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

## Homework Statement

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

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

## Homework Equations

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

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

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

## The Attempt at a Solution

I transform the independdant current source on top first.

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

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

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

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

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

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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.

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

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

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

But how do I find i?

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

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

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

But how do I find i?
Use KVL:

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

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?

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SGT
So we have

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

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

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

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

$$V_x\,=\,\frac{10}{3}\,V\,\approx\,3.33\,V$$ <----- Right?
Seems OK to me.

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!

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)