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,
current will start to flow in
this direction,
this direction,
this direction,
so in the power supply,
the current flows in this
directions.
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,
H two S O 4,
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 S O 4 minus
ions,
and you have copper plus ions
in this solution,
and here you have zinc plus and
you have S O 4 minus.
And as current starts to run,
S O 4 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 S O 4 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 S O 4 minus is going
from the right to the left,
you get fewer S O 4 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 S O 4
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 S O 4 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 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 copper -- copper
sulfate solution,
which H 2 S O 4,
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
difference 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 s- 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 the 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 large 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 D T,
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 down 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 --
e 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, unintelligible 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 can much more light that
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.
N ow, 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.
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
square 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
unintelligible 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 six 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.
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.
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 thousands,
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
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 V 1 and V2,
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 rules,
because a current that goes
around is -- 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
that 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.
normal speed Uh-oh,
look what I did,
I wrote down capital R,
clearly I meant R 1,
there is no capital R in the
whole problem.
Sorry for that,
you should read this as minus I
1, R 1.
recording starts slow,
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.
[inaudible] if I go in this
direction, I 1 is coming towards
me, so I climb up the potential
hill.
So I got 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 that 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
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 th- there's only
current
going in this direction,
one ampere.
And we're recording slows.
normal speed 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 one R one.
Sorry for that.
starts slow,
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.
[inaudible] if I go in this
direction, I 1 is coming towards
me, so I climb up the potential
hill.
So I got 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 that 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
But my method would still work.
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 th- 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 that some of you don't
like it, that's fine.
The reason why I like it is,
I g- 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 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,
unintelligible 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 there 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,
[poit], 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 that must be energy coming
from somewhere.
And so, problem four one,
which you haven't seen yet,
only unintelligible assignment,
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 unintelligible
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.
I 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?
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, unintelligible sound to
all of us.
And now, the sound changes,
you hear change?
And there's the spark!
Once more.
It's running,
spreading, coming up!
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!
unintelligible 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.
