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Mesh and nodal analysis

  1. Dec 25, 2014 #1

    M P

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    1. The problem statement, all variables and given/known data
    Please see a picture
    zad2.png

    2. Relevant equations
    KVL

    3. The attempt at a solution
    V1-I1Z1-(I1-I2)Z4=0
    Z2(I2-I4)+Z5(I2-I3)-(I2-I1)Z4=0
    Z3I3+V2+Z5(I3-I2)=0
    V3+Z2(I2-I4)=0

    Any clues are appreciated..
     
  2. jcsd
  3. Dec 25, 2014 #2

    gneill

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    Staff: Mentor

    When you write your mesh equations make sure to first indicate your assumed current directions, then make sure that you are consistent in summing either potential drops or potential rises. I note, for example, that last term in your second equation seems to be inconsistent. The first two terms look like potential drops due to assumed clockwise mesh currents, while the last term looks like a potential rise. Similarly for your last equation, the currents in the second term don't don't match the assumed mesh current directions if the V3 term is taken to be a potential drop for KVL going clockwise around the loop.
     
  4. Dec 26, 2014 #3

    M P

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    Attached direction of currents as suggested.
     

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  5. Dec 26, 2014 #4

    gneill

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    Staff: Mentor

    Mesh currents should be indicated by curved arrows inside the given loop, not in the wiring itself. Otherwise there can be confusion as to the direction on the shared boarders of loops -- which arrow pertains to which mesh?

    Usually it's convenient to have all the mesh currents be either clockwise or counterclockwise (there's a good reason for this, having to do with a very simple way to write the mesh equations in matrix form by inspection which I'm sure you'll cover in your course at some point).

    You seem to have chosen the following:

    I1 : clockwise
    I2 : clockwise
    I3 : counterclockwise
    I4 : counterclockwise

    That's fine, but now you'll need to write out your mesh equations accordingly.
     
  6. Dec 26, 2014 #5

    M P

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    V1 -(Z1+Z4)J1+Z4J2 =0
    Z4J1-(Z4+Z2+Z5)J2+Z5J3-V3=0
    Z5I2-(Z5+Z3)I3-V2=0
     
  7. Dec 26, 2014 #6

    M P

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    V3+Z2(I4-I2)=0 ??
    [
     
  8. Dec 26, 2014 #7

    gneill

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    Staff: Mentor

    Okay, so you've decided that Z2 is irrelevant to the second loop since its potential difference is constrained by V3. That's a good observation! It reduces the number of loops that you have to deal with. However, it means that Z2 should not appear in the second term of your second equation above. Make that change and try solving the system for the mesh currents.
     
  9. Dec 26, 2014 #8

    M P

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    V1 -(Z1+Z4)J1+Z4J2 =0
    Z4J1-(Z4+Z5)J2+Z5J3-V3=0
    Z5I2-(Z5+Z3)I3-V2=0
    V3+Z2(I4-I2)=0

    Sorry I am a bit confused so I have now 4 equations as above with no Z2 in 2nd equation? does it make sense now??
     
  10. Dec 26, 2014 #9

    gneill

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    Staff: Mentor

    You don't need the fourth equation to find the desired ##I## . As I mentioned above, V3 fixed the potential drop across Z2, and I4 plays no role in determining ##I##.

    If you want to look at it another way, recall that components in parallel share the same potential difference, and their order doesn't matter so long as they remain parallel-conected. So you can swap the locations of Z2 and V3 without altering the circuit behavior at all. That puts V3 in the second loop and removes Z2 from it. The current through Z2 is trivially given by Ohms law since you have the impedance and the potential difference, so no need to include loop ##I4## in your set of equations. The circuit then looks like this:

    Fig1.gif
     
  11. Dec 26, 2014 #10

    M P

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    Words are powerless to express my appreciation :)
     
  12. Dec 26, 2014 #11

    M P

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    Trying to find out if I am on the right track:

    I1 = 0.613 - j26.4 A
    I2 = 9.96 - j50.7 A
    I3 = 26.8 - j20.3 A
     
  13. Dec 26, 2014 #12

    M P

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    sorry missed -0.613 -j26.4 for I1
     
  14. Dec 26, 2014 #13

    gneill

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    Hmm. Those don't look like the values that I'm seeing. What method are you using to solve the system of equations? I know it can be a tough slog to do the complex arithmetic by hand... If you look around online you can find "calculators" that do complex math, which you can use to confirm your steps.
     
  15. Dec 26, 2014 #14

    M P

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    So I calculated V1 and V2 and V3(in rectangular)and I put values instead od (Z) afterwards I used a calculator given to us by tutor and all the values with no number I left as zero, that was result I have achieved. Now I am thinking where did that go wrong?
     
  16. Dec 26, 2014 #15

    gneill

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    I can't comment on what I cannot see. So you'll have to find a way to display your steps.

    This calculator that you were given, is it a matrix-type of solver? If so, perhaps you could post your matrix equation (impedance and voltage matrices).
     
  17. Dec 26, 2014 #16

    M P

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    I think I know what went wrong let me try once again I1= 16.8 - j22.9
    I3= 25.9 - j40.2
    I2= 18.1 - j22.1
     
  18. Dec 26, 2014 #17

    gneill

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    Staff: Mentor

    Yes, those results look much better. You should now be in a position to calculate the desired ##I##.
     
  19. Dec 26, 2014 #18

    M P

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    I cannot thank you enough for helping me:)
     
  20. Dec 26, 2014 #19

    gneill

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    You're very welcome.

    Let us know how the nodal analysis version of the problem goes.
     
  21. Dec 26, 2014 #20

    M P

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    yes I will for sure. to be honest with you it has been a bumpy ride with those questions and long days but you are amazing ...
     
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