I am just a high school student trying to self study, please excuse me if this question sounds silly to you.

I know that current is a product of the speed of electrons and the electron density.When current is increased it either means that the speed of electrons has increased or it means that the number density of the flowing electrons has increased.

I also know that voltage is directly proportional to current and when voltage increases(without no change in the resistance) the current will also increase.

But my question is, when voltage increases does an increase in the speed of electrons contribute for an increase in current or does an increase in electron density contribute for it.

If it isn't that black and white, then in what proportion will each of the two components increase? Does it randomly increase?

Related question:Say the electron density of a circuit that lights a light bulb increases.When this happens what change will we see in the brightness of the light bulb?I know that when the speed of electrons increase the brightness increases but what will happen when the electron density increases?

  • $\begingroup$ see this hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html $\endgroup$ – anna v Mar 3 '19 at 11:30
  • $\begingroup$ "I also know that voltage is directly proportional to current" - please be careful here as this statement is not true in general, i.e, it is (approximately) true only for so-called ohmic conductors. In particular, it is not true for a light bulb filament. $\endgroup$ – Alfred Centauri Mar 3 '19 at 14:39
  • $\begingroup$ Your question is written as if these were general relations, but all of this only holds for resistors. $\endgroup$ – user4552 Mar 3 '19 at 14:44
  • $\begingroup$ @BenCrowell What do you mean by "only holds for resistors". I don't understand. $\endgroup$ – Aditya Bharadwaj Mar 3 '19 at 17:21
  • $\begingroup$ @AlfredCentauri Can you please elaborate on that or suggest links where I can learn the concept required to understand what are saying.Since I really don't know much about electricity and the whole deal, I don't know what you are talking about. $\endgroup$ – Aditya Bharadwaj Mar 3 '19 at 17:26

In a conductive material such as a metal, for all practical purposes, current depends only on the speed of the electrons. The electron density does not change because each metal atom has already given up all of its valence electrons; releasing further electrons would require a very large energy input.

In an insulator or semiconductor, the density of charge carriers may increase during electrical breakdown. This occurs in avalanche diodes, neon lights, lightning bolts, and elsewhere.

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  • $\begingroup$ Can you please answer my second question also. $\endgroup$ – Aditya Bharadwaj Mar 3 '19 at 13:58
  • $\begingroup$ @AdityaBharadwaj Aluminum has a higher electron density than copper, so your second question is equivalent to: what happens if we take all of the wires that connect the light bulb to a power source and change them from copper to aluminum? Intuitively, the answer should be "nothing changes." The reason is that since the light bulb is far more resistive than anything else in the circuit, the amount of current flowing through the circuit is governed by the behavior of the light bulb, not the wires. If the electron density increases, the electron speed decreases proportionately. $\endgroup$ – Thorondor Mar 3 '19 at 21:10

Current is the amount of charge (electrons) passing a point in a wire per unit time. Voltage is the amount of energy in joule in every charge of 1 coulomb moving through the wire. Increase in current translates to increase in speed of electrons moving past our reference point. Electron density in a wire remain relatively constant even at high wire temperature.

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    $\begingroup$ "The amount of coulomb" doesn't make sense (it's like saying "the amount of metre") . You mean the amount of charge. Also "voltage" doesn't "move through the wire". $\endgroup$ – alephzero Mar 3 '19 at 13:42
  • $\begingroup$ I assumed people are going to understand without too much specificity. I'll make amends. $\endgroup$ – TechDroid Mar 4 '19 at 1:14

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