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Equivalent circuit?

  1. Feb 17, 2014 #1
    Just wondering if its ok to re-draw this
    http://postimg.org/image/a074v6cor/ [Broken]

    as this
    http://postimg.org/image/uvtf6f8vv/ [Broken]


    ?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 17, 2014 #2

    BvU

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    Hello, your higness. Not your first post, so you should know better by now. Use the template and don't make me chase links instead of going to sleep. Maybe tomorrow.

    But, out of curiosity: is the central point in the first picture a connection or is I1 = -I3 ?
     
  4. Feb 17, 2014 #3

    lewando

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    Whether the central point is connected or not, the redrawing is not correct. One discrepancy (and there are others): in the original drawing, the 4 Ω and the 6 Ω share a node. In the redrawing, they do not.
     
  5. Feb 17, 2014 #4

    SammyS

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    No


    attachment.php?attachmentid=66728&stc=1&d=1392691615.png

    Is not the same as

    attachment.php?attachmentid=66729&stc=1&d=1392691712.png

    Your circuit would force I3 = -I2

    and I1 = 1 Amp
     

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    Last edited by a moderator: May 6, 2017
  6. Feb 18, 2014 #5
    It would be correct if you add a wire from the point between the 4 and 12 ohm resistors at the bottom to the point between the 6 ohm resistor and the 1 amp current source at the top. You separated the 4 wires coming together in the midpoint into 2 pairs.
     
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