Solving triple mesh, single source circuit using mesh analysis

Discussion in 'Homework Help' started by Connor Spangler, Sep 2, 2015.

  1. Connor Spangler

    Thread Starter New Member

    Sep 2, 2015
    So this problem has me getting ticked, as I've done it a million different ways and followed similar tutorials to the letter and I'm not getting the correct answers.


    I've assigned the current flow as clockwise for all 3. Here are my equations:
    I1: -10 + 220I1 + 390(I1-I2) = 0
    I2: 270I2 + 470(I2-I3) + 390(I1-I2) = 0
    I3: 560I3 + 470(I2-I3) = 0

    These reduce to the following matrices:
    610 -390 0
    390 350 -470
    0 470 90



    Solving (-A-)^-1(-B-)gives me completely incorrect answers (using nodal analysis + LTspice simulation, the correct currents are I1 = -22.65 mA, I2 = -9.6 mA, and I3 = 4.39 mA).

    What am I missing here? Thank you!
    Last edited: Sep 2, 2015
  2. Connor Spangler

    Thread Starter New Member

    Sep 2, 2015
    For some reason it won't let me edit, here's the correct schematic:
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
    Shouldn't the second equation be 270*I2 +470*(I2-I3) +390*(I2-I1) = 0 ?
    Similar problem in the third eq.
  4. WBahn


    Mar 31, 2012
    Sanity check your equations. Since the mesh current for each mesh is flowing the same way through every component along the mesh, the sign of that current must be the same in all terms -- and, conversely, the sign of the other mesh currents from adjacent meshes must have the opposite sign.

    So in the I1 mesh, I1 should be added and I2 should be subtracted. That passes.
    But in the I2 mesh, I2 should be added while both I1 and I3 should be subtracted. That fails the test.

    You also need to learn to track your units. Your first equation, for instance, should be

    I1: -10 V + 220Ω(I1) + 390Ω(I1-I2) = 0