A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this thermistor is 3,6

Question

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this thermistor is 3,650 °C. Find: If the thermistor is then used to measure a particular temperature, and its resistance is measured as 500 Ω, determine the thermistor temperature.

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Neala 4 years 2021-07-22T02:34:59+00:00 1 Answers 92 views 0

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  1. diemkieu
    0
    2021-07-22T02:36:53+00:00
    Reply

    Answer:

    the thermistor temperature = 325.68 \ ^0 \ C

    Explanation:

    Given that:

    A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

    i.e Temperature

    T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K

    Resistance of the thermistor R_1 = 20,000 ohms

    Material constant \beta = 3650

    Resistance of the thermistor R_2 = 500 ohms

    Using the equation :

    R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

    \frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

    Taking log of both sides

    In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

    In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})

    \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})

    \frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}

    {T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}

    Replacing our values into the above equation :

    {T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}

    {T_2} = \frac{1361450}{3650 - 3.6888*373}

    {T_2} = \frac{1361450}{3650 - 1375.92}

    {T_2} = \frac{1361450}{2274.08}

    {T_2} = 598.68 \ K

    {T_2} = 325.68 \ ^0 \ C

    Thus, the thermistor temperature = 325.68 \ ^0 \ C

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