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Homework Help: Please help me solve loop analysis problem

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    http://i.imgur.com/pVIMw.png"

    2. Relevant equations



    3. The attempt at a solution

    I called the top left loop i1, the top right loop i2, the bottom right loop i3, and the bottom left loop i4. I have all of these currents going clockwise in my work. An attempt to do loop analysis has left me with the following equations:

    i4=(2Vx)/(1K)
    i1-i2=2mA
    i3=-4mA
    -i2(4kΩ)=Vx

    With this, I used KVL along the top loop going clcokwise, which got me:

    -2(i1-i4) - 4i1 - 4i2 - 1i2 - 8(i2-i3) = 0
    Solving this got me:
    i2=(-25/44)mA

    Then, V0=(8kΩ)(-25/44 mA)=-1.76V

    Can someone please help me figure it out? When I go and try to check it, my work gets pretty weird....Pretty sure this is wrong. Can someone please help me find my mistake?
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Oct 11, 2011 #2

    gneill

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    Staff: Mentor

    Verify the polarity of Vx.
     
  4. Oct 11, 2011 #3
    I was confused by that.

    Should it be i2(4κΩ)=Vx?

    Redoing my work gets me V0=11V, which looks better. I'll check to see if that answer makes more sense
     
    Last edited: Oct 11, 2011
  5. Oct 11, 2011 #4

    gneill

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    Current through a resistor causes a voltage drop (+ to -) in the direction that the current flows. What's the assumed direction of i2?
     
  6. Oct 11, 2011 #5
    I have all the currents going clockwise.

    I have i2 going from left to right on that resistor.
     
  7. Oct 11, 2011 #6

    gneill

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    So then...
     
  8. Oct 11, 2011 #7
    So then it should be Vx=i2(4κΩ)
     
  9. Oct 11, 2011 #8

    gneill

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    Correct.
     
  10. Oct 11, 2011 #9
    If I use that, I get the following:

    i1=13mA
    i2=11mA
    i3=-4mA
    i4=88mA

    Using these values, I fail to get a correct check using KVL on the top loop. I can't get ƩV=0...
     
  11. Oct 11, 2011 #10

    gneill

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    I get a different value for i2 (I didn't check the other currents) when I solve your equations. You'll have to post your work in detail if we're to see what's happened.
     
  12. Oct 11, 2011 #11
    Actually, I found it myself. Turns out I did the following:

    i1-i2=2mA → i1=2-i2

    Which is wrong :p

    Fixing this error, I got a value of i2=-14.6667=-44/3, if i2 were going clockwise.
     
  13. Oct 11, 2011 #12
    Thanks for all of your help, I really appreciate it!
     
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