= Resonant series RLC response
{title2=$I_{\mathrm{amp}}(\omega_0)=V_0/R$}
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:
$$
i_{\mathrm{ss}}(t)=I_{\mathrm{amp}}\cos(\omega t-\phi),\qquad
I_{\mathrm{amp}}(\omega)=\frac{V_0}{\sqrt{R^2+(\omega L-1/(\omega C))^2}},\qquad
\tan\phi=\frac{\omega L-1/(\omega C)}{R},\quad -\frac\pi2<\phi<\frac\pi2.
$$
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,
$$
i_{\mathrm{ss}}(t)=\frac{V_0}{R}\cos(\omega_0t),\qquad
q_{\mathrm{ss}}(t)=\frac{V_0}{R\omega_0}\sin(\omega_0t).
$$
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>. <University Physics Volume 2 by Samuel J. Ling and others>[OpenStax University Physics 2, §15.3: current amplitude, phase, and resonance]{ref}.
For example, stipulate $L=10\,\mathrm{mH}$, $C=1\,\mu\mathrm F$, $R=20\,\Omega$, and $V_0=1\,\mathrm V$. Then $\omega_0=10^4\,\mathrm{s}^{-1}$, $f_0=\omega_0/(2\pi)\approx1591.55\,\mathrm{Hz}$, and the peak <electric current> <amplitude> is $50\,\mathrm{mA}$. Holding the source <amplitude> fixed is essential when comparing responses at different <frequencies>.
\Image[/rlc-current-resonance.png]
{title=Driven <electric current> <resonance> and the distinct <electric charge> peak}
{description=Left: fixed $L=1\,\mathrm H$, $C=1\,\mathrm F$, $V_0=1\,\mathrm V$, with positive $R=0.5,1,2,4\,\Omega$. Right: at $R=1\,\Omega$, the charge-amplitude peak, free ringing frequency, and current-amplitude peak occur at $\omega/\omega_0\approx0.707,0.866,1$, respectively. Both driven curves are divided by their own peak amplitude.}
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 <amplitude>
$$
q_{\mathrm{amp}}(\omega)=\frac{I_{\mathrm{amp}}(\omega)}{\omega}
=\frac{V_0/L}{\sqrt{(\omega_0^2-\omega^2)^2+4\alpha^2\omega^2}},
$$
whose peak occurs at $\sqrt{\omega_0^2-2\alpha^2}$ only if $\alpha<\omega_0/\sqrt2$. Otherwise it decreases from its zero-frequency limit $CV_0$, 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 $I_{\mathrm{amp}}\sim V_0C\omega$ as $\omega\to0^+$ and $I_{\mathrm{amp}}\sim V_0/(L\omega)$ as $\omega\to\infty$.
The <Q factor> measures selectivity:
$$
Q=\frac{L\omega_0}{R}=\frac{\omega_0}{2\alpha}.
$$
The average <Joule heating> <power> is $RI_{\mathrm{amp}}^2/2$. Its half-maximum <angular frequencies> and their separation are exactly
$$
\omega_\pm=\sqrt{\omega_0^2+\alpha^2}\pm\alpha,\qquad
\Delta\omega=\omega_+-\omega_-=2\alpha=\frac RL=\frac{\omega_0}{Q}.
$$
These 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. https://www.feynmanlectures.caltech.edu/II_22.html#Ch22-S7[Feynman lectures, II.22.7: resonant filters and radio sidebands]{ref}.
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