Question

Two resistance $R_1$ and $R_2$ are made of different materials. The temperature coefficient of the material of $R_1$ is $\alpha$ and that of the material of $R_2$ is $-\beta$. The resistance of the series combination of $R_1$ and $R_2$ does not change with temperature, then the ratio of resistances of the two wires at $0^{\circ }C$ will be:

• A. $\frac{2\alpha }{\beta }$
• B. $\frac{\alpha +\beta }{\alpha -\beta }$
• C. $\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$
• D. $\frac{\beta }{\alpha }$

Answer 1

The correct option is D $\frac{\beta }{\alpha }$

Resistance of wire as temperature changes is
$R_1=R_{01}(1+\alpha \Delta T)$
$R_2=R_{02}(1-\beta .\Delta T)$

$R_{\text{series}}=R_1+R_2=R_{01}+(R_{01}\alpha -R_{02}\beta )\Delta T+R_{02}$
$R_{\text{series}}$ is independent of temperature, if
$R_{01}\alpha -R_{02}\beta =0\Rightarrow \frac{R_{01}}{R_{02}}=\frac{\beta }{\alpha }$