Energy and Coenergy in Magnetic Systems — The Energy Balance

i Learning Objectives

By the end of this chapter you should be able to:

List the destinations of energy supplied to an electromechanical energy conversion device.
State and apply the energy balance equation for a lossless coupling field.
Define field energy \(W_{\text{fld}} = \int i\,\mathrm{d}\lambda\) and identify it as the area to the left of the \(\lambda\)–\(i\) curve.
Define coenergy \(W'_{\text{fld}} = \int \lambda\,\mathrm{d}i\) and identify it as the area beneath the curve.
Prove that \(W_{\text{fld}} + W'_{\text{fld}} = \lambda i\) for any characteristic, linear or not.
Show that the two are equal for a linear system and compute both when the core saturates.
Show that stored energy divides between iron and air gap in proportion to reluctance.
Explain why coenergy is the natural quantity from which to obtain force at constant current.

Section 15-1

Electromechanical Energy Conversion

Electromechanical energy conversion devices convert mechanical energy into electrical energy and vice versa. They are the subject of the rest of this book, and every one of them works through a magnetic field acting as an intermediary.

Electric motors

Driving industrial machines — hammer presses, drilling machines, lathes, shapers, blowers for furnaces — and domestic appliances such as refrigerators, fans, water pumps, toys and mixers.

Generators

Hydro-electric power plants, steam power plants, diesel power plants, nuclear power plants, and in automobiles.

Diagram contrasting generator action and motor action in an electromechanical energy conversion device — Generator action and motor action — the same device run in either direction.

During conversion, the whole of the energy in one form is not converted into the other useful form. The input divides three ways:

Most of the input power is converted into useful output power.
Some of the input power is converted into losses: heat losses \(\left(I^{2}R\right)\) due to the flow of current in the conductors, magnetic losses (hysteresis and eddy-current losses), and friction losses.
A small portion of the input power is stored in the magnetic field of the electromechanical device.

Where the input energy goes. Only the loss branch is irrecoverable.

The stored term is the one this chapter is about. It is the smallest of the three and, in steady state, contributes nothing at all — the field energy is constant, so as much goes in as comes out each cycle. But during changes it is decisive, because a machine produces force precisely by changing its stored field energy as a part moves. Chapters 16 to 18 build the whole theory of force and torque on that observation, and this chapter provides the two quantities they need.

Video · Electromechanical Energy Conversion Devices

1 Worked Example 15.1 — Where the Input Goes

Problem. A motor draws 5.00 kW from the supply. Its copper loss is 250 W, its core loss 180 W, and friction and windage account for 120 W. Find the mechanical output and the efficiency. What contribution does the stored field energy make?

Mechanical output.

\[P_{\text{out}} = 5000 - 250 - 180 - 120 = 4450~\mathrm{W}\]

Efficiency.

\[\eta = \frac{4450}{5000} = 0.890 = 89.0\,\%\]

Stored field energy. In the steady state it makes no contribution at all. The field energy is a fixed quantity — perhaps a joule or two — and once established it neither grows nor shrinks. Energy flows into it and out of it during each cycle in equal measure.

Comment. The stored energy is absent from every efficiency calculation in this book, and it would be easy to conclude it is unimportant. The opposite is true: it is the mechanism by which the other 4450 W crosses the air gap. Without a magnetic field there is no torque, and the whole 5 kW would simply heat the winding. Sections 15-2 onwards give it the precise treatment it needs.

Section 15-2

The Energy Balance Equation

The three destinations of Section 15-1 can be written as an equation. Over any interval,

\[\begin{pmatrix}\text{electrical}\\ \text{energy input}\end{pmatrix} = \begin{pmatrix}\text{mechanical}\\ \text{energy output}\end{pmatrix} + \begin{pmatrix}\text{increase in}\\ \text{field energy}\end{pmatrix} + \begin{pmatrix}\text{energy}\\ \text{lost}\end{pmatrix}\]

The losses are inconvenient but separable. Winding resistance can be taken outside the device as a series resistor; core loss can be represented by a shunt resistance, as Chapter 8 did; friction belongs to the mechanical system. What remains in the middle is a lossless coupling field, and for it the balance is exact:

⚖

The Energy Balance

For a lossless coupling field

\[\mathrm{d}W_{\text{elec}} = \mathrm{d}W_{\text{mech}} + \mathrm{d}W_{\text{fld}}\]

Electrical energy in equals mechanical work done plus the increase in stored field energy. Nothing else. This one equation generates the whole of Chapters 16 to 18.

The electrical term can be written explicitly. The coil's induced EMF is \(e = \mathrm{d}\lambda/\mathrm{d}t\) from Chapter 10, so in a time \(\mathrm{d}t\) the source delivers

\[\mathrm{d}W_{\text{elec}} = e\,i\,\mathrm{d}t = \frac{\mathrm{d}\lambda}{\mathrm{d}t}\,i\,\mathrm{d}t = i\,\mathrm{d}\lambda\]

\[\boxed{\;i\,\mathrm{d}\lambda = \mathrm{d}W_{\text{mech}} + \mathrm{d}W_{\text{fld}}\;}\]

Note what has been assumed and what has not. Nothing here requires the magnetic circuit to be linear, or the inductance to be constant, or the geometry to be fixed. The relation \(\mathrm{d}W_{\text{elec}} = i\,\mathrm{d}\lambda\) is completely general — it follows from Faraday's law alone. That generality is why the energy method succeeds where circuit reasoning becomes unmanageable, and it is why \(\lambda\) rather than \(\Phi\) is the working variable from here on.

Section 15-3

Field Energy

Consider first a device whose mechanical part is clamped, so that no mechanical work is done and \(\mathrm{d}W_{\text{mech}} = 0\). The balance equation then reduces to

\[\mathrm{d}W_{\text{fld}} = i\,\mathrm{d}\lambda\]

and integrating from an unexcited state gives the energy stored in the field:

📗

Field Energy

The area to the LEFT of the λ–i characteristic

\[W_{\text{fld}} = \int_0^{\lambda} i\,\mathrm{d}\lambda\]

Because the integration is with respect to \(\lambda\), the integral sweeps horizontal strips and measures the area between the curve and the vertical (\(\lambda\)) axis.

Two ways of writing the same quantity are worth having. Since \(\lambda = N\Phi\) and \(\mathcal{F} = Ni\),

\[W_{\text{fld}} = \int_0^{\lambda} i\,\mathrm{d}\lambda = \int_0^{\Phi}\mathcal{F}\,\mathrm{d}\Phi\]

so the same area can be read from a \(\lambda\)–\(i\) curve or from a \(\Phi\)–\(\mathcal{F}\) curve, which is simply the magnetisation curve of Chapter 5 rescaled.

Section 15-4

Coenergy

Alongside the field energy sits a second quantity, defined by integrating the other way round.

📘

Coenergy

The area BENEATH the λ–i characteristic

\[W'_{\text{fld}} = \int_0^{i}\lambda\,\mathrm{d}i\]

Now the integration is with respect to \(i\), so the integral sweeps vertical strips and measures the area between the curve and the horizontal (\(i\)) axis.

Coenergy is not a physical energy. No joules are stored in it and no instrument measures it. It is a mathematical device — but a remarkably convenient one, for the reason Section 15-8 explains.

The two areas together make up the whole rectangle of sides \(\lambda\) and \(i\), which gives an identity valid for any characteristic whatever:

🔑

The Fundamental Identity

Energy plus coenergy fills the rectangle

\[W_{\text{fld}} + W'_{\text{fld}} = \lambda i\]

This holds whether the core is linear, saturating, or arbitrarily shaped. It follows from integration by parts, \(\int i\,\mathrm{d}\lambda + \int\lambda\,\mathrm{d}i = \int\mathrm{d}(\lambda i) = \lambda i\), and it is the quickest way to find one quantity once you have the other.

Energy is the area to the left; coenergy is the area beneath. Only for a straight line are they equal.

Section 15-5

The Linear Case

If the magnetic circuit is linear — no saturation — then \(\lambda = Li\) and the characteristic is a straight line through the origin. The two areas are then two triangles of equal size.

Field energy. Substituting \(i = \lambda/L\):

\[W_{\text{fld}} = \int_0^{\lambda}\frac{\lambda}{L}\,\mathrm{d}\lambda = \frac{\lambda^{2}}{2L} = \tfrac{1}{2}Li^{2} = \tfrac{1}{2}\lambda i\]

Coenergy. Substituting \(\lambda = Li\):

\[W'_{\text{fld}} = \int_0^{i}Li\,\mathrm{d}i = \tfrac{1}{2}Li^{2} = \tfrac{1}{2}\lambda i\]

Linear Systems

Numerically equal, conceptually distinct

\[W_{\text{fld}} = W'_{\text{fld}} = \tfrac{1}{2}Li^{2} = \tfrac{1}{2}\lambda i = \frac{\lambda^{2}}{2L}\]

Their sum is \(\lambda i\), as the identity requires. This is the \(\tfrac{1}{2}LI^{2}\) of Chapters 13 and 14, now revealed as one of two quantities that happen to coincide when the curve is straight.

! Equal Is Not the Same as Identical

It is tempting to conclude that coenergy is a redundant concept. It is not. The two are equal numerically for a linear system, but they are functions of different variables: \(W_{\text{fld}}\) is naturally a function of \(\lambda\) and position, while \(W'_{\text{fld}}\) is naturally a function of \(i\) and position.

That distinction is invisible while the geometry is fixed. The moment a part is allowed to move — which is the whole subject of Chapters 16 to 18 — the two give force expressions that differ by a sign and by which variable is held constant. Section 15-8 previews the result.

2 Worked Example 15.2 — A Linear System

Problem. A coil of inductance 0.400 H carries 5.00 A. Find the flux linkage, the field energy, the coenergy, and verify the identity.

Flux linkage.

\[\lambda = Li = (0.400)(5.00) = 2.00~\mathrm{Wb\text{-}t}\]

Field energy.

\[W_{\text{fld}} = \tfrac{1}{2}\lambda i = \tfrac{1}{2}(2.00)(5.00) = 5.00~\mathrm{J}\]

Coenergy.

\[W'_{\text{fld}} = \tfrac{1}{2}Li^{2} = \tfrac{1}{2}(0.400)(25.0) = 5.00~\mathrm{J}\]

Identity check.

\[W_{\text{fld}} + W'_{\text{fld}} = 5.00 + 5.00 = 10.0~\mathrm{J}, \qquad \lambda i = (2.00)(5.00) = 10.0~\mathrm{J} \;\checkmark\]

Comment. Each quantity is exactly half the rectangle, as two equal triangles must be. Whenever a problem gives a constant inductance, energy and coenergy are interchangeable and either may be used — which is why the distinction can be safely postponed until saturation or motion enters the picture.

Section 15-6

The Saturating Case

Real cores saturate, and Chapter 5 showed that this is deliberate — machines are worked past the knee for economy. Once the characteristic bends over, the two triangles become unequal regions and \(W_{\text{fld}} \ne W'_{\text{fld}}\).

Which is larger is easy to see from the figure in Section 15-4. A saturating curve bulges upward, away from the current axis, so it encloses more area beneath it and less to its left:

\[\text{Saturating core:} \qquad W'_{\text{fld}} > W_{\text{fld}}\]

and the gap widens the harder the core is driven. The identity \(W_{\text{fld}} + W'_{\text{fld}} = \lambda i\) continues to hold throughout, so finding either one immediately gives the other.

The dangerous shortcut. Faced with a saturated operating point, it is tempting to define an "apparent inductance" \(L = \lambda/i\) and use \(\tfrac{1}{2}Li^{2}\). This gives neither the energy nor the coenergy — it gives exactly half the rectangle, which is the average of the two, and is wrong for both. Worked Example 15.3 shows the error is not small.

3 Worked Example 15.3 — A Saturating Core

Problem. A magnetic circuit has the characteristic \(i = 2\lambda^{2}\), with \(i\) in amperes and \(\lambda\) in weber-turns. At an operating current of 8.00 A, find the flux linkage, the field energy and the coenergy. Compare with the value that a naive \(\tfrac{1}{2}Li^{2}\) would give.

Flux linkage. From \(i = 2\lambda^{2}\),

\[\lambda = \sqrt{\frac{i}{2}} = \sqrt{\frac{8.00}{2}} = 2.00~\mathrm{Wb\text{-}t}\]

Field energy — integrate \(i\) with respect to \(\lambda\):

\[W_{\text{fld}} = \int_0^{2}2\lambda^{2}\,\mathrm{d}\lambda = \left[\frac{2\lambda^{3}}{3}\right]_0^{2} = \frac{2(8)}{3} = 5.33~\mathrm{J}\]

Coenergy — most easily from the identity:

\[W'_{\text{fld}} = \lambda i - W_{\text{fld}} = (2.00)(8.00) - 5.33 = 16.0 - 5.33 = 10.67~\mathrm{J}\]

Confirming by direct integration with \(\lambda = \sqrt{i/2}\):

\[W'_{\text{fld}} = \int_0^{8}\sqrt{\frac{i}{2}}\,\mathrm{d}i = \frac{1}{\sqrt{2}}\left[\frac{2}{3}i^{3/2}\right]_0^{8} = \frac{2(22.63)}{3\sqrt{2}} = 10.67~\mathrm{J} \;\checkmark\]

The naive calculation. Taking an apparent inductance \(L = \lambda/i = 2.00/8.00 = 0.250\) H:

\[\tfrac{1}{2}Li^{2} = \tfrac{1}{2}(0.250)(64.0) = 8.00~\mathrm{J}\]

Table 15.1 — Three answers to the same question.
Quantity	Value (J)	Error of the naive result
Field energy \(W_{\text{fld}}\)	5.33	overstated by 50 %
Naive \(\tfrac{1}{2}Li^{2}\)	8.00	— neither quantity —
Coenergy \(W'_{\text{fld}}\)	10.67	understated by 25 %
Sum \(= \lambda i\)	16.00	\(\checkmark\) identity holds

Comment. The coenergy here is exactly twice the field energy, and the naive answer of 8.00 J is precisely their average — half the rectangle, as the insight strip warned. A 50 % error in a force calculation is not a rounding matter, and this is why Chapters 16 to 18 are careful to say which quantity is being differentiated and which variable is held constant.

Section 15-7

Where the Energy Is Stored

Chapter 4 asserted that the air gap holds most of a magnetic circuit's stored energy. The energy formulation makes this exact and quantitative.

For a series magnetic circuit the same flux \(\Phi\) passes through every part, and the energy of each part is \(\tfrac{1}{2}S_k\Phi^{2}\). Since \(\Phi\) is common:

🎯

Key Result

Energy divides in proportion to reluctance

\[\frac{W_k}{W_{\text{total}}} = \frac{S_k}{S_{\text{total}}}\]

The part of the circuit with the greatest reluctance stores the greatest share of the energy. Since a millimetre of air routinely outweighs half a metre of iron, the gap holds almost all of it.

This is why every energy-conversion device has an air gap, and why the gap is where the force is developed. It also explains a fact from Chapter 4 that may have seemed arbitrary: a gapped reactor is designed by choosing the gap to store the required energy, with the iron treated as a nuisance rather than a store.

4 Worked Example 15.4 — Iron versus Gap

Problem. A magnetic circuit has 500 turns carrying 2.00 A. The iron path is 399 mm long with cross-section 5 cm² and \(\mu_r = 1200\), and there is a 1.00 mm air gap of the same area. Find the total stored energy and how it divides between iron and gap.

Reluctances (from Worked Example 13.2):

\[S_i = 5.292\times10^{5}, \qquad S_g = 1.592\times10^{6}, \qquad S_{\text{total}} = 2.121\times10^{6}~\mathrm{AT/Wb}\]

Flux.

\[\Phi = \frac{Ni}{S_{\text{total}}} = \frac{(500)(2.00)}{2.121\times10^{6}} = 4.715\times10^{-4}~\mathrm{Wb}\]

Energy in each part.

\[W_{\text{iron}} = \tfrac{1}{2}S_i\Phi^{2} = \tfrac{1}{2}\left(5.292\times10^{5}\right)\left(2.223\times10^{-7}\right) = 0.0588~\mathrm{J}\]

\[W_{\text{gap}} = \tfrac{1}{2}S_g\Phi^{2} = \tfrac{1}{2}\left(1.592\times10^{6}\right)\left(2.223\times10^{-7}\right) = 0.1769~\mathrm{J}\]

\[W_{\text{total}} = 0.0588 + 0.1769 = 0.2358~\mathrm{J}\]

Check against \(\tfrac{1}{2}Li^{2}\). With \(L = N^{2}/S = 250\,000/2.121\times10^{6} = 0.1179\) H,

\[W = \tfrac{1}{2}(0.1179)(4.00) = 0.2358~\mathrm{J} \;\checkmark\]

The split.

\[\frac{W_{\text{gap}}}{W_{\text{total}}} = \frac{0.1769}{0.2358} = 0.750 = 75.0\,\%\]

Comment. Exactly the reluctance ratio, as the key result requires: \(S_g/S_{\text{total}} = 1.592/2.121 = 0.750\). One millimetre of air, occupying a quarter of one percent of the flux path, holds three-quarters of the energy. Reduce the gap and the energy concentrates in a smaller volume; that concentration, and its variation with gap length, is precisely what produces the pull of an electromagnet.

Section 15-8

Why Coenergy Matters

The reason for carrying two quantities becomes clear as soon as the mechanical part is unclamped. Return to the balance equation:

\[\mathrm{d}W_{\text{mech}} = i\,\mathrm{d}\lambda - \mathrm{d}W_{\text{fld}}\]

Now the field energy depends on two variables — the excitation and the position of the moving part. Which variable is held fixed while the part moves determines which quantity is convenient.

Constant flux linkage

If \(\lambda\) is held fixed then \(i\,\mathrm{d}\lambda = 0\), so no electrical energy enters and the mechanical work must come entirely out of the store:

\[f = -\left.\frac{\partial W_{\text{fld}}}{\partial x}\right|_{\lambda}\]

Note the minus sign: the stored energy falls as work is done.

Constant current

If \(i\) is held fixed, working through the algebra with the identity \(W' = \lambda i - W\) gives the companion result:

\[f = +\left.\frac{\partial W'_{\text{fld}}}{\partial x}\right|_{i}\]

A plus sign, and the coenergy rather than the energy.

🔧

Why Coenergy Wins in Practice

Current is what you control; flux linkage is not

Both expressions give the same force — they must, since they describe the same device. But a machine is fed from a supply that fixes its current, not its flux linkage. The constant-current form is therefore the one that matches how problems are posed, and it is \(W'\) that appears in it.

For a linear system this immediately gives \(W' = \tfrac{1}{2}Li^{2}\) and hence the central result of Chapter 18:

\[f = \tfrac{1}{2}i^{2}\frac{\mathrm{d}L}{\mathrm{d}x}\]

Force is produced wherever inductance varies with position. That single sentence is the operating principle of every machine in Parts 2 to 6.

Chapters 16 and 17 apply this to singly and doubly excited systems, and Chapter 18 derives the force and torque expressions properly. What matters here is the structure: energy pairs with flux linkage, coenergy pairs with current, and the two routes must agree.

5 Worked Example 15.5 — A Closing Electromagnet

Problem. An electromagnet has an inductance of 0.0600 H when its armature is 5 mm away and 0.200 H when the armature has closed to 1 mm. The current is held constant at 4.00 A throughout the movement. Find the electrical energy supplied, the change in stored field energy, and the mechanical work done.

Field energy at each position.

\[W_1 = \tfrac{1}{2}L_1i^{2} = \tfrac{1}{2}(0.0600)(16.0) = 0.480~\mathrm{J}\]

\[W_2 = \tfrac{1}{2}L_2i^{2} = \tfrac{1}{2}(0.200)(16.0) = 1.600~\mathrm{J}\]

\[\Delta W_{\text{fld}} = 1.600 - 0.480 = 1.120~\mathrm{J}\]

Electrical energy supplied. At constant current, \(W_{\text{elec}} = \int i\,\mathrm{d}\lambda = i\,\Delta\lambda\), and \(\Delta\lambda = i\,\Delta L\), so

\[W_{\text{elec}} = i^{2}\Delta L = (16.0)(0.200 - 0.0600) = (16.0)(0.140) = 2.240~\mathrm{J}\]

Mechanical work. From the energy balance,

\[W_{\text{mech}} = W_{\text{elec}} - \Delta W_{\text{fld}} = 2.240 - 1.120 = 1.120~\mathrm{J}\]

Check by coenergy. The change in coenergy at constant current is

\[\Delta W'_{\text{fld}} = \tfrac{1}{2}i^{2}\Delta L = \tfrac{1}{2}(16.0)(0.140) = 1.120~\mathrm{J} \;\checkmark\]

exactly the mechanical work, as \(f = \partial W'/\partial x|_i\) predicts.

The Fifty-Fifty Rule

At constant current, half the electrical energy becomes work

Of the 2.240 J supplied, exactly half did mechanical work and half went into increasing the stored field energy. This is not a coincidence of the numbers: for any linear system moving at constant current,

\[W_{\text{elec}} = i^{2}\Delta L, \qquad \Delta W_{\text{fld}} = W_{\text{mech}} = \tfrac{1}{2}i^{2}\Delta L\]

The result is worth remembering as a check on any energy-conversion calculation.

Comment. Note the direction of the force. The inductance increases as the gap closes, so \(\mathrm{d}L/\mathrm{d}x\) is positive in the closing direction and the force acts to close the gap. An electromagnet always pulls its armature towards the position of maximum inductance — that is, minimum reluctance, which is the general rule for all singly excited systems and the subject of Chapter 16.

Section 15-9

Applications

Relays and Contactors

The pull is \(\tfrac{1}{2}i^{2}\mathrm{d}L/\mathrm{d}x\), largest where the gap is smallest — which is why a relay's action accelerates as it closes and why the contacts snap rather than drift.

Reluctance Motors

A rotor with no winding at all develops torque purely because its inductance varies with position. Chapters 16 and 91 exploit exactly the mechanism of Example 15.5.

Solenoid Valves and Actuators

Designed by shaping the plunger so that \(\mathrm{d}L/\mathrm{d}x\) — and therefore force — stays roughly constant over the stroke, rather than rising steeply at the end.

Energy Storage Reactors

Since energy divides in proportion to reluctance, the gap is sized to hold the required joules. This is the design procedure of Worked Example 4.5, now justified.

Magnetic Bearings and Levitation

Force follows from the same coenergy derivative, but the system is inherently unstable — attraction grows as the gap closes — so active control is essential.

Machine Analysis Generally

The energy method handles saturation, saliency and multiple windings without circuit assumptions, which is why it, and not force-on-a-conductor, is the foundation of Parts 2 to 6.

Section 15-10

Summary and Key Formulas

Electromechanical energy conversion devices convert mechanical energy to electrical and vice versa. Input energy divides into useful output, losses (\(I^{2}R\), hysteresis, eddy current, friction) and a small stored portion.
Removing the losses leaves a lossless coupling field for which \(\mathrm{d}W_{\text{elec}} = \mathrm{d}W_{\text{mech}} + \mathrm{d}W_{\text{fld}}\).
From Faraday's law, \(\mathrm{d}W_{\text{elec}} = i\,\mathrm{d}\lambda\) — a completely general result requiring no linearity.
Field energy \(W_{\text{fld}} = \int i\,\mathrm{d}\lambda\) is the area to the left of the \(\lambda\)–\(i\) curve.
Coenergy \(W'_{\text{fld}} = \int\lambda\,\mathrm{d}i\) is the area beneath it. It is not a physical energy but a mathematical convenience.
\(W_{\text{fld}} + W'_{\text{fld}} = \lambda i\) always, linear or not — the quickest route from one to the other.
For a linear system they are numerically equal: \(W = W' = \tfrac{1}{2}Li^{2} = \tfrac{1}{2}\lambda i = \lambda^{2}/2L\).
For a saturating core, \(W' > W\). Using \(\tfrac{1}{2}Li^{2}\) with an apparent \(L = \lambda/i\) gives neither, but their average.
Stored energy divides in proportion to reluctance, so the air gap holds nearly all of it.
Force follows from \(-\partial W/\partial x|_{\lambda}\) or \(+\partial W'/\partial x|_{i}\). The second is preferred because current is what a supply fixes, giving \(f = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\) for linear systems.

Table 15.2 — Formulas introduced in this chapter.
Quantity	Formula	Notes
Energy balance	\(\mathrm{d}W_{\text{elec}} = \mathrm{d}W_{\text{mech}} + \mathrm{d}W_{\text{fld}}\)	lossless coupling field
Electrical input	\(\mathrm{d}W_{\text{elec}} = i\,\mathrm{d}\lambda\)	general — no linearity assumed
Field energy	\(W_{\text{fld}} = \displaystyle\int_0^{\lambda} i\,\mathrm{d}\lambda\)	area left of the curve
Alternative form	\(W_{\text{fld}} = \displaystyle\int_0^{\Phi}\mathcal{F}\,\mathrm{d}\Phi\)	from the \(\Phi\)–\(\mathcal{F}\) curve
Coenergy	\(W'_{\text{fld}} = \displaystyle\int_0^{i}\lambda\,\mathrm{d}i\)	area beneath the curve
Identity	\(W_{\text{fld}} + W'_{\text{fld}} = \lambda i\)	always true
Linear system	\(W = W' = \tfrac{1}{2}Li^{2} = \tfrac{1}{2}\lambda i\)	two equal triangles
In terms of \(\lambda\)	\(W_{\text{fld}} = \dfrac{\lambda^{2}}{2L}\)	linear only
Magnetic-circuit form	\(W_{\text{fld}} = \tfrac{1}{2}S\Phi^{2}\)	Chapters 4 and 13
Energy split	\(\dfrac{W_k}{W_{\text{total}}} = \dfrac{S_k}{S_{\text{total}}}\)	series circuit
Force, constant \(\lambda\)	\(f = -\left.\dfrac{\partial W_{\text{fld}}}{\partial x}\right\|_{\lambda}\)	see Chapter 18
Force, constant \(i\)	\(f = +\left.\dfrac{\partial W'_{\text{fld}}}{\partial x}\right\|_{i}\)	the practical form
Linear force	\(f = \tfrac{1}{2}i^{2}\dfrac{\mathrm{d}L}{\mathrm{d}x}\)	Chapter 18
Constant-current motion	\(W_{\text{mech}} = \Delta W_{\text{fld}} = \tfrac{1}{2}W_{\text{elec}}\)	linear systems

Section 15-11

Common Mistakes

Treating coenergy as a real stored energy. It is an area with the units of energy, nothing more. Only \(W_{\text{fld}}\) is physically stored.
Swapping the two areas. Energy integrates \(i\,\mathrm{d}\lambda\) and lies to the left; coenergy integrates \(\lambda\,\mathrm{d}i\) and lies beneath. The differential tells you which axis you are sweeping along.
Assuming \(W = W'\) for a saturating core. True only for a straight line. For a saturating characteristic \(W'\) is the larger.
Using \(\tfrac{1}{2}Li^{2}\) with an apparent \(L = \lambda/i\) at a saturated point. This gives half the rectangle — the average of the two, and correct for neither.
Forgetting the minus sign in the constant-\(\lambda\) force. With no electrical input, mechanical work must reduce the stored energy.
Believing the two force expressions give different answers. They give the same force; only the bookkeeping differs.
Ignoring the stored term because it is small. It contributes nothing in steady state but everything during motion — it is the conversion mechanism.
Assuming all the electrical input becomes mechanical work. At constant current in a linear system, exactly half does.
Expecting the iron to store the energy. Energy follows reluctance, so it sits in the gap — typically three-quarters or more of the total.
Writing \(\mathrm{d}W_{\text{elec}} = v\,i\,\mathrm{d}t\) for the coupling field. That includes the \(i^{2}R\) loss. The coupling field receives \(e\,i\,\mathrm{d}t = i\,\mathrm{d}\lambda\).

Section 15-12

Chapter Review

Practice Problems

Use \(W + W' = \lambda i\) as a check on every answer — it holds whatever the characteristic.

P15.1 A machine takes 12.0 kW. Copper loss is 480 W, core loss 320 W and friction and windage 200 W. Find the output and the efficiency.

Show answer

\[P_{\text{out}} = 12\,000 - 480 - 320 - 200 = 11\,000~\mathrm{W}\]

\[\eta = \frac{11\,000}{12\,000} = 0.9167 = 91.67\,\%\]
P15.2 A linear coil of 0.250 H carries 6.00 A. Find \(\lambda\), \(W_{\text{fld}}\), \(W'_{\text{fld}}\), and verify the identity.

Show answer

\[\lambda = (0.250)(6.00) = 1.50~\mathrm{Wb\text{-}t}\]

\[W_{\text{fld}} = W'_{\text{fld}} = \tfrac{1}{2}(0.250)(36.0) = 4.50~\mathrm{J}\]

\[W + W' = 9.00~\mathrm{J} = \lambda i = (1.50)(6.00) \;\checkmark\]
P15.3 A saturating circuit obeys \(i = 3\lambda^{2}\). At \(\lambda = 2.00\) Wb-t find the current, the field energy and the coenergy.

Show answer

\[i = 3(2.00)^{2} = 12.0~\mathrm{A}\]

\[W_{\text{fld}} = \int_0^{2}3\lambda^{2}\,\mathrm{d}\lambda = \left[\lambda^{3}\right]_0^{2} = 8.00~\mathrm{J}\]

\[W'_{\text{fld}} = \lambda i - W_{\text{fld}} = 24.0 - 8.00 = 16.0~\mathrm{J}\]
Once again \(W' = 2W\) — a property of this particular square-law characteristic.
P15.4 A series magnetic circuit has \(S_i = 4.00\times10^{5}\) and \(S_g = 1.20\times10^{6}\) AT/Wb, carrying a flux of \(5.00\times10^{-4}\) Wb. Find the energy in each part and the gap's share.

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\[W_{\text{iron}} = \tfrac{1}{2}\left(4.00\times10^{5}\right)\left(2.50\times10^{-7}\right) = 0.0500~\mathrm{J}\]

\[W_{\text{gap}} = \tfrac{1}{2}\left(1.20\times10^{6}\right)\left(2.50\times10^{-7}\right) = 0.150~\mathrm{J}\]

\[W_{\text{total}} = 0.200~\mathrm{J}, \qquad \frac{W_{\text{gap}}}{W_{\text{total}}} = \frac{1.20}{1.60} = 75.0\,\%\]
P15.5 An actuator's inductance rises from 0.0800 H to 0.240 H as it moves, with the current held at 5.00 A. Find the electrical energy supplied, the change in field energy and the mechanical work.

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\[W_{\text{elec}} = i^{2}\Delta L = (25.0)(0.160) = 4.00~\mathrm{J}\]

\[\Delta W_{\text{fld}} = \tfrac{1}{2}i^{2}\Delta L = 2.00~\mathrm{J}, \qquad W_{\text{mech}} = 4.00 - 2.00 = 2.00~\mathrm{J}\]
Half and half, as the fifty-fifty rule requires.
P15.6 A coil stores 3.20 J when carrying 8.00 A. Find its inductance and flux linkage, assuming linearity.

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\[L = \frac{2W}{i^{2}} = \frac{2(3.20)}{64.0} = 0.100~\mathrm{H}\]

\[\lambda = Li = (0.100)(8.00) = 0.800~\mathrm{Wb\text{-}t}\]
Check: \(\tfrac{1}{2}\lambda i = \tfrac{1}{2}(0.800)(8.00) = 3.20\) J \(\checkmark\)
P15.7 Explain, using the shape of the \(\lambda\)–\(i\) curve, why \(W' > W\) for a saturating core. What would make \(W > W'\)?

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A saturating characteristic is concave downward — steep at first, flattening as the iron saturates. It therefore bulges upward, away from the current axis, enclosing a large area beneath it (coenergy) and leaving a smaller area to its left (energy).
The reverse would require a characteristic that is concave upward, in which \(\lambda\) rises ever more steeply with \(i\). No ferromagnetic material behaves that way, since permeability can only fall with increasing flux density, never rise indefinitely. So for real magnetic circuits \(W' \ge W\) always, with equality only for a linear characteristic.
P15.8 Two coupled coils have \(L_1 = 0.400\) H, \(L_2 = 0.300\) H and \(M = 0.200\) H, carrying 3.00 A and 5.00 A with fluxes aiding. Find the stored field energy.

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Using the two-coil result of Chapter 14:
\[W = \tfrac{1}{2}(0.400)(9.00) + \tfrac{1}{2}(0.300)(25.0) + (0.200)(3.00)(5.00)\]

\[W = 1.80 + 3.75 + 3.00 = 8.55~\mathrm{J}\]
With the fluxes opposing it would be \(1.80 + 3.75 - 3.00 = 2.55\) J.
P15.9 A 0.500 H coil carrying 6.00 A is suddenly disconnected. What becomes of the stored energy?

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\[W = \tfrac{1}{2}(0.500)(36.0) = 9.00~\mathrm{J}\]
This energy cannot vanish. As the field collapses it is returned to the circuit, and since the current is being forced to zero very rapidly, \(e = L\,\mathrm{d}i/\mathrm{d}t\) becomes very large. The 9.00 J is dissipated in whatever path the current can find — typically an arc across the opening contacts, which is why freewheel diodes and snubbers are fitted.
Note this is the stored energy of Section 15-1, appearing here as a practical hazard rather than an accounting term.
P15.10 For the electromagnet of Worked Example 15.5, estimate the average force if the armature moves 4.00 mm.

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The mechanical work was 1.120 J over a displacement of \(4.00\times10^{-3}\) m:
\[f_{\text{av}} = \frac{W_{\text{mech}}}{\Delta x} = \frac{1.120}{4.00\times10^{-3}} = 280~\mathrm{N}\]
This is an average. The actual force is far from constant: \(f = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\) and \(L\) rises steeply as the gap closes, so the instantaneous force at 1 mm greatly exceeds that at 5 mm. This is why relays close with a snap rather than a smooth glide.

Multiple-Choice Questions

MCQ 1. For a lossless coupling field, the electrical energy input equals:
(a) mechanical output only (b) field energy only (c) mechanical output plus change in field energy (d) mechanical output minus losses

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(c) \(\mathrm{d}W_{\text{elec}} = \mathrm{d}W_{\text{mech}} + \mathrm{d}W_{\text{fld}}\).
MCQ 2. Field energy is the area:
(a) beneath the \(\lambda\)–\(i\) curve (b) to the left of it (c) the whole rectangle (d) above the \(\lambda\) axis

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(b) to the left of it, since \(W = \int i\,\mathrm{d}\lambda\) sweeps horizontal strips.
MCQ 3. The sum of field energy and coenergy equals:
(a) \(\tfrac{1}{2}\lambda i\) (b) \(\lambda i\) (c) \(2\lambda i\) (d) \(\lambda/i\)

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(b) \(\lambda i\) — the two areas fill the rectangle, whatever the shape of the curve.
MCQ 4. For a linear magnetic system, coenergy compared with field energy is:
(a) larger (b) smaller (c) equal (d) zero

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(c) equal, both being \(\tfrac{1}{2}Li^{2}\). They are equal in value but remain functions of different variables.
MCQ 5. For a saturating core, coenergy is:
(a) larger than the field energy (b) smaller (c) equal (d) negative

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(a) larger. The curve bulges upward, enclosing more area beneath it.
MCQ 6. In a series magnetic circuit the stored energy divides in proportion to:
(a) length (b) permeability (c) reluctance (d) area

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(c) reluctance, since the flux is common and \(W_k = \tfrac{1}{2}S_k\Phi^{2}\).
MCQ 7. Force at constant current is obtained from:
(a) \(-\partial W/\partial x\) (b) \(+\partial W'/\partial x\) (c) \(\partial\lambda/\partial x\) (d) \(\partial i/\partial x\)

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(b) \(+\partial W'/\partial x\) — the coenergy derivative, with a plus sign.
MCQ 8. When a linear device moves at constant current, the fraction of electrical energy converted to mechanical work is:
(a) all of it (b) three-quarters (c) one half (d) one quarter

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(c) one half. The other half increases the stored field energy.
MCQ 9. An electromagnet's armature is attracted towards the position of:
(a) maximum reluctance (b) minimum inductance (c) maximum inductance (d) zero flux

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(c) maximum inductance, equivalently minimum reluctance — the direction in which \(\mathrm{d}L/\mathrm{d}x\) is positive.
MCQ 10. Coenergy is best described as:
(a) energy stored in the iron (b) energy stored in the gap (c) a mathematical quantity with energy units (d) the energy lost as heat

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(c) a mathematical quantity with energy units. Nothing is physically stored as coenergy; it is a convenient function of current and position.

Conceptual Questions

List the destinations of energy supplied to an electromechanical device, and explain why the stored portion is absent from efficiency calculations yet essential to operation.
Derive \(\mathrm{d}W_{\text{elec}} = i\,\mathrm{d}\lambda\) and explain why it requires no assumption of linearity.
Define field energy and coenergy geometrically, and prove that their sum is \(\lambda i\) for any characteristic.
Explain why the two are equal for a linear system, and why the distinction nevertheless cannot be discarded.
Explain why using an apparent inductance \(L = \lambda/i\) at a saturated point gives neither energy nor coenergy, and what it does give.
Show that stored energy divides in proportion to reluctance, and use this to explain why every energy-conversion device has an air gap.
Explain why coenergy rather than energy is the practical quantity for computing force, referring to what a supply actually holds constant.

Looking Ahead

The two quantities are now defined and the balance equation established. Chapter 16 applies them to singly excited magnetic systems — a single winding acting on a movable member, which covers relays, solenoids, contactors and the reluctance motor. The rule glimpsed in Worked Example 15.5, that the moving part is drawn towards maximum inductance, is derived properly there.

Chapter 17 extends to doubly excited systems, where two windings interact and a new torque component appears from the mutual term — the mechanism of every synchronous and DC machine in this book. Chapter 18 then states the general force and torque results, including \(T = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}\theta\) and the separation of alignment torque from reluctance torque. From Chapter 19 the treatment turns to real windings distributed in slots, and Part 2 begins.