At room temperature (27.0 °C) the resistance of a heating element is 100 Ω.

Question:

At room temperature $\left(27.0^{\circ} \mathrm{C}\right)$ the resistance of a heating element is $100 \Omega$. What is the temperature of the element if the resistance is found to be $117 \Omega$, given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$

Solution:

Room temperature, T = 27°C

Resistance of the heating element at TR = 100 Ω

Let T1 is the increased temperature of the filament.

Resistance of the heating element at T1R1 = 117 Ω

Temperature co-efficient of the material of the filament,

$\alpha=1.70 \times 10^{-4} \mathrm{C}^{-1}$

$\alpha$ is given by the relation,

$\alpha=\frac{R_{1}-R}{R\left(T_{1}-T\right)}$

$T_{1}-T=\frac{R_{1}-R}{R \alpha}$

$T_{1}-27=\frac{117-100}{100\left(1.7 \times 10^{-4}\right)}$

$T_{1}-27=1000$

$T_{1}=1027^{\circ} \mathrm{C}$

Therefore, at $1027^{\circ} \mathrm{C}$, the resistance of the element is $117 \Omega$.