advanced31 min studyLesson 245 in Physics

Inductance matrices and magnetic energy

Track winding orientations, signed mutual coupling, reciprocity, and the positivity constraints imposed by magnetic energy.

On this page

What you’ll understand

  • Construct a signed flux-linkage relation for coupled windings.
  • Use dot conventions consistently in mutual-inductance calculations.
  • Check a proposed inductance matrix against reciprocity and nonnegative energy.

Two coils share magnetic energy because the field produced by either current threads both windings. A matrix organizes this coupling, but its off-diagonal entries carry orientation information. A negative mutual inductance can be entirely physical, whereas an inductance matrix allowing negative total stored energy is inconsistent with a passive linear magnetic model. Distinguishing these statements requires defining current arrows, voltage polarities, and flux-linkage directions before doing any algebra.

Flux linkage follows an oriented winding

Flux linkage is the sum of oriented magnetic fluxes through every turn of a winding. If each turn sees the same flux, it is turn count times that single-turn flux. For fixed geometry and a linear material response, each current contributes proportionally to each linkage, defining an inductance matrix measured in henries. Leakage flux reduces coupling between coils without necessarily reducing each coil's self-inductance by the same fraction.

Choose current references entering dotted terminals and voltage drops from dot to undotted terminal. For windings whose dotted current references produce aiding flux, the mutual entry is positive. Reverse one winding's reference direction and its linkage reference consistently: both mutual entries change sign, while the self entries remain unchanged. The sign is therefore a convention tied to orientations, whereas the physical fields and total energy are unchanged by a consistent relabelling.

(λ1λ2)=(L1MML2)(I1I2),vi=dλidt\begin{pmatrix}\lambda_1\\\lambda_2\end{pmatrix}=\begin{pmatrix}L_1&M\\M&L_2\end{pmatrix}\begin{pmatrix}I_1\\I_2\end{pmatrix},\qquad v_i=\frac{d\lambda_i}{dt}
Linkage λ is in Wb, current I in A, and inductances L1, L2, M in H. Voltages use the passive reference convention. The symmetric matrix assumes a reciprocal linear, time-invariant magnetic system with fixed geometry; the circuit voltage drop has the opposite sign to the induced loop emf.↔ Scroll wide equations horizontally.

Energy supplies a consistency test

With negligible resistance, source power is the sum of current times linkage rate. For a constant symmetric inductance matrix this integrates to one half of the current vector contracted with its linkage. The mutual term appears once without a factor one half because two equal cross terms occur in the quadratic form. Dropping it ignores the energy of the shared field and predicts the same energy for aiding and opposing currents.

A passive system's energy cannot be negative for any current vector. Positive self-inductances therefore require the magnitude of mutual inductance to be no larger than the geometric mean of the self entries. Equality is an ideal perfectly coupled limit with a zero-energy current combination; ordinary leakage makes the determinant positive. Reciprocity can fail in a biased nonreciprocal medium, and hysteresis or saturation can invalidate this constant-matrix energy expression even when the circuit still uses the word inductance.

W=12L1I12+MI1I2+12L2I22,M2≤L1L2,k=∣M∣L1L2≤1W=\frac12L_1I_1^2+MI_1I_2+\frac12L_2I_2^2,\qquad M^2\le L_1L_2,\qquad k=\frac{|M|}{\sqrt{L_1L_2}}\le1
W is magnetic energy in J and k is the nonnegative coupling magnitude. These inequalities follow from a positive-semidefinite reciprocal inductance matrix with positive self entries. The sign of M is supplied separately by the winding reference directions.↔ Scroll wide equations horizontally.

Connections change the accessible current combinations

Connecting two windings in series imposes equal physical current magnitudes, but the signed current vector depends on their relative connection. Aiding references give an effective inductance equal to the sum of self terms plus twice the mutual entry. Opposing references subtract that cross contribution. With incomplete coupling the opposing connection generally retains positive inductance because some field energy resides in flux that fails to cancel.

The matrix is a reduced description of a distributed field. It assumes the circuit is small enough for propagation delays to be neglected and the selected current distributions are well defined. Eddy currents can make an inferred inductance frequency dependent; core nonlinearity can make differential inductance differ from linkage divided by current. Check the operating regime before applying a low-frequency constant value to a fast pulse or a saturating transformer.

Opposite signed currents in coupled coils

A reciprocal pair has L1=4.0 mH, L2=9.0 mH, and M=+3.0 mH for its dotted references. At I1=+2.0 A and I2=−1.0 A, find both linkages and stored energy, and check the coupling bound.

  1. The coupling magnitude is k=3/4×9=0.50k=3/\sqrt{4\times9}=0.50. Also L1L2−M2=27 mH2>0L_1L_2-M^2=27\,\mathrm{mH^2}>0, so this matrix is positive definite and passes the passive-energy test.

  2. Multiply each row by the signed currents. The first linkage is 0.004(2)+0.003(−1)=0.005 Wb0.004(2)+0.003(-1)=0.005\,\mathrm{Wb}. The second is 0.003(2)+0.009(−1)=−0.003 Wb0.003(2)+0.009(-1)=-0.003\,\mathrm{Wb}. A negative linkage merely refers to the chosen orientation.

  3. Evaluate the energy terms separately: 12(0.004)(22)=0.008\tfrac12(0.004)(2^2)=0.008 J, 0.003(2)(−1)=−0.0060.003(2)(-1)=-0.006 J, and 12(0.009)(1)=0.0045\tfrac12(0.009)(1)=0.0045 J. Their sum is 0.0065 J, still positive despite the negative mutual contribution.

The result

The linkages are +5.0 mWb and −3.0 mWb, and the stored energy is 6.5 mJ. The coupling is 0.50, safely within the passive bound.

Explore the model

Interactive lab

Interactive lab · function

Energy as a positive quadratic

The function lab plots y=a x². Reset and set both controls to Coefficient a=0.5 and Position x=1; predict y=0.5. For a fresh comparison, Reset and set both controls to Coefficient a=0.5 and Position x=2; predict y=2. Interpret x as a scaled current along a fixed current-vector direction. This illustrates the quadratic scaling of energy without representing a two-coil inductance matrix.

Energy as a positive quadraticThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDoubling all currents at fixed geometry multiplies linear magnetic energy by four. The instrument's tangent and cubic area are mathematical readouts, not linkage or magnetic work.

Check your understanding

For a passive reciprocal linear pair with positive L1 and L2, which proposed mutual inductance is admissible?

Try it yourself

Using the same 4.0 mH, 9.0 mH, +3.0 mH pair, connect the coils in series opposition so I1=I and I2=−I. Find the effective inductance and stored energy at I=3.0 A.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Substitute the constrained current vector into the energy expression and compare it with one half of effective inductance times current squared.

Compare with the solution

Substitution gives W=12(L1+L2−2M)I2W=\tfrac12(L_1+L_2-2M)I^2. The effective inductance is therefore 4+9−6=7.04+9-6=7.0 mH. At 3.0 A its energy is 12(0.007)(9)=0.0315\tfrac12(0.007)(9)=0.0315 J, or 31.5 mJ. It does not vanish because the coupling magnitude is only 0.50. Reversing the connection to series aiding would give 19 mH, which is a different physical current combination rather than a contradictory value for the same connection.

Take it with you

  • Declare winding references before interpreting the sign of mutual inductance.
  • Reciprocal linear coupling produces a symmetric inductance matrix.
  • Nonnegative energy constrains mutual-inductance magnitude and retains leakage energy in opposing connections.

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