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Number of KCL and KVL equations for branch-current method

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  1. Dec 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Find ix and Valpha
    2rdwz7c.jpg


    2. Relevant equations
    KCL
    KVL

    3. The attempt at a solution
    How do I determine the number of KCL and KVL equations? In this problem I am solving, there are 6 unknowns so I need 6 equations. I can write 4 independent KCL equations, 1 KVL at loop 1(to include the Valpha variable) and 1 KVL at loop 4. But these 6 equations result to math error on my solver. Is this because Valpha appeared only once in all equation? Can I also KVL on combined loop 2 and 3 to produce another equation with Valpha?

    TL;DR
    4 KCL equations, KVL at loop 1, KVL at loop 4 = math error
    Should I use 4 KCL, KVL at loop 1, KVL at combined loop 2 and 3?

    Also, in solving circuits using KCL and KVL only(mesh and nodal not allowed yet), is it possible to use different sets of equation to solve circuits?
     
  2. jcsd
  3. Dec 11, 2014 #2

    gneill

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    Staff: Mentor

    How did you write KVL for loop 1? That is, how did you treat the voltage across the ##3 i_x## current source? If you just called the voltage across that whole branch ##v_{\alpha}## then it doesn't take into account the effect of the 10 Ohm resistor in the branch.

    What I would suggest doing is to introduce a new variable, say ##v_x##, as the unknown potential drop across the ##3 i_x## current source. That lets you write an expression for ##v_{\alpha}## in terms of this new voltage and the potential drop across the 10 Ohm resistor. Replace the ##v_{\alpha}## in the controlled current source with this expression, too, so ##v_{\alpha}## disappears for now.

    Then you can write KVL for Loop 1, Loop 4, and around the perimeter of Loop 2+3 (so you don't pass though the 8 A current source and have to introduce yet another unknown voltage). Then write KCL for each node. That should give you a total of 7 equations. With 6 unknowns ##(i_1, i_2, i_3, i_4, i_x, v_x)## you need to pick 6 equations out of the seven.

    When you choose the right one to leave out the resulting matrix will have a determinant that is nonzero. In other words, one of the equations will turn out to not be independent. I find that a reasonable approach is to choose equations so that every column has at least two entries. (Each column represents one of the unknown variables, so you want all of the variables to appear in at least two of the equations).
     
  4. Dec 11, 2014 #3

    The Electrician

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    Gold Member

    What do you mean, "...mesh and nodal not allowed yet.". You are in fact writing mesh equations. If you must avoid mesh (loop) and nodal methods, about the only thing left is to solve branch equations. And, of course, all these methods use KCL and KVL, so saying "solve circuits using KCL and KVL" doesn't imply the use of any particular one of branch, mesh, loop, or nodal methods.
     
  5. Dec 11, 2014 #4

    gneill

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    Staff: Mentor

    Mesh analysis uses virtual mesh currents which do not necessarily correspond to any real current in the circuit. Nodal analysis assumes node potentials with respect to some reference node. If these methods have not yet been introduced then one is left with assigning individual branch currents for writing loop equations (KVL) and summing branch currents at nodes (KCL).

    Usually the "raw" branch current method with KVL loops is introduced first, then the slightly more sophisticated versions of mesh and nodal analysis are built upon them.
     
  6. Dec 12, 2014 #5

    The Electrician

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    Gold Member

    What I'm complaining about is the OP's statement: "Also, in solving circuits using KCL and KVL only(mesh and nodal not allowed yet)". This makes it sound like something more than KCL and KVL is needed for a mesh or nodal solution; does the student think something more is needed? Mesh and nodal analyses can be carried out using only KCL and KVL, so restricting the solution to using only KCL and KVL doesn't rule out those methods.

    I wish the student who asks for help here, and is required to use a particular method to solve a network, would say what that method is. If it's the branch current method, then say he has to use the branch current method. Saying that he has to use KCL and KVL only, doesn't really tell us what method must be used, and certainly doesn't rule out mesh and nodal analysis.

    In addition to helping solve a network, we can also help the student use good descriptive language.

    I wonder if his studies up to this point have made a distinction between a loop and a mesh. Does he know that when he sums the voltages around a loop, he is also summing voltages around a mesh? That's what I meant when I said he is writing "mesh" equations, using the word "mesh" in a general sense, synonymous with "loop" in this case.

    The OP's network is planar, and all the loops are also exterior meshes, so that the "mesh" currents, corresponding to the possibly virtual currents in the formal mesh current analysis method, are all real currents in this case. Not that it's relevant to the branch current method solution of this network, but it's something every student should know about planar networks.
     
  7. Dec 12, 2014 #6

    The Electrician

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    Gold Member

    Note that i4 and ix are linearly dependent. In other words, i4 is carrying the same information as ix. Don't use the KCL equation at node C.

    Write 3 KCL equations at nodes A, B and D.

    Write three KVL equations, around loop 1, loop4, and around the outside of loop 2 and 3. Just keep Vα as Vα.

    The 10 ohm resistor is in series with a current source, so it isn't involved in the solution.
     
  8. Dec 12, 2014 #7
    I see. Why is i4 carrying the same info as ix? I'm a bit confused.

    Also, what I meant was using mesh currents(mesh analysis) and nodal voltages(nodal analysis) were not to be used in this analysis. That's why we had to assign currents for each branch. I'm having a hard time expressing my thoughts in English :)
     
  9. Dec 12, 2014 #8

    The Electrician

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    Gold Member

    If you write the KCL equation for node C, it is "i4 = 8 - 3 ix". If you know i4, you also know ix. i4 is linearly dependent on Ix, and vice versa.

    Have you tried solving the 6 equations I suggested you use?
     
  10. Dec 12, 2014 #9

    Zondrina

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    Homework Helper

    I'd just like to note: You need to apply KCL to ##N-1## nodes, or KVL to ##B - N + 1## loops.

    ##B## are the branches, ##N## are the nodes.

    Many equations can be simplified using the formalisms of nodal and mesh analysis.
     
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