Question

From the given table data of resistane Vs temperature for an RTD the linear approximation of resistance between ${100}^{\circ }C$ & ${130}^{\circ }C$ about a mean temperature of ${115}^{\circ }C$ is
$\begin{array}{cc}\text{Temperature}\left(i{n}^{\circ }C\right)& \text{Resistance}\left(in\Omega \right)\\ 100& 573.40\\ 105& 578.77\\ 110& 584.13\\ 115& 589.48\\ 120& 594.84\\ 125& 600.18\\ 130& 605.52\end{array}$

• A. $589.48[1+0.0180(\theta -115\left)\right]\Omega$
• B. $573.40[1+0.00182(\theta -115\left)\right]\Omega$
• C. $589.48[1+0.00182(\theta -115\left)\right]\Omega$
• D. $605.52[1+0.00182(\theta -115\left)\right]\Omega$

Answer 1

The correct option is C $589.48[1+0.00182(\theta -115\left)\right]\Omega$

${\theta }_1={100}^{\circ }C$, ${\theta }_2={130}^{\circ }C$, ${\theta }_0={115}^{\circ }C$

$R_{{\theta }_1}=573.40\Omega$, $R_{{\theta }_2}=605.52\Omega$, $R_{{\theta }_0}=589.48\Omega$

${\alpha }_1=\frac{1}{R_{{\theta }_0}}\left(\frac{R_{{\theta }_2}-R_{{\theta }_1}}{{\theta }_2-{\theta }_1}\right)=\frac{1}{589.48}\times \left(\frac{605.52-573.40}{130-100}\right)$

${\alpha }_1=0.00182\left(\frac{1}{{}^{\circ }C}\right)$

$R_{\theta }=R_{{\theta }_0}(1+\alpha (\theta -{\theta }_0\left)\right)=589.48[1+0.00182(\theta -115\left)\right]\Omega$