Resistance is given by $\rho L/A$, where $\rho$ is the material constant, $L$ is the length, and $A$ is the area.

Is there any way that this can be derived mathematically, or is the only way experimentally?

Personally, I think experiment is the only way as I do not know how you would get $\rho$ otherwise.

  • $\begingroup$ Intriguingly close to my question physics.stackexchange.com/q/21877 but still fundamentally different in what is being asked. Here, the question is "how can we get this" whereas mine was "I know this is wrong, what are the correct mathematics of the system" $\endgroup$ Apr 16, 2012 at 3:25
  • $\begingroup$ Wait, revision: the equation here is exactly correct under the pertinent assumptions. My question was about a different form with an integral of the inverse area, which introduced the assumption that the change in area is small relative to the length (more-or-less). $\endgroup$ Apr 16, 2012 at 3:37

3 Answers 3


There is actually a student-friendly microscopic model how to derive the real Ohm's law

$$\vec{j} = \sigma \vec{E}.$$

After its derivation you can transform it into the more common form using the answer by Nesp.

The idea goes as following:

We must start with the definition of current:

$$I = \frac{\Delta Q}{\Delta t}.$$

So where does current come from? Current is the result of movement of charged particles in the material. Obviously, current will be proportional to the charge of one particle, the speed of the particle and the total number of particles. Current density $\vec{j}$ can therefore be written as

$$\vec{j} = N q \vec{v}_\text{d},$$

where $N$ is the density of particles, $q$ is the charge of one particle and $\vec{v}_\text{d}$ is the drift speed, that is average speed of particles. I think that this definition is self-explanatory, but it can also be derived more strictly from the second formula.

In material you have certain amount of almost "free" electrons. Those electrons behave like particles in the gas, they are crashing between themselves and into atom cores, bouncing back and forth and there is actually no net movement, average speed and current are zero.

However, if you put some potential on the ends of material, you actually put homogenous electric field in the material, which strength is

$$E = \frac{V}{l}$$

All electrons start accelerating in the direction of the positive potential and you can easily obtain this acceleration using the expression

$$\vec{a} = \frac{\vec{F}}{m} = \frac{q\vec{E}}{m}.$$

So you actually get net movement of electrons. And now comes the beauty of Ohms law. You should ask yourself: If electrons accelerate, how come current (which is proportional to average speed of electrons) isn't becoming larger and larger with the time?

The reason is that electrons keep crashing into atom cores and after those crashes their speed is by average reset back to zero! So let's define some typical time between two crashes $\tau$. The average maximum speed of electrons between two crashes shall be

$$\vec{v}_\text{max} = \vec{a} \tau = \frac{q\vec{E}}{m} \tau.$$

Obviously, average speed between two crashes, which equals drift speed is half of that value.

From the definition of current density you finally obtain

$$\vec{j} = N \frac{q^2 \tau}{2 m} \vec{E}$$

which is Ohm's law.

Therefore - and this is direct answer to your question - conductivity is

$$\sigma = \frac{1}{\rho} = N \frac{q^2 \tau}{2 m}$$

and can be determined by knowing the mass and the charge of electron, density of free electrons in the material and the average time between two crashes.

By the way: these crashes between electrons and atom cores actually adds heat to material (increases temperature), so this microscopic model explains everyhing.

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    $\begingroup$ +1; this answer contains the logic of the Drude model which is the simplest microscopic justification for the Ohm's law, see en.wikipedia.org/wiki/Drude_model $\endgroup$
    – Slaviks
    Apr 16, 2012 at 7:45
  • $\begingroup$ @Pygmalion Thanks for your answer! (I'm right in the middle of revisiting my dogmas) What kills all the charm of Drude model as an intuitive representation for me is the following fact: if you imagine a material as a steeplechase distance, you expect the obstacles to be evenly distributed with the distance (so that material with higher resistance just has higher rate of obstacles). So, when your acceleration (and, thus, average speed) is higher, your average time between crashes should be lower: $x=\frac{a\tau^2}{2}$, $\tau=\sqrt{2x/a}$. But this contradicts the linearity of Ohm's law. $\endgroup$ Jun 9, 2015 at 9:19
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    $\begingroup$ @Bob Unfortunately this is only one of the facts in that simple model that does not fit. My hunch is that proper and complex statistical treatment would solve most of the problems. Of course, as some already pointed out, the only fair treatment is using quantum mechanics. $\endgroup$
    – Pygmalion
    Jun 12, 2015 at 8:12
  • $\begingroup$ R=rho*l/A is experimentally determined. What I mean is that some part of your proof is bound to be an experimentally derived equation which in this case is the relation between j and drift velocity. So, this is just shifting the experimentally derived relation. Nevertheless, your answer is very informative but I still would not count it as proof. $\endgroup$
    – Aniansh
    Mar 4, 2018 at 21:02
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    $\begingroup$ @SNB If cross section varies, then electric field will also vary. And if electric field varies, the drift velocity will also vary. You can post your problem to the stackexchange and someone (me if nobody else) will solve it for you. $\endgroup$
    – Pygmalion
    Jul 10, 2018 at 18:56

The answer is "yes", if you take for granted that $R$ is defined by the relation $\Delta V=IR$. In fact it is derived from (the real) Ohm's Law.

Ohm's law states that, for some materials (the so-called "Ohmic" materials) the current density vector $\vec{J}$ (current per unit area) is parallel to the electric field $\vec{E}$, i.e., $$\vec{J}=\sigma\vec{E}=\frac{1}{\rho}\vec{E}\ \ \ \ \ \ \ (1),$$ where $\sigma=1/\rho$ is the conductivity of the material (which is the inverse of $\rho$, the resistivity), which can be considered a constant for some materials (but is not restricted to be constant in general). From here, consider a material of length $L$ which has two extremes of area $A$ where we apply a potential difference $\Delta V$. Using the definition of the potential difference, it is easy to show that $$|\vec{E}|=\frac{\Delta V}{L}\ \ \ \ \ \ \ (2).$$ On the other hand, we can express the current flowing trough the material, from the definition of current density as $$|\vec{J}|=\frac{I}{A}\ \ \ \ \ \ \ (3).$$ Using, then, the results of equation $(2)$ and $(3)$ on equation $(1)$, we get $$\frac{I}{A}=\frac{1}{\rho}\frac{\Delta V}{L},$$ or, $$\Delta V=I\frac{\rho L}{A}.$$ On the typical relationship, $\Delta V=IR$, then $R=\rho L/A$.

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    $\begingroup$ You mention that you "take for granted" $\Delta V=IR$, but I would take issue in that $R$ can be said to be defined by that equation. Provided the model yields a linear relationship (which it does and has been sufficiently demonstrated) then resistance is just naming the quantity. $\endgroup$ Apr 16, 2012 at 3:29
  • $\begingroup$ @AlanSE: right. However, I just said "take for granted" because this relationship is indeed derived from Ohm's Law. $\endgroup$
    – Néstor
    Apr 16, 2012 at 3:35

One fundamental

Well, if you consider $V=IR$ as fundamental (and not $\bf J=\sigma \bf E$--IMO this is the actual Ohm's law), then you can derive it by what I call "discrete calculus"

Taking the definition of resistivity as "resistance per unit length and unit area":

Take a cuboid of dimensions $L,W,H$ along coordinate axes $x,y,z$. Current flows along $x$. Let $\delta x,\delta y,\delta z${*} be unit elements. We cannot use calculus here as we do not yet know that resistivity is to be multiplied by length but divided by volume.

Now, lets take a horizontal "pillar" at $y,z$, of area $\delta y\delta z$. This pillar is divided into $\frac{L}{\delta x}$ parts (which is equivalent to $L$ parts--since $\delta x$ is a unit quantity--but this gets us a dimensionless constant).

So, we have $\frac{L}{\delta x}$ tiny cubes of resistance $\rho$ (unit dimensions, right)? They are in series, so total resistance is $\rho\frac{L}{\delta x}$. Lets call this $\delta R$

Now, we have $n=\frac{WH}{\delta y\delta y}$ such "pillars", of resistance $\delta R$, all in parallel. Since they have the same resistance, we can just divide $\delta R$ by $n$ to get total resistance.

$$\therefore R=\delta R/n=\frac{\rho L \delta y \delta z}{WH \delta x}$$

We can absorb the $\delta$ terms into $\rho$, since initially we took $\rho$ to be of dimensions of resistance. Now we can just rewrite it to have dimensions of resistivity (which we "did not know" initially). Also, $WH=A$(area).

So, $R=\frac{\rho L}{A}$

Both fundamental

If you consider $\bf J=\sigma \bf E$ or $\rho \bf J= \bf E$ as fundamental as well as $V=IR$, then we can derive the formula:

$E=V/L$, since we're considering uniform cuboid

$J=I/A$ by definition

$\implies \rho I/A=V/L\implies V=I\left(\frac{\rho L}{A}\right)$

Comparing with $V=IR$ we get $R=\frac{\rho L}{A}$


Remember that resistance ad resistivity are sort of arbitrarily defined concepts. Resistance is "ratio of p.d. and current", resistivity is either "resistance of unit area and length", or "ratio of electric field and current density magnitudes".

My first proof sort of tries to do away with as much arbitrariness as possible.

*This is just me having fun--I rarely get to use the quirky lowercase delta :)

  • $\begingroup$ +1 nice work. Now try to derive it from molecular properties and the conductive band of electrons.. $\endgroup$ Apr 16, 2012 at 3:39
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    $\begingroup$ @ja72: Left as an exercise to the reader $\endgroup$ Apr 16, 2012 at 4:31
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    $\begingroup$ @ja72 Are those things sufficient to establish a linear (or otherwise) concept of resistivity in the first place? $\endgroup$ Apr 16, 2012 at 5:46
  • $\begingroup$ @AlanSE: I don't know, but this sounds like an interesting question to be posted in Physics SE. $\endgroup$ Apr 16, 2012 at 12:39
  • $\begingroup$ @ja72: Seems to be slightly different from this--out of scope for this question. You may want to repost with a > related:http://physics.stackexchange.com/q/23813/7433 notice, asking for a derivation from molecular properties. $\endgroup$ Apr 16, 2012 at 12:42

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