I have been asking myself this question for a long time now. Suppose we have two resistors in series connected to a voltage source. Simply put, does the voltage drop on each resistor mean that there are more electrons on one side of the resistor compared to the other side? Because if an analog voltmeter is connected in parallel to that resistor, current will flow through the voltmeter.

  • $\begingroup$ As the answer suggests, electrons have charge, but voltage is the force which causes charge to move, i.e. current. Think of it as a river: there are water molecules everwhere; the density of molecules is the same everywhere. Here gravity rather than voltage makes the current flow. $\endgroup$ Oct 27, 2014 at 11:42
  • $\begingroup$ Here, the drift velocity of the free electrons create the current flow, and thus we have Voltage. As the electrons goes through an electronic appliance the drift velocity of the electrons gets decreased and thus we obtain potential drop. $\endgroup$ Jun 15, 2016 at 18:35

3 Answers 3


Simply put, does the voltage drop on each resistor mean that there are more electrons on one side of the resistor compared to the other side?

Yes, the charge density at one end of the resistor must differ from the other if there is a current through.

Consider, for simplicity, a resistive element of length $L$, area $A$ and resistivity $\rho_r$.

Joined to each end are conductors with area $A$ and resistivity $\rho_c$.

Assume a constant current density of magnitude $J$ along the length of the conductors and resistive element.

The electric field, in the direction of $J$, within the conductors and resistive elements is given by $E_c = J \rho_c $ and $E_r = J \rho_r $ respectively.

Then, at the boundary between the conductor and resistive element, the electric field abruptly changes in value by

$$\Delta E = \pm J\left(\rho_r - \rho_c \right) \approx \pm J\rho_r\;,\quad \rho_c \ll \rho_r$$

This implies a charge density at each end of the resistive element.

See, for example:

enter image description here

For another approach, consider that the slope of the electric potential changes abruptly at the interface.

For a steady current and assuming essentially ideal conductors, the electric potential along the conductors is constant but begins changing with distance inside the resistive element (it must since there is potential difference between the ends of the resistor).

Again, this implies an abrupt change in the electric field at the boundary which requires a charge density at the boundary.

  • $\begingroup$ You said "Yes, the charge density at one end of the resistor must differ from the other if there is a current through." Would this still be true if you had a perfectly uniform circular loop of wire with some nonzero resistance (so you could pick any arbitrary section of it and call that section a 'resistor'), and a current was induced in it by an external magnetic field that was spatially uniform and perpendicular to loop's plane, but changing with time? It seems like the spatial symmetry here would imply the charge density would be uniform throughout the wire, even though current was flowing. $\endgroup$
    – Hypnosifl
    Oct 28, 2014 at 12:11
  • $\begingroup$ Are you defining "resistor" in some more narrow way that excludes some lengths of material with nonzero resistance? If so, what is the element of the definition that excludes my example? What if instead of a loop of wire I imagined a loop of resistors (of the kind you would get from an electronics supplier) connected end-to-end in a loop, would your statement still not be intended to cover this scenario? By symmetry, it seems that in this scenario there would still be current flowing without a charge density difference from one end of each resistor to the other. $\endgroup$
    – Hypnosifl
    Oct 28, 2014 at 12:38
  • $\begingroup$ @Hypnosifl, the context of the question and answer are clear. I have to now assume bad faith on your part. It is now my judgement that you are intentionally interpreting the answer out of context due to some motivation I do not care to know. This is all I shall say about this. $\endgroup$ Oct 28, 2014 at 12:53
  • $\begingroup$ "Bad faith"? That's pretty uncharitable, physicists tend to be interested in defining precisely what conditions must be met for a claim to be true, that's all I'm curious about here. It may be that you are making some implicit assumptions about conditions under which your statement is true, conditions which would be met in the original example but wouldn't be met in mine--if so I'd be interested to see them stated explicitly. Would it be sufficient to say there are no external fields, for example, or would it still be possible to come up with exceptions? $\endgroup$
    – Hypnosifl
    Oct 28, 2014 at 13:00

Try the hosepipe analogy. Voltage is the pressure needed to push the water through the pipe. It's diameter is the resistance, and the amount of water going through it the current.


The reason it's called a voltage "drop" is because there exists an almost exact analogy between voltages and gravitational potential — that is, height.

Imagine electrical current as a real, moving fluid, which is precisely how Ben Franklin viewed it (though he got the difrection of flow wrong!)

The amount of this fluid flowing per second across some cross-section of a path is called amperage, regardless of how quickly or slowly it is moving. Thus the Mississippi River would have an absolutely enormous amperage, even though most of the year its current (notice the parallel use of words!) moves relatively slowly.

What almost always makes water flow over land, though, is differences in the relative height of the water. Without that connected height difference, you just end up with a pond. When the fluid is electricity, such ponds are called static electricity.

Over the length of the Mississippi River for example, there is a drop in height — and thus also of the proportional metric of gravitational potential energy — of several thousand meters. As long as there exists a path over which the water can always flow downhill — which in electrical terms you would call a conduction path — then water will flow from the higher points to the lower points along the river.

Notice, though, that this idea of a drop in height is completely independent of how much water is moving, or even if the water is moving at all. If you dam up the river you can for example create a huge potential drop over a very shot distance, even though no actual flow may be taking place at a given time.

But when you do open up a path, watch out! That big drop over a very short distance means that all the fluid upstream is going to want to move down, and it will give off a lot of potentially destructive energy as it does so. You had best be well prepared to handle it!

A large car battery works in much the same way as such a dam, except that stores the functional (not literal, that would be a capacitor) equivalent of two different "heights" of electrical "fluid."

This "height difference" in a battery is not a real vertical height, so it has its own different name: voltage, which measures the electrical energy potential per unit of electrical fluid. It is a measure of energy stored per unit of fluid, and it very closely parallels the gravitational potential energy of water kept at a higher location.

Essentially every kind of electrical circuit imaginable can be visualized with high accuracy by using this water analogy, though it begins to weaken when magnetic fields are involved.

For example, a low-amperage, high-voltage link corresponds to a tiny stream falling over a high cliff. The energy released per cubic centimeter of water (or per coulomb of electrons) is very high, but both the total flow of water (or amperage of electrons, which is coulombs pet second) and total energy released at the bottom stay low because so little water (so few electrons) are available.

Conversely, a really big 1.5 volt battery can melt a thin wire because it allows the wire to carry every electron it is capable of holding. It's like taking dumping the entire contents of an inflatable outdoor pool into a shallow ditch: the sheer volume of it will do damage, even though the energy released per liter of water is low.

Common electrical components all have simple water-as-fluid analogies. A resistor is like a very rough, convoluted stream bed, e.g. one full of rocks of various sizes, that slows the flow and converts its energy I to heat. As resistance increases the bed eventually becomes impassibe — an insulator.

A capacitor is an elastic membrane that blocks direct flow, but easily allows small back-and-forth motions that do not overstretched its capacity.

Even inductors have a rough equivalent, since for example water flowing in a loop with many coils will have enough kinetic energy to keep flowing if an attempt is made to shut it off. Combine that effect with a membrane-style "water capacitor" and voila, you get a resonator with a tuned frequency.

For rivers, the ocean (usually) is the lowest available height available for extracting water energy, and so corresponds to the electrical concept of "ground." But since it is a difference in height that defines both gravitational potential and voltages, "ground" can also simply be the lowest potential locally available, such as when it is defined as one of the battery terminals of a mobile device.

  • $\begingroup$ Thank you for your answer. If we have to use the water analogy, I could consider a large tank of water at some height above the ground and a water pump on the ground, pumping water into the tank. The tank has a very small hole at the bottom. Is it correct to assume that, right when I turn the pump on, waters starts to fill the tank, increasing the pressure on the "hole", until the flow inside equals to the flow outside? @Alfred Centauri commented that there is indeed a difference in charge density before and after the resistor. $\endgroup$
    – user42768
    Oct 27, 2014 at 19:44
  • $\begingroup$ The pressure in your pump must be high enough to raise a thin column of water, one the same diameter as your "very small hole," up to the top water level of the tank. Once you have that pressure — or voltage — you can "trickle charge" the tank with it, though it may take some time. Something is not quite right about the charge density point you mentioned, though, since for example the instantaneous electron charge density in equal length wire segments on either side of an active resistor is going to be the same. @AlfredCentauri, might you be able to clarify what you meant? $\endgroup$ Oct 27, 2014 at 22:29
  • $\begingroup$ Terry, for a steady current density, the electric field must change abruptly at the boundary between a good conductor and a resistive element. This implies a charge density there. $\endgroup$ Oct 28, 2014 at 0:46
  • $\begingroup$ @AlfredCentauri, ah, no argument there, the electrons must bunch up a bit in the slow stretch, just like cars encountering slow stretch on a highway. User42768, density is not a feature that carries over in the water potential analogy, particularly since we think of water conceptually as an incompresible liquid. $\endgroup$ Oct 28, 2014 at 3:12

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