Ques. about Concept of Thev Analysis

Discussion in 'Homework Help' started by freemindbmx, May 16, 2014.

  1. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    I don't need them to be worked out,I just want to make sure I understand the process using mesh analysis on these circuits for Voc.

    Problem 1) (aka Thev Concept)
    To determine Voc:

    I1:
    9KΩI1=15V
    I1=15V/9KΩ -> Voc=6(15V/9KΩ)=0.0017V

    I didn't include the Dependent Current Controlled Voltage Source and 4Kohm resistor since it isn't a loop that is closed,and doesn't have any voltage across it b/c im treating A and B as an open circuit?Right?
    Problem 2) (aka Thev Concept 2)
    To determine the voltage across A and B, for Voc:

    I1:
    9ΩI1-6ΩI2=16V

    I2:
    -6ΩI1+12ΩI2=0
    I1=4/3A
    I2=-4/3A
    To determine Voc we subtract the difference of B-A=Voc(a,b)

    Voltage at B:
    6Ω*(-4/6)=-4 (I added the 4ohm and 2ohm together since there isn't any current through AB wouldn't this mean that the 4 and 2ohm resistors are in series?
    Voltage at A:
    6Ω(-4/6)=-4

    So B-A= -4-(-4)=0 ? Correct?
     
    Last edited: May 16, 2014
  2. WBahn

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    Mar 31, 2012
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    So if your claim is that the dependent voltage source has no voltage across it, and if the voltage across it is proportional to the current Io, what does that require be true for the current Io?

    Put another way, if it was a 20V battery instead of a dependent voltage source, would you still claim that it has no voltage across it?
     
  3. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    say there was a independent voltage source instead of the dependent voltage source (2000Io) that depends on Io(which is Io=I1-I2. Then depending on the polarity there would be a voltage across A and B right .Would I just write up a KVL to show this relationship like:
    -6ΩI1+4ΩI2?+2000Io=0
     
  4. WBahn

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    Mar 31, 2012
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    Which problem are we talking about? The first problem has a dependent voltage source but no I2 and the second has an I2 but no dependent voltage source.

    This is why you should restrict your thread to ONE problem.

    So please pick ONE of the problems, tell us which one you've picked, and let's get through that problem before even looking at the other one. Okay?
     
  5. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    The first problem,I will post only one from now on. But when trying to figure out the voltage across A and B for Voc, how would we find this?

    would we use the branch with the 4ohm resistor and the 2000Io dependent source with respect to Vab. And write a
    KVL : 4i+2000Io+V(a,b)=0 ??

    I chose the lowercase I to represent the current through the 4ohm resistor (4i).
     
  6. WBahn

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    Mar 31, 2012
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    Even if your equation were structurally correct, your answer would be off by about three orders of magnitude because you don't care about tracking your units. It is not a 4 resistor, it is a 4 kΩ resistor.

    KVL is for the voltage drop around a closed loop. You equation only takes you from point B up to the junction of the 4 kΩ and 6 kΩ resistors.

    Let's call that junction point C and let's call the junction between the 4 kΩ resistor and the dependent voltage source point D as in the following figure.

    [​IMG]

    And it's not enough to say that lowercase I is the current through the resistor since that doesn't indicate which direction the current is flowing. If I gave you the choice of holding your head +3ft relative to the surface of a swimming pool or -3ft relative to the surface of a swimming pool, which would you choose? Before you decide, wouldn't it be nice to know whether I am measuring positive distances upward or downward?

    So I've done in the drawing what you should have done -- defined the current "i", including its polarity.

    Now let's start using the subscript notation like it's intended to be used. If we say "Va" (i.e., a single subscript) then we mean the voltage at point 'a' relative to some agreed upon reference node (i.e, "ground"). If we say "Vab" (i.e., a double subscript) then we mean the voltage at point 'a' relative to the point 'b'. For convenience, we'll use lower case in our subscript notation even though our nodes are uppercase since we can't show proper subscripts (without resorting to tex, which we don't need for this problem).

    So the voltage you are looking for is

    Vab = Va - Vb

    Now, the beauty of subscript notation is that it let's us do the math real easy. For instance, if I have three nodes, X, Y, and Z, then I can say:

    Vxy = Vxz + Vzy because this really means:

    Vxy = Vx - Vy (Left-hand side)

    Vxz + Vzy = (Vx - Vz) + (Vz - Vy) = Vx - Vy (Right-hand side)

    So in our case we can write:

    Vab = Vad + Vdc + Vcb

    Now, in terms of the quantities Io and i, what are

    Vad = ?

    Vdc = ?

    Vcb = ?

    Keep in mind that that "2000Io" needs to be (2000 V/A)Io or, more conveniently, (2 V/mA)Io.
     
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  7. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    Vad = 2000I1

    Vdc = 4KΩi

    Vcb = 6KΩI1

    these would indicate the voltage across them correct? If so I need to ifnd I1 to determine the voltages for Vad & Vcb,to calculate I1:

    I1:
    9KΩI1=15V
    I1=15V/9KΩ

    Vad=2000*(15V/9KΩ)

    Vdc= 4KΩi -> cant we solve for lowercase I by plugging Vad & Vcb in and finding the value of lowercase I? So lowercase I would be 0.0033A so
    Vdc= 4KΩ*(0.0033A)

    Vcb= 6KΩ*(15V/9KΩ)
     
  8. WBahn

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    Mar 31, 2012
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    You have GOT to start paying attention to polarity (and units -- STOP dropping them!).

    Look at the polarity of the dependent voltage source. Which terminal, A or D, is the more positive? If Vad > 0, which terminal, A or D is more positive? If Io>0, is (2 V/mA)*Io) going to be positive or negative. Now, with all of that in mind, what is Vad?

    Remember, by definition, Vad = -Vda.

    Now apply the same reasoning to Vdc and Vcb. If you need to, go back and refresh your understanding of the passive sign convention (although this doesn't really rely on that, but I have a feeling you are very week on those concepts and they are very important).
     
  9. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    Io=I1 (moving in same direction as I1 clockwise when I calculated this).

    Io=(15V/9kΩ)=0.0017A which is > 0.

    (2 V/mA)*Io) -->
    (2V/mA)*0.0017A)= (3.4V/A or .0034V/mA) and this is > then 0.



    Vad= (Va-Vd)= (0V-3.4V/A)= -3.4V

    The voltage potential difference from Va to Vd is -3.4V. Would we say Va is zero? even though it isn't common to our ground? Or do we just use a zero to represent that the voltage at Vd(+) is more positive then Va(-).
     
  10. WBahn

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    "common to our ground"? What do you mean? You have yet to specify a ground. If you want to talk about grounds, you need to pick one of the nodes and declare that to be your reference point (i.e., your "ground").
     
  11. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    Im confused as what to call Va.When calculating Vad.I went back and looked up passive signs and voltage potentials,but since Im saying from A to B, treat it as a open circuit to calculate Voc.There wouldn't be any current through Vad->Vdc normally(if the dependent source wasn't their).But since theirs a dependent source that's related to Io,theres a voltage.

    Say I call my ground to be directly underneath the 6KΩ resistor that shares a node with the (-) end of the independent 15V source and B. What would we label Va to be in reference to this ground when trying to find the voltage potential difference?When calculating Vad=(Va-Vd)
     
  12. WBahn

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    Mar 31, 2012
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    You are confusing current and voltage. What is the voltage across a 12V battery if nothing is connected to it so that there is no current flowing? It's 12V and the fact that no current is flowing is irrelevant.

    Node A is labeled Va regardless of what you call ground. The value of Va will depend on what you call ground, but not that fact that the voltage at Node A is Va which simply means the voltage at Node A relative to ground. Don't make it harder than it is.

    If you choose the ground you indicate, what does that make Vb equal to?

    To find the value of Va (or any other node), simply start from ground and add up the voltage gains as you go from node to node along ANY path that take you from the reference node to Node A (or whatever node you are working with).
     
  13. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    If I chose ground right underneath the 6ohm resistor,as stated earlier. And I add up all the voltages from ground to Node A:

    Va


    (6KΩ*Io)+(-4KΩi)+(-2000Io)=Va ?

    Hypothetical question:
    Say Io was flowing in the opposite direction,up instead of down. Would: (6KΩ*Io) be positive since Io(the current flowing through the 6KΩ resistor) is labeled in the opposite direction and the voltage would come up as a voltage drop in the opposite direction b/c of the direction the current is flowing through it.When adding up the voltage from ground to Node A?
     
  14. WBahn

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    Mar 31, 2012
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    I'll answer your question by working the problem in the detail that I had asked for so that you can see the inner workings.

    Vab = Vad + Vdc + Vcb

    Now let's take each term by itself.

    The voltage across the dependent source is such that the voltage at the positive terminal minus the voltage at the negative terminal is (2V/mA)Io. The positive terminal is Node D and the negative terminal is Node A, thus:

    Vda = (2V/mA)*Io

    Since Vad = -Vda, we have

    Vad = -(2V/mA)*Io

    If 'i', the current in the 4kΩ resistor, is positive, then the voltage at Node C will be greater than the voltage at Node D. Thus

    Vcd = (4kΩ)I

    As before, Vdc = -Vcd, so

    Vdc = -(4kΩ)i

    If 'Io' is positive, then the voltage at Node C will be greater than the voltage at Node B. Thus

    Vcb = (6kΩ)Io

    Bringing these all together, we have

    Vab = Vad + Vdc + Vcb

    Vab = - (2V/mA)*Io - (4kΩ)I + (6kΩ)Io

    Now, this is also equal to Va, but it is important to realize that this is true ONLY because we have picked the reference node such that Vb = 0.

    If you go through the same reasoning, but have Io defined to be in the opposite direction, then you will see that the sign of the third term will change.
     
  15. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    Ahhh...

    So to find the voltage now,we just have to plug in whatever Io is and then solve for 'i' and then plug the value of 'i' into Vab and that's our voltage Vab=Voc.
    For the second problem I posted labeled Thev Concept 2.The wheatstone bridge one.If there isn't any current through A and B for calculating Voc.Then the resistors on both sides are in series.I posted a picture of this.

    Mesh Analysis: Clockwise is my convention,and the resistors I came across will be labeled based on my convention.

    I1:
    11I1-6I2=16V

    I2:
    -6I1-12I2=0

    Solving for I1 and I2 gives I1=2A, I2=1A.
    For calculating Vab.If I chose ground to be directly underneath the 16V source.And with my convention for solving the currents I1,I2.

    B/c of my convention Node A is more positive then Node C.And to represent the voltage between the terminals I have to include the currents that influence it.

    Vac=6(I1-I2)

    Same goes for Vbc,since the gorund I chose and b/c of my convention for calculating current.

    Vbc=6I2
    Vcb=-6I2

    Vab= 6(I1-I2)-6I2..Correct?
     
    Last edited: May 22, 2014
  16. WBahn

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    On the first problem, you know by inspection what the value of 'i' is, correct?

    I'll look at your post regarding the second problem later.
     
  17. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    "on inspection", not sure I know what you mean by this sorry.
     
  18. WBahn

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    It means, "by looking at it."

    Is there a closed circuit in which current through the 4kΩ resistor can flow?

    If not, what does the current in the 4kΩ resistor HAVE to be?

    No calculation required.
     
  19. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    0.

    But lets say hypothetically:
    The dependent source was a current(or)voltage controlled current source. Then we would have to calculate "i' correct. If the current was going through it.
     
  20. WBahn

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    Mar 31, 2012
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    If you had a current source (dependent or independent) in that spot you would have an indeterminate problem that had no solution. On the one hand you would have an open circuit which would allow no current to flow but on the other hand you would have a current source that would produce whatever voltage was required to produce the programmed current flow. One of two things will happen -- the voltage will build until you get an arc across the gap or the voltage will reach its maximum possible voltage and no current will flow. In any case, there is insufficient information in the problem to get a definite answer.
     
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