LC Circuit: Energy Oscillating Between L and C

An LC circuit consists of an inductor $L$ connected to a capacitor $C$. With no resistance, the energy initially stored in the capacitor's electric field (or the inductor's magnetic field) oscillates back and forth indefinitely between the two — exactly analogous to a mass-spring system trading kinetic and potential energy.

The Equations

By Kirchhoff's voltage law, $V_L + V_C = 0$:

$L\,\dfrac{dI}{dt} + \dfrac{Q}{C} = 0$

Differentiating once gives the SHM equation:

$\dfrac{d^{2}Q}{dt^{2}} = -\dfrac{1}{LC}\,Q$

The angular frequency of oscillation is:

$\omega = \dfrac{1}{\sqrt{LC}}$

So the period is $T = 2\pi\sqrt{LC}$ and the frequency is $f = 1/(2\pi\sqrt{LC})$.

Energy Conservation

The total energy:

$E = \dfrac{Q^{2}}{2C} + \dfrac{LI^{2}}{2}$

is constant. The first term (electric, in the capacitor) and the second (magnetic, in the inductor) trade off — when one is at maximum, the other is at zero — but their sum is fixed.

Step-by-Step Solution

Given: $L = 1\;\text{mH} = 10^{-3}\;\text{H}$, $C = 1\;\text{μF} = 10^{-6}\;\text{F}$.

Find: The angular frequency, period, and frequency of oscillation.

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Step 1 — Compute $LC$.

$LC = (10^{-3})(10^{-6}) = 10^{-9}$

Step 2 — Take the square root.

$\sqrt{LC} = \sqrt{10^{-9}} = 10^{-4.5} \approx 3.162 \times 10^{-5}$

Step 3 — Compute the angular frequency.

$\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{3.162 \times 10^{-5}} \approx 31{,}623\;\text{rad/s}$

Step 4 — Compute the period.

$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{31{,}623} \approx 1.987 \times 10^{-4}\;\text{s} \approx 0.199\;\text{ms}$

Step 5 — Compute the frequency.

$f = \dfrac{1}{T} \approx \dfrac{1}{0.000199} \approx 5033\;\text{Hz} \approx 5.03\;\text{kHz}$

So this circuit oscillates at about 5 kHz — well into the audio range. Tuning the values of $L$ and $C$ tunes the frequency:

• Larger $L$ or $C$ → slower oscillation (lower frequency)
• Smaller $L$ or $C$ → faster oscillation (higher frequency)

This is how radio receivers tune to different stations: a variable capacitor changes $C$, sliding the resonant frequency across the radio band.

Step 6 — Energy values.

Suppose the capacitor is initially charged to $Q_0 = 10^{-6}\;\text{C}$. The total energy is:

$E = \dfrac{Q_0^{2}}{2C} = \dfrac{10^{-12}}{2 \times 10^{-6}} = 5 \times 10^{-7}\;\text{J} = 0.5\;\text{μJ}$

That same energy alternates between the capacitor and the inductor, with peak current:

$I_{\max} = \omega\,Q_0 = 31{,}623 \times 10^{-6} \approx 0.0316\;\text{A} \approx 31.6\;\text{mA}$

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Answer:

$\boxed{\;\omega = \dfrac{1}{\sqrt{LC}} = 31{,}623\;\text{rad/s},\quad T \approx 0.199\;\text{ms},\quad f \approx 5.03\;\text{kHz}\;}$

The energy oscillates back and forth between the capacitor (electric) and the inductor (magnetic) at this frequency. With no resistance, the oscillation continues forever.

Try It

• Adjust the capacitance $C$ and inductance $L$ — frequency scales as $1/\sqrt{LC}$.
• Watch the bar graphs of capacitor energy (cyan) and inductor energy (pink) trade off.
• The total bar (green) stays constant — that's energy conservation.
• Try $L = C$ values matching real radio circuits (~ μH and ~ pF) — frequencies in the MHz range.