Prof: Okay,
I left you guys thinking about
inductors last time,
so I should start there.
An inductor is very different
from the resistor when you start
doing circuit theory,
because when you have a
resistor and you connect some
voltage to it,
the current is determined by
this equation.
This is what I call an
algebraic equation.
It's an equation in algebra,
one unknown.
You solve for it by dividing
and you get I at time
t is V at time
t divided by R.
And you can make the network
more complicated,
put a few more resistors,
put a few more there,
a few more here,
resistors within resistors.
It doesn't matter,
because we can always in the
end combine the ones in series
then lump them with the stuff in
parallel,
till the whole thing to the
right of this is one single
effective resistor.
Then you find the current
coming out of the sources as
V divided by that guy.
 
Then every time it comes to a
branch, we sort of know how to
divide it.
 
So resistor circuits are very
easy.
But now when you bring
inductors, things are different,
so I'm just going to tell you a
little more about them.
Here is an inductor.
 
That's the symbol for an
inductor.
It's got some inductance
L measured in henries.
And the implication of that is
the following:
if you have a current going
through this inductor,
there is going to be
necessarily a voltage with this
polarity if the current is
increasing.
And the voltage you need is
LdI/dt.
This is the bottom line.
 
Even if you skip all the other
stuff about conservative forces,
this, that, this is something
you need to know how to do
problems.
 
The difference you notice is
that the relation between
voltage and current is not an
algebraic equation,
but a differential equation.
 
This is called a differential
equation.
Now you don't have to worry,
because I will tell you how to
solve it, because I don't know
if it's a prerequisite or not
for this course.
 
I once had a kid in summer
school who when I started using
partial derivatives,
got very upset and said,
"It's not a prerequisite.
 
Why are you using partial
derivatives?"
So I told him,
"When you come to class,
there's always a danger you'll
learn something new.
You just have to learn to live
with that."
So I'm not going to leave you
in the cold.
If I do something you've not
done, I will tell you how to do
it, but I cannot avoid using
certain notions.
So this one,
even though it's called a
differential equation and so on,
if V is a constant,
we can easily find the
integral.
It will be V over
L times t.
Okay, but now we're going to be
interested in cases where
V is not simply constant;
it could be varying with time
in an arbitrary way.
 
So you take this guy,
the first thing you notice is
that the relation between
current and voltage is given by
derivatives.
 
Second thing is,
when a current flows through a
resistor, whatever energy you
provide is gone in the form of
heat.
 
It has dissipated.
 
The light bulb glows and that's
the end.
With an inductor,
when you drive a current,
you are building a magnetic
field inside the inductor and
there's an energy associated
with the magnetic field.
And that stored energy will be
given back to you later on.
So it's like a capacitor.
 
It takes a lot of work to
charge a capacitor,
because you've got to take
charges from this plate and keep
on piling them in that plate.
 
It's a lot of work,
but then if you connect that to
a bulb and you squeeze the
trigger in your camera,
this charge is giving you the
energy back.
So the inductor and capacitor
are different from the resistor
in two ways: one is,
the voltage across the
capacitor is the charge on the
capacitor divided by C,
which if youlike is the
integral of the current up to
that time divided by C.
 
Again, calculus comes in.
 
So these are the two
differences, the relation
between current and voltage and
voltage is differentiating or
integrating,
and these are energy storing
circuit elements compared to
this one,
where energy is dissipated.
 
So let's start with a simple
problem where I have a voltage
source.
 
I'm going to take a fixed
voltage, V_0.
There is a switch,
there is a resistor R,
and there is this inductor
L.
I'm going to assume the current
is flowing this way,
in which case--I'm sorry.
 
You know, this picture's wrong.
 
What's wrong with this picture?
 
What is wrong?
 
Student:  The switch is
open.
Prof: The switch is
open, okay.
That's what's wrong,
so nothing is flowing yet.
But when I close the switch,
then it's fine.
Now it's flowing.
 
And the question is,
what does it do?
How does it flow?
 
After all, the inductor is a
resistance free wire,
so you may think current =
V_0/R will
start flowing immediately,
but that's wrong,
because if such a current were
flowing,
there'll be an energy
½LI^(2) with that
current in it.
 
And that energy cannot come
from 0 to all of that in no
time.
 
There's no way to do that.
 
Or if you like,
if the current really grows
rapidly or instantaneously from
0 to some value,
that's a current,
dI/dt is infinite here.
dI/dt is infinite,
LdI/dt is infinite.
There's an infinite voltage
needed somewhere in the circuit
and we don't have it,
so that would not happen.
So current will have to be
continuous, start from 0 and it
will have to start climbing.
 
The minute the current starts
climbing up it starts flowing
through this guy.
 
There is the resistance times
current voltage drop across
this.
 
Therefore the amount of voltage
available to drive current
through this one or to increase
the current through this one is
reduced from V_0
to V_0 -
RI.
 
That = LdI/dt.
 
So you can see,
as the current increases,
the motivation to increase the
current decreases,
because this resistor starts
swallowing up more and more of
this voltage.
 
So the growth in current will
be very, very slow,
and we expect that after a very
long time, it will settle down
to some value.
 
What is that value?
 
We can find that value by
saying, "Wait until the
current has stopped
growing."
That means dI/dt =0,
that current will be
V_0/R
and I'm going to give it the
name
I_infinity,
because it will turn out that
you will reach the value only
after infinite time.
 
But you can get very,
very close to that by waiting
some reasonable amount of time,
and part of what we want to do
is, how long should I wait?
 
Suppose I'm settling for 90
percent of maximum.
When does that happen?
 
Is it after 1 second or 10
seconds or 10 hours?
What's going to decide that?
 
Well, once you have the right
equations, you can figure
everything out.
 
So we're going to do that.
 
So basically we are going to
solve the equation
LdI/dt.
 
So let me show you one more
time how it is done,
this equation.
 
You should know how to write it.
 
Start anywhere you like,
go round this guy.
You gain voltage
V_0.
Then go around this guy,
you drop by an amount
RI.
 
You drop by an amount RI.
 
And here you jump from this end
to this end.
Remember, you never go into the
inductor.
That's a bad zone.
 
That's a zone where the
potential is not even defined,
but I've told you multiple
times that even though there are
non conservative things going on
inside this,
once you come outside,
you just find a regular
electrostatic potential
difference = LdI/dt.
So you drop again,
and you come back to where you
start,
the whole thing should add up
to 0,
which means the equation is
LdI/dt RI =
V_0.
That's what we want to solve.
 
So when you have this thing,
as I told you,
solving differential equations
is a matter of guess work.
There is no algorithm.
 
You rely on what you've seen
before and you guess the answer.
You put some free parameters,
you fiddle with them till
everything works.
 
And in this kind of problem,
we can make everything work.
Some problems we cannot solve.
 
We can write the equations but
we cannot solve them.
This is something you're not
used to when you take an
elementary course like this one.
 
But there are situations,
for example,
you know inside the proton
there are these quarks,
and the quarks interact with
each other with entities called
gluons.
 
They're called gluons because
they glue the quarks and form
the protons and neutrons.
 
We know the equations governing
the behavior of the gluons.
We cannot solve them.
 
It's really strange.
 
You have an equation but you
cannot solve it.
It's completely possible,
because certain
equations--every equation has a
solution, but it does not mean
you can write it down
analytically.
So quite often,
one may be on top of the right
answer, but one may not know how
to solve it, so one doesn't even
know if the theory is right.
 
So suppose this was the
equation you got,
but you could not solve it.
 
It happens to be the right
equation.
You don't know what the answer
looks like, you can never be
sure this is the right equation.
 
Similarly when
Mr. Schr�dinger invented his
quantum mechanics,
he wrote an equation that will
tell you the energy levels of
hydrogen.
But luckily,
he was able to solve it.
By solving it,
he was able to show that his
equation gives the energy levels
iand the spectra you expect from
the hydrogen atom.
 
So you have to remember that
not every interesting equation
can be solved,
and I gave you the equation for
these gluons,
which we can write down but
cannot solve.
 
There is no function whose
derivative we cannot take.
There are lots of functions
whose integrals we cannot do.
That's simply asymmetry between
those two.
Here's a very simple function:
e^(−x2)dx from 0
to 19, or some fixed number,
0 to some
x_max.
 
It has a definite value.
 
It's area under the graph,
but no one knows how to write a
formula for this in terms of
x_max,
where it's in closed form.
 
But anyway, this guy we can
trample to death as follows.
First we say at very,
very long times,
the current,
I told you, is
V_0/R.
 
That's when dI/dt is 0.
 
So we're going to write the
actual current in our problem as
the current at infinity some
difference I twiddle.
Standard thing,
you take out the part you know
and solve for the rest of it.
 
So let's take that assumed form
and put it into the differential
equation.
 
So when I do LdI/dt,
this is a number,
V over R.
 
It has no d by dt.
 
So dI/dt becomes
dI˜/dt R
times I˜ R
times
I_infinity =
V_0.
Now you can see why we did that.
 
We did that because now these
two guys cancel and you get
LdI/dt  RI = 0.
 
That means dI˜/dt
is −R/L times
I.
 
So we are saying,
get me a function whose time
derivative looks like the
function itself up to a
constant,
and that's something we know
from high school,
is an exponential function.
So I˜
looks like some constant,
I_0e^(-Rt/L).
 
You can take the derivative of
this guy and out will come
-R/L times the function
itself.
And I˜_0
is completely arbitrary.
It's not determined by this
equation.
There's another useful property
to know.
This is a linear equation.
 
It means if you multiply both
sides by 19, you get another
solution which is 19 times the
old solution.
So the solution's overall scale
is not determined by the
equation.
 
If I give you one solution,
you can multiply it by any
number you like,
it's also a solution,
because if you multiply by a
number,
say 9, the 9 goes into the
derivative,
9 comes here.
 
You can see 9 times I˜
satisfies the same
equation.
 
So we don't know this number
just from the equation,
but we go back to what we know,
which is I of t,
is I_infinity,
which is V_0/R
 I˜_0
e^(−Rt/L).
But I know that at t =
0, this guy should vanish.
t = 0,
this guy should vanish,
because there was no current
when I just threw the switch.
So that tells me that equals 0.
 
That tells me
I˜_0 =
-V_0/R and we
get our final result,
which is I of t =
V_0/R times 1 -
e^(-Rt/L).
 
So at this level,
you are really doing physics
the way we like to do it.
 
We like to study things,
write down some equations and
solve the equations.
 
Then we get a very precise
prediction on what will happen.
Because now we don't have to
guess when the current will come
to 90 percent of its maximum
value.
We can pick any number you
like, because we can now plot
this graph.
 
And let's plot this graph,
I(t) versus t.
At t = 0,
e^(-0) is 1.
You get 1 - 1, you get nothing.
 
At t = infinity,
t to the minus infinity
is 0.
 
Yes?
 
Student:  Over on the
right where it says I˜
sub twiddle = negative
V_0 over R,
should that be 0?
 
Prof: This one?
 
Yeah, right. Thank you.
 
Okay, so now this exponential
has a full strength of 1 at time
equals 0.
 
Goes to 0 at time equals
infinity.
Therefore it takes away
exponentially small stuff at
very large times.
 
And if you want to know roughly
how long should I wait,
the answer is,
the time you should wait is
roughly of order L/R.
 
Because you can write this as
e^(-t)/τ,
where τ is L/R.
 
Whenever we have an
exponential, it will always be e
to the minus a physical quantity
divided by some number that sets
the scale for the physical
quantity.
Exponentials fall exponentially.
 
We know that.
 
But how long should I wait?
 
Well, if you wait t =
τ seconds,
then it will be 1/e,
which is roughly 1 over third
of the starting value and 1
minus that will be roughly 2
thirds of its value.
 
If you wait t = 3τ,
e^(-3) is 1 over 20,
then you'll be what,
95 percent.
So you can calculate everything
you need in this simple problem.
So now let's come to the same
problem.
The switch has been closed for
a long time and the current has
stabilized.
 
So here is the switch.
 
It has been closed.
 
That's my V_0,
that's R.
And the current,
you remember,
is V_0/R.
 
This is L.
 
So now we want to do something
else.
Now I want to open that switch.
 
So what do you think will
happen if you open that switch?
Yes?
 
Student:  The inductor
will resist the changing
current.
 
Prof: But how can it
resist?
Student:  By inducing
the magnetic field.
Prof: It can do what it
wants, but how does it drive a
current through the open
circuit?
Student:  From energy
stored in the magnetic field.
Prof: But in what manner
will the current flow,
given that I've broken the
circuit?
Student:  In the
direction--
Prof: No,
but how will it manage to go
around?
 
It's like removing the bridge,
right?
How will it go around?
 
I agree with everything you
said, but how will it make the
circuit?
 
Yes?
 
Student:  There'll be a
higher concentration of charges
on one side than the other.
 
Prof: Yes,
but how will you really get rid
of the current in a circuit that
seems open?
Have you ever done this?
 
Okay, that's just fine.
 
But we know it has to get rid
of that current,
because otherwise,
what happened to the energy?
It has to go through.
 
So you know what,
you guys don't know what will
happen if I pull that switch.
 
Yes?
 
Student:  Is the current
an electric field?
The change in the magnetic
field will create an electric
field which doesn't need the
closed circuit.
Prof: Right,
no I agree, it will have an
electric field.
 
But how will the electrons flow
when the wire has been
interrupted, is all I'm asking
you?
Yes?
 
Student:  Will the
current oscillate,
go back and forth?
 
Prof: No.
 
Yes?
 
Student:  You will have
a static discharge.
Prof: You will have a
discharge.
You will have an arc.
 
You will have a flash,
and that's when you'll hear a
zip, and then it will jump this
gap.
So whenever you remove the
switch, you have to be careful
that there is not stored energy
somewhere.
You might think,
"Hey, I am pulling the
plug on--"
well, I won't say who.
You're pulling the plug,
okay, and what can it do to me?
Well, it can zap you,
because this is a very
dangerous place to pull the
plug.
So you know what people do?
 
They have another resistor.
 
Let's call this guy
R_1.
This is R_2
and R_2 is a
huge number, 10,000 ohms let's
say.
So most of the time,
R_2 doesn't do
anything,
because when the current comes
here in any situation,
it looks at the inductor and
looks at 10,000 ohms and it
says,
"I'm going this way."
 
So primarily,
it will all go there,
okay?
 
In the end, when the switch has
been closed for a very long time
in our old experiment,
when the current is stabilized,
you remember LdI/dt 0
here.
So there's 0 volts between
these two.
That's the same 0 volt across
R_2.
No current will flow in
R_2.
Everything will flow through
L.
That current in fact will be
V_0 over
R, because this guy has
been cut out of the loop.
This is how it's going.
 
But now if you throw the switch
open, you have given it a path
to discharge.
 
So the inductor will discharge
through the resistor.
The current will continue to
flow in this direction.
Of course, it cannot flow
forever, because the resistor
will burn up the energy,
but you're giving it a path.
You're avoiding--in a way,
if you like,
this is also a path with very,
very high resistance.
The air is a path with very
high resistance.
That means by and large,
there are no free carriers in
the air.
 
But if the two tips get really
charged, just like you said,
there'll be an emf that will
pile up charges.
They're jumping,
they're waiting to jump the
gap.
 
Eventually they'll polarize
air, which is electrically
neutral, into and - parts.
 
Then the - will go one way,
the will go the other way.
You'll recognize that as a
discharge.
But you don't have to worry
about that now,
because R_2
will take up your current.
So now the equation we write is
LdI/dt  RI = 0
because if you start anywhere
and you go around a loop,
you have no change in anything.
 
There's no voltage.
 
And the answer to this one,
R_2,
there I =
I_0e to the
−R_2/L
times t.
So now the current will decay
exponentially.
Again, with the time constant,
R_2 over
L,
which is the 1 over the time
constant,
or the time constant is
L over
R_2.
So you put L in henries,
R in ohms,
you'll get some time.
 
That will give you an idea of
how many times that time you
have to wait before the inductor
is completely discharged.
Well, it's never going to be
fully discharged.
It's going to take forever,
but if you say,
"Look, 1 thousandth of the
original current is safe enough.
How long do I have to
wait?"
you put .001 here and see what
time you get.
So you want
e^(−t/τ) to be
.001.
 
So every big inductor you have
in a circuit you will find has
got a resistor in parallel with
it to take up the energy it has.
So let's do a little energy
calculation.
The energy calculation will be,
I had energy in the inductor in
the beginning,
which is ½
LI_0 ^(2).
 
What happened to it?
 
Well, we know what happened.
 
There's a current in the
circuit, and when the current is
flowing in the circuit,
it is dissipating heat in the
resistor.
 
Therefore the power in the
resistor will be
I^(2)R_2.
 
 I^(2) is
I_0^(2) due to the
minus twice R_2t/L
times another
R_2.
 
Now we integrate this power
from 0 to infinity,
you will get I_0
^(2)R_2.
The integral of e to the
minus something is e to
the minus that divided by that
infinity to 0.
I can promise you that what you
will get will be
L/2R_2.
 
It will give you then 1 over
2LI_0^(2).
So this is not a surprise,
that the stored energy in the
inductor is driving your
current.
So if you want,
one way to store energy,
it's a capacitor,
you charge it and it discharge
when you want the current to
flow.
Or you can have an inductor,
in which a current has been set
up.
 
Then when the power shuts off
in your house and somebody
throws the switch,
that inductor can keep the
light bulb burning,
in fact forever.
It will be infinite time before
the bulb shuts down,
but you won't be able to see
anything after a while,
because this graph is
exponentially falling.
This is a very simple problem
where we can do the math and we
can understand what it's doing.
 
Yes?
 
Student:  The graph,
what is that triangle thing?
Prof: This one here?
 
This is called τ.
 
You mean that symbol?
 
Tau is a unit of time whose
numerical value is
R_2/L and we
can write I as I_0
e^(−t/τ).
 
We always like to write it in
this form,
so that if τ = 100 seconds,
you will have to make many
multiples of 100 seconds before
the current is negligible.
Student:  What are the
units on the axis?
Prof: Oh,
in this graph here?
This is time and this is the
current in the loop.
It starts out at some initial
value.
Okay, now I'm going to do the
next slightly more complicated
circuit, and that is an LC
circuit.
If you have a circuit like
that, it won't do anything.
You can take these guys,
hook them up and wait all day.
Nothing will happen because
it's got no energy.
Why should it do anything?
 
It's just going to sit there.
 
But if you've charged up your
capacitor like this,
and there's an open switch that
you then closed,
then those charges are going to
leave and find their way around
the other side and neutralize
them and the capacitor will
discharge.
 
But you know that once it
discharges, it's not the end of
the story.
 
So why is that?
 
Why doesn't it stop when the
capacitor is discharged?
Yes?
 
Student:  It's kind of
like the inductor has momentum
and it keeps pulling.
 
Prof: The inductor would
have had a current by then,
so the inductor is not going to
suddenly stop having its
current.
 
So it's going to keep driving
more current for a while,
until the capacitor's charged
up the opposite way to an extent
that this fights that one,
that voltage,
and then it will go the other
way.
So it will go back and forth.
 
So let's write down the
equation that tells you how it
goes back and forth.
 
So in other words,
you should draw pictures.
Textbooks have nice pictures
and I won't even try it.
In the beginning,
it may look like--I'm sorry,
in my example,
in the beginning,
there's an electric field here,
nothing in the capacitor.
After a quarter cycle,
there's nothing here,
there's a magnetic field in the
inductor.
Another quarter cycle,
it's back here,
but with the opposite polarity,
and it goes back and forth.
So we can get all that,
by writing the equation
LdI/dt Q/C = 0.
 
So let's write everything in
terms of Q.
Then dQ/dt is I,
because as the current flows
this way, it builds up charge.
 
That's my convention.
 
So I can write it as
Ld^(2)Q/dt^(2) (1/C)Q =
0.
 
Now we have seen exactly this
equation before,
right?
 
We saw the equation for a mass
coupled to a spring the equation
md^(2)x/dt^(2) kx = 0.
 
Now we know what it does.
 
We know it oscillates back and
forth.
And L--this is an SAT
question--L is to
m what 1/C is to
k.
It's very useful to bear in
mind this analogy,
because mathematically,
this equation and this equation
have exactly the same solution.
 
It's no more difficult to solve
this one than that one.
This may involve electric
charges;
that may involve masses.
 
You don't care.
 
So mathematically,
here's an equation:
cow times d^(2) dog over
dt^(2) let's say elephant
times dog = 0,
where dog is a function of
time, has exactly the same
solution.
What does it matter what you
call the unknown variables,
right?
 
But you've got to tell me that
cow and elephant are time
independent and dog is the only
thing that depends on time,
because otherwise the equation
is not the same.
So here m and k don't change,
L and C don't change.
So what's the answer to this
one?
Once again, it's the
differential equation.
You write dQ/dt squared
is − (1/LC) Q and
you're asking yourself,
give me a function which,
when I differentiate twice,
looks like itself.
There's always the exponential.
 
There's also sines and the
cosines and you can take any
combination you like.
 
And the one acceptable solution
is some constant A times
cosine square root of 1 over
LC t. Is that right?
Yes.
 
If you take two derivatives of
Q,
you'll pull one over root
LC each time,
then you'll get a - sine when
cosine becomes - sine and one
more derivative will bring it
back to - cosine.
So this is called the
frequency, which is
1/√LC.
 
That's the analog of square
root of k over m.
Anyway, this will oscillate.
 
It's a mechanical analog--it's
the electrical analog of the
mechanical problem.
 
So I don't want to do too much
of this, because I think you
should be quite familiar with
this, at least,
the oscillatory behavior of
this.
And the analogy is perfect.
 
For example,
to say that I pulled the mass
by 1 meter is to say that
x was given a non-zero
value and dx/dt was 0.
 
Analogous thing here will be
Q was given a non-zero
value and dQ/dt which is
I, is 0.
That means the capacitor was
charged, the inductor was not
carrying any current.
 
Then I let it go.
 
What happens?
 
The problems are identical.
 
Yes?
 
Student:  When you solve
the differential equation,
why did you omit the sine?
 
Prof: Yes,
I should explain to you.
That's an important point.
 
Q(t) can be one number,
cosine
ω_0t.
 
ω_0 is
this guy another number sine
ω_0t.
 
And both are acceptable.
 
And the property of this
differential equation is,
if I have one answer,
cos ωt, and
another possible answer,
sine ωt, then any
constant times cos any constant
times sine is also a solution.
You can verify that.
 
If you have one solution,
Q_1,
a second solution,
Q_2,
you can add the two equations
and you can show
Q_1 Q_2
obeys the same equation.
Not only that,
suppose there's one problem you
solve,
dQ_1/dt
squared is -ω
on R squared
Q_1.
 
There is some function of time
and here's another function of
time, obeying the same equation.
 
Now you can multiply this by a
number A and you can
multiply this by a number
B, take the A
inside the derivative,
B inside the derivative
and add the two sides.
 
On the left hand side you will
find the second derivative of
AQ_1
BQ_2.
Right hand side,
you will find
ω_0^(2)
times AQ_1
BQ_2.
 
That means AQ_1
BQ_2 obeys this
same differential equation.
 
This is called superposing two
solutions.
You've got one solution,
another one,
you can multiply one by a
constant.
It's very important it's a
constant, because that's what
will let you take it inside the
derivative.
Therefore the correct answer
here is really
Acosωd
Bsinωt and you can
ask,
why did you write
Acosω
_0t?
First of all,
that's not a good solution.
I can put a Φ
there which is arbitrary.
And it's not hard to show--let
me call that C if you
like, C.
 
This can be written as
Ccosω
_0t - Φ.
 
In other words,
it's possible to take either a
solution with a cosine and a
sine,
with no extra phases inside,
to another one with one
amplitude and one phase.
 
The phase is the delay in the
cosine.
Anybody here has trouble,
who does not know the details,
I'll be happy to explain why
that's correct.
If you don't know,
you should tell me now.
You can check that if you take
this thing,
make a right angled triangle
with A,
B and C and an
angle Φ here,
then A can be written as
CcosΦ and
B can be written as
CsinΦ,
and that just happens to be
cosωt -
Φ,
this one, some trigonometry.
So the differential equation,
with the second order in time,
will always have two unknown
parameters in it.
You can choose them to be these
numbers A and B,
or this number C and
Φ.
And the equation won't tell you
what they are,
because once you tell me
there's a mass coupled to a
spring and I say,
"Okay, this is the spring
constant this is the mass?"
 
and I ask you,
"Where is the mass right
now?"
 
I don't know.
 
It depends on when you started
it and how you released it.
So I need to know what's called
initial conditions,
which is the value of Q
or the value of x at the
initial time and the value of
the velocity or the current at
an initial time.
 
With those two pieces of
information, I can solve for
A and B.
 
This I think you should know,
this kind of stuff you should
know, so I won't say too much
about that.
All right, so now we do another
problem.
We put an alternating voltage
on these guys.
So this is V_0
cosωt and this
is C and this is
L.
Now we ask, what's the current?
 
This omega is not the natural
frequency of oscillation.
It is some externally given
omega, like 60 hertz from your
power supply,
from your socket.
So that's driving the circuit
and you can ask,
what happens now?
 
The answer will be,
you write the same equation.
You can write Ld^(2)Q/dt^(2)
 (1/C)Q =
V_0cos
ωt.
Now what are you going to do?
 
You have to again guess the
solution.
So I want a function,
Q of t,
so that when I take two
derivatives and add it to some
multiple of itself,
I get something,
something times the cosine.
 
So what should that function
look like?
Yes?
 
Student: 
>
Prof: No,
no, what function Q
of t do you think will
satisfy this equation?
I'm asking for a function of
time which, when I put into
this, has some chance of obeying
the equation.
What's the functional form?
 
Yes?
 
Student:  Cosine times
some constant.
Prof: Right.
 
In other words,
I can take it to be some
constant, C times
cosωt,
period.
 
Not even times Φ.
 
Take this one, see what happens.
 
Now when you take one
derivative, you get
-ωLCcos
ωt.
When you take another
derivative, you get
−ω
^(2)L ^(2)--
I'm sorry, this becomes sine
times cosωt,
and this one is just oh my
god--this is my nightmare.
You can see the nightmare now?
 
How many people see the
nightmare?
Yes?
 
Student:  Constant
C.
Prof: Yes,
so I picked that C to be
the same as this C.
 
So we'll put a hat on this guy
so we can tell them about.
So C˜/C times
C˜ is
V_0
cosωt.
And what this tells you is,
if you take two
derivatives--I'm sorry,
this is so sloppy.
If you take two derivatives
here, you will get
ω^(2).
 
cosω be also be
with the - sign.
Q/C.
 
Okay.
 
You guys buy that?
 
There is no L^(2),
just an L.
That = V_0
cosωt.
Right?
 
Every derivative brings an
ω and there's a net -
sign.
 
So you find out that your
guess, C˜c
osωt is going to
work, provided this equation is
satisfied.
 
And the cosωt can
be canceled.
That's the whole point of doing
this thing.
So then you get C˜
= V_0 -
V_0 divided by
ω^(2)L −
1/C.
 
Or if you like,
-V_0/L divided
by ω^(2) -
ω_0^(2).
Don't worry about the numerator;
look at the denominator.
The denominator says that if
your driving frequency is equal
to the resonant frequency,
C˜ blows up.
That means when a system has
got a resonance--yes?
Student:  Where did the
ω^(2) come from?
Prof: I took two
derivatives--oh,
here.
 
Oh, I'm sorry.
 
Yes, thank you.
 
You know, it's good to do these
things slowly and not rely on
what you heard somewhere.
 
So let me do it for you again.
 
Sorry about that.
 
So this time I take two
derivatives.
Each one brings an
ω.
There is an L,
and the - sine comes because
the cosine becomes - sine.
 
Differentiate one more,
you get - the cosine.
This guy is nothing.
 
It just says "divide my by
C"
and that's equal to that.
 
And then I wrote 1/LC is
ω_0^(2).
Thank you very much.
 
Okay?
 
I was really hung up on the
final result,
which tells you,
you'd better not drive this at
the resonant frequency,
because then current amplitude
will build up indefinitely.
 
That's also true in a swing.
 
If you've got a swing and the
kid's coming back and forth,
and you're reading a newspaper
just pushing the kid,
you've got to push at the right
time to get the best result.
And that kid's not going to fly
off, because there's one more
term in the equation for the
kid, which is friction.
But if you have a frictionless
swing and you're doing this,
you've got to watch out,
because soon there'll be nobody
around,
because the amplitude will keep
growing.
 
Well, these are unrealistic
problems, but I want you to
notice one thing.
 
Here's what I want you to
notice.
The voltage you were given
looked like V_0
cosωt and the
current that you got,
I can obtain by taking the
derivative of this charge,
which is like dQ/dt.
 
That's going to be
C˜ω
sinωt,
and C˜, we can
write as V_0/L
divided by ω^(2) -
ω_0
^(2)sinωt.
This is I.
 
Here's what I want you to
notice.
It looks a lot like Ohm's law,
because the current looks like
voltage divided by some number,
but that's a very big
difference.
 
The voltage is a cosine and the
current is a sine.
That's something I want you to
think about.
The current is not in step with
the voltage, whereas in a
resistor circuit,
the current follows the
voltage.
 
It's the same profile as the
voltage, except you divide by
R.
 
Here one is a cosine,
one is a sine.
That means when one guy is at
maximum, other is at minimum.
It's called out of phase,
in fact out of phase by 90
degrees.
 
Yes?
 
Student:  ________ by
the C that you got,
where did the omega go?
 
Prof: Here?
 
Student:  On the other
side.
When you multiply that through
by the C--
Prof: Oh,
ω did not go,
it's back.
 
Right here. You're right.
 
There's the ω.
 
That came from...
 
 
 
Okay, these problems,
the only thing I'm focusing on
right now,
I would say of all the things I
wrote there,
to be the most important thing
is number one,
this is what I want you to bear
in mind.
 
That's why sometimes I'm not
paying attention to some
constants.
 
What is important to notice is
that this is an equation you can
solve by inspection.
 
Why was it easier to solve it
by inspection?
Because you are trying to get
in the end a cosine to balance
the right hand side,
and you're trying to find a
function who is itself a cosine,
or whose second derivative is a
cosine,
and we know the answer to that
is a cosine,
so we can guess the answer.
Once you guess the answer,
you put it in and you analyze
the solution,
you notice also that the
current and the voltage are not
in step.
One is a cosine.
 
You know what a cosine does,
it does that.
Other is a sine,
so they are out of step.
When one is at maximum,
the other is at minimum and so
on.
 
That means a current as a
function of time is not equal to
the voltage as a function of
time divided by anything.
There is nothing you can divide
a cosine by to turn it into a
sine.
 
Whereas with resistors,
you just divide by R,
you get the current.
 
Now for the more realistic
problem, the realistic problem
has got a capacitor,
an inductor and a resistor and
no other power supply.
 
The equation obeyed by this one
will be Ld^(2)Q/dt^(2)
(that stands for
LdI/dt) RdQ/dt 
Q/C = 0.
 
This is analogous to
md^(2)x/dt^(2)  (I don't
know how you guys wrote this
thing last semester.
It doesn't matter)
γdx/dt  kx =
0.
 
Notice that it's the same form.
 
And I'm going to give this to
you as a homework problem to
analyze the answer to this one.
 
So this describes a problem
where you have a mass and a
spring and some friction on the
table.
This means if you pull it and
let it go, the oscillations will
eventually get damped and it
will die down.
And the general solution for
the case when it's oscillating
will look like some number A
e to the minus some
number αt times cosine
times some frequency
ω't.
 
Let me see.
 
You can always add a phase
Φ,
but I'm not going to do that.
 
If you want,
you can put an extra Φ,
but I'll choose my origin of
time so that I don't have that.
This is what the answer's going
to look like,
where ω' and
α are going to be
controlled by L,
R and C.
That's the thing I don't want
to do in class.
Have you seen this before?
 
Professor Harris tells me
you've seen this last time.
And the hints in the homework
will guide you on how to do
this.
 
You just assume the solution
x at t looks like
some A e to the - (I
don't want to call it
α now)
β times t.
You put it in the equation and
solve for β and
you'll find β
will have a real part and
an imaginary part.
 
And you have to combine the two
to get this answer.
Now that brings me to another
thing, so I don't know how
prepared you guys are for what's
about to happen next,
which is the use of complex
numbers.
So everybody familiar with
complex numbers?
Who doesn't know complex
numbers?
How do you do your taxes?
 
You don't know imaginary
numbers?
So I will tell you.
 
I'll give you a lightning
review.
I'm assuming everybody has seen
them in high school.
I will only tell you the part
you need, but I'm going to
assume that I can use them
fluently.
I don't want to stop every time
and worry about you guys.
So I really need a show of
hands.
Anybody never seen x iy
and x - iy?
And how about
e^(iθ)?
You know that guy?
 
Okay, that's all you need.
 
So I'm going to tell you what
the deal is.
So everyone knows what a
complex number is.
We know i is square root
of -1.
Then we know that we can write
a complex number as x iy
and visualize it in a complex
plane where you measure x
this way and iy is
measured that way.
That's usually called a generic
complex number z.
But now let us find the length
of that--the complex number is a
single point in the complex
plane.
The length of that is square
root of x^(2) y^(2).
So I can also write z as
x divided by square root
of x^(2) y^(2) i
times y divided by square
root of x^(2) y^(2) times
the square root of x^(2)
y^(2).
 
Just done nothing,
just rearranged stuff.
This is going to be called the
modulus of z,
and what is this?
 
This angle is θ
here.
This is really
cosθ
isinθ times
modulus of z.
And that, thanks to this great
identity by Euler,
is e^(iθ). Now I
don't know how much you know
about this great formula that
relates e^(iθ) to
cosine sine.
 
How many people know where it
comes from?
What does it mean to raise
e to a complex power?
You know where it really comes
from?
Yes?
 
Student: 
>
Prof: It comes from
power series.
It turns out you can write a
power series for cosine.
You can write a power series
for sine,
then, for example,
cosine of θ = 1
− θ^(2) over
2 factorial,
θ to the 4
factorial,
or 6 factorial, etc.
 
And you can write a power
series for e^(θ) if
you like,
which is 1 θ
θ^(2)/2.
 
This defines cosθ
in the sense that if you put
θ = Π/2
in this infinite series,
you will get 0.
 
And if you put θ
= Π, you will get
Π -1.
 
In other words,
in spite of this funny looking
form, it really is the same guy.
 
It will oscillate,
it will have zeros,
it will be periodic.
 
None of it is obvious,
but this power series is
numerically equal to this one if
you keep the infinite number of
turns.
 
Likewise, e^(θ) is
defined by this series.
And similarly,
there's a formula for
sinθ.
 
Now once a power series is
defined, you can put e raised to
anything you want there.
 
So you know what I'm going to
put there.
e raised to dog is 1 dog
dog squared.
This is not a joke.
 
This is really true.
 
This is the definition.
 
If someone says,
"How do I raise e
to the power dog?"
 
you do this.
 
This will have all the
properties of the exponential.
It doesn't matter what's in the
exponent.
That's the key.
 
People originally put real
numbers, then they put complex
numbers.
 
Now they put matrices,
operators, anything you want.
e raised to anything,
you formally define to be this
infinite series,
provided the infinite series
converges and gives you a
meaningful answer.
In that sense,
if you put e^(iθ)
here,
and compare the result to
cosθ
isinθ,
it matches, that's all.
 
I'm just going to keep using
that result.
Then you should also know that
for every complex number,
there's a complex conjugate,
which is x - iy.
That means i goes to
-i.
That means z,
the angle θ
will change to
-θ.
And given any complex number,
the real part of z,
which is x,
is z z* over 2 and the
imaginary part of z,
which is y,
is z −
z* over 2i.
This is what you need to know.
 
Every complex number has a real
part and an imaginary part.
In fact, z* looks like
this.
And if you add z z*,
the vertical parts cancel
and you get double the
horizontal part.
That's why you divide by 2.
 
You subtract and divide by
2i, you get y.
Basically, that's all I want
you to know, but I want you to
be able to manipulate them
rapidly enough.
Here's something very useful
about complex numbers.
Let's take a complex number
z_1 which is
mod z_1
e^(iθ1),
and another complex number,
z_2,
which is mod z_2
e^(iθ2).
So what do they look like?
 
Well, z_1
looks like this,
with some angle
θ_1,
and z_2_
may look like that.
It's got some angle
θ_2.
So every complex number has a
length and an angle.
You can think of it two ways.
 
It's got an x part and a
y part,
and a real and an imaginary,
or it's got a modulus and a
phase.
 
That's the Cartesian version of
the number and the polar number
of the number.
 
And the connection between them
comes from Euler's formula.
But now look what happens when
I take the product
z_1z
_2.
I get mod z_1
mod z_2
e^(iθ)_1^(
θ)_2,
because exponentials add when
you multiply them.
Well, this allows you to
immediately guess what the
product is going to look like.
 
The product is going to have a
length equal to the length of
z_1 times
length of z_2.
It's going to have an angle,
I've not done a good picture
here,
which is the sum of the two
angles,
because it's
θ_1
θ_2.
So listen to this statement
very carefully.
When you take a complex number,
say number 1,
and you multiply it by a second
complex number,
you do two things to the first
guy.
You rescale it and you rotate
it.
So two operations are done in
one shot when you use complex
numbers.
 
The mod z_2
rescale is the mod
z_1 and
θ_2 adds
to θ_1.
 
Likewise, if you divide,
if you take
z_1 over
z_2, it is mod
z_1 over mod
z_2 times
e^(iθ)_1 ^(-
θ)_2,
because the exponential
θ_2 is
downstairs.
 
So multiplication by real
numbers is very easy.
You take a real number 4,
you multiply by 8,
you'll get something in the
real axis 8 times longer.
You multiply by -8,
you'll get something back here.
With complex numbers,
you take a number,
you multiply another number,
you rescale and you rotate.
That's what I want you to know.
 
That's going to be very
important for what I'm saying.
Now I come to the problem I
really want to solve,
which is an LCR circuit driven
by an alternating source,
V_0
cosωt.
This is R,
this is C and this is
L.
 
And the equation we have to
solve is L dI/dt
RI 1/C integral
I dt =
V_0
cosωt.
Because Q is the
integral up to time t of the
current,
that's just Q/C. It's
the same equation,
but now, unlike in this
problem, where I had no driving
voltage,
I have a driving voltage,
and I've also written it in
terms of current rather than in
terms of charge.
Because in electrical
engineering, you don't really
watch the charge in the
capacitor.
You look at the current flowing
through the circuit.
So you have to solve this
equation.
The question is,
can you guess the answer?
That's the only way we know.
 
You've got to guess.
 
You're trying to find a
function, I of t,
so that when you differentiate
it, you get a cosine
ωt,
when you integrate it,
you get a cosine
ωt.
So far we can do it.
 
Sine ωt will do it.
 
But when you leave it alone,
then also you get a cosine
ωt. You cannot do
that,
because sine will become a
cosine and sine will become a
sine.
 
In the LC circuit,
when you didn't have this guy,
we were okay,
because if you pick a sine,
this became a cosine and that
also became a cosine and you can
combine them to get a cosine.
 
But you cannot have
dI/dt,
I and integral of I all in one
equation and all satisfied by
trigonometric function.
 
The only time we can guess the
answer is if it looks like
V_0
e^(αt).
Suppose that was the voltage.
 
Then can you guess the answer,
what form the answer will have?
It will also be e^(αt)
times some number,
because then this will look
e^(αt), that will
look like e^(αt),
that will look like
e^(αt).
 
e^(αt) has a great
property that whether you
multiply it,
whether you integrate it,
differentiate it or leave it
alone,
it looks the same.
 
Unfortunately,
no one's interested in this
voltage, because it's growing
exponentially fast,
or if you put a - sign,
it's dying exponentially.
What we really want is this.
 
So the question is,
how do we solve this problem
with cosωt?
 
So we're going to use a certain
trick.
The trick I'm going to use is
the following:
let me take a general case
where this is some function V
of t,
where I'm not even assuming
V is real.
 
I'm not assuming I is
real, but L and R
and C are real.
 
Everything else can be complex
numbers.
Now once you have an equation
and you've found a solution,
if you take the complex
conjugate of both sides,
they will still match.
 
Do you understand that?
 
If two things are equal,
their complex conjugates are
equal,
because this thing is the real
imaginary that balances the real
imaginary on the other side.
Complex conjugate just reverses
the imaginary terms,
so they will still match when
you flip both sides.
So it means if you take L
dI* of dt R times
I* 1/C integral I* of
t' dt' t = V* of
t.
 
So star is the rule by which if
V had a real and
imaginary part,
you flip the imaginary part.
So that solution for V
of t implies a second
equation.
 
In other words,
if the voltage V drives
a current I,
the conjugate of the voltage
drives the conjugate of the
current.
That simply follows from the
equation.
It comes from the fact that
when I conjugated it,
L, R and C are
real.
Do you understand that?
 
Suppose L were imaginary
or had imaginary part,
then I must also put a
conjugate on L.
Then I* obeys a
different equation,
because this had L in
it, this had L* in it.
But if everything is real,
I* obeys the same
equation when you have a
V* driving it.
I'm almost done now.
 
I want you to add the left hand
side to the left hand side and
the right hand side to the right
hand side, and what do I get?
Well, if I write it here,
you may not be able to see it.
Let me put it here then.
 
If you add these two,
you'd get L d/dt of
I I* R
times I I* 1/C
integral of I I* dt =
V V*. This again comes
because the equation is a linear
equation.
You can always add the left
hand side and the right hand
side to get another problem
where the driving voltage is
V V* and the driving
current is I I*,
etc.
 
But I I* is 2 times the
real part of I and V
V* is 2 times the real part
of V.
Therefore this implies that
L d/dt of the real part
of I R plus the
real part of I 1/C
times the integral of the real
part of I = real part of
V.
 
So let me say in words what the
equation means.
If you by luck solve an
equation with a complex
potential V,
and you get a current,
then the answer to the problem
where the potential is only the
real part of the actual
potential you applied,
the current will be the real
part of the answer you got.
That's all I want you to know.
 
If you solve a complex problem
with a driving voltage which is
complex--
which is just a mathematical
fiction,
in real life you don't have
that--then the current that you
get will also be complex.
But its real part will be due
to the real part of your
voltage.
 
And you can also show the
imaginary part of the current
will be due to the imaginary
part of the voltage.
This is just the principle of
superposition.
So here is the trick we are
going to use.
We are going to go back to
original equation,
LdI/dt RI 1/C integral
I dt = V_0cos
ωt,
and we're going to solve a new
problem.
 
We're going to solve a problem
where this is V_0
e^(iωt).
 
And the current for that,
I'm going to call some
I˜.
 
It's a complex current,
because the driving voltage is
a complex voltage.
 
So think of
Ve^(iωt) as the
V in the previous
problem.
The actual problem I was given
had V_0cos
ωt,
but you know that
V_0
e^(iωt) is
V_0
cosωt I
times V_0
sinωt.
Consequently,
this potential has a real part
which is a cosine,
an imaginary part which is a
sine, and the current will have
a real and imaginary part.
The answer to the original
question is the real part of the
answer to this question.
 
This is what you should think
about.
This is what you should
understand.
So I've manufactured a problem
with an unphysical complex
voltage,
and all I know is that at the
end of the day,
if I take the real part of the
current,
I'll get the answer to the real
part of the potential,
which happens to be the actual
potential,
V_0
cosωt.
 
So why would I do all this?
 
Why would I take a problem
that's bad enough with the
cosine and turn it into a
complex exponential?
I think you can sort of guess
what the reason is.
The reason is that it may be
complex, but it's still an
exponential.
 
That means its derivative is
going to look just the same.
That means I can make a guess
that the current I˜
is proportional to
e^(iωt). So you can
make the assumption that
I˜ of t is
some constant I˜
times e^(iωt).
Make that assumption and put it
into this equation and see what
you get.
 
You will find you will get
IωL times
I˜ R times
I˜  1/IωC
times I˜
times e^(Iωt) is
V_0e^(iωt).
 
This is because if you assume
the current is an exponential,
every derivative = an
iω and every integral
= 1/iω, and
multiplying is just multiplying.
So d/dt has got to be
replaced by ω,
and you get this equation
for I˜. That means
you will satisfy the equation,
provided I˜ times
all of this = V .
 
And I'm going to write this as
Z times I˜
= V where Z is the
name for all the guys
multiplying I˜.
 
So I'll start with that here.
 
 
 
Okay.
 
So I wrote iωL
R 1 over iωC
I˜ e^(iωt) is
V_0
e^(iωt).
 
Everybody get that?
 
That's because differentiating
on an exponential is just
multiplying by the exponent.
 
So this is a complex number.
 
R iωL −
i/ωC.
I want you to know that i
in the bottom is like -i
in the top.
 
And this is called the
impedance Z,
it's a complex number.
 
You can visualize the complex
number as follows:
it's got a real part,
which is R.
It's got an imaginary part,
which is ωL −
1/ωC,
and the impedance Z is a
complex number with some modulus
and some phase.
If you know the real and
imaginary parts,
you can construct the modulus
in the phase like that.
If you want,
it's the modulus of Z.
Therefore canceling this,
I find I˜
= V_0/Z.
 
That becomes V_0 over
the modulus of Z times
e^(i)^(Φ),
where Φ is this angle
defined by tanΦ
= ωL −
1/ωC divided by R.
 
So don't worry too much about
writing all of this down,
because this you're going to
find in every book.
There's nothing novel or
different.
So this is the formula for
I twiddle.
Now I itself,
you remember,
= I˜ of 0.
 
I'm sorry, this is
I˜ of 0.
Did I define it that way?
 
Yeah, I should write it--I'm
sorry.
I should write it as
I˜ of 0.
There's a subscript 0 there,
because the full I twiddle has
got an e^(iωt) in
it.
That looks like
V_0 over mod
Z times e^(iωt -
)^(Φ).
But this is not the physical
current, because it's the result
of a complex voltage.
 
The physical current is a real
part of this,
without the twiddle,
that we write as real part of
I˜. Where do I get
the real part?
V_0 and
absolute value of Z are
both real.
 
Real part of e to the
i something is cosine of
something and that's our answer.
 
That's our answer,
the final answer you get.
This tells you that the current
has a magnitude which is
V_0 over the
absolute value of Z,
and it's got a phase
Φ by which it is
behind the driving voltage.
 
The magnitude of Z for
as in any complex number,
if someone says,
"What's the magnitude of
Z?"
 
it is simply square root of
real part squared imaginary part
squared.
 
So I want you to think about
the magic of complex numbers.
Your final answer has a current
which is a cosωt -
Φ.
 
Your driving voltage is a
cosωt.
There is no way you can divide
the voltage by anything to get
that current.
 
But in the complex language,
the complex current is the
complex voltage divided by the
complex impedance.
How does the happen?
 
That happens because impendence
Z has got a magnitude and
a phase,
therefore if you had a voltage
V_0
e^(iωt) and you divide
by mod Z e^(i)
^(Φ) , that
becomes e^(-i)
^(Φ) upstairs.
You're able to change the
magnitude and you're able to
change the phase by dividing.
 
So if you want to take the
voltage and you want to rescale
it and shift its phase,
all in one shot,
you can do it if you divide by
a complex number,
because a complex number will
rescale it and also shift its
phase.
 
And then you take the real part.
 
In other words,
cosωt divided by
nothing will give you
cosωt -
Φ,
but e^(iωt),
when divided by
e^(i)^(Φ),
will in fact give you
e^(iωt -
i)^(Φ).
 
So that phase shift we can
produce only by dividing with a
complex number.
 
So I'll come back and explain
to you a little bit more on how
you do circuit problems using
the complex impedance,
but I think it will be helpful
for those of you who don't know
complex numbers to get use to
this part of it.
 
 
