# Circuit analysis help

#### geft

Joined Dec 8, 2011
19
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.

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#### jegues

Joined Sep 13, 2010
733
Can you show us your work?

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

#### geft

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

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

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

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#### Zazoo

Joined Jul 27, 2011
114
$$I_2 = \frac{V_2-2V_x}{2} + \frac{V_2}{5} = \frac{V_2-2\frac{V_2}{4}}{2} + \frac{V_2}{5} = 0.45V_2$$

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

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