2
$\begingroup$

Backstory: I’m a software engineer just getting into electronics and it seems that everything I’ve ever been told about electricity my whole life is a candy-coated lie. I can’t find consistent logical answers to the most basic of questions and it’s driving me mad!

The kindergarten math V = IR makes sense... unless accounting for conservation of energy, matter, and real laws of physics.

I’m old enough now. I just want to know the truth, even if it hurts.

The effect of voltage on current

  1. A resistor, led, and copper wire walk into a bar
  2. The bar tender serves a 9v battery to share
  3. The electric field is too weak to serve the led directly
  4. The copper wire volunteers to help direct the electric field
  5. The ampacity of the led is too low, so the resistor hops up on the bar in-line with the copper wire so the LED can get a drink
  6. Ohms Law is a lie, but the coulombs are absolutely intoxicating. All hell breaks loose and the resistor catches the whole bar on fire, burns it to the ground, everybody dies, and I lost five dollars.

Pause.

No, wait, there was no fire, that was just my anger at how every explanation I read of this scenario is in direct contradiction to what I thought I knew about conservation of energy and matter.

Contradictions for which I'd like answers

If charge causes the electrical field, then why does the voltage drop across the resistor? The electrons didn’t just magic themselves away. Isn’t the charge the same?

If charge passing though the resistor causes the atoms to enter a lower energy state, thereby releasing IR photons that heat up the place... then where did the extra coulombs go each second?

How come 2x resistance makes my battery lasts (on the scale of) twice as long but at (on the scale of) 1/4 of the power?

If resistance slows the flow of current, shouldn't ALL of the current still be accounted for somewhere on the system? solved: many of the explanations I was reading made it sound as though resistors lowered the current (...from infinity?) by "burning off" the "extra" current, which made no sense and contradicted the idea that the current supply and current drain were equal (Kirchoff's Law, common sense). Hence, the oversimplification of some of what I was reading confused me greatly.

... either my understanding is way off or there’s a well kept secret that few people are sharing (or my Google fu is busted)

$\endgroup$
  • 1
    $\begingroup$ this is entertaining writing, but there are too many questions embedded in it. Can you pare this down or split it into several different questions? $\endgroup$ – niels nielsen Jan 19 at 3:50
  • 1
    $\begingroup$ better still: you need to sit down with a physics guy and a 6-pack of some dry-hopped IPA and a pad of graph paper for as long as it takes to finish off the beer... where are you located? -Niels $\endgroup$ – niels nielsen Jan 19 at 3:52
  • 1
    $\begingroup$ @nielsnielsen I'm in Provo, UT, but I'd happily buy you the beverage of your choosing in exchange for a Skype session. I got rid of the first half of the question and I'll post that as a second question when I'm allowed to in... 38 minutes. $\endgroup$ – CoolAJ86 Jan 19 at 4:10
  • 2
    $\begingroup$ try the hyperphysics microscopic views hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html $\endgroup$ – anna v Jan 19 at 4:51
  • $\begingroup$ Skyping physics- never though of that. Let's do this: see what response your split questions get here and if there are issues remaining afterwards, either I write a dissertation on them for you or we skype. You can also reach me direct via my website, www.nielsenkillowatt.com $\endgroup$ – niels nielsen Jan 19 at 6:54
3
$\begingroup$

What really happens to the electrons in a solid when an electric field is applied is extremely complicated, and depends heavily on the material in question. What's more, the electrons cease to be "electrons" as the elementary particle in vacuum, they become quasiparticles with non well defined velocity and with other strange properties. I'm afraid there's no simple answer to the original question. It has to be at the level of quantum field theory applied to condensed matter. I do not have such a level of understanding (yet at least).

Nevertheless, I can offer a much different insight than the ones already posted, closer to what really happens in a conductor when an electric field is applied. Let's take a simple material as conductor such as an alkali metal. Its atoms/ions form a crystal. An intuitive way to think about the electrons in that solid, is to assume that all the core electrons, i.e. the ones in filled shells, are not free electrons and we can ignore them completely for electrical conduction matters. Only the single valance electron is a free electron. That yields one free electron per atom. All these free electrons behave roughly as in a cold Fermi gas, that is, they have to satisfy Pauli's exclusion principle and their occupation number obeys Fermi-Dirac statistics. Thus:

  • Case when $\vec E = \vec 0$. In that case, the energy of the electrons range from 0 up to about the Fermi energy, $E_F$ (if the temperature is at absolute 0, then it is exactly at the Fermi energy). In k-space (momentum space, not real space), the electron's momentum form a Fermi sphere. Note that this is valid for most alkali metals, but for metals like copper and iron, the shape is not quite spherical. The wavefunction of each electron extends to the crystal sample (they are not localized), and they have velocities ranging from 0 up to the Fermi velocity, which is about two order of magnitude slower than light. But they go in all possible directions and thus the mean velocity is null: there is no current, the drift velocity is 0.

    - Case when $\vec E \neq \vec 0$. What happens when we apply an electric field? Usually ordinary currents have a magnitude which causes a very, very small perturbation to the energy of whole system. Contrarily to what the Drude model assumes, in reality only the electrons near the Fermi surface of the sphere (or simply surface in general) can "feel" or react to the applied electric field. This is due to Pauli's exclusion principle which implies that no two electrons can share the same state. Thus the free electrons that have an energy much lower than $E_F$ cannot increase their energy, since all the states which have an energy slightly above them are already occupied. Therefore the net result of the applied field is to cause the electrons that were moving in the field's direction with momentum near $p_F$, to interact with the field and have their momentum switched in the other direction, with roughly the same magnitude. The fraction of the free electrons that can react to the electric field is of the order of $v_d/v_F$, or about $10^{-4}/10^6 =10^{-10}$. Hence only about one free electron per ten billions will get influenced by the electric field. Mathematically it is equivalent to a shifting of the Fermi surface against the direction of the electric field, by an extremely small amount (because the E field is such a small perturbation). Note that the drift velocity that arises in that free electron model is the same as in Drude's model, but the physics is quite different and proved to be more correct than Drude's.

Just to clear some misconceptions: when one applies an electric field in a conductor, it "travels" at a fraction of the speed of light, roughly about 20% to 80% of light's speed. The electrons that take part in electrical conduction move at speed about two orders of magnitude slower than light, and they are extremely less numerous than the number of free electrons. This yields a drift velocity that matches the one in Drude's model. Note that the number of electrons that can react to an applied electric field does not match the number of electrons that can absorb heat, or take place in heat conduction.

About the resistance (or resistivity): The resistivity is partly due to the scattering of the few free electrons going against the $\vec E$ field that take part in the electrical current. They are scattered by phonons and "put back" in the energy state where they were before the $\vec E$ field was applied. Note that they interact with phonons (a quasiparticle), and defects (like a missing atom in the crystal lattice), among others. The electrons that participate in electrical conduction do not really bump into atoms as Drude's model claims.

$\endgroup$
  • $\begingroup$ If the real answer "has to be at the level of quantum field theory", then how is this "the truth" and not just a slightly deeper lie for children than the one I presented? How do you expect OP to understand terms like "Fermi gas" and "quasiparticles", given the level of knowledge they show in the question? How is the concepts of scattering by phonons different from "bumping into a molecule" at the level of understanding demonstrated by OP? $\endgroup$ – The Photon Jan 19 at 16:05
  • 1
    $\begingroup$ @The Photon the asker exlicitely begs for the truth "even if it hurts". Drude model won't cut it. I agree that our best models are still just another lie to children, but it is still light years closer to the unreachable truth, and as such, posseses value. $\endgroup$ – thermomagnetic condensed boson Jan 19 at 16:44
  • 3
    $\begingroup$ +1 for a pretty good condensation of this model. But if OP wants the "even if it hurts" truth they'll need to start by reading Kittel (or whatever prep materials they need to be able to understand Kittel) and finish by going for a PhD in condensed matter physics. That should hurt quite a bit, I think. $\endgroup$ – The Photon Jan 19 at 16:53
  • $\begingroup$ I agree with you @The Photon! $\endgroup$ – thermomagnetic condensed boson Jan 19 at 17:14
  • $\begingroup$ Intuitively, this sounds much more correct than many of things I’ve read about the flow of electrons, in relation to all of what I’ve read over the past few weeks about electrical physics. Thank you very much for the depth. I also had a feeling that the truth is that we really don’t know, but we’ve got a lot of guesses and observations and some are better than others. I have no idea what Fermi is, but I’ll go Google it and the existing context is nearly sufficient for me to understand as is. I’ll wait until Monday to accept an answer (since so many people have different answers anyway). $\endgroup$ – CoolAJ86 Jan 20 at 7:40
3
$\begingroup$

If charge causes the electrical field, then why does the voltage drop across the resistor? The electrons didn’t just magic themselves away. Isn’t the charge the same?

Crudely, the electric field accelerates an electron. It flies along until it bumps into something (a molecule in the resistor). This causes the electron to slow down (or even fly back the other direction), and the molecule to vibrate.

(Actually the free electrons in the material will mostly be flying around in almost entirely random directions, with only a very slight bias in the direction opposite [because they're negatively charged] the electric field)

Then the electron, still in the electric field starts accelerating again.

The vibration of the molecule gets transferred to other nearby molecules in a random way, which we see at the macro scale as heat.

If charge passing though the resistor causes the atoms to enter a lower energy state, thereby releasing IR photons that heat up the place... then where did the extra coulombs go each second?

There's no atoms changing state or photons here, just electrons being accelerated and then transferring their kinetic energy directly to molecules they run into ("interact with" to use more physics-y language).

How come 2x resistance makes my battery lasts (on the scale of) twice as long but at (on the scale of) 1/4 of the power?

With a fixed source voltage, 2x resistance means 2x less current.

This means 1/2 the power consumed by the resistor, not 1/4.

If resistance slows the flow of current, shouldn't ALL of the current still be accounted for somewhere on the system?

Yes, the current flowing out of the resistor is the same as the current flowing into the resistor. The current is the same at every point around the circuit (out of the battery, into the resistor, out of the resistor, into the LED, out of the LED, and back in to the battery)

$\endgroup$
  • $\begingroup$ Thank you. I'll have a few questions while I digest this. First, I've seen contradictory explanation of the movement of electrons. You seem to be saying that they accelerate, which I've seen many posts say the same. My understanding, however, is that current moves at near the speed of light (Newton's Cradle / marbles style) whereas the electrons barely drift along somewhere in the range centimeters per minute (to our macro view, but exist as a cloud in their energy shell). I don't understand how both could be true. Can you elaborate on that? $\endgroup$ – CoolAJ86 Jan 19 at 5:48
  • 1
    $\begingroup$ @CoolAJ86, they're mostly moving in random directions. But in the presence of a field, they all accelerate opposite that field. This causes a very slight net bias to be moving in one direction. They're also continuously "bumping into" molecules of whatever material they're in, re-randomizing their velocities (explaining why individual electrons don't move along the circuit very quickly at all) But fundamentally if a charged particle is in a field, it will accelerate along that field vector. $\endgroup$ – The Photon Jan 19 at 5:51
  • 1
    $\begingroup$ The "current moving at near the speed of light" is about how I can close a switch in a circuit and suddenly current will be moving around the whole circuit, not just the part near the switch. My explanation is about how the individual electrons move once the steady state is achieved. $\endgroup$ – The Photon Jan 19 at 5:54
  • $\begingroup$ I think the asker wrote he wants to know the truth, not a description of the obsolete and wrong Drude's model $\endgroup$ – thermomagnetic condensed boson Jan 19 at 8:43
  • $\begingroup$ @tttt, I answered it with the goal of giving them a better model than the one they had come up with on their own. If you think you can give them "the truth" in the space of a Stackexchange answer, of course you are welcome to give it a try, and I will be happy to upvote you if you manage to do it. $\endgroup$ – The Photon Jan 19 at 15:57
2
$\begingroup$

In addition to The Photon's excellent answer, the part about charge conservation is formalized in circuit analysis as Kirchoff's laws.

When you look at a circuit as a graph with edges (components) and nodes (mutual connections), the current entering and leaving a node sums to zero (KCL) and the voltage changes as you walk any loop in the graph also sums to zero (KVL).

So, yes, current leaving any part of the circuit is required to come from somewhere else and charge is conserved.

$\endgroup$
  • $\begingroup$ The Photon's answer might be an accurate description of what we knew back in the early 1900's. It's actually quite far from what we know today. $\endgroup$ – thermomagnetic condensed boson Jan 19 at 8:46
1
$\begingroup$

If charge causes the electrical field, then why does the voltage drop across the resistor? The electrons didn’t just magic themselves away. Isn’t the charge the same?

The voltage (of a charge) changes as it goes across an electric field. Just like the potential energy of a mass changes as you lift it in the air. The mass is the same from one place to another, but the potential is different.

The big difference is that gravitational potential near the earth is mostly uniform. There's very little we can do to change it. The electrical potential in a wire changes constantly as the charges rearrange.

When a current goes through a resistor, the charges arrange themselves so that there's an electric field inside. Now when a charge moves through the field, it changes the potential (voltage) from one side to the other.

If charge passing though the resistor causes the atoms to enter a lower energy state, thereby releasing IR photons that heat up the place... then where did the extra coulombs go each second?

Photons are not coulombs. If you slide a block down a ramp, it heats up without the block changing. The photons came from the change in potential as it loses height. In the same way the charges passing through the resistor don't change, but the energy of that charge does change.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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