Resistivity (also known as specific resistance)

• When electrons flow through a wire they experience resistance and lose energy, the furtherer along the wire they flow the more energy they lose, therefore, we can say that the total resistance of a wire is proportional to its length.
• Since the electrons are evenly distributed throughout the wire and since the current is the rate at which a charge passes any point on that wire, we can see that to provide any specific current the electrons in a wider wire will have to flow a shorter distance than electrons in a narrower wire (figure 2.2). 
• We can therefore say that resistance is inversely proportional to the conductor's cross-sectional area. 
• Therefore thicker wires have less resistance per meter and will cause less energy to be lost as heat.

• Putting the previous two concepts together given us:

                                           R~1/a (*proportional)

• where l is the length of wire, a is the cross-sectional area and α means proportional to. The resistivity (ρ) of a material is defined as the resistance between opposite faces of a cube of that material with a given side length. 
• ρ is very small for most conductors and is usually quoted in micro-ohms (μΩ) for a 1 meter cube expressed as μΩm. For example aluminium has a resistivity of 0.0285μΩm. 
• Thus the resistivity of a wire is proportional to the resistivity of the material from which it is made. We can combine resistivity with the previous equation to give:

                                                                    R=ρl / a

• Note that the resistance calculated from this equation will be given in the same units of resistance used for the resistivity (i.e. μΩ). Also l and a must be in the same length units as ρ so that if ρ is in μΩm l must be in m and a must be in m2. Table 2.1 contains the resistivities of some metals.

Example:
Calculate the resistance of 1000m of 16mm2 single annealed copper wire.
From the table: ρ = 17.2μΩmm (since the cross sectional area is given in mm2)
l = 1000m = 10^6m
R=ρl / a = (17.2*10^6)/16 =1075000μΩ=1.075Ω

• All wires and cables have some resistance, therefore there will always be some energy lost and a voltage drop within. Thin wires will get very hot and may burn out if they carry too much current, they may also cause such a large voltage drop that equipment may not function properly. 
• Thick wires will reduce these phenomena however, because they contain more copper they will be considerably more expensive than thin wires. Voltage drops in cables are normally recommended to be no more than 4% of the supply voltage.
• That means a 9.6V drop for a 240V supply or a 4V drop for a 110V supply. Note that you can only measure a voltage drop across a length of cable when a current is flowing in it.