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I need to know the underlying physics of what exactly happens different with the electric field in the resistor than in superconducting wires.

Why is it that when I connect a resistor, potential drops across its (resistor's) ends? Sorry for the silly question but I can't figure it out. I do know, energy will be dissipated and hence the voltage will drop but I'm looking for a more insightful and detailed answer.

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Electric potential is a potential energy just like gravitational potential energy or indeed any other form of potential energy. Specifically, moving one coulomb of charge through an electrical potential of one volt produces (or requires) 1 joule of energy. From your question I guess you're basically happy with this, so the question is really how this energy is dissipated i.e. what happens to that 1 joule of energy?

When you apply a voltage to the conductor you produce a force on the conduction electrons so they accelerate - the potential energy is turned into kinetic energy of the electrons. However conductors are made up from a crystal lattice of atoms/molecules that is randomly vibrating due to thermal energy, and there is a probability that the moving electrons will scatter off this lattice and transfer energy to it. So the electron is slowed down and the magnitude of the lattice vibrations is increased. Increased lattice vibrations mean the conductor is hotter, so the kinetic energy of the electrons has been transferred into thermal energy in the conductor.

And that's what happens to the 1 joule of energy. It's transferred to the conductor and ends up as heat.

Some related issues you might want to look into further: when you cool a conductor you reduce the magnitude of the lattice vibrations and you make it less likely the electron will scatter off the lattice. That's why resistance (usually) decreases with decreasing temperature. The superconducting transition prevents electrons from scattering off the lattice, so they can't transfer energy to it and that's why superconductors have a resistance of zero.

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    $\begingroup$ Thanks! My question is answered. I wanted to up-vote but it seems I require 15 reputation or something to do that even in my own question thread. Thanks again. $\endgroup$ Jul 23, 2014 at 9:47
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    $\begingroup$ Personally I still don't get why resistance modifies voltage, i.e. a difference of potetial energy. As far as I know, jumping down a roof requires only the work of gravity, while getting down the roof using a slope with friction would require additional energy to compensate for the one lost by friction. Now, in each case, the potential energy due to gravity is the same. What part of this analogy fails to represent what goes on in the circuit? Why does resistance modifies the potential energy in the circuit? $\endgroup$
    – Hal
    Jun 15, 2020 at 13:02
  • $\begingroup$ @John Rennie hello .. so do you mean that the electron which enters a high resistive zone in a circuit with certain energy loses some of it inside the resistor due to collisions and leaves that region with a leeser energy. So can we define the potential of the end from which the electron came out as the energy of the electron remaining per unit electronic charge ? $\endgroup$
    – Ankit
    Sep 4, 2021 at 14:00
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In resistors, resistance, current, and voltage (or the voltage drop) are related by Ohm's law: $$ U = RI $$ This law may also be formulated in a more insightful, microscopic form, especially in the simplest microscopic version of Ohm's law (due to Gustav Kirchhoff): $$ \vec J = \sigma \vec E $$ Here, $\vec J$ is the current density, $\sigma$ is the conductivity of the material, and $\vec E$ is the electric field, i.e. the potential (voltage) drop per unit distance (including the direction of the decrease).

First, it's useful to understand why the two Ohm's laws are equivalent. Consider some wire of cross section $A$ and length $L$. Multiply the microscopic law by $A$. The product $A\vec J$ combines to the total current $I$ (it is the density of the current per unit area), so we get $$ I = \sigma A |\vec E |$$ Now, rewrite $|\vec E|$ as the potential drop per unit distance (the electric field is the gradient of the potential), $V/L$, to get $$ I = \frac{\sigma A}{L} V $$ But $\sigma A/L $ is nothing else than the conductance $G=1/R$, the inverse resistance, so this is equivalent to $I = U/R$, the macroscopic Ohm's law.

Now, let's try to understand why the microscopic form of the law is true. $$ \vec J = \sigma \vec E$$ It is true because the electric field (changing electric potential) makes the free electrons (or charge carriers) move. There is some density of the electrons in the material, and the electric field makes them move – because of the electric force $\vec F = m\vec a = q\vec E$ – at some average velocity $\vec v$ which is proportional to $\vec E$ in the linear approximation. The electrons don't accelerate indefinitely because their collisions with the nuclei etc. slows them down. So they reach some "equilibrium speed", much like a skydriver facing the air fraction. By linear algebra, $\sigma \vec E$ is really the only leading $\vec E$-dependent result for $\vec J$ that is mathematically plausible and behaves smoothly for $\vec E\sim 0$.

Nonlinear terms like $\Delta \vec J = \beta |\vec E|^2 \vec E$ may appear and indeed do appear but the point is that for a very small $\vec E$, they are subleading and the leading term doesn't cancel for resistors.

So if you try to explain why Ohm's law holds, it's better to imagine that the voltage drop is the cause while the current is its consequence, not the other way as you tried.

Superconductors don't have any collisions so the electrons wouldn't slow down if an electric field were attached. Consequently, the voltage drop would make huge changes that would completely eliminate the voltage drop at the end, and therefore $V=0$. At any rate, as long as Ohm's law is concerned, you may view superconductors as a simple (and strict) $R\to 0$ or $R=0$ limit of resistors.

Apologies, $U=V$ above. I wasn't consistent about the notation.

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  • $\begingroup$ Thank you for the detailed reply. My question is answered. I wished to up-vote your answer but it seems I require 15 reputation or something, even though this is my own thread. But anyways, thank you for your time. $\endgroup$ Jul 23, 2014 at 9:48
  • $\begingroup$ Please don't worry about that! ;-) $\endgroup$ Jul 23, 2014 at 10:01
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The answer depends a lot on how much into detail you want to go. Superconductivity is a pretty involved phenomenon so I hope I can help you already, by explaining why a resistor causes a potential drop.

A current can be imagined as electrons moving through a conductor. During that movement, they are accelerated due to the electric field. At the same time, they "bump into things", i.e. are scattered. This forces them to slow down and leads to resistance. This can maybe be imagined like pumping fluid through a tissue or something similar. Therefore there are more electrons one one side as on the other which leads to a potential drop.

This of course is a very rough explanation and going into detail, one will discover there is a lot of physics involved.

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It is not a silly question . Potential means energy per unit charge at a point in a field . When a unit charge travels through an e-field, energy is either lost or gained as potential energy . Potential difference is the difference in P.E. between any two points in the field . In a circuit , when a cell is inserted , electric field forms and it is due to this field charges flow(current). When charge flows in the opposite direction of the field inside the cell, energy is stored as PE in the charge . After passing through the cell, it again travels in the direction of electric field thus lossing the stored energy , irrespective of whatever path it takes. When it goes through a resistor, the energy dissipated (as it is going in the direction of field) becomes the heat energy . It can also be said in order to overcome the hinderances(resistance) the charges used their stored energy . Hence after coming out from the resistor, they have less energy . Thus potential drops . Hope this helps.

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I am a hobbyist and have been trying to answer this question for a while, and this what I found. First of all Emf or Electric potential is created by a battery(DC) or Alternator(AC) by separating charges. I.e., forcefully moving -ve charges/electrons to one end called the -ve terminal, and the other end becomes +ve because of the depletion of electrons. Because -ve charges naturally repel each other, and they are kept together by force against their will,their potential rises which can be now be used to drive a current through the circuit where one end are these -ve charges, and the other has atoms of +ve charges ready to receive the -ve charges. This measure of this forced electric potential / pressure is done in volts and called voltage. The higher the voltage the more densely the electrons are packed at the -ve terminal and more work has been done(by the power company) to get this to happen. A single electron from a higher voltage source (say 220v) has greater energy and perhaps more velocity as well than one from a lower voltage source (say 110v). That is why Power is V(voltage) x I(current) and not just I because the higher energy/voltage electrons can do more work and you end up paying more money for it.

Now coming to voltage drop, Electrons always move from a higher potential or -ve terminal to a lower potential or +ve terminal assuming electron flow as opposed to conventional flow. Take a simple circuit a with a single Resistor. One end of R is connected to -ve terminal and the other to +ve terminal through copper wires. Turn on the circuit. Voltage drop happens ONLY when current begins to flow. The drop is noticed from the end of end of R to the +ve positive terminal. Here is why : The copper wire from -ve terminal to R is flooded with electrons and so is at the same potential as the -ve terminal. But because the resistor restricts the current, only some electrons are allowed to go the other side. One the other side,that is from end of R to the +ve terminal, except for the small current that is allowed to flow, there are mostly +ve atoms / charges which means the electron which reaches the +ve side of R does not experience that much pressure from neighboring electrons as it did on the -ve side. That is why the voltage drops.

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What does it mean for a resistor to cause a "potential drop"? This means that the potential of an electron after going through a resistor is lower than the potential of the electron before it went through the resistor. Why is this the case?

First, let us go over what electric potential means. The electric potential of an electron is defined as the energy that would be needed in order to bring this electron to its present location from a distance of infinity. This means an electron has a higher electric potential when it is close to a large amount of other electrons (it would take more energy to overcome the repulsive forces of all these electrons in order to bring our electron to them).

Now, remember, we're trying to explain why electrons that have not yet entered the resistor have a higher potential than electrons that have exited it. It turns out that the reason is that electrons end up being more concentrated on the side of the resistor that they are entering, and less concentrated on the side of the resister that they are exiting. Let's look at why this happens.

Let's take a simple circuit consisting of a battery, a wire, and a resistor. Now, we know the resistor is made of material that is less conductive than the wire, so electrons aren't able to move as quickly in the resistor as they are in the wire.

enter image description here

In the diagram above, the white circles represent the electrons in the wire at their initial state (before the battery is connected). The red circles represent the electrons a moment after the battery is connected to the wire.

When the electron closest to the negative end of the battery moves, it causes the next electron to move as well (because of the repulsive forces between electrons). This creates a cascading effect.

However, electron A in the above diagram cannot move as much (in the same timeframe) as all of the electrons behind it. As a result, electron A will be closer to all the electrons behind it than it was before. Hence, it repels these electrons more strongly. This slightly cancels the repulsive force on these electrons due the negative end of the battery, causing them to slow down.

Now, since the repulsive force an electron exerts on another electron decreases with distance, it is clear that those electrons that are closest to electron A will be slowed down the most and those that are furthest will be slowed down the least. Since the electrons that are in the front (meaning electron A and those that are slightly behind it) are the ones that are moving the most slowly, the electrons will start bunching up a bit near the entrance of the resistor.

Let's look at what happens on the other side of the resistor.

On the other side of the resistor, electron B cannot move as much as the electrons in front of it. Hence, electron B will be farther from these electrons than it was before, so its repulsive force on these electrons decreases. Therefore, these electrons also slow down, since the force with which B is pushing them away is now smaller.

However, the increase in distance from B to the electrons ahead of it will cause a greater decrease in repulsive force for electrons that are further from B. (This can be shown by basic math.) Hence, the electrons that are closest to electron B will slow down the LEAST. Hence, electrons will not bunch up nearly as much on the exiting side of the resistor as they will on the entering side.

Since electrons are more concentrated on the "entering side" of a resistor, an electron will have higher potential when entering the resistor and being among all these close-together electrons than upon exiting the resistor.

I hope that made sense!

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