Is Ohm's Law violating itself?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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=V/R.

I= 15/∞ = 15*(∞^-1)

V=IR

V=I*(∞^1) = 15*(∞^-1)*(∞^1) = 15

.... 15 Volts.

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