How to calculate resistance between two electrodes filled with water

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SUMMARY

This discussion focuses on calculating the electrical resistance between two electrodes submerged in a water solution with known conductivity. The two experimental setups include rectangular plates with a 30 mm gap and a cylindrical configuration with inner and outer diameters of 10 mm and 50 mm, respectively. The resistance formula for the rectangular case is defined as R = ρL/A, where ρ is resistivity, L is the distance between electrodes, and A is the cross-sectional area. The relationship between conductivity and resistivity is clarified, emphasizing that they are inverses of each other.

PREREQUISITES
  • Understanding of electrical resistance and Ohm's Law
  • Familiarity with the concepts of conductivity and resistivity
  • Basic knowledge of geometry related to rectangular and cylindrical shapes
  • Ability to perform integral calculus for complex geometries
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  • Research the relationship between conductivity and resistivity in detail
  • Learn how to apply integral calculus to calculate resistance in cylindrical geometries
  • Explore practical applications of resistance calculations in electrochemistry
  • Investigate the effects of temperature on the conductivity of water solutions
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Researchers, electrical engineers, and students in physics or electrochemistry who are involved in experiments requiring resistance calculations in conductive solutions.

guptasuryakant
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Hello ,

Can someone from this group help me ,how to calculate resistance between two electrodes filled with water with a known conductivity solution.

In my case I have two experimental conditions. First there are two rectangular plates separated with a gap of 30 mm.

In second case this is a cylindrical geometry Innner electrode dia is 10 mm. And outer electrode dia is 50 mm. In between water is filled with a conductivity of 500µ cm/cm.

If some one can answer atleast for one case it will be useful for me.

regards
 
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For the rectangular case, the resistance is

R = \frac{\rho L}{A}

Where A is the crossectional area, and the resistivity is \rho

The calculation is similar for the cylindrical geometry, but you need to set up an integral to do it over small volume elements that approximate the rectangular situation.
 
How I can put the value in place of resistivity

Thanks for your prompt reply

How I can put the value in place of resistivity in above fourmula
 
You said you knew the conductivity of the solution. Probably conductivity and resistivity are just inverses, but check the units to be sure.
 

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