core30 min studyLesson 282 in Physics

RLC transients and resonant response

Distinguish undamped natural frequency, damped ringdown frequency, and response-dependent driven resonance in a series RLC circuit.

On this page

What you’ll understand

  • Classify series RLC transients with damping ratio.
  • Connect initial capacitor charge and inductor current to ringdown.
  • Distinguish current resonance from charge resonance under sinusoidal driving.

A capacitor charged and connected to an inductor does not settle immediately into a sinusoidal steady state. It undergoes a transient determined by its initial charge and current, while resistance removes stored energy. An applied sinusoid introduces a second problem: the response after that transient has decayed. These problems share parameters but can have different characteristic frequencies. Naming the measured quantity is essential whenever someone reports a circuit's resonance.

Charge makes a second-order coordinate

Consider a single series resistor, inductor, and capacitor with constant positive component values. Choose capacitor charge q so that the series current is its time derivative. The loop equation becomes a driven second-order equation for charge, analogous to a mass on a damped spring. Initial charge and initial current are independent data. For an unforced circuit, the resistor dissipates energy at a rate proportional to current squared, so the total capacitor-plus-inductor energy cannot increase.

Dividing by inductance defines the undamped angular frequency and damping ratio. Underdamping gives oscillatory decay, critical damping gives repeated real characteristic roots, and overdamping gives two real decay rates. The damped frequency is smaller than the undamped frequency and exists as an oscillation frequency only in the underdamped regime. A resistor does not simply lower amplitude while leaving the period exactly unchanged; that description is only an approximation when damping is weak.

Lq¨+Rq˙+qC=Vs(t),ω0=1LC,α=R2L,ζ=αω0L\ddot q+R\dot q+\frac qC=V_s(t),\qquad \omega_0=\frac1{\sqrt{LC}},\quad \alpha=\frac R{2L},\quad \zeta=\frac\alpha{\omega_0}
Charge q is in coulombs, source voltage Vs in volts, R in ohms, L in henrys, C in farads, and α and ω₀ in inverse seconds. The model assumes linear lumped elements and series resistance.↔ Scroll wide equations horizontally.

Initial conditions determine the phase

When the source is removed and damping is below critical, charge is an exponentially decaying combination of sine and cosine. If current initially vanishes, the sine coefficient must compensate the derivative of the exponential. Writing only a decaying cosine would generally impose a nonzero initial current. This matters when comparing a charged-capacitor release with a driven sinusoidal fit: their phases are determined by different information and should not be borrowed from one another.

The envelope of charge or voltage decays with rate alpha. Energy is quadratic and its slowly varying envelope decays twice as fast in a weakly damped oscillator. Instantaneous electric and magnetic energies exchange within each cycle, so the exact energy need not be a perfect exponential at every instant for appreciable damping. What is exact is its negative derivative from resistive dissipation. These distinctions prevent a factor-of-two error when extracting a damping rate from a ringdown trace.

q(t)=q0e−αt[cos⁡(ωdt)+αωdsin⁡(ωdt)],ωd=ω02−α2q(t)=q_0e^{-\alpha t}\left[\cos(\omega_dt)+\frac\alpha{\omega_d}\sin(\omega_dt)\right],\qquad \omega_d=\sqrt{\omega_0^2-\alpha^2}
This solution assumes Vs = 0, q(0) = q₀, current q̇(0) = 0, and 0 ≤ α < ω₀. The sine coefficient enforces zero initial current; q̇ is measured in amperes.↔ Scroll wide equations horizontally.

Resonance depends on the response

For a fixed-amplitude sinusoidal voltage source, the series impedance is smallest when inductive and capacitive reactances cancel. The current amplitude therefore peaks at the undamped frequency, independently of resistance in this ideal model. Capacitor voltage and charge have an additional inverse-frequency factor relative to current. Their peak occurs at a lower frequency when damping is sufficiently weak, and disappears as an interior nonzero-frequency peak once the damping ratio reaches the relevant threshold.

A frequency sweep measures steady-state response only if each point allows enough settling or uses an appropriate dynamic measurement model. Comparing that sweep with free decay is a strong diagnostic: ringdown probes characteristic roots, while the swept response probes a specified transfer function. Real parasitic capacitance, frequency dependent resistance, and source impedance can shift all these predictions. A fitted resonance should therefore be reported together with circuit topology, observable, drive convention, and damping assumptions.

ωI,max⁡=ω0,ωq,max⁡=ω02−2α2(ζ<1/2)\omega_{I,\max}=\omega_0,\qquad \omega_{q,\max}=\sqrt{\omega_0^2-2\alpha^2}\quad(\zeta<1/\sqrt2)
These maxima refer to a series RLC driven by a fixed voltage amplitude at variable angular frequency. The charge maximum formula requires weak enough damping; it is distinct from the free-decay frequency ωd.↔ Scroll wide equations horizontally.

Three frequencies from one circuit

A series circuit has L = 1 H, C = 0.25 F, and R = 2 Ω. Find its damping regime, ringdown angular frequency, and the frequencies of maximum driven current and charge.

  1. Compute ω0=1/1×0.25=2 rad/s\omega_0=1/\sqrt{1\times0.25}=2\,\mathrm{rad/s} and α=2/(2×1)=1 s−1\alpha=2/(2\times1)=1\,\mathrm{s^{-1}}. Thus ζ=0.5\zeta=0.5, which is underdamped.

  2. The free-decay oscillation has ωd=4−1=3≃1.73 rad/s\omega_d=\sqrt{4-1}=\sqrt3\simeq1.73\,\mathrm{rad/s}. Its amplitude envelope decays as e−te^{-t} with t in seconds.

  3. The fixed-voltage driven current peaks at 2 rad/s. Charge peaks at 4−2=2≃1.41 rad/s\sqrt{4-2}=\sqrt2\simeq1.41\,\mathrm{rad/s} because the damping ratio is below the charge-peak threshold.

The result

The circuit is underdamped. Its ringdown frequency is 1.73 rad/s, while driven current and charge peak at 2.00 and 1.41 rad/s respectively.

Explore the model

Interactive lab

Interactive lab · oscillator

Compare underdamped and critical releases

For a fresh run Reset and set BOTH Damping ratio ζ to 0.5 and Natural frequency ω₀ to 2; predict an oscillatory decay before Play. Reset again and set BOTH ζ to 1 and ω₀ to 2; predict no oscillation. The unit-mass model starts at x = 1 and velocity = 0. It represents normalized free charge dynamics, not a driven resonance sweep.

Compare underdamped and critical releasesUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe same positive restoring term can give oscillatory or nonoscillatory decay depending on damping.

Check your understanding

For an ideal series RLC circuit driven with fixed voltage amplitude, what peaks at ω₀ = 1/√(LC)?

Try it yourself

An unforced series circuit has L = 2 H, C = 0.125 F, and R = 8 Ω. It starts with q = 1 C and current zero. Classify the decay and give q(t).

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Find α and ω₀, then use the repeated-root solution and impose the initial derivative.

Compare with the solution

Here ω0=1/2(0.125)=2 rad/s\omega_0=1/\sqrt{2(0.125)}=2\,\mathrm{rad/s} and α=8/(2×2)=2 s−1\alpha=8/(2\times2)=2\,\mathrm{s^{-1}}, so damping is critical. The repeated-root solution is (A+Bt)e−2t(A+Bt)e^{-2t}. Initial charge gives A = 1 C, and zero initial current gives B = 2 C/s. Therefore q(t)=(1+2t)e−2t Cq(t)=(1+2t)e^{-2t}\,\mathrm C when t is in seconds. Its derivative is −4te−2t A-4te^{-2t}\,\mathrm A, which starts at zero and never changes sign for positive time.

Take it with you

  • Natural, damped, and driven-response frequencies answer different questions.
  • Both initial charge and current are needed to determine a transient.
  • A fixed-voltage series current peaks where inductive and capacitive reactances cancel.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.