An LC circuit has L = 0.50 H and C = 2.010^-6 F. Find the angular frequency. [2 points]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does energy oscillate between a capacitor and an inductor in an LC circuit?

Topic 13.6 Circuits with Capacitors and Inductors (LC Circuits): model the oscillation of charge and current in an LC circuit and the exchange of energy.

A calculus-based answer to AP Physics C E&M Topic 13.6, covering the differential equation of an LC circuit, the sinusoidal oscillation of charge and current, the angular frequency, and the exchange of energy between the capacitor and inductor.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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Quick answer

An ideal LC circuit (capacitor plus inductor, no resistance) oscillates: charge and current vary sinusoidally with no decay. Combining the capacitor voltage $\dfrac{q}{C}$ and the inductor back-EMF $L\dfrac{dI}{dt}$ with $I = \dfrac{dq}{dt}$ gives $L\dfrac{d^2 q}{dt^2} + \dfrac{q}{C} = 0$ , the simple-harmonic equation $\dfrac{d^2 q}{dt^2} = -\omega^2 q$ with angular frequency $\omega = \dfrac{1}{\sqrt{LC}}$ . With the capacitor starting fully charged, $q(t) = Q_0\cos(\omega t)$ and $I(t) = -Q_0\omega\sin(\omega t)$ . Energy oscillates between the capacitor's electric field ( $\dfrac{q^2}{2C}$ ) and the inductor's magnetic field ( $\tfrac{1}{2}LI^2$ ): when the charge is maximum the current is zero (all energy in the capacitor), and a quarter cycle later the charge is zero and the current is maximum (all energy in the inductor). The total energy is conserved. This is the exact electrical analogue of a mass on a spring.

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What this topic is asking
The differential equation
The oscillation
Energy exchange
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What this topic is asking

The College Board (Topic 13.6) wants you to model an LC circuit: a capacitor and inductor with no resistance. The charge and current oscillate sinusoidally, governed by a second-order differential equation identical in form to the mass-spring oscillator, while energy sloshes back and forth between the electric and magnetic fields.

The differential equation

For a capacitor and inductor in a loop with no resistance, Kirchhoff's loop rule gives

\frac{q}{C} + L\frac{dI}{dt} = 0, \qquad I = \frac{dq}{dt}

Substituting the current yields a second-order differential equation:

L\frac{d^2 q}{dt^2} + \frac{q}{C} = 0 \quad\Longrightarrow\quad \frac{d^2 q}{dt^2} = -\frac{1}{LC}\,q

This is exactly the form $\ddot{q} = -\omega^2 q$ of simple harmonic motion, the same equation a mass on a spring obeys, with $\dfrac{1}{LC}$ playing the role of $\dfrac{k}{m}$ .

The oscillation

The solution, taking the capacitor fully charged at $t = 0$ ( $q = Q_0$ , $I = 0$ ), is

q(t) = Q_0\cos(\omega t), \qquad I(t) = \frac{dq}{dt} = -Q_0\omega\sin(\omega t)

with angular frequency

\omega = \frac{1}{\sqrt{LC}}, \qquad f = \frac{1}{2\pi\sqrt{LC}}, \qquad T = 2\pi\sqrt{LC}

The charge and current are a quarter-cycle out of phase: when one peaks, the other is zero. This natural frequency is the resonant frequency that tuned circuits (radios) exploit.

Energy exchange

The total energy stays constant, shuttling between the two stores:

U_{total} = \frac{q^2}{2C} + \tfrac{1}{2}LI^2 = \frac{Q_0^2}{2C} = \text{constant}

When the capacitor is fully charged, $I = 0$ and all the energy is electric ( $\dfrac{Q_0^2}{2C}$ ). A quarter period later the capacitor is empty, the current is maximum, and all the energy is magnetic ( $\tfrac{1}{2}LI_{max}^2$ ). The cycle repeats forever in the ideal (resistanceless) case.

Frequency and energy of an LC circuit

An LC circuit has $L = 2.0\times10^{-3}$ H and $C = 8.0\times10^{-6}$ F. The capacitor is charged to $Q_0 = 4.0\times10^{-5}$ C. Find the oscillation angular frequency and the maximum current.

step 1 Find the angular frequency

$\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(2.0\times10^{-3})(8.0\times10^{-6})}} = \dfrac{1}{\sqrt{1.6\times10^{-8}}} = \dfrac{1}{1.26\times10^{-4}} = 7.9\times10^3$ rad/s.

step 2 Use energy conservation for the maximum current

When the capacitor is empty, all the energy is in the inductor: $\dfrac{Q_0^2}{2C} = \tfrac{1}{2}LI_{max}^2$ .

step 3 Solve for the maximum current

$I_{max} = Q_0\sqrt{\dfrac{1}{LC}} = Q_0\,\omega = (4.0\times10^{-5})(7.9\times10^3) = 0.32$ A.

step 4 Check the phase

The current is maximum exactly when the charge is zero, a quarter period after the fully charged instant, consistent with $q = Q_0\cos\omega t$ and $I = -Q_0\omega\sin\omega t$ .

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Q1. An LC circuit has $L = 0.50$ H and $C = 2.0\times10^{-6}$ F. Find the angular frequency. [2 points]

Cue. $\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(0.50)(2.0\times10^{-6})}} = \dfrac{1}{1.0\times10^{-3}} = 1.0\times10^3$ rad/s.

Q2. State where the energy is when the current in an LC circuit is maximum. [1 point]

Cue. Entirely in the inductor's magnetic field (the capacitor is uncharged at that instant).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). In an ideal LC circuit, at the instant the capacitor is fully charged, the energy is (A) all in the inductor (B) all in the capacitor (C) split equally (D) zero. Justify your reasoning.

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A 1-point MCQ on energy in an LC circuit. The answer is (B).

When the capacitor is fully charged, the current is momentarily zero (it is reversing), so the inductor's energy $\tfrac{1}{2}LI^2 = 0$ and all the energy is in the capacitor, $\dfrac{Q^2}{2C}$ . A quarter-period later the capacitor is empty and the current is maximum, putting all the energy in the inductor. The trap is (A): that is the opposite instant.

AP 2024 (style)6 marksSection II (FRQ, derivation). An ideal LC circuit has inductance

L

and capacitance

C

; the capacitor starts with charge

Q_0

and no current. (a) Write the loop equation and the differential equation for

q(t)

. (b) State the solution and the angular frequency. (c) Describe how energy moves between the elements over one cycle.

Show worked answer →

A 6-point FRQ deriving LC oscillation.

(a) Differential equation (2 points): loop rule, $\dfrac{q}{C} + L\dfrac{dI}{dt} = 0$ with $I = \dfrac{dq}{dt}$ , so $L\dfrac{d^2 q}{dt^2} + \dfrac{q}{C} = 0$ , that is $\dfrac{d^2 q}{dt^2} = -\dfrac{1}{LC}q$ .
(b) Solution (2 points): simple harmonic, $q(t) = Q_0\cos(\omega t)$ with $\omega = \dfrac{1}{\sqrt{LC}}$ .
(c) Energy (2 points): energy sloshes between the capacitor's electric field ( $\dfrac{q^2}{2C}$ ) and the inductor's magnetic field ( $\tfrac{1}{2}LI^2$ ); when one is maximum the other is zero, and the total stays constant.

Markers reward the second-order differential equation, the cosine solution with $\omega = 1/\sqrt{LC}$ , and the energy exchange.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)