R_1 and R_2 are two resistors made of differe | vedclass

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(D) Let the resistances at $0^{\circ}C$ be $R_{01}$ and $R_{02}$.At temperature $	heta$,the resistances are given by:$R_1 = R_{01}(1 + \alpha 	heta)

Vedclass · Accepted Answer

$\beta/\alpha$

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(D) Let the resistances at $0^{\circ}C$ be $R_{01}$ and $R_{02}$.At temperature $	heta$,the resistances are given by:$R_1 = R_{01}(1 + \alpha 	heta)$$R_2 = R_{02}(1 - \beta 	heta)$The total resistance in series is $R_s = R_1 + R_2$.$R_s = R_{01}(1 + \alpha 	heta) + R_{02}(1 - \beta 	heta)$$R_s = (R_{01} + R_{02}) + 	heta(R_{01}\alpha - R_{02}\beta)$For the total resistance $R_s$ to be independent of temperature $	heta$,the coefficient of $	heta$ must be zero:$R_{01}\alpha - R_{02}\beta = 0$$R_{01}\alpha = R_{02}\beta$Therefore,the ratio of the resistances is $\frac{R_{01}}{R_{02}} = \frac{\beta}{\alpha}$.

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