Electronics Guide

Self-Inductance, Mutual Inductance, and Flux Linkage

A coil of N turns, each linking the same magnetic flux Φ, has a flux linkage Λ = NΦ. In a linear magnetic path the flux linkage is proportional to the current that produces it, and the constant of proportionality is the self-inductance:

L = Λ/i

Faraday's law makes the coil voltage the rate of change of its flux linkage. With the passive sign convention, in which the current enters the terminal marked positive, v = dΛ/dt = L di/dt.

Two Coils

When a second coil sits in the field of the first, part of the flux produced by i₁ links coil 2, and part of the flux produced by i₂ links coil 1. With both currents taken in the directions that make their fluxes add, each flux linkage has two terms:

Λ₁ = L₁i₁ + M₁₂i₂

Λ₂ = M₂₁i₁ + L₂i₂

M₂₁ is the flux linkage in coil 2 per ampere in coil 1, and M₁₂ the flux linkage in coil 1 per ampere in coil 2. In linear media the two are equal. The energy argument later in this article proves it from the circuit side, and Neumann's formula shows it from the field side. The common value M is the mutual inductance, measured in henries.

Windings on a Common Core

A winding of N turns on a magnetic path of reluctance R_m drives a flux Φ = Ni/R_m, so L = N²/R_m, as Inductors and Magnetic Components develops. If every line of flux in the core links both windings, then

L₁ = N₁²/R_m, L₂ = N₂²/R_m, M = N₁N₂/R_m

so M² = L₁L₂ and L₁/L₂ = (N₁/N₂)². With R_m = 2.5 × 10⁶ A/Wb, windings of 100 and 20 turns have L₁ = 4 mH, L₂ = 160 µH, and M = 0.8 mH, exactly √(L₁L₂). Real windings fall short of this. Some flux, the leakage flux, closes through air or the winding space and links only the winding that produced it, which makes M smaller than √(L₁L₂).

The Sign of M

In field theory, M carries a sign that depends on the reference directions chosen for the two currents, and Magnetostatics treats it that way. Circuit analysis instead keeps M positive and records the winding sense on the schematic with dots.

The Dot Convention

A schematic symbol shows neither the direction in which each coil is wound nor how the coils sit relative to each other, yet those details decide whether the flux from one coil adds to or subtracts from the flux in the other. The dot convention encodes them with one mark on one terminal of each coil: currents entering the dotted terminals of two coils produce fluxes that add in the shared magnetic path. Reversing either winding moves its dot to the other terminal.

The Sign Rule

One rule turns the dots into equations. A current entering the dotted terminal of one coil induces a voltage in the other coil, equal to M times the rate of change of that current, with its positive reference at the other coil's dotted terminal.

When each coil's voltage and current follow the passive sign convention, the rule becomes a test on the two current arrows. A current leaving a dotted terminal is the negative of a current entering it, which gives the table.

Worked Example: Applying the Rule

Two coils have L₁ = 20 mH, L₂ = 80 mH, and M = 30 mH.

1. With coil 2 open, i₁ enters the dotted terminal of coil 1 and rises at 100 A/s. Coil 1 shows v₁ = L₁ di₁/dt = 2 V, and coil 2 shows M di₁/dt = 3 V, positive at its dotted terminal.
2. Now let coil 2 carry a current i₂ that leaves its dotted terminal and rises at 25 A/s, while i₁ still rises at 100 A/s. One current enters a dotted terminal and the other leaves one, so the mutual terms are negative: v₁ = 0.020 × 100 − 0.030 × 25 = 1.25 V, positive at the dotted terminal of coil 1.
3. For coil 2, with v₂ positive at the undotted terminal, where i₂ enters: v₂ = 0.080 × 25 − 0.030 × 100 = −1 V. The dotted terminal is 1 V above the undotted one, because the voltage induced by coil 1 outweighs coil 2's own.

Finding the Dots

A battery and an analog DC voltmeter locate the dots on unmarked windings; a digital meter may miss the brief kick. Connect the battery through a switch to one winding and the meter across the other. Closing the switch starts a current rising into the terminal wired to the battery's positive side, so the other winding's terminal of the same polarity goes momentarily positive. If the meter kicks upscale, its positive lead is on that terminal, and the two terminals take the dots. Opening the switch reverses the kick and can induce a voltage far larger than the battery's, so use a low-voltage battery and keep hands off the terminals.

An AC method needs no switching. Join one terminal of each winding, drive one winding with a sine wave, and measure between the two free terminals. If the joined terminals are alike, both dotted or both undotted, the reading is the difference of the two winding voltages; if they are unlike, it is the sum.

More Than Two Coils and Simulation

Windings on one core can share one set of dots, because every winding's polarity can be stated relative to the same core flux. Coils coupled through separate flux paths may need a different mark for each coupled pair, so schematics use dots, squares, triangles, or similar shapes, one shape per pair. SPICE netlists carry the convention in their syntax. In ngspice, a K element names two inductors and a coupling coefficient greater than 0 and at most 1, and the Ngspice User's Manual, version 47, places the dot on the first node listed for each inductor. Three or more inductors are coupled pair by pair, or in one K statement that gives every pair the same coefficient, and ngspice warns when a set of pairwise coefficients is physically impossible. Circuit simulation (SPICE) describes the simulators.

Coupled-Coil Equations in Time and Phasor Form

From here on, unless stated otherwise, both currents enter the dotted terminals and both voltages are positive at the dotted terminals. The first row of the table then gives

v₁ = L₁ di₁/dt + M di₂/dt

v₂ = M di₁/dt + L₂ di₂/dt

These equations hold for any waveform. Winding resistance adds R₁i₁ and R₂i₂ in series. For n coils, the vector of voltages equals an n × n inductance matrix times the vector of current derivatives; the matrix is symmetric, and the energy argument below requires that the energy it implies never be negative.

Laplace Form

Under the Laplace transform, each derivative brings in an initial current:

V₁(s) = sL₁I₁(s) + sMI₂(s) − [L₁i₁(0⁻) + Mi₂(0⁻)]

The bracket is the initial flux linkage Λ₁(0⁻), and V₂(s) takes the matching form. A current already flowing in one coil at t = 0 therefore acts as an initial-condition source in both coils.

Phasor Form

In sinusoidal steady state at angular frequency ω, each derivative becomes multiplication by jω:

V₁ = jωL₁I₁ + jωMI₂

V₂ = jωMI₁ + jωL₂I₂

The product ωM is the mutual reactance. These are the Z-parameter equations of a two-port with Z₁₁ = jωL₁, Z₁₂ = Z₂₁ = jωM, and Z₂₂ = jωL₂, in the form that Two-Port Networks uses. The equal transfer terms mark coupled coils as reciprocal.

Writing the Circuit Equations

Mesh analysis handles coupled coils directly, because the mutual terms need the coil currents, and mesh currents supply them. Nodal analysis does not, because each coil current then depends on the integrals of both coil voltages, and inverting the inductance matrix to express that works only when M² < L₁L₂. SPICE simulators avoid the difficulty with modified nodal analysis, which carries each inductor current as an extra unknown. By hand:

1. Assign a mesh current to each window of the circuit, and express the current through each coil in terms of them.
2. For each coil, note whether that current enters or leaves the dotted terminal.
3. Write Kirchhoff's voltage law around each mesh. Each coil contributes a self term, L times the derivative of its own current, or jωL times the phasor, and a mutual term, M times the derivative of the other coil's current, signed by the dot rule.
4. Use the whole coil current, which may be a difference of mesh currents, in both terms.
5. Solve, then check that the power delivered by the sources equals the power dissipated.

Worked Example: Mesh Analysis

A 10 V RMS source at 1,000 rad/s (159.2 Hz), with an internal resistance of 10 Ω, drives coil 1, and coil 2 feeds a 40 Ω load. The coils are those of the previous example, so ωL₁ = 20 Ω, ωL₂ = 80 Ω, and ωM = 30 Ω. Both dots are at the upper terminals, I₁ and I₂ enter them, and the load sets V₂ = −40I₂.

1. Primary mesh: 10 = (10 + j20)I₁ + j30I₂.
2. Secondary mesh: 0 = j30I₁ + (40 + j80)I₂, so I₂ = −j30I₁/(40 + j80).
3. Substitution: 10 = [10 + j20 + 900/(40 + j80)]I₁ = (14.5 + j11)I₁, because 900/(40 + j80) = 4.5 − j9 Ω.
4. Currents: I₁ = 10/(14.5 + j11) = 0.549 A at a phase of −37.2°, and I₂ = 0.184 A at a phase of 169.4°. The load current, −I₂, is 0.184 A at a phase of −10.6°, flowing out of the dotted terminal of coil 2.
5. Power check: the source delivers 10 × 0.549 × cos 37.2° = 4.38 W. The internal resistance dissipates 0.549² × 10 = 3.02 W and the load 0.184² × 40 = 1.36 W, which together account for all 4.38 W.

The term 4.5 − j9 Ω in step 3 is the secondary circuit as seen from the primary, the reflected impedance treated later in this article.

The Coupling Coefficient

The coupling coefficient compares the mutual inductance with the largest value it can have:

k = M/√(L₁L₂), with 0 ≤ k ≤ 1

The lower limit follows from the circuit convention of a nonnegative M. The upper limit is physical rather than a matter of definition: the next section shows that a value above 1 would let the coils store negative energy. The coils of the worked examples have k = 30/√(20 × 80) = 0.75.

What k Measures

Let a fraction k₁ of the total flux Φ₁₁ produced by coil 1 link coil 2, and a fraction k₂ of the total flux Φ₂₂ produced by coil 2 link coil 1, and assume that each of these fluxes links every turn of each coil it reaches. Then M₂₁ = k₁N₂Φ₁₁/i₁ and M₁₂ = k₂N₁Φ₂₂/i₂. Multiplying the two, and using L₁ = N₁Φ₁₁/i₁ and L₂ = N₂Φ₂₂/i₂, gives

M² = k₁k₂L₁L₂, so k = √(k₁k₂)

The coupling coefficient is the geometric mean of the two flux fractions, and k = 1 means no leakage flux at all. Air-core coils some distance apart couple loosely, with k well below 1. Windings wound closely on a closed core of high permeability bring k very close to 1, where a small change in k means a large change in leakage.

Measuring k

Shorting coil 2 sets V₂ = 0 in the phasor equations, so I₂ = −(M/L₂)I₁, and coil 1 presents an inductance

L_1sc = L₁ − M²/L₂ = L₁(1 − k²)

Readings at coil 1 with coil 2 open and then shorted therefore give k = √(1 − L_1sc/L₁). The test frequency must make each winding's reactance much larger than its resistance and stay well below self-resonance. A flyback transformer that reads 600 µH at its primary with the secondary open and 12 µH with the secondary shorted has k = √(1 − 0.02) = 0.990. Its coupling falls short of perfect by only 1 percent, yet 2 percent of its primary inductance remains in series with the primary. Series connections, described below, give a second method.

Energy Stored in Coupled Coils

Energy delivered to lossless coupled coils stays in their magnetic field, so the stored energy depends only on the final currents, not on the order in which they were established. Raising the currents in two steps shows what that implies.

1. Raise i₁ from 0 to I₁ with coil 2 open. Coil 2 develops a voltage but carries no current, so it absorbs nothing, and coil 1 absorbs ∫L₁i₁ di₁ = ½L₁I₁².
2. Hold i₁ at I₁ and raise i₂ from 0 to I₂. Coil 2 absorbs ½L₂I₂². Coil 1 carries I₁ while its mutual voltage is M₁₂ di₂/dt, so it absorbs M₁₂I₁I₂.

The opposite order gives the same self terms with M₂₁I₁I₂. Because the total cannot depend on the order, M₁₂ = M₂₁ = M, and the energy stored at any instant is

w = ½L₁i₁² + ½L₂i₂² ± Mi₁i₂

The plus sign applies when both currents enter, or both leave, dotted terminals, and the minus sign otherwise. Magnetostatics locates this energy in the field, at a density of B²/(2μ).

Why k Cannot Exceed 1

Stored magnetic energy cannot be negative, because its density is proportional to the square of the flux density. Completing the square in the expression with the minus sign gives

w = ½L₁(i₁ − Mi₂/L₁)² + ½(L₂ − M²/L₁)i₂²

Choosing i₁ = Mi₂/L₁ removes the first term, so the energy stays nonnegative for every pair of currents only if L₂ − M²/L₁ ≥ 0, that is, M² ≤ L₁L₂, or k ≤ 1. The plus sign needs no separate proof, because reversing one current turns it into the minus sign.

At k = 1 the energy can fall to zero with both currents flowing. On a common core, with each inductance proportional to the square of its turns, that happens when N₁i₁ = N₂i₂ with the fluxes opposing: the two magnetomotive forces cancel, the core carries no flux, and nothing is stored. That is the operating condition of the ideal transformer.

Worked Example: Energy and Its Minimum

The coils with L₁ = 20 mH, L₂ = 80 mH, and M = 30 mH carry I₁ = 2 A and I₂ = 0.5 A.

1. Self terms: ½ × 0.020 × 2² = 40 mJ and ½ × 0.080 × 0.5² = 10 mJ.
2. Mutual term: 0.030 × 2 × 0.5 = 30 mJ.
3. Stored energy: 80 mJ with the fluxes aiding and 20 mJ with them opposing.
4. Minimum: with I₁ held at 2 A and the fluxes opposing, the energy is least at I₂ = MI₁/L₂ = 0.75 A, where w = ½L₁(1 − k²)I₁² = 17.5 mJ.

Because k = 0.75 is less than 1, the energy reaches zero only when both currents are zero.

Series and Parallel Connections

Two coupled coils in series carry one current, i. If that current enters both dotted terminals in turn, the connection is series aiding, and the coupled-coil equations give v = (L₁ + M) di/dt + (M + L₂) di/dt. If it enters one dotted terminal and leaves the other, the connection is series opposing, and the mutual terms subtract:

L_aiding = L₁ + L₂ + 2M

L_opposing = L₁ + L₂ − 2M

Series aiding joins the undotted terminal of one coil to the dotted terminal of the other; series opposing joins like terminals. The opposing value is never negative, because it equals (√L₁ − √L₂)² + 2[√(L₁L₂) − M]; the first term is a square, and k ≤ 1 makes the second nonnegative.

Worked Example: Measuring M

An inductance meter reads 4.0 mH and 6.0 mH for two coils measured separately, 13.8 mH with the coils in series, and 6.2 mH with one coil reversed.

1. The larger series reading is the aiding connection. Subtracting the two series equations gives L_aiding − L_opposing = 4M, so M = (13.8 − 6.2)/4 = 1.9 mH.
2. Check: (13.8 + 6.2)/2 = 10.0 mH, which equals L₁ + L₂.
3. Coupling coefficient: k = 1.9/√(4.0 × 6.0) = 0.388.

The measurement also finds the dots: the terminals joined in the aiding connection are unlike.

Parallel Connections

In parallel, both coils share one voltage. Solving the phasor equations for the two coil currents and adding them gives the results in the table, which neglects winding resistance.

The coils of the measurement example give 3.29 mH in parallel aiding and 1.48 mH in parallel opposing. The shared numerator warns of a hazard. It vanishes at k = 1, so perfectly coupled windings on one core form a short circuit when paralleled, unless their turns match and their like terminals are joined. Their induced voltages differ, and in real windings only leakage inductance and winding resistance limit the current that circulates between them.

Identical Coils and the Common-Mode Choke

Two identical coils of inductance L give L(1 + k)/2 in parallel aiding and 2L(1 − k) in series opposing. A common-mode choke exploits the difference. Its two windings carry the supply and return conductors with their dots on the same side, so the load current enters one dotted terminal and leaves the other, the series-opposing case, while noise current flowing the same way in both conductors meets the parallel-aiding case. With L = 10 mH and k = 0.99, common-mode noise meets 9.95 mH, while the load current's loop meets only 2L(1 − k) = 0.2 mH, a leakage inductance that designers often put to use as differential-mode filtering. Common-Mode Filtering covers choke construction and filter design.

The T Equivalent of Coupled Coils

When two coils share a terminal, the coupled pair is a three-terminal network, and three uncoupled inductors reproduce its equations. With both dots away from the common terminal, a T with L₁ − M in series with coil 1's free terminal, L₂ − M in series with coil 2's, and M from their junction to the common terminal has Z₁₁ = jωL₁, Z₂₂ = jωL₂, and Z₁₂ = Z₂₁ = jωM, exactly the coupled-coil equations. Moving one dot to the common terminal reverses the sign of M in the equations and in the T.

The Common-Terminal Condition

The T holds only for coils that share a terminal. Isolated windings can be given one by joining a terminal of each, but only when no other conductive path links the two circuits, so that the new connection carries no current. The T then gives the right currents and voltages within each circuit, but it has discarded the isolation: it says nothing about the voltage between the windings or about current through winding capacitance.

Negative Arms

The coils of the mesh example have M = 30 mH, larger than L₁ = 20 mH, so their T has an arm of −10 mH beside arms of 50 mH and 30 mH. No real inductor has negative inductance, but the arm is harmless in analysis, because only the terminal behavior matters. With the 40 Ω load, the T gives −j10 + j30(40 + j50)/(40 + j80) = 4.5 + j11 Ω at the primary terminals, the value that mesh analysis found. When k is less than 1, coils in the first row of the T-equivalent table also have a Π equivalent, with (L₁L₂ − M²)/(L₂ − M) across coil 1, (L₁L₂ − M²)/(L₁ − M) across coil 2, and (L₁L₂ − M²)/M between the free terminals: 14 mH, −70 mH, and 23.3 mH for these coils. For the second row, replace M with −M. Two-Port Networks derives both forms from the network parameters.

The Linear Transformer and Reflected Impedance

A linear transformer is a pair of coupled coils modeled without the idealizations of the ideal transformer, keeping finite inductances, any coupling coefficient, and winding resistance. Air-core transformers in radio circuits and the coils of wireless power links are analyzed this way. Let a source drive the primary and a load Z_L terminate the secondary. With I₁ and I₂ entering the dotted terminals and V₂ = −Z_LI₂, the winding equations become

V₁ = Z₁₁I₁ + jωMI₂

0 = jωMI₁ + Z₂₂I₂

where Z₁₁ = R₁ + jωL₁ is the impedance of the primary winding and Z₂₂ = R₂ + jωL₂ + Z_L is the total impedance of the secondary loop, load included. The second equation gives I₂ = −jωMI₁/Z₂₂, and substituting it into the first gives the impedance at the primary terminals:

Z_in = V₁/I₁ = Z₁₁ + (ωM)²/Z₂₂

The second term is the reflected impedance, Z_R = (ωM)²/Z₂₂, the secondary circuit as it appears in series with the primary winding. Two-Port Networks reserves Z₂₂ for the winding alone, so its terminated-network formula, Z_in = Z₁₁ − Z₁₂Z₂₁/(Z₂₂ + Z_L), has Z₂₂ + Z_L where this article has Z₂₂; with Z₁₂ = Z₂₁ = jωM, the two results agree.

Signs and Reference Directions

The reflected impedance does not depend on where the dots are or on the reference direction chosen for I₂. Either change reverses the sign of the mutual terms and so the sign of I₂, but M enters Z_R squared. The dots decide only the polarity of the secondary current: with I₁ entering the primary's dotted terminal, the load current, −I₂ = jωMI₁/Z₂₂, leaves the secondary's dotted terminal.

The form of Z_R also fixes the signs of its parts. If R₂₂ and X₂₂ are the resistance and reactance of the secondary loop, Z_R has a resistance of (ωM)²R₂₂/|Z₂₂|², never negative for a passive secondary, because it accounts for the power delivered to the secondary, and a reactance of −(ωM)²X₂₂/|Z₂₂|², opposite in sign to the secondary's. An inductive secondary loop therefore reflects a capacitive reactance. A shorted, resistanceless secondary reflects −jωM²/L₂, leaving the short-circuit inductance L₁(1 − k²) used to measure k. For transients with zero initial currents, the same algebra gives Z_in(s) = Z₁₁(s) − s²M²/Z₂₂(s).

The Secondary Side

Reflection works in both directions. Seen from the load, a source of voltage V_s and impedance Z_s driving the primary becomes a Thevenin equivalent: an open-circuit voltage jωMV_s/(Z_s + Z₁₁), positive at the secondary's dotted terminal, behind R₂ + jωL₂ + (ωM)²/(Z_s + Z₁₁). In the mesh example, that is 13.4 V at a phase of 26.6° behind 18 + j44 Ω, which drives the same 0.184 A through the 40 Ω load.

Worked Example: A Loosely Coupled Resonant Link

Two identical coils, each of 24 µH with 0.1 Ω of winding resistance, face each other with k = 0.2, so M = 4.8 µH. At 100 kHz, ωL = 15.08 Ω and ωM = 3.016 Ω, so (ωM)² = 9.096 Ω². The secondary feeds a 5 Ω load.

1. Untuned secondary: Z₂₂ = 5.1 + j15.08 Ω, so Z_R = 9.096/(5.1 + j15.08) = 0.183 − j0.541 Ω. Only 0.183 Ω of reflected resistance competes with the primary's own 0.1 Ω.
2. Efficiency: input power divides between the primary resistance and the reflected resistance, and the secondary's share divides between R₂ and the load, so η = [0.183/(0.1 + 0.183)] × (5/5.1) = 0.634.
3. Tuned secondary: a series capacitor of 105.5 nF cancels the secondary reactance at 100 kHz, so Z₂₂ = 5.1 Ω and Z_R = 9.096/5.1 = 1.78 Ω, a pure resistance.
4. Efficiency: η = [1.78/(0.1 + 1.78)] × (5/5.1) = 0.928. A matching capacitor on the primary cancels its own 15.08 Ω, and the driver sees 1.88 Ω.

Tuning raises the reflected resistance almost tenfold, because it removes the reactance that had made |Z₂₂| large. With the secondary tuned, varying the load shows that the best efficiency depends only on k²Q₁Q₂, where Q = ωL/R for each coil. The best efficiency is k²Q₁Q₂/[1 + √(1 + k²Q₁Q₂)]², reached at R_L = R₂√(1 + k²Q₁Q₂). Here k²Q₁Q₂ = 910, so the best is 93.6 percent, with a 3.0 Ω load. Inductive power transfer covers such links in practice.

The Ideal Transformer

The ideal transformer is the limit of coupled coils under three assumptions:

• Perfect coupling, k = 1, so no leakage flux.
• Infinite core permeability, so the self- and mutual inductances are infinite and the core needs no current to magnetize it.
• No losses: no winding resistance and no core loss.

The coupled-coil equations show what each assumption contributes. With k = 1 on a common core, the expressions for L₁, L₂, and M in terms of turns give v₁ = (N₁/R_m) d(N₁i₁ + N₂i₂)/dt and v₂ = (N₂/R_m) d(N₁i₁ + N₂i₂)/dt, so the voltage ratio is exactly N₁/N₂ for any currents. The net magnetomotive force N₁i₁ + N₂i₂ equals R_mΦ, and as infinite permeability drives R_m to zero, a finite flux requires it to vanish. With both currents entering and both voltages positive at the dotted terminals,

v₁/v₂ = N₁/N₂ and N₁i₁ + N₂i₂ = 0

A current entering the primary's dotted terminal therefore leaves the secondary's dotted terminal, multiplied by N₁/N₂, and moving one dot reverses the sign in both relations. The power absorbed, v₁i₁ + v₂i₂, is zero at every instant, so the ideal transformer neither dissipates nor stores energy. The model also passes DC, which no real transformer can do.

Impedance Transformation

A load on the secondary sets V₂ = −Z_LI₂. With V₁ = (N₁/N₂)V₂ and I₂ = −(N₁/N₂)I₁,

Z_in = V₁/I₁ = (N₁/N₂)²Z_L

The ratio holds for any load and frequency and for either dot placement, because reversing a dot reverses both signs. The same argument refers any element across the transformer. Moving an impedance from the secondary to the primary multiplies it by (N₁/N₂)², a voltage source by N₁/N₂, and a current source by N₂/N₁, with source polarities set by the dots. Series elements stay in series and shunt elements in shunt. A whole side can be referred this way only if it connects to the other side solely through the transformer.

The reflected-impedance formula agrees. For k = 1 and no resistance, Z_in = jωL₁ + ω²L₁L₂/(jωL₂ + Z_L) reduces to jωL₁ in parallel with (L₁/L₂)Z_L, and L₁/L₂ = (N₁/N₂)². A unity-coupled transformer of finite inductance is thus an ideal transformer with the primary inductance in parallel across its primary; letting that inductance grow without bound removes the shunt.

Worked Example: Matching and Referring

An amplifier stage designed for a 5 kΩ load must drive an 8 Ω loudspeaker, and separately a 100 V RMS source with 50 Ω of internal resistance drives a 2 Ω load through a 5:1 step-down transformer.

1. Turns ratio for the loudspeaker: N₁/N₂ = √(5,000/8) = 25. A 4 Ω loudspeaker on the same transformer reflects 25² × 4 = 2.5 kΩ, half the design load.
2. Referred load for the 5:1 transformer: 5² × 2 = 50 Ω, which matches the source.
3. Primary: I₁ = 100/(50 + 50) = 1 A, and V₁ = 50 V.
4. Secondary: V₂ = 50/5 = 10 V, the load current is 5 × 1 = 5 A, and the load receives 10²/2 = 50 W.
5. Check from the other side: referred to the secondary, the source becomes 100/5 = 20 V behind 50/5² = 2 Ω, which drives 20/(2 + 2) = 5 A.

Per-phase analysis refers impedances through the transformers of balanced three-phase systems in the same way, and Three-Phase Circuits and Power adds the phase shift of delta–wye connections. Power system studies avoid the referral altogether: in the per-unit system, base voltages chosen in the ratio of the rated voltages give each ideal transformer a per-unit turns ratio of 1.

Practical Transformers: Magnetizing and Leakage Inductance

A real transformer departs from every ideal assumption. Its usual equivalent circuit places an ideal N₁:N₂ transformer at the center and adds the elements in the table. The model still assumes a linear core, so saturation lies outside it. Transformers describes construction and types, and Power Transformers finds the parameters of this model from open- and short-circuit tests.

From Coupled Coils to the Model

Without resistances and capacitances, the model has the terminal equations of coupled coils when

L_m = aM, L_lk1 = L₁ − aM, L_lk2 = L₂ − M/a

where a is the turns ratio of the model's ideal transformer. Terminal measurements fix only L₁, L₂, and M, so a is a free choice, and the split of leakage between the windings depends on it. The physical turns ratio, a = N₁/N₂, is the usual choice, because it keeps the elements close to the physical picture in the table. The choice a = M/L₂ puts all of the leakage on the primary, with L_lk1 = L₁(1 − k²) and L_m = k²L₁, a convenient form when one leakage element in series with the primary is wanted.

Magnetizing Inductance

The magnetizing inductance draws a current proportional to the integral of the voltage across it, whether or not the secondary carries load. It sets the low-frequency limit of a signal transformer, where L_m shunts the circuit, and it causes the droop of a pulse transformer. The same volt-seconds set the peak flux, and any DC current in a winding biases the core toward saturation, as Inductors and Magnetic Components explains. In a flyback converter, L_m is the energy-storage element.

Leakage Inductance

Leakage inductance carries the load current. It causes a voltage drop that grows with load, limits short-circuit current, and forms a low-pass filter with the source and load resistances that sets the high-frequency limit of a signal transformer. When a switch interrupts the primary current, the energy stored in the leakage becomes a voltage spike that a clamp or snubber must absorb. Resonant converters instead make the leakage part of the resonant circuit.

Worked Example: Bandwidth of a Line Transformer

A 1:1 transformer couples a 600 Ω source to a 600 Ω load. It has L_m = 2 H and 2.5 mH of leakage in each winding, so as coupled coils L₁ = L₂ = 2.0025 H, M = 2 H, and k = 0.99875.

1. Midband: L_m is nearly an open circuit and the leakage nearly a short circuit, so the load receives half the source voltage.
2. Low-frequency limit: L_m shunts the source and load resistances in parallel, 300 Ω, so the −3 dB frequency is 300/(2π × 2) = 23.9 Hz.
3. High-frequency limit: the total leakage of 5 mH is in series with 1,200 Ω, so the −3 dB frequency is 1,200/(2π × 0.005) = 38.2 kHz. The short-circuit inductance, L₁(1 − k²) = 4.997 mH, measures nearly the same total.
4. Check: the full model gives 23.8 Hz and 38.2 kHz, close to the estimates because the two limits lie more than three decades apart.

A coupling coefficient of 0.99875 sounds nearly perfect, yet its leakage sets the upper band edge. In a real transformer, winding capacitance also resonates with the leakage inductance and shapes the response near that edge.

Coupled Inductors in Power Converters

The magnetic component of a flyback converter is a pair of coupled inductors rather than a transformer in the ideal sense. Its windings are phased so that the output rectifier is reverse biased while the switch conducts. Primary current then builds stored energy, and when the switch opens, that energy drives current out of the secondary. The flyback transformer measured earlier, with 600 µH at its primary, stores ½ × 600 µH × (2 A)² = 1.2 mJ at a peak current of 2 A. Leakage energy does not reach the secondary. If the 12 µH short-circuit reading is taken as the leakage, a dissipative clamp must dispose of at least ½ × 12 µH × (2 A)² = 24 µJ each cycle, 2.4 W at a switching frequency of 100 kHz.

Coupling can also steer ripple current. When two windings carry the same voltage waveform v, positive at both dotted terminals, as the two inductors of a SEPIC converter do to a good approximation, the coupled-coil equations give di₂/dt = v(L₁ − M)/(L₁L₂ − M²). Setting M = L₁, possible when L₂ exceeds L₁, removes the ripple from winding 2 and leaves winding 1 with a ripple set by L₁ alone. Some multiphase buck converters wind their phase inductors on a shared core with inverse coupling, which reduces the ripple current in each phase while keeping the effective inductance low for load steps. Inductor and Reactor Engineering treats these designs, and DC-DC converter topologies describes the circuits that use them.

Unwanted Magnetic Coupling

Every current loop couples to its neighbors through the same M di/dt. A mutual inductance of only 1 nH between two loops on a circuit board, with a current edge of 10 mA/ns in one, induces 10 mV in the other, enough to matter at a sensitive analog input. Crosstalk between adjacent traces combines this inductive coupling with capacitive coupling, as Crosstalk Fundamentals explains, and inductance shared by many switching outputs in one package produces simultaneous switching noise. Mutual inductance depends on geometry, so the remedies are geometric:

• Shrink the loop area of both circuits, above all by routing each return current beside its signal, as over an unbroken ground plane.
• Increase the separation. For small loops far apart compared with their size, mutual inductance falls roughly as the inverse cube of the distance, so doubling the distance cuts it about eightfold.
• Orient coils so that little of one coil's flux threads the other. Two coils whose axes are perpendicular, with one axis also perpendicular to the line joining their centers, ideally have zero mutual inductance by symmetry. Closed-core and shielded inductors also help, because they confine their flux.

The sign rule still applies: reversing a victim loop reverses the sign of the voltage induced in it, which is why, in a uniform field, the voltages induced in the adjacent loops between the crossings of a twisted pair cancel. Coupling Paths and Modes places inductive coupling among the other routes that interference takes.

Summary

Magnetic coupling adds a mutual term, M times the rate of change of the other coil's current, to each coil's voltage, and the dots on the schematic fix its sign. In time, Laplace, or phasor form, the equations are those of a reciprocal two-port. The coupling coefficient k = M/√(L₁L₂) cannot exceed 1, because coupled coils cannot store negative energy. Series connections give L₁ + L₂ ± 2M, and coils with a common terminal reduce to a T of uncoupled inductors. A load on a linear transformer appears at the primary as the reflected impedance (ωM)²/Z₂₂, where Z₂₂ is the impedance of the whole secondary loop. The ideal transformer, with perfect coupling, infinite permeability, and no loss, scales impedance by (N₁/N₂)². A practical transformer is an ideal one surrounded by magnetizing inductance, leakage inductance, and losses. The same equations govern the coupled inductors of power converters and the stray coupling that designers work to suppress.

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