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Finding a specific current in a circuit using Mesh current analysis

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Develop the mesh-current matrix equation for the circuit.
    http://i.imgur.com/tvZA0.png


    2. Relevant equations

    Ʃ[itex]V_{n}[/itex] = 0

    3. The attempt at a solution

    I used Mesh current analysis to find 4 equations, but the equations for the first and second mesh are not related to the 4th. If I make a matrix equation out of these equations I won't get the right answer.

    50I1 - 20I2 - 20 I3 + 0I4 = 12
    -20I1 + 50I2 - 40I3 +0I4 = 0
    -20I1 - 20I2 +50I3 - 10I4 = 0
    0I1 + 0I2 - 10I3 + 50I4 = 0


    The textbook I'm using has a procedure for using Mesh Current:

    Step 1: identify all meshes and assign each of them an unknown mesh current. For convenience, define the mesh currents to be clockwise in direction.

    Step 2: Apply Kirchoff's voltage law to each mesh.

    Step 3: Solve the resultant simultaneous equations to determine the mesh currents.

    The examples in the book only have a maximum of 3 meshes and they always are touching. I'm not sure what to do in this situation.
     
  2. jcsd
  3. Sep 30, 2012 #2
    This problem is similar to the one I'm trying to solve. When I try to solve this using a matrix I get a different answer. I'm confused.

    http://i.imgur.com/U0W9u.png
     
  4. Sep 30, 2012 #3

    gneill

    User Avatar

    Staff: Mentor

    If meshes don't touch then the entries in the matrix for the shared resistances are zero as you've done. That's fine. It just means that the influence of one such mesh on the other will occur though interactions with other intermediary meshes which do touch.

    Your equations look okay except for the third term in the second equation (-40I3). Recheck that one.

    Note that there is a simple procedure for filling out the resistance matrix directly by inspection:
    1. The terms on the diagonal contain the sum of all resistances in the given loop. So, for example, the entry ##a_{22}## would be the sum of all resistances in the loop i2.

    2. The off diagonal terms ##a_{ij}## for ##i \ne j## are minus the sum of the resistances shared by the loops i and j.

    The voltage vector entries are the sum of the voltage sources for each loop (voltages are summed in the direction of the mesh current).

    There is a natural symmetry to the off diagonal entries of the resistance matrix that you should be able to spot, so the matrix can be filled out quickly by inspection.
     
  5. Sep 30, 2012 #4
    Yeah, that is much easier. Thanks for the help.
     
  6. Oct 1, 2012 #5
    The inspection method doesn't work with circuits that have dependent sources, so I tried using Mesh current on this circuit but I end up with wrong answers on all of the currents. I'm pretty sure my algebra is right.

    http://i.imgur.com/nPBYW.png


    25I1 - 10I2 - 5I3 -2.5I = 0
    25I2 - 10I1 - 5I3 = -10
    35I3 -5I1 -5I2 + 2.5I = 0

    I= 10I1-10I2
    2.5I = 25I1-25I2
    -2.5I = 25I2 - 25I1


    -5I1 + 15I2 - 5I3 = 0
    -10I1 + 25I2 - 5I3 = -10
    20I1 - 30 I2 + 35I3 = 0


    I1 = 10/3A correct answer: -4/15
    I2 = 2/3A correct answer: -8/15A
    I3 = -4/3A correct answer: -2/15A
     
    Last edited: Oct 1, 2012
  7. Oct 1, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    Your mesh equations don't look right.

    You can still write them by inspection, although it's a bit trickier. The controlled voltage source appears as a voltage rise in loop 1, and its value is (5/2)(i1 - i2). In the mesh matrix you'd subtract 5/2 from the i1 entry for loop 1, and add 5/2 to the i2 entry for loop 1 (where 5/2 = 2.5, using whole numbers to avoid decimals during the solving process so as to end up with results the same as your given solution). Do the same for loop 3 which also contains the source, only note that there it's a voltage drop rather than a rise, so add 5/2 to the i1 entry and subtract 5/2 from the i2 entry.
     
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