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Potentiometer and EMFs with internal resistance

  1. Jun 12, 2017 #1
    1. The problem statement, all variables and given/known data

    In a potentiometer 99cm of wire of resistance 99ohms is connected with a battery of 50V and 1ohm internal resistance.Find emf of battery giving zero deflection for a length of 13cm.
    2. Relevant equations
    Sadly,the formula I know is only for ideal emfs,which states that:
    If two batteries with emf E1 and E2 obtain null point at L1 and L2 lengths of the potentiometer wire respectively,then : E1/E2=L1/L2
    3. The attempt at a solution
    The question doesn't mention connecting 2 emfs,just 1 case,1setup,hence the formula I know can't be used here.

    Hence,my problem.

    A potentiometer is supposed to compare emfs based on where their null points are obtained,right?Then,how am I supposed to find the emf with just 1 setup(that too batteries with internal resistance)?
    (Maybe I misunderstood the language of the question.Any other interpretations would be useful)
     
  2. jcsd
  3. Jun 12, 2017 #2

    scottdave

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    It says:

    In a potentiometer 99cm of wire of resistance 99ohms is connected with a battery of 50V and 1ohm internal resistance. Find emf of battery giving zero deflection for a length of 13cm.
    So you have a known battery of 50 volts, and a battery of unknown EMF. The question is, do we assume the unknown battery also has 1 ohm internal resistance?
    That would be a decent assumption to go on (and should be stated in your solution steps.
    Since 99cm is 99 ohms, you have 1 ohm/cm so you could just add 1 cm of virtual wire to simulate the internal resistance. You could do this for both sides so that the potentiometer is now 101 cm long, and the setting is at 14 cm.
     
  4. Jun 13, 2017 #3
    I doubt it.The unknown emf needs to be ideal.
     
  5. Jun 13, 2017 #4

    cnh1995

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    I think there are two batteries. One with emf of 50V and internal resistance 1 ohm and the other with unknown emf.
    You are supposed to find the emf of the unknown battery for which the null point is obtained at 13cm.
     
  6. Jun 13, 2017 #5

    scottdave

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    I just re-read your problem and formulas, and realize now how the setup is: 50 volt battery (1 ohm internal) across the entire pot, then the unknown battery is connected via galvometer or ammeter to the tap point, so when there is no current flowing to the tap, the deflection is zero (and it won't matter how much internal resistance).
     
    Last edited: Jun 13, 2017
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