I'm 15 and recently started electronics and I'm having trouble understanding voltage.

I've read so many articles and watched so many videos about voltage and they all give different answers. Some of them say that voltage is like pressure, others say that voltage is is like gravitational potential energy and then some say its a measure of electric field strength. So as you can see, I don't know what to think.

Could someone please explain it to me because I've been trying to find an answer for like 2 months and it's kinda driving me insane :)

And also if voltage is like gravitational potential energy, how does more voltage mean more current?

  • 4
    \$\begingroup\$ If you are planning to learn mathematical circuit analysis soon, it might make sense to understand voltage as "an abstract numerical quantity which is well-modeled by Kirchhoff's Voltage Law in a circuit" without intuition to start. Ultimately, any sort of model has its issues: pressure isn't 100% accurate, integration of electric field over a path is scientifically correct but not very useful. \$\endgroup\$
    – nanofarad
    Aug 21, 2020 at 22:48
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    \$\begingroup\$ When I tried to edit the misspelling of the word 'exactly' (the OP wrote 'exaclty') in the title, I was told a question with exactly the same title already exists. I had to add commas. I don't know if the spelling error by the OP was deliberate to get around that. \$\endgroup\$ Aug 22, 2020 at 8:26
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    \$\begingroup\$ Does this answer your question? What exactly is voltage? \$\endgroup\$ Aug 22, 2020 at 8:30
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    \$\begingroup\$ For the purposes of understanding circuits, electrons are water, voltage is pressure, current is current, pipes are resistors, and Kirchhoff's laws apply. Diodes are check valves, inductance is inertia. Capacitors are a little harder to model with water. Magnetic effects don't correspond at all. \$\endgroup\$ Aug 22, 2020 at 16:32
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    \$\begingroup\$ @TomW it is a mistake to focus only on electrons. If molecules are polar, they will rotate or try to rotate to align with the electrical field. The electrons are pushed in one direction, and the positive part of the molecule is pushed in the other. In a liquid solution, the anions flow one way and the cations flow the other way. \$\endgroup\$
    – mkeith
    Aug 23, 2020 at 0:23

15 Answers 15


I suspect I'm just going to confuse you further, but here goes:

Some of them say that voltage is like pressure, others say that voltage is is like gravitational potential energy and then some say its a measure of electric field strength.

We say that voltage is like pressure, or like gravitational potential energy, because we're trying to draw an analogy to something that you can see or feel (because you can drop a rock on your toe, or feel the pressure in a balloon when you blow it up).

What voltage is gets abstract (hence the analogies). If you have an electron in an electric field, there's a force on it, so it wants to move. If you had a pair of magic tweezers that would let you grab that electron and move it from one spot to another, you'd have to exert force on it -- putting energy into the system -- or it would exert force on you -- taking energy out of the system and delivering it to you.

A volt isn't a measure of the electric field. Volts are a consequence of electric fields, but the electric field is in units of volts per meter. What a volt is is an expression of the amount of energy available per unit of charge. So if you have one Coulomb of charge, and you let that charge flow through something that drops one Volt, then that charge will deliver one Joule of energy to whatever that something is that was dropping one volt.

And also if voltage is like gravitational potential energy, how does more voltage mean more current?

And here our nice analogy breaks down. In this sense voltage is more like pressure in a water pipe.

For all physical things, if you put a voltage across them current will flow -- it may be a lot, it may be minuscule, but current will almost always flow. For most things (there are some exceptions), the more voltage you put on it, the more current will flow.

So in this regard, voltage is like pressure in a water pipe -- more pressure equals more flow, just as more voltage across a resistor equals more current in the resistor. But this is just an analogy. Ultimately, you just have to beat your brain against the physics until everything becomes intuitive, just as you learned that when you let go of something it falls down every time. The difference is that you learned the lesson about dropping things before you were a year old; the voltage lesson comes a bit later in life, so you have to purposely let your brain flex.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Aug 23, 2020 at 20:06

In simple terms, voltage is a measure of energy per unit charge associated with two points in an electric field. But why is it that there is an energy associated with any two points?

To answer this, we need to picture an electric field and its effect on a test charge.

We can picture an electric field as associating a tiny arrow to each point in space. Each arrow in the electric field represents the force that would be felt by an unit of charge if it were placed at that particular point.

Since like charges repel, the arrows point away from a positive charge (as it repels our test charge): Electric field around charge

As the test charge travels through the electric field, it is pushed around and gains or loses energy. If it travels in the same direction of the little arrows in the field, there is work done on the particle, and it gains energy. If it travels opposite to the field, it loses energy instead.

Imagine it like pushing a swing when it's already moving away from you, versus pushing the same swing when it's coming at you. In the first case it is pushed aligned with the direction of motion, accelerating it. In the second, it's pushed opposite to the direction of motion, decelerating it. In a way, you have to add all the contributions from the little arrows along the entire path to calculate the final energy of the swing/test charge.

This adding of arrows is called Line integral, and it involves calculating at each point how much the displacement vector and the field are pointing in the same direction.

A 10V battery is one that generates an electric field such that adding all little arrows from the positive side to the negative side results in a net work of 10 Joules for each unit charge that goes around the circuit.

The electric field looks like this for a wire with uniform electrical resistance everywhere:

Battery E field

Ideally, if there was no resistance, at each cycle our test charge would gain 10 Joules at each loop and speed up forever, but in reality, as the current increases, the energy dissipates more and more in the form of heat.

The test charge may also do work on something else: In LEDs, this electrical energy is converted into luminous form, in motors, mechanical form, and so on.

An important detail to consider is that there may be multiple paths from a point to another. Why should the energy difference not depend on the particular path between the two points?

In the absence of external forces and fields, the electric field is conservative, which implies that the potential difference results the same number no matter what the path.

To see why this is true, imagine that there is a potential of 15V from A to B along the upper path (X), but 5V from A to B along the lower one (Y):

Two paths

Then, if our test charge first goes from A to B through X, and then backwards in the opposite direction through Y, the electric field will do a net work 10 Joules: 15 Joules "downwards" through the field and 5 Joules "upward". (Notice: here I am using "downwards" and "upwards" as an analogy with climbing or going down a gravitational field)

But since the charge is back to the same place it was before, we gained 10 Joules for free! This breaks the law of conservation of energy, unless that energy is being drawn from somewhere else. If there is nothing providing this energy, then all paths are the same potential.

The explanation to the analogies:

Like electric fields, gravitational fields also push things around. Just like in electromagnetic fields, if you go down a gravitational field, the field does work and you gain energy, and this energy can also be used for a variety of purposes by doing work on something else.

In fluids, the force field in question is the pressure differential, which accelerates particles in the direction of reduction of pressure (since there is a force imbalance pointing in that direction)

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    \$\begingroup\$ In general it's associated with two points and a specific path between them. It's only when the field is conservative that voltage depends on the endpoints alone and can thus be expressed as a difference of potential. Anyway, +1 \$\endgroup\$ Aug 24, 2020 at 3:24
  • \$\begingroup\$ @Sredni Vashtar Fixed! \$\endgroup\$ Aug 24, 2020 at 21:18
  • \$\begingroup\$ You went in the opposite direction I had in mind, but it's your answer, so... Let me just point out what I see as an inaccuracy in your addition: you wrote that "in empty space electric field is conservative". Well, no, you can have a nonconservative electric field in a vacuum. All you need is a region of magnetic field variability. And this brings us to the 'direction' I had in mind. If there is no dB/dt the field is conservative and admits a potential, so voltage is also a potential difference. But in general it is not and we have to deal with path dependency and PD is not definable. \$\endgroup\$ Aug 24, 2020 at 22:47
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    \$\begingroup\$ @Sredni Vashtar In my head I pictured empty space as devoid of external objects/forces, but in retrospect it was not obvious what I meant. Thank you for the feedback. \$\endgroup\$ Aug 25, 2020 at 13:23

Some of them say that voltage is like pressure, others say that voltage is is like gravitational potential energy and then some say its a measure of electric field strength.

You haven't asked a question here, but it is like all of those things, if you understand the analogies.

In the gravitational analogy, it would be more accurate to say voltage is like gravitational potential, not gravitational potential energy. For example if you have a hill 10 meters high, the gravitational potential difference between the bottom and the top of the hill is \$(10\ m)(g)\$. This is proportional to the energy you would need to move an object from the bottom to the top of the hill. But you'd need more energy to move a bowling ball than to move a pebble (just like you need more energy to move a bigger charge through an electrical potential difference). And the gravitational potential difference is a defined quantity even if you aren't moving any objects up and down the hill (just like the voltage between two points can be a defined quantity even if there isn't any current flowing between those points).

if voltage is like gravitational potential energy, how does more voltage mean more current?

It's not a bigger difference in voltage per se that produces more current. It's a bigger difference in voltage across a fixed distance (such as the distance between the two terminals of a resistor).

Gravitational potential works the same way: A stream flows faster down a steeper slope, and more slowly where there is less slope.

  • \$\begingroup\$ Beware of confusing "potentials" with the concept "potential energy." They're two completely different things. OP is asking about "potential energy," when he should be asking about the math-concept named "Potentials." (Really, it would have been better if "potentials" had a different name!) Electric potentials are not like energy, instead they're like altitude. The slope of a hill is not a form of energy, but the slope of a hill is called a "potential-gradient." \$\endgroup\$
    – wbeaty
    Aug 22, 2020 at 6:17
  • \$\begingroup\$ @wbeaty, yes, I made that exact point in my second paragraph, then messed it up in the last one. Edited. \$\endgroup\$
    – The Photon
    Aug 22, 2020 at 14:58

Mathematically speaking the voltage is simply the integral of the electric field over a line. (You probably already know what an integral is. Maybe only an integral of a function on an interval. An electric field in the space tells in every point of the space what is the vector force per charge unit. A vector field (and so an electric field) can be integrated on a (curved or straight) line as if it were a function on the interval described by the parameter of the line, where the function is given by the dot product of the vector field and the vector tangent to the line).

Physically speaking, without using any analogy with other branch of physics which can cause confusions and thus remaining in the electric world, it can be three and only three different things:

  1. electric power converted into heat per current unit. It's measured in [W/A]=[V]. It's the phenomenon which is observed when a current flows through a material characterized mainly by a resistance (e.g. a resistor). It is also known with the name of voltage drop.

  2. electric energy stored per charge unit. It's measured in [J/C]=[V]. It's the phenomenon which is observed when a system characterized mainly by a capacitance (e.g. a capacitor) is electrically charged or discharged. It is also known with the name of potential difference.

  3. temporal rate of change of the magnetic flux linkage. It's measured in [Wb/s]=[V]. It's the phenomenon which is observed when a system characterized mainly by an inductance (e.g. a coil) is magnetized or demagnetized. It is also known with the name of emf or electromotive force

You have to sum all these contributions when a system is characterized by a resistance, a capacitance and an inductance at the same time.

  • \$\begingroup\$ Where do you fit an electron acquiring kinetic energy when it is accelerated by an electric field in a vacuum, in your "only three things"? \$\endgroup\$ Aug 22, 2020 at 11:20
  • \$\begingroup\$ For charge-carriers in vacuum under an electric field, case 3 above applies: the electric field integrated on the trajectory of the accelerated electron gives the voltage on that trajectory. This voltage is the temporal rate of change of the magnetic flux generated by the current due to the moving electron. The electron is accelerated so its velocity increases, and the current it creates increases and the magnetic flux produced by such a current increases: the rate of this last increment is the said voltage. \$\endgroup\$
    – trying
    Aug 22, 2020 at 12:23
  • \$\begingroup\$ While your explanation is good, it's likely something a 15 year old cannot comprehend since you dive into multivariate calculus, i.e. talking about line integrals, vector fields, and dot products. \$\endgroup\$
    – user103380
    Aug 22, 2020 at 16:38
  • \$\begingroup\$ @KingDuken yes, I know. I was in doubt whether to talk about the mathematical point of view or not. Anyway I placed the explanation in parentheses, meaning that it's not really important in order to understand the real matter. Anyway I think that a 15 y.o. may know about integral of a function, vectors, and dot product and can at least imagine what's going on on the mathematical side. That said, I have also given an independent physical perspective that cannot be overlooked simply because I have been previously talking about integrals. \$\endgroup\$
    – trying
    Aug 22, 2020 at 17:03
  • \$\begingroup\$ @trying magnetic flux through which surface? Let's say I have a positive charge in 0,0,0 and an electron going from 0,0,1 to 0,0,2 starting from rest along a straight vertical line. \$\endgroup\$ Aug 24, 2020 at 0:31

You can simply compare charged particles to gas molecules: charged particles with the same electrical properties repel each other. When they are closer, they tend to disperse outwards, just as gases expand outward after being compressed. This outward dispersion trend forces charged particles to move outward to form an electric current,This is the voltage. In fact, for a single charged particle, no matter how far away another particle with the same charge is from it, it will be repulsed outwards, but the farther the distance, the smaller the force. The neutral you see is that the number of positive and negative charges is equal, so that the two-point voltage is zero.


Crude analogy: waterfalls.

Voltage is the height of the waterfall.

Current is the amount of water going over the falls.


Electric potential of a point is the amount of work needed to move a unit charge from a zero electric potential point (generally this point is considered to be at infinite distance) to that specific point.

Just like gravitational potential is the amount of work needed to move a unit mass from the zero potential point to that specific point.

The difference of electric potential between two points creates electric field. And this difference is known as potential difference or Voltage.

Let's get back to gravitation for analogy. A mass is bound to move from point with higher gravitational potential(like 5th floor of a building) to a point with lower gravitational potential(ground floor).

Similarly a positive charge is bound to move from the point with higher electric potential to a point with lower electric potential within the electric field.

A train of charge moving in the electric field causes electric current.

Now to address your question about voltage. More potential difference won't mean more current unless charges are riding in the electric field.

But say there are sufficient charges like free electrons in a conductor then more potential difference between two points means more stronger electric field and thus more faster movement of the charges i.e more number of charges passing through a region in the field per unit time, which means more current.

Now to give an analogy with gravitation, consider a waterfall.

In Earth the water will fall faster towards the ground. Hence more water will fall through a certain region of the fall per unit time, hence high water current.

In Moon however, the water will fall slowly, hence less amount of the water will pass though a certain region per unit time, so low water current.


I'm seeing a lot of complicated answers. If you climb up (say 10 meters) you will gain potential energy. As the earth is continuously pulling you toward's it, you have to do work against it. This work will be stored as your potential energy. \begin{equation} E = mgh = 10mg \end{equation}
Now, consider a positive point charge. It will have a field around it. If you want to place a positive 1 C charge inside it, you have to work against the existing field. This work will be called the voltage of that point charge.

Now go back to the 10-meter height case again. You have already gained potential energy. If you jump, you will go towards the surface of the earth (or the reference). As soon as you touch the surface, you will transfer all your energy to the surface (or may create sound, vibration, etc).

Now think yourself as an electron. If I say you have 5-volt potential, it means that you have done some work to gain that potential. And you have always the tendency of going towards reference (or 0 volts). If you compare 'hitting the surface' as resistance, you would clearly see that power is dissipated through it.


Using the water analogy, Voltage is electrical ‘pressure’ (technical term: potential), while current is electrical ‘flow’ of charge.

What makes that pressure? The application of an electric field, which is, a relative difference in charge density from one point to another. For example, a battery, through a chemical process, creates a difference in charge density between its (-) and (+) terminals. Wire a load across this, and the pressure created by the charge difference induces a current, while we measure the difference (electrical pressure) as voltage.

Likewise, static electricity is a build-up (or removal) of charge from an insulated region, that has a potential difference to its neighbors (like thunderclouds vs. the ground below.) When that difference is big enough, the charge finds a path through the air, such as in the form of lightning.

This Q might be helpful to explain how the 'pressure' results in electron flow: Does the voltage difference have an effect on the electrons' speed?


This is a rephrasing from some other answer on an SE site, and it really helped me understand electricity more, as well as a few simple pieces you'll probably use.

If we imagine our wire is a channel through some farmland, we can assign a few variables to voltage and amperage. How big our wire is is related to the size of our channel. Voltage becomes the amount of water in the channel. too much voltage, and the channel overflows, kills the crops and the farmer(your wire melts). too little voltage, and the farmer cant water his crops(your LED won't light up).

Amperage becomes the speed of the water. if the water isn't fast enough, it won't turn the waterwheel and grind the wheat(again, your led won't light up). too fast, and it can shake the building the pieces. But, the farmer can use gears to change the speed and torque(transformer or transistor) and use it to grind his wheat.

This really helped me when I was starting up, and sadly I don't have the link for the original, as it was written a lot better when I first read it. hope you figure it out, good luck!


So, there's the correct explanation and the technically correct explanation. I'll go with the former.

You probably already know about electrostatic forces: equal charges repel, and opposed charges attract. From that, you can imagine that, if you put a bunch of electrons together in a box, they'd seem pretty pissed and would try to escape. If you also happen to have a box with a bunch of protons nearby... these electrons really want to get there.

Voltage is an attempt to quantify how pissed your electrons are. This is very useful because the more pissed they are, the more stuff you can make them do when trying to escape: at 0.1V they'll basically do nothing, at 12V you can start a car (if you have enough of them) and at 10kV they'll zap through air and you'd have trouble containing them.

Keeping this in mind, it's easy to see why more voltage generally results in more current: the more voltage, the more your charges will force their way through whatever you put between them and their desired destination.

Now, what I just said is is pretty fuzzy. "How much electrons want to escape" isn't a very precise idea. And yet, that's really the gist of it. You'll eventually rediscover the precise definition of voltage (and electric potential) if you try to refine this idea. Some food for thought:

  • The concept of voltage should also work for positive charges
  • How do you define "how much X wants to escape"? Force? Escape velocity? Momentum? Energy?
  • Also, escape to where?
  • What happens if the charges can move freely? And what if they can move freely within some bounds? (say, inside a metal sphere, or a metal cilinder, or a metal wire)
  • When we want to store electricity, we buy "batteries", not "electron tanks". What's up with that?

Line up 10 coins in a row on a table in a straight line with their edges touching. Take another coin and flick it with your finger at the end of the line. The first one doesn't move much, but the one at the other end does. The harder you flick that first coin, the more the one at the other end moves, but the movement in the middle is still negligible.

Now make a line of 100 coins and try the same thing. The end coin barely moves. That's because SOME of the energy in your finger flick is absorbed in each coin in the middle; not much on each one, but it adds up to where it affects the end result.

The force you are exerting on the fist coin is the equivalent of "voltage", the movement of the coin at the other end is the "current", the length of the string of coins represents resistance. With no voltage, there is no current. With low resistance (10 coins), current is high but with high resistance (100 coins), the current is low despite the voltage being the same.


Voltage is the potential difference in a circuit.

If a circuit doesn't exist then charge flow cannot occur. There's a ton of voltage on a open circuit but then close a switch and you have a drop in voltage as it begins to equalize and become at the same potential as its surroundings. From our perspective or just mine it will be Earth since that is what charges tend to factor potential difference with. Earth is at one charge whilst everything above the Earth is at another and all wants to mix and mingle to become neutral and stable. Heat transfer is a good analogy.

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    \$\begingroup\$ Joshua, welcome to EE.SE, but this is rather a garbled answer and contains a few inaccuracies and confused thinking. I think you need to study the topic further. \$\endgroup\$
    – Transistor
    Jun 4 at 19:16

electric Voltage is a measure for the work which was applied to seperate positive and negative charges. So voltage U=W/Q, where W ist the work and Q is the charge.

This is a more or less wordly translation from a book in german "Physik für Ingenieure" ISBN 3-540-62442-2, 6th edition, from Elbert Hering, Rolf Martin, Martin Strohrer published by Springer, page 224, chapter 4.1.3

So if you short circuit the seperated charges with 0.000 Ohm resistance, the seperated charges Q will flow together again (current) and you get back the work W.

Thus the analogy with potential energy of a waterfall is a good one - was used in my studies at day one.

  • \$\begingroup\$ Thats cool here - you get downvoted for a 100% correct answer directly taken from a book "phyics for engineers"... \$\endgroup\$
    – schnedan
    Aug 23, 2020 at 20:12
  • \$\begingroup\$ Hi, It's not my downvote, but regarding: "[...] directly taken from a book "phyics for engineers"." - as required by this site rule, when you include something in an answer (e.g. photo, image or text) which isn't your own original work, you need to properly reference (cite) it. Can you please edit your answer, use the "blockquote" symbol > to mark the text from that book, and add a link back to the original web page (or add equivalent details for the book e.g. title, author(s), publisher, edition and page number). Thanks. \$\endgroup\$
    – SamGibson
    Aug 23, 2020 at 21:19
  • \$\begingroup\$ NOT correct. That book is simply wrong. Voltage is not a measure of work. \$\endgroup\$
    – wbeaty
    Aug 25, 2020 at 3:16
  • \$\begingroup\$ Mistaking "Potentials" for work (or for potential energy) is a bad mistake, and very widespread. With a waterfall, "voltage" is like the height of the cliff. If the stream dries up, the waterfall is gone, and the potential energy of water is zero ...the "voltage" or potentials are still there.Voltage is not like a lifted boulder, instead voltage is like the empty sky: it's like "altitude." (In other words, voltage is not associated with the test-charge, instead, voltage is part of the invisible e-field. Voltage is always perpendicular to flux-lines, it's the pattern of equipotential surfaces) \$\endgroup\$
    – wbeaty
    Aug 25, 2020 at 3:19
  • \$\begingroup\$ @wbeaty The work is the work needed to seperate the charges as it is done in any power plant where you using something to propell the generator which seperates charges. That has by no means anything to do with Mistaking "Potentials" for work (or for potential energy). This is what happens in voltage sources, even in chemical voltage sources like batteries, fuel cells, solar pannels,... and it's the same work which is transfered to your voltage sinks like lightblubs, heaters, el. motors, whatever.... if you don't believe me, ask any decent book for professionals or engineers \$\endgroup\$
    – schnedan
    Aug 25, 2020 at 20:06

We can generalize all these analogical quantities in three groups of "something-like" quantities: pressure-like, flow-like and impediment-like quantities. Thus we can formulate a "generalized Ohm's law" and see generalized converters - pressure-like to flow-like, impediment-like to pressure-like, etc.

If we take into account that the water pressure is proportional to the height of the water column, then we can visualize the (drops of) voltages across circuit elements with segments (bars) whose height (length) is proportional to the corresponding voltage. Then, we can visualize the currents in circuits by closed curves (loops) whose thickness is proportional to the magnitude of the corresponding current. Here are some examples:

Decoupling capacitor visualized

Fig. 1 Decoupling capacitor visualized

Dynamic load visualized

Fig. 2 Dynamic load visualized

Differential pair visualized

Fig. 3 Differential amplifier visualized

ECL at low input signal visualized

Fig. 4 ECL gate at low input signal visualized

I apply this technique to every circuit I explain. You can see how I do this in my questions and answers.

See also my Codidact papers about voltage bars, voltage diagrams and current loops.

  • 6
    \$\begingroup\$ If a person who has just started using a computer asks Bill Gates "What is Windows?", what might be the answer? \$\endgroup\$
    – Sadat Rafi
    Aug 23, 2020 at 19:18
  • 1
    \$\begingroup\$ Differential amplifier is not related to this question. \$\endgroup\$ Aug 23, 2020 at 21:11

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