Question

Two conductors have the same resistance at $0^{\circ }C$ but their temperature coefficients of resistance are${\alpha }_1$ and $a_2$ respectively. The temperature coefficient of their series combination is

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

Answer 1

The correct option is B $\frac{{\alpha }_1+{\alpha }_2}{2}$

$$\begin{array}{c}let,\\ R_0=resistanceofbothconductorat{0}^{0}C\\ if{R}_{T1}and{R}_{T2}aretheresistanceofbothconductorat{t}^{0}C\\ then,\\ R_{T1}=R_0(1+a_{1}t)and{R}_{T2}=R_0(1+a_{2}t)\\ and,if{R}_s=resistanceat{t}^{0}Cwhentheyareconnectedinseries.\\ then,\\ R_S=R_{T1}+R_{T2}\\ =R_0(1+a_{1}t)+R_0(1+a_{2}t)\\ =R_0(2+a_{1}t+a_{2}t)\\ =2R_0\left[1+\frac{(a_1+a_2)}{2}t\right]---------\left(i\right)\\ then,\\ R_S={R^{{}^{\prime }}}_0(1+at)-------\left(ii\right)\\ compareequ\left(i\right)and\left(ii\right)\\ weget,\\ {R^{{}^{\prime }}}_0=2R_0\\ \therefore a=\frac{(a_1+a_2)}{2}\\ andsothatthecorrectoptionisB.\end{array}$$