Node analysis with supernode

Discussion in 'Homework Help' started by regexp, Jan 5, 2011.

  1. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Hello,

    In this circuit, i need to find the node voltages for v1 and v2.
    [​IMG]


    I have for V1: 0.1v_{1}+J0.05v_{1} = 0

    and for V2: 0.067V_{2}-J0.2V_{2} = 0

    And V1 = V2 + 10

    Substituting for V1:

    0.1\cdot(v_{2}+10) +J0.05v_{2}+J0.5

    Am i on the wrong track?
     
  2. tyblu

    Member

    Nov 29, 2010
    199
    16
    The question is presented to use loop analysis. Here's a starter:

    Loop 1: I_1 (10 \Omega) + (I_1 - I_2)(j20 \Omega) = 0
    Loop 2: (I_2 - I_1)(j20 \Omega) + (I_2 - I_3)(15 \Omega) = 10V
    Loop 3: (I_3 - I_2)(15 \Omega) + I_3(-j5 \Omega) = 0

    Note: the voltage source symbol is that of a DC source, but the question suggests AC.
     
  3. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Hm, can you really do mesh analysis like that when you have a supernode?
     
  4. tyblu

    Member

    Nov 29, 2010
    199
    16
    No -- supernodes are used in nodal analysis. Transforms to go from one form of analysis to another exist (Thevenin, Norton), but usually only complicate matters. I'm not sure what you're calling a "supernode", here, but I know it as the combination of 2 or more nodes.
     
  5. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0

    \left[ {\begin{array}{cc}<br />
(10+J20) & -J20 & 0 \\<br />
     -J20  & (15+J20) &-15 \\<br />
 0  & -15  &(15-J5)\end{array}}\right] \left[ {\begin{array}{cc}I_{1}\\<br />
I_{2}\\<br />
I_{3}\end{array}}\right]

    Hm, it's easy to see that the middle row becomes zero

    Can you just add the rows
    So that you get:
    10I_{1} -J5I_{3} = 10
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    The red circled arrows do suggest a mesh analysis, but it would be just as reasonable to infer from the marked node voltages V1 and V2 that a node analysis is wanted.

    Your Loop 2 equation has the wrong sign on the right hand side; it should be -10V.

    This gives you the relationship between I1 and I3, but you need V1 and V2.

    The nodal equations would be:

    \left[ {\begin{array}{cc}<br />
\frac{1}{10}+\frac{1}{j20} & \frac{1}{15}+\frac{1}{-j5}  \\ 1  & -1  \end{array}}\right] \left[ {\begin{array}{cc}V_{1}\\<br />
V_{2} \end{array}}\right]=\left[ \begin{array}{cc}0\\10\\ \end{array}\right]

    The second equation (row) is the constraint equation to deal with the fact that the two nodes are connected by a voltage source (they are a supernode).
     
    tyblu likes this.
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