I have been reading a book about electricity which states that: electric current is not the movement of electrons but the "impulse generated when free electrons orderly "jump" from one atom to the other. On some other topic of the book it also states that, electrons lose energy when they move through loads.

I really do not understand how the impulse phenomenon takes place in a conductor. Could anyone elucidate me on the subject?

Secondly, what signs electrons present when they lose energy?


closed as unclear what you're asking by sammy gerbil, M. Enns, Jon Custer, ZeroTheHero, Bill N Aug 17 '17 at 2:09

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    $\begingroup$ Can you provide the title of the book? I may be misinterpreting what you've written, but that sounds like nonsense. The free electrons in a conductor are not bound to any atoms and therefore do not "jump" between them. Electrons lose kinetic energy when they scatter off of the atomic lattice, but they gain it right back due to the electric field which is driving the current. They do lose potential energy in the process, though. $\endgroup$ – J. Murray Aug 15 '17 at 10:45
  • $\begingroup$ Related: physics.stackexchange.com/q/17741/2451 and links therein. $\endgroup$ – Qmechanic Aug 15 '17 at 10:50
  • $\begingroup$ -1. Not clear. Please provide an image of the text which gives this explanation. $\endgroup$ – sammy gerbil Aug 15 '17 at 11:17
  • $\begingroup$ see my answers here physics.stackexchange.com/questions/351517/… and here physics.stackexchange.com/questions/28036/… $\endgroup$ – anna v Aug 15 '17 at 12:33
  • $\begingroup$ @sammygerbil: why should OP provide an unsearchable image of the text when he can simply quote it? I would like to think that most of us would convert the image into text anyway, so the screenshot would be wasted efforts better spent copying the specific lines verbatim (and including the author & title). $\endgroup$ – Kyle Kanos Aug 16 '17 at 9:51

When you connect a battery to a open circuit. Your battery has an excess of electrons at the negative terminal and a deficit at the positive terminal. When your circuit is closed, the excess electrons at the battery negative terminal start diffusing into the wire. Electrons in the wire then start diffusing away from the negative terminal along the wire and the end result is a voltage wave that travels down the wire at a few tenths of the speed of light. Due to this fact, electrons starts accelerating and collectively shifts their position together. Each and every electrons at the negative terminal does not moves completely to the positive terminal, each electron only moves a little bit, but all of them move together, and so the net current is observed.

As the electrons are accelerated by the electric field (also decelerated by resistance), there is change in momentum and hence there is an impulse. That may be stated in that book.


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