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Mesh Analysis

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Find IA, IB, IC, I1, I2, V0, VA

    2. Relevant equations

    3. The attempt at a solution
    (V0/10) = IA-IB

    For IA: 10v + 10(IA+IC) + (IA-IB)=0
    11(IA) - (IB) + 10(IC) = -10v

    For IB: -(IB-IA) - 20(IB+IC) - 10(IB) = 0
    IA - 31(IB) - 20(IC) = 0

    For IC: 10(IA+IC) - 20(IC+IB) - 10(IC) = 0
    10(IA) - 20(IB) - 20(IC) = 0

    I then put these equations into a matrix (Sorry I don't know how to put them into a matrix on here) and used an online matrix calculator to get the currents:

    IA= -0.4A

    So then I plugged these values into this formula:
    (V0/10) = IA-IB
    And got V0= -7.3V

    But when I used v=iR to see if I would get the same thing I got

    v=(-29/55)(10)= -5.27v

    What am I doing wrong?

    Attached Files:

  2. jcsd
  3. Oct 8, 2015 #2


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    Staff: Mentor

    Hi jdawg. Are you able to replace your pdf with a simple jpeg?
  4. Oct 8, 2015 #3


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    Staff: Mentor

    Your term: IA-IB is a current, not a voltage. That won't work for a KVL statement.

    The source in the center of the circuit is a current source. It has a value ##V_o / 10kΩ##, which you'll note is a current value. Lying between two meshes as it is, your instinct should be to form a Supermesh to avoid dealing with its unknown potential change in your KVL equations.

  5. Oct 8, 2015 #4
    Oh ok, I retried the problem using super meshes:

    First I did a super mesh around the whole circuit:
    10+10(IC)-10(IB) = 0
    IC = IB - 1 equation1

    Then I did a second super mesh around meshes IA and IB excluding IC:
    10 - 10(IA+IC) - 20(IC+IB) - 10(IB) = 0
    10(IA) + 30(IB) +30(IC) = 10 equation2

    And then around IC:
    -10(IC) - 10(IC+IA) - 20(IC+IB) = 0
    IA +2(IB) +4(IC) = 0 equation3

    I then plugged equation 1 into equation 2 and equation 3 and IB ended up canceling out. I'm not sure what I did wrong this time.
  6. Oct 8, 2015 #5


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    Staff: Mentor

    You only need two mesh equations plus the supernode constraint equation. Writing a third mesh equation doesn't provide new information, it just duplicates other information from the other two equations. That's why you found things cancelling out.

    So, you previously wrote the required constraint equation relating IA, IB, and IC (via the controlled current supply). Use that in place of one of your mesh equations. Usually the approach is to combine the two loops with the shared current into a supermesh, and write regular mesh equations for any other loops.
  7. Oct 8, 2015 #6
    Ok so I kept my super mesh 2 and IC equations. I'm still a little confused, so now I just write an equation for either loop IA or IB?
  8. Oct 8, 2015 #7


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    Staff: Mentor

    You can't write equations for loops IA or IB because you don't have a potential for the controlled current source. So you combine IA and IB into a supermesh as you've done, and then write mesh equations for what's left: that's loop IC.

  9. Oct 8, 2015 #8
    So now I have IA+3(IB)+3(IC) = 1 from the super mesh and IA+ 2(IB)+4(IC) = 0 for the IC loop. How do I solve for the three variables with only two equations? Sorry for asking so many questions!
  10. Oct 8, 2015 #9


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    Staff: Mentor

    You've forgotten the supermesh constraint equation.
  11. Oct 9, 2015 #10
    Thanks so much for all your help!
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