Q119 / 125 · BITSAT 2016 · Physics
Two resistances at $0^{\circ} \mathrm{C}$ with temperature coefficient of resistance $\alpha_{1}$ and $\alpha_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be
- A$\alpha_{1}+\alpha_{2}$
- B$\frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$
- C$\frac{\alpha_{1}-\alpha_{2}}{2}$
- Correct answerD$\frac{\alpha_{1}+\alpha_{2}}{2}$
Solution
$\begin{aligned} R_{1} &=R_{0}\left(1+\alpha_{1} t\right)+R_{0}\left(1+\alpha_{2} t\right) \\ &=2 R_{0}\left(1+\frac{\alpha_{1}+\alpha_{2}}{2} t\right) \end{aligned}$
Comparing with $R=R_{0}(1+\alpha t)$
$\alpha=\frac{\alpha_{1}+\alpha_{2}}{2}$
Hence, the correct answer is option: 'D'.