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Homework Help: Current Electricity problem

  1. Nov 12, 2006 #1
    The question is:

    A battery drives a current of 3.0 A round a circuit consisting of two 2.0 ohms resistors in parallel. When these resistors are connected in series, the current changes to 1.2A. Calculate:

    a) the e.m.f of the battery

    and

    b) the internal resistance of the battery.


    Here's how far I've got:

    I calculated the effective parallel resistance to 1.0 ohms by doing

    [tex]R=(\frac{1}{2}+\frac{1}{2})^-1[/tex]

    The potential difference in each case is 3V and 4.8V

    So I've put them in a simultaneous equation:

    Let E be the e.m.f, and r be the internal resistance:

    E = 3r + 3
    E = 1.2r + 4.8

    Which can be written as

    3r + 3 = 1.2r + 4.8

    But I can't figure out how to isolate the r values!

    Any help would be greatly appreciated.
     
    Last edited: Nov 12, 2006
  2. jcsd
  3. Nov 12, 2006 #2

    rsk

    User Avatar

    Once you get that far it's just a question of getting your rs on one side and your numbers on the toher....but I don't get teh same equations.


    Apologies - I've mixed up my series and paralle - it should be exactly as you said.

    3r + 3 = 1.2r + 4.8

    So

    3r - 1.2r = 4.8 - 3

    then subs back to find E
     
    Last edited: Nov 12, 2006
  4. Nov 12, 2006 #3
    Aha! Why on earth didn't I think of that... :(

    Anyway, I got the answer now! Thanks!

    3r - 1.2r = 4.8 -3
    1.8r = 1.8
    r=1

    So EMF = 6 and internal resistance = 1

    Thanks!
     
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