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Impossible creation of potential energy

  1. Mar 1, 2004 #1
    Hi folks,

    Sorry for this second post so quick but its bugging me! I have been reconsidering some fundamentals of electrics and have opened a can of worms for myself (thanks Cliff for closing the lid on the previous one). If i consider again a simple closed DC circuit consisting of a wire and a pd applied across it i.e:-


    If i look upon the circuit as above (i know this is an unconventional view but it better illustrates my dilema), a steady current flows in the direction indicated. By definition each coulomb of charge leaving the +10V end of the wire will have lost 10 joules of energy by the time it reaches the 0V end of the wire. If we imagine the wire itself as series of resistors then we can label the pd along the wire as:


    Imagine a coulomb of charge has travelled all the way to were the pd is 1V, we say it has lost 9 joules and can only possibly lose 1 more before being raised to +10 again by the source.
    Now imagine i instantaneously insert a high resistance between the coulomb and the 0V end of the wire. The pd across this resistor would now be tending towards 10V and the coulomb would have to cross this before getting to 0V. This means that the coulomb has lost more than the possible 10V in potential energy???? What am i doing wrong??

    Thanks in advance
  2. jcsd
  3. Mar 1, 2004 #2


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    Voltage is not energy.

    Voltage is essentially constant along a wire.

    No power source can provide its rated voltage through a short (any power supply has an internal resistance that prevents it).

    Transients are transient.

    Your example is of one big, long resistor (such as a light bulb filament), which dissipates energy all along its length.
    Last edited: Mar 1, 2004
  4. Mar 1, 2004 #3


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    The potential drop along the wire is not directly connected to electrons "traveling" the distance, see my post in the other thread. The key word here is POTENTIAL difference. Electrons at the 1v location feel a 1 volt potential those at the 5V point feel a 5V potential. It does not matter where the electrons started it only matters where they are with respect to the total resistance of the circuit. Does this help understanding what happens when you insert a different resistance?
    Last edited: Mar 1, 2004
  5. Mar 2, 2004 #4
    russ_watters: Thanks for the reply. I'm slightly confused when you say that voltage is essentially constant along a wire. What is a piece of wire if not one big long resistor/series of resistors. If its the internal resistance that provides the problem then if we use a 'wire' that is sufficiently high that the internal resistance is negligible would you still say that in the closed circuit with current flowing that the voltage is still essentially constant along a wire?

    Integral: I think i understand were you are coming from, do you mean that by inserting the resistance you simply 're-arrange' the pd across the circuit so you effectively raise the pd of the coulomb again which it can then lose again??
    I keep trying to compare the pd with gravitational potential energy as this is easier to imagine. Could i say that the equivalent of the problem would be to let a ball drop 9m from a height e.g. 10m. and the act of inserting the resistance would be equivalent to me raising the ball again and letting it drop more slowly??
  6. Mar 2, 2004 #5


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    Physical analogies are a tough cookie, its difficult to create a good model that encompasses all the nuances that make up the term 'electricity' without leaving something out. Its surprising how many people misunderstand electricity by taking the "electricity flows like water" metaphor too far and out of context. That said, I'm going to take a stab at your question and analogy.

    If we were to move these experiments in the physical world to water and not air as the medium where the object is traveling, bouyancy would need to be factored in as well as the fluid dynamics of our object traveling through the water to account for losses but potential energy is calculated as the same even if the results are far different.

    Think of it like this: potential energy is calculated with gravity as the source. Well, the battery/power supply here needs to be the source. Normally, we could consider this constant. You're roughly treating the current flow as if it were a changable mass in your PE calculations which isn't very analogous.

    The wire as a dead short, translated loosely into physical terms, could be the center of a black hole or your favorite sub-atomic particle only 10m from the surface of the earth. With either, PE will be secondary to very large 'factors' changing the outcome. :smile: Back to reality, a dead short will do little besides heat up a wire and cook the battery (because the battery is flowing a lot of current across its own internal resistance). The wire will not show a drop from 10V to 0V across it, probably more like 1V or 2V depending on your battery. Regardless, it would be similar to non-linear behavior since other factors take precedence.

    Lets just use a series of resistors to keep things in a linear range. If you suddenly 'swap' a high resistance in place of a lower one, you see a sudden drop in current flow. Changing the current flowing in the circuit doesn't change the voltage at the battery anymore than changing the mass in your PE equasions changes the constant acceleration you use for gravity. See where I'm going with this line of thinking?


    P.S. Someone please correct me if my analogy has too many holes in it.
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