QJEEM24Ph0c83: JEE Main 2024 Physics - Class 12 JEEM24Ph0c83 Solution: Two conductors have the same resistances at but their…

Series :

$\begin{aligned}& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\& 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=\mathrm{R}\left(1+\alpha_1 \Delta \theta\right)+\mathrm{R}\left(1+\alpha_2 \Delta \theta\right) \\& 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=2 \mathrm{R}+\left(\alpha_1+\alpha_2\right) \mathrm{R} \Delta \theta \\& \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}\end{aligned}$

Parallel :

$\begin{aligned}& \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\& \frac{1}{\frac{R}{2}\left(1+\alpha_{\text {eq }} \Delta \theta\right)}=\frac{1}{R\left(1+\alpha_1 \Delta \theta\right)}+\frac{1}{\mathrm{R}\left(1+\alpha_2 \Delta \theta\right)}\end{aligned}$

$\begin{aligned}& \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1}{1+\alpha_1 \Delta \theta}+\frac{1}{1+\alpha_2 \Delta \theta} \\& \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1+\alpha_2 \Delta \theta+1+\alpha_1 \Delta \theta}{\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)} \\& 2\left[\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)\right] \\& =\left[2+\left(\alpha_1+\alpha_2\right) \Delta \theta\right]\left[1+\alpha_{\mathrm{eq}} \Delta \theta\right] \\& 2\left[1+\alpha_1 \Delta \theta+\alpha_2 \Delta \theta+\alpha_1 \alpha_2 \Delta \theta\right]\end{aligned}$

$\begin{aligned}& = \\& 2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta+\alpha_{\mathrm{eq}}\left(\alpha_1+\alpha_2\right) \Delta \theta^2\end{aligned}$

Neglecting small terms

$\begin{aligned}& 2+2\left(\alpha_1+\alpha_2\right) \Delta \theta=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta \\& \left(\alpha_1+\alpha_2\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta \\& \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}\end{aligned}$