3
$\begingroup$

I have heard that Resistance is directly proportional to length and inversely proportional to area of cross-section. Can someone give me a practical explanation for this?

$\endgroup$
4
$\begingroup$

The formula is $$R = \rho \frac{l}{A},$$ where $R$ is the resistance, $l$ the length of the medium current is flowing in and $A$ its cross-sectional area. $\rho$ is the resistivity, a property of the material.

An intuitive way of understanding the dependence on $l$ and $A$ is the following. The longer the wire (increase $l$), the more collisions electrons have to go through in order to get to the end of it. More collisions $\Rightarrow$ larger $R$.

The larger the cross sectional-area $A$, however, and the more electrons you will have flowing at a given distance along the wire -- given a fixed electron surface density in the material, the larger the area the larger the number of electrons. More charge carriers means better conduction of electricity and lower resistance.

I would really like to see a derivation of the formula above though, even a heuristic one.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Well, I guess you could "derive" it in the Drude model (see my post here), where the proportionality of current density $\vec j$ and electric field $\vec E$ is the conductivity $\sigma$ or inverse resistivity $\rho$:

$~~~~~~\vec j = \cfrac{q^2}{m} \tau n \vec E = \sigma \vec E = \cfrac{1}{\rho} \vec E$

Using the current density $|\vec j| = I / A$ as quotient of current $I$ to cross section of your conductor $A$ and the electric field $|\vec E| = U/d$ as quotient of the voltage drop $V$ over a piece of wire of length $d$ one obtains

$~~~~~~\cfrac{I}{A} = \cfrac{1}{\rho} \cfrac{U}{d}$.

The definition of resistance $R = U/I$ than provides the sought relation

$~~~~~R = \rho \cfrac{d}{A}$.

From an engeneering standpoint you could see it as discrete portions of resistance, as a electrical circuit component. Put two in series, you double $d$, and you double the resistance. Put two parallel, well you figure it out youself ;).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your answer really helped me. $\endgroup$ – Sashank Sriram Jul 25 '15 at 7:04
  • $\begingroup$ @t0xic - your last 2 sentences are a real gem. I teach high school physics, and I'll endeavor to use this analogy in the class room. THANKS! $\endgroup$ – David White Aug 8 '15 at 0:39
0
$\begingroup$

The general derivation of Pouillet's Law is given in https://www.academia.edu/1841457/The_Notion_of_Electrical_Resistance

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

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