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Simple Circuit Analysis: Please Help

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Use mesh analysis to find [tex]i_{x}[/tex].

    [​IMG]

    Where [tex]R_{1}[/tex] = 100 ohms , [tex]R_{2}[/tex] = 50, [tex]R_{3}[/tex] =100, R_4 = 220, and [tex]R_{5}[/tex] = 470. [tex]V_{s}[/tex] = 15 V.

    2. Relevant equations

    [tex]V = iR[/tex]
    [tex]\Sigma V=0[/tex]

    3. The attempt at a solution

    First of all, the current orientation of the left ([tex]i_{1}[/tex]) and middle ([tex]i_{x}[/tex]) loops are clockwise, and the right ([tex]i_{2}[/tex]) loop current is counter clockwise.

    Step 1) Finding [tex]i_{1}[/tex] and [tex]i_{2}[/tex]

    [tex]i_{1}[/tex] = 15/100 = .15 A

    [tex]i_{2}[/tex] = 9/100 = .09 A

    Step 2) Mesh Analysis

    [tex]i_{x}*50+(i_{x}+i_{2})*470+(i_{x}-i_{1})*220=0[/tex]

    [tex]i_{x}*50+(i_{x})*470+(.09)(470)+(i_{x})*220+(.15)(220)=0[/tex]

    [tex]740i_{x}+9.3=0[/tex]

    [tex]i_{x} = -.0125 A[/tex]

    *******

    The only thing that is bothering me is that I am getting an negative answer. Is this the correct answer using mesh analysis? Thank you.
     
  2. jcsd
  3. Feb 2, 2009 #2
    I1 is first mesh, I2 second ... (all 3 clockwise)

    sum of voltages around a loop = 0

    -15 + 100I1 + 220I1 - 220I2 = 0
    220I2 - 220I1 + 50I2 +470I2 - 470I3 = 0
    470I3 - 470I2 + 100I3 + 9 = 0

    simplfy and solve using augmented matrix

    I2 = Ix = 0.014371 amps (if numbers are correct)
     
    Last edited: Feb 3, 2009
  4. Feb 2, 2009 #3
    Thanks, but the problem is asking for a mesh analysis.
     
  5. Feb 3, 2009 #4
    ttiger2k7, I re-did the problem using mesh, hope it helps.
     
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