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:

Updated On: Nov 20, 2024
  • \( \alpha_1 + \alpha_2, \quad \frac{\alpha_1 + \alpha_2}{2} \)
  • \( \frac{\alpha_1 + \alpha_2}{2}, \quad \frac{\alpha_1 + \alpha_2}{2} \)
  • \( \alpha_1 + \alpha_2, \quad \frac{\alpha_1 \alpha_2}{\alpha_1 + \alpha_2} \)
  • \( \frac{\alpha_1 + \alpha_2}{2}, \quad \alpha_1 + \alpha_2 \)
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The Correct Option is B

Solution and Explanation

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:

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

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

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

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