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CIRCUIT ANALYSIS: Use source transformation to find the Voltage Vo

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Use source transformation to find the voltage [itex]V_x[/itex] in the circuit below.

    [​IMG]


    2. Relevant equations

    [tex]V_S\,=\,i_S\,R[/tex]

    [tex]i_S\,=\,\frac{V_S}{R}[/tex]

    KCL, KVL, v = i R, super-node?


    3. The attempt at a solution

    I transform the independdant current source on top first.

    [tex]V_S\,=\,(3\,A)\,(10\Omega)\,=\,30\,V[/tex]

    [​IMG]


    Then I transform the V.C.C.S. on the right.

    [tex]V_S\,=\,(2\,V_x)\,(10\Omega)\,=\,20\,V_x[/tex]

    [​IMG]


    Then I combine the two resistors in series.

    [​IMG]


    Here I am stuck, how do I proceed?
     
  2. jcsd
  3. Jan 27, 2007 #2

    SGT

    User Avatar

    Since all your elements are in series, the current is the same.
    What is Vx in terms of this current?
    Replace this value in the controlled source and equate.
     
  4. Jan 27, 2007 #3
    [tex]v\,=\,i\,R[/tex]

    [tex]V_x\,=\,i\,(8\Omega)\,=\,8\,i[/tex]

    [tex]20\,V_x\,=\,20\,(8\,i)\,=\,160\,i[/tex]

    But how do I find i?
     
  5. Jan 28, 2007 #4

    SGT

    User Avatar

    Use KVL:

    -40 + 8i - 30 + 160i = 0
     
  6. Jan 28, 2007 #5
    So we have

    [tex]168\,i\,=\,70[/tex]

    [tex]i\,=\,\frac{70}{168}\,A[/tex]

    Now use the [itex]i[/itex] to find the [itex]V_x[/itex].

    [tex]V_x\,=\,I\,R\,=\,\left(\frac{70}{168}\,A\right)\,(8\Omega)\,=\,\frac{10}{3}\,V[/tex]

    [tex]V_x\,=\,\frac{10}{3}\,V\,\approx\,3.33\,V[/tex] <----- Right?
     
    Last edited: Jan 28, 2007
  7. Jan 28, 2007 #6

    SGT

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    Seems OK to me.
     
  8. Feb 20, 2010 #7
    Why aren't we adding the voltage across he 20Ohm resister? The equation should read -40 + 8i - 30 + 20i + 160i = 0 right?
     
  9. Oct 23, 2010 #8
    yup very rite, we do need to include that 20i, otherwise the answer is wrong!
     
  10. Apr 14, 2013 #9
    The Correct answer is :-

    Using KVL (Clockwise) for Loop, we get....

    -40 + Vx - 30 + 20i+ 20Vx = 0

    21Vx + 20i = 70. // i = Vx/8

    21Vx + 20(Vx/8) = 70

    Vx ( 21 + 20/8) = 70

    Vx = 2.97 Approx (3v)
     
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