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
$\therefore \quad$ 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 $R_{s0} = R_{0} + R_{0} = 2R_{0}$
$\therefore \quad2R_{0} \left(1+ \alpha _{s} \theta \right) = 2R_{0}+R_{0} \theta\left(\alpha_{1}+ \alpha _{2} \right)$
or$\quad\alpha_{s} = \frac{\alpha _{1}+ \alpha _{2}}{2}$
for parallel combination,$\quad R_{p} = \frac{R_{1}R_{2}}{R_{1}+R_{2}}$
$R_{p0} \left(1+\alpha_{p}\theta\right) = \frac{R_{0} \left(1+ \alpha _{1} \theta \right)R_{0} \left(1+ \alpha _{2} \theta \right)}{R_{0} \left(1+ \alpha _{1} \theta \right)+R_{0} \left(1+ \alpha _{2} \theta \right)}$
where, $R_{p0} = \frac{R_{0}R_{0}}{R_{0}+R_{0}} = \frac{R_{0}}{2}$
$\therefore \quad\frac{R_{0}}{2}\left(1+ \alpha_{p} \theta \right) = \frac{R^{2}_{0}\left(1+\alpha_{1} \theta+\alpha_{2} \theta +\alpha _{1}\alpha _{2} \theta \right)}{R_{0}\left(2+\alpha _{1} \theta +\alpha _{2} \theta \right)}$
as $\quad\alpha _{1}$ and $\alpha _{2}$ are small quantities
$\therefore \quad\alpha _{1} \, \alpha _{2}$ is negligible
or $\quad\alpha_{p} = \frac{\alpha _{1} + \alpha _{2}}{2+\left(\alpha _{1} + \alpha _{2}\right)\theta} = \frac{\alpha _{1} + \alpha _{2}}{2}\left[1+\left(\alpha _{1} + \alpha _{2}\right)\theta\right]$
as$\quad\left(\alpha _{1} + \alpha _{2}\right)^{2}$ is negligible
$\therefore \quad\alpha_{p} = \frac{\alpha _{1} + \alpha _{2}}{2}$