Question

Two conductors have the same resistances at $0^\circ \text{C}$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients for their series and parallel combinations are:

• A. $\alpha_1 + \alpha_2, \quad \frac{\alpha_1 + \alpha_2}{2}$
• B. $\frac{\alpha_1 + \alpha_2}{2}, \quad \frac{\alpha_1 + \alpha_2}{2}$
• C. $\alpha_1 + \alpha_2, \quad \frac{\alpha_1 \alpha_2}{\alpha_1 + \alpha_2}$
• D. $\frac{\alpha_1 + \alpha_2}{2}, \quad \alpha_1 + \alpha_2$

Answer 1

Answer: (B)

$\frac{\alpha_1 + \alpha_2}{2}, \quad \frac{\alpha_1 + \alpha_2}{2}$

Answer 2

Series:

$R_{\text{eq}} = R_1 + R_2$

$2R(1 + \alpha_{\text{eq}} \Delta \theta) = R(1 + \alpha_1 \Delta \theta) + R(1 + \alpha_2 \Delta \theta)$

$2R(1 + \alpha_{\text{eq}} \Delta \theta) = 2R + (\alpha_1 + \alpha_2)R \Delta \theta$

$\alpha_{\text{eq}} = \frac{\alpha_1 + \alpha_2}{2}$

Parallel:

$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{\pi}{2} \frac{1}{1 + \alpha_{\text{eq}} \Delta \theta} = \frac{1}{R(1 + \alpha_1 \Delta \theta)} + \frac{1}{R(1 + \alpha_2 \Delta \theta)}$

$\frac{2}{1 + \alpha_{\text{eq}} \Delta \theta} = \frac{1}{1 + \alpha_1 \Delta \theta} + \frac{1}{1 + \alpha_2 \Delta \theta}$

$\frac{2}{1 + \alpha_{\text{eq}} \Delta \theta} = \frac{1 + \alpha_2 \Delta \theta + 1 + \alpha_1 \Delta \theta}{(1 + \alpha_1 \Delta \theta)(1 + \alpha_2 \Delta \theta)}$

$2[(1 + \alpha_1 \Delta \theta)(1 + \alpha_2 \Delta \theta)] = [2 + (\alpha_1 + \alpha_2) \Delta \theta][1 + \alpha_{\text{eq}} \Delta \theta]$

$2 \left[1 + \alpha_1 \Delta \theta + \alpha_2 \Delta \theta + \alpha_1 \alpha_2 \Delta \theta^2 \right] = 2 + 2(\alpha_1 + \alpha_2) \Delta \theta + (\alpha_1 + \alpha_2) \Delta \theta$

Neglecting small terms:

$2 + 2(\alpha_1 + \alpha_2) \Delta \theta = 2 + 2 \alpha_{\text{eq}} \Delta \theta + (\alpha_1 + \alpha_2) \Delta \theta$

$(\alpha_1 + \alpha_2) \Delta \theta = 2 \alpha_{\text{eq}} \Delta \theta$

$\alpha_{\text{eq}} = \frac{\alpha_1 + \alpha_2}{2}$