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Will a liquid resistor have different values with simple DC voltage versus pulsed DC voltage? And do voltage levels affect the resistance of a liquid resistor?

For example, if test a liquid resistor with simple DC supply, will it have same resistance as when tested with pulsed DC?

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  • \$\begingroup\$ If you know what happens for a "normal" resistor, you can assume the same for liquid resistor. Because it's nothing special..in the end, it's just a resistor which has linear characteristics. \$\endgroup\$
    – Mitu Raj
    Nov 25, 2021 at 20:32
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    \$\begingroup\$ Define what you mean by "liquid resistor". Nearly all substances that show resistivity have a linear current to voltage characteristic (certainly nearly any salt in water does). But Ohm's law is just an empirical law -- it's not based on fundamental physics, and not everything has a nice linear current to voltage characteristic. If your circuit counts on some random liquid having well-behaved current-voltage properties it's up to you to double-check that assumption. \$\endgroup\$
    – TimWescott
    Nov 25, 2021 at 20:41
  • \$\begingroup\$ Do you mean to use it in sensor applications? As in a variable resistance depending on tilt & co? \$\endgroup\$ Nov 25, 2021 at 20:43
  • \$\begingroup\$ What sort of "liquid resistor"? Metal such as Ga or Hg? Molten copper? Ionic liquid? \$\endgroup\$ Nov 25, 2021 at 21:29
  • \$\begingroup\$ What is that liquid resistor? I probably need it. Are you talking about "impedance and pulse response" of liquid, apart from solid and gas? \$\endgroup\$
    – jay
    Nov 25, 2021 at 23:24

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It depends on the liquid.

If it's a liquid metal like mercury, gallium, molten solder or sodium, then it will behave just like a normal solid resistor (save for possibly some magnetohydrodynamic flow in either self or external magnetic fields, which will likely be negligible if not intended) which is the same for AC, DC, pulsed DC etc.

If it's a heterogenous fluid, like an aqueous solution or minerals disolved in a molten salt, then it depends on the chemistry of the solution and electrodes. Polarisation, that is a chemical concentration gradient in response to a current flow, near the electrodes will usually increase the voltage drop across that part of the solution, so looking like an increase in resistance. This will occur with DC, and will increase as the current continues to flow, resulting in high and changing readings. This is why AC is usually specified for measuring the resistance of a solution.

If the solution is compatible with the electrodes, for instance copper electrodes and a copper sulphate solution, then at low currents polarisation will be less significant. However impurities can cause problems, and practical electroplating solutions contain all sorts of other chemicals (technology, practical know-how, 'secret sauce') to make the process work better.

There are many cases where measurements of aqueous solutions often have their own field of practical knowledge, in addition to just 'measuring with AC', for instance moisture levels in soil, or heart waveform electrodes.

When the solution is very weak, 'plain' water for instance, the effect can be the other way, that electrolysis of the electrodes puts orders of magnitude more solute ions into the fluid, increasing the conductivity.

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  • \$\begingroup\$ Can you elaborate more on Copepr sulphate solution with copper electrodes encapsulated in some insulation \$\endgroup\$ Nov 30, 2021 at 14:10
  • \$\begingroup\$ @MuhammadMuzammilNawaz One way to make high power or high voltage resistors is to use copper sulphate solution (CuSO4(aq)) with copper electrodes. For a high power resistor, two plates of copper in a bucket of solution. You can alter the conductivity by the concentration of the solution, the area and spacing of the plates. For a high voltage resistor (so physically long) that has an accurately calculable ability to handle pulse power, you can use a plastic tube filled with CuSO4aq, stoppered at the ends with copper rod or tube for the electrodes, avoiding gas bubbles in the solution \$\endgroup\$
    – Neil_UK
    Nov 30, 2021 at 14:29

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