JEE Main 2024 — Thermal Properties of Matter Question with Solution

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

For Series Combination:

$R_{eq}=R_1+R_2$

$\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=R\left(1+{\alpha }_1∆T\right)+R\left(1+{\alpha }_2∆T\right)$

$\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=2R+\left({\alpha }_1+{\alpha }_2\right)R∆T$

$\Rightarrow {\alpha }_{eq}=\frac{{\alpha }_1+{\alpha }_2}{2}$

For Parallel Combination:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{{\displaystyle \frac{R}{2}\left(1+{\alpha }_{eq}∆T\right)}}=\frac{{\displaystyle 1}}{{\displaystyle R\left(1+{\alpha }_1∆T\right)}}+\frac{{\displaystyle 1}}{{\displaystyle R\left(1+{\alpha }_2∆T\right)}}$

$\Rightarrow \frac{2}{1+{\alpha }_{eq}∆T}=\frac{{\displaystyle 1}}{{\displaystyle 1+{\alpha }_1∆T}}+\frac{{\displaystyle 1}}{{\displaystyle 1+{\alpha }_2∆T}}$

$\Rightarrow \frac{2}{1+{\alpha }_{eq}∆T}=\frac{1+{\alpha }_2∆T+1+{\alpha }_1∆T}{\left(1+{\alpha }_1∆T\right){\displaystyle \left(1+{\alpha }_2∆T\right)}}$

$\Rightarrow 2\left[\left(1+{\alpha }_1∆T\right)\left(1+{\alpha }_2∆T\right)\right]=\left[2+\left({\alpha }_1+{\alpha }_2\right)∆T\right]\left[1+{\alpha }_{eq}∆T\right]$

$\Rightarrow 2\left[1+{\alpha }_1∆T+{\alpha }_2∆T+{\alpha }_1{\alpha }_2∆T\right]=2+2{\alpha }_{eq}∆T+\left({\alpha }_1+{\alpha }_2\right)∆T+{\alpha }_{eq}\left({\alpha }_1+{\alpha }_2\right)∆T^2$

Neglecting small terms

$2+2\left({\alpha }_1+{\alpha }_2\right)∆T=2+2{\alpha }_{eq}∆T+\left({\alpha }_1+{\alpha }_2\right)∆T$

$\Rightarrow \left({\alpha }_1+{\alpha }_2\right)∆T=2{\alpha }_{eq}∆T$

$\Rightarrow {\alpha }_{eq}=\frac{{\alpha }_1+{\alpha }_2}{2}$