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

Solution

Physics sir · 4 steps

Write each resistance as R₀(1 + α t), combine them, then read off the new α.

1. Set up. At temperature t, the two resistances are R₁ = R₀(1 + α₁ t) and R₂ = R₀(1 + α₂ t). For any combination, find the effective α by writing its resistance in the form R_(0,eq)(1 + α_(eq) t).
2. Series. Rₛ = R₁ + R₂ = 2R₀ + R₀(α₁ + α₂)t = 2R₀(1 + (α₁ + α₂)/2t). The resistance at 0°C is 2R₀, so αₛ = (α₁ + α₂)/2.
3. Parallel. Add conductances. Since α t is very small, 1/R₀(1 + α t) ≈ 1/R₀(1 - α t). Adding the two conductances and flipping the sum back the same way gives the result below.

   1/Rₚ ≈ 2/R₀(1 - (α₁ + α₂)/2t) ⇒ Rₚ ≈ R₀/2(1 + (α₁ + α₂)/2t)

4. So. Both combinations have the coefficient (α₁ + α₂)/2. With equal resistors, each one contributes half of the total resistance in either arrangement. The answer is (b).

The idea to keep. The coefficient of a combination is the average of the individual coefficients, weighted by resistance. With equal resistances it is the plain average, in series and in parallel alike.

Answer from the NTA final answer key. Solution by PrepWiser.