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Voltmeter reading in a parallel circuit with two batteries

  1. Sep 30, 2016 #1
    1. The problem statement, all variables and given/known data
    In the circuit below the batteries have negligible internal resistance and the voltmeter V has a very high resistance. What would be the reading of the voltmeter?

    ccd53ec942ce.jpg

    Answer: 9.6 V.

    2. The attempt at a solution
    I used the Kirchhoff's rule: the current that flows from 10 V is I1, current that flows to the voltmeter is (I1 - I3) and the current that flows into the 8 V is I3. So I got 10 - 8 = 4 I3 + 1 I1 and 10 = RV (I1 - I3) + 1 I1. But this doesn't look like it can be solved. We have R which I don't what to do with it. Any ideas please?
     
  2. jcsd
  3. Sep 30, 2016 #2

    gneill

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    Staff: Mentor

    The voltmeter has a high resistance (ideally infinite resistance) so it passes no current. What does that tell you about the relationship between ##I_1## and ##I_3##?
     
  4. Sep 30, 2016 #3
    That I1 - I3 = 0?

    I also calculated: I = V / R = 10 / 1 = 10 A and so V1 = 10 * 1 = 10 V and I = 8 / 4 = 2 A, so V2 = 8 V.
     
  5. Sep 30, 2016 #4

    gneill

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    Staff: Mentor

    Sure, or in other words, ##I_1 = I_3##. So there's really only one current:
    upload_2016-9-30_8-44-46.png
    No, you need to take the entire loop into consideration in order to determine the current. Write KVL around the loop and solve for the current.
     
  6. Sep 30, 2016 #5
    Don't we get 10 - 8 = 4 * I + 1 * I → 2 = 5 * I → I = 0.4 A?
     
  7. Sep 30, 2016 #6

    gneill

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    Staff: Mentor

    Looks good.
     
  8. Sep 30, 2016 #7
    And then we have V = I R + 0.4 * 1 = 0.4 V or 0.4 * 4 = 1.6 V. And then either 10 - 0.4 = 9.6 V or 8 + 1.6 = 9.6 V. Should be correct.
     
  9. Sep 30, 2016 #8

    gneill

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    Staff: Mentor

    Yes.
     
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