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The voltage across a voltmeter Olymp 09

  1. Jun 7, 2014 #1
    ImageUploadedByPhysics Forums1402164069.874746.jpg

    The question I'm asking about is Olymp 09

    My attempt : 2.36 V

    Vb1 / (R + r1 + r2) = 2 / 2.5 = 0.8 A
    Vb2 / (R + r1 + r2) = 5 / 2.5 = 2 A

    The reading of the voltmeter = the voltage across r1 + the voltage across r2 and R = (0.8)(2.2) + (2)(0.3)
    = 2.36 V

    I'm not sure of my answer, and even if it's correct I'm not completely convinced how is my solution is right.
     
  2. jcsd
  3. Jun 7, 2014 #2

    rude man

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    The voltmeter sits across a 2V battery. What do you suppose it will read?
     
  4. Jun 7, 2014 #3

    mfb

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    Not 2 V, as we have to consider the internal resistances.

    @ElmorshedyDr: What is meant with "in the opposite circuit [...]"?
    I don't understand your approach to convert those currents to voltages across the 0.3 Ohm resistor.
     
  5. Jun 7, 2014 #4

    rude man

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    Right. Didn't see that.
     
  6. Jun 8, 2014 #5
    So the answer is voltmeter reading = VB + V across the internal resistance = 2 + (1.2)(0.3) = 2.36 V ??
     
  7. Jun 8, 2014 #6

    rude man

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    Correct.
     
  8. Jun 8, 2014 #7

    NascentOxygen

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    Your working is basically correct, but your technique is not what I was expecting. Can you explain the method you have used?
     
  9. Jun 8, 2014 #8

    But I'm still confused, why can't we just calculate the total voltage and the total resistance and then find the total current ad multiply it by the internal resistance of B1 to find the reading, since it is connected in parallel with internal resistance of the battery which would be 0.36V
     
  10. Jun 8, 2014 #9

    mfb

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    Sure, and the result is the same.
     
  11. Jun 8, 2014 #10

    NascentOxygen

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    That can't be correct because the voltmeter is not connected directly across the internal resistance of the battery. The voltmeter reading is the sum of two voltages---you can see how it also captures the battery's EMF.

    The best we can do is use the voltmeter to measure the terminal voltage of the battery, which means it's the scalar addition of two potentials that we measure here.

    I'm stll interested in your explanation of the first method you showed. It is correct, BTW.
     
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