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GILBERT STRANG: This
video is about one

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of the key applications
of ordinary differential

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equations to electrical flow,
flow of currents in a network.

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And so I drew a network,
a very simple network.

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It's just called an RLC loop.

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It's only got one loop, so
it's a really simple network.

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The R stands for
resistance to the flow.

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The L stands for an inductance.

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And the C is the.

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Capacitance those are
the three elements

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of a simple linear constant
coefficient problem associated

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with one loop.

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And then there is a
switch, which I'll close,

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and the flow will begin.

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And there is a voltage
source, so like a battery,

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or maybe let's make this
alternating current.

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So the voltage source
will be some voltage

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times an e to the i omega t.

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So we're going to have
alternating current.

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And the question is,
what is the current?

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We have to find the
current, I. So the current

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is I of t going around the loop.

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And we saw our
differential equation

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will have that unknown I of
t, rather than my usual y.

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I'm going to use I for current.

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Again, this is an RLC loop that
everybody has to understand,

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as in electrical engineering.

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So I'm going to have a second
order differential equation.

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Well, you'll see what
that equation is.

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So you'll remember Ohm's law.

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That the voltage is the
current times the resistance.

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So this gives me a voltage
across the resistor.

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If the current is I and
the resistance is R,

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then the voltage drop from
here to here is I times R.

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So that's that term.

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But now I have also my
current is changing with time.

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This is alternating current.

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It's going up and down.

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So the current is also going
through the inductance.

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And there, the voltage
drop across the inductance

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has this form.

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The derivative of the
current comes into it.

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And in the capacitance,
which is building up charge,

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the integral of the
current comes in.

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So there, that's the
physical equation that

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expresses this
voltage law, which

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says that around a closed
loop-- this is a closed,

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loops are closed-- add to 0.

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So I have four terms, and
they combine to give us 0.

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So there's an equation
I'd like to solve.

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And how am I going to
solve that equation?

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By the standard
idea which applies

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when I have constant
coefficients

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and I have a pure
exponential forcing term.

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I look for a solution that is
a multiple of that exponential,

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right?

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The solution to
differential equations

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with constant
coefficients, if they

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have an exponential
forcing, then

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the solution is I
equals some, shall

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I say W e to the i omega t.

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Some multiple of the source
gives me the solution

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to that differential equation.

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Well, it's actually a
differential integral equation.

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I can make it a more familiar
looking differential equation

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by taking the derivative
of every term.

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Suppose I do that.

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Suppose I take the derivative
of every term, just to make

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it look really familiar.

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That would be L
times I double prime.

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Taking the derivative
of the derivative.

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This would be RI prime.

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The derivative of the integral
would be just I itself.

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So I'd have 1 over
C I. And I would

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have the derivative here, I
omega V e to the I omega t.

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So it's just a
standard second order

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constant coefficient linear
differential equation.

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And in fact, if you are
a mechanical engineer,

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you would look at
that and say, well, I

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don't know what L, R,
and 1 over C stand for.

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But I know that I should
see the mass, the damping,

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and the stiffness there.

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So we have a complete parallel
between two important fields

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of engineering, the electric
engineering with L, R,

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and 1 over C, mechanical
engineering with M,

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B for damping, and
K for stiffness.

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And actually, that parallel
allowed analog computers--

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which came before
digital computers

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and lost out in
that competition.

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An analog computer was just
solving this linear equation

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by actually imposing the voltage
and measuring the current.

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So an analog computer
actually solved the equation

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by creating the model
and measuring the answer.

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But we're not creating
an analog computer here.

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We're just doing
differential equations.

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So why don't I figure
out what that W is.

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So what am I going to do?

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As always, I have this equation.

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I have a pure exponential.

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I look for a solution
of that same form.

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I plug it in.

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And I get an equation for W.

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That's exactly what I'll
do on the next board.

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I'll put W e to the I
omega t into this equation

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and find W. Let's do it.

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Maybe I'll bring
that down just a hair

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and I'll do it here where
you can watch me do it.

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So I have L times
the derivative.

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So I have L. The
derivative will bring down

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an I omega L. Everything is
going to multiply W and match

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V. When I put this
into the equation,

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the derivative is an I omega
L W e to the I omega t,

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and it's matching V
e to the I omega t.

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Now, what happens when I put
I in for that second term,

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R. I just get an R. R times
W times e to the I omega t.

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No problem.

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And now finally, a 1
over C. The integral.

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The integral of the
exponential brings down--

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let me put it in the denominator
neatly-- I divide by I omega

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when I integrate e
to the I omega t.

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I have a division by I omega.

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That's it.

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That's it.

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Those are the three terms that
come-- times W, the unknown.

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This is to find.

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And of course, we
find it right away.

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We find W is V
over-- and now we're

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seeing this I
omega L plus R. Oh,

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let me combine the I omegas.

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Combine the real part
and the imaginary part.

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The real part is R. And
the imaginary part is I

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omega L minus 1 over I omega C.

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Straightforward.

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And that has a name.

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That is the resistance.

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But when there's
also terms coming

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from an inductance
and a capacitance,

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then the whole thing is
called the impedance.

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So this whole thing,
this whole denominator,

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is called the complex impedance.

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Believe me, all these
ideas are so important.

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There's a whole vocabulary here.

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But you see, we've done
exactly the same thing

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for other constant
coefficient equations.

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We just called the coefficient
A, B, C. Or maybe M, B, K.

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And now we have slightly
different letters,

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but we don't have
a new idea here.

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The idea is this 1 over, that
1 over the impedance, that

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will be the transfer function,
which multiplies the source

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to give the complex number
W. And W is a complex number.

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I have to now think about that.

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And that impedance
is always called Z.

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So we now have a new letter
for the important quantity that

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shows up in the
denominator there.

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And again, its real
part is the resistance.

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Its imaginary part
comes from L and C.

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So we can easily see
how large-- what's

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the size of that impedance?

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What's going to be the
magnitude of this current?

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We want the size of that number.

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V is the size of the voltage.

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Here is the size
of the impedance.

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And the answer will
give us the size of W.

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I'm using size or
magnitude to say

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that when I only
do magnitudes, you

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won't be seeing the phase lag.

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So complex numbers,
like this complex number

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has a magnitude which
we're about to write down.

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And also it has a
phase lag that tells us

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how much is in
the imaginary part

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and how much is
in the real part.

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But the magnitude is easy.

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What's the magnitude
of a complex number?

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It's the real part squared and
the imaginary part squared.

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Oh, that should have been
a plus there, I think.

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I don't know how
it became a minus.

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It will become a minus,
so I was thinking

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if I put the I up there.

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Let me show you what I'm saying.

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The imaginary part
is omega L minus 1

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over omega C. What I'm saying
is that if I put the I up there,

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then 1 over I is minus I.
That's the brilliant step I just

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took there.

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So all that squared.

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Are you OK with that?

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It's the real part squared,
which is the resistance.

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And this combination
gives the imaginary part.

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We square that.

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That's maybe called
the reactants.

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And the sum of those squares
is the impedance squared,

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the magnitude.

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So we have essentially
successfully solved

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a second order constant
coefficient single equation

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for the current.

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What to do now.

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Just let me add a
little bit more.

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Maybe just a comment.

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That video was about one loop.

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When I told Dr. Mohler that
one of the applications, one

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of the real applications
in this series of videos

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would be an RLC
circuit, his reply

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was an RLC circuit is
not an application, not

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a realistic application.

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One loop.

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So how do we proceed
with a full scale circuit

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with many nodes, many resistors,
many conductors, many edges?

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Well, we have a big
decision to make.

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And that's the comment
I want to make.

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They have a choice.

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They can use Kirchoff's
current law at the nodes

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and solve for the
voltages at those nodes.

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Or they can do as
we did for one loop,

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use Kirchoff's voltage
law around that one loop

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which said that the
currents in the loop

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gave a total voltage
drop adding to 0.

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So we solve the current
equation for the unknown I. This

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is what we did for one loop.

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My message is just for a big
system, this is the winner.

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So writing down the
equations in terms

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of Kirchoff's current law,
that the currents-- we

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get the nodal
picture, the picture

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with an equation for every
node instead of the picture

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for an equation for every loop.

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Because it's not so easy to see
which are the loops to consider

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and which loops are
combinations of other loops.

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The linear algebra
is the question.

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And the linear algebra,
to get the loop picture

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independent and clear,
is more difficult

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than the node picture.

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The node picture with
the unknown voltages,

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V at the nodes, is the good one.

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And the matrix that comes into
that is the incidence matrix.

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It connects nodes and edges.

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It says how the network
is put together.

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And that matrix, I'll study with
a little bit of linear algebra.

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So that comes in a later video.

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If you look for
incidence matrices,

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you'll see probably two
videos about those very, very

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important and
beautiful matrices.

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Thank you.