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 $1.7\times {10}^{-4}$ ${}^o{C}^{-1}$).

• A. $999.9$ ${}^{o}C$
• B. $1005.3$ ${}^{o}C$
• C. $1020.2$ ${}^{o}C$
• D. $1037.1$ ${}^{o}C$

Answer 1

Answer: (D)

$1037.1$ ${}^{o}C$

Here, $R_0=99\Omega ,T_0={27}^o$C
$R_T=116\Omega ;\alpha =1.7$ ${10}^{-4}$ ${}^o{C}^{-1}$
$\therefore R_T=R_0[1+\alpha (T-T_0\left)\right]$

$\therefore \frac{R_T}{R_0}-1=\alpha (T-T_0)\Rightarrow \frac{116}{99}-1=\alpha (T-T_0)$

$T-T_0=\frac{1}{\alpha }\left[\frac{116-99}{99}\right]=\frac{17}{99\alpha }=\frac{1}{1.7\times {10}^{-4}}\times \frac{17}{99}$

$\therefore T-T_0=\frac{{10}^5}{99}={1010.10}^o$C

$\Rightarrow T=1010.1+T_0=1010.1+27=1037.1$ ${}^{o}C$.