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Circuit Analysis - Source Transformation

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Find Vx using source transformation
    attachment.php?attachmentid=41205&stc=1&d=1322126072.jpg

    2. Relevant equations
    V=IR
    KVL

    3. The attempt at a solution
    After doing source transformation on the three pairs of current sources and resistors...
    attachment.php?attachmentid=41207&stc=1&d=1322126953.jpg

    KVL:
    -6+2I-Vx+4I-3+3I+5I-2=0
    14I-Vx-11=0
    but Vx=5I (is it? i'm not so sure about this)
    14I-5I-11=0
    1I=11
    I=1.222 A

    Vx=-5I=-5(1.222)
    Vx=-6.11 V

    Did I do the source transformation right? Also, I'm confused with the value of Vx. Is it -5I or just 5I? Because in our class, our convention is that when a current source is transformed into a voltage source, the part where the arrow head in the current source becomes the positive terminal for the voltage source. But since we're only dealing with a resistor here, I just followed our convention for the KVL (the voltage across 5 ohms is positive since the mesh current enters the positive terminal)
     

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    • ckt2.jpg
      ckt2.jpg
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    Last edited: Nov 24, 2011
  2. jcsd
  3. Nov 24, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Everything looks fine except for your last two lines! Vx is in fact 5I, since its specified polarity indicates that it's a drop in the same direction as your current. So Vx = 6.11 V.
     
    Last edited: Nov 24, 2011
  4. Nov 24, 2011 #3
    thank you very much for the verification sir!
     
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