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Homework Help: Mesh law in a resistor circuit with current and voltage sources

  1. Feb 17, 2014 #1
    Hello everyone. Hope somebody can help.

    1. The problem statement, all variables and given/known data
    We are given the following circuit.
    http://puu.sh/70uaQ.png [Broken]
    We need to find the currents i1, i2 and i3 by explicitly using the mesh law.

    2. Relevant equations
    There are no relevant equations.

    3. The attempt at a solution
    So, I draw these 4 meshes on the circuit.
    Let's call iA the current in mesh A, iB the current in mesh B, and so on.
    Let's also call v_I0 the voltage at I0 and v_I1 the voltage at I1. For each current source, we assume the current enters at + and leaves at -.

    Considering we already know the current in mesh D, there is no need to build an equation for that one. We only need to build the equations for meshes A, B and C.

    We know that iD = 4A and that iC - iA = 2A, so that iC = 2 + iA.

    I ended up with those equations.

    Mesh A
    12 = 2 * (iA) + 2 * (iA - iB) - v_I0

    Mesh B
    -6 = 2 * (iB - 4) + 2 * (iB - iA) + 2 * iB

    Mesh C
    24 = 1 * iC + v_I0 + 1 * (iC - 4)

    After some algebra, and by substituing iC = 2 + iA, we end up with 3 equations and 3 unknowns in iA, iB and v_I0.

    I end up finding iA = 6.875A, iB = 2.625A and v_I0 = 10.25V, so iC = 8.875.

    Therefore, the currents that we want to know are :

    i1 = iA = 6.875A
    i2 = iA - iB = 4.25A
    i3 = iD = 4A

    The problem I have is that I used an online tool called DoCircuits to simulate the circuit, and by the result of the run, I would be suppose to have :
    i1 = -4.375A, i2 = 5.25A and i3 = 4A
    (Link to the circuit : http://www.docircuits.com/my-circuit/7691/2)

    I therefore think there must be something I did wrong...

    Thanks to anyone who helps me. :)
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 18, 2014 #2
    The voltage across 2(iB+4), not 2(iB-4).
  4. Feb 18, 2014 #3
    Can I ask why ? The way I've put the currents (all clockwise), iB meets iD at R5 in opposite directions, no ? Is it wrong to put the current iD in that direction ? If I take iB+4, would that mean that I also have to take iC+4 and not iC-4 at R3 ?

    And it does not change the fact that I would still have a positive current for iA, when I'm getting a negative one with the sim... Unless I just put the ammeter in the wrong direction too.
  5. Feb 18, 2014 #4


    User Avatar

    Staff: Mentor

    PinguZaide, have you been introduced to the concept of the supermesh yet? If so, you can avoid having to introduce a potential difference variable for the current supply.

    Also take note that the current dource in loop D flows contrary to your defined loop current direction, so iD should be negative.
  6. Feb 18, 2014 #5
    So I tried by switching the direction of the current so that iD is not negative, and found out I made an algebra mistake somewhere, and now it works. :)

    And no I have not been introduced to the concept of supermesh, but I heard of it. Is it simply using a mesh that runs around two or more "independant" mesh? I know my solution now works, but I'm curious as to where I should have put that "supermesh"?

    Thanks for your input guys. :)
  7. Feb 18, 2014 #6


    User Avatar

    Staff: Mentor

    The idea of the supermesh is to write a loop (KVL) equation for a loop that doesn't include the current source, yet includes the components of the loops that it borders. This avoids introducing the current source potential difference as an unknown, but still allows you to "touch" all the components of the circuit. For the circuit under discussion, the following path in green would satisfy that requirement:


    Note that you still write the loop equation using the original mesh currents. The constraint equation relating the individual loop current to the current source is the only additional equation you need, so in this case, iC - iA = 2.

    Attached Files:

    • Fig2.gif
      File size:
      18.4 KB
  8. Feb 18, 2014 #7
    I see. I will remember that for further circuit analysis that I will need to do. Thanks a lot, gneill.
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