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Thévenin's Equivalent Circuit

  1. Apr 11, 2012 #1
    As seen from a and b.
    tC3DW.jpg

    My teacher did the norton's equivalent in the class, and now for studying purposes i was trying to get to the Thévenin's one. However it seems my equations are somehow wrong, and i can't figure out why.

    The Eq resistance is 8||2, which is 1.6 ohms.

    I tried to apply Nodal analysis first

    WNqBl.jpg

    so,
    for node V1: [tex]\frac{20-V_1}{3}=\frac{V_1-V_3}{2}[/tex]
    for node V2: [tex]\frac{V_1-V_3}{2}=6+\frac{V_2-0}{5}=[/tex]
    The relationship between two voltages is also known, which is [tex]V_2=V_3+10[/tex]

    which gives V1=2 ; V2=0 ; V3=-10.

    V1-V2 should indeed be 2, because from my teachers resolution, Vab = 2V. I don't get is how can V2 be 0? Then there can't be another voltage drop when it reaches the ground (0V)

    Also, then i tried mesh analysis:
    UKA3A.jpg

    For mesh 1[tex]20+10=3I_1+2I_1+5(I_1+I_2)[/tex]
    For mesh 2 [tex]0=5(I_2-I_1)+v[/tex]
    [tex]I_2=6[/tex]

    Where v is the voltage drop across the current source.

    It gives I1 = 6 and v (which is equal to the V2 of the nodal analysis) = 0 again.
     
  2. jcsd
  3. Apr 11, 2012 #2
    My answer is that :
    V2 is 0V because 6A current source short 5Ω resistor.
    Also notice that (20V + 10V) / ( 3Ω + 2Ω ) = 6A

    So from superposition point of view we have this situation

    attachment.php?attachmentid=46116&stc=1&d=1334158488.png

    V2' = 30*R2/(R1 +R2) = 15V

    And

    V2'' = - (6A * 5/(10)) * 5 = - 15V

    and V2 = V2' + V2'' = 15 + (- 15V) = 0V
     

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  4. Apr 11, 2012 #3
    Thanks :) it means the current source is just discharging current. Like i had a current source between two earth connections.

    Another question:

    In nodal analysis, when calculating a current entering / leaving a node we do [tex]\frac{V_x-V_y}{R}[/tex] where [tex]V_x > V_y[/tex] My question is: what if we don't know which of the two nodes has higher voltage when writting the equations? Or should it always be possible to know ?
     
  5. Apr 12, 2012 #4
    You don't need to know which of the two nodes has higher voltage.
    All you need to do is assume that one of a voltage nodes has a higher voltage than the other node. And you can treat all nodes individually and assume that the node your currently examining has higher voltage.

    For example for this circuit

    attachment.php?attachmentid=46151&stc=1&d=1334238395.png

    We can assume Vin > Va > Vout

    And we can write nodal equation

    [tex]\frac{(Vin-Va)}{R1}=\frac{Va}{R3}+\frac{(Va-Vout)}{R2}[/tex]

    [tex]\frac{(Va-Vout)}{R2}=\frac{Vout}{R4}[/tex]

    Or we can assume when we examining Va node that Va has the highest voltage.
    And this means that all current flow-out (leave) from the Va node.

    [tex]\frac{(Va-Vin)}{R1}+\frac{Va}{R3}+\frac{(Va-Vout)}{R2} = 0 [/tex]

    And the same think we can do for Vout

    [tex]\frac{(Vout-Va)}{R2}+\frac{Vout}{R4} = 0 [/tex]

    Simply I assume that all the current leaving the node. If so this means that the voltage at this node has to be be the highest.

    http://www.wolframalpha.com/input/?...+++(A+-+B)/100+=+0+,+B/100+++(B+-+A)/100+=+0+
     

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    Last edited: Apr 12, 2012
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