1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mesh analysis

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Use mesh current analysis to determine I1

    2. Relevant equations
    Loop 1

    47i1+j100(i1+i2)-12=0
    (47+j100)i1+j100i2=12

    Loop 2
    -j75i2+j100(i1+i2)-10=0
    j100i1+j25i2=10

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations

    j100i1+j25i2=10 (-4)
    -j400i1-j100i2=-40

    (47+j100)i1+j100i2=12
    -j400i1+(47+j100)i1=-38
    j400i1-i1(47+j100)=38
    (-47-j300)i1=38

    Is it correct?
     

    Attached Files:

    • f1.jpg
      f1.jpg
      File size:
      24.8 KB
      Views:
      97
    Last edited by a moderator: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    A slight problem with the voltage value indicated above.
     
  4. Nov 13, 2013 #3
    40-12=28
    Well spotted
     
  5. Nov 13, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Right. Now finish, isolating i1 and writing it in a standard form.
     
  6. Nov 13, 2013 #5
    Thanks.I decided to use matrices to solve this problem.I hope matrices is correct
     

    Attached Files:

    • m1.jpg
      m1.jpg
      File size:
      13 KB
      Views:
      67
  7. Nov 14, 2013 #6
    For i1 I have -0.0557<84.63deg

    Is it correct?
     
  8. Nov 14, 2013 #7

    gneill

    User Avatar

    Staff: Mentor

    No, it doesn't match what I find.
     
  9. Nov 14, 2013 #8
    (47+j500)i1=-28
    i1=(-28)/(47-j500)=(-28)/(sqroot(472+5002))tan -1(500/47)=
    -28/502.2<84.63deg=-0.05557<84.63deg
     
  10. Nov 14, 2013 #9

    gneill

    User Avatar

    Staff: Mentor

    Show your algebra beginning from where you left off here in your first post:

    ##j400 i_1 - (47 + j100) i_1 = 28##
     
  11. Nov 14, 2013 #10
    For mesh A
    -12+47i1+j100(i1+i2)=0
    47+j100i1+j100i2=12
    (47+j100)i1+j100i2=12

    For mesh B
    -10-j752+j100(i1+i2)=10
    j100i1+j25i2=10

    Therefor
    (47+j100)i1+j100i2=12
    j100i1+j25i2=10 (x(-4))

    (47+j100)i1+j400i1+j100i2-j100i2=12-40
    (47+j100)i1-j400i1=-28
    47i1+j100i1-j400i1=-28
    (47-j300)i1=-28
     
  12. Nov 14, 2013 #11
    i1=-0.922<80.09deg

    hope is correct
     
    Last edited: Nov 14, 2013
  13. Nov 14, 2013 #12

    gneill

    User Avatar

    Staff: Mentor

    The final expression is okay.

    The final expression is okay.

    I think you tried to subtract the mesh B equation from the mesh A equation, but didn't change the sign of the i1 term.
     
    Last edited: Nov 14, 2013
  14. Nov 14, 2013 #13
    j100i1+j25i2=10 multiply by (-4)
    -j400i1-j100i2=-40
    then I add equation for mesh A to B
    (47+j100)i1-j400i+j100i2-j100i2=12-40
    +j100i2-j100i2=0
    Therefor
    (47+j100)i1-j400i1=-28
    47i1+j100i1-j400i1=-28
    47i1-j300i1=-28
    (47-j300)i1=-28
     
  15. Nov 14, 2013 #14

    gneill

    User Avatar

    Staff: Mentor

    Okay, that looks better. Continue.
     
  16. Nov 14, 2013 #15
    i1=-=-28/303.65<80.096=-0.0922<80.096

    As well should I mention that I have used KVL to solve this problem
     
  17. Nov 14, 2013 #16

    gneill

    User Avatar

    Staff: Mentor

    You should incorporate the sign of the numeric part into the angle since the numeric part is the magnitude of the complex value, and magnitudes are always positive (absolute) values. Your angle appears to be off by one degree... typo?

    Yes, that was clear from your equations.
     
  18. Nov 14, 2013 #17
    i1=-28/(47-j300)=-28/sqr (472-3002)tan-1(-300/47)=-28/303.65<80.096=0.093<-81.10deg

    10=j100i1+j25i2 =100x1<90x0.093<-81.10+j25i2
    10=9.3<(90-81.10)+j25i2
    10=9.3cos(9)+j9.3sin(9)=9.19+j1.439+j25i2
    10=9.19+j1.439+j25i2
    -0.81+j1.439=-j25i2
    1.651<-0.94/-j25i2=1.651<-0.94/25<90=0.066<-90.94
     
    Last edited: Nov 14, 2013
  19. Nov 14, 2013 #18

    gneill

    User Avatar

    Staff: Mentor

    The angle you've found for the denominator doesn't reflect the sign of the argument of the arctan function. Note that the "-" is associated with the numerator of the argument, -300. So the angle should end up in the [STRIKE]third[/STRIKE] fourth quadrant. Also the angle value is off by a degree (typo?).

    Next, the negative sign on the numerator (-28) has disappeared. It should get rolled into the final angle... think of the numerator as 28 ∠ ±180°. Or, perhaps more easily, make the numerator positive to begin with (multiply top and bottom by -1 before you begin).
     
    Last edited: Nov 14, 2013
  20. Nov 14, 2013 #19
    Calculated differently:

    (-28/47-j300)(47+j300)/(47+j300)=(-1316-j8400)(2209-j290000)=(-1316-j8400)/92209
    (-1316/92209)-(j8400/92209)=-0.0142-j0.091=0.092<98.86deg
     
  21. Nov 14, 2013 #20

    gneill

    User Avatar

    Staff: Mentor

    Good, but the angle should be negative. The two components of the current are both negative, putting the angle in the third quadrant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Mesh analysis
  1. Mesh analysis (Replies: 2)

  2. Mesh Analysis (Replies: 1)

  3. Mesh analysis (Replies: 25)

  4. Mesh Analysis (Replies: 9)

Loading...