Electrical & Electronics Engineering Laboratory

Chapter 6
Magnetics, Transformers and Electrical Machines

Bachelors Program in Electrical Engineering
Electrical & Electronics Engineering Laboratory   •   Bachelor

Magnetics, Transformers and Electrical Machines

Introduction

Magnetic circuits, transformers and rotating machines convert energy through magnetic fields. Their laboratory characterization connects electromagnetic theory with measurable quantities such as voltage, current, power, speed, torque, losses, efficiency and regulation. Because direct internal measurement is usually impossible, equivalent circuits and standardized tests are used to infer machine parameters.

Reliable characterization requires more than recording instrument readings. Connections, phase quantities, temperature, instrument accuracy, operating limits and assumptions must all be documented. This chapter develops the principal models and laboratory methods for magnetic circuits, single- and three-phase transformers, DC machines, induction motors and synchronous machines.

Learning Objectives

After completing this chapter, students should be able to:

  • calculate flux, reluctance, magnetomotive force and stored magnetic energy;
  • interpret saturation, hysteresis and eddy-current loss;
  • determine transformer equivalent-circuit parameters from open- and short-circuit tests;
  • evaluate single- and three-phase transformer regulation, losses and efficiency;
  • extract DC-machine parameters and separate important loss components;
  • characterize induction-motor equivalent circuits and torque–speed behavior;
  • apply open-circuit, short-circuit and load tests to synchronous machines;
  • conduct machine tests safely and report uncertainty and limitations.

1. Magnetic Circuits

1.1 Flux, magnetomotive force and reluctance

A magnetic circuit provides a path for magnetic flux. For a uniform section carrying approximately uniform flux,

Φ = BA

where Φ is magnetic flux in webers (Wb), B is flux density in teslas (T), and A is cross-sectional area in square metres (m²).

Ampère’s law relates magnetic field intensity to current:

∮ H·dl = NI

Here H is magnetic field intensity in A/m, N is the number of turns, I is current in amperes, and NI is magnetomotive force, or mmf, in ampere-turns.

For a linear, uniform section of length l and permeability μ,

ℜ = l/(μA), Φ = NI/ℜ

where ℜ is reluctance in A-turns/Wb and μ = μ0μr. The analogy with Ohm’s law is useful, but permeability is often nonlinear and depends on flux density, temperature and magnetic history.

Reluctances in series add when the same flux passes sequentially through each section. Parallel reluctances require flux division and must be treated as a magnetic network. Leakage and fringing make simple lumped models approximate.

Magnetic circuit with an air gap air gap g N turns Φ core window
Magnetic core with a coil, an air gap, and flux circulating around the core through the gap
Figure: Magnetic core with a coil, an air gap, and flux circulating around the core through the gap

Figure: Flux circulates around the magnetic core and crosses the physical air gap; it does not pass directly across the core window.

1.2 Air gaps and fringing

For a core and one air gap carrying the same useful flux,

NI = Φ(ℜc + ℜg)

with

ℜc = lc/(μcAc), ℜg ≈ g/(μ0Ag,eff)

The effective gap area Ag,eff may exceed the physical core area because of fringing. Since air has much lower permeability than ferromagnetic material, the gap often dominates total reluctance.

1.3 Inductance and stored energy

For a linear magnetic circuit with negligible leakage,

L = NΦ/I = N2/ℜ

The stored field energy is

Wm = ½LI2

This expression applies when L is constant. In nonlinear materials, incremental energy must be found from the appropriate integral of flux linkage λ against current.

1.4 Saturation and magnetic losses

The B–H curve becomes nonlinear near saturation. A small increase in flux then requires a large increase in magnetizing current. Alternating excitation produces:

  • Hysteresis loss, associated with repeated traversal of the B–H loop;
  • Eddy-current loss, caused by induced currents in the core;
  • Excess loss, associated with dynamic domain-wall effects.

Laminated, high-resistivity electrical steel reduces eddy currents. Ferrites are advantageous at high frequencies. Core-loss models based only on frequency and peak flux are empirical and must be used within the material data range.

Worked Example 1: Gapped Magnetic Core

A core has mean iron length 0.40 m, cross-sectional area 400 mm², relative permeability 2000, a 1.0 mm air gap and a 500-turn winding. Neglect fringing and leakage. Find the current required for 0.50 T.

The area is A = 400 × 10-6 m2, so

Φ = BA = 0.50(400 × 10-6) = 2.00 × 10-4 Wb

Using μ0 = 4π × 10-7 H/m,

ℜc = 0.40/[(4π × 10-7)(2000)(400 × 10-6)]
= 3.98 × 105 A·turn/Wb
ℜg = 0.001/[(4π × 10-7)(400 × 10-6)]
= 1.99 × 106 A·turn/Wb

Thus

NI = Φ(ℜc + ℜg) = 477.5 A·turn

and I = 477.5/500 = 0.955 A. The gap contributes about 83% of the reluctance, illustrating why gap dimensions must be measured carefully.

2. Transformers

2.1 Ideal and practical behavior

For an ideal transformer with sinusoidal steady-state excitation,

V1/V2 = N1/N2 = a, I1/I2 = 1/a

where a is the turns ratio. A load impedance referred from secondary to primary is

Z'L = a2ZL

A practical transformer includes winding resistance, leakage reactance, magnetizing current and core loss. Referred to one side, the approximate equivalent circuit contains series parameters Req and Xeq, plus shunt parameters Rc and Xm.

Transformer equivalent circuit referred to the primary V₁R_c X_mR_eq jX_eqZ′_L all quantities referred to the primary
Approximate transformer equivalent circuit with shunt excitation branch, series impedance, and a load impedance element
Figure: Approximate transformer equivalent circuit with shunt excitation branch, series impedance, and a load impedance element

Figure: Approximate transformer equivalent circuit. The referred load Z′L is an impedance element connected across the output terminals, not a short circuit.

2.2 Open-circuit and short-circuit tests

In an open-circuit test, rated voltage is applied, usually to the low-voltage winding, while the other winding is open. The wattmeter reading approximates core loss because no-load copper loss is normally small.

For single-phase measurements V0, I0 and P0,

Ic = P0/V0, Im = √(I02 - Ic2)
Rc = V02/P0, Xm = V0/Im

In a short-circuit test, one winding is shorted through suitably rated conductors and a reduced voltage is applied to circulate rated current. Core loss is small because the flux is low. For single-phase data,

Zeq = Vsc/Isc, Req = Psc/Isc2
Xeq = √(Zeq2 - Req2)

The parameters are referred to the energized side. Winding resistance varies with temperature, so test temperature should be recorded.

2.3 Three-phase parameter extraction

For a balanced three-phase transformer, total measured power is divided by three and a per-phase equivalent circuit is used. Conversion from line to phase quantities depends on the connection:

ConnectionPhase voltagePhase current
Star (Y)Vph=VL/√3Iph=IL
Delta (Δ)Vph=VLIph=IL/√3

For open-circuit data,

P0,ph = P0/3, Rc = Vph2/P0,ph
Ic = P0,ph/Vph, Xm = Vph/√(Iph2-Ic2)

For short-circuit data,

Zeq = Vsc,ph/Isc,ph, Req = (Psc/3)/Isc,ph2

The connection used during each test must be known. A line-voltage ratio is not automatically the per-phase turns ratio when the two windings have different Y/Δ connections.

A practical three-phase procedure is:

  1. Verify nameplate voltage, vector group, tap position and grounding arrangement.
  2. Measure winding resistance before energization.
  3. Perform the open-circuit test at rated frequency, increasing voltage gradually.
  4. Record all three line currents; large inequality may indicate supply imbalance or winding problems.
  5. De-energize, discharge and establish a rated short circuit.
  6. Apply low voltage gradually until rated current flows.
  7. Record line voltage, line current, total three-phase power and winding temperature.
  8. Convert readings to phase values before extracting parameters.

Worked Example 2: Three-Phase Transformer

A balanced 30 kVA, 400/200 V, Y–Y transformer gives the following results:

  • open-circuit test on the 200 V side: 200 V line, 5.0 A line, 600 W total;
  • short-circuit test on the 400 V side: 36.0 V line, 43.3 A line, 900 W total.

For the open-circuit test,

Vph = 200/√3 = 115.47 V, P0,ph=200 W
Ic = 200/115.47 = 1.732 A
Im = √(5.02-1.7322)=4.690 A

Thus, on the low-voltage side,

Rc = 115.472/200 = 66.67 Ω
Xm = 115.47/4.690 = 24.62 Ω

The per-phase turns ratio is 2, so referred to the high-voltage side, R'c=266.7 Ω and X'm=98.5 Ω.

For the short-circuit test,

Vsc,ph=36/√3=20.785 V
Zeq=20.785/43.3=0.480 Ω
Req=(900/3)/(43.32)=0.160 Ω
Xeq=√(0.4802-0.1602)=0.453 Ω

At rated current and 0.8 lagging power factor, approximate voltage regulation is

VR ≈ I(Reqcosφ + Xeqsinφ)/Vrated,ph

where sinφ=0.6. Therefore,

VR ≈ 43.3[0.160(0.8)+0.453(0.6)]/230.94 = 0.0749

or 7.49%. This approximation neglects the shunt current and second-order phasor effects.

At rated 0.8 power factor, output is 24.0 kW. Taking core loss as 600 W and rated copper loss as 900 W,

η = 24000/(24000+600+900)=94.12%

2.4 Regulation and efficiency

The exact percentage regulation is

VR = (|VNL|-|VFL|)/|VFL| × 100%

with primary voltage and frequency held constant. Regulation may be negative for sufficiently leading loads.

Transformer efficiency is

η = Pout/(Pout+Pcore+Pcu+Pstray)

At fractional load x, copper loss is approximately x2Pcu,rated if resistance is treated as constant. Maximum efficiency occurs approximately when variable copper loss equals constant loss.

3. DC Machines

3.1 Electromechanical relations

For a DC machine,

E = keΦω

and

Te = ktΦIa

where E is internally generated emf, Φ is useful flux per pole, ω is angular speed, Ia is armature current and T is electromagnetic torque. In consistent SI units, the numerical torque and emf constants are equal.

For a motor,

V = E + IaRa + Vb

where Ra is total armature-circuit resistance and Vb is the total brush contact drop. For a generator, the current-dependent drops are subtracted from generated emf to obtain terminal voltage.

3.2 Parameter extraction and laboratory tests

TestMain resultImportant limitation
Four-wire resistance testArmature and field resistanceCorrect resistance to operating temperature
No-load testRotational and core-loss estimateStray-load loss is not represented
Direct load testTorque, speed, efficiency and regulationRequires a rated mechanical load
Swinburne testPredicted shunt-machine efficiencyAssumes approximately constant rotational loss
Hopkinson testFull-load heating and efficiency of two similar machinesRequires two coupled machines
Retardation testRotational-loss variation with speedNeeds known or estimated inertia

Armature resistance should be measured with a low DC voltage while preventing rotation. A four-terminal method reduces lead and contact error. Because resistance changes with temperature, copper winding resistance may be corrected using

R2/R1 = (T2+235)/(T1+235)

where temperatures are in °C; 235 °C is a commonly used copper temperature constant. The applicable standard and conductor material should be checked.

For a shunt motor at no load,

Pin,0=VIL,0, Ia,0=IL,0-If

An estimate of core plus mechanical loss is

Prot ≈ Pin,0-VIf-Ia,02Ra-VbIa,0

The result is valid near the tested voltage and speed. Core and windage losses are not strictly constant over wide operating ranges.

Under load,

Pout=Pin-Pfield-Pa,cu-Pbrush-Prot-Pstray

where Pa,cu=Ia2Ra. If stray-load loss is omitted, the calculated efficiency is optimistic.

Worked Example 3: DC Shunt-Motor Losses

A 240 V shunt motor has Ra=0.40 Ω, field current 1.0 A and total brush drop 2.0 V. At no load, line current is 3.5 A. At load, line current is 30 A and speed is 1200 r/min.

At no load, Ia,0=2.5 A. Therefore,

Prot=240(3.5)-240(1)-2.52(0.40)-2(2.5)=592.5 W

At load, Ia=29 A and

E=240-29(0.40)-2=226.4 V

The losses are armature copper loss 336.4 W, field loss 240 W, brush loss 58 W and estimated rotational loss 592.5 W. Neglecting additional stray-load loss,

Pout=7200-336.4-240-58-592.5=5973.1 W
η=5973.1/7200=82.96%

At 1200 r/min, ω=2π(1200)/60=125.66 rad/s, giving

Tshaft=5973.1/125.66=47.53 N·m

4. Three-Phase Induction Motors

4.1 Slip and equivalent circuit

The synchronous speed is

ns = 120f/P

where f is frequency in hertz and P is the number of poles. Motor slip is

s=(ns-nr)/ns

The per-phase equivalent circuit contains stator resistance R1, stator leakage reactance X1, magnetizing branch Rc∥jXm, and referred rotor branch

R'2/s + jX'2

The air-gap power, rotor copper loss and converted mechanical power satisfy

Pr,cu=sPg, Pconv=(1-s)Pg

These relations apply to balanced steady-state operation using the fundamental-frequency equivalent circuit.

4.2 Torque–speed behavior

Using the Thevenin equivalent seen by the rotor, electromagnetic torque is

Te = [3|Vth|2(R'2/s)]/[ωs((Rth+R'2/s)2+(Xth+X'2)2)]

where ωs is mechanical synchronous angular speed. Near synchronous speed, torque is approximately proportional to slip. The maximum-torque slip increases with rotor resistance, while the idealized maximum torque is comparatively insensitive to rotor resistance.

4.3 No-load, DC and blocked-rotor tests

  1. DC resistance test: Apply low DC voltage between terminals. Convert the measured line-terminal resistance to per-phase stator resistance according to the winding connection.
  2. No-load test: Run at rated voltage and frequency. Input power supplies stator copper, core, friction and windage losses. Voltage variation can help separate approximately voltage-dependent and mechanical losses.
  3. Blocked-rotor test: Lock the rotor securely, apply reduced voltage and raise current only to the specified value. The test determines total series resistance and leakage reactance at the test frequency.

For balanced per-phase blocked-rotor quantities,

Zbr=Vph/Iph, Rbr=(Pbr/3)/Iph2
Xbr=√(Zbr2-Rbr2)

Approximately, Rbr=R1+R'2 and Xbr=X1+X'2. Splitting leakage reactance equally is a simplifying assumption, not a universal physical law. Rotor skin effect also makes blocked-rotor parameters differ from normal-slip parameters.

Motor output and efficiency are measured by

Pout=Tshaftωr, η=Pout/Pin

where balanced three-phase input power may be obtained from a power analyzer or the two-wattmeter method.

5. Synchronous Machines

5.1 Operating principles

A synchronous machine operates at synchronous speed. For a cylindrical-rotor machine neglecting armature resistance, the approximate three-phase real-power transfer is

P = 3VEf/Xs · sinδ

where V and Ef are per-phase rms magnitudes, Xs is synchronous reactance and δ is the torque angle. This relationship is a steady-state, unsaturated approximation and must not be used beyond the stability region.

In a synchronous motor, field current controls reactive power. Underexcitation generally produces lagging stator current; overexcitation can produce leading current. A plot of armature current against field current at constant mechanical load is called a V-curve.

5.2 Open- and short-circuit tests

For a synchronous generator:

  • The open-circuit characteristic records terminal voltage against field current at rated speed.
  • The short-circuit characteristic records armature current against field current with the terminals shorted through rated test equipment.

At a common field current, the synchronous-impedance method estimates

Zs=Eoc,ph/Isc,ph, Xs=√(Zs2-Ra2)

The method tends to predict conservative regulation because saturation differs between open- and short-circuit conditions. Potier or ASA methods can provide improved separation of leakage reactance and armature reaction.

Before a synchronous generator is connected to a live bus, phase sequence, voltage magnitude, frequency and phase angle must match. Incorrect synchronization can produce destructive torque and current transients.

6. Comparative Testing and Measurement Practice

EquipmentPrincipal testsParameters or outcomes
Magnetic coreB–H loop, inductance, core-loss testPermeability, saturation, loss
TransformerResistance, open circuit, short circuit, loadEquivalent circuit, efficiency, regulation
DC machineResistance, no load, load, retardationLoss separation, torque, speed, efficiency
Induction motorDC, no load, blocked rotor, loadEquivalent circuit, slip, torque–speed curve
Synchronous machineResistance, OCC, SCC, load/V-curveSynchronous impedance, regulation, reactive behavior

Good laboratory practice includes recording instrument ranges, calibration status, ambient and winding temperatures, connection diagrams, repeated observations and estimated uncertainty. Efficiency obtained by subtracting two large measured powers may have greater relative uncertainty than direct torque–speed output measurement.

7. Applied Case: Energy-Efficient Motor Replacement

Background. This is an illustrative industrial case. A facility operates an older four-pole, 50 Hz induction motor for 3000 hours annually. A candidate replacement is tested at 400 V.

Problem. Determine the candidate motor’s operating efficiency and estimate annual energy savings at an equivalent shaft-output demand of 6.9 kW.

Analysis. During the candidate’s load test, line current is 13.8 A, power factor is 0.82, speed is 1460 r/min and shaft torque is 48.0 N·m.

Pin=√3(400)(13.8)(0.82)=7.84 kW
Pout=2π(1460/60)(48.0)=7.34 kW

Thus η=7.34/7.84=93.6%. Synchronous speed is 1500 r/min, so

s=(1500-1460)/1500=0.0267

or 2.67%.

The older motor consumes 8.20 kW while supplying 6.90 kW, giving 84.1% efficiency. Assuming the candidate maintains 93.6% efficiency near that load, its required input is

Pnew=6.90/0.936=7.37 kW

Outcome. Estimated annual saving is

(8.20-7.37)(3000)=2490 kWh/year

approximately. A purchase decision should also consider duty cycle, harmonics from variable-speed drives, starting torque, enclosure, bearing currents and maintenance cost.

Learning Points. Nameplate efficiency should not replace duty-specific measurement. Equal-output comparison is essential, and uncertainty in torque, power and operating hours should accompany any financial claim.

8. Safety, Standards and Professional Responsibility

Machine laboratories involve hazardous voltage, high fault current, rotating shafts, heat and stored mechanical energy. Essential controls include:

  • lockout and verification before rewiring;
  • guarded shafts and couplings;
  • current-limited supplies and correctly rated protection;
  • secure rotor locking during blocked-rotor tests;
  • gradual voltage application with an emergency disconnect;
  • prohibition of loose clothing near rotating equipment;
  • discharge and earthing of isolated windings where required;
  • stopping a test when temperature, vibration or current limits are exceeded.

IEC and IEEE standards define test methods, temperature corrections, efficiency classes and permissible tolerances. Reporting selective readings or ignoring unfavorable results is professionally unethical. Test data may affect equipment safety, energy contracts and regulatory compliance.

9. Contemporary Developments and Limitations

Modern laboratories increasingly use digital power analyzers, automated dynamometers, thermal imaging and vibration monitoring. Finite-element analysis improves magnetic-field and loss prediction, while digital twins combine equivalent circuits with live measurements. Machine-learning systems can identify bearing, insulation and rotor faults, but their predictions require representative training data and independent validation.

Equivalent circuits remain approximations. Saturation, harmonics, temperature, skin effect, stray-load loss, supply imbalance and inverter switching can make parameters operating-point dependent. A defensible report therefore states test conditions and does not present extracted parameters as universal constants.

Chapter Summary

Magnetic circuits are characterized by flux, mmf and reluctance, with air gaps frequently dominating total reluctance. Transformers are represented by shunt excitation and series leakage parameters obtained from open- and short-circuit tests. Correct three-phase extraction requires conversion between line and phase quantities according to winding connection.

DC-machine testing combines resistance, no-load and load measurements to estimate copper, brush, field and rotational losses. Induction-motor tests determine stator, magnetizing and rotor parameters and support analysis of slip, air-gap power and torque–speed behavior. Synchronous-machine OCC and SCC tests provide an approximate synchronous impedance and voltage-regulation model. Across all tests, safety, temperature correction, uncertainty and model limitations are integral to technically and ethically valid conclusions.

Key Terms

Magnetic flux: Surface integral of magnetic flux density, measured in webers.

Flux density: Magnetic flux per unit normal area, measured in teslas.

Magnetomotive force: Magnetic driving quantity NI, measured in ampere-turns.

Reluctance: Opposition of a magnetic path to flux.

Saturation: Region in which permeability falls and large H increases produce small B increases.

Hysteresis loss: Core energy loss caused by cyclic magnetization.

Leakage reactance: Reactance associated with flux that does not link both windings.

Voltage regulation: Change in terminal voltage between specified loading conditions.

Open-circuit test: Test used primarily to determine excitation behavior and core loss.

Short-circuit test: Reduced-voltage test used to determine series impedance and copper loss.

Armature reaction: Effect of armature-produced magnetic field on the main field.

Back emf: Internally induced motor voltage opposing the applied voltage.

Slip: Difference between synchronous and rotor speed divided by synchronous speed.

Blocked-rotor test: Induction-motor test with the rotor mechanically stationary.

Air-gap power: Electromagnetic power transferred from stator field to rotor.

Synchronous reactance: Lumped reactance representing leakage and armature-reaction effects.

Torque angle: Angular displacement associated with synchronous-machine power transfer.

V-curve: Synchronous-motor armature current plotted against field current at constant load.

Review Questions

Short Answer Questions

  1. Why does an air gap often dominate magnetic-circuit reluctance?
  2. What quantities are obtained from a transformer open-circuit test?
  3. Why must three-phase line quantities be converted according to winding connection?
  4. Define induction-motor slip.
  5. What conditions must be matched before synchronizing a generator to a bus?

Long Answer Questions

  1. Derive the reluctance model of a series magnetic circuit containing a core and air gap, and discuss fringing and saturation.
  2. Explain the complete procedure for open- and short-circuit testing of a three-phase transformer.
  3. Develop a loss and efficiency model for a DC shunt motor using no-load and loaded measurements.
  4. Explain induction-motor parameter extraction using DC, no-load and blocked-rotor tests.
  5. Compare synchronous-impedance, Potier and direct-load approaches to synchronous-generator regulation.

Analytical / Application Questions

  1. A three-phase transformer is tested in delta on one side. Given line readings, determine its per-phase excitation and series parameters and refer them to the other winding.
  2. Design a laboratory procedure to obtain the efficiency–load curve of a DC motor, including uncertainty and temperature corrections.
  3. Given induction-motor equivalent-circuit parameters, calculate starting torque, maximum-torque slip and rated-load efficiency, then evaluate the effect of supply-voltage reduction.

Multiple Choice Questions with Answers

Question 1

A magnetic core has flux density 0.8 T and cross-sectional area 250 × 10-6 m2. What is the flux?

A. 0.20 mWb

B. 3.20 mWb

C. 20 mWb

D. 200 mWb

Correct Answer: A. 0.20 mWb

Question 2

Why does a small air gap often dominate the reluctance of a ferromagnetic magnetic circuit?

A. The air gap has a much smaller permeability than the core

B. The air gap carries no magnetic flux

C. The core area becomes zero at the gap

D. The air gap eliminates leakage flux completely

Correct Answer: A. The air gap has a much smaller permeability than the core

Question 3

A transformer has turns ratio a=N1/N2=4. A 5 Ω secondary load referred to the primary is:

A. 1.25 Ω

B. 20 Ω

C. 80 Ω

D. 100 Ω

Correct Answer: C. 80 Ω

Question 4

Which transformer test primarily determines equivalent series resistance and leakage reactance?

A. Open-circuit test

B. Short-circuit test

C. Polarity test

D. Insulation-resistance test

Correct Answer: B. Short-circuit test

Question 5

A balanced star-connected winding has a line voltage of 400 V. Its phase voltage is approximately:

A. 133 V

B. 231 V

C. 400 V

D. 693 V

Correct Answer: B. 231 V

Question 6

A transformer supplies 24 kW while its total losses are 1.5 kW. Its efficiency is approximately:

A. 88.2%

B. 90.6%

C. 94.1%

D. 96.0%

Correct Answer: C. 94.1%

Question 7

A 240 V DC motor draws 20 A armature current through a 0.5 Ω armature resistance and has a total brush drop of 2 V. Its back emf is:

A. 228 V

B. 230 V

C. 248 V

D. 252 V

Correct Answer: A. 228 V

Question 8

A four-pole, 50 Hz induction motor runs at 1440 r/min. What is its slip?

A. 2%

B. 4%

C. 6%

D. 8%

Correct Answer: B. 4%

Question 9

In a balanced induction motor, rotor copper loss is related to air-gap power by:

A. Pr,cu=(1-s)Pg

B. Pr,cu=Pg/s

C. Pr,cu=sPg

D. Pr,cu=s2Pg

Correct Answer: C. Pr,cu=sPg

Question 10

Which set of conditions must be matched before connecting a synchronous generator to a live bus?

A. Resistance, reactance, torque and efficiency

B. Voltage magnitude, frequency, phase sequence and phase angle

C. Field resistance, shaft power, slip and temperature

D. Core loss, copper loss, speed regulation and inertia

Correct Answer: B. Voltage magnitude, frequency, phase sequence and phase angle

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