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A charged capacitor connected to a resistor simply discharges and stops. Connect it to an inductor instead and something different happens: the charge flows out, flows back in with the opposite sign, and keeps going. This is an LC oscillation. This post explains why it happens, shows the spring analogy that makes every formula easy to remember, and works one circuit through with numbers.
Which exams include LC oscillations?
Check this first, because it has changed. The JEE (Advanced) 2026 brochure lists "RC, LR, LC and LCR (in series) circuits with d.c. and a.c. sources" in its physics syllabus. The JEE (Main) 2026 syllabus, the NEET (UG) 2026 syllabus and the CBSE 2026-27 physics curriculum list the series LCR circuit, resonance and power, but not LC oscillations as a separate topic. The current NCERT Alternating Current chapter ends with power and transformers and has no LC oscillation section. If you are preparing for JEE Advanced or a physics olympiad, learn it fully. For the other exams it is still a good way to understand resonance, since ω₀ = 1/√(LC) is the same number. Check your own year's syllabus.
Why does the charge keep oscillating?
You need two energy formulas, both in NCERT. A capacitor with charge q stores U = q²/2C in its electric field. An inductor with current i stores U = ½Li² in its magnetic field, and NCERT points out that this looks like ½mv², with L playing the role of mass.
Here is one cycle in words:
At t = 0 the capacitor holds charge Q₀ and the current is zero. All the energy is electric.
The capacitor discharges through the inductor. The inductor opposes the growing current, so it builds up gradually.
When the capacitor is empty, the current is at its maximum. All the energy is magnetic.
The inductor now opposes the falling current, so current keeps flowing and charges the capacitor with the opposite sign.
The process reverses, and after one full period the circuit is back where it started.
The MIT 8.02 notes on inductance write the total energy as q²/2C + ½Li² and set its time derivative to zero. That gives d²q/dt² = −q/(LC), the equation of simple harmonic motion, with q(t) = Q₀ cos ω₀t, i(t) = I₀ sin ω₀t, ω₀ = 1/√(LC) and I₀ = ω₀Q₀.
How does it compare with a mass on a spring?
The MIT notes set the two systems side by side. For a spring, energy is ½mv² + ½kx² and ω = √(k/m). Match the terms:
q (charge) corresponds to x (displacement).
i = dq/dt corresponds to v = dx/dt.
L corresponds to m: both resist changes in motion.
1/C corresponds to k: a small capacitor is like a stiff spring.
q²/2C corresponds to ½kx², and ½Li² corresponds to ½mv².
If you have done SHM problems, you already know how this circuit behaves. Our guide to simple harmonic motion problems covers the mechanical side.
Worked example: one LC circuit, all the numbers
A 20 μF capacitor is charged to 10 V and then connected across a 50 mH inductor with negligible resistance. Find the frequency, the energy, the maximum current, and the time at which the energy is shared equally.
Angular frequency. ω₀ = 1/√(LC) = 1/√(0.050 × 20 × 10⁻⁶) = 1/√(1.0 × 10⁻⁶) = 1000 rad/s. So f = 1000/2π ≈ 159 Hz and T = 2π/1000 ≈ 6.28 ms.
Initial charge and energy. Q₀ = CV = 20 × 10⁻⁶ × 10 = 2.0 × 10⁻⁴ C. U = ½CV² = ½ × 20 × 10⁻⁶ × 100 = 1.0 × 10⁻³ J.
Maximum current. I₀ = ω₀Q₀ = 1000 × 2.0 × 10⁻⁴ = 0.20 A. Check with energy: ½LI₀² = ½ × 0.050 × 0.040 = 1.0 × 10⁻³ J. It matches.
First time the capacitor is empty. q = Q₀ cos ω₀t = 0 when ω₀t = π/2, so t = T/4 ≈ 1.57 ms.
Equal sharing. Electric energy is U cos²ω₀t, so it equals half the total when cos²ω₀t = ½, at ω₀t = π/4. That is t = T/8 ≈ 0.785 ms. Then q = Q₀/√2 ≈ 1.41 × 10⁻⁴ C and i = I₀/√2 ≈ 0.141 A.
Notice that the electric energy goes from maximum to zero and back to maximum in half a period. Energy oscillates at twice the frequency of the charge.
What changes in a real circuit?
Real wires and coils have resistance, which turns some energy into heat on each swing. The MIT notes treat this as the RLC circuit, the electrical form of a damped oscillator: the amplitude dies away, and for R = 0 you get back the undamped ω₀ = 1/√(LC). Driving that circuit with an AC source gives resonance, which is covered in our post on LCR resonance and the quality factor.
Which mistakes cost marks?
Using f where ω is meant. ω₀ = 1/√(LC) is in rad/s; divide by 2π for hertz.
Saying the energy frequency equals the charge frequency. Energy repeats every T/2.
Putting maximum current at maximum charge. They are a quarter period apart: i is largest when q = 0.
Forgetting the √2 at equal sharing. Half the energy means q = Q₀/√2, not Q₀/2.
Mixing units. Convert μF and mH to F and H before you calculate.
Charge and current a quarter cycle apart are easier to grasp when you can watch them. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Day 1: Write the spring–circuit matching list from memory and derive ω₀ = 1/√(LC) from energy conservation.
Day 2: Three numerical problems like the worked example, changing L and C each time.
Day 3: Sketch q, i, electric energy and magnetic energy against time on one set of axes.
Day 4: Two past JEE (Advanced) questions on LC or LCR circuits, timed.
Today, redo the worked example with C four times larger and predict the new period before calculating it.
