Question

The reisistance of thermistors A & B both at 298K is $1k\Omega$ and the ratio of their resistance at 398K is 0.75. The difference in the material constants of thermistor A & B (in K) is _________?

• A. 341

Answer 1

The correct option is A 341
The resistance of thermistor at any temperature T is expressed as:-

$R_T=R_0\text{exp}\beta (\frac{1}{T}-\frac{1}{T_0})$

Where $R_T\text{\&}{R}_0$ are resistances at temperature $T\text{\&}{T}_0$ and $\beta$ is materials constant.

Hence,
$R_A\left(T\right)=R_0\text{exp}{\beta }_A(\frac{1}{T}-\frac{1}{T_0})$
and
$R_B\left(T\right)=R_0\text{exp}{\beta }_B(\frac{1}{T}-\frac{1}{T_o})$

$\text{Their Ratio =}\frac{R_A}{R_B}\left(T\right)=0.75$

$\therefore 0.75=\frac{exp{\beta }_A(\frac{1}{T}-\frac{1}{T_o})}{exp{\beta }_B(\frac{1}{T}-\frac{1}{T_o})}$

$\Rightarrow 0.75=exp({\beta }_A-{\beta }_B)[\frac{1}{T}-\frac{1}{T_o}]$

$\Rightarrow \text{ln}\left(0.75\right)=\text{ln}\left[exp\right({\beta }_A-{\beta }_B\left)(\frac{1}{T}-\frac{1}{T_o})\right]$

$\Rightarrow -0.2876=({\beta }_A-{\beta }_B)(\frac{1}{T}-\frac{1}{T_o})$

$\Rightarrow ({\beta }_A-{\beta }_B)=-\frac{0.2876}{(\frac{1}{T}-\frac{1}{T_o})}$

$=\frac{0.2876}{(\frac{1}{T_0}-\frac{1}{T})}$

$\text{In the given problem T = 398 K \&}{T}_0\text{= 298 K}$

$\therefore ({\beta }_A-{\beta }_B)=\frac{0.2876}{(\frac{1}{298K}-\frac{1}{398K})}=341$