# Why does a Resistor cause a potential drop?

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.

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.

• 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. – SilverSlash Jul 23 '14 at 9:47
• 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? – torito verdejo Jun 15 at 13:02

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.

• 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. – SilverSlash Jul 23 '14 at 9:48
• Please don't worry about that! ;-) – Luboš Motl Jul 23 '14 at 10:01

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.

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.