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:
- \( \alpha_1 + \alpha_2, \quad \frac{\alpha_1 + \alpha_2}{2} \)
- \( \frac{\alpha_1 + \alpha_2}{2}, \quad \frac{\alpha_1 + \alpha_2}{2} \)
- \( \alpha_1 + \alpha_2, \quad \frac{\alpha_1 \alpha_2}{\alpha_1 + \alpha_2} \)
- \( \frac{\alpha_1 + \alpha_2}{2}, \quad \alpha_1 + \alpha_2 \)
The Correct Option is B
Approach Solution - 1
To determine the temperature coefficients for the series and parallel combinations of two conductors having the same resistances at \( 0^\circ \text{C} \), and temperature coefficients \( \alpha_1 \) and \( \alpha_2 \), we start by understanding the basic concept of temperature coefficients of resistance.
The resistance \( R_T \) of a conductor at a temperature \( T \) is given by the formula:
\(R_T = R_0(1 + \alpha \Delta T)\)
where \( R_0 \) is the resistance at the reference temperature (often \( 0^\circ \text{C} \)), \( \alpha \) is the temperature coefficient of resistance, and \( \Delta T = T - T_0 \).
- For the series combination of the two conductors:
The total resistance in series, \( R_s \), is:
\(R_s = R_1 + R_2 = R_0(1 + \alpha_1 \Delta T) + R_0(1 + \alpha_2 \Delta T)\) \(= 2R_0 + R_0(\alpha_1 + \alpha_2) \Delta T\)
Thus, the effective temperature coefficient, \( \alpha_s \), for the series is:
\(\frac{R_s - 2R_0}{2R_0 \Delta T} = \frac{\alpha_1 + \alpha_2}{2}\)
- For the parallel combination of the two conductors:
The total resistance in parallel, \( R_p \), can be found using:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_0(1 + \alpha_1 \Delta T)} + \frac{1}{R_0(1 + \alpha_2 \Delta T)}\)
For small values of \( \alpha \Delta T \), this simplifies to:
\(R_p = \frac{R_0}{1 + \frac{\alpha_1 + \alpha_2}{2} \Delta T}\)
Thus, the effective temperature coefficient, \( \alpha_p \), for the parallel combination is also:
\(\frac{\alpha_1 + \alpha_2}{2}\)
Therefore, the correct answer for the respective temperature coefficients for their series and parallel combinations is:
Option: \( \frac{\alpha_1 + \alpha_2}{2}, \quad \frac{\alpha_1 + \alpha_2}{2} \)
Approach Solution -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:
\[ 2 + 2(\alpha_1 + \alpha_2) \Delta \theta = 2 + 2 \alpha_{\text{eq}} \Delta \theta + (\alpha_1 + \alpha_2) \Delta \theta \]
\[ (\alpha_1 + \alpha_2) \Delta \theta = 2 \alpha_{\text{eq}} \Delta \theta \]
\[ \alpha_{\text{eq}} = \frac{\alpha_1 + \alpha_2}{2} \]
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