Impedance for a RLC Circuit

Discussion in 'Homework Help' started by LockednLoaded, Jun 28, 2016.

  1. LockednLoaded

    Thread Starter New Member

    May 10, 2015
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    I've been having real trouble determining the Impedance (Magnitude) for the following RLC circuit. My end goal is to retrieve the total current of the circuit. I'm not even too sure if I'm going about getting the total Impedance the right way, or if a different approach is needed since it is a Series-Parallel RLC Circuit. I'd appreciate any help or guidance to point me in the right direction.

    Firstly, I collected the Resistance total (140.65 ohms), Total Inductance (.3 mH) as well as the Total Capacitive (1.18*10^-8) of the circuit.
    Then, using my values from RT / Xt / CT, I found the Inductive Reactance (47.1 H) and Capacitive Reactance (540) values.
    Finally, I tried using the RLC Impedance formula that I had handy using my magnitude values though I didn't reach a correct answer.
    [​IMG]
     
  2. DGElder

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    Apr 3, 2016
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    Inductance in series add just like resistors in series not like resistors in parallel.

    Reactance is not in units of Henrys, it is in units of ohms.

    Also it is a series RLC circuit, not series parallel.

    Correct those errors and show all your work: how you determine the complex impedance, current magnitude and phase..
     
    Last edited: Jun 28, 2016
  3. DGElder

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    Apr 3, 2016
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    shteii01, why are you doing his homework for him?
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    You assume I did it right.

    The worst part is that my trusty calculator needs new batteries so I had to fire up the MatLab to do the problem.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Yeah, that result in post 3 doesnt look right, but i'll wait a while to discuss it further.

    The result in post 1 looks ok for the resistance, but not for the L or C.
     
    Last edited: Jun 28, 2016
  6. DGElder

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    Apr 3, 2016
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    "You assume I did it right."

    I made no assumption. If you were correct you short circuit the OP's learning opportunity and maybe give him easy credit he didn't earn. If your answer was wrong it could be misleading and may cause the OP more confusion. As it turns out the later is the case.
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Ok. I know where I made mistake. The picoFarad I did as 10^-9, when it should be 10^-12.
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I redid the calculations.

    OP. The key to doing this problem is to convert the inductors and capacitors into "resistors". Inductor (L) is replaced with jwL or j*(2*pi*f)*L. Capacitor (C) is replaced with \frac{1}{jwC} or \frac{1}{j*2*pi*f*C}. Once you do all these substitutions, you have a circuit where all the elements are "resistors". The two inductors become two "resistors" in series. The two capacitors become two "resistors" in parallel. You already did the two resistors in parallel so you know their equivalent resistance. Once you are done simplifying the circuit, you will have one "resistor" that has complex representation.

    In case you don't recall: w=2*pi*f
    The f is given to you, it is 25 kHz.

    On the general note. In the course of my education, I found reactances to be completely USELESS. Don't even bother finding them, it is waste of time.
     
  9. DGElder

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    Apr 3, 2016
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    "The key to doing this problem is to convert the inductors and capacitors into "resistors". .........The two inductors become two "resistors" in series. The two capacitors become two "resistors" in parallel. ....... Once you are done simplifying the circuit, you will have one "resistor" that has complex representation........
    On the general note. In the course of my education, I found reactances to be completely USELESS. Don't even bother finding them, it is waste of time."




    No, no and.... wait for it....................... NO! :eek:

     
    Last edited: Jun 28, 2016
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Fine, fine.
    Capacitors in parallel add up.
    Inductors in series add up.
    Yes, that is simpler way of doing it. Then convert them to impedances and you got three "resistors" in series, a resistor + inductor impedance + capacitor impedance.

    The OP said they needed to find the current. The impedance of the circuit, that you and I are fighting over, is just an intermediate step to get the current. The frequency is given, for a reason or at least I think it is given for a reason. The current will be complex, it will have a real part and an imaginary part.
     
  11. WBahn

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    Mar 31, 2012
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    Your calculation for the effective resistance of the two resistor is correct. Carrying intermediate results to five sig figs is not unreasonable, but if you are doing to do so then you should round it properly in which case you should have 140.66 Ω. Very minor nit, so don't worry about it, just try to round values properly to whatever level of resolution you decide to record them.

    Inductances in series add just like resistors (and they combine in parallel just like resistors, too). You apparently combined them using the parallel resistor formula. But notice here that the result would have been .33333 mH if you were recording it to five sig figs. By recording it to only one sig fig, you are potentially rendering the five sig figs that you recorded the resistance to meaningless.

    The capacitance total is numerically correct, but you neglected to include the units. Always, always, ALWAYS include units (unless the value is unitless). So that's 11.8 nF. While technically you should have 11.800 nF to have five sig figs, it's an intermediate result and so there's no compelling need to actually write down trailing zeros.

    Now, we already know that 0.3 mH is incorrect, but your 47.1 Ω (it should be Ω and not H -- these are NOT the same) comes from using 0.3 mH. What if you had used 0.33333 mH? You would have gotten an inductive reactance of 52.360 Ω. That's a very noticeable difference. Do you see how it renders recording the resistance to five sig figs pointless?

    Your capacitive reactance is only recorded to two sig figs, once again rendering the five sig figs of the resistance a waste of time. You also need to include the units, which are again Ω. If you want to record things to five sig figs, then it should be 539.51 Ω. Now, I'm assuming that your text/instructor has you work with both inductive and capacitive reactances as positive values and then you have a bunch of formulas that have things like (X_l - X_c) in them. If so, that's a shame, but you need to stick with whatever is used by whoever is doing the grading. The far better way to do it is to have a reactance be a reactance with inductive reactances being positive and capacitive reactances being negative (and the even better way is to work with impedances, but those involve complex numbers and arithmetic that you might not be familiar with).

    You then say that you tried to plug these into the handy, dandy RLC Impedance formula but you didn't get the correct answer. Well, since you didn't tell us what formula you used (there's no universal "RLC Impedance formula" that I'm aware of) nor what value you got, there's not much guidance I can give you.
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Universal "RLC Impedance formula"...

    Isnt it sqrt(R^2+X^2) or something like that?
    Which boils down to sqrt(R^2+(xL-xC)^2).
    And of course:
    xL=w*L
    xC=1/(w*C)
    So when xL cancels xC we have xL-xC=0 and then just Z=R.

    I dont use it myself i use complex math for everything.
     
  13. WBahn

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    The fact that you use complex math for it instead of that formula means that that formula is the "universal" formula.

    The better form of that formula is simply

    |Z| = sqrt(R² + X²)

    X_L = wL
    X_C = -1/(wC)

    When you have X_C being positive and using (X_L - X_C), you set yourself up for problems and limit yourself to only being able to work problems in small chunks and for actual values instead of parametrically. Just consider that if you don't know the component values, or if you don't know the frequency, you don't know if the result is capacitive or inductive, which means you don't know if you need to add it or subtract it to the reactance of the next component, which means you can't work with them symbolically (without having a increasingly byzantine decision tree that has to be followed once you have the component values and/or frequency. But if you have capacitive reactances as being negative, then it doesn't matter since you always add them.
     
  14. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    "The fact that you use complex math for it instead of that formula means that that formula is the "universal" formula."
    I am afraid i dont understand what you mean here. How does the fact that i use complex math for computing something the reason that some other formula is 'universal'?

    Actually, (xL-xC)^2 is the most common form found in literature, especially for these simple cases of one of each type of element. That might be because xC is most often defined as simply xC=1/wC not xC=-1/wC. I guess it is all too easy to remember to subtract capacitive reactances especially when there is only one. And of course this goes for the resulting phase angle calculation too, which is also a bit ambiguous but still often used (ie atan() instead of atan2() ). Also, the form (xL-xC)^2 appears from right out of the calculation for a circuit with a single L and C.

    So in some literature we are, like it or not, going to see 1/wC and in others -1/wC.

    This is almost like your creative use of the double "that", where the second one can be dropped leaving just one 'that' and the meaning not changed :)
    I believe myself that the single 'that' is the more correct one, but the double use is also acceptable so we cant really say it is wrong.

    Side note:
    Hey notice the other thread was closed (ha ha). Nice :)
    I dont think i was ever so happy to see a thread close.
     
  15. dannyf

    Well-Known Member

    Sep 13, 2015
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    you will need to simplify it into three "impedance" first. You then need to realize that impedance is a vector (=resistance with non-zero phase angle). So when you add vector's together, you will need to consider their phase angle.
     
  16. WBahn

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    Mar 31, 2012
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    I agree. It is just a shame that we hobble people with having to track names around (is this reactance value sitting in front of me capacitive or inductive) instead of being able to just let the math track it nice and seamlessly.

    For instance, consider an inductor in series with a parallel combination of an inductor and a capacitor. You don't know the values of any of the components yet or the operating frequency, but you want to come up with a single expression for the total reactance. So you combine the parallel components and come up with a formula for it. But now how to you combine that with the series inductor? Do you add it to the inductor's reactance, or subtract it? You don't know, because you don't know if the parallel components are going to be inductive or capacitive. So you have a mess on your hands -- and that's just with three components. But if you let capacitive reactance be negative, then you can just combine any reactance of any type using the exact same formulas that you use for resistors and just let the math take care of itself all the way through.

    To see how bad this gets, consider a Wheatstone bridge in which each of the five elements is either a capacitor or an inductor, but you don't know which yet. Find a formula for the total Thevenin equivalent reactance as seen by the bridging element that you can evaluate once you know the reactances (a positive value and whether it is inductive or capacitive) of each of the four components.
     
  17. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Do they even give a formula for that? I am not sure, but i guess you mean this works better:
    parallel(L,-C)

    than this:
    parallel(L,C)

    I have to agree, and maybe that's why i turned exclusively to using complex math a long, long time ago :)
    The only time this comes up is when we see elementary questions on impedance.
    I also see a lot of 'short cuts' being talked about in three phase systems. If they used used complex math it would make the whole procedure one complete procedure for any problem.

    I just wish complex math was taught sooner than later.
     
  18. WBahn

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    Mar 31, 2012
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    Not quite sure I follow. What is "that" that you are referring to.

    If capacitive reactances are negative, then you don't need any special formula. If you have X1 and X2, the parallel combination of them is

    X12 = (X1·X2)/(X1+X2)

    So if these are in series with a third reactance, the total is

    Xt = X3 + X12 = X3 + (X1·X2)/(X1+X2)

    At no point am I talking about negative capacitance.

    I wish they taught complex math more, too, but we have to keep in mind that a lot of folks that work with reactances have little or know algebra skills at all, either, and that is the bigger handicap. They rely on page after page of formulas, most of which are simple algebraic manipulations of the same relationship, such as all the formulas on that wheel that is always being posted here.

    As for the three phase systems, assuming I know the kind of short cuts you are referring to, those are separate from whether complex math is used. They exploit the symmetry of a balanced three-phase system to significantly reduce the complexity of the circuit that has to be analyzed. This is particularly the case when you can transform a three-phase circuit into an equivalent single-phase circuit that is then analyzed using complex impedances. If you are talking about different types of short-cuts, please feel free to correct me.
     
  19. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Sorry, i had a lot going on today. What i meant to type was:
    parallel(xL,-xC)

    was better than:
    parallel(xL,xC)

    I am only concerned about the parallel part right now because that is the formula they dont always give.
     
  20. WBahn

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    Mar 31, 2012
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    Okay, let's just consider the parallel combination of an inductor and a capacitor.

    If define capacitive reactance as being negative, then the formula is ultra simple and ultra familiar:

    X = (xL·xC)/(xL + xC)

    The math very cleanly tracks whether it is inductive or capacitive. If X>0, it's inductive, if X<0 it's capacitive.

    If reactance is always positive, then you have to complicate things quite a bit and in a very non-intuitive way.

    Let's first look at how you add them in series.

    You can't just say

    X = xL - xC

    because if xC > xL you have a negative reactance, which is undefined since both inductive and capacitive reactances are positive. So you have to go:

    if (xL > xC), then X = xL - xC (inductive), else X = xC - xL (capacitive).

    See how much more complicated just the series formula is when all reactances are positive?

    But for the parallel case it is:

    if (xL > xC), then X = (xL·xC)/(xL - xC) (capacitive), else X = (xL·xC)/(xC - xL) (inductive).

    The (for most people working at this level) non-intuitive part is that if the inductive reactance is larger than the capacitive reactance, then the overall reactance is NOT inductive, but rather capacitive. If you think about it just a bit, however, this is not surprising since the effective resistance of two parallel resistors is dominated by the smaller resistance, so it shouldn't be surprising that the effective capacitance of two parallel reactors is dominated by the smaller reactance.

    But notice that this is for a parallel circuit containing an inductive reactance and a capacitive reactance. In general you also have to allow for two inductive reactances and two capacitive reactances, so given two (unknown) reactances X1 and X2, your generic formula for combining them becomes even more convoluted:

    IF:

    xL = wL
    xC = 1/(wC)

    then to find the effective reactance of two reactances in parallel:

    if (X1 and X2 are both inductive)
    ---- X = (X1·X2)/(X1 + X2) (inductive)
    else if (X1 and X2 are both capacitive)
    ---- X = (X1·X2)/(X1 + X2) (capacitive)
    else if (X1 is inductive and X2 is capacitive)
    ---- if (X1 > X2)
    ---- ---- X = (X1·X2)/(X1 - X2) (capacitive)
    ---- else
    ---- ---- X = (X1·X2)/(X2 - X1) (inductive)
    else if (X1 is capacitive and X2 is inductive)
    ---- if (X1 > X2)
    ---- ---- X = (X1·X2)/(X1 - X2) (inductive)
    ---- else
    ---- ---- X = (X1·X2)/(X2 - X1) (capacitive)
    else
    ---- X = infinity

    And that's just for two reactances in parallel. Now when you go to add another reactance in series it REALLY becomes a nightmare.

    Personally, if I didn't know about complex numbers, I'd find the following SO much better:

    IF:

    xL = wL
    xC = -1/(wC)

    then to find the effective reactance of two reactances in parallel:

    X = (xL·xC)/(xL + xC)

    But, hey, that's just me.

    EDIT: Typo corrected -- thanks for pointing it out, The Electrician.
     
    Last edited: Jul 1, 2016
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