This post is my best effort to seek assistance on a topic which is quite vague to me, so that I am struggling to formulate my questions. I hope that someone will be able to figure out what it is I'm trying to articulate.

If we have a circuit with a resistor, we speak of the voltage drop across the resistor.

I understand all of the calculations involved in voltage drop (ohm's law, parallel and series, etc.). But what I seek is to understand on a conceptual level what voltage drop is. Specifically: what is the nature of the change that has taken place between a point just before the resistor and a point just after the resistor, as the electrons travel from a negatively to a positively charged terminal.

Now as I understand it, "voltage" is the force caused by the imbalance of charge which causes pressure for electrons to travel from a negatively charged terminal to a positively charged terminal, and "resistance" is a force caused by a material which, due to its atomic makeup, causes electrons to collide with its atoms, thus opposing that flow of electrons, or "current". So I think I somewhat understand voltage and resistance on a conceptual level.

But what is "voltage drop"? Here's what I have so far:

  • Voltage drop has nothing to do with number of electrons, meaning that the number of electrons in the atoms just before entering the resistor equals the number of atoms just after

  • Voltage drop also has nothing to do with the speed of the electrons: that speed is constant throughout the circuit

  • Voltage drop has to do with the release of energy caused by the resistor.

Maybe someone can help me understand what voltage drop is by explaining what measurable difference there is between points before the resistor and points after the resistor.

Here's something that may be contributing to my confusion regarding voltage drop: if voltage is the difference in electrons between the positive terminal and the negative terminal, then shouldn't the voltage be constant at every single point between the positive terminal and the negative terminal? Obviously this is not true, but I'd like to get clarification as to why.

Perhaps I can clarify what I'm trying to get at with the famous waterwheel analogy: we have a pond below, a reservoir above, a pump pumping water up from the pond to the reservoir, and on the way down from the reservoir, the water passes through a waterwheel, the waterwheel being analogous to the resistor. So if I were to stick my hand in the water on its way down from the reservoir, would I feel anything different, depending on whether I stuck my hand above or below the waterwheel? I hope that this question clarifies what it is I'm trying to understand about voltage drop.

EDIT: I have read and thought about the issue more, so I'm adding what I've since learned:

It seems that the energy which is caused by the voltage difference between the positive and negative terminals is used up as the electrons travel through the resistor, so apparently, it is this expenditure of energy which is referred to as the voltage drop.

So it would help if someone could clarify in what tangible, empirical way could we see or measure that there has been an expenditure of energy by comparing a point on the circuit before the resistor and a point on the circuit after the resistor.

EDIT # 2: I think at this point what's throwing me the most is the very term "voltage drop".

I'm going to repeat the part of my question which seems to be still bothering me the most:

"Here's something that may be contributing to my confusion regarding voltage drop: if voltage is the difference in electrons between the positive terminal and the negative terminal, then shouldn't the voltage be constant at every single point between the positive terminal and the negative terminal? Obviously this is not true, but I'd like to get clarification as to why."

In other words, whatever takes place across the resistor, how can we call this a "voltage drop" when the voltage is a function of the difference in number of electrons between the positive terminal and negative terminal?

Now I've been understanding the word drop all along as "reduction", and so I've been interpreting "voltage drop" as "reduction in voltage". Is this what the phrase means?

Since I've read that voltage in all cases is a measurement between two points, then a reduction in voltage would necessarily require four different points: two points to delineate the voltage prior to the drop and two points to delineate the voltage after the drop, so which 4 points are we referring to?

Perhaps a more accurate term would have been "drop in the potential energy caused by the voltage" as opposed to a drop in the voltage?

EDIT # 3: I think that I've identified another point which has been a major (perhaps the major) contribution to the confusion I've been having all along, and that is what I regard as a bit of a contradiction between two essential definitions of voltage.

When we speak of a 1.5V battery, even before it is hooked up to any wiring / switches / load / resistors / whatever, we are speaking of voltage as a function of nothing other than the difference in electric charge between the positive and negative terminals, i.e the difference in excess electrons between the two terminals.

Since there is a difference in number of electrons only in reference to the terminals, I therefore have been finding it confusing to discuss voltage between any other two points along the circuit -- how could this be a meaningful issue, since the only points on the circuit where there is a difference in the number of electrons is at the terminals -- so how can we discuss voltage at any other points?

But there is another definition of voltage, which does make perfect sense in the context of any two points along a circuit. Here we are speaking of voltage in the context of Ohm's law: current/resistance. Of course, in this sense, voltage makes sense at any two points, and since resistance can vary at various points along the circuit, so clearly voltage can vary at different points along the circuit.

But, unlike the first sense of voltage, where the voltage is a result of the difference in electrons between the terminals, when we speak of voltage between two points along the circuit, say, between a point just before a resistor and a point just after the resistor, we are not saying that there any difference in number of electrons between these two points.

I believe that it is this precise point which has been the main source of my confusion all along, and that's what I've been trying to get at all along. And this is what I've been struggling to ask all along: okay, in a battery, you can tell me that there is a voltage difference between the two terminals, meaning that you can show me, tangibly and empirically, that the atoms at the positive terminal have a deficit of electrons, and the atoms at the negative terminal have a surplus of electrons, and this is what we mean by the voltage between the two, then I can understand that.

But in contrast, I accept that there is voltage (I/R) between a point just before a resistor and just after a resistor -- but can you take those two points, the one before the resistor and the one after the resistor, and show me any measurable qualitative difference between the two? Certainly there is no difference between the number of electrons in the atoms of those two points. In point of fact, I believe that there is no measurable difference between the two points.

Ah, now you'll tell me that you can show me the difference between the two points: you'll hook up a voltmeter to the two points, and that shows the voltage between them!

Sure, the voltmeter is telling us that something has happened between the two points. But the voltmeter does not tell us anything inherent in the points themselves -- unlike the two terminals of a battery, where there is an inherent difference between the two points: one has more excess electrons than the other -- that is a very inherent, concrete difference.

I guess what we can say is that the electrons travelling at a point just before the resistor are travelling with more energy than the electrons travelling at a point just after the resistor. But is there any way of observing the difference in energy other than a device that simply tells us that the amount of energy has dropped between the two points?

Let me try another way: we could also hook up a voltmeter to the two battery terminals, and the reading would indicate that there is voltage between the two terminals. And if I would ask you yes, but what is it about those two points that is causing that voltage, you could then say, sure: look at the difference in electrons between the two points -- that is the cause for the reading of the voltmeter.

In contrast, when we hook up the voltmeter to the points just before and after the resistor, and the reading indicates a voltage between the two terminals. But in this case if I would now ask you the same question: yes, but what is it about those two points that is causing the voltage, I'm not sure if you'd have an answer.

I think this crucially fundamental difference between the two senses of voltage is generally lost in such discussions.

  • 2
    $\begingroup$ Your three points are very good ones. To make your third one more specific: "The voltage drop has to do with the energy released PER electron caused by the resistor." $\endgroup$
    – jeffdk
    Mar 5, 2013 at 18:25
  • $\begingroup$ It might help to look at the operation of a voltmeter. If you attempted to measure the voltage in a wire on two points at the same side of a resistor you would see no voltage indication. If you chose points on opposite sides of the resistor, would see a voltage difference. If you wanted to see the voltage in a wire relative to ground, you would put one probe on the wire and the other on a conductive point outside the circuit. As pointed out in several answers, the voltage is a measure of potential, nothing more. $\endgroup$
    – Freedom
    Mar 8, 2013 at 2:02
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    $\begingroup$ I'll just point out one misconception I noticed throughout your question: voltage has nothing to do with the number of electrons (that's what current is). Voltage is the energy per electron. It has the units of joules per coulomb, where a coulomb is just a way of counting electrons. The same number of electrons leave the resistor as enter it (the current out is the same as the current in); but they have less energy going out than they did coming in, since they've give up some energy as heat. $\endgroup$ Mar 9, 2013 at 22:04
  • $\begingroup$ "unlike the two terminals of a battery, where there is an inherent difference between the two points: one has more excess electrons than the other -- that is a very inherent, concrete difference." That is not really correct. When the battery is drained it is no longer true. $\endgroup$ Mar 9, 2013 at 22:48
  • 1
    $\begingroup$ In the waterwheel analogy: the thing you can notice is that the water after the wheel is lower than the water before the wheel. $\endgroup$
    – user253751
    Jun 1, 2015 at 4:52

15 Answers 15


Perhaps I can clarify what I'm trying to get at with the famous waterwheel analogy

99 years ago, Nehemiah Hawkins published what I think is a marginally better analogy:

enter image description here
Fig. 38. — Hydrostatic analogy of fall of potential in an electrical circuit.

Explanation of above diagram

  • In this diagram, a pump at bottom centre is pumping water from right to left.
  • The water circulates back to the start through the upper horizontal pipe marked a-b
  • The height of water in the vertical columns C,m',n',o',D indicates pressure at points a,m,n,o,b
  • The pressure drops from a to b due to the resistance of the narrow return path
  • The pressure difference between a and b is proportional to the height difference between C and D


  • Pump = Battery
  • Water = Electric charge carriers
  • Pressure = Voltage
  • Vertical Pipes = Voltmeters
  • pipe a-b = Resistor (or series of four resistors)


  • A "particle" of water at a has a higher potential energy than it has when it reaches b.

There is a pressure drop across a "resistive" tube.

Voltage (electric potential) is roughly analogous to water pressure (hydrostatic potential).

If you could open a small hole at points a,m,n,o,b in the tube and hold your finger against the hole, you would be able to feel the pressure at those points is different.

The potential at some point is the amount of potential energy of a "particle" at that point.

it would help if someone could clarify in what tangible, empirical way could we see or measure that there has been an expenditure of energy by comparing a point on the circuit before the resistor and a point on the circuit after the resistor.

  1. Purchase a 330 ohm 1/4 watt resistor and a 9V PP3 battery
  2. Place the resistor across the battery terminals
  3. Place your finger on the resistor.
  4. Wait.
  • $\begingroup$ Unfortunately, I don't really follow this diagram. Perhaps if you explained how it works... $\endgroup$
    – oyvey
    Mar 6, 2013 at 20:36
  • $\begingroup$ @oyvey: See updated answer. $\endgroup$ Mar 7, 2013 at 11:23
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    $\begingroup$ Sorry, even with your explanation, I find the water pump analogy difficult both to follow on its own and to relate to electricity. $\endgroup$
    – oyvey
    Mar 7, 2013 at 21:19
  • $\begingroup$ "Place your finger on the resistor". But read the part of my post that you cited just prior: "comparing a point on the circuit before the resistor and a point on the circuit after the resistor." $\endgroup$
    – oyvey
    Mar 7, 2013 at 21:20
  • $\begingroup$ @oyvey: My problem was that the only way I know to measure flow of energy into, and out of the resistor involves using voltmeters and ammeters which is somewhat empirical but involves a somewhat circular argument as more resistors are introduced. Viz measure V & I at each end, multiply V.I to get Joules/sec at each end, subtract to get "expenditure of energy". I worried you would find this equally unsatisfying. $\endgroup$ Mar 7, 2013 at 21:53

"voltage" is the force caused by the imbalance of charge which causes pressure for electrons to travel from a negatively charged terminal to a positively charged terminal,

Nope, voltage is not a force. Voltage is a difference in potential energy per unit charge. More precisely: electric potential is the potential energy per unit charge (just like $gh$ is the gravitational potential energy per unit mass), and a voltage (a.k.a. voltage difference a.k.a. voltage drop) is a difference in electrical potential between two points.

The actual value of electric potential at any point has no physical meaning; only its difference relative to the electrical potential at some other point, i.e. the voltage, is meaningful or measurable. This means the whole idea of voltage is inherently bound to a choice of two points. There's no measurement you can make at a single point only that will tell you anything about voltage or electric potential. However, if you have two points, you can determine the voltage between them by pushing a unit charge from one point to the other and measuring how much work it takes (or gives). This is how we can establish voltages in a circuit with resistive elements: move a charge through the circuit from one point to another and see how much energy needs to be put in to get it there.

The reason it takes energy is fundamentally complicated, having to do with quantum mechanical effects, but as a rough classical model, you could say that the electrons lose energy from colliding with the atoms and molecules of the resistive material, and you need to put in enough energy to make up for those losses.

  • $\begingroup$ "Nope, voltage is not a force." From the website allaboutcircuits.com: 'The force motivating electrons to "flow" in a circuit is called voltage.' $\endgroup$
    – oyvey
    Mar 7, 2013 at 21:01
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    $\begingroup$ @oyvey that website is wrong. Voltage is absolutely not a force. $\endgroup$
    – David Z
    Mar 7, 2013 at 22:36
  • 1
    $\begingroup$ I think that the website and I are using "force" in a much looser way than you are. $\endgroup$
    – oyvey
    Mar 8, 2013 at 11:47
  • 13
    $\begingroup$ Perhaps so. In that case I would suggest that, at least for purposes of this question, you use "force" to mean what it actually means in physics, rather than whatever looser definition you may have in mind. $\endgroup$
    – David Z
    Mar 8, 2013 at 19:18
  • $\begingroup$ @DavidZaslavsky, you're right. But if you subject the electron to an electric field (the field being caused by a potential difference), the electron will feel a force according to the strength and sign of the field, and that force is what causes the electron to move. Which I think is in the same spirit as what oyvey is trying to say. I just think that the terminology, though important, can be given some leeway in this case, because it's not what is being asked about? $\endgroup$ Mar 10, 2013 at 4:43

Voltage (Joule per coulomb) is a measure of electric potential energy gained by a positive test charge, or the work done in moving a positive test charge from infinity to a point in a positive electric field. This energy gained is due to the conservative electrostatic force between charge. When a charge gains potential it naturally does work equal to its gained energy to restore its neutral position, by this definition infinity.

Potential difference is a measure of the voltage between two points in an electric field. Potential difference is potential relative to a fixed point in an electric field instead of infinity. Potential difference should not be confused as the difference in charge between two terminals. The difference is what causes a voltage (energy gained by a test charge in an electric field). Unless there is a difference in charge, no field can be established and hence there is no potential.

In a circuit, a battery provides the difference in potential between two terminals by doing work against the electric field, to give electrons potential energy. Once these charges get this potential energy, they naturally do work to get to their neutral positions by changing this potential energy to kinetic energy which drives current through the circuit. But since the electrons travel through a medium they lose some of this energy as heat due to collision with atoms or molecules. A measure of the energy lost/expended between two points in a circuit is known as potential drop. The drop in potential increases with resistance.

Due to the law of conservation of energy, the sum of all voltage drops must equal the applied voltage as stated by kirchoff's second law. Hence, there will be a greater voltage drop across a load in a circuit if the total resistance of the circuit is lower.


I'm not a physicist, rather a simple engineering student, and I believe your confusion has confused me, too. That's great, because it made me look for the answers myself. This is just my personal guess (though grounded in some fact), but it makes a lot of sense to me, and I hope it makes sense to you, too.

Bear with me if this is a long answer.

First off, when we speak of the difference in the amount of electrons just before and just after a resistor, you're right: there is fundamentally no difference in the number of electrons. We know this because all the electrons (current) that comes in must be equal to the number of electrons (current) that comes out. This is one of Kirchoff's laws. Moreover, we know that matter doesn't just disappear and it isn't being converted into energy, either.

Of course, the second of Kirchoff's laws tell us that the sum of all the voltages in a circuit must be zero (so, in a simple circuit, the initial voltage from the battery minus all the voltage drops from all the resistors is zero). That tells us that something must already be happening within those resistors that has nothing to do with the battery. So that's a good start.

Then, we should also understand that current and voltage are not the same thing. That is, if you already have a flowing current, that current does not have its own voltage. You might need voltage to jumpstart the flow of current, but once it's already flowing, the current exists as the jumping of one electron across ions, and each electron is only a single point -- it doesn't have voltage because it's only one point. More importantly, current flows uniformly, so there is no voltage that is being generated as it does flow. Whether it is just before or just after the resistor, the current is the same there, and it produces no voltage.

Here's a drawing I made for you:

Drawing of current

I drew it as it really is, against the convention. Notice that, despite there being a pre-existing voltage difference between the two batteries, the distribution of electrons is uniform. So, if you tried to take a voltage difference between the first electron and the fourth electron, you get a voltage difference of zero, even if they are flowing, because their distribution will continue being uniform. This, I think, is the equivalent of measuring the voltage across a wire with no resistor.

Because I think it will be useful in analyzing what happens across the resistor, I'd like to point out that the battery is the initial source of voltage. It's made so that one end is much more electronegative than the other, and that one pole has a large supply of electrons whereas the other has a large deficit. So, if you link the two ends together, the electrons flow from the less electronegative end (with more electrons) to the more electronegative end (with less electrons). In the process, the difference in electrons from pole to pole equalizes. Moreover, the chemical reactions happening inside the battery make it so that the poles (anode for negative, cathode for positive) degrade both cathode and anode -- unless it's a rechargeable battery, in which case this reaction is reversible.

But that doesn't quite spell out why there is a voltage drop across the load. If the voltage difference exists between the poles, then shouldn't the same voltage difference exist all throughout the circuit? The answer seems to be yes... but only if there is no load.

So, we have to go back and ask, what is a voltage drop, and where does it come from?

I think your question really is, what happens to the energy at a molecular level? It doesn't just disappear. When the voltage drops, what does it mean? If the same amount of electrons enter the resistor and then leave it, doesn't their energy stay the same?

First, we know that if their energies are the same, then the voltmeter will tell us there is no voltage. We also know that, if their energies are the same, there is no difference between the two points. And finally, if their energies are the same, we know that no energy was lost, and therefore, no work was taken out of the system.

But we do know that work was taken out of the system, and we can see the voltmeter giving us a reading. But if the number of electrons is the same across the resistor, then there must be energy being lost. But, where did that energy come from?

We go back to what voltage is -- it is the amount of energy per unit charge. One unit of charge is the electron. So voltage is just the measure of how much energy each electron carries. There are two separate energies in the electron -- its rest energy and its momentum energy. Its rest energy is constant, but its momentum energy is definitely not.

An electron carries energy by its orbital and shell. That is, the more energy an electron has, the higher up it goes into its shell. I think this is because it has more kinetic energy, so it tends to move faster. Moving faster causes it to occupy a larger volume, and so for the nucleus to keep the electron, there must be more pull to keep it in place. I'm imagining it's like the orbit of a planet around the sun, where wider orbits signify more total energy in the system. The orbitals are (I'm guessing, but fairly certain) caused by the discrete nature of energy -- photons are the most irreducible form of energy, so an electron can only jump between orbitals by discrete amounts.

So, inside the battery, the electrons feel a potential difference between the different poles of the battery. This potential difference imparts energy into the electron, causing it to jump to a higher orbital. It then travels through the atoms in the wire until it reaches the resistor.

Inside the resistor, energy is taken out of the electron. The electron jumps to a smaller orbital, and in the process it releases a photon. This release of a photon means it has given off energy. The photon, having energy, is converted to work and waste.

The electrons, now occupying a smaller orbital, exits the resistor with less energy.Because their energy is smaller, their voltage drops. And this, I think, is the reason why it is called a voltage drop.

This answer raises the question, though -- how does a voltmeter detect this voltage drop, if it occurs in the orbital shells of an atom? And the answer is interesting: it doesn't. At least, not directly.

From what I can read, what voltmeters do is measure the deflection of a pointer against a spring. The pointer, in turn, is moved by the repulsive forces of the electrons on a pivoting wheel inside the voltmeter. And the repulsive forces is, of course, directly proportional to the amount of energy inside the electrons. Because we know the equations of energy, all this is calibrated. What the voltmeter actually measures is not the voltage itself, but rather the force that the voltage difference exerts on the pointer of the voltmeter.

I can't say much more on how the voltmeter works. I'm frankly not so sure if that's true, but I know that the moving current with a certain voltage must work in some way like I just said.

When the electron comes back home to the cathode (because we're thinking in terms of actual electrons moving, rather than the conventional electron holes moving), its voltage will not be zero, but rather will be equal to the energy per eletron in the cathode. Kirchoff's law tells us that at every load, the voltage difference decreases due to the voltage drop. Because the electrons still have energy inside the cathode, this doesn't mean the voltage is zero there (causing electrons to have no energy, which intuitively makes them have no momentum, which can't be true). It just means that the energy when the electron comes back is now the same as the energy at the cathode, so there is no more potential difference and the journey of the electron stops.

If I've made any glaring errors here (and I'm bound to have, as most of this is me guessing and trying to make sense of what I know), please feel free to edit me. And if I mis-explained something, please point out where in a comment, so I and everybody else can see where it's wrong and why. Thank you!

EDIT: I realize I used voltage in two ways here: the difference in potential between two points, and the energy per unit charge. One definition requires two points to be defined, whereas one requires only one point.

I think we can reconcile this by imagining that there is the same voltage everywhere. Each electron has exactly the same amount of energy. In such a case, there is no voltage difference. There is no potential difference, meaning there is no gradient which will cause the electrons to move.

So I guess I'm saying, voltage can be a point quantity, whereas the potential difference, or voltage difference, requires two points to be defined. Certainly, this is consistent with the definitions of voltage (and as I remember in class, voltage by itself is a scalar, so it only requires itself to be defined, not another point; the gradient or difference is what requires another point).

  • $\begingroup$ "One definition requires two points to be defined, whereas one requires only one point." There is an implicit definition of some other set of points which mark $V=0$ (AKA "ground"). They really are the same usage. $\endgroup$ Mar 10, 2013 at 6:57
  • $\begingroup$ Thanks. I realize that now (the voltage at that one point is just the work required to pull it from infinity, right?), but in conceptual usage it sounded like only one point. $\endgroup$ Mar 10, 2013 at 10:00

Voltage is an electrical potential (relative some arbitrary value called "ground").

What it mean is that if I take an electron from ground and move it to a point with voltage $V$ it requires that I do work $W = V e$ (here $e$ is the magnitude of the charge on the electron) because there was a force $\vec{F} = q\vec{E}$ due to the electrical field, $\vec{E}$ along path and $W = \int \vec{F} \cdot d\vec{s}$, conversely moving the electron from a point of potential $V$ to a point at the potential of ground gets back the same amount of energy.

This is exactly like the gravitational potential energy of mass $m$ relative it's position on the floor. $V$ is analogous to $gh$ in introductory mechanics (where the work of lifting the mass is $W = mgh$ and the force is $F_g = -mg\hat{z}$).

The thing to note is that $V$ is a property of a particular position at a particular time, and if you look at two points along a circuit and find that the one "further along" has lower potential then you can say that the potential "dropped" by $V_1 - V_2$ between points 1 and 2.

That is the whole meaning of "voltage drop", but it does not explain the microscopic physics that are responsible for the change. Like many other things in physics it is easier to get a handle on the physics if you will just agree to accept the meaning of the symbols and vocabulary before you begin. When you understanding matures it will be clear that definitions are internally consistent and useful for performing calculation.


I think what your getting at is a question about the energy associated with the electron shell states of a material before a resistor and after a resistor. So in this sense one can think in terms of relative ionization in two different materials. Good thought experiments in this regard are to consider how a battery works, and then consider how a transistor works.

In a battery, (as you probably already know) the potential energy is stored chemically, and it is the interaction of two materials when there is an ion bridge that allows for the reaction to occur and give off energy. The important piece of course being the bridge that completes the circuit and allows for a flow of ions. So in any case, there are higher entropy states (more energetically favorable) available to the system, and flow of electricity can be seen as a response to achieve the more energetically favorable conditions. So a voltage drop can be understood to not only involve a change of energy, but also a change in entropy as well.

The second example, the transistor, one of the layers is doped so that there is natural bias in the distribution of the electrons. This bias can be used as a resistance to current flow, and in most cases transistors are used as switches and changing the voltage in the "gate" allows one to control current flow. Again, this is a change in the distribution of the electrons in the "shells" of the relevant ions.

A voltage drop then is viewed as a change in charge, or ionization across a resistor. It is a direct control of the ionization that allows for flow.

This brings up the point that one of the surest ways to determine if a circuit design is faulty is to look for instances in the diagrams where there are missing grounds. Current will not flow if there is no ground which would cause a more energetically favorable condition in the circuit.

If this is helpful, I can expand, but I think several of us are struggling with understanding what you are after.


The question now appears to be whether one can tell a difference between two points across the resistor other than using a voltmeter? What can you tell me about the difference between those two points?

It helps to understand a little about Energy, Power, Voltage, Current and Charge.

In its simplest mathematical definition, Energy is the product of Power and time.


where time is actually an interval so that

$$E=P(t_2 - t_1) $$

Power is most simply defined as :

$$P = \dfrac{QV}{(t_2 - t_1)}$$

Where $Q$ is charge and $V$ is voltage

Charge is actually very well defined as

More abstractly, a charge is any generator of a continuous symmetry of the physical system under study. When a physical system has a symmetry of some sort, Noether's theorem implies the existence of a conserved current. The thing that "flows" in the current is the "charge", the charge is the generator of the (local) symmetry group. This charge is sometimes called the Noether charge.

Thus, for example, the electric charge is the generator of the U(1) symmetry of electromagnetism. The conserved current is the electric current.

In the case of local, dynamical symmetries, associated with every charge is a gauge field; when quantized, the gauge field becomes a gauge boson. The charges of the theory "radiate" the gauge field. Thus, for example, the gauge field of electromagnetism is the electromagnetic field; and the gauge boson is the photon.

Sometimes, the word "charge" is used as a synonym for "generator" in referring to the generator of the symmetry. More precisely, when the symmetry group is a Lie group, then the charges are understood to correspond to the root system of the Lie group; the discreteness of the root system accounting for the quantization of the charge.

It should be added that charge is quantized in the real world, and electrons carry a fundamental unit of charge, however, it is not necessarily electrons that are flowing when we talk about a circuit, rather it is the more abstract conserved quantity of charge that is flowing. The electrons may move in the circuit, but the physical distance they are moving will be very small. The charge is moving, and since charge by itself is effectively massless, it can move very rapidly in the circuit, however it is still fundamentally quantized.

Looking back to the Power equation:

$$P = \dfrac{QV}{(t_2 - t_1)} $$

One can see that since energy is the product of power and time, one can derive energy as the simple product of charge and voltage.

$$E = QV$$

So the difference of energy between two points is

$$E_2 - E_1 = (QV)_2 - (QV)_1$$

However charge is conserved, which means that the number of fundamental charges associated with a current flowing between two points must be the same at the beginning and end of the flow. Therefore we can assume $Q$ is constant and we have:

$$E_2 - E_1 = Q(V_2 - V_1)$$

This tells us that the change in energy between two points is directly proportional to the change of voltage (or drop in voltage) between the two points. Here we see the similarity of this equation to the equation for the potential energy in gravitational field. If we define weight as:

$$\text{Weight} = F_g = mg$$

where $m$ is mass and $g$ is the acceleration due to gravity, then the change of energy between two points in a gravitational field is:

$$E_2 - E_1 = F_g(h_2 - h_1)$$

where $h$ is height.

Here the similarity of the two equations should be evident.

Classically the units of charge (coulomb - C) are given as the product of unit of capacitance (Farad - F) and the units of voltage (V).

$$1C = 1F \times 1V $$

Where capacitance is simply a proportionality constant relating charge and voltage, which is more clear in the time varying equation:

$$i(t) = C \dfrac{dv(t)}{dt} = Cv'(t)$$

The energy (or equivalently work) emitted by a resistor over time is:

$$W = E = \int_{t_1}^{t_2} i(t)v(t) dt$$

Substituting we have

$$W = E = \int_{t_1}^{t_2} Cv'(t)v(t) dt$$

As we have shown above the voltage is analogous to height (or position), so $v'(t)$ would be analogous to velocity (time varying position).

If we look at the integral for mechanical work (where Force and velocity (s'(t)) are in one dimension):

$$W = E = \int_{t_1}^{t_2} F(t) s'(t) dt$$

One could regroup the electrical version as:

$$W = E = \int_{t_1}^{t_2} Cv(t)v'(t) dt$$

This suggest that the quantity $Cv(t)$ is analogous to force (but it is not force since force is a mechanical term and the units here are different, the relationship here is shown so that an analogy can be understood). Charge can also be written as

$$Q(t) = Cv(t)$$

So our integral becomes

$$W = E = \int_{t_1}^{t_2} Q(t)v'(t) dt$$

Charge again can be considered a constant since it must be conserved, so we can write:

$$W = E = Q\int_{t_1}^{t_2} v'(t) dt$$

and since

$$v(t) + const = \int_{t_1}^{t_2} v'(t) dt$$

We can write

$$W = E = Q[v(t_2) - v(t_1)]$$

So again we find that work or energy is proportional to a change in voltage with respect to some variable, where the proportionality constant is the charge.

This is enshrined in the definition of the energy as an electronvolt, where

Historically, the electron volt was devised as a standard unit of measure through its usefulness in electrostatic particle accelerator sciences because a particle with charge q has an energy E=qV after passing through the potential V; if q is quoted in integer units of the elementary charge and the terminal bias in volts, one gets an energy in eV.

The point of all the above is to illustrate the following:

  1. Voltage is not a force, it is a potential analogous to height in a gravitational field.
  2. By means of analogy, if voltage is a potential, then charge would be the analogy to force, but it is not "force" as used in physics, since that is a mechanically defined term and is equivalent to mass times acceleration.
  3. Charge is quantized, and it is also a conserved quantity.

So we can readily relate a change in voltage to a change in energy. So when there is a voltage drop across a resistor, there is a change in energy. For the resistor this energy is usually given off as heat.

If there is no change in voltage, then their will be no change in energy. So in a circuit, the energy contained in the wire before the resistor is greater than the energy contained in the wire after the resistor. Again, this is reflected in a change in the configurations accessible to the electrons in the wire.

I hope this is helping. Let me know if we need to expand more.


Consider a circuit with just a battery and a loop of wire. The wire has ideally zero resistance, that means the Voltage between any two points on the wire will be zero! $V=IR= I\times0=0$ $^1$

Now consider the circuit you mention (i.e. with a resistor). The resistor will cause a decrease in current flow - following ohms law which you know so well. What does that mean? Well the voltage across the resistor must be equal to the that of the source. I.e. $|V_{\textrm{resistor}}|=IR=|V_{\textrm{battery}}|$. Note that ideally, no voltage drop is present if you probe the voltage across any segment of wire (just as before).

Why is it called a voltage drop? The sum of all voltages in a loop must yield zero. So the technically we would have $V_{\textrm{battery}} +V_{\textrm{resistor}}=0$. (See Kirchhoff's circuit laws, specifically Kirchhoff's Voltage Law)

Speaking of 'voltage drops' is just an easy way to say Power from the source battery is going somewhere - as you pointed out "Release of energy". So in more complicated circuits you may keep track of things in terms of voltage drops across the various components.

Edit: Further example. Consider now a circuit with two resistors in series. The voltage drop across both in total must be as we discussed before but the voltage drop across each individual resistor will be weighted by its value.


Voltage drops:



$^1$ In real circuits, wires have a really low resistance.

  • $\begingroup$ The poster said that he/she understood the simple calculations that go along with voltages in circuits "(ohm's law, parallel and series, etc.)". Since your answer just restates the simplest of these with the typical black-box-don't-worry-about-the-physics type of answer, this is not answering the question asked at all. I believe the OP would like an explanation of the actual physics behind differences in voltages when current passes through simple circuit components. Such as the energy lost to collisions with the material of the resistor when electrons pass through and so on. $\endgroup$
    – user44430
    Mar 5, 2013 at 21:49
  • $\begingroup$ user444320: absolutely, this post simply restates the formulae I'm already familiar with from all of the usual basic electricity texts, and doesn't address the issue I asked at all. $\endgroup$
    – oyvey
    Mar 6, 2013 at 20:34
  • $\begingroup$ I take it back. I re-read this post (actually, I've been re-reading all of them several times) and it was somewhat helpful. My apologies. Fire: can you please read my remarks in edit # 2? $\endgroup$
    – oyvey
    Mar 9, 2013 at 12:00
  • $\begingroup$ I might need to think about this more but I've read Edit 2 a few times. Certainly a battery works by a separation of charge, but when in a circuit I think one must be careful about thinking about things in terms of "difference in number of electrons". Its all about flow (current). The Voltage is the potential difference which allows flow to occur (think gravitational potential difference allowing an object to fall from high to low). What if something impedes this motion? In our classical mechanics example we could introduce a viscous liquid or pinball maze to divert flow. $\endgroup$
    – Fire
    Mar 9, 2013 at 21:37
  • $\begingroup$ (Part II) Energetically we could talk about an effective potential which now takes into account this new resistive element. It mightn't be a bad idea to think about voltage like this. I can't get around it - but the basis of this concept is $V=IR$ which is why my answer focused on it. Note it is the relationship between a flow, impedance and potential difference. One needs to think about the entire circuit at once, macroscopically - I don't know what happens individual electrons. Once the battery is connected, practically instantly the circuit will reach a steady state of flow. $\endgroup$
    – Fire
    Mar 9, 2013 at 21:39

The "drop" comes from the analogy of current being the flow of water and each difference in height that makes the water flow is a drop = a voltage difference.

So voltage drop is just a difference in voltage across a component that makes a current flow.

  • 1
    $\begingroup$ Sorry, but this does not help me at all. $\endgroup$
    – oyvey
    Mar 7, 2013 at 9:35

I trust that you've read the Wikipedia pages on voltage and voltage drops.

Anyhow, this isn't particularly rigorous but it helps with the intuition. A resistor, as the name implies, tries to resist current flow through it. What that really means is that there are less "free" electrons in the material to help with the flow of current. If the electrons are tightly bound to the atom, they tend to not want to move, so there's more resistance to current flow.

A voltage difference is the electric potential difference between two points on the circuit, and the current flows in a direction in which the potential difference can be minimized. So when current is flowing through a resistor (Note: In a circuit, the wires are usually assumed to be of zero resistance), it finds it hard(er) to flow across the resistor, but there's still an energy flowing "into" the resistor. And we all know that energy has to be conserved at all costs.

So what effectively happens is that some of this energy is lost when the current flows across the resistor, either because it spent energy in trying to get those tightly bound electrons to leave their atoms or in the form of heat. This means that on the other side of the resistor, there's been some energy lost, which really means that the voltage at that point is lower than at the point before the resistor. So there's less "push" for the electron to get to the side with lower potential, because of the loss of energy.

  • $\begingroup$ Kitchi, can you please see what I wrote in "Edit # 2" and see if it clarifies what I'm trying to get at? $\endgroup$
    – oyvey
    Mar 9, 2013 at 5:41
  • $\begingroup$ @oyvey - Does this help clarify this to some extent for you? Or do you need me to elaborate ore on why there is a potential difference between the two points? The analogy with the battery can be extended to this case, if modified slightly. There is a difference in voltage between the two ends. But in your question it seems like you think that the difference in voltage in a battery is significant, but here it isn't... I'm a little confused about your position on this question. $\endgroup$
    – Kitchi
    Mar 9, 2013 at 18:10

Voltage is just a way of talking about energy per unit charge. What a batter guarantees is that charge entering from the negative terminal will be raised in potential energy (through an electrochemical process) to a specified higher energy at the positive terminal.

What does it mean to give charges energy? It's the same as saying that a bowling ball has more energy at rest at a height of 10 feet than it does at rest on the ground. A battery pushes the charge that comes in from the negative terminal against the electric force, thus giving those charges energy. The energy given is proportional to the amount of charge. Talking about voltage, then, is just a convenient way of talking about a battery in terms of one quantity: how much energy it bestows per unit charge.

The reason a battery can do this is the chemical reaction going on within. In a battery, the electric field points opposite the direction of positive current. If there were only this E-field alone, movement of current from the negative to positive terminals would not be possible, but the chemical reaction allows it.

Conversely, a resistor is an object whose electric field points along the direction of current, so that currents moving through the resistor lose electrical energy.

Here's an analogy for you: a battery is an escalator. Put a ball at the base of an escalator, and the ball will go to a higher gravitational potential due to the escalator's influence, even though gravity always points down. Note that the escalator is able to do this because it provides energy from some other source--in this case, it is powered--rather than from the gravitational field. This is how it is able to do work against gravity.

A resistor, on the other hand, is like a way from the top floor to the base of the escalator.

Batteries do not produce any difference in the number of charge-carriers at the two terminals. Similarly, you can feed a constant rate of objects to an escalator and it produces no difference in the number of objects between the top and ground floors. The current of such objects is taken to be steady.

For this reason, there is no fundamental difference between the voltage measured across a battery and the voltage measured across a resistor. They both measure electrical energy per unit charge flowing through the meter. Yes, charges flow through a voltmeter. A voltmeter does not measure potential across two points as much as the change in energy per unit charge from the input to the output of the voltmeter. In this sense, adding a voltmeter to a circuit fundamentally changes the circuit--it does not strictly tell you about the original circuit--but you can quantify the error in induces. In general, you can only use a voltmeter reliably when it has much greater resistance than what you're trying to measure across.


I didn't read all the responses, so hopefully this helps the OP or anyone who stumbles upon this a little. I think an example may help understand what is going on and help shed light on what the real question is as I see it.

Consider the same situation but with gravity. Imagine you have a mass at height h. Now move by hand that mass to a height 0. It still has the same velocity as before, but now it has less energy. This energy loss is in the form of potential energy. Where did it go? Your body did a work on the mass and probably heated up. What is different about the mass before and after moving it? Nothing really physically obvious, except in relationship to your REFERENCE POINT. So in some sense the OP is right, there is no physical difference when you take a look at the electron before and after.

However, that is not to say there is no difference. The OP needs to remember that potentials have NO MEANING by themselves, by which I mean, the meaning comes only in relationship to two points (a difference). So you can only tell the difference in the electron by checking two points and seeing there is a difference which is why a voltmeter uses two probes at the same time. This would be instead of use 1 probe checking the electron before the resistor,remove, and then touch after the resistor, and see what you see as being different.

Now consider the same situation without a resistance (my hand in gravity, resistor in circuit.) If it were in a field, the electron would have sped up. What keeps the electron from speeding up is the resistance in the circuit. So instead of speeding up, the energy gets transfered to heat.


The first thing to note is that you must never use the word voltage unless you are sure that you and your audience are aware of the fact that is shorthand for “a difference in voltage” or as you have used “voltage drop”.
I would prefer to use the term “potential difference”.

All this has to do with energy and a type of energy that a system has which makes it capable of doing work which is called potential energy.

The water circuit is a very good analogue of an electrical circuit but let me modify it a bit in the following way.

enter image description here

In diagram 1 here is no friction and $A$ and $B$ are “pumps” with built in taps.
All of the taps are open but none of the pumps are working.
What happens next? – Nothing, as the system is in its lowest gravitational potential energy state.
Does that mean that the water has no gravitational potential energy ie cannot do any work.
No. You can say it has gravitational potential energy because it could do work when falling onto the bench top which is below the apparatus but it is not allowed to do that in this case.

Close tap $B$ and let the weight attached to pump $A$ fall.
What happens? – In diagram 2 the water level in tube $a$ goes up and the water level in tube $c$ goes down

Pump $A$ is doing work on the water.
How is it doing that? - There is a weight attached to the pump’s shaft and the weight is falling.
So what is it that makes the water flow? - The falling weight losing gravitational potential energy and so the pump does work on the water.
The water in tube $a$ is gaining gravitational potential energy and the water in tube $c$ is losing gravitational potential energy.
There has been a conversion of the gravitational potential energy of the weight to the gravitational potential energy of the water and there is a net gain in the gravitational potential energy of the water.

When the difference in water level between tubes $a$ and $c$ reaches a certain value the pump stops and tap $A$ closes.
Tap $B$ is now opened.
Water flows through pump $B$ from tube $a$ to tube $c$ raising the weight attached to pump $B$ until the water levels in tube $a$ and tube $c$ are the same.
Pump $B$ has done work and the gravitational potential energy of the water has been converted into the gravitational potential of the weight.

In diagram 3 a third pump and tap $C$ have been added.
All the taps are open with pump $A$ maintaining a constant difference in water level between tube $a$ and tube $c$.
Pump $A$ is continually doing work and the net result is the increase of the gravitational potential of the water going from tube $c$ to tube $a$ and the decrease of the gravitational potential energy of the falling weight.
Pump $B$ is continually doing work and the net result is the increase of the gravitational potential energy of the falling weight attached to it and the decrease of the gravitational potential of the water going from tube $a$ into tube $b$.
Pump $C$ is continually doing work and the net result is the increase of the gravitational potential energy of the falling weight attached to it and the decrease of the gravitational potential of the water going from tube $b$ into tube $c$.

The essence of your question is “What is different about the water in tube $a$ and the water in tube $b$?”
The answer is “The water in tube $a$ can do more work than the water in tube $b$.”
The water in tube $a$ lifts the weights attached to both pump $b$ and pump $c$ whereas the water in tube $b$ only lifts the weight attached to pump $C$.

Now to translate all this into electrical terms.
To make it easier just assume that in my electrical circuit the mobile charge carrier are positive as I find it easier to visualise positive charge wanting to falling down a (potential) hill as opposed the electrons wanting to climb up a (potential) hill.

enter image description here

In diagram 4 pump $A$ is battery $A$ and pumps $B$ and $C$ are motors $B$ and $C$.
The battery is a device which raises the electrical potential energy of the mobile charge carriers when they move from node $c$ to node $a$ (the increase in water level).
This comes from a chemical reaction in the battery and a decrease in chemical potential energy (the falling weight.

Motors $B$ and $C$ are devices which do work raising weights at the expense of a decrease in the electrical potential energy of the mobile charge carriers passing through them.

Rather than use the term “change in potential energy between two points” all the time it is easier to use the term “change in potential energy between two points per unit charge” and even simpler still “potential difference between two points” or “potential difference” or your term “voltage drop”.

So when you have a voltage drop across a component it is telling you that the amount of work the mobile charge carriers can do on leaving the component is reduced (drops) compared with the amount of work the mobile charge carrier can do before entering the component.

The amount of work (joules, J) which can be done by unit charge (1 coulomb, 1C) of mobile charge carriers between two points, the potential difference between those two points, is measured in joules per coulomb which is the volt.

Diagram 5 is a simple circuit.
The voltage drop across the 3$\Omega$ resistor is 3V.
Starting at node $a$ and finishing at node $c$ 1 coulomb of mobile charge carriers will produce 5 joules of heat whereas if they started at node $b$ and reached node $c$ they would only have generated 2 joules of heat.
So one coulomb of mobile charge carriers capacity to do work has fallen from 5 joule at node $a$ to 2 joules at node $b$ a drop of 3 joules per coulomb which is a voltage drop of 3 volts.


I believe that a simple example will be the best way to answer your question.

Let's start with a 10v voltage source (battery, power supply, generator, etc.- the type does not matter). You then connect 10, 1 ohm, resistors in series across the supply. With this arrangement, we know there is a 1 amp current flow through each resistor, so the voltage drop across each resistor is 1v.
If you now take your voltmeter and place one lead at the bottom of the bottom resistor (#10) and the other lead to the top of the top resistor (#1) you will read 10v. You now take the top lead and move it to the point below the first resistor, it will now read 9v. If you continue this process, you will read 8v, 7v... 1v, 0v. As can be seen, each resistor has caused a reduction in voltage equal to the voltage drop across it.
It should be noted that each resistor is consuming 1W, giving a total of 10W, which is the same power that the supply is delivering.

If we now replace the resistors with only one (10 ohm), although we no longer have access to the intermediate points, the situation is the same. The voltage drops occur inside the resistor in proportion to its length (assuming linear distribution). If the resistor is 10 cm long, each cm would drop 1v.
The fact that a resistor causes a reduction in voltage, proportional to the current flowing through it, is the reason it is called a voltage drop (across the resistor).


When the battery is turned on, a wave propagates within the circuit to determine the current of the sytem.At this point, the current is inhomogeneous withing the circuit(but this dissapears almost at the speed of light). Let's suppose there are also resistances of different resistivity connected to this circuit. The electrons within the circuit are a part of the same system, thus, if one slows down, the rest will slow down too. The "voltage drop", at this point affects the whole system at the same time. The loss of potential is calculated in the initial turn on, and the current is thus established. The reason why this drop of voltage equals to the V of the battery is because all is calculated according to this wave which is launched from the battery.


oyvey I can understand your confusion.I am just a student but I can make an attempt to clear the confusion using an analogy.

Consider two playgrounds (the two negative terminals of two different batteries 2V and 3V).Children in the first playground(2V) are weak and lazy.When asked to run through a pathway the number of children(electrons) passing through a point in the pathway is less.Children in the second playground(3V) are strong and active.When asked to run through a pathway we will see more number of children passing through a point in the pathway than in the previous case because they are strong and active.

This I think very well explains Ohm's Law that current is directly proportional to voltage and also clears your confusion hopefully.

  • 2
    $\begingroup$ I don't think this is a good analogy, because all electrons are the same. There are no lazy or weak electrons. $\endgroup$
    – DavePhD
    Apr 25, 2014 at 12:36
  • $\begingroup$ But each coulomb coming out of a 2V battery carries 2J energy and that coming out of a 3V battery carries 3J energy,hence weak and strong coulombs $\endgroup$
    – copernicus
    Apr 28, 2014 at 9:31

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