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

The temperature coefficient of resistivity for a linear thermistor is
`1.38*10^(-3)^(\circ)C^(-1)`

Step-by-step 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=(R-R_(0))/(R_(0)\Delta T)`

Put the value into the formula

`\alpha=(275-75)/(275*(525-0))`

`\alpha=1.38*10^(-3)^(\circ)C^(-1)`

Hence, The temperature coefficient of resistivity for a linear thermistor is
`1.38*10^(-3)^(\circ)C^(-1)`

Answer 2

5.08 x 10^-3 /°C

Step-by-step 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 =(R_(t)-R_(0))/(R_(0)\Delta T)`

`\alpha =(275-75)/(75* 525)`

α = 5.08 x 10^-3 /°C

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