# Resistance of an object with arbitrary shape

Suppose to have a conductor with a non-trivial geometry, like the one in the picture:

The shape does not really matter for this question, as long as you suppose to have at least one "hole" in the $$z-x$$ plane and the sectional area is not constant along the $$y$$ direction.

I want to calculate the effective resistance of this conductor, that can be supposed to be homogeneous with conductivity $$\sigma$$, between the two extreme points on the $$y$$ axis.

Normally, to calculate the resistance of a material, I would proceed in the following way:

1. Solve numerically the Laplace equation $$\nabla^2V=0$$ with boundary conditions:

• $$V(x,y=0)=1$$ and $$V(x,y=L)=0$$
• $$\nabla V \cdot \hat n=0$$ at every other boundary
2. Integrate the Ohm's law $$I=\int \mathbf J \cdot d\mathbf S = \int \sigma \mathbf E\cdot d\mathbf S =-\int \sigma \nabla V \cdot d\mathbf S = \Delta V/R$$

3. Solve with respect to $$R$$

If the section $$A$$ was constant this algorithm would give you the well-known formula $$R=\rho\frac{L}{A}$$, but it's clear that in this case is not appliable since the section is different at different $$y$$ points. How is then possible to calculate $$R$$?

Without any holes, I think the formula

$$R=\int_0^LdR=\int_0^L\rho\frac{dy}{A(y)}$$

should work, but I have the feeling that if I have a bifurcation things become more complicated.

• Even without holes, the last formula doesn't work. – Massimo Ortolano Mar 18 '19 at 20:15
• @MassimoOrtolano It's possible, I haven't proved it, it just sounded reasonable. Can you please explain why it doesn't work? – Alessandro Zunino Mar 18 '19 at 20:30
• I have a numerical model of current flow in a conductor with such a geometry. I can calculate the resistance. – Alex Trounev Mar 18 '19 at 20:37
• @AlexTrounev This is fantastic! Would you mind sending me the details of your model? – Alessandro Zunino Mar 18 '19 at 20:39
• I don't have much time, but the point is that your equation can be thought to be derived in an axisymmetric problem from $\boldsymbol{E} = \rho\boldsymbol{J}$, with $\boldsymbol{J}$ uniform along the cross section depending only on the reciprocal of $A(y)$ to keep the current constant along the axis. But such a field doesn't meet the boundary conditions, and so the equation is a good approximation only for very slowly varying $A(y)$, but not for a general shape. – Massimo Ortolano Mar 18 '19 at 21:09

It's necessary only to use your numerical results to calculate the total current entering or leaving one of the connections. At each discretized point at the boundary, calculate the gradient of the voltage normal to the connection and divide it by the material's resistivity. Since you've set the voltage drop to 1, you can take the reciprocal of the sum of these values weighted by boundary area to determine the object's electrical resistance.

For improved accuracy, you could average the current at both connections (the values should be identical in theory, but some discrepancy will arise in a numerical solution).

The same approach can be used to find the total thermal resistance of an object by replacing the voltage by temperature and the resistivity by the reciprocal of the thermal conductivity.

• I'm not sure to understand: you suggest to choose any section that does not contains holes to calculate the current $I$, because there is alredy the electric field (solution of the laplace equation) that is going to take in account the geometry of the conductor? – Alessandro Zunino Mar 18 '19 at 21:17
• You should specifically choose the boundary where you set $V=0\,\mathrm{V}$ (or the boundary where you set $V=1\,\mathrm{V}$) because you know for sure that the current density vector will be perpendicular to that surface. (You could also choose the middle because of symmetry.) – Chemomechanics Mar 18 '19 at 21:32
• Sorry, maybe it's a stupid question, but why at the boundaries with fixed $V$ I should be sure that $\mathbf J$ is perpendicular to the surface? – Alessandro Zunino Mar 18 '19 at 21:59
• Not stupid at all—this result arises because (1) the voltage gradient is perpendicular to any contour of constant voltage and (2) you're implicitly assuming that the resistivity is isotropic, so the current flux points in the same direction as the electric field. – Chemomechanics Mar 19 '19 at 0:33
• That is very nice, I think this settles the issue. Can I ask you a little off-topic question? Which software or programming language do you suggest to numerically solve the Laplace equation? – Alessandro Zunino Mar 19 '19 at 16:43

Using a numerical model of current flow in a conductor with such a geometry, we can calculate the resistance. The Laplace equation is solved, with the boundary conditions formulated by the author. It will be enough to calculate the total current in the initial and in the final section (for comparison). In fig. The distribution of potential and current density is shown at $$\sigma =1$$. In this case $$I=-\int {\sigma \nabla V \vec {dS}} =0.133\sigma \delta$$, $$\delta$$ - is thickness.

• Why the current should be calculated in the initial and final section? – Alessandro Zunino Mar 18 '19 at 22:01
• @AlessandroZunino Because the total current $I=const$. – Alex Trounev Mar 18 '19 at 22:03
• Yes, the total charge is conserved for sure, but what I am asking is if I can choose any section arbitrarily or if the initial and final ones are privileged. Chemomechanics, in his answer, suggests that I should specifically choose the boundary where I set $𝑉=0\,\mathrm V$ (or the boundary where I set $𝑉=1\,\mathrm V$) because I should know for sure that the current density vector will be perpendicular to that surface, but I do not understand why. – Alessandro Zunino Mar 18 '19 at 22:07
• @AlessandroZunino You can choose any section. Just the initial and final cross sections are convenient for integration in this case. – Alex Trounev Mar 18 '19 at 22:12
• Why are they convenient? – Alessandro Zunino Mar 18 '19 at 22:20