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LC Circuits

Key idea: Derive the LC oscillator equation, use ω = 1/√(LC) and T = 2π√(LC), and solve charge/current/energy questions with examples.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • define self-inductance as the ratio of the e.m.f. induced in an electrical circuit / component to the rate of dI change of current causing it and use V = L to solve problems dt
  • show an understanding that mutual inductance is the tendency of an electrical circuit / component to oppose a change in the current in a nearby electrical circuit / component
  • show a qualitative understanding that dielectric materials enhance capacitance, and that dielectric breakdown can occur when the electric field is sufficiently strong (knowledge of the quantitative modification of electric fields in matter through the permittivity is not required)
  • show a qualitative understanding that ferromagnetic materials enhance inductance and that this enhancement is non-linear especially near saturation (knowledge of the quantitative modification of magnetic fields in matter through the permeability is not required)
  • Apply U = ½LI² to the energy stored in an inductor.
  • Derive the expression U = ½LI² for the potential energy stored in an inductor by considering the work done on charges.
  • solve problems using the formulae for the combined inductance of two or more inductors in series and in parallel
  • solve problems involving circuits with resistors, inductors, and sources of constant e.m.f. (includes solving first-order differential equations) [RL series circuits with constant e.m.f. source]
  • solve problems involving circuits with inductors and capacitors only (includes solving second-order differential equations) [LC series circuits without e.m.f. source]
  • solve problems involving circuits with resistors, inductors and capacitors only (candidates are not expected to solve the general second-order differential equations, though they can be asked to verify and use particular solutions). [RLC series circuits without e.m.f. source]

An LC series circuit (no resistor, no e.m.f. source) can undergo electromagnetic oscillations: energy swaps between the capacitor’s electric field and the inductor’s magnetic field. Mathematically, it behaves like SHM with angular frequency: ω = 1/(square root of LC)

1. Definitions (Must Know)

  • Capacitor charge: q(t) (C).
  • Current convention: choose positive current so that I(t) = dq/dt, where q is the charge on the selected capacitor plate.
  • Capacitor voltage: V_C = q/C
  • Inductor voltage: V_L = LdI/dt = Ld²q/dt²
  • Natural (angular) frequency (LC): ω = 1/(square root of LC)
  • Period: T = 2π/ω = 2π square root of LC
  • Symbols used in this lesson: L (H), C (F), q (C), Q maximum charge (C), I (A), Iₘₐₓ maximum current (A), t (s), ω (rad s⁻¹), T (s), U_C (J), U_L (J).

2. Key Ideas (What Earns Marks)

  • In an ideal LC circuit, the total energy is conserved: U = q²/2C + 1/2 LI² = constant

  • KVL gives a second-order ODE for q(t) that matches SHM: d²q/dt² + (1/LC)q = 0

  • Solutions are sinusoidal: q(t) = Q cos(ω t + φ)

    I(t) = -ω Q sin(ω t + φ)

3. Detailed Explanations

A. Deriving the LC differential equation (no source)

Apply KVL around the loop (sign convention consistent): V_L + V_C = 0 Substitute: Ld²q/dt² + q/C = 0 Divide by L: d²q/dt² + (1/LC)q = 0 Compare with SHM form x double dot + ω² x = 0, so: ω² = 1/LC ⇒ ω = 1/(square root of LC)

B. What oscillates (and what is out of phase)

  • q(t) and V_C(t) are in phase (since V_C = q/C).
  • I(t) = dq/dt is π/2 out of phase with q(t).
  • Energy swaps:
    • capacitor energy: U_C = (1/2)q²/C
    • inductor energy: U_L = 1/2 LI² When q is maximum, I = 0 (energy purely in capacitor). When I is maximum, q = 0 (energy purely in inductor).
Four quarter-cycle states showing a charged capacitor, maximum inductor current, oppositely charged capacitor, and maximum reverse current
In the ideal model, total energy is constant. It is entirely electric at maximum charge and entirely magnetic at maximum current.

4. Common Mistakes

  • Forgetting I = dq/dt and writing I = q/t.
  • Using V_C = CV (wrong); it is V_C = q/C.
  • Dropping the second derivative: LC circuits require d²q/dt².
  • Treating the oscillations as decaying without a resistor (ideal LC does not decay).

5. Exam Tips

  • Start from KVL and decide your state variable (q is usually cleanest).
  • Quote ω = 1/square root of LC and T = 2π square root of LC once you recognise “ideal LC”.
  • Use phase reasoning: “current leads charge by 90°” (or “I is derivative of q”).

6. Worked Examples

Modelled example 1

Find the natural frequency and period

Core

Problem

An LC circuit has L = 0.50 H and C = 8.0 μF. Find ω and T.
Study the worked solution

  1. Find the LC time scale

    Method

    square root of LC = 2.0 × 10⁻³ s.

    Reason

    Convert microfarads before multiplying.

    Working

    square root of ((0.50)(8.0 × 10⁻⁶)) = 2.0 × 10⁻³ s

  2. Find angular frequency

    Method

    ω = 500 rad s⁻¹.

    Reason

    ω = 1/square root of LC.

    Working

    ω = 1/(2.0 × 10⁻³) = 500 rad s⁻¹

  3. Find period

    Method

    T = 1.3 × 10⁻² s.

    Reason

    T = 2π/ω.

    Working

    T = 2π(2.0 × 10⁻³) = 1.3 × 10⁻² s

Guided practice 2

Maximum current from initial charge (energy method)

About 6 min

Problem

A capacitor initially holds charge Q and is connected to an ideal inductor L. Derive Iₘₐₓ.

Try this before viewing the solution

Hints

Hint 1: compare two extreme states
Initially energy is electric; at maximum current it is magnetic.
View solution step by step

  1. Write the initial energy

    Method

    U = Q²/(2C).

    Reason

    Initially current is zero and capacitor charge is maximum.

    Working

    U_C = Q²/(2C)

  2. Write maximum-current energy

    Method

    U = (1/2)LIₘₐₓ².

    Reason

    At maximum current the ideal capacitor is momentarily uncharged.

    Working

    U_L = (1/2)LIₘₐₓ²

  3. Equate and solve

    Method

    Iₘₐₓ = Q/square root of LC = ω Q.

    Reason

    Total energy is conserved.

    Working

    Q²/(2C) = LIₘₐₓ²/2

Common misconception 3

Find Q from Iₘₐₓ (reverse of Example B)

Find and correct the mistake

Learner claim

For L = 0.40 H, C = 10 μF and Iₘₐₓ = 1.5 A, a learner divides by square root of LC to find Q. Explain the rearrangement error and calculate Q.

Try this before viewing the solution

Correct rearrangement

View solution step by step

  1. Find the time scale

    Method

    square root of LC = 2.0 × 10⁻³ s.

    Reason

    Convert 10 μF to 10 × 10⁻⁶ F.

    Working

    square root of ((0.40)(10 × 10⁻⁶)) = 2.0 × 10⁻³ s

  2. Rearrange correctly

    Method

    Q = 3.0 × 10⁻³ C.

    Reason

    Iₘₐₓ = Q/square root of LC implies multiplication by the time scale.

    Working

    Q = (1.5)(2.0 × 10⁻³) = 3.0 × 10⁻³ C

Examiner practice 4

Energy at maximum current

4 marks

Examination question

An ideal LC circuit has Qₘₐₓ = 2.0 × 10⁻³ C, C = 5.0 μF and L = 0.20 H. Find Iₘₐₓ. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Capacitor energy

    1 mark

    Method

    U = 0.40 J.

    Reason

    At maximum charge all ideal-circuit energy is electric.

    Working

    U = Q²/(2C) = 0.40 J

  2. Equate energies

    1 mark

    Method

    (1/2)LIₘₐₓ² = U.

    Reason

    At maximum current all energy is magnetic.

    Working

    Iₘₐₓ = square root of (2U/L)

  3. Substitute

    1 mark

    Method

    Iₘₐₓ = square root of (2(0.40)/0.20).

    Reason

    Use SI quantities.

    Working

    Iₘₐₓ = square root of 4

  4. Report

    1 mark

    Method

    Iₘₐₓ = 2.0 A.

    Reason

    Take the positive magnitude.

    Working

    Iₘₐₓ = 2.0 A

Challenge 5

Phase relationship: when is current maximum?

Minimal support

Independent transfer

An ideal LC circuit has q(t) = Q cos(ω t). When does current first reach maximum magnitude after t = 0?

Try this before viewing the solution

Hints

Hint 1: differentiate charge
Current is dq/dt.
View solution step by step

  1. Find current

    Method

    I = -ω Q sin(ω t).

    Reason

    Current is the time derivative of capacitor charge.

    Working

    I = dq/dt

  2. Maximise magnitude

    Method

    | sin(ω t)| = 1.

    Reason

    The amplitude factor ω Q is constant.

    Working

    ω t = π/2 first after zero

  3. State the time

    Method

    t = π/(2ω).

    Reason

    This is one quarter of the oscillation period.

    Working

    t = T/4

7. Mind Stretchers

Mind stretcher 1: Why oscillations need both L and CExtension

Why doesn’t a circuit with only a capacitor (no inductor) “oscillate” once discharged?

Answer

You need an inductor to provide “inertia” for current change (magnetic energy storage). A capacitor alone can store electric energy, but without an inductor there’s no mechanism for current to keep flowing and reverse the charge repeatedly.

Mind stretcher 2: What changes when you add a small resistance?Extension

Real circuits always have some resistance. Compared to an ideal LC circuit, what qualitative changes do you expect if you add a small series resistor?

Answer

You get damping: the oscillations decay in amplitude with time because energy is dissipated as heat in the resistor. The oscillation frequency becomes slightly lower than the ideal 1/square root of LC value.

8. Optional/Enrichment: Real LC circuits (damping)

In practice, resistance is never exactly zero, so real LC oscillations decay. The damped case is treated in the RLC analysis lesson:

Next step

Continue to RLC Circuits Analysis to add resistive energy loss and classify the resulting response.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027