Resistivity is defined in units of \$\Omega \times \textrm{cm}\$. I don't conceptually understand what is meant by the unit.

If it was \$\Omega / \textrm{cm}\$, that would be easy to understand - a certain number of ohms for every centimeter. How can one understand \$\Omega \times \textrm{cm}\$?

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    \$\begingroup\$ Hint: It's really \$\Omega\frac{{\rm cm}^2}{\rm cm}\$. \$\endgroup\$
    – The Photon
    May 21, 2020 at 1:15
  • 2
    \$\begingroup\$ Don't forget conductivity, resistivity's twin brother, which is 1/resistance. It's often constructive to flip into thinking in terms of conductivity. \$\endgroup\$ May 22, 2020 at 15:47

5 Answers 5


Imagine a block of material with a uniform density to it. Something like this:

enter image description here

The material also has, let's say, a uniform "resistivity" to it.

Now, suppose we cover the entire face pointed at by the arrow, and the face opposite to it that we cannot see, by plating them with silver (which is very conductive.) We then measure the resistance between these two silvered faces on opposite ends using an ohmmeter. There will be some value for that in Ohms.

Now, let's consider three modifications:

  1. Suppose we doubled the length. Here, since the silvered faces touched by the ohmmeter have the same area as before, but are further apart, we should expect that the resistance we'd measure between the opposite X faces would double.
  2. Suppose we doubled the height. Here, since the silvered faces touched by the ohmmeter have doubled in area but are the same distance apart as before, we should expect that the resistance we'd measure between the opposite X faces would be cut in half.
  3. Suppose we doubled the width. Here, since the silvered faces touched by the ohmmeter have doubled in area and are the same distance apart as before, we should again expect that the resistance we'd measure between the opposite X faces would be cut in half.

So, we postulate the following about the resistance we'd measure:

  • \$R\propto \text{Length}\$
  • \$R\propto \frac1{\text{Width}}\$
  • \$R\propto \frac1{\text{Height}}\$
  • \$\therefore R\propto \frac{\text{Length}}{\text{Width}\:\cdot\:\text{Height}}\$

Now, if we call the length, \$L\$, the width, \$W\$, and the height, \$H\$, and introduce a constant of proportionality, we can say:

$$R=\rho \cdot \frac{L}{W\cdot H}$$

Let's now express the above only looking at the SI dimensions:

$$\begin{align*}\Omega=\rho \cdot \frac{\text{m}}{\text{m}^2}, &&\therefore \rho=\Omega\cdot\frac{\text{m}^2}{\text{m}}=\Omega\cdot\text{m}\end{align*}$$

Just simple dimensional analysis.

  • 22
    \$\begingroup\$ My car's fuel economy has units of area (say 8.1 liters/100km), which is roughly AWG 28. Physical interpretation is a thin cylinder of gasoline consumed as the car moves forward. \$\endgroup\$ May 21, 2020 at 3:14
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    \$\begingroup\$ @SpehroPefhany Yeah. And torque has units of Joules but really is Newton metres and isn't a scalar, but a vector besides. But this doesn't injure what I wrote. \$\endgroup\$
    – jonk
    May 21, 2020 at 4:22
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    \$\begingroup\$ Not at all (+1), torque having units of work is a good one. \$\endgroup\$ May 21, 2020 at 4:23
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    \$\begingroup\$ @SpehroPefhany Torque should probably be read as Joules per radian, I suppose. Radians are unitless -- but still important despite it. Dimensional analysis has a lot of subtle corners. We spent many many hours in my high school physics class applying it scaled modeling problems (like, you find you need crazy-minded fluid characteristics for a scale waterfall where you first imagined just using water only to find out you can't.) \$\endgroup\$
    – jonk
    May 21, 2020 at 4:43
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    \$\begingroup\$ @elile Twice the conductance. Two equally conductive paths driven by the same EMF (volts per meter) will have twice the conductance of just one of them, separately. The resistance is therefore half as much. (That is, without introducing other problems. It's always possible to hypothesize edge-case scenarios which may complicate such questions. But these are very rarely encountered in practice. Usually, they only happen when scientists intentionally create those circumstances to see what then happens.) \$\endgroup\$
    – jonk
    May 21, 2020 at 8:38

To understand this, you must first know that resistivity is basically the total number of resistance per unit length AND cross sectional area.

$$ \frac{\Omega}{\textrm{cm}} \times \textrm{cm}^2 = \Omega \times \textrm{cm} $$


  • \$\Omega / \textrm{cm}\$: value of resistance per unit length
  • \$\textrm{cm}^2\$: cross sectional area
  • \$\begingroup\$ If you say that the bigger the cross section is, the higher the resistance is, that's incorrect imho... \$\endgroup\$
    – Déjà vu
    May 21, 2020 at 12:41

Another way to think about this repeats essentially the same dimensional analysis as what jonk wrote above, but it starts from Ohm's law which can be written more generally as:

$$J = \frac{E}{\rho}$$

where \$J\$ is the current density, \$E\$ is the electric field and \$\rho\$ is the resistivity. This is always true while \$V=IR\$ is actually rarely true. However, if we keep it simple and consider the rectangular prism that jonk describes above, we can consider the material to be isotropic (meaning the resistivity is the same in all directions), and we have:

$$J = \frac{I}{A} = \frac{E}{\rho}$$

where \$I\$ is the current above and \$A\$ is the cross sectional area. This can simply be rearranged:

$$\rho = \frac{E\times A}{I}$$ Looking at the RHS and doing SI units analysis (fudging dimensional analysis a bit) gives:

$$ \require{cancel} \frac{[\frac{V}{\cancel{m}}][m^{\cancel{2}}]}{[\frac{C}{s}]}= \frac{V}{Amp}\cdot m = \Omega\cdot m$$

Here we have used the usual units of volts per meter for electric field, and coulombs per second for amperes. The best way to think about resistivity or conductivity is that it translates an external electric field into a current density inside of a material with free charge carriers.

In electromagnetic theory, units are sometimes highly confusing and it's better to focus on what the quantity means through fundamental equations. As food for thought, consider that in Gaussian units, the resistivity is measured in seconds! You could rationalize that as a time required to travel unit length in response to an applied field, etc, but I still think it's better to stick with the fundamentals.


So if \$\rho\$ is given as \$1.6 \mu\Omega-\text{cm}\$ (copper)

If you consider a strip 1cm long and 1 cm wide, it's the thickness of the strip in cm to make it \$1 \mu\Omega\$ in resistance.

  • \$\begingroup\$ There's a typo, either in the first sentence or in the second. \$\endgroup\$
    – vu2nan
    May 21, 2020 at 4:18
  • \$\begingroup\$ @vu2nan how so? \$\endgroup\$ May 21, 2020 at 4:20
  • \$\begingroup\$ My apologies! I stand corrected. The thickness of that strip would certainly be 1.6cm. \$\endgroup\$
    – vu2nan
    May 21, 2020 at 4:38

If we start with restistance as relating to a particular thing, say a length of wire of length 10cm. It is measured in Ohms.

Now, the wire manaufacturer will probably specify the wire as 99% copper, with a specificed resistance in Ω/cm. That is a slightly more abstract concept. If we put two resistors in series, we know the resistance is doubled.

For a physicist the interest is more about the properties of 98% pure copper in general. So far we have a quantity that depends on the gauge of the wire. To a good approximation the resistance is inversely proportional to the area: a good way to visualize that is to imagine we have a stranded cable, so a thicker cable is just more strands in parallel. This gets us Ω.cm.

It's not easy to visualize units like that. One thing that might help would be to think of the resistance between two faces of a cube of copper. It would decrease as the cube grows. Maybe it's easier to think of the conductivity of a material having units of S/cm (where S = Siemens aka mhos)


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