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Parallel resistors and ammeter problem

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A student mistakenly connects a perfect ammeter (i.e one of negligible resistance), in parellel with one of the two of the two identical resistors. The ammeter reads of 0.40 A.
    What would the reading have been if he had correctly connected the ammeter in series with two resistors? The cell also has negligible resistance.

    http://upload.lhurgoyf.net/gal/Other/physics0908.JPG
    2. Relevant equations

    [itex] \Delta Q = I\Delta T[/itex]
    Kirchhoff's first law

    3. The attempt at a solution

    At this stage in the book, only the two relevant equations/law have been discussed. The answer is 0.20A but i'm not exactly sure why and how I would get this answer. It seems that it can only possibly be relevant to Kirchhoff's, the sum of the currents entering a junction must equal the sum leaving the junction.

    The current in between the two resistors (before the junction) must be 0.40A then, so it seems that the reading on the ammeter should be 0.40A. I don't think I understand the concept properly, a clue would be good.
     
  2. jcsd
  3. Feb 3, 2009 #2
    Have you considered the current in the first resistor as shown in the diagram?
     
  4. Feb 3, 2009 #3
    You're making this too complicated; as you suspected, you're not seeing the concept. You need only Ohm's Law at this point.
     
  5. Feb 3, 2009 #4
    [itex]I=\frac{V}{R}[/itex]

    [itex]\therefore I \propto \frac{1}{R}[/itex]

    [itex] I_1 = \frac{V}{R} = 0.40[/itex]

    [itex] I_2 = \frac{V}{2R} = \frac{I_1}{2} = 0.20[/itex]
     
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