Question

The resistance of a heating element is $99\,\Omega$ at room temperature. What is the temperature of the element if the resistance is found to be $116\,\Omega$ ?
(Temperature coefficient of the material of the resistor is $1.7 \times 10^{-4}\,{}^{\circ}C^{-1}$)

• A. $999.9\,{}^{\circ}C$
• B. $1005.3\,{}^{\circ}C$
• C. $1020.2\,{}^{\circ}C$
• D. $1037.1\,{}^{\circ}C$

Answer 1

Answer: (D)

Here, $R_{0}=99\,\Omega$, $T_{0}=27^{\circ}C$
$R_{T}=116\,\Omega$;
$\alpha=1.7\,10^{-4}\,{}^{\circ}C^{-1}$
$\therefore R_{T}=R_{0}\,\left[1+\alpha\left(T-T_{0}\right)\right]$
$\therefore \frac{R_{T}}{R_{0}}-1=\alpha\left(T-T_{0}\right)$
$\Rightarrow \frac{116}{99}-1=\alpha\left(T-T_{0}\right)$
$T-T_{0}=\frac{1}{\alpha}\left[\frac{116-99}{99}\right]$
$=\frac{17}{99\,\alpha}=\frac{1}{1.7 \times 10^{-4}} \times \frac{17}{99}$
$\therefore T-T_{0}=\frac{10^{5}}{99}=1010.10\,{}^{\circ}C$
$\Rightarrow T=1010.1+T_{0}=1010.1+27$
$=1037.1\,{}^{\circ}C$