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Electrons and energy expended in a circuit

  1. Dec 8, 2015 #1
    Now, this is probably a very noob kind of question, but I am just going to say I just don't know how to answer this following series of questions.

    If we have a circuit, with a battery and bulb and the wires connected per normal, how do we physically explain the expenditure of energy in terms of the electrons of the current? Is the energy transferred from the electrons to the filaments in the bulb? If so, what kind of energy is expended? Kinetic energy of the electrons? If so, why doesn't the electrons reduce speed? Then would not the current be reduced? I remember playing around with a electric circuit simulator that seemed to imply that the electrons were "pushing" against one another...but I would like to have a more concise (mathematically, if possible) understanding of this conceptual problem I have.

    I understand current as being an average of the electrons passing through a cross-sectional area of the wire per unit of time....I hope that this is not the reason that I cannot comprehend this problem.
     
  2. jcsd
  3. Dec 8, 2015 #2

    cnh1995

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    Homework Helper

    If you refer some good physics book (e.g Halliday-Resnick), you'll come across the mechanism of electrical conduction in the conductors. In brief, when current flows, the electrons move in zigzag manner, continuously colliding with the vibrating lattices. However, the resultant or net velocity of the electrons is almost constant (That needs a pictorial explanation.) Due to collision, a part of their kinetic energy is converted into heat (like friction in mechanical analogy). Now this "part" will be dependent on resistance 'R' of the material and kinetic energy of electrons i.e. it will be proportional to square of velocity (v2). Now velocity v will be proportional to current I flowing through the wire. So, according to Joule's law, heat generated in the wire (or bulb) will be,
    H=I2Rt, where t is the time for which the current is passed.
     
  4. Dec 8, 2015 #3

    jbriggs444

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    Cnh1995 does not explicitly mention it, but there is an electric field throughout the filament. The electrons lose energy as they interact with the lattice. They gain energy as they move through and are propelled by the field. That is how they can maintain a constant average drift velocity even though they are delivering energy to the lattice.

    If you chase this back, the electric field is there because the battery imposes a fixed potential difference between one end of the circuit and the other. The charges on the wires, filaments, resisters and what not will automatically adjust themselves so as to distribute that potential into an equilibrium such that current flow is constant throughout the circuit. [If the current flow were not constant, charge would build up somewhere. The result would be an electric field opposing the build-up. This works to automatically restore an equilibrium]

    The electric field produced by a small amount of charge is very large. It takes only a tiny redistribution of charge to attain an equilibrium electric field. Because it is so tiny, this redistribution may not be mentioned in introductory materials. Instead, one is expected to take the behavior of electrical components and circuits as a given.
     
  5. Dec 8, 2015 #4

    cnh1995

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    Right. I forgot to mention this key part. Thanks a lot..:smile:
     
  6. Dec 8, 2015 #5
    Thanks so much for the replies jbriggs444 and cnh1995! I still don't understand this fully, but I now know what is lacking in my understanding...thanks again!
     
  7. Dec 8, 2015 #6

    Dale

    Staff: Mentor

    Just to add to what has already been said, there are basically three things to consider when looking at energy in EM: energy density, energy flow, and work. In a DC circuit the energy density is constant in time, so only the flow and work is important.

    In a circuit like you describe the energy of the battery does "work" on the fields, the fields transport energy to the filament, and the energy of the fields then do work on the filament. Once the EM energy has done work on one part of the matter, like the electrons, then it can be transfered to other parts of the matter, like the lattice, as described above by the others.
     
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