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Today, while drinking some water from a \$500mL\$ bottle, I started reading the info about the water and found out that the conductivity (\$\sigma\$) at \$25°\$C is \$147.9\mu S/cm\$. So it came to my attention that maybe I could calculate the resistance of the water bottle, from top to bottom. After some measuring, I found out that the bottle can be approximated as a cylinder with \$18cm\$ height and \$3cm\$ base radius.

So we can do the following: \$R_{eq} = \frac{\rho L}{A}\$, where \$\rho = \frac{1}{\sigma}\$ is the resistivity, \$L\$ is the bottle's height and \$A\$ is the base area. By doing this, I got \$R_{eq} \simeq 4.3k\Omega\$.

Then, I bought a new full bottle, made a hole on it's bottom (of course avoiding leakages) and measured the resistance (with a digital multimeter) from this hole to the "mouth", at first making it so that only the tip of the probes touches water. The measured resistance was really high, ranging from \$180k\Omega\$ to even \$1M\Omega\$ depending on how deep in water I positioned the probes.

Why is the measured resistance so different from what I calculated? Am I missing something? Is it possible at all to use a bottle of water as a resistor?

Edit #1: Jippie pointed out that I should use electrodes with the same shape as the bottle. I used some aluminum foil and it actually worked! Except this time I measured ~\$10k\Omega\$ and not the \$4.3k\Omega\$ I calculated. One thing I was able to notice while lighting a LED with water as a resistor was that the resistance was slowly growing over time. May this phenomenon be explained by the electrolysis that happens while DC current travels through water (the electrodes slowly get worse because of ion accumulation at their surfaces)? This would not happen for AC current, right?

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    \$\begingroup\$ Water conductivity will have an awful lot to do with the ionic content of the water. \$\endgroup\$
    – Scott Seidman
    Oct 16, 2014 at 17:57
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    \$\begingroup\$ Of course, but I imagined that the conductivity stated at the bottle would be enough to calculate the resistance. \$\endgroup\$
    – Thiago
    Oct 16, 2014 at 18:04
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    \$\begingroup\$ Interestingly, I've used a water bottle as a resistor before to test my power supply. It has excellent specific heat and can take a lot of energy before it heats up. The down side is that unless you plan to work with alternating current, electrolysis turn your water bottle into a hydrogen bomb! \$\endgroup\$
    – fuzzyhair2
    Oct 16, 2014 at 19:27
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    \$\begingroup\$ @fuzzyhair2 A hydrogen bomb is not just a mixture of oxygen and hydrogen :-) \$\endgroup\$
    – user1844
    Oct 16, 2014 at 19:51
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    \$\begingroup\$ The resistance of pure pH 7 water is quite high, but virtually anything dissolved in it will lower its resistance dramatically. On the other hand, virtually all conductive electrode materials react electrolytically with water, and, for DC systems, an insulating oxide layer will develop on one electrode. \$\endgroup\$
    – Hot Licks
    Oct 16, 2014 at 20:34

7 Answers 7

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The formula you use is valid for a certain area, but the size of your probes is nowhere near the area you used in your calculation. If you want a closer approximation, you'll have to use electrodes similar in size as the area you calculated the water column for, one flat on top, one flat at the bottom.

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  • \$\begingroup\$ So my approximation would be better if I use such electrodes at the top and at the bottom? Would it be fine to use them with simple wires soldered? Would the electrodes generate considerable capacitance? \$\endgroup\$
    – Thiago
    Oct 16, 2014 at 18:13
  • \$\begingroup\$ You get capacitance when you have a dielectric. Water isn't a dielectric, as it conducts. There won't be capacitance as the charge from one plate can travel through the water to the other plate. \$\endgroup\$
    – Majenko
    Oct 16, 2014 at 18:22
  • \$\begingroup\$ I will try it and add the results later. \$\endgroup\$
    – Thiago
    Oct 16, 2014 at 18:33
  • \$\begingroup\$ Tried the electrodes and it actually got a lot better. More info on the post. \$\endgroup\$
    – Thiago
    Oct 16, 2014 at 20:29
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    \$\begingroup\$ Sure, it may have a high dielectric, but that doesn't mean it will hold a charge between two plates when it conducts that charge between the plates equalizing the charge. If you want to use water as a dielectric you have to insulate the plates from the water, as you do when you want to use the capacitative method of probing water depth, or soil moisture content, etc. \$\endgroup\$
    – Majenko
    Oct 17, 2014 at 9:34
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I agree with @jippie.

For instance, take this cross-section of a good old-fashioned carbon rod resistor:

enter image description here

You notice the wires don't just stick into the carbon rod - instead they attach to metal plates the same diameter as the carbon rod.

The same with a more modern carbon film resistor:

enter image description here

Here the wires attach to nickel caps which connect with the carbon tube right around its circumference, not just at one point.

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    \$\begingroup\$ The carbon film is cut into a spiral pattern that wraps around the ceramic. So it does mostly just make contact over a small area. \$\endgroup\$
    – George Herold
    Oct 16, 2014 at 18:26
  • \$\begingroup\$ Yes, but it'll still make contact with all that area at the end, not just a single small point where the wire is connected. The important thing is that the connection is the entire size of the resistive element, whatever that be, not just a point on that resistive element. \$\endgroup\$
    – Majenko
    Oct 16, 2014 at 18:40
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As Jippie already pointed out, one of the issues is that your electrodes were much smaller than what your calculations assumed. They seem to assume the entire top and bottom areas of the cylinder will be the electrodes.

However, the resistivity of "water" varies widely. Very very pure and deionized water has very high resistivity. The resistivity of any real water you likely have access to is all about what impurities are in it. Even tiny amounts can make a large difference to resistivity.

Another issue for making a resistor from water is that there will be electrolisys at the electrodes. With no impurities and inert electrodes (like graphite), you will get hydrogen released at one electrode and oxygen at the other. With impurities and chemically active electrodes, lots of things can happen. For example, if you electrolyze salt water, you will in part get chlorine gas. Most metals will corrode at one end of the other if used as electrodes.

Water simply isn't a good substance to make resistors out of.

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    \$\begingroup\$ In the old days big barrels of salty water with copper plates that could be submerged in them, were used to control electric motors at the carnaval. So they actually were used as a sort of resistor. \$\endgroup\$
    – jippie
    Oct 16, 2014 at 20:26
  • \$\begingroup\$ I saw a setup much like that at a factory a few years ago, used during startup of a big plastic film extruder or something like that. \$\endgroup\$
    – brhans
    Oct 16, 2014 at 23:26
  • \$\begingroup\$ Early stage lighting setups sometimes used water for dimmers. As Olin says water by itself is not very useful- there was salt or acid added to the water to greatly increase the conductivity. See this for example. \$\endgroup\$
    – Spehro Pefhany
    Oct 17, 2014 at 1:25
  • \$\begingroup\$ Nice that you pointe doit that ion make changes on water conductivity. +1 \$\endgroup\$
    – RawBean
    Oct 23, 2014 at 20:29
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I've tried to measure the conductivity of water a few times with a DMM without much luck... or reproducible results. (using big flat probes.) Reading this, http://en.wikipedia.org/wiki/Conductivity_(electrolytic)

I think the problem may be DC electrolysis in the water/ probe ends. Now I'll have to try it AC some day!

Edit addition: (Friday Fun.)
So I was motivated to measure the resistance of water.
I put some 1/2 inch diameter SS posts in a plastic tub with ~1" of Buffalo tap water in the bottom. (A picture and data are here.)

Signals from a function generator where sent through the probes to an opamp TIA. (R = 1 k ohm) I moved the probes around an got ~ 1k ohm of resistance (See TEK000). Then I stuck the probes into a DMM (resistance scale). The resistance changed rapidly at first (starting at ~3k ohm) then slowly rose up to ~50k Ohm, at which point the DMM auto ranged and went to ~300k Ohm and then the resistance dropped to ~200k Ohm.

I then played some, Looked at step response, changed voltage drive amplitude.
(again data is in dropbox link)

I then sprinkled a pinch of salt. The resistance dropped quickly to ~100 Ohms (closer 150) Trying to measure with a DMM the resistance was 40 k Ohm!

The time constant was a lot faster with salt in the water.

To measure the resistance of water you need to do it AC with a frequency that is faster than the time constant of the water. (Time constant of the water changes with electrolyte concentration.)

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  • \$\begingroup\$ @Thiago, my pleasure. I did this years ago with a DMM and could never figure out why it didn't work. Your question stimulated me to figure it out. (Do try it AC... a much different number than the DMM gives.. try adding some salt with the DMM.) \$\endgroup\$
    – George Herold
    Oct 20, 2014 at 12:49
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I did my high school physics project on the DC conductivity of pure water (32 years ago) and found that increasing the current decreased the resistance linearly at first and then quite dramatically, the former and latter possibly caused by electrolysis at the electrodes (as mentioned by Olin Lathrop) causing ionization, the opposite of what you have found.

Hydrogen and oxygen gas at the electrodes will reduces their conductive surface area, increasing resistivity, but the hydrogen and oxygen travelling to each of the electrodes will conduct electricity, so you may have reverse/competing effects that may depend upon the shape and size of the electrodes. Perhaps my electrodes were large enough to discount the former effect (reduction in surface area) leaving only the latter.

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You need to measure resistance of water using an AC current. You measure the AC voltage across the electrodes and the AC current going through the water and divide out to get the effective resistance. The size of the electrodes will absolutely also affect the effective resistance. Measuring with a DC ohmmeter using point contact electrodes (lead tips) will always give you a higher than calculated resistance. All kinds of weird things happen at the electrode-water interface. There are many papers written on the subject.

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What you are missed in calculations is the temperature coefficient to correct the temperature changes if it is other than 25 degr C. For most applications has a value of a 2% per degr Celcius.

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  • \$\begingroup\$ There is no way this accounts for the huge difference between prediction and measurement. \$\endgroup\$
    – Chris Stratton
    Oct 17, 2014 at 1:18
  • \$\begingroup\$ Who is mention about a huge difference? Just remind what is missed. By downvote you mean we have to ignore the temperature coefficient at all....Really interesting!!! \$\endgroup\$
    – GR Tech
    Oct 17, 2014 at 21:14
  • \$\begingroup\$ The original error was a factor of 41 from expected. After some changes, it's about a factor of two. Your temperature model can explain neither. \$\endgroup\$
    – Chris Stratton
    Oct 17, 2014 at 21:20

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