We have often talked
about power supplies, which are devices which maintain
a constant potential difference. Here, we have such a power supply,
potential difference V, this be the plus side
and this be the minus side. I'm going to connect this,
I have a resistor here, R and as a result of this, a current
will start to flow in this direction, this direction,
this direction, so in the power supply,
the current flows in this direction. Through the resistance,
the current flows in this direction. In what direction is the electric field? The electric field always runs
from plus to minus potential. So right here, in this resistor,
the electric field is in this direction, from plus to minus. But inside the supply it must also go
from plus to minus. And so inside the supply,
the electric field is in the direction that opposes the current. So some kind of a pump mechanism
must force the current to go inside the supply,
against the electric field. A boulder does not, all by itself,
move up a hill. And so something is needed
to push it. You remember, with the VandeGraaff,
we were spraying charge onto a belt and then we rotated the belt and
the belt forces the charge into the dome. It had to overcome the repelling force
of the dome. So work had to be done. So the energy must come
from somewhere. And in the case
of the VandeGraaff, it was clearly the motor
that kept the belt running. In the case of the Wimshurst it was I
who turned the crank, so I did work. In the case of common batteries,
the ones that you buy in the store, it is chemical energy
that provides the energy. And I will discuss now with you and demonstrate
a particular kind of chemical energy, which is one whereby we have a zinc
and a copper plate in a solution. So we have here, H2SO4
and we have here, zinc plate and we have here
a copper plate. This side will become positive
and this side will become negative. You will get a potential difference between
these two plates. To understand that really
takes quantum mechanics, this goes beyond
this course. But the potential difference that you get
is normally something around one volt. The secret, really, is not necessarily
in the solution, because if you take
two conductors, two different conductors
and you touch them, metal on metal, there will also be
a potential difference. So let's look at his now
in some more detail. We have here a porous barrier that the ions
can flow freely from one side to the other. I reconnect them here, with a resistor
and so a current is now flowing. A current is flowing in this direction,
through the resistor, from the plus side of the battery,
to the minus side, that means inside the battery,
the current is flowing like this and the electric field, here,
is in this direction, from plus to minus, but also inside the battery, the electric field
must be from plus to minus, so you see again,
as we saw here, that the electric field
is in the opposite direction of the current. You will have here SO4 minus ions,
and you have copper plus ions in this solution and here you have zinc plus
and you have SO4 minus. And as current starts to run,
SO4 minus ions, which are now the current-carrier
inside this battery, is going from the right-- they're
going from the right to the left. Now why would SO4 minus ions travel through
an electric field that opposes them? That opposes their motion? And they do that
because in doing so, they engage in
a chemical reaction which yields more energy
than it costs to climb the electric hill. And while a current
is flowing, while the SO4 minus is going
from the right to the left you get fewer SO4 minus ions here,
this liquid here remains neutral, so copper plus must disappear. And it precipitates onto
this copper bar. So it is like copper-plating. On this side, you get an increase
of SO4 minus, therefore you also must get
an increase of zinc plus, because again, this liquid there
remains neutral and that means that some of the zinc
is being dissolved, so you get an increase
in the concentration of the zinc. So the charge carriers inside this battery,
the SO4 minus ions, travel through this barrier and they go
from here to here, so they travel through an electric field
that opposes their motion. And this happens at the expense
of chemical energy. Now, when the copper solution
becomes very dilute, because all the copper
has been plated onto the copper, and when this becomes concentrated
zinc plus, then the battery stops and now what you can do, you can
run a current in the opposite direction, so you can run a current,
now, in this direction, you can force a current to run
with another external power supply and now the chemical reactions
will reverse, so now, copper will go back
into the solution, so it will dissolve and now the zinc will
be precipitated onto the zinc and so now, if you do this long enough,
you can run the battery again the way it is here. A car battery is exactly
this kind of battery, except that you have lead and lead oxide
instead of zinc and copper, but you also have sulfuric acid,
like you have here and a nickel-cadmium battery is well-known,
you can charge that, too, those are the ones that are
readily available in the stores, you can run your flashlights
with these nickel-cadmium batteries. The symbol for battery that we will be using
in our circuits is this, this is the positive side
and this is the negative side, this is a symbol that symbolizes
that we are dealing with a-- with a battery. So let this point be B and let this point be A
and here, we have a resistor R. So we have a current going, the current is
going in this direction, a current I. This could be a light bulb,
could be your laptop, could be a hair dryer, whatever,
that you supply. If this R is not there, that means that the
resistance is infinitely large, that means that the current
that is running is zero, then the voltage that we would measure
over this battery, which is V B minus V A-- for which
I will simply write down, V of the battery-- that voltage we call a curled E,
which stands for EMF, which is electromotive force. I will show you that later. If I put a resistance R in here,
which is not infinitely large, then a current will start to flow, but now,
we should never forget, that between the points A and B, invisible to the human eye, there is always an internal
resistance which I call little r of i, and so if a current starts to flow,
it goes, not only through capital R, but it will also go through this little r,
and so, according to Ohm's Law, the EMF is now I times the external
resistance plus the internal one. The voltage that you would measure between
point B and A is now going to change. That voltage, according to Ohm's Law,
is I R and so it's also the EMF
minus I times r of i. And you see it's a little lower
than the EMF. And the reason is
this internal resistance here. If I shorted out this battery, stupid thing
to do, but if I make R equal zero-- so I take the battery
and I just short it out-- then, the maximum current that I can
draw then-- so R is now zero, so you can see that the maximum I
that you can get is E divided by r of i-- and V of B, the voltage that
you would measure now, between point B and A
goes to zero. It doesn't mean that there is
no current running, but it means that
between these points, your potential difference
goes down to zero. Shorting out a battery, of course,
is not a very smart thing to do. You can put batteries in series and thereby getting
a higher potential difference-- this is the negative side
and this is the positive, I have an independent one
negative positive, and an independent one,
negative, positive, each one, with an EMF E-- and I can connect the positive side of one
with the negative side of the other, just a conducting wire, and the positive side of this
with the negative side of the other, and now the potential difference
between these two points-- is now three E, open circuit,
if I don't draw any current. If I draw a current, then, of course, I have
to deal with the internal resistance. I'm going to build with you a copper-zinc
battery of the kind that we just discussed. You see it here. Here's the cupper-- copper sulfate solution,
with H2SO4 and-- here are my plates,
this is my zinc plate and this is my copper plate and you are going to see
the voltage displayed, I think, over there,
that is correct-- there's no potential difference now,
because they're not in place yet and so here comes my-- my zinc
and here comes my copper and they go into the solution
and you see about one volt. In general these potential differences
are of that order of one volt. Oh point nine five. So now what I will do,
I'm going to create a double one, so I have two independent-- batteries, I have here one whereby I have
copper and zinc-- and I have another one
whereby I have copper and zinc, and I'm going to connect this one
and you will see now that the EMF will double. If we're ready for that-- this is my second one,
it's going to be completely independent, so here comes the other two plates, make sure that I have the copper
and the zinc not confused-- there we go-- and now you should see twice
the potential. And you do see that. It's open circuit,
there is no current running. Well, there is a minute little small current
running through the volt meter that you see. But that's so small that that's--
can always be ignored. And you see you get double the EMF. Now what I will do-- so I have, now,
about two volts between these two plates, two batteries in series, I now have
a little light bulb here and I'm going to
turn on the light bulb. And now what you will see is that the voltage
that you measure right here-- that's all you can do,
you can only measure the voltage-- at the plates of the battery-- that now this voltage will drop, because of the internal
resistance of the battery, in addition you will see
some light, but that's really not
my objective. For those of you
who are sitting close, you can see this light bulb
going to be lit. So I do this now, I can see the light bulb,
a little bit of light and notice that the voltage
goes down. And so this value that you measure now,
V of B, is now lower than the one point nine volts
because of this term. You lose inside the battery
through the internal resistance, you lose there
potential difference. All right? So let's take this out, because
this produces a lot of-- a lot of smelly fumes. OK. That's fine. If a charge moves from point A to point B,
and here the potential is V A and here the potential is V B
and a charge d q moves-- and let's suppose, for simplicity,
that V B minus V A-- yes, let's make V A larger than V B,
that's just a little bit easier to think in that--
in those terms. It's not necessary,
of course. So let's make V A
larger than V B. So the electric field is from A to B. And I move charge from A to B,
then the electric field is doing work. And the work that the electric field is doing,
d W, is the charge times
the potential difference, which is V A minus V B. This work can be positive,
if the charge is positive, it can be negative
if the charge is negative because we have assumed that
this is positive, in this case. I can now do something that you shouldn't
tell your math teachers, but physicists do it all the time,
we divide by dt and now we say, "Aha!
What we have on the left side is now work, per unit time.
That's power. Joules per second." So this, now, is power. d q d t is current,
how many Coulombs per second flow. So this is current, I. And the potential difference,
I simply call-- I use that symbol, V. So you see now here that the power
delivered by a power supply-- is the current that it produces
times the potential difference. And this is independent of Ohm's Law,
this always holds. If you also include Ohm's Law,
if you can use it-- last time, we discussed
the limitations of Ohm's Law-- but if you can use it,
and V equals I R, then of course, you can also write down for the power,
that it is I squared R, and it is also V squared
divided by R. Power is joules per second, but we write,
for that, joules per second, we write for that, watts, just a capital W,
but it's named after the physicist Watt. So we always express the power
in terms of watts. So suppose we have a resistance R
and we run a current through it-- this is the resistance--
and we run a current I through it and let us take an example,
that the current I is one ampere and that the resistance
is one hundred ohm. Then the power which is dissipated in this
resistor has to be provided by your battery, that power P is now
one hundred watts. I square R,
if you want to use this. If it is two amperes-- and you don't change
the resistance, then it becomes
four hundred watts. Because it's I square R. I doubles, the power
and four times higher. Now, this energy is dissipated
in the form of heat and if it gets hot enough,
then maybe you can produce light, that's the idea behind
a light bulb. The filament into-- in a tungsten
incandescent light bulb becomes very high-- twenty five hundred degrees Centigrade,
maybe even three thousand-- not so high, of course,
that the tungsten melts-- and so, you begin
to see light. So, for instance, a hundred watt light bulb--
oh, work on here-- so if we have a hundred watt light bulb
in your dormitory-- and the voltage is hundred ten volts,
you just plug it into the wall, then the current that will run
is about oh point nine amperes. P equals V I. This product must be a hundred. And then the resistance that you have
is about a hundred and twenty ohms. V equals I R. So even though it's quite hot, a light bulb,
the amount of light that it produces is, in general, not more than twenty percent
of this power. It's not a very efficient thing,
an incandescent light bulb. A fluorescent tube is much better. So if you have a forty watt
fluorescent tube, you get about four times
more light then you get out of a forty watt
incandescent bulb. If we take your heaters
that you have in your dormitories-- typically two kilowatts,
but let's make it twenty two hundred watts, because that divides nicely
through a hundred and ten volts-- so then you would have twenty
amperes-- that's a lot of amperes. If the dormitory is very old,
chances are that your fuses will go. Twenty amperes is more than
many houses can handle. But nowadays, I think most outlets are good
for twenty five amperes or so. But not for much more. And so, now you have a resistor in your heater
which is about five point five ohms. Just to give you a feeling
for some numbers. Now, you want heat
out of your heater, and you want light
out of your light bulb, so you want to keep the temperature
of your heater modest, not so high that you get
a lot of light. If you make it two thousand
or twenty five hundred degrees, then you would get a lot of light
out of your heater. And so suppose that half of that power
would come out in terms of light and you turn on
your heater it night, would be like having a thousand-watt
light bulb in your dormitory, you don't want that. So how do you do--
what do you do now? Well, you simply keep the temperature about,
maybe, thousand degrees Centigrade, it gets a little red-hot,
very little light is produced and how do you keep the temperature low?
Well, you could cool it with air, some of these heaters
have fans that cool them. Or you just make the resistance,
you do both, very large, huge surface area of the resistance,
not small, but large and so now they have a large surface area
so they can radiate their heat and so the temperature
remains low. So if you look at your-- things that you
have at home, you have light bulbs, forty to two hundred watts,
your toaster, maybe three hundred watts. Your cooking plates and your heaters,
something like two kilowatts, TV, a few hundred watts, your electric toothbrush probably
only four watts, very modest. Your own body produces about hundred watts
heat-- of course that's energy. You have a very large
surface area, so you don't get nearly as hot
as a hundred-watt light bulb, because your surface area is large, so you only have a modest
ninety-eight degrees Fahrenheit, unless you're running a fever. So you don't produce any light,
because you're not hot enough for that, so you produce infrared radiation
and that's very noticeable. You hold someone
in your arms, the good feeling is,
you feel the body heat. That's the infrared radiation. That radiates at about a hundred
joules per second. Hundred watts. An electric blanket
is only fifty watts. So a partner is about twice as effective
as an electric blanket. Maybe also more fun. [laughter] The power delivered
by a battery-- is the current that the battery delivers
times E, which is this EMF. And so when a current
is running, it is I squared times the sum
of the two resistances. The external one
plus the internal one. We can never bypass that. The heat that is produced in the external
one is I squared capital R, but the heat that is produced inside the battery
is i squared little r, you can't avoid that. And so if you make R zero,
by shorting out the battery, then you get a current which is the maximum
current that you can get, which is the EMF
divided by r of i-- so you've killed the capital R,
it's zero now-- and so-- so you get a power
which is the maximum power which is now E squared
divided by r of i. I maximum squared
times r of i. It's the same thing, it is the maximum
current squared times r of i, and all of that comes
out inside the battery. Nothing comes out outside it. If you have a nine-volt Duracell battery,
the ones that we're all so familiar with, so then E is about nine volts,
the EMF, the internal resistance of such a battery
is about two ohms, and so the maximum current that
you can ever get out of a Duracell battery would be about four and a half amperes,
that is I Max, would be about four and a half
amperes and so P Max would be
about forty watts. So if you take a nine-volt battery
and you short it out, then the battery
should get warm, because all that heat, all that forty
watts is generated inside your battery. The value for V of B that you measured
would go down to zero, if you really could short it out,
with a resistor which has zero resistance. Now it's, of course, a pretty stupid thing
to do, to short out a battery, but it's not dangerous. Forty watts, the thing gets
a little warm, big deal. So let's do it. So I have, here, the voltage,
that you can see, that we measure,
a nine-volt Duracell battery, I have the battery here. And you can read it here. I hope the decimal point is in there,
but it's about nine point six volts. And now I am going to do something stupid,
but, again, it's not dangerous-- I'm going to take my car keys
and i'm going to short out the battery. So simply connect point A
with point B and so the voltage that you're going
to see is going-- maybe not go to zero, exactly, because
my key may not have zero resistance, but it goes very low-- and what you cannot experience
is something that I can, that this battery
will get hot. These forty watts will be generated
inside here. It is possible, though,
that when the battery gets hot, that the internal resistance
may even go up a little because, remember that resistance
goes up when temperature goes up, in which case, the power will go down,
so it may not be the full forty watts. But I can assure you that I can feel
this thing getting warm. So let me short it out, now. I'm doing this now. And you read the voltage,
I can see it, too here-- oh, it's always not so easy
with a key to do that-- here, it's very low,
hey, look at that. It's about a tenth of a volt
and I feel this thing getting hot. It's really warming up, now. And so I'm ruining this battery. This is a terrible thing to do,
batteries don't like that. But, when I take off the external resistance,
some of that may come back. It may not be permanently damaged,
and you see, it's already eight and a half volts. So there's no way that you can start a car
with a nine-volt Duracell battery, because you just can't get the current
that you need for your starter engine. Your starter motor needs
a few hundred amperes. If you take a car battery,
that's about twelve volts. It has a very low internal resistance,
of about one-fiftieth of an ohm. So that means that the maximum current that
you can draw, if you short-circuit it, would be something like
six hundred amperes. And so the maximum power,
if you were so stupid to short-circuit it, that would all be generated
inside the battery, would be something like
seven kilowatts. If you ever work on your car-- make sure that you never drop accidentally
the wrench that you're using onto the battery. Because if you did, then inside the battery,
about seven kilowatts, seven thousand joules per second,
are going to be produced in terms of heat and the sulfuric acid is going to boil,
the case may melt and that's no good. Not only is that stupid,
but it's also very dangerous. So let's do it. [laughter] I have here, this battery-- and I have here, the wrench. Just in case. I'm going to short out that battery
and as I do that, you will clearly see
that the battery doesn't like it. I will be very careful not to hold on
to this wrench too long, because it would weld onto it, actually,
it can weld on to it and stay there, the current is so high,
it can go up to six hundred amperes, that it can weld onto it
and then you can't get it off any more. In case that happens,
I will walk out of here. [laughter] And I advise you
to do the same. You ready? OK, I go now. You see? That's what happens. A very high current. And when you do this
too often to batteries, they're not going to live very long,
they don't like it. But I wasn't joking when I said,
when you work on the car, that you should avoid this,
because I have seen it happen, that wrenches actually welded
onto the terminals. Your electric company
charges you for energy, they don't care about the power,
how many joules you use second, but they care about
how much energy you're using. So they will charge you, then,
for joules, you think. That's energy. However, if you look at your bill,
you're being charged for kilowatt-hours. Well, a kilo is thousand
and an hour is thirty six hundred seconds, so the units of energy,
for which they charge you, is this in joules. Two thousand watts. Cooking plates, you run for two hours,
that is four kilowatt-hours. They will probably charge you ten cents per
kilowatt-hour-- for that same amount of money, you could run your hundred-watt
light bulb for forty hours. Again, that would be the same
four kilowatt hours-- or, you could brush your teeth
with your electric toothbrush for about one thousand
hours. Now I want to take a look with you at a network
which consists of resistors and batteries. And this is the kind of stuff
that you see on homework assignments, and, perhaps, on exams. And so now, we start out with a--
a very modest circuit, here we have a resistance R 1-- here we have a resistor R 2-- and here R 3-- and then we put
a battery in here and we put the plus side, say, on the left,
this the plus side, a minus side and let the potential difference
of this one be V 2. It's really the EMF, but I will ignore any
kind of internal resistance of the batteries, it's completely negligible
in this problem. And here I put also a battery,
let this be the negative side and this be the positive side and let the potential difference
be V 1. And so, imagine that you know
V 1, V 2, R 1, R 2, and R 3. But what I'm going to ask you is, "What
is I 1, what is I 2 and what is I 3?" I want the magnitude
and I want the direction. When you look at this,
it's by no means obvious that the current in this resistor
will be to the right or to the left, it's by no means obvious,
it depends on the-- on the values of V 1 and V 2
and on the resistances. The basic idea behind solving these problems
are in what we call Kirchoff's rules. Kirchoff's first rule is that
the closed loop integral over a closed loop of E dot d l
is zero. We've seen that before. I don't know why Kirchoff
gets the credit for this. This always is the case when we're dealing
with conservative fields. When you start at a particular point,
you go around E dot d l, you're back at the same potential
where you were before, so this must be zero, as long as you deal
with conservative fields. So that's his first rule. And you can do this
closed loop anywhere. You can even do it here. It would still be zero. You can do it here...
also zero. You can do it there. No matter where you do it,
that closed loop integral must be zero. And then there is
the second Kirchoff's rule, and that is what we call
charge conservation. If there is a steady-state situation, then,
independent of which junction you go to, the current that flows in
must flow out. Can't have a pile-up of charge. That's the second rule. And I gave you a problem, three seven,
to work out, and you can look in the book
how that is done. However, I'm going to work on this
with you in a slightly different way than the book is doing it, which I,
personally, like better. But it may confuse you. So I warn you in advance, you may not want
to use my method at all. What I do is the following. I say, OK, I assume that there is
a closed loop current here, I 1. And that there is a closed loop current here,
I 2. Whether I make them clockwise,
or counterclockwise, is unimportant. I could have chosen one clockwise, the other
counterclockwise, unimportant. However, once I choose a direction,
it has consequences, as you will see. And that's all that's running. One current like this
and one current independently like that. If I assume that, then I have automatically--
automatically, I am obeying the second rule, because a current that goes around--
there's charge conservation, right? There's no charge piling up. So the second rule of Kirchoff
is already obeyed. So now I go to the first one-- and I can start, now,
at any point in that circuit, and go around-- I can go around clockwise,
I can go around counterclockwise, it makes no difference
as long as I return to the same point, that integral E dot d l
must be zero. I'm returning at the same potential. What is the integral of E dot d l
in going from point one to point two? Well, that's the potential difference
between point one and point two. And so let us start here
and let us go around-- and we have to adopt
a certain convention, namely-- if we go up in potential
and we go down in potential. Again, you're free to choose
the sign convention. But I would say, when I go up in potential,
I give that a plus sign, when I go down in potential,
I give that a minus sign. So I start here. I could have started there, I could have
started there, makes no difference, as long as I don't start here,
that makes no sense. So I start here
and I go around like this. So right here,
I go down in potential, V 1. So I get minus V 1. Now I go with current I 1 in the direction,
from left to right, so that means that the potential here
must be higher than there. V equals I R. Potential here must be higher than there. So I go down in potential,
so I get minus I 1, R 1. Now I go through R 3. This current, I 1,
is going down, so this has a higher potential
than here, so I go down in potential,
so I get minus I 1 times R 3. But I have, independently,
a current I 2 which is now coming towards me
when I go down. And so if it comes
towards me, that current would give me
an increase in potential. This would have to have
a higher potential than this, for this current to do this. So now I climb up the potential hill,
so I get now plus I 2 times R 3 [recording slows down] Uh-oh, look what I did,
I wrote down I 1 R, there is no capital R
in the whole problem. I clearly meant I 1 R 1,
so read minus I 1 R 1, sorry for that. [recording speeds up to normal] I'm back where I was, because these wires
have no resistance. And so I'm back where I am,
so this is zero. One equation with two unknowns,
I 1 and I 2. So now,
let's go around this one. We can go clockwise,
we can go counterclockwise, makes no difference. Let's start here-- and I go in this direction,
once around. So now, I go through R 3 and this current I 2
is running in this direction, so I go down in potential. So I get minus I 2 times R 3. But current I 1 is coming
towards me. See, if I go in this direction,
I 1 is coming towards me, so I climb up
the potential hill. So I get plus
I 1 times R 3. Now I go through R 3 in this direction,
current I 2 is also in this direction and so this must have
a higher potential than this, so I go downhill
in potential, so I have minus I 2 times R 2. I come down here--
ah, here's a battery. And it goes up in potential. So I get plus V 2-- and that's zero. Two equations with two unknowns. I can solve for I 1
and I can solve for I 2. So I 1 and I 2 pop out. Let us assume that I 1 is positive,
I find a positive value. It means, it's really in this direction. Let's suppose that I 1 is negative. I find minus three amperes. Well, it means that I 1 is in this direction,
big deal. And so the whole operation
is sign sensitive. And the same is true here. If I two is positive,
it means it's in this direction. If I 2 is negative,
then it's in that direction. How about I 3 now? Well, let us assume
that I 1 is plus three amperes and that you find
that I 2 is plus one ampere. That's possible, right? You have two equations, two unknowns
and these are the answers. So three amperes goes like this...
[wssshhht] down and one ampere comes up. Well, it's clear then, that I 3
is three minus one, is plus two. Another way of looking at it is,
three amperes come in at this juncture, I 2 is one ampere, so one ampere goes through,
so two must go down. That's really Kirchoff's second rule. If I 1 were plus one ampere
and I 2 was also plus one ampere, then I 3 will be zero. No current would flow through I 3
but my method would still work. I find one ampere going down
and one ampere going up, so there's no-- no current
going through R 3-- there's only current going in
this direction, one ampere. And so, you have to recognize then,
that I 3 is I 1 minus I 2, which is really application, then,
of Kirchoff's second rule. I like this idea
of a closed loop current. I know some of you don't like it,
it's fine. The reason why I like it is, I always end up in this case
with two equations with two unknowns I solve for I 1, I solve for I 2-- and then the third one
comes out in a natural way by just thinking, "Ah, one current
goes in this direction and the other goes
in that direction." But if you prefer the method
that the book will present to you, you get three equations
with three unknowns-- and you get I 1, I 2 and I 3,
right at the start you get an I 3, you see I don't even start off
with an I 3 It's not there,
it comes in later. So the choice is yours. Now I want to entertain you for the last six
minutes with something amazing, something that is truly amazing. And it is a form of a battery
that is mind-boggling. And the battery is right here, on my--
my left, on your right. It is a battery that produces an enormous
potential difference, ten, twenty kilovolts-- you see a schematic here
on the transparency, you have a bucket of water,
here on the top and you have glass and the bucket
of water is hiding behind here-- it's not that because we hide it from you,
but that's the best place to be-- and you see plastic tubing coming down
and the water can run out on the right, and it can run out on the left. It runs out here, there is a, uh--
some paint can, no top and no bottom. And you see this paint can here,
it's completely open. There's a letter A. And there's another paint can on the right,
there's a letter B. It's a conducting can. And this is also a conducting can. And this water runs into another
conducting trash can and this water also runs
into a conducting trash can. And now comes the key point,
that this conductor here, A, is connected through
a conducting wire with C and the conductor B,
the paint can, is connected with a conducting wire
to this trash can D. You let the water run for a while--
and you will see, between these two points here,
sparks. Even when the points are as far apart as,
say, five millimeters, when you're talking about
at least a potential difference of something like ten, fifteen
thousand volts-- [spark] you will see the sparks. And you wait--
see another spark. And you wait--
and you see another spark. So this is a power supply. And there must be energy
coming from somewhere. And so, problem four one,
which you haven't seen yet, on your fourth assigment is asking you how this works. I will demonstrate it today
and I will come back to it later. The way it works is actually quite subtle,
but I want you to think about it. It's a remarkable battery,
a remarkable power supply. As the water starts running, I want to draw your attention
to the fact that you can almost anticipate
when the start-- when the spark occurs, because the water, at the very last,
is beginning to spread. It doesn't come out any more,
just like a narrow cylinder, but it begins to spread. And then comes the spark. And then it goes back
to running normally, and then slowly, in time,
it will spread, and then comes the spark. So let us get it going,
have some light here, Marcos and Bill spent a lot of time
getting this going-- Marcos, do we have--
are my lights the way you want them? You're happy with that. There, you see the two bowls,
which are really here-- and let's first look
at the sparks, so I will start the water
running now. Let's just be patient a little bit. And let's see where [when?]
we see in [a?] spark. Keep -- ah! Did you see one?
Did you see the spark? Oh, you were not looking. Man, I'm paying for this. Look at the, uh,
look at the two bowls. Give it some time again. Have to charge up. Oh, I can already anticipate,
it's coming up, it's coming up, dah! Did you see it? Ten, fifteen thousand volts. Let's give it a little bit more time and
then we'll take a look at the water flow, which I can see, I'm close
but we can make you see the water flow. Look again. Ah, it's coming up-- ah! Did you see it?
I could see it coming up. I can make you listen by having
my microphone near the water, and you can hear
this water running [sound of water running] familiar sound to all of us. And now, the sound changes,
you hear change? [spark] And there's the spark! Once more-- It's running-- spreading-- coming up! [spark] Yah! Amazing, isn't it?
I can make you see this water. Just stay there. We have one and a half minutes
left. So now you can see the water. You happy with the light, Marcos?
You can improve on it. So look at the water. Ah! It was just spreading already, you can't see the spark and the water
at the same time. See, the water's running, now,
normally It's going to spread slowly-- I will tell you when I see
the spark here, but it's already--
I can almost predict when it happens. The water is spreading now, coming up shortly--
yah! I saw the spark. And you immediately see
the water go like this. I want you to think about it
and explain this. This is one of the most remarkable things
I've ever seen in my life.
