I am looking for the derivation of the Ohm's Law i.e., V is directly proportional to I. Can someone help me with it?

  • $\begingroup$ On a measured function $U=f(R,I)$ Ohms law correspond to the tangential function $U=R\cdot I$. $\endgroup$ – Lehs Jul 20 '15 at 15:54
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    $\begingroup$ For a simple theory behind Ohm's law, see e.g. the Drude model at Wikipedia or this Phys.SE post. $\endgroup$ – Qmechanic Jul 20 '15 at 18:13
  • $\begingroup$ Yes, see physics.stackexchange.com/q/40907 $\endgroup$ – Thomas Jul 21 '15 at 4:22
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    $\begingroup$ A simple answer is that except in an oversimplified linear response model, it isn't even true. The characteristic curve $U(I)$ for some material can be quite nonlinear and depends on the particular properties of the material. So the reasoning is simply: everything looks like a straight line if you zoom in enough. Drude model is a pretty standard "idealized" material model, but basically, you should look at Ohm's Law as an empirical statement. $\endgroup$ – orion Jul 21 '15 at 10:51

Ohm's Law is not a construct which can be derived. It is essentially a generalized observation. It is only useful for a few materials (conductors and medium resistivity), and even then virtually all of those materials show deviations from the ideal, such as temperature coefficients and breakdown voltage limits.

Rather, Ohm's Law is an idealization of the observed behavior of these materials. As the saying goes, "All models are wrong. Some models are useful." In this case, Ohm's Law is extraordinarily useful, but that doesn't make it universal. Semiconductors, for instance, do not follow Ohm's Law in any large sense, and look how widespread their use is.

As originally discovered and formulated, there was a great deal of wishful thinking involved. There was no understanding of the forces involved, and there was no real definition, for instance, of voltage or current. Nonetheless, it was determined that a self-consistent set of values was possible (you can define different battery chemistries as producing specific voltages, and get consistent behavior of galvanometers - as long as you're willing to accept experimental error). Over time, standards were set and more objective measures discovered, such as the quantity of electrons in a coulomb, so that a current of 1 amp can be unambiguously measured) Eventually, a very good understanding of the behavior of electrons (and holes) in conductors was reached, and that understanding is generally, for a wide range of useful conditions, expressible as Ohm's Law.

But it is not derived.

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    $\begingroup$ I thought we could start from the Maxwell's equations and derive the Ohm's law. Atleat this is the impression I had from Anant Agarwal of MIT. $\endgroup$ – quantum231 Jul 21 '15 at 0:05
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    $\begingroup$ @quantum231 - Of course you can - see tOxic's answer. However, that's not how it happened, and it is not accurate for a wide variety of materials. As a matter of fact, it's not accurate even for "regular" conductors under high current conditions, or insulators under high voltage. $\endgroup$ – WhatRoughBeast Jul 21 '15 at 0:24
  • $\begingroup$ i don't think it is Ohm's Law across a PN junction. but one can derive Ohm's Law, at the lumped-element level from a model of resistivity in a material like a metal. and you can derive resistivity of a material from certain assumptions about the movement of free electrons in that material. $\endgroup$ – robert bristow-johnson Jul 21 '15 at 7:01
  • $\begingroup$ Similarly, you could say Maxwell's Laws are an idealization, and don't hold at scales of QED or General Relativity. I don't see how that's a useful answer, however. $\endgroup$ – BlueRaja - Danny Pflughoeft Jul 21 '15 at 7:08
  • $\begingroup$ Note that the Coulomb's definition is based on the Ampere, not vice versa (as the answer implies). Ampere is actually one of the SI base units. $\endgroup$ – Geier Jul 21 '15 at 11:37

You could start from Drude in zero magnetic field, that equates the derivative of the momentum $\vec p$ by the electrostatic force $\vec F_{el} = q \vec E$ as a product of charge $q$ and electric field $\vec E$ minus a scattering term (with time constant $\tau$; compared to Newtons second law that does not feature the latter, crystal term):

$~~~~~~\dot {\vec p} = q {\vec E} - \cfrac{\vec p}{\tau}$

The stationary solution ($\dot {\vec p} = 0$) utilizing the current density $\vec j$

$~~~~~~\vec j = n q \vec v$

as the product of carrier denstiy $n$, charge $q$ and carrier velocity $\vec v$ entails

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

which is nothing but the linear relationship between current (density) "$j = \cfrac{I}{A}$" and electric field (potential gradient) "$E = \cfrac{U}{d}$" that is stated by Ohm.

  • $\begingroup$ ..and you get a contradiction, because characteristic time $\tau$ between collisions can not be considered constant for different velocities - it decreases with increasing velocity, if you imagine the obstacles as a steeplechase distance. :) $\endgroup$ – Boris Burkov Jul 22 '15 at 9:34
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    $\begingroup$ Fundamentally, the Drude formulation is a model and as such, the formula is as approximate as Ohm's law itself. But it's a good way to derive it. Just keep in mind they're both (Drude's and Ohm's) laws that don't apply in every possible case. $\endgroup$ – legrojan Jul 22 '15 at 9:55
  • $\begingroup$ Is there a reference for experimental data that confirms the Drude model in the sense that $R=md/(n\tau q^2 A)$ is confirmed within some empirical bounds? This would require a specification of $\tau$ that is to some extend independent of the measurement result of $R$ when only varying the other observables ($nq,m,...$). $\endgroup$ – exchange Nov 20 '17 at 17:00
  • $\begingroup$ Some more notes on the Drude model: colorado.edu/physics/phys4340/phys4340_sp09/notes/… $\endgroup$ – exchange Nov 20 '17 at 17:16

In my opinion, the mathematical equation we call Ohm's Law is best taken not as a “law”, a fact about the universe, but as the definition of resistance.

$$R \overset{\mathrm{def}}{=} \frac{V}{I}$$

Given this definition of the quantity $R$, we can then make (as other answers have mentioned) the empirical observation that many materials have approximately constant $R$ (which we call being ohmic) and therefore $R$ is a useful quantity to have defined.

But if there is something to derive, then it is the answer to “why do we observe approximately constant $R$?”, not $V = IR$.


Ohm's law isn't fundamental and holds true only under certain conditions, like constant temperature for example. However, there is a simple way to think about it. Imagine the flow of massive objects through a wide water pipe. This is like a current. The water pressure causes the objects to flow quickly, that's your voltage. If the pipe is narrow then the objects can't flow as quickly. Also, the objects can be slowed down as they hit the sides of the pipe which isn't perfectly smooth. This is the resistance. Now if you take the water pressure (voltage) and divide by a resistance to flow, you get the rate at which objects flow through the pipe. Dividing by a large resistance means less flow.

Think of it as V/R = I instead of V = IR Division is easier to grasp mentally. As the resistance gets smaller the current goes up and vice versa.

  • $\begingroup$ Uh... how does the water-flow analogy imply that $U \propto I$? (That is actually valid for a purely laminar flow, but who says it makes sense to assume high viscosity for the water-flow model?) $\endgroup$ – leftaroundabout Jul 21 '15 at 13:03

Ohm's Law actually follows the definition of power, current and voltage.

Let's begin by defining power $P$, current $I$ and voltage $U$ as $P = \displaystyle \lim_{\Delta t \to 0} \frac{E}{\Delta t}$, $I = \displaystyle \lim_{\Delta t \to 0} \frac{Q}{\Delta t}$ and $U = \frac{E}{Q}$.

We then find for a constant current $I$ with a constant voltage $U$ that the work done during the time period $\Delta t$ is the work done per charge times the charge passing a certain point during $\Delta t$, i.e. the work done $E = \left(\displaystyle \lim_{\Delta t \to 0} \frac{Q}{\Delta t} \cdot \Delta t \right) \cdot \frac{E}{Q} = U \cdot \Delta t \cdot I$.

Remember that this was the work done during a time period, more specifically $\Delta t$, and we can thus say that the power during that time period is $\frac{U \cdot \Delta t\cdot I}{\Delta t} = P = U \cdot I$.

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    $\begingroup$ But P = U•I is not Ohm's law, and in fact it applies in non-Ohmic situations. Ohm's law says U ~ I, so P ~ I^2. $\endgroup$ – pwf Jul 21 '15 at 5:40

protected by Qmechanic Jul 21 '15 at 16:52

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