About Thevenin Theorem

Discussion in 'Homework Help' started by BrillianceLin, Oct 18, 2007.

  1. BrillianceLin

    Thread Starter New Member

    Oct 18, 2007
    Hi, I got a question in my assignment about Thevenin Theorem, question shown below (picture shown in attachment).
    Find the Thevenin equivalent circuit for the following circuit, where Is=2 with 0 phrase angle, R=4ohm, ZL=3j(impedance from inductance) and ZC=-3j((impedance from capacitor)
    In the picture, the round thing on top is current source Is, the curly thing on top of left hand side is a inductance ZL, and a Resistor across the drawing, the one at the bottom is capacitor ZC.
    My solution is use Is multiply by R then I can get Thevenin voltage, which is
    Vth= 2 X 4 with 0 phrase angle=8V with 0 phrase angle.(all angles are in degrees)
    and I tried to find the Thevenin impedance by doing following calculation
    Rth= ZC + ZL||R= -3j + (48j+36)/25= 1.8 with phrase angle 36.8698765 (This is my professor said it is wrong)
    He said I suppose to replace this current source with an open circuit. Thus, the Rth is ZC+R= 5 with phrase angel -90. This part confuse me a very much. Because this is a current source, means it provide voltage at the same time. Thus, I think should short the current source instead of open it. But he said this is WRONG idea.
    Ok, I went home and search on Wiki. All Thevenin Theorem examples are basic on voltage source. And I did not aware any other source can provide example for this current source business thing. (Which make sense, there is no such thing can provide current without providing voltage) Until I read my text book (Electrical Engineering Principles and Applications 4th edition by Allan R. Hambley Department of Electrical and Computer Engineering of Michigan Technology University ISBN-13: 978-0-13-198922-1) page 86 exercise 2.22 with a correct answer only, and I can only get the "right" number by open the current source. After this question, there 3 paragraphs sort of explain what happen in the question, it says" If a network contains no dependent sources, there is an alternative way to find the Thevenin resistance. First, we zero the sources in the network. In zeroing a voltage source, we reduce its voltage to zero. A voltage source with zero voltage is equivalent to a short circuit.
    In zeroing a current source, we reduce its current to zero. By definition, an element that always carries zero current is an open circuit. Thus, to zero the independent sources, we replace voltage sources with short circuits and replace current sources with open circuits.
    (I omit the last paragraph, since is relate to voltage source)"
    These three paragraphs seem clear enough. Then I tried to apply this "THEORY" in to the example shown on Wiki (http://en.wikipedia.org/wiki/Thevenin's_theorem) I did a little bit modification. I replace the known value of the power source (15V) with output 3.75mA (which exactly the current output from the source). I did this modification because I ask,:"Is this power source a voltage source?" the answer is obvious, "yes". And I ask myself: "Is this power source a current source?" "It provide current, then I guess it should be a current source." So I think this modification did not change the circuit. Now I can start working on this modify question. I use the current to calculate the open circuit voltage, Which is 7.5V. This is the Thevenin equivalent voltage. So far so good. And I try to find out the Thevenin equivalent resistance. So, by definition, I should open the current source, and stand on the other side of circuit and look back for total resistance. Since R4 on an open circuit it will not being consider as a part calculation. I get, Thevenin equivalent resistance is Rth=R1+R2+R3=3000 ohm. Now I stuck here. I got 3000ohms, but the example shows 2000ohms. Both of them are Thevenin equivalent resistance, so they should be equal. I appreciate some one can give me a hand to tell me which part I did wrong, in both my assignment and my modify example question. And I am looking forward some one can tell me more about this Thevenin Theorem and why it will work. Plus, why a current source can not be a voltage source and a voltage source can not be a current source.
    Thanks for every reply this thread.
  2. milee


    Sep 20, 2007
    I was trying to view the diagram but it seems blur...can u re attach the diagram???
  3. BrillianceLin

    Thread Starter New Member

    Oct 18, 2007
    I can not manage to focus the diagram. hope you can understand this one.
    iC.... ..R
    ZL =impedance of inductance ZC= impedance of capacitor
    IC= alternative current source R= resistor
  4. chuckey

    Well-Known Member

    Jun 4, 2007
    1. A current source, can be regarded as a voltage source with a high impedannce in series with it. For example a current source of 1mA, could be regarded as a voltage of 1kV with a resistor of 1M ohm inseries with it, now in your circuit an extra 1Mohm would make little difference to your answer, if it did, use 1MV with a resistor of 1G ohm in series with it. Now can you see why the current source should be open circuited?. Basically in the extreme, its an infinite voltage with an infinite resistor in series with it , but the ratio of the two are known?
    For thevenin, the circuit is voltage drop R/ZL+R at I, with an output impedance of L in parrallel with R with C in series.
    Its sometime helpful to think of a simple example, i.e. a 12V batt with two 1K ohm resistors in series across it. At the joint of the two resistors the voltage is 12/2 = 6V the output impedance is of two 1Ks in parallel = 1K/2 = 500 ohms.