# How to compute the resistance of a nonuniform cylinder with varying resistivity?

The generally quoted formula foe resistance is

$$$$R = \rho \ell/A$$$$

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?

• Turn the object into a variety of 'resistors' in series and parallel. Commented Feb 12 at 21:25
• I thought that would have been done and published Commented Feb 12 at 21:27
• FYI, this is not a homework. I have removed that tag. Commented Feb 12 at 21:56
• 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. Commented Feb 12 at 22:55

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, $$$$\nabla\cdot\mathbf{J}=-\nabla\cdot\left(\sigma(r,z)\nabla\phi\right) =-\nabla\sigma(r,z)\cdot\nabla\phi-\sigma(r,z)\Delta\phi=0$$$$ is the equation to solve. This equaition differs from the pure laplace equation. Numerical solver, for example finite element method, is preferable.