Bridge type circuit (Electric DC current) : Thevinin circuit

Discussion in 'Homework Help' started by picopaul, Jun 25, 2015.

  1. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    Hi there,
    I got this question from a student I was helping. This circuit is a WheatStone bridge type circuit.
    The image of the circuit is show here below.
    These types of circuits are usually solved with Delta-Y or Delta-star, etc , kinds of techniques.
    And I solved it this usual way.

    BUT, I have always been wondering if this kind of circuit can be solved by using just Thevinin Theorem and using the regular Kirchoffs laws for series and parallel resistors:
    Original circuit: Circuit01.png

    Note: this circuit has inductors, but just replace all inductors with resistors and replace the henry with ohms.

    Next I did some circuit reduction and re-drew the circuit to a form that is a Wheatstone Bridge type of circuit.
    I made a drawing of this reduced circuit:

    CircuitReduced.png
    We want to find Vth and Rth, between points A-B.
    Based on my analysis, Va-b = Vth = Vsource = 120 volts.

    The difficult part is finding Rth.
    Using Thevinin Theorem, one must take the voltage source and short circuit it.
    Then try to used Kirchoff rules to find the equivalent resistance between points A-B.

    Without using Delta-Y type of transformation, I am not sure how to reduce these resistor just with Kirchoffs rules.

    My question is: is it even possible to get equivalent resistance with Thevinin approach.
    Or is it only possible via the Delta-Y type of transformations.

    In the Electrical Engineering circuit courses I took as an undergrad, they "Never Proved" the theorems such as Thevinin, Norton and Delta-Y transformations, never showed the derivations or proofs for these.
    They just showed how to use them, and presented these as the Holy Bible, to be accepted as Holy Facts, not to be questioned, just used.

    Due to this way of teaching in Engineering, one does not have the proper knowledge to understand when a certain approach is to be applied and when not.

    Hope someone can help.
    Regards,
    P
     
  2. WBahn

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    Mar 31, 2012
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    This is not a reassuring thought. More on that later.

    Kirchhoff's laws are fundamental and are not "for series and parallel resistors". They apply to ANY circuit (provided you are dealing with conservative electric fields, which is usually the case).

    I don't see that anything has been reduced, other than combining the series resistors in each branch. It is no more nor no less in the form of a Wheatstone Bridge circuit than it was to begin with.

    So what's the problem? You have a short circuit in parallel with a bunch of resistors. What's the effective resistance of R1 and R2 if one of them is zero ohms?

    You never HAVE to use delta-wye conversions. They are merely a tool that CAN make some circuits easier to analyze. You can always fall back on the fundamentals of KVL/KCL and techniques such as mesh current analysis (MCA) and node voltage analysis (NVA) which are nothing more than formalized ways of applying KVL and KCL.

    The first thing you need to consider is what, in this context, is meant by "solving" the problem? Find ALL of the voltages and currents in each branch/component? Or finding the voltage across the bridge elements?

    If you are trying to avoid delta-wye conversions, then your approach needs to break up the part of the circuit that would lead you to use a delta-wye conversion. Your choice hasn't accomplished that.

    A few quick observations about the circuit.

    1) Do you see that the 30Ω resistor and the bridge circuit are completely isolated from each other by the voltage source? Thus you can analyze those parts of the circuit completely separately. That resistor will have 4A of current flowing downward in it. Done. Move on.

    2) The Thevenin resistance at A and B is 0Ω because you are looking into an ideal voltage source.

    A much better choice for your Thevenin terminals is to choose the bridging components (the 8Ω resistor in your modified circuit) as your load.


    This is very sad and, frankly, you got ripped off by your school if this is the case. Was this, by change, a for-profit school? There are certainly some good for-profit schools (as well as quite a few bad non-profit schools), but there is a very high correlation between being for-profit and being bad.

    And that's because all they taught you to do was to imitate a monkey imitating an engineer. And now you are in the position of only being able to teach other students how to imitate monkeys.

    At least you are trying to identify and overcome your deficiencies -- my hat's off to you on that score because that's the first step that so few people in your shoes are willing to take.

    I would recommend getting a good Circuits I text and working your way through it step by step making sure you understand how to derive each and every single equation and technique in there. The ONLY things you should accept as given are KVL, KCL, and the constitutive (defining) relations for a resistor, capacitor, and inductor (and that's only because you should have been through the derivation of these in Physics II -- if you don't know where these come from, start with a Physics II (Intro E&M text)).
     
  3. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    Hi there, thanks for your very detailed analysis. In terms of school, this is a school in Canada, will not say which one. But they were very bad teaching, on purpose, because they take in a lot of students, and want to then weed out many. So they purposely did not teach the students well, for the sake of weeding out and then making money off of them to re-take the courses as a second change to get back in the program, so they can make a profit.
    SO the profit making technique goes like this: Say you only want to graduate 100 engineers at the end of 4 years.
    SO you take in 500 in first year, then weed every year a certain amount, until you get only 100 students at the end, and to NOT teach the material very well.

    I learned how to think in the Sciences, went to a different and good school for that.
    But the other Engineering school, they wanted to teach you more to be a Technician than an better Engineering.
    SO I got taught to be a Monkey in Engineering!
    Since I did my science degree first, and saw how things were done at this Engineering place, I knew they were ripping me off, but basically had no choice, once I got there. There was ZERO support there to learn!

    BUT, on to the question here, I also restrict myself to not using Node or Mesh analysis either.

    I will need some time to look at you very detail post here.
    Thanks for your great support.

    P
     
    Last edited: Jun 25, 2015
  4. WBahn

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    You don't need to teach bad in order to weed out -- quite the opposite, in fact. By teaching good and holding the standards high, you weed out the students that can't handle the material and educate the ones that can. But for-profits often take an attitude close to what you describe. They don't want the students that can't meet high standards to go away, they want them to retake the course several times thinking that they are capable of getting through it eventually and to convince them of that you need to have low standards. But that then means that, as you identified, you need to NOT teach the material well. Now, having said that, I will say that most of the teachers in any classroom, be it for-profit or nonprofit, are going to do their best to teach the material as well as they can given both their abilities and the abilities of the students. In a typical for-profit, neither the teachers nor the students are adequately prepared for their role.

    If you don't want to do Node or Mesh, then just use KVL/KCL.

    The first step is to annotate your diagram with branch currents.
     
  5. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    Not sure how to make a boxed area as you have, i think it is referred to as a blockquote.
    If I have short-circuited the source and want to use KCL & KVL (for resistors in this case), how does one conclude that one of the resistors or more are zero ohms. These are all closed paths that I can see, and if a source was connected then there would be current flowing thru all the resistors.

    The only time I would see a resistor becoming zero is if there was a current source that one would "open circuit" in this kind of analysis. And the resistor near this circuit would be hanging in free space not part of any closed path.

    So having difficulty still trying to see what you are saying with the particular sentence I am quoting you here.

    P
     
    Last edited: Jun 25, 2015
  6. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    So, just to reply to your recent post. When the voltage source is operational, there will be current flowing thru all branches, I believe. I need to do another diagram and make labels, to label the Resistors as R1, R2, R3,... etc
    then one can refer to the currents in each resistor as I1, I2, I3,... etc.

    P
     
  7. WBahn

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    To encode a block quote you surround the quoted text with [QUOTE] and [/QUOTE] tags.
     
  8. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    Thanks earlier for letting me know that KVL and KCL are universal, and any circuit(given the condition you mentioned) can be analyzed from the ground up. Which means that all theses Network theorems must have been derived from the basis of KVL and KCL.
     
  9. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    Ok, thanks for that!
     
  10. WBahn

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    It the voltage source that is short-circuits that is zero ohms. What's the resistance of a short circuit?

    The reason that you replace a voltage source with a short circuit is because what you are really doing is setting its output to 0V. Such a voltage source will source or sink whatever current is necessary in order to maintain 0V across it's terminals. Well, this is a pretty good description of a short circuit, so a 0V ideal voltage source and an ideal short circuit are interchangeable.

    Similarly, the reason that you replace a current source with an open circuit is because what you are really doing is setting it's output to 0A. Such a current source will produce whatever voltage is necessary in order to maintain 0A through it. Well, this is a pretty good description of an open circuit, so a 0A ideal current source and an open circuit are interchangeable.
     
  11. WBahn

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    EXACTLY!
     
  12. WBahn

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    Close. If I tell you that I1 is 3A that only tells you the magnitude of the current in R1. But a current flowing left to right in R1 is VERY different from a current flowing right to left. Signs matter -- big time.

    Your diagram needs to make it unambiguously clear what the polarities of ALL of the voltages and currents are that you are going to use in your analysis.
     
  13. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    Hi,
    Yes, I am familiar with Open circuit and Short circuit.
    Now you are referring to labeling directions of the current, and you are saying that this is paramount.

    But when I was taught Thevinin analysis, as in this case only one voltage source, once you find out Vth,
    then to find Rth, you just need to know which resistors have a current going thru them, and not the directions, because you are left with a bunch of resistors and you need to use some techniques to reduce them to one equivalent resistor.
    I am trying to follow this path of logic.

    Not sure if we are following the same ideas here.
    Let me know if I am wrong with my approach to doing Thevinin.

    P
     
  14. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    Hi,
    In terms of current direction, you will see in the first diagram, that there is a star symbol on every circuit element. The way I interpret that star symbol, is that that is where current enters the element and the other side of the element that does not have a star, would be where the current would exit the element.
    I am assuming Conventional Current Flow (positive flow that is). It is quite obvious from the voltage source +- symbols in the original figure.

    Also, I forgot to mention this, once one figures out Vth and Rth, then one can find the current I coming from the voltage source. So this is the aim.
    P
     
    Last edited: Jun 25, 2015
  15. WBahn

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    Not unless you know the resistance of the load, which with the say you set up your Thevenin terminals means that you still have that troublesome bridge as part of your load.
     
  16. WBahn

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    Are there times when you only need the magnitude? Sure. Just as there are times when you only need the direction.

    But current has both a magnitude and a direction and if you get sloppy about that you will come to grief far more often than is necessary. We've seen it time and time again here and I've seen it in real work engineering projects on more than one occasion.
     
  17. picopaul

    Thread Starter New Member

    Jun 25, 2015
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    So the funny thing is, this question wants you to find out the equivalent resistance about the points A-B.
    Usually questions of this type in textbooks would put a resistor load across A-B. But I guess that the 30 ohm resistor to the far right, is in parallel to the bridge circuit and is across the points A-B. I think I saw a practical implementation of something like this a while back when I helped a Mechanical engineer with a Strain Gauge device and its circuit representation.

    And yes, that is what makes this circuit difficult because the Bridge is part of the load!

    Also from the textbook problems that I have done in the past in terms of Thevinin problems, I only recall doing problems where we only had to know that there was some current in a resistor once one shorted and or opened source to find Rth. And Never incorporated direction of current.
    So this could be how the Engineering school gave a very slanted view of Thevinin theorem, and in this circuit course, we used a very custom book created by the Electrical Engineering department, and was only used in that Electrical Eng program at that University. SO we did not use the standard textbooks that were out there used at other Universities.
     
    Last edited: Jun 25, 2015
  18. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You may be looking for this method:
    http://www.hallikainen.com/rw/theory/theory6.html
     
  19. WBahn

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    I would say that this pretty clearly is looking for the effective resistance (impedance in the original circuit) as seen by the 120V DC source.

    If you do not want to use delta-wye transformations, then probably the easiest way to find the equivalent resistance is to apply a test voltage, Vtest (and 120V will do just fine), use KVL/KCL to find the resulting current, Itest, and then recognize that, by definition of what an equivalent resistance is, Req=Vtest/Itest.

    In general, if you only have one current then it usually doesn't matter what the direction is. But as soon as you have two or more currents, direction matters because at some point you are going to have to do some math on them and you need to know whether they add or subtract and that is completely dictated by their relative directions.

    This is probably not a good sign. While I have problems with just about every EE text out there (because I haven't found one yet that properly tracks units), the fact remains that a text that is used by many universities and has had several editions published has been pretty thoroughly vetted. Your university when out on its own because it didn't feel its students were up to a standard text and/or figured that it could make that much more money by forcing students to by its text.
     
  20. WBahn

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    That's definitely one of the classic ways of doing it. And it involves a very useful trick that is often and easily overlooked -- namely adding a second voltage source in parallel with the first, moving it onto the other side of the circuit, and then realizing that no current will flow in the connecting wire so that it can be severed. Do this breaks up the bridge circuit.
     
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