This question is related to the electric potential of a circuit point, not the voltage between two points.

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Here is a sample circuit diagram with \$V_{circuit}=5V\$ and \$ R_{circuit}=5 \Omega \$.
If we neglect the voltage meter of the left-hand side, \$I_{circuit}=1A\$ and \$V_{1,2}=5V\$.

Let's consider a voltage meter that is connected to the ground and point 1.
If we assume the electric potential of the ground as zero, what will be the electric potential of the point 1?
In other words, what is the absolute voltage of the point 1, \$V_{1}\$?

The circuit itself is valid if \$V_{1,2}\$ is maintained as a constant value \$5 V\$.
Therefore, every combination of \$V_{1,ground}=\alpha+5 \$ and \$ V_{2,ground}=\alpha\$ is valid.

Is there any theory that can help to determine the value of \$\alpha\$?
Or, if it is impossible to determine the value of \$ \alpha \$, does it at least remain a constant value during the circuit operation?

P.S. This question is originated from the pipe network analysis.
Pipe network is usually analyzed with a similar manner to the electric circuit.
The absolute pressure of each pipe point can introduce a safety problem, however, there seems like no theory to determine the absolute pressure, either.

  • 4
    \$\begingroup\$ There's no such thing as an absolute voltage. But perhaps you are interested in something else? For example, the electric field intensity at a point is the gradient of the electric potential at that point, negated: \$E=-\nabla V\$. Voltmeters don't measure this, though. In your picture, if I'm gathering it right and the left side schematic is floating, then your voltmeter is likely to read zero volts, as its own impedance has just galvanically connected that node it touches to earth ground. Of course, if the left side circuit isn't floating but otherwise has a galvanic connection to earth... \$\endgroup\$
    – jonk
    Aug 8, 2021 at 6:24
  • 2
    \$\begingroup\$ The "absolute voltage" of the point V1 is 0, because of the voltmeter's connection. If the voltmeter was instead connected between ground and point V2, then V2's "absolute voltage" would be 0. If there was no path from V1 and V2 to ground, there would be no way to define the voltage. Note that in the wikipedia Electric potential article you linked, it mentions the electric potential field is relative to a "reference point", i.e. 0 volts i.e. ground. Electrical voltage does not work quite like hydraulic pressure, "water model" is a flawed analogy. \$\endgroup\$
    – MarkU
    Aug 8, 2021 at 6:30
  • \$\begingroup\$ Thank you everyone for nice explanations despite my poor question quality, particularly the circuit diagram with a voltage meter. A gauge pressure meter for fluid normally measures the pressure with respect to the vacuum or free air. I've tried to display a similar thing but seems like it resulted in a kind of nonsense. I want to choose both answers as accepted ones but I can only choose one; I would choose the second answer with respect to his effort to provide a nice answer. Thank you again for all. \$\endgroup\$
    – J. Choi
    Aug 8, 2021 at 8:13
  • \$\begingroup\$ Although you say "This question is related to the electric potential of a circuit point, not the voltage between two points.", isn't what you are asking the voltage between GND and Point 1? So if a real Voltmeter is connected it is 0 (since any access charges will flow through the Voltmeter to the GND). If you mean the Voltmeter has infinite resistance, then you can get any value. Charges might evaporate to the air, or (charged) cosmic particle can be captured by the circuit. \$\endgroup\$
    – lalala
    Aug 9, 2021 at 8:15

4 Answers 4


Your question assumes there is such a thing as absolute potential. It requires that, somewhere in the universe there is a place "X" where electric charges have no potential energy at all. In other words, there exists no other place in the entire universe that charges could move to, where they they have less potential energy than they do at "X".

The only reason an electric charge would move (that movement being called "current"), is because there's a point where it would have lower potential energy, and to get there they don't have to climb a potential "hill" first.

Think of it like a ball on a hillside. As long as the ball is sitting on a slope towards a lower gravitational potential energy (otherwise known as "downwards") it will be accelerated in the direction where the downward slope is greatest, and will only settle when it arrives at a point where the slope in all directions is upwards.

There may be a valley next door whose lowest point is below the ball's current position, but if it has to climb another hill to get into that valley and settle there instead, the only way it can do that is if it gets pushed up and over the hill separating them, by some external source of energy.

There's always a valley somewhere else, whose lowest point is lower than where it is right now. How low can the ball get? Sea level? But some land is below sea level. What about the bottom of the Kola Bore Hole? What about the centre of the earth? Supposing the ball could reach the centre of the Earth without damage, what about the center of the Sun, would the ball have less potential energy there? You see why it's not reasonable to say that potential energy, and by extension voltage, can have some absolute value. It's always relative to some other place.

You told us to neglect the voltmeter. In doing so, we assume that point 1 is physically disconnected, electrically isolated from Earth. In the absence of any means by which a charge at point 1 or 2 or anywhere in the circuit can travel to Earth, what you have created is a huge hill between those two locations. You've also relinquished any say you have in the question of which of the two valleys either side of the hill is "deeper", because there are other influences such as capacitive coupling to nearby mains wiring, which have suddenly become far more significant factors shaping the "terrain".

Sure, the Earth may have a lower potential, or it may have a greater potential than point 1, but you'll never know. In making the measurement, by inserting a voltmeter, you provide a route for charges to move between those locations, effectively removing the hill between them. Measurement changes the voltage "terrain". By joining the two locations electrically, do we raise the Earth's potential, or do we lower the circuit's potential? It's all relative, and the question is moot. In the end, all you need to know is now the Earth and the point 1 have been joined.

In your circuit, neglecting the voltmeter, the only thing you can say that has any meaning, is that point 1 is five volts higher in potential than point 2. Any other statement referring to the Earth or anywhere else ouside the circuit is meaningless (finite electrical resistance of the air notwithstanding) .

When you put the voltmeter in place, you join the Earth to point 1, providing a path that charges can use to move between them. However, will any charges make that journey? Is there an potential difference to motivate those charges to flow? In the absence of any other forces at work (like electromagnetic radiation, noise, and so on), the simple answer is no. There's no reason for any charge to move from point 1 to Earth, or the other way.

If you want charges to move between Earth and point 1 you have to make them move! You have to explicitly impose a potential difference, so that charges at one place find themselves on a potential "slope" that they can "roll" down. You have to do what you did to move charges through your 5Ω resistor, use a voltage source to impose a potential difference across it. If there's no such voltage source, between Earth and point 1, just the voltmeter, then there's no potential difference across the voltmeter, and the potential at Point 1 is the same as the potential of Earth.

Edit: In the parlance of somebody who teaches electricity to 10 year-olds, where generally you must show them that there needs to be a closed loop for current to flow "around", that last paragraph amounts to saying that the addition of a voltage source across the voltmeter creates just such a loop. Current is now permitted to flow through the voltmeter, and the presence of a voltage source in that loop motivates the charges to move. It's quite naive, though, because it lacks many elements of the full picture.

Edit2: Where the pressure analogy is concerned, there is such a thing as zero pressure, it's the vacuum, and therefore you may quote pressures relative to the vacuum if you wish. Typically though, that's not very useful, given that locally flow is dependent on the difference in pressure between two points, and has nothing to do with the vacuum, per se. I have always found the analogy between water and electricity to be wanting in a several respects, this being one of them.

  • \$\begingroup\$ Great answer +1. Just to be annoying, we can choose (as astrophysicists do for the energy of planetary orbits) a sphere of infinite radius as the datum, so that: \$V_r=-\int_\infty^r E\cdot \text{d}l\$. Knowing the path independence principle and taking charge \$q^{'}\$ placed centrally, a point charge voltage out as \$V_p=-\int_\infty^r \left[\hat{r}\frac{q}{4\pi\,\epsilon\, r^2} \right]\cdot\hat{r}\text{d}r=+\frac{q}{4\pi\,\epsilon\, r}\$. And that's also dimensionally correct. But none of this would have helped the OP! ;) A point voltage can technically be defined. It's just mostly useless! \$\endgroup\$
    – jonk
    Aug 8, 2021 at 15:15
  • \$\begingroup\$ @jonk: Thanks. I wrote a whole section on how two independent, closed groups of charges technically exist in each other's electric fields, and therefore have a finite combined potential energy, and finite, calculable potential difference with some third distance point in empty space, but I scrapped it, because when I read it I confused myself! \$\endgroup\$ Aug 8, 2021 at 16:04

In a real circuit there are resistances and capacitances that come into play. Voltmeters may have relatively low input resistance compared to leakages, perhaps 10M\$\Omega\$. So the voltage reading would be close to zero.

If we imagine a world without significant leakages but still consider capacitance to ground, the isolated circuit may well have some charge relative to ground, so there could be a voltage which could be detected by an electrometer.

The voltage V is q/C where q is the charge in coulombs and C is the capacitance.

An effect of that is that the voltage will increase if you move the circuit further away from ground, reducing the capacitance, since the charge has nowhere to go. The energy in the capacitance also increases. You may be able to say where that energy comes from.


We define α, we don't determine it.

We either define it explicitly by saying α=some value, or implicitly by defining some other point not on the circuit as having a value (usually the point is something ground-like or infinitely distant, and the value 0), and then measuring it.

The whole point of a potential is it's measured with respect to a reference.


electric potential = voltage between the measurement point and a reference ground; is the integral of the electric field


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