LC Oscillations
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Direct answer
Connect a charged capacitor across an inductor and close the loop: charge sloshes back and forth at the natural frequency f = 1/(2π√(LC)) — the electrical twin of a mass on a spring, with q playing displacement, 1/C the spring constant and L the mass. Energy alternates entirely between the capacitor's electric field (q²/2C at maximum charge q0) and the inductor's magnetic field (½LI² at maximum current), the total q0²/2C staying constant in the ideal circuit. The governing equation L d²q/dt² + q/C = 0 is SHM verbatim, so every result from the oscillations chapter transfers directly — including the amplitude logic and the energy split at intermediate charges.
What you must remember
- Frequency relations: ω = 1/√(LC); f = 1/(2π√(LC)); T = 2π√(LC) — larger L or C swings slower.
- The analogy table: q ↔ x; I = dq/dt ↔ v; L ↔ mass (inertia); 1/C ↔ spring constant; magnetic energy ½LI² ↔ kinetic ½mv²; electric energy q²/2C ↔ potential ½kx².
- The two extremes: fully charged capacitor — all energy electric, current zero (the turning point); fully discharged capacitor — all energy magnetic, current maximum at I0 = ωq0.
- Energy conservation: E = q0²/2C = ½LI0², constant in the ideal circuit; at q = q0/2 the electric share is E/4 and the magnetic 3E/4 — the same quadratic split as SHM.
- Time behaviour: q = q0 cos ωt and I = q0ω sin ωt — charge and current stay a quarter period out of phase.
- Why real oscillations decay: circuit resistance dissipates energy as heat, and the loop radiates some as electromagnetic waves; sustaining oscillations needs an amplifier-driven tank circuit.
- Tuning logic: the LC tank selects its resonant frequency in radio receivers — the practical application NCERT attaches to the idea.
One circuit from start to steady slosh
Charge a 5 μF capacitor to 12 V: q0 = CV = 60 μC, storing E = ½CV² = ½ × 5 × 10^-6 × 144 = 3.6 × 10^-4 J. Connect it to L = 20 mH. The natural frequency: ω = 1/√(LC) = 1/√(10^-7) ≈ 3160 rad/s, f ≈ 500 Hz. The peak current follows from energy conservation: ½LI0² = 3.6 × 10^-4 gives I0 = ωq0 ≈ 0.19 A. Follow the cycle: at t = 0 the capacitor holds everything; a quarter period later it is empty, the inductor carries 0.19 A and all 3.6 × 10^-4 J; at the half period the capacitor is charged to 12 V with reversed polarity. Check the intermediate logic — when q = q0/2, the electric share is a quarter of E, magnetic the rest — and every SHM energy instinct is doing double duty.
Where NEET sets the trap
The factor of 2π is the most stolen mark: ω = 1/√(LC) and f = 1/(2π√(LC)) both appear as options, and the question's wording ("frequency" versus "angular frequency") decides. The analogy question is near-annual: which quantity plays the role of mass? Inductance L — inertia against current change; and 1/C plays the spring's stiffness, so a stiffer (smaller) capacitor raises the frequency. Energy items test I0 = ωq0, derivable from ½LI0² = q0²/2C, and the q0 = CV0 step that starts every numerical. Assertion items: charge and current are not in phase (quarter period apart); total energy is constant only in the ideal circuit; real tank circuits die out without feedback amplification. NEET keeps this block light — usually one concept or a one-step numerical — which makes these exact traps the whole game.
Frequently asked questions
What is the oscillation frequency of an LC circuit?
f = 1/(2π√(LC)); for L = 20 mH and C = 5 μF, about 500 Hz (ω ≈ 3160 rad/s).
In the LC-SHM analogy, what plays the mass and the spring?
Inductance L plays the mass (inertia) and 1/C the spring constant; charge q oscillates exactly like displacement x.
Where is the energy when the capacitor is fully discharged?
Entirely in the inductor's magnetic field, ½LI_max², at the instant current peaks — the total q0²/2C never changes in an ideal circuit.
Why do LC oscillations die out in real circuits?
Resistance dissipates the energy as heat and the loop radiates part as electromagnetic waves; only an amplified tank circuit sustains oscillations.
What is the maximum current in terms of the initial charge?
I_max = ωq0 = q0/√(LC), obtained by equating ½LI_max² with q0²/2C.