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First construct the periodic response required in the introductory clause. At ${\omega }_0^2=1/(LC)$, the terms $L{I}_0^{\prime \prime }+I_0/C$ cancel for a sinusoid, leaving $R{I}_0^{\prime }={\omega }_0\cos ({\omega }_{0}t)$. ThusThis resonant series RLC response dissipates energy through Joule heating. Over one period,The Q factor, using the maximum magnetic energy $L{I}_M^2/2$, is consequently

The Q factor measures stored energy relative to loss: $Q=2\pi E/(\text{energy lost per cycle})$, or $Q={\omega }_{0}E/P_{\mathrm{loss}}$ in the weak-loss limit. For viscous damping $m\ddot{x}+b\dot{x}+kx=0$, write $\gamma =b/(2m)$, ${\omega }_0=\sqrt{k/m}$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The conventional oscillator parameter isIn free ringdown, fit the amplitude envelope $A(t)=A_0{e}^{-\gamma t}$. Its decay time is ${\tau }_A=1/\gamma$, whereas the cycle-averaged energy envelope decays on ${\tau }_E=1/(2\gamma )$; thus $Q={\omega }_0{\tau }_A/2={\omega }_0{\tau }_E$. Equivalently the logarithmic decrement between same-sign peaks is $\delta =\ln (A_n/A_{n+1})=2\pi \gamma /{\omega }_d$. It gives $Q\simeq \pi /\delta$ for weak damping, or the exact parameter relation $Q=\sqrt{1/4+(\pi /\delta {)}^2}$. Fit multiple peaks above the noise floor, measure the damped period, and verify an exponential envelope rather than assuming all amplitude loss is viscous.

For a driven measurement, allow transients to settle and scan frequency at fixed force amplitude. Near resonance and at high $Q$, the response is Lorentzian in displacement power, with half-power points at displacement amplitudes $1/\sqrt{2}$ of the maximum. Their angular-frequency separation gives $Q\simeq {\omega }_0/\Delta \omega$. Halving the amplitude is not a half-power criterion.

The absorbed-power bandwidth of a viscously damped oscillator can be found exactly. Sinceits maximum is at $\Omega ={\omega }_0$, and the half-power frequencies satisfy $|k-m{\Omega }^2|=b\Omega$. HenceFor broad resonances, displacement and absorbed-power peaks differ, so the measured observable must be specified. Frequency-dependent drive, backgrounds or several overlapping modes can otherwise masquerade as a changed Q factor.

Drive a series RLC circuit with electric potential difference $V(t)=V_0\cos (\omega t)$, where $V_0>0$ and $\omega >0$. After the free transient decays, the electric current has the same angular frequency as the source:For fixed positive $R,L,C,V_0$, the electric current amplitude is largest exactly at $\omega ={\omega }_0=1/\sqrt{LC}$. The two electrical reactance contributions cancel there, leaving only the drop from electrical resistance. In particular,Every solution approaches this periodic response because its difference from the displayed solution is a decaying free response. The peak remains at ${\omega }_0$ even for an overdamped RLC response; free ringing is not required for a maximum of driven electric current. ^[ref].

For example, stipulate $L=10\mathrm{mH}$, $C=1\mu \mathrm{F}$, $R=20\Omega$, and $V_0=1\mathrm{V}$. Then ${\omega }_0=10^4{\mathrm{s}}^{-1}$, $f_0={\omega }_0/(2\pi )\approx 1591.55\mathrm{Hz}$, and the peak electric current amplitude is $50\mathrm{mA}$. Holding the source amplitude fixed is essential when comparing responses at different frequencies.

Three frequencies must be distinguished. The free underdamped RLC response rings at ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$. Driven electric current peaks at ${\omega }_0$. Driven electric charge has amplitudewhose peak occurs at $\sqrt{{\omega }_0^2-2{\alpha }^2}$ only if $\alpha <{\omega }_0/\sqrt{2}$. Otherwise it decreases from its zero-frequency limit $C{V}_0$, with no positive-frequency peak. The different peak locations follow because electric current is the time derivative of electric charge, multiplying a sinusoidal amplitude by $\omega$. The limiting responses are $I_{\mathrm{amp}}\sim V_{0}C\omega$ as $\omega \to 0^{+}$ and $I_{\mathrm{amp}}\sim V_0/(L\omega )$ as $\omega \to \infty$.

The Q factor measures selectivity:The average Joule heating power is $R{I}_{\mathrm{amp}}^2/2$. Its half-maximum angular frequencies and their separation are exactlyThese follow by setting $|\omega L-1/(\omega C)|=R$. In the numerical example, $Q=5$ and $\Delta \omega =2000{\mathrm{s}}^{-1}$. Reducing $R$ makes a taller, narrower electric current peak; varying $C$ tunes the selected frequency. Measuring the drop $v_R=Ri$ gives a band-pass filter that suppresses very low and very high frequencies while passing a range near resonance. This is useful for selecting a radio signal, though the range must be broad enough to pass its modulation. ^[ref].

A spark gap can rapidly discharge a charged capacitor through an inductor. After excitation, the stored energy runs down through Joule heating, radiation and other losses. A key held down can permit repeated discharges: each discharge starts a damped train of rapid oscillations. The spark repetition frequency, the electrical oscillation frequency within each train, and the rate of keyed symbols are different quantities. Repetition does not make each train a maintained sinusoid.

An original lumped model replaces the conducting spark gap and other losses by a constant total electrical resistance $R>0$ in an isolated series RLC circuit. For electric charge $q$ on the capacitor, the Kirchhoff cycle law givesFor $Q>1/2$, its characteristic roots are $-\alpha \pm \mathrm{i}{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$, proving the stated damped form. The electric current $\dot{q}$ has the same exponential decay scale. Its wave envelope falls by $\exp (-2\pi \alpha /{\omega }_d)$ per cycle; the energy envelope falls twice as fast in the exponent. A maintained sinusoid instead requires continuing input power balancing losses.

With $L=5\mathrm{mH}$, $C=2\mathrm{nF}$ and $R=100\Omega$, the model gives $f_0={\omega }_0/(2\pi )=50.329\mathrm{kHz}$, Q factor $Q=15.811$, amplitude time constant $1/\alpha =100\mu \mathrm{s}$, and damped frequency ${\omega }_d/(2\pi )=50.304\mathrm{kHz}$. These are chosen component values. A real spark gap switches nonlinearly, an antenna radiates, and coupling exchanges energy between circuits; one constant-$R$ free decay does not model the complete transmitter. Rapid damping also broadens the frequency spectrum, which helps explain why short trains make sharp receiver selection difficult.