Physics - Heat & Thermo Question with Solution | TestHub

Let $R_0$ be the initial resistance of both conductors $\therefore$ At temperature $\theta$ their resistance will be, $R_1=R_0\left(1+{\alpha }_1\theta \right)$ and $R_2=R_0\left(1+{\alpha }_2\theta \right)$ for, series combination, ${\mathrm{R}}_{\mathrm{s}}={\mathrm{R}}_1+{\mathrm{R}}_2$ $R_{s0}\left(1+{\alpha }_s\theta \right)=R_0\left(1+{\alpha }_1\theta \right)+R_0\left(1+{\alpha }_2\theta \right)$ where ${\mathrm{R}}_{\mathrm{s}0}={\mathrm{R}}_0+{\mathrm{R}}_0=2{\mathrm{R}}_0$ $\therefore 2{\mathrm{R}}_0\left(1+{\alpha }_s\theta \right)=2{\mathrm{R}}_0+{\mathrm{R}}_0\theta \left({\alpha }_1+{\alpha }_2\right)$ or ${\alpha }_s=\frac{{\alpha }_1+{\alpha }_2}{2}$ for parallel combination, $$R_p=\frac{R_1{R}_2}{R_1+R_2}$$ where, $R_{p0}=\frac{R_0{R}_0}{R_0+R_0}=\frac{R_0}{2}$ $\therefore \frac{{\mathrm{R}}_0}{2}\left(1+{\alpha }_{\mathrm{p}}\theta \right)=\frac{{\mathrm{R}}_0^2\left(1+{\alpha }_1\theta +{\alpha }_2\theta +{\alpha }_1{\alpha }_2\theta \right)}{{\mathrm{R}}_0\left(2+{\alpha }_1\theta +{\alpha }_2\theta \right)}$ as ${\alpha }_1$ and ${\alpha }_2$ are small quantities $\therefore {\alpha }_1{\alpha }_2$ is negligible or ${\alpha }_{\mathrm{p}}=\frac{{\alpha }_1+{\alpha }_2}{2+\left({\alpha }_1+{\alpha }_2\right)\theta }=\frac{{\alpha }_1+{\alpha }_2}{2}\left[1-\left({\alpha }_1+{\alpha }_2\right)\theta \right]$ as ${\left({\alpha }_1+{\alpha }_2\right)}^2$ is negligible $\therefore {\alpha }_p=\frac{{\alpha }_1+{\alpha }_2}{2}$