SCIENCE · INVESTIGATE AND EXPLAIN

Use both initial conditions in an LC oscillation

Choose one capacitor plate and let q be its signed charge; define i = dq/dt as current toward that plate.

Learning outcome: Derive both state functions and check their initial values and derivatives. Retain several phase snapshots with electric, magnetic and total energy. Contrast a general initial state with the initially charged special case and propose independent evidence for a hypothetical decaying envelope.

Starting recap

Differentiate linked flux, preserve circuit state and test energy and independent-calibration constraints.

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Try

L = 2 H, C = 1 F, q0 = 1 C and i0 = 1 A give omega = 1/sqrt(2), charge amplitude sqrt(3) C, peak current sqrt(3/2) A and total energy 1.5 J. At a quarter period, q = sqrt(2) C and i = −1/sqrt(2) A: this is not a pure magnetic-energy instant because the initial current was not zero.

Open the experiment

Predict

In the ideal LC model, at an instant when capacitor charge is zero but total energy is positive, what follows?

Your prediction and reason

Experiment

Derive both state functions and check their initial values and derivatives. Retain several phase snapshots with electric, magnetic and total energy. Contrast a general initial state with the initially charged special case and propose independent evidence for a hypothetical decaying envelope.

Solve the LC initial-value problem

Worked teaching model · synthetic values · Ideal isolated LC circuit with capacitance 1 F and initial signed current i = dq/dt = +1 A. Current reference is toward the selected positively labeled plate. No resistance, radiation or source; signed charge may reverse.

Inductance: 2 H Initial signed capacitor charge: 1 C

Reset keeps your prediction, captures and practice history.

State and topology reference

This original schematic defines one closed loop; it is not an apparatus wiring instruction. Exact element connections and event conditions are stated below. Component sizes are schematic.

The capacitor and inductor form one isolated loop; there is no source or resistor.
The q label marks the selected left capacitor plate. The arrow is toward that plate, so i = dq/dt.
L d²q/dt² + q/C = 0. Reversing the chosen plate reverses q and i together, not the nonnegative energy.

Signed capacitor charge

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Signed capacitor charge: numerical coordinates

time (s)	charge (C)
0	1
0.148096	1.14235
0.296192	1.27218
0.444288	1.38807
0.592384	1.48876
0.74048	1.57313
0.888577	1.64027
1.03667	1.68944
1.18477	1.7201
1.33286	1.73191
1.48096	1.72474
1.62906	1.69869
1.77715	1.65401
1.92525	1.59122
2.07335	1.51099
2.22144	1.41421
2.36954	1.30194
2.51763	1.1754
2.66573	1.03598
2.81383	0.885212
2.96192	0.724745
3.11002	0.556338
3.25811	0.381835
3.40621	0.203149
3.55431	0.0222369
3.7024	-0.158919
3.8505	-0.338333
3.99859	-0.51404
4.14669	-0.684116
4.29479	-0.846696
4.44288	-1
4.59098	-1.14235
4.73908	-1.27218
4.88717	-1.38807
5.03527	-1.48876
5.18336	-1.57313
5.33146	-1.64027
5.47956	-1.68944
5.62765	-1.7201
5.77575	-1.73191
5.92384	-1.72474
6.07194	-1.69869
6.22004	-1.65401
6.36813	-1.59122
6.51623	-1.51099
6.66432	-1.41421
6.81242	-1.30194
6.96052	-1.1754
7.10861	-1.03598
7.25671	-0.885212
7.4048	-0.724745
7.5529	-0.556338
7.701	-0.381835
7.84909	-0.203149
7.99719	-0.0222369
8.14529	0.158919
8.29338	0.338333
8.44148	0.51404
8.58957	0.684116
8.73767	0.846696
8.88577	1

Signed current toward chosen plate

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Signed current toward chosen plate: numerical coordinates

time (s)	current (A)
0	1
0.148096	0.920609
0.296192	0.831132
0.444288	0.732549
0.592384	0.625939
0.74048	0.512472
0.888577	0.39339
1.03667	0.269998
1.18477	0.143648
1.33286	0.0157238
1.48096	-0.112372
1.62906	-0.239238
1.77715	-0.363482
1.92525	-0.483743
2.07335	-0.598705
2.22144	-0.707107
2.36954	-0.807762
2.51763	-0.899566
2.66573	-0.981516
2.81383	-1.05271
2.96192	-1.11237
3.11002	-1.15985
3.25811	-1.19461
3.40621	-1.21629
3.55431	-1.22464
3.7024	-1.21958
3.8505	-1.20115
3.99859	-1.16956
4.14669	-1.12516
4.29479	-1.06843
4.44288	-1
4.59098	-0.920609
4.73908	-0.831132
4.88717	-0.732549
5.03527	-0.625939
5.18336	-0.512472
5.33146	-0.39339
5.47956	-0.269998
5.62765	-0.143648
5.77575	-0.0157238
5.92384	0.112372
6.07194	0.239238
6.22004	0.363482
6.36813	0.483743
6.51623	0.598705
6.66432	0.707107
6.81242	0.807762
6.96052	0.899566
7.10861	0.981516
7.25671	1.05271
7.4048	1.11237
7.5529	1.15985
7.701	1.19461
7.84909	1.21629
7.99719	1.22464
8.14529	1.21958
8.29338	1.20115
8.44148	1.16956
8.58957	1.12516
8.73767	1.06843
8.88577	1

Calculated values under the declared model
Quantity	Value	Unit
Angular frequency	0.707107	rad/s
Charge amplitude	1.73205	C
Peak current magnitude	1.22474	A
Total energy	1.5	J
Charge after quarter period	1.41421	C
Current after quarter period	-0.707107	A
Energy residual at sampled phase	2.22045e-16	J

Both initial charge and current determine phase; energy oscillates between electric and magnetic storage without dissipation.

With i = dq/dt, the isolated loop gives L d²q/dt² + q/C = 0. Its solution retains both initial conditions and omega = 1/sqrt(LC). Maximum current occurs when charge passes through zero, not when the energy disappears. No resistor or source is present in this ideal model.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Both initial charge and current determine phase; energy oscillates between electric and magnetic storage without dissipation.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define charge and current references

Choose one capacitor plate and let q be its signed charge; define i = dq/dt as current toward that plate. A positive q does not require positive i: the selected plate may be discharging. The ideal closed circuit contains only L and C, with no source, resistance or radiation loss. Its loop equation is L d²q/dt² + q/C = 0.

Derive phase from the initial state

The angular frequency is omega = 1/sqrt(LC). For q(0) = q0 and i(0) = i0, q(t) = q0 cos(omega t) + (i0/omega) sin(omega t). Differentiating gives i(t) = −q0 omega sin(omega t) + i0 cos(omega t). A cosine-only solution silently assumes i0 = 0. The charge amplitude is sqrt[q0² + (i0/omega)²], and the peak current magnitude is omega times that amplitude.

Track energy through a zero crossing

Total energy is q²/(2C) + L i²/2. Differentiation and the loop equation give zero time derivative. At q = 0, all energy is magnetic and current magnitude is maximal; at i = 0, all energy is electric and charge magnitude is maximal. A zero instantaneous charge therefore does not mean an unenergized circuit. Charge and current are separated in phase, and signed quantities should not be confused with nonnegative energies.

Compare the special initially charged case

For q0 = C V0 and i0 = 0, the maximum current is V0 sqrt(C/L), directly from energy. After a quarter period, q = 0 and i = −q0 omega in the chosen toward-plate reference. After half a period, charge reverses sign with zero current. These statements depend on the explicit initial phase; with nonzero i0 the same clock times need not mark energy extrema.

Test the idealization rather than fitting forever

A persistent decrease in the envelope would contradict the ideal isolated lossless LC model, although it could arise from resistance, radiation, probe loading or time-dependent components. One decaying trace alone does not identify which mechanism is responsible. Retain a proposed independent comparison instead of relabeling a damped record as an exact ideal solution.

Worked original case

Investigation

Changed-case practice

Try a new practice case

An ideal LC circuit has L = 4 H and C = 1 F. What is angular frequency in rad/s?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and boundaries

Original teaching checked 31 October 2026 against current official Unit 13 and the relevant OpenStax discussion. No source exercises or diagrams are reproduced. All records and apparatus are virtual, supplied or generated; no real wiring, switching, magnetic braking or electrical measurement is requested. Public feedback and retained work do not establish independent mastery, supervised practical participation or AP authorization.

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Public practice checks a specific claim. It does not establish independent mastery.

Which orientation, initial state, symmetry or independent calibration assumption would change this induction inference?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Preserve inductor current when the closed path changes Next: Separate LC components with independent amplitude evidence

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