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Coefficient of linear expansion of material of resistor is A.its temperature coefficient of resistivity and resistance are Ap and Ar respectively , then what is the relation between A Ap and Ar?

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by NajiChacko

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by NajiChacko

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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.

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

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.