Help with Thevenin's Theorem!

Discussion in 'Homework Help' started by tonlika, Jun 25, 2011.

  1. tonlika

    Thread Starter New Member

    May 22, 2011
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    I have to find the current across V2 using Thevenin's theorem.

    Below in the attachments, draft2.jpg is the circuit and draft5.jpg is the Thevenin equivalent.

    R2 and R3 are in series and give a 6 Ohm resistance.
    R1 is shorted out. R4 is in parallel with R(23) so 1/Rt = 1/R(23) + 1/R4
    1/6 + 1/6 = 2/6 which gives Rt the total resistance equal to 3 Ohm.

    Then I write the nodal voltage equations:

    Va/2 + (Va-V1)/4 = 0
    Vb/6 + 4 + (Vb-V1)/2 = 0
    (V1-Va)/4 - 4 + (V1-Vb)/2 = 0

    which give:

    3*Va - V1 = 0
    4*Vb - 3*V1 = -24
    -Va - 2*Vb + 3*V1 = 16

    and solving the system/matrix for Va, Vb and V1 gives approximately:
    Va = 1.1
    Vb = -3.4
    V1 = 3.4

    The Thevenin voltage equivalent should be Va-Vb = 1.1 - (-3.4) = 4.5 Volts

    So the current over V2 should be: I = (Vt + V2)/Rt = (4.5 + 8)/3 = 4.17 Amps

    But when I do the Ltspice simulation the current is shown to be -6 A.

    Can you please tell me what I'm doing wrong?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Last edited: Jun 25, 2011
    tonlika likes this.
  3. tonlika

    Thread Starter New Member

    May 22, 2011
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  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Do you mean this version?
    Now we have one supernode so the equation will look like this

    For Vb

    \frac{Vb}{2}+\frac{Vb-V2}{6}+4 = 0

    And for V2 + Va

    \frac{Vb-V2}{6} = \frac{Va}{4}

    And

    Va = V2+12V
     
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  5. tonlika

    Thread Starter New Member

    May 22, 2011
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    Yes, this version.
    Thank you so much! I got it!
     
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