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ECE802 – Lec09 [00:00~10:31] When positive charges move in this direction, then, for definition, we say the current goes in this direction. When negative goes in this direction, we also say the current goes in that direction. That’s just our convention. If I apply a potential difference over a conductor, then I am going to create electric field in that conductor. And the electrons, there are free electrons in the conductor. They can move, but the ions can not move. Because they are frozen into the solid, into the crystal. And so when the current flows into the conductor is always the electrons that responsible for the current. The electrons fill the electric field. And then the electrons try make the electric field zero, but they can not succeed because we keep the potential difference over the conductor. Often, there is a linear relationship between currents and potential. In which case we talk about Ohm’s Law. Now I will try to derive Ohm’s Law in very cruel way, cool man’s version, and not really 100% causal. It requires quantum mechanics which is beyond this course, but I will do the job that still gets the interest inside into Ohm’s Law. If I start off with a conductor, for instance, copper (Cu), at room temperature 300 degree kelvin (300K). 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. It’s in all directions. It is in chaotic motion. It is a thermo motion. It is due to the temperature. The time between collisions……time between the collisions……and this is the collision of the free electrons with the atoms, is approximately…… I called it t, is about 3*10–14 seconds. No surprise because the speed is abnormally high. And the number of free electrons in copper per cube meter…… I call that number n, is about 1029. There is about one free electron for every atom. So it would turn to the 1029 cube meter. So now imagine that I apply a potential difference, a piece of copper, for any conductor for that matter, then the electrons will experience a force F, which is the charge of the electrons that is my little e times the electric field E that I was creating because I apply a potential difference. I realized that the force and the electric field are in opposite direction for the electrons. But that is the detail. I am interesting in magnitude itself only. And so now these electrons will experience an acceleration, which is the force F divide to the electric mass of the electron me. And so they will pick up the speed v in the collision, which we called the drift velocity vd, which is a times t. This is 801 meterial. and so a equals F divide by me, F is eE, so we get e times E divide by the mass of the electrons times t. And that is the drift velocity. When the electric field E goes up, the drift velocity goes up, so the electrons move faster in the direction opposite to the current. If the time between collision gets larger and the acceleration last longer. So also they pick up a large speed and that is into a different pleasing. If we take a specific case and I take, for instance, a copper, and I applied over the ……over the wire. Lets say the wire has a length of 10 meters. I apply a potential difference. I called delta V , ∆V, but I could have just said a V. I applied the potential difference of 10 volts. Then the electric field inside the conductor now is about 1 volt per meter. And so I can calculate now for that specific case. I can calculate what the drift velocity will be. So the drift velocity for those free electrons would be the charge of the electrons, which will is 1.6*10–19 coulomb. The E field is 1 so I can forget about that. t is 3*10–14 as long as I am in room temperature. And the mass of the electron, me, is about 10–30 kilograms. And so if I didn’t slip up, I found this is 5*10–3 meters per second, which is half centimeter per second. So imagine, due to the thermo motion, these free electrons move with a million meters per second. Due to this electric field, they only advanced along the wire slowly, like a snail with a speed and an average of half of centimeters per second. And this goes very much again due to my annual tuition. And this is the way it is. Even the turtle will go faster then these electrons. To go along a 10 meter wire will take half hour. Something that you never thought of that it will take a half hour for these electrons to go along the wire if you apply potential difference of 10 volts a copper wire 10 meters long. Now I want to massage this further and see whether we can somehow squeeze out Ohm’s Law, which is linear relation with 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 V over the wire, plus here and minus there, potential V. So I will get the current in this direction that is our definition of current, going from plus to minus. The electrons e, of course are moving in this direction, with the drift velocity. And so the electric field in here which is in this direction. The electric field E is approximately V divide by l, potential difference V divide by distance l. In 1 second, these free electrons will move from left to right over a distance vd meters. So if I rake any cross section through this wire anywhere, I can calculate how many electrons pass through that cross section in 1 second. In 1 second, the volume that passed through here, the volume is vd times A. But the number of the free electrons per cube meter is called n. So this is now the number of the free electrons that passed per a second through any cross section. 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 is the detail. If I now substitute the drift velocity, which we have ……here!! I substitute that in there, then I find the current ……I got a e2, the charge squared. I got n, I got t, I got downstairs the mass of the electron, me. And then I get A times electric field E. Because I have here the electric field E. When you look at this here, it really depends only on its properties of the my substance for a given temperature. And we give that a name. We called this s, which is called conductivity. If I calculate for copper the conductivity at room temperature. That is very easy. Because I have given you what the n is. On the blackboard the n is attend to 1029, and you know what the t is at room temperature, 3*10–14. So, for copper at the room temperature, you will find about 108. You will see more value for s later on during this course. This is in SI unit.
[10:32~20:36] I can massage this a little bit further because E is V divided by l, and so I can write now that the current is that s times A times V divided by l. I can write it down a little bit differently. I can say V, therefore equals l divided sA times I. And now you are staring at ohm’s law whether you like it or not. Because this is what we called the resistance, capital R. We often write down r, for 1 over s, and r is called the resistivity. So either one or two you can write down V equals I*R, and this R then……is either l divided by sA, or l times r, lets make a nice r, divided by A. That is the same thing. The unit for the resistance R is volt per amp here, (V/I), what we called that ohm. And so the unit for [R] is [W]. And so you want to know what the unit for r and s is that following it immediately from equations. The unit for r is that is 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 physic of on the Tomas scale, which is all by himself. It’s interesting. If you look at the resistance, you see this proportional with the length of your wire to each provide current. Think about this as water try to go through a pipe. If you make the pipe longer, the resistance goes up. So that is very intuitive basic. Notice that you have an A down stair. That means if the pipe is wider, larger cross section is also easier for the current to flow, easy for the water to flow (resistance is lower). So that is also quite reason. Ohm’s Law also often holds for insulators. They are not conductors even though I have derived in here. For conductors which has these free electrons. So now I want to make it comparison between very good conductors and very good insulators. Well, I will start off with a chunk of material, called sectial area A. Lets take 1mm by 1mm, so A is 10–6 square meters. S o here I have a chunk material. And the length for that material l is 1 meter. Put the potential difference over there, plus there and minus here. Current I will start to flow in this direction, electrons e will flow in this direction. And the question now is what is the resistance of this chunk of material? Well, that is easy. You take this equation. You know l and A, if I tell you what the s is, then you can immediately calculate what the resistance is. So lets take first, a good conductor. Silver (Au) and Gold (Ag) and Copper (Cu) are very good conductors. They will have values for s, 108, which is calculated for common. You have seen it, in front of your eyes. So that means the r will be 10–8, 1 over s. And so in this particular case, since A is 10–6, the resistance R is simply 106 times r because l is 1 meter. So it is very easy. The resistance here, R, is 10–2 ohms, 1 over 100 ohms. For this material, it works for copper. Lets now take a very good insulator. Glass is the example, Cours, Auxilen, very good insulators. Now s, the conductivity is extremely low. It is in very somewhere from 10–12 to the 10–16. So r now, the resistivity, something like 1012 to 1016. And if I take 1014, just like after grab the middle of those number, then you will find that R now is 1020 ohms. One with 20 zeros. That is an abnormal resistance. So you see the difference 22 odd makes the difference between a good conductor and a good insulator. And if I make this potential difference over the wire, if I make that 1 volt, and if I apply ohm’s law, V equals to I*R, then I can also calculate the current that is going to flow. If I*R is 1, then the current here is a 100 amp here, and the current here is 10–20 amps here, insignificant current, 10–20 amps here. I first want to demonstrate to you that ohm’s law sometimes holds. I will do the demonstration by the voltage supply, put the V in here and we change the voltage, the matter of few seconds from 0 to 4 volts. This is the plus side, this is the minus side. I have connected here to the resister which is 50 ohms. We use this symbol for resister and here is the current meter. And the current meter has negative resistancy. You can ignore that. And I am going to show you the oscilloscope. We never discuss the oscilloscope. But maybe we will in the future. I am going to show you there the projected. The voltage is going from 0 to 4 versus the current. So we start from here by the time we reach 4 volts. Then we will have reached the current, 4 divide by 50 according to ohm’s law. I will write down just 4 divide by 50 amps here which is 0.8 amps here. And if ohm’s law holds, then you will find a straight line. That is 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 lets do that!! Lets take a look at that! You are going to see that there and I have to change my lights so that you can get a good shot at it. Oh! It is already going!! So you see horizontally we have the current and vertically we have the voltage. So it takes about a second to go from 0 to 4. So this is going from 0 to 4 volts. So you see the current is beautifully linear. Yea……I blocking it……oh……no. That is my reflection. That is interesting. (student laughing) Ohm’s Law doesn’t allow for that! So you see how beautiful 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 s is a strong function of the temperature. If you increase the temperature, then the time t between collisions goes down because the speed of these free electrons goes up. It is a very strong function of temperature. And so if t goes down, then clearly what will happened that the conductivity will go down. And that means r will go up. And so you get more resistance. And so you when heat the substance. The resistance goes up. Higher temperature higher resistance.
[20:37~30:14] So the moments that the resistance are becomes a function of the temperature. I called that the total break down of V equals I*R. Total break down of ohm’s law. If you looking at your book, they say……oh…no! no! no! That is not a break down. You just have to adjust the resistance for different temperature. Wow… yeah! That is 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 current the higher the temperature. So now you get the ratio V divide by I which is no longer constants. It becomes a function of the current. That’s end up with the ohm’s law. And so I want to show you that if I do the same experiment that I did here, but I replace this by a light bulb of 50 ohms. It is a very small light bulb. Resistance when it is hot is 50 ohms. When it is cold, it is 7 ohms. So Rcold of the light bulb is roughly 7 ohms I believed, but I know that when it is hot, it is very closed to the 50 ohms. I think it is a little lower. What you are expect it now? Well you expect now when the resistance is low in the beginning you get this, and then the resistance goes up and you are going to get this. I might end up a little bit higher current because I think the resistance is a little lower then 50 ohms. And if you see a curve like this, that is not linear anymore. So that is the end of the ohm’s law. And that is what I want to show you now. So all I do is here I have this a little light bulb. For those of you who sat closed, they can actually see the light bulb starts glowing but that is not important. I really want you to see that V versus I is no longer linear. There you go!! And you see every time this light bulb go on, heats up! And during the heating up, the resistance increases. And this is end of the ohm’s law for this light bulb at least. But it was fine for another resistor but it was not fine for this light bulb. There is another way that I can show you that the ohm’s law is not always doing so well. I have a 125 volts power supply. So V is 125 volts. This is the potential difference. And I have the light bulb you see it here. That is the light bulb. The resistance of the light bulb, Rcold, I believe it is 25 ohms, and the hot, Rhot, is about 250 ohms. Huge difference!! So if the resistance …… if I take the cold resistance then I will get 5 amps here. By the time that the bulb is hot I will only get half amps (2.5 amps) here. So huge difference. I want to show you the gain of the oscilloscope is the current as the function of time. When you switch on the light bulb, you will expect if ohm’s law holds. And when you switch on the current……well, you switch on the voltages I shall say, then you see this. Then this is end up 5 amps here. And they will stay there. That is the whole idea. Namely that the voltage divided by the current remains constant. However what you are going to see is like this. The current goes up, then the resistance goes down……oh…no……then the resistance goes up, when the current goes up, the resistance goes up, and therefore the current will go down and with the level of and the level which substantial below this. So you are looking at there and you are staring at the break down of the ohm’s law. So that is what I want to show you now. So here we need a 125 volts and there is the light bulb. And when I through this switch, you will see the pattern of the current versus the time. You will only see it at once and then it freeze with the oscilloscope. So turn this off. So look closely now……There is!! Forget these ripples you see on it. That has to do with the way that produce the 125 volts. And so you see here the horizontally time……(the time between 2 adjacent vertical lines is 20 mini-seconds)……and so indeed very early on the current search on to the very high value (and then the filament heats up), and so the resistance goes up on the light bulb, and the current just goes back again (from the far left to the far right on the screen is about 200 milliseconds, that is about 2/10 of seconds). And here you get the current level is way lower then what you get it there. That’s the break down of the ohm’s law. It is actually very nice that the resistance heats goes up with the light bulb when the temperature goes up because supposedly it will be the other way around. Suppose you turn on a light bulb, and the resistance will 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 even goes further down. That means the current even goes further up. And so what does it means every time your turn on the light bulb, it will right in front of your eyes destructed itself. That is not happening. It is the other way around. So in the way it is fortuned that the resistance goes up when the light bulb get hot. All right, lets now be a little bit more quantitative. On some networks of resistors and we will have you do few problems like that. Whereby, we just assume easily that ohm’s law holds. Another words we will always assume that the value for the resistance is that we give you will not change. So we will assume that the heat that is produced will not play any equal in role. So we will just use ohm’s law for now and you can’t use it if we will be more specific about that. So suppose I have here between point A and point B, suppose I have two resisters, R1 and R2. And suppose I apply a potential difference between A and B. And 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 resister along which I will call V1, and what is the potential difference over the second resister which I called V2. Very straight forward question. Well you apply now ohm’s law. And so between A and B there are two resisters 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 R1 plus R2, V=I*(R1+R2). Suppose these two resisters were the same. They have the same length, the same cross section area. If you put two in series, you have twice the length. Well, so twice the length. Remember, resistance is linear proportional with the length of the wire, and so you add them up! So now you know R1 and you know R2, and 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 along, so then you get that the V1 equal I times R1, V1 = I*R1. So now you have the voltage over this resistor. And of course V2, must be the current I times R2, V2=I*R2. And so you have solved the problem. All the questions that I asked you, you have! The answers, too. We could now have a slightly different problem.[30:15~40:25] Hereby we have the point A is here, but now we have the resistor here which is R1, and we have here R2. This is the point B. This is R2. And the potential difference is V, and that is again is 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 will go through the resistor one, and what is the current that could go through resistor two, and I will allow you to use ohm’s law. So now you say~~ Ah~ha! The potential difference from A to B go in this loop that potential difference is V, that is given. So V must now be I1*R1, V=I1*R1, that is the ohm’s law for this upper branch. But, of course, you can also go the lower branch. So, the same V, is also I2 times R2, V=I2*R2. But what every the current comes in here must spread up between these two. Think about it is water. You can not get rid of this charges. The number of charges per second that flow into this juncture continue on. And so I, the total current, is I1 plus I2. And so now you see you have all the ingredients that you need to solve for the current I, for the current I1 and for the current I2. And you can turn this into the industry. You can make it extremely complicated network of the resistors. And if you’re in course six, you should love it. I don’t like it at all! So you don’t need to worry about you are going to get a complicated resistor networks from me. But in course six, you are going to see a lot of them. They are going to throw them, stuff them down into your throat. The conductivity of the 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, r for air, rair, cold, dry, at one atmosphere……Cold means the temperature we have outside, is about 4*1013. That is the resistivity of air. It is about what is this room, maybe a little lower because the temperature is a little bit 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 have this lighting, its step-lit came down, and we created a channel full of ions and electrons, had a very low resistivity, and 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 the air go up tremendously. Not only the electrons move but also the ions now will start to move. And the way I am going to do this is I am going to put charge on the electroscope. Oh! That is no so good! No hard down. I am going to put the charge on the electroscope. And you will see that the conductivity of the air is so poor that will stay there for hours. And then what I will do……I will create ions in the facility of the electroscope. So lets first put some charge on the electroscope. I have here a glass rod, and I will put some charge on it. Ok! That’s a lot of charge. And the air is quite dry. Conductivity is very very small. And so the charge can not go up through the air to the surrounding to the earth. And now we are going to create an ion there by heating it up. And I decide to do that with a candle because the candle is very romantic as we all known. So here I have this candle. Look up how well the charge is holding it. And here is my candle. And now I will bring the candle 20 centimeters from the electroscope. Look at it, look at this!! It is already going! Now is about from 15 centimeters away. I will take my candle away and it will stop again. So with all due to the fact that I am ionize the air there. Creating free electrons as well as ions that they both participate now. In the current and the charge can flow away from electroscope to the earth because the conductivity now is so much higher. I stop it again and it stops. 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 a clean distil water. I mean clean water. I don’t mean the stuff that you get in the cambridge. That along the means that the stuff that is in the Charles River. I mean clean water that has a pH of seven, ph=7. That means one out of 107 of the water molecule is ionized, H+ and OH– . The conductivity by the way is not the result of the free electrons. But it is really the result of these H+ and OH– ions. It is one of the case is whereby nothing. The electrons are major responsibility for the current. If I add 3% of salt in terms of weight. Then all that salt will ionize. So you get salting plus, Cl+, and Cl– ions. You increase the number of ions by enormous factor. And so the conductivity will shoot up by a factor of 300,000 or up to a million. Because you increase the ion by that amount. And so it is not surprise then for you that the conductivity of sea water is million times higher. Think about million times higher then the conductivity of the distil water. And I would like to give you the number for water. So this is the distil water, rdist water, that is about 2*105 ohm-meters, (Wm). That is the resistivity, 2*105 ohm-meters. I have here a bucket of distil water. I will make a drawing for you on the black board there. So here is the bucket of distil water and in there is a cut of plate, another cut of plate. And here is the light bulb. And this will go straight to the outlet, shueet! Stick in! On the 10 volts. This light bulb is 800 ohms resistances, 800W, when it is hot. You see the light bulb here. You can calculate what this resistance is between these two plates. That is easy! You all have the tools now. If you know the distances, is about 20 centimeters. And you know the surface area of the plate because remember the resistance inverse proportional with the A. You have to take that into account, and you take the resistivity of water into account. It is a trivial calculation. You can calculate what resistance of this proportion here and I found that this resistance here is about 2 mega-ohms, 2MW, 2 million ohms. . [40:26~49:16] So when I plug this into the wall. The current that will flow is extremely low because it has to go through 800 ohms and the 2 mega-ohms. So you won’t see any thing. Your light bulb will not show any light. But now if I put salt in here, if I really manage to put 3% in weight salt in here, then this 2 mega-ohms will go down to 2 ohms, million times less. So now the light bulb will be happy like a climb high type because 2 ohms here plus the 800 ohms here is not significant. And this is what I want to demonstrate to you now. Enormous important of increasing ions. I increase ions here by heating the air. Now I am going to increase the ions by adding salt. So the first thing that I would like to do, is I will stick this in here, there is the light bulb, and I make a daring prediction that you will see nothing. There you go! Nothing. Is 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. This water is such a huge resistance that the current is too low. Well, lets add some………not papper……add some salt. (student laughing) Yeah, there are salt in there. It is about as much as I will put into my egg this morning. Stir a little. Ha~Hey! Look at that! Is that amazing? When I bring them closed together, it will come even brighter. Because l is now smaller……the distance is smaller. Bring them far apart (the light bulb will be gloomy). It is amazing. Just a tiny mini little bit of salt. About as much as I used on my egg that along………what the hell………lets put it everything in there. (student laughing) That is a little bit more put in the everything. Then, of course, you got almost down to the 2 ohms and the light bulb will just burning normally. But even with that little bit of sale you saw the huge difference. My body is a fairly good conductor. Yours too. We all came out of to sea. So we are all almost of water. And therefore, when we do the experiment, the little charge like the Van de Graaff, bitting a student (see the 802, lecture-1), (then we have to insulated outselves very carefully), putting the glass plate on the plastic tool to prevent that the charge runs down to the earth. In fact, the resistance……my resistance between my body and the earth is large dictated by the soul of my shoe. Not by my body. Not by my skin. But if you look at my shoes, then you get something like this, and it has a certain thickness. And this, maybe 1 centimeter. This now is l in my calculation for the resistance because current may flow in this direction. So that is l. Well, how large is my foot? Lets say it is 1 foot long. No point imply. And lets say it is about 10 centimeters wide. So you can calculate what the surface area, A, is. You know what the l is. And if you know now what the resistivity for my shoe………I can make a roughly gauss………I look up the material and I found out that the resistivity is about 1010. So I can now calculate what the resistance is in this direction. And I found that the resistance then putting it into the numbers, is about 2*109 ohms. Well, you will say……WOW……oh! It is 4 actually. No big deal! 4 billion ohms. So you will say that is enormous resistance. Well, first of all, I walk in on two feet, not only one. So if I will be standing in one in whole lecture, it would be probably 4 billion ohms. But if I have two feet on the ground, it is really 2 billion ohms. You will say, well, that is still extremely large. Well, it may looks large but it really isn’t because all the experiment that we are doing here in building 26100, you are dealing with a very small amount of charge. Even if you take the Van de Graaff. The Van de Graaff say has the 200 thousand volts, 200 kv. And lets assume that my resistance is 2*109 ohms, two feet on the ground. So when I touch the Van de Graaff, the current that will flow, according to the ohm’s law, will be 100 micro-amps ,100μa , here. That means in 1 second, I can take 100 micro-coulomb of the Van de Graaff. The Van de Graaff has only 10 micro-coulomb. So the resistance of 4 billion or 2 billion ohms is way too low for this experiment that we have been doing it in building 26100. And that is why we used these plastic stool and we used these glass plates in order to measure that the current is not draining off the charge that we need for the experiment. I want to demonstrate that to you that indeed, even with my shoes on. That means even with my 2 billion ohms resistances to the ground, that it will be very difficult for me for instance to keep charge on the electroscope. I am going to put the charge on this electroscope. By scuffing my feet……But since I keep my shoes on and I am not standing on the glass plate, the charge will flow through to me. You can apply ohm’s law. And you will see as I do this, I am scuffing my feet now that I can only keep that the electroscope as long as I keep my feet scuffing. With the moment that I stop the scuffing, it is 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 2 billion ohms. Like along if I take my shoes off. I apologize for that. If now I scuff, you can’t even get any charge on the electroscope. Because now the resistance is so ridiculously low. I don’t even have 2 billion ohms. I can’t even put any charge on the electroscope. It is always very difficult for us to do this experiment unless we isulate us very well. And if some how the weather is a little damn (dry). We get very thin of water onto our tools, and then the current can flow off just through this very thin layers of water. That is why we always like to do this experiment in winter, so that the conductivity of the air is very low. No water anywhere. Did you see a slide of rubbery. I have scuff my feet across the rug and I am armed with the static charge and overall your money, or I will touch your nose! This person either never took 802, or he is wearing a very very special shoes. (student laughing!) See you on Wednesday! End …… Lecture 09 Last Modified 6/5/06 11:54 AM |
