# Circuit energy evaluation

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.

• 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. Commented Jan 21, 2016 at 9:14