Question

Two wires of resistance $R_1$ and $R_2$ have temperature co-efficient of resistance ${\alpha }_1$ and ${\alpha }_2$ respectively. These are joined in series. The effective temperature co-efficient of resistance is

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

Answer 1

The correct option is C $\frac{{\alpha }_1{R}_1+{\alpha }_2{R}_2}{R_1+R_2}$

Suppose at $t^{\circ }C$ resistances of the two wires becomes $R_{1t}$ and $R_{2t}$ respectively and equivalent resistance becomes $R_t$. In series grouping $R_t=R_{1t}+R_{2t}$, also $R_{1t}=R_1(1+{\alpha }_{1}t)$ and $R_{2t}=R_2(1+{\alpha }_{2}t);R_t=R_1(1+{\alpha }_{1}t)+R_2(1+{\alpha }_{2}t)=(R_1+R_2)+(R_1{\alpha }_1+R_2{\alpha }_2)t=(R_1+R_2)[1+\frac{R_1{\alpha }_1+R_2{\alpha }_2}{R_1+R_2}t]$. Hence effective temperature co-efficient is $\frac{R_1{\alpha }_1+R_2{\alpha }_2}{R_1+R_2}$.