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(D) Let $R_0$ be the initial resistance of both conductors at $0\,^{\circ}C$.At temperature $	heta$,their resistances are given by:$R_1 = R_0(1 + \al

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

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(D) Let $R_0$ be the initial resistance of both conductors at $0\,^{\circ}C$.At temperature $	heta$,their resistances are given by:$R_1 = R_0(1 + \alpha_1 	heta)$$R_2 = R_0(1 + \alpha_2 	heta)$For a series combination,the total resistance $R_s$ is the sum of individual resistances:$R_s = R_1 + R_2$Let $\alpha_s$ be the temperature coefficient of the series combination. Then:$R_{s0}(1 + \alpha_s 	heta) = R_1 + R_2$Since $R_{s0} = R_0 + R_0 = 2R_0$,we have:$2R_0(1 + \alpha_s 	heta) = R_0(1 + \alpha_1 	heta) + R_0(1 + \alpha_2 	heta)$$2R_0 + 2R_0 \alpha_s 	heta = 2R_0 + R_0 	heta(\alpha_1 + \alpha_2)$Dividing both sides by $2R_0 	heta$:$\alpha_s = \frac{\alpha_1 + \alpha_2}{2}$

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 coefficient of their series combination is

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