Nodal Analysis with voltage source but no resistor?

Discussion in 'Homework Help' started by Charger1, Jun 5, 2016.

  1. Charger1

    Thread Starter New Member

    Jun 5, 2016
    2
    0
    V1 = 20V, R1 = R3 = 10 ohms, R2 = 20 Ohms, Current source = 2Ix. Solving for Ix and I1 using nodal analysis.

    I set up the bottom as my reference node, and have node 1 connected to the voltage source and node two connected to the current source. My KCL for node 2 is

    (V2-V1)/R2 - V2/R3 +2Ix

    Where V1 and V2 are the voltages of node 1 and 2 respectively, with V1 being set to 20V because of the voltage source. However on Node 1 I've got V1/20 for the downward branch an (V1-V2)/20 for the right branch, but with no resistor on the left I'm not sure how to treat the current coming from the voltage source. If I work it through ohms law I end up with infinite current, and that simply doesn't feel right. What am I forgetting to apply?

    Thanks in advance.
     
  2. DGElder

    Member

    Apr 3, 2016
    345
    85
    First off, you show your current source with an independent source symbol but you notate the current as dependent on Ix. Then I don't see that you have identified Ix location. Second issue, this statement: "However on Node 1 I've got V1/20 for the downward branch" should be "......V1/10......"; should it not?

    As to your issue with the voltage source, you need to ask yourself what are your trying to do? You are trying to setup KCL nodal equations to figure out the voltages on node 1 and node 2, i.e. where V1 and V2 are your 2 variables to be solved in 2 equations. But the node 1 voltage is already known as it is established by the 20V voltage source. So you only need to solve for node 2 using KCL. After finding V2 you can find all voltage and currents with simple application of ohms law.

    Because of the voltage source you can't use KCL on node 1, unless you create variable Is, the current from the voltage source. This replaces V1, which is a known constant, with Is in your node 1 KCL equation. This combined with the node 2 equation you have two equations in two unknowns, but one is a voltage and the other is a current.

    I hope this helps.
     
    Last edited: Jun 5, 2016
    Charger1 likes this.
  3. Charger1

    Thread Starter New Member

    Jun 5, 2016
    2
    0
    Looking at the original circuit again you are right(an you're also correct about my goof on node one), and there was an another Ix, going from node one into node two, which means that the current going from node two to the reference node has to be 3Ix(as I have Ix entering from the left an 2Ix entering from the right). This gives me a couple useful equations

    Ix = (V2-20)/20
    3Ix = V2/10

    From there some algebra gets me V2 = 12v, and Ix = 400 mA. I1 on the other hand should just be 20/10, or 2 amps. While not necessary for my homework a nodal equation for n1(putting in new unknown Is) gives me

    Is - 20/10 - 12/20 = 0 => Is = 2.6A

    Which gives me values for everything in the system.

    Thank you DG. I think that does it, unless I messed up a sign somewhere.
     
Loading...