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Step-by-Step Solution
Step 1: Express Individual Resistances at Temperature $t$
Let $R_0$ be the resistance of each conductor at $0^\circ\!C$.
If the temperature is raised by an amount $\Delta t$, then the resistances are:
$R_1 = R_0 \bigl[1 + \alpha_1\,\Delta t\bigr], \quad R_2 = R_0 \bigl[1 + \alpha_2\,\Delta t\bigr].$
Step 2: Find the Equivalent Temperature Coefficient in Series
When these two resistors are connected in series, their combined resistance $R_{\text{series}}$ is:
$R_{\text{series}} = R_1 + R_2
= R_0 \bigl[1 + \alpha_1 \,\Delta t\bigr] + R_0 \bigl[1 + \alpha_2 \,\Delta t\bigr]
= R_0 \bigl[2 + (\alpha_1 + \alpha_2)\,\Delta t\bigr].
$
Factor out $2 R_0$:
$R_{\text{series}} = 2\,R_0 \Bigl[\,1 + \dfrac{\alpha_1 + \alpha_2}{2}\,\Delta t\Bigr].$
By comparing this form $2\,R_0 \,[\,1 + \alpha_{\text{eq,series}}\, \Delta t\,]$ to the above expression, we identify:
$\alpha_{\text{eq,series}} = \dfrac{\alpha_1 + \alpha_2}{2}.$
Step 3: Find the Equivalent Temperature Coefficient in Parallel
For the parallel combination, the total resistance $R_{\text{parallel}}$ satisfies:
$\dfrac{1}{R_{\text{parallel}}}
= \dfrac{1}{R_1} + \dfrac{1}{R_2}
= \dfrac{1}{R_0 \bigl[1 + \alpha_1\,\Delta t\bigr]} \;+\; \dfrac{1}{R_0 \bigl[1 + \alpha_2\,\Delta t\bigr]}.
$
After algebraic simplification (combining fractions and comparing the resulting form to $ \tfrac{1}{R_{\text{parallel}}} \;=\; \tfrac{1}{\,\tfrac{R_0}{2}\,\bigl(1 + \alpha_{\text{eq,parallel}} \,\Delta t\bigr)} }$), one finds that the effective temperature coefficient for the parallel combination is also:
$\alpha_{\text{eq,parallel}} = \dfrac{\alpha_1 + \alpha_2}{2}.$
Final Answer
Therefore, the respective temperature coefficients of the series and parallel combinations are both
$\dfrac{\alpha_1 + \alpha_2}{2}.$
