GILBERT STRANG: This
video is about one
of the key applications
of ordinary differential
equations to electrical flow,
flow of currents in a network.
And so I drew a network,
a very simple network.
It's just called an RLC loop.
It's only got one loop, so
it's a really simple network.
The R stands for
resistance to the flow.
The L stands for an inductance.
And the C is the.
Capacitance those are
the three elements
of a simple linear constant
coefficient problem associated
with one loop.
And then there is a
switch, which I'll close,
and the flow will begin.
And there is a voltage
source, so like a battery,
or maybe let's make this
alternating current.
So the voltage source
will be some voltage
times an e to the i omega t.
So we're going to have
alternating current.
And the question is,
what is the current?
We have to find the
current, I. So the current
is I of t going around the loop.
And we saw our
differential equation
will have that unknown I of
t, rather than my usual y.
I'm going to use I for current.
Again, this is an RLC loop that
everybody has to understand,
as in electrical engineering.
So I'm going to have a second
order differential equation.
Well, you'll see what
that equation is.
So you'll remember Ohm's law.
That the voltage is the
current times the resistance.
So this gives me a voltage
across the resistor.
If the current is I and
the resistance is R,
then the voltage drop from
here to here is I times R.
So that's that term.
But now I have also my
current is changing with time.
This is alternating current.
It's going up and down.
So the current is also going
through the inductance.
And there, the voltage
drop across the inductance
has this form.
The derivative of the
current comes into it.
And in the capacitance,
which is building up charge,
the integral of the
current comes in.
So there, that's the
physical equation that
expresses this
voltage law, which
says that around a closed
loop-- this is a closed,
loops are closed-- add to 0.
So I have four terms, and
they combine to give us 0.
So there's an equation
I'd like to solve.
And how am I going to
solve that equation?
By the standard
idea which applies
when I have constant
coefficients
and I have a pure
exponential forcing term.
I look for a solution that is
a multiple of that exponential,
right?
The solution to
differential equations
with constant
coefficients, if they
have an exponential
forcing, then
the solution is I
equals some, shall
I say W e to the i omega t.
Some multiple of the source
gives me the solution
to that differential equation.
Well, it's actually a
differential integral equation.
I can make it a more familiar
looking differential equation
by taking the derivative
of every term.
Suppose I do that.
Suppose I take the derivative
of every term, just to make
it look really familiar.
That would be L
times I double prime.
Taking the derivative
of the derivative.
This would be RI prime.
The derivative of the integral
would be just I itself.
So I'd have 1 over
C I. And I would
have the derivative here, I
omega V e to the I omega t.
So it's just a
standard second order
constant coefficient linear
differential equation.
And in fact, if you are
a mechanical engineer,
you would look at
that and say, well, I
don't know what L, R,
and 1 over C stand for.
But I know that I should
see the mass, the damping,
and the stiffness there.
So we have a complete parallel
between two important fields
of engineering, the electric
engineering with L, R,
and 1 over C, mechanical
engineering with M,
B for damping, and
K for stiffness.
And actually, that parallel
allowed analog computers--
which came before
digital computers
and lost out in
that competition.
An analog computer was just
solving this linear equation
by actually imposing the voltage
and measuring the current.
So an analog computer
actually solved the equation
by creating the model
and measuring the answer.
But we're not creating
an analog computer here.
We're just doing
differential equations.
So why don't I figure
out what that W is.
So what am I going to do?
As always, I have this equation.
I have a pure exponential.
I look for a solution
of that same form.
I plug it in.
And I get an equation for W.
That's exactly what I'll
do on the next board.
I'll put W e to the I
omega t into this equation
and find W. Let's do it.
Maybe I'll bring
that down just a hair
and I'll do it here where
you can watch me do it.
So I have L times
the derivative.
So I have L. The
derivative will bring down
an I omega L. Everything is
going to multiply W and match
V. When I put this
into the equation,
the derivative is an I omega
L W e to the I omega t,
and it's matching V
e to the I omega t.
Now, what happens when I put
I in for that second term,
R. I just get an R. R times
W times e to the I omega t.
No problem.
And now finally, a 1
over C. The integral.
The integral of the
exponential brings down--
let me put it in the denominator
neatly-- I divide by I omega
when I integrate e
to the I omega t.
I have a division by I omega.
That's it.
That's it.
Those are the three terms that
come-- times W, the unknown.
This is to find.
And of course, we
find it right away.
We find W is V
over-- and now we're
seeing this I
omega L plus R. Oh,
let me combine the I omegas.
Combine the real part
and the imaginary part.
The real part is R. And
the imaginary part is I
omega L minus 1 over I omega C.
Straightforward.
And that has a name.
That is the resistance.
But when there's
also terms coming
from an inductance
and a capacitance,
then the whole thing is
called the impedance.
So this whole thing,
this whole denominator,
is called the complex impedance.
Believe me, all these
ideas are so important.
There's a whole vocabulary here.
But you see, we've done
exactly the same thing
for other constant
coefficient equations.
We just called the coefficient
A, B, C. Or maybe M, B, K.
And now we have slightly
different letters,
but we don't have
a new idea here.
The idea is this 1 over, that
1 over the impedance, that
will be the transfer function,
which multiplies the source
to give the complex number
W. And W is a complex number.
I have to now think about that.
And that impedance
is always called Z.
So we now have a new letter
for the important quantity that
shows up in the
denominator there.
And again, its real
part is the resistance.
Its imaginary part
comes from L and C.
So we can easily see
how large-- what's
the size of that impedance?
What's going to be the
magnitude of this current?
We want the size of that number.
V is the size of the voltage.
Here is the size
of the impedance.
And the answer will
give us the size of W.
I'm using size or
magnitude to say
that when I only
do magnitudes, you
won't be seeing the phase lag.
So complex numbers,
like this complex number
has a magnitude which
we're about to write down.
And also it has a
phase lag that tells us
how much is in
the imaginary part
and how much is
in the real part.
But the magnitude is easy.
What's the magnitude
of a complex number?
It's the real part squared and
the imaginary part squared.
Oh, that should have been
a plus there, I think.
I don't know how
it became a minus.
It will become a minus,
so I was thinking
if I put the I up there.
Let me show you what I'm saying.
The imaginary part
is omega L minus 1
over omega C. What I'm saying
is that if I put the I up there,
then 1 over I is minus I.
That's the brilliant step I just
took there.
So all that squared.
Are you OK with that?
It's the real part squared,
which is the resistance.
And this combination
gives the imaginary part.
We square that.
That's maybe called
the reactants.
And the sum of those squares
is the impedance squared,
the magnitude.
So we have essentially
successfully solved
a second order constant
coefficient single equation
for the current.
What to do now.
Just let me add a
little bit more.
Maybe just a comment.
That video was about one loop.
When I told Dr. Mohler that
one of the applications, one
of the real applications
in this series of videos
would be an RLC
circuit, his reply
was an RLC circuit is
not an application, not
a realistic application.
One loop.
So how do we proceed
with a full scale circuit
with many nodes, many resistors,
many conductors, many edges?
Well, we have a big
decision to make.
And that's the comment
I want to make.
They have a choice.
They can use Kirchoff's
current law at the nodes
and solve for the
voltages at those nodes.
Or they can do as
we did for one loop,
use Kirchoff's voltage
law around that one loop
which said that the
currents in the loop
gave a total voltage
drop adding to 0.
So we solve the current
equation for the unknown I. This
is what we did for one loop.
My message is just for a big
system, this is the winner.
So writing down the
equations in terms
of Kirchoff's current law,
that the currents-- we
get the nodal
picture, the picture
with an equation for every
node instead of the picture
for an equation for every loop.
Because it's not so easy to see
which are the loops to consider
and which loops are
combinations of other loops.
The linear algebra
is the question.
And the linear algebra,
to get the loop picture
independent and clear,
is more difficult
than the node picture.
The node picture with
the unknown voltages,
V at the nodes, is the good one.
And the matrix that comes into
that is the incidence matrix.
It connects nodes and edges.
It says how the network
is put together.
And that matrix, I'll study with
a little bit of linear algebra.
So that comes in a later video.
If you look for
incidence matrices,
you'll see probably two
videos about those very, very
important and
beautiful matrices.
Thank you.
