3
$\begingroup$

I have been thinking for a while about what really causes voltage drop and how to explain it in terms of what the electric and magnetic fields do. So I've been reading a lot of posts here and in other places, but I could not really connect the dots to what happens to an electron at each point in the circuit, especially before and after the resistor. Either the explanations were too mathy or were lacking a little more in-depth explanation, which is really what I am looking for. I am confortable with math, but it doesn't explain to me the phenomena, and doesn't really help visualize the interactions at that quantum level. It just wants me to accept that the whole thing is a black box, but hey, according to the formula, this is what you get in the end!

I will try to explain what I know so far. Some other questions ([Q]) may pop along the way and I will try to list them for you to correct me if you think that's where i'm making a wrong assumption or I got something totally wrong. If it's not relevant to the main question, skip it.

So let's say we have a battery and a resistor in a circuit.

Here is what I have so far:

  1. When we complete the circuit, the electric field guided by the surface charges of the wire, propagates through and around the wire to the negative terminal and establishes a potential difference between the positive and negative terminal. If the wire were a perfect conductor, there would be no current.

    • [Q] If there is a potential difference between the leads of the battery, why there has to be a voltage drop somewhere in the circuit (wire, resistor, load etc.) in order for the current to be flowing? In a perfect conductor, shouldn't the electrons be constantly accelerating therefore infinte current? or at least a current proportional to the total number of electrons available in the wire?
  2. Because of the electric field, all charges start to move. And it's not that the first electron hit by the field pushes the next and so on, but because each electron is affected by the field with the same force magnitude.

  3. Now because the electrons are moving, in a stationary surface charge's frame of reference, all the electrons appear closer (Lorentz's transformation), so there is a magnetic field created around the wire, perpendicular to the current flow, which acts upon all the surface charges at that point.

  4. Because the electron is moving, some of it's potential energy has been converted into kinetic energy, and the rest of its potential energy is stored in the magnetic field and the electric field.

  5. Let's say an electrons passes through the resistive material. It collides with the impurities (atoms, other electrons) transferring it's kinetic energy to the atom or electron, which in turn vibrates, releasing this vibrational energy into heat and light.

    • [Q] Now because the electron has lost some KE, it's going a bit slower, so there will be a charge build-up. And because it's going slower, the magnetic field intensity should decrease, because the relativistic effects should have lower effect on the surface charges. Now what happens to the energy stored in the magnetic field, is it transfered to the electron to accelerate it again to keep the average velocity constant or it passes the energy to the electric field, which in turn accelerates the electron so on and so forth?
  6. Now what are we really measuring when we hook up a voltmeter to measure voltage across the resistor. Is is the strength of the magnetic field? electric field? both? neither?

    • [Q] From electrostatics we know that if we bring a test charge at distance d in an uniform electric field, then it's potential will be equal to qEd, and if we move it further away from the field it will have less potential energy. So voltage (which is the diference in potential energy at these points) is a function of distance, then a voltage drop would mean that the charges somehow got further apart in the field or the field has been distorted so the flux density at a point after the resistor is lower ... ?

If anyone could point the flaws in all of this, help me understand what really happens, or point me to some books, that would be great.

$\endgroup$
0
$\begingroup$
  • Q1 If there is a potential difference between the leads of the battery, why there has to be a voltage drop somewhere in the circuit (wire, resistor, load etc.) in order for the current to be flowing? In a perfect conductor, shouldn't the electrons be constantly accelerating therefore infinte current? or at least a current proportional to the total number of electrons available in the wire?
  • A1: As you know, we got Ohm's law, $U = R \cdot I$. Now, we take the limit $R \to 0\Omega$, while $U=const$. You see, that the current $I$ must diverge, $I\to \infty$. So, if we assume that the battery is a perfect voltage supply, the current must diverge. Alternatively, we can ask the question which voltage the battery supplies, if the current is limited by some value $I_{max}$. Now we get $U\to 0 V$ if $R\to 0\Omega$. So somehow we have to decide, which picture we like to use. For me, I prefer the first one, because I like to think of a battery as a perfect voltage supply.
  • Q2: ... I don't know.
  • Q3: I take this as your question: What are we really measuring when we hook up a voltmeter to measure voltage across the resistor?
  • A3: There is not only one working principle but many working principles exists. Have a look here. I think, at least some of their names are self explaining. So I just copy them here: Parmanent Magnet Moving coil Voltmeter, Moving Iron Voltmeter, Electro Dynamometer Type Voltmeter, Rectifier Type Voltmeter, Induction Type Voltmeter, Electrostatic Type Voltmeter, Digital Voltmeter.
$\endgroup$
0
$\begingroup$

Q1: for an ideal battery with 0 internal resistance, closing the circuit with perfectly conducting wire will result in infinite current. If the battery has internal resistance, as real ones do, all the voltage drop will happen in this internal resistor and thus heat the battery until it is ruined (this could even be a fire hazard). The reason there is a voltage drop is because there is an electric field across some distance. Inside a perfect conductor, there is no field and therefore no voltage drop. Q2: the same current flows through the resistor as through the perfectly conducting wire. The charges do lose energy to collisions within the resistor, but they are re-acceleration again. Q3: the most simple volt meter allows charges to accumulate on two metal pieces which then repel each other against some restoring force such as gravity. They repel electrostatically, so it is the electric potential or voltage associated with the electric field which balances the potential difference associated with the resorting force. That is, the voltage causes a proportional increase in the average height of the metal pieces. Obviously there are other methods and modern equipment is more advanced.

I'm not sure your discussion of what the magnetic fields are doing is entirely necessary for the conceptual understanding you appear to be seeking here, since you haven't brought up inductance. Inductance is related to how much momentum gets stored in the magnetic field. Changing momentum takes time, so adding inductance to a circuit causes a lag after a change in voltage while the magnetic field "spins up" to its new equilibrium value. For simple ideal circuits, inductance is often neglected wherever possible and current is treated as if it hits equilibrium instantly after flipping the switch.

$\endgroup$
0
$\begingroup$

[Q] If there is a potential difference between the leads of the battery, why there has to be a voltage drop somewhere in the circuit (wire, resistor, load etc.) in order for the current to be flowing?

I love this question! Here's how I explain it to my freshmen: Imagine a "potential" you're more familiar with: gravity. When you go up an escalator (which has a potential difference $mg\Delta h$), in order to get back to the bottom, you have to lose potential energy $mg\Delta h$ some way or another (usually by walking down stairs). You also don't keep all that energy as kinetic once you get to the bottom, right? Your shoes & joints create friction (analogous to a resistor) shed the energy so you approach the escalator at ~0 speed again.

Now for a superconductor, with almost 0 resistance, it will keep increasing the current until the tiny resistance accounts for the voltage drop. Imagine, perhaps, an escalator and frictionless slide that keeps speeding you up until the air resistance gives you terminal velocity.

Now because the electron has lost some KE, it's going a bit slower, so there will be a charge build-up. And because it's going slower, the magnetic field intensity should decrease, because the relativistic effects should have lower effect on the surface charges. Now what happens to the energy stored in the magnetic field, is it transfered to the electron to accelerate it again to keep the average velocity constant or it passes the energy to the electric field, which in turn accelerates the electron so on and so forth?

It's not like the current decreases once it hits a resistor, imagine this as a bumper to bumper traffic jam on a single lane highway. They can all drive fast together, but if something slows some of them down it slows all of them down.

In the case that you could dial the resistance up or down, certainly if you were to increase the resistance quickly it would cause a change in the magnetic field which would induce an electric field to accelerate the charges as in Faraday's law. But this is the same for any change in current, not unique to changing resistance.

From electrostatics we know that if we bring a test charge at distance d in an uniform electric field, then it's potential will be equal to qEd, and if we move it further away from the field it will have less potential energy. So voltage (which is the diference in potential energy at these points) is a function of distance, then a voltage drop would mean that the charges somehow got further apart in the field or the field has been distorted so the flux density at a point after the resistor is lower ... ?

Circuits are really a special case that can't be treated with electrostatics. $V = qEd$ doesn't work well to describe circuits because electrons can't move freely past one another, so the effects of the electron in front of and behind of the one you're looking at are really what determines the $\vec{E}$ field at that point. But that is a messy, unhelpful way of looking of the special case that is electric circuits.

$\endgroup$
-1
$\begingroup$

I believe your misunderstanding on all of these points is coming from the core idea of electricity being the flow of electrons. When the theories electromagnetism were developed in the 1800's, the electron had not been discovered yet, so you have to keep that in mind when trying to explain these ideas with the electron.

By trying to apply a simplistic explanation of electricity given in high school to Quantum Mechanics, you end up running into some problems where the conceptualisation doesn't match up with real life.

Properly understanding electricity and electromagnetism cannot at all be done by just thinking about the electron, so start by read about electric potential and some history on electromagnetism

$\endgroup$

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.