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 $\mathrm{\Delta }T$ 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:
$R={\displaystyle \frac{\rho l}{A}}$
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 $R={\displaystyle \frac{\rho l}{A}}$. 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,
$R_0={\displaystyle \frac{{\rho }_0{l}_0}{A_0}}$
Now let’s increase the temperature with $\mathrm{\Delta }T$ then the value will change,i.e.,
$R_0=R_0(1+{\alpha }_R\mathrm{\Delta }T)$
$$\Rightarrow {\rho }_0={\rho }_0(1+{\alpha }_{\rho }\mathrm{\Delta }T)$$
$\Rightarrow l_0=l_0(1+\alpha \mathrm{\Delta }T)$
$\Rightarrow A_0=A_0(1+2\alpha \mathrm{\Delta }T)$

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