Is this website correct? (direction of DC current in basic example)

Discussion in 'Homework Help' started by Amenably_Amendable, Jan 21, 2008.

  1. Amenably_Amendable

    Thread Starter Member

    Jan 21, 2008
    13
    0
    If you look at this website: http://www.ibiblio.org/kuphaldt/electricCircuits/DC/DC_10.html


    Search for the word "downstream", you'll see a basic DC circuit with 3 unknown currents. There is a node in the middle that I called Vx. Using the equations:

    -I1 + I2 -I3 = 0
    I1 = (Vx-28)/4
    I2 = -Vx/2
    I3 = (Vx-7)

    We get:
    -(Vx-28)/4 -Vx/2 - (Vx-7) = 0
    -Vx + 28 - 2Vx -4Vx + 28 = 0
    -7Vx = -56
    Vx = 8

    I1 = (Vx-28)/4 = -5A
    I2 = -Vx/2 = -4A
    I3 = (Vx-7) = -1A

    BUT if you scroll down on this site and look at their solution, they found I1 and I2 to be positive. Who is correct, me or this website? Thanks! :eek:
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    It just means the original assumptions about current direction were wrong. Try working the problem assuming that all currents are positive and entering the node. The currents with negative signs are actually leaving the node instead of entering the node.
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
  4. kotophey

    New Member

    Jan 20, 2008
    3
    0
    You may wanna check your calculations as
    From your listing

    -I1 + I2 -I3 = 0

    and answers are

    I1 = (Vx-28)/4 = -5A
    I2 = -Vx/2 = -4A
    I3 = (Vx-7) = -1A

    say substitute I1 and I3 in which gives

    5 + I2 + 1 = 0

    I2 = -6A.
     
  5. Amenably_Amendable

    Thread Starter Member

    Jan 21, 2008
    13
    0
    Sorry..

    I1 = (Vx-28)/4 = -5A
    I2 = -Vx/2 = -4A
    I3 = (Vx-7) = 1A


    So I have the exact opposite answer than they got.. how does that work?
     
  6. Amenably_Amendable

    Thread Starter Member

    Jan 21, 2008
    13
    0
    Here is the schamatic comparing my results to the website's
     
  7. kotophey

    New Member

    Jan 20, 2008
    3
    0
    Just think of Vx being smaller then sources ( thats why currents are in those directions) Now your equations will turn to
    I1 = (28-Vx)/4
    I2 = -Vx/2
    I3 = (7-Vx)

    That should solve the problem.
     
  8. Amenably_Amendable

    Thread Starter Member

    Jan 21, 2008
    13
    0
    It shouldn't matter what direction I initially assume the currents to be flowing in. anyway, someone on Yahoo Answers helped me with this.
     
Loading...