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NTC thermistor in a circuit

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  1. Oct 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A student is provided with a 2 V cell, a lamp, a switch and a thermistor with a negative temperature coefficient of resistance. The lamp, which is in series with the cell as in Figure 1, lights immediately the switch is turned on.

    eb7e837356dc.jpg

    Explain why (a) when the thermistor is connected in series with the lamp, as in Figure 2, and the switch is turned on, the lamp lights up slowly, and (b) if the lamp is omitted, as in Figure 3, and the switch is turned on, the cell is soon destroyed by overheating.

    2. The attempt at a solution
    We have a NTC thermistor, whose resistance decreases with the increase in temperature. I'm still not sure whether I got the usage of this apparatus. It is used in order not to create large voltage rises for the devices like lamps. Is it correct?

    (a) If my understanding of this devide is corrrect, than the lamp lights slowly because the thermistor slowly passes the current from the 2 V cell.

    (b) If the resistance of the thermistor decreases with the increase in its temperature, then over time it will have no resistance and the battery will ovearheat itself?

    Where am I wrong / correct?
     
  2. jcsd
  3. Oct 7, 2016 #2

    cnh1995

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    Homework Helper

    Before the battery gets overheated, what will happen to the thermistor itself?
    Right. Can you say why?
     
  4. Oct 7, 2016 #3
    Shouldn't it become an ammeter? An ammeter has no resistance, and also will be a thermistor. No?

    Because the thermistor is designed to keep other devices from getting too much voltage?
     
  5. Oct 7, 2016 #4

    cnh1995

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    Homework Helper

    Yes. That is correct. I actually misread the OP and thought the thermistor itself will get burned before the battery because of increased temperature, but we can't say that without knowing more about the thermistor.
    Initially, thermistor resistance will be in series with the lamp, which will create a voltage divider and the bulb will get a reduced voltage. So it will light up slowly.
     
  6. Oct 7, 2016 #5
    (a) question:
    Alright, so for the first question let's sum up. The resistance of the thermistor will be in series with the lamp and will divide the voltage between the thermistor and the lamp and therefore the lamp will get a reduced voltage. And since the thermistor loses it's resistance over the temperature, the lamp will get more and more voltage until it will have a 100 % light.

    (b) question:
    The cell will be destroyed by overheating because the thermistor will loose its resistance with the increase in temperature and therefore it will become an ammeter subsequently. That's the answer?

    And also is something like this correct: since the current will only go through the battery so this is the reason why the cell will be destroyed?
     
  7. Oct 7, 2016 #6

    gneill

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    Staff: Mentor

    It can't actually "become an ammeter". It won't suddenly grow a meter movement and test leads. It may share an important characteristic with ammeters, a low resistance, but that's as far as its "ammeterness" goes. A better comparison would be that it becomes more like a wire or small value resistor.
     
  8. Oct 7, 2016 #7
    And the battery will be destroyed by overheating because it has no resistance in the circuit?
     
  9. Oct 7, 2016 #8

    gneill

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    Staff: Mentor

    Yes. As the load resistance gets smaller the current in the circuit rises, and so too must the power dissipated by the battery's internal resistance. This heats the battery. If the load resistance were to go to zero, the only limiting factor for the current would be battery's own internal resistance. As the battery gets hotter it's chemical reactions quicken, causing more current, more heat, and so on until the battery destroys itself.
     
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