Mutual Inductance

Start with a single coil. A current \(i_1\) in coil 1 produces a flux \(\phi_1\), and Faraday's law gives the voltage across it:

Self-inductance is the constant of proportionality between a coil's own flux linkage and its own current — Set 16.

Bring a second coil nearby. Only part of \(\phi_1\) passes through it:

The linked portion induces a voltage in coil 2 even though no current flows there:

This is the whole phenomenon. A voltage appears across terminals connected to nothing, because flux — not charge — carries the influence.

The coupling coefficient measures what fraction of the flux is shared:

Relating \(k\) to \(M\). Working through both directions and multiplying:

Why \(M_{12} = M_{21}\). This is not obvious — the coils may have wildly different sizes and turn counts. The reason is energy. Bring the two currents to their final values by two different routes:

The stored energy of a lossless system depends only on the final state, not on how it was reached. Hence \(M_{12} = M_{21} \equiv M\) — an instance of the reciprocity that Set 21, Problem 17 established for any bilateral network.

The complete pair of equations:

The \(\pm\) is the only thing left undetermined, and geometry rather than algebra fixes it — Problem 2.

The problem the dots solve. A circuit diagram shows how coils are connected but not how they are wound. Reversing the winding direction of one coil reverses the sign of every mutual term — and no amount of information about the connections can reveal it.

What a dot means. Dots mark terminals such that

Equivalently: currents entering dotted terminals produce fluxes that add. The dots are a two-dimensional summary of a three-dimensional winding fact.

The sign rule in the form used throughout this set:

Only the relative orientation matters, which is why moving both dots to the opposite ends of both coils changes nothing.

A reliable procedure:

The commonest error is deciding the sign from the appearance of the diagram rather than from the assigned current directions. If a mesh current is later found negative, the sign of \(M\) does not change — the assignment was a bookkeeping choice, not a physical claim.

The symmetry check. Whatever sign is chosen, it must be the same in both equations:

A mesh matrix with \(Z_{12} \ne Z_{21}\) is always an error for a passive coupled circuit, since \(M_{12} = M_{21}\) was proved in Problem 1. Problem 15 shows a published example where this check catches a defect.

Determining the dots experimentally. Apply a DC source to coil 1 through a switch and watch a voltmeter on coil 2 at the instant of closing:

The current enters the dotted terminal of coil 1 by construction, so the dot on coil 2 is wherever the induced voltage is positive.

The energy stored in a coupled pair, taking the worst case where the mutual term subtracts:

Passivity requires \(w \ge 0\) for every possible pair of currents — a passive network cannot store negative energy, since that would mean energy could be extracted from it indefinitely.

Treat it as a quadratic in \(i_1\):

A quadratic \(ax^2+bx+c\) with \(a > 0\) is non-negative for all \(x\) exactly when its discriminant is non-positive:

What happens at each value of \(k\), taking \(L_1 = 4\ \text{H}\), \(L_2 = 9\ \text{H}\) so \(\sqrt{L_1L_2} = 6\):

At \(k = 1\) there is a particular current combination that stores exactly zero energy — the fluxes cancel completely. Beyond it, the stored energy goes negative, which no passive arrangement of coils can do.

The current combination that stores nothing. At \(k = 1\) the quadratic has a repeated root:

With \(L_1 = 4\), \(L_2 = 9\), \(i_2 = 1\): \(i_1 = 1.5\ \text{A}\), giving \(w = \tfrac12(4)(2.25) + \tfrac12(9)(1) - 6(1.5)(1) = 4.5+4.5-9 = 0\) ✓. The two fluxes are equal and opposite, so the core carries no net field at all.

Why \(k = 1\) is unattainable in practice. It requires every flux line from one coil to link the other. Real windings have leakage — flux that escapes into the air around the core — so even a good iron-cored transformer reaches only \(k \approx 0.998\). That small shortfall is the leakage inductance, and it is what limits a transformer's short-circuit current and sets its voltage regulation.

The mutual inductance:

The two currents:

Both in phase, so both fluxes vary together — and with both entering the dots, the mutual term adds.

The derivatives at \(t = 0\):

The primary voltage:

The mutual term contributes 27% of the total — substantial even at \(k = 0.6\) and with the secondary current only a quarter of the primary.

The instantaneous currents at \(t = 0\):

The stored energy, with the mutual term positive:

The mutual term is 35% of the total — larger than the secondary's own self-energy. Coupling contributes to stored energy quite out of proportion to the smaller current.

Check the coupling is physical:

Always worth a glance — a stated \(M\) exceeding \(\sqrt{L_1L_2}\) is the fastest way to detect a misprinted problem.

The mutual inductance:

Both currents enter the dots, so the fluxes aid and the mutual energy term is positive:

All three terms happen to be equal here — a coincidence of the chosen numbers, but a convenient reminder that the mutual term can be as large as either self term.

Steady currents, yet energy is stored. No voltage is induced — \(di/dt = 0\) — but the magnetic fields exist and hold energy. Interrupting either current would release it, which is why opening an inductive circuit produces an arc.

If one coil were reversed, the mutual term would subtract:

A factor of three difference in stored energy from reversing one winding, with every component value unchanged. This is the quantitative statement of Problem 2's point about dots carrying real information.

Verify against the passivity bound. With the mutual term subtracting, the minimum possible energy is

Here \(k = 0.5\), well inside the bound, and 16 J is comfortably positive. Had \(k\) been 1 with these currents:

Exactly zero — the fluxes cancel completely, which is Problem 3's limiting case appearing with concrete numbers.

Derive it first. In series both coils carry the same current \(i\), and the total voltage is the sum:

The factor of two is the point: the mutual term appears once in each coil's equation, so it enters the total twice.

Series aiding — the current enters both dotted terminals, fluxes add:

Series opposing — one coil reversed, fluxes oppose:

A factor of 2.33 between them, from an identical pair of coils differently connected. The two are not small perturbations of the uncoupled value \(L_1+L_2 = 10\ \text{H}\) — they straddle it symmetrically:

Can \(L_{eq}\) go negative? The opposing case requires

and since \(M \le \sqrt{L_1L_2}\) by Problem 3:

So it is never negative — the passivity bound of Problem 3 guarantees it. Equality occurs only when \(k = 1\) and \(L_1 = L_2\), giving \(L_{eq} = 0\) exactly: two identical perfectly coupled coils in opposition behave as a short circuit at every frequency.

The practical use — a bifilar winding. Winding two identical wires together and connecting them in opposition gives \(k \approx 1\) and hence almost zero inductance, producing a genuinely non-inductive resistor. The same trick in reverse, connected aiding, doubles the inductance for the same wire length.

In parallel the coils share a voltage, not a current — so the derivation must start from the two branch equations and eliminate the currents:

Taking the aiding case; the opposing case replaces \(M\) by \(-M\) throughout.

Subtract to relate the two currents:

Substitute into the first equation and form \(\mathbf{V}/(\mathbf{I}_1+\mathbf{I}_2)\):

Note the sign in the denominator is opposite to the series case — a frequent source of error, and worth checking against a limiting case every time.

The check. With \(M = 0\) both formulas reduce to

the familiar product-over-sum for uncoupled inductors in parallel ✓.

Evaluating:

Aiding increases the parallel inductance and opposing decreases it — the same direction as the series case, though the algebra looks reversed.

Why the numerator is \(L_1L_2-M^2\). This is the determinant of the inductance matrix:

Problem 3 showed this determinant is non-negative, vanishing exactly at \(k = 1\). So perfectly coupled coils in parallel have zero equivalent inductance in either orientation — the same degenerate behaviour as the series-opposing case, arrived at differently.

The two measurements are the aiding and opposing connections of Problem 6:

Subtract to isolate \(M\):

Add to obtain the sum of the self-inductances:

So with \(L_1 = 4\), \(L_2 = 6\ \text{H}\) — recovering Problem 6's coils exactly.

The coupling coefficient:

Moderate coupling — typical of air-cored coils placed close together but not interleaved.

Why two series tests do not suffice alone. They give \(M\) and \(L_1+L_2\), which is two facts for three unknowns. A third measurement is needed — usually \(L_1\) alone, with the second coil open-circuited:

Any three of these four determine everything, and the fourth then serves as a consistency check — a redundancy worth using.

The method also identifies the dots. Whichever connection gives the larger reading is the aiding one, so the terminals joined in that connection are opposite in dot status: a current entering one coil's dot leaves the other's. This is a purely electrical way to establish the dot markings, needing only an inductance meter rather than the switching test of Problem 2.

The reactances at \(\omega = 2000\ \text{rad/s}\):

Convert the sources to phasors, with cosine as the reference:

The coupled equations. From the dot markings one current enters a dot and the other the undotted end, so the mutual term is subtractive:

The same sign in both — the symmetry required by \(M_{12} = M_{21}\).

Solve the first for \(\mathbf{I}_1\):

Note \(-j16.97 \times -j2 = j^2(33.94) = -33.94\) — a real number, because two imaginary quantities multiply to a real one.

Substitute into the second:

Back to the time domain:

Note the two contributions to \(\mathbf{V}_2\). The mutual term gives −9.32 V and the self term +20 V, partly cancelling to leave 10.68 V. Had the dots been the other way round they would have added to 29.32 V — nearly three times as much, from the same components.

The mesh equations, with the mutual term appearing off-diagonal in both rows:

Check the symmetry immediately: \(Z_{12} = Z_{21} = j\) ✓. This is the first thing to verify in any coupled mesh problem.

Solve by Cramer's rule. The determinant:

Rationalise:

Second-quadrant angle — the real part is negative and the imaginary part positive, so \(\tan^{-1}\) must be taken with care.

The output voltage:

Where the mutual term went. Notice that \((j)^2 = -1\) increased the determinant's real part from −14 to −13. The mutual coupling has changed the network's response even though mesh 2 contains no source — it is the sole reason any current flows there at all. Setting \(M = 0\) gives \(\mathbf{I}_2 = 0\) exactly, and no output.

Sanity check on the coupling. With mesh reactances of \(j6\) and \(j3\), if these were the coil reactances alone then

— light coupling, comfortably below 1 and therefore physical. Whenever mesh reactances are given, this check costs one line.

Transform the current source first. A \(2\angle0°\) A source in parallel with \(-j1\ \Omega\) becomes a voltage source in series with the same impedance — Set 22, Problem 11:

Source transformation applies unchanged here, because it concerns only the source branch and not the coupling.

The mesh equations after the transformation:

Symmetric off-diagonals ✓. Mesh 2's self-impedance is \(2 - j1 + j4 = 2+j3\), combining the capacitor and the coil.

Solving the pair:

The output voltage:

A caution about transforming coupled branches. The transformation was applied to a branch containing no coupled coil. Attempting to transform a source in series with one of the coupled inductors is not valid in the ordinary way — the coil's voltage depends on the other mesh's current, so it is not a simple impedance:

Why the sign of the transformed source matters. Mesh 2's right-hand side is \(+j2\), not \(-j2\) — the polarity follows from which way the original current source pushed relative to the assumed mesh direction. Getting this wrong changes \(\mathbf{I}_2\) substantially without producing anything obviously wrong, which is why the transformed circuit should be sketched rather than done in the head.

The test-source method. Drive the port with 1 V at \(\omega = 1\ \text{rad/s}\), find the resulting input current, and take the ratio — Set 22, Problem 8's technique applied to a purely inductive network.

Eliminate the second loop current:

Note the two \(j\) factors cancel, leaving a purely real ratio — the two loop currents are in phase or antiphase, as they must be in a lossless network.

Substitute into the first equation:

The input impedance and hence the inductance:

Purely imaginary with a positive imaginary part, so the network is net inductive — as it must be, since it contains only inductors.

Note what the coupling did. Without the mutual term the second loop would carry no current and the input would see \(j8\), giving \(L_{eq} = 8\ \text{H}\). The coupling has reduced the equivalent inductance to 7.111 H:

This is the reflected-impedance formula of Problem 13 in its simplest, purely inductive form — a shorted secondary always lowers the primary inductance.

Setting \(\omega = 1\) is a convenience, not a restriction. Because every impedance is proportional to \(\omega\), the ratio \(\mathbf{Z}/j\omega\) is frequency-independent for a purely inductive network. Working at \(\omega = 1\) makes the reactances numerically equal to the inductances and removes a layer of arithmetic.

Derive first. The two coupled equations with the secondary loop closed on itself:

From the second, \(\mathbf{I}_2 = -j\omega M\mathbf{I}_1/\mathbf{Z}_{22}\). Substituting:

The sign works out positive because \((j\omega M)^2 = -(\omega M)^2\). Note the mutual term's sign disappears — it is squared, so the dot orientation does not affect the input impedance at all.

Find the operating frequency from the load:

The mutual inductance and all reactances:

The secondary loop impedance:

The reflected impedance:

Note the sign inversion. The secondary is inductive (\(+j120\)), but what it reflects into the primary is capacitive (\(-j6.14\)). Dividing by a complex number conjugates its angle, so an inductive load always reflects as capacitive and vice versa.

The input impedance:

The reflected resistance of 3.07 Ω represents real power delivered to the secondary's 60 Ω — power crossing the air gap with no conducting path.

Two limiting cases worth knowing:

Transform the source:

The mutual reactance, computed directly in reactance terms:

Working in reactances avoids ever computing \(\omega\) or the inductances — legitimate because \(k\) is dimensionless and the \(\omega\) factors cancel inside the square root.

The mesh equations:

Symmetric ✓. Mesh 1's self-impedance combines the resistance, the capacitor and the coil: \(50 - j30 + j20 = 50 - j10\).

Solving:

Check the coupling is physical:

Recovering the stated \(k = 0.6\), which confirms the mutual reactance was computed correctly.

The current in mesh 2 exists only because of the coupling. There is no source in that loop, and no conducting path to mesh 1 — the entire 1.5 A is induced magnetically. Setting \(k = 0\) gives \(\mathbf{I}_2 = 0\), and the network falls apart into two unconnected pieces.

Apply the symmetry check first, as the Method Recap requires:

The matrix is not symmetric. For a network of resistors, inductors, capacitors and mutual inductance this is impossible. Problem 1 proved \(M_{12} = M_{21}\) from energy conservation, and Set 21, Problem 17 showed the mesh matrix of any bilateral network must be symmetric. There are no dependent sources here. The equations as printed contain an error.

Solving them anyway does reproduce the quoted answer:

So the published answer follows correctly from the published equations — the defect is upstream, in forming them.

Locating the error. One of the two mutual terms is wrong. The two repairs give quite different answers:

Both repairs are physically valid — each gives \(k \le 1\) against \(\sqrt{(10)(3)} = 5.48\). Without the original figure the choice cannot be settled from the equations alone.

Which is more likely. A dropped digit is a commoner typographical accident than a spurious one, so \(-j\mathbf{I}_1\) is probably a mistyping of \(-j4\mathbf{I}_1\). On that reading:

Confirm via reflected impedance: \(\mathbf{Z} = (1+j10) + 16/(2+j3) = 1+j10+2.462-j3.692 = 3.46+j6.31\) ✓, agreeing with Problem 13's formula.

Why this matters more than the arithmetic. The published answer is internally consistent — it solves its own equations correctly, has a plausible magnitude, and a plausible angle. Nothing about the number reveals the problem. Only the structural check on the matrix catches it:

The apparent discrepancy. With \(k = 0.5\) and \(\omega L_1 = \omega L_2 = 10\):

Yet the off-diagonal term is \(j10\), twice the mutual reactance. This looks like an error — and is not.

The resolution: an off-diagonal term has two sources. When two meshes share a branch and are magnetically coupled, both contribute:

For \(Z_{12} = j10\) with \(\omega M = 5\), the shared branch must contribute \(+j5\), so \(\mathbf{Z}_{\text{shared}} = -j5\) — a capacitor.

Confirm against the diagonals. If mesh 1 contains 4 Ω, the shared \(-j5\) capacitor and its coil \(j10\):

Both diagonals reproduced exactly. The circuit is entirely consistent — the coil reactances are 10 Ω each, the shared capacitor is \(-j5\ \Omega\), and the mutual reactance is 5 Ω.

Solving the system:

\(\mathbf{I}_3\) is the current in the shared capacitor, and it is larger than either mesh current because the two are nearly antiphase.

The inductances, from \(\omega L = 10\) at \(\omega = 1000\):

Instantaneous currents at \(t = 2\ \text{ms}\), where \(\omega t = 2\ \text{rad} = 114.59°\):

The angle must be converted from radians: \(\omega t\) is in radians while the phase is in degrees, and mixing them is a standard error.

The stored energy, with the mutual term subtracting (opposite dot orientations):

The mutual term is positive here despite the minus sign in the formula, because the two currents have opposite signs at this instant. Energy is a function of the instantaneous currents, and their signs matter.

Every pair couples, so with three coils there are three mutual terms, each with its own sign fixed by its own pair of dots.

Coil by coil, summing each coil's self term and the mutual contributions it receives from the other two:

Coil 2's contribution is negative. That is permitted — an individual coil's share of the total can be negative when its mutual terms outweigh its self-inductance. Only the total must be positive.

The direct formula, in which each mutual term appears twice:

Agreeing with the coil-by-coil route. The factor of two arises exactly as in Problem 6 — each mutual pair contributes to both of its coils.

Check every coupling coefficient:

All below 1, so the arrangement is physically realisable. Coils 1 and 3 couple most weakly, which is expected if coil 2 sits between them.

The general rule for \(n\) coils in series:

The system, with all reactances evaluated at \(\omega = 720\ \text{rad/s}\):

Symmetry check: \(Z_{12} = Z_{21}\), \(Z_{13} = Z_{31}\), \(Z_{23} = Z_{32}\) ✓ — all three pairs match, so the matrix is properly assembled.

Reading the structure. Each off-diagonal combines a shared-branch impedance with a mutual term, as Problem 16 explained:

The tiny \(Z_{13}\) is characteristic of a chain: mesh 1 couples to 2, and 2 to 3, but 1 and 3 are separated and interact only through the intermediate stage.

Note that \(Z_{22}\) is purely imaginary — \(j0.77183\), with no resistance. Mesh 2 contains only reactive elements, which is consistent with it being a coupling stage rather than a source or load loop.

Solving the 3×3 system:

Back to the time domain. The source was given as a sine, so the answer is stated as a sine to match:

Consistency of reference is what matters — Set 22, Problem 3. A cosine reference would give \(i_2 = 3.132\cos(720t - 32.78°)\), which is the same waveform.

A plausibility note. The mesh-1 impedance has magnitude \(|1.8+j0.72| = 1.94\ \Omega\), so an 8 V source would drive about 4 A there if the other meshes were absent. A mesh-2 current of 3.1 A is of the right order — the coupling is strong enough to transfer most of the action into the second loop.

The claim. A coupled pair with one terminal in common behaves identically to three uncoupled inductors in a T:

Verify by comparing impedance matrices. The coupled pair has

For the T, mesh 1 traverses \((L_1-M)\) and the shunt \(M\), mesh 2 traverses \((L_2-M)\) and the same shunt:

All four entries match, so the two networks are indistinguishable at their terminals.

The numerical case:

Three ordinary uncoupled inductors reproducing a coupled pair exactly.

Arms may be negative. If \(M > L_1\) the primary arm is negative — which is perfectly acceptable as a mathematical equivalent:

A negative inductance cannot be constructed, but it can appear in an equivalent circuit — exactly as negative resistances appear in Thévenin equivalents of active networks (Set 22, Problem 9). The T is a calculating device, not a construction plan.

Why this is worth having. Once decoupled, every technique of Parts 1 and 2 applies without restriction:

The one requirement is a common terminal between primary and secondary. Where the two sides must remain electrically isolated — a real transformer's whole purpose — the T-equivalent cannot be used, and the coupled equations or an ideal-transformer model must be retained. That restriction leads directly into Set 27.

Take the limit \(k \to 1\), so \(M = \sqrt{L_1L_2}\). With the secondary open-circuited (\(\mathbf{I}_2 = 0\)):

Bring in the turns. Inductance goes as the square of the turns for coils on a common core:

The transformer voltage ratio, obtained from mutual inductance alone with no further assumption than perfect coupling.

The current ratio follows from power conservation in a lossless element:

Voltage up means current down, in exact proportion — which is why transmission at high voltage carries small currents and small losses.

Impedance transformation, combining the two:

A load appears at the primary divided by the turns ratio squared — which makes a transformer a matching network in the sense of Set 22, Challenge C1, and one that works over a wide band rather than at a single frequency.

What the ideal model omits:

The first two are the important ones, and both are visible already in this set — leakage is the shortfall of \(k\) from 1, and magnetising current is what flows in Problem 13's open-circuit case.

What this set established, in summary: