Courses for Kids
Free study material
Offline Centres

Two resistances ${{R}_{1}}$ and ${{R}_{2}}$ are made of different materials. The temperature coefficient of the material of ${{R}_{1}}$ is $\alpha $and that of material of ${{R}_{2}}$ is $-\beta $. The material of ${{R}_{1}}$ and ${{R}_{2}}$will not change with temperature if $\dfrac{{{R}_{1}}}{{{R}_{2}}}$ equal to
(A) $\dfrac{\alpha }{\beta }$
(B) $\dfrac{\alpha +\beta }{\alpha -\beta }$
(C) $\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{2\alpha \beta }$
(D) $\dfrac{\beta }{\alpha }$

Last updated date: 01st Mar 2024
Total views: 341.1k
Views today: 6.41k
IVSAT 2024
341.1k+ views
Hint: The temperature coefficient of a material is the relative change of its physical properties with the change in temperature. If the material of the two resistances will not change then the initial sum of temperature coefficients should equal the final sum after time t.

Complete answer:
According to the question, let us consider two resistances ${{R}_{1}}$ and ${{R}_{2}}$ made of different materials. Their temperature coefficients are, $\alpha $and $-\beta $ respectively.
Resistance and temperature coefficient relation is given by the following equation,
$R={{R}_{0}}(1+\alpha \Delta t)$
Where, ${{R}_{0}}$ is initial or reference temperature at temperature ${{T}_{0}}$ and $\Delta t$is $T-{{T}_{0}}$.
In this case, the resistance will not change so, using the above formula, we can write,
  & {{R}_{1}}+{{R}_{2}}={{R}_{1}}(1+\alpha T)+{{R}_{2}}(1-\beta T) \\
 & \Rightarrow {{R}_{1}}\alpha T={{R}_{2}}\beta T \\
 & \Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\beta }{\alpha } \\
Thus, the material of ${{R}_{1}}$ and ${{R}_{2}}$will not change with temperature if $\dfrac{{{R}_{1}}}{{{R}_{2}}}$ equal to $\dfrac{\beta }{\alpha }$.

Hence, the correct answer is option (D).

Additional Information:
Temperature coefficients of materials have useful applications in electric and magnetic properties of material. Normally, the value of temperature coefficient lies between $-2$ and $3$.
The resistance and temperature coefficient relation ($R={{R}_{0}}(1+\alpha \Delta t)$) is a linear approximation, if temperature coefficient of material does not vary too much with temperature, i.e. $\alpha \Delta t\langle \langle 1$
The temperature coefficient of elasticity i.e. the elastic modulus of elastic materials decreases with the increase in temperature.
Temperature coefficient has a dimension that is reverse of temperature. Unit of temperature coefficient is $\kappa^{-1}$(per Kelvin). Sometimes thermal coefficient is expressed as ppm/$^{0}C$ or ppm/K, ppm is parts per million.

The resistance and temperature coefficient relation ($R={{R}_{0}}(1+\alpha \Delta t)$) is important to remember for solving similar question, however this relation is a linear approximation and is only applicable for small temperature differences ($\Delta t$) between initial and final temperature changes.
Recently Updated Pages
Trending doubts