-1
$\begingroup$

This question already has an answer here:

The electrons that carry the electric energy are in wires or in the battery ? battery provides a potential difference but from where electrons flow to make for example a bulb light. If they flow from wires then how a path is established in order electrons to go through wire over and over ?

$\endgroup$

marked as duplicate by Qmechanic Dec 25 '18 at 17:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

The electrons that carry the electric energy

Electrons do not carry the energy in typical electric circuits. The energy is carried in the fields. The motion of the electrons is very small, so if energy were carried by the electrons then it would take a long time for a light to turn on in a DC circuit and they would never turn on in an AC circuit.

In contrast, the fields propagate at an appreciable fraction of the speed of light, corresponding with the nearly instantaneous lighting of a light bulb in either an AC or a DC circuit. The power flux through the fields is given by the Poynting vector $S=\frac{1}{\mu_0} E \times B$

https://en.m.wikipedia.org/wiki/Poynting_vector

The currents go through both the batteries and the wires and the fields go through and around the entire circuit.

$\endgroup$
  • $\begingroup$ How are energies carried in fields? Is there anything I can refer? $\endgroup$ – Karthik V Dec 25 '18 at 14:52
  • $\begingroup$ I added a bit explaining the Poynting vector which describes the energy flux in the fields $\endgroup$ – Dale Dec 25 '18 at 15:08
1
$\begingroup$

Electrons are in a "free moving state" and they just need a kick to start moving in the entire circuit which consists of both battery and wire. There is a very special bond called "metallic bond" that allows this flexibility.

$\endgroup$
  • 2
    $\begingroup$ you should elaborate why you think ac has less energy loss than dc, after all dissipation is $I(t)^2R$. $\endgroup$ – hyportnex Dec 25 '18 at 12:03
  • 1
    $\begingroup$ The second paragraph doesn't seem to relate to the question. How about just cutting it? This is a fine answer if you just keep the first paragraph. $\endgroup$ – Ben Crowell Dec 25 '18 at 17:02
  • 1
    $\begingroup$ @jimandr it's not true that AC has fewer energy losses than DC. It is only stated as such when you compare the energy loss occuring when you step up or down an AC voltage (which is achieved via transformer). This has lower energy losses compared to a DC voltage step up or down, that cannot be done by a simple transformer - needs to be done by semiconductors, resulting in poorer efficiency. $\endgroup$ – Karthik V Jan 6 at 13:40
  • $\begingroup$ ok removed the paragraph regarding energy losses. $\endgroup$ – jim andr Jan 9 at 10:31
-1
$\begingroup$

Electrons from everywhere. To make it easier to understand, a battery has got chemical reactions running that sends electrons out through the negative terminal of the battery.

As you will know, an electron can only move that freely if it is placed in an electric field (generated by some potential difference - presence of charges). So, in this case it goes to the positive terminal where the reaction is in need of these electrons. There is more to it, but this is the main point.

These electrons are not that free to make it individually to the other terminal in that fraction of time. These electrons have random velocities before the electric field came into play, but in random directions in the wire. That is, say that the number of electrons that went right is the exact same as those that came from the right.

It is due to the application of the electric field that the electrons (in the wire) move around. But they don't move at those lightning speeds, but instead collide with each other, the atoms, and move about. However, the energy has been transmitted all that distance quickly.

$\endgroup$
  • $\begingroup$ This answer assumes a variant of the obsolete Drude model is a representation of reality. $\endgroup$ – t t t t Dec 25 '18 at 12:49
  • $\begingroup$ Could you please elaborate? $\endgroup$ – Karthik V Dec 25 '18 at 14:27
  • $\begingroup$ Add an answer if you can $\endgroup$ – Karthik V Dec 25 '18 at 14:27
  • $\begingroup$ I elaborate a bit more. The last paragraph, as is, is wrong. The very few free (quasi)electrons that account for the electrical current move at speeds around Fermi velocity, i.e. about 2 order of magnitude slower than light speed in vacuum. Also, they do not collide with each other and atoms (or ions). They interact with phonons, and between each other although the electron-electron interaction can be mostly neglected in most cases. $\endgroup$ – t t t t 2 days ago
  • $\begingroup$ I mentioned "a variant of the Drude model" because in Drude's model the electrons are like in a classical ideal gas, so they do not interact with each other. However they do collide with atoms/ions, as you mention. So your model is a variant of the Drude model, with the addition of interacting electrons. $\endgroup$ – t t t t 2 days ago

Not the answer you're looking for? Browse other questions tagged or ask your own question.