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.
     
    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.
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