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Why does the current take this path?

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    (see attached file)

    2. Relevant equations
    Kirchhoff's laws

    3. The attempt at a solution
    (see attached file) I only included the work for measurement 1 since measurement 2 follows from that.

    I got both voltage values correct. However the paths of the current flow was incorrect for both measurements. (See the attached file for solution) It specifically states that "Arrows to show that the current is only in the left hand loop i.e. not through R3." are needed, but I also included arrows for the RH loop.

    My question: Why is there no current through the RH loop i.e. through R3? Is it because the voltmeter in that loop has an infinite resistance? If yes does this always apply?
     

    Attached Files:

  2. jcsd
  3. Feb 25, 2017 #2

    BvU

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    Direct answer to your question: With V = IR and R 'infinite' you get I = 0

    [edit] a bit more:
    In itself that isn't wrong at all. It's just that the value of the current comes out zero.
     
  4. Feb 25, 2017 #3

    BvU

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    Still more: from the wording 'ideal voltmeter' you can guess that the answer is no. But it's almost true for nowadays digital meters. Old analog meters had a coil in a magnet and needed a little current (e.g. 50 ##\mu##A full scale) which gave them a finite internal resistance (20 k##\Omega##/V in the example) that you had to take into account seriously.
     
  5. Feb 25, 2017 #4
    Would I normally show the path of the current in a loop with an ideal voltmeter even though I=0, I just feel like the current would still travel down that path. The solution specifying that there shouldn't be any arrows at all seems odd to me...
     
  6. Feb 25, 2017 #5

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    Consequence of the fact that the voltmeter must be considered ideal. Solution has an arrow and 0.5 A at left, nothing at right. An arrow and 0 A at the right to me would be a correct answer.
     
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