The generally quoted formula foe resistance is

\begin{equation} R = \rho \ell/A \end{equation}

some special cases are easy to solve. For example the case where the current flowing along the z-axis and $\rho$ or $A$ only depend on $z$ can be easily solved and is commonly mentioned in introductory textbooks.

Usually, the resistance is calculated by dividing the voltage (V) by the current (I) according to Ohm's law (R = V/I). However, consider the case when neither the voltage nor the current are known but the resistivity $\rho(r, z)$ is known.

How to compute the resistance when $\rho$ depends on both $r$ and $z$ i.e. $\rho(r,z)$ is an arbitrary but continuous function?

  • $\begingroup$ Turn the object into a variety of 'resistors' in series and parallel. $\endgroup$
    – Jon Custer
    Commented Feb 12 at 21:25
  • $\begingroup$ I thought that would have been done and published $\endgroup$
    – wander95
    Commented Feb 12 at 21:27
  • $\begingroup$ FYI, this is not a homework. I have removed that tag. $\endgroup$
    – wander95
    Commented Feb 12 at 21:56
  • $\begingroup$ Note also this tip about working with conductances in parallel, for radial variation in resistivity. But if the resistivity changes strongly with both $r$ and $z$, these approaches don't work because the current is no longer moving in the $z$-direction. Consider reviewing Resistivity of an object with arbitrary shape, which involves the same complicating issue. Summary: A numerical solution is probably required. $\endgroup$ Commented Feb 12 at 22:55

2 Answers 2


Resistance is a property of a component with linear relationship between current density $\mathbf{J}$ and electric field $\mathbf{E}$. In such cases one can show that there will be linear relationship between the voltage drop ($V$) between two points on that component and the current ($I$) flowing between them.

The general equations that govern this situation are (isotropic case):

$$ \begin{align} \mathbf{J}=\sigma\mathbf{E}&=-\sigma\boldsymbol{\nabla}\phi \\ \nabla^2\phi &= 0\\ \end{align} $$

Where $\phi$ is the electrostatic potential. You would solve them and then get relationship between voltage and current. Below is an example of a simple solution. In general solution may not be that simple, but it will give you what you are after - linear coefficient that relates voltage and current.

Simple solution for a cylindrical segment of a conductor

The resistance equation you have quoted is a simple solution of the equation I gave above.

In your case you could arrive at it by choosing a cylinder with axis of symmetry along z-axis, cross-section area $A=\pi R^2$, height $l$ and its centre at the origin. We choose the boundary conditions to be $\phi=0$ at the bottom face of the cyllinder $-\phi=V$ at the top face of the cylinder and $\partial_r\phi=0$ at the sides of the cylinder (no current electric field perpendicular to cyllinder walls).

One then finds, in cyllindrical coordinates with symmetry $$ \nabla^2\phi=\frac{1}{r}\partial_{r}\left(r\partial_r\phi\right)+\partial_{zz}\phi=0 $$

Assuming a more general $\partial_r\phi=0$ which means no current is flowing side-ways, only along the length of the cyllinder, we get the solution $\phi=-\frac{V}{l}\cdot \left(z-l/2\right)$.

Current ($I$) is then the current density, $\mathbf{J}=-\sigma\boldsymbol{\nabla}\phi=\mathbf{\hat{z}}\sigma\cdot V/l$, flowing through the cross-section of the cyllinder, so $I=\pi R^2 \mathbf{\hat{z}}.\mathbf{J}=A\sigma \cdot V/l$. Finally, introducing $\rho=1/\sigma$:

$$ \frac{V}{I}=l\rho/A $$


Since $\sigma(=1/\rho)$ is not constant in the cylinder, that is, \begin{equation} \nabla\cdot\mathbf{J}=-\nabla\cdot\left(\sigma(r,z)\nabla\phi\right) =-\nabla\sigma(r,z)\cdot\nabla\phi-\sigma(r,z)\Delta\phi=0 \end{equation} is the equation to solve. This equaition differs from the pure laplace equation. Numerical solver, for example finite element method, is preferable.


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