# CIRCUIT ANALYSIS: Use source transformation to find the Voltage Vo

1. Jan 27, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

Use source transformation to find the voltage $V_x$ in the circuit below.

2. Relevant equations

$$V_S\,=\,i_S\,R$$

$$i_S\,=\,\frac{V_S}{R}$$

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

3. The attempt at a solution

I transform the independdant current source on top first.

$$V_S\,=\,(3\,A)\,(10\Omega)\,=\,30\,V$$

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

$$V_S\,=\,(2\,V_x)\,(10\Omega)\,=\,20\,V_x$$

Then I combine the two resistors in series.

Here I am stuck, how do I proceed?

2. Jan 27, 2007

### 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. Jan 27, 2007

### VinnyCee

$$v\,=\,i\,R$$

$$V_x\,=\,i\,(8\Omega)\,=\,8\,i$$

$$20\,V_x\,=\,20\,(8\,i)\,=\,160\,i$$

But how do I find i?

4. Jan 28, 2007

### SGT

Use KVL:

-40 + 8i - 30 + 160i = 0

5. Jan 28, 2007

### VinnyCee

So we have

$$168\,i\,=\,70$$

$$i\,=\,\frac{70}{168}\,A$$

Now use the $i$ to find the $V_x$.

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

$$V_x\,=\,\frac{10}{3}\,V\,\approx\,3.33\,V$$ <----- Right?

Last edited: Jan 28, 2007
6. Jan 28, 2007

### SGT

Seems OK to me.

7. Feb 20, 2010

### ms1234

Why aren't we adding the voltage across he 20Ohm resister? The equation should read -40 + 8i - 30 + 20i + 160i = 0 right?

8. Oct 23, 2010

### bigb_bilal

yup very rite, we do need to include that 20i, otherwise the answer is wrong!

9. Apr 14, 2013

### tajiknomi

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)