JEE Main 2024 — Current Electricity Question with Solution

Series :

$$\begin{array}{rl}& {\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_1+{\mathrm{R}}_2\\ & 2\mathrm{R}\left(1+{\alpha }_{\mathrm{eq}}\Delta \theta \right)=\mathrm{R}\left(1+{\alpha }_1\Delta \theta \right)+\mathrm{R}\left(1+{\alpha }_2\Delta \theta \right)\\ & 2\mathrm{R}\left(1+{\alpha }_{\mathrm{eq}}\Delta \theta \right)=2\mathrm{R}+\left({\alpha }_1+{\alpha }_2\right)\mathrm{R}\Delta \theta \\ & {\alpha }_{\mathrm{eq}}=\frac{{\alpha }_1+{\alpha }_2}{2}\end{array}$$

Parallel :

$$\begin{array}{rl}& \frac{1}{R_{\text{eq }}}=\frac{1}{R_1}+\frac{1}{R_2}\\ & \frac{1}{\frac{R}{2}\left(1+{\alpha }_{\text{eq }}\Delta \theta \right)}=\frac{1}{R\left(1+{\alpha }_1\Delta \theta \right)}+\frac{1}{\mathrm{R}\left(1+{\alpha }_2\Delta \theta \right)}\end{array}$$

$$\begin{array}{rl}& \frac{2}{1+{\alpha }_{\mathrm{eq}}\Delta \theta }=\frac{1}{1+{\alpha }_1\Delta \theta }+\frac{1}{1+{\alpha }_2\Delta \theta }\\ & \frac{2}{1+{\alpha }_{\mathrm{eq}}\Delta \theta }=\frac{1+{\alpha }_2\Delta \theta +1+{\alpha }_1\Delta \theta }{\left(1+{\alpha }_1\Delta \theta \right)\left(1+{\alpha }_2\Delta \theta \right)}\\ & 2\left[\left(1+{\alpha }_1\Delta \theta \right)\left(1+{\alpha }_2\Delta \theta \right)\right]\\ & =\left[2+\left({\alpha }_1+{\alpha }_2\right)\Delta \theta \right]\left[1+{\alpha }_{\mathrm{eq}}\Delta \theta \right]\\ & 2\left[1+{\alpha }_1\Delta \theta +{\alpha }_2\Delta \theta +{\alpha }_1{\alpha }_2\Delta \theta \right]\end{array}$$

$$\begin{array}{rl}& =\\ & 2+2{\alpha }_{\mathrm{eq}}\Delta \theta +\left({\alpha }_1+{\alpha }_2\right)\Delta \theta +{\alpha }_{\mathrm{eq}}\left({\alpha }_1+{\alpha }_2\right)\Delta {\theta }^2\end{array}$$

Neglecting small terms

$$\begin{array}{rl}& 2+2\left({\alpha }_1+{\alpha }_2\right)\Delta \theta =2+2{\alpha }_{\mathrm{eq}}\Delta \theta +\left({\alpha }_1+{\alpha }_2\right)\Delta \theta \\ & \left({\alpha }_1+{\alpha }_2\right)\Delta \theta =2{\alpha }_{\mathrm{eq}}\Delta \theta \\ & {\alpha }_{\mathrm{eq}}=\frac{{\alpha }_1+{\alpha }_2}{2}\end{array}$$