Resistivity

All materials have some resistance to the flow of charge
As free electrons move through a metal wire, they collide with ions which get in their way
As a result, they transfer some, or all, of their kinetic energy on collision, which causes electrical heating

Free electrons collide with ions which resist their flow

Since current is the flow of charge, the ions resisting their flow causes resistance
Resistance depends on the length of the wire, the cross-sectional area through which the current is passing and the resistivity of the material

$R = \frac{ρ L}{A}$

The resistivity equation shows that:
- The longer the wire, the greater its resistance
- The thicker the wire, the smaller its resistance

The length and width of the wire affect its resistance

Resistivity is a property that describes the extent to which a material opposes the flow of electric current through it
It is a property of the material, and is dependent on temperature
Resistivity is measured in Ω m

The higher the resistivity of a material, the higher its resistance
This is why copper, with its relatively low resistivity at room temperature, is used for electrical wires — current flows through it very easily
Insulators have such a high resistivity that virtually no current will flow through them

Two electrically-conducting cylinders made from copper and aluminium respectively.

Their dimensions are shown below.

WE - resistivity question image, downloadable AS & A Level Physics revision notes

Copper resistivity = 1.7 × 10^-8 Ω m

Aluminium resistivity = 2.6 × 10^-8 Ω m

Which cylinder is the better conductor?

Step 1: The better conductor will have lower resistance

Step 2: Resistance is calculated from

$R = \frac{ρ L}{A}$

Step 3: Resistance of the copper cylinder

$A = π \times r^{2} = π \times {(\frac{d}{2})}^{2} = π \times {(\frac{5 \times 10^{- 3}}{2})}^{2} = 2.0 \times 10^{- 5} m^{2}$

$R = \frac{1.7 \times 10^{- 8} \times 8 \times 10^{- 3}}{2.0 \times 10^{- 5} m^{2}} = 6.8 \times 10^{- 6} Ω$

Step 4: Resistance of the aluminium cylinder

$A = π \times r^{2} = π \times {(\frac{d}{2})}^{2} = π \times {(\frac{10 \times 10^{- 3}}{2})}^{2} = 7.9 \times 10^{- 5} m^{2}$

$R = \frac{2.6 \times 10^{- 8} \times 16 \times 10^{- 3}}{7.9 \times 10^{- 5} m^{2}} = 5.3 \times 10^{- 6} Ω$

Step 5: Resistance of aluminium cylinder < resistance of copper cylinder

The aluminium cylinder is the better conductor

You won’t need to memorise the value of the resistivity of any material, these will be given in the exam question.
Remember if the cross-sectional area is a circle e.g. in a wire, it is proportional to the diameter squared. This means if the diameter doubles, the area quadruples causing the resistance to drop by a quarter.

Resistivity (CIE A Level Physics)