Confirmation with exercises on thévenin (part one)

Discussion in 'Homework Help' started by jorgemotalmeida, Jun 3, 2008.

  1. jorgemotalmeida

    Thread Starter Member

    Oct 23, 2007
    23
    2
    Hi


    This is not really a homework. I'm trying to grasp fully the Thévenin theorem.

    I will give you an exercise that I managed resolved almost fully, with 3 doubts.
    I attached 3 files HERE: the solution provided by the site (from Austin - Texas);
    the part one of my solution for Rth (with one doubt highlighted in red); and the other part two of my solution with mesh analysis... (they used the nodal analysis).

    The exercise is in the PART ONE of my solution (file name: exercise2_1part.jpg), there I have one doubt concerning the reason of taking off the resistor 5ohms...

    The part two is just to show and confirm if the mesh analysis is correct. I'm pretty sure it is . :) (file name: exercise2_2part.jpg )

    And for final, see the solution they provided... I also have two doubts highlighted... where it comes the Ib = Vb - 1.6?? And
    why in the final they subtract the Ve ?? See Ve = 1.61 - 1.01

    PS I used the comma as decimal point. :)
     
  2. mrmount

    Active Member

    Dec 5, 2007
    59
    7
    Part 1: The 5 ohm resistor can be taken off because it is in open circuit (one end is free).

    Part 3: If you rearrange the KCL at node E, you get Ib= Ve-1.
    whereas Ve is at -0.6v lesser than Vb. (see the original problem, a 0.6v source is connected between B and E).
    So, Ib= Vb-0.6-1
    = Vb-1.6
    And this also explains your second question in part 3.
     
  3. jorgemotalmeida

    Thread Starter Member

    Oct 23, 2007
    23
    2
  4. mrmount

    Active Member

    Dec 5, 2007
    59
    7
    Again, it is the same... For determining Rth, you will have to look back at the effective resistance from terminals A and B. So assuming that the current enters the node A, the only possible paths are 60 ohms and 40 ohms in series with 20 ohms. No current passes through either 10 ohm or 30 ohm resistors. So two 60 ohms in parallel is the effective resistance.

    I hope you understand how the value for Va is derived. Va is the value of voltage between A and ground. To find Vab all you have to do use voltage divider formula.

    Or alternately, you can apply KCL to node B.

    Va/80 = Vb/20.
    Solving for Vb, you get 15V.

    So Vab= Va-Vb = 60-15 = 45v.
     
  5. jorgemotalmeida

    Thread Starter Member

    Oct 23, 2007
    23
    2
    thanks! Now it is all clarified. :D
     
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