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Help with mesh analysis, please

  1. Oct 6, 2005 #1
    I need help with the following mesh problem. Doing the source conversion screws me up. I really need to get this problem correct and would like to check my answer with somebody who is more competant than myself at mesh analysis.
    Please help asap
    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. Oct 6, 2005 #2
    re

    There is 4 loops, 2 with current sources and 1 with a voltage source, and one with I.

    Label each loop with I1, I2 and I3 with a clockwise direction. You know the current sources already establish the current in the loops.
     
  4. Oct 6, 2005 #3
    The way that the instructor has us solving these is to convert the current sources to voltage sources. Is that not neccesary? If not what are the equations? I do not know how to do it with the current sources. We have been told that they must be voltage sources.
    Thanks
     
  5. Oct 6, 2005 #4

    berkeman

    User Avatar

    Staff: Mentor

    I agree with waht. The circuit is presented with lots of current info, so I would stay with Ii as the middle vector. If your instructor wants you to go to the extra work of redrawing the circuit with voltage sources, then whatever.

    BTW, you've labeled the capacitor as 3 Ohms. Do you mean -j3 Ohms at the frequency of interest? (or I guess 3<-90 in the notation in the drawing)
     
  6. Oct 6, 2005 #5
    yes you are right it is 3<-90 or -j3. My instructor wants us to solve mesh for voltage sources so that is how I have to do it to get credit.
    The mesh equations I came up with after converting sources are.
    I the first curretn source 4a<0 is V1, the second current source is V2 and the voltage source is V3
    V1= 4A<0*j2=8V<90
    V2= 2A<45*4=8V<45
    V3= 3V<90
    Mesh equation
    (j2+4+2)I1 -(2)I2 = V1 - V2
    -(2)I1 + (2 - j3)I2 = V3

    (6+j2)I1 -(2)I2= 6.12V<157.5
    -(2)I1 +(2 - j3)I2 = 3V<90
    Simultaneous equation solved:
    I1= 1.41301<143.804
    I2= 1.44114<172.344

    I through the resistor=I1-I2
    (1.41301<143.804) - (1.44114<172.344) = .70405A<65.855
    Is this correct?
     
  7. Oct 6, 2005 #6
    re

    ok here it is, I2 should be -2<45 A.

    Converting to voltage sources is not the way to do meesh analysis but I showed you convert it anyway.
     
    Last edited: Apr 8, 2007
  8. Oct 6, 2005 #7
    Ok thanks for your help but now I am really connfused. I redrew the circuit after I converted the voltage sources to show the change. I have not seen the method you use that is not what any of the professors use for Mesh. I can almost make sense of what you are saying but I use the simultaneous equations function on my TI-86 to calculate the results. You way looks to me like it would require 4 simultaneous equations. I would rather do it the way the prof. wants to see it but I also want to know the easier way to do it also.
    Thanks
     
  9. Oct 7, 2005 #8
    ok so is it
    (4<0)I1 +(2<45)I2 +(2-j3)I3 -(2)I = 3<90 I3
    (j2)I1 -(4)I2 +2-j3)I3 +(6+j2)I = 0 I
    (4<0)I1 +(4)I2 +(3<90)I3 -(4)I = 2<45 I2
    (j2) +(2<45)I2 +(3<90)i3 -(j2)I =4<0 I1
     
  10. Oct 7, 2005 #9

    mezarashi

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    Homework Helper

    I agree with waht on the method to do Mesh Analysis. You establish "meshes" or currents around each loop; that is, you assume there are currents I, I1, I2, and I3 going around the loops as waht has drawn in his diagram. There are 4 unknowns. If you do a KVL around these loops, you will have 4 equations. 4 equations, 4 unknowns, question solved.
     
  11. Oct 7, 2005 #10
    re

    Yea, you have 4 equations with 4 unknowns, but you already know 2 variables I1 = 4<0 and I2 = -2<45 so the equation simplifies to 2 unkown variable I and I3


    I1 = 4<0
    I2 = -2<45

    I3*(2-3j) - I*(2) = 3<90

    I3*(2) - I*(6+2j) = (4<0 * 2j) + (-2<45 * 4)

    I don't have my calc, so maybe you could run the numbers.
     
  12. Oct 7, 2005 #11
    I1 = 4<0
    I2 = -2<45

    I3*(2-3j) - I*(2) = 3<90

    I3*(2) - I*(6+2j) = (4<0 * 2j) + (-2<45 * 4)

    Would the 3<90 not be -3<90 since the mesh currents are going clockwise which would be current flowing into the positive and out the negative making it -3<90?
     
  13. Oct 7, 2005 #12
    re

    yes it was

    I3*(2-3j) - I*(2) - 3<90 = 0

    hence, I3*(2-3j) - I*(2) = 3<90
     
  14. Oct 7, 2005 #13
    Ok so the result was:
    I3=.343<-128.5
    I=.944<-47.28
    I1=4<0
    I2=-2<45
    So since the two equations that were solved were not accounted for would I the =I+I1+(I2)-I3
    I=.944<-47.28 + 4<0 + (-2<45) - .343<-128.5
    =3.9<-28.135
    ?
     
  15. Oct 11, 2005 #14
    Solved

    Well I was able to solve the problem. I am not sure why our instructors insist on source conversion but I guess it makes it simpler since I do not understand otherwise. The correct matrix is as follows for source conversion.

    V1=4<0*j2=8<90
    V2=4<45*4=8<45

    (6+j2)I1 -(2)I2 I1 V1-V2
    (-2)I1 +(2-j3) I2 V3


    I=I1-I2=1.47<133

    I would be interested too see how to do it with the four loop equations if someone wants to write out the simultaneous equations and solve for I to see if it matches. According to the instructor my answer above is correct.
     
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