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(A) Let the resistance of both conductors at $0\,^{\circ}C$ be $R_0$. The resistance at temperature $\Delta t$ is given by $R = R_0(1 + \alpha \Delta

Vedclass · Accepted Answer

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

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(A) Let the resistance of both conductors at $0\,^{\circ}C$ be $R_0$. The resistance at temperature $\Delta t$ is given by $R = R_0(1 + \alpha \Delta t)$.For the series combination: $R_s = R_1 + R_2 = R_0(1 + \alpha_1 \Delta t) + R_0(1 + \alpha_2 \Delta t) = 2R_0(1 + \frac{\alpha_1 + \alpha_2}{2} \Delta t)$. Comparing this with $R_s = 2R_0(1 + \alpha_s \Delta t)$,we get $\alpha_s = \frac{\alpha_1 + \alpha_2}{2}$.For the parallel combination: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_0(1 + \alpha_1 \Delta t)} + \frac{1}{R_0(1 + \alpha_2 \Delta t)}$. Using the binomial approximation $(1+x)^{-1} \approx 1-x$,we get $\frac{1}{R_p} \approx \frac{1}{R_0}(1 - \alpha_1 \Delta t + 1 - \alpha_2 \Delta t) = \frac{2}{R_0}(1 - \frac{\alpha_1 + \alpha_2}{2} \Delta t)$.Thus,$R_p \approx \frac{R_0}{2}(1 + \frac{\alpha_1 + \alpha_2}{2} \Delta t)$. Comparing this with $R_p = \frac{R_0}{2}(1 + \alpha_p \Delta t)$,we get $\alpha_p = \frac{\alpha_1 + \alpha_2}{2}$.

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

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