Question

You are using as 2252 ohm thermistor beta=3940 and you measure a resistance of 672.5 ohms what temperature degree celcius you are measuring

Answer 1

Given that,

Resistance $R_{th}=2252\ \Omega$

Resistance $R_{0}= 672.5\ \Omega$

$\beta=3940$

Suppose. Temperature is

$T_{0}=45^{\circ}C$

$T_{0}=318\ K$

We need to calculate the temperature

Using formula of resistance

$R_{th}=R_{0} e^{\beta(\dfrac{1}{T_{0}}-\dfrac{1}{T})}$

$T=\dfrac{\beta}{ln(\dfrac{R_{th}}{R_{0}\times e^{\dfrac{-\beta}{T_{0}}}})}$

Where, $R_{0}$ = resistance

$R_{th}$ =thermistor resistance

$T_{0}$ = temperature

Put the value into the formula

$T=\dfrac{3940}{ln(\dfrac{2252}{672.5\times e^{\dfrac{-3940}{318}}})}$

$T=289.73\ K$

$T=16.73^{\circ}C$

Hence, The temperature is 16.73°C.