Answer check to challenging problem

Discussion in 'Homework Help' started by ihaveaquestion, Sep 20, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
    0
    This is a challenging problem to me, so I'd like to make sure my thinking is correct for my way of solving:

    http://img24.imageshack.us/img24/88/imguu.jpg

    http://img23.imageshack.us/img23/8479/img0001ty.jpg

    Given the original circuit drawn at the top (in the first link), I am asked to find the voltage vo by using Thevenin's theorem. So the strategy is to find Vth, find Rth, then use voltage division to find vo, i.e. vo = Vth(2) / Rth + 2.

    I do normal nodal analysis to find Vth = V2 in the first link.

    On the second link I find Rth by replacing the 2ohm resistor by a 1V voltage source, shorting the original voltage source, and do nodal analysis again. In this circuit with the excitation 1v voltage, I know v2 = 1v... so I find out what V1 is... the reason for all of this is because I want to determine what the current through my excitation voltage source is so I can determine Rth by Vs / i1 (labeled in my work). Once I find out what V1 is I do KCL at v2... I'm wondering if the directions of the currents at v2 matters... I said the current coming in from the left into V2 was v1-v2/j2... I calculate i1 (current through excitation Vs) and calculate Rth...

    I finally do my voltage division with my Thevenin equiv..

    what do you guys think?

    Thanks in advance.
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Replace the capacitors and inductor with their equivalent impedance. Remove the 2R resistor and find Vth. Then replace the voltage source by a short circuit and the current source by an open circuit and find the impedance (Rth) looking into the terminals across the 2R resistor (without 2R connected).
     
  3. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
    0
    This is shown in my work.

    The current source is dependent, can't open circuit it. This is why I applied the 1v source in my work.
     
  4. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
    0
    As an FYI - you don't need to check all my numbers, just want confirmation that my procedure and way of thinking is correct.
     
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