Physics · Electrodynamics

LC circuit oscillations and Thomson’s formula

Q: How do you calculate the frequency of an LC circuit?

For a free ideal circuit, $f_0=\frac1{2\pi\sqrt{LC}}$. Convert L and C to henry and farad before multiplying. With 10 mH and 100 nF their product is $10^{-9}\,\mathrm s^2$, giving about 5.033 kHz and a period of 198.692 μs. Angular frequency is two pi times frequency in hertz. Increasing inductance or capacitance under otherwise equal conditions lengthens the period. This result says nothing about amplitude or successful starting: without initial charge, initial current or excitation the circuit remains at rest. In a real damped or driven circuit you must additionally specify which frequency is being discussed.

Q: What happens physically in an LC circuit?

With an initially charged capacitor and zero current, all energy first occupies the electric field. After a quarter-period charge is zero and all energy is magnetic in the inductor. Continuing current then charges the capacitor with reversed polarity. After half a period the energy is electric again, but charge is negative rather than positive. Only after a full period do charge and current return together to their initial state. Energy fractions contain squares and therefore repeat after half a charge period. Current sign follows its chosen reference arrow. In this article the arrow points into the selected plate, making current negative during the first discharge.

Q: How do you rearrange the Thomson formula for C or L?

First square the period equation and then isolate the required component: $C=\frac{T_0^2}{4\pi^2L}$, or $L=\frac{T_0^2}{4\pi^2C}$. The frequency form gives $C=\frac1{4\pi^2f_0^2L}$. Frequency enters squared rather than merely linearly in the denominator. At 1 MHz and 100 μH the ideal result is about 253.3 pF. Back-substitution should recover the target frequency; higher frequency at fixed inductance requires lower capacitance. A real setup uses effective total capacitance including additional capacitances. A measurement probe can therefore shift frequency and distort a naive inference of inductance from the observed period.

Q: What does the LC circuit have in common with the spring pendulum?

The evolution equations have the same mathematical structure. Signed charge corresponds to displacement, its time-derivative current to velocity, inductance to mass and reciprocal capacitance to spring stiffness. Magnetic inductor energy therefore plays the role of the spring oscillator’s kinetic energy; electric capacitor energy plays the elastic role. This mapping assumes linear ideal systems and does not turn henry into kilogram. Greater capacitance means less voltage for the same charge and hence weaker restoring action, comparable to a softer spring. Initial conditions and signs must survive the translation. Mechanical damping maps simply to electrical resistance only when the selected circuit topology supports the corresponding parameter mapping.

Q: Why does a real LC circuit not oscillate forever?

Real losses remove energy from a free circuit. In the simple series RLC model, $\frac{\mathrm dE}{\mathrm dt}=-Ri^2$. An oscillatory damped response exists only below the critical series resistance $R_{\mathrm{crit}}=2\sqrt{\frac LC}$. Its angular frequency is $\omega_d=\sqrt{\frac1{LC}-\left(\frac R{2L}\right)^2}$; the shift is small only under weak damping. Critical and overdamped responses are not periodic. Additional losses or measurement loading need further model assumptions. Sustained oscillation requires energy input, such as a suitable drive or feedback. A prescribed drive frequency must not be equated with free damped natural frequency. A passive circuit with neither initial energy nor a source does not start oscillating by itself.

Thomson’s formula gives the period of free undamped oscillation in an ideal LC circuit. Initial energy sets amplitude rather than ideal frequency.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Create flashcards or a quiz I already have an account

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Formula

Calculate

T = 2 π L \cdot C

T in s · L in H (henry) · C in F (farad)

Variables & units

Variables & units
Symbol	Meaning	Unit
$T$	Period of the electromagnetic oscillation	s
$L$	Inductance of the coil	H
$C$	Capacitance of the capacitor	F

LaTeX

T = 2\pi \sqrt{L \cdot C}

QUANTA STUDYquanta-study.de

Original quantitative curves over t/T₀ from 0 to 1: upper q/Q=cos(2πt/T₀), i/Imax=−sin(2πt/T₀); lower EC/E=cos² and EL/E=sin². At ¼ period q=0 and i=−1; at ½ period q=−1 and i=0. Example 8 mH, 50 nF, 12 V: T₀=125.664 μs. — Free ideal LC circuit with initially positive charge and zero current. Positive current enters the selected plate. Charge and current are separated by a quarter-period; the two energy fractions repeat after half a charge period and always sum to one.

LC circuit frequency calculator

Find the period and natural frequency of an ideal LC circuit using Thomson's formula, or the matching coil or capacitor, without damping.

Find

Inductance L

Inductance of the coil, e.g. 8 mH; an iron core makes it much larger.

Capacitance C

Capacitance of the capacitor, e.g. 50 nF; a tuning capacitor in a radio has a few hundred pF.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for f, substitute, evaluate

1. Rearrange for f
$f=\frac{1}{2\pi\sqrt{LC}}$
2. Substitute values (in SI units)
- L = 8 mH = 0.008 H
- C = 50 nF = 5 · 10⁻⁸ F
$f=\frac{1}{2\pi\sqrt{0.008\cdot 5\times10^{-8}}}=7957.75$
3. Result
f = 7.958 kHz

Result: Frequency f

f = 7.958 kHz

With L = 8 mH and C = 50 nF, f = 7.958 kHz.

Quick tips

Four times the capacitance or four times the inductance halves the frequency, because f depends on 1/√(L · C).
Estimate: work out √(L · C) first; H · F gives s². For 8 mH and 50 nF, √(4 · 10⁻¹⁰ s²) = 2 · 10⁻⁵ s, so T ≈ 6.3 · 2 · 10⁻⁵ s ≈ 0.13 ms.

Assumptions and limits

Undamped ideal LC oscillator with constant L > 0 and C > 0. T = 2π√(LC), f = 1/T. Damping, component losses and nonlinear effects are excluded.

Derivation & background

Energy in a free ideal circuit alternates between the capacitor’s electric field and the inductor’s magnetic field. Signed charge and current are separated by a quarter-period; their energy fractions repeat after half a charge period. Real losses damp the circuit. Only underdamped free series RLC circuits have an oscillatory response; its frequency shift is small only for weak damping. Sustained oscillation with losses needs energy input, for example a drive or suitable feedback.

Exam blueprint

Validity range

Free ideal lumped LC circuit with positive constant L and C, without resistance, extra stores or drive. A nontrivial oscillation requires initial charge or current. A free series RLC circuit has a damped oscillation frequency only for $R<2\sqrt{(\frac{L}{C})}$ ; drive frequency and resonance response are separate statements.

Derivation steps

The closed free loop couples signed capacitor charge and continuous inductor current into harmonic motion.

1Select a capacitor plate: $u_C=\frac qC$ . Define positive current into it, hence $i=\frac{\mathrm dq}{\mathrm dt}$ .
2After disconnecting the source the passive loop budget is $u_C+u_L=0$ .
3With $u_L=L\frac{\mathrm di}{\mathrm dt}$ obtain $L\ddot q+\frac qC=0$ .
4Divide by L and identify harmonic form: $\ddot q+\omega_0^2q=0$ , $\omega_0=\frac1{\sqrt{LC}}$ .
5Initial charge and current set amplitude and phase. If both vanish, the free circuit remains at rest.
6A full state repetition spans two pi of phase: $T_0=\frac{2\pi}{\omega_0}=2\pi\sqrt{LC}$ , $f_0=\frac1{T_0}$ .

Rearrangements

Natural frequency

f = \frac{1}{2\pi \sqrt{L C}}

Smaller L or C means a higher frequency.

Capacitance

C = \frac{1}{4\pi^2 f^2 L}

This is how a tuning capacitor for station selection is sized.

Inductance

L = \frac{T^2}{4\pi^2 C}

From a measured period and a known capacitance.

Task variant

An LC circuit ( $L=100\,\mathrm{\mu H}$ ) is to oscillate at 1 MHz. What capacitance is needed?

$C=\frac{1}{4\pi^2f_0^2L}=\frac{1}{4\pi^2(10^6\,\mathrm{s^{-1}})^2(100\times10^{-6}\,\mathrm H)}\approx253.3\,\mathrm{pF}$

Compute T for $L=25\,\mathrm{mH}$ and $C=40\,\mathrm{\mu F}$ .

$T_0=2\pi\sqrt{LC}=2\pi\sqrt{(25\times10^{-3}\,\mathrm H)(40\times10^{-6}\,\mathrm F)}\approx6.283\,\mathrm{ms}$

Common mistakes

Substituting mH, µF or nF directly.

Convert to henry and farad first, otherwise the result is off by powers of ten.

Confusing frequency f and angular frequency ω.

$\omega_0=\frac1{\sqrt{LC}}$ , but $f_0=\frac{\omega_0}{2\pi}$ ; the factor 2π matters.

Forgetting the square root when solving for L or C.

Square T: L and C enter under the root, so rearrange quadratically.

Exam context

Exam tasks couple the Thomson formula with energy arguments ( $\frac12CU^2=\frac12LI^2$ ), the spring-pendulum analogy and the tuning of receiver circuits.

Formula cluster

Electromagnetic oscillations

Electrical analogue of the mechanical pendulum, the basis of radio engineering.

C = Q/U

Capacitance of a Capacitor

U_ind = −L·(dI/dt)

Self-Induction of a Coil

T = 2π·√(m/k)

Spring Pendulum (Oscillation Period)

T = 2π·√(l/g)

Simple Pendulum (Period)

Worked example

For $L=10\,\mathrm{mH}$ and $C=100\,\mathrm{nF}$ , $LC=10^{-9}\,\mathrm s^2$ . Thus $T_0\approx198.692\,\mathrm{\mu s}$ and $f_0\approx5.033\,\mathrm{kHz}$ .

Applications

Ideal natural frequency and tuning of LC storage pairs, free oscillations and the spring analogy; loaded or driven filters require further parameters.

How do you calculate with Thomson formula for an LC circuit?

Here is how to work through a typical Thomson formula for an LC circuit (T = 2π·√(L·C)) task step by step:

1
Task
An LC circuit ( $L=100\,\mathrm{\mu H}$ ) is to oscillate at 1 MHz. What capacitance is needed?
Solution path
$C=\frac{1}{4\pi^2f_0^2L}=\frac{1}{4\pi^2(10^6\,\mathrm{s^{-1}})^2(100\times10^{-6}\,\mathrm H)}\approx253.3\,\mathrm{pF}$
2
Task
Compute T for $L=25\,\mathrm{mH}$ and $C=40\,\mathrm{\mu F}$ .
Solution path
$T_0=2\pi\sqrt{LC}=2\pi\sqrt{(25\times10^{-3}\,\mathrm H)(40\times10^{-6}\,\mathrm F)}\approx6.283\,\mathrm{ms}$

Understand and apply the equation

Thomson’s formula gives the timescale of an ideal free LC circuit. Oscillation draws on initial energy exchanged between the capacitor and inductor. Period, charge direction and energy evolution are connected; damping and driven operation require additional statements.

Energy changes its store

A precharged capacitor is connected to an inductor in a closed loop. Initially its voltage drives current. Once the capacitor is discharged, the current already established in the inductor cannot simply vanish: it charges the capacitor with reversed polarity. This repeated exchange explains oscillation before we calculate its timing.[1][3]

Thomson’s formula requires constant positive L and C, negligible losses and a free circuit with no drive. Initial conditions set amplitude and phase. No oscillation starts without initial charge, initial current or energy input.[1][3]

Why does an initially empty ideal LC circuit not oscillate by itself?

\begin{aligned}\omega_0&=\frac1{\sqrt{LC}}\\T_0&=2\pi\sqrt{LC}\\f_0&=\frac1{2\pi\sqrt{LC}}\end{aligned}

Positive constant inductance and capacitance, negligible resistance and no connected driving source.[1]

A closed circuit containing an inductor and capacitor is a pair of energy stores. The formula specifies timing, not initial amplitude. If charge and inductor current are both initially zero with no source, the system remains at rest. In the model a precharged capacitor or an existing inductor current can start oscillation. Switching initial conditions must match the actual circuit.[3]

“Ideal” means linear time-independent components without dissipation or additional storage. Leads are electrically short enough for a lumped circuit. Real instruments may load the circuit. The ideal period can be useful when those effects remain small; neglecting them is an assumption, not a conclusion of the formula.[1]

How is Thomson’s formula derived from the loop balance?

q denotes signed charge on the selected capacitor plate; $u_C=\frac qC$ . Positive current i enters that plate, so $i=\frac{\mathrm dq}{\mathrm dt}$ . After source disconnection, capacitor and inductor form a closed series loop. With compatible passive voltage references their voltages sum to zero, fixing the restoring sign.[1]

The restoring action is visible before calculating: positive charge produces a voltage which reduces that charge in a free circuit. Positive charge therefore requires a negative second charge derivative. The positive charge term in the differential equation produces precisely this restoring effect. The opposite sign would give a growing response instead of an oscillation.[1][3]

\begin{aligned}u_C+u_L&=0\\\frac qC+L\frac{\mathrm di}{\mathrm dt}&=0\\L\frac{\mathrm d^2q}{\mathrm dt^2}+\frac qC&=0\\\frac{\mathrm d^2q}{\mathrm dt^2}+\omega_0^2q&=0\end{aligned}

In the last line $\omega_0^2=\frac1{LC}$ ; this is a differential equation rather than merely a static voltage budget.[3]

\begin{aligned}q(t)&=q_i\cos(\omega_0t)+\frac{i_i}{\omega_0}\sin(\omega_0t)\\i(t)&=-\omega_0q_i\sin(\omega_0t)+i_i\cos(\omega_0t)\end{aligned}

Initial values $q_i=q(0)$ and $i_i=i(0)$ determine the unique free motion.[3]

One full period is reached when charge and current together return to their initial state. The cosine requires a phase advance of two pi. Period is therefore two pi divided by angular frequency. Reporting angular frequency directly as frequency in hertz omits that distinction.[1]

How do charge, current and energy change during a period?

\begin{aligned}E_C&=\frac{q^2}{2C}=\frac12Cu_C^2\\E_L&=\frac12Li^2\\E&=E_C+E_L=\mathrm{constant}\end{aligned}

E is total energy; uC remains a signed voltage.[1]

Starting with positive capacitor charge and zero current places all energy in the electric store. After a quarter-period charge vanishes and current reaches its negative maximum in our reference direction. After a half-period charge reverses and current is again zero. Energy is then electric again although the full state has not yet returned to its start.[1]

\begin{aligned}q(t)&=Q\cos(\omega_0t)\\i(t)&=-\omega_0Q\sin(\omega_0t)\\\frac{E_C}{E}&=\cos^2(\omega_0t)\\\frac{E_L}{E}&=\sin^2(\omega_0t)\end{aligned}

Special case of an initially charged capacitor with $i(0)=0$ ; the two energy fractions add to one.[1]

Charge and current are offset by a quarter-period. Their squared energy fractions repeat after half the charge period. Energy does not become negative when charge or current reverses direction. The normalised energy fractions in the following diagram show the exchange between stores. Their positive values indicate neither charge polarity nor current direction. Energy stored in the capacitor depends quadratically on voltage magnitude in the linear model; an energy fraction therefore cannot reveal polarity.[1][4]

Energy repeats before the complete state

QUANTA STUDYquanta-study.de

Initially the capacitor is charged and current is zero. At

\frac{T_0}{4}

all energy is magnetic; at

\frac{T_0}{2}

electric again but with reversed capacitor polarity. Energy fractions have half the state period.[1]

Free with a Quanta account. Your download stays ready after registration.

Calculating LC frequency, period, inductance and capacitance

\begin{aligned}[LC]&=\mathrm H\,\mathrm F=\mathrm s^2\\C&=\frac{T_0^2}{4\pi^2L}=\frac1{4\pi^2f_0^2L}\\L&=\frac{T_0^2}{4\pi^2C}=\frac1{4\pi^2f_0^2C}\end{aligned}

Henry times farad is second squared; convert mH, μH, μF, nF or pF to SI before substitution.[1]

The square root gives characteristic scaling: Four times the capacitance at fixed inductance doubles period and halves frequency. Doubling initial voltage instead doubles current amplitude and quadruples energy in the linear model, leaving timing unchanged. These changes must not be confused.[1]

\frac{\Delta f_0}{f_0}\approx-\frac12\left(\frac{\Delta L}{L}+\frac{\Delta C}{C}\right)

First-order approximation for small relative changes; use the exact square-root factor for larger changes.[3]

Calculate charge exchange and tuning capacitance

A precharged capacitor starts the circuit

Given: Ideal: $L=8.0\,\mathrm{mH}$ , $C=50\,\mathrm{nF}$ , $u_C(0)=12\,\mathrm V$ , $i(0)=0$ . Source disconnected before free evolution; positive current enters the initially positive plate.
Find: Find period, frequency, total energy, peak-current magnitude and the state at $\frac{T_0}{8}$ .

Approach: Initially zero inductor current fixes the cosine start. Energy conservation gives peak-current magnitude.
$\begin{aligned}\omega_0&=\frac1{\sqrt{LC}}\\T_0&=\frac{2\pi}{\omega_0}\\E&=\frac12CU_0^2\\I_{\mathrm{max}}&=U_0\sqrt{\frac CL}\end{aligned}$
Substitution: Convert both prefixes before multiplying and distinguish frequency from angular frequency.
$\begin{aligned}LC&=(8.0\times10^{-3})(50\times10^{-9})\,\mathrm s^2\\&=4.0\times10^{-10}\,\mathrm s^2\\\omega_0&=5.0\times10^4\,\mathrm{s^{-1}}\\T_0&\approx125.664\,\mathrm{\mu s}\\f_0&\approx7.958\,\mathrm{kHz}\end{aligned}$
Calculate energy and current amplitude from the initial data.
$\begin{aligned}E&=\frac12(50\times10^{-9}\,\mathrm F)(12\,\mathrm V)^2\\&=3.60\,\mathrm{\mu J}\\I_{\mathrm{max}}&=(12\,\mathrm V)\sqrt{\frac{50\times10^{-9}\,\mathrm F}{8.0\times10^{-3}\,\mathrm H}}\\&=30.0\,\mathrm{mA}\end{aligned}$
At one eighth of a period phase is π/4; the two stores contain equal energy.
$\begin{aligned}q\!\left(\frac{T_0}{8}\right)&=CU_0\cos\frac\pi4\approx0.4243\,\mathrm{\mu C}\\i\!\left(\frac{T_0}{8}\right)&=-I_{\mathrm{max}}\sin\frac\pi4\approx-21.21\,\mathrm{mA}\\E_C&=E_L=1.80\,\mathrm{\mu J}\end{aligned}$

Result: The circuit oscillates at about 7.958 kHz. At 15.708 μs the plate remains positively charged but its signed charging current is negative.

Check: Check: $\frac12L(30\,\mathrm{mA})^2=3.60\,\mathrm{\mu J}$ matches initial energy. At $\frac{T_0}{8}$ both 1.80 μJ fractions sum to 3.60 μJ. Negative current reduces positive plate charge.

Tune an ideal circuit to 250 kHz

Given: Target $f_0=250\,\mathrm{kHz}$ , nominal $L=220\,\mathrm{\mu H}$ . Then suppose actual inductance is 10% larger and actual capacitance 5% larger than calculated. No additional losses.
Find: Find required capacitance and actual frequency for this specified deviation case.

Approach: Rearrange quadratically for C; both component factors act together under the square root.
$\begin{aligned}C_{\mathrm{nom}}&=\frac1{4\pi^2f_0^2L_{\mathrm{nom}}}\\\frac{f_{\mathrm{actual}}}{f_0}&=\frac1{\sqrt{1.10\cdot1.05}}\end{aligned}$
Substitution: Use the exact factor rather than just the linearised tolerance formula.
$\begin{aligned}C_{\mathrm{nom}}&=\frac1{4\pi^2(250\times10^3\,\mathrm{s^{-1}})^2(220\times10^{-6}\,\mathrm H)}\\&\approx1.8422\,\mathrm{nF}\\f_{\mathrm{actual}}&=\frac{250\,\mathrm{kHz}}{\sqrt{1.155}}\\&\approx232.621\,\mathrm{kHz}\end{aligned}$

Result: About 1.842 nF is required nominally. The specified same-direction deviations lower frequency by about 6.952% to 232.621 kHz.

Check: Check: Direct substitution of $L=242\,\mathrm{\mu H}$ and $C\approx1.93431\,\mathrm{nF}$ gives the same frequency. The first-order −7.5% estimate gets the direction right but overestimates magnitude here.

What do resistance, damping and driving change?

\begin{aligned}L\ddot q+R\dot q+\frac qC&=0\\\alpha&=\frac R{2L}\\\omega_d&=\sqrt{\frac1{LC}-\alpha^2}\\R_{\mathrm{crit}}&=2\sqrt{\frac LC}\end{aligned}

Frequency ωd describes only the underdamped free series RLC circuit: R<Rcrit.[2]

Underdamping retains an oscillatory component with a decaying envelope. Critical or overdamping leaves no periodic free oscillation, so a Thomson period is not the actual repeat time of the response. Frequency shift is small only for weak damping. The word “slightly” therefore requires comparison with critical resistance. Impedance of a series RLC circuit instead describes the sinusoidally driven response; the requested resonance value must match the quantity being observed.[2][5]

For $L=8.0\,\mathrm{mH}$ and $C=50\,\mathrm{nF}$ critical series resistance is 800 Ω. With $R=20\,\Omega$ , free damped frequency is about 7.955 kHz rather than 7.958 kHz. At $R=1000\,\Omega$ the same circuit is overdamped. These numbers apply specifically to the series circuit; a parallel-loss topology needs a different damping mapping.[2]

\frac{\mathrm dE}{\mathrm dt}=-Ri^2

In the passive free series circuit energy is lost as Joule heat; charge and current still require consistent signs.[2]

Distinguish free frequency from drive and measurement

A tuned receiver uses L and C together but is excited by an external signal and affected by loads. Drive frequency is initially set by the source. Thomson gives a reference value for the ideal storage pair, not the complete amplitude or bandwidth response of an arbitrary filter. Sustained oscillation with real losses requires energy input; suitable feedback is one option, not a universal requirement for every circuit.[2]

For a freely decaying circuit compare several peaks of the same type. A full period lies, for example, between successive positive voltage maxima, not between a positive and a negative maximum. Probe capacitance shifts frequency. Ignoring that loading would give an incorrect inductance inferred from the measured period.[3]

Explore further: the spring-oscillator analogy

\begin{aligned}m\ddot x+kx&=0\\L\ddot q+\frac qC&=0\\m&\leftrightarrow L,\quad k\leftrightarrow\frac1C\\x&\leftrightarrow q,\quad\dot x\leftrightarrow i\end{aligned}

The analogy concerns the structure of the evolution equation, not identical physical units.[3]

The inductor mathematically plays the role of inertia; reciprocal capacitance plays the role of restoring stiffness. A larger capacitor has less voltage for the same charge and therefore weaker restoring action, like a softer spring. This gives a qualitative check: larger capacitance increases period. Units such as henry and kilogram are not equated. The period of a mass on a spring provides the corresponding mechanical model with mass and spring stiffness.[1][6]

Explore further: probe loading

A probe lengthens the measured period

Given: An ideal free circuit has $L=8.00\,\mathrm{mH}$ and $C=50.0\,\mathrm{nF}$ . An ideal measurement probe adds $C_p=10.0\,\mathrm{nF}$ in parallel with the capacitor; introduce no other losses here.
Find: Find measured frequency and the error when inferring L while ignoring probe capacitance.

Approach: parallel capacitances add. Inference must use the same actual capacitance.
$\begin{aligned}C_{\mathrm{total}}&=C+C_p\\f_p&=\frac1{2\pi\sqrt{L(C+C_p)}}\\L_{\mathrm{wrong}}&=\frac1{4\pi^2f_p^2C}\\&=L\frac{C+C_p}{C}\end{aligned}$
Substitution: additional capacitance is one fifth of the original value.
$\begin{aligned}C_{\mathrm{total}}&=60.0\,\mathrm{nF}\\f_p&\approx7.26440\,\mathrm{kHz}\\T_p&\approx137.658\,\mathrm{\mu s}\\L_{\mathrm{wrong}}&=(8.00\,\mathrm{mH})\frac{60.0}{50.0}\\&=9.60\,\mathrm{mH}\end{aligned}$

Result: The probe lowers frequency from about 7.958 to 7.264 kHz. Inference using only 50.0 nF would overestimate L by 20%.

Check: Frequency ratio is $\sqrt{\frac{50}{60}}\approx0.912871$ ; period grows by its reciprocal. The actual inductor has not changed; the error comes from overlooked capacitance.

Five exercises with different reasoning paths

Calculate · Foundation

An ideal free circuit has $L=20\,\mathrm{mH}$ and $C=2.0\,\mathrm{\mu F}$ . Find period and frequency.

Show hint

LC has unit s².

Learning goal: Convert SI prefixes for a complete period calculation.

Show solution and calculation

Approach: Convert components to henry and farad and evaluate Thomson period.
Results are about 1.257 ms and 795.8 Hz.

Check: Period times frequency equals one.

\begin{aligned}T_0&=2\pi\sqrt{(20\times10^{-3})(2.0\times10^{-6})}\,\mathrm s\\&\approx1.25664\,\mathrm{ms}\\f_0&=\frac1{T_0}\approx795.775\,\mathrm{Hz}\end{aligned}

[1]

Rearrange · Application

Given $L=330\,\mathrm{\mu H}$ , what ideal capacitance gives $f_0=80\,\mathrm{kHz}$ ?

Show hint

Square the frequency equation before isolating C.

Learning goal: Rearrange a target frequency quadratically for capacitance.

Show solution and calculation

Approach: Reciprocal squared target frequency determines C at fixed L.
The result is about 11.994 nF.

Check: Substitution into the original frequency form returns 80 kHz; frequency requires the positive root.

C=\frac1{4\pi^2(80\times10^3\,\mathrm{s^{-1}})^2(330\times10^{-6}\,\mathrm H)}\approx11.994\,\mathrm{nF}

[1]

Interpret · Application

An ideal circuit starts with positive charge Q and zero current. At $\frac{T_0}{2}$ all energy is again in the capacitor. Has the initial state returned? State q, i and both energy fractions.

Show hint

Energy contains squared charge and therefore loses its sign.

Learning goal: Distinguish a full charge period from energy repetition.

Show solution and calculation

No: Charge is −Q and current is zero; electric energy is E and magnetic energy zero.
The initial state additionally requires the same charge polarity and returns only at T₀.
Energy fractions alone cannot distinguish the two polarities.

\begin{aligned}q\!\left(\frac{T_0}{2}\right)&=-Q\\i\!\left(\frac{T_0}{2}\right)&=0\\E_C&=E\\E_L&=0\end{aligned}

[1]

Error analysis · Application

For $L=8.0\,\mathrm{mH}$ and $C=50\,\mathrm{nF}$ someone reports 50 kHz because $\frac1{\sqrt{LC}}=50000\,\mathrm{rad/s}$ . Locate and correct the mistake.

Show hint

How much phase corresponds to one full repetition?

Learning goal: Distinguish angular frequency from frequency through the missing 2π factor.

Show solution and calculation

The value 50000 is angular frequency in radians per second, not cycles per second.
A complete cycle spans two pi radians.
Frequency is therefore about 7.958 kHz and period about 125.664 μs.

Check: The original 50 kHz value is too large by a factor of two pi.

f_0=\frac{\omega_0}{2\pi}=\frac{50000\,\mathrm{s^{-1}}}{2\pi}\approx7957.75\,\mathrm{Hz}

[1]

Transfer · Application

An ideal circuit with $L=8.0\,\mathrm{mH}$ and $C=50\,\mathrm{nF}$ starts at $q(0)=0$ and $i(0)=+20\,\mathrm{mA}$ . What peak voltage does the selected plate reach, and when is its first positive maximum?

Show hint

Initial energy is in the inductor; positive current increases plate charge.

Learning goal: Transfer initially magnetic energy to peak capacitor voltage.

Show solution and calculation

Approach: Equate initial magnetic energy to maximum electric energy.
Positive initial current first charges the selected plate, so its first positive voltage peak occurs after a quarter-period.
It reaches 8.0 V at about 31.416 μs.

Check: Each store at its respective maximum contains 1.60 μJ.

\begin{aligned}U_{\mathrm{max}}&=i_i\sqrt{\frac LC}=8.0\,\mathrm V\\t_{\mathrm{first}}&=\frac{T_0}{4}\approx31.416\,\mathrm{\mu s}\\E&=\frac12L i_i^2=1.60\,\mathrm{\mu J}\end{aligned}

[1][3]

In brief

Thomson’s formula sets timing of an ideal free LC circuit. Initial conditions determine whether and how it oscillates. Charge and current have a full state period; energy fractions already repeat in half that period. Losses, probes and drive need additional model elements.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made Thomson rearrangements, initial energy, full-state period, damping, spring analogy and existing probe loading accessible through questions; linked capacitor energy, driven RLC impedance and the spring oscillator in context.
October 1, 2026Added an original DE/EN explanation with two checked examples and five distinct solved exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

Cite this reference page

Quanta. Thomson formula for an LC circuit. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/schwingkreis-thomson

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Frequently asked questions about Thomson formula for an LC circuit

How do you calculate the frequency of an LC circuit?+

For a free ideal circuit, $f_0=\frac1{2\pi\sqrt{LC}}$ . Convert L and C to henry and farad before multiplying. With 10 mH and 100 nF their product is $10^{-9}\,\mathrm s^2$ , giving about 5.033 kHz and a period of 198.692 μs. Angular frequency is two pi times frequency in hertz. Increasing inductance or capacitance under otherwise equal conditions lengthens the period. This result says nothing about amplitude or successful starting: without initial charge, initial current or excitation the circuit remains at rest. In a real damped or driven circuit you must additionally specify which frequency is being discussed.

What happens physically in an LC circuit?+

With an initially charged capacitor and zero current, all energy first occupies the electric field. After a quarter-period charge is zero and all energy is magnetic in the inductor. Continuing current then charges the capacitor with reversed polarity. After half a period the energy is electric again, but charge is negative rather than positive. Only after a full period do charge and current return together to their initial state. Energy fractions contain squares and therefore repeat after half a charge period. Current sign follows its chosen reference arrow. In this article the arrow points into the selected plate, making current negative during the first discharge.

How do you rearrange the Thomson formula for C or L?+

First square the period equation and then isolate the required component: $C=\frac{T_0^2}{4\pi^2L}$ , or $L=\frac{T_0^2}{4\pi^2C}$ . The frequency form gives $C=\frac1{4\pi^2f_0^2L}$ . Frequency enters squared rather than merely linearly in the denominator. At 1 MHz and 100 μH the ideal result is about 253.3 pF. Back-substitution should recover the target frequency; higher frequency at fixed inductance requires lower capacitance. A real setup uses effective total capacitance including additional capacitances. A measurement probe can therefore shift frequency and distort a naive inference of inductance from the observed period.

What does the LC circuit have in common with the spring pendulum?+

The evolution equations have the same mathematical structure. Signed charge corresponds to displacement, its time-derivative current to velocity, inductance to mass and reciprocal capacitance to spring stiffness. Magnetic inductor energy therefore plays the role of the spring oscillator’s kinetic energy; electric capacitor energy plays the elastic role. This mapping assumes linear ideal systems and does not turn henry into kilogram. Greater capacitance means less voltage for the same charge and hence weaker restoring action, comparable to a softer spring. Initial conditions and signs must survive the translation. Mechanical damping maps simply to electrical resistance only when the selected circuit topology supports the corresponding parameter mapping.

Why does a real LC circuit not oscillate forever?+

Real losses remove energy from a free circuit. In the simple series RLC model, $\frac{\mathrm dE}{\mathrm dt}=-Ri^2$ . An oscillatory damped response exists only below the critical series resistance $R_{\mathrm{crit}}=2\sqrt{\frac LC}$ . Its angular frequency is $\omega_d=\sqrt{\frac1{LC}-\left(\frac R{2L}\right)^2}$ ; the shift is small only under weak damping. Critical and overdamped responses are not periodic. Additional losses or measurement loading need further model assumptions. Sustained oscillation requires energy input, such as a suitable drive or feedback. A prescribed drive frequency must not be equated with free damped natural frequency. A passive circuit with neither initial energy nor a source does not start oscillating by itself.

Quanta exam set

Curated exam set for "Thomson formula for an LC circuit":

Question (front)

Which formula describes Thomson formula for an LC circuit?

Answer in your set

Question (front)

How do you rearrange Thomson formula for an LC circuit for Natural frequency?

Answer in your set

Question (front)

Which common mistake happens with Thomson formula for an LC circuit?

Answer in your set

+ 7 more cards: units, variables, derivation, example, exam task

The formula set contains 10 cards. After importing, review them and start practicing; your ratings determine later review intervals.

Create flashcards or a quiz I already have an account

Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §14.5 Oscillations in an LC Circuit

More formulas: Electricity and magnetism

F = k_e \cdot \frac{|q_1 q_2|}{r^2}

Coulomb's Law

\vec{F} = q \cdot (\vec{v} \times \vec{B})

Lorentz Force

\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}

Faraday’s law of induction

\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

Gauss's Law for Electrostatics

Spaced-repetition apps

Compare scheduling workflows

Study evidence

Read methods, results and limits

Compare flashcard apps

FSRS support and study features

Create a formula exam set

From Essential in Quanta

Retain Thomson formula for an LC circuit for exams

Create a curated FSRS exam set for T = 2π·√(L·C): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Create flashcards or a quiz Already registered? Continue directly

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

LC circuit oscillations and Thomson’s formula

Variables & units

LC circuit frequency calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electromagnetic oscillations

Worked example

Applications

How do you calculate with Thomson formula for an LC circuit?

Energy changes its store

Why does an initially empty ideal LC circuit not oscillate by itself?

How is Thomson’s formula derived from the loop balance?

How do charge, current and energy change during a period?

Energy repeats before the complete state

Calculating LC frequency, period, inductance and capacitance

Calculate charge exchange and tuning capacitance

A precharged capacitor starts the circuit

Tune an ideal circuit to 250 kHz

What do resistance, damping and driving change?

Distinguish free frequency from drive and measurement

Explore further: the spring-oscillator analogy

Explore further: probe loading

A probe lengthens the measured period

Five exercises with different reasoning paths

An ideal free circuit has L=20 mHL=20\,\mathrm{mH}L=20mH and C=2.0 μFC=2.0\,\mathrm{\mu F}C=2.0μF. Find period and frequency.

Given L=330 μHL=330\,\mathrm{\mu H}L=330μH, what ideal capacitance gives f0=80 kHzf_0=80\,\mathrm{kHz}f0​=80kHz?

An ideal circuit starts with positive charge Q and zero current. At T02\frac{T_0}{2}2T0​​ all energy is again in the capacitor. Has the initial state returned? State q, i and both energy fractions.

For L=8.0 mHL=8.0\,\mathrm{mH}L=8.0mH and C=50 nFC=50\,\mathrm{nF}C=50nF someone reports 50 kHz because 1LC=50000 rad/s\frac1{\sqrt{LC}}=50000\,\mathrm{rad/s}LC​1​=50000rad/s. Locate and correct the mistake.

An ideal circuit with L=8.0 mHL=8.0\,\mathrm{mH}L=8.0mH and C=50 nFC=50\,\mathrm{nF}C=50nF starts at q(0)=0q(0)=0q(0)=0 and i(0)=+20 mAi(0)=+20\,\mathrm{mA}i(0)=+20mA. What peak voltage does the selected plate reach, and when is its first positive maximum?

In brief

Sources for this explanation

Frequently asked questions about Thomson formula for an LC circuit

Subject references and further reading

More formulas: Electricity and magnetism

Retain Thomson formula for an LC circuit for exams

An ideal free circuit has $L=20\,\mathrm{mH}$ and $C=2.0\,\mathrm{\mu F}$ . Find period and frequency.

Given $L=330\,\mathrm{\mu H}$ , what ideal capacitance gives $f_0=80\,\mathrm{kHz}$ ?

An ideal circuit starts with positive charge Q and zero current. At $\frac{T_0}{2}$ all energy is again in the capacitor. Has the initial state returned? State q, i and both energy fractions.

For $L=8.0\,\mathrm{mH}$ and $C=50\,\mathrm{nF}$ someone reports 50 kHz because $\frac1{\sqrt{LC}}=50000\,\mathrm{rad/s}$ . Locate and correct the mistake.

An ideal circuit with $L=8.0\,\mathrm{mH}$ and $C=50\,\mathrm{nF}$ starts at $q(0)=0$ and $i(0)=+20\,\mathrm{mA}$ . What peak voltage does the selected plate reach, and when is its first positive maximum?