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Why is power volts * current?

  1. Apr 15, 2014 #1
    I don't have an intuitive understanding of this. To me it seems that
    power should be related to just current, since this is the measurement
    of how the charge is flowing. Volts and current are inextricably related,
    true? So then why P=VI instead of P=I?

    Thank you
  2. jcsd
  3. Apr 15, 2014 #2
    I find it useful to analyse the dimensions (units) of these things.

    Power is measured in Joules per second. [itex]\frac{J}{s}[/itex]

    Volts is measured in Joules per coulomb. [itex]\frac{J}{C}[/itex]

    Current is measured in Coulombs per second [itex]\frac{C}{s}[/itex]

    [itex]\frac{J}{C}[/itex] x [itex]\frac{C}{s}[/itex] = [itex]\frac{J*C}{C*s}[/itex] = [itex]\frac{J}{s}[/itex] which are the same units as power!
  4. Apr 15, 2014 #3


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    This is puzzling. If P=I, why would we invent another name to designate the same thing?

    And this is before we point out that I Is current, while P is power. They are not the same animal. So why would you think they are equal to each other?

  5. Apr 15, 2014 #4


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    It's analogous to a classical water wheel. The same amount of water flowing through the wheel at the same rate will do more work and deliver more power if it drops through a greater height than a smaller height; the height difference is analogous to the voltage drop.

    In fact, the English words that we use to describe electricity ("current" and voltage "drop") come from this analogy. The physics of water flow was well understood by the scientists studying electrical phenomena.
  6. Apr 15, 2014 #5


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    Power is ##\frac{E}{t}##
    Voltage is ##\frac{E}{Q}##
    Current is ##\frac{Q}{t}##
    So ##\frac{E}{Q} \times \frac{Q}{t}## gives ##\frac{E}{t}## which is equal to power.
    EDIT:BOAS beat me to it
  7. Apr 15, 2014 #6
    What about the power generated by a waterfall, say on a turbine at the bottom. Is it dependent only on the mass of water which flows over the top in a given time?

    EDIT: Got in very late
  8. Apr 15, 2014 #7
    Thank you for the helpful answers. I can see the mathematical relationship
    thanks to your formulas, but I will still strive to obtain an intuitive understanding
    if you will be patient with me.

    Dropping water from a greater height results in more velocity when it hits the wheel. For the analogy, increasing current, dQ/dt, increases the power. What does the voltage have to do with it?
  9. Apr 15, 2014 #8


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    Voltage is a measure of potential energy; if you multiply the voltage by the amount of charge it is exactly the potential energy of that amount of charge, and is equal to the amount of work that can be done.

    The current is charge per second ... so voltage x current = (voltage x charge)/time = work/time = power.
  10. Apr 15, 2014 #9
    Using the waterfall analogy you can view voltage as the height of the fall, while current as the amount of water. I.e. one drop wouldn't cause much power even if it fell from very high (low current, high voltage) and the same could be said for a large amount of water that merely falls from very low height (high current, low voltage). But when you have a large amount of water falling from high (high current, high voltage) on the other hand, this will create a large amount of power. Hope this helps!
  11. Apr 15, 2014 #10


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    Not necessarily. In the most efficient waterwheel designs (turbines are subtly different, but that's an irrelevant digression here) the top of the wheel is level with the water behind the dam, so the water doesn't speed up before it hits the wheel. Instead, the force on the wheel comes completely from the weight of the water sitting on its buckets, and the water is moving at a constant speed determined by the speed of the wheel, throughout its entire drop.

    In this case, if a given mass flows through the water wheel each second, the amount of work done in that second will be that mass times the drop (##W=Fd##). Double the drop, keep the flow the same and I'll get twice the work out per unit time. And...

    When I speak of "a given mass flows each second", I'm talking about the quantity dM/dt where M is the amount (mass) of the water, and that is exactly analogous to your dQ/dt where Q is the amount of charge.

    Note that I can get the same dM/dt by moving a large mass of water slowly or a smaller mass faster - focusing on the speed instead of the total amount moved per unit time will just confuse you. In the electric current case, you'll notice that we almost never need to think about the speed of the charge carriers, although I can get the same dQ/dt by moving a small number of charge carriers through the circuit very quickly, or a larger number more slowly.
  12. Apr 16, 2014 #11


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    Let's do some math and I think you'll have a better understanding.

    As you've said, power is P=IV. So let's consider two examples.

    First, consider a circuit where the voltage is 10 volts and current is 1 amp. Per Ohm's law this requires a resistance of 10 ohms. The power is 10 watts.

    Next, a circuit where the voltage is 20 volts and the current is still 1 amp. Again, using Ohm's law gives us a resistance of 20 ohms. The power is 20 watts.

    So in the 1st example we have 1 amp flowing through 10 ohms of resistance while in the 2nd we have 1 amp flowing through 20 ohms of resistance. In order to force the same amount of current through twice the resistance we are required to perform twice the work on each charge, requiring twice the voltage to do so.

    Finally, remember that power is a rate of energy transfer. In our circuits, all of this energy is being converted to heat, so when the power is doubled by increasing the voltage, we can heat something up twice as fast.

    To use the water analogy, imagine a water powered winch lifting a 10 kg object. Doubling the pressure of the water (analogous to voltage) will let you lift a 20 kg object at the same speed as the 10 kg.
  13. Apr 16, 2014 #12


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    Voltage is an electric potential. and a change in voltage between two points in a circuit or two points in a field is related to the intensity of the electrical field which is the force per unit charge (note the force is usually position dependent, with a few exceptions like the field between two charged plates or the ideal field from an infinite plane or disc).

    Current is the amount of charge that flows past a point per unit time.

    netwon = N = kg m / s^2
    volt V = (kg m^2 / A s^3)
    ampere A = 1 coulomb of charge flow across a point / second

    So voltage x current is similar to force x speed, (kg m^2 / A s^3) x (A) = N (m/s) = force x speed, and power = force x speed, in this case (kg m^2 / s^3) = N (m/s) = watt.
  14. Apr 19, 2014 #13
    Thank you for all the answers
  15. Apr 19, 2014 #14
    ^^ This along with you username leads me to believe that your main issue is an unholy love of current.
  16. Apr 29, 2014 #15
    I am still confused. How can different amounts of potential energy be associated with the same amount of charge? Voltage does not increase the *velocity* of the charge, right? So why does dQ/dt have more potential energy when more voltage was applied to induce the dQ/dt, when the velocity of the charges nor the amount of charge hasn't changed?

    Voltage and current are inextricable by ohm's law. You change one, you change the other. So how can they be multiplied to produce power.

    I am really not getting it.

    Edit: Maybe my biggest problem is that I don't understand how a charge can have a variable amount of energy
  17. Apr 29, 2014 #16


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    How can the same amount of mass or volume have different amounts of potential energy?
    One thing doesn't have anything to do with the other. Similarly, the force you apply to an object affects how far it goes (per unit time), yet we still multiply force and distance to get energy.
  18. Apr 29, 2014 #17
    For mass and volume, when you increase their potential energy, the result of transforming that into kinetic energy is higher velocity. Or the amount of mass or volume was increased to give more kinetic energy. Charge has only one velocity right? And from what I gather from this thread, the same exact amount of charge can have two different amounts of energy.
  19. Apr 29, 2014 #18


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    No. Electric charges can have any velocity from 0 up to (but less than) the speed of light.

    Either you are confusing electric charge with electromagnetic radiation, or you have the wrong idea that when electric current flows in a wire, the electrons move very fast. Actually, for "normal" currents in copper wires, the electrons only move at about 1 meter/second, or slower.But if you push an electron into one end of a wire, a different electron pops out of the other end almost instantaneously - just like if one drop of water flows into a long pipe, a different drop of water comes out of the other end almost instantaneously, independent of the speed of the water flowing in the pipe.
  20. Apr 29, 2014 #19


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    Voltage is a difference in electric potential. Electric potential is the electric potential energy a point charge would have at that location in space. Two points at the same electric potential have no voltage between them and electric charges at these points would experience no force.

    Consider a negative test charge located halfway between a positive and negative electrode. The test charge will be repelled by the negative electrode and attracted by the positive electrode. This means that it has potential energy and will experience a force that accelerates it, performing work on it and transforming the potential energy into kinetic energy. In other words, there is a difference in electric potential between the point where the test charge is at, and a point closer to or further away from either of the electrodes. Closer to the positive electrode, a test charge will not have as much potential energy since it can't accelerate for very long before colliding with the electrode. When placed near the negative electrode, a test charge will have a much greater distance to accelerate and will reach a much higher velocity before colliding with the positive electrode and thus has a greater amount of potential energy. The amount of potential energy the test charge has is known as electric potential, and the difference between the electric potential between two points is known as the voltage. (Electric potential is actually a little more complicated, but for this example all you need to know is that it is a measure of potential energy. I'm mentioning this because I want you to know there is a reason "electric potential" is being used instead of just going with "electric potential energy")

    If the electrodes have a higher negative and positive charge, then our test charge will experience a greater force, meaning that it has more potential energy than when the electrodes were at a lower charge. It will be accelerated faster than before, reach a higher velocity, and have more kinetic energy when it impacts the positive electrode. As you can see, this greater force results in more potential energy, which then results in a greater electric potential (electric potential energy) and a greater difference in electric potential between two points, aka voltage.

    If we take these concepts and apply them to a circuit, we see that a higher voltage causes the charges to have more energy. Since we are constantly applying a force to the charges, the charges are constantly being accelerated. The difference lies in the fact that charges in a conductor aren't able to simply accelerate to any arbitrary speed. They constantly interact with ions and other electrons and transfer some of their energy through collisions, magnetic forces, etc. In other words, applying a voltage performs work on the charges, which in turn perform work on the circuit as a whole by heating it up and interacting with the load.

    A higher voltage, aka a greater difference in electric potential, means that the force applied to the charges is greater and thus more work is performed on the charges themselves and by the charges on the circuit. The amount of work performed can be found by multiplying the voltage by the current, since the current is the number of charges and the voltage is the force applied to them.

    Think back to our test charge example. If we double the voltage then the force on the test charge is twice what it was before. Twice the force means twice the work performed on the charge (Since work = force x displacement). But remember that power is work over time. Our test charge, experiencing twice the force, undergoes twice the acceleration, meaning that it takes less time to reach the positive electrode. Since we did more work AND took less time, the power isn't simply twice what it was before, it's actually four times what it was before.

    So if we take a circuit with a set resistance, then doubling the voltage doubles the current and the total power quadruples.

    Make sense?
  21. Apr 29, 2014 #20


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    Forgot to add:

    In a conductor we are dealing with a great many mobile charges, not just one. Applying a voltage causes these charges to move. Each charge takes a certain amount of energy to move a set distance through a set resistance. Increasing the voltage causes the charges to move further and faster.

    Consider a circuit with 1 ohm of resistance and 1 volt applied to it. The current is 1 amp, or 6x1018 charges per second passing through a point in the circuit. Each charge is constantly having work performed on it, and in turn performing work on the circuit and losing whatever energy it gained as heat. Power is work over time, and since amps is measured in charges per second, power is the work done to these charges in that second. In this case total power is 1 watt.

    Now, let's increase the voltage to 2 volts. Current is now 2 amps, or 12x1018 charges passing through a point in the circuit per second. We can think of this as each charge passing through twice the distance (twice the displacement) than before.

    So each charge has double the force applied over twice the displacement, meaning that total work on each charge is quadruple what it was before. (Remember, work = force x displacement) Since this work took place over 1 second, power is also 4x what it was with 1 volt and we end up with 4 watts of power.
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