Temperature Dependence of Resistivity: Overview, Questions, Preparation

Physics Current Electricity 2023

Updated on Sep 28, 2023 17:24 IST

Temperature dependence of resistivity is an important concept in the Current Electricity NCERT Class 12 Physics Chapter. It describes how the resistivity of a material changes with temperature. The behaviour can be different for different types of materials, but in general, there are two main categories: conductors and semiconductors.

Temperature Dependence of Resistivity – Metals and Conductors

In metals, as temperature increases, the resistivity typically also increases. This is because in metals, electrical conduction is primarily due to the movement of free electrons. When the temperature rises, the lattice structure of the metal vibrates more vigorously. These lattice vibrations, or phonons, scatter the free electrons, making it more difficult for them to move through the material. As a result, the resistance increases.

The relationship between resistivity (ρ) and temperature (T) for metals is often described by the empirical equation:

ρ(T) = ρ₀[1 + α(T - T₀)]

Where:

ρ(T) is the resistivity at temperature T

ρ₀ is the resistivity at a reference temperature T₀

α is the temperature coefficient of resistivity (a positive constant)

T₀ is the reference temperature

As mentioned, various materials exhibit varying degrees of sensitivity to temperature changes. Certain materials, such as Nichrome (an alloy of nickel, iron, and chromium), as well as Manganin and Constantan, show a minimal sensitivity of resistivity to changes in temperature. Because of this characteristic, these materials are commonly employed in the production of wire-wound standard resistors. This choice is made because their resistance values remain remarkably stable even when subjected to variations in temperature.

Temperature Dependence of Resistivity – Semiconductors

In semiconductors, the behaviour is quite different. As temperature increases, the resistivity typically decreases. This is because in semiconductors, the electrical conduction is primarily due to the movement of charge carriers (either electrons or holes) across the energy band gap.

With increasing temperature, more electrons gain enough thermal energy to jump from the valence band to the conduction band, creating more charge carriers and reducing resistivity. We can qualitatively explain this temperature-dependent resistivity based on our derivation.

ρ= 1/σ = m/ne2τ

ρ is the resistivity of the material

n is the number of free electrons per unit volume

τ is the average time between collisions

According to this equation, the resistivity of a material is determined by several factors, including the number (n) of free electrons per unit volume and the average time (τ) between collisions. As we elevate the temperature, the average velocity of the electrons, which serve as current carriers, increases, leading to more frequent collisions. Consequently, the average collision time (τ) decreases with rising temperature.

In the case of metals, the number of free electrons (n) is not significantly influenced by temperature. Therefore, the decrease in the value of t with increasing temperature leads to an observed increase in resistivity. However, for insulators and semiconductors, the number of free electrons (n) increases with temperature. This increase more than compensates for any reduction in τ. As a result, for such materials, resistivity decreases with temperature.

FAQs on Temperature Dependence of Resistivity

Q. At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^–4 °C–1.

A. Let T be the room temperature and R be the resistance at room temperature and T1 be the required temperature and R1 be the resistance at that temperature and α be the coefficient of resistor.

We have:

T = 27°C, R = 100 Ω, T1 = ?, R1 = 117 Ω, α = 1.70 X 10-4 ° C-1

We know the relation of α can be given as

α = R1 -R / R (T2-T)= (117-100) / 100(T1-27) = 1.70 X 10-4

or ( T1 - 27) = (117-100) / 100 X 1.70 X 10-4 = 1000

T1 = 1000 +27 = 1027° C

Q. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^–7 m^2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

A. Given, length of the wire, l = 15 m

Uniform cross section, A = 6.0 X 10-7m2

Resistance measured, R = 5.0 Ω

From the relation of

R = ρ l/A, where ρ is the resistivity of the material, we get

ρ = AR/l = 6.0 X 10-7X 5 / 15 = 2 X 10-7  Ωm.

Q. A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

A . Let us assume R = 2.1 Ω, T = 27.5 °C, R1 = 2.7 Ω, T1 = 100 , α = resistivity of silver

We know the relation of α can be given as

α = (R1 - R) /R ( T1 -T) = 2.7 - 2.1 / 2.1(100-27.5) = 3.94 X 10-3 ° C-1

Q. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^–4 °C^–1.

A Given

Supply voltage, V = 230 V

Initial current drawn, I = 3.2 A

Final current drawn, I1 = 2.8 A

Room temperature, T = 27.0 °C

Steady temperature, T1 = ?

From Ohm’s law, we get initial resistance, R = V/l = 230/3.2 Ω = 71.875 Ω

Final resistance, R1 = V/l1 = 230/2.8 Ω = 82.143 Ω

From the relation of  α = R1 -R/ R (T1-T), where α is the temperature coefficient of resistance, we get

1.7 X 10-4 = 82.143 - 71.875 / 71.875 (T1-27) 

T1-27 = 840.34

T1 = 867.34°C

Therefore the steady temperature of heating element required is 867.34 °C

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