{title2=$Lq''+Rq'+q/C=V(t)$}

In a <series RLC circuit>, the same <electric current> $i=q'$ flows through the <resistor> with <electrical resistance> $R$, the <inductor>, and the <capacitor>. Choose $q$ as the signed <electric charge> on the <capacitor> plate into which positive $i$ flows. With terminal drops measured in that direction, <Kirchhoff's voltage law> gives
$$
v_R=Ri,\qquad v_L=Li',\qquad v_C=q/C,\qquad
Lq''+Rq'+q/C=V(t).
$$
Here $V(t)$ is the applied <electric potential difference>; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfies
$$
q''+2\alpha q'+\omega_0^2q=0,\qquad
\alpha=\frac{R}{2L},\qquad \omega_0=\frac{1}{\sqrt{LC}}.
$$
The <angular frequency> $\omega_0$ is the value for lossless free <oscillation>; $\alpha$ is the decay parameter. The three possibilities are <underdamped RLC response> for $\alpha<\omega_0$, <critically damped RLC response> for $\alpha=\omega_0$, and <overdamped RLC response> for $\alpha>\omega_0$. https://openstax.org/books/university-physics-volume-2/pages/14-6-rlc-series-circuits[OpenStax University Physics 2, §14.6: free RLC response]{ref}.

For a fair comparison, start each circuit with the same <electric charge> $q(0)=q_0$ and zero <electric current> $q'(0)=0$. Set $\tau=\omega_0t$ and $a=\alpha/\omega_0$; then $x=q/q_0$ obeys $d^2x/d\tau^2+2a\,dx/d\tau+x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and <electric charge>.

\Image[/rlc-free-damping.png]
{title=Free <series RLC circuit> responses with identical <initial conditions>}
{description=Left: <electric charge> for <underdamped RLC response>, <critically damped RLC response>, and <overdamped RLC response>. Right: total stored <energy>, which decreases in all three cases even when <electric charge> oscillates. The normalization is $\omega_0=1$, $q_0=1$, $q'(0)=0$.}

The stored <energy> is the sum of <magnetic energy> and <capacitor energy>:
$$
E=\frac12Li^2+\frac{q^2}{2C}.
$$
Differentiating and using the circuit equation gives the complete <power> balance,
$$
E'=i\left(Li'+\frac qC\right)=V(t)i-Ri^2.
$$
In the free response, $E'=-Ri^2\leq0$: <Joule heating> accounts exactly for the lost <energy>. For the plotted <initial conditions>, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau)^2$. Since every free solution tends to zero for $R>0$,
$$
\int_0^\infty Ri(t)^2\,dt=E(0).
$$
In the lossless limit $R=0$, <energy> stays constant and shuttles between the <capacitor> and <inductor>. Thus an <oscillation> of <electric charge> and a decrease of total <energy> are compatible statements.

The analogy with a <damped harmonic oscillator> is exact within these ideal models: <electric charge> corresponds to displacement, <electric current> to <velocity>, <inductance> to <mass>, $1/C$ to <spring constant>, and $R$ to the coefficient of a resisting force proportional to <velocity>. The two stored <energy> terms play the roles of <kinetic energy> and elastic <potential energy>. https://www.feynmanlectures.caltech.edu/I_23.html#Ch23-S3[Feynman lectures, I.23.3: the oscillator analogy]{ref}.

Differentiating the driven equation also gives $Li''+Ri'+i/C=V'$. When using this equation to solve for <electric current>, retain the original constraint $Li'(0)+Ri(0)+q(0)/C=V(0)$: the initial <electric charge> fixes the initial rate of change of <electric current>.


 Series RLC circuit (source code)