Question

Digital thermometers often make use of thermistors, a type of resistor with resistance that varies with temperature more than standard resistors. Find the temperature coefficient of resistivity for a linear thermistor with resistances of 75.0 V at 0.00°C and 275 V at 525°C.

Answer 1

Answer:

The temperature coefficient of resistivity for a linear thermistor is $1.38\times10^{-3}^{\circ}C^{-1}$

Explanation:

Given that,

Initial temperature = 0.00°C

Resistance = 75.0 Ω

Final temperature = 525°C

Resistance = 275 Ω

We need to calculate the temperature coefficient of resistivity for a linear thermistor

Using formula for a linear thermistor

$R=R_{0}(1+\alpha\Delta T)$

$R=R_{0}+R_{0}\alpha\Delta T$

$\alpha=\dfrac{R-R_{0}}{R_{0}\Delta T}$

Put the value into the formula

$\alpha=\dfrac{275-75}{275\times(525-0)}$

$\alpha=1.38\times10^{-3}^{\circ}C^{-1}$

Hence, The temperature coefficient of resistivity for a linear thermistor is $1.38\times10^{-3}^{\circ}C^{-1}$

Answer 2

Answer:

5.08 x 10^-3 /°C

Explanation:

Let the temperature coefficient of resistivity is α.

resistance at 0°C , Ro = 75 ohm

Resistance at 525°C, Rt = 275 ohm

the formula for the temperature coefficient of resistivity

$\alpha =\frac{R_{t}-R_{0}}{R_{0}\Delta T}$

$\alpha =\frac{275-75}{75\times 525}$

α = 5.08 x 10^-3 /°C

thus, the temperature coefficient of resistivity is 5.08 x 10^-3 /°C.