1. Oct 11, 2011

### ee_ess_ee-kid

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

http://i.imgur.com/pVIMw.png"

2. Relevant equations

3. The attempt at a solution

I called the top left loop i1, the top right loop i2, the bottom right loop i3, and the bottom left loop i4. I have all of these currents going clockwise in my work. An attempt to do loop analysis has left me with the following equations:

i4=(2Vx)/(1K)
i1-i2=2mA
i3=-4mA
-i2(4kΩ)=Vx

With this, I used KVL along the top loop going clcokwise, which got me:

-2(i1-i4) - 4i1 - 4i2 - 1i2 - 8(i2-i3) = 0
Solving this got me:
i2=(-25/44)mA

Then, V0=(8kΩ)(-25/44 mA)=-1.76V

Last edited by a moderator: Apr 26, 2017
2. Oct 11, 2011

### Staff: Mentor

Verify the polarity of Vx.

3. Oct 11, 2011

### ee_ess_ee-kid

I was confused by that.

Should it be i2(4κΩ)=Vx?

Redoing my work gets me V0=11V, which looks better. I'll check to see if that answer makes more sense

Last edited: Oct 11, 2011
4. Oct 11, 2011

### Staff: Mentor

Current through a resistor causes a voltage drop (+ to -) in the direction that the current flows. What's the assumed direction of i2?

5. Oct 11, 2011

### ee_ess_ee-kid

I have all the currents going clockwise.

I have i2 going from left to right on that resistor.

6. Oct 11, 2011

### Staff: Mentor

So then...

7. Oct 11, 2011

### ee_ess_ee-kid

So then it should be Vx=i2(4κΩ)

8. Oct 11, 2011

### Staff: Mentor

Correct.

9. Oct 11, 2011

### ee_ess_ee-kid

If I use that, I get the following:

i1=13mA
i2=11mA
i3=-4mA
i4=88mA

Using these values, I fail to get a correct check using KVL on the top loop. I can't get ƩV=0...

10. Oct 11, 2011

### Staff: Mentor

I get a different value for i2 (I didn't check the other currents) when I solve your equations. You'll have to post your work in detail if we're to see what's happened.

11. Oct 11, 2011

### ee_ess_ee-kid

Actually, I found it myself. Turns out I did the following:

i1-i2=2mA → i1=2-i2

Which is wrong :p

Fixing this error, I got a value of i2=-14.6667=-44/3, if i2 were going clockwise.

12. Oct 11, 2011

### ee_ess_ee-kid

Thanks for all of your help, I really appreciate it!