Need Help with Nodal KCL labeling

Discussion in 'Homework Help' started by freemindbmx, Apr 24, 2014.

  1. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    The problem wants me to find the voltages V1 and V2 using nodal analysis and then determine Ix.I get stuck on V1 when trying to label the KCL that includes the 12 ohm resistor, and I getstuck with V2.It seems that V2 shares the same node as the 3A independent source and the 12 ohm,I just don't know how to label this.
     
  2. WBahn

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    Mar 31, 2012
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    Just like you have an Ix for the current in the 12Ω resistor. Define and label two currents (I1 and I2 work) for the currents in the two 6Ω resistors.
     
  3. shteii01

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    Feb 19, 2010
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    Label it V3. Use it to solve stuff.
     
  4. WBahn

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    Label what V3? There are three nodes. One is ground and the other two are already labeled V1 and V2.
     
  5. shteii01

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    Feb 19, 2010
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    I see 4 nodes. Ground, V1, V2, that is three. The last node does not have a label. I don't really care what OP uses for the label for it, but if they want to solve the problem, they will have to use it and they will have to call it something.
     
  6. WBahn

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    Well, your eyes are a lot better than mine because I still only see three nodes.

    [​IMG]

    I see a green node that is ground, a red node that is labeled V1, and a blue node that is labeled V2.

    Please show me where this fourth node is, because I just don't see it.
     
  7. freemindbmx

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    Mar 5, 2014
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    The book labeled the V1 and V2 for me,but im alittle confused.Say for instance with V1 and the 12 ohm= (V1-V2)/12 and as for the branch that shares the same node with V2, the 3A source and 12ohm.Should I just say its 3A for one KCL and (V2-V1)/12 for the other KCL that is applied to node V2 , since they share the same node,denoted by the straight line.

    My Nodal KCLS are: (current leaving is positive, and current entering is negative as my convention)
    V1=
    -2A+(V1/6)-4A+(V1-V2)/12=0

    V2=
    4A+(V2/6)-3A+(V2-V1)/12=0

    Let me know if I need to change them.


    On a sidenote,if I was allowed to use mesh,the first mesh on the very left(2a source and 6ohm resistor),the KVL would be just 2A correct? Since this mesh already has a current source that dictates the characteristics of that KVL.
     
    Last edited: Apr 25, 2014
  8. Jony130

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    Feb 17, 2009
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    Your nodal KCL equation looks good.
     
  9. shteii01

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    Ok. I see it now. Thank you.
     
  10. freemindbmx

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    Mar 5, 2014
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    another quick question,I posted a new nodal problem,if I could get a confirmation of my KCL's thatd be great.

    V1=

    (V1-21)/5+V1/10+(V1-V2)/5=0

    V2=

    (V2-V1)/5+V2/10+(V2+10.5)/5=0

    the independent voltage source that has its polarities switch is the only problem I have with this nodal problem,I just want to make sure on my analysis.I just add it together instead of substracting because of the source polarity?Or would it seem more reasonable to source transform the 21v source and label my ground at the negative end of the 10.5 source?Trying to make sure I conceptually understand it.
     
  11. Jony130

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    Once again your nodal equation looks good for me.
     
  12. WBahn

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    Mar 31, 2012
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    You did it right.

    Just keep in mind what Nodal analysis is and you will be fine. You are simply summing up the currents leaving the node and setting them equal to zero.

    For a resistance connected between the node being analyzed and another node, the current leaving the node is (Vnode-Vothernode)/Rbetween. In the case of the right-most branch, Vothernode is -10.5V.

    I STRONGLY encourage you to start using and tracking your units. Most mistakes you make will affect your units and thus you can catch them at the point they are made or shortly thereafter. Also, don't use the equals sign as you do since it is misleading. Instead, use something like a colon. Here is a possible format for you to consider:


    V1: (V1-21V)/5Ω + V1/10Ω + (V1-V2)/5Ω = 0
    V2: (V2-V1)/5Ω + V2/10Ω + (V2+10.5V)/5Ω=0

    With just a bit of practice you will be able to write down the normalized equations by inspection, which is one of the key strengths of nodal and mesh analysis, as follows:

    V1: V1(1/5Ω + 1/10Ω + 1/5Ω) - V2(1/5Ω) = 21V/5Ω
    V2: -V1(1/5Ω) + V2(1/5Ω + 1/10Ω + 1/5Ω) = -10.5V/5Ω

    At this point you can multiply both sides of both equations by 10Ω to get:

    V1: V1(5) - V2(2) = 42V
    V2: -V1(2) + V2(5) = -21V

    From here, of course, the solution is very straightforward. Multiplying the first equation by 2 and the second by 5, we get

    V1: V1(10) - V2(4) = 84V
    V2: -V1(10) + V2(25) = -105V

    Adding them we get

    V2(21) = -21V

    to yield V2 = -1V

    Notice how the units worked out on their own. If they hadn't, I would have KNOWN that I had made a mistake and that the answer was wrong. The same is true at each step along the way. I know in my node equations that each term must have the same units in order to be able to add them together. So if I forget to include the resistance on the right hand side (a very common mistake) I would have terms on the left that have units of current and terms of the right that have units of voltage. Since current is not equal to voltage, I would have known it was wrong AT THAT POINT and could spend thirty seconds fixing it instead of wasting ten minutes solving a set of equations that are guaranteed to yield the wrong answer.

    Now, while tracking units will catch most mistakes, it obviously can't catch them all. I could easily have simply screwed up the math along the way. So you want to always ask if your answers make sense and also check your results. One of the nice things about most engineering problems, even in the real world, is that you can verify the correctness of the answers from the answers themselves. You do this by assuming that your answer is correct and treating it as a given and then seeing if it is consistent with the rest of the problem. So if V2 = -1V, I can find the current flowing in the rightmost two resistors immediately and, summing them, get the current flowing in the resistor bridging V1 and V2, which let's me find that voltage drop and, knowing V2, I can find V1. That let's me find the current in the leftmost 10Ω resistor and, summing that with the bridging resistor current, the current in the leftmost 5Ω resistor. Knowing that and the voltage at V1, I can calculate the voltage of the leftmost source and if it comes out anything other than 21V I have a problem. In fact, when I first did this I had a problem. So I went back and looked at my work and, sure enough, found a simple math error. But because I was methodical with my equations, it was a matter of a few seconds to correct them and get a new answer. Which I then verified the same way and this time got that the leftmost source was, indeed, 21V. Along the way, I also found that V1=8V and then verified this by using my original equations to solve for V1 and getting the same result.

    Another way to check your work is to simply solve the problem using a different approach such as mesh current or superposition.

    If you start tracking your units and checking your work, you will be amazed at how quickly your skills will improve and how much your grades will go up.
     
  13. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    wow, thank you.I need to get into the habit of labeling the units correctly.And I didn't even think about using Inspection. As of right now, I know I need more practice analyzing circuits using these techniques. The Text Book doesn't give enough at times lol.But another problem,I posted below.Two things come to mind,first either source transform the voltage source(6v) and the 2ohm resistor in series shunting it,then use Inspection(Conductance), or show the relationship of the nodes in nodal analysis.

    It seems theres only two extraordinary nodes,on bothsides of the 3A current source.

    So by nodal analysis(I'll use inspection to check my answer or mesh if needed).

    V1:
    (V1-6V)/2Ω + V1/8Ω + V1/8Ω = -3A

    V2:
    V2/8Ω + V2/8Ω = 7A

    /*Just added the two current sources and moved to right side for V2.Current leaving is positive,as my convention.
    I can solve it,but I just want to get the hang of writing down these relationships. I will try your way of using the answers I get and relizing whats in series/parallel to plug and chug for Voltages/Currents.
     
    Last edited: Apr 27, 2014
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