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Question: Coefficient of linear expansion of material of resistor is \(\alpha \) .Its temperature coefficient ...

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

Explanation

Solution

First, we will assume the complete system at T0{T_0} temperature. And then increase the temperature with ΔT\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=ρlAR = \dfrac{{\rho l}}{A}
Where, RR is the resistance, ρ\rho is the resistivity of the material, ll is the length of the conductor and AA is the area of the cross section.

Complete step by step answer:
We know that R=ρlAR = \dfrac{{\rho l}}{A}. Let’s assume at T0{T_0} temperature, Resistance is R0{R_0} ,Resistivity is ρ0{\rho _0}, l0{l_0} is the length and A0{A_0} is the area. So, at T0{T_0} temperature the value of resistance will be,
R0=ρ0l0A0{R_0} = \dfrac{{{\rho _0}{l_0}}}{{{A_0}}}
Now let’s increase the temperature with ΔT\Delta T then the value will change,i.e.,
R0=R0(1+αRΔT){R_0} = {R_0}(1 + {\alpha _R}\Delta T)
⇒ρ0=ρ0(1+αρΔT)\Rightarrow {\rho _0} = {\rho _0}(1 + {\alpha _\rho }\Delta T)
⇒l0=l0(1+αΔT)\Rightarrow {l_0} = {l_0}(1 + \alpha \Delta T)
⇒A0=A0(1+2αΔT)\Rightarrow {A_0} = {A_0}(1 + 2\alpha \Delta T)

Not putting together in formula, we get,
R0(1+αRΔT)=ρ0(1+αρΔT)l0(1+αΔT)A0(1+2αΔT){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,

\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 $.