Thermal Properties Of Matter Question 2
Question 2 - 2024 (31 Jan Shift 1)
Two conductors have the same resistances at $0^{\circ} \mathrm{C}$ but their temperature coefficients of resistance are $\alpha_{1}$ and $\alpha_{2}$. The respective temperature coefficients for their series and parallel combinations are :
(1) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$
(2) $\frac{\alpha_{1}+\alpha_{2}}{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$
(3) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$
(4) $\frac{\alpha_{1}+\alpha_{2}}{2}, \alpha_{1}+\alpha_{2}$
Show Answer
Answer: (2)
Solution:
Series:
$\mathrm{R}{\mathrm{eq}}=\mathrm{R}{1}+\mathrm{R}_{2}$
$2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=\mathrm{R}\left(1+\alpha_{1} \Delta \theta\right)+\mathrm{R}\left(1+\alpha_{2} \Delta \theta\right)$
$2 \mathrm{R}\left(1+\alpha_{\text {eq }} \Delta \theta\right)=2 \mathrm{R}+\left(\alpha_{1}+\alpha_{2}\right) \mathrm{R} \Delta \theta$
Parallel :
$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$
$\frac{1}{\frac{R}{2}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)}=\frac{1}{\mathrm{R}\left(1+\alpha_{1} \Delta \theta\right)}+\frac{1}{\mathrm{R}\left(1+\alpha_{2} \Delta \theta\right)}$
$2\left[\left(1+\alpha_{1} \Delta \theta\right)\left(1+\alpha_{2} \Delta \theta\right)\right]$
$=\left[2+\left(\alpha_{1}+\alpha_{2}\right) \Delta \theta\right]\left[1+\alpha_{\mathrm{eq}} \Delta \theta\right]$
$2\left[1+\alpha_{1} \Delta \theta+\alpha_{2} \Delta \theta+\alpha_{1} \alpha_{2} \Delta \theta\right]$
$=2+2 \alpha_{\text {eq }} \Delta \theta+\left(\alpha_{1}+\alpha_{2}\right) \Delta \theta+\alpha_{\text {eq }}\left(\alpha_{1}+\alpha_{2}\right) \Delta \theta^{2}$
Neglecting small terms
$2+2\left(\alpha_{1}+\alpha_{2}\right) \Delta \theta=2+2 \alpha_{\text {eq }} \Delta \theta+\left(\alpha_{1}+\alpha_{2}\right) \Delta \theta$
$\left(\alpha_{1}+\alpha_{2}\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta$
$\alpha_{\text {eq }}=\frac{\alpha_{1}+\alpha_{2}}{2}$
