LC oscillations
Charge that swings to and fro
This topic is beyond the current NCERT syllabus, but it is a beautiful idea and it is useful for competitive exams. What happens when a charged capacitor is connected across an inductor, with no source at all?
At the start, the capacitor holds a charge q m, and the current is zero. All the energy is in the electric field between the plates.
Close the switch. The capacitor starts to discharge through the coil. The inductor opposes the change, so the current builds up gradually, not all at once.
When the capacitor is completely discharged, the current is at its largest. Now all the energy is in the magnetic field of the coil.
The inductor does not let the current stop suddenly. It keeps the charge flowing, and charges the capacitor again, but with the opposite polarity.
Then the whole process runs backwards, and the capacitor returns to its starting charge. The charge keeps swinging to and fro: these are LC oscillations.
Now the equation. Round the loop, q over C equals L di by dt. The current drains the capacitor, so i equals minus dq by dt. This gives d²q by dt² plus q over LC equals zero.
Compare it with simple harmonic motion: d²x by dt² plus ω₀² x equals zero. It is the same equation, with q in place of x and ω₀² equal to 1 over LC.
So q equals q m cos ω₀t, with ω₀ equals 1 over √(LC). The natural frequency is ν₀ equals 1 over 2π√(LC). The current, i equals q m ω₀ sin ω₀t, is greatest whenever q is zero.
Energy and the spring analogy
Watch the energy. The electric energy is U E equals q² over 2C, which is q m² over 2C, times cos² ω₀t.
The magnetic energy is U B equals ½ L i². Using L ω₀² equals 1 over C, this becomes q m² over 2C, times sin² ω₀t.
Add them: cos² plus sin² is 1. So the total energy stays fixed at q m² over 2C. It just passes back and forth between the capacitor and the coil.
Put a block on a spring next to the circuit. Pull the block out and let go. Its displacement x behaves exactly like the charge q, and its velocity v like the current i.
Mass m plays the part of L: both resist a change in motion. The spring constant k plays the part of 1 over C. Kinetic energy ½ m v² matches ½ L i², and spring energy ½ k x² matches q² over 2C.
Example: a 30 microfarad capacitor is connected to a 27 millihenry inductor. ω₀ equals 1 over √(27 × 10⁻³ × 30 × 10⁻⁶), which is 1 over √(8.1 × 10⁻⁷), about 1.1 × 10³ radians per second.
That is a frequency of about 177 hertz. If the initial charge is 6 millicoulomb, the total energy is q² over 2C: 36 × 10⁻⁶ over 60 × 10⁻⁶, which is 0.6 joule.
With no resistance, the energy at any later time is still 0.6 joule. It is only shared differently between the two fields.
A real circuit always has some resistance, which turns energy into heat, and it also radiates a little. So the oscillations slowly die away: they are damped.
Try it: LC oscillations
- This is the example: 30 μF and 27 mH. The motion is slowed down: one real period (5.65 ms) takes 4 s on screen.
- Watch the energy bars: the blue electric bar and the orange magnetic bar trade places, but the total stays at 0.6 J.
- The block on the spring moves exactly like the charge, and its velocity arrow matches the current arrow.
- Make C four times bigger: the period doubles, since T = 2π√(LC). Then tick “Add a small resistance” and watch the swings die away.
Natural frequency
A 10 mH inductor is connected across a charged 1 μF capacitor. What is the frequency ν₀ of the LC oscillations?
- About 1.59 kHz
- 10 kHz
- 100 MHz
- About 159 Hz
Answer
A. ω₀ = 1/√(LC) = 1/√(10⁻² × 10⁻⁶) = 1/√(10⁻⁸) = 10⁴ rad/s. Then ν₀ = ω₀/2π = 10⁴/6.28 ≈ 1.59 kHz. (10 kHz forgets to divide ω₀ by 2π; 100 MHz uses 1/LC without the square root.)
Sharing the energy
A 30 μF capacitor, initially holding 6 mC, oscillates with an inductor (no resistance). The total energy is 0.6 J. How much energy is in the inductor at the moment the charge on the capacitor is 3 mC?
- 0.15 J
- 0.30 J
- 0.45 J
- 0.60 J
Answer
C. At q = 3 mC, U_E = q²/2C = (3 × 10⁻³)²/(2 × 30 × 10⁻⁶) = 9 × 10⁻⁶/60 × 10⁻⁶ = 0.15 J. The total is fixed at 0.6 J, so U_B = 0.6 − 0.15 = 0.45 J. Half the charge leaves only a quarter of the electric energy, because U_E depends on q², not on q (so 0.30 J is wrong).
Summary
- Equation of LC oscillations:
d²q/dt² + q/LC = 0 - Charge and current:
q = q_m cos ω₀t, i = q_m ω₀ sin ω₀t - Natural frequency:
ω₀ = 1/√(LC), ν₀ = 1/(2π√(LC)) - Total energy:
q²/2C + ½Li² = q_m²/2C
- Extra topic, beyond the current NCERT syllabus. A charged capacitor connected to an inductor makes the charge oscillate with no source.
- Energy passes back and forth between the electric field of the capacitor and the magnetic field of the inductor; without resistance the total stays constant.
- It is the electrical twin of a block on a spring: q ↔ x, i ↔ v, L ↔ m, 1/C ↔ k.
- Real circuits have resistance (and radiate), so the oscillations are damped.