Introduction to Electrical Machines

An electrical machine is a device that converts energy between electrical and mechanical form, using a magnetic field as the intermediate store. The direction of conversion gives the machine its name:

• Generator — mechanical energy in, electrical energy out. A turbine shaft turns; current flows out of the terminals.

• Motor — electrical energy in, mechanical energy out. Current flows into the terminals; the shaft turns and drives a load.

The word convert deserves care. Energy is never created in a machine and never destroyed in one; it is only moved from one form to another, with an unavoidable fraction diverted into heat along the way. That statement is not a platitude — it is the energy-balance equation that Chapter 15 will turn into the central analytical tool of the subject:

Read for a motor, every term is as written. Read for a generator, the first and second simply change sign. This single equation is the reason one machine can do both jobs.

Because the same hardware supports flow in either direction, every rotating machine is reversible in principle. A DC motor driven above its no-load speed becomes a DC generator. An induction motor driven above synchronous speed becomes an induction generator — which is exactly how many wind turbines feed the grid. Whether reversal is practical depends on the excitation and control arrangements, not on the physics.

Energy can be stored in an electric field as readily as in a magnetic one, and electrostatic motors do exist. So why does every machine of any consequence use a magnetic field? The answer is a matter of energy density, and it is worth computing rather than asserting.

The energy stored per unit volume in each kind of field is

Each field is limited by a breakdown mechanism. Air breaks down at an electric field of about \(3 \times 10^{6}~\mathrm{V/m}\), and there is no practical way around it in a machine air gap. A magnetic circuit is limited instead by saturation of the iron, which for ordinary silicon steel sets in somewhere around \(1.5 \text{ to } 1.8~\mathrm{T}\). Substituting the two limits:

Two further practical advantages reinforce the choice. First, magnetic flux can be concentrated by iron, whose relative permeability of a few thousand funnels the flux into precisely the air gap where it is wanted, while there is no comparable material for guiding electric flux. Second, a magnetic circuit is driven by current, which is easy to supply and control, whereas an electrostatic machine requires kilovolts across a millimetre gap — with all the insulation, contamination and safety problems that implies.

Two effects, both known since the 1830s, do all the work. They are not independent: each is a consequence of the same electromagnetic field, and in a machine they always occur together.

Both effects are always present. This is the point students most often miss. A generator being driven produces current; that current, sitting in the field, produces a force \(BIl\) that opposes the motion — which is exactly why the turbine has to do work. A motor drawing current produces motion; that motion induces an EMF \(Blv\) that opposes the applied voltage — the back EMF, which is why a motor does not draw infinite current. In both cases the opposition is Lenz's law doing its bookkeeping, and in both cases it is the mechanism by which energy actually crosses the air gap.

A transformer converts AC electrical energy at one voltage level to AC electrical energy at another. It has no mechanical port and no moving part, so strictly it is not an electromechanical converter at all. Yet it belongs firmly in a machines course, for three reasons.

• It uses the same physics. Faraday's law, a laminated iron core, copper windings, hysteresis and eddy-current loss, leakage flux, magnetising current — every concept in transformer theory reappears unchanged in the induction motor.

• It is the induction motor with the rotor locked. This is not an analogy but a fact: the standard equivalent circuit of an induction motor is the transformer equivalent circuit with one resistance made slip-dependent. Chapter 64 does exactly this substitution, and the blocked-rotor test of Chapter 65 is literally a short-circuit test.

• It makes the power system possible. Generation at 11 kV, transmission at 400 kV and distribution at 415 V exist only because voltage can be changed almost losslessly. Without the transformer, AC would have lost to DC in the 1890s and the modern grid would not exist.

For the ideal transformer, voltage transforms up in the ratio of turns while current transforms down in the same ratio, so that \(V_1 I_1 = V_2 I_2\) — power in equals power out. Chapter 41 derives this from the EMF equation and Chapter 44 adds the losses that make a real transformer fall short of it.

Machines are classified along three cuts that you should be able to apply to any machine you meet: static or rotating, DC or AC, and within AC, synchronous or asynchronous.

The synchronous/asynchronous distinction is the one that carries the most physics. A synchronous machine has a rotor field locked in step with the stator's rotating field — it turns at exactly \(120f/P\) or not at all. An asynchronous (induction) machine has no independently excited rotor field; the rotor field is induced by relative motion, so the rotor must slip behind the stator field for any torque to exist. That single structural difference produces every behavioural difference between the two families: starting torque, power-factor control, speed regulation, and stability all follow from it.

A machine's rating is a promise from the manufacturer: operate within these numbers and the machine will reach a steady temperature that its insulation can survive for the expected service life. A rating is therefore, at bottom, a thermal statement dressed up in electrical units. Understanding that removes most of the mystery from nameplate conventions.

Insulation class and machine life. The permissible temperature is set by the weakest organic material in the winding. The standard classes are:

Every machine in this book has the same four families of loss. Learning them once here saves learning them five more times later.

• Copper (\(I^{2}R\)) loss. Resistive heating in the windings. Varies as the square of the load current, so it is small at light load and dominant at overload. Also called variable loss for this reason.

• Iron (core) loss. Hysteresis plus eddy currents in the laminated core. Depends on flux density and frequency, both of which are essentially fixed by the supply, so this loss is nearly constant from no load to full load. Chapters 6 and 7 treat the two components in detail.

• Mechanical loss. Bearing friction, brush friction where brushes exist, and windage — the power spent stirring the air. A function of speed, hence constant for a constant-speed machine.

• Stray load loss. Everything not captured above: flux harmonics, skin effect, eddy currents in the frame and end plates. Difficult to compute, conventionally taken as about 1 % of output, and never zero.

Efficiency is then simply

Electric machines are so thoroughly embedded in ordinary life that they are usually noticed only when they fail. A short inventory is a useful corrective.

Ninety-two chapters is a long journey, and it helps to see the shape of it before setting out. The book is built in six parts, and the order is deliberate: each part supplies exactly the tools the next one consumes.

1. Part 1 · Principles of Energy Conversion (Ch 1–19). Magnetic circuits, materials and losses; Faraday's and Lenz's laws; inductance; and finally the energy–coenergy argument that produces force and torque from a stored-energy function. Nothing here is a machine, and everything here is in every machine.
2. Part 2 · DC Machines (Ch 20–39). The first machine studied, because the commutator makes the field–armature geometry static and therefore easy to visualise. Construction, windings, the EMF equation, armature reaction, generator and motor characteristics, speed control, braking and testing.
3. Part 3 · Transformers (Ch 40–56). Energy conversion with the mechanics removed. The EMF equation, phasor diagrams, the equivalent circuit, OC and SC tests, regulation and efficiency, three-phase connections, and the per-unit system that makes power-system calculation tractable.
4. Part 4 · Induction Machines (Ch 57–70). The rotating magnetic field, slip, the torque–slip curve, power flow, the equivalent circuit inherited directly from Part 3, testing, the circle diagram, starting and speed control. The most widely used machine in existence.
5. Part 5 · Synchronous Machines (Ch 71–81). Alternators and synchronous motors: the EMF equation and winding factors, armature reaction, voltage regulation by three methods, two-reaction theory for salient poles, synchronising, V-curves and hunting.
6. Part 6 · Single-Phase and Special Machines (Ch 82–92). Why a single-phase motor will not start itself and what is done about it; then the electronically commutated machines — stepper, BLDC, PMSM, switched reluctance — that now dominate precision drives and electric vehicles.

How to study it. Read a chapter's objectives first, then the derivations, then attempt the worked examples with the solution covered. The practice problems at the end of each chapter are graded from routine to demanding; the answers are hidden behind a disclosure so that you can genuinely attempt them. Where a chapter builds on an earlier result, the cross-reference is given — following those links backwards when something does not sit right is far more productive than re-reading the same paragraph.

This book is self-contained, but no single treatment suits every reader. The following are the standard references, and each has a distinct character worth knowing.

• Chapman — the most readable introduction, strong on physical explanation and modern power-electronic context. Best first book if the subject feels abstract.

• Nagrath & Kothari — rigorous and comprehensive, with the generalised-machine treatment. The standard Indian university reference.

• P. C. Sen — excellent on the machine–drive interface; the natural bridge to a power-electronics course.

• Bimbhra — the deepest analytical treatment, particularly on two-reaction theory and transients. Consult when a derivation here feels compressed.

• Theraja and Mehta — very large solved-problem collections. Use them for drill once the theory is secure.

• An electrical machine converts energy between electrical and mechanical form through a magnetic coupling field. Generator and motor action are the same process in opposite directions, and every rotating machine is reversible in principle.

• The magnetic field is used because its practical energy density exceeds that of an electric field by about four orders of magnitude, and because iron can concentrate flux into the air gap where the energy is wanted.

• Almost all field energy sits in the air gap, not in the iron, which is why the gap is the seat of torque production.

• The two governing effects, \(e = Blv\) and \(F = BIl\), are always both present. Back EMF in a motor and opposing torque in a generator are the mechanism of energy transfer, not a nuisance.

• A transformer converts between voltage levels with no moving part and requires AC. It supplies the equivalent-circuit machinery reused by the induction motor.

• Machines classify as static/rotating, DC/AC, and synchronous/asynchronous. The synchronous–asynchronous split — whether the rotor field is separately excited or induced — explains nearly every behavioural difference in the AC family.

• A rating is a thermal promise. Motors are rated in kW because they deliver shaft power; transformers and alternators in kVA because their losses depend on voltage and current separately, not on power factor.

• Losses fall into copper (variable), iron (constant), mechanical and stray. Efficiency peaks where variable loss equals constant loss.

• "A motor and a generator are different machines." They are the same machine with the energy flowing the other way. Whether a given unit will actually work in reverse is a question about excitation and control, not about the physics of conversion.

• "Back EMF is a loss." It is the opposite. Back EMF times armature current is the power converted to mechanical form. A motor with no back EMF converts nothing and simply dissipates \(V^2/R_a\) as heat — which is precisely the condition at standstill, and precisely why starting current is dangerous.

• "Energy is stored in the iron." Because \(w_m = B^2/2\mu\) and iron's permeability is thousands of times that of air, the iron stores almost nothing. The air gap holds the energy; the iron merely steers the flux to it.

• "The kVA rating of a transformer is just its kW rating rounded up." They differ by the power factor, which the manufacturer cannot know. A 100 kVA transformer supplies 100 kW at unity pf and 60 kW at 0.6 pf, running equally hot in both cases.

• "A rated motor can be loaded to its rating indefinitely under any conditions." The rating assumes a stated ambient temperature (usually 40 °C), altitude (usually below 1000 m) and duty type. Derate for hot climates, high altitude, or frequent starting.

• "Efficiency is highest at full load." It is highest where the variable loss equals the constant loss, which most designers place appreciably below full load — often near 75 %.

• "A synchronous motor is just an induction motor that runs faster." A synchronous motor runs at exactly \(N_s\) under all loads and produces no starting torque unaided. An induction motor must slip to produce any torque at all. They are structurally different machines.

• "Power factor is the machine's problem, not mine." A poor power factor raises current for the same useful power, which raises \(I^2R\) loss everywhere upstream, requires larger cables and transformers, and attracts a tariff penalty.

• Confusing \(P\) for power with \(P\) for pole number in \(N_s = 120f/P\). Unfortunate but universal notation; context always resolves it, and pole number is always an even integer.

Work these before opening the answers. Where a power balance is available, use it as a check.

1. P1.1 A motor delivers 7.5 kW at 960 rev/min. Find the shaft torque.

   Show answer

   \[\omega = \frac{2\pi(960)}{60} = 100.5~\mathrm{rad/s}, \qquad T = \frac{7500}{100.5} = 74.6~\mathrm{N\cdot m}\]

   Or directly, \(T = 9.55(7500)/960 = 74.6~\mathrm{N\cdot m}\) \(\checkmark\)

2. P1.2 A machine draws 25 kW and delivers 22.5 kW. Find its efficiency and total loss. If the constant loss is 1.0 kW, what is the variable loss at this load?

   Show answer

   \[\eta = \frac{22.5}{25} = 0.900 = 90.0\,\%, \qquad P_{\text{loss}} = 2.5~\mathrm{kW}\]

   Variable loss \(= 2.5 - 1.0 = 1.5~\mathrm{kW}\). Since variable loss already exceeds constant loss, the machine is operating past its peak-efficiency point.

3. P1.3 Find the synchronous speed of a 6-pole machine on a 50 Hz supply and on a 60 Hz supply. What pole number gives 3000 rev/min at 50 Hz?

   Show answer

   \[N_s = \frac{120(50)}{6} = 1000~\mathrm{rev/min}; \qquad \frac{120(60)}{6} = 1200~\mathrm{rev/min}\]

   \[P = \frac{120 f}{N_s} = \frac{120(50)}{3000} = 2~\text{poles}\]

   Two poles is the minimum, so 3000 rev/min is the highest synchronous speed available at 50 Hz.

4. P1.4 Find the energy stored per cubic metre of air gap at 1.2 T, and compare it with the same volume of iron of \(\mu_r = 2500\) at the same flux density.

   Show answer

   \[w_{\text{gap}} = \frac{(1.2)^{2}}{2(4\pi\times10^{-7})} = \frac{1.44}{2.513\times10^{-6}} = 5.73\times10^{5}~\mathrm{J/m^{3}}\]

   \[w_{\text{iron}} = \frac{(1.2)^{2}}{2(2500)(4\pi\times10^{-7})} = \frac{5.73\times10^{5}}{2500} = 229~\mathrm{J/m^{3}}\]

   The iron stores 2500 times less per unit volume — confirming that the air gap is where the field energy, and therefore the torque, lives.

5. P1.5 A DC generator has a generated EMF of 250 V, an armature current of 40 A and an armature resistance of 0.25 \(\Omega\). Find the terminal voltage, the output power and the armature copper loss.

   Show answer

   For a generator the resistive drop subtracts:

   \[V = E - I_a R_a = 250 - (40)(0.25) = 240~\mathrm{V}\]

   \[P_{\text{out}} = V I_a = (240)(40) = 9600~\mathrm{W}, \qquad P_{cu} = (40)^{2}(0.25) = 400~\mathrm{W}\]

   Check: converted power \(EI_a = 10\,000\) W \(= 9600 + 400\) \(\checkmark\)

6. P1.6 A 22 kW, 415 V, three-phase motor has an efficiency of 91 % and a power factor of 0.88. Find the line current and the reactive power drawn.

   Show answer

   \[P_{\text{in}} = \frac{22\,000}{0.91} = 24\,176~\mathrm{W}\]

   \[I_L = \frac{24\,176}{\sqrt{3}(415)(0.88)} = \frac{24\,176}{632.5} = 38.2~\mathrm{A}\]

   \[S = \sqrt{3}(415)(38.2) = 27\,472~\mathrm{VA}, \qquad Q = S\sin\phi = 27\,472 \times 0.475 = 13.0~\mathrm{kVAr}\]

7. P1.7 A 100 kVA transformer supplies a load at 0.8 power factor lagging. What active power does it deliver at full rating? If the load power factor were corrected to 0.95, how much more active power could the same transformer carry?

   Show answer

   \[P = S\cos\phi = 100(0.8) = 80~\mathrm{kW}; \qquad P' = 100(0.95) = 95~\mathrm{kW}\]

   An extra 15 kW — an 18.75 % increase in useful throughput — from the same iron and copper, at the same temperature. This is the commercial case for power-factor correction stated in one line.

8. P1.8 An 8-pole induction motor on a 50 Hz supply runs at 720 rev/min. Find the synchronous speed, the slip, and the frequency of the rotor currents.

   Show answer

   \[N_s = \frac{120(50)}{8} = 750~\mathrm{rev/min}, \qquad s = \frac{750-720}{750} = 0.040 = 4.0\,\%\]

   \[f_r = s f = 0.040 \times 50 = 2.0~\mathrm{Hz}\]

   The very low rotor frequency is why rotor core loss is negligible in normal running — a result Chapter 60 develops properly.

9. P1.9 A 5 kW motor of efficiency 88 % runs 8 hours a day for 300 days a year. Find the annual input energy and the energy wasted as heat. What annual saving would a 94 %-efficient replacement give at ₹8.00 per kWh?

   Show answer

   \[E_{\text{out}} = 5 \times 8 \times 300 = 12\,000~\mathrm{kWh}\]

   \[E_{\text{in}} = \frac{12\,000}{0.88} = 13\,636~\mathrm{kWh} \quad\Rightarrow\quad \text{waste} = 1636~\mathrm{kWh}\]

   \[E_{\text{in}}' = \frac{12\,000}{0.94} = 12\,766~\mathrm{kWh}, \qquad \Delta E = 13\,636 - 12\,766 = 870~\mathrm{kWh}\]

   At ₹8.00 per kWh the replacement saves 870 × ₹8.00 = ₹6960 per year — on a motor that costs only a few times that to buy.

10. P1.10 The same 4 kW output is required from a 2-pole and from an 8-pole 50 Hz induction motor, running at 2900 and 730 rev/min respectively. Compare the torques, and comment on the physical size of the two machines.

    Show answer

    \[T_2 = \frac{9.55(4000)}{2900} = 13.2~\mathrm{N\cdot m}, \qquad T_8 = \frac{9.55(4000)}{730} = 52.3~\mathrm{N\cdot m}\]

    \[\frac{T_8}{T_2} = \frac{52.3}{13.2} = 3.97 \approx 4\]

    The ratio is simply the inverse speed ratio. Since torque scales roughly with rotor volume, the 8-pole machine is about four times larger and heavier for the same output — which is why high-torque, low-speed drives are usually built as a fast motor plus a gearbox rather than as a many-pole machine.