Question

Two resistors with temperature coefficients of resistance ${\alpha }_1$ and ${\alpha }_2$ have resistances $R_1$ and $R_2$ at $0^{\circ }C$. What is the temperature coefficient of the compound resistor consisting of the two resistors connected in parallel?

• A. $\frac{R_1{\alpha }_1+{\alpha }_2{R}_2}{R_1+R_2}$
• B. $\frac{{\alpha }_1+{\alpha }_2}{2}$
• C. $\frac{2{\alpha }_1+{\alpha }_2}{{\alpha }_1+{\alpha }_2}$
• D. $\frac{{\alpha }_1{R}_2+{\alpha }_2{R}_1}{R_1+R_2}$

Answer 1

The correct option is D

$\frac{{\alpha }_1{R}_2+{\alpha }_2{R}_1}{R_1+R_2}$

When connected in parallel,
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
Differentiating both sides,
$\frac{-dR}{R^2}=\frac{-d{R}_1}{R_1^2}+\frac{-d{R}_2}{R_2^2}\Rightarrow \frac{\alpha RdT}{R^2}=\frac{{\alpha }_1{R}_{1}dT}{R_1^2}+\frac{{\alpha }_2{R}_{2}dT}{R_2^2}\Rightarrow \frac{\alpha }{R}=\frac{{\alpha }_1}{R_1}+\frac{{\alpha }_2}{R_2}\Rightarrow \alpha =R(\frac{{\alpha }_1}{R_1}+\frac{{\alpha }_2}{R_2})=\frac{R_1{R}_2}{R_1+R_2}\left(\frac{{\alpha }_1{R}_2+{\alpha }_2{R}_1}{R_1{R}_2}\right)\therefore \alpha =\frac{{\alpha }_1{R}_2+{\alpha }_2{R}_1}{R_1+R_2}$