University Physics II · Inductance and AC Circuits · 11.7
LC Oscillations & Series RLC Resonance
Charge sloshing between a capacitor and an inductor is a harmonic oscillator: L is the mass, 1/C the stiffness. Resistance drains it; a source tuned to the natural frequency replaces the loss until only R limits the current.
Build the model
Connect the measurement to the mechanism.
An inductor stores energy in a magnetic field, a capacitor in an electric field, and neither can dump its store instantly. Wire them in a loop and the energy shuttles between them: charge drains from C, current builds in L, and the collapsing current recharges C the other way. The loop equation L d²q/dt² + q/C = 0 is a = −ω²x in disguise, with ω₀ = 1/√(LC), so the circuit is a harmonic oscillator whose mass is L and whose spring constant is 1/C.
Add resistance and the free oscillation decays at rate R/2L; drive the loop instead and X(L) = ωL and X(C) = 1/(ωC) cancel exactly at ω₀, leaving Z = R and the largest current the source can push. One dimensionless number, Q = ω₀L/R = (1/R)√(L/C), governs both pictures: it is the number of radians of free oscillation in one energy-decay time, it fixes the fractional bandwidth Δω/ω₀ = 1/Q, and it is the factor by which the voltage across L or C exceeds the source's.
- Simple definition
- An LC loop oscillates as energy trades between magnetic and electric storage at ω₀ = 1/√(LC). Resistance damps the free oscillation and, when the loop is driven, sets the sharpness Q = ω₀L/R of the current peak at ω₀.
- Example
- L = 10.0 mH with C = 1.00 µF rings at ω₀ = 1.00 × 10⁴ rad s⁻¹, f₀ = 1592 Hz. Add R = 10.0 Ω and Q = 10.0: the free ring decays to 1/e in 3.2 cycles, and the driven peak is 159 Hz wide.
a = −ω²x in circuit dress: L is the mass, 1/C the spring constant.
L in H, C in F · f₀ = ω₀/2π · 10.0 mH with 1.00 µF gives 1592 Hz
Each store peaks when the other is empty, and the two peaks are equal.
Ideal loop: U constant · I(max) = ω₀ q(max)
Amplitude decays with 2L/R, energy with L/R — 2.0 ms and 1.0 ms here.
Underdamped while R < 2√(L/C); here that critical value is 200 Ω
The peak sits where the reactances cancel, not where R is small.
Z in Ω · Z is minimum and equal to R at ω₀, for any R
±45° at the half-power points, → ±90° far off; slope 2Q/ω₀ at ω₀.
φ > 0 means the current lags the source · φ = 0 at ω₀
Also the L or C voltage gain at resonance, and 2π × stored/lost per cycle.
Dimensionless · Δω is the full width at half power
The ideal loop is a harmonic oscillator
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.
Resistance makes the ring decay
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 + φ), with 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.
Driving the loop: impedance and the current peak
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 free loop rings at. 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.
Phase: what the current does relative to the source
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.
Bandwidth and quality factor
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.
Transient plus steady state, and where the model stops
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.
Change one variable at a time
Make the relationship visible.
Slide R and the curve broadens but the peak never moves — only L (or C) can shift it, because ω₀ = 1/√(LC) while the width is Δω = R/L.
RESONANCE f₀1592 Hz
QUALITY FACTOR Q10.0
HALF-POWER WIDTH Δf159 Hz
FREE RING TO 1/e3.2 cycles
Live interpretationRESONANCE f₀: 1592 Hz. QUALITY FACTOR Q: 10.0. HALF-POWER WIDTH Δf: 159 Hz. FREE RING TO 1/e: 3.2 cycles
Catch the common trap
Explain before calculating.
A series RLC circuit with L = 10.0 mH, C = 1.00 µF and R = 10.0 Ω is driven by a source of fixed amplitude and variable frequency; the current peaks at 1592 Hz with a half-power width of 159 Hz. The 10.0 Ω resistor is replaced by 20.0 Ω. What happens to the peak frequency and to the width?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 25 mH coil is connected across a 400 nF capacitor charged to 12 V, with negligible resistance. Find the ring frequency and the peak current.
- LC = (25 × 10⁻³ H)(4.00 × 10⁻⁷ F) = 1.00 × 10⁻⁸ s², so ω₀ = 1/√(LC) = 1.00 × 10⁴ rad s⁻¹.
- f₀ = ω₀/2π = 1.00 × 10⁴ ÷ 6.283 = 1.59 × 10³ Hz.
- The charge starts at its maximum: q(max) = CΔV = (4.00 × 10⁻⁷ F)(12 V) = 4.8 × 10⁻⁶ C.
- A quarter cycle later the capacitor is empty and the whole store is current: I(max) = ω₀q(max) = (1.00 × 10⁴ s⁻¹)(4.8 × 10⁻⁶ C) = 4.8 × 10⁻² A.
- Check by energy: ½CΔV² = ½(4.00 × 10⁻⁷ F)(12 V)² = 28.8 µJ, and ½LI(max)² = ½(0.025 H)(0.048 A)² = 28.8 µJ — the same joules, moved from the field between the plates to the field in the coil.
Answerω₀ = 1.00 × 10⁴ rad s⁻¹ (f₀ = 1.59 kHz); I(max) = 48 mA
MediumThe same L and C are put in series with R = 15 Ω and driven by a source of amplitude 12 V, swept in frequency. Find the quality factor, the half-power width, and the amplitude of the voltage across the inductor at resonance.
- At resonance X(L) = X(C), so Z = R and I(max) = ΔV(max)/R = 12 V ÷ 15 Ω = 0.80 A.
- Q = (1/R)√(L/C) = (1/15 Ω)√(0.025 H ÷ 4.00 × 10⁻⁷ F) = 250 Ω ÷ 15 Ω = 16.7.
- Δω = R/L = 15 Ω ÷ 0.025 H = 600 rad s⁻¹, so Δf = 600 ÷ 2π = 95.5 Hz — a fraction 95.5/1592 = 0.060 of f₀, which is 1/Q as it must be.
- X(L) at resonance is ω₀L = (1.00 × 10⁴ s⁻¹)(0.025 H) = 250 Ω, so ΔV(L) = I(max)X(L) = (0.80 A)(250 Ω) = 2.0 × 10² V.
- That is Q × 12 V. The capacitor carries the same 200 V, 180° out of phase, so the pair cancels in the loop sum while each reading is real across its own component.
AnswerQ = 16.7; Δf = 95.5 Hz; ΔV(L) = 2.0 × 10² V, Q times the source amplitude
HardA receiver front end must peak at 909 kHz and pass a 9.0 kHz band between its half-power points, using a 220 µH coil. Find the tuning capacitance, the required Q and the total series resistance, then say how far down a station 90 kHz away sits.
- The peak sets C: ω₀ = 2π(909 × 10³ Hz) = 5.711 × 10⁶ rad s⁻¹, and C = 1/(ω₀²L) = 1 ÷ [(5.711 × 10⁶ s⁻¹)²(2.20 × 10⁻⁴ H)] = 1.39 × 10⁻¹⁰ F = 139 pF.
- The width sets Q: Q = f₀/Δf = 909 kHz ÷ 9.0 kHz = 101.
- Q sets the allowed loss: Δω = R/L gives R = 2πLΔf = 2π(2.20 × 10⁻⁴ H)(9.0 × 10³ Hz) = 12.4 Ω — and that is every loss in the loop, winding and dielectric included, not just a soldered-in resistor.
- Selectivity: 90 kHz off tune is 2Q(δf/f₀) = 2(101)(90/909) = 20 half-widths away, so the response is 1/√(1 + 20²) ≈ 0.05 of the peak.
AnswerC = 139 pF, Q = 101, R = 12.4 Ω; a station 90 kHz off comes through at about 5% of full response