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Question:

Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are α₁ and α₂. The respective temperature coefficients of their series and parallel combinations are nearly

α₁+α₂/2, α₁+α₂/2

α₁+α₂/2, α₁+α₂

α₁+α₂, α₁+α₂/2

α₁+α₂, α₁α₂/(α₁+α₂)

Solution:

Let R₀ be the initial resistance of both conductors.

At temperature θ their resistance will be,
R₁ = R₀(1 + α₁θ) and R₂ = R₀(1 + α₂θ)

for series combination,
Rₛ = R₁ + R₂
Rₛ₀(1 + αₛθ) = R₀(1 + α₁θ) + R₀(1 + α₂θ)
where Rₛ₀ = R₀ + R₀ = 2R₀.
2R₀(1 + αₛθ) = 2R₀ + R₀θ(α₁ + α₂)
or αₛ = (α₁ + α₂)/2

for parallel combination,
Rₚ = R₁R₂/(R₁ + R₂)
Rₚ₀(1 + αₚθ) = R₀(1 + α₁θ)R₀(1 + α₂θ)/[R₀(1 + α₁θ) + R₀(1 + α₂θ)]
where, Rₚ₀ = R₀R₀/(R₀ + R₀) = R₀/2
R₀/2(1 + αₚθ) = R₀²(1 + α₁θ + α₂θ + α₁α₂θ²)/R₀(2 + α₁θ + α₂θ)
as α₁ and α₂ are small quantities α₁α₂ is negligible
or αₚ = (α₁ + α₂)/2 + (α₁ + α₂)θ = (α₁ + α₂)/2[1 - (α₁ + α₂)θ]
as (α₁ + α₂)² is negligible.
αₚ = (α₁ + α₂)/2