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Homework Help: Electric circuit with resistors in series

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data

    A battery , ammeter with resistance , 1 ohm and a resistor are connected in series .A voltmeter of 500 ohm is connected parallel to the resistor . The actual resistance of R is 100 ohm , what is the ratio of (the resistance R determined)/(the actual resistance of R) ?

    2. Relevant equations



    3. The attempt at a solution

    Let the emf of the battery be E and resistance of resistor , R

    pd across R = [R/(1+R)]E

    I(500)= [R/(1+R)]E

    and that leads me to nowhere .
     
    Last edited: Mar 27, 2010
  2. jcsd
  3. Mar 27, 2010 #2

    rl.bhat

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    Homework Helper

    Re: electriccircuit

    When the voltmeter is connected across the unknown resistance, what is the equivalent resistance R' of the parallel combination?
    Then find PD across R' and the ratio of R'/R.
     
  4. Mar 27, 2010 #3
    Re: electriccircuit

    ok , the effective resistance , R'=500R/(500+R) and the potential difference across the unknown resistor is

    [500R/(501R+500)]E

    but thats not the resistance .
     
  5. Mar 27, 2010 #4
    Re: electriccircuit

    The effective resistance R' is 500R/(550+R) but the potential difference is ER'/(R'+1). (Don't forget the ammeter resistance.)

    So now you know the potential difference across R' and the current through it. Can you calculate the determined resistance?
     
  6. Mar 27, 2010 #5
    Re: electriccircuit

    For the effective resistance , i got 500R/(500+R) , i don see where is the another 50 coming ..

    and the pd across the uknown resistor ,

    using the concept of potential divider , [500R/(500+R)]/[(500R/(500+R))+1]

    =500R/(501R+500)

    maybe i am wrong here , could yuo point me to my mistake ? and also i do not know the current flowing through the unkown resistor .
     
  7. Mar 27, 2010 #6

    rl.bhat

    User Avatar
    Homework Helper

    Re: electriccircuit

    If you do the experiment to determine the resistance , you use an ammeter of very small resistance and a voltmeter of very high resistance. Ammeter measures the current and voltmeter measures the voltage across the resistance. Then R = V/I. This is the actual resistance. If the voltmeter is of lower resistance, then the measured resistance is smaller than the actual resistance.
    But I can't understand what you mean by
    what is the ratio of (the resistance R determined)/(the actual resistance of R) ?

    Probably the given data is not sufficient to determine the above ratio.
     
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