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I have very poor understaning of electricity and usually go for "water analogy" (or Ohm's law) but it seem to have no use here:

I need to calculate a resistance of a metal disk with given diameter, thikness and conductivity but I'm also trying to understand what "actually" will go on?

Will current go in a thin channel of "least resistance"? Will it "go broadly" through all of the disk? My intuitiun tells me that it will matter what points of disk are actually in contact with the wires but why?

What is the "normal" way to calculate resistance of a complex body?

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  • \$\begingroup\$ You have not said how you are connecting to the disk. face to face is different from edge to center is different from point on edge to point on edge is different from chord on edge to cord on edge... Usually it's calculus if you really need to compute it awkwardly. Bulk resistivity, cross sectional area, slice up the area into dx. Someone with less rust and more mathjax can lay that out in proper form. \$\endgroup\$ – Ecnerwal Feb 28 '16 at 1:13
  • \$\begingroup\$ @Ecnerwal right, I was expecting some kind of integral coming up :) And I'm hoping to improve my intuition so it would be nice to get some kind of common solution. \$\endgroup\$ – Amomum Feb 28 '16 at 1:20
  • \$\begingroup\$ Ohms law has lot of use here - if you have a real physical disk with real physical connections, you run some current and measure a voltage, or vice versa, rather than dusting off the calculus book; It's much faster. Current will not go in a narrow channel - it will flow though all the "parallel resistors" made by all the various paths that go from one connection to the other. "Water analogy-wise" (I once really wanted to build a water 555) a circular puddle is your disk, fill it with sand (coarser or finer) for bulk resistivity. \$\endgroup\$ – Ecnerwal Feb 28 '16 at 1:44
  • \$\begingroup\$ There is something called skin effect which applies to higher frequency currents. It causes the current to concentrate in the outer (skin) area of the conductor. For DC or low frequency, the highest current density will be in the straight line from one contact point to the other, and fall off as you move away from that line, but exactly how it falls off, I don't really know. I think it would be easier to measure the resistance than calculate it. If you have the disk. \$\endgroup\$ – mkeith Feb 28 '16 at 2:48
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    \$\begingroup\$ The "parallel resistor" that is the shortest path is also the one with the least resistance, so the most current flows in it. There is no disagreement here. You can sort of do "taking a picture of it" by using stainless steel (or polished regular steel) shim stock in air, and letting the oxidation colors show the varying temperatures (you'd have to stop at whatever point you think the colors got interesting, but it takes its own thermal picture.) \$\endgroup\$ – Ecnerwal Feb 29 '16 at 2:56
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Once you get beyond approximations from using \$R=\frac{\rho L}{A}\$, the math gets pretty hairy to do it in closed form. The current flow depends on the electric field, which depends on the shape (boundary conditions), to varying degrees (the resistance from 1mm probes through a sheet of 3mm aluminum doesn't change much as you change the shape of the aluminum past a couple tens of mm).

I think the most reasonable approach for a complex structure would be to use a software package such as COMSOL and calculate it numerically. Or build a physical model from a material of known and consistent resistivity and test it (especially practical if it can be reduced to a 2D model). For simple situations it might be possible to mesh it manually and create a matrix of resistors, and then use SPICE to get a solution (when your only tool is a hammer..)

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