Question

Two wires of resistance $R_{1}$ and $R_{2}$ have temperature coefficient of resistance $\alpha_{1}$ and $\alpha_{2}$, respectively. These are joined in series. The effective temperature coefficient of resistance is

• A. $\frac{\alpha_{1}+\alpha_{2}}{2}$
• B. $\sqrt{\alpha_{1} \alpha_{2}}$
• C. $\frac{\alpha_{1} R_{1}+\alpha_{2} R_{2}}{R_{1}+R_{2}}$
• D. $\frac{\sqrt{R_{1} R_{2} \alpha_{1} \alpha_{2}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}$

Answer 1

Answer: (C)

$R_{t_{1}}=R_{1}\left(1+\alpha_{1} t\right)$ and $R_{t_{2}}=R_{2}\left(1+\alpha_{2} t\right)$
Also $R_{e q .}=R_{t_{1}}+R_{t_{2}} $
$\Rightarrow R_{e q}=R_{1}+R_{2}+\left(R_{1} \alpha_{1}+R_{2} \alpha_{2}\right) t$
$\Rightarrow R_{e q}=\left(R_{1}+R_{2}\right)\left\{1+\left(\frac{R_{1} \alpha_{1}+R_{2} \alpha_{2}}{R_{1}+R_{2}}\right) \cdot t\right\}$
So $\alpha_{e f f}=\frac{R_{1} \alpha_{1}+R_{2} \alpha_{2}}{R_{1}+R_{2}}$