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Calculating the voltage over a resistance with and without voltmetre

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Here's a little drawing I made with all the info. In case my drawing is unclear. R1 is 100 kilo omhs
    R2 is 100 kilo omhs as well. The current is 36,5 uA over the digital Ampere meter , And 0,4 Volt over the analog volt meter

    2. Relevant equations

    U= I x R
    In a parallel currentflow U1=U2=U3
    In a serieflow I1=I2=I3

    3. The attempt at a solution

    I'm completely baffled by this one. They're asking for the voltage of the resistance, but which one? Do they want me to add them together and are we even allowed to do that?
    So R1+R2=R3 Which is 200k omh, but we still don't know the current over the resistance.
    But we know the voltage over the voltage meter which is 0,4 Volt so the voltage over R3 is 0,4 V as well. Using I=U/R The current over R3 Is 0,000002 A. So the voltage of the resistance with voltmeter is 0,4 V, but how do you calculate it when you leave the voltmeter out of the picture?
     

    Attached Files:

  2. jcsd
  3. Sep 29, 2008 #2

    Defennder

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    Homework Helper

    I think one way you can solve this is by interpreting the ammeter as a current source of that given value, and the 0.4V as a voltage source.
     
  4. Sep 30, 2008 #3
    So the voltage over the resistance with voltmeter is 0,4 V and without 7,3 V (200k omhx36,5uA), but then the voltage over the resistance is larger than the battery??
     
  5. Nov 20, 2008 #4
    Hey!

    Your drawing doesn't make any sense. None of those numbers follow Ohms law and you would never insert an ammeter in series with a voltmeter like that.

    What is the exact wording of the question?

    Remember you can figure out any series or parallel circuit using Ohms law...

    Voltage (V) = Current (I) x Resistance (R)

    or

    Current (I) = Voltage (V) / Resistance (R)

    or

    Resistance (R) = Voltage (V) / Current (I)


    See what I mean? Following Ohms law, your numbers are meaningless...

    AdeptRapier
     
  6. Nov 20, 2008 #5

    berkeman

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    Staff: Mentor

    I agree that the diagram is wrong. Could you please check the problem statement again and clarify? The only way they'd put an ammeter in series with the voltmeter in a circuit question like that, would be if the input resistance of the voltmeter were specified...
     
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