Nepal Engineering Council · Electrical Engineering · Chapter 8
Utilization of Electrical Energy
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185 questions in 6 syllabus topics.
8.1 Illumination
32 questions · AElE0801
1. What is the SI unit of luminous flux?
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Think of the quantity that counts the total light output of a lamp.
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Answer: A. Lumen
Luminous flux is the total light energy emitted per second, judged by the eye's response, and is measured in lumens. Candela is intensity, lux is illuminance.
2. One candela of luminous intensity is equivalent to:
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Intensity relates to flux in a cone of directions.
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Answer: D. One lumen per steradian
Luminous intensity is flux per unit solid angle in a given direction, so 1 cd = 1 lm/sr.
3. The total solid angle subtended at the centre of a sphere by its whole surface is:
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Use surface area of a sphere divided by r².
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Answer: B. 4π steradian
Solid angle = area/r² = 4πr²/r² = 4π sr.
4. A lamp radiates a uniform luminous intensity of 100 cd in all directions. Its total luminous flux is about:
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Multiply intensity by the total solid angle around the lamp.
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Answer: B. 1257 lm
Flux = 4π × I = 4π × 100 ≈ 1257 lm.
5. A lamp of 200 cd hangs 4 m directly above a point on a table. The illuminance at that point is:
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Apply the inverse square law.
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Answer: C. 12.5 lux
E = I/d² = 200/4² = 12.5 lux (inverse square law, normal incidence).
6. A lamp of 400 cd (towards the point) is mounted 3 m above the floor. The horizontal illuminance on the floor at a point 4 m horizontally away from the foot of the lamp is:
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Combine the inverse square law with the cosine law.
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Answer: D. 9.6 lux
d = √(3²+4²) = 5 m, cos θ = 3/5. E = I cos θ / d² = 400 × 0.6 / 25 = 9.6 lux.
7. A 9 W LED lamp gives 900 lm. Its luminous efficacy is:
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Efficacy is output light per input watt.
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Answer: D. 100 lm/W
Efficacy = lumens/watts = 900/9 = 100 lm/W.
8. The filament of a general lighting incandescent lamp is made of tungsten mainly because tungsten has:
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The filament must glow white-hot without melting.
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Answer: C. A very high melting point and low vapour pressure
Tungsten melts at about 3400 °C and evaporates slowly, so the filament can run hot enough to emit useful visible light.
9. In a tungsten-halogen lamp, the halogen cycle:
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Think about what normally blackens an ordinary bulb.
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Answer: C. Returns evaporated tungsten to the filament and keeps the bulb clear
Evaporated tungsten combines with halogen near the hot bulb wall and the compound breaks down at the filament, redepositing tungsten and preventing blackening.
10. In a fluorescent tube, most of the visible light comes from:
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Mercury vapour discharge mostly emits invisible radiation.
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Answer: A. The phosphor coating excited by ultraviolet from the mercury discharge
The low-pressure mercury discharge emits mainly UV (253.7 nm); the phosphor coating on the tube converts it to visible light.
11. In a conventional fluorescent lamp circuit, the choke (ballast):
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A discharge lamp cannot limit its own current.
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Answer: B. Gives a high voltage surge to strike the tube and then limits current
When the starter opens, the collapsing choke current induces a high voltage to strike the arc; afterwards the choke limits the arc current because the discharge has negative resistance.
12. The capacitor connected across the supply terminals of a fluorescent lamp fitting is meant to:
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The choke draws lagging reactive current.
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Answer: C. Improve the power factor of the fitting
The choke makes the fitting strongly inductive (low lagging power factor); a shunt capacitor supplies the reactive power and raises the power factor.
13. The stroboscopic effect of fluorescent lighting in a workshop can be reduced by:
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Spread the light pulses out in time.
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Answer: A. Connecting adjacent fittings to different phases of the three-phase supply
Light from lamps on different phases peaks at different instants, so the combined light fluctuates much less. High-frequency electronic ballasts also help.
14. The electronic ballast of a CFL typically runs the lamp at a frequency of about:
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Higher frequency allows small magnetic parts and no flicker.
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Answer: B. Tens of kilohertz
CFL ballasts rectify the supply and invert it to roughly 20–50 kHz, which raises efficacy, removes visible flicker and allows a tiny ballast.
15. An LED produces light by:
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It is a diode operated with forward bias.
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Answer: D. Electroluminescence from carrier recombination in a forward-biased p-n junction
In a forward-biased junction, electrons recombine with holes and release energy as photons whose wavelength depends on the band gap.
16. Most white LEDs used for general lighting produce white light by:
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Phosphor conversion of a short-wavelength chip.
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Answer: A. A blue LED chip coated with a yellow phosphor
Blue light from an InGaN chip partly excites a yellow phosphor (e.g. YAG:Ce); the blue + yellow mixture is seen as white.
17. Which of the following light sources has the lowest luminous efficacy?
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Which source converts most of its input into heat?
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Answer: C. General lighting incandescent lamp
Incandescent lamps give roughly 10–17 lm/W since most input becomes heat; CFLs give about 50–70, T5 tubes around 90–100 and good LEDs over 100 lm/W.
18. An office 10 m × 8 m needs an average illuminance of 300 lux. Utilisation factor = 0.6, maintenance factor = 0.8 and each luminaire gives 3000 lm. The number of luminaires required is:
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Use the lumen method and round up.
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Answer: D. 17
N = E·A/(Φ·UF·MF) = 300 × 80/(3000 × 0.6 × 0.8) = 16.7, rounded up to 17.
19. A hall of 100 m² has 20 lamps of 2000 lm each. With utilisation factor 0.5 and maintenance factor 0.8, the average illuminance on the working plane is:
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Lumen method in reverse: total useful lumens over area.
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Answer: C. 160 lux
E = N·Φ·UF·MF/A = 20 × 2000 × 0.5 × 0.8/100 = 160 lux.
20. In illumination design, the utilisation factor (coefficient of utilisation) is the ratio of:
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It compares useful light to light produced.
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Answer: A. Lumens reaching the working plane to the total lumens emitted by the lamps
UF accounts for light lost to absorption by walls, ceiling and fittings before it reaches the working plane.
21. The maintenance factor used in lighting design accounts for:
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What happens to a lighting installation over months of use?
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Answer: B. Fall in illuminance due to dirt on fittings and ageing of lamps
Maintenance factor (inverse of depreciation factor) allows for lumen depreciation and dirt accumulation over time.
22. A lighting scheme in which 90–100 % of the light is directed upward towards the ceiling and reflected down is called:
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The ceiling becomes the light source.
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Answer: A. Indirect lighting
In indirect lighting the ceiling acts as the effective source, giving soft, glare-free but less efficient illumination.
23. For a luminaire with a maximum space-to-mounting-height ratio of 1.5 mounted 3 m above the working plane, the maximum spacing between luminaires is:
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Multiply the ratio by the height above the working plane.
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Answer: D. 4.5 m
Spacing = SHR × mounting height = 1.5 × 3 = 4.5 m.
24. Which measure helps reduce direct glare from luminaires in an office?
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Hide the bright lamp from the eye.
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Answer: D. Using luminaires with louvres or diffusers giving proper cut-off
Shielding the lamp with louvres/diffusers keeps high-luminance parts out of the normal field of view.
25. In flood lighting calculations, the waste light factor accounts for:
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Flood beams rarely fit the target exactly.
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Answer: B. Light falling outside the edges of the area to be lit
Some of the beam spills beyond the target surface; the waste light factor (typically about 1.2–1.5) increases the lumens required to allow for this spill.
26. Street lamps are spaced 30 m apart on a 10 m wide road. For an average illuminance of 10 lux with utilisation factor 0.4 and maintenance factor 0.8, the lumen output required per lamp is about:
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Each lamp lights one span of road.
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Answer: D. 9375 lm
Area per lamp = 30 × 10 = 300 m²; Φ = E·A/(UF·MF) = 10 × 300/(0.4 × 0.8) = 9375 lm.
27. Which lamp has very high luminous efficacy but gives nearly monochromatic yellow light with very poor colour rendering?
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Think of the yellow sodium D-line.
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Answer: B. Low-pressure sodium vapour lamp
Low-pressure sodium lamps emit almost entirely at 589 nm, giving efficacy up to about 150–200 lm/W but almost no colour discrimination.
28. For a wide road with heavy traffic, street lighting poles are usually arranged:
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Uniformity across a wide carriageway.
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Answer: A. On both sides of the road, opposite or staggered
One-side lighting is adequate only for narrow roads; wide roads need luminaires on both sides (opposite or staggered) for uniform illuminance.
29. The polar curve of a luminaire shows:
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It is plotted in polar coordinates of angle.
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Answer: B. Luminous intensity in different directions in a plane
A polar curve plots candle power (intensity) against angle, showing how the luminaire distributes light.
30. A lamp gives a total luminous flux of 1000 lm. Its mean spherical candle power (MSCP) is about:
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Divide total flux by the full-sphere solid angle.
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Answer: A. 79.6 cd
MSCP = total flux/4π = 1000/12.566 ≈ 79.6 cd.
31. Replacing a 60 W incandescent lamp with a 9 W LED lamp used 5 hours a day saves how much energy in 30 days?
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Power difference times total hours.
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Answer: C. 7.65 kWh
Saving = (60 − 9) W × 5 h × 30 = 7650 Wh = 7.65 kWh.
32. A typical recommended average illuminance for general office work is about:
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Much more than a street, much less than a surgical table.
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Answer: C. 300–500 lux
Lighting codes (e.g. CIBSE, IS 3646) recommend roughly 300–500 lux for general office tasks; 5–10 lux suits streets and thousands of lux only precision tasks.
8.2 Electrical design and estimation
31 questions · AElE0802
33. The electrical layout (wiring) plan of a building mainly shows:
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It is drawn over the architectural floor plan.
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Answer: D. Locations of light points, sockets, switches, distribution boards and circuit runs on the floor plan
The layout drawing marks every point, switch, board and wiring route on the architectural plan using standard symbols; it is the basis of the estimate.
34. A single-line diagram of a three-phase installation:
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The name describes how the phases are drawn.
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Answer: D. Represents all three phases by one line with standard symbols for equipment
A single-line (one-line) diagram simplifies a balanced three-phase system to one line, showing sources, transformers, switchgear, cables and loads.
35. The first step in preparing an estimate for the electrical installation of a building is to:
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You cannot count materials before knowing where things go.
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Answer: B. Study the building plan and mark the locations of all points and loads
Quantities of wire, conduit and accessories can only be worked out after the layout of points and circuits is fixed on the plan.
36. A schedule of materials (bill of quantities) in an electrical estimate normally lists:
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It must allow the cost to be computed item by item.
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Answer: B. Item description, quantity, unit, rate and amount
The BOQ itemises every material and work item with quantity, unit rate and amount, which add up to the estimated cost.
37. Which wiring system gives the best mechanical protection and neat appearance for modern residential and commercial buildings?
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The wiring is hidden inside the structure.
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Answer: A. Concealed conduit wiring
Cables drawn through conduits embedded in walls/slabs are protected from damage and moisture, can be rewired and are invisible.
38. Cleat wiring is generally suitable for:
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Cheap, quick and easily removed.
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Answer: D. Temporary installations
Cleat wiring is cheap and quick to install and remove, but has poor protection and appearance, so it is used for temporary work.
39. A single-pole switch controlling a lamp must be connected in the:
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When off, the fitting should not be live.
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Answer: B. Phase (line) conductor
Switching the phase ensures the lamp holder is dead when the switch is off, which is safer for lamp replacement.
40. As per the IEC harmonised colour code, the neutral and protective earth conductors are respectively:
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Brown is a phase colour.
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Answer: C. Blue and green-yellow
IEC 60445 colours: phase brown/black/grey, neutral blue, protective earth green-and-yellow.
41. For protection of persons against electric shock, a residual current circuit breaker (RCCB) usually has a rated residual operating current of:
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It must act well below the current that causes heart fibrillation.
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Answer: A. 30 mA
30 mA RCCBs trip before shock current through the body becomes dangerous; 100–300 mA devices are used for fire protection only.
42. An MCB whose magnetic (instantaneous) trip operates between 5 and 10 times its rated current is a:
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Recall the B, C, D trip bands.
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Answer: C. Type C MCB
IEC 60898: Type B trips at 3–5 In, Type C at 5–10 In, Type D at 10–20 In.
43. A 3 kW single-phase resistive water heater is fed at 230 V. The most suitable standard MCB rating for its circuit is:
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Find the current, then the next standard size above it.
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Answer: C. 16 A
I = 3000/230 ≈ 13 A, so the next standard rating above it, 16 A, is chosen; 10 A would trip and 63 A would not protect the wiring.
44. A 2.5 mm² copper cable (ρ = 0.0172 Ω·mm²/m) feeds a single-phase load 30 m away carrying 10 A. The voltage drop in the circuit (go and return) is about:
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Remember the current flows out and back.
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Answer: D. 4.13 V
R = ρ × 2L/A = 0.0172 × 60/2.5 = 0.413 Ω; drop = 10 × 0.413 = 4.13 V (about 1.8 % of 230 V).
45. A building has a connected load of 20 kW and a demand factor of 0.6. Its maximum demand is:
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Demand factor is always 1 or less.
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Answer: D. 12 kW
Maximum demand = demand factor × connected load = 0.6 × 20 = 12 kW.
46. The individual maximum demands of a group of consumers add up to 50 kW, while the maximum demand of the whole group is 40 kW. The diversity factor is:
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It is never less than 1.
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Answer: C. 1.25
Diversity factor = sum of individual maximum demands / maximum demand of group = 50/40 = 1.25.
47. A flat has 22 light and fan points. If each lighting sub-circuit may have at most 10 points, the minimum number of lighting sub-circuits is:
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Round up, not down.
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Answer: A. 3
22/10 = 2.2, rounded up to 3 sub-circuits.
48. An estimate has material cost Rs 2,00,000. Labour is taken as 20 % of material cost and contingency as 5 % of (material + labour). The total estimated cost is:
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Apply contingency on the subtotal, not on material alone.
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Answer: B. Rs 2,52,000
Material + labour = 2,40,000; contingency 5 % = 12,000; total = 2,52,000.
49. A balanced three-phase industrial load of 50 kW at 0.8 power factor lagging is supplied at 400 V. The line current is about:
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Three-phase power uses √3 and the power factor.
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Answer: A. 90.2 A
I = P/(√3 V cos φ) = 50000/(1.732 × 400 × 0.8) ≈ 90.2 A.
50. A factory load is 40 kW at 0.8 power factor. The kVA rating required of the supply transformer (ignoring margin) is:
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Apparent power is larger than real power at lagging pf.
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Answer: C. 50 kVA
S = P/cos φ = 40/0.8 = 50 kVA.
51. A house has 10 lamps of 40 W, 5 fans of 60 W, 4 light-plug sockets taken as 100 W each and 2 power sockets taken as 1000 W each. The total connected load is:
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Add every category, including both power sockets.
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Answer: B. 3100 W
400 + 300 + 400 + 2000 = 3100 W.
52. A PVC cable has a current rating of 100 A under standard conditions. With a grouping factor of 0.8 and an ambient temperature factor of 0.9, its derated current carrying capacity is:
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Apply both correction factors together.
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Answer: A. 72 A
Derated capacity = 100 × 0.8 × 0.9 = 72 A; correction factors multiply.
53. The main purpose of earthing the metal enclosures of electrical equipment is to:
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Think of what happens when a live wire touches the casing.
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Answer: C. Keep exposed metal near earth potential and provide a low-impedance fault path so protection operates
On an insulation failure, fault current flows through the earth path, the fuse/MCB/RCD operates, and touch voltage on the enclosure stays low.
54. Layers of charcoal and salt are placed around a pipe or plate earth electrode in order to:
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Lower resistance needs moist, conductive soil.
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Answer: B. Lower the soil resistivity around the electrode by retaining moisture
Salt makes the soil more conductive and charcoal holds moisture, both reducing earth resistance.
55. Two earth electrodes, each of 10 Ω resistance, are placed far enough apart that they do not interact and are connected in parallel. The combined earth resistance is about:
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Parallel resistors.
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Answer: A. 5 Ω
Well-separated electrodes act as parallel resistances: 10 × 10/(10 + 10) = 5 Ω.
56. The resistance of an earth electrode is usually measured with:
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It needs auxiliary spikes driven into the ground.
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Answer: B. An earth tester using the fall-of-potential method
An earth tester injects current through an auxiliary current electrode and measures voltage with a potential electrode (fall-of-potential/three-point method).
57. In Nepal, the National Building Code (NBC) is given legal force for building construction by the:
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It is a construction law, not a power-sector law.
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Answer: D. Building Act, 2055 BS
The Building Act 2055 (1998) provides for preparation and enforcement of the National Building Code; the other acts deal with the power sector.
58. Nepal's National Building Code NBC 207 deals with:
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It is the code in the series that concerns electrical engineers.
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Answer: B. Electrical design requirements for public buildings
NBC 207 covers electrical design requirements for public buildings; seismic design is NBC 105 and sanitary/plumbing is NBC 208.
59. Nepal Standards (NS) for electrical products such as cables and switches are issued by:
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Standards and measurement body.
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Answer: D. Nepal Bureau of Standards and Metrology
NBSM is the national standards body and issues NS marks; NEA is a utility, DoED handles licensing, and NEC registers engineers.
60. The IEC 60364 series of standards deals with:
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The standard used for building wiring rules.
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Answer: A. Low-voltage electrical installations
IEC 60364 covers design, erection and verification of LV installations; symbols are IEC 60617, transformers IEC 60076 and machines IEC 60034.
61. The standard low-voltage supply provided by NEA to consumers is:
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Line voltage = √3 × phase voltage.
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Answer: A. 230 V single-phase and 400 V three-phase at 50 Hz
Nepal uses 50 Hz with 230 V phase-to-neutral and 400 V line-to-line on LV distribution.
62. A lightning protection system for a building essentially consists of:
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Intercept, conduct, dissipate.
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Answer: C. Air terminals, down conductors and an earth termination system
Air terminals intercept the stroke, down conductors carry the current, and the earth termination dissipates it into the ground.
63. In an industrial complex, a motor control centre (MCC) is:
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It centralises motor switching and protection.
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Answer: C. An assembly of starters and protective devices for many motors fed from a common busbar
An MCC groups motor starters, contactors, overload relays and breakers in one enclosure with common busbars, simplifying distribution and maintenance.
8.3 Tariff schemes
31 questions · AElE0803
64. The basic objective of an electricity tariff is to:
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A utility must stay financially viable.
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Answer: C. Recover the cost of supplying energy plus a reasonable return on investment
A tariff must recover fixed and running costs of generation, transmission and distribution and give the utility a fair return.
65. Which of the following is a desirable characteristic of a tariff?
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A good tariff rewards behaviour that helps the system.
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Answer: D. It should encourage consumers to improve load factor and power factor
A good tariff is simple, fair, gives a proper return and gives incentives that help the system, such as higher load factor and power factor.
66. The annual fixed charges of a power supply undertaking depend mainly on:
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What decides how much plant must be built?
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Answer: C. The maximum demand (installed capacity)
Interest, depreciation and similar fixed charges arise from the capital cost of plant, which is sized for the maximum demand; running charges depend on units generated.
67. A main drawback of the simple (uniform) rate tariff, which charges a fixed rate per kWh, is that it:
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It looks only at units, not at demand.
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Answer: A. Does not distinguish between consumers with different load factors
Two consumers using the same kWh pay the same, although the one with higher maximum demand (lower load factor) costs the utility more.
68. A domestic consumer's tariff is: first 20 units at Rs 4 per unit, next 30 units at Rs 7 per unit and remaining units at Rs 9 per unit. The energy charge for 100 units is:
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Bill each slab separately.
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Answer: C. Rs 740
20 × 4 + 30 × 7 + 50 × 9 = 80 + 210 + 450 = Rs 740.
69. A block-rate tariff in which the rate per unit increases for higher consumption blocks is mainly used to:
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Higher use costs more per unit.
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Answer: B. Encourage energy conservation by heavy users
Inverted (increasing) block rates keep basic consumption cheap while making extra units costlier, which discourages waste.
70. A two-part (Hopkinson demand) tariff consists of:
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Fixed costs relate to demand, running costs to energy.
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Answer: B. A charge per kW of maximum demand plus a charge per kWh of energy
Two-part tariff: total = a × kW (max demand) + b × kWh; the demand part recovers fixed costs and the energy part running costs.
71. An industry is billed Rs 200 per kW of maximum demand per month plus Rs 8 per kWh. In a month its maximum demand is 50 kW and it uses 15,000 kWh. The monthly bill is:
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Add the demand and energy charges.
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Answer: D. Rs 1,30,000
Demand charge 200 × 50 = 10,000; energy charge 8 × 15,000 = 1,20,000; total Rs 1,30,000.
72. A disadvantage of the two-part tariff for small consumers is that:
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Look at the part of the bill that does not depend on kWh.
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Answer: B. They must pay the demand charge even if they use very little energy
The fixed (demand) component is payable regardless of consumption, which is a burden on small users, and a demand meter is required.
73. A three-part (Doherty) tariff consists of:
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Two-part tariff plus one more fixed element.
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Answer: B. A fixed customer charge, a maximum demand charge and an energy charge
Three-part tariff = a (fixed per consumer) + b × kW + c × kWh; the fixed part covers metering, billing and service costs.
74. Under a three-part tariff of Rs 100 per month fixed + Rs 150 per kW of maximum demand + Rs 7 per kWh, a consumer with 10 kW maximum demand and 2000 kWh consumption pays:
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There are three components to add.
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Answer: B. Rs 15,600
100 + 150 × 10 + 7 × 2000 = 100 + 1500 + 14,000 = Rs 15,600.
75. A factory has a maximum demand of 80 kW at 0.8 power factor and is charged Rs 250 per kVA of maximum demand per month. Its monthly demand charge is:
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Convert kW to kVA first.
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Answer: A. Rs 25,000
kVA = 80/0.8 = 100 kVA; demand charge = 100 × 250 = Rs 25,000.
76. Which tariff gives a consumer a direct financial incentive to improve power factor?
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kVA rises as power factor falls.
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Answer: B. A tariff with demand charge based on maximum kVA
At the same kW, lower pf means higher kVA and a higher bill, so improving pf reduces the kVA demand charge.
77. A consumer uses 3600 kWh in a 30-day month with a maximum demand of 10 kW. The monthly load factor is:
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Average load over 720 hours, divided by the peak.
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Answer: C. 0.50
Load factor = average load/maximum demand = (3600/720)/10 = 5/10 = 0.5.
78. For a consumer on a two-part tariff, improving the load factor (same maximum demand) results in:
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Same demand charge, more units.
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Answer: D. A lower average cost per kWh
The fixed demand charge is spread over more units, so the overall cost per kWh falls.
79. A consumer on a tariff of Rs 300 per kW of maximum demand per month plus Rs 6 per kWh has a maximum demand of 20 kW and a load factor of 0.5 in a 30-day month. The average cost per kWh is about:
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Find the energy from load factor first.
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Answer: A. Rs 6.83
Energy = 20 × 0.5 × 720 = 7200 kWh; bill = 6000 + 43,200 = 49,200; average = 49,200/7200 ≈ Rs 6.83/kWh.
80. The main purpose of a time-of-day (TOD) tariff is to:
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Price signals that vary with the clock.
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Answer: A. Encourage consumers to shift load from peak hours to off-peak hours
TOD tariffs charge more per kWh in peak hours and less in off-peak hours, flattening the system load curve.
81. Under a TOD tariff, an industry uses 1000 kWh at peak (Rs 12/kWh), 3000 kWh at normal (Rs 9/kWh) and 2000 kWh at off-peak (Rs 5/kWh). The energy charge is:
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Multiply each period's energy by its rate.
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Answer: A. Rs 49,000
12,000 + 27,000 + 10,000 = Rs 49,000.
82. Tariff A charges Rs 10 per kWh. Tariff B charges Rs 1000 per month plus Rs 8 per kWh. Both give the same monthly bill at a consumption of:
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Set the two bills equal.
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Answer: C. 500 kWh
10x = 1000 + 8x gives x = 500 kWh; above this Tariff B is cheaper.
83. The low-priced first block (lifeline rate) in a domestic electricity tariff is mainly intended to:
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Who uses only the first few units each month?
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Answer: D. Keep basic electricity affordable for low-consumption households
A small first block at a low (often subsidised) rate covers essential needs such as lighting, so poor households can afford electricity, while higher slabs pay more.
84. A maximum demand meter with a 30-minute integration period records 30 kWh in its highest half-hour of the month. The maximum demand registered is:
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Demand is average power over the interval.
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Answer: C. 60 kW
Demand = energy in the interval/interval length = 30 kWh/0.5 h = 60 kW.
85. A flat-demand tariff, which charges according to the connected load (e.g., number of lamps) without metering, is best suited to:
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Load and hours are known in advance.
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Answer: B. Street lighting and similar fixed, predictable loads
Where the load and its hours of use are fixed and known, metering can be avoided and the bill based on connected load.
86. Domestic consumers of NEA are billed with a minimum charge that depends on the meter's ampere capacity and with energy charges that:
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Nepal uses increasing block slabs for households.
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Answer: C. Increase in rate for higher consumption slabs
NEA's domestic tariff has categories by meter capacity (e.g. 5 A, 15 A, 30 A, 60 A) and slab-wise energy rates that rise with consumption.
87. In Nepal, electricity tariffs are at present approved by the:
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An independent regulator.
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Answer: B. Electricity Regulatory Commission
The ERC, formed under the Electricity Regulatory Commission Act 2074, determines and approves retail tariffs of NEA and other distributors.
88. The Electricity Regulatory Commission of Nepal was established under the:
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The regulator has its own Act, enacted recently.
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Answer: C. Electricity Regulatory Commission Act, 2074 BS
The ERC Act 2074 (2017) created the ERC, taking over tariff fixation from the earlier Electricity Tariff Fixation Commission.
89. Under the Electricity Act, 2049 BS, a licence is NOT required for generating electricity up to:
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The threshold is one thousand kilowatts.
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Answer: D. 1 MW
The Act exempts generation up to 1000 kW from licensing; the promoter only needs to inform the prescribed authority.
90. Survey and generation licences for hydropower projects in Nepal are generally issued by the:
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A government department under the energy ministry.
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Answer: D. Department of Electricity Development
DoED, under the Ministry of Energy, Water Resources and Irrigation, processes and issues licences under the Electricity Act 2049; ERC regulates tariffs and NEA is a utility.
91. Private hydropower projects in Nepal are developed under generation licences on a build-own-operate-transfer (BOOT) basis. When the licence term ends, the project is:
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The last letter of BOOT stands for Transfer.
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Answer: B. Handed over to the Government of Nepal
Under BOOT the developer builds, owns and operates the plant during the licence term and recovers its investment from power sales; at the end of the term the project is transferred to the Government of Nepal (in working condition), as provided under the Electricity Act 2049.
92. The detailed rules for implementing the Electricity Act, 2049 BS are given in the:
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It was issued one year after the Act.
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Answer: A. Electricity Regulation, 2050 BS
The Electricity Regulation 2050 (1993) was framed under the Electricity Act 2049 to implement licensing, safety and other provisions.
93. The Nepal Electricity Authority (NEA) was established under the:
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NEA has its own Act from the 1980s.
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Answer: A. Nepal Electricity Authority Act, 2041 BS
NEA was created in 2042 BS (1985) under the NEA Act 2041 by merging the earlier electricity department, corporation and boards.
94. A maximum demand meter registers:
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It averages over an interval.
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Answer: A. The highest average power over a fixed demand interval in the billing period
Demand meters average power over an interval (commonly 15 or 30 minutes) and record the highest such value, so brief starting surges are not counted.
8.4 Electric drives and motor selection
30 questions · AElE0804
95. In an electric drive system, the function of the power modulator (converter) is to:
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It sits between the source and the motor.
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Answer: C. Regulate the flow of power from the source to the motor
The power modulator (rectifier, chopper, inverter, starter etc.) shapes voltage, current and frequency supplied to the motor as commanded by the control unit.
96. In a group drive, a major disadvantage is that:
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Many machines depend on one motor.
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Answer: D. A fault in the single motor stops all the machines driven by it
In a group drive one motor drives several machines through a line shaft; its failure halts all of them and the shafting wastes power.
97. A drive in which several motors drive different parts of the same machine, such as the hoist, trolley and bridge of an overhead crane, is called a:
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One machine, many motors.
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Answer: C. Multi-motor drive
Multi-motor drives use separate motors for separate motions of one machine, e.g. cranes and rolling mills.
98. Which of the following load torques is an active load torque?
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Which torque could drive the motor by itself?
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Answer: B. Torque due to gravity on the load of a hoist
Active torques (gravity, compression, deformation) can drive the motor and keep their sign when the direction reverses; passive torques such as friction and windage always oppose motion.
99. For a centrifugal fan or pump, the load torque varies approximately as:
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Power for fans varies as the cube of speed.
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Answer: D. The square of the speed
Fan/pump torque ∝ N², so the power required ∝ N³.
100. Loads such as coilers and winders, in which torque varies inversely with speed, are called:
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Multiply torque by speed.
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Answer: D. Constant-power loads
If T ∝ 1/N, then P = Tω is constant.
101. A hoist lowering a heavy load at steady speed, with the motor developing torque upward to hold the load back, is operating in:
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Speed is negative but motor torque still points upward.
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Answer: B. Quadrant IV (reverse braking)
Taking hoisting as positive speed, lowering has negative speed while motor torque remains positive (upward): negative speed, positive torque = quadrant IV, a braking mode.
102. An equilibrium operating point of a motor-load system is steady-state stable if, at that point:
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A small speed rise must produce a decelerating torque.
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Answer: A. dT_L/dω > dT_m/dω
If speed rises slightly, load torque must exceed motor torque to bring it back: the load torque curve must have the greater slope.
103. A motor delivers 100 N·m at 1450 rpm. The mechanical power output is about:
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Convert rpm to rad/s.
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Answer: C. 15.2 kW
P = Tω = 100 × (2π × 1450/60) = 100 × 151.8 ≈ 15.2 kW.
104. A 7.5 kW motor runs at 960 rpm at full load. Its full-load torque is about:
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T = P/ω with ω in rad/s.
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Answer: B. 74.6 N·m
T = P/ω = 7500/(2π × 960/60) = 7500/100.5 ≈ 74.6 N·m.
105. A pump delivers 0.05 m³/s of water against a total head of 20 m with a pump efficiency of 75 %. The motor output power required is about:
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Hydraulic power divided by efficiency.
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Answer: A. 13.1 kW
P = ρgQH/η = 1000 × 9.81 × 0.05 × 20/0.75 ≈ 13.1 kW.
106. A hoist raises a mass of 1000 kg at 1 m/s. If the mechanical efficiency of the hoist is 80 %, the motor power required is about:
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Force × speed, divided by efficiency.
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Answer: D. 12.3 kW
P = mgv/η = 1000 × 9.81 × 1/0.8 ≈ 12.26 kW.
107. A motor of moment of inertia 0.5 kg·m² drives a load of 10 kg·m² through a gear that reduces speed by 5:1. The equivalent inertia referred to the motor shaft is:
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Load inertia is divided by the square of the gear ratio.
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Answer: B. 0.9 kg·m²
J_eq = J_m + J_L/a² = 0.5 + 10/25 = 0.9 kg·m², where a = 5 is the speed ratio.
108. A motor operates a repeating 60 s cycle: 100 N·m for 10 s, 50 N·m for 20 s and no load for 30 s. Ignoring changes in cooling, the RMS (equivalent) torque is:
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Square, average over the whole cycle, then root.
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Answer: D. 50.0 N·m
T_rms = √[(100² × 10 + 50² × 20 + 0)/60] = √(150,000/60) = √2500 = 50 N·m.
109. A drive has total inertia 2 kg·m² and a constant accelerating torque of 40 N·m. The time to accelerate from rest to 1500 rpm is about:
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T = J dω/dt with ω in rad/s.
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Answer: C. 7.9 s
ω = 2π × 1500/60 = 157.1 rad/s; t = Jω/T = 2 × 157.1/40 ≈ 7.85 s.
110. A conveyor needs a pull of 2000 N at a belt speed of 2 m/s. With a drive efficiency of 90 %, the motor rating required is about:
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Force times speed, then allow for efficiency.
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Answer: A. 4.4 kW
P = Fv/η = 2000 × 2/0.9 ≈ 4.44 kW.
111. A motor has a final steady temperature rise of 60 °C and a heating time constant of 60 min. Starting cold, its temperature rise after 60 min of continuous full load is about:
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After one time constant the rise reaches about 63 % of final.
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Answer: D. 37.9 °C
θ = θ_f(1 − e^(−t/τ)) = 60(1 − e^(−1)) = 60 × 0.632 ≈ 37.9 °C.
112. The heating time constant of an electric motor is the time taken for its temperature rise to reach:
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Evaluate 1 − e^(−1).
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Answer: A. 63.2 % of its final steady value
With θ = θ_f(1 − e^(−t/τ)), at t = τ the rise is 1 − e⁻¹ = 0.632 of final.
113. A motor rated for continuous duty (S1) is used on short-time duty (S2) with long rest periods. For the short operating period it can deliver:
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It does not have time to reach final temperature.
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Answer: A. More than its continuous rating without overheating
Because the operating time is shorter than the time needed to reach final temperature, the motor can carry overload for that period and cool during rest.
114. The motor traditionally preferred for traction and crane drives needing very high starting torque and speed that falls with load is the:
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Field winding carries the armature current.
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Answer: C. DC series motor
Torque of a DC series motor ∝ I² (before saturation), giving very high starting torque, and its speed adjusts naturally with load.
115. For constant-speed drives such as pumps, fans and compressors without speed control, the most commonly selected motor is the:
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Rugged, cheap and almost maintenance-free.
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Answer: D. Three-phase squirrel-cage induction motor
Squirrel-cage motors are rugged, cheap, need little maintenance and run at nearly constant speed.
116. A slip-ring induction motor is preferred over a squirrel-cage motor when the drive needs:
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What can you add through the slip rings?
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Answer: C. High starting torque with low starting current, using rotor resistance
External rotor resistance raises starting torque and limits starting current; also gives limited speed control, useful for cranes and hoists.
117. A large synchronous motor is often chosen for a big compressor drive because it:
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Over-excitation gives a useful side benefit.
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Answer: A. Runs at constant speed and can improve the plant power factor
A synchronous motor runs at constant speed and, when over-excited, draws leading current, correcting the plant power factor.
118. In variable-frequency (V/f) control of an induction motor below base speed, the ratio V/f is kept constant in order to:
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Flux is proportional to voltage divided by frequency.
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Answer: B. Keep the air-gap flux constant
Air-gap flux ∝ V/f; constant V/f keeps flux and torque capability constant and avoids saturation at low frequency.
119. Regenerative braking of an induction motor occurs when:
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The machine becomes a generator.
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Answer: B. The rotor is driven above synchronous speed and power is returned to the supply
Above synchronous speed slip becomes negative; the machine works as a generator, feeding energy back to the supply.
120. Plugging (reverse current braking) of an induction motor is done by:
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Reverse the rotating field.
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Answer: A. Interchanging two supply phases while the motor is running
Reversing the phase sequence reverses the rotating field; the large opposing torque stops the motor quickly, but current and losses are very high.
121. In dynamic (rheostatic) braking, the kinetic energy of the drive is:
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The energy is not reused.
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Answer: B. Dissipated as heat in a braking resistor
The motor acts as a generator loaded by a resistor, so the stored energy is burnt in the resistor.
122. A motor runs at 1440 rpm and must drive a machine at 288 rpm. The required gear reduction ratio is:
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Divide motor speed by load speed.
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Answer: B. 5 : 1
Ratio = 1440/288 = 5, a 5 : 1 reduction from motor to load.
123. A 4-pole, 50 Hz induction motor runs with 4 % slip. Its rotor speed is:
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Start from synchronous speed.
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Answer: C. 1440 rpm
N_s = 120f/P = 1500 rpm; N = N_s(1 − s) = 1500 × 0.96 = 1440 rpm.
124. The IP rating (e.g., IP55) of a motor enclosure specifies:
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Ingress Protection.
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Answer: A. Degree of protection against ingress of solid objects and water
IP code: first digit gives protection against solids/dust, second against water; it guides motor selection for the site environment.
8.5 Electric heating
31 questions · AElE0805
125. Which of the following is an advantage of electric heating over fuel-fired heating?
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Think cleanliness and control.
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Answer: A. Accurate temperature control with no products of combustion
Electric heating is clean, needs no flue, allows precise and uniform temperature control and high efficiency, though energy cost may be higher.
126. In direct resistance heating:
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Where is the I²R loss produced?
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Answer: C. Current is passed through the charge itself, which acts as the resistance
Examples are salt-bath furnaces and electrode boilers, where the charge (or bath) carries the current and heats directly.
127. An electric oven in which heat produced in nichrome elements reaches the charge by radiation and convection uses:
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The current does not pass through the charge.
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Answer: C. Indirect resistance heating
The current flows only in the heating elements; the charge is heated indirectly by transfer from them.
128. A desirable property of a material for resistance heating elements is:
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You want a compact element whose power stays steady when hot.
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Answer: A. High resistivity and a low temperature coefficient of resistance
High resistivity keeps the element short; low temperature coefficient keeps power steady; high melting point and oxidation resistance give long life.
129. Nichrome, widely used for heating elements up to about 1150 °C, is an alloy of:
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The name gives it away.
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Answer: A. Nickel and chromium
Typical nichrome is about 80 % nickel and 20 % chromium; it resists oxidation and has high resistivity.
130. A non-metallic heating element suitable for furnace temperatures of about 1400 °C in air is made of:
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A ceramic-like conducting compound.
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Answer: B. Silicon carbide
Silicon carbide (globar) rods work in air up to about 1500 °C, beyond the range of nickel-chromium alloys.
131. Three identical heating elements are switched from delta to star on the same three-phase supply. The power in star is:
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Power varies as voltage squared.
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Answer: A. One-third of the power in delta
In star each element gets V/√3 instead of V, so power per element falls to (1/√3)² = 1/3.
132. A heating element is rated 2 kW at 230 V. Its hot resistance is about:
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Use R = V²/P.
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Answer: C. 26.45 Ω
R = V²/P = 230²/2000 = 26.45 Ω.
133. A 3 kW immersion heater with 90 % efficiency heats 50 kg of water from 15 °C to 85 °C (c = 4186 J/kg°C). The time taken is about:
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Find heat in kWh, allow for efficiency, then divide by power.
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Answer: B. 1.51 h
Heat = 50 × 4186 × 70 = 14.65 MJ = 4.07 kWh; input = 4.07/0.9 = 4.52 kWh; time = 4.52/3 ≈ 1.51 h.
134. A 10 kW resistance oven takes 1 hour to heat a charge that absorbs 30 MJ. The oven efficiency is about:
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Convert MJ to kWh (1 kWh = 3.6 MJ).
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Answer: D. 83.3 %
Useful heat = 30 MJ = 8.33 kWh; input = 10 kWh; efficiency = 8.33/10 = 83.3 %.
135. If the temperature of a radiating heating element is raised from 1000 K to 1200 K (surroundings much cooler), the heat radiated increases by a factor of about:
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Stefan–Boltzmann law uses the fourth power.
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Answer: C. 2.07
Radiated power ∝ T⁴ (Stefan's law): (1200/1000)⁴ = 1.2⁴ ≈ 2.07.
136. In a direct arc furnace used for steel making:
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The charge is part of the electric circuit.
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Answer: B. The arc is struck between the electrodes and the charge, so current flows through the charge
In direct arc furnaces the charge is part of the arc circuit, giving intense heat suitable for steel melting.
137. An indirect arc furnace, often of the rocking type, is mainly used for melting:
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The charge is not in the arc circuit.
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Answer: D. Non-ferrous metals such as brass and copper
The arc between two electrodes heats the charge by radiation; rocking spreads heat and protects the lining. It suits non-ferrous metals.
138. Induction heating of a metal charge is based on:
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Transformer action without contact.
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Answer: D. Eddy current and hysteresis losses induced by an alternating magnetic field
The charge acts as a short-circuited secondary; induced eddy currents (and hysteresis in magnetic materials) heat it.
139. Coreless induction furnaces are operated at higher frequencies (hundreds of Hz to several kHz) mainly because:
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Induced emf is proportional to frequency.
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Answer: A. Without an iron core, high frequency is needed to induce enough eddy current heating in the charge
With weak magnetic coupling, induced emf and eddy loss must be raised by increasing frequency.
140. A limitation of the core-type (e.g., Ajax-Wyatt) induction furnace is that:
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The charge itself is the secondary winding.
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Answer: C. A closed loop of molten metal must be kept to start the next heat
The charge forms the secondary loop; a continuous molten ring is needed, so it cannot start from cold solid charge pieces easily and is used for continuous operation.
141. If the frequency of the current in an induction heater is increased four times, the depth of current penetration (skin depth) in the charge becomes:
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Depth varies with the inverse square root of frequency.
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Answer: D. Half
Skin depth δ ∝ 1/√f, so 4 × f gives δ/2.
142. Surface hardening of steel gear teeth is best done by:
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Use the skin effect.
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Answer: B. High-frequency induction heating
High frequency confines the induced current to a thin surface layer, heating it quickly while the core stays cool.
143. Dielectric heating is used for heating:
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The material is placed between capacitor plates.
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Answer: A. Non-conducting materials such as wood, plastics and glue
In dielectric heating an insulating material between electrodes is heated by dielectric losses in a high-frequency field.
144. In dielectric heating, if the applied voltage is doubled with frequency unchanged, the heat produced becomes:
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Power depends on V squared.
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Answer: C. Four times
Power P = ωCV² tan δ, so P ∝ V²; doubling V quadruples power.
145. Domestic microwave ovens typically operate at a frequency of about:
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Gigahertz, not megahertz.
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Answer: B. 2.45 GHz
Microwave ovens use the 2.45 GHz ISM band, which is readily absorbed by water molecules in food.
146. Infrared heating with special lamps is commonly used for:
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It heats surfaces, not bulk.
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Answer: A. Drying paint and varnish on surfaces
Infrared radiation heats surfaces quickly, ideal for drying painted or varnished surfaces.
147. The main purpose of an electrical energy audit in an industry is to:
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It ends with recommendations to save energy.
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Answer: B. Find where energy is used and wasted and identify cost-effective saving measures
An energy audit measures and analyses consumption to find losses and recommend conservation measures with their payback.
148. A plant draws 100 kW at 0.7 power factor lagging. The capacitor kVAR needed to raise the power factor to 0.95 lagging is about:
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Difference of the tangents of the two angles.
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Answer: A. 69.2 kVAR
Q = P(tan φ₁ − tan φ₂) = 100(1.020 − 0.329) ≈ 69.2 kVAR.
149. A fan's speed is reduced to 70 % of rated using a variable-speed drive. Its power demand (ignoring drive losses) falls to about:
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Affinity law: power varies as the cube of speed.
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Answer: C. 34 % of rated
Fan power ∝ N³: 0.7³ = 0.343, i.e. about 34 %.
150. A 15 kW (output) motor running 4000 h per year is replaced by one of 92 % efficiency instead of 88 %. The annual energy saved is about:
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Compare input powers, not efficiencies directly.
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Answer: B. 2960 kWh
Inputs: 15/0.88 = 17.05 kW and 15/0.92 = 16.30 kW; saving 0.74 kW × 4000 h ≈ 2960 kWh.
151. If the power factor of a feeder carrying a fixed kW load at constant voltage is improved from 0.7 to 1.0, the I²R loss in the feeder becomes about:
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Current falls in proportion to the power factor rise.
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Answer: C. 49 % of the original
I ∝ 1/pf, so loss ∝ 1/pf²: (0.7/1.0)² = 0.49.
152. Flow from a pump is controlled by partly closing a throttle valve. The most effective energy conservation measure is to:
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Remove the wasteful pressure drop.
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Answer: B. Control flow by varying the pump speed with a variable-frequency drive
Throttling wastes energy across the valve; reducing speed cuts power roughly as the cube of speed.
153. In demand-side management, valley filling means:
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Fill the low parts of the load curve.
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Answer: D. Building up load during off-peak periods, e.g. night-time storage heating or EV charging
Valley filling adds load in low-demand periods, improving load factor and utilisation of plant.
154. A heat pump saves energy for space heating compared with a resistance heater because:
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It moves heat rather than making it.
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Answer: D. It moves heat from the surroundings, giving a coefficient of performance greater than 1
A heat pump delivers several kWh of heat per kWh of electricity (COP typically 2–4) by pumping heat from outside air or ground.
155. Which of the following is an energy conservation measure in building lighting?
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Light only when and where it is needed.
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Answer: D. Occupancy sensors and daylight-linked dimming
Switching or dimming lights when spaces are empty or daylight is sufficient reduces lighting energy without loss of service.
8.6 Electric traction
30 questions · AElE0806
156. A major disadvantage of electric traction compared with diesel traction is:
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Think of what must be built along the route.
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Answer: C. High initial cost of the overhead supply system and dependence on it
Electric traction is clean, has high starting torque and can regenerate, but needs costly fixed electrification and stops if supply fails.
157. A diesel-electric locomotive is an example of:
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It carries its own power source but drives the wheels electrically.
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Answer: A. Self-contained electric traction, where a diesel engine drives a generator feeding traction motors
It carries its own prime mover; the electric transmission gives smooth, flexible control of the traction motors.
158. The track electrification system most widely adopted for new main-line railways today is:
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Fed straight from the public grid frequency.
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Answer: A. 25 kV, 50 Hz single-phase AC overhead
25 kV, 50 Hz AC can be fed directly from the public grid, needs fewer substations and lighter overhead conductors, so it is the standard for new main lines.
159. Tramways in cities are usually supplied with:
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Low-voltage DC, using the rails for return.
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Answer: C. 600–750 V DC through an overhead wire with rail return
Low-voltage DC is safe for street running and suits DC series motors; the running rails serve as the return path.
160. A trolley bus needs two overhead contact wires because:
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What does a tram use for current return?
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Answer: B. Its rubber tyres cannot provide a return path through rails
Unlike a tram, a trolley bus runs on the road, so both supply and return must be via overhead wires collected by two trolley poles.
161. In a 25 kV AC locomotive that uses DC series traction motors, the AC supply is:
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AC for distribution, DC for the motors.
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Answer: B. Stepped down by an on-board transformer and rectified to DC
Such rectifier locomotives combine the advantages of AC distribution with the good traction characteristics of DC series motors.
162. The current collector used on high-speed electric trains with overhead catenary is the:
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A spring-loaded hinged frame on the roof.
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Answer: D. Pantograph
A pantograph keeps steady contact with the contact wire at high speed; trolley poles and bow collectors are for low-speed trams/trolley buses.
163. The DC series motor has traditionally been preferred for traction because:
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Consider torque versus current for a series field.
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Answer: B. It gives high starting torque and its speed falls naturally as load rises
Torque ∝ I² at starting gives strong acceleration; the falling speed-torque characteristic suits traction and shares load between motors.
164. Modern electric locomotives and EMUs mostly use:
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Power electronics plus a rugged AC motor.
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Answer: A. Three-phase induction motors fed from VVVF inverters
Inverter-fed squirrel-cage motors are rugged, low-maintenance, give smooth speed control and easy regenerative braking.
165. With two identical DC series motors, series-parallel starting raises the theoretical starting efficiency from 50 % (plain rheostatic starting) to about:
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Less energy is wasted in the starting resistance.
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Answer: D. 66.7 %
With plain rheostatic starting half the supply energy is lost in the starting resistance (efficiency 50 %). With series-parallel starting the resistance loss falls to one third of the input energy, so starting efficiency = 2/3 ≈ 66.7 %.
166. Field weakening (field diverter) on a DC series traction motor is used to:
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Speed is inversely proportional to flux.
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Answer: C. Increase the speed above that at full field
Reducing flux raises speed (N ∝ E/φ), giving higher running speeds after series-parallel transition.
167. During coasting on the speed-time curve of a train:
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No power, no braking.
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Answer: A. Power is switched off and the train runs on its momentum with slowly falling speed
Coasting saves energy; speed falls slowly due to train resistance before braking begins.
168. A train covers 2 km between stations in 3 minutes of running time and stops for 30 s at each station. Its schedule speed is about:
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Include stop time.
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Answer: B. 34.3 km/h
Schedule speed = distance/(running + stop time) = 2/(3.5/60) ≈ 34.3 km/h (average speed would be 40 km/h).
169. For the same run, the schedule speed of a train compared with its average speed is:
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Which one includes the time standing at stations?
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Answer: B. Always lower, because it includes stop time
Average speed uses running time only; schedule speed adds station stop time, so it is smaller.
170. A train accelerates uniformly at 2 km/h/s to 60 km/h. The distance covered during acceleration is:
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Area of the triangle under the speed-time curve.
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Answer: D. 0.25 km
Time = 60/2 = 30 s; distance = ½ × (60/3600 km/s) × 30 = 0.25 km.
171. A train accelerates at 1.8 km/h/s for 25 s from rest. Its speed at the end of acceleration is:
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Speed = acceleration × time, in matching units.
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Answer: D. 45 km/h
V = αt = 1.8 × 25 = 45 km/h.
172. A train of 200 t with 10 % allowance for rotating masses (effective mass 220 t) accelerates at 1.5 km/h/s on level track. The tractive effort for acceleration alone is about:
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1 km/h/s = 0.2778 m/s².
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Answer: B. 91.7 kN
F = 277.8 × We × α = 277.8 × 220 × 1.5 ≈ 91,670 N, using effective mass (83.3 kN would ignore rotating masses).
173. The tractive effort required to overcome gravity for a 300 t train on a 1 % up gradient is about:
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Gradient of 1 % means a rise of 1 m in 100 m.
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Answer: C. 29.4 kN
F = W g sin θ ≈ 300,000 × 9.81 × 0.01 = 29.4 kN (98.1 N per tonne per % gradient).
174. A 300 t train has a specific train resistance of 50 N per tonne. The tractive effort to overcome train resistance is:
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Multiply by the mass in tonnes.
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Answer: C. 15 kN
F = 300 × 50 = 15,000 N = 15 kN.
175. A locomotive exerts a tractive effort of 50 kN at 36 km/h. The power at the wheels is:
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Convert km/h to m/s.
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Answer: A. 500 kW
P = F × v = 50,000 × 10 m/s = 500 kW.
176. A locomotive has an adhesive weight of 80 t and the coefficient of adhesion is 0.25. The maximum tractive effort it can exert without slipping is about:
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Friction-like limit: μ times the weight on driving wheels.
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Answer: B. 196 kN
F_max = μ × W × g = 0.25 × 80,000 × 9.81 ≈ 196 kN.
177. Sand is dropped on the rails in front of the driving wheels in order to:
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Grip between wheel and rail.
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Answer: A. Increase the coefficient of adhesion and prevent wheel slip
Wet or oily rails reduce adhesion; sanding increases it so more tractive effort can be applied without slipping.
178. On long down gradients, the most energy-efficient form of electric braking is:
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Which method recovers energy?
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Answer: D. Regenerative braking
Regenerative braking returns the energy of the descending train to the supply instead of wasting it as heat.
179. Specific energy consumption of a train is usually expressed in:
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Energy per unit mass per unit distance.
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Answer: A. Wh per tonne-km
It is the energy needed to move one tonne of train mass over one kilometre.
180. For a given maximum speed, the specific energy consumption of a train generally decreases when:
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Each stop throws away kinetic energy.
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Answer: D. The distance between stops increases
Fewer stops mean less kinetic energy wasted in braking per kilometre travelled.
181. An elevator car weighs 800 kg and has a rated load of 1000 kg. If the counterweight balances the car plus 50 % of rated load, the counterweight mass is:
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Car mass plus a fraction of the rated load.
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Answer: C. 1300 kg
Counterweight = 800 + 0.5 × 1000 = 1300 kg.
182. An elevator has an unbalanced mass of 500 kg (full load minus counterweight effect) and travels at 1.5 m/s. With an overall efficiency of 80 %, the motor power required is about:
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Only the unbalanced mass needs lifting power.
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Answer: A. 9.2 kW
P = mgv/η = 500 × 9.81 × 1.5/0.8 ≈ 9.2 kW.
183. Modern traction elevators are usually driven by:
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Smooth speed control with power electronics.
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Answer: B. AC induction or permanent-magnet motors with VVVF drives
VVVF drives give smooth acceleration, accurate floor levelling, low starting current and regenerative operation.
184. In an elevator, the overspeed governor:
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A safety device, not the speed controller.
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Answer: C. Trips the safety gear to grip the guide rails if the car exceeds a set speed
It is a safety device independent of the drive; on overspeed it actuates safety gear clamping the car to the rails.
185. Starting of a DC series tramway motor is commonly done by:
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DC motors cannot use star-delta.
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Answer: D. Series resistance combined with series-parallel control of the motors
Resistances limit current at starting; switching the motors from series to parallel reduces the energy wasted in resistors.