Relation between linear expansion and temprature co-efficiet of resistance
Answers
we will find the relation for small temperature changes.
resistance R = ρ L / A
coefficient of linear expansion = α
length of conductor: L = L₀ ( 1 + α ΔT) ΔL = α L₀ ΔT
β = coefficient of expansion in area : = 2 α
Area of cross section: A = A₀ (1+ 2 α ΔT) ΔA = 2α A₀ ΔT
Resistivity ρ = ρ₀ (1 + Aρ ΔT) Δρ = ρ₀ Aρ ΔT
Resistance R = R₀ (1 + Ar ΔT) : ΔR = Ar R₀ ΔT
If ΔA = 2 α ΔT is very small then, and for small ΔT,
R₀ = ρ₀ L₀ / A₀
R = ρ₀ (1 + Aρ ΔT) L₀ (1 + α ΔT) / [ A₀ (1 + 2 α ΔT) ]
= (ρ₀ L₀ / A₀) (1 + Aρ ΔT) (1 + α ΔT) (1 - 2 α ΔT)
= R₀ (1 + Aρ ΔT) (1 - α ΔT) ignoring the 2 α² ΔT² term
= R₀ [ 1 + (Aρ - α) ΔT ] ignoring the Aρ α ΔT² term
Ar = (Aρ - α)
If the conductor does not expand, then the value of coefficient of resistance is same as the coefficient of resistivity. But as the area of cross section increases with temperature, the coefficient decreases by α.