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A simple Physics question regarding Ohm's law. The resistivity of the conductor is inversely proportional to the area of the conductor. I would like to have a simple explanation for this.

How does resistivity increase if the area decreases in a conductor?

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    $\begingroup$ "The resistivity of the conductor is inversely proportional to the area of the conductor." Your answer is not correct. $\endgroup$
    – Farcher
    Commented Mar 14, 2019 at 15:14

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You mean: The resistance of a conductor is inversely proportional to the area. [Resistivity is a property of the material of which the conductor is made, and is independent of the size or shape of the conductor.]

You can imagine making up a cylindrical conductor from two strands of semicircular cross-section (running lengthways). If each strand has a resistance $R_{semi},$ then the complete conductor will have a resistance $R$ of $\frac{1}{2}R_{semi}$ because the strands are in parallel. [It doesn't matter that the strands are in contact all along and not just at the ends, because no current will flow between them.]

So a wire of twice the cross-sectional area (but the same length and material) has half the resistance.

You can easily extend the argument to show that resistance is inversely proportional to cross-sectional area (for a given length and material).

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  • $\begingroup$ While resistivity is an intrinsec property of the material, if the dimensions are small enough then ballistic conduction might take over and the resistivity will certainly be different than for a larger sample of the exact same material. $\endgroup$
    – untreated_paramediensis_karnik
    Commented Mar 14, 2019 at 19:19
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    $\begingroup$ I judged that the question-asker was concerned with resistances of everyday pieces of wire and not with very small specimens of conductor. $\endgroup$
    – Philip Wood
    Commented Mar 14, 2019 at 19:26
  • $\begingroup$ I understand. I was just pointing out that your sentence " [Resistivity is a property of the material of which the conductor is made, and is independent of the size or shape of the conductor.]" is not strictly true. My remark is very informative for the curious physics student who wants to dig deeper into this matter. $\endgroup$
    – untreated_paramediensis_karnik
    Commented Mar 14, 2019 at 20:44
  • $\begingroup$ Also for the curious student: DC is assumed. But in AC applications the skin effect may become important. In the limiting case, resistance is then proportional to circumference (radius). $\endgroup$
    – user137289
    Commented Mar 14, 2019 at 21:13
  • $\begingroup$ @thermomagneticcondensedboson, it wouldn't make sense to even talk about resistivity for a sample small enough that those effects matter. $\endgroup$
    – The Photon
    Commented Mar 14, 2019 at 21:54
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Think of the analogy of water flowing in a pipe. Water flows through a pipe more easily the greater the cross sectional area. Similarly, current will flow through a conductor more easily the greater the cross sectional area of the conductor.

However, it is the “resistance” of the conductor not its “resistivity” that varies inversely with the cross sectional area of the conductor. The “resistivity” of a conductor is a material property of the conductor.

The resistance of a conductor is given by

$$R=\frac {ρL}{A}$$

Where

$R$ = conductor resistance in Ohms

$ρ$ = the resistivity of the conductor material in Ohm-meter

$L$ = the length of the conductor in meters

$A$ = the cross sectional area of the conductor in square meters.

Hope this helps

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  • $\begingroup$ Poor analogy. The flow of a liquid in a pipe is not constant over its cross section (in the case of laminar flow it is parabolic profile from zero at the walls which results in the Poiseuille $R^4$ law). But in an electrical wire, current density is constant over its cross section (in DC applications). $\endgroup$
    – user137289
    Commented Mar 14, 2019 at 20:08
  • $\begingroup$ @Pieter I never said the flow of a liquid in a pipe is constant over its cross section. Don't put words in my mouth. I'm going to restate this one more time. $\endgroup$
    – Bob D
    Commented Mar 14, 2019 at 20:17
  • $\begingroup$ @Pieter Fluid motion is really complex. Your example is one model. As any other, it's not always valid. Anyways, the fact that the velocity profile varies along the cross-section doesn't mean that the net effect is what the answerer says. $\endgroup$
    – FGSUZ
    Commented Mar 14, 2019 at 20:33
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I suppose you are concerned with homogeneous and isotropic ohmic conductors in the diffusive regime. As others have pointed out already, the (specific) resistivity is a material property independent of the dimensions of the specimen.

We are now going to think about how resistance depends on the cross sectional area of the conductor perpendicular to the direction of current flow. Suppose we apply a constant dc voltage to the conductor of cross sectional area $A$ and resistance $R_A$. Then the current flowing through it is given according to Ohm's law by $$ I_A = \frac{U}{R_A}.$$ This current is homogeneously distributed across the cross sectional area of the conductor. This property will now help us to draw the right conclusions.

If we reduce the cross sectional area of the same conductor to $a$ (say, we cut the rest of the conductor away), but keep the voltage $U$ the same, the current will be reduced in proportion to the area reduction, i.e., $$ I_a = \frac{a}{A}I_A = \frac{a}{A}\frac{U}{R_A}.$$ At the same time, the current $I_a$ is, according to Ohm's law, $$ I_a = \frac{U}{R_a},$$ where $R_a$ is the resistance of the resistor with reduced cross-sectional area. Comparing the two equations for $I_a$ we see that $$ R_a = \frac{R_AA}{a}.$$ Since in our problem, the product $R_AA$ is simply a constant given by the conductor with the initial cross section $A$, you see that the resistance $R_a$ of the conductor with reduced cross sectional area $a$ is inversely proportional to $a$.

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