LC Oscillations & Series RLC Resonance

Put a charged capacitor across an inductor and the loop rule gives L di/dt + q/C = 0. With i = dq/dt that is L d²q/dt² = −q/C, which is a = −ω²x with ω₀ = 1/√(LC): L plays the mass, 1/C the spring constant, q the displacement, i the velocity. Nothing dissipates, so U = q²/(2C) + ½Li² is constant while the two terms trade places every quarter cycle — all electric when the current is zero, all magnetic when the charge is zero. Take L = 10.0 mH and C = 1.00 µF, charged to ΔV = 5.0 V. Then ω₀ = 1/√(1.00 × 10⁻⁸ s²) = 1.00 × 10⁴ rad s⁻¹, f₀ = 1592 Hz, and the stored energy is ½CΔV² = 12.5 µJ. The current peaks at I(max) = ω₀q(max) = (10⁴ s⁻¹)(5.0 µC) = 50 mA, and ½L I(max)² = 12.5 µJ — the same energy, now entirely in the coil's field. The ratio that sets that current, √(L/C) = 100 Ω, is the loop's characteristic impedance.

Real loops have resistance, and the loop rule becomes L d²q/dt² + R dq/dt + q/C = 0 — the damped-oscillator equation, with R sitting in the velocity term. While R < 2√(L/C) the solution is a decaying sinusoid, i = I₀e^(−Rt/2L) cos(ω′t + θ), with θ set by how the loop was started and ω′ = √(ω₀² − (R/2L)²). Two rates matter. The amplitude envelope decays with time constant 2L/R; the energy, going as amplitude squared, decays with L/R. With R = 10.0 Ω added to the loop above, R/2L = 500 s⁻¹, so the amplitude falls to 1/e in 2.0 ms — 3.2 cycles — and the energy in 1.0 ms. The frequency shift is slight: R is a twentieth of the critical value 2√(L/C) = 200 Ω, so ω′ = ω₀√(1 − 0.05²) = 9987 rad s⁻¹, 0.13% below ω₀. Raise R to 200 Ω and ω′ reaches zero: the charge returns to zero without overshooting, in the shortest possible time. Beyond that the loop is overdamped and crawls back.

Now drive the series loop with ΔV(t) = ΔV(max) cos ωt. Once the transient has died the current is sinusoidal at the source's frequency, not at ω₀, and phasor addition gives its amplitude: I(max) = ΔV(max)/Z with Z = √(R² + (X(L) − X(C))²), X(L) = ωL, X(C) = 1/(ωC). R is fixed, but the reactive part changes sign: below ω₀ the capacitor dominates and X(L) − X(C) is negative, above ω₀ the inductor dominates. Only at ω₀ = 1/√(LC) do they cancel exactly, leaving Z = R — the smallest impedance the loop can present, at the same frequency the loss-free loop rings at, not at the damped ω′. That is resonance, and where it sits does not depend on R. How sharp is it? Drive our circuit 10% high, at ω = 1.10 × 10⁴ rad s⁻¹: X(L) = 110 Ω, X(C) = 90.9 Ω, so Z = √(10² + 19.1²) = 21.6 Ω and the current is R/Z = 46% of its resonant value. A 10% frequency error has cost more than half the amplitude.

The phase follows the same reactance difference: tan φ = (X(L) − X(C))/R, with φ the angle by which the source voltage leads the current. Far below resonance the loop is essentially a capacitor and φ → −90°, so the current leads; far above it is essentially an inductor and φ → +90°, so the current lags. At ω₀ the reactances cancel, φ = 0, and current and source voltage rise and fall together: the source sees a pure resistance. The useful markers on either side are the frequencies where |X(L) − X(C)| = R, which give φ = ±45°; for our circuit those are 1514 Hz and 1673 Hz. The whole swing from −45° to +45° happens across that span, so a low-Q circuit turns its phase lazily while a high-Q circuit snaps through — the slope at ω₀ is dφ/dω = 2L/R = 2Q/ω₀. Circuits that lock onto a frequency track that slope, because near the peak the amplitude is flat while the phase is steepest.

Define the half-power points as the two frequencies where the average power in R is half its peak value. Power goes as I², so they are where I = I(max)/√2, hence where Z = √2 R, hence where |X(L) − X(C)| = R. Solving gives a full width Δω = R/L, which does not contain C, and a sharpness Q = ω₀/Δω = ω₀L/R = (1/R)√(L/C). For our circuit Δω = 10.0/0.0100 = 1000 rad s⁻¹, so Δf = 159 Hz, the band runs 1514–1673 Hz about f₀ = 1592 Hz, and Q = 10.0. Q then reappears everywhere. The free ring lasts Q/π = 3.2 cycles before its amplitude falls to 1/e. Q equals 2π × (energy stored)/(energy lost per cycle). And at resonance the inductor voltage is X(L)I = (100 Ω)(ΔV/10 Ω) = 10 ΔV, with the capacitor voltage equal and opposite: feed 5.0 V in and 50 V appears across each reactive component.

The driven solution is the steady sinusoid plus the decaying natural oscillation, with the initial conditions fixing how much transient there is. It dies at the rate the free ring dies, 2L/R = 2Q/ω₀, so a resonator takes of order Q cycles to settle: our Q = 10 loop needs a few milliseconds, and a Q = 10⁴ quartz resonator thousands of cycles. Narrow band and slow response are one trade seen twice. As for the practical limits, R here is every loss, not the resistor you soldered in — winding resistance and skin effect, capacitor ESR and dielectric loss, core loss in a ferrite, plus whatever the source and load add, so a loaded Q sits below the unloaded value. Tolerances move the target: 5% on L and 5% on C shift f₀ by up to 5%, half a bandwidth at Q = 10. And the model is lumped. Saturate a ferrite core and L falls with current, bending the curve over; go high enough in frequency and the coil's own stray capacitance resonates it.