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Mesh analysis and 3 simultanious equations please help!

  1. Sep 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Apply mesh analysis to determine the magnitude and phase of the currents in the circuit of FIGURE 4, given the values in TABLE A

    Vs = 20 Vrms
    Z1 = 0 - j5 Ω
    Z2 = 5 + j5 Ω
    Z3 = 5 + j5 Ω
    Z4 = 0 - j10 Ω
    Z5 = 10 + j0 Ω

    2. Relevant equations

    KVL
    Simultanious equations

    3. The attempt at a solution

    I first highlighted three appropriate mesh's to use in the calculation, these are shown in the attachment. I know that these are appropriate mesh's to use.

    I then calculated an equation for each circuit:

    Green:
    = E - I1Z1 - (I1-I3)Z3 = 0
    = E - (Z1+Z3)I1 +Z3I3 = 0
    = 20 - (5+j0)I1 + (5+j5)I3 = 0

    Pink:
    = E - I2Z2 - (I2-I3)Z4
    = E - (Z2+Z4)I2 - (I2-I3)Z4 = 0
    = 20 - (5-j5)I2 + (0-j10)I3 = 0

    Orange:
    = -I3Z5 - (I3+I2)Z4 - (I3-I1)Z3 = 0
    = Z3I1 - Z4I2 - (Z3+Z4+Z5)I3 = 0
    = (5+j5)I1 - (0-j10)I2 - (15-j5)I3 = 0


    What I need is someone to confirm that what I have done so far is correct. And if so I am stuck at how to solve the equations to find the current and phase. I have tried to be as clear as possible (My paper workings out aren't so...) Thank you in advance
     

    Attached Files:

  2. jcsd
  3. Sep 17, 2013 #2

    gneill

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    Staff: Mentor

    Are you meant to be applying mesh analysis or just KVL/KCL? I ask because the equations you've written don't employ mesh currents but use branch currents instead.

    For KCL/KVL purposes, your Green and Orange equations look fine but take a closer look at the KCL for current through Z4 in the Pink equation.
     
  4. Sep 17, 2013 #3
    I see my mistake now. I wrote the second line of pink completely incorrectly, I must have missed this in my proof reading.
    The current I2 and I3 are in the same direction also and should therefore be added rather than subtracted, correction below:

    Pink:
    = E - I2Z2 - (I2+I3)Z4
    = E - (Z2+Z4)I2 - Z4I3 = 0
    = 20 - (5-j5)I2 - (0-j10)I3 = 0

    The question says apply mesh analysis, but in my guidance material it shows to calculate using KVL, hence me thinking they were the same thing.
    My guidence says, "Once you have three equations, they may be solved conventionally using substitution or elimination, or more readily using matrices."
    I don't fully understand matrices as of yet so was trying to solve conventionally first.
    Any further help would be appreciated.
     
  5. Sep 17, 2013 #4

    gneill

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    Staff: Mentor

    Yes, that looks better.

    Mesh analysis also employs KVL, but rather than assigning unknown currents to branches they are assigned to entire loops. These are called the "mesh currents". Equations are then written using KVL with those mesh currents, then solved by whatever method you choose for simultaneous equations. The current in any given component can then be found by summing the mesh currents that pass through it (components that border two loops will carry two mesh currents, one for each loop).

    Matrix methods are particularly nifty for mesh analysis because you can write down the matrix equation by inspection of the circuit, no splitting and joining currents or worrying about current directions involved. Cramer's Rule is a handy way to solve a matrix equation.
     
  6. Sep 18, 2013 #5
    From what I understand I now have to solve the simultaneous equations. My lecturer provided me with the attached 'complex matrix by decomposition' calculator, so I assume I should be using this to calculate the mesh currents. I have filled it out how I assume it is supposed to be done, however I get strange answers. Should I be using this calculator? If so how do I use it correctly to find the currents and phase?

    I assume that the conversion to polar form gives me the phase angle, but have I filled out the calc correctly? Because I get I3=0A (Or could this be true as the Orange mesh 'I3' does not contain a voltage source?)
     

    Attached Files:

    Last edited: Sep 18, 2013
  7. Sep 18, 2013 #6
    Actually I now think this is a bridge network. Meaning that current will not flow for I3 because Z2 and Z3 are equal. If Z2 and Z3 were not equal then there would be a current for I3.

    Am I correct?
     
  8. Sep 18, 2013 #7
    This last comment is incorrect as for it to be balanced Z1/Z3 = Z2/Z4. So I3 should not be 0.

    What have I done wrong?
     
  9. Sep 18, 2013 #8

    gneill

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    Staff: Mentor

    No, I don't think that's correct. I calculate a non-zero current for I3. You might find the circuit topology more clear if you rearrange the layout to avoid the crossed-over wires.

    My system won't open your attached spreadsheet; it's looking for a password.
     
  10. Sep 18, 2013 #9
    I have made a pdf copy, the grey boxes are the editable boxes. The ones I filled out as per my equations.

    I have also tried re-arranging the circuit but it doesn't really help me.
     

    Attached Files:

  11. Sep 18, 2013 #10

    gneill

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    Staff: Mentor

    It looks to me like you've made a few sign errors in transposing your equation coefficients to the spreadsheet. For example, if I take your Orange equation to divide into real and imaginary parts:

    (5+j5)I1 - (0-j10)I2 - (15-j5)I3 = 0

    (5*I1 + 0*I2 -15*I3) + j(5*I1 + 10*I2 + 5*I3)

    Compare the above with what you've entered.
     
  12. Sep 20, 2013 #11
    Okay, I understand what I have done wrong now.
    I have now entered the info correctly into the calc, which gives me the following for I:

    I1 = 2.15E+00 +j1.23E+00
    I2 = 7.69E-01 +j1.54E-01
    I3 = -3.08E-01 +j1.54E+00

    Now all I need to do to find the phase angle is convert to polar form. Thanks gneill!
     
  13. Sep 20, 2013 #12

    gneill

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    Staff: Mentor

    Those results look fine, and I see the same values for the currents.

    Keep in mind though that, strictly speaking, you haven't applied mesh analysis in solving the problem. Whether or not this is going to "bite" you depends upon what you need to hand in and how the problem is graded.
     
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