Question

The resistance of a heating element is 99 $\rho$ at room temperature. What is the temperature of the element if the resistance is found to be $\frac{r_{1}}{r_{2}}\frac{\rho}{2}$

(Temperature coefficient of the material of the resistor is $\frac{r_{2} - r_{1}}{r_{1}r_{2}}\frac{\rho}{4\pi}$ )

• A. 999.9C
• B. 1005.3C
• C. 1020.2C
• D. 1037.1C

Answer 1

Answer: (D)

1037.1C

Answer 2

: Here, $T_{0} = 27^{o}C$

}{\alpha = 1.71{0^{- 4}}^{o}C^{- 1}}$$ $$\therefore R_{T} = R_{0}\lbrack(1 + \alpha(T - T_{0})\rbrack$$ $$\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\lbrack \frac{116 - 99}{99} \right\rbrack = \frac{17}{99o} = \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$$