Mesh Analysis Labeling Quesiton

Discussion in 'Homework Help' started by freemindbmx, May 6, 2014.

  1. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    I have two HW problems that I have some questions about. I have to use Mesh analysis techniques to solve for Ix in both separate problems. Im not asking to solve for Ix,I just need help labeling the KVL's for the Loops in question.

    Problem 1) (posted below as problem 1)

    I don't know how to label I3's KVL, I know I2 and I4 are a super mesh, but I cant get the right answer. I know I1 is just the 2A independent source, but for I3 im just lost.

    Problem 2)

    Same kinda of problem with labeling the KVL, for I2 and I3.For 3 would it just be the 2A source, same goesfor I2, I just don't know how to label these.

    Any help would be great.
     
  2. WBahn

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    Mar 31, 2012
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    There's a recent thread on this exact same problem. See if you can track it down and take a look at it, then post your best effort and we can go from there.
     
  3. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    after looking around, you can "apply a KVL on a closed branch to calculate the voltage drops in a conservative field" :p Is what I found. So say for problem one, if we are calculating a voltage drop across elements. In loop three. The voltage across the 6ohm resistor would be 3A*6=18V and the KVL would be(I think): -6I2 + 18v = 0
     
  4. WBahn

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    Mar 31, 2012
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    On what basis are you claiming that the current in the 6Ω resistor is 3A?
     
  5. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    Ive never had a problem like this before in class. The only problems I had that were similar were when we knew for sure the independent current source was the only one effecting the elements in the loop/node. But for this one im at a loss. B/C at first I was trying to visualize the 6ohm and 3A source in parallel hooked up end to end, and thought that I3 would be just the 3A source.
     
  6. WBahn

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    Mar 31, 2012
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    The 6Ω resistor and the 3A source ARE in parallel, but being in parallel only constrains the voltage across each to be the same. It says nothing about the current through each having to be the same -- that constraint would apply only if they were connected in series.

    I3 is due only to the 3A source -- though you need to pay attention to the polarity of each; but I3 is not the only current flowing in the 6Ω resistor, is it?
     
  7. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    the way the 6ohm resistor is set up, it looks like it has I2 going through it as well.But trying to label the direction in which current flows through isn't clear. Should I label the signs (+,-) on the 6ohm to set up my convention? So when I write the supermesh equation to represent the 6ohm resistor's current:

    6Ω(I2-3A)
     
  8. WBahn

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    Mar 31, 2012
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    You get to pick the reference direction for your polarities, then you just have to be consistent in using them.

    The easiest way to do this is to decide, once and forever, whether you are going to sum the voltage drops or the voltage gains as you go around a mesh in the direction of the mesh current for that mesh. After that, always do it the same way. Personally, I always sum up the voltage drops around the mesh because this results in the voltage term for the mesh current for each mesh being positive and the terms for the adjacent mesh currents being negative.

    The thing that trips people up is that the polarity of the voltage across each resistor is not fixed -- it depends on which mesh is being analyzed. So you can't label the polarity on the diagram. Well, you could, but then you would just be doing plain old KVL around a bunch of loops instead of taking advantage of the formalisms that mesh current analysis provides.
     
  9. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    ahhh, so just writ down the polarities and keep the same throughout the whole process of analyzing the circuit.I got ya.... I think.

    So:

    I1= 2A
    I2+I4(Supermesh)= 6Ω(2A-I2)+6Ω(I3-3A)+12ΩI3=0
    AUX Equaiton: 4A=I4-I2
    I3=3A
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Remember what I said about being careful about the polarities.

    If I3 is 3A, then that means that there are 3A flowing in the direction given by the polarity indication of I3. So is this 3A flowing upward or downward through the right-most current source? What direction is the actual current in the source flowing?

    On a notation note, don't use the equals sign unless you mean it. In your supermesh equation you are saying that I2+I4 is equal to something which is then equal to zero. Instead, use a colon or an arrow or some other non-mathematical notation.

    I1: I1= 2A
    I2,I4: 6Ω(2A-I2)+6Ω(I3-3A)+12ΩI3=0
    I3: I3 = 3A
    AUX Equation: 4A=I4-I2

    (NOTE: I'm just reexpressing what you have, not correcting it.)

    You might ask yourself why you are multiplying 12Ω by I3.

    You are making progress.
     
  11. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    I meant to say 12ΩI4,

    I1: I1= 2A
    I2,I4: 6Ω(2A-I2)+6Ω(I3-3A)+12ΩI4=0
    I3: I3 = -3A (since my mesh's direction is clockwise, and 3A is going in the opposite direction)

    AUX Equation: 4A=I4-I2

    And thanks for the help, I try to look through the forums and see which problems I might be able to help on. To try and repay the favors :)
     
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