• Support PF! Buy your school textbooks, materials and every day products Here!

CIRCUIT ANALYSIS: Use source transformation to find the Voltage Vo

  • Thread starter VinnyCee
  • Start date
489
0
1. 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]


2. Homework 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]

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?
 
Last edited by a moderator:

Answers and Replies

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.
 
489
0
[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?
 
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
 
489
0
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:
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.
 
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?
 
yup very rite, we do need to include that 20i, otherwise the answer is wrong!
 
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)
 

Related Threads for: CIRCUIT ANALYSIS: Use source transformation to find the Voltage Vo

Replies
6
Views
886
  • Last Post
5
Replies
105
Views
9K
  • Last Post
Replies
24
Views
3K
Top