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Hi I'm an electrical engineer but have not seen this problem before. I'mn trying to measure/calculate the resistance of a liquid.

Background: I'm working on an electronic device that can come in contact with a liquid. Unintentional, can be avoided but is not considered a big deal. It's easy to clean and does not corrode/hurt the print.

What I noticed is that it misbehaves when in contact with the liquid. I've found that the liquid sometimes was conductive enough to cause a small bias current on a transistor, introducing a large bias on the output. Just enough to break the function.

Problem: I want to calculate the resistance of this liquid. But I've not found a solution for it yet. For instance I can measure the conductivity but to me it's not straight forward going from Siemens/cm to Ohm over a given distance. I assume the measured conductance is a from a 3D resistor network of unknown value.

Example: Glass tapwater 22 deg C, measured at 264uS/cm. I use a cheap Chinese TDS&EC so this measurement might be off. R= 1/S or R = about 3k8 at 1cm.

What would the resistance be at 2mm? Or at 1mm? How can I find this value?

I'm not looking for a solution to my circuit, there are many easy fixes. I want to understand the conductivity/resistance problem first.

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    \$\begingroup\$ We know Dk is 80 and I assume this behaves like a a high impedance. Normally resistivity is take geometrically as the volume divided by the area to get Ohm-m result. If you are getting something different, you may be seeing a chemical reaction or a slow reactance or some ionic contamination. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 28 at 20:31
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Resistivity is measured in (Ω⋅m). But you should think of the single distance measurement as "area over distance". Area over distance is m^2/m, which reduces to m. So resistivity goes up linearly with distance, and down linearly with cross-sectional area.

So if your tap water is 264µS/cm ~= 3800Ω.cm, then there is 3800Ω of resistance for every cm of distance through a water column with a cross-sectional area of one square cm. Double the length of the water column and the resistance doubles. Double the cross sectional area of the water column and the resistance halves.

If you stick with a water path that has a cross sectional area of one square centimetre, then if the path reduces to 2mm long, the resistance will be 760Ω. At 1mm it will be 380Ω.

Of course, practically speaking it would be very hard to control the cross-sectional area, and even if you could there's likely to be all sorts of edge effects as you start to get down to smaller distances. But at least that should help you understand resistivity.

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  • \$\begingroup\$ yes conductivity = S/m and wire resistance = Ω/m but "The SI" unit of electrical resistivity = ohm-meter (Ω⋅m). which may be more practical for a glob of liquid. R=ρL/A but this may not define gap as a capacitor leakage Rp \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 28 at 22:10
  • \$\begingroup\$ Oops! Yes, thank you. Fixed Ω/m -> Ω⋅m. \$\endgroup\$ – Heath Raftery Sep 28 at 22:17
  • \$\begingroup\$ Thank you for the input. I'll consider a water column as described to be the upper bound for my calculations. At the moment I've not got a good lower bound yet. I have to take in mind the other answers and comments: we are doing chemistry instead of pure resistive measurements. \$\endgroup\$ – JWRM22 Oct 6 at 16:41
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Hmm inside a liquid charge don't move as electrons but as ions. The resistance of the liquid changes with change in the applied voltage.What i try to say is that the resistance may not be ohmic.

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  • \$\begingroup\$ Yes, and for applied volts greater than about 0.5V, you start doing chemistry. VERY non-ohmic. \$\endgroup\$ – wbeaty Sep 29 at 8:30
  • \$\begingroup\$ Thank you for the answer. It was not exactly what I was looking for but I'll keep it in mind. \$\endgroup\$ – JWRM22 Oct 6 at 16:43
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te resistance between two pieces of metal in a liquid depends ---- on the topology

Lets assume the liquid forms a thin sheet, and that thin sheet lies between two parallel copper PCB traces. Assume the traces are 1mm apart, and run parallel for 10cm (100mm); assume the liquid sheet resistance is 1 Mega Ohm per square (for any size square, it turns out) because the sheet is very thin or the conductive is very low.

What is the resistance between the two parallel traces? there are 100mm/1mm = 100 squares.

Thus the resistance thru the sheet of liquid is 1Mega Ohh / 100 == 10,000 ohms.

What about other topologies? Consider this

schematic

simulate this circuit – Schematic created using CircuitLab

In this 2_D grid, we have a point entry for current, spreading out and around to end up on the left side at a BAR.

Can we model this easily?

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