Electronic – Conductivity of liquids (3D resistor network)

Regard it as a transmission line problem and from that, we know that the characteristic impedance is: -

\$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}\$

It's not too tricky to prove this - see my answer here

So for DC this reduces to \$Z_0 = \sqrt{\dfrac{R}{G}}\$

R is series resistance per unit length and G is parallel conductivity per unit length. For example if R is 1 ohm per metre and G is 1 micro siemen per metre then the characteristic impedance is 1000 ohms.

Looks like I'll have to do the proof. Imagine one section of the line having R series ohms and 1/G parallel Mohms. If this section is repeated to infinity the impedance looking into the 1st section is the same impedance as if the 1st section is discarded and you looked into the 2nd section.

From this simple and straightforward observation you can say: -

\$Z_{IN} = R + \dfrac{1}{G} || Z_{IN}\$. In other words this: -

^{simulate this circuit – Schematic created using CircuitLab}

So, \$Z_{IN} = R + \dfrac{\frac{Z_{IN}}{G}}{Z_{IN} + \frac{1}{G}} \$

\$Z_{IN} = R + \dfrac{Z_{IN}}{1+Z_{IN}\cdot G}\$

\$Z_{IN} + Z_{IN}^2\cdot G = R + Z_{IN}\cdot G\cdot R + Z_{IN} = R(1 + Z_{IN}\cdot G) + Z_{IN}\$

As the sections of cable are made infinitely small, \$Z_{IN}\cdot G\$ becomes an insignificant term (right hand side of the equation) hence we are left with: -

\$Z_{IN}^2\cdot G = R\$ or \$Z_{IN} = \sqrt{\dfrac{R}{G}}\$

Not an easy problem to solve without modeling software. You have probably noticed that water has a value of resistivity rather than resistance, stated in ohm-meters. This is because the ratio of voltage to current is proportional to the distance and inversely proportional to the cross section of the water.

Take a look at the two images below and you can begin to see the problem. The voltage distribution is shown in figure 12.4 and the current density in 12.5. The voltage varies between any two points in the current path, and the current density would also vary if the graph had higher resolution. Then consider that this is a model of a thin disc, and so there is a third dimension that would be present in real life to add complexity. The boundary might be conductive, which would change everything. Finally, the surface area of the conductor in the water will make a difference - the model uses point sources.

Your best bet is to place the water in a vessel of known geometry with electrodes of know geometry and do some modeling or some empirical testing. Bare copper may form a non-conductive oxide layer.