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(D) Let the resistance of the series combination at temperature $T$ be $R_s(T)$.Initially,at temperature $T_0$,the total resistance is $R_s = R_1 + R_

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$\frac{\beta}{\alpha}$

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(D) Let the resistance of the series combination at temperature $T$ be $R_s(T)$.Initially,at temperature $T_0$,the total resistance is $R_s = R_1 + R_2$.At a temperature $T = T_0 + \Delta T$,the resistances become $R_1' = R_1(1 + \alpha \Delta T)$ and $R_2' = R_2(1 - \beta \Delta T)$.The new series resistance is $R_s' = R_1' + R_2' = R_1(1 + \alpha \Delta T) + R_2(1 - \beta \Delta T)$.For the total resistance to remain unchanged,$R_s' = R_s$,which implies:$R_1 + R_2 = R_1 + R_1 \alpha \Delta T + R_2 - R_2 \beta \Delta T$.Subtracting $(R_1 + R_2)$ from both sides,we get:$0 = R_1 \alpha \Delta T - R_2 \beta \Delta T$.$R_1 \alpha \Delta T = R_2 \beta \Delta T$.Dividing both sides by $R_2 \alpha \Delta T$,we get:$\frac{R_1}{R_2} = \frac{\beta}{\alpha}$.

Two resistances $R_1$ and $R_2$ are made of different materials. The temperature coefficient of the material of $R_1$ is $\alpha$ and of the material of $R_2$ is $-\beta$. The resistance of the series combination of $R_1$ and $R_2$ will not change with temperature,if $R_1/R_2$ equals

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