Why V=IR is not Ohm’s Law, and why that matters

I have lost count of the number of times I have heard people say that Ohm’s Law is [latex]V=IR[/latex].  I have seen it in lots of electrical training materials.

This post is to explain that Ohm’s Law and [latex]V=IR[/latex] are not the same thing, and that the difference matters; thinking they are the same is likely to lead to at least two misconceptions, which I will describe.

Having said all that, I have just done an internet search for “what is Ohm’s law?” and the top 8 results all say Ohm’s Law is [latex]V=IR[/latex]. So I’m rather nervous now about the amount of anger that might come this post’s way… Still, if I lack the courage, I can always take it down again!

What is Ohm’s Law?

Ohm’s Law says that the current ([latex]I[/latex]) through an electrical conductor is directly proportional to the voltage ([latex]V[/latex]) across it. Mathematically we can write that statement as [latex]I \propto V[/latex].

The relationship between [latex]I[/latex] and [latex]V[/latex] can be shown in a simple graph called a voltage-current characteristic. A V-I characteristic for a component obeying Ohm’s Law (like a resistor) would look like any other proportional relationship – a straight line through the origin, as below.

So far, so good.

What is V=IR?

This formula is the definition of electrical resistance (often stated as [latex]R=\frac{V}{I}[/latex], but it’s the same thing, just rearranged). The SI units of the quantities voltage, current and resistance are volts (V), amperes (A) – amps for short, and ohms (Ω), respectively. So if a voltage of 10 V causes a current of 2 A in a resistor, then its resistance is 5 Ω.

Why do Ohm’s Law and V=IR look the same?

Any proportional relationship, like the [latex]I \propto V[/latex] of Ohm’s Law, can be turned into an equation (with an equals sign) using a ‘constant of proportionality’. Doing this to our statement of Ohm’s Law gives this:

[latex]I=constant \times V[/latex]

And if we say that the constant of proportionality is [latex]1/R[/latex], then we end up with [latex]I=V/R[/latex], which is just [latex]V=IR[/latex] rearranged! So Ohm’s Law and [latex]V=IR[/latex] do look the same at first sight.

Next, let’s return to our resistor, in which a 10 V voltage causes a 2 A current…

The resistance of the resistor when the current is 2 A is 5 ohms, as we saw before. That is shown as the red point in the graph above. (The graph is a straight line, because we know resistors obey Ohm’s Law.)

When the current is higher, what is the resistance? Let’s look at the orange point:

[latex]R=\frac{V}{I}=\frac{20 \mathrm{V}}{4 \mathrm{A}}=5 \Omega[/latex]

So the resistance is still 5 Ω. In other words, whatever the current through the resistor, the resistance is 5 Ω. It is constant. Just like the constant of proportionality in [latex]I=constant \times V[/latex].

So it’s very natural to think that Ohm’s Law and [latex]V=IR[/latex] are the same thing. It just happens not to be true…

So what’s the problem?

The V-I characteristic for an old-style filament light bulb provides a useful route to seeing the problem. Here is one…

[latex]I[/latex] is not proportional to [latex]V[/latex] (the graph is not a straight line). So Ohm’s Law does not hold. But [latex]V=IR[/latex] does still apply at every point on the graph. It’s just that the values of resistance at each point on the graph are different. We can see this by calculating the resistance at the red and orange points on this graph:

Red: [latex]R=\frac{V}{I}=\frac{5 \mathrm{V}}{3 \mathrm{A}}=1.67 \Omega[/latex]

Orange: [latex]R=\frac{V}{I}=\frac{20 \mathrm{V}}{6 \mathrm{A}}=3.33 \Omega[/latex]

The first misconception, then, from thinking Ohm’s Law and [latex]V=IR[/latex] are the same, is:

“when Ohm’s Law doesn’t apply, then [latex]V=IR[/latex] doesn’t apply either.”

You would then think you can’t use [latex]V=IR[/latex] for a light bulb, for example. It is true that Ohm’s Law doesn’t apply in the case of a light bulb. But [latex]V=IR[/latex] does. A light bulb just has a resistance that changes with current, unlike the constant resistance of a resistor.

What’s the other problem?

I said there were two misconceptions. Here’s the other one. It’s a bit more subtle, but very common.

When you have a proportional relationship, say [latex]y \propto x[/latex], which is the same as [latex]y=constant \times x[/latex], the constant of proportionality is equal to the gradient of the graph. For example, in the graph [latex]y=3x[/latex], the gradient is equal to 3 (at all points on the graph, because the line is straight).

If you think Ohm’s Law is [latex]V=IR[/latex] you may well think the gradient of a V-I characteristic gives you [latex]1/R[/latex].

Let’s return to our resistor.

The gradient of the V-I characteristic has the value (4/20 = 0.2), and that does indeed equal [latex]1/R[/latex] (you can see this from 1/5 = 0.2). But you don’t need to think about resistance in terms of a gradient; you just read off values for [latex]V[/latex] and [latex]I[/latex]. For a straight line graph the two approaches are equivalent.

In fact, many people draw V-I characteristics ‘the other way round’, with [latex]I[/latex] on the [latex]x[/latex]-axis. To do so breaks a common convention of graphs, in which the independent variable is plotted on the [latex]x[/latex]-axis. Voltage causes current, so it makes more sense to plot [latex]V[/latex] on the [latex]x[/latex]-axis. The only reason I can think to ‘swap the axes’ is because for the straight line graph of the resistor, the gradient then gives you [latex]R[/latex] directly, rather than [latex]1/R[/latex].

All of this makes the second misconception very persuasive. The second misconception is:

“gradient = resistance, or 1/resistance, depending on which way round the axes are.”

But that only works for components that obey Ohm’s Law. It’s not a general rule. For our light bulb, if we try to calculate resistance from the gradient of the V-I characteristic, we will not get the right answer.

At a current of 6 A, the gradient of the graph has the value 0.08. If we try to calculate resistance from using [latex]1/R[/latex], we get a value of 12.5 Ω. That is not the right answer for the resistance. It isn’t the right method for calculating resistance. The right method is to read off the values of [latex]V[/latex] (20 V) and [latex]I[/latex] (6 A), and apply [latex]V=IR[/latex]. We did that before and got a value of 3.33 Ω (the right answer).

Here are three diagrams found from the first page of an internet image search of ‘voltage-current characteristics’.

I think they are misleading. Yes, the gradient happens to give you the right result, because the V-I characteristic in question obeys Ohm’s Law. But I think you could draw the conclusion from the diagrams that using the gradient to calculate resistance is a general technique. And it isn’t.

(There is a quantity equal to the gradient – it is called the differential resistance. But it isn’t the resistance [latex]R[/latex]. Differential resistance [latex]dV/dI[/latex] and resistance [latex]V/I[/latex] have the same value when [latex]I \propto V[/latex], but not otherwise. The usefulness of differential resistance is beyond the scope of this post.)

Interestingly, I looked at UK exam board specifications for 16 year-olds, and they don’t mention Ohm’s law at all. This startlingly simple approach cures all the problems above. All you need is [latex]V=IR[/latex], and the knowledge that for some components [latex]R[/latex] is constant, and for others it changes with current. Then it would never occur to anyone to calculate a gradient, or to think that sometimes [latex]V=IR[/latex] doesn’t apply.

If you still don’t think the difference between Ohm’s Law and [latex]V=IR[/latex] matters, then this will have seemed like the most pedantic post on the web, and I apologise…