Mesh Analysis help with voltage sources

Discussion in 'Homework Help' started by rambam, Oct 18, 2015.

  1. rambam

    Thread Starter New Member

    Oct 18, 2015
    This is my first quarter in university and taking intro to Electrical Engineering. I been doing really good so far with all the problems I've encounter, but this one I just can't solve.
    I am having a hard time finding a relationship for I_o, I simply don't see it.
    I obtained 3 equations from the 3 meshes:

    Mesh i_1: 2(i_2 - i_1) + 8(i_2 - i_3) - 10I_o =0
    Mesh i_2: -16 + 4(i_1 - i_3) + 2(i_1 - i_2)= 0
    Mesh i_3: 4(i_i - i_3) + 8(i_2 - i_3) + 6I_o = 0
    (I_o = 2i_2 - i_3 ) <--- I think that this is wrong or one of the mesh equations is wrong?

    I keep getting the wrong answer, which according to the book as stated in the image is -4A.
  2. WBahn


    Mar 31, 2012
    Which of the mesh currents is/are flowing through the upper-left branch (the same one that Io is flowing in)?

    Your equations don't make much sense.

    Your equation for Mesh i_1 includes an 8 Ω resistor. Where is there an 8 Ω resistor in that mesh? This equation seems to be for Mesh i_2. Similarly, your equation for Mesh i_2 seems to be for Mesh i_1.

    Your equation for Mesh i_3 is apparently summing up the voltage gains as you go clockwise around the mesh, except that the last term is a voltage drop.

    Why do you use Io in the last term for Mesh i_3? What would you have used if I_o had not been involved in the circuit? Would you have used (2i_2 - i3), since that is what you claim I_o is on the next line.
  3. rambam

    Thread Starter New Member

    Oct 18, 2015
    Yes you are right. I labeled them incorrectly sorry for the confusion.

    Mesh 1: -16+4(i_1-i_3)+2(i_1-i_2) =0
    6i_1 -2i_2-4i_3 =16
    Mesh 2: 8(i_2-i_3)-10(2i_2-i_3)+2(i_2-i_1) =0
    Mesh 3: 6(2i_2-i_3)+8(i_3-i_2)+4(i_3-i_1) = 0
    -4i_1+4i_2+6i_3 = 0
    The current flowing through the 6Ω resistor is I_o, thus I have to use the relation ( I_o= 2i_2 - i_3 ) I found for it and substitute.
    By applying what I found I was able to obtain the correct answer for I_o= -4A. Where i_2= -2/7 and i_3 = 24/7.
    Thank you
  4. WBahn


    Mar 31, 2012
    Do any of these mesh equations pass a sanity check? For instance, what if I3 were forced to zero increasing the 6 Ω resistor to infinity?

    Where is this relation (I_o= 2i_2 - i_3) that you "found" coming from? Does it make any sense at all? If I_o is the current flowing through the 6 Ω resistor, then what is the current flowing through that resistor in terms of i3 (which, not to put too fine a point on it, is the ONLY mesh current flowing in that resistor).

    Write your mesh equations in terms of the mesh currents. Use control parameters such as I0 only where required -- namely in those terms for which they act as control parameters.

    Then look at your diagram and write control currents in terms of mesh currents -- you should be able to do this by inspection.
  5. RBR1317

    Active Member

    Nov 13, 2010
    My copy of "Fundamentals of Electric Circuits" by Alexander & Sadiku explains (see attached extract) that a branch current is the algebraic sum of the mesh currents in the branch.

    So how is the I_o branch current the algebraic sum of the mesh currents in the branch?