Resistance And Resistivity | Physics - Lumen Learning

Material and Shape Dependence of Resistance

The resistance of an object depends on its shape and the material of which it is composed. The cylindrical resistor in Figure 1 is easy to analyze, and, by so doing, we can gain insight into the resistance of more complicated shapes. As you might expect, the cylinder’s electric resistance R is directly proportional to its length L, similar to the resistance of a pipe to fluid flow. The longer the cylinder, the more collisions charges will make with its atoms. The greater the diameter of the cylinder, the more current it can carry (again similar to the flow of fluid through a pipe). In fact, R is inversely proportional to the cylinder’s cross-sectional area A.

Figure 1. A uniform cylinder of length L and cross-sectional area A. Its resistance to the flow of current is similar to the resistance posed by a pipe to fluid flow. The longer the cylinder, the greater its resistance. The larger its cross-sectional area A, the smaller its resistance.

For a given shape, the resistance depends on the material of which the object is composed. Different materials offer different resistance to the flow of charge. We define the resistivityρ of a substance so that the resistance R of an object is directly proportional to ρ. Resistivity ρ is an intrinsic property of a material, independent of its shape or size. The resistance R of a uniform cylinder of length L, of cross-sectional area A, and made of a material with resistivity ρ, is

[latex]R=\frac{\rho L}{A}\\[/latex].

Table 1 gives representative values of ρ. The materials listed in the table are separated into categories of conductors, semiconductors, and insulators, based on broad groupings of resistivities. Conductors have the smallest resistivities, and insulators have the largest; semiconductors have intermediate resistivities. Conductors have varying but large free charge densities, whereas most charges in insulators are bound to atoms and are not free to move. Semiconductors are intermediate, having far fewer free charges than conductors, but having properties that make the number of free charges depend strongly on the type and amount of impurities in the semiconductor. These unique properties of semiconductors are put to use in modern electronics, as will be explored in later chapters.

Table 1. Resistivities ρ of Various materials at 20º C

Material	Resistivity ρ ( Ω ⋅ m )
Conductors
Silver	1. 59 × 10−8
Copper	1. 72 × 10−8
Gold	2. 44 × 10−8
Aluminum	2. 65 × 10−8
Tungsten	5. 6 × 10−8
Iron	9. 71 × 10−8
Platinum	10. 6 × 10−8
Steel	20 × 10−8
Lead	22 × 10−8
Manganin (Cu, Mn, Ni alloy)	44 × 10−8
Constantan (Cu, Ni alloy)	49 × 10−8
Mercury	96 × 10−8
Nichrome (Ni, Fe, Cr alloy)	100 × 10−8
Semiconductors[1]
Carbon (pure)	3.5 × 105
Carbon	(3.5 − 60) × 105
Germanium (pure)	600 × 10−3
Germanium	(1−600) × 10−3
Silicon (pure)	2300
Silicon	0.1–2300
Insulators
Amber	5 × 1014
Glass	109 − 1014
Lucite	>1013
Mica	1011 − 1015
Quartz (fused)	75 × 1016
Rubber (hard)	1013 − 1016
Sulfur	1015
Teflon	>1013
Wood	108 − 1011

Example 1. Calculating Resistor Diameter: A Headlight Filament

A car headlight filament is made of tungsten and has a cold resistance of 0.350 Ω. If the filament is a cylinder 4.00 cm long (it may be coiled to save space), what is its diameter?

Strategy

We can rearrange the equation [latex]R=\frac{\rho L}{A}\\[/latex] to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section.

Solution

The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in [latex]R=\frac{\rho L}{A}\\[/latex], is

[latex]A=\frac{\rho L}{R}\\[/latex].

Substituting the given values, and taking ρ from Table 1, yields

[latex]\begin{array}{lll}A& =& \frac{\left(5.6\times {\text{10}}^{-8}\Omega \cdot \text{m}\right)\left(4.00\times {\text{10}}^{-2}\text{m}\right)}{\text{0.350}\Omega }\\ & =& \text{6.40}\times {\text{10}}^{-9}{\text{m}}^{2}\end{array}\\[/latex].

The area of a circle is related to its diameter D by

[latex]A=\frac{{\pi D}^{2}}{4}\\[/latex].

Solving for the diameter D, and substituting the value found for A, gives

[latex]\begin{array}{lll}D& =& \text{2}{\left(\frac{A}{p}\right)}^{\frac{1}{2}}=\text{2}{\left(\frac{6.40\times {\text{10}}^{-9}{\text{m}}^{2}}{3.14}\right)}^{\frac{1}{2}}\\ & =& 9.0\times {\text{10}}^{-5}\text{m}\end{array}\\[/latex].

Discussion

The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits.