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Kirchoff's Rules

  1. Mar 5, 2012 #1
    Problem:
    Two automobile batteries are connected in parallel to power a wheelchair. If each of the batteries has electromotive force 12V and internal resistance r=0.025Ohms, and the the wheelchair motor has resistance R=1Ohms, find the current provided to the motor.

    Attempt at solution:
    So I have emf=12 V, r = 0.025 ohms, and R = 1 ohms...

    I set up the current routes such that
    I1 + I2 = I3
    -I1r1 + emf - I3R = 0
    -I2r2 + emf - I3R = 0

    Then when solving for I3, I get r(I2 - I1) = I3 *made one "r" since both are same*

    Where have I gone wrong in setting up the equations?
     
  2. jcsd
  3. Mar 5, 2012 #2

    gneill

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    Can you show how you tried to solve the equations? Just by looking at them I can see that adding your two loop equations together would be promising.
     
  4. Mar 6, 2012 #3
    Actually I'm not even certain now, I think I got tangled between equations yesterday...

    I tried again, just adding both like you said, and ended up with
    -I1r1 - I2r2 + 2emf - 2I3R = 0

    Factored out the r, both I's equal I3,

    Then after moving I3 to one side and solving for it, I eventually got I3 = 11.9, but the answer is supposed to be 11.5. Thoughts?
     
  5. Mar 6, 2012 #4

    gneill

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    Your answer looks better. I find I3 = 11.85 A.
     
  6. Mar 6, 2012 #5
    Yup, I just rounded. So was there a mistake in my equations? I'm not that sure they're right since the algebra seems to be fine..
     
  7. Mar 6, 2012 #6

    gneill

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    Your equations look fine. Could be whoever set the problem rounded down to the nearest .5A (for unknown reasons).
     
  8. Mar 6, 2012 #7
    I'll have to figure what went wrong there.

    I'm confused as to how to set up equations for series (same question but instead of parallel) though, would it be the same equations but signs flipped?
     
    Last edited: Mar 6, 2012
  9. Mar 6, 2012 #8

    gneill

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    No, you'd have a single equation for a single loop.
     
  10. Mar 6, 2012 #9
    So would the equation be
    I1r1 + I2r2 + 2emf = I3R?

    I'm just trying to think that the current would just accumulate across the batteries, right?
     
  11. Mar 6, 2012 #10

    gneill

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    If the batteries and motor are connected in series then there is a single loop and only one current. So change all the various current variables to one variable name.

    The same amount of current flows through each component in a series connection.
     
  12. Mar 6, 2012 #11
    Right, current would be one variable, so what about this equation:
    -Ir + emf - Ir +emf - IR = 0

    then I would just have to find I?
     
  13. Mar 6, 2012 #12

    gneill

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    Yes, that should do it.
     
  14. Mar 6, 2012 #13
    Thanks for your help!
     
  15. Mar 6, 2012 #14

    gneill

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    No problem!
     
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