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Homework Help: Supermesh Analysis

  1. Feb 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the power supplied by the 3-A current source in the network in the Figure using loop analysis:
    Figure: http://i.imgur.com/OcDUuhX.jpg

    2. Relevant equations
    KVL applied to 1 mesh and 1 super mesh, 1 source equation

    3. The attempt at a solution
    -To start off, I defined all my currents to go in the counter-clockwise direction.
    -From left to right, I labeled the loop currents as IX, I2, I3, & I4 (I4=3A)
    -Next, I constructed my source equation:
    -For my IX loop, I wrote down the following:
    IX*4 + 50 + (IX-I2)*4 = 0
    -Considering there is a current source between I2 and I3, I wrote down a super mesh equation:
    (I2-IX)*4 + (I3-3)*5 = 0

    So, that's 3 equations and 3 unknowns, but I know one of my equations is incorrect because solving that system results in ridiculous currents, on the scale of ~20A.

    Where did I go wrong? Thank you.
  2. jcsd
  3. Feb 16, 2014 #2


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    Staff: Mentor

    I would simply say 'let the voltage across the 5 Ohm resistor be V, then use KCL that the sum of the currents into that long top node be zero. Only 2 unknowns are involved, V and Ix.
  4. Feb 16, 2014 #3
    Hey Oxygen, here are my results:

    KCL @ top node: -3+V/5-2X+V/4+(V-50)/4=0

    Relation between V and X: X=(V-50)/4

    Solving this gives me that V=-47.5v, multiplying that by the 3A current source gives me 142.5W which is incorrect. Is my KCL equation incorrect or my V/X relationship?
  5. Feb 16, 2014 #4


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    Staff: Mentor

    I suggested using the sum of currents entering the long top node = 0

    You chose to use the sum of currents leaving the top node = 0
    which is equally legitimate, except you got one sign wrong:
    BTW, "long top" node is too informal for project submisions and examinations :smile:
  6. Feb 16, 2014 #5

    The Electrician

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    Gold Member

    Is this really homework?

    Solving a problem by more than one method is a good way to verify the solution, but what about the restriction given in the problem statement:

    "Find the power supplied by the 3-A current source in the network in the Figure using loop analysis"
  7. Feb 16, 2014 #6


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    Staff: Mentor

    I don't think you went wrong at all! Your loop equations look fine. Currents on the order of ~20 A should be okay. Show your results and we can confirm.
  8. Feb 16, 2014 #7


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    Staff: Mentor

    You raise an interesting point there, Electrician. :redface: Reading the instructions is always recomended. Thank you for pointing out my oversight. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]

    In which case OP's working is all in order, as far as it goes. Just substitute for I3 in the last two equations.

    Checking the answer by solving using an alternative method is always a good idea, too, as you indicate.
    Last edited by a moderator: May 6, 2017
  9. Feb 16, 2014 #8
    Thanks for your help everyone! I consider myself to be pretty good with nodal analysis (my preferred method), but I'm really trying to understand mesh analysis because both methods are incredibly powerful. (Note to Electrician: Yes, this is homework for an introductory 200 level E E course. This is the first E E course most engineers take at my school)

    Anyways, putting my equations into Wolfram Alpha to solve the matrix:
    2X=W-Y, X*4 + 50 + (X-W)*4 = 0, (W-X)*4 + (Y-3)*5 = 0
    (Here I made IX=X, I2=W, & I3=Y)

    W=-36.25, X=-24.37, & Y=12.50

    These should all be in Ampere's correct?
    So, to find the power through the 3A current source, I need to find the voltage drop across the 4ohm resistor on the left. (Current labeled IX), so 24.37*4= 97.5.

    So, the voltage drop between the 50v source and the IX resistor is 100v. Which means that the voltage on the other side of the resistor is -47.5v. So, since the rest of the circuit elements are in parallel, -47.5*3= -142.5w.

    Question: I put 142.5w as my answer (That was incorrect), because I'v-
    Wow, well, I just realized it's -142.5w. That was the correct answer. Thanks for all your help everyone!

    Though, I have a question for you, Oxygen, I also think it's important to know how to figure out circuit multiple ways confidently, so you mentioned that: "KCL @ top node: -3+V/5-2X+V/4+(V-50)/4=0" was incorrect because I made my last current positive. I'm a little confused here. I made it positive because I thought IX was leaving the node, as shown in the figure.

    So, my question is: Let's say in my example I chose, again, for all the currents to leave the node. I know that the two current sources are going into the node, so they're negative, and I don't know the current through the lower two resistors, so I left those to be positive, but I figured the left-most resistor leaving the node to also be positive, coincidentally aligning with the direction of IX in the picture.

    How do you know IX is going into the node? Thank you!!! Edit: Actually, are you sure that's correct, because I just realized that I get V to be equal to -47.5 when I make IX leave the node, which is the voltage I got using loop analysis?

    (Yes, I know "long top node" is definitely inappropriate for formal stuff, thank goodness it's just homework :D)
    Last edited: Feb 16, 2014
  10. Feb 16, 2014 #9


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    Staff: Mentor

    Stay strictly to the polarities and directions marked, and the answer appears in one step:

    With the nomenclature of the figure, top node voltage Vn = 50 + 4.Ix

    Vn = 50 + 4(-24.37) = -47.48

    Back to nodal analysis, mark throughout node voltages and current directions, then sum the currents into each node. The lower node I assumed to be ground node, of zero volts, and the top is at V volts. Current down through the 5 ohm is (V-0)/5. Current up through the 5 ohm is given by (0-V)/5, because for current to flow up it means the lower end must be at the higher potential (current flows from a point of higher potential to a point of lower potential).

    When summing currents leaving the node, you involve the current flowing down through the 5 ohm.
    Last edited: Feb 16, 2014
  11. Feb 17, 2014 #10
    Hey again Nascent! Thanks for your help again, my equation -3+V/5-2X+V/4+(V-50)/4=0 you listed as incorrect because of the last term +(V-50)/4. I understand how the current through the 5ohm and 4ohm resistors are flowing downwards (V-0)/5 & (V-0)/4, but I don't understand why (V-50)/4 should be negative. Wouldn't that mean I am saying the current through the 4ohm resistor connected to the 50v is flowing *into* the node? Doesn't leaving it positive mean it's leaving the node, in the direction of IX in the figure?
  12. Feb 17, 2014 #11


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    Staff: Mentor

    Ah, your equation is correct as you wrote it, with +(V-50)/4

    I see in my working I'd used +(50-V)/4 and this requires the negative sign.When I saw you hadn't changed the sign there although you'd changed to current leaving I flagged it as an error. Apologies, the error was mine. (Keeping one eye on the olympics and not paying full attention here, it seems.)
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