Solveeit Logo

Question

Question: The resistance of a conductor at \(20^\circ C\) is 3.15ohm and at \(100^\circ C\) is 3,74ohm. Determ...

The resistance of a conductor at 20C20^\circ C is 3.15ohm and at 100C100^\circ C is 3,74ohm. Determine the temperature coefficient of resistance of the conductor. What will be the resistance of the conductor at 0C{0^\circ }C ?

Explanation

Solution

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 20C20^\circ C and the resistance of a conductor at 100C100^\circ C and then we will equate these two equations and find the value of α\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 0C{0^\circ }C .

Complete step by step answer
Given: The resistance of a conductor at 20C20^\circ C is 3.15 Ω\Omega and at 100C100^\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. RT=R0[1+αT]{R_T} = {R_0}\left[ {1 + \alpha T} \right] as the resistance varies with the temperature.
Let the resistance of a conductor at 20C=R20=3.15Ω20^\circ C = {R_{20}} = 3.15\Omega and
the resistance of a conductor at 100C=R100=3.75Ω100^\circ C = {R_{100}} = 3.75\Omega .
We use RT=R0[1+αT]{R_T} = {R_0}\left[ {1 + \alpha T} \right] , where RT{R_T} is the resistance at temperature TCT^\circ C , R0{R_0} is the resistance at temperature at 0C{0^\circ }C and α\alpha is the temperature coefficient of resistivity .
Now R20=3.15=R0(1+20α){R_{20}} = 3.15 = {R_0}(1 + 20\alpha ) and
R100=3.75=R0(1+100α)\Rightarrow {R_{100}} = 3.75 = {R_0}(1 + 100\alpha )
Dividing the above two equations we get R0(1+20α)R0(1+100α)=3.153.75\dfrac{{{R_0}(1 + 20\alpha )}}{{{R_0}(1 + 100\alpha )}} = \dfrac{{3.15}}{{3.75}}
Now after solving the above equation we get α=0.0025C1\alpha = 0.0025^\circ {C^{ - 1}} , now we put the value of α\alpha in the equation R20=3.15=R0(1+20α){R_{20}} = 3.15 = {R_0}(1 + 20\alpha ) to find the value of R0{R_0} .
We get R0=3.15(1+20α){R_0} = \dfrac{{3.15}}{{(1 + 20\alpha )}} , after putting the value of α\alpha we get:
R0=3.15(1+20×0.0025)=3.151.05=3Ω\Rightarrow {R_0} = \dfrac{{3.15}}{{(1 + 20 \times 0.0025)}} = \dfrac{{3.15}}{{1.05}} = 3\Omega .
So, we get R0=3Ω{R_0} = 3\Omega i.e. the resistance of the conductor at 0C{0^\circ }C .

Note
The general rule i.e. resistivity increases with increasing temperature in conductors and at 20C20^\circ C is 3.15 Ω\Omega decreases with increasing temperature in insulators as we can see that at 100C100^\circ C it is 3.75 ohm, at 20C20^\circ C it is 3.15 Ω\Omega and at 0C{0^\circ }C it is 3Ω3\Omega .