I know there has been asked a similar question about this before but I still have some issues to understand the whole voltage and current thing. ( Current without Voltage and Voltage without Current? and Is Ohm's Law violating itself?)

Ohm's law states: $$V = R \times I$$

But what came to my mind is: if either the resistance or the current is zero the voltage will be zero, too (according to ohms law):

$$V = R \times 0 = 0$$ and $$V = 0 \times I = 0$$

I'm not sure if Ohm's law is applicable here. Please correct me if I'm completely wrong or missing something obvious!

So lets say I have a single power supply with a 5V fixed output:

I'm going to measure the voltage with a typical \$10\text{M}\Omega\$ input resistance: enter image description here

Since there is a "resistor" as a load, current will flow:

$$V = R \times I$$ $$I = \frac{V}{R}$$ $$I = \frac{5\text{V}}{10\Omega} = 500\text{nA}$$

But what if I disconnect the probes? Now we are at a point where it gets really confusing for me:

There will still be a "load" on the supply, since: air is not a perfect isolator so it must have electrical resistance: (I'm assuming \$1\text{G}\Omega\$ here)

enter image description here

Same as before, current would theoretically be:

$$I = \frac{V}{R} = \frac{5\text{V}}{1\text{G}\Omega} = 5\text{nA}$$

So my question is: does this small amount of current really flow? Because if not there will be no voltage across the two terminals according to Ohm's law! This is really confusing and I don't think current will flow because it simply sounds really unlikely.

  • \$\begingroup\$ In the circuit you have shown, yes current does flow through the 1GOhm resistor. As to the broader question, "does Ohm's Law apply when the resistance is infinite?", personally I'd waft that away by saying Ohm's law only applies to circuits - an (ideal) battery standing on its own does not constitute a circuit - but there are plenty more answers to this on the the question you have already linked to. \$\endgroup\$
    – peterG
    Apr 23, 2015 at 22:56
  • \$\begingroup\$ Even this tinny current may be it is significant for an extremilly thin wire in room temperatures. \$\endgroup\$
    – GR Tech
    Apr 24, 2015 at 3:14
  • \$\begingroup\$ If you can train your electrons to jump from one terminal to another, current will flow. If they can train fleas, someone should be able to train electrons. Seriously, no current flows. I = 5V / ∞Ω = 0A. On the other side, take a large current and a circuit breaker. Open up the circuit breaker contacts and a spark will be formed as current jumps contacts. The current that is flowing does not want to stop. \$\endgroup\$ Apr 24, 2015 at 17:17

6 Answers 6


Your estimate is off by several orders of magnitude. Wikipedia gives the resistivity of air as being around \$10^{16}\ \Omega \cdot m\$. I'd guess an actual resistance between two points would be at least on the order of teraohms. Assuming \$1\ T\Omega\$, that gives a current of 5 picoamps, which is far too small to measure easily. As pointed out in an answer to another EE.SE question, the material the battery is made of is probably a better conductor than air.

To actually figure out what's going on in extreme situations, you need a more detailed model of the materials involved. How many electrons and/or ions are available for conduction? An ideal dielectric (insulator) has no free electrons, but a real dielectric might. What's the strength of the electric field? If you have a 40 kilovolt voltage source, you can rip apart air molecules, creating lots of free electrons! A less extreme example would be a vacuum tube, which "conducts" through empty space \$(R = \infty)\$ using electrons liberated from a piece of metal.

Ohm's law is an approximation that works for many materials at low voltages, frequencies, and temperatures. But it is far from a complete description of electrodynamics and physical chemistry, and should not be treated as such.

To answer your question more directly, regardless of whether a tiny current flows through the air, there can definitely be a voltage between the terminals. Voltage is another way of describing the electric field. Wherever there is an electric field, there is a voltage difference, even in a vacuum with no matter at all! HyperPhysics shows what this looks like.

Specifically, the gradient of the voltage field gives you the magnitude and direction of the electric field:

$$\vec E = -\nabla V$$

I don't know whether a tiny current actually flows through the air, but hopefully now you have a better appreciation for the physics of the situation. :-)

  • 2
    \$\begingroup\$ Sidenote for the reader: The last equation is valid only for magnetostatics. In electrodynamics, \$ \mathbf E = - \nabla V - \dfrac{\partial \mathbf A}{\partial t} \$, where \$ \mathbf A \$ is the magnetic (vector) potential, and \$ V \$ is the well-known electric (scalar) potential. \$\endgroup\$
    – alejnavab
    Jun 28, 2020 at 16:06

The voltage across the (ideal) battery is independent of the current through. That is to say, the battery is not an ohmic device and thus, does not 'obey' Ohm's law.

In other words, the voltage across the (non-zero) resistance is fixed by the battery; that voltage is given and is independent of Ohm's law.

Since the voltage across the resistance is fixed, the current through is determined by Ohm's law.

Thus, for an (ideal) open circuit (the limit as \$R \rightarrow \infty\$), the current through is zero but the voltage across is fixed by the battery voltage.

In summary, the voltage across the resistance (in this ideal circuit) is not determined by Ohm's law, it is determined by the battery. When the resistance is 'infinite', the current through is zero by Ohm's law.

Note that there is difficulty if we allow the resistance to go to zero. In the ideal case, the current is unbounded. However, this isn't physical. A physical battery cannot supply unlimited current (there is an effective internal resistance) and so, to model this, we add a small resistance in series with the battery.


When you have a fixed voltage and unknown current, you should re-state Ohm's law this way:

$$ I = \frac{V}{R} $$

Since V is constant and finite, then you can see that

$$\lim_{R\to{}\infty}I = \lim_{R\to{}\infty}\frac{V}{R} = 0$$.

So even though R is never quite infinite, Ohm's law still shows that the more you increase R, the closer I gets to zero.

  • \$\begingroup\$ I know that, but does that current really flow? \$\endgroup\$
    – Marco
    Apr 24, 2015 at 18:04
  • \$\begingroup\$ @d3l, Sure. In the case of an air gap, the current is more likely in the form of a few ions drifting one way or the other in the air between the electrodes rather than free electron flow. \$\endgroup\$
    – The Photon
    Apr 24, 2015 at 18:19

Open circuit means infinite resistance.

So: $$ V = I\times R = 0 \times \infty$$ and

$$ 0 \times \infty$$ is not defined.
See: Why is Infinity multiplied by Zero not an easy Zero answer?


Ohm's Law confusion — can there be voltage without current?

Yes. The wall outlet has voltage all the time, but current only flows if the circuit is closed with something that has a resistance.


Ohm's law applies only for the current flow through a resistor and the voltage across it.When there is no current flow Ohm's law is not at all applicable.Ohm's law is not a relation between unrelated voltage and current. There is a difference between source voltage and voltage drop.( I am going to shock you all! Whole electrical engineering is wrong! There is no flow of electrons in a pure conductor. People thought of heat as flow of caloric fluid and now have realized it is wrong.So is current flow! Thinking in terms of mechanics is an age-old bad habit. OK. A capacitor is a dielectric bound on two sides by conductors.Dielectric is an insulator. How does current flow through it? Engineers will say the capacitive current is 'displacement' current! So there is a 'current' not flowing?!) How do I back-up my 'theory'? Physics now says electron has no shape and volume. "To be more accurate, an electron is neither literally a traditional wave nor a traditional particle, but is instead a quantized fluctuating probability wavefunction."

  • \$\begingroup\$ Welcome to EE.SE, but this seems to be a rant rather than an answer. \$\endgroup\$
    – Transistor
    Aug 28, 2019 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.