Nodal analysis and Super Nodes

Discussion in 'Homework Help' started by Alustriel, Apr 26, 2009.

  1. Alustriel

    Thread Starter Member

    Apr 1, 2009
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    Hey everyone, I've been having some problems doing these questions from my work book. :(]
    I1/Is = 8mA , Find V1.
    R1 = 75 kΩ : R2 = 6.5 kΩ : R3 = 4.5 kΩ

    For this question, I kind of figured it requires nodal analysis, I did R3//R3 + R2 simplifying the right side of the circuit. Taking the top left hand corner node as the reference node, I got I1 + R1 - R2 - (R3//R3 + R2) = 0, is this equation correct? I can't get a second equation though. Or is there some other way of doing this?



    VS1 = 11 V : VS2 = 6 V
    R1 = 19 kΩ : R2 = 9 kΩ : R3 = 4 kΩ

    This next question is a little bit more complicated. I'm guessing it requires the use of the supernode for which I did V = 11v and V = 6v, how do I continue from here? I'm required to find Vo.

    Thanks in advanced. :)

    Edit : took down pics cuz i'm running out of bandwidth >.<
     
    Last edited: Apr 27, 2009
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Taking the first network, you should be aware that the current entering and leaving the node of interest must sum to zero. This understanding would then lead to the equation that follows.

    I_1\ -\ \frac{V_1}{R_2}\ -\ \frac{V_1}{R_2+R_3||R_3}\ =\ 0

    From this equation you have enough information to calculate V_1.

    hgmjr
     
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    The same problem can be tackled using Millman's Theorem as follows.

    V\small_1\ =\ \frac{\large 0.008}{\large \frac{1}{R_2}\ +\ \frac{1}{R_2\ +\ R_3||R_3}}

    hgmjr
     
    Last edited: Apr 26, 2009
  4. Alustriel

    Thread Starter Member

    Apr 1, 2009
    14
    0
    Hey there, thanks for the quick reply, I think I'm almost getting the first question. :) But for the second question, I haven't really learned Millman's Theorem so I'll go with the other method. Thanks.

    Anyone with ideas on question two please post up :p it really is a killer, I've ask a few of my friends and all of them can't do it. >.<
     
  5. vvkannan

    Active Member

    Aug 9, 2008
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  6. hgmjr

    Moderator

    Jan 28, 2005
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    You may want to take the approach of applying Thevenin's Equivalent for _{V\normalsize_{s1}\ , R_2\ ,} and _{R_3}.

    This will simplify the overall network and give you a network that is more familiar in its form.

    hgmjr
     
  7. Alustriel

    Thread Starter Member

    Apr 1, 2009
    14
    0
    For question 2,

    I named the top left node on the first row V1 while top middle node on first row V2 while the first node on the 2nd row is named V4 and so on and so forth.

    V1 = Vs1 = 11v
    V2 = V3 + Vs2

    Super node : Ix+Iy+Iz = 0 assuming all the currents are going out of the circuit.
    Ix = (V2-V1)/R3
    Iy = (V2-V4)/R2
    Iz = (V3-V5)/R3

    How do I continue from here?

    EDIT: (V1-V3)/R1 for that part on top...
     
    Last edited: Apr 27, 2009
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
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    You shouldn't have a 4th node. Let the - terminal on Vo be your reference node; you don't give the reference node a designation normally. Then your 3 nodes can be as you have designated them, left to right V1, V2 and V3.

    So right away you have V1 = Vs1

    and V2 - V3 = Vs2

    Finally, you have:

    (V2-V1)/R3 + (V3-V1)/R1 + (V2)/R2 + (V3)/R3 = 0

    You get this last one by treating V2 and V3 as if they were just one node (a supernode, so to speak).

    Now, can you solve these 3 simultaneous equations?
     
  9. jasperthecat

    Member

    Mar 26, 2009
    20
    0
    Any takers for the Superposition theorem ?

    No simultaneous equations

    A few resistors in parallel

    2 Sources

    2 Voltage dividers
     
  10. Alustriel

    Thread Starter Member

    Apr 1, 2009
    14
    0
    Thanks for all your help guys, finally managed to solve all 20 questions. xD
     
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