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Best choice of KVL and KCL in a circuit

  1. Oct 6, 2012 #1
    Hi everyone! Since this is very confusing for me, i wanted to know what is the best choice to take in choosing the kvl and kcl.
    The questions are ?

    How many KVL and KCL i should take ?? ( best number, without redundance)
    And is there a way to see if a kvl or kcl is linear dependent on another equation and i could leave the linear equation ?
     
  2. jcsd
  3. Oct 6, 2012 #2

    tiny-tim

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    hi nebbione! :smile:
    technically, the number of independent loops is the minimum number of wires that you'd need to cut to remove all loops

    in practice, it's usually the obvious number …

    eg the letter "B" has three loops, but only any two are independent (and KVL for two of them will add to make KVL of the third, so choose the two most convenient ones!) :wink:

    (KCL: i think you need a separate equation for each node)
     
    Last edited: Oct 6, 2012
  4. Oct 6, 2012 #3
    Ok but can you explain me better.

    I mean, i now know that the number of KVL is the minumum number of wires that i would cut to have no loops.

    But i didn't understand about KCL.
    Can you make me some examples or explain better ?

    Anyway thank you!
     
  5. Oct 6, 2012 #4
    Almost.

    For each loop you cut you can obtain one independent equation.
    The 'number' is the number of independant equations you can obtain by cutting all the loops.

    As, I'm sure you know, you need as many independent equations as unkwnowns to solve the system.

    KCL is simply the law of conservation of charge at any point.
    "What goes in must come out."

    So if we apply it at any junction (node) we have

    Sum of all currents entering the junction = Sum of all currents leaving the junction.

    So you can obtain as many equations as there are nodes, but they will not all be independent.

    You can also mix KVL (loop equations) and KCL (node equations) to obtain solutions.
     
  6. Oct 6, 2012 #5
    Yes but i mean in a circuit analysis i should only take the KVL ?
    I remember thath i should construct a matrix that is containing both KCL and KVL is that right or wrong ?
    In this case wht's the number of KCL i should write ?
     
  7. Oct 6, 2012 #6
    I think it would be a good idea to understand what you are doing before you start to write matrices.

    What are you going to do with those matrices?
     
  8. Oct 6, 2012 #7
    i write matrices to solve for current and voltage for each branch and solving the system
     
  9. Oct 6, 2012 #8
    Sorry, i reflected about it, and my question is :

    ASSUMPTION:
    If i know that in a circuit there are n=4 nodes and b=6 branches, we can get (n-1) independent node equations and [b-(n-1)] independent loop equations, so in this case i will have 3 independent node equations and 3 independent loop equations.

    So my question is how can i choose independent equations ? In the beginning i don't know which of the equations are independent ? How can i get to know which are independent ??
     
  10. Oct 6, 2012 #9

    tiny-tim

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    i think the only problem is with the loop equations …

    it should be fairly obvious

    eg with 3 independent loops, the circuit will look like three boxes side-by-side

    you could choose the left-hand pair, the right-hand pair, and any other loop except the centre box :wink:

    (you'd always be safe if you chose the "smallest" loops, but that might not be the most convenient!)
     
  11. Oct 6, 2012 #10
    what do you mean by right-hand pair and left hand pair ?
     
  12. Oct 6, 2012 #11

    tiny-tim

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    the outline of the two boxes on the right, and the outline of the two boxes on the left :wink:
     
  13. Oct 6, 2012 #12
    Do you realise why you cannot solve using KCL alone?
     
  14. Oct 7, 2012 #13
    yes i know that i must use kvl and kcl the question was which are independent of all kcl and kvl. i 've understood about kcl but have some problems sometimes to choose the correct kvl, but i think i've understood now
     
  15. Oct 7, 2012 #14
    Well I hope so, I'm only trying to help.

    Just remember that Kirchoff's theorems are only a means to an end.

    They provide a formal and efficient way to analyse some circuit networks. There are other methods as well.

    Analysing circuit networks is really about solving a system of simultaneous equations.

    To do this you have to

    1) Have the same number of equations as unknowns
    2) Include at least one instance of each of the required unknowns in at least one of these equations.
    3) Ensure that none of the equations must be a simple multiple of any other equation.
    4) Ensure that you cannot form any of the equations by a simple combination of the others.

    The answer to your original question and the others asked in this thread lie in knowing and applying this.

    KCL cannot be used by itself because it contains no voltage terms so you cannot fully solve a network with any number of KCL equations.
    You can obtain voltage drops from Ohm's law, but that does not tell you generator (battery) voltages.

    The answer to the question which equations to choose is that you choose a set of those which contains all the unknowns you want to solve for (you may be lucky, you may have to solve for others as well along the way).

    Some of the equations in your set may be KVL, some may be KCL, some may be Ohm's. That is fine.

    It is sometimes possible (small correction Tim:wink:) to pick linearly dependent (conditions 3 & 4) equations from both the possible KVL and KCL set. The maths will soon let you know if you have done this as the system is then insoluble.

    Often KVL leads to fewer equations so many prefer it however judicious use of KCL can significantly reduce the calculation effort.
    Spotting where this can be done comes with practice.
     
  16. Oct 7, 2012 #15

    es1

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    Is this true? Do you have an example of a circuit that requires KVL to be solved? I totally agree when solving circuits by hand using both can save a lot of work and time but I didn't think using both was strictly required.

    I ask because I was under the impression spice only uses KCL when solving circuits. But admittedly, it does it numerically. So maybe that matters somehow... Just curious.

    http://www.emcs.org/acstrial/newsletters/summer09/HowSpiceWorks.pdf
     
  17. Oct 7, 2012 #16
    hello. es1

    KCL is an equation containing only currents.

    Can you tell me how to solve it for voltage?

    I would be most impressed.
     
  18. Oct 7, 2012 #17

    es1

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  19. Oct 8, 2012 #18
    @es1

    I can solve circuits quite adequately, thank you.

    However you seemed to doubt my statement that an equation involving only currents cannot be solved for voltages.

    If this is not your doubt what was your real question?
     
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