Question

A metal wire of diameter 4 mm and length 100 m has a resistance of $0.408\Omega$ at ${10}^{\circ }C$ and $0.508\Omega$ at ${120}^{\circ }C$. Find the value of
a. temperature coefficient of resistance
b. its resistance at $0^{\circ }C$
c. its resistivities at $0^{\circ }C$ and ${120}^{\circ }C$

Answer 1

Given that $r=4/2mm=2mm=2\times {10}^{-3}m,\ell =100m$,
$t_1={10}^{\circ }C,t_2={120}^{\circ }C,R_{t_1}=0.408\Omega ,R_{t_2}=0.508\Omega$
i. Temperature coefficient of resistance is given by
$\alpha =\frac{R_{t_2}-R_{t_1}}{R_{t_1}(t_2-t_1)}=\frac{0.508-0.408}{0.408(120-10)}=2.2\times {10}^{-3}{C}^{-1}$
ii. We know that $R_{t_1}=R_0(1+\alpha t_1)$
or $R_0=\frac{R_{t_1}}{1+\alpha t_1}=\frac{0.408}{1+2.2\times {10}^{-3}\times 10}=\frac{0.408}{1.022}=0.4\Omega$
iii. Resistivity at $0^{\circ }C$ is given by
$\rho =\frac{R_{0}A}{\ell }=\frac{R_0\pi r^2}{\ell }$
$=\frac{0.4\times 3.14\times (2\times {10}^{-3}{)}^2}{100}\Omega m=5.02\times {10}^{-8}\Omega m$
Resistivity at ${120}^{\circ }C$ is given by
${\rho }_{20}={\rho }_0(1+\alpha t)=5.02\times {10}^{-8}\left(\right(l+2.2\times {10}^{-3}\times 120)\Omega m$
$=5.02\times {10}^{-8}\times 1.264\Omega m=6.35\times {10}^{-8}\Omega m$