2
$\begingroup$

A Comsol study using frequency sweep on electric current physics yields only current density as accessible variables.

I understand the underlying equation used is Ohm's law, i.e.

$$\mathbf{J} = \sigma \mathbf{E}$$

where $\mathbf{E}$ is the electric field, $\mathbf{J}$ is the current density, and $\sigma$ is the conductivity.

Why is it that they do not implement $V = R I$ directly.

What is the computational difficulty in its implementation ?

$\endgroup$
  • 3
    $\begingroup$ They represent different models, the first equation describes a continuum model, where you have a value of current density and electric field. This model would let you analyze what is happening inside your domain. In the second model you have a lumped version, i.e., a mascrocopic version of it. A model where you are considering the overall behavior of your system and not the internal details. Since Comsol uses FEM... I suppose that is natural to expect the first version, although you can obtain things like the current by integration. $\endgroup$ – nicoguaro Nov 29 '16 at 21:48
2
$\begingroup$

They are saying the same thing, at a fundamental level - it's just that the implementation is a bit different. The first equation you noted is valid throughout the system because it solves for the intrinsic values throughout the system, such as the local current density and local electric potential.

The second equation you mentioned is a lumped-value equation, that is technically valid throughout the system, but by the way you wrote it, I'm guessing that $\mathbf{V}$, $\mathbf{R}$, and $\mathbf{I}$ have units of voltage ($V$), resistance ($\Omega$), and current ($I$), respectively. The question is then, voltage where? Current where? Resistance where?

When modeling a continuum physics problem it's beneficial to reduce everything into its intrinsic properties instead of extrinsic properties, which means that you end up working with scale-independent quantities.

Now back to your first equation:

$$ \mathbf{J} = \sigma \mathbf{E} $$ is the same as $$ \mathbf{E} = \frac{1}{\sigma} \mathbf{J} $$

It's good to remember that conductivity ($\sigma$, with units Siemens/meter) is just the reciprocal of resistivity ($\rho$), which has units of $\Omega \cdot m$. Thus

$$ \mathbf{E} = \rho \mathbf{J} $$

This is more or less the valid local equation that an FEM solver would solve at each node in the problem domain, and which most resembles your global equation of Ohm's law.

For a more comprehensive answer to how resistivity relates to conductivity, resistance, and conductance, check out the wiki article on exactly this topic:

https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.