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Simple KCL help

Discussion in 'Electronic Basics' started by R.Spinks, Jun 6, 2005.

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  1. R.Spinks

    R.Spinks Guest

    I seem to be unable to properly calculate the proper voltage for what I've
    called Node B. Would someone mind helping me discover where I've turned a
    wrong corner here. Schematic below (top right corner and top of 3Vx
    source -- ie. I drew an ideal wire -- it's the same node : Node B):


    2i
    .--------.
    |dependent
    |current | Node B
    ------------ source +----------------------------,|
    | | | ,-´ |
    | '--------' -´ |
    | ,´ |
    | 5 ohm 2 ohm ,-´ |
    | ___ ___ ,-´ |
    Node A-----|___|--------|___|--------,|´ |
    | ,-´ | |
    | -Vx + ,-´ | |
    | ,-´ | |
    | ,-´ | |
    | ,-´ | |
    .---|----. |´ i | |
    | Indep. | /+\ .-. .-.
    10 A | current| ( ) 3Vx | | | 5 ohm | | 1 ohm
    | source | \-/ \|/| | | |
    | | | '-' '-'
    '---|----' | | |
    | | | |
    | | | |
    | | | |
    | | | |
    | | | |
    | | | |
    |--------------|-----------------|-----------------
    - |
    |
    GND
    (created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)

    Here's the math I did:

    5ohm || 1 ohm = 5/6 ohm
    5ohm in series 2 ohm = 7 ohm
    i=Vb / 5 ohm
    Vb=3Vx

    KCL @ A:
    10+((Vb-Va)/7) - 2i =0
    10+(Vb/7)-(Va/7)-2Vb/5=0
    Vb(1/7 - 2/5) - 1/7 * Va = -10
    -0.25714Vb -0.14285Va = -10 (I simulated the problem and this part is
    correct)

    KCL @ B:
    2i - ((Vb-Va)/7 - (6/5)*Vb = 0
    (2/5)Vb - (Vb/7) + (Va/7) - (6/5)*Vb =0
    (2/5 - 1/7 - 6/5)Vb = -1/7 Va
    Va = 6.6Vb (this is incorrect based on my simulation)

    I expect the answer to be voltage a node b to be 30v and at node a to be 16
    volts. What have I done incorrectly? Thanks.
     
  2. Which term accounts for the current due to the dependent voltage source?

    Jonathan
     
  3. R.Spinks

    R.Spinks Guest

    If you mean additional current contributions (due to the 3Vx voltage source)
    in the legs I've already accounted for in the above equation -- then... I'm
    not sure I know what to do with it. How should I account for it in the KCL
    equation since it's a voltage and not a current.
     
  4. Don't write KCL at a node whose voltage is already fixed by a voltage
    source. You can't, because you don't know the current through the voltage
    source, and you don't need to, because you already know that Vb = 3*Vx.

    Your KCL at A is fine, so you have one equation:

    -0.25714Vb -0.14285Va = -10

    Now Vb = 3*Vx, and Vx is the drop across the 5 ohm resistor, which is equal
    to 5 ohms times the current through that branch:

    Vx = 5*((Vb - Va)/7)
    Vb/3 = 5/7*(Vb - Va)
    0.3333*Vb = 0.7143*Vb - 0.7143*Va
    -0.3810Vb + 0.7143Va = 0

    Now you have two equations in two variables. Solve to get

    Vb = 30 V
    Va = 16 V

    which is what you were hoping for.

    Jonathan
     
  5. R.Spinks

    R.Spinks Guest

    Ah, very good. I kept thinking I had to use KCL for both nodes. That makes
    much more sense now. Thanks.
     
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