Question

Two conductor have the same Resistance at 0°c but their temprature coefficients of resistance are α₁ and α₂ .the respective temprature coefficients of their series and parallel combinations are nearly [AIEEE 2011] ​

Answer 1

(a) : Let R0 be the resistance of both conductors at 0°C. Let R1 and R2 be their resistance at t°C. Then R1 = R0(1 + α1t) R2 = R0(1 + α2t) Let Rs is the resistance of the series combination of two conductors at t°C.

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Answer 2

Answer:

The respective temprature coefficient of their series and parallel combinations are nearly α₁+α₂ /2 , and α₁+α₂ /2

Explanation:

Let R₀ be the initial resistance of both conductors .

∴At temprature θ , their resistance will be

R₁ = R₀ (1+α₁θ)

and R₂ = R₀(1+ α₂θ)

For series combination .

Rₛ = R₁ + R₂

Rₛₒ = (1+αₛθ) = R₀ ( 1+α₁θ) + R₀ (1+α₂θ)

where, Rₛₒ = R₀+ R₀ = 2R₀

∴ 2R₀ (1+αₛθ) = 2R₀ + R₀ θ ( α₁ +α₂ )

or, αₛ= α₁ + α₂ / 2

For parallel combination ,

Rₚ = R₁R₂ / R₁ + R₂

Rₚₒ (1+αₚθ ) = R₀ (1+α₂θ )+ R₀ (1+α₂θ) /R₀(1+α₁θ) + R₀(1+α₂θ )

where, Rₚₒ = R₀R₀ /2R₀ = R₀/2

∴ R₀/2(1+αₚθ ) = R²(1+α₁θ + α₂θ + α₁α₂θ²) /Rₒ(2+α₁θ + α₂ θ )

as, α₁ and α₂ is negligible .

or, αₚ = α₁ + α₂ /2 + (α₁ + α₂) θ

= α₁ + α₂ /2 [(1- (α₁ +α₂ /2)θ ]

∴ αₚ = α₁ + α₂ /2