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Supermesh analysis w/ dependent sources

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Use mesh analysis to find the power delivered by the current-controlled voltage source in the circuit in the Figure: http://i.imgur.com/LUXYFO5.jpg

    2. Relevant equations
    1 super mesh equation, 1 source equation, 1 mesh equation, KVL

    3. The attempt at a solution
    To start off, I made all of my currents flow counter-clockwise and I labeled them, starting from the bottom left loop, then bottom right, then upper left, then upper right loop as: W, X, Y & Z, respectively.

    Next, I drew a super mesh equation around Y and Z (Upper left and upper right). My super mesh equation:

    6*Y-X+8*Z=0

    Next, I wrote down my source equation:

    3=Z-Y

    Finally, I did a regular loop around X (Bottom right):

    6*X+X+32*(X+W/8)=0 (Note, I added W/8 instead of subtracted because I defined all my currents to go counter-clockwise, except, the current coming from the voltage-controlled current source is going the opposite way, so it would technically be 32*(X-(-W/8)). I'm not sure if this assumption is correct. I think my mistake may be there.

    Next, since W is an unknown, I need one more equation. Since W is defined as the voltage drop across the upper left resistor, I wrote the following:

    W=6*Y

    Now, here I am also unsure whether or not this is correct. Since Y is entering the negative terminal of the resistor, or at least I've defined it to be, would that make W=6*-Y?

    So, my two questions are whether W/8 is positive or negative in 6*X+X+32*(X(+/-)W/8) and whether Y should be negative or positive in W=6*(+/-)Y. (As well as my mesh loops, not quite sure I completely have the hang of mesh analysis)

    Thanks for all of your help again!
     
  2. jcsd
  3. Feb 17, 2014 #2

    NascentOxygen

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    Staff: Mentor

    The 6Y is a voltage
    The -X is a current http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken]
    The 8Z is a voltage. Why aren't these three terms all voltages?

    The dependent voltage source has a gain of 15 volts/ampere. Therefore its voltage is 15X. Correct?

    Right.
     
    Last edited by a moderator: May 6, 2017
  4. Feb 18, 2014 #3
    Hey Nascent, that was my error. I don't know why, but I kept ignoring the 15. My equations should be

    Supermesh: 6*Y-15*X+8*Z=0
    Source: 3=X-Y
    Voltage-controlled current source: W=6*7
    Bottom right mesh: 6*X+X+32*(X+W/8)=0

    Do these look correct?
     
  5. Feb 18, 2014 #4

    NascentOxygen

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    Staff: Mentor

    i) right
    ii) no, it's 3=Z-Y
    iii) how did you get 6*7
    iv) wrong

    W is not a voltage drop; W is a current. (EDIT: you noticed W = -Vx/8 but messed up when trying to use this. You have the right idea, but are making careless mistakes.)
     
    Last edited: Feb 18, 2014
  6. Feb 18, 2014 #5
    Thanks for all your help Oxygen, I just noticed I switched up my loop variables when I redid some of the equations, namely ii), which should be 3=Z-Y
    For iii), that should be W=Y*6
    For iv) 6*X+15*X+32*(X-W/8)=0

    Now, why is it 32*(X-W/8) and not plus? I defined my W current to go the opposite direction of X, but since I see the current source, don't I just make W negative? Or should I always just follow my loop direction designation and assume the math will work out in the end? (Just that W will be a negative value?)

    When I solve the system, I get the following: (With 32*(X-W/8))
    W=-19.972, X=-1.5079, Y=-3.398, Z=-0.3298
     
  7. Feb 18, 2014 #6
    I think this should be:

    6*X+15*X+32*(X-W)=0

    The current in the 32Ω resistor is just X-W.
     
  8. Feb 18, 2014 #7

    NascentOxygen

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    Please explain why W = Y*6
     
  9. Feb 18, 2014 #8
    I assume this question is actually directed to rms5643 rather than to me.
     
  10. Feb 18, 2014 #9
    Using Ohm's law V=IR, the voltage drop across the 6 ohm resistor is the current, Y, times the resistance, 6 ohms. I am attempting to use W as Vx in the figure and account for the 1/8 in my X mesh equation since it does not appear anywhere else.

    Would it be more appropriate to assign W=(Y*6/8)?
     
  11. Feb 18, 2014 #10

    NascentOxygen

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    Staff: Mentor

    It would be.
     
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