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I was reading about the quality factor of a series RLC circuit. The quality factor is defined as

$$2\pi*\frac{\text{maximum energy stored}}{\text{total energy lost per period}}$$

Here maximum energy stored is calculated at the resonant frequency. My doubt is, is there any other frequency at which the maximum energy stored in the circuit is greater than that at resonance?

I have done the following calculations.

Energy in storage at any instant, $$ \mathrm{Energy}(t)=\frac{1}{2}*\textrm{L}i_{L}^2(t)+\frac{1}{2}*\textrm{C}v_{C}^2(t)$$

$$\text{Let source voltage}=v_{m}sin\omega t$$

After some calculations

Energy in storage at any instant = $$\frac{1}{2}*\frac{v_{m}^{2}}{[R^{2}+(\omega L-\frac{1}{\omega C})^2]}*[L*sin^2(\omega t-tan^{-1}\phi )+\frac{1}{\omega ^{2}C}*cos^2(\omega t-tan^{-1}\phi )]$$

At resonance, this reduces to a constant value, $$ \mathrm{Energy}(t)=\frac{1}{2}*\frac{v_{m}^2*L}{R^2}$$

This is the maximum energy stored at resonance.The question is, is there any other radian frequency at which the function ' Energy(t) ' peaks greater than that at resonance?

Voltages across R, L, C & I versus frequency

Image source: Boylestad, Robert L. Introductory Circuit Analysis. Pearson Education.

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  • \$\begingroup\$ In case the chat room is inaccessible now that I deleted my answer, it's now clear that the maximum peak voltage across the capacitor does indeed peak at slightly below true LC resonance. I found this through simulation. Sorry for my misguided answer. \$\endgroup\$
    – Andy aka
    Commented May 21, 2023 at 18:23
  • \$\begingroup\$ @Andyaka . It's ok. \$\endgroup\$ Commented May 21, 2023 at 18:31

2 Answers 2

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The question is, is there any other radian frequency at which the function ' Energy(t) ' peaks greater than that at resonance?

I was wrong in my "now deleted" answer and, I should have been less hasty in posting it. For a simple RLC series circuit you can look at the voltage across the capacitor just as you would when designing a simple 2nd order low-pass filter like this: -

enter image description here

If my brain had worked immediately, I'd have posted this picture from my basic website: -

enter image description here

  • The maximum peak amplitude is: \$\dfrac{1}{2\zeta\sqrt{1-\zeta^2}}\$
  • This occurs at a frequency of: \$\omega_n\sqrt{1-2\zeta^2}\$

There is a proof further down that linked page. For the peak inductor energy, this occurs at true resonance (\$\omega_n\$) because the reactances cancel and, current becomes a maximum.

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  • \$\begingroup\$ This is my conclusion: For a particular frequency source, maximum energy is stored in the circuit at the moment when capacitor voltage peaks and inductor current is zero (except at resonance frequency where energy stored at any moment is constant). Capacitor voltage peak is maximum for frequency $$\omega_n\sqrt{1-2\zeta^2}$$ as shown in the frequency response above. Therefore, maximum energy storage occurs at $$\omega_n\sqrt{1-2\zeta^2}$$. Is this correct? \$\endgroup\$ Commented May 24, 2023 at 13:13
  • \$\begingroup\$ @LelouchYagami that's what my answer says; maximum capacitor voltage occurs at that frequency (slightly lower than resonance). \$\endgroup\$
    – Andy aka
    Commented May 24, 2023 at 13:31
  • \$\begingroup\$ If we consider the dual of this circuit, i.e., parallel RLC circuit, does the current through the inductor determine at what frequency maximum energy storage occurs? \$\endgroup\$ Commented May 24, 2023 at 14:22
  • \$\begingroup\$ Maybe ask that as a brand new question @LelouchYagami <-- without going into the math I believe it's simpler i.e. peak voltage coincides exactly at resonance. \$\endgroup\$
    – Andy aka
    Commented May 24, 2023 at 14:43
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For a overdamped series RLC, at 0 Hz (DC).

That's when energy is stored just in the capacitor. There is no current, so no power is wasted in the R and L.

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    \$\begingroup\$ This will be the case if the RLC circuit is overdamped. However, when the circuit is underdamped, there will be "peaking" of the capacitor voltage, as shown in AndyAka's answer. \$\endgroup\$ Commented May 21, 2023 at 20:20
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    \$\begingroup\$ Thank you. I edited my answer accordingly. \$\endgroup\$ Commented May 21, 2023 at 21:22

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