8
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ see this hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html $\endgroup$
    – anna v
    Mar 3, 2019 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. $\endgroup$ Mar 3, 2019 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, 2019 at 14:44
  • $\begingroup$ @BenCrowell What do you mean by "only holds for resistors". I don't understand. $\endgroup$
    – whae
    Mar 3, 2019 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. $\endgroup$
    – whae
    Mar 3, 2019 at 17:26

3 Answers 3

10
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Can you please answer my second question also. $\endgroup$
    – whae
    Mar 3, 2019 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, 2019 at 21:10
  • $\begingroup$ @Thorondor, The idea (that increasing voltage increases proportionately the speed of conducting electrons) is easily comprehensible in DC circuits. But will this hold true also in AC circuits? Let's say, 50 Hz. Doubling the voltage gets electrons speed doubled. But would this not translate to current frequency of 100 Hz (bcoz the electrons would now traverse the same area -- say it's a load of 1 ohm resistor -- twice per unit of time) while maintaining 50 Hz for voltage? $\endgroup$
    – Roy Closa
    Feb 20, 2021 at 8:55
  • $\begingroup$ @RoyClosa in an AC circuit, if you double the voltage, the electrons move twice as far each cycle (which still isn't a very long distance). $\endgroup$
    – Thorondor
    Feb 20, 2021 at 17:55
1
$\begingroup$

Increasing the voltage applied to a circuit of a given resistance will increase the current flow. That flow is defined in electrons per second past a point. So increasing the voltage increases the speed of the electron flow. The number of electrons free to flow is a constant for a material. For Copper that is one electron per atom.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
4
  • 3
    $\begingroup$ "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, 2019 at 13:42
  • $\begingroup$ I assumed people are going to understand without too much specificity. I'll make amends. $\endgroup$
    – TechDroid
    Mar 4, 2019 at 1:14
  • $\begingroup$ Thank you, Everyone else in the entire Internet is trying to define current and voltage with a water analogy (just leading to more questions, like definition of "pressure"). You are the first one I find who tells it in a more understandable way. $\endgroup$
    – simon
    Jul 21, 2020 at 20:52
  • $\begingroup$ @ TechDroid Let's have a DC circuit of 10-vdc source and a simple load of 10-ohms resulting in a current of 1 amp, i.e, a flow of 6.25E18 electrons per second. By your logic, if the voltage source is increased to 20v, the same 1 Coulomb of electrons would circulate but 2x the speed. This implies that the 6.25E18 electrons gets 10-v during 1st circulation, and another 10-v during 2nd circulation to deliver a total of 20-v in same time of 1-sec. BUT this translate to the same 10 volts drop in the 1-ohm resistor (instead of 20-v) each time 1-C charges circulates. Are we missing something? $\endgroup$
    – Roy Closa
    Feb 20, 2021 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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