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CIRCUIT ANALYSIS: Use source transformation to find the Voltage Vo

  • Thread starter VinnyCee
  • Start date
  • #1
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Homework Statement



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

http://img207.imageshack.us/img207/5150/chapter4problem24ij9.jpg [Broken]


Homework Equations



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

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

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


The Attempt at a Solution



I transform the independdant current source on top first.

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

http://img441.imageshack.us/img441/4210/chapter4problem24part2be2.jpg [Broken]


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

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

http://img255.imageshack.us/img255/9743/chapter4problem24part3kh6.jpg [Broken]


Then I combine the two resistors in series.

http://img101.imageshack.us/img101/5192/chapter4problem24part4bx8.jpg [Broken]


Here I am stuck, how do I proceed?
 
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Answers and Replies

  • #2
SGT
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.
 
  • #3
489
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[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?
 
  • #4
SGT
[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?
Use KVL:

-40 + 8i - 30 + 160i = 0
 
  • #5
489
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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:
  • #6
SGT
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}\right)\,(8\Omega)\,=\,\frac{10}{3}\,V[/tex]

[tex]V_x\,=\,\frac{10}{3}\,V\,\approx\,3.33\,V[/tex] <----- Right?
Seems OK to me.
 
  • #7
1
0
Use KVL:

-40 + 8i - 30 + 160i = 0
Why aren't we adding the voltage across he 20Ohm resister? The equation should read -40 + 8i - 30 + 20i + 160i = 0 right?
 
  • #8
yup very rite, we do need to include that 20i, otherwise the answer is wrong!
 
  • #9
1
0
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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