thevenin equivalent example problem

Discussion in 'General Electronics Chat' started by PG1995, Jun 5, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi :)

    In the following linked pictures I have marked the sections which I have found more troublesome with numbers (1, 2, 3, 4, 5). The aim is to find Thevenin equivalent. Please help me. It would be really kind and nice of you. Thank you

    1: http://img843.imageshack.us/img843/4292/norton1.jpg
    2: http://img847.imageshack.us/img847/2450/norton2h.jpg

    At "1" in link #1, it says (Vo - 0)/2. It is for the branch in which Ix is flowing. Ix is flowing upward - from the ground toward Vo which simply means that Vo is at lower potential than the ground. Therefore, it should be (in my view): (0 - Vo)/2. Do you see the problem? I understand that if we do that the nodal equation will also change.

    At "2" it says Vo = -4 V. What does this mean? In what situation does the current flows from ground to some other point (at -ve potential) in real world?

    At "3" having Rth = -4 ohm is completely incomprehensible for me. Please help me with it. I know the author gives some explanation. But one ohm resistance is equivalent one volt used per coulomb. So having a -ve sign with resistance doesn't make sense because it's directionless quantity.

    At "4", can we take the direction of current CCW instead of CW? (To see what is author doing here in this circuit please have a look on link #2).

    At "5", why has the author used -4i when he has used +ve sign for +9i?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    How about showing us your efforts on one of the problems as a starter?

    hgmjr
     
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    What would you imagine to be the meaning of the fact that the results for Rth is negative?

    hgmjr
     
  4. crazyengineer

    Member

    Dec 29, 2010
    156
    2
    1) For node equations, you must pay attention to the direction of the current and the convention you use.
    2) The only times I can think of which you have a negative voltage is where you apply your voltage. From inspection, it seems like the voltage controlled source is greater in magnitude then the 10v source and in the example is wired negative to positive, so there's a negative potential at vo (which I'm assuming is at the top of the 2ohm resistor).
    3) There's a controlled voltage source. You cannot use resistor reduction since you cannot turn off a voltage controlled source, so they just divided the voltage by the current
    4) Yes, as long as your consistent in the rules you use (e.g going from negative to positive in a loop equation results in a "positive" voltage drop)
    5) In single current mesh equations, Resistors are passive components, so you just write down the resistance value times the current following through it.

    Hope this helps
     
  5. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Sir, :)

    I have already shown you the work I have done. I was trying to understand Thevenin theorem and an encountered a little weird sample problem, thought about it but couldn't reach anywhere. These things and the questions I ask might seem easy and silly to you because you are presumably an expert in your field. For example, this is first time in my life I have read anywhere that a resistance could be -ve. I do believe there would also be first time for you too when you read such thing(s) and it took you some time and help to understand such stuff.

    Best regards
    PG

    _____________________________________________________________

    Let me rephrase some of my misunderstandings which are hampering my understanding...

    Please have a look on the linked diagrams:
    http://img810.imageshack.us/img810/1671/imgef.jpg

    1:
    The nodal equation used by the author in link #1 in my first post is the one for the Diagram 1. Right?

    But in the circuit diagram Ix is flowing upward as shown in Diagram 2. So, in my humble opinion (which I'm almost sure is wrong :) ) the equation I have written below the Diagram 2 should be applicable because both Io and Ix are flowing toward the same junction.

    Do you see the problem I'm having? How can we assume some other direction for the current when we have explicitly been told the direction of the current (which is "Ix" in this case)? Yes, in the case of the branch contain 4 ohm the current will flow downward toward the ground.

    2:
    The Vo has been found to be -4V. Vo is equivalent to Vab ( where, Vab = Va - Vb ). So Vo = -4V simply tells us that Va is at lower potential for the circuit shown.

    3:
    Okay, the Rth = -4 ohm.

    As the author says the -ve sign tells us that the circuit is supplying power.

    (1) P = VI, (2) P=I^2.R, (3) P=V^2/R

    So, for the power to be -ve the R should take -ve values because otherwise for (2) and (3) power can never be -ve because I^2 and V^2 can never be negative. Perhaps, when the current I flows into the resistor R from lower potential point and comes out of the higher potential point the resistor is said to be -ve (and perhaps this is only done for the same of mathematical computations!). Generally, the current I enters the resistors from the point of higher potential.

    Thank you very much for all your help and time.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    If you look closely at the equation 4.10.1, you will notice that the equation has been written consistent with the assumption that all currents leaving the node are positive and currents entering the node is negative. That means that all of the terms have been assumed consistently. Notice that the 1A source that was added as an aide to deriving the solution has been given a negative sign even though it is clearly drawn flowing into the node. As long as one is consistent with their assumptions of current direction going into a solution, then the sign of the solution will work out in the end.

    hgmjr
     
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