Understanding Resistivity and Current Relationships: Explained Simply"

[JohnGano]

So, resistivity goes like $$\rho = R \frac{A}{l}$$ (where A = cross sectional area, l = length)

I solved for resistance and got $$R = \rho \frac{l}{A}$$

If $$R = \rho \frac{l}{A} = \frac{V}{I}$$ would the current correspond to the ampacity of the material? Or, if you were to replace R in the first equation with V/I, what current value would be used in determining the resistivity of a material?

This is not a homework question. I just read about resistivity and now I'm trying to figure this out.

 

[cepheid]

Or, if you were to replace R in the first equation with V/I, what current value would be used in determining the resistivity of a material?

No, even if you were to do that to the equation, it would not mean that 'I' somehow determines 'rho'. The resistivity is an inherent property of the material. So, what that equation would be saying is:

"Given the resistivity of this particular material, and the geometry of the particular chunk of it that I have here, if I apply voltage 'V' across this chunk, a current 'I' through it will result."

The resistivity is determined by the nature of the material at the atomic scale. For instance, metals are chemical elements that easily lose their outermost electrons. As a result, a metal tends to contain a 'sea' of electrons that are not bound to any of the atomic nuclei in the lattice structure. As a result, they are able to flow freely in response to applied electric fields, and hence metals make good electrical conductors (they have a high conductivity, which means a low resistivity -- the two are merely reciprocals). In contrast, materials that are bad conductors (high resistivity) tend not to have many (or any) free electrons.

The point of resistivity is that, unlike resistance, it is a property of the material itself that does not depend on the specific amount of it that you have (kind of like density). In order to determine the resistance of a specific piece of that material, you need to know both its resistivity and the geometry of that piece.