Understanding Electrical Resistivity and Conductivity: Definitions, Formulas, and Solved Problems

Electrical Resistivity & Conductivity

1. Electrical Resistivity

You can verify by doing an experiment that the resistance of any conductor ‘R’ is directly proportional to the length of the conductor ‘L’ and is inversely proportional to its area of cross section ‘A’.

$$ R = \rho \frac{L}{A} $$

Where, ρ (rho) is a constant, called as electrical resistivity or specific resistance of the material of the conductor.

From the above equation, we can write:

$$ \rho = \frac{RA}{L} $$

If L = 1 m, A = 1 m² then, from the above equation ρ = R

Hence, the electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section. Its unit is ohm metre (Ω m).

Electrical resistivity of a conductor is a measure of the resisting power of a specified material to the passage of an electric current. It is a constant for a given material.

2. Conductance and Conductivity

Conductance of a material is the property of a material to aid the flow of charges and hence, the passage of current in it. The conductance of a material is mathematically defined as the reciprocal of its resistance (R). Hence, the conductance ‘G’ of a conductor is given by:

$$ G = \frac{1}{R} $$

Its unit is ohm^–1. It is also represented as ‘mho’.

The reciprocal of electrical resistivity of a material is called its electrical conductivity ($\sigma$).

$$ \sigma = \frac{1}{\rho} $$

Its unit is ohm^–1 metre^–1. It is also represented as mho metre^–1. The conductivity is a constant for a given material. Electrical conductivity of a conductor is a measure of its ability to pass the current through it. Some materials are good conductors of electric current. Example: copper, aluminium, etc. While some other materials are non-conductors of electric current (insulators). Example: glass, wood, rubber, etc.

Conductivity is more for conductors than for insulators. But, the resistivity is less for conductors than for insulators. The resistivity of some commonly used materials is given in Table 4.2.

Table 4.2: Resistivity of some materials

Solved Problem

The resistance of a wire of length 10 m is 2 ohm. If the area of cross section of the wire is 2 × 10^–7 m², determine its (i) resistivity (ii) conductance and (iii) conductivity.

Solution:

Given: Length, L = 10 m, Resistance, R = 2 ohm and Area, A = 2 × 10^–7 m²

(i) Resistivity ($\rho$):

$$ \rho = \frac{R \times A}{L} = \frac{2 \times (2 \times 10^{-7})}{10} = \frac{4 \times 10^{-7}}{10} = 4 \times 10^{-8} \, \Omega \text{ m} $$

(ii) Conductance (G):

$$ G = \frac{1}{R} = \frac{1}{2} = 0.5 \, \text{mho} $$

(iii) Conductivity ($\sigma$):

$$ \sigma = \frac{1}{\rho} = \frac{1}{4 \times 10^{-8}} = 0.25 \times 10^{8} \, \text{mho m}^{-1} $$