Two conductors have the same resistances at 0 ° C but their...

Change in resistance due to temperature is given by, ${R}^{'} = R\left(1 + \alpha ∆ T\right)$

For Series Combination:

${R}_{eq} = {R}_{1} + {R}_{2}$

$⇒ 2R\left(1 + {\alpha}_{eq} ∆ T\right) = R\left(1 + {\alpha}_{1} ∆ T\right) + R\left(1 + {\alpha}_{2} ∆ T\right)$

$⇒ 2R\left(1 + {\alpha}_{eq} ∆ T\right) = 2R + \left({\alpha}_{1} + {\alpha}_{2}\right)R ∆ T$

$⇒ {\alpha}_{eq} = \dfrac{{\alpha}_{1} + {\alpha}_{2}}{2}$

For Parallel Combination:

$\dfrac{1}{{R}_{eq}} = \dfrac{1}{{R}_{1}} + \dfrac{1}{{R}_{2}}$

$⇒ \dfrac{1}{\dfrac{R}{2}\left(1 + {\alpha}_{eq} ∆ T\right)} = \dfrac{1}{R\left(1 + {\alpha}_{1} ∆ T\right)} + \dfrac{1}{R\left(1 + {\alpha}_{2} ∆ T\right)}$

$⇒ \dfrac{2}{1 + {\alpha}_{eq} ∆ T} = \dfrac{1}{1 + {\alpha}_{1} ∆ T} + \dfrac{1}{1 + {\alpha}_{2} ∆ T}$

$⇒ \dfrac{2}{1 + {\alpha}_{eq} ∆ T} = \dfrac{1 + {\alpha}_{2} ∆ T + 1 + {\alpha}_{1} ∆ T}{\left(1 + {\alpha}_{1} ∆ T\right)\left(1 + {\alpha}_{2} ∆ T\right)}$

$⇒ 2\left[\left(1 + {\alpha}_{1} ∆ T\right)\left(1 + {\alpha}_{2} ∆ T\right)\right] = \left[2 + \left({\alpha}_{1} + {\alpha}_{2}\right) ∆ T\right]\left[1 + {\alpha}_{eq} ∆ T\right]$

$⇒ 2\left[1 + {\alpha}_{1} ∆ T + {\alpha}_{2} ∆ T + {\alpha}_{1}{\alpha}_{2} ∆ T\right] = 2 + 2{\alpha}_{eq} ∆ T + \left({\alpha}_{1} + {\alpha}_{2}\right) ∆ T + {\alpha}_{eq}\left({\alpha}_{1} + {\alpha}_{2}\right) ∆ {T}^{2}$

Neglecting small terms

$2 + 2\left({\alpha}_{1} + {\alpha}_{2}\right) ∆ T = 2 + 2{\alpha}_{eq} ∆ T + \left({\alpha}_{1} + {\alpha}_{2}\right) ∆ T$

$⇒ \left({\alpha}_{1} + {\alpha}_{2}\right) ∆ T = 2{\alpha}_{eq} ∆ T$

$⇒ {\alpha}_{eq} = \dfrac{{\alpha}_{1} + {\alpha}_{2}}{2}$