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Combining inductors

Derive series and parallel equivalent inductance from shared current or voltage, then check the assumptions behind each rule.

  • GCE A-Level H3 Physics 2027

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An equivalent inductance describes the terminal relation V = L_eq dI/dt of a network. First identify what the components share: the same current in series, or the same terminal voltage in parallel. The rules below assume ideal constant-inductance components with negligible mutual magnetic coupling.

Derive the series rule

The same current I passes through each component. Define all voltage drops in that current direction. Kirchhoff’s voltage law gives

V = V₁ + V₂ = L₁dI/dt + L₂dI/dt.

Comparing with V = L_eq dI/dt,

L_eq = L₁ + L₂.

For more components, add all inductances. The equivalent is greater than any individual positive inductance.

Shared current in series; shared voltage in parallelTwo uncoupled ideal inductors connect from terminal A to terminal B. In series, the same current passes through both, so terminal voltage is the sum of their inductive drops. In parallel, both branches have the same voltage, so the total rate of change of current is the sum of the branch rates. Current arrows define positive directions from A to B; voltage is A relative to B.
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Ideal, uncoupled inductors with constant L. The arrows define positive current directions; A and B are the two terminals of each network.

Guided practice 1

Equivalent inductance (series)

About 4 min

Problem

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

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Hints

Hint 1: same branch current

For uncoupled series inductors, inductances add directly.

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  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

Derive the parallel rule

Both branches share terminal voltage V, while total current is I = I₁ + I₂. Thus

dI/dt = dI₁/dt + dI₂/dt = V/L₁ + V/L₂.

Comparing with dI/dt = V/L_eq gives

1/L_eq = 1/L₁ + 1/L₂.

For two branches this is L_eq = L₁L₂/(L₁ + L₂), smaller than either positive branch inductance. Extra branches allow a greater change of total current for the same applied voltage.

Spot the mistake 2

Equivalent inductance (parallel)

About 5 min

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.

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Correct combination

Show 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

Check your understanding 1: Compare the changes in branch current

Two ideal, uncoupled inductors have 0 < L₁ < L₂ and share the same positive voltage V in parallel. A learner says, “Their currents must increase at the same rate because their voltages are equal.”

Which branch current increases faster? Explain why the equivalent inductance is smaller than either branch inductance even though the current changes are unequal.

Show answer

The smaller inductor has the larger current gradient: dI₁/dt = V/L₁ is larger than dI₂/dt = V/L₂. Equal voltage does not imply equal current gradient when the inductances differ.

The total current gradient is the sum of both positive branch gradients, so it is greater than either alone at the same applied voltage. From L_eq = V/(dI/dt), that larger total gradient means a smaller equivalent inductance. The argument assumes positive, constant inductances and negligible magnetic coupling.

Check the assumptions and initial state

Significant magnetic coupling adds mutual-induction terms, so these simple sums no longer apply. Real winding resistance can also affect the network’s terminal behaviour.

For ideal parallel inductors, the voltage law describes the change in total current. Separate initial branch currents can also contain stored energy or a circulating current that the equivalent’s total current alone does not describe. The usual initially unenergised-network model avoids that ambiguity.

Try it yourself 3

A combined network

Minimal support

Problem

The three inductors are connected as shown. They are ideal and uncoupled, and initially carry zero current. A voltage of 3.0 V is then applied across A–B, with A positive relative to B.

Identify which inductors carry the same current and which complete paths share the applied voltage. Find the equivalent inductance and the initial rate of change of total current from A to B.

An uncoupled inductor networkTerminal A connects to the 0.20 H inductor, then to point M, then through the 0.40 H inductor to terminal B. A separate wire from A passes through the 0.30 H inductor to B. Nothing else connects at M. A is 3.0 V above B. The upper and lower current arrows both point from A towards B.
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Schematic: all inductors are ideal, have negligible magnetic coupling and initially carry zero current. Arrows define positive current directions.

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Hints

Hint 1: reduce the inner branch
Add the inductances in the series branch before combining it with the other branch.
Show solution step by step

  1. Reduce the series branch

    Method

    The branch has inductance 0.60 H.

    Reason

    No current can leave at M, so the 0.20 H and 0.40 H inductors carry the same current and are in series.

    Working

    Lₛ = 0.20 + 0.40 = 0.60 H

  2. Combine the parallel branches

    Method

    The network equivalent is 0.20 H.

    Reason

    The complete upper path and the 0.30 H path both connect A to B, so they are in parallel. The applied voltage is across the whole upper path, not across each upper inductor separately.

    Working

    1/L_eq = 1/0.60 + 1/0.30 = 5.0 H⁻¹

  3. Find the total current gradient

    Method

    dI/dt = 15 A s⁻¹.

    Reason

    Use the network’s equivalent terminal voltage law.

    Working

    dI/dt = V/L_eq = 3.0/0.20 = 15 A s⁻¹

Common mistakes

  • Adding parallel inductances directly. Direct addition is the series rule.
  • Giving the full applied voltage to each inductor in a series branch.
  • Assuming equal voltages across parallel inductors mean equal current gradients.
  • Using the simple series and parallel rules when the coils are magnetically coupled.
Syllabus and review details