Question

The resistance of a thermistor at ${25}^{\circ }C$ is found to be $1250\Omega \pm 5\%$.The material constant $\beta =3800$. The magnitude of maximum error (kelvin) in measurement of temperature $T-T_2$. If resistance R at tempeatrue $T_1$ is measured to be $2000\Omega$

• A. 286.29
• B. 1.111
• C. 28.62
• D. 11.11

Answer 1

The correct option is B 1.111

$\text{Resistor of thermistor is given by,}$

$R_T=R_{o}exp\left\{\beta [\frac{1}{T}-\frac{1}{T_o}]\right\}$

$Given,T_o={25}^{\circ }c=298K,R_o=1250\Omega and{R}_T=2000\Omega and\beta =3800$

$\therefore T=287.407K$

$\text{But it is given that}$

$R_o=1250\Omega \pm 5\%$

$So,R_o=1250\Omega +62.5=1312.5\Omega$

$\therefore T_1=288.471K$

$Andfor{R}_o=1250-62.5=1187.5\Omega$

$T_2=286.296K$

$\therefore \text{Maximum error}\Delta T=287.407-286.296K$

$=1.111K$