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Half power frequency divides by sqrt(2) instead of 2 why?

  1. Aug 16, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    1. The problem statement, all variables and given/known data
    Problem is to find the half power frequencies of a RLC circuit.
    Z = √(R2 + (WL-1/WC)2)

    2. Relevant equations

    Power(P) = V2/Z
    Pmax = V2/R <---because Z min = R

    3. The attempt at a solution
    P/2 = V2/2R = V2/Z
    Z=2R
    (R2 + (WL-1/WC)2) = 4R2
    (WL-1/WC)2 - 3R2 = 0
    W2-3WR/L-1/LC = 0
    Solve for w to get half power frequencies.

    Book gets : W2+-WR/L -1/LC = 0

    Cant figure out why Pmax is divided by sqrt(2) instead of 2
     
    Last edited: Aug 16, 2016
  2. jcsd
  3. Aug 16, 2016 #2

    Twigg

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    Did you make sure to convert from amplitudes to RMS values of voltage and/or current?
     
  4. Aug 16, 2016 #3
    To convert V2/R to VRMS2/R you divide V by 1/√2
    It would square, and become power = V2/2R
    P/2 = V2/4R
    Trying to get P/2 = V2/R√2
    I dunno... interesting tip though. 10 min messing with it and I am still not seeing it though.
     
  5. Aug 16, 2016 #4

    Twigg

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    The issue isn't RMS/peak conversions, sorry for the confusion. I missed this the first time I read your post, but ##P \neq \frac{V^{2}}{Z}##. The correct relationship is ##P = RI^{2} = R\frac{V^{2}}{Z^{2}} = R\frac{V^{2}}{R^{2} + (X_{L} - X_{C})^{2}}##. I can't tell what the book is doing from the information given though. Can you show their steps and refer us to the book and page?
     
  6. Aug 16, 2016 #5

    cnh1995

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    Is there resonance involved? Is that the complete problem statement? Check out this thread if it is any help.
    https://www.physicsforums.com/posts/5372988/
     
  7. Aug 17, 2016 #6

    NascentOxygen

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    Only if Z is a pure resistance AND the full voltage V is across that pure resistance Z.

    For your series RLC circuit, at resonance, Z is purely real, and equal to R Ohms.
    Power = power in the resistance = I2⋅R

    Off resonance, at a frequency where X = R, the value of Z has increased to magnitude ##\sqrt{2}{R}## so current is reduced by a factor ##\frac 1 {\sqrt2}##
    At this reduced current, what now is the power in the circuit's series resistance?
     
  8. Aug 17, 2016 #7
    Schaums outlines Electric Circuits 5th edition Q12.5

    Here is the complete information :
    For the series RLC circuit shown in fig. 12-36, find the resonant frequency ##W_{0}=2πf_{0}##. Also obtain the half power frequencies and the bandwidth β. V(w), R=100, L = .5H, C=0.4u (V(w) is a variable. I dont care about the numbers I only care about the derivation because I want to understand this...)

    At resonance, ##Z_{in}(w)=R## and ##w_{0} = \frac{1}{\sqrt{LC}}##
    ##w_{0} = \frac{1}{\sqrt{0.5(0.4x10^-6)}}=2236.1rads## ##f_{0}=\frac{w_{0}}{2π}=355.9 Hz##

    The power formula ##P = I_{eff}^{2}R = \frac{V_{eff}R}{|Z_{in}^{2}|}## shows that ##P_{max} = \frac{V_{eff}^{2}}{R}##, which is achieved at ##w=w_{0}##, and that ##P= \frac{1}{2}P_{max}## when ##|Z_{in}|^{2}=2R^{2}##; that is, when ##wL - \frac{1}{WC} = +-R## or ##w^{2} +- \frac{R}{L}w - \frac{1}{LC} = 0##

    corresponding to the upper sign, there is a single real positive root ##w_{h} = \frac{R}{2L} + \sqrt{\frac{R}{2L}^{2} + \frac{1}{LC}} = 2338.3## rad/s or ##f_{h} = 372.1Hz##

    and corresponding to the lower sign, the single real positive root ##w_{l} = -\frac{R}{2L} + \sqrt{\frac{R}{2L}^{2} + \frac{1}{LC}} = 2138.3## rad/s or ##f_{l}=340.3## Hz

    Another thing I dont get is why the +- in the quadratic formula is ##+-B+\frac{\sqrt{B^{2}-4AC}}{2A}## instead of ##B+-\frac{\sqrt{B^{2}-4AC}}{2A}## I know it has to do with that +- R but I guess it just becomes arbitrary when you have a +- B term? even though doesnt it change the answers? Still, my main question is figuring out once and for all that blasted ##\sqrt{2}## ! This took like 25 minutes to type up! So ill read the other replys after a break!
     
    Last edited: Aug 17, 2016
  9. Aug 19, 2016 #8

    BvU

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    Quiet in this thread ...

    Re: "why the +- in the quadratic formula is ##\ \ \displaystyle \frac {\pm B+\sqrt{B^{2}-4AC}}{2A}##."

    You were looking at ##
    |Z_{in}|^{2}=2R^{2} ## which is satisfied if ##
    w^{2} +\frac{R}{L}w - \frac{1}{LC} = 0## and also if ##
    w^{2} - \frac{R}{L}w - \frac{1}{LC} = 0##.

    That's two quadratic equations. If there is a resonance, each has two real solutions, one positive ##\omega## and one negative. You only want the positive.
     
  10. Aug 22, 2016 #9
    Interesting. Is there a theory behind why to only take the positive roots that I can google?
     
  11. Aug 22, 2016 #10
    When you say "Only if Z is a pure resistance AND the full voltage V is across that pure resistance Z.", Can you elaborate a bit... I am trying to find the Z where ##P=\frac{1}{2}Pmax##. Max power is where the impedance is pure resistance and the frequency is resonance. So if power = ##\frac{V^2}{R}## at resonance, there should be a answer for Z where that power is 1/2. So voltage, resistance, everything is known but Z in the equation ##\frac{V^2}{2R} = \frac{V^2}{Z}##. So if I am trying to find the value for Z where it would be equal to half of pure resistance. Can you elaborate a bit on why thats wrong because I am staring blankly lol

    Then you said : "Off resonance, at a frequency where X = R" <--- (do you mean AT resonance, X = R ?), "the value of Z has increased to magnitude ##\sqrt{2}R##"

    At resonance, ##Z=\sqrt{R^2 + (WL-\frac{1}{WC})^2} = R^2##. So, how does ##Z=R^2## become ##\sqrt{2}R## ?

    I see though if I take for face value and plug ##P=IV = \frac{I}{\sqrt{2}}V = \frac{V^2}{\sqrt{2}R}## it works out...
    But at Vmax P=ImaxVmax, Z=R. Why does ##I = \frac{1}{\sqrt{2}}Imax##? One of my assumptions is probably wrong I know, since I am getting the problem wrong.
     
  12. Aug 22, 2016 #11

    BvU

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    No. Your driving frequency is positive, so the response frequency is positive as well.
     
  13. Aug 22, 2016 #12

    NascentOxygen

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    I mean move away from resonance until you find a frequency where reactance X equals resistance R, in magnitude. As you know, precisely at resonance, in a series circuit the total reactance = 0 ohms.

    The equation ##\require{enclose}\enclose{box}[mathcolor="white",mathbackground="yellow"]{\color{black}{power=\frac{V^2}Z}}## is correct only where Z is resistance (no part of Z can be reactance in that equation), and V must be the voltage measured across that resistance (and not across the whole RLC circuit).
     
  14. Aug 23, 2016 #13

    ehild

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    Your power formula is wrong. Z is the magnitude of the impedance. The power is P=V I cos(θ), where V and I mean the rms values of voltage and current, and θ is the phase lag of the current behind the voltage. Now, you know that I=V/Z, and the resistance is the real part of the impedance, R=Zcos(θ). Therefore ##P=\left(\frac{V^2}{Z}\right)\left(\frac{R}{Z}\right)=\frac{V^2R}{Z^2}##
    You get the same formula by using the fact that AC power is dissipated only on resistors. The power is P=I2R, and I=V/Z, so ##P=\frac {V^2R}{Z^2}##. At resonance, Z=R, so Pres=V2/R.
     
  15. Aug 27, 2016 #14
    God thanks dood I am finally seeing it now.
     
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