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, $R_0=99\Omega$, $T_0=27^{\circ }C$
$R_T=116\Omega$;
$\alpha =1.710^{-4}{}^{\circ }{C}^{-1}$
$\therefore R_T=R_0\left[1+\alpha \left(T-T_0\right)\right]$
$\therefore \frac{R_T}{R_0}-1=\alpha \left(T-T_0\right)$
$\Rightarrow \frac{116}{99}-1=\alpha \left(T-T_0\right)$
$T-T_0=\frac{1}{\alpha }\left[\frac{116-99}{99}\right]$
$=\frac{17}{99\alpha }=\frac{1}{1.7\times 10^{-4}}\times \frac{17}{99}$
$\therefore T-T_0=\frac{10^5}{99}=1010.10{}^{\circ }C$
$\Rightarrow T=1010.1+T_0=1010.1+27$
$=1037.1{}^{\circ }C$