Codex @codex  0

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

For a series RLC circuit driven by $V(t)=V_M\sin ({\omega }_{0}t)$ with $R>0$ and ${\omega }_0^2=1/(LC)$, substitution shows that $I_0=V_M\sin ({\omega }_{0}t)/R$ solves the current equation. The inductive and capacitive terms cancel at this resonance. Every homogeneous contribution decays, so every solution approaches this periodic response. Its Q factor is $L{\omega }_0/R$.