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Mesh Analysis / Sign Convention - Conceptual Problem

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data

    I am having a little trouble using mesh analysis. I'm trying to stick rigorously to the passive sign convention, as I'm told that this is sensible.

    Consider the circuit diagram attached, where I have added the mesh currents in orange. In red are the polarities for the components where it is obvious - of course, using the passive sign convention.

    attachment.php?attachmentid=64166&d=1385058228.jpg

    How do I assign the polarities for the other components, i.e. the 6V voltage source and the 5 and 1 ohm resistors? For example, the 6V source cannot be assigned a polarity that satisfies the passive sign convention for both mesh 1 and mesh 3.

    As it turns out, assigning positive to the top and negative to the bottom gives the correct answer (after the mesh analysis), but I'm not sure how you can tell.
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2013 #2

    gneill

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    Staff: Mentor

    Potential drops in the direction of the current flow. Since mesh currents are treated as separate currents in shared components (like the 5 and 1 ohm resistors in your example), there's a separate potential drop associated with each current.

    attachment.php?attachmentid=64170&stc=1&d=1385059766.gif

    You need to account for both changes in potential when you write the equation for a loop.
     

    Attached Files:

  4. Nov 21, 2013 #3
    Thank you gneill, that makes sense. However, I still get the incorrect equations when I try to analyse my circuit, so I guess I'm still missing something.

    attachment.php?attachmentid=64173&d=1385063770.jpg

    I get, for mesh 1, 2 and 3 respectively,

    [itex]2i_1+5(i_1-i_2)+6=0[/itex]
    [itex]\Rightarrow7i_1-5i_2=-6[/itex]

    [itex]12+(i_2-i_3)+5(i_2-i_1)=0[/itex]
    [itex]\Rightarrow 5i_1-6i_2+i_3=12[/itex]

    [itex]4i_3+6+(i_3-i_2)=0[/itex]
    [itex]\Rightarrow -i_2+5i_3=-6[/itex]

    Not sure what I'm doing wrong. :(
     

    Attached Files:

  5. Nov 21, 2013 #4

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, for one thing, in loops 1 and 3, the 6V source has different polarities simultaneously. This can only lead to confusion.

    http://en.wikipedia.org/wiki/Mesh_analysis

    It's best to pick one polarity and adjust the sign of the voltage according to the assumed direction of the current when writing your mesh equations. There is no rule that says voltages must always be positive.
     
  6. Nov 21, 2013 #5
    Okay, but that raises the question: how do I know which way to assign the polarity to the 6V source? I could either assign it to match up with the mesh 1 current (which gives the correct answer) or the mesh 3 current (which gives the incorrect answer).
     
  7. Nov 21, 2013 #6

    gneill

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    Staff: Mentor

    Voltage sources are not passive components like resistors. They have a fixed polarity which you must use. A voltage source always produces the same potential difference across its terminals regardless of the current flowing though it.
     
  8. Nov 21, 2013 #7
    Got it, thanks!

    For the record, the book I'm using doesn't use the passive sign convention. So I've got into the habit of ignoring their labelling and redoing it. So in this example I changed the labelling of the voltage sources!
     
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