### Resistance of a rectangular block

The resistance of a block/wire depends on two factors: they are the cross sectional area and the length.
The relationship is given by the following formula:
R = ρ l/A
The longer the wire/block, the greater the resistance.
The thinner the wire/block, the greater the resistance.
The constant, ρ, in this case is called resistivity. It depends only on the substance.
E.g.

MetalResistivity - Ωm
Silver1.59x10-8
Nickel6.99x10-7
Iron1.0x10-7
Copper1.68x10-8
Tungsten5.69x10-8

E.g.1

The dimensions of a metal cuboid are 6cm, 4cm and 2cm respectively. Find its resistance, if the resistivity is 1.2 X 10-8Ωm and the current enters through the smallest surface.
R = ρl/A
A = 8 x 10-4m2
l= 6 x 10-2m
R = 1.2 X 10-8 x 6 x 10-2 / 8 x 10-4
R = 9 X 10-7Ω

E.g.2

The radius of a wire is 2cm and the length is 8cm. If the resistivity is 3x10-6

, find the resistance of the wire.

A = πr2 = 3.142*22*10-4 = 12.562*10-4
R = ρl/A = 3X10-6*8*10-2/12.562*10-4 = 1.9x10-4Ω.

E.g.3

The resistance of a wire is 4Ω. It is folded up and then twisted in the middle. What is its new resistance?

When it is folded up and then twisted, the cross sectional area gets doubled and length gets halved.
So, 4 = ρl/A for the first wire
R = ρ(l/2)/2A for the second wire
4/R = 4
R = 1Ω.