# Circuit analysis help

Discussion in 'Homework Help' started by geft, Dec 25, 2011.

1. ### geft Thread Starter New Member

Dec 8, 2011
19
0
I'm having a hard time finding V2 in terms of I2. I've tried various forms of KCL and KVL but I can't seem to find the right answer. I hope someone can enlighten me.

• ###### Untitled.png
File size:
41.5 KB
Views:
37
Last edited: Dec 25, 2011
2. ### jegues Well-Known Member

Sep 13, 2010
735
43
Can you show us your work?

Then we can further diagnose what the problem is and point you in the right direction.

3. ### geft Thread Starter New Member

Dec 8, 2011
19
0
$I_2 = \frac{V_2-2V_x}{2} + \frac{V_2}{5} = \frac{V_2-2\frac{V_2}{4}}{2} + \frac{V_2}{5} = 0.45V_2$

$Z_{22} = \frac{V_2}{I_2} = \frac{V_2}{0.45V_2} = 2.222$

This is clearly wrong since the answer is supposed to be 1.111 ohm.

Last edited: Dec 25, 2011
4. ### Zazoo Member

Jul 27, 2011
114
43
My node equation is:
$I_2 = \frac{V_2-(-2V_x)}{2} + \frac{V_2}{5} = \frac{V_2+2V_2(\frac{1}{1+4})}{2} + \frac{V_2}{5} = 0.9V_2$

Current direction was chosen as leaving the node for both branches.
Note the polarity on the dependent voltage source.
Also, Vx can be replaced by the expression for a 1Ω and 4Ω voltage divider (making it 1/5 V2)

This gives 1.111ohms when used in the second equation.

geft likes this.