LC circuit and Thomson formula

An LC circuit is an idealized loop made of a capacitor of capacitance $C$ and an inductor of inductance $L$, whose resistance is taken to be zero.

When a charged capacitor is connected across the inductor, the capacitor's electric field weakens while the current grows: the electric-field energy $\dfrac{q^{2}}{2C}$ converts into the magnetic-field energy $\dfrac{Li^{2}}{2}$. When the capacitor is fully discharged the current is maximal, and the inductor keeps it flowing, recharging the capacitor with the opposite polarity.

This produces free harmonic oscillations of charge $q(t) = q_m\cos(\omega t + \varphi_0)$ and current $i(t) = q'(t)$ with natural angular frequency $\omega = \dfrac{1}{\sqrt{LC}}$. Hence Thomson's formula for the period:

$$T = 2\pi\sqrt{LC}, \qquad \nu = \frac{1}{T} = \frac{1}{2\pi\sqrt{LC}}.$$

The formula follows from the equation $q'' = -\dfrac{1}{LC}q$, obtained from energy conservation or from Kirchhoff's loop rule. The period increases with $L$ or $C$; in a real circuit with resistance $R$ the oscillations decay, but for $R \ll 2\sqrt{L/C}$ Thomson's formula remains a good approximation to the natural frequency.

An inductor with inductance $L = 0.10$ H is connected to a capacitor with capacitance $C = 10$ µF. The resistance is negligible. Find the period $T$ and the frequency $\nu$ of the free oscillations.

The tuning circuit of a radio receiver must be set to a frequency $\nu = 500$ kHz. The capacitance is $C = 100$ pF. What inductance $L$ should be chosen?

A capacitor with capacitance $C = 4.0$ µF was charged to a voltage $U_0 = 10$ V and connected to an inductor with inductance $L = 1.0$ mH (zero resistance). Using energy conservation, find the maximum current $I_m$ and the period $T$. How do $T$ and $I_m$ change if the capacitance is increased 4 times while $L$ and $U_0$ stay the same?

Self-study plan in a simulator. A real low-resistance circuit needs an oscilloscope and gives rapidly decaying oscillations, and working with mains voltage is dangerous, so perform the observations in a virtual circuit (PhET "LC circuit", Falstad, or similar), using only low-voltage parameters.

1. Build the loop: capacitor, inductor, switch; keep the resistance as small as possible.
2. Set $L = 0.1$ H and $C = 10$ µF, charge the capacitor and start the oscillations.
3. From the graph of $q(t)$ (or $u_C(t)$) measure the period between two neighbouring maxima and compare it with $T = 2\pi\sqrt{LC} \approx 6.3$ ms; record the discrepancy.
4. Increase $L$ 4 times: first predict the new period, then check that it grew 2 times.
5. Decrease $C$ 4 times: check that the period decreased 2 times.
6. State the conclusion about the dependence $T \propto \sqrt{LC}$ and write it down.

Do not connect your circuit to a 220 V mains supply and do not use high voltages.

To show the circuit layout, the energy exchange between the capacitor and the inductor, and the phase shift between charge and current over one period.

Diagram: a capacitor $C$ and an inductor $L$ in a closed loop, with an arrow showing the positive current direction. Below it, three graphs share the time axis $t$ over $0 \ldots T$: $q(t)$ is a cosine starting at the maximum $q_m$; $i(t)$ is a sine that is zero at that moment (a shift of $\pi/2$); the energies $W_C = q^{2}/(2C)$ and $W_L = Li^{2}/2$ are plotted so that their sum stays constant. Vertical lines mark the instants $T/4$, $T/2$, $3T/4$, $T$.

In a medium-wave receiver a coil wound on a ferrite rod gives $L = 1{,}0\ \mathrm{mH}$, while the tuning knob turns a variable capacitor whose capacitance runs from $18\ \mathrm{pF}$ to $100\ \mathrm{pF}$. Thomson's formula $f=\frac{1}{2\pi\sqrt{LC}}$ gives $f=1{,}19\ \mathrm{MHz}$ at $C=18\ \mathrm{pF}$ and $f=0{,}50\ \mathrm{MHz}$ at $C=100\ \mathrm{pF}$ — exactly the medium-wave broadcast band. A station at $900\ \mathrm{kHz}$ becomes audible only when the knob has brought $C$ to resonance.

This exercises the period formula of the circuit, $T=2\pi\sqrt{LC}$: the natural frequency is fixed by the product $LC$, so with a fixed $L$ the whole tuning range is set by the two extreme values of $C$. At resonance the energy is exchanged twice per period between the capacitor's electric field, $\tfrac{1}{2}CU^2$, and the coil's magnetic field, $\tfrac{1}{2}LI^2$, and the current in the loop is at its maximum.

A beach detector's search coil has $L=1{,}0\ \mathrm{mH}$ and its capacitor $C=100\ \mathrm{nF}$, so the circuit formula gives $f=15{,}9\ \mathrm{kHz}$ and $T=63\ \mu\mathrm{s}$. When the coil passes 5 cm above a coin, eddy currents in the metal raise the effective inductance by about $1\%$; since $T\propto\sqrt{L}$, the period grows by $0{,}5\%$ and the frequency drops by $80\ \mathrm{Hz}$, from $15{,}9$ to $15{,}8\ \mathrm{kHz}$ — the electronics turn that shift into an audible tone.

The detector is an LC tank that a transistor keeps oscillating: by Thomson's formula the frequency reads out the coil's inductance, because $f\propto 1/\sqrt{L}$, so a target is detected as a shift $\Delta f/f=-\tfrac{1}{2}\,\Delta L/L$, not through any change of capacitance. The energy exchange $\tfrac{1}{2}CU^2\leftrightarrow\tfrac{1}{2}LI^2$ is what keeps the oscillation running.

Under the 4 mm glass-ceramic plate of an induction hob lies a flat coil with $L=100\ \mu\mathrm{H}$, wired with a capacitor $C=0{,}40\ \mu\mathrm{F}$; the natural frequency is $f=25{,}2\ \mathrm{kHz}$ and the period $T=40\ \mu\mathrm{s}$. The transistors switch at exactly this frequency and sustain about $10\ \mathrm{A}$ in the coil; that current corresponds to a voltage amplitude $U=I\sqrt{L/C}=158\ \mathrm{V}$ across the capacitor. The eddy currents induced by this field heat the base of a 1.5 kg steel pan holding a litre of water.

The electronics drive the tank at the natural frequency fixed by Thomson's formula, so the reactive energy $\tfrac{1}{2}CU^2=\tfrac{1}{2}LI^2$ shuttles between capacitor and coil instead of being dissipated. The same equality of energies yields the amplitude relation $U=I\sqrt{L/C}$ between the largest current and the largest voltage in the tank.