# I Current flow model through a resistance field

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1. Aug 18, 2016

### akyriazis

Hey everyone.

I thought about this problem the other day, but I don't know what theory I need to solve this problem.

Given a 2D rectangle with a known, varying resistance, and a known voltage drop across the length, how would one calculate the current density at each point on the surface. By intuition, the current should in general go in the direction of least resistance, but I also think there is probably a mechanism to prevent current from all flowing through the same place. Is this the right thinking?

I also would like to be able to numerically calculate this, but how to do this will be much clearer when I have an idea of the theory.

Is anyone able to point me in the right direction?

2. Aug 18, 2016

### BvU

Hello Akyriazis,

The theory you need is a kind of localized form of Ohm's law. Resistance is an over-all property of a piece of material. And current through your rectangle an over-all result.
Resistivity and current density are the local equivalents. The first is a scalar ( a number, depending on position ) and the second is a vector (with a magnitude and components - in your case x- and y components).
Check out the link and see if you can write down the differential equations and the boundary conditions for your case.

3. Aug 18, 2016

### Khashishi

If the resistivity is constant, the problem is equivalent to solving Laplace's equation with Dirichlet boundary conditions. If it isn't constant, you have an equation that looks like this:
$\nabla \cdot \mathbf{J} = 0$
$\mathbf{J} = \sigma \nabla{V}$
$\nabla{\sigma} \cdot \nabla{V} + \sigma \nabla^2 V = 0$
$\sigma$ is the reciprocal of the resistivity.
This is a generalized Laplace equation. It reduces to Laplace's equation for $\nabla{\sigma}=0$
Unless you can exploit some kind of symmetry, you'll probably need numerical methods to solve this.

4. Aug 18, 2016

### lychette

It might help you to look at Van der Pauw technique for calculating resistivity for arbitrary shapes.