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Current in a loop (Kirchhoff's Laws)

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the following circuit:
    gian1940.gif

    Calculate the currents I1, I2, and I3
    The given resistances are:
    r: 1.05 ohm
    R1: 11.7 Ohm
    R2: 10.5 Ohm
    R3: 7.20 Ohm
    R4: 14.3 Ohm
    R5: 18.8 Ohm

    The given emf's are:
    E1 = E2: 9.0 V
    E3: 3.0 V

    2. Relevant equations
    We know Kirchhoff's laws for current and voltage. Namely, Iin = Iout for any junction, and voltage around any loop must = 0.


    3. The attempt at a solution
    I attempted to use Kirchhoff's voltage laws around the big loop (all the way around the outside), the upper (top) loop, and the lower (bottom) loop. Then, I put these into a matrix, and tried to use the reduced row echelon form (rref) to solve for the currents.

    Big loop:
    E3 - I3r - I3R5 - I1R1 + E1 - I1r - I1R3 - I3R4 = 0

    This is equivalent to: I1(-R1 - r - R3) + I2(0) + I3(-r - R5 - R4) = -E3 - E1

    Top loop:
    E2 - I2r - I2R2 - I1R1 + E1 - I1r - I1R3 = 0

    Equivalent to: I1(-R1 - r - R3) + I2(-r - R2) + I3(0) = E2 - E1

    Bottom loop:
    E3 - I3r - I3R5 + I2R2 + I2r - E2 - I3R4 = 0

    Equivalent to: I1(0) + I2(R2 + r) + I3(-r - R5 - R4) = -E3 + E2

    Resultant matrix
    Code (Text):

    [(-R1 - r - R3)       0      (-r - R5 - R4)  ]  [ I1 ]      [ -E3 - E1  ]
    [(-R1 - r - R3)  (-R2 - r)       0            ]  [ I2 ]  =  [  E2 - E1  ]
    [       0         (R2 + r)  (-r - R5 - R4)   ]  [ I3 ]      [ -E3 + E2 ]
     
    However, the rref of this matrix ends up with a row of all 0's (implying infinite solutions). Therefore, I'm assuming there is a mistake in how I went around the loops and setup the equations. Can anyone help?
     
  2. jcsd
  3. Mar 12, 2009 #2
    Aha! Solved it. I needed one more equation, which I neglected.

    1 -1 -1 = 0 (I1 = I2 + I3)

    with this, the matrix can be solved :)
     
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