Potential Energy of a Battery

Discussion in 'General Electronics Chat' started by sjgallagher2, Apr 29, 2013.

  1. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    So I just learned that the potential energy of an object U is the energy due to the position of the body or the arrangement of the particles, measured in Joules here.

    Voltage is E = U/q where U is potential energy and q is charge in coulombs. Knowing this, apparently the potential energy between the two ends of a 1.5V battery can be found like so:
    1.5V = U/1.602x10^-19 = 2.4x10^-19J
    So hold on a sec, does that mean that the battery has a potential energy of 2.4x10^-19? Like, no matter what resistance, or work is being performed, the potential energy of a 1.5V source will always be that? When talking about potential energy in a circuit, what exactly are we measuring? The wikipedia definition above isn't doing anything for me, and my book doesn't explain potential energy, just voltage.
  2. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    Oh, apparently there's a thing called electric potential energy, looking into that. Any help is still welcome, and encouraged
  3. hexreader

    Active Member

    Apr 16, 2011
    You seem to be assuming that your battery contains a single electron (or proton).

    ... or at least, has a charge difference equivalent to one electron charge.
  4. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    This was just an example from the book Practical Electronics for Inventors and it goes like this:
    To be honest, none of that makes too much sense to me, I think I have a poor mental concept of this potential energy business :O
  5. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    Sorry to bombard with questions, is 1 coulomb a measure of the amount of electrons on a surface? Like, 1C = 6.28x10^18 electrons on a given surface? And since I have one electron that would mean the charge is 1.602x10^-19 C, or 1 electron. Then voltage between this electron and that one is E=U/q. Now why in the above passage is the q only one electron? Why not two, or the number of electrons between the ends, or anything... if that makes sense
  6. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    Okay, so: I don't understand the base example much, but if the charge is that of 1 electron (positive for examples sake) then that means that I=1.602x10^-19/t and in one second that's 1.602X10^-19A? But in a real circuit with a current of say 0.1A, the charge is: 0.1 = C/1 = 0.1C or 6.28x10^20 electrons! That might not have value since I just need the charge huh. So, snap 0.1C into V=U/q and you get V=U/0.1
    Now if the voltage of the battery is 1.5V and it's producing 0.1A then we get: 1.5=U/0.1, or 0.15J (per second). Is that right? Well let me check the book. YES it is, hell yes, okay thanks hexreader for your input, I've solved this problem I think. Before the battery was producing an extremely small currrent, gotcha. :)
    Last edited: Apr 29, 2013
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
    It would actually be true to say that a battery has chemical potential energy.

    Probably the best way to regards potential energy of any sort is to think of it as energy that is not be used or expended at the moment, but can be accessed by some means.

    A battery works by converting the energy in a chemical reaction into electrical energy.

    Electrical energy may or may not be potential energy depending upon circumstances.

    Energy is often called 'the capacity to do mechanical work' This means it is changed into another form and is then no longer potential.

    Electrical energy may be transformed into heat energy by passing electrical current through a resistance.

    It may be transformed into mechanical energy (work) in an electric motor.

    There are, of course, mathematical equations to describe precisely how much energy is involved in each case. But I suggest you get the underlying ideas straight first.

    Does this help?
    Last edited: Apr 30, 2013
    screen1988 likes this.
  8. WBahn


    Mar 31, 2012
    Note what this is (and is not) saying. It is saying that a voltage (which is ALWAYS a measure of potential energy between two points) is a measure of the potential energy PER UNIT CHARGE done on that charge as it travels between those two points.

    Huh? If you have x = y = z, then isn't x = z?

    Get over the sloppy math!

    Tracking your units would also help.

    1.5V = U/-1.602x10^-19C/e-

    U = 1.5V*-1.602x10^-19C/e- = -2.4x10^-19J/e-

    Since 1V = 1J/C

    Thus, in moving from the positive terminal to the negative terminal of a battery, an electron has -2.4x10^-19J of work done on it. Since this is negative, the work is actually being done on the battery increasing it's potential energy. But electrons actually flow from negative to positive and so the work done on EACH electron flowing in that direction is 2.4x10^-19J.

    You are partly correct. The amount of work done PER ELECTRON is 2.4x10^-19J for a 1.5V voltage source now matter what resistance is involved. But when you say "or work", you are wrong because the 2.4x10^-19J IS the work that is performed.

    We are measuring the potential to perform work, where work is a force through a distance. In this case, it is the force needed to move an electron from one terminal of the battery to the other integrated (summed) over the distance travelled by the electron.
    screen1988 likes this.