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Electric power

In this video David derives the formula to find the power used by a resistor. Created by David SantoPietro.

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  • piceratops tree style avatar for user karoomoogon124
    I got a general question about electric power:
    At school, I have been told that the power loss in A.C transmission is minimized when voltage is maximized and current is minimized, but why is it so? Isn`t power proportional to both voltage and current? Please this is really worrying me. Thanks!
    (23 votes)
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    • purple pi purple style avatar for user APDahlen
      Hello Karoomoogon124,

      You are very close to the answer:

      1) Power lines have resistance. There will be a power loss calculated as I^2R. By inspection, this loss can be reduced by either reducing the resistance or reducing the current. Reducing the resistance is not economical as more conductor would be required along with the added cost of supporting the transmission line...

      2) There is a electrical component you need to understand. The backbone of our AC transmission system is the transformer. This allows us to step the voltage up or down as desired. A transformer is very efficient (high 90%). For our discussion let's assume the efficiency is 100%. Consequently power in equals power out.

      3) Using a transformer let's take 100 V from our generator and use a transformer to step the voltage up by a factor of 10. Since our ideal transformer is 100% efficient the current has been reduced by a factor of 10. This must be so as P = IV.

      4) We send the high voltage but low current through the transmission lines. At the other end we use a transformer to step the voltage back down.

      Your teacher is correct a high voltage low current is the way to transmit electrical power. Since the resistance of the power line is fixed we get the greatest economy when the current is low and the voltage high. The transformer is the key to the entire process.

      Keep going - there are many fascinating levels to this topic waiting for your discovery.


      (49 votes)
  • blobby green style avatar for user Ash Court
    This lesson quickly gets ahead of itself when done after the first few in "Ohm's law and circuits with resistors" He very quickly runs through definitions using symbols that haven't been explained yet...Or maybe I'm just stupid haha
    (8 votes)
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  • aqualine ultimate style avatar for user Hafsa Kaja Moinudeen
    To make an electrical appliance work, why is it necessary that it gets x amount of energy per second(watts)? Would it still work if I were to give the same amount of energy over a longer period on time or a shorter period of time?
    (5 votes)
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  • piceratops tree style avatar for user Rahul Kunte
    Does more current mean faster moving electrons or is it more magnitude of electrons?? David mentions at that the electrons can't go faster because the current must be same
    (6 votes)
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  • piceratops ultimate style avatar for user Hari
    Why does David show the positive charges to be flowing through the resistor? Don't the negative charges (electrons) flow from the negative terminal to the positive terminal but not the positive charges?

    (3 votes)
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    • starky ultimate style avatar for user KLaudano
      Yes, the electrons are the particles moving to produce the current, however, this was not known at the time current was defined. It was thought that positive charges moved to create current, so the current direction is the opposite of the movement of the electrons. Since the definition of current had already become convention around the world, it was never corrected.
      (5 votes)
  • starky tree style avatar for user {Rayeed}^3
    It says that a resistor is measured by watts , like it says if , the power of a resistance is 20 watts , then it converts 20 joules of energy into something else in 1 sec.

    Now my question is why is a resistor measured in watts and not in Ohms ?
    (2 votes)
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    • male robot hal style avatar for user Andrew M
      The resistance of the resistor is not measured in watts, it's measured in Ohms.
      If you want to put a very high voltage, and therefore a lot of current, through a resistor, and you know it's going to get hot (because that's what resistors do, they dissipate energy), then you can imagine that you might need a different version of, say, a 10 ohm resistor to carry 10 amps than you would if you wanted it to carry 100 milliamperes. Therefore there are different power ratings of different resistors. "This is a 10 ohm resistor that can handle 100 Watts" is different from "this is a 10 ohm resistor that can handle half a watt". The first one will be big and fat. The second one will be much smaller.
      (4 votes)
  • orange juice squid orange style avatar for user santhosh prabahar
    In the equation P=(I^2)R, power dissipated is directly proportional to the resistance of the resistor. But in the equation P=(V^2)/R, power dissipated is inversely proportional to the resistance of the resistor. This seems contradicting! How can this contradiction be solved?
    (2 votes)
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  • male robot johnny style avatar for user Gregory
    Why is V2 - V1? Shouldnt it be V1 -V2 since U = Va - Vb? Does it matter?
    (3 votes)
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  • aqualine seedling style avatar for user Priyesha
    Major doubt: if there was no resistance, but the charges won't speed up and nor would they heat up the non existent resistor, where would the potential energy go?
    Please answer asap.
    (2 votes)
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    • male robot hal style avatar for user Charles LaCour
      You have created a nonrealistic situation. If you have an ideal voltage source that you connect its poles with a 0 resistance conductor you will get infinite current. A more realistic voltage source has a limit on its current which is usually depicted as an internal resistance in the voltage source. It is this internal resistance that release the energy as heat.
      (3 votes)
  • blobby green style avatar for user Jujube
    so is V1 and V2 basically an electric potential energy?
    (2 votes)
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Video transcript

- [Instructor] So if you've ever run current through a resistor, you might've noticed that that resistor warms up. And in fact if you run too much current through the resistor that resistor can get so hot it can burn you when you touch it, so you have to be careful. So what I'm saying is when you have current flowing through a resistor it warms up, and we want to explain why. Conceptually, why does current moving through a resistor heat it up, and is there way a to calculate exactly how much that current would heat up that resistor over a given amount of time? There is a way to calculate it. We'll derive that in this video. But first we should just explain conceptually why is it that current moving through a resistor heats up the resistor. And so we'll explain that with this current. Now the way I've got it drawn here, notice that I've got these positive charges. Positive charges don't actually move through a wire, but physicists always pretend like they do because positives moving one way through a wire is equivalent to negatives moving the other way through the wire, and if you use the positive description, even though it's technically incorrect, you don't have to deal with all those negative signs. It's easier to deal with, and it's equivalent, so we may as well use it. But you can go through this whole description with negative charges moving the other way, and it works just as well. You just have to be very careful with the negatives, and it kind of obscures the conceptual meaning 'cause it's hiding behind a bunch of negative signs. So we'll just use these positive charges, but know that it's really negatives going the other way. So why do positive charges flowing through a resistor cause that resistor to heat up? Well here's why. So we know that when current flows through a resistor, there's a voltage across that resistor. In other words, between this point here and this point here, there's a difference in electrical potential. So there's a voltage across that resistor. Technically I'm gonna call it delta V 'cause it's a difference in electric potential. In other words, V on this side, the electric potential on this first side of the resistor is gonna have a different value from the electric potential on this second side of the resistor. So why does this matter? It matters 'cause this final side where the charges end up, is gonna have a lower electrical potential than the beginning. So these positive charges are gonna be moving from a high potential region to a low potential region. And that means they're gonna be changing their potential energy, so these charges have electric potential energy, and if they go from a region of higher electric potential to a region of lower electric potential, they've started with more potential energy, electrical potential energy than they end with. So they're decreasing their electric potential energy. And in case that's confusing, remember that the definition of electric potential is the amount of electric potential energy per charge. So to just put a number in here so it's not so abstract, let's say V two was two joules per coulomb. That would mean for every coulomb of charge at that point V two, there'd be two joules of electric potential energy. And if V one is at a higher electric potential, maybe this is at six joules per coulomb, that would mean over here at this position V one, for every coulomb of charge there'd be six joules of electric potential energy. So as these charges move through the resistor, they're gonna be decreasing their electric potential energy, and so the obvious question is, well where does that energy go? If these charges are decreasing their electric potential energy, where's that potential energy going? My first guess is that they'd increase their kinetic energy because I'd remember then on Earth if you drop a ball and it decreases its potential energy, its gravitational potential energy, we know that when it decreases its gravitational potential energy and falls down, it increases its kinetic energy, it just speeds up. So the decrease in gravitational potential energy just corresponds to an increase in kinetic energy of that object. And so maybe that's happening over here. Maybe as these charges lose potential energy, they speed up, but that can't happen. Remember the current on one side of a resistor has to be the same as the current on the other side. These charges don't speed up. They're losing potential energy, but they don't speed up. This is a little counterintuitive. We're used to things speeding up when they lose potential energy, but these charges aren't going to speed up. What they do is they just heat up the resistor. So as these charges fly through this resistor, they strike the atoms and molecules in this lattice structure of this solid. So this resistor's made out of atoms and molecules, and as these charges flow through here, and again it's really electrons flowing the other way, but as the charges flow through, same idea. They strike the atoms and molecules, they transfer energy into them. And as they pass through in their wake, they leave a resistor that's hotter, at a higher temperature. Which means these atoms are jiggling around more than they were before. And since they're oscillating more than they were before, they're jiggling, they've got more energy, the temperature of this resistor increases. So these charges, rather than keeping all the energy for themselves, they actually just spread it out over that resistor as they pass through, and they spread it out in the form of heat, or thermal energy. And they emerge with basically the same kinetic energy that they started with, so this change in potential energy, electrical potential energy, corresponds to an increase in thermal energy of this resistor. And that's why the resistors heat up. But is there a way to calculate exactly how much this resistor will heat up? How much energy it's gonna gain per time? There is, we just have to use the definition of power. So we know the definition of power is the work per time, or since work is the change in energy, or the energy transferred, we can just write this as the amount of energy this resistor's gaining per time. What we want is a formula that tells us how much energy are these charges depositing in the resistor per time. Well this energy gained by the resistor is coming from the loss of potential energy of these charges. So these charges are losing potential energy. They're losing electric potential energy, and that electric potential energy's turning into thermal energy, so the thermal energy this resistor gains is just equal to the amount of electric potential energy that these charges lose. So I can just rewrite this. I can just say that the power is gonna be equal to the change in electrical potential energy of these charges per time. And so I'll just continue down here. Power's gonna be equal to, how do we find the change in electric potential energy? Well remember, potential is defined to be the potential energy per charge. So that means the electric potential energy is just the charge times the electric potential. So if I want to find delta U, I can just say that that's gonna be U when they emerge, U two minus U one, and this is a way we can find the U values. So the U at two, since it's Q times V, is just gonna be the charge at two times V two. And then the U at one, so we'll do minus the U at one, is the charge at one. But that's the same charge. Whatever charge enters this resistor has to exit it, so it'd be the charge at one times the V at one. This is the change in electric potential energy. So I could rewrite this. I can pull out a common factor of Q in this expression right here. Then we get that the power's going to be equal to this common factor of Q times V two minus V one. So that's delta U, and that's what we're plugging in right here for delta U. Delta U is just the difference in these Q times V values. And then we still have to divide by time since we're talking about a power. But what is V two minus V one? That's simply the voltage across this resistor. Delta V is V two minus V one. So I could rewrite this. I could just say that this is Q times delta V, the voltage, across that resistor, divided by the time it took for the charge to pass through that resistor. And now something magical happens, check this out. So we got power equals, I've got charge the past through the resistor, divided by the time that it took for that charge to pass through the resistor, but charge per time is just the definition of current. So we get this beautiful formula. If I just factor out this Q over T, I get Q over T times delta V, the voltage across the resistor. But Q over T is just the current, so I get that the power is going to equal the current through that resistor times the voltage across that resistor. And this is the formula for electrical power. This tells you how many joules of thermal energy are being created in that resistor per time. So the units are joules per second, or in other words, watts. Because what this is telling you is the amount of thermal energy generated per second. If the power value came out to be 20 watts, that would mean there'd be 20 joules of thermal energy generated every second, which is a really useful thing to know. This formula's extremely useful when you want to figure out how much energy's gonna be used by a light bulb, or a toaster, or a TV, or whatever electronic device you want to use. This tells you how much energy that it's going to turn into either thermal energy, or light, or sound, or whatever other kind of energy that it's converting that electric potential energy into. 'Cause notice, we never really assumed that this was thermal energy. We just called it E, and then we said that whatever energy got transformed came from the change in potential energy of those charges, and that's always gonna be true. It's always gonna be electric potential energy converting into something, whether it's heat, or light, or sound. So this doesn't just work for resistors. It works for almost all electrical components that turn electric potential energy into some other kind of energy. And you could rewrite this in different forms. Sometimes you'll see it a different way. There's this form, and then I'm just gonna copy this. I'm gonna show you there's a couple other forms you could see this in. Let me clean this up, we'll put this formula down here. This gives us the power, but we know from Ohm's law that delta V is equal to I R. So if both of these formulas are true, I can plug delta V as I R. So I could take this I R, I could plug it in for delta V, and I get an alternate expression for the power used by electrical component. I get that the power's gonna be I times I times R, which is just I squared times R. So this one might be more useful if you've got a situation where you don't know the voltage, but you happen to know the current and the resistance. But there's one more. I could've solved this Ohm's law formula for I, and I'd get that I equals delta V over R. And then if I plug delta V over R in for I up here, I'd get an alternate expression for the power. I'd get that the power equals delta V over R times delta V. It's just gonna be delta V squared. The voltage across that resistor squared divided by the resistance of that resistor. And this formula might be more useful if you know the voltages and the resistance, but you don't know the current. So depending on what you know, these can get you the power used by a resistor. They're all equivalent. They will all give you the correct and the same power. It's just a matter of what's more convenient for the actual problem that you're dealing with. So recapping, when current passes through a resistor, it converts electrical potential energy into thermal energy, and you can calculate the amount of electrical potential energy converted per second using current times voltage, current squared times the resistance, or voltage squared divided by the resistance.