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Energy in an Inductor

Key idea: Derive and use the energy stored in an inductor U = 1/2 L I^2, and practise exam-style energy and current calculations.

  • 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 inductor stores energy in its magnetic field when current flows. In H3, you derive and use: U = 1/2 LI²

1. Definitions (Must Know)

  • Inductor voltage (ideal): V = LdI/dt
  • Power delivered to a component: P = VI.
  • Energy stored in an inductor: U (J).
  • Symbols used in this lesson: U (J), L (H), I (A), V (V), t (s), R (Ω), τ (s), E (V).

2. Key Ideas (What Earns Marks)

  • Energy is stored when current is being built up (non-zero dI/dt).
  • The stored energy depends on I²: doubling I quadruples U.
  • The derivation is a standard “work done on charges” chain: P = VI ⇒ dU = VI dt plus V = L dI/dt.

3. Detailed Explanations

A. Derivation of U = 1/2 LI²

Instantaneous power into the inductor: P = VI So the incremental energy supplied is: dU = P dt = VI dt

Use the inductor relation V = LdI/dt: dU = (LdI/dt)I dt = LI dI

Integrate from zero current to a final current I, using x as the dummy variable: U = ∫₀^I Lx dx

U = L[1/2 x²]₀^I

Therefore, U = 1/2 LI².

This derivation assumes L is constant. A saturating ferromagnetic core can make L current-dependent, so the simple expression then needs a more careful model.

4. Common Mistakes

  • Writing U = LI (wrong); it must scale as I².
  • Forgetting that the derivation assumes an ideal inductor (no resistive losses inside the inductor).
  • Mixing up energy in an inductor (1/2 LI²) with energy in a capacitor (1/2 CV²).

5. Exam Tips

  • If asked to “derive by considering work done on charges”, start with P = VI and dU = VI dt. Substitute V = L dI/dt to obtain dU = LI dI, then integrate.
  • Use consistent SI units: L in H, I in A gives U in J.

6. Worked Examples

Modelled example 1

Energy stored at a given current

Core

Problem

An inductor has L = 0.30 H and carries I = 2.0 A. Find its stored energy.
Study the worked solution

  1. Apply the energy relation

    Method

    U = 0.60 J.

    Reason

    Magnetic energy depends linearly on L and quadratically on I.

    Working

    U = (1/2)LI² = (1/2)(0.30)(2.0)² = 0.60 J

Guided practice 2

Current needed for a target energy

About 5 min

Problem

A 0.50 H inductor must store 1.0 J. Find the required current magnitude.

Try this before viewing the solution

Hints

Hint 1: isolate the squared current
Start from 2U = LI².
View solution step by step

  1. Rearrange

    Method

    I = square root of (2U/L).

    Reason

    Current is squared in the energy relation.

    Working

    I² = 2U/L

  2. Evaluate

    Method

    I = 2.0 A.

    Reason

    Use the positive root for current magnitude.

    Working

    I = square root of (2(1.0)/0.50) = 2.0 A

Common misconception 3

Comparing two inductors

Find and correct the mistake

Learner claim

A learner says B stores more energy because L_B = 0.80 H exceeds L_A = 0.20 H. Given I_A = 3.0 A and I_B = 1.5 A, explain what the comparison overlooks.

Try this before viewing the solution

Energy comparison

View solution step by step

  1. Calculate A

    Method

    U_A = 0.90 J.

    Reason

    Use A’s own inductance and current.

    Working

    U_A = (1/2)(0.20)(3.0)² = 0.90 J

  2. Calculate B

    Method

    U_B = 0.90 J.

    Reason

    B’s fourfold inductance is offset by its current being halved and therefore I² being quartered.

    Working

    U_B = (1/2)(0.80)(1.5)² = 0.90 J

Examiner practice 4

Energy change when current doubles

3 marks

Examination question

For fixed L, current rises from I to 2I. Derive the factor change in stored energy. [3 marks]

Try this before viewing the solution

View solution step by step

  1. State dependence

    1 mark

    Method

    U = (1/2)LI².

    Reason

    L is fixed.

    Working

    U ∝ I²

  2. Form ratio

    1 mark

    Method

    U_2I/U_I = (2I)²/I².

    Reason

    Common (1/2)L factors cancel.

    Working

    U_2I/U_I = ((1/2)L(2I)²)/((1/2)LI²)

  3. Conclude

    1 mark

    Method

    Energy increases by factor 4.

    Reason

    Squaring the factor of two gives four.

    Working

    2² = 4

Challenge 5

Energy stored in a series RL circuit at a given time

Minimal support

Independent transfer

An RL circuit has R = 4.0 Ω, L = 2.0 H and E = 12 V applied at t = 0. Find the inductor energy at t = 0.50 s.

Try this before viewing the solution

Hints

Hint 1: find the transient current first
Calculate τ = L/R and I_∞ = E/R before using I = I_∞(1-e^(-t/τ)).
View solution step by step

  1. Find circuit scales

    Method

    τ = 0.50 s and I_∞ = 3.0 A.

    Reason

    These set the exponential growth curve.

    Working

    τ = 2.0/4.0 = 0.50 s, I_∞ = 12/4.0 = 3.0 A

  2. Find current at one time constant

    Method

    I = 1.9 A.

    Reason

    Here t/τ = 1.

    Working

    I = 3.0(1-e⁻¹) = 1.9 A

  3. Find stored energy

    Method

    U = 3.6 J.

    Reason

    Use the instantaneous current in U = (1/2)LI².

    Working

    U = (1/2)(2.0)(1.9)² = 3.6 J

7. Mind Stretchers

Mind stretcher 1: Where does the energy go when current decreases?Extension

In an RL circuit, when the switch is opened, the current decays. Where does the inductor’s stored energy go?

Answer

It is transferred to other parts of the circuit (typically dissipated as thermal energy in resistive elements, or delivered back to the source/other components depending on the circuit). The inductor produces an induced e.m.f. that drives current while releasing stored magnetic energy.

Mind stretcher 2: Why an inductor can “kick” a large voltageExtension

An inductor stores energy as 1/2 LI². Yet the voltage across it is V = L dI/dt.

Explain how the same stored energy can produce a very large voltage when a switch opens.

Answer

When the circuit opens, the inductor tries to keep current flowing, so dI/dt can become very large in magnitude (rapid current change). Since V = L dI/dt, that produces a large induced voltage even though the energy is limited; the large voltage typically occurs over a very short time as the stored energy is transferred/dissipated.

8. Optional/Enrichment: Energy Density and Field View

At more advanced levels, energy can be described as being stored in the magnetic field throughout space. H3 circuit questions usually do not need the field-energy density approach; 1/2 LI² is the required tool here.

Next step

Continue to RL Circuits to see how the source supplies this stored energy while current rises exponentially.

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