Question

Two conductors have the same resistance at $0^{\circ }C$ but their temperature coefficients of resistance are ${\alpha }_1$ and ${\alpha }_2$. The respective temperature coefficients of their series and parallel combinations are nearly

• A. $\frac{{\alpha }_1+{\alpha }_2}{2},{\alpha }_1+{\alpha }_2$
• B. ${\alpha }_1+{\alpha }_2,\frac{{\alpha }_1+{\alpha }_2}{2}$
• C. ${\alpha }_1+{\alpha }_2,\frac{{\alpha }_1{\alpha }_2}{{\alpha }_1+{\alpha }_2}$
• D. $\frac{{\alpha }_1+{\alpha }_2}{2},\frac{{\alpha }_1+{\alpha }_2}{2}$

Answer 1

Answer: (D)

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