Class average on the exam was fifty-five.
Three homework problems were on the exam.
I'm a little puzzled by the fringe field problem.
That was the one homework problem that was graded.
It was graded on your homework, and I went back to your
homework scores and they were ninety percent.
You all did extremely well on that problem.
But on the exam, it was only forty percent.
And so perhaps this is telling you something.
And maybe it's telling me something too.
If you turn in a correct solution for homework,
it's not very useful if you do not understand what you wrote
down. The understanding,
of course, is what matters, not that you somehow,
one way or another, get a correct solution.
If I had to judge you only on the basis of exam one and two,
forgetting the quizzes, forgetting the homework,
then those with eighty or lower would fail the course,
as of now. But of course those who have
between eighty and ninety are by no
means home free. They are still in the danger
zone. And anyone who has ninety-five
or even a hundred, I cannot guarantee you that you
will pass the course. The depends,
of course, on how you will do in the future,
the third exam and on the final.
Today I want to discuss with you RC circuits.
We already discussed RL circuits.
Now we get RC circuits. I have here a battery
with an EMF V zero. And I have here a switch
connecting to a capacitor. And here a resistor.
And I close the loop. When I have the switch in this
position, the capacitor is going to charge up.
I get a current I going in this direction.
If I call this point A here and this point P and this point S to
make sure that we are on the same wavelength,
I will call the potential over the capacitor,
I will call that V A minus V P. The potential over the resistor
is of course always I times R, is then V of P minus V of S.
The question now is what is the potential over the capacitor
doing as a function of time and what is the current doing as a
function of time as I charge up this capacitor.
And your intuition will help you a great deal without any
fancy differential equations. It is clear that at T equals
zero, the capacitor is not charged.
There is no charge on the capacitor.
And it will take time to charge the capacitor.
So at T equals zero, you expect that the potential
over the capacitor is zero. If you make T a little larger
than zero, you are going to charge up this capacitor,
and so the potential over the capacitor will go up,
and therefore the current will go down.
And if you wait long enough -- we call that infinitely long --
then the capacitor will be fully charged.
It will have the potential V zero of
the battery, and then the current has become
zero. No current is flowing anymore
if the battery -- if the capacitor is fully charged.
And so this capacitor is going to charge up,
this becomes positive, and this becomes negative.
And so you can construct a, a plot whereby you plot as a
function of time here the potential over the capacitor --
we haven't used any differential equations yet.
You know that if you wait long enough you will reach that value
V zero. And it's going to build up like
this, asymptotically reach that value.
And if C is very large, then the current will be more
like this, and if C is very small, then it will go much
faster, of course, small C.
The current as a function of time.
In the beginning, the current will be high,
but ultimately the current will die down, when the capacitor is
fully charged. So you expect something like
this. So this you can do without any
differential equations. Let's now do it the correct
way. The closed loop integral of E
dot DL happens to be zero, which will make Mr.
Kirchoff very happy.
The closed loop integral of E dot DL in this circuit is zero.
Mr. Faraday is happy and Mr.
Kirchoff is happy. There is no magnetic flux
change here; we don't have self-inductances.
The electric field inside the capacitor is in this direction,
from plus to minus. The electric field in the
resistor is in this direction, the current is flowing in this
direction. And this battery,
which has this as the positive side and this as the negative
side, inside the battery the electric field is in this
direction. So if I start at point A and I
go around the circuit, then E and DL are in the same
direction if I go from A to P, and so I get plus V of C,
that's the E dot DL from A to P.
Then I go through the resistor. Again, E and DL are in the same
direction, so I get plus I times R.
Then I come through the battery.
Now the electric field is opposing the direction in which
I go, so I get minus V zero. And that is zero.
So that's my differential equation.
And I can write it differently, because I know that I,
the current, is dQ dT, Q being the charge on
the capacitor. And it is only if that number
changes, if the capacitor is either charging up or
discharging, is there a current flowing.
And in addition I know that V of C is Q divided by C.
That's the definition of capacitance.
And so I can write down for this V of C, I can write down Q
divided by C. For I I can write down dQ dT.
So I get R times dQ dT minus V zero equals zero.
And this is a differential equation in Q.
We have seen an identical differential equation.
It was not in Q, it was in I,
but it had of course a completely similar solution.
And the solution to this differential equation is
actually quite simple. I will put it on this board.
Q -- so this is going to be Q as a function of time --
is V zero C times one minus E to the minus T over R C.
If I take the derivative of this Q then I have I,
because I is dQ dT. So I, which is dQ dT then
becomes V zero times C, I get a minus sign,
I get another minus sign, then I get one over R C,
and then I get E to the minus T divided by R C.
So the two minus signs eat each other up and I lose one C here.
And so I get that the current as a function of time is V zero
divided by R times E to the minus T divided by R C.
So that's the curve at the bottom.
And the potential over the capacitor is now very simple,
because the potential over the capacitor is Q divided by C.
And I already have Q here, so I simply have to divide this
C out. So I get V zero times one minus
E to the minus T over R C. And so that is the upper curve.
And so we can now make a small table and we can look at various
values for T. Maybe I should do that here on
the blackboard, because I don't want to erase
anything yet. So we have T here,
we have I, and we have V of C. And when T is zero,
you go to your equation of I, so this is one.
So you have a current V zero divided by R.
And you V C -- V C of C is zero.
You can see that. If T is zero,
you get one minus one -- oh sorry, I have to go here.
You get one minus one, so you see indeed that the
potential over the capacitance is still zero.
If you wait long enough, then the current must go to
zero. That exponential function goes
to zero if you wait long enough. And your potential over the
capacitance then reaches V zero, which is exactly consistent
with our solution. If you wait a time RC,
that's called the time constant of this circuit.
In the case of the current, it's also called the decay time
of the circuit. Then of course your current is
one over E times V zero divided by R.
And one over E is a roughly point three seven,
oh point three seven. So your current is down to
thirty-seven percent of what it was at the beginning.
And after a time R C, the potential over the
capacitor is one minus one over E times V zero,
and that is then about oh point seven three -- uh,
six three, oh point six three, sixty-three percent.
In other words, if I go back here to my plot
and if I draw a line at time R C, then this value here is
thirty-seven percent of the maximum and this
value here is then sixty-three percent of the maximum.
So it's in -- the solution is rather obvious,
very intuitive. And the R C times can vary an
enormous amount, as you can imagine,
depending upon the values for R and C.
If we have here an R and a C and we want to know what the R C
time is, convince yourself that the product of R and C indeed
has units of seconds.
Remember we had L over R before, which had also units of
seconds. R C also has units of seconds.
If you have R equals one ohm and C is one microfarad,
then the R C time is only a microsecond.
But if you have R equals hundred megaohms and you have
this one millifarad, then this is ten to the five
seconds, which is longer than a day.
And that would mean that it would
take you even three days to reach ninety-five percent of V
zero. After three days,
you would still have only ninety-five percent of the
potential difference of the capacitor of the maximum value
that you can get. Now what I want to do is I make
a change here. I have here a conducting wire.
I call this position one of the switch.
And I'm going to put the switch in this position,
position two. And I can do that without any
danger. Remember, I waited until this
capacitor was fully charged. There was no current running.
And so when no current was running, I can quietly take the
switch and put it in this position.
And what is going to happen now is of course this side is
positively charged and this is negatively charged,
so now you're going to get a current which is going to run
counterclockwise, in opposite direction.
And what is going to happen is the
capacitor is discharging now. And the resistor will dissipate
the energy that is in the capacitor.
The one half C V squared energy stored in the capacitor is going
to be dissipated in the resistor in terms of I squared R,
in terms of heat. And if you wait long enough,
the current will become zero. So it should be obvious what's
going to happen. If I return to this curve here,
if I redefine my time equals zero,
and if this is the moment that I put the switch in position
two, then I expect that the capacitor will discharge.
You get a curve like this. And I expect that the current,
which now becomes negative -- it reverses direction and I call
that negative. And so the current will come
like this. And if you wait long enough,
of course, the current will aga- again become zero.
You have discharged the capacitor.
So if you want the formal solution,
you have to go back to the differential equation.
And you take this term out, because it's not there.
And now you have to solve this differential equation again,
which is now utterly trivial. And I would like you to solve
that differential equation. You couldn't have an easier
one. I will give you the solution to
I as a function of time, and you then will come up with
this part. I as a function of time is
exactly the same as this except with a minus
sign, provided that I call this T equals zero.
So I redefine the zero time. And so I as a function of time
is the equation that you have here, but now with a minus sign.
So you get an exponential change again,
but the current has flipped over.
I can demonstrate this to you. I have a electronic switch,
so I go between one and two -- every four milliseconds I throw
the switch. And so what I have is,
as a function of time, what we call a square wave.
So this is my battery. And this time here,
from here to here, is eight milliseconds.
This is time. And so this is that value V
zero. And the values that I have
chosen for R and C,
I will give them to you. The value for V zero here is
one volt, and here of course it's zero.
The value that I have chosen for R is six kiloohms.
And the value I have for C I think is oh point one
microfarad. Yes, that's what it is.
And there's a reason why I've chosen these values.
This is oh point one microfarads.
And so the R C time is six times ten to the minus four
seconds. So this is point six
milliseconds, which is substantially smaller
than four. And so you can expect that the
capacitor becomes almost fully charged in these four
milliseconds. That's why I chose the R C time
substantially smaller than four milliseconds.
And my setup is such that I can show you the,
um, the input, the driving voltage of the
battery. I can show you this.
I can show you then how the capacitor is charged and how the
capacitor is discharged with the time constant R C.
And also how the current goes through the system.
So I can show you also this curve.
And then I can change the capacitor
so that the R C time changes. I can also change the
resistance. So let me change first the
lights, so that you get high quality for your money.
You already see there the switching voltage between one
volt and zero. And now I'm going to show you
the voltage over the capacitor, and I will take the other one
out. And here you see indeed exactly
the image that we discussed. So you see early on the
charging of the capacitor, and it reaches effectively the
maximum value. You know, you don't have to
wait infinitely long. That's always -- we say
infinitely long, but that's -- clearly if you
wait three or four, five times R C then you're
almost there. And then here I'm going to
switch it back to position two, and you see it's discharging.
And then it's charging up again, and discharging.
And now I can also show you the current.
You see it on the same plot as the battery -- as the capacitor
is charging, you see the current is high.
But by the time that the capacitor is fully charged,
the current is zero. But the moment that I throw the
switch to position two, the current flips direction,
becomes negative, and then as the capacitor
discharges and heat is dissipated in the resistor,
ultimately the current again will become zero.
What I can do now is to increase my capacitance by,
for instance, a factor of five.
But that would change my R C time to three milliseconds.
So now there is no way that the battery -- not the battery,
no, there is no way that the capacitor can become fully
charged in the four milliseconds that it has.
And you will see that indeed if I
make this five, point five microfarads,
oh point five microfarads, then you see that the capacitor
has not enough time to get fully charged, and so here you begin
to discharge. Notice also that the current
doesn't reach to zero, for the same reason that the
capacitor doesn't get fully charged.
So it never reaches the point that the current becomes zero.
I'm already switching here to position two,
and the current then flips over but never becomes zero.
All right, so let's go back to point one value.
There it is. This is easier than LR circuits
for the reason that we don't have to deal with Faraday's Law.
We don't have a non-conservative field.
And so it's easier to imagine things, because we're dealing
with Kirchoff's Law. So potential differences are
uniquely defined and don't depend on the path,
whereas with Faraday's Law they do depend on the path.
So now I want to change to a different subject,
which is transformers. Transformers play a very
important role in our lives. Transformers are key in many
instruments that you use at home but
also key in getting the energy all the way from the power
station to us, as you will see.
A full understanding of transformers is not easy.
It's extremely complicated. It's really more like an
engineering problem than a physics problem.
I will give you a simplified version, in which I leave a lot
of details out, but the basic idea will be
there. Here I have a coil,
and I call this coil the primary.
This coil has N1 windings and has a self-inductance L1.
And I put here a voltmeter, which I call V1.
It's always in that circuit. And a current I1 is running
through the coil and returning. And there is a teeny weeny
little current I1 -- I put a little I there -- running
through this voltmeter. Insignificantly small,
so I just call this also I1, because this I1 is so small.
But it gives me, it allows me to always monitor
V1. So that's the primary.
Now we have a secondary, which is wound in such a way
that there is a magnetic flux coupling between the two.
This is the secondary. It has N1 -- N2 windings.
Self-inductance is L2. And I put here a voltmeter V2.
The current I2 is flowing here. This is I2, here is the
consumer, this is where you are maybe.
And I2 is flowing back through the coil.
But here is this teeny weeny little current I2 flowing so
that I can monitor the value of V2.
What I can do now is I can write down the closed loop
integral of E dot DL for this closed circuit here.
And I have to apply, of course, Faraday's Law,
because I now have a changing magnetic flux.
They're always AC. Transformers always require
alternating current. And so if I start here and I go
through the self-inductor, I know there is no electric
field in the self-inductor, so that contribution of E dot
DL is zero. Then I arrive here at this
current, which is opposing me. I go in this direction,
but the E field is opposing me, and so I get minus V1.
And that now, according to Faraday's Law,
since I went in the direction of the current through the
self-inductor, equals minus L1 DI1 DT.
And I'll tel- tell you what the simplification is that I made,
because I want to be honest with you about where I am rigid
and where I am not. here is really a second term
here, which is the mutual inductance between the two
coils. And there should really be a
term here in terms of that mutual inductance capital M
times DI2 DT. And I leave it out here.
I leave it out because the final result in most cases is
not affected by it. But I want you to know that
strictly speaking, this equation is
simplified. I know that this is the E
induced in side one. And now I do the same closed
loop integral through the coil, through the voltmeter back to
the coil. I start here.
The current I2 is in the same direction as the direction that
I move. So I have plus V2.
Then I go through the self-inductor.
There's no electric field in the self-inductor,
because there's no electric field in a wire that has no
resistance. And that now,
since I went in the direction of the current,
is minus L2 DI2 DT. And again, I bury the M
contribution. And this now equals the EMF
induced in the secondary. But clearly this is also N1
times D phi B DT, with a minus sign.
This is the magnetic flux change through one loop in the
primary. The surface of one loop in the
primary sees a magnetic flux going through it which is
changing with time, and that value D phi B DT is
through one loop. But I have N1 loops,
so I have an N1 there. And so in the secondary,
I have N2 times the same D fee by DT
if I have perfect magnetic coupling between the two loops,
which is not always the case. And so now you see that I have
that V2 over V1, if I take the magnitude,
is now simply N2 over N1. That is an amazing result.
It tells you that you can make V2 on the secondary side way
larger than V1. We call that transforming up,
by making N2 larger than N1. But you can also make it lower
-- we call that a step-down transformer -- by making N2
smaller than N1. When I write down capital V
here you should think of that. It's an alternating current,
so you have cosine omega Ts in there, or sines omega T.
You should think of this perhaps as the maximum value
possible. But of course that's not
important in the concept.
But there are always cosine omega Ts all over the place.
The power station puts the electricity, so to speak,
on the line at three hundred thousand volts.
We discussed that earlier, why they do that,
that's such high voltage. So they have a transformer from
their generator to transform the potential up to a very high
value, to three hundred thousand volts.
When it arrives at Boston, it's stepped down to twelve
kilovolts. And so there are transformers,
which you see, huge transformers,
which have a ratio N1 over N2 of twenty-five.
So N1 is twenty-five times larger than N2.
So I step it down to twelve kilovolts.
But you don't want twelve kilovolts at your home.
You want hundred ten, hundred twenty volts at home.
So when you look at these power poles outside people's homes,
you see again transformers in these power poles,
which bring it down from twelve kilovolts to about hundred
twenty volts. So there you have a ratio N1
over N2 of about one hundred. To calculate the ratios of the
current, I1 on I2, is way more tricky.
There are some big ifs. And one big if is that R in the
primary -- in the secondary side, is much,
much smaller than omega L. That is one if.
The second if is that no energy is lost in terms of eddy
currents. Very often these two coils are
coupled through an iron core, and then you get,
get eddy currents in the iron core, then you lose energy.
Also, do you not always have ideal flux coupling.
So it is not always true that the magnetic flux through one
coil in the primary is the same as the one through one coil in
the secondary. But if all of that were the
case, so if R is much,
much less than omega L and if there is no energy lost in eddy
currents and if there is perfect flux coupling,
then you can show that whatever power is delivered on the
primary side is going to be consumed on the secondary side.
And if that is true, if the power delivered on the
primary side, which is I1 V1,
if that is the same as the power consumed on the secondary
side, then that must be I2 V2.
But if this is true, which is only under those
conditions, and this is also correct, then of course you will
find now that I2 divided by I1 is N1 divided by N2,
and let's put here magnitudes again, so that we are not
worried about possible minus signs.
And then you look at this, you see that you can generate,
if you want to, in the secondary an enormously
high current by simply having a large ratio, N1 to N2.
And I will demonstrate that today, that indeed if you meet
those conditions that you can indeed do that.
I2 over I1, that's fine, V2 over V1, that's fine.
So I will do two demonstrations.
The first thing I want to demonstrate is that if I make --
oh, that's beautiful. I have to erase something here
-- oh, actually, I could work on this board here
for a while. I have here a demonstration
whereby you see here the primary coil.
You see it right in front of you.
It has two hundred and twenty windings.
So N1 is two hundred and twenty.
And V1 is hundred and ten volts, and it is AC,
has to be. The frequency is sixty hertz.
It's just what we get out of the plug here,
like you get it in your dormitory.
That means omega is about three hundred and seventy-seven.
It's two pi times F. If I give you really the
correct value for V1, it's really the maximum value
for V1 times the cosine of omega T.
And this value is really hundred ten times the square
root of two. I've mentioned that earlier
in my lectures. The voltmeter,
however, are cir- are designed in such a way that they only
show you hundred and ten. And so we call that hundred and
ten volts. I can now make the secondary
just one winding. I have a wire that I re- ro-
wrap around once. And if indeed that equation
holds, which it will, I expect now that V2 will be
about oh point five volts, namely the hundred ten volts
divided by two twenty, which is the ratio primary over
secondary. And then I will put four
windings around, and then you expect about two
volts. And you will see it there right
in front of you. This is where you're supposed
to see the reading. I can't see it.
So I have to trust you when you give me those values.
OK, so here is my primary coil, and here is my secondary.
This is my secondary coil. And the way I have this
demonstration is that this is completely open.
So I only show you immediately the value of V2.
I'll show you this value. It has a huge value of a
resistor, several megaohms. So for sure I don't meet
this condition. But that condition is not
necessary for this to hold. That condition is necessary for
this to hold, not for this to hold.
So I have no fears that I am going to be embarrassed.
So I'm going to put one -- this is very long wire.
And I'm going to put one loop around there.
I'm now powering this solenoid. There's an iron core in there.
Sixty hertz, hundred and ten volts.
And there we go, one loop around it.
I measure oh point four eight volts.
That's not bad. Now I put two,
three, four around it. What do I see now?
I see one point nine nine six. You may not see the six.
Very close to two. What I will do now,
I will move it up a little, and when I move it up a little,
the secondary, the flux coupling is not ideal
anymore, and so you expect that V2 will go down.
Just move it up a little along the primary, and you see the
value goes down. Because now the flux coupling
is not ideal. And if I bring it further out,
the flux coupling becomes hopeless.
And when I bring it here, the flux coupling is so poor,
but I still get oh point three volts, by the way.
So that's the first part, to show you that this is indeed
quite accurate. Now I want to do a second part,
and that is more interesting. I'm going to try to convince
you that I can get a pr- secondary current which is huge,
maybe even one thousand amperes.
But now I have to take special precautions.
I have to do the experiment in a very different way.
And the way I'm going to do this experiment now is with a
secondary which is specially designed to
have an enormously low value of resistance, because I want to
approach this situation so that this approximately holds.
We have a very thick copper secondary, almost half an inch
thick. It's like so.
This is copper. And I put here a nail,
an iron nail. And the resistance of this iron
nail is about four times ten to the minus four ohms.
The self-inductance of this ring, purely geometry,
has nothing to do with the kind of material I have,
is easy to estimate. I made a calculation and I
found about five times ten to the
minus seven henry. So omega times L2 -- I have to
multiply this by three hundred and seventy-seven -- becomes
about two times ten to the minus four ohms.
Even though R2 is not much, much smaller than omega L2,
they are now beginning to be comparable.
My I1, the current that I drive in the primary,
is going to be twenty amperes.
I do not expect that I2 will be two hundred and twenty times
larger than I1, because my -- I use the same
two hundred and twenty windings for the coil,
and this is just one winding. So I have the same ratio N1
over N2 that I had before in that experiment,
when I did N2 equals one. I don't expect that this is
going to be two hundred and twenty times
twenty amperes, which would be about four
thousand four hundred amperes, because I am really not in the
domain that R is much much smaller than omega L.
But I probably get five hundred to a thousand amperes.
And if my current I2 becomes five hundred to a thousand
amperes, you can calculate what I square R2 is,
because you know R2. That's all in that nail.
And that comes out to be about hundred to four hundred watts.
You better believe that that that nail is going to be
red-hot, is going to glow, and probably melt.
And this is the idea behind so-called induction ovens.
This is purposely done to get a very high temperature.
It's also done in welding to get very high currents.
And so this second part, the demonstration with the same
primary, I want to make it dark so that you can see the glowing
of that nail. And I will also show you,
of course, first the -- so here is the coil, and here is my
secondary. This is that very thick copper.
And then you see the nail here. Make sure there is no current.
My God, I almost shoved it over there.
[laughter]. Thank goodness that I realized
that there was current. So the current is now not going
through the solenoid, and so here is the,
the, the secondary now. Put some pieces of wood under
there so that you can see it better.
And so now if I power the primary, I can't tell you
exactly what the current is through the secondary,
but it's probably somewhere in the vicinity between five
hundred and one thousand amperes.
It will glow. It may take one minute to melt.
It may not melt at all. It may take three seconds.
Unpredictable, because we don't know exactly
the resistance of that nail. So I will tell you when I do
it. I will count down.
Three, two, one, zero.
It's already gone. That was fast.
So you saw it glowing, and you can actually see that
the nail melted. So you've seen an example now
where the current is immensely high.
So now I want to discuss with you another practical
application, which has immediate consequences,
or applications I should say, in your cars.
Spark plugs. Cars have coils,
and coils are run by a car battery, which is DC,
it's not AC. But you got to get high voltage
to get a spark going in spark
plug. And that's done in a very
clever way. So here is my twelve volt
battery. There is always some
resistance, of course, in, in the circuit.
If you have a coil, you always have some final --
finite resistance. And then here is a coil,
and here is a switch. And this has N1 windings,
has a certain self-inductance, the same situation that we had
before. And then I have here a
secondary, where I have N2 windings, which is way,
way larger than N1. I close the switch.
A current is going to build up. And the time constant is L over
R seconds -- we discussed that before,
it's going to build up. And now I'm going to do
something extremely cruel. I'm going to open this circuit,
open the switch. Imagine what is -- what I am
doing. I'm cutting instantaneously the
current. The current is going happily,
reaches a maximum value after several L over R seconds,
and now I *wssht* open it up. So I get a huge value for DI1
DT. I1 is now the current
in the primary. The current,
of course, will die down on a time scale L over R which is the
new value for R. And the new value for R is the
fact that I open it up. I make the resistance
infinitely large here. And if the resistance is
infinitely large here, my L over R time goes to zero.
So that makes you see indeed why the DI1 DT becomes enormous.
With a twelve volt battery, you can easily get several
hundred volts EMF now, because the DI1 DT will of
course create an EMF, an induced EMF in that circuit,
because you get a D phi DT. You get an enormous change in
the magnetic flux here, which is directly coupled to
the current. And that induced EMF could be
several hundred volts. But now look at the secondary.
The secondary has an N2 which is way larger than N1.
So the induced EMF in the secondary is very roughly N2
over N1 times the induced EMF in the primary.
That was the ratio that we had earlier, the V2 over V1,
remember was N2 over N1. So now I get in the secondary
an absolutely horrendous potential difference.
I could get up to a million volts if I wanted.
In your cars, that is not necessary.
It is enough in your cars that you get here something like ten
kilovolts. And that will comfortably give
you a spark over your spark plug.
And so what is happening with your car without you realizing
it, there is a circuit that closes and opens and closes and
opens and closes and opens, and it's only when you open it
that you get this pathetic high voltage
here and that the spark flies over.
And if you run your engine at three thousand RPM and you have
a four cylinder, that happens two hundred times
per second. But it is the breaking that
does it. And I can demonstrate this to
you. We have a very special spark
plug. Maybe it should not be called
spark plug. It's a beautiful device.
You'll see it all the way over there.
And we have there a situation whereby we don't even know what
N2 over N1 is. This is ancient work of art,
probably built in the nineteenth century or maybe
early twenty century. We think that N2 over N1 is at
least several thousand, but let's say it is ten to the
third. But it could be way higher than
that, we don't know. And we run it with a car
battery, just like your car,
twelve volts. We let the current build up in
the primary, just like we have here, and then in a very cruel
way we open the primary and we create, in the secondary,
up to three, four, five hundred thousand
volts. And then what you will see,
you will see a spark fly over here, at three hundred,
four, five hundred thousand volts.
You can draw a spark over a distance of ten centimeters
easily, because all you have to do is get
above three million volts per meter, which is the breakdown
electric field. And this is what I want to show
you now. And for that I also have to
make it a little dark, because if you see,
want to see sparks of course, you want to keep it,
make it a little dark. So this is the way we are going
to do it. Maybe you want to take a look
at that beautiful coil first. It's there.
So this is the, uh, the whole setup.
Here you see the car battery there.
It's just twelve volts. And here you see the,
the secondary. This is the open end of the
secondary, those two little balls that I have there.
And all of this is enclosed. We cannot measure very much.
And so now I will change the lights for you.
OK, I'm now running a current through the primary.
And that current is about -- oh, we don't even know what the
current is. I can't tell you.
All I can tell you is now I will open up the current in the
primary, so -- boy, I didn't mean to kill myself,
but I did in the process. OK, I open up the switch,
and every time that I open it you see there this enormous
voltage at the secondary. This is the way that your car
coil works, and that you get sparks
in your spark plug. That is an enormous potential
difference there. We can also set this in such a
way that it opens and closes without my having to do it
manually. And then you see,
you get an idea of how your spark
plug works. But that's not the reason why I
showed it to you. This is such a wonderful piece
of engineering. And a lot of research was done
in the early part of the twenty century with instruments like
this. It was named after the person
who invented this, which is Ruhmkorff,
and when I was a student we always referred to this
instrument as a Ruhmkorff. Well, I'll let it go so when
you walk out you can come close. But be very careful.
You're dealing there with several
hundred thousand volts. OK, have a good weekend.