Question

Two conductors have the same resistance at $0^{\circ }C$ but their temperature coefficient of resistance are ${\alpha }_1$ and ${\alpha }_2$. The respective temperature coefficient of their series and parallel combinations are nearly:

• A. $\frac{{\alpha }_1+{\alpha }_2}{2},{\alpha }_1+{\alpha }_2$
• B. ${\alpha }_1+{\alpha }_2,\frac{{\alpha }_1+{\alpha }_2}{2}$
• C. $\frac{{\alpha }_1+{\alpha }_2}{3},{\alpha }_1+{\alpha }_2$
• D. $\frac{{\alpha }_1+{\alpha }_2}{2},\frac{{\alpha }_1+{\alpha }_2}{2}$

Answer 1

Answer: (D)

$\frac{{\alpha }_1+{\alpha }_2}{2},\frac{{\alpha }_1+{\alpha }_2}{2}$

Let the resistance at $0^{o}C$ be $R$ and at temperature $T$ be $R_{1,T}$ and $R_{2,T}$ respectively

For series combination, net resistance at $T$ is:

$R_{s,T}=R_{1,T}+R_{2,T}$

$R_s(1+{\alpha }_{s}T)=R(1+{\alpha }_{1}T)+R(1+{\alpha }_{2}T)$

$2R(1+{\alpha }_{s}T)=2R(1+\frac{{\alpha }_1+{\alpha }_2}{2}T)$

${\alpha }_s=\frac{{\alpha }_1+{\alpha }_2}{2}$

For parallel combination, net resistance at $T$ is:

$\frac{1}{R_{p,T}}=\frac{1}{R_{1,T}}+\frac{1}{R_{2,T}}$

$\frac{1}{R_p(1+{\alpha }_{p}T)}=\frac{1}{R(1+{\alpha }_{1}T)}+\frac{1}{R(1+{\alpha }_{2}T)}$

$\frac{2}{R(1+{\alpha }_{p}T)}=\frac{2+({\alpha }_1+{\alpha }_2)T}{R(1+({\alpha }_1+{\alpha }_2)T+{\alpha }_1{\alpha }_2{T}^2)}$

$1+{\alpha }_{p}T=1+\frac{({\alpha }_1+{\alpha }_2+2{\alpha }_1{\alpha }_{2}T)T}{2+({\alpha }_1+{\alpha }_2)T}$

Since temperature coefficient of resistance is small (order of ${10}^{-3}{/}^{o}C$), the terms $2{\alpha }_1{\alpha }_{2}T$ in the numerator and $({\alpha }_1+{\alpha }_2)T$ in denominator can be ignored.

Using the above approximation,

${\alpha }_p\approx \frac{{\alpha }_1+{\alpha }_2}{2}$