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Internal Resistance of a Battery

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data
    The potential difference across the terminals of a battery is 8.20 V when there is a current of 1.54 A in the battery from the negative to the positive terminal. When the current is 3.56 A in the reverse direction, the potential difference becomes 9.40 V. What is the internal resistance r of the battery?

    2. Relevant equations

    V_ab = E - Ir

    3. The attempt at a solution

    So I thought that it would make no difference if the charge was reversed. So I just plugged the values into the equations, so since I have two equations and two variables I could solve for r.

    8.56 = E - 1.54r -> E = 8.56 + 1.54r
    9.40 = E - 3.56r -> E = 9.40 + 3.56r -
    0 = -.84 - 2.02r
    r = -.416

    Obviously the resistance can't be negative. What am I doing wrong, does it have something to do with the reversing part of the problem?
  2. jcsd
  3. Oct 25, 2007 #2


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    In your second equation, remember that the direction of the current is reversed. (Just like in a pre-Star Wars science-fiction movie: "Reverse the polarity!")
  4. Oct 25, 2007 #3
    I don't understand why that makes a difference, because won't the change in potential be the same? So what's negative the potential or the amps??? I'm confused.
  5. Oct 26, 2007 #4


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    You are asked to reverse the direction of the current through the battery. In the first part, the current flows from the negative to positive terminal within the battery, so you may write

    8.20 V = E (in V) - (1.54 A) · r (in ohms) ,

    as you did.

    But in the second part, we are told that the 3.56 A current is being applied in the opposite direction (positive to negative terminal inside the battery), so the sign of the current must be switched.
  6. Oct 26, 2007 #5
    Thanks. I guess I need to go back and read up on charges deff. of a Coulomb and such. I knew the potential would be the same.
  7. Oct 26, 2007 #6


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    You should get a small value for the battery's internal resistance. (It's typically a fraction of an ohm in such problems.)
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