By the end of this chapter you should be able to:
Distinguish the aiding and opposing connections of two coupled coils.
Derive and apply \(L_T = L_1 + L_2 \pm 2M\) for the series connection.
Use the two series readings to find \(M\) from \(M = (L_a - L_o)/4\) — the standard laboratory method.
Derive and apply the parallel formulas, and explain why the \(2M\) signs appear reversed compared with series.
Evaluate the special cases \(M = 0\), equal coils, and perfect coupling.
Explain bifilar winding and why it produces a nearly non-inductive component.
State where the energy supplied to a coil goes, and compute \(W = \tfrac{1}{2}LI^{2}\).
Compute the stored energy of two coupled coils, \(W = \tfrac{1}{2}L_1I_1^{2} + \tfrac{1}{2}L_2I_2^{2} \pm MI_1I_2\).
Two Coupled Coils
Consider two coils magnetically coupled, having self-inductances of \(L_1\) and \(L_2\) respectively, and a mutual inductance of \(M\) henry.
The two coils, in an electrical circuit, may be connected in different ways giving different values of resultant inductance.
That last point is worth pausing on. Two resistors in series always give \(R_1 + R_2\), whatever way round they are wired. Two coupled coils do not behave so simply: reversing the connections of one of them changes the answer, sometimes dramatically, because it reverses the direction in which that coil's flux threads the other.
Four connections are therefore possible — series and parallel, each with the fields aiding or opposing — and this chapter works through all four. The series pair turns out to be more than a curiosity: taking two readings and subtracting them is the standard way of measuring \(M\), a quantity that cannot be measured directly at all.
Inductances in Series
The two coils may be connected in series in the following two ways.
When their fields — or mmfs — are additive, that is, their fluxes are set up in the same direction. In this case the inductance of each coil is increased by \(M\):
Also called the cumulative connection.
When their fields — or mmfs — are subtractive, that is, their fluxes are set up in opposite directions. In this case the inductance of each coil is decreased by \(M\):
Also called the differential connection.
In series the same current \(i\) flows through both coils. The total voltage is the sum of four contributions — each coil's self-induced EMF, and the EMF mutually induced in each by the other:
\(M\) appears twice because the coupling works in both directions at once: coil 1 induces an EMF in coil 2 and coil 2 induces one in coil 1, and by the symmetry of Section 13-4 both use the same \(M\). Reversing one coil changes the sign of both mutual terms together, giving \(-2M\).
Problem. Two coupled coils have \(L_1 = 0.6\) H, \(L_2 = 0.4\) H and \(M = 0.3\) H. Find the total inductance in both series connections, and the coefficient of coupling.
Fields aiding.
Fields opposing.
Coefficient of coupling.
Comment. The two connections differ by a factor of four, from the same two coils with nothing changed but a pair of wires. Note also the check: \(M = 0.3\) H is below \(\sqrt{L_1L_2} = 0.49\) H, as Chapter 13 requires. Had the problem quoted \(M = 0.6\) H, the data would have been impossible — and the opposing connection would have given a negative inductance, which is the giveaway.
Measuring Mutual Inductance
Mutual inductance cannot be measured directly: there is no instrument with two terminals that reads \(M\). But the two series connections give it away immediately.
Subtracting eliminates the self-inductances entirely:
And adding them gives the sum of the self-inductances as a bonus:
Two measurements with an ordinary inductance bridge, and a subtraction. This is how \(M\) is determined for every coupled pair in a laboratory.
If either \(L_1\) or \(L_2\) is also measured separately, the other follows from the sum, and then \(k = M/\sqrt{L_1L_2}\) completes the picture. Three measurements characterise the pair fully.
Problem. Two coupled coils connected in series read 2.1 H in one connection and 0.9 H when one coil is reversed. Coil 1 alone measures 0.8 H. Find \(M\), \(L_2\) and the coefficient of coupling.
Identify the connections. The larger reading is always the aiding one, so \(L_a = 2.1\) H and \(L_o = 0.9\) H.
Mutual inductance.
Sum of self-inductances.
Coefficient of coupling.
Comment. Every quantity describing the pair has been obtained from three bridge readings and some arithmetic. A coupling of 0.40 tells you at once that these are air-cored or loosely coupled coils — a transformer on a closed iron core would read above 0.99, as Table 13.2 showed.
Inductances in Parallel
The two coils may be connected in parallel in the following two ways.
When the fields — or mmfs — produced by them are in the same direction:
When the fields — or mmfs — produced by them are in the opposite direction:
Compare carefully with the series result. In series, aiding gives \(+2M\). In parallel, aiding gives \(-2M\) — in the denominator.
Since a smaller denominator means a larger inductance, the aiding connection still produces the larger total inductance in both cases. The physics is consistent; only the algebra looks inverted, because parallel combination inverts everything. Memorising "aiding means plus" will get the parallel case wrong every time — remember instead that aiding always gives more inductance, and read the formula accordingly.
Take the aiding case. The two coils share the same voltage \(v\) but carry different currents \(i_1\) and \(i_2\), with \(i = i_1 + i_2\). Each coil's voltage has a self and a mutual term:
Equating the two expressions and collecting terms:
The total current's rate of change is therefore
and substituting back into the second voltage equation,
Dividing the last two results, the common factor \((L_1-M)^{-1}\,\mathrm{d}i_2/\mathrm{d}t\) cancels:
Reversing one coil replaces \(M\) by \(-M\) throughout. Since \(M^{2}\) is unchanged, only the denominator's sign flips, giving the opposing result.
Problem. For the coils of Example 14.1 (\(L_1 = 0.6\) H, \(L_2 = 0.4\) H, \(M = 0.3\) H), find the total inductance in both parallel connections. Compare with the value the coils would give if they were not coupled at all.
Common numerator.
Fields in the same direction (aiding).
Fields in opposite directions.
If uncoupled (\(M = 0\)), the ordinary parallel rule applies:
| Connection | \(L_T\) (H) | Relative to uncoupled |
|---|---|---|
| Series, aiding | 1.600 | 1.60 × the series sum |
| Series, uncoupled | 1.000 | reference |
| Series, opposing | 0.400 | 0.40 × |
| Parallel, aiding | 0.375 | 1.56 × the parallel value |
| Parallel, uncoupled | 0.240 | reference |
| Parallel, opposing | 0.0938 | 0.39 × |
Comment. Four wiring options span a range of 17:1 from the same pair of coils. Note that in both the series and parallel cases the aiding connection gives the larger value, exactly as the alert box promised — even though one formula has \(+2M\) and the other \(-2M\).
Special Cases and Limits
Three limiting cases are worth checking, both as a memory aid and as a sanity test on any formula you have just written down.
No coupling (\(M = 0\)). All four formulas collapse to the familiar rules for uncoupled elements:
Any formula that fails this test has been written down wrongly.
Two identical coils (\(L_1 = L_2 = L\), so \(M = kL\)). The four results simplify beautifully:
| Connection | \(L_T\) | At \(k = 0\) | At \(k = 1\) |
|---|---|---|---|
| Series, aiding | \(2L(1+k)\) | \(2L\) | \(4L\) |
| Series, opposing | \(2L(1-k)\) | \(2L\) | 0 |
| Parallel, aiding | \(\tfrac{1}{2}L(1+k)\) | \(L/2\) | \(L\) |
| Parallel, opposing | \(\tfrac{1}{2}L(1-k)\) | \(L/2\) | 0 |
Perfect coupling with identical coils (\(k = 1\)). The series-opposing result becomes exactly zero. Two identical, perfectly coupled coils wired in opposition produce no net flux at all and behave as a pure resistance. This is not a mathematical curiosity — it is the working principle of the bifilar winding.
Problem. A precision resistor is wound bifilar from two identical half-windings, each of inductance 0.5 H, connected in opposition. Find the residual inductance if (a) the coupling is perfect, and (b) the coupling is 0.98. Compare with the aiding connection in case (b).
(a) Perfect coupling, \(k = 1\). Then \(M = kL = 0.5\) H and
The winding is perfectly non-inductive. Equivalently, from Table 14.2, \(2L(1-k) = 2(0.5)(0) = 0\).
(b) Realistic coupling, \(k = 0.98\). Now \(M = (0.98)(0.5) = 0.49\) H:
The aiding connection, for comparison.
Comment. The ratio between the two connections is
so simply reversing one half of the winding reduces the inductance by a factor of about a hundred. Note how sensitive the opposing connection is: a 2 % shortfall in coupling leaves 2 % of the uncoupled inductance behind. This is why bifilar windings are laid down with the two strands twisted together — every millimetre of separation costs coupling, and therefore costs performance.
Energy Stored in a Magnetic Field
When some electrical energy is supplied to a coil, it is spent in two ways.
A part of it is spent to meet the \(I^{2}R\) loss, which is dissipated in the form of heat and cannot be recovered.
The remaining part is used to create the magnetic field around the coil and is stored in the magnetic field. When this field collapses, the stored energy is released by the coil and is returned to the circuit.
The energy stored in the magnetic field is given by the expression:
The distinction between the two parts matters more than it may appear. The \(I^{2}R\) term is gone for good; the \(\tfrac{1}{2}LI^{2}\) term is merely on loan. It comes back when the current falls — which is why interrupting an inductive circuit produces an arc, and why the freewheel diodes and snubbers of Chapter 11 exist to give the returning energy a harmless path.
Chapter 13 showed this quantity has three equivalent forms, \(\tfrac{1}{2}LI^{2} = \tfrac{1}{2}\lambda I = \tfrac{1}{2}S\Phi^{2}\). From Chapter 15 onward it becomes the primary tool of the subject.
Two coupled coils. When both coils carry current, a third term appears, representing the energy of their interaction:
The sign is positive when the fluxes aid and negative when they oppose — the same rule as for the series connection, and for the same reason.
Setting \(I_1 = I_2 = I\) recovers the series result: \(W = \tfrac{1}{2}\left(L_1 + L_2 \pm 2M\right)I^{2} = \tfrac{1}{2}L_TI^{2}\), as it must.
A useful consequence: since stored energy can never be negative for any pair of currents, the mutual term cannot dominate the self terms. Working that condition through gives \(M^{2} \le L_1L_2\) — the same limit Chapter 13 obtained from \(k \le 1\), now derived from energy conservation instead of geometry.
Problem. The coils of Example 14.1 (\(L_1 = 0.6\) H, \(L_2 = 0.4\) H, \(M = 0.3\) H) carry \(I_1 = 4\) A and \(I_2 = 3\) A. Find the stored energy for both field directions. Then verify the series result by setting both currents to 4 A.
Self-energy terms.
Mutual term.
Total energy.
Verification against the series result. With \(I_1 = I_2 = 4\) A and fields aiding,
Comment. The same two coils carrying the same two currents store 10.2 J or 3.0 J depending purely on which way one of them is connected — a factor of 3.4, with no change to any physical quantity except a pair of wires. The energy difference is exactly twice the mutual term, and it is the energy that would be released or absorbed if one coil were reversed while the currents were held constant.
Applications
The \(M = (L_a - L_o)/4\) method of Section 14-3 is the standard laboratory determination, requiring nothing but an inductance bridge and a pair of reversed leads.
Precision resistors, current shunts and AC bridge standards are wound so their two halves oppose, cancelling inductance while keeping resistance.
Getting the connection the wrong way round is the same error as choosing the opposing series connection — which is why transformers are polarity-tested before paralleling (Chapter 51).
Multi-phase buck converters and SEPIC circuits use deliberately coupled inductors, where the aiding or opposing choice sets the ripple current and the transient response.
Two windings on one core, arranged so that the wanted differential current sees the opposing connection (low inductance) while unwanted common-mode current sees the aiding one (high inductance).
Rotating one coil relative to another varies \(M\) continuously between \(+M\) and \(-M\), giving a smoothly variable inductance — an elegant use of the same four connections.
Summary and Key Formulas
Two magnetically coupled coils connected in different ways give different resultant inductances — unlike resistors, the wiring order matters.
Series, fields aiding: each coil's inductance is increased by \(M\), giving \(L_T = L_1 + L_2 + 2M\).
Series, fields opposing: each coil's inductance is decreased by \(M\), giving \(L_T = L_1 + L_2 - 2M\).
\(M\) appears twice because the coupling acts in both directions at once, with the same \(M\) each way.
Subtracting the two series readings gives \(M = (L_a - L_o)/4\) — the standard laboratory method. Adding them gives \(L_1 + L_2\).
Parallel: \(L_T = (L_1L_2 - M^{2})/(L_1 + L_2 \mp 2M)\), with the minus sign for aiding. The signs look reversed but the physics is not: aiding always gives the larger inductance.
At \(M = 0\) all four formulas reduce to the ordinary series and parallel rules — a good check.
For identical perfectly coupled coils in opposition, \(L_T = 0\): the principle of the bifilar winding.
Energy supplied to a coil goes partly to irrecoverable \(I^{2}R\) heat and partly to the magnetic field, where \(W = \tfrac{1}{2}LI^{2}\) is stored and later returned.
For two coupled coils, \(W = \tfrac{1}{2}L_1I_1^{2} + \tfrac{1}{2}L_2I_2^{2} \pm MI_1I_2\).
| Connection | Formula | Notes |
|---|---|---|
| Series, aiding | \(L_T = L_1 + L_2 + 2M\) | cumulative |
| Series, opposing | \(L_T = L_1 + L_2 - 2M\) | differential |
| Series, uncoupled | \(L_T = L_1 + L_2\) | \(M = 0\) |
| Parallel, aiding | \(L_T = \dfrac{L_1L_2 - M^{2}}{L_1 + L_2 - 2M}\) | note the minus |
| Parallel, opposing | \(L_T = \dfrac{L_1L_2 - M^{2}}{L_1 + L_2 + 2M}\) | note the plus |
| Parallel, uncoupled | \(L_T = \dfrac{L_1L_2}{L_1 + L_2}\) | \(M = 0\) |
| Mutual inductance | \(M = \dfrac{L_a - L_o}{4}\) | measurement method |
| Sum of self-inductances | \(L_1 + L_2 = \dfrac{L_a + L_o}{2}\) | from the same two readings |
| Identical coils, series | \(L_T = 2L(1 \pm k)\) | zero at \(k=1\), opposing |
| Identical coils, parallel | \(L_T = \tfrac{1}{2}L(1 \pm k)\) | zero at \(k=1\), opposing |
| Energy, one coil | \(W = \tfrac{1}{2}LI^{2}\) | stored and returned |
| Energy, two coils | \(W = \tfrac{1}{2}L_1I_1^{2} + \tfrac{1}{2}L_2I_2^{2} \pm MI_1I_2\) | + for aiding |
Common Mistakes
Writing \(L_1 + L_2 + M\) instead of \(+2M\). The coupling acts both ways, so \(M\) is counted twice.
Assuming "aiding means plus" in the parallel formula. Aiding uses \(-2M\) in the denominator. Remember instead that aiding always gives the larger inductance.
Forgetting \(M^{2}\) in the parallel numerator. It is \(L_1L_2 - M^{2}\), not \(L_1L_2\). Only when \(M = 0\) do the two agree.
Dividing the difference by 2 instead of 4. \(L_a - L_o = 4M\), because each reading already contains \(2M\) with opposite sign.
Mixing up which reading is which. The larger series reading is always the aiding one.
Accepting a negative total inductance. If \(L_1 + L_2 - 2M < 0\) then \(M > \sqrt{L_1L_2}\) and the data is impossible.
Applying series formulas to a parallel circuit. The currents differ in parallel, so \(\mathrm{d}i_1/\mathrm{d}t \ne \mathrm{d}i_2/\mathrm{d}t\) and the derivation is quite different.
Believing a bifilar resistor has exactly zero inductance. It has \(2L(1-k)\), small but finite, and sensitive to how tightly the strands are laid.
Treating stored energy as a loss. Only the \(I^{2}R\) part is lost; the field energy is returned when the field collapses.
Omitting the mutual term from the energy. With both coils energised, \(\pm MI_1I_2\) can be a large fraction of the total, as Example 14.5 showed.
Chapter Review
Check every answer against \(M \le \sqrt{L_1L_2}\) — a negative inductance means the data is impossible.
P14.1 Two coupled coils have \(L_1 = 0.25\) H, \(L_2 = 0.15\) H and \(M = 0.12\) H. Find both series inductances and the coefficient of coupling.
Show answer
\[L_a = 0.25 + 0.15 + 2(0.12) = 0.640~\mathrm{H}, \qquad L_o = 0.25 + 0.15 - 2(0.12) = 0.160~\mathrm{H}\]\[k = \frac{0.12}{\sqrt{(0.25)(0.15)}} = \frac{0.12}{\sqrt{0.0375}} = \frac{0.12}{0.1936} = 0.620\]P14.2 Two coils in series read 1.8 H in one connection and 0.6 H when one is reversed. Find \(M\) and \(L_1 + L_2\).
Show answer
Note that \(L_1\) and \(L_2\) individually cannot be found from these two readings alone — a third measurement is needed.\[M = \frac{1.8 - 0.6}{4} = 0.300~\mathrm{H}, \qquad L_1 + L_2 = \frac{1.8 + 0.6}{2} = 1.20~\mathrm{H}\]P14.3 Two coils of 0.5 H and 0.32 H have a mutual inductance of 0.3 H. Find \(k\).
Show answer
\[k = \frac{0.3}{\sqrt{(0.5)(0.32)}} = \frac{0.3}{\sqrt{0.16}} = \frac{0.3}{0.400} = 0.750\]P14.4 Coils of 0.8 H and 0.5 H with \(M = 0.4\) H are connected in parallel. Find the inductance for both field directions and for no coupling.
Show answer
\[L_1L_2 - M^{2} = 0.40 - 0.16 = 0.240\]\[L_{\text{aiding}} = \frac{0.240}{1.3 - 0.8} = \frac{0.240}{0.500} = 0.480~\mathrm{H}\]\[L_{\text{opposing}} = \frac{0.240}{1.3 + 0.8} = \frac{0.240}{2.100} = 0.114~\mathrm{H}\]\[L_{\text{uncoupled}} = \frac{0.40}{1.3} = 0.308~\mathrm{H}\]P14.5 Two identical coils of 0.2 H each have \(k = 0.75\). Find both series inductances.
Show answer
\[M = kL = (0.75)(0.2) = 0.150~\mathrm{H}\]\[L_a = 2L(1+k) = 2(0.2)(1.75) = 0.700~\mathrm{H}, \qquad L_o = 2(0.2)(0.25) = 0.100~\mathrm{H}\]P14.6 A bifilar resistor is wound from two 0.36 H halves with \(k = 0.99\). Find the residual inductance and compare with the aiding connection.
Show answer
\[M = (0.99)(0.36) = 0.3564~\mathrm{H}\]\[L_o = 0.72 - 2(0.3564) = 0.72 - 0.7128 = 0.00720~\mathrm{H} = 7.20~\mathrm{mH}\]Improving \(k\) from 0.98 to 0.99 halves the residual inductance — every increment of coupling counts.\[L_a = 0.72 + 0.7128 = 1.4328~\mathrm{H}, \qquad \frac{L_a}{L_o} = 199\]P14.7 Coils of 0.5 H and 0.3 H with \(M = 0.2\) H carry 6 A and 4 A. Find the stored energy for both field directions.
Show answer
\[\tfrac{1}{2}L_1I_1^{2} = \tfrac{1}{2}(0.5)(36) = 9.00~\mathrm{J}, \qquad \tfrac{1}{2}L_2I_2^{2} = \tfrac{1}{2}(0.3)(16) = 2.40~\mathrm{J}\]\[MI_1I_2 = (0.2)(6)(4) = 4.80~\mathrm{J}\]\[W_{\text{aiding}} = 16.20~\mathrm{J}, \qquad W_{\text{opposing}} = 6.60~\mathrm{J}\]P14.8 Three uncoupled inductors of 0.40 H, 0.60 H and 0.12 H are connected first in series and then in parallel. Find the total inductance in each case.
Show answer
\[L_{\text{series}} = 0.40 + 0.60 + 0.12 = 1.12~\mathrm{H}\]\[\frac{1}{L_{\text{parallel}}} = \frac{1}{0.40} + \frac{1}{0.60} + \frac{1}{0.12} = 2.500 + 1.667 + 8.333 = 12.50\]With no coupling the rules are identical to those for resistors.\[L_{\text{parallel}} = \frac{1}{12.50} = 0.0800~\mathrm{H}\]P14.9 Show that all four coupled-coil formulas reduce to the standard uncoupled results when \(M = 0\).
Show answer
Series: both \(L_1 + L_2 \pm 2(0) = L_1 + L_2\).
Parallel: both \(\dfrac{L_1L_2 - 0}{L_1 + L_2 \mp 0} = \dfrac{L_1L_2}{L_1+L_2}\).With no coupling, the two connections become indistinguishable, which is exactly what the graph in Section 14-5 shows at \(k = 0\): all four curves meet in pairs. This is the quickest check on a half-remembered formula.
P14.10 A designer needs the largest possible inductance from two coils with \(L_1 = L_2 = 0.3\) H and \(k = 0.9\), and separately the smallest. Which connections should be used, and what are the values?
Show answer
Largest — series aiding:\[M = (0.9)(0.3) = 0.270~\mathrm{H}\]Smallest — parallel opposing:\[L_T = 2L(1+k) = 2(0.3)(1.9) = 1.140~\mathrm{H}\]A ratio of 76:1 from the same two coils. Series aiding is always the maximum and parallel opposing always the minimum, for any \(k > 0\).\[L_T = \tfrac{1}{2}L(1-k) = \tfrac{1}{2}(0.3)(0.1) = 0.0150~\mathrm{H}\]
MCQ 1. Two coupled coils in series with fields aiding have a total inductance of:
(a) \(L_1+L_2\) (b) \(L_1+L_2+M\) (c) \(L_1+L_2+2M\) (d) \(L_1L_2+2M\)Show answer
(c) \(L_1+L_2+2M\) — each coil's inductance is increased by \(M\).MCQ 2. Mutual inductance is found from two series readings as:
(a) \((L_a-L_o)/2\) (b) \((L_a-L_o)/4\) (c) \((L_a+L_o)/4\) (d) \(L_a-L_o\)Show answer
(b) \((L_a-L_o)/4\), since the difference is \(4M\).MCQ 3. Of the two series readings, the larger corresponds to:
(a) fields opposing (b) fields aiding (c) either (d) no couplingShow answer
(b) fields aiding. This is how the connections are identified in practice.MCQ 4. For two coupled coils in parallel with fields aiding, the denominator is:
(a) \(L_1+L_2+2M\) (b) \(L_1+L_2-2M\) (c) \(L_1L_2-M^{2}\) (d) \(L_1+L_2\)Show answer
(b) \(L_1+L_2-2M\). The minus sign gives a smaller denominator and hence the larger inductance — still consistent with aiding meaning more.MCQ 5. Two identical perfectly coupled coils in series opposing have a total inductance of:
(a) \(4L\) (b) \(2L\) (c) \(L\) (d) zeroShow answer
(d) zero — the principle of the bifilar winding.MCQ 6. A bifilar winding is used to make a resistor that is:
(a) non-inductive (b) highly inductive (c) temperature-stable (d) high-voltage ratedShow answer
(a) non-inductive. The two halves oppose, cancelling almost all the inductance.MCQ 7. If \(L_1 + L_2 - 2M\) comes out negative, this means:
(a) the coils oppose (b) the data is impossible (c) \(k = 0\) (d) the coils are in parallelShow answer
(b) the data is impossible, since it requires \(M > \sqrt{L_1L_2}\) and hence \(k > 1\).MCQ 8. The energy supplied to a coil that can be recovered is:
(a) \(I^{2}R\) (b) \(\tfrac{1}{2}LI^{2}\) (c) both (d) neitherShow answer
(b) \(\tfrac{1}{2}LI^{2}\). The \(I^{2}R\) part is dissipated as heat and cannot be recovered.MCQ 9. The energy of two coupled coils carrying currents in aiding directions is:
(a) \(\tfrac{1}{2}L_1I_1^{2}+\tfrac{1}{2}L_2I_2^{2}\) (b) with \(+MI_1I_2\) added (c) with \(-MI_1I_2\) added (d) with \(+2MI_1I_2\) addedShow answer
(b) with \(+MI_1I_2\) added. Note this is \(M\), not \(2M\) — the factor of 2 appears only when both currents are equal and the \(\tfrac{1}{2}\) is factored out.MCQ 10. The connection giving the smallest inductance from two coupled coils is:
(a) series aiding (b) series opposing (c) parallel aiding (d) parallel opposingShow answer
(d) parallel opposing — parallel always reduces, and opposing reduces further.
Explain why two coupled coils give different inductances depending on how they are wired, whereas two resistors do not.
Derive \(L_T = L_1 + L_2 + 2M\) from the voltage equation, explaining clearly why \(M\) appears twice.
Show how two inductance-bridge readings yield both \(M\) and \(L_1 + L_2\), and explain why a third measurement is needed to separate \(L_1\) from \(L_2\).
The parallel formula uses \(-2M\) for the aiding connection while the series formula uses \(+2M\). Explain why this is not a contradiction.
Describe how a bifilar winding is made and why it produces a nearly non-inductive resistor. What limits how close to zero the inductance can be brought?
Distinguish the two destinations of energy supplied to a coil, and explain the practical consequence of one being recoverable.
Show that requiring the stored energy of two coupled coils to be non-negative for all currents leads to \(M^{2} \le L_1L_2\), and comment on obtaining the same limit two different ways.
Part 1's treatment of induction is complete. Chapters 10 to 12 established Faraday's law, its direction and its two families; Chapters 13 and 14 turned the constants of proportionality into physical quantities and showed how to measure and combine them.
From Chapter 15 the subject changes method. Energy and coenergy become the primary tools, and the \(\tfrac{1}{2}LI^{2}\) of Section 14-6 is generalised to systems whose inductance varies with position. That single generalisation is the key: if \(L\) depends on where a moving part sits, then moving that part changes the stored energy, and the rate of change of stored energy with displacement is the force. By Chapter 18 that idea produces \(F = \tfrac{1}{2}I^{2}\,\mathrm{d}L/\mathrm{d}x\) and its rotational counterpart, and every machine in Parts 2 to 6 becomes a special case of one result.