double supermesh problem

Discussion in 'Homework Help' started by hunterage2000, Dec 31, 2010.

  1. hunterage2000

    Thread Starter Active Member

    May 2, 2010
    400
    0
    Im having problems figuring out the equations for the attached supermesh problem I've not seen many examples of these type of problems so not sure what I'm doing.

    What I tried to do was:

    supermesh 1

    (R1+R2+R3)m1 - R3m3 = Vs1 - Vs2 - R2Is1 - R3Is1

    supermesh 2

    R3m2 + (R3+R4+R5)m3 = Vs2 + Vs3 + R4Is2 + R5Is2

    Can someone tell me whats wrong?
     
  2. kira12a8

    New Member

    Dec 31, 2010
    2
    1
    With mess current method, you can ONLY use it with independant loops, you can't use it with loops that contain another loop inside it. Wish you good luck man :D
     
  3. hunterage2000

    Thread Starter Active Member

    May 2, 2010
    400
    0
    So you would reduce the circuit so their is just one superloop at the right hand side. Thats what I've done.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I've attached a solution using mesh analysis. There are many ways to do this. I haven't specifically used a supernode technique.
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Using the designations for the mesh currents and the node voltages given by t_n_k in his attachment, we can set up a matrix formulation for the problem. There are two supermeshes, a left one and a right one. With currents in mA and resistances in k ohms, we have:

    \left[ \begin{array}{4}-1 & 1 & 0 & 0 \\5 & 15 & -10 & 0 \\0 & 0 & -1 & 1 \\0 & -10 &  10 & 55 \end{array}\right]\left[ \begin{array}{1}I_A\\I_B\\I_C\\I_D \end{array}\right]=\left[ \begin{array}{1}5\\-5\\\frac{16}{10}\\17\end{array}\right]

    Rows 1 and 3 are the constraint equations, and rows 2 and 4 are the supermesh equations.

    Solving gives:

    I_A = \frac{-541}{120}
    I_B = \frac{59}{120}
    I_C = \frac{-61}{60}
    I_D = \frac{7}{12}

    You get all 4 mesh currents at once.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Using Millman's Theorem allows one to form the formula for the voltage across RL by inspection.

    V_{R_L} =\ \left{ \left[ \Large \frac{\frac{(V_1+R_1I_1)}{R_1+R_2}\ +\ \frac{V_2}{R_3}\ +\ \frac{(I_2(R_4+R_L)\ -\ V_3)}{(R_4+R_L)}}{\frac{1}{R_1+R_2}\ +\ \frac{1}{R_3}\ +\ \frac{1}{R_4+R_L}}\right]\ +\ V_3\ \right}\ \Large *\ \frac{R_L}{(R_4+R_L)}

    I plugged in the circuit values and I came up with the same value for the voltage across RL as t_n_k.

    hgmjr
     
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