Question

Coefficient of linear expansion of material of resistor is $\alpha $ .Its temperature coefficient of resistivity and resistance are ${\alpha _\rho }$ and ${\alpha _R}$ respectively, then correct relation is
A. ${\alpha _R} = {\alpha _\rho } - \alpha $
B. ${\alpha _R} = {\alpha _\rho } + \alpha $
C. ${\alpha _R} = {\alpha _\rho } + 3\alpha $
D. ${\alpha _R} = {\alpha _\rho } - 3\alpha $

Answer 1

Hint:First, we will assume the complete system at ${T_0}$ temperature. And then increase the temperature with $\Delta T$ and simplify the obtained equation and with the help of binomial expansion we will reduce the term and hence we will get the required solution.

Formula used:
$R = \dfrac{{\rho l}}{A}$
Where, $R$ is the resistance, $\rho $ is the resistivity of the material, $l$ is the length of the conductor and $A$ is the area of the cross section.

Complete step by step answer:
We know that $R = \dfrac{{\rho l}}{A}$. Let’s assume at ${T_0}$ temperature, Resistance is ${R_0}$ ,Resistivity is ${\rho _0}$, ${l_0}$ is the length and ${A_0}$ is the area. So, at ${T_0}$ temperature the value of resistance will be,
${R_0} = \dfrac{{{\rho _0}{l_0}}}{{{A_0}}}$
Now let’s increase the temperature with $\Delta T$ then the value will change,i.e.,
${R_0} = {R_0}(1 + {\alpha _R}\Delta T)$
\[\Rightarrow {\rho _0} = {\rho _0}(1 + {\alpha _\rho }\Delta T)\]
$\Rightarrow {l_0} = {l_0}(1 + \alpha \Delta T)$
$\Rightarrow {A_0} = {A_0}(1 + 2\alpha \Delta T)$

Not putting together in formula, we get,
${R_0}(1 + {\alpha _R}\Delta T) = {\rho _0}(1 + {\alpha _\rho }\Delta T)\dfrac{{{l_0}(1 + \alpha \Delta T)}}{{{A_0}(1 + 2\alpha \Delta T)}}$
Now, simplifying,
\[{R_0}(1 + {\alpha _R}\Delta T) = {\rho _0}\dfrac{{{l_0}}}{{{A_0}}} \times \dfrac{{(1 + {\alpha _\rho }\Delta T)(1 + \alpha \Delta T)}}{{(1 + 2\alpha \Delta T)}} \\
\Rightarrow {R_0}(1 + {\alpha _R}\Delta T) = {R_0} \times \dfrac{{(1 + {\alpha _\rho }\Delta T)(1 + \alpha \Delta T)}}{{(1 + 2\alpha \Delta T)}} \\
\Rightarrow (1 + {\alpha _R}\Delta T) = (1 + {\alpha _\rho }\Delta T)(1 + \alpha \Delta T){(1 + 2\alpha \Delta T)^{ - 1}} \\ \]
Now, reducing with the help of binomial expansion, we will get
\[(1 + {\alpha _R}\Delta T) = (1 + {\alpha _\rho }\Delta T)(1 + \alpha \Delta T){(1 + 2\alpha \Delta T)^{ - 1}} \\
\Rightarrow (1 + {\alpha _R}\Delta T) = (1 + {\alpha _\rho }\Delta T)(1 + \alpha \Delta T)(1 - 2\alpha \Delta T) \\ \]
Here we have neglected the last term as the multiplication of $\alpha $ and $ - 2\alpha $ is very small.
\[\Rightarrow (1 + {\alpha _R}\Delta T) = (1 + {\alpha _\rho }\Delta T)[1 + \alpha \Delta T - 2\alpha \Delta T] \\
\Rightarrow (1 + {\alpha _R}\Delta T) = (1 + {\alpha _\rho }\Delta T)(1 - \alpha \Delta T) \\
\Rightarrow 1 + {\alpha _R}\Delta T = 1 - \alpha \Delta T + {\alpha _\rho }\Delta T - {\alpha _\rho }\alpha \Delta T \\ \]
Here we have neglected the last term as the multiplication of \[{\alpha _\rho }\alpha \] is very small.
\[\Rightarrow 1 + {\alpha _R}\Delta T = 1 - \alpha \Delta T + {\alpha _\rho }\Delta T \\
\therefore {\alpha _R} = {\alpha _\rho } - \alpha \]
So, the correct relation is \[{\alpha _R} = {\alpha _\rho } - \alpha \] .

Hence the correct option is A.

Note:If the conductor does not expand, then the value of coefficient of resistance is same as the coefficient of resistivity. But as the area of the cross section increases with temperature, the coefficient decreases by $\alpha $.