Nepal Engineering Council · Electrical Engineering · Chapter 2
Electrical Machines
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181 questions in 6 syllabus topics.
2.1 Magnetic circuits
30 questions · AElE0201
1. In a magnetic circuit, which quantity plays the role that resistance plays in an electric circuit?
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Think of the magnetic version of Ohm's law: flux = mmf / ?
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Answer: A. Reluctance
Reluctance S = l/(µA) opposes the establishment of flux, just as resistance opposes current; flux = mmf / reluctance.
2. An iron ring has a mean length of 0.5 m, cross-section 4 cm² and relative permeability 1000. Its reluctance is approximately
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Convert cm² to m² before using S = l/(µ0 µr A).
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Answer: B. 9.95 × 10⁵ A/Wb
S = l/(µ0 µr A) = 0.5 / (4π×10⁻⁷ × 1000 × 4×10⁻⁴) ≈ 9.95 × 10⁵ A/Wb.
3. A 200-turn coil carries 2 A on a core of reluctance 4 × 10⁵ A/Wb. The self-inductance of the coil is
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Inductance depends on turns squared divided by reluctance.
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Answer: D. 0.1 H
L = N²/S = 200² / (4×10⁵) = 0.1 H (the flux is NI/S = 1 mWb, and L = NΦ/I = 0.1 H).
4. The SI unit of magnetic flux linkage is
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Flux linkage is turns multiplied by flux.
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Answer: C. Weber-turn
Flux linkage λ = NΦ, so its unit is weber-turn (dimensionally the weber).
5. A 100-turn coil links a flux of 2 mWb. The flux linkage is
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Multiply turns by flux in webers.
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Answer: C. 0.2 Wb-turn
λ = NΦ = 100 × 2×10⁻³ = 0.2 Wb-turn.
6. The energy stored in an inductor of 0.5 H carrying a steady current of 4 A is
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Use the half-L-I-squared formula.
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Answer: A. 4 J
W = ½LI² = 0.5 × 0.5 × 16 = 4 J.
7. If the number of turns of a coil is doubled while the magnetic core and its dimensions remain the same, its inductance becomes
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Inductance is proportional to a power of N.
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Answer: D. Four times
L = N²/S; with S unchanged, doubling N multiplies L by 2² = 4.
8. Two coils have self-inductances 0.4 H and 0.9 H with a coefficient of coupling of 0.5. Their mutual inductance is
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M equals k times the geometric mean of the self-inductances.
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Answer: A. 0.3 H
M = k√(L1L2) = 0.5 × √0.36 = 0.5 × 0.6 = 0.3 H.
9. Two coils of 2 H and 3 H with mutual inductance 1 H are connected in series so that their fluxes aid each other. The equivalent inductance is
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Series aiding adds twice the mutual inductance.
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Answer: C. 7 H
Series aiding: L = L1 + L2 + 2M = 2 + 3 + 2 = 7 H.
10. In a magnetic circuit with an iron core and a small air gap, most of the magnetic field energy is stored in
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Compare B²/(2µ) for a high-µ and a low-µ region carrying the same flux density.
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Answer: D. The air gap
Energy density is B²/(2µ); since µ of air is far lower than that of iron for the same B, the air gap stores most of the energy.
11. The magnetic energy density in an air gap where the flux density is 1 T is about
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Energy density = B²/(2µ0).
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Answer: C. 3.98 × 10⁵ J/m³
w = B²/(2µ0) = 1 / (2 × 4π×10⁻⁷) ≈ 3.98 × 10⁵ J/m³.
12. The mmf needed to establish a flux density of 1 T across an air gap of 1 mm is approximately
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Find H in air from B/µ0, then multiply by the gap length.
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Answer: A. 796 AT
H = B/µ0 = 1/(4π×10⁻⁷) ≈ 7.96×10⁵ A/m; mmf = H × l = 7.96×10⁵ × 10⁻³ ≈ 796 AT.
13. Which class of materials has a relative permeability slightly less than 1?
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These materials are weakly repelled by a magnet.
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Answer: B. Diamagnetic
Diamagnetic materials (e.g. copper, bismuth) have a small negative susceptibility, so µr is slightly below 1.
14. Above the Curie temperature, a ferromagnetic material becomes
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It loses its domain structure but still responds weakly to a field.
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Answer: C. Paramagnetic
Thermal agitation destroys the domain alignment above the Curie temperature, and the material behaves paramagnetically.
15. A material suitable for making permanent magnets should have
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It must stay magnetised and resist being demagnetised.
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Answer: B. High retentivity and high coercivity
A permanent magnet must keep a large residual flux (retentivity) and resist demagnetisation (coercivity), giving a wide hysteresis loop.
16. The core of a power transformer is usually made of
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The material must be magnetically soft with low losses along the rolling direction.
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Answer: D. Cold-rolled grain-oriented silicon steel
CRGO silicon steel has high permeability, a narrow hysteresis loop and raised resistivity, giving low core loss.
17. Ferrites are preferred for cores at high frequencies mainly because they have
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Which loss grows fastest with frequency, and what limits it?
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Answer: D. Very high electrical resistivity
Ferrites are ceramic and nearly insulating, so eddy current losses remain small even at high frequency.
18. The area enclosed by the hysteresis loop of a magnetic material represents
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Integrating H dB over a cycle gives a quantity with units of J/m³.
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Answer: A. Energy lost per cycle per unit volume
The loop area (∮H dB) is the energy dissipated as hysteresis loss in one cycle per unit volume.
19. By the Steinmetz relation (Ph ∝ f·Bmax^1.6), if Bmax is increased by 20% at constant frequency, the hysteresis loss increases by about
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Raise 1.2 to the Steinmetz exponent.
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Answer: D. 34%
Ratio = 1.2^1.6 ≈ 1.34, i.e. an increase of about 34%.
20. If the lamination thickness of a core is halved (all else unchanged), the eddy current loss becomes
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Eddy loss varies with a power of thickness.
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Answer: B. One quarter
Eddy current loss is proportional to the square of lamination thickness, so (½)² = ¼.
21. The portion of the B-H curve where a large increase in H produces only a small increase in B is called
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The domains have nearly all lined up.
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Answer: B. Saturation
In saturation almost all domains are aligned, so B increases only at the rate µ0 for further increases in H.
22. A coil of 500 turns links a flux that falls uniformly from 4 mWb to zero in 0.02 s. The average emf induced is
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Use Faraday's law with the rate of change of flux.
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Answer: A. 100 V
e = N dΦ/dt = 500 × 0.004 / 0.02 = 100 V.
23. A conductor 0.4 m long moves at 10 m/s at right angles to a uniform field of 0.5 T. The emf induced in it is
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Dynamically induced emf is B·l·v.
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Answer: B. 2 V
e = Blv = 0.5 × 0.4 × 10 = 2 V.
24. A straight conductor 0.5 m long carrying 20 A lies perpendicular to a uniform field of 0.8 T. The force on it is
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Force on a current-carrying conductor is B·I·l.
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Answer: B. 8 N
F = BIl = 0.8 × 20 × 0.5 = 8 N.
25. The emf induced in a transformer winding is an example of
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Nothing in a transformer physically moves.
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Answer: B. Statically induced emf
In a transformer the conductors are stationary and the flux linking them changes with time, so the emf is statically induced.
26. Lenz's law states that the direction of an induced emf is such that it
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Think of energy conservation and the minus sign in Faraday's law.
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Answer: C. Opposes the change of flux that produces it
Lenz's law, a consequence of energy conservation, gives the minus sign in e = -N dΦ/dt: the induced current opposes the cause.
27. Fleming's right-hand rule is used to find the direction of
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Right hand is associated with generators.
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Answer: D. Induced emf in a generator conductor
The right-hand (generator) rule relates motion, field and the direction of induced emf; the left-hand rule gives motor force.
28. Fringing in an air gap of a magnetic circuit causes
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Flux spreads out as it crosses the gap.
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Answer: C. An increase in the effective area of the gap
Flux lines bulge outward at the gap edges, so the effective gap area is larger and its reluctance is slightly lower.
29. Relative permeability of a material is defined as the ratio of
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It is a dimensionless comparison with vacuum.
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Answer: A. Its absolute permeability to the permeability of free space
µr = µ/µ0; it is dimensionless. B/H gives absolute permeability, and M/H is susceptibility.
30. The flux linkage of a linear magnetic system is λ = Li. The co-energy of the system equals its stored energy because
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Look at the two areas on either side of the λ-i curve.
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Answer: A. The λ-i characteristic is a straight line
For a linear λ-i curve the areas above and below the line are equal, so co-energy = energy = ½Li².
2.2 Transformers
31 questions · AElE0202
31. In a core-type transformer,
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Ask which one encloses the other.
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Answer: D. The windings surround a considerable part of the core
In core type the windings encircle the limbs; in shell type the core surrounds the windings and the central limb carries the full flux.
32. Shell-type transformers are generally preferred for
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Which type offers easier insulation for high voltage, and which suits the other case?
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Answer: D. Low-voltage, high-current applications
Shell type gives better mechanical support and lower leakage for heavy-current low-voltage windings; core type gives more room for high-voltage insulation.
33. The function of the conservator in an oil-immersed transformer is to
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Oil volume changes as the transformer heats and cools.
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Answer: B. Accommodate expansion and contraction of oil with temperature
The conservator is a small tank above the main tank that keeps it full of oil while oil volume changes with temperature, reducing the oil surface exposed to air.
34. The breather of a transformer contains silica gel to
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What harms insulating oil that comes in with the air?
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Answer: D. Absorb moisture from the air entering the conservator
As oil contracts, air is drawn in through the breather; silica gel dries it so moisture does not degrade the oil's dielectric strength.
35. The explosion vent on a transformer tank is provided to
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It is a safety device for abnormal conditions.
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Answer: A. Relieve excessive internal pressure during a severe internal fault
A thin diaphragm at the end of the vent bursts under high pressure from a heavy fault, protecting the tank from rupture.
36. Bushings in a transformer are used to
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Live leads must pass through an earthed metal wall.
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Answer: A. Bring winding terminals out through the earthed tank with insulation
A bushing is an insulated conductor passing through the tank so that the live leads are insulated from the earthed tank.
37. Transformer oil serves mainly to
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It does two jobs: electrical and thermal.
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Answer: B. Provide insulation and carry heat away from windings and core
Mineral transformer oil is both an insulating medium and a coolant that transfers heat to the tank and radiators.
38. The emf induced in a winding of 200 turns on a 50 Hz transformer with a maximum core flux of 0.01 Wb is about
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Use the transformer emf equation with Φm.
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Answer: C. 444 V
E = 4.44 f N Φm = 4.44 × 50 × 200 × 0.01 = 444 V (rms).
39. A 2200/220 V single-phase transformer has 500 turns on the HV winding. The number of turns on the LV winding is
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The voltage ratio equals the turns ratio.
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Answer: D. 50
N2 = N1 × V2/V1 = 500 × 220/2200 = 50.
40. A 100 kVA, 11000/400 V single-phase transformer has full-load primary and secondary currents of about
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Current = VA / V on each side.
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Answer: B. 9.09 A and 250 A
I1 = 100000/11000 ≈ 9.09 A; I2 = 100000/400 = 250 A. The HV side carries the smaller current.
41. The no-load current of a transformer lags the applied voltage by nearly 90° (low power factor) because
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Compare the two components of the no-load current.
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Answer: A. Its magnetising component is much larger than its core-loss component
No-load current I0 consists of a large reactive magnetising component Iµ and a small in-phase core-loss component Iw, so the power factor is typically 0.1-0.2.
42. When the load on a transformer increases, the mutual flux in its core
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The applied voltage fixes the flux.
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Answer: C. Remains nearly constant
Since V1 ≈ 4.44 f N1 Φm and V1 is fixed, Φm stays nearly constant; the extra primary current neutralises the secondary mmf.
43. A secondary resistance of 0.02 Ω on a transformer with turns ratio N1/N2 = 10 has an equivalent value referred to the primary of
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Impedances are referred by the square of the turns ratio.
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Answer: C. 2 Ω
R2' = R2 (N1/N2)² = 0.02 × 100 = 2 Ω.
44. The open-circuit test on a transformer is used mainly to determine
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With rated voltage applied and no load current, what loss dominates?
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Answer: A. Core (iron) loss and the shunt branch parameters
At rated voltage with one side open, the current is small, so the wattmeter reads essentially core loss; V, I0 and P give R0 and X0.
45. The short-circuit test on a transformer is normally performed
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Only a small voltage is needed to circulate rated current.
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Answer: A. On the HV side with the LV side shorted, at reduced voltage
Rated current is circulated at a small voltage (about 5-10% of rated) applied to the HV side, which carries the smaller current, with the LV side shorted.
46. In a short-circuit test, the readings are 500 V, 10 A and 1000 W. The equivalent reactance referred to the side of measurement is about
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Find Z and R first, then X by Pythagoras.
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Answer: D. 49 Ω
Z = 500/10 = 50 Ω, R = 1000/10² = 10 Ω, X = √(50² - 10²) ≈ 49 Ω.
47. Sumpner's (back-to-back) test needs two identical transformers and is used to
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It reproduces full-load conditions while drawing only the losses.
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Answer: C. Determine efficiency and temperature rise under full-load conditions without loading them
Connected back-to-back, the transformers draw only losses from the supply while carrying full-load current and flux, so heat run and efficiency can be found.
48. A 100 kVA transformer has an iron loss of 1 kW and a full-load copper loss of 2 kW. Its efficiency at full load and 0.8 pf is about
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Efficiency = output / (output + iron loss + copper loss).
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Answer: C. 96.4%
Output = 100 × 0.8 = 80 kW; losses = 3 kW; η = 80/83 ≈ 96.4%.
49. The efficiency of a transformer is maximum when
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Equate the variable loss to the constant loss.
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Answer: C. Copper loss equals iron loss
Differentiating efficiency with respect to load current gives the condition: variable (copper) loss = constant (iron) loss.
50. A transformer has an iron loss of 1 kW and a full-load copper loss of 4 kW. Maximum efficiency occurs at
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Find the load at which the copper loss falls to the iron loss.
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Answer: B. 50% of full load
Load fraction x = √(Pi/Pcu,fl) = √(1/4) = 0.5.
51. Hysteresis loss in a transformer core can be reduced by
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Hysteresis loss is a material property.
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Answer: B. Using a material with a narrow hysteresis loop such as silicon steel
Hysteresis loss depends on the loop area of the material; silicon steel has a narrow loop. Thin laminations reduce eddy loss instead.
52. A 50 Hz transformer is operated at 60 Hz with the same applied voltage. Its eddy current loss
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Express Bmax in terms of V and f, then substitute.
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Answer: C. Remains about the same
Bmax ∝ V/f, and eddy loss ∝ (Bmax f)²; with V fixed, Bmax f is constant, so eddy loss is unchanged (hysteresis loss falls).
53. The approximate percentage voltage regulation of a transformer with 1% resistance and 4% reactance at full load, 0.8 pf lagging, is
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Use %R·cosφ + %X·sinφ for lagging power factor.
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Answer: D. 3.2%
Regulation ≈ %R cosφ + %X sinφ = 1 × 0.8 + 4 × 0.6 = 3.2%.
54. A three-phase, 500 kVA, 11 kV/415 V transformer has an HV full-load line current of about
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Use the three-phase power formula with line voltage.
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Answer: A. 26.2 A
I = S/(√3 VL) = 500000/(√3 × 11000) ≈ 26.2 A.
55. In a Dyn11 distribution transformer, the LV line voltage
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Each hour on the clock is 30°; read 11 o'clock relative to 12.
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Answer: C. Leads the HV line voltage by 30°
The clock number 11 means the LV phasor is at the 11 o'clock position relative to HV at 12, i.e. it leads by 30°.
56. Three single-phase 100 kVA transformers in delta-delta supply a balanced load. If one is removed and the bank is run in open delta (V-V), the maximum load it can supply without overloading is about
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Open-delta capacity is not simply two-thirds of the bank.
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Answer: B. 173 kVA
Open-delta capacity = √3 × rating of one unit = √3 × 100 ≈ 173 kVA (57.7% of the original 300 kVA).
57. Which condition is essential (not merely desirable) for parallel operation of two three-phase transformers?
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Which mismatch would cause a short circuit between the secondaries?
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Answer: A. Same phase displacement (vector group) and phase sequence
Different phase displacements or phase sequences cause heavy circulating currents; equal kVA or X/R ratios only improve load sharing.
58. Two transformers rated 500 kVA and 250 kVA with identical per-unit impedances and voltage ratios share a 600 kVA load. The load on the 500 kVA unit is
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Equal per-unit impedance means sharing in proportion to rating.
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Answer: A. 400 kVA
With equal per-unit impedances, load divides in proportion to ratings: 600 × 500/750 = 400 kVA.
59. A single-phase load of 160 kW at 0.8 power factor lagging is to be supplied. The minimum standard transformer rating required is
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Transformer ratings are in kVA, not kW.
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Answer: B. 200 kVA
Required kVA = kW / pf = 160/0.8 = 200 kVA. Transformers are rated in kVA because their losses depend on current and voltage, not pf.
60. The permissible temperature rise of a transformer is limited mainly by
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What deteriorates fastest when a transformer runs hot?
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Answer: D. The thermal class of its winding insulation and oil
Insulation ageing accelerates with temperature, so the temperature rise limits are set to protect the insulation (and oil).
61. Buchholz relay in a transformer is fitted
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It detects gas that rises from the tank towards the conservator.
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Answer: B. In the pipe between the main tank and the conservator
The Buchholz relay is a gas-actuated relay in the connecting pipe; gas from incipient faults collects in it and operates its float.
2.3 DC machines
30 questions · AElE0203
62. In a DC generator, the commutator acts as a
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The coil emf alternates, yet the output is DC.
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Answer: B. Mechanical rectifier converting the alternating armature emf to DC at the brushes
The emf induced in each armature coil is alternating; the commutator reverses coil connections at the right instant so the brush voltage is unidirectional.
63. The yoke of a DC machine serves to
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It is the outer frame of the machine.
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Answer: D. Provide mechanical support and carry the magnetic flux between poles
The yoke is the outer frame; it supports the poles and provides the return path for the main flux.
64. The armature core of a DC machine is laminated in order to reduce
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Laminations break up circulating current paths.
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Answer: B. Eddy current loss
The armature rotates in the field, so its core experiences alternating flux; thin insulated laminations limit eddy currents.
65. A 4-pole, lap-wound DC generator has 500 armature conductors and a flux of 0.02 Wb per pole. At 1200 rpm its generated emf is
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For lap winding, the number of parallel paths equals the number of poles.
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Answer: B. 200 V
E = PΦZN/(60A) = 4 × 0.02 × 500 × 1200/(60 × 4) = 200 V (A = P = 4 for lap winding).
66. A 4-pole DC generator with a simplex lap winding generates 200 V. If it is rewound as a simplex wave winding with the same number of conductors, flux and speed, its emf becomes
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Count the parallel paths in a simplex wave winding.
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Answer: A. 400 V
Wave winding has A = 2 instead of 4, so E doubles to 400 V (while current capacity halves).
67. A DC shunt generator delivers 50 A at 220 V. The shunt field resistance is 110 Ω and armature resistance 0.1 Ω. The generated emf is (neglect brush drop)
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Armature current = load current + shunt field current.
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Answer: D. 225.2 V
Ish = 220/110 = 2 A, Ia = 50 + 2 = 52 A, E = 220 + 52 × 0.1 = 225.2 V.
68. For a self-excited DC shunt generator to build up voltage, which of the following is necessary?
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Think of the conditions for voltage build-up starting from a tiny voltage.
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Answer: B. Residual magnetism in the poles and field resistance below the critical value
Build-up needs residual flux, correct field connection (aiding the residual flux), and field circuit resistance less than the critical resistance at that speed.
69. The main effect of armature reaction in a DC generator is
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The armature's own mmf acts across the main field.
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Answer: D. Distortion and weakening of the main field flux with a shift of the magnetic neutral axis
Armature mmf cross-magnetises the field, distorting it, shifting the MNA in the direction of rotation (in a generator) and, with saturation, reducing the net flux.
70. Compensating windings in large DC machines are placed
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They must follow the armature current to cancel its mmf under the poles.
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Answer: C. In slots on the pole faces, connected in series with the armature
Compensating windings in the pole faces carry armature current in the opposite direction to neutralise armature reaction under the poles.
71. Interpoles (commutating poles) are provided in DC machines mainly to
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They act on the coils being short-circuited by the brushes.
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Answer: D. Improve commutation by neutralising reactance voltage in the commutating zone
Interpoles, in series with the armature, induce an emf in the coils undergoing commutation that cancels the reactance voltage, reducing sparking.
72. Which DC generator is most suitable for arc welding?
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Welding needs voltage to fall sharply as current rises.
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Answer: A. Differentially compounded generator
Differential compounding gives a steeply drooping characteristic, so current is limited when the arc is short-circuited.
73. A DC series generator is used as a booster in DC lines mainly because
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The line drop rises with current; what should the booster do?
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Answer: D. Its terminal voltage rises with load current over the working range
In the rising part of its characteristic, a series generator's voltage increases with current, compensating line drop proportional to load.
74. A DC generator delivers 10 kW with total losses of 1 kW. Its efficiency is about
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The input is the output plus the losses.
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Answer: A. 90.9%
η = output/(output + losses) = 10/11 ≈ 90.9%.
75. The efficiency of a DC machine is maximum when
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Split the losses into constant and load-dependent parts.
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Answer: C. Variable copper loss equals constant losses
As for transformers, efficiency peaks when the load-dependent armature copper loss equals the constant (iron, mechanical, shunt field) losses.
76. Swinburne's test on a DC shunt machine determines
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It is a no-load, indirect test.
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Answer: C. No-load (constant) losses, from which efficiency at any load is predicted
Swinburne's test runs the machine as a motor at no load to measure constant losses; efficiency at other loads is then calculated.
77. A 230 V DC shunt motor draws an armature current of 40 A. With an armature resistance of 0.5 Ω, the back emf is
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In a motor the armature drop is subtracted from the supply voltage.
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Answer: C. 210 V
Eb = V - IaRa = 230 - 40 × 0.5 = 210 V.
78. A 220 V DC shunt motor (Ra = 0.5 Ω) runs at 1000 rpm taking an armature current of 20 A. If the load increases so that Ia = 40 A with flux constant, the new speed is about
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With constant flux, speed is proportional to back emf.
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Answer: B. 952 rpm
Eb1 = 220 - 10 = 210 V, Eb2 = 220 - 20 = 200 V; N2 = 1000 × 200/210 ≈ 952 rpm.
79. A DC motor has a back emf of 200 V and armature current of 50 A at 1000 rpm. The electromagnetic torque developed is about
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Convert rpm to rad/s and divide gross mechanical power by it.
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Answer: B. 95.5 N·m
T = EbIa/ω = 200 × 50 / (2π × 1000/60) ≈ 95.5 N·m.
80. A 220 V DC motor has an armature resistance of 0.5 Ω. If switched directly on the supply at standstill, the armature current would be
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At the moment of starting there is no back emf.
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Answer: A. 440 A
At standstill Eb = 0, so Ia = 220/0.5 = 440 A, which is why a starter is required.
81. A 220 V DC shunt motor with Ra = 0.5 Ω has a rated armature current of 20 A. The external starting resistance needed to limit the starting current to twice rated current is
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Subtract the armature resistance from the total required.
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Answer: B. 5.0 Ω
Total R = 220/40 = 5.5 Ω; external R = 5.5 - 0.5 = 5.0 Ω.
82. A DC series motor should never be started without load because
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Speed is inversely related to flux, and flux depends on load current here.
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Answer: C. Its speed becomes dangerously high at light load
In a series motor flux ∝ current; at no load the current and flux are very small, so speed (∝ Eb/Φ) rises to dangerous values.
83. Below magnetic saturation, if the current of a DC series motor is doubled, its torque becomes
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In a series motor flux follows the armature current.
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Answer: C. Four times
T ∝ ΦIa and Φ ∝ Ia (unsaturated), so T ∝ Ia²; doubling current gives four times torque.
84. DC series motors are commonly used for
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Which application needs very high starting torque?
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Answer: B. Electric traction and cranes
Series motors give very high starting torque and speed that falls with load, ideal for traction, hoists and cranes.
85. A cumulatively compounded DC motor is suitable for loads such as
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These loads have sudden heavy torque peaks.
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Answer: A. Punching presses and shears with flywheels
Cumulative compounding gives high starting torque with a defined no-load speed; with a flywheel it suits intermittent heavy loads like presses and shears.
86. The speed of a DC shunt motor can be raised above its rated (base) speed by
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Speed is inversely proportional to flux.
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Answer: A. Inserting resistance in the shunt field circuit
Field weakening reduces flux, and N ∝ Eb/Φ, so speed rises above base; armature resistance or voltage control gives speeds below base.
87. The armature resistance (rheostatic) method of speed control of a DC shunt motor is
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Consider where the dropped voltage's energy goes.
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Answer: A. Wasteful, as considerable power is lost in the control resistance
Speed is reduced below base by dropping voltage across series resistance; the I²R loss in it makes this method inefficient, and speed varies with load.
88. The Ward-Leonard system controls the speed of a DC motor by
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It uses a dedicated generator to feed the motor's armature.
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Answer: D. Varying the armature voltage supplied from a separately driven generator
A motor-generator set supplies a variable armature voltage to the motor, giving smooth, wide-range speed control in both directions.
89. The no-volt release coil in a three-point starter of a DC shunt motor
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It is in series with the field in a three-point starter.
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Answer: C. Returns the starter arm to OFF if supply fails or the field circuit opens
The hold-on coil is in series with the shunt field; loss of supply or field current releases the arm to the OFF position.
90. In a DC motor, the torque developed is directly proportional to
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Recall the torque equation derived from EbIa = Tω.
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Answer: D. The product of flux per pole and armature current
T = (PZ/2πA) Φ Ia, so torque ∝ ΦIa.
91. In a DC shunt motor, if the field circuit suddenly opens while running on light load, the motor will
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Think about N ∝ Eb/Φ with Φ almost zero.
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Answer: A. Tend to race to a dangerously high speed
Only residual flux remains, so the speed must rise greatly to develop the required back emf, which is dangerous (hence field-failure protection).
2.4 Synchronous machines
30 questions · AElE0204
92. Salient-pole rotors are normally used in alternators driven by
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Many poles are needed at low speed.
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Answer: A. Low-speed hydraulic turbines
Low-speed prime movers need many poles to produce 50 Hz, which is practical only with large-diameter, short salient-pole rotors.
93. Cylindrical (non-salient) rotors are used in turbo-alternators mainly because
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Consider the mechanical stresses at 3000 rpm.
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Answer: D. They withstand the large centrifugal forces at high speed and give lower windage loss
At 1500/3000 rpm a smooth, small-diameter, long rotor is mechanically robust and quieter, with low windage loss.
94. In large synchronous generators the armature winding is placed on the stator and the field on the rotor because
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Which circuit carries the large output power?
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Answer: C. High-voltage armature is easier to insulate and connect when stationary, and the low-power field needs only two slip rings
Stationary armature avoids sliding contacts for the high-voltage, high-current output; the DC field needs only low-power slip rings (or none, if brushless).
95. The field winding of a synchronous machine is supplied with
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The rotor poles must be of fixed polarity.
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Answer: D. Direct current
The rotor field winding is excited with DC to produce a fixed pole pattern that rotates with the rotor.
96. In a brushless excitation system, the main field winding is fed through
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No brushes means the rectifier must rotate with the field.
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Answer: C. A rotating rectifier supplied by a shaft-mounted AC exciter
An AC exciter with its armature on the shaft feeds rotating diodes that supply the main field directly, eliminating brushes and slip rings.
97. A static excitation system obtains its field power from
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It has no rotating exciter at all.
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Answer: A. The generator terminals through a transformer and controlled thyristor rectifier fed via slip rings
Static excitation takes power from the machine terminals, rectifies it with thyristors and feeds the field through slip rings, giving a fast response.
98. The synchronous speed of a 4-pole alternator generating 50 Hz is
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Use Ns = 120f/P.
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Answer: C. 1500 rpm
Ns = 120f/P = 120 × 50/4 = 1500 rpm.
99. A hydro-generator runs at 250 rpm to produce 50 Hz. The number of poles is
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Rearrange Ns = 120f/P.
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Answer: B. 24
P = 120f/N = 120 × 50/250 = 24.
100. A 6-pole alternator running at 1000 rpm generates a frequency of
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Use f = PN/120.
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Answer: A. 50 Hz
f = PN/120 = 6 × 1000/120 = 50 Hz.
101. An alternator has 100 turns per phase, a flux of 0.05 Wb per pole and a winding factor of 0.95. At 50 Hz the rms emf per phase is about
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Include the winding factor in the emf equation.
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Answer: D. 1055 V
E = 4.44 f Φ Nph Kw = 4.44 × 50 × 0.05 × 100 × 0.95 ≈ 1055 V.
102. The pitch factor of a winding whose coils span 150 electrical degrees is about
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kp = cos(α/2), where α is the angle by which the coil is short of full pitch.
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Answer: D. 0.966
Short-pitched by 30°, kp = cos(30°/2) = cos 15° ≈ 0.966.
103. A three-phase winding has 3 slots per pole per phase with a slot angle of 20 electrical degrees. Its distribution factor is about
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Use kd = sin(mβ/2) / (m sin(β/2)).
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Answer: B. 0.960
kd = sin(mβ/2)/(m sin(β/2)) = sin 30°/(3 sin 10°) ≈ 0.960.
104. Short-pitched coils are used in alternator armatures mainly to
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Think about waveform quality.
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Answer: B. Reduce harmonics in the emf waveform and save copper in end connections
Short pitching slightly reduces the fundamental but strongly suppresses selected harmonics and shortens the end connections.
105. An alternator's open-circuit voltage is 480 V and its full-load terminal voltage is 400 V at the same excitation and speed. Its voltage regulation is
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Regulation is referred to the full-load voltage.
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Answer: B. 20%
Regulation = (E0 - V)/V × 100 = (480 - 400)/400 × 100 = 20%.
106. At a certain field current, a star-connected alternator gives 400 V (line) on open circuit and 100 A on short circuit. The synchronous impedance per phase is about
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Use per-phase voltage for a star connection.
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Answer: A. 2.31 Ω
Zs = (open-circuit phase voltage)/(short-circuit current) = (400/√3)/100 ≈ 2.31 Ω.
107. The synchronous impedance (EMF) method of finding alternator regulation is called pessimistic because it
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It overestimates the effect of saturation-free reactance.
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Answer: B. Gives a value higher than the actual regulation
Zs is found from unsaturated short-circuit conditions and applied to a saturated machine, overestimating the drop and the regulation.
108. Armature reaction in an alternator supplying a purely inductive (zero pf lagging) load is
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Lagging current produces an mmf that opposes the main field axis.
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Answer: C. Demagnetising
At zero pf lagging the armature mmf is directly opposite the field mmf, so it weakens the field and lowers terminal voltage.
109. A generator with a 4% speed droop, rated at 50 Hz, has its frequency fall from no load to full load by
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Apply the droop percentage to rated frequency.
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Answer: A. 2 Hz
Droop is the frequency change from no load to full load as a percentage of rated: 0.04 × 50 = 2 Hz.
110. Which condition is NOT required for synchronising an alternator to the bus bars?
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Which item has nothing to do with matching waveforms?
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Answer: A. Equal kVA ratings of the incoming and running machines
Synchronising requires equal voltage, frequency, phase sequence and phase angle; kVA ratings need not be equal.
111. In the dark-lamp method of synchronising, the correct instant to close the switch is when
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Zero voltage across the lamps means the two voltages match.
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Answer: D. All the lamps are dark
With lamps across corresponding phases, zero voltage across them (all dark) means the machine and bus voltages are in phase.
112. An alternator connected to an infinite bus has its field excitation increased while the turbine input is unchanged. The result is that
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Which quantity does the prime mover control, and which does excitation control?
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Answer: D. Its reactive power output changes while active power stays the same
An infinite bus fixes voltage and frequency; excitation controls reactive power (and pf), while active power is set by the prime mover input.
113. To increase the active power delivered by an alternator connected to an infinite bus, one must
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Power must come from somewhere mechanically.
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Answer: B. Increase the prime mover input (steam or water supply)
Active power output depends on mechanical input; more input increases the load angle δ and the power P = EV sinδ/Xs.
114. A cylindrical-rotor generator has E = 1.2 pu, V = 1.0 pu and Xs = 1.0 pu (resistance neglected). When delivering 0.6 pu active power, its load angle is
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Use P = EV sinδ / Xs.
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Answer: D. 30°
P = EV sinδ/Xs, so sinδ = 0.6 × 1.0/(1.2 × 1.0) = 0.5 and δ = 30°.
115. The main losses in a synchronous generator include
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List every place where power is dissipated in the machine.
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Answer: B. Stator copper, core, friction and windage, and excitation losses
Synchronous machine losses are armature I²R, core (hysteresis + eddy), mechanical (friction, windage) and field excitation losses; there is no commutator.
116. A three-phase synchronous motor is not self-starting because
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The rotor sees poles sweeping past it rapidly in alternate directions.
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Answer: C. The average torque on the stationary rotor from the rotating field is zero
The stator field rotates at synchronous speed; the heavy rotor cannot follow, so it gets alternating pulls in opposite directions with zero average torque.
117. A common method of starting a synchronous motor is
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The rotor pole faces carry short-circuited bars.
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Answer: C. Starting it as an induction motor using damper windings, then applying DC excitation
Damper (amortisseur) bars act as a squirrel cage; the motor runs up near synchronous speed, then DC excitation pulls it into synchronism.
118. An over-excited synchronous motor running on no load is used as a
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Over-excitation makes the motor draw leading current.
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Answer: C. Synchronous condenser for power factor improvement
Over-excited, it draws leading current and supplies reactive power to the system, acting like a capacitor bank.
119. The V-curves of a synchronous motor are plots of
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The shape comes from a minimum at unity power factor.
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Answer: A. Armature current against field current at constant load
Armature current is minimum at unity pf (normal excitation) and rises for under- or over-excitation, giving a V shape.
120. Damper windings in synchronous machines help to
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Relative motion induces currents in them.
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Answer: B. Suppress hunting (rotor oscillations) and assist starting of motors
Induced currents in the damper bars oppose any relative motion between rotor and field, damping oscillations; they also provide starting torque.
121. Synchronous motors are preferred for applications that require
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The name gives away the speed behaviour.
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Answer: A. Constant speed independent of load, with possible power factor correction
They run at exactly synchronous speed for any load within their pull-out limit and can be over-excited to improve plant power factor (e.g. large compressors).
2.5 Three phase induction motors
30 questions · AElE0205
122. The rotor of a squirrel-cage induction motor consists of
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The name describes its shape.
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Answer: A. Bars short-circuited at both ends by end rings
Copper or aluminium bars lie in rotor slots and are permanently shorted by end rings, forming a cage.
123. The main advantage of a slip-ring (phase-wound) induction motor over a squirrel-cage motor is that
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What can be connected through the slip rings?
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Answer: B. External resistance can be added to the rotor circuit for high starting torque and limited starting current
Slip rings give access to the rotor winding, so external resistance improves starting torque and limits current, and can give some speed control.
124. Rotor bars of squirrel-cage motors are usually skewed in order to
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Think about the interaction between stator and rotor teeth.
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Answer: C. Reduce magnetic noise and prevent cogging
Skewing smooths out slot harmonics, reducing magnetic hum and the tendency of the rotor to lock (cog) with stator teeth.
125. The air gap of an induction motor is kept as small as mechanically possible in order to
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Air requires a lot of magnetising ampere-turns.
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Answer: D. Reduce the magnetising current and improve power factor
The air gap demands most of the magnetising mmf; a small gap lowers magnetising current and so raises the power factor.
126. The rotor of an induction motor rotates because
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It works like a transformer with a rotating secondary.
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Answer: B. Currents induced by the rotating stator field interact with that field
The rotating field cuts the rotor conductors, induces currents, and the resulting force drags the rotor in the direction of the field.
127. An induction motor cannot run at synchronous speed because
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What drives the rotor current?
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Answer: D. There would be no relative motion, hence no induced rotor current or torque
At synchronous speed the rotor conductors do not cut the field, so emf, current and torque all become zero.
128. A 4-pole, 50 Hz induction motor runs at 1440 rpm. Its slip is
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Find the synchronous speed first.
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Answer: A. 4%
Ns = 1500 rpm; s = (1500 - 1440)/1500 = 0.04.
129. The frequency of rotor current of a 50 Hz induction motor running with 4% slip is
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Rotor frequency is the slip times supply frequency.
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Answer: B. 2 Hz
f2 = s f = 0.04 × 50 = 2 Hz.
130. A 6-pole, 50 Hz induction motor runs with 5% slip. Its speed is
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Compute synchronous speed, then reduce by the slip.
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Answer: B. 950 rpm
Ns = 120 × 50/6 = 1000 rpm; N = 1000 × (1 - 0.05) = 950 rpm.
131. A 50 Hz induction motor has a full-load speed of 960 rpm. The number of poles is
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Running speed is a little below the synchronous speed.
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Answer: B. 6
The nearest synchronous speed above 960 rpm is 1000 rpm, which corresponds to 120 × 50/1000 = 6 poles.
132. The air-gap power of an induction motor is 10 kW and the slip is 4%. The rotor copper loss and gross mechanical power are
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Air-gap power splits in the ratio s : (1 - s).
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Answer: C. 400 W and 9.6 kW
Rotor copper loss = s Pag = 0.04 × 10 = 0.4 kW; mechanical power = (1 - s) Pag = 9.6 kW.
133. The standstill rotor emf of an induction motor is 100 V per phase. At a slip of 5%, the rotor emf per phase is
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Rotor emf is proportional to the relative speed.
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Answer: D. 5 V
E2s = s E2 = 0.05 × 100 = 5 V.
134. In the per-phase equivalent circuit of an induction motor, the mechanical load is represented by a resistance of
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Split R2'/s into the actual rotor resistance plus a load term.
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Answer: C. R2'(1 - s)/s
R2'/s = R2' + R2'(1 - s)/s; the first part is the rotor copper loss, and the second represents gross mechanical power.
135. Neglecting stator impedance, the slip at which an induction motor develops maximum torque is
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Maximum torque occurs when rotor resistance equals the slip-dependent rotor reactance.
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Answer: C. R2/X2
Differentiating the torque expression gives maximum torque when R2 = sX2, i.e. s = R2/X2.
136. Increasing the rotor resistance of a slip-ring induction motor
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Look at which terms R2 appears in.
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Answer: D. Increases the slip at maximum torque but leaves the maximum torque unchanged
Tmax ∝ 1/(2X2) is independent of R2, while sm = R2/X2 increases; this is used to raise starting torque.
137. In the low-slip region, the torque-slip characteristic of an induction motor is approximately
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At small slip, which term in the denominator dominates?
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Answer: B. Linear, with torque proportional to slip
For small s, sX2 is negligible compared with R2, so T ∝ s/R2, a straight line.
138. For maximum starting torque in a slip-ring induction motor, the rotor resistance per phase (referred) should equal
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Set the slip for maximum torque equal to 1.
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Answer: B. The standstill rotor reactance per phase
Maximum torque at s = 1 requires sm = R2/X2 = 1, i.e. R2 = X2.
139. The torque of an induction motor is proportional to the square of the supply voltage. If the voltage falls by 10%, the torque at a given slip falls to
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Square the voltage ratio.
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Answer: A. 81% of its original value
T ∝ V²: (0.9)² = 0.81.
140. A star-delta starter reduces the starting line current and starting torque of a motor, compared with direct-on-line starting, to
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Torque goes as voltage squared.
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Answer: C. One-third of their DOL values
Each winding gets 1/√3 of line voltage; phase current falls by 1/√3, line current by 1/3 (vs delta DOL) and torque by 1/3.
141. An autotransformer starter with a 60% tap gives a starting torque, as a fraction of the DOL starting torque, of
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Torque varies as the square of the applied voltage.
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Answer: D. 0.36
Motor voltage is 0.6 V, so torque ∝ (0.6)² = 0.36 of the DOL value; line current is also reduced to 0.36.
142. The direction of rotation of a three-phase induction motor can be reversed by
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What decides the direction of the rotating field?
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Answer: A. Interchanging any two supply leads
Swapping two phases reverses the phase sequence, so the rotating field and the motor reverse.
143. The no-load test on an induction motor gives
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With very small slip, what does the input power supply?
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Answer: A. Core and mechanical losses, and the magnetising branch parameters
At no load the slip is tiny, rotor current is negligible, and the input covers stator copper, core, friction and windage losses.
144. In a blocked-rotor test, the per-phase readings are 100 V, 10 A and 600 W. The equivalent reactance per phase is
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Find Z and R, then X.
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Answer: C. 8 Ω
Z = 100/10 = 10 Ω, R = 600/10² = 6 Ω, X = √(10² - 6²) = 8 Ω.
145. The blocked-rotor test of an induction motor is analogous to the
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With s = 1, the load resistance in the equivalent circuit becomes zero.
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Answer: A. Short-circuit test of a transformer
With the rotor locked, the 'secondary' is shorted and only a reduced voltage is applied, as in a transformer short-circuit test.
146. A 4-pole induction motor is fed from a variable-frequency drive at 40 Hz. Its synchronous speed is
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Synchronous speed is proportional to frequency.
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Answer: C. 1200 rpm
Ns = 120 × 40/4 = 1200 rpm.
147. In V/f control of an induction motor, the ratio V/f is kept constant in order to
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The emf equation links flux, voltage and frequency.
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Answer: B. Keep the air-gap flux nearly constant
Air-gap flux ∝ V/f; keeping it constant avoids saturation at low frequency and preserves torque capability.
148. Which speed control method is applicable only to slip-ring induction motors?
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Which method needs access to the rotor winding?
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Answer: A. Rotor resistance control
Adding external resistance to the rotor requires access through slip rings; it is not possible in a cage motor.
149. Pole-changing speed control is commonly used with
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The rotor must work with any number of poles.
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Answer: A. Squirrel-cage motors, giving a few fixed speeds
A cage rotor adapts to any stator pole number, so changing stator connections gives discrete speeds (e.g. 2:1).
150. When an induction machine is driven above synchronous speed by a prime mover while connected to the supply, it
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What is the sign of the slip above synchronous speed?
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Answer: D. Operates as an induction generator with negative slip
Above Ns the slip is negative; the torque reverses and the machine feeds active power to the supply (while still drawing reactive power).
151. The phenomenon of a squirrel-cage motor running stably at about one-seventh of its synchronous speed is called
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It is caused by a space harmonic.
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Answer: D. Crawling
Crawling is caused by the 7th space harmonic field, which can create a stable operating point near Ns/7.
2.6 Single phase induction motor
30 questions · AElE0206
152. A single-phase induction motor with only its main winding energised is not self-starting because
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Resolve the pulsating field into two rotating components.
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Answer: A. The single-phase winding produces a pulsating field with zero net starting torque
A pulsating field can be resolved into two equal fields rotating in opposite directions; at standstill their torques cancel.
153. The theory commonly used to explain the operation of a single-phase induction motor is the
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The pulsating field is split into two equal parts.
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Answer: A. Double revolving field theory
Double revolving field theory splits the pulsating mmf into forward and backward rotating fields of half amplitude.
154. Once a single-phase induction motor is started in one direction by hand with the auxiliary winding absent, it will
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After rotation begins the two field torques are no longer equal.
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Answer: A. Continue to run in the direction in which it was started
Once the rotor moves, the forward-field torque exceeds the backward-field torque, so the motor keeps running in that direction.
155. A 4-pole single-phase induction motor on a 50 Hz supply runs at 1425 rpm. Its slip with respect to the forward field is
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Compute the synchronous speed for 4 poles at 50 Hz.
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Answer: C. 0.05
Ns = 1500 rpm; s = (1500 - 1425)/1500 = 0.05.
156. A single-phase induction motor runs with a forward slip of 0.05. Its slip with respect to the backward rotating field is
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Measure the rotor speed relative to a field rotating the other way.
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Answer: D. 1.95
Backward slip = 2 - s = 2 - 0.05 = 1.95.
157. For a 50 Hz single-phase induction motor running at a forward slip of 0.05, the frequency of rotor currents produced by the backward field is
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Multiply the backward slip by the supply frequency.
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Answer: A. 97.5 Hz
Rotor frequency due to the backward field = (2 - s) f = 1.95 × 50 = 97.5 Hz.
158. In a resistance split-phase motor, the auxiliary (starting) winding has
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Its current must be closer in phase to the voltage than the main current.
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Answer: B. High resistance and low reactance compared with the main winding
The auxiliary winding uses fewer turns of thinner wire, making its current more in phase with the voltage than the main winding current, creating a phase split.
159. The centrifugal switch in a split-phase motor
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The starting winding is not designed for continuous duty.
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Answer: C. Disconnects the auxiliary winding at about 70-80% of synchronous speed
Once the motor has accelerated, the centrifugal switch opens the starting winding circuit, which is rated only for short-time duty.
160. If the starting winding of a split-phase motor is not disconnected after starting (switch stuck closed), the most likely result is
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What duty is the auxiliary winding designed for?
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Answer: C. The starting winding overheats and may burn out
The thin, high-resistance auxiliary winding is short-time rated; continuous current overheats it.
161. In a split-phase motor the main winding current lags the voltage by 40° and the auxiliary winding current lags by 15°. The phase difference between the two currents is
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Both currents lag; subtract the angles.
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Answer: D. 25°
Phase difference = 40° - 15° = 25°.
162. Starting torque of a two-winding single-phase motor is proportional to Im·Ia·sinα, where α is the angle between winding currents. For the same currents, changing α from 25° (split-phase) to 80° (capacitor start) increases starting torque by a factor of about
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Take the ratio of the sines, not of the angles.
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Answer: B. 2.33
Ratio = sin 80°/sin 25° = 0.985/0.423 ≈ 2.33.
163. A capacitor-start induction motor develops higher starting torque than a resistance split-phase motor mainly because
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Starting torque depends on sinα.
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Answer: A. The capacitor makes the phase difference between winding currents nearly 90°
The series capacitor makes the auxiliary current lead, giving a phase split approaching 90° and hence much larger starting torque.
164. A 2.5 µF capacitor used in a 50 Hz ceiling fan has a reactance of about
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Use Xc = 1/(2πfC) with C in farads.
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Answer: B. 1273 Ω
Xc = 1/(2πfC) = 1/(2π × 50 × 2.5×10⁻⁶) ≈ 1273 Ω.
165. Ceiling fans in homes commonly use a
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The capacitor stays in circuit all the time and there is no centrifugal switch.
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Answer: D. Permanent split-capacitor motor
A permanent split-capacitor (PSC) motor runs continuously with its capacitor in circuit, giving quiet operation, good pf and easy speed control.
166. Compared with a capacitor-start motor, a capacitor-start capacitor-run motor has
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The extra capacitor stays connected while running.
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Answer: B. Better running power factor and efficiency, with quieter operation
A run capacitor stays in circuit to improve running pf, efficiency and smoothness, while a larger start capacitor is cut out after starting.
167. A refrigerator compressor, which must start against a significant load, is usually driven by a
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Which single-phase motor has the highest starting torque among these?
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Answer: B. Capacitor-start induction motor
Capacitor-start motors give high starting torque (often 3-4.5 times full-load torque) for compressors and pumps.
168. The direction of rotation of a capacitor-start motor can be reversed by
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Interchanging the supply leads reverses both windings at once.
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Answer: D. Reversing the connections of either the main or the auxiliary winding
Reversing one winding reverses the phase relationship between the winding currents and hence the direction of the rotating field.
169. In a shaded-pole motor, the shading coil is
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It is a single turn of copper on part of the pole.
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Answer: B. A short-circuited copper ring around part of each pole
A copper band shorted on itself surrounds a portion of each salient pole; induced currents delay the flux in that portion.
170. A shaded-pole motor rotates in the direction from
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The flux in the shaded part is delayed.
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Answer: C. The unshaded part of the pole to the shaded part
Flux in the shaded portion lags that of the unshaded portion, so the field sweeps from the unshaded to the shaded part.
171. A disadvantage of the shaded-pole motor is
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Think about the permanently shorted copper rings.
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Answer: D. Very low starting torque and low efficiency
Continuous loss in the shading rings and weak phase split give low efficiency (often 5-35%) and low starting torque.
172. Shaded-pole motors are typically used in
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They are only suitable for very light starting loads.
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Answer: A. Small table fans, hair dryers and exhaust fans
They suit very small, low-torque loads where simplicity and cost matter more than efficiency.
173. The direction of rotation of a standard shaded-pole motor
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Direction is built into the pole construction.
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Answer: C. Cannot easily be reversed without changing the shading arrangement
Direction is fixed by the physical position of the shading rings; reversing requires turning the stator or using extra shaded windings.
174. A universal motor is essentially a
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It works on both AC and DC because field and armature reverse together.
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Answer: C. Series-wound commutator motor that runs on both AC and DC
Field and armature are in series, so both reverse together on AC and torque remains unidirectional; laminated field cores keep losses low.
175. Which appliance commonly uses a universal motor?
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Look for an appliance that needs very high speed.
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Answer: B. Mixer-grinder
Universal motors give high speed and high torque per weight, suiting mixers, drills and vacuum cleaners.
176. The synchronous speed of a 16-pole single-phase ceiling fan motor on a 50 Hz supply is
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Use Ns = 120f/P.
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Answer: A. 375 rpm
Ns = 120f/P = 120 × 50/16 = 375 rpm; the fan runs slightly below this.
177. The speed of a ceiling fan is most commonly controlled by
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A fan load's torque rises steeply with speed.
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Answer: D. Reducing the voltage applied to the motor with a regulator
Lower applied voltage lowers torque (∝ V²), so the fan settles at a lower speed with higher slip; electronic (triac) or capacitor-type regulators are used.
178. If a fan regulator reduces the applied voltage to 80% of rated, the motor torque at a given slip falls to about
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Torque varies as the square of applied voltage.
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Answer: D. 64% of its original value
Induction motor torque ∝ V²: 0.8² = 0.64.
179. The speed of a universal motor can be conveniently controlled by
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Its speed is not tied to supply frequency.
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Answer: B. Varying the applied voltage, e.g. with a triac or series resistance
Speed of a series motor depends on applied voltage; phase-angle (triac) control is common in drills and mixers.
180. Compared with a three-phase induction motor of the same rating, a single-phase induction motor generally has
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Think about the effect of the backward rotating field.
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Answer: A. Lower efficiency and lower power factor
The backward field causes extra losses and pulsating torque, so single-phase motors are bigger, less efficient and have poorer pf.
181. The torque produced by a single-phase induction motor while running
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The forward and backward fields rotate in opposite directions at the same speed.
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Answer: C. Pulsates at twice the supply frequency
Interaction of forward and backward fields gives a torque component at double supply frequency, causing vibration and noise.