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Inductance

Key idea: Learn self-inductance and mutual inductance, use V = L dI/dt, and combine inductors in series/parallel with worked 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

An inductor opposes changes in current. When its current changes, self-induction produces an e.m.f. whose direction opposes that change. In magnitude, |E_L| = L|dI/dt|

1. Definitions (Must Know)

  • Self-inductance, L (H): the ratio of the magnitude of the induced e.m.f. in a circuit or component to the magnitude of the rate of change of current producing it.
  • Induced e.m.f. across an inductor (magnitude): V = L|dI/dt| (use the sign convention in circuit equations to show “opposes the change”.)
  • Mutual inductance (qualitative): a changing current in one circuit changes the magnetic flux linking a nearby circuit and induces an e.m.f. there. Its direction is set by Lenz’s law.
  • Unit: 1 H = 1 V s A⁻¹.
  • Symbols used in this lesson: L (H), V (V), I (A), t (s), M mutual inductance (H).

2. Key Ideas (What Earns Marks)

  • Inductors oppose changes in current, not steady current. If dI/dt = 0, then V = 0 across an ideal inductor.
  • Bigger L means a bigger induced e.m.f. for the same dI/dt.
  • For ideal, uncoupled inductors, the combination rules match resistor rules:
    • series: add directly,
    • parallel: add reciprocals.

Quick reference:

SetupKey resultWhen to use
Ideal inductorV = LdI/dtTransients / changing current
Steady currentdI/dt = 0 ⇒ V = 0Long time after switching
Inductors in seriesL_eq = L₁ + L₂ + …Same current through each
Inductors in parallel1/L_eq = 1/L₁ + 1/L₂ + …Same voltage across each

Self-inductance and mutual inductance

Two separate coils are shown side by side. A labelled arrow shows changing current in the primary. Two labelled curved arrows show changing magnetic flux linking both coils, and an arrow at the secondary identifies the induced effect.

Two nearby coils with changing primary current, linked changing magnetic flux and an induced effect in the secondary coilTwo nearby coils with changing primary current, linked changing magnetic flux and an induced effect in the secondary coil
Self-induction occurs in the current-carrying circuit itself; mutual inductance describes the induced effect in a nearby magnetically linked circuit.
View figure data
Topology and causal annotations for the coupled coils
PartMeaning
Primary coilChanging current I₁ produces changing magnetic flux and a self-induced e.m.f.
Magnetic linkChanging flux links the primary and nearby secondary coil
Secondary coilThe linked changing flux produces a mutually induced e.m.f.

3. Detailed Explanations

A. What V = LdI/dt means physically

If the current is increasing, the induced e.m.f. acts to oppose the increase (it acts like a “back e.m.f.”). If the current is decreasing, the induced e.m.f. acts to oppose the decrease (it “supports” the current).

That is why inductors smooth current changes in transient circuits.

B. Series and parallel inductors

For inductors in series: L_eq = L₁ + L₂ + …

For inductors in parallel: 1/L_eq = 1/L₁ + 1/L₂ + …

These relations are examinable in H3 and apply when mutual magnetic coupling between the inductors is negligible. Coupled coils require extra mutual-inductance terms.

C. Mutual inductance (what you need to know)

If current in coil 1 changes, coil 2 experiences an induced e.m.f. The syllabus only requires a qualitative understanding: this induced effect tends to oppose the change that caused it.

The syllabus requires this effect qualitatively. Unless a question supplies an additional model, do not assume a numerical mutual-inductance equation is needed.

4. Common Mistakes

  • Using V = LI (wrong); it is proportional to dI/dt, not I.
  • Forgetting that L is in henries (H), not ohms.
  • Treating the inductor as a “voltage source” with a fixed e.m.f.; the induced e.m.f. depends on how fast the current changes.
  • Mixing up series/parallel rules (series adds; parallel adds reciprocals).
  • Applying the uncoupled combination formulae to coils that share significant magnetic flux.

5. Exam Tips

  • Always write the rate of change explicitly: dI/dt ≈ (Δ I)/(Δ t) when data is given.
  • If a question says “steady state” (long time after switching), use dI/dt = 0 ⇒ V_L = 0 for an ideal inductor.
  • When combining inductors, compute L_eq first, then plug into circuit equations.

6. Worked Examples

Modelled example 1

Induced e.m.f. from a changing current

Core

Problem

An inductor has L = 0.40 H. Its current increases uniformly from 0 A to 3.0 A in 0.50 s. Find the induced-e.m.f. magnitude.
Study the worked solution

  1. Find the current gradient

    Method

    |dI/dt| = 6.0 A s⁻¹.

    Reason

    A uniform change allows the derivative to be found from Δ I/Δ t.

    Working

    |dI/dt| = (3.0-0)/0.50 = 6.0 A s⁻¹

  2. Apply self-inductance

    Method

    |E| = 2.4 V.

    Reason

    The induced-e.m.f. magnitude is L|dI/dt|.

    Working

    |E| = (0.40)(6.0) = 2.4 V

Guided practice 2

Equivalent inductance (series)

About 4 min

Problem

Uncoupled inductors 0.20 H and 0.50 H are connected in series. Find L_eq.

Try this before viewing the solution

Hints

Hint 1: same branch current
For uncoupled series inductors, inductances add directly.
View solution step by step

  1. Add the inductances

    Method

    L_eq = 0.70 H.

    Reason

    The same current passes through both series components and mutual coupling is excluded.

    Working

    L_eq = 0.20 + 0.50 = 0.70 H

Common misconception 3

Equivalent inductance (parallel)

Find and correct the mistake

Learner claim

A learner directly adds 0.30 H and 0.60 H for two uncoupled parallel inductors. Explain why the rule is wrong and find L_eq.

Try this before viewing the solution

Correct combination

View solution step by step

  1. Use the parallel rule

    Method

    1/L_eq = 5.00 H⁻¹.

    Reason

    The same voltage appears across both parallel branches.

    Working

    1/L_eq = 1/0.30 + 1/0.60 = 5.00

  2. Invert

    Method

    L_eq = 0.20 H.

    Reason

    The equivalent is the reciprocal of the summed reciprocal.

    Working

    L_eq = 1/5.00 = 0.20 H

Examiner practice 4

Steady current means zero inductor voltage

3 marks

Examination question

An ideal 0.80 H inductor carries a steady 2.0 A current. Find its voltage and explain why the non-zero current does not imply a voltage. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Interpret steady

    1 mark

    Method

    dI/dt = 0.

    Reason

    Steady current is constant in time.

    Working

    I = constant

  2. Calculate voltage

    1 mark

    Method

    V_L = 0 V.

    Reason

    An ideal inductor responds to current change, not current magnitude.

    Working

    V_L = L(dI/dt) = (0.80)(0) = 0

  3. State the distinction

    1 mark

    Method

    A non-zero steady current can flow with zero ideal-inductor voltage.

    Reason

    V = LI is not the inductance law.

    Working

    I ≠ 0, dI/dt = 0

Challenge 5

Mutual inductance reasoning

Minimal support

Independent transfer

A switch makes the current in coil 1 increase while some of its flux links nearby coil 2. Explain what causes the e.m.f. in coil 2 and what fixes its direction.

Try this before viewing the solution

Hints

Hint 1: follow the causal chain
Current change → magnetic-field change → linked-flux change.
View solution step by step

  1. Identify the changing quantity

    Method

    The increasing primary current changes the magnetic field and linked flux through coil 2.

    Reason

    Only changing linked flux induces an e.m.f.

    Working

    dI₁/dt ≠ 0 ⇒ dΦ₂/dt ≠ 0

  2. Apply induction

    Method

    Faraday induction produces an e.m.f. in coil 2.

    Reason

    The secondary loop experiences changing magnetic flux.

    Working

    E₂ ∝ -dΦ₂/dt

  3. Set direction

    Method

    Any resulting current opposes the linked-flux change.

    Reason

    Lenz’s law determines the opposition; terminal polarity also depends on winding orientation.

    Working

    induced effect opposes the cause

7. Mind Stretchers

Mind stretcher 1: “Opposes the change” but current still changesExtension

An RL circuit is switched on. The current starts at 0 and rises towards a final value.

If the inductor “opposes changes in current”, why does the current change at all?

Answer

The inductor’s induced e.m.f. opposes the change, but the source e.m.f. provides a net driving effect. The result is a gradual change rather than an instantaneous jump.

Mind stretcher 2: Why parallel inductors reduce the equivalent inductanceExtension

Resistors in parallel give a smaller R_eq. Inductors in parallel also give a smaller L_eq.

Give a short physical intuition for why putting inductors in parallel makes the overall circuit “less inductive”.

One good intuition

In parallel, the same applied voltage is shared across each inductor, but the current can split between branches. For a given rate of change of total current, the induced voltage needed can be smaller because each branch can contribute part of the changing current, so the effective opposition to change (inductance) is reduced.

8. Optional/Enrichment: Flux Linkage (Only If Given)

Some treatments define self-inductance via flux linkage λ = NΦ and L = λ/I. You do not need this formalism unless a question explicitly introduces it.

Next step

Continue to Dielectrics and Ferromagnetic Materials for the qualitative material limits that affect real capacitors and inductors.

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