Two resistances,each equal to R_0 at 0,^oC,ha | vedclass

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(D) Let the resistances at $0\,^oC$ be $R_0$ for both resistors.At temperature $t$,the resistances are $R_1 = R_0(1 + \alpha_1 t)$ and $R_2 = R_0(1 +

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

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(D) Let the resistances at $0\,^oC$ be $R_0$ for both resistors.At temperature $t$,the resistances are $R_1 = R_0(1 + \alpha_1 t)$ and $R_2 = R_0(1 + \alpha_2 t)$.When connected in series,the equivalent resistance $R_{eq}$ is $R_1 + R_2$.$R_{eq} = R_0(1 + \alpha_1 t) + R_0(1 + \alpha_2 t) = 2R_0 + R_0(\alpha_1 + \alpha_2)t$.$R_{eq} = 2R_0 \left( 1 + \frac{\alpha_1 + \alpha_2}{2} t \right)$.Since the equivalent resistance at $0\,^oC$ is $R'_{0} = 2R_0$,we have $R_{eq} = R'_{0}(1 + \alpha_{eq} t)$.Comparing the two expressions,the equivalent temperature coefficient is $\alpha_{eq} = \frac{\alpha_1 + \alpha_2}{2}$.

Two resistances,each equal to $R_0$ at $0\,^oC$,have temperature coefficients of resistance $\alpha_1$ and $\alpha_2$. When they are joined in series,the temperature coefficient of the equivalent resistance is:

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