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I P=U*I: Why do electrons drop ALL energy e*U in the circuit?

  1. Oct 30, 2016 #1
    Hi.

    Consider a simple circuit consisting of a voltage source ##U## and a load with resistance ##R##, e.g. a lamp or a motor. The current is given by ##I=U/R##. The number of electrons passing the circuit per second is ##n=I/e##. The power consumed by the load is calculated by
    $$P=U\cdot I=U\cdot e\cdot n=\Delta E\cdot n\enspace,$$
    where ##\Delta E=U\cdot e## is the (kinetic) energy an electron would gain travelling from the negative to the positive pole of the power source if there was no load.

    In this computation, we assume the electron gives all its energy to the load and has kinetic energy zero when it arrives at the plus pole of the power source. But why is that? Why can't the electron maybe only lose half its energy to the load and still have kinetic energy when it enters the battery?
     
  2. jcsd
  3. Oct 30, 2016 #2

    Dale

    Staff: Mentor

    The kinetic energy of the electrons is pretty much irrelevant for any device other than a particle accelerator.

    However, in a bigger picture view of the question, you should not think of energy being stored in individual electrons to be dropped off elsewhere later. The energy is in the fields.
     
    Last edited: Oct 30, 2016
  4. Oct 30, 2016 #3
    But how does the electron know this? Why does it "think" while passing the lamp "I need to give all my energy to the filament such that I arrive at the battery at rest"?
     
  5. Oct 30, 2016 #4

    Dale

    Staff: Mentor

    The electron doesn't know that. Neither does the electron carry energy to give to a load. The fields carry the energy.
     
  6. Oct 30, 2016 #5

    Dale

    Staff: Mentor

    Consider what happens when you turn on a light switch. The change in the fields moves at a little less than the speed of light. By contrast, the electrons move at about a mm/s. How quickly does the energy get from the source to the light? Is it something that happens nearly at the speed of light or closer to a mm/s?
     
  7. Nov 1, 2016 #6
    Ok, then I think I need a different derivation of ##P=U\cdot I##. Because the equation I wrote in #1 starting from the right is basically the derivation I learnt in school, i.e. looking at the energy difference of an electron (or any other current carrier) between the poles of a voltage source and then counting how many of them pass the circuit per second.

    How can ##P=U\cdot I## be derived alternatively without looking at individual charges and make assumptions about how much energy they drop in the circuit?
     
  8. Nov 1, 2016 #7

    Dale

    Staff: Mentor

    In terms of circuit theory it should not be derived, it should simply be defined. Then with KVL and KCL you can show that it leads to energy conservation.

    The place where you would derive it would be in electromagnetism, after Poynting's theorem is introduced. I like the treatment here, in chapter 11, especially 11.3

    http://web.mit.edu/6.013_book/www/book.html
     
  9. Nov 6, 2016 #8

    David Lewis

    User Avatar
    Gold Member

    The current drawn by a motor will be less than V/R unless the motor shaft is prevented from turning. Also, the voltage across the load should be used, not the voltage of the source.
     
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