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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.
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Notice, I continue to follow
this convention where whenever I
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label voltages and currents for
an element, I will show the
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current going into the positive
terminal of the element
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variable, OK,
after element variable voltage.
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So here, I have V nought and I
nought.
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Let me pause here for five
seconds and show you a point of
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confusion that happens
sometimes.
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Often times,
people confuse between what is
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called the variable that is
associated with the element
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versus the element value.
OK, notice that here,
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capital V nought is the voltage
that this voltage source
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provides, while this name here,
v nought, is simply a variable
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that we've used to label the
voltage across that element.
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So, similarly,
I can label v1 as the voltage
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across the resistor,
and i1 is the current flowing
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through the resistor.
So this method of labeling,
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where I follow the convention,
that the current flows into the
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positive terminal is called the
associated variables discipline.
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I was trying to use the word
discipline in situations where
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you have a choice,
OK, and of a variety of
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possible choices,
you pick one as the convention.
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OK, so here,
as a convention,
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we use the associated variables
discipline, and use that method
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to consistently label the
unknown voltages and currents in
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our circuits.
OK, so let me continue the
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labeling here,
v4, i4, i3, v3 here,
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and v2 and i2,
v5, and i5.
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I think that's it.
So, I've gone ahead and labeled
262
00:20:22 --> 00:20:27
all my unknowns.
So each of these voltages and
263
00:20:27 --> 00:20:35
currents are the voltages and
currents associated with each of
264
00:20:35 --> 00:20:40
the elements.
And my goal is to solve for
265
00:20:40 --> 00:20:44
these.
OK, so in terms of our solution
266
00:20:44 --> 00:20:49
here, let's follow the method
that I outlined for you.
267
00:20:49 --> 00:20:54
So, as the first step I am
simply going to go ahead and
268
00:20:54 --> 00:21:00
write down all the element VI
relationships.
269
00:21:00 --> 00:21:05
OK, so as a first step,
I'm going to go ahead and write
270
00:21:05 --> 00:21:12
down all the VI relationships.
So, can someone yell out for me
271
00:21:12 --> 00:21:16
the VI relationship for the
voltage source?
272
00:21:16 --> 00:21:20
OK, good.
So, v0 is capital V nought,
273
00:21:20 --> 00:21:25
that is that the variable V
nought is simply equal to the
274
00:21:25 --> 00:21:29
voltage, v0.
Similarly, I can write the
275
00:21:29 --> 00:21:32
others.
v1 is i1, R1.
276
00:21:32 --> 00:21:36
v2 is i2, R2,
and so on.
277
00:21:36 --> 00:21:39
OK, and I have one,
two, three, four,
278
00:21:39 --> 00:21:44
five, six elements.
So, I will get six such
279
00:21:44 --> 00:21:48
equations.
Step two, I'm going to go ahead
280
00:21:48 --> 00:21:52
and write KCL for the nodes in
my system.
281
00:21:52 --> 00:21:57
So, let me start with node A.
So, for node A,
282
00:21:57 --> 00:22:05
let me take as positive the
currents going out of the node.
283
00:22:05 --> 00:22:10
So, I get i nought flowing out,
plus i1 flowing out,
284
00:22:10 --> 00:22:16
plus i4 flowing out,
and they must sum to zero for
285
00:22:16 --> 00:22:20
node A.
Then, I can go ahead and do the
286
00:22:20 --> 00:22:24
other nodes, let's say,
for example,
287
00:22:24 --> 00:22:28
I do node B.
For node B, I have i2 going
288
00:22:28 --> 00:22:31
out.
That's positive,
289
00:22:31 --> 00:22:36
i3, and i1 is coming in,
so I get -i1 equals zero.
290
00:22:36 --> 00:22:39
OK, so I have one,
two, three, four,
291
00:22:39 --> 00:22:44
I have four nodes.
OK, so I would get four
292
00:22:44 --> 00:22:47
equations.
It turns out that the fourth
293
00:22:47 --> 00:22:53
equation is not independent.
You can derive it from the
294
00:22:53 --> 00:22:56
others.
So, I get three independent
295
00:22:56 --> 00:23:01
equations out of this.
I can then write KVL.
296
00:23:01 --> 00:23:05
And for KVL,
I just go down my loops here.
297
00:23:05 --> 00:23:10
And let me go through this
first loop here in this manner.
298
00:23:10 --> 00:23:15
OK, and a simple trick that I
use, you have to be incredibly
299
00:23:15 --> 00:23:20
careful when you go through this
in keeping your minuses and
300
00:23:20 --> 00:23:23
pluses correct.
Otherwise you can get
301
00:23:23 --> 00:23:26
hopelessly muddled.
Once you label it,
302
00:23:26 --> 00:23:32
you need to be sure that you
get all your minuses and pluses
303
00:23:32 --> 00:23:34
correct.
So, for KVL,
304
00:23:34 --> 00:23:38
what I'd like to do is,
let's say I start at C,
305
00:23:38 --> 00:23:40
and from C I'm going to go to
A.
306
00:23:40 --> 00:23:44
For A I go to B,
and from B I'm going to come
307
00:23:44 --> 00:23:46
back to C.
OK, that's how I traverse my
308
00:23:46 --> 00:23:49
loop.
And, the trick that I'm going
309
00:23:49 --> 00:23:52
to follow is,
as my finger walks through that
310
00:23:52 --> 00:23:57
loop, I'm going to label the
voltage as the first sign that I
311
00:23:57 --> 00:24:02
see for that voltage.
OK, so I'm going to start with
312
00:24:02 --> 00:24:05
C, and I go up.
I start by punching into the
313
00:24:05 --> 00:24:08
voltage source element,
and then punch into it,
314
00:24:08 --> 00:24:10
I hit the minus sign for the V
nought.
315
00:24:10 --> 00:24:14
OK, so I'm just going to write
down minus V nought,
316
00:24:14 --> 00:24:18
plus then I go through and as I
come up to A and go down to B,
317
00:24:18 --> 00:24:21
I punch to the plus sign of the
V1.
318
00:24:21 --> 00:24:24
So, that's plus V1.
And then I punch into the plus
319
00:24:24 --> 00:24:26
sign of the V2,
and so I get plus V2,
320
00:24:26 --> 00:24:30
and that is zero.
OK, good.
321
00:24:30 --> 00:24:33
So, that matches what you have
in your notes as well.
322
00:24:33 --> 00:24:36
So, this is the first equation.
Similarly, I can go through my
323
00:24:36 --> 00:24:40
other loops and write down
equations for each of the loops.
324
00:24:40 --> 00:24:43
OK, and the convention that I
like to follow is as I go
325
00:24:43 --> 00:24:45
through the loop,
I write down as a sign for the
326
00:24:45 --> 00:24:49
voltage the first sign that I
counter for that element.
327
00:24:49 --> 00:24:51
OK, you can do the exact
opposite, if you want,
328
00:24:51 --> 00:24:54
just to be different.
But, as long as you stay
329
00:24:54 --> 00:24:57
consistent, you'll be OK.
All right, so in the same
330
00:24:57 --> 00:25:00
manner here, there are four
loops that I can have,
331
00:25:00 --> 00:25:03
so four equations.
Again, one of them turns out to
332
00:25:03 --> 00:25:08
be dependent on the others.
So I end up getting three
333
00:25:08 --> 00:25:12
independent equations.
So, I get a total of 12
334
00:25:12 --> 00:25:15
equations.
I get 12 equations.
335
00:25:15 --> 00:25:20
There are six elements,
OK, voltage source,
336
00:25:20 --> 00:25:24
and five resistors.
So, there are six unknown
337
00:25:24 --> 00:25:27
voltages, and six unknown
currents.
338
00:25:27 --> 00:25:33
So, I have 12 equations,
and 12 unknowns.
339
00:25:33 --> 00:25:38
OK, I can take all of the
equations and put them through a
340
00:25:38 --> 00:25:42
big crank, and sit there and
grind.
341
00:25:42 --> 00:25:47
And if I was really cruel,
I'd give this as a homework
342
00:25:47 --> 00:25:53
problem, and have you grind,
and grind, and grind until you
343
00:25:53 --> 00:25:57
get your six voltages and six
currents.
344
00:25:57 --> 00:26:01
OK, it works.
OK, so you get 12 equations,
345
00:26:01 --> 00:26:07
and this method just works.
However, notice that this is
346
00:26:07 --> 00:26:10
quite a grubby method.
It's quite grungy.
347
00:26:10 --> 00:26:14
I get 12 equations,
and it's quite a pain even for
348
00:26:14 --> 00:26:18
a simple circuit like this.
However, suffice it to say that
349
00:26:18 --> 00:26:22
this fundamental method is one
step away from Maxwell's
350
00:26:22 --> 00:26:25
equations, simply works.
OK?
351
00:26:25 --> 00:26:28
So what you'll do is the rest
of this lecture,
352
00:26:28 --> 00:26:33
I'll introduce you to a couple
more methods.
353
00:26:33 --> 00:26:40
One is an intuitive method,
and another one called the node
354
00:26:40 --> 00:26:46
method is a little bit more
formal, but is much more,
355
00:26:46 --> 00:26:51
I guess, terse Than the KVL KCL
method.
356
00:26:51 --> 00:26:56
Method 2.
So the relevant section to read
357
00:26:56 --> 2.4.
in the course notes is section
358
2.4. --> 00:27:02
359
00:27:02 --> 00:27:06
One of the things that I will
be stressing this semester is
360
00:27:06 --> 00:27:08
intuition.
What you'll find is that as you
361
00:27:08 --> 00:27:11
become EECS majors,
and so on, and go on,
362
00:27:11 --> 00:27:15
or if you talk to your TAs or
your professors and so on,
363
00:27:15 --> 00:27:19
you will find that very rarely
do they actually go ahead and
364
00:27:19 --> 00:27:21
apply the formal methods of
analysis.
365
00:27:21 --> 00:27:25
OK, by and large,
engineers are able to look at a
366
00:27:25 --> 00:27:28
circuit and simply by
observation write down an
367
00:27:28 --> 00:27:31
answer.
And usually in the past,
368
00:27:31 --> 00:27:35
what we have tried to do is
kind of ignore that process and
369
00:27:35 --> 00:27:37
told our students,
look, we teach you all the
370
00:27:37 --> 00:27:40
formal methods,
and you will develop your own
371
00:27:40 --> 00:27:44
intuition and be able to do it.
What we'll try to do this term
372
00:27:44 --> 00:27:47
is try to stress the intuitive
methods, and try to show you how
373
00:27:47 --> 00:27:51
the intuitive process goes,
so you can very quickly solve
374
00:27:51 --> 00:27:54
many of these circuits simply by
inspection.
375
00:27:54 --> 00:27:57
OK, so this method that I'm
going to show you here is one
376
00:27:57 --> 00:28:02
such an intuitive method.
And I'll call it element
377
00:28:02 --> 00:28:08
combination tools.
OK, for many simple circuits,
378
00:28:08 --> 00:28:14
you can solve them very quickly
by applying this method.
379
00:28:14 --> 00:28:18
The components of this method
are these.
380
00:28:18 --> 00:28:24
I learned about how to compose
a bunch of elements.
381
00:28:24 --> 00:28:27
So, let's say,
for example,
382
00:28:27 --> 00:28:32
I have a set of resistors,
R1 through RN,
383
00:28:32 --> 00:28:38
in series.
OK, you can use KVL and KCL to
384
00:28:38 --> 00:28:44
show that this is equivalent to
a single resistor whose value is
385
00:28:44 --> 00:28:48
given by the sum of the
resistances.
386
00:28:48 --> 00:28:54
OK, so if I have resistors in
series, then effectively it's
387
00:28:54 --> 00:28:59
the same as if there was a
single resistor whose value is
388
00:28:59 --> 00:29:07
the sum of all the resistances.
OK, you can look at the course
389
00:29:07 --> 00:29:11
notes for a proof for derivation
of this fact.
390
00:29:11 --> 00:29:16
Similarly, if I have
resistances in parallel,
391
00:29:16 --> 00:29:20
so let me call them
conductances.
392
00:29:20 --> 00:29:24
A conductance is the reciprocal
of a resistance.
393
00:29:24 --> 00:29:31
If resistance is measured in
ohms, conductance is measured in
394
00:29:31 --> 00:29:36
mhos, M-H-O-S.
OK, so that's the conductance
395
00:29:36 --> 00:29:39
is G1, G2, and G3.
And effectively,
396
00:29:39 --> 00:29:44
this is the same as having a
single conductance whose
397
00:29:44 --> 00:29:49
effective value is given by the
sum of the conductances.
398
00:29:49 --> 00:29:53
OK, the conductances in
parallel add,
399
00:29:53 --> 00:30:00
and resistances in series add.
Similarly, for voltage sources,
400
00:30:00 --> 00:30:06
if I have voltage sources in
series, then they are tantamount
401
00:30:06 --> 00:30:10
to the sum of the voltages.
And similarly,
402
00:30:10 --> 00:30:14
for currents,
if I have currents in parallel,
403
00:30:14 --> 00:30:19
then they can be viewed as a
single current source,
404
00:30:19 --> 00:30:24
whose currents are the sum of
the individual parallel
405
00:30:24 --> 00:30:30
currents.
So, let's do a quick example.
406
00:30:30 --> 00:30:36
So let's do this example.
So, let's say I have a circuit
407
00:30:36 --> 00:30:41
that looks like this,
and three resistances.
408
00:30:41 --> 00:30:47
And let's say all I care about
is the current,
409
00:30:47 --> 00:30:50
I, that flows through this
wire.
410
00:30:50 --> 00:30:54
All I care about is that
current.
411
00:30:54 --> 00:31:02
Of course, you can go ahead and
write KVL and KCL.
412
00:31:02 --> 00:31:06
You will get four equations,
and there are four unknowns.
413
00:31:06 --> 00:31:09
And you can solve it.
But, I can apply my element
414
00:31:09 --> 00:31:12
combination rules,
and very quickly figure out
415
00:31:12 --> 00:31:16
what the current I is,
using the following technique.
416
00:31:16 --> 00:31:19
So, what I can do is,
I can, first of all,
417
00:31:19 --> 00:31:23
take this circuit.
And, I can compose these two
418
00:31:23 --> 00:31:27
resistances and show that the
circuit is equivalent as far as
419
00:31:27 --> 00:31:30
this current,
I, is concerned to the
420
00:31:30 --> 00:31:33
following circuit,
R1.
421
00:31:33 --> 00:31:39
And I take the sum of the two
conductances,
422
00:31:39 --> 00:31:46
OK, and that comes out to be
R1, R2, R3, R2 plus R3.
423
00:31:46 --> 00:31:54
And then, I can further
simplify it, and I get a single
424
00:31:54 --> 00:32:01
resistance, whose value is given
by R1 plus R2,
425
00:32:01 --> 00:32:06
R3, R3.
OK, I'm just simplifying the
426
00:32:06 --> 00:32:08
circuit.
Now, from this circuit,
427
00:32:08 --> 00:32:11
I can get the answer that I
need.
428
00:32:11 --> 00:32:15
I is simply the voltage,
V, divided by R1 plus.
429
00:32:15 --> 00:32:20
OK, so in situations like this
where I'm looking for a single
430
00:32:20 --> 00:32:25
current, I can very quickly get
to the answer by applying some
431
00:32:25 --> 00:32:28
of these element combination
rules.
432
00:32:28 --> 00:32:34
And, I can get rid of having to
go through formal steps.
433
00:32:34 --> 00:32:36
So, in general,
whenever you encounter a
434
00:32:36 --> 00:32:41
circuit, by and large attempt to
use intuitive methods to solve
435
00:32:41 --> 00:32:43
it.
And go to a formal method only
436
00:32:43 --> 00:32:47
if some intuitive method fails.
Even in your homework,
437
00:32:47 --> 00:32:51
by and large,
the homeworks are not meant to
438
00:32:51 --> 00:32:54
be grungy.
OK, if you find a lot of grunge
439
00:32:54 --> 00:32:57
in your homework,
just remember you're probably
440
00:32:57 --> 00:33:04
not using some intuitive method.
OK, so just be cautious.
441
00:33:04 --> 00:33:11
All right, so let me go on to
the third method of circuit
442
00:33:11 --> 00:33:19
analysis, and the third method
is called the node method.
443
00:33:19 --> 00:33:28
So, the node method is simply a
specific application of the KVL
444
00:33:28 --> 00:33:36
KCL method and results in a
much, much more compact form of
445
00:33:36 --> 00:33:42
the final equations.
If there's one method that you
446
00:33:42 --> 00:33:46
have to remember for life,
then I would say just remember
447
00:33:46 --> 00:33:48
this method.
OK, the node method is a
448
00:33:48 --> 00:33:50
workhorse of the easiest
industry.
449
00:33:50 --> 00:33:55
OK, if there's one method that
you want to consistently apply,
450
00:33:55 --> 00:33:58
then this is the one to
remember.
451
00:33:58 --> 00:34:01
So, let me quickly outline for
you to method,
452
00:34:01 --> 00:34:04
and then work out an example
for you.
453
00:34:04 --> 00:34:09
The first step of the node
method will be to select a
454
00:34:09 --> 00:34:14
reference or a ground node.
This is the symbol for a ground
455
00:34:14 --> 00:34:16
node.
The ground node simply says
456
00:34:16 --> 00:34:21
that I'm going to denote
voltages at that point to be
457
00:34:21 --> 00:34:26
zero, and measure all my other
voltages with reference to that
458
00:34:26 --> 00:34:30
point.
So, I'm going to select a
459
00:34:30 --> 00:34:36
ground node in my circuit.
Second, I want to label the
460
00:34:36 --> 00:34:41
remaining voltages with respect
to the ground node.
461
00:34:41 --> 00:34:48
So, label voltages for all the
other nodes with respect to the
462
00:34:48 --> 00:34:52
ground node.
Next, write KCL for each of the
463
00:34:52 --> 00:34:57
nodes write KCL.
OK, but don't write KCL for the
464
00:34:57 --> 00:35:02
ground node.
Remember, if you have N nodes,
465
00:35:02 --> 00:35:07
the node equations will give
you N-1 independent equations.
466
00:35:07 --> 00:35:12
So, write KCL for the nodes,
but don't do so for the ground
467
00:35:12 --> 00:35:15
node.
Then, solve for the node
468
00:35:15 --> 00:35:18
voltages.
So, let's say when we label
469
00:35:18 --> 00:35:21
voltages.
I want to be labeling them as E
470
00:35:21 --> 00:35:26
something or the other.
So, solve for the unknown node
471
00:35:26 --> 00:35:31
voltages.
And then, once I know all the
472
00:35:31 --> 00:35:37
voltages associated with the
nodes, I can then back solve for
473
00:35:37 --> 00:35:41
all the branch voltages and
currents.
474
00:35:41 --> 00:35:47
OK, once I know all the node
voltages, I can then go ahead
475
00:35:47 --> 00:35:52
and figure out all the branch
voltages and the branch
476
00:35:52 --> 00:35:56
currents.
So, let's go ahead and apply
477
00:35:56 --> 00:36:02
this method, and work out an
example.
478
00:36:02 --> 00:36:05
Again, remember,
if there's one method that you
479
00:36:05 --> 00:36:08
should remember,
it's the node method.
480
00:36:08 --> 00:36:11
OK, and when in doubt,
consistently apply the node
481
00:36:11 --> 00:36:15
method and it will work whether
your circuit is linear or
482
00:36:15 --> 00:36:20
nonlinear, if the resistors are
built in the US or the USSR it
483
00:36:20 --> 00:36:23
doesn't matter.
OK, the node method will simply
484
00:36:23 --> 00:36:26
work, linear or nonlinear,
OK?
485
00:36:26 --> 00:36:30
So, what I'm going to do is I'm
going to build a circuit that's
486
00:36:30 --> 00:36:36
my old faithful.
It's our old faithful,
487
00:36:36 --> 00:36:45
plus I'll make it a little bit
more complicated by adding in
488
00:36:45 --> 00:36:52
the current source.
So, let's go have some fun.
489
00:36:52 --> 00:36:59
Let's do this.
So here's my voltage source,
490
00:36:59 --> 00:37:04
as before.
OK, what I'll do is for fun,
491
00:37:04 --> 00:37:13
add a current source out there.
And, you can convince
492
00:37:13 --> 00:37:20
yourselves that if you go ahead
and apply the KVL KCL method,
493
00:37:20 --> 00:37:25
it'll really be a mess of
equations.
494
00:37:25 --> 00:37:28
OK, so R1, R3,
R4, R2, R5.
495
00:37:28 --> 00:37:36
OK, so let's follow our method
and just plug and chug here.
496
00:37:36 --> 00:37:39
So let's apply the first step.
I select a ground node.
497
00:37:39 --> 00:37:42
It's a reference node from
which I'll measure all my other
498
00:37:42 --> 00:37:44
voltages.
OK, now without knowing
499
00:37:44 --> 00:37:48
anything about the node method,
try to use intuition as to
500
00:37:48 --> 00:37:51
which node you should choose as
a ground node.
501
00:37:51 --> 00:37:55
Remember, you want to label the
ground node with the voltage
502
00:37:55 --> 00:37:58
zero, and measure all the other
voltages with respect to that
503
00:37:58 --> 00:38:02
node.
OK, a usual trick is to pick a
504
00:38:02 --> 00:38:07
node which has the largest
number of elements connected to
505
00:38:07 --> 00:38:11
it as the ground node.
OK, and in particular,
506
00:38:11 --> 00:38:16
you will find out later it's
useful to pick a node in which
507
00:38:16 --> 00:38:20
all your voltage sources,
the maximum number of your
508
00:38:20 --> 00:38:23
voltage sources are also
connected.
509
00:38:23 --> 00:38:27
OK, so in this instance,
I'm going to choose this as my
510
00:38:27 --> 00:38:32
ground node.
OK, that's my first step.
511
00:38:32 --> 00:38:38
I chose that as my ground node.
And I'm going to label that as
512
00:38:38 --> 00:38:41
having a voltage zero.
Second step,
513
00:38:41 --> 00:38:47
I'll label voltages of the
other branches with respect to
514
00:38:47 --> 00:38:52
the ground node.
OK, so what I'll do is add this
515
00:38:52 --> 00:38:55
node here.
So I'm going to label that
516
00:38:55 --> 00:39:00
voltage E1.
These are my unknowns.
517
00:39:00 --> 00:39:05
Remember, node method,
because my node voltages are my
518
00:39:05 --> 00:39:09
unknowns.
So, I label this as E1.
519
00:39:09 --> 00:39:14
I label this one as my unknown
voltage, E2.
520
00:39:14 --> 00:39:19
What about this one here?
Is that voltage unknown?
521
00:39:19 --> 00:39:23
No.
I know what the voltage is
522
00:39:23 --> 00:39:28
because I know that this node is
at a voltage,
523
00:39:28 --> 00:39:33
V0, higher than the ground
node.
524
00:39:33 --> 00:39:38
OK, notice that to go from here
to here, I directly go through a
525
00:39:38 --> 00:39:42
voltage source.
And so, this node has voltage
526
00:39:42 --> 00:39:45
V0.
And I'll simply write down V0.
527
00:39:45 --> 00:39:50
OK, try to simplify the number
of steps that you have to go
528
00:39:50 --> 00:39:55
through, so directly go ahead
and write down the voltage,
529
00:39:55 --> 00:39:58
V0, for that node.
What I will also do,
530
00:39:58 --> 00:40:02
is for convenience,
I'm going to write down,
531
00:40:02 --> 00:40:08
I'm going to use conductances.
So I'm going to use GI in the
532
00:40:08 --> 00:40:12
place of one by RI,
and write down a bunch of node
533
00:40:12 --> 00:40:14
equations.
OK, so step one,
534
00:40:14 --> 00:40:19
I've chosen my ground node.
Step two, I've labeled my node
535
00:40:19 --> 00:40:23
voltages, E, OK?
I've done that with two of my
536
00:40:23 --> 00:40:27
steps.
Now, let me go ahead and --
537
00:40:27 --> 00:40:41
538
00:40:41 --> 00:40:44
OK, so let me go ahead and
apply step three.
539
00:40:44 --> 00:40:50
And, step three says go ahead
and apply KCL for each of the
540
00:40:50 --> 00:40:54
nodes at which you have an
unknown node voltage.
541
00:40:54 --> 00:40:59
And then that will give you
your equations.
542
00:40:59 --> 00:41:02
So let me start by applying KCL
at E1.
543
00:41:02 --> 00:41:06
So, let me write KCL at E1.
I do one more thing.
544
00:41:06 --> 00:41:09
Notice, I don't have any
currents there.
545
00:41:09 --> 00:41:14
OK, so how do I write KCL?
KCL simply says the sum of
546
00:41:14 --> 00:41:18
currents into a node is zero
again, remember,
547
00:41:18 --> 00:41:23
by the lump matter discipline.
So, if I don't have currents in
548
00:41:23 --> 00:41:28
there, so the trick that I adopt
is that to write KCL,
549
00:41:28 --> 00:41:34
I use the node voltages,
and implicitly substitute for
550
00:41:34 --> 00:41:37
the node voltages,
divide by the elemental the
551
00:41:37 --> 00:41:41
resistance, for instance,
so I take the node voltages,
552
00:41:41 --> 00:41:44
and divide by the resistance,
get the current.
553
00:41:44 --> 00:41:48
OK, so I implicitly apply
element relationships to get the
554
00:41:48 --> 00:41:51
node currents.
So, the example that make it
555
00:41:51 --> 00:41:55
clear, so I take node E1 and,
again, for currents going out
556
00:41:55 --> 00:42:00
I'm going to assume to have,
to be positive.
557
00:42:00 --> 00:42:05
So, the current going up is E1
minus V nought,
558
00:42:05 --> 00:42:09
divide by R1,
so I multiplied by the G1.
559
00:42:09 --> 00:42:16
That's the current going up.
Plus, the current going down is
560
00:42:16 --> 00:42:22
E1 minus zero where the ground
node potential is zero,
561
00:42:22 --> 00:42:28
G2, OK, plus the current that
is going to resistor R3,
562
00:42:28 --> 00:42:35
which is simply E1 minus E2,
divide by R3.
563
00:42:35 --> 00:42:37
So, E1 minus E2,
divide by R3,
564
00:42:37 --> 00:42:40
or multiplied by G3 is equal to
zero.
565
00:42:40 --> 00:42:44
OK, see how I got this?
This is simply KCL,
566
00:42:44 --> 00:42:49
but to get my currents,
I simply take the differences
567
00:42:49 --> 00:42:54
of voltages across elements,
and divide by the element of
568
00:42:54 --> 00:42:58
resistance, and I get the
currents.
569
00:42:58 --> 00:43:01
OK, so I can similarly write
KCL at E2.
570
00:43:01 --> 00:43:06
So, at KCL at E2,
again, let me go outwards.
571
00:43:06 --> 00:43:14
So, the current going up is E2
minus V nought multiplied by G4.
572
00:43:14 --> 00:43:22
The current going left is E2
minus E1 divided by R3 or
573
00:43:22 --> 00:43:29
multiplied by G3.
The current going down is E2
574
00:43:29 --> 00:43:37
minus zero multiplied by G5.
And, the current going down is
575
00:43:37 --> 00:43:40
-I1.
OK, you've got to be careful
576
00:43:40 --> 00:43:46
with your polarities here.
So all the currents going out
577
00:43:46 --> 00:43:50
sum to zero.
And here are the currents that
578
00:43:50 --> 00:43:56
are going out at this point.
So what I do next is I can move
579
00:43:56 --> 00:44:01
the constant terms to the
left-hand side and collect my
580
00:44:01 --> 00:44:07
unknowns.
So, let me write them out here.
581
00:44:07 --> 00:44:14
So, let's say I get E1 here,
OK, and from this equation,
582
00:44:14 --> 00:44:20
I have a V nought,
G1, which comes out here.
583
00:44:20 --> 00:44:26
So, minus V nought G1 comes
over to the other side.
584
00:44:26 --> 00:44:34
And, let me collect all the
values that multiply E1.
585
00:44:34 --> 00:44:39
So I get, G1 is one example.
I have G2, and I have G3.
586
00:44:39 --> 00:44:43
And then, for E2,
I have minus G3.
587
00:44:43 --> 00:44:49
OK, so I'll simply express this
as the element voltages
588
00:44:49 --> 00:44:55
multiplied by some terms in
parentheses, and I put my
589
00:44:55 --> 00:44:59
external sources on the right
hand side.
590
00:44:59 --> 00:45:06
Similarly, I go ahead and do
the same thing here.
591
00:45:06 --> 00:45:10
In this instance,
let me move my sources to the
592
00:45:10 --> 00:45:13
right.
So, I get I1 coming out there,
593
00:45:13 --> 00:45:16
and I get V nought G4 coming
out there.
594
00:45:16 --> 00:45:21
By the way, I just want to
mention to you that if you're
595
00:45:21 --> 00:45:25
looking to fall asleep,
this is a good time to do so
596
00:45:25 --> 00:45:30
because as soon as I write down
these two equations,
597
00:45:30 --> 00:45:36
OK, from now on it's nap time.
There's nothing new that you're
598
00:45:36 --> 00:45:40
going to learn from here on.
It's just Anant Agarwal having
599
00:45:40 --> 00:45:43
fun at the blackboard,
pushing symbols around.
600
00:45:43 --> 00:45:46
So, once you write down these
two node equations,
601
00:45:46 --> 00:45:48
the rest of it is just grubby
math.
602
00:45:48 --> 00:45:52
So, let me just have some fun.
So let me just go ahead and do
603
00:45:52 --> 00:45:54
that.
So, I moved my voltages and
604
00:45:54 --> 00:45:58
currents to the other side.
And let me collect all the
605
00:45:58 --> 00:46:01
coefficients for E1 here.
So, E1 minus G3,
606
00:46:01 --> 00:46:04
and that's it,
I guess.
607
00:46:04 --> 00:46:08
OK, and then I'll do the same
for E2.
608
00:46:08 --> 00:46:12
So, I get G4,
and I get G3,
609
00:46:12 --> 00:46:18
and I get G5.
OK, so notice here that I have
610
00:46:18 --> 00:46:22
two equations,
and two unknowns.
611
00:46:22 --> 00:46:29
OK, the two equations are on
the right hand side,
612
00:46:29 --> 00:46:35
I have some voltages and
currents which are my dry
613
00:46:35 --> 00:46:43
voltages and dry currents.
OK, so actually this is getting
614
00:46:43 --> 00:46:46
quite boring.
I'm going to pause here,
615
00:46:46 --> 00:46:52
and talk about something else.
So, you can take this and you
616
00:46:52 --> 00:46:56
can put it in matrix form,
so I've done that for you on
617
00:46:56 --> 00:47:00
page ten.
It's all matrix form.
618
00:47:00 --> 00:47:03
Yeah, I know that.
You can use any technique to
619
00:47:03 --> 00:47:07
solve it, use algebraic
techniques, use linear algebraic
620
00:47:07 --> 00:47:09
methods to solve it,
use a computer,
621
00:47:09 --> 00:47:11
whatever you want.
And, computers,
622
00:47:11 --> 00:47:15
when computers analyze
circuits, they write down these
623
00:47:15 --> 00:47:18
equations, and deal with solving
matrices.
624
00:47:18 --> 00:47:21
So, when you take the linear
algebra across,
625
00:47:21 --> 00:47:25
how many people here have taken
a linear algebra class?
626
00:47:25 --> 00:47:30
How many people here have heard
of Gaussian elimination?
627
00:47:30 --> 00:47:34
How can more people have heard
of Gaussian elimination than
628
00:47:34 --> 00:47:37
took a linear algebra class?
Well anyway,
629
00:47:37 --> 00:47:42
so now you know why you took
those linear algebra classes.
630
00:47:42 --> 00:47:47
And so, if I just collected
these into matrix form --
631
00:47:47 --> 00:48:06
632
00:48:06 --> 00:48:09
OK, so I just simply expressed
those two equations in linear
633
00:48:09 --> 00:48:12
algebraic form,
and here's my column vector of
634
00:48:12 --> 00:48:15
unknowns, and you can apply any
of the techniques you've learned
635
00:48:15 --> 00:48:17
in linear algebra to solve for
this.
636
00:48:17 --> 00:48:20
Gaussian elimination works.
OK, and in computer,
637
00:48:20 --> 00:48:23
people doing research in
computer techniques,
638
00:48:23 --> 00:48:26
or solving such equations
simply deals with huge equations
639
00:48:26 --> 00:48:28
like this, building computer
programs that,
640
00:48:28 --> 00:48:33
given equations like this,
can go ahead and solve them.
641
00:48:33 --> 00:48:36
OK, so let me stop here and
reemphasize that what you've
642
00:48:36 --> 00:48:39
done is made a huge leap from
Maxwell's equations to using the
643
00:48:39 --> 00:48:43
lump matter discipline to KVL
and KCL, which ended up giving a
644
00:48:43 --> 00:48:46
simple algebraic equation to
solve, and not having to worry
645
00:48:46 --> 00:48:49
about partial differential
equations that were the form of
646
00:48:49 --> 48:52
Maxwell's equations.