8.02x – Lect 9 – Electric Currents, Resistivity, Conductivity, Ohm’s Law

8.02x – Lect 9 – Electric Currents, Resistivity, Conductivity, Ohm’s Law

When positive charges move
in this direction, then, per definition we say the current goes
in this direction. When negative charges
go in this direction, we also say the current
goes in this direction, that’s just our convention. If I apply a potential difference
over a conductor, then I’m going to create an electric field
in that conductor. And the electrons, there are
free electrons in a conductor, they can move,
but the ions cannot move, because they are frozen into the solid,
into the crystal. And so when a current
flows in a conductor, it’s always the electrons
that are responsible for the current. The electrons feel
the electric field, and then the electrons try
to make the electric field zero, but they can’t succeed, because we keep the potential difference
over the conductor. Often, there is a linear relationship
between current and the potential, in which case,
we talk about Ohm’s Law. Now, I will try to derive Ohm’s Law
in a very crude way, a poor man’s version and not
really one hundred percent kosher, it requires quantum mechanics,
which is beyond the course– beyond this course– but I will do a job that still gives us
some interesting insight into Ohm’s Law. If I start off with a conductor,
for instance copper, at room temperature,
three hundred degrees Kelvin, the free electrons in copper
have a speed, an average speed of about
a million meters per second. So this is the average speed
of those free electrons, about a million
meters per second. This in all directions. It’s a chaotic motion. It’s a thermal motion,
it’s due to the temperature. The time between collisions,
time between the collisions, and this is a collision of the free electron
with the atoms– is approximately,
I call it tau, is about three times ten
to the minus fourteen seconds. No surprise because the speed
is enormously high. And the number of free electrons
in copper– per cubic meter– I call that number n– is about ten to the twenty-nine. There’s about one free electron
for every atom. So we get twen– ten to the twenty-nine
free electrons per cubic meter. So now imagine that I apply
a potential difference– piece of copper,
or any conductor for that matter, then the electrons will
experience a force which is the charge
of the electron, that’s my little e times the
electric field that I’m creating, because I apply
a potential difference. I realize that the force and the electric field
are in opposite directions for electrons, but that’s a detail, I’m interested in the
magnitudes only. And so now these electrons
will experience an acceleration, which is the force divided
by the mass of the electron and so they will pick up a speed,
between these collisions, which we call the drift velocity,
which is a times tau. It’s just 801. And so a equals F divided by m e. F is e E [misspoken A], so we get e times E
divided by the mass of the electrons, times tau. And that is the–
the drift velocity. When the electric field goes up,
the drift velocity goes up, so the electrons move faster in the direction opposite
to the current. If the time between collisions gets larger,
they– the acceleration lasts longer, so also, they pick up a larger speed,
so that’s intuitively pleasing. If we take a specific case and I take,
for instance copper and I apply over the– over a wire, let’s say the wire
has a length of 10 meters, I apply a potential difference
I call delta V, but I could have said just V,
I apply there a potential difference of ten volts, then the electric field,
inside the conductor now, is about one volt per meter. And so I can calculate now,
for that specific case, I can calculate what the drift velocity
would be. So the drift velocity of those free electrons
would be the charge of the electron, which is one point six times
ten to the minus nineteen coulombs. The E field is one,
so I can forget about that. Tau is three times ten to the minus fourteen,
as long as I’m room temperature and the mass of the electron
is about ten to the minus thirty kilograms. And so, if I didn’t slip up, I found that
this is five times ten to the minus three meters per second, which is
half a centimeter per second. So imagine,
due to the thermal motion, these free electrons move
with a million meters per second. But due to this electric field,
they only advance along the wire– slowly, like a snail, with a speed on average
of half a centimeter per second. And this goes very much against your
and my own intuition, but this is the way it is. I mean, a turtle would go faster
than these electrons. To go along a ten-meter wire
would take half hour. Something that you never thought of. That it would take a half hour for these electrons
to go along the wire if you apply potential difference of ten volts,
copper, ten meters long. Now, I want to massage
this further and see whether we can somehow
squeeze out Ohm’s Law, which is the linear relation
between the potential and the current. So let me start off with a wire– which has a cross-section A and it has a length l and I put a potential difference
over the wire, plus here and minus there,
potential V, so I would get a current in this direction,
that’s our definition of current, going from plus to minus. The electrons, of course, are moving in this
direction, with the drift velocity. And so the electric field in here,
which is in this direction, that electric field is approximately
V divided by l, potential difference
divided by distance. In one second, these free electrons will move
from left to right over a distance v d meters. So if I make any cross-section
through this wire, anywhere, I can calculate how many electrons pass
through that cross-section in one second. In one second, the volume that
passes through here, the volume is v d times A– but the number of free electrons
per cubic meter is called n, so this is now the number of free electrons that passes, per second,
through any cross-section. And each electron has a charge e
and so this is the current that will flow. The current, of course, is in this direction,
but that’s a detail. If I now substitute the drift velocity,
which we have here, I substitute that in there
but then I find that the current– I get a e squared,
the charge squared, I get n, I get tau, I get downstairs,
the mass of the electron and then I get A
times the electric field E. Because I have here,
this electric field E. When you look at this here, that really depends
only on the properties of my substance, for a given temperature. And we give that a name. We call this sigma,
which is called conductivity. Conductivity. If I calculate, for copper, the conductivity,
at room temperature, that’s very easy, because I’ve given you what n is, on the blackboard there,
ten to the twenty-nine, you know what tau is at room temperature,
three times ten to the minus fourteen, so for copper, at room temperature,
you will find about ten to the eighth. You will see more values for sigma
later on during this course. This is in SI units. I can massage this a little further,
because E is V divided by L, and so I can write now that the current
is that sigma times A times V divided by l. I can write it down a little bit differently,
I can say V, therefore, equals l divided by sigma A,
times I. And now, you’re staring at Ohm’s Law,
whether you like it or not, because this is what we call
the resistance, capital R. We often write down rho for one over sigma
and rho is called the resistivity. So either one will do. So you can also write down–
you can write down V equals I R and this R, then, is either
L divided by sigma A, or L times rho– let me make it
a nicer rho — divided by A. That’s the same thing. The units for resistance R is volts per ampere,
but we call that ohm. And so the unit for R is ohm. And so if you want to know
what the unit for rho and sigma is, that follows immediately
from the equations. The unit for rho
is then ohm-meters. So we have derived the resistance here
in terms of the dimensions– namely, the length
and the cross-section– but also in terms of the physics
on an atomic scale, which, all by itself,
is interesting. If you look at the resistance, you see it is proportional
with the length of your wire through which you drive a current. Think of this as water trying
to go through a pipe. If you make the pipe longer,
the resistance goes up, so that’s very intuitively pleasing. Notice that you have A downstairs. That means if the pipe is wider,
larger cross-section, it’s also easier
for the current to flow, it’s easier for the water
to flow. So that’s also
quite pleasing. Ohm’s Law, also, often holds for insulators,
which are not conductors, even though I have derived it here
for conductors, which have
these free electrons. And so now, I want to make a comparison
between very good conductors and very good insulators. So I’ll start off with a–
a chunk of material, cross-sectional area A– let’s take it one millimeter
by one millimeter– so A is ten to the minus six
square meters. So here I have a chunk
of material– and the length of that material, l,
is one meter. Put a potential difference over there,
plus here and minus here. Current will start to flow in this direction,
electrons will flow in this direction. The question now is, what is the resistance
of this chunk of material? Well, very easy. You take these equations, you know l and A,
so if I tell you what sigma is, then you can immediately calculate
what the resistance is. So let’s take, first,
a good conductor. Silver and gold and copper
are very good conductors. They would have values for sigma,
ten to the eight, we just calculated for copper,
you’ve seen in front of your own eyes. So that means rho would be
ten to the minus eight, it’s one over sigma. And so in this particular case,
since A is ten to the minus six, the resistance R is simply
ten to the sixth times rho. Because l is one meter. So it’s very easy– resistance here, R,
is ten to the minus two ohms. One-hundredth of an ohm. For this material
if it were copper. Let’s now take
a very good insulator. Glass is an example. Quartz, porcelain,
very good insulators. Now, sigma, the conductivity,
is extremely low. They vary somewhere from ten to the minus
twelve to ten to the minus sixteen. So rho, now, the resistivity, is something like
ten to the twelve to ten to the plus sixteens and if I take ten to the fourteen,
just I gra– have to grab– a number– then you’ll find that R, now,
is ten to the twenty ohms. A one with twenty zeros. That’s an enormous resistance. So you see the difference–
twenty-two orders of magnitude difference between a good conductor
and a good insulator. And if I make this potential difference
over the wire, if I make that one volt, and if I apply Ohm’s Law, V equals I R,
then I can also calculate the current that is going to flow. If I R is one, then the current here
is hundred amperes and the current here is
ten to the minus twenty amperes, an insignificant current,
of ten to the minus twenty amperes. I first want to demonstrate to you
that Ohm’s Law sometimes holds, I will do a demonstration, whereby you have a voltage supply–
I put a V in here– and we change the voltage
in a matter of a few seconds from zero to four volts. This is the plus side,
this is the minus side, I have connected it here
to a resistor which is fifty ohms– we use this symbol for a resistor–
and here is a current meter. And the current meter
has negligible resistance, so you can ignore that. And I’m going to show you
on an oscilloscope, we’ve never discussed
an oscilloscope, but maybe we will
in the future. I’m going to show you,
they are projected, the voltage, which goes from zero to four,
versus the current. And so it will start here
and by the time we reach four volts then we would have reached
a current of four divided by fifty, according to Ohm’s Law,
I will write down just four divided by fifty amperes,
which is 0.08 amperes. And if Ohm’s Law holds, then you
would find a straight line. That’s the whole idea about Ohm’s Law,
that the potential difference, linearly proportional to the current. You double the potential difference,
your current doubles. So let’s do that, let’s take a look at that,
you’re going to see that there– and I have to change my lights
so that you get a good shot at it– oh, it’s already going. So you see, horizontally, we have the current
and vertically, we have the voltage. And so it takes about a second
to go from zero to four– so this goes from zero
to four volts– and you’ll see that the current
is beautifully linear. Yes, I’m blocking it– oh, no, it’s my
reflection, that’s interesting. Ohm’s Law doesn’t allow for that. So you see how beautifully
linear it is. So now, you may have great confidence
in Ohm’s Law. Don’t have any confidence
in Ohm’s Law. The conductivity sigma is a strong
function of the temperature. If you increase the temperature, then the
time tau between collisions goes down, because the speed of these
free electrons goes up. It’s a very strong function of temperature. And so if tau goes down, then clearly,
what will happen is that the– conductivity will go down. And that means rho will go up. And so you get more resistance. And so when you heat up a substance,
the resistance goes up. A higher temperature,
higher resistance. So the moment that the resistance R
becomes a function of the temperature, I call that a total breakdown of V equals I R,
a total breakdown of Ohm’s Law. If you look in your book, they say, “Oh, no,
no, no, that’s not a breakdown, you just have to adjust the resistance
for a different temperature.” Well, yes, that’s an incredible poor man’s way
of saving a law that is a very bad law. Because the temperature itself
is a function of current, the higher the current
the higher the temperature. And so now, you get a ratio,
V divided by I, which is no longer constant. It becomes a function
of the current. That’s the end of
Ohm’s Law. And so I want to show you that if I do the
same experiment that I did here, but if I replace this by a light bulb of fifty ohms
— it’s a very small light bulb, resistance when it is hot is fifty ohms,
when it is cold, it is seven ohms. So R cold of the light bulb is roughly
seven ohms, I believe, but I know that when it is hot,
it’s very close to the fifty ohms. Think it’s a little lower. What do you expect now? Well, you expect now, that when
the resistance is low in the beginning, you get this and then when the resistance
goes up, you’re going to get this. I may end up a little
higher current, because I think the resistance
is a little lower than fifty ohms. And if you see a curve like this,
that’s not linear anymore. So that’s the end of Ohm’s Law. And that’s what I want
to show you now. So, all I do is, here I have
this little light bulb– for those of you
who sit close, they can actually see that light bulb–
start glowing, but that’s not important,
I really want you to see that V versus I is no longer linear,
there you go. And you see, every time you see this
light bulb go on, it heats up, and during the heating up, it, um,
the resistance increases. And it’s the end of Ohm’s Law,
for this light bulb, at least. It was fine for the other resistor,
but it was not fine for this light bulb. There is another way
that I can show you that Ohm’s Law is not always
doing so well. I have a hundred twenty-five volt
power supply, so V is hundred and twenty-five
volts– this is the potential difference–
and I have a light bulb, you see it here, that’s the light bulb– the resistance of the
light bulb, cold, I believe, is twenty-five ohms and hot, is about
two hundred and fifty ohms. A huge difference. So if the– resistance, if I take the cold
resistance, then I would get five amperes, but by the time that the bulb is hot,
I would only get half an ampere. It’s a huge difference. And what I want to show you,
again with the oscilloscope, is the current
as a function of time. When you switch on a light bulb,
you would expect, if Ohm’s Law holds, that when you switch on the voltage–
that you see this. This is then your five amperes. And that it would stay there. That’s the whole idea. Namely, that the voltage divided by
the current remains a constant. However, what you’re going
to see is like this. The resistance goes up and then therefore
the current will go down. And will level off at a level which is
substantially below this. So you’re looking there– you’re staring
at the breakdown of Ohm’s Law. And so that’s what I want
to show you now. Ehm– so, here we need
a hundred and twenty five volts– and there is the light bulb, and when I throw this switch, you will see the pattern
of the current versus time– you will only see it once and then
we freeze it with the oscilloscope– turn this off–
so look closely, now. There it is. Forget these little ripple
that you see on it, it has to do with the way that we produce
the hundred and twenty five volts. And so you see here,
horizontally, time. The time between two adjacent
vertical lines is twenty milliseconds. And so, indeed, very early on, the current
surged towards– to a very high value and then the filament heats up and so
the resistance goes up, the light bulb, and the current just
goes back again. From the far left to the far right on the
screen is about two hundred milliseconds. That’s about two tenths of a second. And here you get a current level which is
way lower than what you get there. That’s a breakdown of Ohm’s Law. It is actually very nice
that resistances go up with light bulbs when
the temperature goes up. Because, suppose it were
the other way around. Suppose you turn on a light bulb
and the resistance would go down. Light bulb gets hot, resistance goes down,
that means the current goes up. Instead of down,
the current goes up. That means it gets hotter. That means the resistance
goes even further down. That means the current
goes even further up. And so what it would mean is that
every time you turn on a light bulb, it would, right in front of your eyes,
destruct itself. That’s not happening. It’s the other way around. So, in a way, it’s fortunate that the resistance
goes up when the light bulbs get hot. All right. Let’s now be a little bit more qualitative
on some networks of resistors and we’ll have you do
a few problems like that, whereby we just will assume,
naively, that Ohm’s Law holds. In other words,
we will always assume that the values for the resistances that
we give you will not change. So we will assume
that the heat that is produced will not play
any important role. So we will just use Ohm’s Law,
for now, and if you can’t use it,
we will be very specific about that. So suppose I have here,
between point A and point B, suppose I have two resistors,
R one and R two. And suppose I apply a potential difference
between A and B, that this be plus and this be minus
and the potential difference is V. And you know V, this is known,
I give you V, I give you this resistance
and I give you that one. So I could ask you now, “What
is the current that is going to flow?” I could also ask you, then, “What is the
potential difference over this resistor alone?” — which I will call V one and “What is the potential difference
over the second resistor?” which I call V two. Very straightforward question. Well, you apply, now,
Ohm’s Law and so between A and B,
there are two resistors, in series. So the current has to go
through both and so the potential difference V,
in Ohm’s Law, is now the total current
times R one plus R two. Suppose these two resistors
were the same, they had the same length,
same cross-sectional area. If you put two in series,
you have twice the length. Well, so, twice the length,
remember, resistance is linearly proportional
with the length of a wire and so you add them up. So now you know R one
and you know R two, you know V,
so you already know the current, very simple. You can also apply Ohm’s Law, as long as it
holds, for this resistor alone. So then you get that V one
equals I times R one, so now you have the voltage
over this resistor and of course, V two must be the
current I times R two. And so you have solved
your problem. All the questions that I asked you,
you have the answers to. We could now have a
slightly different problem, whereby point A is here,
but now we have a resistor here, which is R one and we have
here, R two. This is point B
and this is R two. And the potential difference is V,
that is, again, given and now I could ask you, “What,
now, is the current that will flow here?” And then I can also ask you, “What is
the current that would go through one– resistor one and what is the current
that could go through resistor two?” And I would allow you
to use Ohm’s Law. So now you say, “Aha! The potential difference
from A to B going this route, that potential difference,
is V, that’s a given.” So V must now
be I one times R one. That’s Ohm’s Law,
for this upper branch. But, of course, you can also go
the lower branch. So the same V is also I two
times R two. But whatever current comes in here
must split up between these two think of it as water. The number of charges that flow
into this juncture continue on and so I, the total current,
is I one plus I two. And so now, you see,
you have all the ingredients that you need to solve
for the current I, for the current I one
and for the current I two. And you can turn this
into an industry, you can make extremely complicated
networks of resistors– and if you were in course six,
you should love it– I don’t like it at all, so you don’t have to worry about it,
you’re not going to get very complicated resistor net–
networks from me– but in course six,
you’re going to see a lot of them. They’re going to throw them–
stuff them down your throat. The conductivity of a substance
goes up if I can increase
the number of charge carriers. If we have dry air
and it is cold, then the resistivity of cold, dry air
at one atmosphere– so rho for air, cold,
dry, one atmosphere– cold means temperature
that we have outside– it’s about four times
ten to the thirteen. That is the resistivity of air. It is about what it is in this room,
maybe a little lower, because the temperature
is a little higher. If I heat it up, the air, then
the conductivity will go up. Resistivity will go down, because now, I create oxygen and
nitrogen ions by heating up the air. Remember when we had this lightning,
the step leader came down and we created a channel
full of ions and electrons, that had a very low resistivity,
a very high conductivity. And so what I want to demonstrate to you,
that when I create ions in this room, that I can actually make the conductivity
of air go up tremendously. Not only will the electrons move,
but also the ions, now, will start to move. And the way I’m going to do that is,
I’m going to put charge on the electroscope– [cracking sound] oh, that is not so good–
no harm done. I’m going to put charge
on the electroscope and you will see that
the conductivity of air is so poor that it will stay there
for hours. And then what I will do, I will create ions
in the vicinity of the electroscope. But let’s first put some charge
on the electroscope. I have here a glass rod and I’ll put some
charge on it. OK, that’s a lot of charge. And, uh, the r- the air is quite dry,
conductivity is very, very small and so the charge cannot go off
through the air to the surroundings, to the earth. But now I’m going to create ions there
by heating it up, and I decided to do that with a candle,
because a candle is very romantic, as we all know. So here I have this candle–
look how well the charge is holding, eh? And here’s my candle. And I will bring the candle– oh, maybe twenty
centimeters from the electroscope. Look at it, look at it,
already going. It’s about fifteen centimeters
away. I’ll take my candle away
and it stops again. So it’s all due to the fact
that I’m ionizing the air there, creating free electrons
as well as ions, and they both participate now
in the current and the charge can flow away
from the electroscope through the earth, because the conductivity now
is so much higher. I stop again and it stops. You see in front of your eyes
how important the temperature is, in this case, the presence
of the ions in the air. If I have clean, distilled water,
I mean, clean water. I don’t mean the stuff
that you get in Cambridge, let alone did I mean the stuff
that is in the Charles River, I mean clean water,
that has a pH of seven. That means one out of ten to the seven
of the water molecules is ionized, H plus and O H minus. The conductivity, by the way,
is not the result of the free electrons, but is really the result of these H plus
and O H minus ions. It’s one of the cases whereby not the–
the electrons are a maj–
the major responsibility for the current. If I have add three percent of salt,
in terms of weight, then all that salt will ionize, so you get sodium plus
and C L minus ions, you increase the number of ions
by an enormous factor. And so the conductivity will soar up
by a factor of three hundred thousand, or up to a million, because you increase
the ions by that amount. And so it’s no surprise then, for you,
that the conductivity of seawater is a million times higher, think about it,
a million times higher– than the conductivity
of distilled water. And I would like to give you
the number for water, so this is distilled water, that is about two times
ten to the fifth ohm-meters. That is the resistivity, two times ten
to the five oh-meters. I have here a bucket
of distilled water. I’ll make a drawing for you
on the blackboard there. So here is a bucket of distilled water
and in there, is a copper plate and another copper plate,
and here is a light bulb, and this will go straight
to the outlet [wssshhht], stick it in,
hundred ten volts. This light bulb has eight hundred
ohm resistance when it is hot. You see the light bulb here. You can calculate what this resistance is
between the two plates, that’s easy,
you have all the tools now. If you know the distance,
it’s about twenty centimeters and you know the surface area
of the plates, because remember, the resistance
is inversely proportional with A, so you have to take that
into account– and you take the resistivity
of water into account, it’s a trivial calculation, you can calculate what
the resistance is of this portion here. And I found that this resistance here
is about two megaohms. Two million ohms. So, when I plug this into a wall,
the current that will flow is extremely low, because it has to go through
the eight hundred ohms, and through the
two megaohms. So you won’t see anything,
the light bulb will not show any light. But now, if I — put salt in here, if I really manage to put
three percent in weight salt in here, then this two megaohm will go down
to two ohms, a million times less. So now, the light bulb will be happy like
a clam at high tide, because two ohms here,
plus the eight hundred, the two is insignificant. And this is what I want to–
to demonstrate to you now, the enormous importance
of increasing ions. I increased ions here
by heating the air, now i’m going to increase the ions
by adding salt. And so the first thing that I will do is,
I will stick this in here. There’s the light bulb. And I make a daring prediction
that you will see nothing. There we go. Nothing. Isn’t that amazing? You didn’t expect that, right?
Physics works, you see nothing. If I take the plates out and touch them
with each other, what will happen? There you go. But this water has such a huge resistance
that the current is too low. Well, let’s add some– not pepper–
add some salt. Yes, there’s salt in there. It’s about as much as I would put
on my eggs in the morning– stir a little– ah, hey,
look at that. Isn’t that amazing? And when I bring them closer together,
it will become even brighter, because L is now smaller,
the distance is smaller. I bring them farther apart,
it’s amazing. Just a teeny, weeny little bit of salt,
about as much as I use on my egg, let alone– what the hell,
let’s put everything in there– that’s a [unintelligible]
I put everything, then, of course, you go almost down
to the two ohms, and the light bulb
will be just burning normally. But even with that little bit of salt,
you saw the huge difference. My body is a fairly good conductor–
yours too, we all came out of the sea– so we are almost all water– and therefore,
when we do experiments with little charge, like the VanderGraaf,
beating a student, then we have to insulate ourselves
very carefully, putting glass plates under us,
or plastic stools, to prevent that the charge
runs down to the earth. In fact, the resistance, my resistance
between my body and the earth is largely dictated by the soles of my shoe,
not by my body, not by my skin. But if you look at my soles,
then you get something like this and it has a certain thickness
and this, maybe one centimeter. This, now, is l in my calculation
for the resistance, because current may flow
in this direction, so that’s l. Well, how large is my foot? Let’s say it’s one foot, long–
no pun implied– and let’s say it’s about
ten centimeters wide. So you can calculate what the surface area
A is, you know what l is and if you know now what the resistivity is
for my sole, I can make a rough guess, I looked up the material and I found that
the resistivity is about ten to the tenth. So I can now calculate what the resistance
is in this direction. And I found that that resistance then,
putting in the numbers, is about ten billion ohm. And you will say, “Wow!”
Oh, it’s four, actually. Well, big deal. Four billion ohm. So you will say, “That’s an
enormous resistance!” Well, first of all,
I’m walking on two feet, not on one, so if I would be standing one
the whole lecture, it would probably
be four billion, but if I have two feet on the ground,
it’s really two billion. You will say, “Well,
that’s still extremely large!” Well, it may look large,
but it really isn’t, because all the experiments
that we are doing here in twenty six one hundred, you’re dealing with very small
amounts of charge. Even if you take the VanderGraaff– the VanderGraaff, say, has
two hundred thousand volts– and let’s assume that my resistance
is two times ten to the nine ohms, two feet on the ground. So when I touch the VanderGraaff,
the current that would flow, according to Ohm’s Law,
would be hundred microamperes. That means, in one second, I can take hundred
microcoulombs of the VanderGraaff, but the van der Graff has only
ten microCoulombs on it. So the resistance of four billion or two billion
ohms is way too low for these experiments that we have been doing
in twenty six one hundred, and that’s why we use
these plastic stools and we use these glass plates
in order to make sure that the current is not
draining off the– the charge that we need
for the experiments. I want to demonstrate that to you,
that, indeed, even with my shoes on– that means, even with my two billion ohm
resistance to the ground– that it will be very difficult for me,
for instance, to keep charge
on an electroscope. I’m going to put charge on this electroscope
by scuffing my feet. But, since I keep my– I have my shoes on,
I’m not standing on the glass plate, the charge will flow through me. You can apply Ohm’s Law. And you will see that as I do this–
I’m scuffing my feet now– that I can only keep that electroscope charged
as long as I keep scuffing. But the moment that I stop scuffing,
it’s gone. Start scuffing again, that’s fine,
but the moment that I stop scuffing, it goes off again. Even though this resistance is something like
two billion ohms. Let alone if I take my shoes off. I apologize for that. If now I scuff, I can’t even get any charge
on the electroscope, because now, the resistance
is so ridiculously low, I don’t even have
the two billion ohms, I can’t even put any charge
on the electroscope. It’s always very difficult for us
to do these experiments unless we insulate ourselves
very well. And if, somehow,
the weather is a little damp, then the conductivity of many materials
can increase That’s why we always like to do
these experiments in winter, so that the conductivity of the air
is very low, no water anywhere. Here you see a slide of a robbery. I have scuffed my feet across the rug,
and I am armed with a static charge. Hand over all your money,
or I’ll touch your nose. This person either
never took 802, or he is wearing very,
very, special shoes. See you on wednesday.

Posts created 29718

100 thoughts on “8.02x – Lect 9 – Electric Currents, Resistivity, Conductivity, Ohm’s Law

  1. Sir , I am in search of the exact reason for the fact that electrons move opposite to the current (direction)
    thanks !!

  2. Hi Sir

    when you are doing scuffing experiment , at that time you are wearing shoes , then your shoes are rubbed against woolen cloth so your shoes gets charged right , not your feet ? ( i.e because your feet is not exerting any friction with the cloth )

  3. 1 Consider a 6V power supply , ( i.e positive terminal of the battery is at +3V in potential and negative terminal is at -3V of the battery ). If we are connecting the battery to a resistor R ( i.e closed circuit with battery and resistor ) then the electrons ( One coulomb of negative charge will be having -3V potential ) will gain all the 6V potential in the resistor and the each One coulomb of charge will be at +3 V potential right Sir ?


    1. V = 6V , R = 1 OHM , I = 6 AMPS , n = 37.5 X 10^18 ELECTRONS

    2. V = 6V , R = 3 OHM , I = 2 AMPS , n = 18.75 X 10^18 ELECTRONS

    3. V = 6V , R = 6 OHM , I = 1 AMPS , n = 6.25 X 10^18 ELECTRONS

    As we are increasing the resistance in the circuit , then the current is decreasing , means number of electrons that are producing the current are decreasing and potential of each electron will increases right Sir ?

  4. when you are doing scuffing experiment , at that time you are wearing shoes , then your shoes are rubbed against woolen cloth so your shoes gets charged right , not your feet ? ( i.e because your feet is not exerting any friction with the cloth )

  5. Hello dear professor

    I don't know ,what i am going to explain you, discovered or not.

    if we increase the voltage , increase the current.. obvious, but what the deal with conducter at particular time. means, what is the chang in current with respect to conducter length??

    i have tried to find on Google but fail , maybe I didn't find with correct word. i tryed to solve it , i got some weird equation.
    I hope you understood my sentence ,if not please tell i will try in different ways.

    thank you

  6. Sir suppose there are two resistors A and B. sir we have connected these two resistors in series. sir why the current has to be same through both resistor

  7. sir suppose there is battery of emf 2volt. its internal resistances is 1ohm. and we have applies variable external resistance to the terminal of battery. if we change external resistance the current given by the battery is always constant or current changes according to the magnitude of current

  8. suppose i have a BATTERY AND RESISTOR. THE resistor has two ends A AND B. sir i have connected the end A of the –
    resistor to positive terminal of batter through cell and negative terminal of battery to end B of the resistor through conducting wire. sir between end A AND +VE terminal of the battery there is no resistance and no potential difference. hence accordig to ohms law i= v upon resistance. i =0 upon 0 = infinite.hence there will be infinite current in conducting wire between positive terminal and end A. but in resistor the resistance is R and potential difference is v hence according to ohms law there will be an finite current. is am i right sir.

  9. sir is it means current changes according to the magnitude of potential difference applied between ends of resistor and resistance of resistor. now is am I right

  10. Sir I calculate power of the bulb with three formula VI , V2R and through I2R. the power of the bulb through VI is not same through I2R. Is it calculation mistake only

  11. sir suppose I have connected positive and and -ve terminal of battery through bulb by super conducting wires. sir I applied ohms law between any two of point of conducting wires. sir potential difference between two points will be zero and resistance is also zero. hence I=0/0 =infinite. sir now I apply ohms law between two points of bulb. there is potential difference between these point and there is resistance .hence there will be current= V/R. sir is it means that current in conducting wire is different than current in bulb.

  12. sir larger is the drift velocity of electron larger is the current. sir no suppose I have joined 2ohm, 3ohm resistor in series with cell. when electron firstly passes through the 3ohm resistance they have some drift velocity suppose v. now when electron passes through 2ohm resistance they have to face less resistance as compare when they passes through 3ohm resistance. As now resistance faced by electron is less hence drift velocity of electron must be higher than v. hence current must be higher that's why I think that current must not be same in resistors joined . SIR please clear this doubts.

  13. Sir suppose there is bulb. When it is cold its resistance is low than when it is hot. As we know power of the bulb=i2R. Therefore power of bulb is higher when bulb is at room temperature and its power decreases when it becomes hot. It means power of the bulb decreases as its temperature increases. Sir it means that when we switch on the bulb its power must be high and after some time its power must be less. Sir than why not bulb show much glow when we switch on the bulb and why not its glow decreases after switch on the bulb because temperature is increases hence resistance must be increases but current decrease hence power must be decreases Sir.
    Thanks Sir

  14. sir when we switch on the bulb its resistance is less than after sometime. sir it means current is higher when we switch on the bulb. sir it means power decreasing after switch on the bulb because resistance increase s. sir than why bulb does not show much glow we we switch on the bulb

  15. sir I a simple question from my practical life. sir as the power of the bulb is decreaseing when we switch on our home bulb because current is decreasing. sir than why light by the bulb is not decreasing asw we see in our home bulb

  16. SIR a steady current is flowing in a metallic conductor of uniform cross-section where electric field will be more in larzer area or smaller area. sir please give answer with reason

  17. sir suppose there is metallic conductor which have non uniform cross-section area. current is flowing through it. is electric field changes or remain same throughout the conductor. sir please explain in easy language

  18. Respected sir, At 23:00 , you explained the ohms law when temperature changes with the help of graph!
    How could the curve is plotted below for higher temperature for same voltage? However if it has nothing to do with the straight line then I have no problem!

  19. Professor, I am having a tough time visualising the E field inside the wire. At 4:49 Why is the E field inside the whole conductor a constant (1 V/m). Since the length of the conductor is 10m, shouldn't the magnitude of the E field inside the wire at different distances (from the origin of the E field) be different? Where is the origin of this E field actually?

  20. What does scuffing your feet do exactly? Does it increase the resistivity of your feet, slowing the current, and allowing the charge to stay on the thing? If so, why does the resistivity increase? Is the temperature of your soles changing?

  21. Where to download ur lecture notes in PDF format, the links in the video represents assignment and solutions only. I'm really very big fan of ur teaching. Thank you sir…

  22. Hello sir! In the case of our day-to-day electric devices, where do the electrons come from and where do they go later?

  23. sir suppose I have a bullb which is attached with 220volt supply and current is 10 ampere. sir as we know that in bulb electrical energy is converted in light energy and heat energy. sir total output power can be find out by three formula V2/R and I2R or VI. sir how we can find out amount of heat energy .

  24. Wow……you sir are AMAZING. I mostly had no idea what you were talking and writing about but as a 17 yr employed automotive technician that uses an oscilloscope literally daily, u were directly describing ptc and ntc style thermistors. A negative temperature coefficient thermistor or variable resistor is the most commonly style engine temperature sensor used in automobiles. Temperature up, resistance down. The lab scope is also used for issues pertaining to variable relector, hall effect…whatever sensors controlled by a transistor driver ( fuel injectors, ignitions coils, output solenoids, etc…..) Again….have no clue what you are saying but it is addicting and badass. I could listen to you all day.

  25. Question : It is desired to supply a current of 2A through a Resistance of 10 ohm . As many as 20 cells are provided , each of 2V and internal Resistance 0.5 ohm . Two Friends David and John try their hand on the Problem . David succeeds but John fails . Answer the Following questions : a) Justify the set up of David.b) What might have gone wrong with John when gets 1.4A current in the External Circuit…Sir , Please tell What will be the Answer of part (a) and (b)

  26. How does the distilled water vs. salt water affect on corrosion? which one is better choice example in ferrite metal pipes to extend lifetime?

  27. So professor, at 4:45 we are assuming that the E field is same throughout the copper wire, right? Otherwise we couldn't be saying V=EL (where L is the total length of the wire and V is the voltage difference between the two ends of the wire), right?

  28. Sir, I have two questions,
    First, you said that resistivity increases with increase in temperature but when you heated up the air the electroscope showed more deviation which means resistivity of air decreased and both these statements contradict.
    Second, can you please explain the part that why the 2 billion resistance of the shoe is very low for the experiments you carry.

  29. Sir can you please explain the part that why the 2 billion resistance of the shoe is very low for the experiments you carry?


  31. Sir , why is earth potential zero ? is it zero even when potential at infinity is taken to be zero ? Many books say earth is a source of huge negative charge , then should not charge go through our body when we touch the earth ? and if earth is a source of neg charge why should neg charge still able to flow to earth ?

  32. Can you please tell me know what lecture in 8.01 covers drift velocity? I do not understand what tor (T) means.
    Thank you

  33. Thank you Professor Lewin for all the great lectures and resources you provide! You are a true inspiration!!

  34. Sir I have a doubt

    Inside a conductor, there are atoms..And in atoms electrons are always revolving (never stops)..So why current is not there always?

  35. Sir, since current is defined as the rate of transfer of charge from one side of the area to the other side ,so 'i' must depend on area also but the equation is just i=dq/dt ,how?

  36. Hello Sir , At 1:12 you say that the derivation of Ohm's law you show in the lecture is a very 'crude derivation' & we need to go into quantum mechanics for a more accurate one ? But why is the derivation not perfect as I think the math is completely fine . Is it because we don't assume the temperature dependence of Resistivity while deriving Ohm's Law by this method ?

  37. Temperature of a substance is due to the vibrational and kinetic energy now if the electrons are made to go this slow then how come the temperature don't drop.

  38. Temperature of a substance is due to the vibrational and kinetic energy now if the electrons are made to go this slow then how come the temperature don't drop.

  39. If the drift velocity is almost negligible compared to the velocity due to thermal motion, why does temperature increase with current? Thanks!

  40. Hello,
    I cannot understand why I would be constant throughout the circuit when considering the series of resistors.
    I mean, after going through the resistor, the current should increase or decrease, not stay the same. Right ?
    Is it because of charge density of the material, and that therefore if localy a change in speed takes place, it will spread to junctions ?

  41. Sir, why current is taken alway equal along resistor?
    Is electric field or drifting of electron is responsible for current? If current is energy isn't it used by resistors?

  42. Sir if resistance is proportional to temperature as we saw in bulb that as it goes hotter and hotter resistance increase and thus current decreases so how come current is proportional to temperature

  43. Sir as you said it few minutes after the starting of video that temperature is proportional to current isn't the bulb violating it

  44. Sir I have a doubt. We calculated drift velocity as Vd=eET/m .But while calculating current through unit area where the charge travelled a "distance Vd", we substituted the value of drift velocity in the place of distance. Can you pls clarify my doubt sir……

  45. Dear Dr. Lewin: I have greatly enjoyed your lectures. My father was a professor of electrical engineering at VJTI in Mumbai. I have a Ph.D. in Mechanical engineering. Today at the age of 70, I still enjoy clarifying my fundamentals. Your down to earth and experiment based teaching is lucid and profound.
    My recent calibrations: Light travels 30 cm or 1 ft in 1 nanosecond. So a 1 GHz computer cannot have its primary memory more than 6 inches away!!

    1 coulomb charges 1 m apart have a Force of 1 million metric tons!! That is 100m sided cube of water or 1 cubic football field of water!!

  46. Professor❤ I'm feeling stressful and Sad😞 , This thing make me feel Failure that I will at Home for 1 more Year – I feel like crying too. Even though I got Best rank in my City in JEE Advance – I'm at Home and Everyone is getting admission somewhere. Sometime I also had Bad thoughts:(. What can I do to Overcome this ?

  47. Hi Dr. Lewin. In terms of the drift velocity is it understood that this is concurrent with the electron motion ? They are still bouncing all over and hitting atoms while also meandering in the direction the voltage field is pushing them?

  48. Why do resistors in series have the same current and different potential difference and resistors in parallel have the same potential difference and different current?

  49. It doesn't matter where the resistor is placed in a circuit,it affects the total current passing in a circuit,i.e. connecting two resistors in series and connecting a resistor having th equivalent resistance of both the resistors has the same effect.But,how is this possible?The current can only get reduced after it flows through the circuit,right?

  50. Question about problem 3.1(b) – you say that since charge cannot flow into or out of the circled part of the circuit, that the charge must be symmetrically distributed. But doesn't that also mean that there is no current through that part of the circuit? If so, how does the current "resume" to the right of this portion of the circuit? There must be sparks going between the plates as they reach saturation, at which point there is a current going through the circled portion… No?

  51. I start getting why the MIT is so prestigeous. Professors like you are capable of perfectly explaining something to the students without confusing them. I am currently studying electrical engineering at the Technical University in Vienna and here are the students the ones, that have to understand stuff out of books instead of listening of such great lectures. i guess everyone has it a bit different and their way of learning may vary, but the finishing pount is the same for everyone. thank you for sharing your knowledge in your unique way.

  52. Hi Professor Walter, thanks for the fantastic lectures.

    I have a question: What would happen if you would touch de van de graaf while wearing your shoes?

    The charge of the VDG (the one you commonly used during your lectures) is only 10uC, so I can't figure out what would happen. I could come up with the follow (probably wrong) hypothesis:

    I assumed that you would experience a current from the VDG to ground, which would be limited by the VDG charge. The "tricky" part is that since the current flow is I = dQ/dt and Ohm's law gives us 100mA, it means the VDG can't provide enough current to keep that up unless it would get charged almost instantly. My (probably incorrect) conclusion is that therefore you would just end up feeling a initial shock, and if you could hold on the sphere after the shock, you would end up acting as a path for the current between the VDG and ground. Since the moving belt inside the VDG would start inducing more charge on the sphere, you would experience a (very low) current flow by the induced charge. All of this is assuming that the initial current could cause some muscle contraction but wouldn't harm you any further.

    Where did I go wrong?

    Once again thanks for the fantastic lectures, they are really unmatched. They make me love physics again and have been an excellent entertainment source every single day since I found out about them.

    Thank you!

  53. I have a question Prof. At the end, when you take your shoes off, you have very low resistance, so the current should be higher. But the electroscope shows no deflection. This implies that no charge flows through it. Where am I getting it wrong?

  54. Professor I have a question…what would those SPECIAL shoes( worn by thief in the end) look like in order to make a working static gun from his finger?

  55. teacher,I have some asks.
    #Does the velocity of electron increase when they travel through the circuit?I think it tries to increse but can't because of small collision period.So,in everywhere the kinetic energy converts into heat energy.Then why we consider the potential difference just over a load and why potential difference is very significant there?
    #What should be the case in a superconductor?

  56. Hello Professor,please provide the exact definition of series and parallel combination of resistors.

  57. Hello sir!
    At 6:17 you derived the speed of electrons in that conductor. Then how come current flows within just the flick of a switch?
    Thank you, sir.

  58. Why is the drift velocity = a * Tau ( at 3:37) ?
    If it is the average speed of the electron, shouldn't it be 1/2 * a * Tau ?
    I mean, a * Tau should be the maximal velocity, just befor another colision, not the average…
    Where is the problem in my reasoning?

  59. Sir , why the current density has to be uniform over the cross section of a conductor ? I read the answer in purcell but i didnot get it .

  60. Thank you so much for your lectures dear Prof. Walter Lewin. They are very inspiring. Please let me show some doubts about the last demonstration of the lecture:
    1) The charge detected by the electroscope is due to a bulk conduction process through the soles or a surface conduction process over the soles, the shoes,… (or both processes)?
    2) The charge detected by the electroscope could be an induced one by the charge created is the low surface of the soles during the scuffing? I know that a opposite charge is generated in the pad by the scuffing, but perhaps it can easily go to earth and have a net charge in the soles that could cause an induced charge in your body.
    3) The electroscope null response when you stop the scuffing is due to the fact that the charge goes to earth through your body with the shoes. Do you think that can play a role the charge recombination in the interface between the sole and the pad?

    4) It could be interesting and informative to repeat the experiment but placing bellow the pad a thick plate of teflon or fused silica to prevent the charge leakage to earth. Perhaps the electroscope do not discharge even when you stop the scuffing. Or may be the electroscope is discharged due to the recombination mentioned in 3) even when the leakage to earth is not posible.
    Thank you so much.

  61. Sir, do electrons really move during current or is it our assumption. Because if an electron moves from its position then…. The configuration of the atom collapse. Can you please explain what is actually happening.

  62. i have one meter copper wire with 0.1 meter square cross sectional area … if I apply only 1.5 volts voltage i am going to get 9375000 amperes current ..it tastes very bitter isn't it ??

  63. Sir when a resistance(AB) is added through a conducting wire across a battery, the potential difference between one end of the battery(say +) and one end of the resistance (A) is zero. But inspite of that there is a current flow between the two.why

  64. Hi Sir,
    Does increase in voltage will increase the speed of the electrons or numbers of electrons in a conductor ?

  65. I am a retired EE and CS engineer, having gotten my EE degree in the 60s. That being said, I am learning more from your videos, some of which I forgot, some I wasn't taught and some I never really fully understood. You obviously love your job and do a great job teaching. Right now, I am watching your videos selectively to get a better grasp of EM wave propagation, transmission lines and antennas. Thank you for posting these classes Professor Lewin! By the way, you probably should use less salt on your eggs. I know I have to. 🙂
    I wonder what would happen if you poured some Gatorade into the distilled water. I think the answer is easy. I have to avoid the stuff because of the salt content.

  66. This is the answer to our lack of education problems. Get our most brilliant minds and make their knowledge free and universal.

  67. Professor, thank you for your excellent lectures.
    I wanted to ask a question:
    If the drift velocity of electrons is that slow as mentioned in the lecture, how current travels so fast then?

  68. showing the (non-linear) time regime impuls of resistivity is one thing, stating "not to trust Ohm's law" is a bit reductive (not to say blunt) … we realized quite a lot trusting it (as we did with Newton's imperfect laws for instance) … oh and though it's fortunate that resistance goes up with temperature, it's also necessary from a physical standpoint no?

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