Question

The resistance of a wire at room temperature \[{30^\circ }C\] is found to be \[10\Omega \] . Now to increase the resistance by \[10\% \] , the temperature of the wire must be: [The temperature coefficient of resistance of the material of the wire is 0.002 per Celsius]
(A) \[{36^\circ }C\]
(B) \[{83^\circ }C\]
(C) \[{63^\circ }C\]
(D) \[{33^\circ }C\]

Answer 1

Hint In this question, we are given the initial conditions of the wire. We know that when a wire is heated, it experiences gain in resistance. This gain in resistance should be equal to \[10\% \] . We can also infer that if the original resistance was R, the new resistance should be 1.1R.
Use the following relation
 ${R_T} = {R_0}(1 + \alpha T)$

Complete step by step solution
According to the question
Temperature coefficient of resistance of the wire,
\[\alpha = 0.002{/^\circ }C\]
Resistance at temperature,
\[{30^\circ }C = {R_{30}} = 10\Omega \]
We know,
${R_T} = {R_0}(1 + \alpha T)$
Putting the value of \[{R_{30}}\] , \[\alpha \] and T in above equation we have
$10 = {R_0}(1 + 30\alpha )$ … (i)
After \[10\% \] increase in resistance the new resistance becomes
$10 + 10 \times \dfrac{{10}}{{100}} = 11\Omega $
Again, for the new resistance we have
$11 = {R_0}(1 + \alpha T)$ … (ii)
Dividing equation (ii) by (i), we get
$\dfrac{{11}}{{10}} = \dfrac{{{R_0}(1 + \alpha T)}}{{{R_0}(1 + 30\alpha )}}$
\[{R_0}\] gets cancelled out as it’s the same material and same reference temperature.
Or, $\dfrac{{11}}{{10}} = \dfrac{{(1 + \alpha T)}}{{(1 + 30\alpha )}}$
Or, \[11(1 + 30\alpha ) = 10(1 + \alpha T)\]
Or, \[11 + 330\alpha = 10 + 10\alpha T\]
Or, \[1 + 330\alpha = 10\alpha T\]
Or, \[\dfrac{{1 + 330\alpha }}{{10\alpha }} = T\]
Putting the value of \[\alpha \] , we get
 \[T = \dfrac{{1 + 330 \times 0.002}}{{10 \times 0.002}}\]
Or, $T = {83^ \circ }C$ (Option B)
Option B is correct answer

Note The above calculation is made with the consideration that the reference temperature is the same. Although in case of different reference temperature the above formula becomes,
 ${R_T} = {R_0}(1 + \alpha (T - {T_0}))$
Where, \[{R_T}\] = Conductor resistance at temperature “T”.
\[{R_0}\] = Conductor resistance at reference temperature.
\[\alpha \] = Temperature coefficient of resistance of the material of the conductor.
T = Conductor temperature in degree Celsius.
\[{R_0}\] = Reference temperature in degree Celsius.
Resistance of a conductor increases with the increasing temperature because the chance of electrons colliding with protons increases. Resistance decreases with decreasing temperature. However, in some cases when the temperature is decreased beyond a critical temperature its resistance becomes zero, such materials are called superconductors. The critical temperature for mercury is 4.15 K.