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Archived Power factor at half power points for RLC circuit

  1. May 12, 2004 #1
    Can anyone help me with this homework problem? I worked it through, but the answer i get doesn't agree with the answer the book gives. I'll include my work, and if anyone can help, I'll be eternally grateful, at least for a while.

    The Problem:

    In a series RCL circuit the dissapated power drops by a factor of two when the frequency of the generator is changed from the resonant frequency to a nonresonant frequency. The peak voltage is held constant while this change is made. Determine the power factor of the curcuit at nonresonant frequency.​

    My Work:

    [tex]P_r = 2 P_{nr}[/tex]
    [tex]P = I * V * \cos \phi[/tex]
    [tex]\phi = \arctan \frac{X_L - X_C}{R} [/tex]
    At resonant frequency, [itex]X_L - X_C = 0[/itex]
    [tex]I * V * \cos ( \arctan \frac{X_L-X_C}{R} ) = 2 * I * V \cos (\arctan\frac{X_L-X_C}{R})[/tex]
    [tex]I * V * \cos ( \arctan (0)) = 2 * I * V \cos (\arctan\frac{X_L-X_C}{R})[/tex]
    [tex]I * V * \cos (0) = 2 * I * V \cos (\phi)[/tex]
    [tex] 1 = 2 \cos (\phi)[/tex]
    [tex]\cos \phi = \frac{1}{2}[/tex]

    This seems correct to me, but the book says the answer is [itex]\frac{\sqrt {2}}{2}[/itex]

    Am I mistaking the "Factor of two" part of the problem? Does this really mean that the power is squared?


    If anybody can help, thanks. On a sidenote, LaTeX is very nice, if a bit tedious to write in.
     
    Last edited by a moderator: May 14, 2004
  2. jcsd
  3. Feb 9, 2016 #2

    gneill

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    Staff: Mentor

    A complete solution is offered.

    At resonance the current is in phase with the voltage source, and the power factor is 1. In that case the power dissipated is simply

    ##P_o = \frac{V^2}{R}~~~##where ##R## is the circuit resistance and ##V## the fixed voltage.

    We are looking for the power factor for the situation where the dissipated power drops by a factor of two. That is,

    ## P = \frac{P_o}{2} = \frac{V^2}{2R}##

    Off-resonance there will be a net reactance, X. We won't worry about the individual reactances of the inductance and capacitance, only their net effect. Let:

    ##Z = R + j X## be the impedance at the off-resonance frequency.

    Then the complex power (apparent power) for the circuit is given by:

    ##p = V~I^* = V ⋅ \frac{V}{(R + jX)^*} = V^2 \frac{R + jX}{R^2 +X^2}##

    where ##I^*## means the complex conjugate of ##I##.

    Now, the real power dissipated is the real component of the complex power. Setting this equal to our desired value:

    ##\frac{V^2}{2R} = V^2 \frac{R}{R^2 +X^2}##

    Solving for the reactance we find that ##X = R##. This means that the impedance has a value ##Z = R (1 + j)##, which has a phase angle of 45°. The current, which is V/Z will have an angle of -45°. The cosine of -45° is ##\frac{\sqrt{2}}{2}##, so that is the power factor.

    Note that the reactance could be positive or negative (net inductive or net capacitive reactance), and that the power factor will be the same since ##~~~cos(-45°) = cos(45°) = \frac{\sqrt{2}}{2}##.
     
  4. Feb 9, 2016 #3
    Here's my two cents..
    At resonance,
    Pr=VIr..(cosΦ=1)
    At non-resonant frequency,
    Pnr=VInrcosΦ
    But Pr=2Pnr
    So,
    VIr=2VInrcosΦ
    ∴Ir=2InrcosΦ
    ∴V/R=2*(V/Z)cosΦ
    ...(R remains same in both the cases)
    ∴Z/R=2cosΦ
    Since R/Z is cos Φ, we can directly write,
    2cos2Φ=1
    ∴cosΦ=1/√2=0.7071
    (=√2/2 as given in the answer)
     
  5. Feb 10, 2016 #4

    gneill

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    Staff: Mentor

    Nice. That works. It's great to see variations!
     
  6. Aug 22, 2016 #5
    Why do you set ##I=\frac{V}{R-JX}## instead of ##\frac{V}{R+JX}## In other words, why use the complex conjugate and not Z as it is?
     
  7. Aug 22, 2016 #6

    gneill

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    Staff: Mentor

    Because the complex power is given by ##P = V I^*##. See, for example: An Introduction to Electric Circuit Theory by G. Williams, section 5.8. Since the V that we are using here is purely real, the way to form the complex conjugate is to change the sign of the imaginary term in the impedance in the denominator.
     
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