Impedance of an electrode

In summary, the conversation discusses the calculation of the impedance of an electrode submerged in sea water using a ring shaped counter electrode. The approach and details presented in the conversation are incorrect and the assumption that current returns to a single edge is unphysical. A finite-element or similar simulation package is recommended for solving 3D geometries.
  • #1
toniojesusde
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[Moderator note: This is an old thread from 2010. But it sounds like a good question, so I moved it here and bumped it to the top.]

Hello, this is my first post to PF. I've been trying to find out my answer in similar posts (resistance of a disk, ...) with no luck. So I would really appreciate some help. Here's the thing:

I am trying to calculate the impedance of an electrode submerged in sea water, using a ring shaped counter electrode (the attached images depicts the system).

In order to find out a value I decompose the cylinder in diferential surfaces dS, and calculate the resistance for each element using R_dS=R_sx/(2∙π∙r∙dx)
Where R_sx can be R_s1 (for those elements of the electrode that overlap with the cathode) or R_s2 (for the rest of the elements)

R_s1=ρ∙d, being ρ the resistivity of the electrolyte and d the distance between the cathode and the anode (electrode).
R_s2=ρ∙√(d^2+x^2 ), being x the distance in the x-axis between the right border of the cathode and the considered diferential element.

to calculate total resistance i do the inverse of the sum of the inverses of all the diferential resistances. (that a tongue twister! :tongue2:)

Is this whole solution valid? I assume that all the current lines of the elements separated from the cathode go from the electrode to the same point (right border) of the cathode, is that the real situation? any suggestions for improving the model? Thanks a lot guys, you are the best, keep up with the hard work.
 

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  • #3
Unfortunately both the approach and the details are incorrect. Given a potential difference V, one must solve for the spatially varying current density in order to arrive (via integration) at total current and, finally, at R=V/I. The assumption that current returns to a single edge (the drawing doesn't make much sense, furthermore, it doesn't look like a ring) is unphysical. On the left edge of the drawing, a great deal of current will flow vertically between cathode and anode, for instance.

The number of configurations that can be solved analytically is very small. Conformal mapping is useful for certain 2D configurations and is covered in older texts (see the excellent book by E. Weber, Electromagnetic Theory: Static Fields and Their Mapping, or Smythe, Static and Dynamic Electricity.) For 3D geometries like the one above, the best approach is to use a finite-element or similar simulation package.
 
  • Informative
Likes anorlunda

1. What is impedance of an electrode?

Impedance of an electrode refers to the measure of resistance to the flow of electrical current through the electrode. It is a combination of resistance, capacitance, and inductance and is measured in ohms.

2. How does impedance affect electrode performance?

Higher impedance can lead to a weaker signal and potential signal loss. It can also cause heating and tissue damage in electrodes used for medical purposes. Lower impedance generally results in better electrode performance.

3. What factors can affect the impedance of an electrode?

The material and size of the electrode, as well as the electrolyte solution used, can affect impedance. Other factors include the surface area of the electrode and the distance between the electrode and the tissue.

4. Why is it important to measure impedance in electrode-based systems?

Measuring impedance can provide information about the quality and functionality of the electrode. It can also help in detecting any changes or damage to the electrode, which can affect the accuracy and reliability of the system.

5. How is impedance of an electrode measured?

Impedance is measured using an impedance analyzer or a multimeter. The electrode is connected to the device, and a small AC current is passed through it. The device then measures the voltage drop and calculates the impedance using Ohm's law.

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