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Stuck on a single loop circuit, two emfs and resistor

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data
    ε[1]=12v
    internal resistance[1]=.016Ω
    ε[2]=12v
    internal resistance[2]=.012Ω

    some unknown resistance 'R' in a stand alone resistor.

    a) what value of 'R' will make the terminal-to-terminal potential difference of one of the batteries zero?
    b) which battery would it be?
    (picture below shows circuit)
    2. Relevant equations
    R = V/i

    3. The attempt at a solution

    I put together this equation:
    ε[2] - ir[2] + ε - ir[1] - iR = 0
    and continued to simplify to this:
    i(.028 + R) = 24
    The problem is I don't have a way to find 'i'...
    Any ideas? This problem is marked as a three dot (harder difficulty) in my book, so I'm guessing there's an assumption I'm missing.

    Thanks in advance,
    Caleb
     

    Attached Files:

  2. jcsd
  3. May 3, 2012 #2
    What would short circuit current be, i.e. R = 0?
     
  4. May 3, 2012 #3

    rude man

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    Gold Member

    Why solve for i?

    What is the equation for voltage across E1 to be zero? E2?

    Could you come up with 2 equations with 2 unknowns (i and R) for either of those?
     
  5. May 3, 2012 #4
    Alright, I think I got it.

    I set V1 = ε1 - ir1 = 0
    and got ε1 = ir1
    and also
    ε2 = ir2

    Subbing 'i' into the equation I had simplified to above produces, for emf #1, I get R to equal .004Ω (which is what the back of the book says). Doing the same for emf #2, I get -.004Ω.

    My only question now is about the negative. Doesn't a negative value on a resistor mean it encourages the flow of charge? Wouldn't that mean that the battery set to zero potential could be either one, just depending on the sign of R? (The book says battery one is the answer to part b.)

    Thanks,
    Caleb
     
  6. May 4, 2012 #5

    rude man

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    For you, there is no such thing as a negative resistor, certainly not a dc negative resistor. So your only valid solution is for E1.

    You've run into this sort of thing many times before, even in high school. For example, Pythagoras says c^2 = a^2 + b^2. But that equation has two solutions for c, and you know one of them is nonsensical.
     
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