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Nodal Analysis

  1. Jul 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Refer Attachment

    2. Relevant equations

    3. The attempt at a solution
    I have drawn the current direction in green. Are there any rules regarding the direction of the current?
    I am at a bit of a loss regarding the horizontal current through the middle. The best I could come up with is: (V1-V3)/6-V3/3 that is through the 5i voltage source and (V1-V3)/6-V3/3-V2/4 through the 10V voltage source.
    For the 10V supernode I have the following equation from KCL: 6V1+6V2+4V3=0. KVL: V1=10+v2.
    For the 5i supernode I have the following equation from KVL:V3-V2-5i=0 and i=V1/2.
    Therefore V3-V2-5V1/2=0.
    Using these equations I get V1=3.85V, V2=-6.15V and V3=3.45V.

    Any help would be appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Jul 20, 2011 #2

    The Electrician

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    Gold Member

    The equations you have attributed to KVL: V1=10+v2
    and: KVL:V3-V2-5i=0

    aren't really KVL equations. KVL says that the sum of the voltages around a closed loop add up to zero. You didn't traverse a closed loop to get those equations. You need those equations, but they are constraint equations rather than KVL equations.

    You should probably just treat the 3 nodes (V1, V2 and V3) as a single supernode rather than as two separate supernodes.

    You already have 2 equations (the two constraint equations) and you need a third. The currents from the 3 nodes to ground must add to zero, so you just need to form a node equation for the supernode using KCL: V1/2 + V2/4 + V3/3 = 0

    Notice that you don't need to form an equation involving the upper 6 ohm resistor; the voltages at the three nodes are completely independent of that resistor (because it's connected across a couple of voltage sources which are in series). However, once you have the node voltages you can calculate the current in the top 6 ohm resistor.
  4. Jul 21, 2011 #3
    Thanks for that Electrician. Joining the two voltage sources into a super duper node makes things much easier.
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