Maxwell-Wein bridge

Discussion in 'Homework Help' started by SilverKing, Mar 8, 2015.

  1. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    Hi everyone,

    I was asked to determine the radian frequency (ω) in the following Maxwell-Wein bridge diagram:
    [​IMG]
    Where R1=500 kΩ, C1=0.01 μF, R2=5 kΩ, R3= 15 kΩ, Lx=0.75 H and Rx=0.15 kΩ.

    My Answer:
    Since:
    Zx Z1 = Z2 Z3 (Balance) --> (1)
    Z1=R1/(1+ωR1C1) --> (2) //Was edited from Z1=R1/1+ωR1C1
    Z2=R2 --> (3)
    Z3=R3 --> (4)
    Z4=Rx+ωLx --> (5)

    From (1) we get:
    Z1=(Z2 Z3)/Zx ---> (6)

    Putting (6) in the left-hand side of (2) [While defining the unkowns using the rest of the equations]:
    (75*10^6)/(0.15*10^3+ω*0.75) = (500*10^3)/(1+ω*0.05*10^-3)

    Solving for ω we get:
    ω=0 rad/sec

    Is that OK?

    EDIT: Equation (2) was edited
     
    Last edited: Mar 8, 2015
  2. WBahn

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    Mar 31, 2012
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    Always, always, ALWAYS ask yourself if the answer makes since.

    At zero frequency, what will the voltage be on the left side of the bridge? What will the voltage be on the right side of the bridge? Will the bridge be balanced?

    Always, always, ALWAYS track your units.

    What are the units on R1/1 + ωR1C1?

    Don't you think that there should be some phase angles involved in this somewhere?
     
  3. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    That's why I posted the question in the first place.

    Since there is L, there should be ω (frequency), but no matter how I tried, I couldn't get a value for ω other than zero. But if the frequency is zero, then we will end up with a DC source not AC.

    BTW, I didn't understand what you meant by "units', you meant like ohm, volt?
     
  4. WBahn

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    Having a DC source doesn't necessarily mean that it isn't a solution. But look at your system and ask whether a DC source IS a solution.

    And, yes, I'm talking about ohms, volts, and the like.

    You are adding two terms and you can't do that unless they have identical units. Just like you can't add (1 lb)+(1 kg) to get (2 something). You also can't add (1 cm)+(1 kg) and get anything meaningful at all.

    So if you want to add R1/1 to ωR1C1, they must have the same units. What are the units of (R1/1)? What are the units of (ωR1C1). Do they match?

    Now, if you MEANT that to be R1/(1 + ωR1C1) then you need to say so by using parentheses to override the standard order of operations (multiplication/division before addition/subtraction, remember?).

    You also can't ignore the phase relationships. Think of your household wall outlets (assuming you are in the U.S. or other country that uses split-phase power). Your household line voltages come in two sets that are both the same magnitude and 180° of out phase. If you put two lines from the same set across your stove element it will sit there and look stupid, but if you put one line from each set across it you can cook your roast for dinner. Phase counts.
     
    Last edited: Mar 8, 2015
  5. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    I know that I'm dealing with an AC source here, I don't understand what you said about DC source being a solution. Do you mean to convert the whole circuit to some equivalent circuit?

    Sorry, my bad. I meant to write R1/(1 + ωR1C1) in the first place. ^^"


    Unfortunately, my lecturer in this course (Electrical Circuits II) said that this is course out of his profession, so he taught us the simplest subjects in this course. We don't use complex numbers at all!

    From what you said, I understood that there should be voltage difference to turn on any electrical elemet in the house, right?
    So, in my case, there should difference in phase between voltage source and the voltage across the impedances? How does that help me if I don't know the voltage source magnitude? And How does that affect ω?
     
  6. WBahn

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    You were actually the one that proposed a solution that involved zero frequency (i.e., DC). What I'm saying is that just because there is an inductor and/or a capacitor does NOT mean that zero frequency is NOT a solution. It is a trivial matter to adjust R1 to a value that will result in a balanced bridge at DC. The question is whether or not that is the case for THIS circuit with THESE values. Even if it is (and, for the bridge to balance, it actually HAS to be the case), this is what is known as the "trivial" or the "uninteresting" solution, though in practice this would let you find the unknown resistance of the inductor, Rx.
    You might seriously consider going to a better school (assuming this is a university engineering program). You are being set up for failure because you will not be prepared to understand the material in your future courses.

    Let's use a really crude analogy that might get the gist of things across.

    Imagine taking a clear sealed tube that has some water in it (not full, maybe half full). Now grab each end of the tube in one of your hands and hold it out in front of you so that the tube is at chest height. Now hold one hand still and move the other hand up and down (above and below) the other hand and notice how the water sloshes back and forth. This is a poor visualization of an AC current flowing in the tube when one hand (the stationary one) is tied to ground and the other is a hot wire (the line voltage). Now move both hands up and down together. Notice that there is very little movement of the water within the tube since both ends of the tube are always at the same height. This would be the case if you put the stove element across two wires from the same line voltage. But now move your hands up and down but in opposite direction and notice that you get the most intense sloshing yet. This corresponds to putting the stove element across two wires from the opposite line voltages. The other difference between the two is the phase angle of the motion (i.e., where your hand was and what direction it was moving when the 'clock' started).

    In your bridge circuit you have a similar situation. The voltage on the left of the bridge is like an up and down motion of your left hand and the voltage on the right is like an up and down motion of your right hand. The reactances, resistances, and frequencies interact to determine how big the motion is and what it's phase is. You are trying to find a frequency that results in the motions being at the same amplitude and same phase.

    If you can use complex impedance, then finding the solution is very easy. If you can't, then what I would recommend would be finding the expression for the voltage (amplitude and phase) at the left side of the bridge and then doing the same for the right side. In order to balance, both the amplitude AND the phase of these two voltages have to be equal.

    Once you slug through this, you should actually come up with a non-intuitive result regarding the frequency that you need to drive the circuit at -- but it's a conclusion that is very handy in practice.
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I'd want my money back I paid for that course.
     
  8. WBahn

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    In spades!

    Stuff like this really irks me. Yes, people are going to get tasked to teach courses that are out of their comfort zone and, at times, even courses that they have no prior background in. Hell, I'm teaching one of those right now! But that just means that the instructor has the obligation to knuckle down that much harder and stay ahead of the students and learn the material well enough to teach it at a level appropriate for the course. I've been in that situation several times -- most instructors have -- and it actually has a silver lining in that if you are just learning the material it places you in a position to be struggling with the same concepts and stumbling blocks that many of the students are going to be struggling with. But to adopt the attitude that you are just going to teach the simplest topics and completely ignore major portions of the course is inexcusable.
     
  9. JoeJester

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    And those instructors that don't put forth the effort, don't deserve the compensation. My wife, many years ago took a course titled "Statistics using Excel" or something similar, but Statistics and Excel were explicitly linked in the course title. The professor stated on day one that he wasn't using Excel because he didn't know how to use it. The interim Chair hired that professor and tried to talk my wife into keeping the course, but she got her money back from that false advertising.

    Some can't fake it till they make it, especially without having the requisite knowledge ... mostly because their effort is non-existent. I'll bet the pass rate is high in the non-profession taught related courses.

    I guess with the advent of the "internet", the students can get their questions answered and not "bother" their professors with mundane questions.
     
  10. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    So, the "correct" answer in my case is the one my lecturer wants. -__-"

    I cannot move from this school -- at least not right now -- but I'm doing my best by teaching myself and asking here and there.


    I think I could imagine the water flow as the current flow through the circuit.

    One thing I need to be sure of: by the right side and the left side of the bridge, you mean Z1 and Z2 (as the right side), and Z3 and Zx (as the left side)?
    Can guide me to some website, textbook where I could learn how to determine complex impedance?
     
  11. WBahn

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    Any introductory electrical engineering circuits text should cover that. Phasors and complex impedance are usually introduced in Circuits I and then the concepts extended to transform methods in Circuits II.
     
  12. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    Thank you for your efforts. I'll do my best to learn that and came back with another answer if I could, but until then, ω=0 will be the answer although it's not "interesting" answer.
     
  13. WBahn

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    It's probably not an answer that is going to get you much, if any, credit. You can use DC to find the inductor resistance, but it does you no good in terms of finding the inductance itself.
     
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