How do we solve this super mesh question?

Discussion in 'Homework Help' started by saaketh, Sep 13, 2012.

  1. saaketh

    Thread Starter New Member

    Sep 13, 2012
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    0
    We've been taught the super mesh method of solving circuits in class and we've been given an assignment. I'm stuck on this one question which I know is a super mesh, but I don't know what to do with the current source on the left limb. If it had a resistor in parallel I would have been able to convert it into a voltage source but it has a resistor in series. Also, I cannot ignore the branch as the current source isn't in between two meshes. It is important that we solve this question using the mesh analysis method. Please help me out. A couple of hints would suffice.

    Thank You,
    Saaketh

    [​IMG]
     
  2. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    Try placing the common ground reference at the negative terminal of Vx. This resolves the problem into two node KCLs.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    Do you HAVE to use a supermesh, or can you just use normal mesh current analysis?

    So you've got three mesh currents, right? One of those is for the left-most mesh. Say it is I1 and is assumed to be going counter clockwise. Given the current source in the left branch, which is an outside branch and therefore ONLY has I1 flowing through it, what does I1 absolutely HAVE to be?

    Now you have a mesh current for the top right mesh (call that I2) and one for the lower right mesh (call that I3). The current source that I1 and I3 share forces a relationship between them that provides you with a second equation, but that adds another unknown (Vx).

    Now do the mesh equation for I2 the way you always would.

    Now right the expression for Vx in terms of the mesh currents.

    Note that the coefficient for the dependent current source is not 0.1, it is 0.1 seimens (recipricol resistance).
     
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  4. saaketh

    Thread Starter New Member

    Sep 13, 2012
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    Thanks a lot Wbahn! Really understood it!

    is Vx= 3(I2-I3) or 3(I3-i2)??
    I think its the first one because current passes from right to left through the resistor right?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    That depends on the direction of your mesh currrents. I defined them as gong counter-clockwise, though that was an editorial goof. I meant to have them going clockwise, which is almost always the direction that I use.

    Regardless of which direction you choose, I2 will pass through the 3 ohm resistor in one direction and I3 will pass through it in the opposite direction, which is why the two terms have opposite sign. The voltage polarity for Vx is for a positive current flowing from left-to-right, so whichever mesh current is doing that is added while the other is subtracted.
     
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  6. saaketh

    Thread Starter New Member

    Sep 13, 2012
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    Thank You so much! Helped me out a lot! :D
     
  7. ChrisChemist116

    Active Member

    Mar 13, 2009
    78
    1
    Maybe i came here a bit late, but i suggest you try to read Charles Alexander and Matthew Sadiku's "Fundamentals of Electric Circuits" (refer Chapter 3, Analysis methods), that book saved my day tons of times, plus it covers much of the topics i guess you're looking to learn, their examples are pretty easy to understand and maths used by the authors aren't out of this world (if you know what i mean). If you're looking to master analysis in a fast and easy way that's my advice. Good luck! :)
     
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