Consider the following circuit...


simulate this circuit – Schematic created using CircuitLab

Now suppose the resistor has infinite resistance. Then obviously the current through the resistor will be zero. Now if we apply Ohm's law to this situation then the voltage drop across the resistor will be zero (since the current through the resistor is zero). So it means that the points A and B are at the same potential. But that's not possible since a resistor with infinite resistance will drop all the voltage across it. Isn't it? So is Ohm's law violating itself?

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    \$\begingroup\$ Why do you think that infinity times zero equals zero? A capacitor also has infinite resistance, but we say that it has 15V across it. \$\endgroup\$
    – Dave Tweed
    Feb 5, 2014 at 13:14
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    \$\begingroup\$ You need to remember that Zero * Infinite is an indeterminate form. So the result is not zero. \$\endgroup\$
    – Oceanic815
    Feb 5, 2014 at 13:16
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    \$\begingroup\$ Mathematically \$ \infty \cdot 0 \$ is not defined and can be anything since \$ \infty \$ is not a number but a concept of something being able to increase without bounds. In this particular case \$ \infty \cdot 0 = 15\$ as that's the voltage source you have used. \$\endgroup\$ Feb 5, 2014 at 15:50
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    \$\begingroup\$ Your "issue"is NOT with understanding Ohm's law but with understanding an aspect of mathematics. Using a circuit as a tool to investigate how dealing with infinity works just confuses things. | As a demonstration that Ohm's law works for real resistor value try 10^6 Ohms, then 10^9, then 10^12 etc You can asymptote to infinity for any finite resistor value this way and get a sensible answer. \$\endgroup\$
    – Russell McMahon
    Feb 5, 2014 at 16:18
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    \$\begingroup\$ If the resistor has infinite resistance, then it becomes an insulator, which may also react as a capacitor or another component other than a resistor. You may replace the resistor with an insulator and continue your circuit analysis to complete this problem. Note that if you do, in fact, come across an "infinite ohm resistor" for arbitrary voltages and currents, then you should not store it near your perpetual motion machine or flux capacitor, as it may result in local temporal instability. \$\endgroup\$
    – Adam Davis
    Feb 5, 2014 at 19:33

5 Answers 5


You are confused about what the concept of infinity means. Infinity isn't a number that can ever actually measure a quantity of something, like resistance, because it's not a real number. As Wikipedia aptly puts it:

In mathematics, "infinity" is often treated as if it were a number (i.e., it counts or measures things: "an infinite number of terms") but it is not the same sort of number as the real numbers.

When we talk about an "infinite" resistance, what we are really considering is this: as the resistor gets arbitrarily large, what does something (current, voltage, etc) approach?

For example, we can say that as the resistance gets arbitrarily large, current gets arbitrarily small. That is, it approaches zero:

$$ \lim_{R\to\infty} \frac{15\mathrm V}{R} = 0\mathrm{A} $$

That's not the same as saying the current is zero. We can't ever increase R all the way to infinity, so we can't ever decrease current to zero. We can just get arbitrarily close. That means you can't now do this:

$$ \require{cancel} \cancel{0\mathrm A \cdot \infty \Omega = ?}$$

This is a bit of a mathematical contradiction by most definitions of infinity, anyhow. Most numbers, when multiplied by an arbitrarily large number, approach infinity. But, anything multiplied by zero is zero. So when you multiply zero by an arbitrarily large number, what do you get? I haven't a clue. Read more about it on Mathematics.SE: Why is Infinity multiplied by Zero not an easy Zero answer?

You could ask, as the current becomes arbitrarily small, what does the resistance approach?

$$ \lim_{I\searrow 0} \frac{15\mathrm V}{I} = \infty \Omega $$

However, if you look closely, you will notice that if \$I = 0\$, then you are dividing by zero, which is your hint you are approaching something that can't happen. This is why we must ask this question as a one sided limit.

Leaving the realm of mathematics, and returning to the realm of electrical engineering, what do you really get if you remove the resistor from that circuit, and leave it open? What you have now is more like this circuit:


simulate this circuit – Schematic created using CircuitLab

C1 represents the (extremely small) capacitance between the two wires that aren't connected. Really, it was there all along but wasn't significant until the resistance went away. See Why aren't wires capacitors? (answer: they are) and everything has some capacitance to everything else.

  • \$\begingroup\$ I find it confusing that infinity * 0 != 0. I get that some mathematical concepts are just made up to be useful, but why would anyone define infinity such that infinity * 0 != 0? Zero of anything is zero. \$\endgroup\$
    – Clonkex
    Apr 13, 2023 at 2:34

It's best to consider Ohm's Law to be i = V/R to get an idea of which parameter is the dependent one when you have an ideal voltage source. In this case, your V is 15, your R is infinite, which makes your i = 0. No laws have been violated here.

When resistance is infinite, there no longer needs to be current flow for a potential difference. In steady state, it acts like an open circuit. You would do the same math for infinite resistance in steady state DC that you would for a capacitor. The voltage source is driving the potential difference across the resistor but the circuit is producing no current.

  • \$\begingroup\$ Which parameter is "the dependant one" just depends on the circuit and which things we happen to be allowing to change in a particular circumstance. What if instead of V1 we had a current source? Using your i=V/R I could make a similar argument about how Ohm's law is being violated. So is Ohm's law i=V/R, or is it V=iR? If you want to pick just one, I can pick a circuit that shows it's "impossible". \$\endgroup\$
    – Phil Frost
    Feb 5, 2014 at 16:01
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    \$\begingroup\$ Good point. It was mostly just a way to approach the problem from a conceptual standpoint. As pointed out in comments above, I think the main hangup is with the mathematics involved in infinities. \$\endgroup\$
    – scld
    Feb 5, 2014 at 16:36
  • \$\begingroup\$ Ohm's law shouldn't be applied to ideal shortcircuits or ideal open-circuits. In your answer, there's a contradiction when you apply Ohm's law: from this law, we know \$v \propto i\$, and in your example you got \$v = 15 \text{ V}\$ and \$i=0 \text{ A}\$, which violates Ohm's law because the current isn't proportional to voltage. \$\endgroup\$
    – alejnavab
    Jun 28, 2020 at 17:47

Here's how I like to think of it: if the resistor has infinite resistance, just replace the resistor with an open circuit:


simulate this circuit – Schematic created using CircuitLab

Now, do we want to say that there is no voltage drop between A and B? We can actually get rid of the wires, if we assume that the wires are ideal and don't have any resistance:


simulate this circuit

So the claim is that this schematic is equivalent to your circuit if the resistance of the resistor is infinite. Or very large, as it is in (unionized) air.


In DC circuits (during transient mode it is not a DC circuit per se) there are only 4 types of elements:

  • Power sources (voltage/current)
  • Connections (R=0)
  • Gaps (R=\$ \infty \$)
  • Resistors (R=arbitrary_number)

You are not solving abstract mathematical problem, but rather engineering problem, therefore your equations come from circuit, but not the other way.

When resistance of resistor becomes infinity it is no longer a resistor, but rather a gap - disconnected circuit - you adjust the circuit accordingly.
Applied to your example this means complete elimination of resistor. We can safely remove dangling wires, which leaves us with the source and points A and B become source poles.
Therefore, potential between A and B is source voltage. All the voltage drops on the resistor that is now gone.

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    \$\begingroup\$ It's usually not a good idea to answer old answered questions, unless you have a very good answer. In this case there is already a very good answer and the question is quite elementary. \$\endgroup\$
    – clabacchio
    Mar 3, 2014 at 14:54

This can be solved with powers of infinity. Discard the assumption that infinity is "forever" and consider it as being the "highest physically possible value" but unknown, represented as the everyday variable ∞ such that ∞^1 = ∞, ∞^-1=1/∞, etc.


I= 15/∞ = 15*(∞^-1)


V=I*(∞^1) = 15*(∞^-1)*(∞^1) = 15

.... 15 Volts.

Don't use this in school; they'll banish you for heresy. ;)


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