Stumped on a resistance problem

  1. Hello! I have tried for roughly 3 hours now with no success on this one problem. My professor assigns webassign's (homework submittal process) for just review...which isn't graded or anything and I can't seem to get this one

    [I cannot imitate the emf symbol which looks like a capital script E, so I have subsituted that symbol with 'E'.]

    Here is the problem:
    In the figure below, the resistances are R1 = 1.3, R2 = 1.5 , and the ideal batteries have emfs E1 = 2.0 V, and E2 = E3 = 4.5 V.


    The problem asks for the following:
    1.) What is the current through battery 1, 2 and 3
    2.) and What is the potential difference Va - Vb ?

    My answers were:
    Batteries 1, 2, 3, = .74 A, .373 A, .373 A
    Potential Difference = 3.941 V

    Some of the equations I manipulated to get values:
    Vb - Va = E2 - IR2 = R1 + (2R1)(2I)
    I = (e2 - e1 / 4R1 + R2)

    I did the problem in the book which was similar except it had different numbers, and I got that one right. But this one comes out wrong.
    Can you guys tell me what Im doing wrong, and if you are able to solve the problem, what values did you get, and more importantly, how did you get them?

    Last edited: Oct 13, 2005
  2. jcsd
  3. Can't see your image 'cause it's pending there any way you can just host it on imageshack and post up a link??
  4. Ah, sorry about that big man, I wasn't aware that it required authorization. Here you go:


  5. vanesch

    vanesch 6,189
    Staff Emeritus
    Science Advisor
    Gold Member

    The trick is to call the potential difference between a and b, V.
    This leads, for each branch, to an equation:
    I1 = (V - E1) / 2 R1
    I2 = (V - E2) / R2
    I3 = (V - E3) / 2 R1
    (that's the total potential difference per branch over the total resistance over each branch).
    We have one further equation (Kirchhoff): I1 + I2 + I3 = 0.
    This system of 4 equations has 4 unknowns: V, I1, I2 and I3.
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