Two conductors have the same resistances at 0 but their temperature coefficients of resistance are α₁ and α₂. The respective temperature coefficients for their series and parallel combinations are: (a) α₁ + α₂, (α₁α₂)/(α₁+α₂) (b) (α₁+α₂)/(2), (α₁+α₂)/(2) (c) α₁ + α₂, (α₁+α₂)/(2) (d) (α₁+α₂)/(2), α₁ + α₂

Answer: (b). Series: Rs = R₀(1+α₁ t) + R₀(1+α₂ t) = 2R₀(1 + (α₁+α₂)/(2)t), so αs = (α₁+α₂)/(2). Parallel: 1/Rp = 1/R₀[(1+α₁ t)⁻¹ + (1+α₂ t)⁻¹] ≈ 2/R₀(1 - (α₁+α₂)/(2)t) So Rp ≈ R₀/2(1 + (α₁+α₂)/(2)t), giving αp = (α₁+α₂)/(2). The answer is (b).

Two conductors have the same resistances at 0 but their temperature… | JEE Main 2024 (31 Jan, Shift 1)

Q10JEE Main 2024 · 31 JAN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature

Two conductors have the same resistances at $0^\circ\text{C}$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$ . The respective temperature coefficients for their series and parallel combinations are:

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Answer: (b)

Series: $R_s = R_0(1+\alpha_1 t) + R_0(1+\alpha_2 t) = 2R_0\left(1 + \tfrac{\alpha_1+\alpha_2}{2}t\right)$ , so $\alpha_s = \tfrac{\alpha_1+\alpha_2}{2}$ .
Parallel: $\dfrac{1}{R_p} = \dfrac{1}{R_0}\left[(1+\alpha_1 t)^{-1} + (1+\alpha_2 t)^{-1}\right] \approx \dfrac{2}{R_0}\left(1 - \tfrac{\alpha_1+\alpha_2}{2}t\right)$
So $R_p \approx \tfrac{R_0}{2}\left(1 + \tfrac{\alpha_1+\alpha_2}{2}t\right)$ , giving $\alpha_p = \tfrac{\alpha_1+\alpha_2}{2}$ .
The answer is (b).

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