# Why does voltage drop across a resistor?

I don't get why there would be a voltage across a resistor as that would mean electrons pile up on 1 side of the resistor and as far as I know, that's not the case. My assumption is that voltage is dropping all over the circuit and the resistor just happened to be between a potential difference in the circuit. Someone please explain me!

• They don't exactly pile up. It's that the wires leading to the resistor have such an ocean of conduction band electrons. The resistor doesn't have nearly so many. So most of the voltage appears across the resistor. The wire only needs a very slight, gradual change in its surface charges (a very slight gradation - ergo slight voltage differential) in order to impel some electrons through the resistor. Basically, the voltage difference appears almost entirely across the resistor, impels some electrons through it and almost no gradation appears across the wire, which can supply so many so easily.
– jonk
Commented Apr 21, 2021 at 7:20
• You're thinking of current drop. If there was a current drop the electrons would pile up. That's why there isn't a current drop. Commented Apr 21, 2021 at 10:09

electrons pile up on 1 side of the resistor

That is entirely correct. And there is a "shortage" of electrons on the opposite side.

as far as I know, that's not the case.

It is the case.

Except in the case of a super-conductor, electrons interact with atoms which compose the conductor in a way that transfers energy and momentum to those atoms, and gives a random motion the electron. To use a simple model, an electron collides with an atom, and ricochets off in a random direction. The difference between the electron's initial momentum and final momentum is transferred to the atom, and the difference between the electron's initial energy and final energy is also transferred to the atom. While energy and momentum are conserved by such an event in our model, the flow of current is not. The "forward" motion of the charged electron, is replaced by the "forward" motion of a neutral atom. (Of course in a very short time, the atom will collide with another atom, but we will not concern ourselves with this at the moment). So these collections in our model create two effects. One they randomize motion, turning the organized motion of a current into heat. Second, they stop the flow of charge. If there were no electric field to start accelerate the original or a new electron, current would come to a stop there.

But we know that in a circuit with no branches and with an applied electromotive force current doesn't just stop at one place, but is uniform throughout the circuit? How does that happen? If the electrons flow did just stop when they collided with atoms, as in our model, then charge would pile up. But the piled up charge would create an electric field. This field would accelerate electrons away from it, on both sides of the pile-up. This would both reduce the current into the pile-up area, and increase the current leaving the pile-up area. Net accumulation stops when current into an area equals the net current leaving that area are equal. In equilibrium, the algebraic sum of all currents into a "node" (really anywhere) is 0. This is Kirchhoff's Current Law.

So where might electrons pile up? What is it's distribution? For this we need Maxwell's/Heaviside's Equations. (The equations we actually use today were first formulated by Heaviside based upon work done by Maxwell). The first equation we will use is

$$\nabla \times E = \frac{\partial B}{\partial t}$$

This says that the divergence of E is equal to the rate of change of the magnetic field with respect to time. We will assume that there is no time varying magnetic field in our model, so that the divergence of E is 0.

$$\nabla \times E = 0$$

Whenever a vector field has a divergence of 0, then that vector field is equal to the gradient of a scalar potential field. In our case, we will call that potential field $$\V\$$

$$E = \nabla V$$

$$\V\$$ is what we think of a the "voltage" at a point, when that voltage is well defined. $$\E\$$ is proportional to the force on a charge (in the case of no magnetic field), and is what causes a charge to accelerate.

In order for there to be a uniform current in a section of a resistive conductor, there needs to be a constant $$\E\$$ throughout that section. But then we turn to another Maxwell/Heaviside Equation

$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$

$$\\rho\$$ in this equation is net charge density, and $$\\epsilon_0\$$ is just a constant. This equation tells us then wherever there is a uniform E field ($$\\nabla\cdot E =0\$$) there is no net charge density. In other words, in a conductor with a uniform E field, there is no net charge density. Charges only build up at points where the E field changes.

But we already saw that in a resistive conductor with a uniform linear resistance, there is a constant E field, so in such a section of a resistive conductor, there is no charge accumulation. All the charge accumulation occurs at the interfaces where linear resistance changes.

My assumption is that voltage is dropping all over the circuit

That is also true. But remember that $$\E = \nabla V\$$. The voltage can drop all along the circuit, but $$\E\$$ remains uniform as long as the rate of voltage drop per distance is uniform.

• Note that some equations need a minus sign: ∇×E=-∂B/∂t and E=-∇V
– Bart
Commented Apr 23, 2021 at 10:44

If the voltage across a resistor is non-zero, then the current is non-zero. This means that some charge carriers are dissipating energy, in the form of heat (neglecting radiation). Since energy must be conserved, the dissipated energy came from somewhere, and it was a reduction in the potential energy of the charge carriers. Notice how this corresponds with the formal terminology in which voltage is called "electrostatic potential".

Alternatively, an ideal wire is one in which the charge carriers don't lose any energy when moving along the wire. Carriers moving along resistor do lose electric potential energy by imparting kinetic energy elsewhere.

Think of voltage as potential energy in the electrons. Electrons repel each other and this force is very, very strong so it only takes a tiny, tiny change in distance between them to result in a massive change in potential energy, which is why you do not see "piling".

Electrons on one end of a resistor have less energy than those at the other end because they have expended some of their potential energy traveling through the resistor. Hence the voltage drop: you are measuring difference in potential energy of the electrons between each end of the resistor.

Voltage is always measured between two points; A single point in a circuit does not have a voltage. Knowing this, think very carefully about what you mean when you said:

My assumption is that voltage is dropping all over the circuit and the resistor just happened to be between a potential difference in the circuit.

• What I meant by that voltage was the voltage between a specific point in the circuit and the negative terminal. Won't electrons lose their potential energy anyway, because they're moving in the circuit instead of staying fixed at a certain potential difference from the negative terminal? Please clarify, sir :D Commented Apr 21, 2021 at 14:14
• Put your black probe on the negative terminal of the power supply. We'll call this the point of lowest potential energy. Then move your red probe to the positive terminal of the power supply and measure how much potential energy the electrons start off with. Then move the red probe around the circuit following the circuit measuring voltages as you go. The voltage relative to the negative terminal drops as you go, because the electrons are losing potential energy. Commented Apr 21, 2021 at 19:01

A resistor is a two terminal circuit element that enforces Ohm's law (V = I x R). All uniform solid materials (except superconductors) impede the flow of electrons to some extent. Even a copper wire has resistance. This is very similar to friction except at the electron level. As electrons are forced through a resistor, energy is converted to heat inside the resistor because of this "friction."

The resistivity of a material depends on the atomic structure and how many free electrons are available. In some materials, it may not be free electrons but some other mobile charge. But let's just stick with electrons for now.

If there are lots of highly mobile electrons in a metal, it will have low resistivity, and a small voltage will produce a relatively large current. If there are few mobile electrons, then it will have higher resistivity, so a comparatively higher voltage will be needed to produce the same current. So it is not so much that electrons pile up on one side of a resistor. It is just that within the resistor itself, the electrons have a hard time moving. So some voltage is required to force the electrons through. The resistive region is not concentrated in a single point. It is distributed over 3 dimensions inside the resistor element.

In my opinion, the best way to understand resistors at the circuit level is just that they enforce Ohm's law. But it is good to be curious about the dynamics of electrons inside conductors and resistors too, I guess.

• Charge builds up on the interface where a circuit changes linear resistivity, rather than being distributed uniformly throughout the resistor. Within (straight) sections of conductor (or resistor) of uniform resistivity, there is a uniform E field. Hence there cannot be a distributed (net) charge within them. Commented Apr 21, 2021 at 11:18

There is, in fact, an accumulation of charge, particularly at the interface between the conducting wires and the resistor. Whether you want to call this 'piling' is a question of semantics.

Something has to push the charge carriers through the resistor, and that something is an electric field. The electric field is extremely small in the conducting leads (zero in the case of an ideal conductor), and charge accumulates around the resistor in such a way as to create that electric field that pushes the charge through the resistor. As the electrons travel through the resistor, there is a drag-like force from their interactions with atoms, and so they are steadily dissipating energy that they are picking up from the local electric field created by those charges.

Ultimately, it is the voltage source that drives all this, but charges redistribute themselves to satisfy the requirement that the electric field in the conducting wire is extremely small (again, zero in the case of an ideal conductor), and to satisfy the local conditions inside the resistor that $$\\mathbf{J}=\sigma \mathbf{E}\$$ at all points inside that resistor.

• A minor quibble. In an ideal conductor, E field is not 0 when current is changing. Some people incorrectly state that an E field cannot exist in an ideal conductor. It can and does, but not with current in a steady state. Commented Apr 21, 2021 at 12:15
• Yes, I agree with you. I think that introducing this idea at this level of the discussion would cause confusion, but from a physics standpoint, your point is correct. Commented Apr 21, 2021 at 12:18

You see a potential difference (voltage drop) across a resistor because the potential at the terminal of the resistor connected to the positive terminal of the power source has a higher potential than at the other terminal connected to the negative terminal of the power source. The reason for the drop in potential is simple, let me explain. When two ends of a metal is connected to a power-source an electric field is setup in the metal where the field lines start from the end of the metal connected to the positive terminal of the power source and end the the end connected to the negative terminal of the power source. This set-up electric field causes the free electrons to gain energy and move from the one end of the metal to the other end then from there across the power-source through the wires. Now, as these free-elctrons move inside the metal they collide inelastically with the stationary atoms and ions present in the metal. During the inelastic collision and due to it the elections loose some of their energy. Further as the electrons move they encounter more atoms and ions and collide with them causing further lose in energy of the electrons. And by the time the electrons reach the other end of the metal they would have lost a significant amount of energy. Since there are many free-electrons moving in the metal there will be a collective loss in energy. This collective loss in energy of the electrons as they move from one point to another in a metal is what is called as a voltage drop. This means the hindrance (resistance) caused by the stationary atoms to the movement of electrons causes voltage drop.

So, remember- "resistance in a conductor causes voltage drop across the conductor."

Also, Voltage drop (V) in a conductor is related to the resistance (R) in the conductor when a current (I) flows through the conductor. This relationship is given by the Ohm's law as:

V = I*R.

And coming to why there is no voltage drop across the entire circuit?
There is.
There is a voltage drop across any two nodes along the circuit because any circuit consists of resistive elements only. You can measure the voltage drop by connecting the red probe of a voltmeter to say node A on the circuit and the black probe to the ground and note the potential (say V1) displayed by the voltmeter. Then connect the red probe to say node B on the circuit while the black probe stays connected to ground. Note the potential shown by the voltmeter (say V2). Now subtract V2 from V1 you will get the voltage drop across the part of the circuit between node A and node B. Now the reason as to why we talk about voltage drop across resistors but don't talk about voltage drop across the entire circuit is simply because resistors, nichrome wires, etc. have a significant amount of resistance where as other components usually don't have a significant amount of resistance. That's the exact reason why we use a resistor to cause electrical resistance and not use any other component.