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Homework Help: Kirchoff's rules in circuits

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Use Kirchoff's rules to determine I1, I2 and I3 for the following circuit: https://chip.physics.purdue.edu/protected/Prelab221img/e4pp2.jpg

    E1 = emf = 9 V
    r = internal resistance = 1.5 ohms
    R1 = 5 ohms
    R2 = 15 ohms
    R3 = 22.75 ohms

    Find I1, I2, I3

    2. Relevant equations

    I(R1+R2) = V and Req = R1 + R2+ ... for resistors in series
    I = emf/R1 + emf/R2 + ... and 1/Req = 1/R1 + 1/R2 + ... for resistors in parallel
    Krichoff's junction rule Iin-Iout = 0

    3. The attempt at a solution
    I found I1 to be .741 A by using I=V/R with R being the calculated total resistance: (1.5 + 1/(1/22.75+ 1/20) = 12.14

    Now I've been using various combinations of the equations above to get the other two I's but I can't seem to get a correct answer, can someone point me in the right direction?

    Thanks in advance
  2. jcsd
  3. Feb 12, 2007 #2


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    What exactly were you doing?
    Note that I1 = I2 + I3
  4. Feb 12, 2007 #3
    So if I1 = I2 + I3, shouldn't I2 = I1 - I3 = .741 - 9/22.75 = .345?
  5. Feb 12, 2007 #4


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    Yup, thats correct.
    Last edited: Feb 12, 2007
  6. Feb 12, 2007 #5
    DARN IT, I got .345 for I2 when I first tried the problem, but when I submitted it (online) it was wrong, so I posted here. But when I submit it for I3, it is correct (and .396 is correct when I submit that for I2). I think they must've gotten switched. Thanks for the help though, ranger.
  7. Feb 12, 2007 #6


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    :rofl: I see what you mean. When I double checked them with the current divider rule, I noticed the same thing. Its just a small mix up, thats all.
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