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Different emf across the same resistance

  1. Jun 1, 2008 #1

    I'm having some difficulties understanding some elementary physics concept.

    When different emf is supplied to a resistor of the same resistance, the p.d. across changes in different scenarios. By definition, p.d. is the amount of electrical energy converted to other forms of energy per unit charge. Why does the same resistance converts different amounts of energy? Can anyone kindly come up with an analogy? Because I can't find any explanation to this. And why does the charge-carriers want to lose all the energy given to them by the source?
  2. jcsd
  3. Jun 1, 2008 #2
    The same resistance dissipates different electric energies because the electric currents are different. Haven't studied Ohm's law?

    [tex]\Delta V = R*i[/tex]

    i is the current.

    It's not that the charge-carriers want to lose all the energy given to them by the source, it's that they lose it because of the material' resistance: this acts slowing down to a constant (average) speed the charge-carriers put in motion from the electric field. At constant speed no work is done on them, so all the electric work have to go to the conductor (as heat) and you have completely dissipated the electric energy.
  4. Jun 2, 2008 #3
    A lager emf will maintain a larger pd across the resistor. This means that the decrease in electrical potential energy (EPE) of the charge carriers is greater (as they pass through the resistor), and so by conservation of energy there is a greater energy transfer (to internal thermal energy) to the resistor. This also answers your last point - the charge carriers lose EPE via the potential drop and this lost EPE appears as thermal energy.
  5. Jun 11, 2008 #4
    My problem is, why must all the emf supplied by different cells be used to maintain a larger pd across the resistor? Why can't the pd across a resistor remain constant while you change the emf cell?

    I do know the actual facts; it's that I somehow can't seem to accept them as I can hardly find any analogies that can satisfy me..

    Thanks for any replies.
  6. Jun 11, 2008 #5
    The pd (V) across a resistor is set and maintained by the emf source *not* the resistor. Imagine a simple circuit consisting of just an emf source and a resistor (R). A given emf (E) will produce a certain pd (V) across R. The value of R will then determine the current (I). If you increase the emf, V *must* also increase (by conservation of energy) and so too will I.
    The key point is the first sentence. Hope this helps :smile:
  7. Jun 14, 2008 #6
    "p.d." = potential difference, a measure of voltage?
    "p.d." = power disipation, a measure of power?
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