Question

It is desired to make a long cylindrical conductor whose temperature coefficient of resistivity at ${20}^{\circ }C$ will be close to zero. If such a conductor is made by assembling alternate disks of iron and carbon, Find the ratio of the thickness of a carbon disk to that an iron disk. (For carbon, $\rho =3500\times {10}^{-8}\Omega m$ and $\alpha =-0.50\times {10}^{-3}{}^{\circ }{C}^{-1}$ for iron, $\rho =9.68\times {10}^{-8}\Omega$ and $\alpha =6.5\times {10}^{-3}{}^{\circ }{C}^{-1})$.

• A. $0.36$
• B. $0.036$
• C. $1.0$
• D. $2.0$

Answer 1

Answer: (B)

$0.036$

Change in the resistance of the conductor on increasing its temperature,
$R-R_0=R_0{\alpha }_{av}(T-T_0)$
The disks will be effectively in series, so we will add the resistances to get the total. Looking only at one pair of disk, we have
$R_c+R_i=R_{0c}\left({\alpha }_c\right(T-T_0)+1)+R_{0i}\left({\alpha }_i\right(T-T_0)+1)$

$=R_{0c}+R_{0i}+(R_{0c}{\alpha }_c+R_{0i}{\alpha }_i)(T-T_0)$.
This equation will only be constant if the coefficient for the term $(T-T_0)$ vanishes.
Then $R_{0c}{\alpha }_c+R_{0i}{\alpha }_i=0$,
but $R=\rho L/A$, and the disks have the same cross sectional area, so

$L_c{\rho }_c{\alpha }_c+L_i{\rho }_i{\alpha }_i=0$

or $\frac{L_c}{L_i}=\frac{{\rho }_i{\alpha }_i}{{\rho }_c{\alpha }_c}=-\frac{9.68\times {10}^{-8}\times 6.5\times {10}^{-3}}{3500\times {10}^{-8}\times (-0.50\times {10}^{-3})}=0.036$.