What is an LC circuit?
An LC circuit (also called an oscillating circuit or tank circuit) has just two parts: a capacitor C and an inductor (coil) L, joined in a loop.
- A capacitor stores energy in an electric field between its plates. Energy = q² / (2C).
- An inductor stores energy in a magnetic field inside the coil. Energy = ½ L i².
If we charge the capacitor and then connect it to the coil, the charge does not just flow once and stop. It rushes back and forth again and again. These are free electromagnetic oscillations: nothing outside pushes them; the circuit swings on its own energy.
Energy conversion in the circuit
Follow one cycle, step by step:
- t = 0: capacitor full (q = Q₀), current zero. All energy is electric: Q₀² / (2C).
- t = T/4: capacitor empty (q = 0), current biggest (i = I₀). All energy is magnetic: ½ L I₀².
- t = T/2: capacitor full again, but with opposite sign. Current zero. All energy electric again.
- t = 3T/4: current biggest again, in the opposite direction.
- t = T: back to the start.
Why does the current not stop when the capacitor is empty? Because a coil opposes any change in current (self-induction). The falling current makes an emf that keeps charge moving, and this charges the capacitor the other way.
With no resistance the total energy stays constant: q²/(2C) + ½ L i² = Q₀²/(2C) = ½ L I₀². So I₀ = Q₀ / √(LC).
This is just like a mass on a spring: charge q acts like displacement, current i like velocity, L like mass, and 1/C like the spring constant.
Thomson formula: period and frequency
The charge changes like a cosine: q = Q₀ cos(ω₀t), and the current is i = −Q₀ω₀ sin(ω₀t).
The angular frequency is ω₀ = 1 / √(LC). So the Thomson formula for the period is
T = 2π√(LC), and the frequency is f = 1 / (2π√(LC)).
- L in henry (H), C in farad (F), T in seconds (s), f in hertz (Hz).
- 4 times L (or C) → 2 times T. T grows as the square root.
- The energy in each part swings twice as fast as the charge: energy has period T/2.
Where it comes from: the voltage across C (q/C) and the voltage across L (L di/dt) must add to zero round the loop, which gives d²q/dt² = −q/(LC). This is the same equation as simple harmonic motion with ω² = 1/(LC).
Damped, forced and self-sustained oscillations
Damped oscillations
Real wires have resistance R. Each cycle some energy turns into heat (I²R). The amplitude of charge falls step by step. This is a damped oscillation. Small R: many slow-dying swings. Very large R: no swings at all, the capacitor just slowly empties.
Forced oscillations and resonance
If an AC source keeps pushing the circuit at frequency f, the circuit oscillates at that frequency (forced oscillations). The current is biggest when the driving frequency equals the natural frequency 1/(2π√(LC)). This is resonance. Radio tuning uses it.
Self-oscillating systems
A self-oscillating system makes steady oscillations by itself, from a DC source. It has three parts: the LC circuit (sets the frequency), an energy source (battery), and a valve such as a transistor that lets energy in at the right moment of each cycle, using feedback from the circuit. The energy added each cycle equals the energy lost as heat, so the amplitude stays constant. A pendulum clock works the same way.
Key formulas and definitions
- T = 2π√(LC) (Thomson formula)
- f = 1 / (2π√(LC)), ω₀ = 1 / √(LC)
- q = Q₀ cos(ω₀t), i = −I₀ sin(ω₀t), I₀ = ω₀Q₀
- Electric energy U_E = q² / (2C), magnetic energy U_B = ½ L i²
- Total energy U = Q₀² / (2C) = ½ L I₀² = constant (no resistance)
Worked examples
1. An LC circuit has L = 4 mH and C = 10 µF. Find its period.
T = 2π√(LC) = 2π√(4×10⁻³ × 10×10⁻⁶) = 2π√(4×10⁻⁸) = 2π × 2×10⁻⁴ = 1.26×10⁻³ s ≈ 1.26 ms.
2. Find the frequency of the same circuit.
f = 1 / T = 1 / 1.26×10⁻³ ≈ 796 Hz.
3. The capacitance is made 9 times bigger. What happens to the frequency?
f ∝ 1/√C. √9 = 3, so the frequency becomes one third of its old value.
4. A 2 µF capacitor is charged to 100 V and connected to a coil. Find the total energy of the oscillations.
Q₀ = CV = 2×10⁻⁶ × 100 = 2×10⁻⁴ C. U = Q₀²/(2C) = (2×10⁻⁴)² / (4×10⁻⁶) = 4×10⁻⁸ / 4×10⁻⁶ = 0.01 J. (Same as ½CV² = ½ × 2×10⁻⁶ × 10⁴ = 0.01 J.)
5. In the circuit above, L = 0.5 H. Find the largest current.
All energy becomes magnetic: ½ L I₀² = 0.01 J. I₀² = 0.02 / 0.5 = 0.04, so I₀ = 0.2 A.
6. A radio has a coil of 1 µH. What capacitance tunes it to 100 MHz?
C = 1 / (4π²f²L) = 1 / (4 × 9.87 × (10⁸)² × 10⁻⁶) = 1 / (3.95×10¹¹) ≈ 2.5×10⁻¹² F = 2.5 pF.
7. At some moment the electric energy equals the magnetic energy. What is the charge then, in terms of Q₀?
Each is half the total: q²/(2C) = ½ × Q₀²/(2C). So q² = Q₀²/2 and q = Q₀/√2 ≈ 0.71 Q₀. This happens at t = T/8.
Common mistakes
- Forgetting to convert units: mH → ×10⁻³ H, µF → ×10⁻⁶ F, pF → ×10⁻¹² F before using T = 2π√(LC).
- Thinking T ∝ LC. It is T ∝ √(LC): four times C gives only two times T.
- Thinking the current stops when the capacitor is empty. At that moment the current is at its maximum.
- Saying energy has the same period as the charge. Energy is never negative, so it repeats every T/2.