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If I have just a battery, the terminals are separated by air, a very good insulator. We can say, practically, that no current flows.

However, air has a very large, but finite resistivity. Wikipedia gives a range of \$1.3\cdot 10^{16}\$ to \$3.3\cdot 10^{16} \Omega \mathrm m\$, which I'm sure varies by pressure, temperature, humidity, pollutants, and so on.

From that, can we calculate the resistance between the battery terminals, knowing the dimensions of the battery, and assuming that there is an infinite space of air around the battery? What's the math?

Could this be extended for calculating the resistance between two points in an arbitrary material, of infinite volume, with a known resistivity? The practicality of obtaining an infinite volume of a thing aside, I'm wondering how resistivity relates mathematically to idealized volumes of things that aren't extruded blocks of things. (That is, not \$R = \rho L / A\$.)

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  • \$\begingroup\$ I think it is more complicated than the resistivity (or conductivity) of solid volumes. Air conductivity has dynamic function and mainly it depends of ion number concentration and ion mobility, and can vary considerably from ~2-100 fSm-1, so you can not ignore geometry. Measuring the degree of decrease in charge for example of an electrically charged pieces of Teflon, we can measure ion concentration, and those air conductivity. \$\endgroup\$ – GR Tech Feb 7 '14 at 22:14
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Just a note: the resistance between two points in space is highly dependent on the geometry of the space. Even things far away affect the electromagnetic properties of things close by, which means the approximation isn't very good except for wide open spaces.

Ok, so to get the resistance in the setting you mentioned you also need to factor the current emitter itself. It makes no sense to ask the resistance between two points if the full geometry of the problem isn't specified (and this includes the battery).

In practice (or is it theory?), this means you have to e.g. model your two points as charged spheres with a constant +I/-I into them and solve Maxwell's equations. If you shrank you spheres too much, the current would inevitably be concentrated in a too small area, and you get rapidly increasing resistance. In other words, the resistance between any two points assuming pointwise sources is infinite -- which shows mathematically that asking "what is the resistance of the air" is incomplete -- you need to specify everything.

Edit: I forgot to add that because of the linearity of Maxwells equations in simple media with uniform conductivity like the models I exemplified, the final resistance is going to be simply proportional to the conductivity (a simple subdivision argument will convince you of this), so you can compare different (linear/homogeneous/etc) conductors directly.

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  • \$\begingroup\$ @PhilFrost Good catch, for some reason I had \$R=\rho*L/A\$ in my head. \$\endgroup\$ – almightyon Feb 7 '14 at 18:30
  • \$\begingroup\$ So this is an interesting thought, but why would infinitely small points result in an unlimited resistance? If you look at the grid of resistors in Andy's answer, I can pick two points there and there is some finite resistance. Does this not hold if we make the grid infinite, and 3-dimentional? \$\endgroup\$ – Phil Frost Feb 7 '14 at 18:36
  • \$\begingroup\$ Never mind, because this is already answered on physics.SE and you would seem to be correct. The explanation that made sense to me is actually in the comments: "You can't inject a current in a point without creating a divergent potential." \$\endgroup\$ – Phil Frost Feb 7 '14 at 18:47
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It's more likely that the surfaces of the battery will form a better conduction path than the surrounding air. For instance, the surface resistivity of Nylon 66 is about \$10^{11}\$ ohms per square. See this document for other common plastics.

Given the formula for volume resistivity (R = \$\rho \cdot L \over A\$) you could calculate what an equivalent volume of air would introduce "in parallel" with the surface resistance of your battery enclosure.

However, all the notes that I find on recommendations for measuring surface resistivity suggest that the air must not be humid when taking measurements as this will lead to errors. Given that non-humid air doesn't cause a significant error when making surface resistivity measurments it's probably reasonable to assume that the dominant conducting material around the terminals of your battery is the surface of the battery's material and not the air.

So, I would say that it is an interesting but fruitless exercise to calculate the resistance of air.

How would you calculate the resistance of a volume of air given that the resistance between two points may be X?

enter image description here

See also Phil Frost's attempt at humour in one of his comments LOL

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    \$\begingroup\$ An interesting point about the surface resistance, but I don't think \$R=\rho L/A\$ is useful here. What's \$L\$ or \$A\$ for an infinite volume of air? \$\endgroup\$ – Phil Frost Feb 6 '14 at 20:36
  • \$\begingroup\$ I am naive on this topic, but isn't knowing the resistance of air important for determining the distance at which to separate high voltage transmission lines? (As such, isn't there some method or formula already in use?) \$\endgroup\$ – JYelton Feb 6 '14 at 20:59
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    \$\begingroup\$ @JYelton - I think the breakdown voltage of air and its volume resistivity are seperate issues. When an insulator holds a wire from the pylon, that insulator does its best to maximize its surface area by its undulating shape - this kind of implies to me that the surface resistance is likely to be a much more dominant effect than the resitance of the surrounding air but, again this is breakdown voltage rather than ohmic resistance. \$\endgroup\$ – Andy aka Feb 6 '14 at 21:30
  • \$\begingroup\$ @PhilFrost I did think about this and concluded (rightly or wrongly) that the dominant conductance offered by air is the air most local to the surface of the insulator. I guess it could be modelled like an infinitely large cube of X ohm resistors forming square shapes - what's the resistance if you probe two nodes near the centre - it won't be zero and it won't be X but my math aint as good as it used to be. \$\endgroup\$ – Andy aka Feb 6 '14 at 21:34
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    \$\begingroup\$ @SpehroPefhany xkcd.com/356 \$\endgroup\$ – Phil Frost Feb 6 '14 at 21:44
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You could calculate the resistance of a rectangular block of any material from the formula R = \$ \rho \cdot L \over A\$, where L is the length of the block and A is the cross-sectional area (all in consistent units, obviously). That's assuming two conducting sheets at each end.

Actually calculating the resistivity between two electrodes of arbitrary shape immersed in a sea of conductor would probably have to be done by a numerical method such as using a field solver.

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    \$\begingroup\$ What about calculating the resistance between to magic, ideal, infinitely small electrodes? Surely this is a step towards the real case where there's a battery, and wires, and other stuff. \$\endgroup\$ – Phil Frost Feb 6 '14 at 20:48
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Doesn't quite make sense to look at it this way. You need to break up the air into a comparmental model, and use numerical methods to solve it (unless you can see a nice closed form solution. It might make more sense to think of it as a current or voltage field in a distributed resistivity. This is the way to thing about two electrodes in biological tissue. Might check out http://www.cmu.edu.cn/jcyxy/upl_files/20081122184243806.pdf

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