Respuesta :
Answer:
The temperature coefficient of resistivity for a linear thermistor is [tex]1.38\times10^{-3}^{\circ}C^{-1}[/tex]
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
[tex]R=R_{0}(1+\alpha\Delta T)[/tex]
[tex]R=R_{0}+R_{0}\alpha\Delta T[/tex]
[tex]\alpha=\dfrac{R-R_{0}}{R_{0}\Delta T}[/tex]
Put the value into the formula
[tex]\alpha=\dfrac{275-75}{275\times(525-0)}[/tex]
[tex]\alpha=1.38\times10^{-3}^{\circ}C^{-1}[/tex]
Hence, The temperature coefficient of resistivity for a linear thermistor is [tex]1.38\times10^{-3}^{\circ}C^{-1}[/tex]
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
[tex]\alpha =\frac{R_{t}-R_{0}}{R_{0}\Delta T}[/tex]
[tex]\alpha =\frac{275-75}{75\times 525}[/tex]
α = 5.08 x 10^-3 /°C
thus, the temperature coefficient of resistivity is 5.08 x 10^-3 /°C.
