the laws of motion,
the discrete laws of motion

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break down if your objects begin
moving at the speed of light.

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OK, the same token here,
our lump circuit abstraction

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breaks down if we approach the
speed of light.

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And there are follow on courses
that talk about waveguides and

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other distributed analysis
techniques that deal with

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signals that travel close to
speeds of light.

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OK, so with that,
let me go on to talking about

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method one of circuit analysis.
This is called the basic KVL

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KCL method.
So just based on those two

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simple algebraic relations,
I can analyze very interesting

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and complicated circuits.
The method goes as follows.

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So, let's say our goal is,
given a circuit like this,

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our goal is to solve it.
OK, in this course,

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we will do two kinds of things:
analysis and synthesis.

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Analysis says,
given a circuit,

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OK, what can you tell me about
the circuit?

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OK, so we'll solve existing
circuits for all the voltages

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and currents,
voltages across elements,

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and currents through those
elements.

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Synthesis says,
given a function,

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I may ask you to go and build
circuits.

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OK, so for analysis here,
we can apply this method that I

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want to show you.
And the idea here is that,

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given a circuit like this,
let us figure out all the

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voltages and currents that are a
function of the way these

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elements are connected.
So, the basic KVL and KCL

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method has the following steps.
The first step is to write down

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the element VI relationships.
OK, right down the element VI

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relationships for all the
elements.

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The second step is write KCL
for all the nodes,

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and the third step is to write
KVL for all the loops in the

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circuit.
That's it.

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Just go ahead and write down
element rules,

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KVL, and KCL,
and then go ahead and solve the

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circuit.
So, what we'll do,

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we'll do an example,
of course.

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But, just as a refresher,
we've looked at a bunch of

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elements so far,
and for the resistor,

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the element relation says that
V is pi R, where R is the

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resistance of the element here.
For a voltage source,

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V is equal to V nought.
That's the element

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relationship.
And for a current source,

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the element is the relation is,
"i" is simply the current

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flowing through the element.
OK, so these are some of the

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simple element rules for the
devices that the current source,

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voltage source,
and the resistor.

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So let's go ahead and solve
this simple circuit.

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And what I'll do is go ahead
and solve the circuit for you.

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OK, if you turn to page five of
your notes, I'm going to go

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ahead and edit the circuit here.
You can scribble the values on

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your notes on page five.
OK, so as a first step of my

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KVL KCL method,
I need to write down all my

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element VI relationships.
So, before I do that,

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let me go ahead and label all
the voltages and currents that

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are unknowns in the circuit.
So, let me label the voltages

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and currents associated with the
voltage source as here.

231
00:18:22,000 --> 00:18:26,000
Notice, I continue to follow
this convention where whenever I

232
00:18:26,000 --> 00:18:31,000
label voltages and currents for
an element, I will show the

233
00:18:31,000 --> 00:18:35,000
current going into the positive
terminal of the element

234
00:18:35,000 --> 00:18:40,000
variable, OK,
after element variable voltage.

235
00:18:40,000 --> 00:18:42,000
So here, I have V nought and I
nought.

236
00:18:42,000 --> 00:18:47,000
Let me pause here for five
seconds and show you a point of

237
00:18:47,000 --> 00:18:49,000
confusion that happens
sometimes.

238
00:18:49,000 --> 00:18:53,000
Often times,
people confuse between what is

239
00:18:53,000 --> 00:18:57,000
called the variable that is
associated with the element

240
00:18:57,000 --> 00:19:01,000
versus the element value.
OK, notice that here,

241
00:19:01,000 --> 00:19:07,000
capital V nought is the voltage
that this voltage source

242
00:19:07,000 --> 00:19:12,000
provides, while this name here,
v nought, is simply a variable

243
00:19:12,000 --> 00:19:17,000
that we've used to label the
voltage across that element.

244
00:19:17,000 --> 00:19:21,000
So, similarly,
I can label v1 as the voltage

245
00:19:21,000 --> 00:19:26,000
across the resistor,
and i1 is the current flowing

246
00:19:26,000 --> 00:19:30,000
through the resistor.
So this method of labeling,

247
00:19:30,000 --> 00:19:36,000
where I follow the convention,
that the current flows into the

248
00:19:36,000 --> 00:19:44,000
positive terminal is called the
associated variables discipline.

249
00:19:44,000 --> 00:19:48,000
I was trying to use the word
discipline in situations where

250
00:19:48,000 --> 00:19:50,000
you have a choice,
OK, and of a variety of

251
00:19:50,000 --> 00:19:54,000
possible choices,
you pick one as the convention.

252
00:19:54,000 --> 00:19:56,000
OK, so here,
as a convention,

253
00:19:56,000 --> 00:20:00,000
we use the associated variables
discipline, and use that method

254
00:20:00,000 --> 00:20:04,000
to consistently label the
unknown voltages and currents in

255
00:20:04,000 --> 00:20:09,000
our circuits.
OK, so let me continue the

256
00:20:09,000 --> 00:20:13,000
labeling here,
v4, i4, i3, v3 here,

257
00:20:13,000 --> 00:20:16,000
and v2 and i2,
v5, and i5.

258
00:20:16,000 --> 00:20:22,000
I think that's it.
So, I've gone ahead and labeled

259
00:20:22,000 --> 00:20:27,000
all my unknowns.
So each of these voltages and

260
00:20:27,000 --> 00:20:35,000
currents are the voltages and
currents associated with each of

261
00:20:35,000 --> 00:20:40,000
the elements.
And my goal is to solve for

262
00:20:40,000 --> 00:20:44,000
these.
OK, so in terms of our solution

263
00:20:44,000 --> 00:20:49,000
here, let's follow the method
that I outlined for you.

264
00:20:49,000 --> 00:20:54,000
So, as the first step I am
simply going to go ahead and

265
00:20:54,000 --> 00:21:00,000
write down all the element VI
relationships.

266
00:21:00,000 --> 00:21:05,000
OK, so as a first step,
I'm going to go ahead and write

267
00:21:05,000 --> 00:21:12,000
down all the VI relationships.
So, can someone yell out for me

268
00:21:12,000 --> 00:21:16,000
the VI relationship for the
voltage source?

269
00:21:16,000 --> 00:21:20,000
OK, good.
So, v0 is capital V nought,

270
00:21:20,000 --> 00:21:25,000
that is that the variable V
nought is simply equal to the

271
00:21:25,000 --> 00:21:29,000
voltage, v0.
Similarly, I can write the

272
00:21:29,000 --> 00:21:32,000
others.
v1 is i1, R1.

273
00:21:32,000 --> 00:21:36,000
v2 is i2, R2,
and so on.

274
00:21:36,000 --> 00:21:39,000
OK, and I have one,
two, three, four,

275
00:21:39,000 --> 00:21:44,000
five, six elements.
So, I will get six such

276
00:21:44,000 --> 00:21:48,000
equations.
Step two, I'm going to go ahead

277
00:21:48,000 --> 00:21:52,000
and write KCL for the nodes in
my system.

278
00:21:52,000 --> 00:21:57,000
So, let me start with node A.
So, for node A,

279
00:21:57,000 --> 00:22:05,000
let me take as positive the
currents going out of the node.

280
00:22:05,000 --> 00:22:10,000
So, I get i nought flowing out,
plus i1 flowing out,

281
00:22:10,000 --> 00:22:16,000
plus i4 flowing out,
and they must sum to zero for

282
00:22:16,000 --> 00:22:20,000
node A.
Then, I can go ahead and do the

283
00:22:20,000 --> 00:22:24,000
other nodes, let's say,
for example,

284
00:22:24,000 --> 00:22:28,000
I do node B.
For node B, I have i2 going

285
00:22:28,000 --> 00:22:31,000
out.
That's positive,

286
00:22:31,000 --> 00:22:36,000
i3, and i1 is coming in,
so I get -i1 equals zero.

287
00:22:36,000 --> 00:22:39,000
OK, so I have one,
two, three, four,

288
00:22:39,000 --> 00:22:44,000
I have four nodes.
OK, so I would get four

289
00:22:44,000 --> 00:22:47,000
equations.
It turns out that the fourth

290
00:22:47,000 --> 00:22:53,000
equation is not independent.
You can derive it from the

291
00:22:53,000 --> 00:22:56,000
others.
So, I get three independent

292
00:22:56,000 --> 00:23:01,000
equations out of this.
I can then write KVL.

293
00:23:01,000 --> 00:23:05,000
And for KVL,
I just go down my loops here.

294
00:23:05,000 --> 00:23:10,000
And let me go through this
first loop here in this manner.

295
00:23:10,000 --> 00:23:15,000
OK, and a simple trick that I
use, you have to be incredibly

296
00:23:15,000 --> 00:23:20,000
careful when you go through this
in keeping your minuses and

297
00:23:20,000 --> 00:23:23,000
pluses correct.
Otherwise you can get

298
00:23:23,000 --> 00:23:26,000
hopelessly muddled.
Once you label it,

299
00:23:26,000 --> 00:23:32,000
you need to be sure that you
get all your minuses and pluses

300
00:23:32,000 --> 00:23:34,000
correct.
So, for KVL,

301
00:23:34,000 --> 00:23:38,000
what I'd like to do is,
let's say I start at C,

302
00:23:38,000 --> 00:23:40,000
and from C I'm going to go to
A.

303
00:23:40,000 --> 00:23:44,000
For A I go to B,
and from B I'm going to come

304
00:23:44,000 --> 00:23:46,000
back to C.
OK, that's how I traverse my

305
00:23:46,000 --> 00:23:49,000
loop.
And, the trick that I'm going

306
00:23:49,000 --> 00:23:52,000
to follow is,
as my finger walks through that

307
00:23:52,000 --> 00:23:57,000
loop, I'm going to label the
voltage as the first sign that I

308
00:23:57,000 --> 00:24:02,000
see for that voltage.
OK, so I'm going to start with

309
00:24:02,000 --> 00:24:05,000
C, and I go up.
I start by punching into the

310
00:24:05,000 --> 00:24:08,000
voltage source element,
and then punch into it,

311
00:24:08,000 --> 00:24:10,000
I hit the minus sign for the V
nought.

312
00:24:10,000 --> 00:24:14,000
OK, so I'm just going to write
down minus V nought,

313
00:24:14,000 --> 00:24:18,000
plus then I go through and as I
come up to A and go down to B,

314
00:24:18,000 --> 00:24:21,000
I punch to the plus sign of the
V1.

315
00:24:21,000 --> 00:24:24,000
So, that's plus V1.
And then I punch into the plus

316
00:24:24,000 --> 00:24:26,000
sign of the V2,
and so I get plus V2,

317
00:24:26,000 --> 00:24:30,000
and that is zero.
OK, good.

318
00:24:30,000 --> 00:24:33,000
So, that matches what you have
in your notes as well.

319
00:24:33,000 --> 00:24:36,000
So, this is the first equation.
Similarly, I can go through my

320
00:24:36,000 --> 00:24:40,000
other loops and write down
equations for each of the loops.

321
00:24:40,000 --> 00:24:43,000
OK, and the convention that I
like to follow is as I go

322
00:24:43,000 --> 00:24:45,000
through the loop,
I write down as a sign for the

323
00:24:45,000 --> 00:24:49,000
voltage the first sign that I
counter for that element.

324
00:24:49,000 --> 00:24:51,000
OK, you can do the exact
opposite, if you want,

325
00:24:51,000 --> 00:24:54,000
just to be different.
But, as long as you stay

326
00:24:54,000 --> 00:24:57,000
consistent, you'll be OK.
All right, so in the same

327
00:24:57,000 --> 00:25:00,000
manner here, there are four
loops that I can have,

328
00:25:00,000 --> 00:25:03,000
so four equations.
Again, one of them turns out to

329
00:25:03,000 --> 00:25:08,000
be dependent on the others.
So I end up getting three

330
00:25:08,000 --> 00:25:12,000
independent equations.
So, I get a total of 12

331
00:25:12,000 --> 00:25:15,000
equations.
I get 12 equations.

332
00:25:15,000 --> 00:25:20,000
There are six elements,
OK, voltage source,

333
00:25:20,000 --> 00:25:24,000
and five resistors.
So, there are six unknown

334
00:25:24,000 --> 00:25:27,000
voltages, and six unknown
currents.

335
00:25:27,000 --> 00:25:33,000
So, I have 12 equations,
and 12 unknowns.

336
00:25:33,000 --> 00:25:38,000
OK, I can take all of the
equations and put them through a

337
00:25:38,000 --> 00:25:42,000
big crank, and sit there and
grind.

338
00:25:42,000 --> 00:25:47,000
And if I was really cruel,
I'd give this as a homework

339
00:25:47,000 --> 00:25:53,000
problem, and have you grind,
and grind, and grind until you

340
00:25:53,000 --> 00:25:57,000
get your six voltages and six
currents.

341
00:25:57,000 --> 00:26:01,000
OK, it works.
OK, so you get 12 equations,

342
00:26:01,000 --> 00:26:07,000
and this method just works.
However, notice that this is

343
00:26:07,000 --> 00:26:10,000
quite a grubby method.
It's quite grungy.

344
00:26:10,000 --> 00:26:14,000
I get 12 equations,
and it's quite a pain even for

345
00:26:14,000 --> 00:26:18,000
a simple circuit like this.
However, suffice it to say that

346
00:26:18,000 --> 00:26:22,000
this fundamental method is one
step away from Maxwell's

347
00:26:22,000 --> 00:26:25,000
equations, simply works.
OK?

348
00:26:25,000 --> 00:26:28,000
So what you'll do is the rest
of this lecture,

349
00:26:28,000 --> 00:26:33,000
I'll introduce you to a couple
more methods.

350
00:26:33,000 --> 00:26:40,000
One is an intuitive method,
and another one called the node

351
00:26:40,000 --> 00:26:46,000
method is a little bit more
formal, but is much more,

352
00:26:46,000 --> 00:26:51,000
I guess, terse Than the KVL KCL
method.

353
00:26:51,000 --> 00:26:56,000
Method 2.
So the relevant section to read

354
00:26:56,000 --> 00:00:02,400
in the course notes is section

355
00:27:02,000 --> 00:27:06,000
One of the things that I will
be stressing this semester is

356
00:27:06,000 --> 00:27:08,000
intuition.
What you'll find is that as you

357
00:27:08,000 --> 00:27:11,000
become EECS majors,
and so on, and go on,

358
00:27:11,000 --> 00:27:15,000
or if you talk to your TAs or
your professors and so on,

359
00:27:15,000 --> 00:27:19,000
you will find that very rarely
do they actually go ahead and

360
00:27:19,000 --> 00:27:21,000
apply the formal methods of
analysis.

361
00:27:21,000 --> 00:27:25,000
OK, by and large,
engineers are able to look at a

362
00:27:25,000 --> 00:27:28,000
circuit and simply by
observation write down an

363
00:27:28,000 --> 00:27:31,000
answer.
And usually in the past,

364
00:27:31,000 --> 00:27:35,000
what we have tried to do is
kind of ignore that process and

365
00:27:35,000 --> 00:27:37,000
told our students,
look, we teach you all the

366
00:27:37,000 --> 00:27:40,000
formal methods,
and you will develop your own

367
00:27:40,000 --> 00:27:44,000
intuition and be able to do it.
What we'll try to do this term

368
00:27:44,000 --> 00:27:47,000
is try to stress the intuitive
methods, and try to show you how

369
00:27:47,000 --> 00:27:51,000
the intuitive process goes,
so you can very quickly solve

370
00:27:51,000 --> 00:27:54,000
many of these circuits simply by
inspection.

371
00:27:54,000 --> 00:27:57,000
OK, so this method that I'm
going to show you here is one

372
00:27:57,000 --> 00:28:02,000
such an intuitive method.
And I'll call it element

373
00:28:02,000 --> 00:28:08,000
combination tools.
OK, for many simple circuits,

374
00:28:08,000 --> 00:28:14,000
you can solve them very quickly
by applying this method.

375
00:28:14,000 --> 00:28:18,000
The components of this method
are these.

376
00:28:18,000 --> 00:28:24,000
I learned about how to compose
a bunch of elements.

377
00:28:24,000 --> 00:28:27,000
So, let's say,
for example,

378
00:28:27,000 --> 00:28:32,000
I have a set of resistors,
R1 through RN,

379
00:28:32,000 --> 00:28:38,000
in series.
OK, you can use KVL and KCL to

380
00:28:38,000 --> 00:28:44,000
show that this is equivalent to
a single resistor whose value is

381
00:28:44,000 --> 00:28:48,000
given by the sum of the
resistances.

382
00:28:48,000 --> 00:28:54,000
OK, so if I have resistors in
series, then effectively it's

383
00:28:54,000 --> 00:28:59,000
the same as if there was a
single resistor whose value is

384
00:28:59,000 --> 00:29:07,000
the sum of all the resistances.
OK, you can look at the course

385
00:29:07,000 --> 00:29:11,000
notes for a proof for derivation
of this fact.

386
00:29:11,000 --> 00:29:16,000
Similarly, if I have
resistances in parallel,

387
00:29:16,000 --> 00:29:20,000
so let me call them
conductances.

388
00:29:20,000 --> 00:29:24,000
A conductance is the reciprocal
of a resistance.

389
00:29:24,000 --> 00:29:31,000
If resistance is measured in
ohms, conductance is measured in

390
00:29:31,000 --> 00:29:36,000
mhos, M-H-O-S.
OK, so that's the conductance

391
00:29:36,000 --> 00:29:39,000
is G1, G2, and G3.
And effectively,

392
00:29:39,000 --> 00:29:44,000
this is the same as having a
single conductance whose

393
00:29:44,000 --> 00:29:49,000
effective value is given by the
sum of the conductances.

394
00:29:49,000 --> 00:29:53,000
OK, the conductances in
parallel add,

395
00:29:53,000 --> 00:30:00,000
and resistances in series add.
Similarly, for voltage sources,

396
00:30:00,000 --> 00:30:06,000
if I have voltage sources in
series, then they are tantamount

397
00:30:06,000 --> 00:30:10,000
to the sum of the voltages.
And similarly,

398
00:30:10,000 --> 00:30:14,000
for currents,
if I have currents in parallel,

399
00:30:14,000 --> 00:30:19,000
then they can be viewed as a
single current source,

400
00:30:19,000 --> 00:30:24,000
whose currents are the sum of
the individual parallel

401
00:30:24,000 --> 00:30:30,000
currents.
So, let's do a quick example.

402
00:30:30,000 --> 00:30:36,000
So let's do this example.
So, let's say I have a circuit

403
00:30:36,000 --> 00:30:41,000
that looks like this,
and three resistances.

404
00:30:41,000 --> 00:30:47,000
And let's say all I care about
is the current,

405
00:30:47,000 --> 00:30:50,000
I, that flows through this
wire.

406
00:30:50,000 --> 00:30:54,000
All I care about is that
current.

407
00:30:54,000 --> 00:31:02,000
Of course, you can go ahead and
write KVL and KCL.

408
00:31:02,000 --> 00:31:06,000
You will get four equations,
and there are four unknowns.

409
00:31:06,000 --> 00:31:09,000
And you can solve it.
But, I can apply my element

410
00:31:09,000 --> 00:31:12,000
combination rules,
and very quickly figure out

411
00:31:12,000 --> 00:31:16,000
what the current I is,
using the following technique.

412
00:31:16,000 --> 00:31:19,000
So, what I can do is,
I can, first of all,

413
00:31:19,000 --> 00:31:23,000
take this circuit.
And, I can compose these two

414
00:31:23,000 --> 00:31:27,000
resistances and show that the
circuit is equivalent as far as

415
00:31:27,000 --> 00:31:30,000
this current,
I, is concerned to the

416
00:31:30,000 --> 00:31:33,000
following circuit,
R1.

417
00:31:33,000 --> 00:31:39,000
And I take the sum of the two
conductances,

418
00:31:39,000 --> 00:31:46,000
OK, and that comes out to be
R1, R2, R3, R2 plus R3.

419
00:31:46,000 --> 00:31:54,000
And then, I can further
simplify it, and I get a single

420
00:31:54,000 --> 00:32:01,000
resistance, whose value is given
by R1 plus R2,

421
00:32:01,000 --> 00:32:06,000
R3, R3.
OK, I'm just simplifying the

422
00:32:06,000 --> 00:32:08,000
circuit.
Now, from this circuit,

423
00:32:08,000 --> 00:32:11,000
I can get the answer that I
need.

424
00:32:11,000 --> 00:32:15,000
I is simply the voltage,
V, divided by R1 plus.

425
00:32:15,000 --> 00:32:20,000
OK, so in situations like this
where I'm looking for a single

426
00:32:20,000 --> 00:32:25,000
current, I can very quickly get
to the answer by applying some

427
00:32:25,000 --> 00:32:28,000
of these element combination
rules.

428
00:32:28,000 --> 00:32:34,000
And, I can get rid of having to
go through formal steps.

429
00:32:34,000 --> 00:32:36,000
So, in general,
whenever you encounter a

430
00:32:36,000 --> 00:32:41,000
circuit, by and large attempt to
use intuitive methods to solve

431
00:32:41,000 --> 00:32:43,000
it.
And go to a formal method only

432
00:32:43,000 --> 00:32:47,000
if some intuitive method fails.
Even in your homework,

433
00:32:47,000 --> 00:32:51,000
by and large,
the homeworks are not meant to

434
00:32:51,000 --> 00:32:54,000
be grungy.
OK, if you find a lot of grunge

435
00:32:54,000 --> 00:32:57,000
in your homework,
just remember you're probably

436
00:32:57,000 --> 00:33:04,000
not using some intuitive method.
OK, so just be cautious.

437
00:33:04,000 --> 00:33:11,000
All right, so let me go on to
the third method of circuit

438
00:33:11,000 --> 00:33:19,000
analysis, and the third method
is called the node method.

439
00:33:19,000 --> 00:33:28,000
So, the node method is simply a
specific application of the KVL

440
00:33:28,000 --> 00:33:36,000
KCL method and results in a
much, much more compact form of

441
00:33:36,000 --> 00:33:42,000
the final equations.
If there's one method that you

442
00:33:42,000 --> 00:33:46,000
have to remember for life,
then I would say just remember

443
00:33:46,000 --> 00:33:48,000
this method.
OK, the node method is a

444
00:33:48,000 --> 00:33:50,000
workhorse of the easiest
industry.

445
00:33:50,000 --> 00:33:55,000
OK, if there's one method that
you want to consistently apply,

446
00:33:55,000 --> 00:33:58,000
then this is the one to
remember.

447
00:33:58,000 --> 00:34:01,000
So, let me quickly outline for
you to method,

448
00:34:01,000 --> 00:34:04,000
and then work out an example
for you.

449
00:34:04,000 --> 00:34:09,000
The first step of the node
method will be to select a

450
00:34:09,000 --> 00:34:14,000
reference or a ground node.
This is the symbol for a ground

451
00:34:14,000 --> 00:34:16,000
node.
The ground node simply says

452
00:34:16,000 --> 00:34:21,000
that I'm going to denote
voltages at that point to be

453
00:34:21,000 --> 00:34:26,000
zero, and measure all my other
voltages with reference to that

454
00:34:26,000 --> 00:34:30,000
point.
So, I'm going to select a

455
00:34:30,000 --> 00:34:36,000
ground node in my circuit.
Second, I want to label the

456
00:34:36,000 --> 00:34:41,000
remaining voltages with respect
to the ground node.

457
00:34:41,000 --> 00:34:48,000
So, label voltages for all the
other nodes with respect to the

458
00:34:48,000 --> 00:34:52,000
ground node.
Next, write KCL for each of the

459
00:34:52,000 --> 00:34:57,000
nodes write KCL.
OK, but don't write KCL for the

460
00:34:57,000 --> 00:35:02,000
ground node.
Remember, if you have N nodes,

461
00:35:02,000 --> 00:35:07,000
the node equations will give
you N-1 independent equations.

462
00:35:07,000 --> 00:35:12,000
So, write KCL for the nodes,
but don't do so for the ground

463
00:35:12,000 --> 00:35:15,000
node.
Then, solve for the node

464
00:35:15,000 --> 00:35:18,000
voltages.
So, let's say when we label

465
00:35:18,000 --> 00:35:21,000
voltages.
I want to be labeling them as E

466
00:35:21,000 --> 00:35:26,000
something or the other.
So, solve for the unknown node

467
00:35:26,000 --> 00:35:31,000
voltages.
And then, once I know all the

468
00:35:31,000 --> 00:35:37,000
voltages associated with the
nodes, I can then back solve for

469
00:35:37,000 --> 00:35:41,000
all the branch voltages and
currents.

470
00:35:41,000 --> 00:35:47,000
OK, once I know all the node
voltages, I can then go ahead

471
00:35:47,000 --> 00:35:52,000
and figure out all the branch
voltages and the branch

472
00:35:52,000 --> 00:35:56,000
currents.
So, let's go ahead and apply

473
00:35:56,000 --> 00:36:02,000
this method, and work out an
example.

474
00:36:02,000 --> 00:36:05,000
Again, remember,
if there's one method that you

475
00:36:05,000 --> 00:36:08,000
should remember,
it's the node method.

476
00:36:08,000 --> 00:36:11,000
OK, and when in doubt,
consistently apply the node

477
00:36:11,000 --> 00:36:15,000
method and it will work whether
your circuit is linear or

478
00:36:15,000 --> 00:36:20,000
nonlinear, if the resistors are
built in the US or the USSR it

479
00:36:20,000 --> 00:36:23,000
doesn't matter.
OK, the node method will simply

480
00:36:23,000 --> 00:36:26,000
work, linear or nonlinear,
OK?

481
00:36:26,000 --> 00:36:30,000
So, what I'm going to do is I'm
going to build a circuit that's

482
00:36:30,000 --> 00:36:36,000
my old faithful.
It's our old faithful,

483
00:36:36,000 --> 00:36:45,000
plus I'll make it a little bit
more complicated by adding in

484
00:36:45,000 --> 00:36:52,000
the current source.
So, let's go have some fun.

485
00:36:52,000 --> 00:36:59,000
Let's do this.
So here's my voltage source,

486
00:36:59,000 --> 00:37:04,000
as before.
OK, what I'll do is for fun,

487
00:37:04,000 --> 00:37:13,000
add a current source out there.
And, you can convince

488
00:37:13,000 --> 00:37:20,000
yourselves that if you go ahead
and apply the KVL KCL method,

489
00:37:20,000 --> 00:37:25,000
it'll really be a mess of
equations.

490
00:37:25,000 --> 00:37:28,000
OK, so R1, R3,
R4, R2, R5.

491
00:37:28,000 --> 00:37:36,000
OK, so let's follow our method
and just plug and chug here.

492
00:37:36,000 --> 00:37:39,000
So let's apply the first step.
I select a ground node.

493
00:37:39,000 --> 00:37:42,000
It's a reference node from
which I'll measure all my other

494
00:37:42,000 --> 00:37:44,000
voltages.
OK, now without knowing

495
00:37:44,000 --> 00:37:48,000
anything about the node method,
try to use intuition as to

496
00:37:48,000 --> 00:37:51,000
which node you should choose as
a ground node.

497
00:37:51,000 --> 00:37:55,000
Remember, you want to label the
ground node with the voltage

498
00:37:55,000 --> 00:37:58,000
zero, and measure all the other
voltages with respect to that

499
00:37:58,000 --> 00:38:02,000
node.
OK, a usual trick is to pick a

500
00:38:02,000 --> 00:38:07,000
node which has the largest
number of elements connected to

501
00:38:07,000 --> 00:38:11,000
it as the ground node.
OK, and in particular,

502
00:38:11,000 --> 00:38:16,000
you will find out later it's
useful to pick a node in which

503
00:38:16,000 --> 00:38:20,000
all your voltage sources,
the maximum number of your

504
00:38:20,000 --> 00:38:23,000
voltage sources are also
connected.

505
00:38:23,000 --> 00:38:27,000
OK, so in this instance,
I'm going to choose this as my

506
00:38:27,000 --> 00:38:32,000
ground node.
OK, that's my first step.

507
00:38:32,000 --> 00:38:38,000
I chose that as my ground node.
And I'm going to label that as

508
00:38:38,000 --> 00:38:41,000
having a voltage zero.
Second step,

509
00:38:41,000 --> 00:38:47,000
I'll label voltages of the
other branches with respect to

510
00:38:47,000 --> 00:38:52,000
the ground node.
OK, so what I'll do is add this

511
00:38:52,000 --> 00:38:55,000
node here.
So I'm going to label that

512
00:38:55,000 --> 00:39:00,000
voltage E1.
These are my unknowns.

513
00:39:00,000 --> 00:39:05,000
Remember, node method,
because my node voltages are my

514
00:39:05,000 --> 00:39:09,000
unknowns.
So, I label this as E1.

515
00:39:09,000 --> 00:39:14,000
I label this one as my unknown
voltage, E2.

516
00:39:14,000 --> 00:39:19,000
What about this one here?
Is that voltage unknown?

517
00:39:19,000 --> 00:39:23,000
No.
I know what the voltage is

518
00:39:23,000 --> 00:39:28,000
because I know that this node is
at a voltage,

519
00:39:28,000 --> 00:39:33,000
V0, higher than the ground
node.

520
00:39:33,000 --> 00:39:38,000
OK, notice that to go from here
to here, I directly go through a

521
00:39:38,000 --> 00:39:42,000
voltage source.
And so, this node has voltage

522
00:39:42,000 --> 00:39:45,000
V0.
And I'll simply write down V0.

523
00:39:45,000 --> 00:39:50,000
OK, try to simplify the number
of steps that you have to go

524
00:39:50,000 --> 00:39:55,000
through, so directly go ahead
and write down the voltage,

525
00:39:55,000 --> 00:39:58,000
V0, for that node.
What I will also do,

526
00:39:58,000 --> 00:40:02,000
is for convenience,
I'm going to write down,

527
00:40:02,000 --> 00:40:08,000
I'm going to use conductances.
So I'm going to use GI in the

528
00:40:08,000 --> 00:40:12,000
place of one by RI,
and write down a bunch of node

529
00:40:12,000 --> 00:40:14,000
equations.
OK, so step one,

530
00:40:14,000 --> 00:40:19,000
I've chosen my ground node.
Step two, I've labeled my node

531
00:40:19,000 --> 00:40:23,000
voltages, E, OK?
I've done that with two of my

532
00:40:23,000 --> 00:40:27,000
steps.
Now, let me go ahead and --

533
00:40:41,000 --> 00:40:44,000
OK, so let me go ahead and
apply step three.

534
00:40:44,000 --> 00:40:50,000
And, step three says go ahead
and apply KCL for each of the

535
00:40:50,000 --> 00:40:54,000
nodes at which you have an
unknown node voltage.

536
00:40:54,000 --> 00:40:59,000
And then that will give you
your equations.

537
00:40:59,000 --> 00:41:02,000
So let me start by applying KCL
at E1.

538
00:41:02,000 --> 00:41:06,000
So, let me write KCL at E1.
I do one more thing.

539
00:41:06,000 --> 00:41:09,000
Notice, I don't have any
currents there.

540
00:41:09,000 --> 00:41:14,000
OK, so how do I write KCL?
KCL simply says the sum of

541
00:41:14,000 --> 00:41:18,000
currents into a node is zero
again, remember,

542
00:41:18,000 --> 00:41:23,000
by the lump matter discipline.
So, if I don't have currents in

543
00:41:23,000 --> 00:41:28,000
there, so the trick that I adopt
is that to write KCL,

544
00:41:28,000 --> 00:41:34,000
I use the node voltages,
and implicitly substitute for

545
00:41:34,000 --> 00:41:37,000
the node voltages,
divide by the elemental the

546
00:41:37,000 --> 00:41:41,000
resistance, for instance,
so I take the node voltages,

547
00:41:41,000 --> 00:41:44,000
and divide by the resistance,
get the current.

548
00:41:44,000 --> 00:41:48,000
OK, so I implicitly apply
element relationships to get the

549
00:41:48,000 --> 00:41:51,000
node currents.
So, the example that make it

550
00:41:51,000 --> 00:41:55,000
clear, so I take node E1 and,
again, for currents going out

551
00:41:55,000 --> 00:42:00,000
I'm going to assume to have,
to be positive.

552
00:42:00,000 --> 00:42:05,000
So, the current going up is E1
minus V nought,

553
00:42:05,000 --> 00:42:09,000
divide by R1,
so I multiplied by the G1.

554
00:42:09,000 --> 00:42:16,000
That's the current going up.
Plus, the current going down is

555
00:42:16,000 --> 00:42:22,000
E1 minus zero where the ground
node potential is zero,

556
00:42:22,000 --> 00:42:28,000
G2, OK, plus the current that
is going to resistor R3,

557
00:42:28,000 --> 00:42:35,000
which is simply E1 minus E2,
divide by R3.

558
00:42:35,000 --> 00:42:37,000
So, E1 minus E2,
divide by R3,

559
00:42:37,000 --> 00:42:40,000
or multiplied by G3 is equal to
zero.

560
00:42:40,000 --> 00:42:44,000
OK, see how I got this?
This is simply KCL,

561
00:42:44,000 --> 00:42:49,000
but to get my currents,
I simply take the differences

562
00:42:49,000 --> 00:42:54,000
of voltages across elements,
and divide by the element of

563
00:42:54,000 --> 00:42:58,000
resistance, and I get the
currents.

564
00:42:58,000 --> 00:43:01,000
OK, so I can similarly write
KCL at E2.

565
00:43:01,000 --> 00:43:06,000
So, at KCL at E2,
again, let me go outwards.

566
00:43:06,000 --> 00:43:14,000
So, the current going up is E2
minus V nought multiplied by G4.

567
00:43:14,000 --> 00:43:22,000
The current going left is E2
minus E1 divided by R3 or

568
00:43:22,000 --> 00:43:29,000
multiplied by G3.
The current going down is E2

569
00:43:29,000 --> 00:43:37,000
minus zero multiplied by G5.
And, the current going down is

570
00:43:37,000 --> 00:43:40,000
-I1.
OK, you've got to be careful

571
00:43:40,000 --> 00:43:46,000
with your polarities here.
So all the currents going out

572
00:43:46,000 --> 00:43:50,000
sum to zero.
And here are the currents that

573
00:43:50,000 --> 00:43:56,000
are going out at this point.
So what I do next is I can move

574
00:43:56,000 --> 00:44:01,000
the constant terms to the
left-hand side and collect my

575
00:44:01,000 --> 00:44:07,000
unknowns.
So, let me write them out here.

576
00:44:07,000 --> 00:44:14,000
So, let's say I get E1 here,
OK, and from this equation,

577
00:44:14,000 --> 00:44:20,000
I have a V nought,
G1, which comes out here.

578
00:44:20,000 --> 00:44:26,000
So, minus V nought G1 comes
over to the other side.

579
00:44:26,000 --> 00:44:34,000
And, let me collect all the
values that multiply E1.

580
00:44:34,000 --> 00:44:39,000
So I get, G1 is one example.
I have G2, and I have G3.

581
00:44:39,000 --> 00:44:43,000
And then, for E2,
I have minus G3.

582
00:44:43,000 --> 00:44:49,000
OK, so I'll simply express this
as the element voltages

583
00:44:49,000 --> 00:44:55,000
multiplied by some terms in
parentheses, and I put my

584
00:44:55,000 --> 00:44:59,000
external sources on the right
hand side.

585
00:44:59,000 --> 00:45:06,000
Similarly, I go ahead and do
the same thing here.

586
00:45:06,000 --> 00:45:10,000
In this instance,
let me move my sources to the

587
00:45:10,000 --> 00:45:13,000
right.
So, I get I1 coming out there,

588
00:45:13,000 --> 00:45:16,000
and I get V nought G4 coming
out there.

589
00:45:16,000 --> 00:45:21,000
By the way, I just want to
mention to you that if you're

590
00:45:21,000 --> 00:45:25,000
looking to fall asleep,
this is a good time to do so

591
00:45:25,000 --> 00:45:30,000
because as soon as I write down
these two equations,

592
00:45:30,000 --> 00:45:36,000
OK, from now on it's nap time.
There's nothing new that you're

593
00:45:36,000 --> 00:45:40,000
going to learn from here on.
It's just Anant Agarwal having

594
00:45:40,000 --> 00:45:43,000
fun at the blackboard,
pushing symbols around.

595
00:45:43,000 --> 00:45:46,000
So, once you write down these
two node equations,

596
00:45:46,000 --> 00:45:48,000
the rest of it is just grubby
math.

597
00:45:48,000 --> 00:45:52,000
So, let me just have some fun.
So let me just go ahead and do

598
00:45:52,000 --> 00:45:54,000
that.
So, I moved my voltages and

599
00:45:54,000 --> 00:45:58,000
currents to the other side.
And let me collect all the

600
00:45:58,000 --> 00:46:01,000
coefficients for E1 here.
So, E1 minus G3,

601
00:46:01,000 --> 00:46:04,000
and that's it,
I guess.

602
00:46:04,000 --> 00:46:08,000
OK, and then I'll do the same
for E2.

603
00:46:08,000 --> 00:46:12,000
So, I get G4,
and I get G3,

604
00:46:12,000 --> 00:46:18,000
and I get G5.
OK, so notice here that I have

605
00:46:18,000 --> 00:46:22,000
two equations,
and two unknowns.

606
00:46:22,000 --> 00:46:29,000
OK, the two equations are on
the right hand side,

607
00:46:29,000 --> 00:46:35,000
I have some voltages and
currents which are my dry

608
00:46:35,000 --> 00:46:43,000
voltages and dry currents.
OK, so actually this is getting

609
00:46:43,000 --> 00:46:46,000
quite boring.
I'm going to pause here,

610
00:46:46,000 --> 00:46:52,000
and talk about something else.
So, you can take this and you

611
00:46:52,000 --> 00:46:56,000
can put it in matrix form,
so I've done that for you on

612
00:46:56,000 --> 00:47:00,000
page ten.
It's all matrix form.

613
00:47:00,000 --> 00:47:03,000
Yeah, I know that.
You can use any technique to

614
00:47:03,000 --> 00:47:07,000
solve it, use algebraic
techniques, use linear algebraic

615
00:47:07,000 --> 00:47:09,000
methods to solve it,
use a computer,

616
00:47:09,000 --> 00:47:11,000
whatever you want.
And, computers,

617
00:47:11,000 --> 00:47:15,000
when computers analyze
circuits, they write down these

618
00:47:15,000 --> 00:47:18,000
equations, and deal with solving
matrices.

619
00:47:18,000 --> 00:47:21,000
So, when you take the linear
algebra across,

620
00:47:21,000 --> 00:47:25,000
how many people here have taken
a linear algebra class?

621
00:47:25,000 --> 00:47:30,000
How many people here have heard
of Gaussian elimination?

622
00:47:30,000 --> 00:47:34,000
How can more people have heard
of Gaussian elimination than

623
00:47:34,000 --> 00:47:37,000
took a linear algebra class?
Well anyway,

624
00:47:37,000 --> 00:47:42,000
so now you know why you took
those linear algebra classes.

625
00:47:42,000 --> 00:47:47,000
And so, if I just collected
these into matrix form --

626
00:48:06,000 --> 00:48:09,000
OK, so I just simply expressed
those two equations in linear

627
00:48:09,000 --> 00:48:12,000
algebraic form,
and here's my column vector of

628
00:48:12,000 --> 00:48:15,000
unknowns, and you can apply any
of the techniques you've learned

629
00:48:15,000 --> 00:48:17,000
in linear algebra to solve for
this.

630
00:48:17,000 --> 00:48:20,000
Gaussian elimination works.
OK, and in computer,

631
00:48:20,000 --> 00:48:23,000
people doing research in
computer techniques,

632
00:48:23,000 --> 00:48:26,000
or solving such equations
simply deals with huge equations

633
00:48:26,000 --> 00:48:28,000
like this, building computer
programs that,

634
00:48:28,000 --> 00:48:33,000
given equations like this,
can go ahead and solve them.

635
00:48:33,000 --> 00:48:36,000
OK, so let me stop here and
reemphasize that what you've

636
00:48:36,000 --> 00:48:39,000
done is made a huge leap from
Maxwell's equations to using the

637
00:48:39,000 --> 00:48:43,000
lump matter discipline to KVL
and KCL, which ended up giving a

638
00:48:43,000 --> 00:48:46,000
simple algebraic equation to
solve, and not having to worry

639
00:48:46,000 --> 00:48:49,000
about partial differential
equations that were the form of

640
00:48:49,000 --> 00:48:52,000
Maxwell's equations.