Question

The resistance of a conductor at $20^\circ C$ is 3.15ohm and at $100^\circ C$ is 3,74ohm. Determine the temperature coefficient of resistance of the conductor. What will be the resistance of the conductor at ${0^\circ }C$ ?

Answer 1

Hint
We will first use the formula of dependence of resistance with temperature formula of dependence of resistance with temperature to find the temperature coefficient of resistance of the conductor.
We will find two equation for resistance of a conductor at $20^\circ C$ and the resistance of a conductor at $100^\circ C$ and then we will equate these two equations and find the value of $\alpha $ i.e. the temperature coefficient of resistivity.
After finding it we will put this into one of the equations for resistance of a conductor and find the resistance of the conductor at ${0^\circ }C$ .

Complete step by step answer
Given: The resistance of a conductor at $20^\circ C$ is 3.15 $\Omega $ and at $100^\circ C$ is 3.75 ohm.
To find the temperature coefficient of resistance of the conductor we use the formula of dependence of resistance with temperature i.e. ${R_T} = {R_0}\left[ {1 + \alpha T} \right]$ as the resistance varies with the temperature.
Let the resistance of a conductor at $20^\circ C = {R_{20}} = 3.15\Omega $ and
the resistance of a conductor at $100^\circ C = {R_{100}} = 3.75\Omega $ .
We use ${R_T} = {R_0}\left[ {1 + \alpha T} \right]$ , where ${R_T}$ is the resistance at temperature $T^\circ C$ , ${R_0}$ is the resistance at temperature at ${0^\circ }C$ and $\alpha $ is the temperature coefficient of resistivity .
Now ${R_{20}} = 3.15 = {R_0}(1 + 20\alpha )$ and
 $\Rightarrow {R_{100}} = 3.75 = {R_0}(1 + 100\alpha )$
Dividing the above two equations we get $\dfrac{{{R_0}(1 + 20\alpha )}}{{{R_0}(1 + 100\alpha )}} = \dfrac{{3.15}}{{3.75}}$
Now after solving the above equation we get $\alpha = 0.0025^\circ {C^{ - 1}}$ , now we put the value of $\alpha $ in the equation ${R_{20}} = 3.15 = {R_0}(1 + 20\alpha )$ to find the value of ${R_0}$ .
We get ${R_0} = \dfrac{{3.15}}{{(1 + 20\alpha )}}$ , after putting the value of $\alpha $ we get:
 $\Rightarrow {R_0} = \dfrac{{3.15}}{{(1 + 20 \times 0.0025)}} = \dfrac{{3.15}}{{1.05}} = 3\Omega $ .
So, we get ${R_0} = 3\Omega $ i.e. the resistance of the conductor at ${0^\circ }C$ .

Note
The general rule i.e. resistivity increases with increasing temperature in conductors and at $20^\circ C$ is 3.15 $\Omega $ decreases with increasing temperature in insulators as we can see that at $100^\circ C$ it is 3.75 ohm, at $20^\circ C$ it is 3.15 $\Omega $ and at ${0^\circ }C$ it is $3\Omega $ .