Question

Resistance of a wire at temperature ${t^ \circ }\,C$ is $R = {R_0}\left( {1 + at + b{t^2}} \right)$ Here, ${R_0}$ is the temperature at ${0^ \circ }\,C$ . Find the temperature coefficient of resistance at temperature $t$ is
(A) $\dfrac{{a + 2bt}}{{1 + at + b{t^2}}}$
(B) $\dfrac{{a + 2b}}{{1 + a{t^2} + bt}}$
(C) $\dfrac{{a + 2bt}}{{1 + a{t^2} + bt}}$
(D) $\dfrac{{2bt}}{{1 + a{t^2} + bt}}$

Answer 1

Use the below formula of the temperature coefficient. Substitute the given resistance of the wire at ${t^ \circ }\,C$ and differentiate the obtained equation with respect to the temperature. The simplification of the differentiated equation gives the answer.
Useful formula:
The formula of the temperature coefficient is given by
$T = \dfrac{1}{{{R_t}}}\dfrac{{dR}}{{dt}}$
Where $T$ is the temperature coefficient of the resistance, ${R_t}$ is the temperature at ${t^ \circ }\,C$ and $T = \dfrac{{dR}}{{dt}}$ is the rate of change of the temperature with respect to the time.

Complete Step by Step Solution:
It is given that the
Resistance of the wire at the temperature ${t^ \circ }\,C$ is, $R = {R_0}\left( {1 + at + b{t^2}} \right)$
The is the temperature at ${0^ \circ }\,C$ is ${R_0}$
The temperature coefficient is the term that is used in which the resistance may increase or decrease the temperature in a given material. Depending on that it may be positive or negative in nature.
Using the formula of the temperature coefficient,
$T = \dfrac{1}{{{R_t}}}\dfrac{{dR}}{{dt}}$
Substituting the value of the resistance of the wire in the above formula,
$T = \dfrac{1}{{{R_0}\left( {1 + at + b{t^2}} \right)}}\dfrac{{d\left( {{R_0}\left( {1 + at + b{t^2}} \right)} \right)}}{{dt}}$
By simplifying the above step, we get
$T = \dfrac{1}{{\left( {1 + at + b{t^2}} \right)}}\dfrac{{d\left( {\left( {1 + at + b{t^2}} \right)} \right)}}{{dt}}$
By differentiating the above equation,
$T = \dfrac{{a + 2bt}}{{\left( {1 + at + b{t^2}} \right)}}$
Hence the value of the temperature coefficient obtained as $\dfrac{{a + 2bt}}{{\left( {1 + at + b{t^2}} \right)}}$.

Thus, the option (A) is correct.

Note: Remember that the differentiating of the constant is zero. And the differentiating of the constant multiplied by a variable is the constant. The differentiating of the square of the variable yields twice the variable. This is the concept that is obtained in the differentiation of the equation.