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Go easy on me if you find this question stupid, just another high schooler here.

So here's what I am thinking.A battery "pushes" electrons through a circuit due to the electric potential difference between its positive and negative terminals.So, basically the battery takes the chemical energy stored in it and gives it to the electron in the conducting wire (say).This way the electrons gain some kinetic energy and move towards the positive terminal.Now lets put a bulb in this circuit.Since the bulb has a high resistance some of the kinetic energy of the electrons changes into heat and light.Therefore the electrons coming out of the bulb should have less energy than electrons that enter the bulb.In other words the current in the wire on one side of the bulb is more than that of the other side.Right?

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  • $\begingroup$ Think of electrons in the wire as being like cars in a railroad train. There is a tiny amount of play from one car to the next, but on average, they must all move together at the same rate. $\endgroup$
    – Solomon Slow
    Commented Aug 29, 2019 at 12:20
  • $\begingroup$ If you've ever stood beside the tracks when a long train started to move, then you'll know that the "news" that the train has started to move can take a few seconds to travel along the whole length of the train. That's because of that "tiny amount of play" in the coupling from one car to the next. Electrons in a wire are "coupled" by electromagnetic forces. When you close the switch in an electric circuit, the "news" that current is flowing must travel around the circuit, from one electron to the next, as an electromagnetic wave that travels almost at the speed of light. $\endgroup$
    – Solomon Slow
    Commented Aug 29, 2019 at 12:23
  • $\begingroup$ @SolomonSlow That's a nice analogy. But by the way, is it really close to the speed of light? I genuinely thought that it was approximately the speed of sound, but now that I think about it, it seems that it cannot be and it must largely depend on the material. Do you know more about this? $\endgroup$
    – cinico
    Commented Aug 29, 2019 at 12:31
  • $\begingroup$ @cinico, That's a bit outside my area of competence. I know that the simplest mathematical models you will find describe the propagation of electric signals along a transmission line, which you can read about on Wikipedia. Unfortunately, that article describes it from the viewpoint of an electrical engineer, and so it does not speak of photons or valence-band electrons or any of that. But yeah, electronic signals on a wire propagate at almost-but-not-quite the speed of light in vacuum. $\endgroup$
    – Solomon Slow
    Commented Aug 29, 2019 at 12:54

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No.

If you look at the definition of current, or what an Ampere is, you will realize that we are talking about charge per time (note that it's not a measure of energy). The charge of an electron never changes. Thus, if the amount of current is the number of electrons per unit time, and the electrons don't just evaporate, then the number of electrons that enter the bulb must be the same number of electrons that exit the bulb. In other words, "what goes in must come out".

The electrons do lose energy into the bulb, but this is what we call a voltage drop between the terminals of the lamp.

P.S. this assumes, of course, that we are talking about steady state conditions.

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  • $\begingroup$ An ampere is the rate of flow of charges.The number of electrons don't change, therefore the charge doesn't change.However the speed of these electrons change due to loss of energy.This change in speed reduces the rate of flow of charges or in other words reduces the current.I know that this doesn't happen but I don't know the reason why.I couldn't find the reason in your answer.Please provide a more in depth explanation. $\endgroup$
    – whae
    Commented Aug 29, 2019 at 12:30
  • $\begingroup$ @AdityaBharadwaj No, the speed of the electrons does not change.They simply are at different energy levels. If the speed would go down while exiting the bulb, then you would have less electrons coming out (per second) which would mean that the electrons that were coming in had to be going somewhere else. $\endgroup$
    – cinico
    Commented Aug 29, 2019 at 12:36
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    $\begingroup$ Of course, with a hot filament you could very well emit a few electrons into the vacuum, never to come back. An unintentional vacuum tube, if you will. Without trying to, it won't be enough to notice. $\endgroup$
    – Jon Custer
    Commented Aug 29, 2019 at 13:14
  • $\begingroup$ @JonCuster true! $\endgroup$
    – cinico
    Commented Aug 29, 2019 at 13:51

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