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
$(a) \frac{\alpha_1 + \alpha_2}{2} \quad (b) \sqrt{\alpha_1 \alpha_2} \\ (c) \frac{\alpha_1 R_1 + \alpha_2 R_2}{R_1 + R_2} \quad (d) \frac{\sqrt{R_1 R_2 \alpha_1 \alpha_2}}{\sqrt{R_1^2 + R_2^2}}$
Text Solution
Verified by ExpertsC
$\mathbf{R}_{t_1} = \mathbf{R}_1 (1 + \alpha_1 t) \text{ and } \mathbf{R}_{t_2} = \mathbf{R}_2 (1 + \alpha_2 t)$
Also $R_{eq.} = R_{t_1} + R_{t_2}$
$\Rightarrow R_{eq} = R_1 + R_2 + (R_1 \alpha_1 + R_2 \alpha_2) t$
$\Rightarrow R_{\text{eq}} = (R_1 + R_2) \left\{ 1 + \left(\frac{R_1 \alpha_1 + R_2 \alpha_2}{R_1 + R_2}\right) \cdot t \right\}$ So $\alpha_{\text{eff}} = \frac{R_1 \alpha_1 + R_2 \alpha_2}{R_1 + R_2}$
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