In the simplest circuit made of a capacitor and a coil, the charge, current and voltage change periodically — these are electromagnetic oscillations. They closely resemble mechanical oscillations of a mass on a spring, so the formulas look alike too. Radio and TV receivers, mobile phones and every radio transmitter depend on an oscillating (LC) circuit.
§1. The LC circuit and what happens in it
An LC circuit is a closed circuit of a capacitor (C) and a coil (L). If the capacitor is charged and connected to the coil, the following happens:
t = 0: the capacitor is fully charged, no current — all the energy is in the electric field: W = q_max²/(2C).
The capacitor discharges through the coil; because of self-induction the current grows gradually, not instantly.
t = T/4: the charge is zero, the current is maximum — all the energy is in the magnetic field: W = LI_max²/2.
Self-induction keeps the current flowing, and the capacitor recharges with the opposite polarity (t = T/2).
The process repeats in reverse, and at t = T the circuit is back to its initial state.
Why does the current not stop when the charge reaches zero? The self-induced EMF in the coil opposes any change of current, so the coil plays the role of inertia (mass) in mechanics. Just as a load passing through equilibrium does not stop, the current keeps flowing and recharges the capacitor.
Mechanical analogy: charge q ↔ displacement x, current i ↔ velocity v, inductance L ↔ mass m, 1/C ↔ spring constant k.
§2. Thomson’s formula
In an ideal circuit (resistance R = 0) the oscillations are harmonic:
L in henries (H), C in farads (F), T in seconds (s), ν in hertz (Hz).
Compare with the spring pendulum T = 2π√(m/k).
The period is proportional to √L and √C: making the capacitance 4 times larger doubles the period.
Worked example
L = 1 mH, C = 10 μF. Period and frequency?
LC = 10⁻³·10⁻⁵ = 10⁻⁸ s², √(LC) = 10⁻⁴ s.
T = 2π·10⁻⁴ ≈ 6.28·10⁻⁴ s.
ν = 1/T ≈ 1.6·10³ Hz = 1.6 kHz.
§3. Graphs of charge, current and voltage
Besides Thomson’s formula, problems often give the equation of oscillation and ask you to “read off” quantities. The general form is
q = q_max·cos(ωt), u = q/C = U_max·cos(ωt), i = q′ = −ωq_max·sin(ωt)
The capacitor voltage oscillates in phase with the charge: U_max = q_max/C.
The current leads the charge by a quarter of a period (π/2): when the charge is greatest the current is zero, and when the charge is zero the current is greatest.
Angular frequency ω (rad/s) and ordinary frequency ν (Hz) are linked by ω = 2πν = 2π/T.
On graphs q(t) and u(t) are cosine curves and i(t) is a negative sine curve. It is just like a mechanical pendulum: at maximum displacement the speed is zero, and passing through equilibrium the speed is greatest.
Worked example
The charge in a circuit varies as q = 2·10⁻⁵·cos(500t) (C), C = 10 μF. Find the period, frequency, inductance and the maximum current and voltage.
From the equation ω = 500 rad/s, q_max = 2·10⁻⁵ C.
T = 2π/ω = 6.28/500 ≈ 0.0126 s, ν = 1/T ≈ 79.6 Hz.
From ω² = 1/(LC): L = 1/(ω²C) = 1/(500²·10⁻⁵) = 0.4 H.
I_max = ωq_max = 500·2·10⁻⁵ = 0.01 A, U_max = q_max/C = 2·10⁻⁵/10⁻⁵ = 2 V.
Energy check: CU²/2 = 10⁻⁵·4/2 = 2·10⁻⁵ J and LI²/2 = 0.4·10⁻⁴/2 = 2·10⁻⁵ J, equal, so correct.
§4. Conservation of energy
In an ideal circuit the total energy is constant:
CU_max²/2 = LI_max²/2 = Cu²/2 + Li²/2
Hence I_max = U_max·√(C/L). The energy "flows" between the electric and magnetic fields with period T/2 — it oscillates twice as fast as the charge.
Worked example
A 2 μF capacitor charged to 100 V is connected to a coil with L = 0.5 H. Energy of the circuit and maximum current?
W = CU²/2 = 2·10⁻⁶·10⁴/2 = 0.01 J.
I_max = √(2W/L) = √(0.02/0.5) = √0.04 = 0.2 A.
Check: U√(C/L) = 100·√(4·10⁻⁶) = 100·2·10⁻³ = 0.2 A.
§5. Damped, forced and self-sustained oscillations
A real circuit has resistance: energy turns into Joule heat and the amplitude decreases — damped oscillations.
Under an external alternating EMF the circuit performs forced oscillations. When the driving frequency equals the circuit’s natural frequency the amplitude rises sharply — resonance.
Self-sustained oscillations are undamped because energy is topped up from a source at the right moments. A transistor oscillator works this way: the transistor acts as a switch that feeds energy into the circuit every period.
Worked example
What capacitance makes a circuit with L = 0.1 H resonate at 50 Hz?
ω = 2π·50 ≈ 314 rad/s; the resonance condition is ω² = 1/(LC).
C = 1/(ω²L) = 1/(314²·0.1) ≈ 1/9860 ≈ 1.0·10⁻⁴ F = 100 μF.
How fast the oscillations die away depends on the circuit’s resistance: the larger R, the faster the amplitude falls, and with a very large resistance there are no oscillations at all; the capacitor simply discharges. The shape of the resonance curve depends on resistance too: with small R the resonance is sharp and tall, so the circuit separates nearby frequencies well. This is the basis of a receiver’s “selectivity”: the wanted station comes in loudly while stations on neighbouring frequencies are hardly heard.
§6. Tuning a radio receiver
The receiver’s circuit is tuned to resonance with the wanted station’s frequency (usually with a variable capacitor). Wavelength and frequency are related by:
ν = c/λ, λ = cT = 2πc√(LC)
So λ ~ √C and ν ~ 1/√C.
Worked example
A receiver is tuned to 6 MHz, but a 200 m wave must be received. How should the capacitance change?