Electrical Machines¶

Exam-ready answer

A transformer is a static electromagnetic device that transfers AC power from one circuit to another at the same frequency, usually changing voltage and current, by mutual induction. There is no conductive connection between isolated primary and secondary windings.

Construction and operation¶

In a core-type transformer the windings surround two limbs; in a shell-type transformer the core surrounds much of both windings. Applying sinusoidal \(V_1\) causes a small magnetising current to establish core flux \(\phi\). By Faraday's law,

The flux links both windings and induces \(E_1\) and \(E_2\). When a load is connected, secondary current \(I_2\) produces opposing ampere-turns; the primary draws additional \(I_1\) so core flux remains nearly constant and power is transferred magnetically.

EMF equation derivation¶

Let \(\phi=\phi_m\sin\omega t\) Wb, where \(\phi_m\) is maximum flux per turn. Flux changes from \(0\) to \(\phi_m\) in one quarter-cycle, \(T/4=1/(4f)\) s. Hence average induced EMF per turn during that quarter-cycle is

The form factor of a sine wave is \(1.11\), so RMS EMF per turn is \(1.11(4f\phi_m)=4.44f\phi_m\). Therefore

For an ideal transformer,

\(k>1\) is step-up and \(k<1\) is step-down.

Losses, efficiency and regulation¶

and maximum efficiency occurs when variable copper loss equals constant core loss. Voltage regulation at specified power factor is

An open-circuit test measures core loss and shunt parameters; a short-circuit test measures full-load copper loss and series impedance.

Worked design check: for \(f=50\,\text{Hz}\), \(N_1=500\) and \(\phi_m=3\,\text{mWb}\),

For a \(110\,\text{V}\) secondary, \(N_2=N_1E_2/E_1=500(110/333)\approx165\) turns. If the no-load and full-load secondary voltages are \(115\,\text{V}\) and \(110\,\text{V}\), regulation is \((115-110)100/110=4.55\%\).

Transformers provide efficient voltage conversion, impedance transformation and galvanic isolation in grids, rectifiers, telecom power plants and instrumentation.

Practice target: 16–18 minutes; draw and label the construction, reproduce the quarter-cycle derivation, and include losses, efficiency and regulation.

Exam-ready answer

Part A — Construction and working [7 marks]¶

A DC generator converts mechanical input into DC electrical output by Faraday's law of electromagnetic induction. When an armature conductor cuts magnetic flux, an EMF \(e=Blv\) V is induced; its direction follows Fleming's right-hand rule. Each rotating coil generates alternating internal EMF, while the split copper commutator acts as a mechanical rectifier, reversing the coil connection at the brushes so terminal voltage is unidirectional.

The prime mover rotates the armature at \(N\) rpm. Conductors under successive north and south poles cut flux in opposite directions, so their induced EMFs reverse every half-turn. The commutator simultaneously reverses the brush connection to each coil, maintaining one brush positive and the other negative. With a closed load, generated EMF drives current from the positive brush through the load and back to the negative brush.

Generators may be separately excited, shunt, series or compound according to field connection. Armature reaction distorts the main field and poor commutation causes brush sparking; interpoles and compensating windings improve both effects.

Part B — EMF equation derivation [3 marks]¶

Let \(P\) be poles, \(\phi\) flux per pole in webers (Wb), \(Z\) total armature conductors, \(N\) speed in revolutions per minute (rpm), and \(A\) parallel armature paths.

Step 1 — Flux cut: one conductor cuts \(P\phi\) Wb in one revolution.

Step 2 — Conductor EMF: time for one revolution is \(60/N\) s, so average EMF per conductor is

Step 3 — Series conductors: each parallel path has \(Z/A\) conductors in series. Therefore generated brush EMF is

Thus \(E_g\propto\phi N\) for a fixed machine.

Numerical check: a four-pole lap generator has \(Z=480\), \(\phi=20\,\text{mWb}=0.02\,\text{Wb}\), and \(N=1200\,\text{rpm}\). Since lap winding gives \(A=P=4\),

The result has the expected proportionality: doubling speed or flux doubles voltage. DC generators remain important for teaching energy conversion, welding, excitation and specialised DC supplies, although alternator-rectifier systems now dominate general generation.

Practice target: 16–18 minutes; allocate about 11–12 minutes to the labeled construction/operation and 5–6 minutes to the four-line EMF derivation and numerical.

Exam-ready answer

A DC motor converts DC electrical energy into mechanical rotation. A current-carrying conductor of active length \(l\) metres in magnetic flux density \(B\) tesla experiences force

when conductor, field and force are mutually perpendicular; direction follows Fleming's left-hand rule. Forces on opposite armature sides form a torque, and the commutator reverses conductor current at the correct instant so torque remains unidirectional.

As the armature rotates it also acts as a generator. The induced back EMF opposes the applied voltage by Lenz's law:

At starting, \(N=0\) and \(E_b=0\), so \(I_a=V/R_a\) can be destructive because armature resistance is small; a starter or electronic current limiter is required.

The gross converted mechanical power is \(P_m=E_bI_a\) W. Since angular speed \(\omega_m=2\pi N/60\) rad/s,

Self-regulation: increased shaft load slows the motor, reducing \(E_b\); armature current then rises, producing more torque. A lighter load causes the reverse sequence. A series motor must never be run unloaded because reduced flux can cause dangerous overspeed.

Worked check: a \(220\,\text{V}\) motor with \(R_a=0.5\,\Omega\) draws \(I_a=20\,\text{A}\) at \(1000\,\text{rpm}\). Then \(E_b=220-(20)(0.5)=210\,\text{V}\), converted power is \(E_bI_a=4.20\,\text{kW}\), and

Back EMF therefore limits running current and gives the motor its automatic load response; the torque and speed equations guide drive selection and control.

Practice target: 8–9 minutes; show the force law, back-EMF loop equation, self-regulation chain and one torque check.

Exam-ready answer

Both machines use the rotating magnetic field produced by balanced three-phase stator currents. Its synchronous speed is

where \(f\) is supply frequency in hertz (Hz) and \(P\) is number of poles.

In a three-phase induction motor, the stator field cuts short-circuited squirrel-cage bars or a wound rotor, inducing rotor EMF and current. Their interaction with the rotating field produces torque. Relative motion is essential for induction, so rotor speed \(N\) must remain below \(N_s\) in motor operation. Slip and rotor-current frequency are

It is self-starting because at standstill \(s=1\) and the three-phase field already rotates.

In a synchronous motor, a DC-excited or permanent-magnet rotor is brought near \(N_s\) and its poles lock magnetically to the stator field. It then runs exactly at \(N_s\) with zero steady-state slip. It is not inherently self-starting; damper bars, a pony motor or a variable-frequency drive provides starting torque.

Numerical comparison: at \(f=50\,\text{Hz}\) and \(P=4\), \(N_s=120(50)/4=1500\,\text{rpm}\). An induction motor running at \(1440\,\text{rpm}\) has

A four-pole synchronous motor on the same supply runs at exactly \(1500\,\text{rpm}\). Overexcitation lets it supply leading reactive power as a synchronous condenser, whereas the induction motor is selected when ruggedness, simple starting and low cost dominate.

Practice target: 8–9 minutes; calculate \(N_s\) and slip, then reproduce at least five comparison rows.