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Ohm's Law, Circuits, and Wires confusion

  1. Jul 12, 2014 #1
    I can do circuit problems, but conceptually I cannot figure out what is going on from a basic physics perspective, with respect to the following:

    Say you have a circuit with a battery (emf = V) and a resistor (resistance = R).

    Then, electrons on one side of the resistor are in a constant potential, and electrons on the other side of the resistor are at a different constant potential. Since E = grad V, there is no force on the electrons in the perfectly conducting wire (on either side of the resistor). We know the battery is supplying an EMF and that the electrons are losing thermal energy through the resistor. Yet they cannot lost any kinetic energy in the direction of motion, since the current has to be the same on both sides of the wire. So my understanding is that just after the battery is turned on, electrons flow unimpeded to the resistor, then they slow down as they encounter resistance. That then slows down the electrons behind them (Coulomb interaction) and speeds up the electrons in front of them. This ends up working out to a constant velocity.

    But with all these electric interactions in play, I don't see how the electrons in the top wire are actually at a constant potential V from the battery relative to the electrons in the bottom wire, since they are already being slowed due to the resistor. I am phrasing this poorly but I am very confused.

    Thanks!
     
  2. jcsd
  3. Jul 12, 2014 #2

    phinds

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    Electron flow is exactly like a bicycle chain. It flows in all parts of a circuit equally at all times. Any slowing or speeding of the flow happens throughout, not localized as you are positing.
     
  4. Jul 12, 2014 #3
    I guess I am asking *why* that is the case. What is physically happening, in this idealized process? (In the wires and in the resistor--I don't really care how the emf is generated).
     
  5. Jul 12, 2014 #4

    phinds

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    Ah, well that's a question for one of our many physicists. I'm an engineer. I can tell you a few things about how stuff works at a macro level, but not at the level you're interested in.
     
  6. Jul 12, 2014 #5

    UltrafastPED

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    Resistance is easy to analyze for a circuit, but the physics is not obvious!i

    The most important thing to recognize is that while the individual electrons are travelling at quite high speeds,their directions are essentially random; the imposed EMF generates a current is due to a "drift velocity", which is quite slow. I wrote up the details for a semi-classical analysis for an electrical engineering circuits class; see: https://www.physicsforums.com/showthread.php?t=740567#15

    A more complex (and accurate!) analysis requires quantum mechanics, but you can study that when you take a course in condensed matter physics.
     
  7. Jul 13, 2014 #6
    This is roughly true, but not exactly. If you model the wire as a continuous chain of very small resistors, does that answer you question?
     
  8. Jul 13, 2014 #7
    Thanks for these replies!

    rigetFrog, if I think of it that way, then it does make sense.

    UltrafastPED, that helps too. What confuses me is this (and I couldn't figure out the answer from your pdf):


    1. Electrons drift through the wire from the battery to the resistor at velocity vd. They are in a region of constant potential, so they experience no force.

    2. The electrons then drift through the resistor at velocity vd. Here they are in a region of non-constant potential, so they experience a force, but they still will not accelerate, merely continue at velocity vd. However, they have lost energy (=V/q) each. That has to be thermal energy, since their kinetic energy is unchanged.

    3. The electrons drift back to the negative terminal of the battery at velocity vd.

    Is that a correct description?

    If not, where did I go awry? If yes, how do the electrons in (1) and (3) "know" to move at velocity vd, which is effectively determined only by (2).
     
  9. Jul 13, 2014 #8

    davenn

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    Hi abcdcba

    firstly,
    point 3 is incorrect

    electrons LEAVE the negative terminal of the battery and drift through the circuit

    Note again what rigetFrog said to you

    The whole wire has resistance ( ie. it is a resistor) it just may not have the same amount of resistance of fixed lumped resistor placed somewhere in the circuit.

    cheers
    Dave
     
  10. Jul 14, 2014 #9
    The sign error was carelessness on my part only. And I get what you are saying about different materials, etc., but am still thrown off in my understanding of the current. It has to be constant throughout the circuit or else we would be violating conservation of charge, or charge would be building up. Can charge not build up because that is not how the material properties of conductors/insulators work? Or is there some better explanation?
     
  11. Jul 14, 2014 #10

    Delta²

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    There is an initial stage before the steady state (steady state=state in which the velocity Vd is constant) during which the velocity Vd is "determined". During this stage the electrons in (1) and (3) dont know how fast or slow they should move so they arrive at the start of the resistor or leave from the end of resistor with different speeds, and in general the velocity of the electrons is not equal throughout the circuit. This has as a result a charge build up throughout the whole circuit. Only when this charge buildup has such a distribution that the potential difference across the ends of the resistor is V, and the potential in the perfectly conducting wires is equal across their length (so that the electrons do not accelerate or deccelerate while traveling inside them) then the circuit comes to steady state. (Potential depends to charge distribution, this follows from Gauss's law).
     
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