Nepal Engineering Council · Electrical Engineering · Chapter 7
Transmission and Distribution Lines
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180 questions in 6 syllabus topics.
7.1 Transmission lines
31 questions · AElE0701
1. Which of the following is NOT a main component of an overhead transmission line?
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Think of which item belongs inside a cable, not on a tower.
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Answer: A. Bedding
Bedding is a layer inside an underground cable that cushions the metallic sheath from the armour. Overhead lines consist of conductors, supports, cross arms, insulators, earth wire and protective devices.
2. Compared with an overhead line of the same rating, an underground cable generally has
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Consider the materials and civil work needed to bury a conductor.
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Answer: D. a much higher initial cost
Cables cost several times more than overhead lines because of insulation, sheathing, armouring and trenching. They have higher capacitance (more charging current), lower inductance, and faults are harder to locate.
3. Underground cables are preferred over overhead lines mainly in
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Where is space, safety and aesthetics more important than cost?
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Answer: D. densely populated urban areas where safety and appearance matter
Cables are safe, unaffected by weather and invisible, which suits congested cities despite their higher cost; overhead lines are cheaper and easier to tap and extend.
4. The correct order of layers in an armoured underground cable, from the conductor outwards, is
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Insulation must touch the core; serving is always outermost.
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Answer: A. core, insulation, metallic sheath, bedding, armouring, serving
Insulation surrounds the core, the metallic (lead/aluminium) sheath keeps out moisture, bedding cushions the sheath, armouring gives mechanical protection, and serving protects the armour.
5. The main purpose of the bedding layer in an underground cable is to
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It sits between two metallic layers.
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Answer: A. protect the metallic sheath from corrosion and mechanical damage by the armouring
Bedding (jute or fibrous material impregnated with compound) lies between the metallic sheath and the armour and protects the sheath from corrosion and from injury by the armouring.
6. Belted (three-core paper-insulated) cables are generally used only for voltages up to about
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Belted construction is the oldest and lowest-voltage design.
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Answer: D. 11 kV
Belted cables suffer tangential stresses across the paper layers at higher voltages, so they are limited to about 11 kV (occasionally up to 22 kV). Screened or pressure cables are used above that.
7. The main reason belted cables are unsuitable for higher voltages is that
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Paper is strong across its thickness but weak along its layers.
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Answer: C. tangential electric stress along the paper layers causes leakage currents and local heating
In a belted cable the field is not purely radial; the tangential component acts along the laminations where the paper is weak, causing leakage, heating and eventual breakdown.
8. In an H-type (Hochstadter) screened cable, the metallic screen around each core is provided to
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Screening removes one particular direction of stress.
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Answer: C. make the electric stress purely radial in each core's insulation
Each core's metallised screen is earthed through the sheath, so the field in each core's dielectric is radial, eliminating tangential stress; it also aids heat dissipation.
9. Oil-filled and gas-pressure cables are used for high voltages chiefly because they
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What forms in impregnated paper during heating and cooling cycles?
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Answer: A. eliminate voids in the dielectric, preventing ionisation
Voids in solid-type cables ionise at high stress and gradually destroy the insulation; oil or gas under pressure fills or suppresses these voids.
10. The maximum continuous conductor operating temperature normally permitted for XLPE-insulated power cables is
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It is 20 °C higher than the usual PVC rating.
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Answer: D. 90 °C
XLPE is rated for 90 °C continuous (about 250 °C short-circuit), which is higher than PVC (70 °C) and gives XLPE cables a higher current rating.
11. A single-core cable is 2 km long. The conductor radius is 1 cm, the inner radius of the sheath is 2.5 cm and the insulation resistivity is 4.5 × 10¹² Ω·m. The insulation resistance is about
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Use R = ρ ln(R/r) / (2πl) with l in metres.
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Answer: B. 328 MΩ
R = ρ ln(R/r)/(2πl) = 4.5×10¹² × ln 2.5 /(2π × 2000) = 4.5×10¹² × 0.916/12566 ≈ 3.28×10⁸ Ω = 328 MΩ.
12. A 1 km length of a cable has an insulation resistance of 500 MΩ. The insulation resistance of 4 km of the same cable is
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Longer cable means more parallel leakage paths.
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Answer: A. 125 MΩ
Insulation resistance is inversely proportional to length (leakage paths are in parallel), so 500/4 = 125 MΩ.
13. A single-core cable has conductor diameter 2 cm and sheath inside diameter 5 cm. If the conductor-to-sheath voltage is 20 kV, the maximum electric stress in the dielectric is about
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Use the radius, not the diameter, in g_max = V/(r ln(R/r)).
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Answer: D. 21.8 kV/cm
g_max = V/(r ln(R/r)) = 20/(1 × ln 2.5) = 20/0.916 ≈ 21.8 kV/cm, occurring at the conductor surface.
14. In a single-core cable, the electric stress in the insulation is maximum
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Stress varies inversely with radius.
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Answer: D. at the surface of the conductor
Stress g = V/(x ln(R/r)) is inversely proportional to the distance x from the centre, so it is highest at x = r, the conductor surface.
15. A single-core cable must work at 50 kV (peak) to earth with a maximum permissible stress of 40 kV/cm (peak). The most economical conductor diameter is
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At the optimum, ln(D/d) = 1.
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Answer: B. 2.5 cm
For minimum overall diameter, ln(D/d) = 1, which gives d = 2V/g_max = 2 × 50/40 = 2.5 cm (and D = e × d ≈ 6.8 cm).
16. In capacitance grading of a cable, the dielectric with the highest permittivity is placed
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Grading aims to bring down the stress where it is naturally highest.
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Answer: B. nearest to the conductor
Stress is inversely proportional to permittivity × radius; putting the highest-permittivity material where the radius is smallest makes the stress more uniform.
17. In intersheath grading of a cable, the intersheaths are
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The grading here is done by fixing voltages, not by changing material.
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Answer: A. metallic layers kept at suitable intermediate potentials between core and sheath
Intersheaths are thin metallic cylinders in a homogeneous dielectric, held at chosen potentials (e.g. via transformer tappings) so that the stress is distributed more evenly.
18. A 5 km single-core cable has a dielectric of relative permittivity 3.5 and a sheath-to-conductor diameter ratio of 2.5. Its capacitance is about (ε₀ = 8.854 × 10⁻¹² F/m)
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C = 2πε₀εr l / ln(D/d).
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Answer: C. 1.06 µF
C = 2πε₀εr l / ln(D/d) = 2π × 8.854×10⁻¹² × 3.5 × 5000 / ln 2.5 ≈ 1.06 × 10⁻⁶ F.
19. Each core of a three-phase 11 kV, 50 Hz cable has a capacitance of 1.06 µF to the (earthed) sheath. The charging current per phase is about
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Use the phase voltage, not the line voltage.
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Answer: B. 2.12 A
I = Vph ωC = (11000/√3) × 2π × 50 × 1.06×10⁻⁶ ≈ 6351 × 3.33×10⁻⁴ ≈ 2.12 A.
20. The Murray loop test is used to
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It is a bridge test that needs a healthy companion core.
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Answer: C. locate earth faults and short-circuit faults in underground cables
The Murray loop is a Wheatstone-bridge method that uses a sound conductor to form a loop and locates the distance to an earth or short-circuit fault.
21. Which of the following is NOT a method of laying underground cables?
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One option is borrowed from overhead line insulators.
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Answer: A. Pin-type system
Cables are laid directly in trenches, drawn into ducts/conduits (draw-in system) or laid in troughing filled with bitumen (solid system). Pin type refers to an overhead insulator.
22. The main advantage of the draw-in system of laying cables is that
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Think of what the ducts allow you to do later.
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Answer: D. cables can be repaired, replaced or added without re-excavating the route
Ducts are laid once with manholes at intervals, so cables can later be pulled in or out; however, its initial cost is high and heat dissipation is poorer than direct laying.
23. The effective AC resistance of a conductor due to skin effect is greater when
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Skin effect is an AC phenomenon that grows with size.
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Answer: C. the frequency and the conductor diameter are larger
Skin effect pushes current towards the surface; it increases with frequency, conductor diameter and permeability, and is absent with DC.
24. A three-phase line has conductors of radius 1 cm placed 2 m apart (equilateral). Taking surface irregularity factor 0.85, air density factor 1 and g₀ = 21.1 kV/cm (rms), the disruptive critical voltage to neutral is about
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Use the natural log of D/r with r in cm.
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Answer: C. 95 kV
Vc = m₀ g₀ δ r ln(D/r) = 0.85 × 21.1 × 1 × 1 × ln 200 ≈ 17.94 × 5.30 ≈ 95 kV (rms, phase).
25. At a site with barometric pressure 70 cm of mercury and temperature 30 °C, the air density factor δ = 3.92b/(273 + t) is about
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b in cm of Hg, t in °C.
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Answer: B. 0.91
δ = 3.92 × 70/(273 + 30) = 274.4/303 ≈ 0.91. Lower air density reduces the corona inception voltage, which matters for lines in hilly areas.
26. Which of the following is an ADVANTAGE of corona on overhead lines?
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Think of surges from lightning or switching.
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Answer: B. It reduces the steepness of travelling surge waves
Corona increases the effective conductor diameter and dissipates energy during a surge, reducing its steepness; its drawbacks are power loss, radio interference and ozone production.
27. Corona loss on an EHV overhead line can be reduced by
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Corona depends on the field gradient at the conductor surface.
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Answer: C. using bundled conductors or larger-diameter conductors
Bundling or larger diameter lowers the surface field gradient, raising the critical disruptive voltage and reducing corona.
28. A single-phase line has two conductors each of radius 0.5 cm spaced 1.5 m apart. The loop inductance is about (use r' = 0.7788r)
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Loop inductance includes both conductors.
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Answer: C. 2.38 mH/km
L(loop) = 0.4 ln(D/r') mH/km = 0.4 × ln(150/0.3894) = 0.4 × 5.95 ≈ 2.38 mH/km. 1.19 mH/km is the inductance of one conductor only.
29. A three-phase line has equilaterally spaced conductors of radius 1 cm, 2 m apart. Its capacitance to neutral is about
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C per phase = 0.0556/ln(D/r) µF/km.
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Answer: B. 0.0105 µF/km
C = 2πε₀/ln(D/r) = 0.0556/ln(200) µF/km = 0.0556/5.30 ≈ 0.0105 µF/km. Using log₁₀ instead of ln gives the wrong 0.0242.
30. Transposition of the conductors of a three-phase overhead line is done mainly to
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It is about the symmetry of the three phases.
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Answer: D. balance the inductance and capacitance of the three phases
With unsymmetrical spacing the phase inductances differ; transposing the conductors over the route equalises the average parameters (and also reduces interference with nearby communication lines).
31. The Ferranti effect, in which the receiving-end voltage exceeds the sending-end voltage, occurs mainly in
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Charging current through series inductance.
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Answer: A. long lines and cables at light load or no load
At light load the capacitive charging current flowing through the line inductance produces a voltage rise toward the receiving end; it is pronounced in long lines and in cables because of their large capacitance.
7.2 Transmission lines circuit selection
30 questions · AElE0702
32. The main reason for transmitting bulk power at high voltage is that, for the same power,
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P = √3 V I cos φ.
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Answer: D. the line current is lower, so I²R loss and conductor material are reduced
For a given power, current is inversely proportional to voltage, so I²R loss falls as 1/V² and a smaller conductor cross-section can be used.
33. For the same power, distance, power factor and percentage line loss, raising the transmission voltage from 66 kV to 132 kV changes the volume of conductor material required to about
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Volume ∝ 1/V².
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Answer: A. 25 % of the original
Conductor volume required is inversely proportional to V² (for fixed percentage loss), so doubling V gives (1/2)² = 1/4 = 25 %.
34. A line delivers a fixed power using the same conductor and power factor. If the voltage is raised from 11 kV to 33 kV, the I²R loss becomes about
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Loss ∝ I² and I ∝ 1/V.
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Answer: A. 11.1 % of the original
Current falls to 1/3, so I²R loss falls to (1/3)² = 1/9 ≈ 11.1 %.
35. Which of the following INCREASES as the transmission voltage is increased?
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Think of what must withstand the voltage.
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Answer: D. Cost of insulators, towers, transformers and switchgear
Higher voltage needs more insulation, larger clearances and taller supports, and costlier terminal equipment; this offsets the conductor saving and limits the economic voltage.
36. The most economical transmission voltage for a line is the one at which
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It is a trade-off between two opposing cost trends.
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Answer: A. the total cost of conductors, insulation, supports and terminal equipment is minimum
Raising voltage reduces conductor cost but increases insulation, support and equipment cost; the economic voltage minimises the sum of all these costs.
37. Using Still's formula V = 5.5 √(L/1.6 + P/100) kV (L in km, P in kW), the economical line voltage for transmitting 20 000 kW over 80 km is about
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Divide km by 1.6 and kW by 100 before taking the root.
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Answer: C. 87 kV
V = 5.5 √(80/1.6 + 20000/100) = 5.5 √(50 + 200) = 5.5 × 15.81 ≈ 87 kV; the next higher standard voltage would then be chosen.
38. According to Kelvin's law, the most economical conductor cross-section is the one for which
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Two annual costs, one rising and one falling with area.
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Answer: D. the annual interest and depreciation on the conductor capital cost equals the annual cost of energy lost
Kelvin's law equates the variable part of annual charges (interest and depreciation on conductor cost, ∝ area) with the annual cost of I²R energy loss (∝ 1/area).
39. A limitation of Kelvin's law for conductor size selection is that it does NOT take into account
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The law considers only costs.
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Answer: A. voltage drop, corona and mechanical strength requirements
Kelvin's law is purely economic; the resulting size must still be checked for voltage regulation, corona, current rating and mechanical strength.
40. If the spacing between the conductors of an overhead line is increased, then
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Look at where D appears in L and C formulas.
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Answer: A. the inductance increases and the capacitance decreases
L ∝ ln(D/r') and C ∝ 1/ln(D/r), so larger spacing D raises inductance and lowers capacitance.
41. Why must the spacing between conductors be increased for longer spans?
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Think of conductors swinging in the wind.
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Answer: D. Larger sag allows greater swinging in wind, increasing the risk of conductors clashing
Sag grows with the square of span; a sagging conductor swings more in wind, so phase spacing must be increased to avoid clashing and flashover.
42. The maximum permissible temperature for Class B insulation is
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Class B lies between E and F.
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Answer: C. 130 °C
Thermal classes: Y 90 °C, A 105 °C, E 120 °C, B 130 °C, F 155 °C, H 180 °C, C above 180 °C.
43. Insulating materials used in electrical equipment are classified into thermal classes (Y, A, E, B, F, H, C) according to their
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The class letters correspond to temperatures.
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Answer: B. maximum permissible operating temperature
Thermal classification groups insulating materials by the highest temperature at which they can operate continuously without undue deterioration.
44. Which of the following is an inorganic solid insulating material?
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Which one is a mineral?
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Answer: A. Mica
Mica is a mineral (inorganic) insulator able to withstand high temperatures; paper, PVC and rubber are organic.
45. A desirable combination of properties for an insulating material used in high-voltage equipment is
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It should withstand voltage without heating.
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Answer: D. high dielectric strength and low dielectric loss
A good insulator should have high dielectric strength, high resistivity, low dielectric loss (tan δ), good mechanical strength and be non-hygroscopic.
46. SF₆ gas is widely used in circuit breakers and gas-insulated substations mainly because it
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Think about electron attachment.
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Answer: C. has high dielectric strength and is strongly electronegative
SF₆ has about 2–3 times the dielectric strength of air at the same pressure and captures free electrons (electronegative), which helps quench arcs; it is non-flammable and inert.
47. A transmission line is usually treated as a short line, with its capacitance neglected, when its length is less than about
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Medium lines start where short lines end.
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Answer: B. 80 km
Lines up to about 80 km (and usually below about 20 kV) are short lines; about 80–200 (or 250) km are medium lines; longer ones are long lines.
48. For a three-phase short line, voltage regulation is defined as
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Regulation is referred to the full-load receiving voltage.
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Answer: A. (sending-end voltage − receiving-end voltage)/receiving-end voltage at full load
%VR = (V_NL − V_FL)/V_FL × 100; for a short line the no-load receiving voltage equals Vs, so %VR = (Vs − Vr)/Vr × 100.
49. A three-phase short line delivers 2 MW at 11 kV (line), 0.8 pf lagging. Its resistance and reactance per phase are 2 Ω and 3 Ω. The voltage regulation is about
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Work per phase with Vr = 11000/√3.
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Answer: D. 7.05 %
I = 2×10⁶/(√3 × 11000 × 0.8) = 131.2 A; Vr = 6351 V/phase; Vs = |6351 + 131.2(0.8 − j0.6)(2 + j3)| ≈ 6799 V; %VR = (6799 − 6351)/6351 ≈ 7.05 %.
50. A three-phase line delivers 2 MW at 11 kV, 0.8 pf lagging, with a line current of 131.2 A and a resistance of 2 Ω per phase. The transmission efficiency is about
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Remember all three phases have I²R loss.
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Answer: B. 95.1 %
Loss = 3I²R = 3 × 131.2² × 2 ≈ 103.3 kW; η = 2000/(2000 + 103.3) ≈ 95.1 %. Taking only one phase's loss would wrongly give 98.3 %.
51. A single-phase short line supplies 50 A at 0.8 pf lagging with receiving-end voltage 2200 V. The total loop resistance and reactance are 1 Ω and 2 Ω. The approximate sending-end voltage is
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Approximate drop = I(R cos φ + X sin φ) for lagging pf.
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Answer: C. 2300 V
Vs ≈ Vr + I(R cos φ + X sin φ) = 2200 + 50(1 × 0.8 + 2 × 0.6) = 2200 + 100 = 2300 V.
52. Negative voltage regulation (receiving-end voltage higher than sending-end voltage) in a short line occurs when the load power factor is
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The sign of the reactive drop reverses for capacitive loads.
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Answer: B. leading, with X sin φ greater than R cos φ
With leading pf the drop is approximately I(R cos φ − X sin φ), which becomes negative when X sin φ > R cos φ.
53. A short line has R = 2 Ω and X = 3 Ω per phase. The load power factor at which the voltage regulation is approximately zero is
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Set R cos φ − X sin φ = 0.
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Answer: C. 0.832 leading
Zero regulation when R cos φ = X sin φ (leading), i.e. tan φ = R/X = 2/3, so cos φ = 3/√13 ≈ 0.832 leading.
54. A medium line represented by the nominal-π model has series impedance Z = 10 + j40 Ω and total shunt admittance Y = j4 × 10⁻⁴ S. Its A constant is about
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A = 1 + YZ/2.
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Answer: B. 0.992
A = 1 + YZ/2 = 1 + (j4×10⁻⁴)(10 + j40)/2 = 1 + (−0.016 + j0.004)/2 = 0.992 + j0.002.
55. In the nominal-T representation of a medium transmission line,
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The letter T shows the shape.
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Answer: C. half the series impedance is placed on each side, with the total shunt admittance at the middle
Nominal T: Z/2 – Y – Z/2. Nominal π: Y/2 at each end with Z in between.
56. A 132 kV (line) medium line in nominal-π form has a shunt admittance of 4 × 10⁻⁴ S at each end. The charging current drawn by the receiving-end shunt branch at rated voltage is about
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Use the phase voltage with the admittance of one end.
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Answer: B. 30.5 A
I = V_ph × Y/2 = (132000/√3) × 4×10⁻⁴ = 76210 × 4×10⁻⁴ ≈ 30.5 A.
57. For a short transmission line, the ABCD constants are
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With no shunt branch, sending and receiving currents are equal.
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Answer: D. A = D = 1, B = Z, C = 0
Neglecting shunt admittance, Vs = Vr + Z Ir and Is = Ir, giving A = 1, B = Z, C = 0, D = 1.
58. For any passive, linear, bilateral two-port network representing a transmission line, the ABCD constants satisfy
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It is the determinant of the transmission matrix.
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Answer: C. AD − BC = 1
Reciprocity of a passive bilateral network gives the determinant AD − BC = 1.
59. A 400 kV line has a surge impedance of 400 Ω. Its surge impedance loading (SIL) is
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SIL = (line voltage)² / Zc.
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Answer: C. 400 MW
SIL = V_L²/Zc = (400 kV)²/400 Ω = 400 MW.
60. A load of fixed kW is supplied over a line at fixed voltage. If the load power factor is improved from 0.8 to 1.0, the line I²R loss becomes
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Loss ∝ 1/(V² cos²φ).
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Answer: B. 64 % of the original
I ∝ 1/cos φ, so loss ∝ 1/cos²φ: (0.8/1.0)² = 0.64, i.e. 64 % of the original loss.
61. Increasing the system voltage of a line while keeping the same conductor and load (in kW) generally
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Current falls as voltage rises.
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Answer: D. improves both efficiency and percentage voltage regulation
Lower current reduces I²R loss (better efficiency), and the voltage drop is a smaller percentage of the higher voltage (better regulation).
7.3 Mechanical design of overhead line
30 questions · AElE0703
62. A conductor weighing 6 N/m is strung between level supports 200 m apart with a horizontal tension of 12 000 N. The sag is
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For level supports, S = wl²/8T with l the span.
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Answer: B. 2.5 m
S = wl²/(8T) = 6 × 200²/(8 × 12000) = 240000/96000 = 2.5 m.
63. A conductor has an ultimate tensile strength of 30 000 N and weighs 6 N/m. With a factor of safety of 2 and a level span of 250 m, the sag is about
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First find the working tension from the safety factor.
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Answer: C. 3.13 m
Working tension T = 30000/2 = 15000 N; S = wl²/(8T) = 6 × 250²/(8 × 15000) = 3.125 m.
64. For level supports and the same conductor tension, if the span length is doubled, the sag becomes
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Sag ∝ (span)².
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Answer: D. four times
S = wl²/(8T), so sag is proportional to the square of the span: 2² = 4.
65. If the working tension of a conductor on a given level span is reduced by 20 %, the sag
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Sag is inversely proportional to tension.
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Answer: A. increases by 25 %
S ∝ 1/T, so the new sag = S/0.8 = 1.25 S, an increase of 25 %.
66. A conductor weighs 6 N/m; ice loading adds 4 N/m and the wind load is 8 N/m. The effective loading per metre is about
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Vertical loads add; wind acts horizontally.
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Answer: B. 12.8 N/m
w_e = √[(w + w_i)² + w_w²] = √(10² + 8²) = √164 ≈ 12.8 N/m; the vertical loads add directly and the horizontal wind load adds at right angles.
67. When an overhead conductor is subjected to wind load, the sag
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The resultant force is not vertical.
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Answer: D. lies in an inclined plane at an angle tan⁻¹[w_w/(w + w_i)] to the vertical
The resultant of the vertical (weight + ice) and horizontal (wind) loads is inclined, so the conductor sags in that inclined plane; the vertical sag is S cos θ.
68. A conductor of diameter 2 cm is covered with a radial ice thickness of 1 cm. Taking ice density 915 kg/m³ and g = 9.81 m/s², the weight of ice per metre is about
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Ice cross-section = π t (d + t).
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Answer: C. 8.46 N/m
Ice area = π t (d + t) = π × 0.01 × 0.03 = 9.42×10⁻⁴ m²; weight = 915 × 9.42×10⁻⁴ × 9.81 ≈ 8.46 N/m (0.86 kg/m is the mass, not the weight).
69. A span of 300 m is between supports differing in height by 6 m. The conductor weighs 8 N/m with a tension of 20 000 N. The lowest point of the conductor is at a horizontal distance from the LOWER support of
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x₁ = l/2 − Th/(wl).
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Answer: B. 100 m
x₁ = l/2 − Th/(wl) = 150 − (20000 × 6)/(8 × 300) = 150 − 50 = 100 m.
70. A line must have a minimum ground clearance of 7 m, and the maximum sag (at the highest temperature) is 4.2 m. The minimum height of the conductor attachment point on level supports is
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The lowest point of the conductor must still clear the ground.
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Answer: C. 11.2 m
Attachment height = minimum ground clearance + maximum sag = 7 + 4.2 = 11.2 m.
71. A conductor on a level span of 200 m has a sag of 2.5 m. The length of the conductor in the span is about
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L = l + 8S²/(3l).
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Answer: B. 200.08 m
L = l + 8S²/(3l) = 200 + 8 × 2.5²/(600) = 200 + 0.083 = 200.08 m.
72. For an overhead line conductor, the sag is maximum and the tension minimum
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Metals expand when hot.
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Answer: A. at the highest temperature without ice or wind
At high temperature the conductor expands, so its length increases, sag increases and tension decreases; clearance must be checked for this condition.
73. A stringing chart for an overhead line shows
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It is used while erecting the conductor.
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Answer: D. sag and tension of the conductor against temperature
Stringing charts let linemen set the correct sag and tension for the ambient temperature at the time of stringing.
74. For transmission lines of 132 kV and above, the supports normally used are
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Long spans and heavy conductors need the strongest support.
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Answer: A. steel lattice towers
Steel lattice towers are strong, allow long spans and large clearances, and are used for EHV lines; poles are used for distribution voltages.
75. The main disadvantage of wooden poles as line supports is
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Wood is a natural organic material.
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Answer: D. limited life due to decay and the need for preservative treatment
Wooden poles are cheap and light with some insulating property, but they rot at the ground line and are attacked by insects, so their life is limited.
76. The main function of a cross arm on a line support is to
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Think of what is fixed on the cross arm.
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Answer: A. carry the insulators and maintain the required spacing between conductors
Cross arms (galvanised steel or wood) hold the insulators and conductors at the required horizontal separation.
77. For the same length and resistance, an aluminium conductor compared with a copper conductor (Al conductivity ≈ 61 % of Cu, densities 2.7 and 8.9 g/cm³) is
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Compare area ratio × density ratio.
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Answer: D. larger in cross-section but about half the weight
Area ratio = 1/0.61 ≈ 1.64; weight ratio = 1.64 × 2.7/8.9 ≈ 0.50. Aluminium needs a bigger but lighter conductor.
78. In an ACSR conductor, the steel strands mainly provide
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Steel is stronger, aluminium conducts better.
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Answer: A. high tensile strength, allowing longer spans with less sag
The central galvanised steel core provides mechanical strength; the outer aluminium strands carry most of the current.
79. Pin-type insulators are generally economical only up to about
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Above this voltage suspension strings take over.
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Answer: D. 33 kV
Beyond about 33 kV a pin insulator becomes too large, heavy and costly; suspension strings are used for higher voltages.
80. An important advantage of suspension (disc) insulators over pin insulators for high voltages is that
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Think of the modular construction.
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Answer: C. the insulation level can be increased simply by adding more discs
Each disc is designed for a modest voltage, so a string of the required number can be used for any voltage, and a damaged disc can be replaced alone.
81. Strain insulators are used on overhead lines
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Where does the conductor pull hardest?
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Answer: B. at dead ends, sharp corners and long river crossings
Where the conductor tension must be taken by the insulator, as at terminations, angles and long crossings, suspension discs are mounted horizontally as strain insulators.
82. Shackle insulators are mainly used
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They are the small spool-shaped ones.
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Answer: A. on low-voltage distribution lines, either vertically or horizontally
Shackle (spool) insulators are used on LV distribution lines and service connections and can serve as small strain insulators.
83. In a string of suspension insulators, the voltage across the disc nearest to the line conductor is
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Shunt capacitance adds current at each junction.
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Answer: D. the highest of all discs
Shunt capacitance from each link to the earthed tower makes the current through the discs increase toward the line end, so the disc nearest the conductor has the highest voltage.
84. A string of 3 identical discs has a shunt capacitance (link to tower) equal to 0.1 times the self-capacitance of each disc. The string efficiency is about
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Apply KCL at each link junction: V₂ = V₁(1 + m).
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Answer: B. 86.8 %
V₂ = 1.1V₁; V₃ = V₂ + 0.1(V₁ + V₂) = 1.31V₁; total = 3.41V₁; η = 3.41V₁/(3 × 1.31V₁) ≈ 86.8 %.
85. A 33 kV three-phase line uses strings of 3 discs. If the shunt-to-self capacitance ratio is 0.1, the voltage across the disc nearest the line is about
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Work with the phase voltage and the 1 : 1.1 : 1.31 shares.
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Answer: C. 7.32 kV
Phase voltage = 33/√3 = 19.05 kV; shares are V₁ : V₂ : V₃ = 1 : 1.1 : 1.31 (sum 3.41), so V₃ = 19.05 × 1.31/3.41 ≈ 7.32 kV.
86. A string of 4 discs has 30 kV across it, and the disc nearest the conductor has 9 kV across it. The string efficiency is
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String efficiency = total voltage/(number of discs × voltage on line-end disc).
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Answer: C. 83.3 %
η = V/(n × V_max) = 30/(4 × 9) = 0.833 = 83.3 %.
87. A guard ring fitted around the line end of a suspension string improves the string efficiency by
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It supplies the charging current that would otherwise flow through the lower discs.
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Answer: A. adding capacitance between the line and the disc links, which equalises the voltage distribution
The guard ring, connected to the line, introduces capacitance between each link and the line conductor that cancels the charging current to the tower, making the voltage distribution more uniform.
88. In capacitance grading of a suspension string, the discs are chosen so that
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More current with the same voltage needs more capacitance.
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Answer: C. the disc nearest the line conductor has the largest capacitance
Since the line-end disc carries the most current, giving it the largest capacitance (Xc smallest) equalises the voltage across all discs.
89. A jumper on an overhead transmission line is
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It carries current across a support where the conductor is dead-ended.
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Answer: B. a short conductor loop that electrically connects the conductors on either side of a tension (dead-end) tower
At tension towers the conductor is terminated on strain strings on each side; the jumper loop bridges them electrically while keeping clearance to the tower.
90. Stockbridge dampers are fitted on overhead line conductors to
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Vibration from wind blowing steadily across the conductor.
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Answer: A. reduce aeolian vibration caused by steady low-speed winds
A Stockbridge damper (two weights on a messenger cable clamped near the support) absorbs the energy of high-frequency, low-amplitude aeolian vibration, preventing fatigue at clamps.
91. Galloping of overhead conductors is best described as
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Contrast it with aeolian vibration.
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Answer: D. low-frequency, large-amplitude oscillation, usually of ice-coated conductors in wind
Galloping occurs when ice gives the conductor an aerofoil shape; wind then produces large, slow oscillations. Aeolian vibration is the high-frequency, low-amplitude kind.
7.4 Electrical loads
30 questions · AElE0704
92. Irrigation pumping load is classified as
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Who uses irrigation pumps?
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Answer: D. agricultural load, which is largely seasonal and motor-dominated
Irrigation pumps form agricultural load: mostly induction motors, used in particular seasons and hours, giving a low annual load factor.
93. Typical residential (domestic) load on a distribution feeder peaks
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When are most people at home with lights on?
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Answer: D. in the evening, mainly due to lighting, cooking and appliances
Domestic demand rises sharply in the evening when lighting, cooking and entertainment loads coincide, giving residential feeders a low load factor.
94. Which category of load usually has the highest load factor?
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Load factor is high when demand is steady all day.
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Answer: A. Continuous-process industrial load
Process industries running 24 hours a day draw nearly constant power, so their average demand is close to their maximum demand.
95. For a street lighting load, the demand factor is
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Are any lamps left off when the street lights come on?
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Answer: B. close to unity, because all lamps operate together
All street lights are switched on together, so maximum demand ≈ connected load, giving a demand factor near 1.
96. Which of the following loads is a major source of harmonic currents in a distribution system?
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Look for non-linear loads.
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Answer: A. Variable-frequency drives and switch-mode power supplies
Power-electronic loads draw non-sinusoidal current, injecting harmonics; resistive heaters and lamps are linear loads.
97. For small voltage changes, a resistive heater behaves as a
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P = V²/R for a fixed R.
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Answer: D. constant-impedance load, its power varying as the square of voltage
A fixed resistance draws P = V²/R, so it is a constant-impedance load; induction motors behave nearer to constant-power loads.
98. A consumer has a connected load of 100 kW and a maximum demand of 60 kW. The demand factor is
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Maximum demand divided by connected load.
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Answer: C. 0.60
Demand factor = maximum demand/connected load = 60/100 = 0.60 (it is normally less than 1).
99. Connected load of a consumer is
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It is found from nameplates, not meters.
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Answer: B. the sum of the continuous ratings of all the equipment connected to the supply
Connected load is the total nameplate rating of all apparatus installed; maximum demand is usually lower because not all equipment runs together at full load.
100. The maximum demand indicated by an MD meter is
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Short inrush peaks are not what it records.
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Answer: A. the highest average demand over a specified demand interval such as 15 or 30 minutes
MD meters integrate demand over fixed intervals (commonly 15 or 30 min) and record the largest interval average, so brief starting surges are ignored.
101. A system has a maximum demand of 50 MW and an annual load factor of 45 %. The energy generated in a year is about
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Average demand = load factor × maximum demand.
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Answer: B. 197 GWh
Energy = LF × MD × 8760 h = 0.45 × 50 MW × 8760 h = 197 100 MWh ≈ 197 GWh.
102. A daily load curve shows 20 MW for 6 h, 50 MW for 12 h and 30 MW for 6 h. The average load is
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Find energy first, then divide by 24 h.
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Answer: C. 37.5 MW
Energy = 20×6 + 50×12 + 30×6 = 900 MWh; average = 900/24 = 37.5 MW. Simply averaging the three levels (33.3 MW) ignores their durations.
103. Three consumers have individual maximum demands of 40 kW, 30 kW and 50 kW, and their combined maximum demand is 100 kW. The diversity factor is
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Sum of individual MDs over the group MD.
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Answer: C. 1.2
Diversity factor = sum of individual maximum demands/maximum demand of the group = 120/100 = 1.2.
104. A higher diversity factor in a power system results in
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Peaks that do not coincide add up to less.
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Answer: D. lower installed capacity needed and lower cost per unit of energy
When consumers' peaks occur at different times, the combined maximum demand is smaller, so less plant capacity is needed for the same consumers.
105. Coincidence factor is
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It is the inverse of a ratio that is ≥ 1.
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Answer: D. the reciprocal of the diversity factor
Coincidence factor = group maximum demand/sum of individual maximum demands = 1/diversity factor, so it is ≤ 1.
106. A power station of 80 MW installed capacity has a maximum demand of 50 MW and a load factor of 45 %. Its plant capacity factor is about
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Average load divided by installed capacity.
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Answer: A. 28.1 %
Average load = 0.45 × 50 = 22.5 MW; plant capacity factor = 22.5/80 = 28.1 % (= load factor × utilisation factor = 0.45 × 0.625).
107. A 40 MVA substation transformer has a maximum demand of 30 MVA. Its utilisation factor is
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Maximum demand over rated capacity.
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Answer: C. 0.75
Utilisation factor = maximum demand/rated capacity = 30/40 = 0.75.
108. A load duration curve is obtained by
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The time axis is not chronological.
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Answer: A. arranging the loads of a period in descending order of magnitude
The load duration curve shows loads arranged from highest to lowest, giving the time for which each load level is equalled or exceeded; its area still equals the energy.
109. The area under a daily load curve represents
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Power × time.
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Answer: B. the energy consumed during the day
The load curve plots power against time, so its area is the energy (e.g. MWh) for the day; dividing by 24 h gives the average load.
110. Which type of power plant is best suited to supply the peak portion of the load?
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Peaks need fast response.
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Answer: D. Storage hydro or gas-turbine plant that can start and change output quickly
Peak loads last short periods and need quick start and fast ramping; storage hydro and gas turbines do this, while nuclear and large thermal units suit base load.
111. Long-term load forecasting (5–20 years ahead) is mainly used for
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Building a power plant takes years.
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Answer: B. planning new generation, transmission and distribution capacity
Long-term forecasts guide expansion planning and investment; short-term forecasts (hours to a week) are used for operation and scheduling.
112. The most important variable affecting short-term (hourly to weekly) load forecasts is usually
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What changes from day to day?
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Answer: A. weather, such as temperature
Over hours or days population and GDP hardly change, but temperature and other weather conditions strongly influence demand.
113. The end-use method of load forecasting estimates demand by
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It works upwards from appliances.
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Answer: D. combining the number of appliances or processes with their typical energy use
End-use models build demand bottom-up from appliance stock, ownership levels and usage per appliance for each consumer category.
114. The econometric method of load forecasting relates electricity demand to
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'Econo-' gives the clue.
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Answer: A. economic and demographic variables such as GDP, population and price
Econometric models use statistical (regression) relationships between demand and explanatory variables like income, GDP, population and tariff.
115. A region's peak demand is 100 MW and is expected to grow at 7 % per year (compounded). The forecast peak after 5 years is about
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Use P = P₀ (1 + r)ⁿ.
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Answer: C. 140 MW
P = 100 × 1.07⁵ = 100 × 1.4026 ≈ 140 MW. Simple (non-compound) growth would give 135 MW.
116. At a steady annual load growth of 7 %, the peak demand doubles in about
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Solve 1.07ⁿ = 2.
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Answer: C. 10.2 years
n = ln 2/ln 1.07 = 0.693/0.0677 ≈ 10.2 years (close to the rule of 70: 70/7 = 10).
117. Peak loads for five successive years were 50, 54, 58, 62 and 66 MW. By linear trend extrapolation, the expected peak load in year 7 is
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Find the yearly increment.
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Answer: C. 74 MW
The load rises by 4 MW per year; year 5 is 66 MW, so year 7 = 66 + 2 × 4 = 74 MW.
118. A feeder has a load factor of 0.5. Using the empirical relation LLF = 0.3 LF + 0.7 LF², the loss load factor is
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Substitute LF = 0.5 into the formula.
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Answer: B. 0.325
LLF = 0.3 × 0.5 + 0.7 × 0.25 = 0.15 + 0.175 = 0.325.
119. A high load factor for a power system is desirable because it
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Fixed costs are shared by more units.
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Answer: B. reduces the cost per unit of energy generated
With a high load factor the plant capacity is used more fully, so fixed charges are spread over more kWh and the cost per unit falls.
120. Spatial (small-area) load forecasting is mainly concerned with predicting
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Think of a map of the service area.
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Answer: A. where in the service area the load will grow, for siting substations and feeders
Distribution planning needs to know the location as well as the amount of future load, to place substations and feeders.
121. The reserve capacity of a generating station is
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It is spare capacity at the time of peak.
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Answer: D. installed capacity minus maximum demand
Reserve capacity = plant capacity − maximum demand; it covers outages, maintenance and load growth.
7.5 Distribution systems
30 questions · AElE0705
122. The main disadvantage of a radial distribution system is that
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There is only one supply path.
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Answer: A. a fault on the feeder interrupts supply to all consumers beyond it
A radial feeder has only one path, so any fault cuts off all downstream consumers; the far-end consumers also suffer the largest voltage fluctuation.
123. A radial distribution system is most appropriate for
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Simplest and cheapest system.
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Answer: D. low-load-density areas where low cost is more important than reliability
Radial systems are the simplest and cheapest, suiting rural or low-density areas; ring and network systems are used where reliability is critical.
124. In a ring main distribution system,
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Think of a loop.
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Answer: A. each distribution transformer can be fed from two directions
The ring main forms a closed loop, so each substation has two supply paths; a faulty section can be isolated and supply maintained from the other side.
125. Compared with a radial system, a ring main system gives
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Two paths share the load.
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Answer: B. better reliability and less voltage fluctuation at consumer terminals
Two supply paths improve continuity, and since the load is shared by two paths the voltage drop and fluctuations are smaller.
126. An interconnected (network) distribution system is characterised by
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More than one source feeds the network.
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Answer: A. the network being fed from several generating stations or substations
In an interconnected system, the network is supplied from multiple sources, giving the highest reliability and the ability to share reserve, but at the highest cost.
127. In distribution system terminology, a feeder is designed mainly on the basis of
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Its current is the same along its whole length.
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Answer: D. current-carrying capacity, since it has no tappings
A feeder connects the substation to a distribution point without tappings, so its current is the same throughout and its size is chosen by current capacity.
128. A distributor is designed mainly on the basis of
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Consumers are tapped along its length.
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Answer: D. voltage drop, since consumers along it need voltage within limits
Loads are tapped along the distributor, so the statutory limit on voltage variation at consumer terminals governs its size.
129. Standard low-voltage supply in Nepal is
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Nepal uses 50 Hz.
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Answer: B. 230 V single-phase and 400 V three-phase, 50 Hz
Nepal's LV system is 400/230 V, 50 Hz, three-phase four-wire; the frequency in Nepal is 50 Hz.
130. In a three-phase four-wire distribution system with a phase voltage of 230 V, the line voltage is about
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Line voltage = √3 × phase voltage in star.
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Answer: C. 400 V
V_L = √3 × V_ph = 1.732 × 230 ≈ 398 V ≈ 400 V.
131. A three-phase four-wire system supplies unity-power-factor single-phase loads drawing 10 A, 20 A and 30 A in the three phases (balanced voltages). The neutral current is about
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Add the currents as phasors 120° apart.
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Answer: B. 17.3 A
The neutral current is the phasor sum: 10∠0° + 20∠−120° + 30∠120° gives |In| = √(10² + 20² + 30² − 10×20 − 20×30 − 30×10) = √300 ≈ 17.3 A.
132. The chief advantage of the three-phase four-wire secondary distribution system is that it
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Two different voltages from one system.
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Answer: A. provides both 400 V for three-phase motors and 230 V for single-phase loads
Line-to-line voltage supplies three-phase loads while line-to-neutral voltage supplies lighting and domestic single-phase loads.
133. If the neutral conductor of a three-phase four-wire LV system breaks while the loads are unbalanced,
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The load star point is no longer held at earth potential.
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Answer: B. the phase voltages become unequal and the lightly loaded phase may get dangerous overvoltage
Without the neutral, the star point of the loads floats; voltages divide according to load impedances, so the lightly loaded (high-impedance) phase sees a voltage approaching line voltage.
134. Distribution transformers supplying a three-phase four-wire LV network are normally connected
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The LV side needs a neutral.
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Answer: A. delta on the HV side and star on the LV side, with the LV neutral earthed
Delta–star (e.g. Dyn11) provides an LV neutral for single-phase loads and the delta winding allows triplen harmonic currents to circulate.
135. A single-phase two-wire distribution circuit is mainly used for
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One phase and one return.
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Answer: D. domestic and small commercial loads, using one phase and the neutral
Individual houses and small shops are usually supplied single-phase (phase and neutral) at 230 V.
136. A single-phase two-wire distributor 300 m long supplies 20 A at its far end. Each conductor has a resistance of 0.5 Ω/km. The voltage drop (resistive) is
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Current flows out and back.
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Answer: C. 6 V
Loop length = 2 × 0.3 = 0.6 km; loop resistance = 0.6 × 0.5 = 0.3 Ω; drop = 20 × 0.3 = 6 V.
137. A 40 kW three-phase load at 400 V (line) operates at 0.8 power factor. The line current is about
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I = P/(√3 V_L cos φ).
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Answer: C. 72.2 A
I = P/(√3 V_L cos φ) = 40000/(1.732 × 400 × 0.8) ≈ 72.2 A.
138. A two-wire DC distributor AB is fed at A. Loads of 50 A, 30 A and 20 A are taken at 200 m, 400 m and 500 m from A. The loop resistance is 0.2 Ω per km. The voltage drop at the far end is
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Use current-moments about the feeding point.
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Answer: B. 6.4 V
Drop = r × Σ(I × distance) = 0.2×10⁻³ × (50×200 + 30×400 + 20×500) = 0.2×10⁻³ × 32000 = 6.4 V.
139. A 300 m distributor is fed at both ends at equal voltages and has loads of 60 A at 100 m and 40 A at 200 m from end A. The current supplied from end A is about
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Take moments of the loads about end B.
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Answer: C. 53.3 A
I_A = Σ I(L − x)/L = (60 × 200 + 40 × 100)/300 = 16000/300 ≈ 53.3 A; end B supplies 46.7 A, and the minimum voltage occurs at the 60 A load.
140. A 400 m two-wire distributor fed at one end is uniformly loaded at 1 A per metre. The loop resistance is 0.2 Ω/km. The voltage drop at the far end is
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Uniform load acts like the total load at mid-length.
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Answer: C. 16 V
Drop = i r l²/2 = 1 × 0.2×10⁻³ × 400²/2 = 16 V, i.e. the same as the total load (400 A) concentrated at the mid-point.
141. In a uniformly loaded distributor fed at one end, the total voltage drop is equal to that produced by
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Where is the 'centre of gravity' of a uniform load?
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Answer: B. the whole load concentrated at the mid-point of the distributor
Drop = irl²/2 = (il)(rl/2), i.e. total current times the resistance of half the length.
142. Which type of support is least likely to be used for an LV (400/230 V) secondary distribution line in a town?
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Towers are used for long spans and high voltages.
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Answer: A. Steel lattice tower
LV lines use poles (wooden, steel tubular, RCC/PCC); lattice towers are used for high-voltage transmission lines.
143. Compared with steel tubular poles, pre-stressed cement concrete (PCC) poles
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Concrete is durable but heavy.
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Answer: D. have long life and little maintenance but are heavy to transport
Concrete poles do not rust or rot and need little maintenance, but their weight makes transport and erection in difficult terrain harder.
144. Aerial bundled cable (ABC) is used for LV distribution mainly because it
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The conductors are insulated.
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Answer: C. reduces electricity theft by hooking and faults due to trees and contact
ABC consists of insulated phase conductors twisted around a messenger neutral; insulation makes hooking difficult and avoids faults from branches or contact.
145. When selecting conductors for an LV secondary distribution line, which factor(s) must be considered?
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Thermal, electrical and mechanical requirements all apply.
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Answer: D. All of the above
The conductor must carry the load current without overheating, keep voltage drop within limits, and be strong enough for the span and weather loads.
146. The main purpose of a stay (guy) wire on a distribution pole is to
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Think of forces that do not cancel.
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Answer: A. balance the unbalanced horizontal pull of the conductors at dead-end and angle poles
Where conductor tensions do not cancel (terminal, angle or tee-off poles), a stay counteracts the resultant horizontal force and prevents the pole from bending or tilting.
147. A terminal pole carries a resultant horizontal conductor pull of 5 kN at the stay attachment point. The stay wire makes an angle of 30° with the pole. The tension in the stay wire is about
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Horizontal component = T sin θ (θ from the pole).
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Answer: D. 10.0 kN
The horizontal component of stay tension must equal the pull: T sin 30° = 5 kN, so T = 5/0.5 = 10 kN.
148. A stay (guy) insulator is fitted in the stay wire of a distribution pole so that
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Who may touch the lower part of the stay?
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Answer: C. the lower part of the stay stays dead if a live conductor touches the upper part
The egg-shaped stay insulator separates the upper and lower parts of the stay wire, protecting people near ground level if a live conductor falls on the stay.
149. In a stay set, the turnbuckle (or stay bow/adjuster) is used to
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It can be tightened or loosened.
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Answer: B. adjust the tension in the stay wire
Tension in the stay is set and re-adjusted using the turnbuckle or adjustable stay clamp; the stay rod and anchor plate secure it in the ground.
150. For the same power, distance, losses and maximum voltage to earth, a three-phase system needs less conductor material than a single-phase two-wire system mainly because
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Think about return conductors.
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Answer: A. the three phases share the load so less current per conductor is needed overall
In a balanced three-phase system the return currents cancel and each conductor carries a smaller current for the same power, so the total conductor volume needed is less than for single-phase.
151. Service mains are the conductors that
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It is the last link to the customer.
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Answer: D. connect the distributor to the consumer's premises
Feeders run from substation to distribution points, distributors carry supply along streets, and service mains connect the distributor to individual consumers' terminals.
7.6 Voltage regulation and power factor correction
29 questions · AElE0706
152. Voltage regulation in a distribution system is important mainly because
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Think of lamps and motors at the consumer end.
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Answer: D. the performance and life of consumer equipment such as lamps and motors depend on supply voltage
Low voltage reduces light output and makes motors draw more current and overheat; high voltage shortens the life of lamps and appliances, so voltage must be held within limits.
153. For a three-phase feeder with resistance R and reactance X per phase supplying P (W) and Q (var) at line voltage V, the approximate line voltage drop is
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Real power pairs with resistance, reactive with reactance.
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Answer: D. (PR + QX)/V
ΔV ≈ (PR + QX)/V; the reactive power flowing through the reactance is often the dominant cause of voltage drop.
154. An 11 kV feeder with R = 2 Ω and X = 4 Ω per phase supplies a load of 2 MW and 1.5 Mvar. The approximate voltage drop is about
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Use ΔV ≈ (PR + QX)/V with V = 11 kV.
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Answer: C. 909 V
ΔV ≈ (PR + QX)/V = (2×10⁶ × 2 + 1.5×10⁶ × 4)/11000 = 10×10⁶/11000 ≈ 909 V (about 8.3 %).
155. A 500 kvar shunt capacitor bank is connected at the end of an 11 kV feeder whose reactance is 4 Ω per phase. The approximate rise in voltage at that point is
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The capacitor's reactive power flows back through the feeder reactance.
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Answer: B. 182 V
ΔV ≈ Qc X/V = 500×10³ × 4/11000 ≈ 182 V.
156. An on-load tap changer (OLTC) on a power transformer
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'On-load' is the key.
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Answer: A. changes the turns ratio without interrupting the load current
OLTCs use transition resistors or reactors so that tapping can change under load without breaking the circuit or short-circuiting a winding section.
157. An 11/0.4 kV distribution transformer with an off-circuit tap changer is fed from a supply that is 5 % low (10.45 kV). To restore 400 V on the LV side, the tap should be set to
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Fewer primary turns give more secondary volts per primary volt.
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Answer: D. reduce the HV winding turns by 5 %
LV voltage = V_HV × N_LV/N_HV; with 0.95 of the HV turns, 10.45 kV/(0.95 × 11 kV) × 400 = 400 V.
158. A booster transformer or step voltage regulator is used in a distribution system to
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Think of a long rural feeder.
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Answer: A. raise the voltage at an intermediate point along a long feeder
A booster or feeder voltage regulator, connected in series at a point along the feeder, adds a controllable voltage to compensate for the drop in the line.
159. A line drop compensator, used with the automatic voltage regulator of a tap-changing transformer, enables the transformer to
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It estimates the drop along the line.
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Answer: B. maintain a constant voltage at a remote load point instead of at its own terminals
The compensator models the feeder's R and X and subtracts the estimated drop from the measured voltage, so the regulator acts on the voltage at the load centre.
160. A drawback of shunt capacitors for voltage support is that their reactive power output
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Q = V²ωC.
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Answer: A. falls as the square of the voltage, so they help least when voltage is low
Q = V²ωC, so a 10 % fall in voltage reduces the capacitor output by about 19 %, just when support is most needed.
161. The voltage boost given by a series capacitor in a distribution feeder
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The capacitor is in the current path.
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Answer: D. is automatically proportional to the load current
A series capacitor cancels part of the line reactance; its voltage I Xc rises and falls with the load current, so the compensation is self-regulating.
162. A synchronous condenser is
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It is a rotating machine.
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Answer: C. an over-excited synchronous motor running without mechanical load to supply reactive power
Running over-excited, a synchronous machine draws leading current (supplies vars) and its output can be varied smoothly by adjusting the field current.
163. A static VAR compensator (SVC) controls reactive power mainly by using
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It is a static (no moving parts) device.
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Answer: B. thyristor-controlled reactors and thyristor-switched capacitors
An SVC combines thyristor-controlled reactors and switched capacitors to vary its reactive output quickly and continuously.
164. The power factor of an induction motor is lowest when it is
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Magnetising current hardly changes with load.
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Answer: A. running at light load or no load
At light load the magnetising current is nearly the same as at full load while the active current is small, so the power factor is very low.
165. Which of the following is NOT a consequence of operating a system at low power factor?
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Current = P/(V cos φ).
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Answer: D. Reduced line current for the same kW
For a given kW, a lower power factor means a larger current, which increases kVA rating, I²R losses and voltage drop.
166. A three-phase load draws 50 kVA at 40 kW. Its reactive power is
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Use the power triangle.
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Answer: B. 30 kvar
Q = √(S² − P²) = √(50² − 40²) = 30 kvar (power factor 0.8).
167. A 100 kW load operates at 0.7 power factor lagging. The capacitor rating needed to raise the power factor to 0.95 lagging is about
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Qc = P(tan φ₁ − tan φ₂).
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Answer: C. 69.2 kvar
Qc = P(tan φ₁ − tan φ₂) = 100(1.0202 − 0.3287) ≈ 69.2 kvar.
168. A capacitor bank of 69.2 kvar is to be delta-connected on a 400 V, 50 Hz three-phase supply. The capacitance required per phase is about
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In delta each capacitor gets the full line voltage.
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Answer: B. 459 µF
Per phase Q = 23.07 kvar at 400 V across each capacitor: C = Q/(V²ω) = 23070/(400² × 314.16) ≈ 459 µF.
169. For the same total kvar, capacitors connected in delta compared with star require
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Compare V² across each capacitor.
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Answer: D. one-third of the capacitance per phase, but rated for line voltage
Q per phase ∝ V²C; in delta each capacitor sees √3 times the star voltage, so only one-third the capacitance is needed, though at a higher voltage rating.
170. A load of constant kW at constant voltage has its power factor raised from 0.7 to 0.95. The reduction in line I²R loss is about
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Loss ∝ I² and I ∝ 1/cos φ.
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Answer: C. 45.7 %
Loss ∝ 1/cos²φ: new/old = (0.7/0.95)² = 0.543, a reduction of about 45.7 % (the current falls by 26.3 %).
171. Improving the power factor of a 100 kW load from 0.7 to 0.95 releases supply capacity of about
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Compare kVA = kW/pf before and after.
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Answer: A. 37.6 kVA
kVA before = 100/0.7 = 142.9; after = 100/0.95 = 105.3; capacity released ≈ 37.6 kVA.
172. A consumer pays Rs 200 per kVA of maximum demand per year, and capacitors cost Rs 60 per kvar per year (interest and depreciation). The most economical power factor is about
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sin φ (economic) = capacitor cost per kvar / tariff per kVA.
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Answer: C. 0.954
Most economical pf: sin φ₂ = y/x = 60/200 = 0.3, so cos φ₂ = √(1 − 0.09) ≈ 0.954.
173. Power factor is not usually corrected all the way to unity because
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Consider diminishing returns.
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Answer: A. the extra capacitor cost near unity outweighs the small further saving and risks leading pf at light load
Near unity, each extra kvar gives very little reduction in kVA, so the cost is not justified; fixed capacitors may also over-compensate at light load.
174. To obtain the greatest reduction in distribution losses, power-factor correction capacitors should be installed
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Reactive current should travel the shortest path.
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Answer: C. as close to the inductive loads as possible
Capacitors at the load cancel reactive current in all upstream conductors and transformers, so losses are reduced over the whole path.
175. Fixed capacitors sized for full load, left connected at light load, can cause
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Too many vars for the load.
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Answer: B. leading power factor and a rise in voltage
At light load the reactive demand falls but the capacitor output remains, so the system becomes over-compensated, leading pf and over-voltage result.
176. An automatic power factor correction (APFC) panel improves power factor by
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Steps of capacitors are switched automatically.
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Answer: A. switching capacitor steps in or out according to the measured reactive demand
An APFC relay monitors pf or kvar and switches capacitor steps through contactors, keeping pf near the target as load varies.
177. A phase advancer improves the power factor of
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It is used with slip-ring induction motors.
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Answer: D. an induction motor, by supplying its rotor with exciting ampere-turns at slip frequency
A phase advancer (AC exciter on the motor shaft) feeds the rotor circuit so the magnetising current need not come from the supply, improving the motor's pf.
178. Detuned (series) reactors are connected with power-factor correction capacitors in installations with non-linear loads mainly to
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Capacitors and inductance can resonate.
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Answer: C. avoid resonance between the capacitors and the system at harmonic frequencies
Capacitors can resonate with the network inductance near a harmonic frequency, amplifying harmonic currents; a series reactor tunes the combination below the dominant harmonic.
179. Shunt reactors are installed on long EHV lines and cable systems mainly to
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They are the opposite of shunt capacitors.
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Answer: B. absorb excess reactive power and limit voltage rise at light load
At light load, line charging generates excess vars and raises voltage (Ferranti effect); shunt reactors absorb these vars.
180. Utilities penalise consumers with low power factor mainly because
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Equipment is rated in kVA.
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Answer: A. the utility must provide larger kVA capacity and bear more losses for the same kW
Low pf increases current and kVA demand for the same real power, requiring larger generators, transformers and lines and causing more losses, which the utility recovers through penalties or kVA tariffs.