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    Figure 1 shows a circuit including a thermistor T in series with a 2.2 kilo-ohm resistor R. The emf of the power source is 5V with negligible internal resistance.



    b) The output voltage is 1.8V which the thermistor is 20 degrees. Calculate the thermistor resistance at this temperature.
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    (Original post by blankboi)
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    Figure 1 shows a circuit including a thermistor T in series with a 2.2 kilo-ohm resistor R. The emf of the power source is 5V with negligible internal resistance.



    b) The output voltage is 1.8V which the thermistor is 20 degrees. Calculate the thermistor resistance at this temperature.
    Have you attempted to do this question? Where are you stuck?

    Show your working and tell us what you are having difficulty with and we can help from there.

    Hint: Potential divider.
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    (Original post by uberteknik)
    Have you attempted to do this question? Where are you stuck?

    Show your working and tell us what you are having difficulty with and we can help from there.

    Hint: Potential divider.

    i first worked out the current by doing (1.8/(2.2x10^3) and this gave me 8.18x10^-4 A. I then used this current and did 3.2/8.18x10^-4 and this gave me 3911 ohms which was the incorrect answer (unless the book is wrong). I'm not sure how else I could solve this question
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    (Original post by blankboi)
    i first worked out the current by doing (1.8/(2.2x10^3) and this gave me 8.18x10^-4 A.
    Correct. You used ohms law and worked out the current flowing in the 2200 ohm resistor:

    I = V/R = 1.8/2200 = 8.18x10-4A

    (Original post by blankboi)
    I then used this current and did 3.2/8.18x10^-4 and this gave me 3911 ohms which was the incorrect answer (unless the book is wrong). I'm not sure how else I could solve this question
    Correct.

    Because the resistor and thermistor are in a series circuit, the current must be the same at all points in that series path. i.e. The current must also be 8.18x10-4A flowing through the thermistor.

    The p.d. developed across the thermistor is (5 - 1.8) = 3.2V

    Which means the resistance of the thermistor @20oC must be:

    V/I = 3.2/8.18x10-4 = 3911 ohms.


    I agree with your answers. What answer does the book give? It does indeed seem to be in error.
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    the book gives this working:

    Pd across R = 5.0 – 1.8 V = 3.2 VThermistor resistance = (1.8 V / 3.2 V) × 2.2 kΩ= 1.2 kΩ
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    (Original post by blankboi)
    the book gives this working:

    Pd across R = 5.0 – 1.8 V = 3.2 VThermistor resistance = (1.8 V / 3.2 V) × 2.2 kΩ= 1.2 kΩ
    Have you drawn the circuit diagram correctly and shown the 1.8V developed across the correct component?

    The text book answer you quoted implies that the 1.8V is developed across the thermistor and not the 2200 ohms resistor.

    The simple test is that the p.d's must always be in exactly the same ratio as the resistances where the 1.8V must be developed across the smaller resistance and 3.6V developed across the larger resistance.
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    oh so the book used this method?: (V1/V2 = R1/R2)

    the diagram is drawn correctly and i copied the question word for word :hmmmm:
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    (Original post by blankboi)
    oh so the book used this method?: (V1/V2 = R1/R2)

    the diagram is drawn correctly and i copied the question word for word :hmmmm:
    I got the same answer as above, im pretty sure the answer in the book must be wrong. Here's my working:

    Find the current in the series circuit:
    V = IR
    1.8 = I x 2200
    1.8/2200 = I
    I = 0.000818A

    Find the voltage in the Thermistor:
    Sum of EMF = Sum of P.D
    5 - 1.8 = 3.2V

    V = IR
    3.2 = 0.000818 x R
    3.2/0.000818 = R
    R = 3911.98 Ω

    Resistance of thermistor at that temperature is 3912 Ω (to 4 s.f.)
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    (Original post by blankboi)
    oh so the book used this method?: (V1/V2 = R1/R2)

    the diagram is drawn correctly and i copied the question word for word :hmmmm:
    Yes they used the ratio method.

    It does appear that the text book is in error.
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    thanks
 
 
 
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