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Resistivity and conductivity Circuits Physics Khan Academy.mp3 | You probably know that if you hook up a battery of voltage V to a resistor of resistance R, then you'll get a certain amount of current, and you can determine how much current flows here by using Ohm's Law, and remember, Ohm's Law says that the voltage across a resistor equals the current through that resistor times the resistance of that resistor. So this pretty much gives you a way to define resistance. The resistance of this resistor is defined to be the amount of voltage applied across it divided by the amount of current through it. And this is good, we like definitions because we want to be sure that we know what we're talking about. That's the definition of resistance. Remember, it has units of Ohms. But be careful, don't fall into the trap of thinking about this the way some people do. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And this is good, we like definitions because we want to be sure that we know what we're talking about. That's the definition of resistance. Remember, it has units of Ohms. But be careful, don't fall into the trap of thinking about this the way some people do. Some people think, oh, okay, if I want a bigger resistance, I'll just increase the voltage because that'll give me a bigger number up top. Bigger resistance, nah, it doesn't work that way. If you increase the voltage, well, you're gonna increase the current. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | But be careful, don't fall into the trap of thinking about this the way some people do. Some people think, oh, okay, if I want a bigger resistance, I'll just increase the voltage because that'll give me a bigger number up top. Bigger resistance, nah, it doesn't work that way. If you increase the voltage, well, you're gonna increase the current. And this ratio is gonna stay the same. The resistance is a constant. At this resistor, if you're not changing the material makeup or size or dimensions of this resistor, this number that is the resistance is a constant if it's truly an ohmic material. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | If you increase the voltage, well, you're gonna increase the current. And this ratio is gonna stay the same. The resistance is a constant. At this resistor, if you're not changing the material makeup or size or dimensions of this resistor, this number that is the resistance is a constant if it's truly an ohmic material. So ohmic materials maintain a constant resistance regardless of what voltage or current you throw at them. It'll just be constant. Yeah, if you throw too much current or voltage, the thing will burn up. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | At this resistor, if you're not changing the material makeup or size or dimensions of this resistor, this number that is the resistance is a constant if it's truly an ohmic material. So ohmic materials maintain a constant resistance regardless of what voltage or current you throw at them. It'll just be constant. Yeah, if you throw too much current or voltage, the thing will burn up. I don't suggest you do that. So there's an operating range here. But if you're within that range, this resistance, this number, this number of Ohms is a constant. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Yeah, if you throw too much current or voltage, the thing will burn up. I don't suggest you do that. So there's an operating range here. But if you're within that range, this resistance, this number, this number of Ohms is a constant. It stays the same no matter what voltage or current you put through it. So we define it by talking about voltage and current, but it doesn't even really depend on that. If you really wanna change this ratio, this number that comes out here for the resistance, you need to change something about the resistor itself, its size, what it's made out of, its length, its shape. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | But if you're within that range, this resistance, this number, this number of Ohms is a constant. It stays the same no matter what voltage or current you put through it. So we define it by talking about voltage and current, but it doesn't even really depend on that. If you really wanna change this ratio, this number that comes out here for the resistance, you need to change something about the resistor itself, its size, what it's made out of, its length, its shape. So let's figure out how to do that. If we take this resistor, imagine taking this resistor, bringing it into the shop, what is it gonna look like? Well, for simplicity's sake, let's say we just have a perfectly cylindrical resistor. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | If you really wanna change this ratio, this number that comes out here for the resistance, you need to change something about the resistor itself, its size, what it's made out of, its length, its shape. So let's figure out how to do that. If we take this resistor, imagine taking this resistor, bringing it into the shop, what is it gonna look like? Well, for simplicity's sake, let's say we just have a perfectly cylindrical resistor. So this is the wire going into one end. This is your resistor. It's a cylinder, let's say. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Well, for simplicity's sake, let's say we just have a perfectly cylindrical resistor. So this is the wire going into one end. This is your resistor. It's a cylinder, let's say. And then here's the wire going out of the other end. So this is the blown up version of this resistor. One thing it could depend on is the length. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | It's a cylinder, let's say. And then here's the wire going out of the other end. So this is the blown up version of this resistor. One thing it could depend on is the length. So the length of this resistor could affect the resistance of this resistor. Another thing it could depend on is the area of this front part here, this cross-sectional area. It's called the cross-sectional area because that's the direction that the current's heading into. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | One thing it could depend on is the length. So the length of this resistor could affect the resistance of this resistor. Another thing it could depend on is the area of this front part here, this cross-sectional area. It's called the cross-sectional area because that's the direction that the current's heading into. This current is heading into that area there, like a tunnel, and it comes out over here. Now this is full. This isn't hollow. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | It's called the cross-sectional area because that's the direction that the current's heading into. This current is heading into that area there, like a tunnel, and it comes out over here. Now this is full. This isn't hollow. This is made up of some material. Maybe it's a metal or some sort of carbon compound or a semiconductor, but it's a solid material right here that the current flows into and then flows out of. So what would happen if we made this resistor longer? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | This isn't hollow. This is made up of some material. Maybe it's a metal or some sort of carbon compound or a semiconductor, but it's a solid material right here that the current flows into and then flows out of. So what would happen if we made this resistor longer? Let's say we start changing some of these variables and we increase the length of this resistor. Well, now this current's gotta flow through a longer resistor. It's gotta flow through this resistor for more of this path. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | So what would happen if we made this resistor longer? Let's say we start changing some of these variables and we increase the length of this resistor. Well, now this current's gotta flow through a longer resistor. It's gotta flow through this resistor for more of this path. And it makes sense to me to think that the resistance is gonna increase. If I increase the length of this resistor, then the resistance is gonna go up. How about the area, this cross-sectional area? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | It's gotta flow through this resistor for more of this path. And it makes sense to me to think that the resistance is gonna increase. If I increase the length of this resistor, then the resistance is gonna go up. How about the area, this cross-sectional area? Let's say I increase this area and make it a wider, larger diameter cylinder. Well, it makes sense to me to think that now the current's got more room to flow through, essentially. There's a bigger area through which this current can flow. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | How about the area, this cross-sectional area? Let's say I increase this area and make it a wider, larger diameter cylinder. Well, it makes sense to me to think that now the current's got more room to flow through, essentially. There's a bigger area through which this current can flow. It's not as restricted. That means the resistance should go down. And if we try to put this in a mathematical formula, what that means is if I increase the length, R should depend on the length. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | There's a bigger area through which this current can flow. It's not as restricted. That means the resistance should go down. And if we try to put this in a mathematical formula, what that means is if I increase the length, R should depend on the length. It turns out it's directly proportional to the length. If I double the length of a resistor, I get twice the resistance. But area, if I increase the area, I should get less resistance because there's more room to flow. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And if we try to put this in a mathematical formula, what that means is if I increase the length, R should depend on the length. It turns out it's directly proportional to the length. If I double the length of a resistor, I get twice the resistance. But area, if I increase the area, I should get less resistance because there's more room to flow. So then over here in this formula, my area has got to go on the bottom. The resistance of the resistor is inversely proportional to this cross-sectional area. But there's one more quantity that this resistance could depend on, and that's what the material is actually made of. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | But area, if I increase the area, I should get less resistance because there's more room to flow. So then over here in this formula, my area has got to go on the bottom. The resistance of the resistor is inversely proportional to this cross-sectional area. But there's one more quantity that this resistance could depend on, and that's what the material is actually made of. So the geometry determines the resistance as well as what the material's made of. Some materials just naturally offer more resistance than others. Metals offer very little resistance, and nonmetals typically offer more resistance. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | But there's one more quantity that this resistance could depend on, and that's what the material is actually made of. So the geometry determines the resistance as well as what the material's made of. Some materials just naturally offer more resistance than others. Metals offer very little resistance, and nonmetals typically offer more resistance. So we need a way to quantify the natural resistance a material offers, and that's called the resistivity. And it's represented with a Greek letter rho. And the bigger the resistivity of a material, the more it naturally resists the flow of current through it. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Metals offer very little resistance, and nonmetals typically offer more resistance. So we need a way to quantify the natural resistance a material offers, and that's called the resistivity. And it's represented with a Greek letter rho. And the bigger the resistivity of a material, the more it naturally resists the flow of current through it. To give you an idea of the numbers here, the resistivity of copper, well, that's a metal. It's gonna be small. It's like 1.68 times 10 to the negative eighth. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And the bigger the resistivity of a material, the more it naturally resists the flow of current through it. To give you an idea of the numbers here, the resistivity of copper, well, that's a metal. It's gonna be small. It's like 1.68 times 10 to the negative eighth. We'll talk about the units in a second. But the resistivity of something like rubber and insulator is huge. It can be on the order of 10 to the 13th. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | It's like 1.68 times 10 to the negative eighth. We'll talk about the units in a second. But the resistivity of something like rubber and insulator is huge. It can be on the order of 10 to the 13th. So there's a huge range of possible values as you go from metal conductor to semiconductor to insulator. Huge range of possible resistivities. And this is the last key here. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | It can be on the order of 10 to the 13th. So there's a huge range of possible values as you go from metal conductor to semiconductor to insulator. Huge range of possible resistivities. And this is the last key here. This is the last element in this equation. The resistivity goes right here. So the bigger the resistivity, the bigger the resistance. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And this is the last key here. This is the last element in this equation. The resistivity goes right here. So the bigger the resistivity, the bigger the resistance. That makes sense. And then it also depends on these geometrical factors of length and area. So here's a formula to determine what factors actually change the resistance of a resistor. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | So the bigger the resistivity, the bigger the resistance. That makes sense. And then it also depends on these geometrical factors of length and area. So here's a formula to determine what factors actually change the resistance of a resistor. The resistivity, the length, and the area. So what are the units of resistivity? Well, I can rearrange this formula and I can get that the resistivity equals the resistance times the area of the resistor divided by the length. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | So here's a formula to determine what factors actually change the resistance of a resistor. The resistivity, the length, and the area. So what are the units of resistivity? Well, I can rearrange this formula and I can get that the resistivity equals the resistance times the area of the resistor divided by the length. And so that gives me units of ohms times meters squared, because that's area, divided by meters. And so I end up getting ohms. One of these meters cancels out. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Well, I can rearrange this formula and I can get that the resistivity equals the resistance times the area of the resistor divided by the length. And so that gives me units of ohms times meters squared, because that's area, divided by meters. And so I end up getting ohms. One of these meters cancels out. Ohms times meters. Those are the units of these resistivities, ohm meters. But how do you remember this formula? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | One of these meters cancels out. Ohms times meters. Those are the units of these resistivities, ohm meters. But how do you remember this formula? It's kind of complicated. I mean, is area on top? Is length on bottom? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | But how do you remember this formula? It's kind of complicated. I mean, is area on top? Is length on bottom? Hopefully you can remember why those factors affected it. But sometimes students have a hard time remembering this formula. One of my previous students from a few years ago figured out a way to remember it. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Is length on bottom? Hopefully you can remember why those factors affected it. But sometimes students have a hard time remembering this formula. One of my previous students from a few years ago figured out a way to remember it. He thought this looked like replay. So this R is like R, and this equal sign kind of looks like an E, and the row kind of looks like a P. And the L looks like an L, and then the A looks like an A, and it kind of looks like replay. There's a missing Y here, but every time I think of this formula, I think of it as the replay formula, because my former student Mike figured out this mnemonic, and it's handy. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | One of my previous students from a few years ago figured out a way to remember it. He thought this looked like replay. So this R is like R, and this equal sign kind of looks like an E, and the row kind of looks like a P. And the L looks like an L, and then the A looks like an A, and it kind of looks like replay. There's a missing Y here, but every time I think of this formula, I think of it as the replay formula, because my former student Mike figured out this mnemonic, and it's handy. I like it. So thank you, Mike. And since we're talking about resistivity, it makes sense for us to talk about conductivity, electrical conductivity. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | There's a missing Y here, but every time I think of this formula, I think of it as the replay formula, because my former student Mike figured out this mnemonic, and it's handy. I like it. So thank you, Mike. And since we're talking about resistivity, it makes sense for us to talk about conductivity, electrical conductivity. Now, the resistivity gives you an idea of how much something naturally resists current, and the conductivity tells you how much something naturally allows current. So they're inversely proportional, and if you're thinking it might be this easy, it is. The resistivity is just equal to 1 over the electrical conductivity, and the symbol we use for electrical conductivity is sigma. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And since we're talking about resistivity, it makes sense for us to talk about conductivity, electrical conductivity. Now, the resistivity gives you an idea of how much something naturally resists current, and the conductivity tells you how much something naturally allows current. So they're inversely proportional, and if you're thinking it might be this easy, it is. The resistivity is just equal to 1 over the electrical conductivity, and the symbol we use for electrical conductivity is sigma. So this Greek letter sigma is the electrical conductivity. And rho, the resistivity, is just 1 over sigma, the electrical conductivity. And vice versa, sigma is just going to equal 1 over the resistivity, because if something's a great resistor, it's a bad conductor. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | The resistivity is just equal to 1 over the electrical conductivity, and the symbol we use for electrical conductivity is sigma. So this Greek letter sigma is the electrical conductivity. And rho, the resistivity, is just 1 over sigma, the electrical conductivity. And vice versa, sigma is just going to equal 1 over the resistivity, because if something's a great resistor, it's a bad conductor. And if something's a great conductor, it's a bad resistor. So these things are inversely proportional. They're like two peas in a pod. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And vice versa, sigma is just going to equal 1 over the resistivity, because if something's a great resistor, it's a bad conductor. And if something's a great conductor, it's a bad resistor. So these things are inversely proportional. They're like two peas in a pod. If you know one of these, you know the other. All right, if this all seems a little bit too abstract still, there's a nice analogy you can make to water. We saw that a resistor depended on a few things like the resistivity. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | They're like two peas in a pod. If you know one of these, you know the other. All right, if this all seems a little bit too abstract still, there's a nice analogy you can make to water. We saw that a resistor depended on a few things like the resistivity. The bigger the resistivity, the bigger the resistance. And we saw that the bigger the length of the resistor, the larger the resistance. And if you divide by the area of the resistor, it shows that the resistance is inversely proportional to the area of the resistor. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | We saw that a resistor depended on a few things like the resistivity. The bigger the resistivity, the bigger the resistance. And we saw that the bigger the length of the resistor, the larger the resistance. And if you divide by the area of the resistor, it shows that the resistance is inversely proportional to the area of the resistor. So let's make an analogy to water. Let's say you had, instead of electrons flowing through a wire, instead of the wire, let's say you had a tube, a pipe, that water could flow through. So instead of electrons, you've got water flowing through a pipe. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And if you divide by the area of the resistor, it shows that the resistance is inversely proportional to the area of the resistor. So let's make an analogy to water. Let's say you had, instead of electrons flowing through a wire, instead of the wire, let's say you had a tube, a pipe, that water could flow through. So instead of electrons, you've got water flowing through a pipe. Different pipes are going to offer different amounts of resistance to the water flowing through that pipe. What would affect it? Well, imagine you had a constriction in this pipe. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | So instead of electrons, you've got water flowing through a pipe. Different pipes are going to offer different amounts of resistance to the water flowing through that pipe. What would affect it? Well, imagine you had a constriction in this pipe. If this pipe got constricted, it'd be harder for the water to flow. You'd find that it resists the flow of water more because of this constriction. And what would it depend on? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Well, imagine you had a constriction in this pipe. If this pipe got constricted, it'd be harder for the water to flow. You'd find that it resists the flow of water more because of this constriction. And what would it depend on? Well, the smaller this area of the constriction, the larger the resistance. And that agrees with what we have up here. If you have a really small area, you're dividing by a small number. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And what would it depend on? Well, the smaller this area of the constriction, the larger the resistance. And that agrees with what we have up here. If you have a really small area, you're dividing by a small number. And when you divide by a small number, you get a big number. That'd be a big resistance. So that makes sense. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | If you have a really small area, you're dividing by a small number. And when you divide by a small number, you get a big number. That'd be a big resistance. So that makes sense. Also, the length. If you increase the length of this constriction, the water will have a harder time flowing. There's manuals for plumbers. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | So that makes sense. Also, the length. If you increase the length of this constriction, the water will have a harder time flowing. There's manuals for plumbers. And you can look it up. There's a key to determine if your pipe is a certain length. You're going to need more pressure over here. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | There's manuals for plumbers. And you can look it up. There's a key to determine if your pipe is a certain length. You're going to need more pressure over here. So the smaller the constriction in terms of its area, and the longer it is, the more pressure you need back here. The pressure is like the source of the battery. So instead of a battery providing a voltage to this circuit, you'd have something offering pressure to get the water flowing. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | You're going to need more pressure over here. So the smaller the constriction in terms of its area, and the longer it is, the more pressure you need back here. The pressure is like the source of the battery. So instead of a battery providing a voltage to this circuit, you'd have something offering pressure to get the water flowing. And just like a battery, what matters is the difference in electric potential. What matters for the pressure here is the change in pressure between one point in the system and another point in the system. So that makes sense. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | So instead of a battery providing a voltage to this circuit, you'd have something offering pressure to get the water flowing. And just like a battery, what matters is the difference in electric potential. What matters for the pressure here is the change in pressure between one point in the system and another point in the system. So that makes sense. A longer constriction means more resistance. A smaller area means more resistance. What would this resistivity be analogous to? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | So that makes sense. A longer constriction means more resistance. A smaller area means more resistance. What would this resistivity be analogous to? Well, it would be kind of like what the pipe is made out of. If this pipe has a rough inner surface, the water wouldn't flow as smoothly. You would get a greater resistance, regardless of how long it is or what the area is. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | What would this resistivity be analogous to? Well, it would be kind of like what the pipe is made out of. If this pipe has a rough inner surface, the water wouldn't flow as smoothly. You would get a greater resistance, regardless of how long it is or what the area is. Just the natural built-in effect of the pipe itself is what the resistivity would depend on, just like up here. The resistivity depends on what the material is made out of. The resistivity of this pipe depends on what this pipe is made out of, at least the inner wall. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | You would get a greater resistance, regardless of how long it is or what the area is. Just the natural built-in effect of the pipe itself is what the resistivity would depend on, just like up here. The resistivity depends on what the material is made out of. The resistivity of this pipe depends on what this pipe is made out of, at least the inner wall. So hopefully this analogy makes this formula seem a little more intuitive. But just in case, let's do an example. Let's get rid of all this. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | The resistivity of this pipe depends on what this pipe is made out of, at least the inner wall. So hopefully this analogy makes this formula seem a little more intuitive. But just in case, let's do an example. Let's get rid of all this. Let's say you got this question. How much resistance would be offered by 12 meters of copper wire with a diameter of 0.01 meters if copper has a resistivity of 1.68 times 10 to the negative eighth? Now what units does resistivity have? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Let's get rid of all this. Let's say you got this question. How much resistance would be offered by 12 meters of copper wire with a diameter of 0.01 meters if copper has a resistivity of 1.68 times 10 to the negative eighth? Now what units does resistivity have? Turns out resistivity has units of ohm meters, so ohms times meters. Well, let's try this out. We've got to use our formula. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Now what units does resistivity have? Turns out resistivity has units of ohm meters, so ohms times meters. Well, let's try this out. We've got to use our formula. Remember, replay. So R equals rho L over A. The resistivity we have right here, 1.68 times 10 to the negative eighth. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | We've got to use our formula. Remember, replay. So R equals rho L over A. The resistivity we have right here, 1.68 times 10 to the negative eighth. Notice how small this is. This is hardly anything at all. Copper is a great conductor. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | The resistivity we have right here, 1.68 times 10 to the negative eighth. Notice how small this is. This is hardly anything at all. Copper is a great conductor. It's a terrible resistor. It lets electrons flow through it like a charm. All right, so the length, that's pretty easy. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Copper is a great conductor. It's a terrible resistor. It lets electrons flow through it like a charm. All right, so the length, that's pretty easy. The length is 12 meters. Notice we're asking, what's the resistance of the wire itself? Now there's not really a quote unquote resistor in here, but every piece of wire is going to offer some resistance. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | All right, so the length, that's pretty easy. The length is 12 meters. Notice we're asking, what's the resistance of the wire itself? Now there's not really a quote unquote resistor in here, but every piece of wire is going to offer some resistance. And this formula applies just as well to a piece of wire as it does to a resistor. So the length of this wire is 12 meters, and the diameter is 0.01. What do we do with that? |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | Now there's not really a quote unquote resistor in here, but every piece of wire is going to offer some resistance. And this formula applies just as well to a piece of wire as it does to a resistor. So the length of this wire is 12 meters, and the diameter is 0.01. What do we do with that? Well, we need the area. Remember the cross-sectional area. And the area of a circle is pi R squared. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | What do we do with that? Well, we need the area. Remember the cross-sectional area. And the area of a circle is pi R squared. So the area down here is going to be pi times, not 0.01 squared, that's the diameter. We need the radius. We need to take half of this, so 0.005 meters squared. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | And the area of a circle is pi R squared. So the area down here is going to be pi times, not 0.01 squared, that's the diameter. We need the radius. We need to take half of this, so 0.005 meters squared. And if you calculate all this, you get a resistance of 0.0026 ohms. Hardly anything, but there is some resistance. And if this is gonna have an effect on a very delicate experiment, you've gotta take that into account. |
Resistivity and conductivity Circuits Physics Khan Academy.mp3 | We need to take half of this, so 0.005 meters squared. And if you calculate all this, you get a resistance of 0.0026 ohms. Hardly anything, but there is some resistance. And if this is gonna have an effect on a very delicate experiment, you've gotta take that into account. If you get this really long, the longer it is, the more resistance it has. That could affect your system, but typically it doesn't matter too much. The copper wire, electrons flow through that like water, like it's not even there, because the resistance is so small. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | And so they all change such that the initial pressure times the volume divided by the initial temperature is equal to the final pressure times volume divided by the final temperature. Assuming that you're not changing the energy of the system. And we'll do more of that later. And the other thing you should remember is that pressure times volume is equal to n, where n is the number of moles. Moles is like a number like dozen, but moles is a huge number, 6 times 10 to the 23 times r. r was the universal gas constant. That's 8.21, I think, times 8.31. This is 8.31 joules per mole Kelvin times the temperature. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | And the other thing you should remember is that pressure times volume is equal to n, where n is the number of moles. Moles is like a number like dozen, but moles is a huge number, 6 times 10 to the 23 times r. r was the universal gas constant. That's 8.21, I think, times 8.31. This is 8.31 joules per mole Kelvin times the temperature. And remember, just to be safe, always convert to Kelvin first. So let's see if we can do a problem that I can make up on the fly of this situation. Let's say I have a balloon. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | This is 8.31 joules per mole Kelvin times the temperature. And remember, just to be safe, always convert to Kelvin first. So let's see if we can do a problem that I can make up on the fly of this situation. Let's say I have a balloon. And the volume of the balloon is, let's say it's 1 meter cubed, so this is a big balloon. That's fairly large if you imagine a cubic meter. So the volume is a cubic meter. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | Let's say I have a balloon. And the volume of the balloon is, let's say it's 1 meter cubed, so this is a big balloon. That's fairly large if you imagine a cubic meter. So the volume is a cubic meter. Let's say the pressure is equal to, I don't know, let's say there's not a lot of pressure in it. So the pressure is equal to 5 pascals, and that's newtons per meter squared. And let's say we're at a reasonably warm temperature. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | So the volume is a cubic meter. Let's say the pressure is equal to, I don't know, let's say there's not a lot of pressure in it. So the pressure is equal to 5 pascals, and that's newtons per meter squared. And let's say we're at a reasonably warm temperature. So temperature is equal to 20 degrees Celsius. So my question to you is, and let's say that balloon is filled with helium. So my question to you is, how many molecules of helium do I have in the balloon? |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | And let's say we're at a reasonably warm temperature. So temperature is equal to 20 degrees Celsius. So my question to you is, and let's say that balloon is filled with helium. So my question to you is, how many molecules of helium do I have in the balloon? Well, let's just substitute into the equation. So we have pressure, which is 5, and I'll actually write the units, I never do it, but you should. And you should always do it in an exam. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | So my question to you is, how many molecules of helium do I have in the balloon? Well, let's just substitute into the equation. So we have pressure, which is 5, and I'll actually write the units, I never do it, but you should. And you should always do it in an exam. 5 newtons per meter squared times volume, 1 meter cubed, is equal to my number of moles, n, times the universal gas constant, 8.31 joules per mole Kelvin, times temperature. And remember, I can't repeat this enough, always convert the temperature to Kelvin. So whatever our Celsius temperature is, add 273. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | And you should always do it in an exam. 5 newtons per meter squared times volume, 1 meter cubed, is equal to my number of moles, n, times the universal gas constant, 8.31 joules per mole Kelvin, times temperature. And remember, I can't repeat this enough, always convert the temperature to Kelvin. So whatever our Celsius temperature is, add 273. So add 273 to that, you get 293 Kelvin. So let's see, I get 5 times 1, and meter squared, meters cubed, this cancels out, this just becomes a meter. Newton meters joules, 5 joules, is equal to n moles times 8.31 joules per mole Kelvin. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | So whatever our Celsius temperature is, add 273. So add 273 to that, you get 293 Kelvin. So let's see, I get 5 times 1, and meter squared, meters cubed, this cancels out, this just becomes a meter. Newton meters joules, 5 joules, is equal to n moles times 8.31 joules per mole Kelvin. Well, this Kelvin and this Kelvin cancel out, so 8.31 times 293. So let's see, 8.31 times 293 is equal to 2434.83. So let me write that down, times 2434.83 joules per mole. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | Newton meters joules, 5 joules, is equal to n moles times 8.31 joules per mole Kelvin. Well, this Kelvin and this Kelvin cancel out, so 8.31 times 293. So let's see, 8.31 times 293 is equal to 2434.83. So let me write that down, times 2434.83 joules per mole. And so to get the number of moles, we just divide both sides of this equation by that, and the unit should work out, so you get 5. So n, switch colors to ease the monotony, n is equal to 5 joules times 1 over that, 1 over 2434.83. And then since we're dividing by this, this flips, moles per joule, and so of course, this joule cancels out with this joule, so we just have to divide 5 by this, and we'll get the number of moles. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | So let me write that down, times 2434.83 joules per mole. And so to get the number of moles, we just divide both sides of this equation by that, and the unit should work out, so you get 5. So n, switch colors to ease the monotony, n is equal to 5 joules times 1 over that, 1 over 2434.83. And then since we're dividing by this, this flips, moles per joule, and so of course, this joule cancels out with this joule, so we just have to divide 5 by this, and we'll get the number of moles. So let's take the inverse of what I had there, times 5. So I get 0.002 moles. So this equals 0.0021 moles. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | And then since we're dividing by this, this flips, moles per joule, and so of course, this joule cancels out with this joule, so we just have to divide 5 by this, and we'll get the number of moles. So let's take the inverse of what I had there, times 5. So I get 0.002 moles. So this equals 0.0021 moles. Now that might seem like a small number to you, but let's figure out how many molecules that is. So we already said, we knew that, let me make some space free, write some Avogadro's number down, so I can get rid of all of this stuff. OK, now I have space. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | So this equals 0.0021 moles. Now that might seem like a small number to you, but let's figure out how many molecules that is. So we already said, we knew that, let me make some space free, write some Avogadro's number down, so I can get rid of all of this stuff. OK, now I have space. So Avogadro's number, did I even say what Avogadro's number is? Avogadro's number is the number of molecules per mole, it's that number. So number Avogadro is equal to 6.022 times 10 to the 23 molecules per mole. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | OK, now I have space. So Avogadro's number, did I even say what Avogadro's number is? Avogadro's number is the number of molecules per mole, it's that number. So number Avogadro is equal to 6.022 times 10 to the 23 molecules per mole. So the top is molecules, the bottom is mole. I know you can't read that. So if I have 0.021 moles, how many molecules do I have? |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | So number Avogadro is equal to 6.022 times 10 to the 23 molecules per mole. So the top is molecules, the bottom is mole. I know you can't read that. So if I have 0.021 moles, how many molecules do I have? Well, I just multiply that, 0.0021 times how many moles per molecule, right, because this is moles, let me write that on moles, times Avogadro's number molecules per mole, I know that's molecules, this is mole. Maybe I should write the whole thing, molecules per mole, so then the moles will cancel out. And Avogadro's number is 6.022 times 10 to the 23, let's just remember that, and let's just multiply that times 0.0021. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | So if I have 0.021 moles, how many molecules do I have? Well, I just multiply that, 0.0021 times how many moles per molecule, right, because this is moles, let me write that on moles, times Avogadro's number molecules per mole, I know that's molecules, this is mole. Maybe I should write the whole thing, molecules per mole, so then the moles will cancel out. And Avogadro's number is 6.022 times 10 to the 23, let's just remember that, and let's just multiply that times 0.0021. It equals 0.0126 times 10 to the 23, right, we have to remember that, molecules. Let's see, this is 0.0126. That's the same thing as 1.26 times 0.01, right, and then of course times 10 to the 23. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | And Avogadro's number is 6.022 times 10 to the 23, let's just remember that, and let's just multiply that times 0.0021. It equals 0.0126 times 10 to the 23, right, we have to remember that, molecules. Let's see, this is 0.0126. That's the same thing as 1.26 times 0.01, right, and then of course times 10 to the 23. And what's 0.01? That's 10 to the negative 2, right? 10 to the negative 1 is 0.1, so this is 10 to the negative 2. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | That's the same thing as 1.26 times 0.01, right, and then of course times 10 to the 23. And what's 0.01? That's 10 to the negative 2, right? 10 to the negative 1 is 0.1, so this is 10 to the negative 2. So then we get 1.26, 10 to the negative 2 times 10 to the negative 3, we add the exponents, times 10 to the 21st power, so roughly 1, 2, 6, and then another 19 0's, or roughly 1 followed by 21 0's, is how many molecules of, in this case, helium, we had in the balloon. So not too difficult, the hard part is really just remembering Avogadro's number, remembering the universal gas constant is 8.31 per joules per mole Kelvin, remembering to always convert your temperature to Kelvin, and then just making sure all your units match up. I mean, sometimes it might be tricky, they might give volume in liters, and you have to, especially in this case, you have to convert it to meters cubed before you do it, or they might give pressure in atmospheres, or bars, and then you should know the conversion and convert it to pascals, or newtons per meter squared. |
Thermodynamics part 5 Molar ideal gas law problem Physics Khan Academy.mp3 | 10 to the negative 1 is 0.1, so this is 10 to the negative 2. So then we get 1.26, 10 to the negative 2 times 10 to the negative 3, we add the exponents, times 10 to the 21st power, so roughly 1, 2, 6, and then another 19 0's, or roughly 1 followed by 21 0's, is how many molecules of, in this case, helium, we had in the balloon. So not too difficult, the hard part is really just remembering Avogadro's number, remembering the universal gas constant is 8.31 per joules per mole Kelvin, remembering to always convert your temperature to Kelvin, and then just making sure all your units match up. I mean, sometimes it might be tricky, they might give volume in liters, and you have to, especially in this case, you have to convert it to meters cubed before you do it, or they might give pressure in atmospheres, or bars, and then you should know the conversion and convert it to pascals, or newtons per meter squared. But other than that, it's just substituting and just doing the hairy math and the scientific notation. Anyway, hopefully that was vaguely clarifying. See you in the next video. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | Let's say you had two charges, positive eight nanocoulombs and negative eight nanocoulombs, and instead of asking what's the electric field somewhere in between, which is essentially a one-dimensional problem, we're gonna ask what's the electric field up here at this point P? Now this is a two-dimensional problem because if we wanna find the net electric field up here, the magnitude and direction of the net electric field at this point, we approach it the same way initially. We say, all right, each charge is gonna create a field up here that goes in a certain direction. This positive charge creates a field up here that goes radially away from it, and radially away from this positive at point P is something like this. So I'll call this electric field blue E because it's created by this blue positive charge. And this negative charge creates its own electric field at that point that goes radially into the negative, and radially into the negative is gonna look something like this. So I'll call this electric field yellow E because it's created by the yellow electric field. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | This positive charge creates a field up here that goes radially away from it, and radially away from this positive at point P is something like this. So I'll call this electric field blue E because it's created by this blue positive charge. And this negative charge creates its own electric field at that point that goes radially into the negative, and radially into the negative is gonna look something like this. So I'll call this electric field yellow E because it's created by the yellow electric field. So far, so good. but now things get a little weird. Look, these fields aren't even pointing in the same direction. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | So I'll call this electric field yellow E because it's created by the yellow electric field. So far, so good. but now things get a little weird. Look, these fields aren't even pointing in the same direction. They're lying in this two-dimensional plane, and we want to find the net electric field. So what we have to do in these 2D electric field problems is break up the electric fields into their components. In other words, the field created by this positive charge is gonna have a horizontal component, and that's gonna point to the right. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | Look, these fields aren't even pointing in the same direction. They're lying in this two-dimensional plane, and we want to find the net electric field. So what we have to do in these 2D electric field problems is break up the electric fields into their components. In other words, the field created by this positive charge is gonna have a horizontal component, and that's gonna point to the right. And I'll call that blue EX, because it was the horizontal component created by the blue positive charge. And this electric field is gonna have a vertical component. That's gonna point upward. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | In other words, the field created by this positive charge is gonna have a horizontal component, and that's gonna point to the right. And I'll call that blue EX, because it was the horizontal component created by the blue positive charge. And this electric field is gonna have a vertical component. That's gonna point upward. I'll call that blue EY. And similarly, for the electric field this negative charge creates, it has a horizontal component that points to the right. We'll call that yellow EX. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | That's gonna point upward. I'll call that blue EY. And similarly, for the electric field this negative charge creates, it has a horizontal component that points to the right. We'll call that yellow EX. And a vertical component, but this vertical component points downward. I'll call that yellow EY. So what do we do with all these components to find the net electric field? |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | We'll call that yellow EX. And a vertical component, but this vertical component points downward. I'll call that yellow EY. So what do we do with all these components to find the net electric field? Typically what you do in these 2D electric problems is focus on finding the components of the net electric field in each direction separately. So we divide and conquer. We're gonna ask, what's the horizontal component of the net electric field, and what's the vertical component of the net electric field? |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | So what do we do with all these components to find the net electric field? Typically what you do in these 2D electric problems is focus on finding the components of the net electric field in each direction separately. So we divide and conquer. We're gonna ask, what's the horizontal component of the net electric field, and what's the vertical component of the net electric field? And then once we know these, we can combine them using the Pythagorean theorem if we want to, to get the magnitude of that net electric field. But we're kind of in luck in this problem. There's a certain amount of symmetry in this problem. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | We're gonna ask, what's the horizontal component of the net electric field, and what's the vertical component of the net electric field? And then once we know these, we can combine them using the Pythagorean theorem if we want to, to get the magnitude of that net electric field. But we're kind of in luck in this problem. There's a certain amount of symmetry in this problem. And when there's a certain amount of symmetry, you can save a lot of time. What I mean by that is that both of these charges have the same magnitude of charge. And because this point P lies directly in the middle of them, the distance from the charge to point P is gonna be the same as the distance from the negative charge to point P. So both of these charges create an electric field at this point of equal magnitude. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | There's a certain amount of symmetry in this problem. And when there's a certain amount of symmetry, you can save a lot of time. What I mean by that is that both of these charges have the same magnitude of charge. And because this point P lies directly in the middle of them, the distance from the charge to point P is gonna be the same as the distance from the negative charge to point P. So both of these charges create an electric field at this point of equal magnitude. The fields just point in different directions. And what that means is that this positive charge will create an electric field that has some vertical component upward of some positive amount. We don't know exactly how much that is, but it'll be a positive number, because it points up. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | And because this point P lies directly in the middle of them, the distance from the charge to point P is gonna be the same as the distance from the negative charge to point P. So both of these charges create an electric field at this point of equal magnitude. The fields just point in different directions. And what that means is that this positive charge will create an electric field that has some vertical component upward of some positive amount. We don't know exactly how much that is, but it'll be a positive number, because it points up. And this negative charge is gonna create an electric field that has a vertical component downward, which is gonna be negative, but it's gonna have the same magnitude as the vertical component of the blue electric field. In other words, the field created by the positive charge is just as upward as the field created by the negative charge is downward. So when you add those up, when you add up these two vertical components to find the vertical component of the net electric field, you're just gonna get zero. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | We don't know exactly how much that is, but it'll be a positive number, because it points up. And this negative charge is gonna create an electric field that has a vertical component downward, which is gonna be negative, but it's gonna have the same magnitude as the vertical component of the blue electric field. In other words, the field created by the positive charge is just as upward as the field created by the negative charge is downward. So when you add those up, when you add up these two vertical components to find the vertical component of the net electric field, you're just gonna get zero. They're gonna cancel completely, which is nice, because that means we only have to worry about the horizontal components. These will not cancel. How come these don't cancel? |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | So when you add those up, when you add up these two vertical components to find the vertical component of the net electric field, you're just gonna get zero. They're gonna cancel completely, which is nice, because that means we only have to worry about the horizontal components. These will not cancel. How come these don't cancel? Because they're both pointing to the right. If one was pointing right and the other was left, then the horizontal components would cancel, but that's not what happens here. These components combine to form a total component in the x direction that's larger than either one of them. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | How come these don't cancel? Because they're both pointing to the right. If one was pointing right and the other was left, then the horizontal components would cancel, but that's not what happens here. These components combine to form a total component in the x direction that's larger than either one of them. In fact, it's gonna be twice as big, because each charge creates the same amount of electric field in this x direction because of the symmetry of this problem. So we've reduced this problem to just finding the horizontal component of the net electric field. To do that, we need the horizontal components of each of these individual electric fields. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | These components combine to form a total component in the x direction that's larger than either one of them. In fact, it's gonna be twice as big, because each charge creates the same amount of electric field in this x direction because of the symmetry of this problem. So we've reduced this problem to just finding the horizontal component of the net electric field. To do that, we need the horizontal components of each of these individual electric fields. If I can find the horizontal component of the field created by the positive charge, that's gonna be a positive contribution to the total electric field, since this points to the right. And I'd add that to the horizontal component of the yellow electric field, because it also points to the right. Even though the charge creating that field is negative, the horizontal component of that field is positive, because it points to the right. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | To do that, we need the horizontal components of each of these individual electric fields. If I can find the horizontal component of the field created by the positive charge, that's gonna be a positive contribution to the total electric field, since this points to the right. And I'd add that to the horizontal component of the yellow electric field, because it also points to the right. Even though the charge creating that field is negative, the horizontal component of that field is positive, because it points to the right. So if I can get both of these, I will just add these up, and I'd get my total electric field in the x direction. So how do I get these? How do I determine these horizontal components? |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | Even though the charge creating that field is negative, the horizontal component of that field is positive, because it points to the right. So if I can get both of these, I will just add these up, and I'd get my total electric field in the x direction. So how do I get these? How do I determine these horizontal components? Well, to get the horizontal component of this blue electric field, I first need to find what's the magnitude of this blue electric field. We know the formula for that. I'll write it over here. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | How do I determine these horizontal components? Well, to get the horizontal component of this blue electric field, I first need to find what's the magnitude of this blue electric field. We know the formula for that. I'll write it over here. The magnitude of the electric field is always kq over r squared. So for this blue field, we'll say that E is equal to nine times 10 to the ninth, and the charge is eight nanocoulombs. Nano means 10 to the negative ninth. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | I'll write it over here. The magnitude of the electric field is always kq over r squared. So for this blue field, we'll say that E is equal to nine times 10 to the ninth, and the charge is eight nanocoulombs. Nano means 10 to the negative ninth. And then we divide by the r, but what's the r in this case? It's not four or three. Remember, the r in that electric field formula is always from the charge to the point you're trying to determine the electric field at. |
Net electric field from multiple charges in 2D Physics Khan Academy.mp3 | Nano means 10 to the negative ninth. And then we divide by the r, but what's the r in this case? It's not four or three. Remember, the r in that electric field formula is always from the charge to the point you're trying to determine the electric field at. So r is this. This distance is r. So how do we figure out what this is? Well, we're kind of in luck. |
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