Thevenin Problem

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

  1. beeorz

    Thread Starter New Member

    Sep 20, 2009

    To solve for Vth, can I do a simply KVL on the outer loop or do I need to do a mesh/loop analysis?
  2. hgmjr

    Retired Moderator

    Jan 28, 2005
    There is no shortcut. The analysis of this circuit will need to take all components into account.

  3. Accipiter

    New Member

    Sep 20, 2009
    If I were you, I would use superposition and voltage/current division. That's a whole lot faster than nodal/loop analysis. If you replace the current source with an "open", then you can solve for Vo in terms of the 10V source. Then, put the current source back, and short out the voltage source. Solve for Vo in terms of the current source alone. Then, add the two results. Once you get the hang of it, you can solve circuits like these in a fraction of the time it would normally take with nodal/loop analysis.
  4. beeorz

    Thread Starter New Member

    Sep 20, 2009

    I have been trying this problem many different ways. My I can only see solving for Vth via mesh/loop analysis then possibly cramer's rule. However, I am very out of practice. This is my work - can you guys check to see if I am doing anything wrong or if I am even doing anything right lol..

  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    Your first equation for mesh 1 is:

    -5(3) - 1(Inct) = 0

    but this should be:

    -5(3) - 1(Inct) = (voltage across 3A source)

    The sum of the voltages across the two resistors isn't zero.

    The equation for mesh 1 should be simply:

    I1 = -3

    Substitute that into your equation for mesh 2 and you should be able to get I2.