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Nodal & Mesh Matrix Equations By Inspection.

  1. May 17, 2007 #1
    My circuit analysis professor showed us a method of nodal/mesh analysis today that allows us to write the matrix equations for either nodal/mesh analysis just by inspection of the circuit.

    Is anyone familiar with this method or ever heard of it?

    The circuits that he used this technique on were all resistive and there were only voltage sources present.
    Last edited: May 17, 2007
  2. jcsd
  3. May 17, 2007 #2
    I've heard of it and seen it, but it is not very intuitive and has a lot of rules to remember. If you do it the good and honest way you will make less mistakes, and things will make more sense.

    If you are really dying to see it, I could piece something together.
  4. May 17, 2007 #3
    Yes, Please do!

    I actually like solving nodal/mesh equations the old fashioned way, but the professor is emphasizing this method. He said that he gets paid to show us things that are not in the textbook otherwise "we could stay home and read the d**n book and not come to class"
  5. May 17, 2007 #4
    Do you know of a website that demonstrates this method and lists the rules? I've googled for it but I don't come up with anything.
  6. May 17, 2007 #5
    I don't know of any websites that have the method.

    Node Voltage:
    This is pretty easy to do by inspection, you just look at the node and anything nearby nodes are going to be related by (1/R_node)v_node (make sure to consistently give them similar signs) and the node that you are on will be the sum of all the resistances^-1 around it. Work through an example to confirm this idea. The matrix follows easily from the equations by inspection.

    Mesh Current:
    This is the tricky one. The mesh equations have a symmetrical pattern that is similar to the coefficient symmetry in node equations. Say you have two meshes, mesh A and mesh B: the coefficients of i_b in the and i_a are nagatives of the resistance common to meshes A and B. So the coefficients of i_a and i_b are the sum of the resistances in the meshes A and B.

    In other words, the voltage across resistances in mesh A involves:
    1) i_a times the sum of the resistances in mesh A
    2) -i_b times the sum of the resistances common to mesh A and mesh B, and similar terms for any other mesh adjacent to mesh A

    This is not really all that intuitive, but you can sort of see why it would work.
  7. May 17, 2007 #6
    Just as a note these break down very easily. Any dependent sources kill these methods. Multiple voltage or current sources, such that supernodes or supermeshes are needed, will also destroy the method.
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