I saw in a documentary that given a circuit, the electron comes out of the negative side of the battery with some electrical potential energy and upon passing through a resistor it gives up all its energy to the resistor, thus transferring energy from the battery to the resistor. Upon giving all its energy up it returns back to the positive terminal and the process starts all over again. My question is that when the electron as given up all its energy to the resistor, should it not have some energy for it to travel back to the positive terminal? I understand that the said explanation is highly simplistic, but even so, my textbooks on circuit analysis also more or less use the same analogy and go on to state that all the energy of the battery is converted to work. Could someone clarify how this energy distribution is understood and what assumptions are in place.

PS: I also read about lumped matter abstraction and didn't quite get it. Further if we were to remove the resistor, wouldn't the electron travel back to the positive terminal with no loss of energy? I don't think that is even possible.

Pls give a detailed explanation. This is giving me nightmares and I can't seem to progress.

  • $\begingroup$ The energy is not in kinetic energy of electrons but in potential energy of the electric field. The force of this field on the electron's charge accelerates electrons only for microscopic distance before they collide with atoms in the resistor, which converts their kinetic energy into heat. If there is no resistor in the circuit and the electrons are allowed to move in vacuum, then all of the potential energy would, indeed, be converted into kinetic energy. The electrons would then collide with the positive electrode, which would get hot and, at high enough voltage, would release x-rays. $\endgroup$ – CuriousOne Jan 21 '16 at 9:14

You don't need to supply any extra energy to return electron to + ve plate . Electron already have enough KE even after passing all the acquired energy by battery to resistor. Never forget that actual speed of elections is very huge. When these electrons were in battery they already had enough KE. When you applied resistor it just increased its energy. Electrons now flow in wire with huge velocity, but then comes resistance which reduce its speed to average drift velocity. They interact with ions and loose their energy gained by electric field.But they still have huge KE. Now, through diffusion they easily reach positive plate.

  • $\begingroup$ The velocity of electrons in conductors is very small (fractions of a mm/s). $\endgroup$ – CuriousOne Jan 21 '16 at 9:15
  • $\begingroup$ No, that's drift veolcity . Actual velocity is 3/2kt. $\endgroup$ – Anubhav Goel Jan 21 '16 at 9:28
  • $\begingroup$ For the problem of the OP only the drift velocity matters. The free electron gas model is meaningless for this application. At the very least you need something like a Drude model. $\endgroup$ – CuriousOne Jan 21 '16 at 9:31
  • $\begingroup$ Hey, I added diffusion in my answer so that OP knows how electron can reach positive plate even after losing its some of KE. $\endgroup$ – Anubhav Goel Jan 21 '16 at 9:36
  • $\begingroup$ I know that you know the correct solution, but you are leaving the OP with the wrong physical intuition (that electron velocity is large) while mentioning the very effect that makes effective velocity small at the very end of your post. The OP can not know that diffusion basically negates what you just told him. $\endgroup$ – CuriousOne Jan 21 '16 at 9:41

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