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Mesh Method and Node Method Check

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Given the following circuit using mesh's method to solve for the power of the dependent source.
    Using the same circuit and using the node voltage method solve for the power of the dependent source.

    Circuit:
    5nlpw5.png

    2. Relevant equations
    Noting that I have to implement a super mesh for the dependent source to solve the unknown currents. I have arrived to the following equations

    [itex]Vx = 4I_1[/itex]

    [itex]3Vx = I_2 - I_1[/itex]

    The Mesh equation:

    [itex]-4I_1-12I_2+30-8I_1-20 = 0[/itex]

    Simplifying:

    [itex]-12I_1-12I_2+10 = 0[/itex]

    3. The attempt at a solution
    Solving the simulatenous equations by doing various substitutions of

    [itex]Vx = 4I_1[/itex]

    [itex]3Vx = I_2 - I_1[/itex]

    By solving the mesh equations for [itex]I_1 = \frac{10}{12} - I_2[/itex]

    I have arrived at the following solutions

    [itex]I_1 = \frac{5}{84}[/itex]

    [itex]I_2 = \frac{65}{84}[/itex]

    [itex]Vx = \frac{5}{21}[/itex]

    [itex]3Vx = \frac{5}{7}[/itex]

    Checking by Kirchoff's voltage law to see if all voltages within the outter loop it does equate to 0.

    I need a check to see if this is right and I have no idea how to properly implement the node voltage method because the equations do not seem to come out right. I think my trouble is properly putting the correct polarity/sign convention of all devices

    Help! Thanks!
     
    Last edited: Feb 18, 2010
  2. jcsd
  3. Feb 18, 2010 #2
    Sigh bump....
     
  4. Feb 18, 2010 #3
    Dammit! I need this checked by tomorrw morning!
     
  5. Feb 18, 2010 #4

    berkeman

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    Staff: Mentor

    Watch the attitude please. The PF is a volunteer help site, so you won't always get help in a timely fashion. The right way to check your problem is for you to do it the two ways the problem asks you -- if the answers are the same, then you probably got it right.

    Can you be more specific about what part of the KCL equation writing is bothering you?
     
  6. Feb 18, 2010 #5
    Well I am not entirely sure if my sign convention for all devices is correct. And which or how many nodes to pick when implementing the node method and what to use when using KCL for node for the middle branch, would I just put the current down that branch as 3Vx, and I just want to see if my first solutions using Mesh is correct
     
  7. Feb 19, 2010 #6

    berkeman

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    Staff: Mentor

    On your KVL work, I'm not sure I understand why you wrote an equation around the whole outside loop, instead of writing the 2 equations for the left and right loops. Also, be sure to label the +/- assumed voltage drop direction on R3, so you can keep it consistent in later equations.

    For the KCL, I would put ground at the bottom of R4, and write equations for nodes on each side of the V1 source, and the top node at the top end of the dependent current source.
     
  8. Feb 19, 2010 #7
    Thanks
    Im in my originating equations I believe when doing super mesh you "pretend" as if the current source in a common branch is there hence the equation for my mesh.

    Also thanks for the tips on the node method, ill try that out
     
  9. Feb 19, 2010 #8

    berkeman

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    Staff: Mentor

    To be honest, I'm not familiar with the super mesh concept. I'd just write the two loop equations for the KVL. But I hardly ever use KVL, so I'm probably not a great help on that. The KCL should be fine, and a good check against your answer for the KVL.
     
  10. Feb 19, 2010 #9
    Thanks for help, so do I label the polarities of the devices based on the direction of each loop I have chosen or based of the direction of the device?

    I believe super mesh is used for a dependent source or any device shared between a common branch of currents.

    Also would I put a node on top of V2?
     
  11. Feb 19, 2010 #10

    The Electrician

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    Gold Member

    You've got a sign error here:

    The Mesh equation:

    [itex]-4I_1-12I_2+30-8I_1-20 = 0[/itex]

    It should be:

    [itex]-4I_1-12I_2-30-8I_1+20 = 0[/itex]

    And you have a sign error here:

    [itex]Vx = 4I_1[/itex]

    [itex]3Vx = I_2 - I_1[/itex]

    You should have:

    [itex]Vx = -4I_1[/itex]

    [itex]3Vx = I_2 - I_1[/itex]

    Combine them to get:

    [itex]-12I_1 = I_2 - I_1[/itex]

    Now if you solve your simultaneous equations, you should get:

    I1 = 1/12
    I2 = -11/12

    From which the voltage at the top middle node = 19 volts, assuming the bottom of R4 is the reference, or ground.

    For the nodal method, I would move R2 to a position between R1 and the middle node; then the negative terminal of V1 is grounded and the whole thing is easier to conceptualize.
     
    Last edited: Feb 19, 2010
  12. Feb 19, 2010 #11
    Thanks!
    I always have a problem with getting the right sign convention here, I've been working on this for multiple hours trying to find a comprehendable explainable solution to my mess. Being that the original direction gets confused with the direction I am trying to take. How do you know which convention of the device is correct, I can never seem to get them correct. Also one minor detail left.

    To find the power of the dependent source can I simply use the fact that

    [itex]P_ds = I^2R[/itex]

    Which R is the resistance of 6 ohms
     
  13. Feb 19, 2010 #12

    The Electrician

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    Gold Member

    You know the current in R4, so you can calculate the voltage across R4. Then the voltage across the dependent source is (19 - voltage across R4). Then the power delivered by the dependent source is (voltage across the source)*(current through the source).
     
  14. Feb 19, 2010 #13
    I got -1A for the current dependent source, does that seem right?
     
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