Question

Two wires $A$ and $B$, having resistivity ${\rho }_A=3\times {10}^{-5}\Omega m$ and ${\rho }_a=6\times {10}^{-5}\Omega m$ of same cross section area are pointed together to form a single wire. If the resistance of the joined wire does not change with temperature, then find the ratio of their lengths given that temperature coefficient of resistivity of wires $A$ and $B$ are ${\alpha }_A=4\times {10}^{-5}{/}^{\circ }C$ and ${\alpha }_B=-4\times {10}^{-6}{/}^{\circ }C$. Assume that mechanical dimensions do not change with temperature.

Answer 1

we know that , $\rho ={\rho }_0(1+\alpha \Delta T)$ and $R=\frac{\rho l}{A}$

and when rods are combined end to end it behaves as series combination

so, combined resistance will be : $R=R_1+R_2$

this gives , $R=\frac{{\rho }_1{l}_1}{A}+\frac{{\rho }_2{l}_2}{A}$

using equation of roh, we get : $R=\frac{1}{A}\left({\rho }_1{l}_1\right(1+{\alpha }_1\Delta T)+{\rho }_2{l}_2(1+{\alpha }_2\Delta T\left)\right)$

$=\frac{1}{A}({\rho }_1{l}_1+{\rho }_2{l}_2+({\rho }_1{l}_1{\alpha }_1+{\rho }_2{l}_2{\alpha }_2\left)\Delta T\right)$

Given that this resistance does not depend on temperature therefore coefficient of temperature difference will be zero , i.e,

${\rho }_1{l}_1{\alpha }_1+{\rho }_2{l}_2{\alpha }_2=0$

This gives : $\frac{l_1}{l_2}=\frac{-{\rho }_2{\alpha }_2}{{\rho }_1{\alpha }_1}$

$=\frac{6\times {10}^{-5}\times 4{\times 10}^{-6}}{{3\times 10}^{-5}\times 4\times {10}^{-5}}$

$=0.2$