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Finding internal resistance

  1. Aug 8, 2010 #1
    1. The problem statement, all variables and given/known data
    http://img408.imageshack.us/img408/6537/screenshot20100808at134.png [Broken]

    2. Relevant equations



    3. The attempt at a solution
    I redrew the circuit as follow:

    http://img293.imageshack.us/img293/6247/screenshot20100808at135.png [Broken]

    Is this correct (or even useful in the first place)?

    Anyway when switch S1 is closed, the total resistance = rx + ry, the internal resistances of the batteries.

    But when both switches are closed, I figured that the voltage increased by 0.02V, to 0.07V. However I am not sure this is the voltage across which components because the millimeter is connected in a rather strange way.

    I'm lost from here onwards. I'm mostly troubled by the placement of the millivoltmeter I guess.

    Am I on the right track? Any alternative method?

    The answer is (A) by the way.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 8, 2010 #2

    ehild

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    The ideal millivoltmeter has infinitely high resistance.

    ehild
     
  4. Aug 8, 2010 #3
    okay i was thinking about that. when switch 1 is closed there is no current, and when both switches are closed the entire circuit system is basically a series circuit because no current flows though the millivoltmeter.

    but i don't know what is the implication of this :tongue: what does the 5.0mV mean? what does the change of 2.0mV mean? any clue?
     
  5. Aug 8, 2010 #4

    ehild

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    When S2 is open the branch wit R just does not exist. Draw this arrangement. It might help. The voltage of what does the voltmeter read?

    ehild
     
  6. Aug 8, 2010 #5
    alright. so 5.0mV would be the potential difference between the two cells. Since there is no current flowing this is equal to the emf difference between the two cells, correct? so i can effectively treat X and Y as if it is a single cell with emf 5.0mV, right?

    when s2 is closed, will the voltage increase by 2.0mV or decrease by 2.0mV? i know there is now current flowing in the branch with S2 but can you elaborate how exactly does this affect the millivoltmeter reading?
     
  7. Aug 8, 2010 #6
    Oh okay i think i get it. So initially the voltmeter reading of 5.0mV is

    [tex]E_{x} - E_{y}[/tex] where Ex and Ey is the emf of X and Y.

    Now after closing the circuit with R, current flows through the internal resistor of X and the voltmeter reading of 3.0mV is

    [tex]V_{x} - E_{y}[/tex] where Vx is the terminal p.d. of X, which is lower than Ex its emf.

    The difference of these two is basically

    [tex]E_{x} - V_{x} = Ir[/tex] where r is the internal resistance of battery

    So 2mV = 5.0(r) and r is 0.004 ohm.

    Am I correct?
     
  8. Aug 8, 2010 #7

    ehild

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    Perfect, except two minor errors: The first voltage reading is 50 mV and the other one differs from this by 20 mV.

    From 2 mv =r (5 A) --->r = 0.0004 ohm, but you get the correct result if you use the original 20 mV. :biggrin:

    ehild
     
  9. Aug 8, 2010 #8
    oh okay that's more of a typo :tongue:

    thanks for your help!!
     
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