Nepal Engineering Council · Electrical Engineering · Chapter 6
Power System Protection
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188 questions in 6 syllabus topics.
6.1 Fuses, isolators and reactors
32 questions · AElE0601
1. Which metal is most widely used as the fuse element in HRC cartridge fuses?
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Think of a precious metal with the lowest resistivity that does not oxidise.
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Answer: D. Silver
Silver has very low resistivity, does not oxidise, and its vapour does not form a conducting deposit, so its fusing characteristic stays stable.
2. The filling powder packed around the element inside an HRC cartridge fuse is usually
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The filler must be a good insulator that cools the arc; conducting powders are ruled out.
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Answer: A. Quartz sand (silica)
Pure quartz sand absorbs the arc energy and reacts with the metal vapour to form a high-resistance glassy mass (fulgurite) that quenches the arc.
3. The fusing factor of a fuse (minimum fusing current divided by current rating) is always
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Should a fuse blow at its own rated current?
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Answer: A. Greater than 1
A fuse must carry its rated current indefinitely without melting, so the minimum fusing current is higher than the rating, making the ratio greater than 1.
4. A fuse has a current rating of 20 A and a minimum fusing current of 30 A. Its fusing factor is
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Divide the minimum fusing current by the rating.
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Answer: B. 1.5
Fusing factor = minimum fusing current / current rating = 30 / 20 = 1.5.
5. A fuse wire of 0.5 mm diameter melts at 10 A. Using Preece's law (fusing current ∝ d^1.5), a wire of the same material with 1.0 mm diameter will melt at about
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Raise the diameter ratio to the power 1.5.
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Answer: C. 28.3 A
I2 = I1 × (d2/d1)^1.5 = 10 × 2^1.5 = 28.3 A.
6. In fuse terminology, the 'cut-off current' is
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It is the peak of the current that the fuse actually lets through.
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Answer: D. The maximum instantaneous current actually reached before the fuse element melts
When a current-limiting fuse melts before the first peak of fault current, the highest current reached is called the cut-off current; it is lower than the prospective peak.
7. The total operating time of a fuse is the sum of
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First the element has to melt, then something has to be extinguished.
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Answer: B. Pre-arcing (melting) time and arcing time
Operating time = pre-arcing time (from fault start until the element melts) + arcing time (from melting until the arc is extinguished).
8. The time-current characteristic of a fuse is
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Heating depends on I²t.
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Answer: D. Inverse: the larger the current, the shorter the melting time
Heat generated is proportional to I²t, so a higher current melts the element sooner, giving an inverse time-current characteristic.
9. A main drawback of a semi-enclosed rewirable (kit-kat) fuse compared with an HRC fuse is
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Think about what an untrained person might do when the wire blows.
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Answer: A. Low and uncertain breaking capacity, and its wire can be replaced by a wrong size
Rewirable fuses have low breaking capacity, their element deteriorates by oxidation, and a wrong gauge of wire can easily be fitted, so protection is unreliable.
10. An HRC fuse is described as 'current limiting' because it
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Compare the cut-off current with the prospective peak current.
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Answer: C. Clears the fault before the fault current reaches its first prospective peak
On heavy faults the HRC element melts within the first quarter cycle, so the let-through current is much lower than the prospective peak.
11. A disadvantage of HRC cartridge fuses is that
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Ask what happens to the fuse itself once it has done its job.
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Answer: A. They must be replaced after each operation
HRC fuses are fast, reliable and have high breaking capacity, but the cartridge cannot be reused and must be replaced after it operates.
12. Drop-out (expulsion) fuses are typically used for
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Its operation is visible from the ground.
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Answer: D. Protecting pole-mounted distribution transformers and overhead feeders
The expulsion drop-out fuse is mounted outdoors on poles; when it blows, the carrier swings down, giving a visible indication and isolation.
13. Fuses used to protect power semiconductor devices such as thyristors must have
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Semiconductors have very low thermal capacity.
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Answer: B. Very fast operation and a low let-through I²t
Semiconductor devices have very small thermal capacity, so their fuses must clear within a fraction of a cycle and limit I²t below the device's withstand value.
14. A 3 kW, 230 V single-phase heater is to be protected by a fuse. Which is the smallest of these standard fuse ratings that will carry its full-load current continuously?
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Find I = P/V first.
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Answer: C. 16 A
Full-load current = 3000 / 230 ≈ 13.0 A, so the next standard rating above it, 16 A, is chosen.
15. A fuse has a pre-arcing I²t of 400 A²s. If a constant fault current of 2000 A flows, the pre-arcing time is
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Divide the I²t value by the square of the current.
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Answer: A. 0.1 ms
t = I²t / I² = 400 / (2000²) = 1 × 10⁻⁴ s = 0.1 ms.
16. An isolator (disconnector) in a substation is designed to be operated
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Does an isolator have any means of extinguishing an arc?
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Answer: B. Only under no-load conditions
An isolator has no arc-quenching arrangement, so it is opened only after the circuit breaker has interrupted the current.
17. Before maintenance on a feeder, which switching sequence is correct?
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Current must be interrupted by the device that can handle an arc.
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Answer: C. Open the circuit breaker, then open the isolators, then close the earth switch
The breaker interrupts the current; the isolators then provide a visible open point; finally the earth switch grounds the isolated section for safety.
18. The main reason isolators cannot replace circuit breakers is that isolators
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What is needed to break a current-carrying circuit?
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Answer: A. Have no arc-quenching means
Isolators only provide visible isolation; without arc-quenching chambers they cannot safely interrupt load or fault current.
19. Electrical or mechanical interlocking between an isolator and its circuit breaker is provided to
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It protects the isolator from carrying out an operation it is not designed for.
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Answer: D. Prevent the isolator from being operated while the breaker is closed
Interlocking ensures that the isolator can be opened or closed only when the associated breaker is open, so the isolator never makes or breaks current.
20. The earthing switch fitted with a line isolator is used to
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It is closed only after the line is dead.
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Answer: D. Discharge trapped charge and ground the isolated line before work
After the line is isolated, the earth switch connects it to ground to drain trapped charge and to protect workers from accidental energisation or induced voltages.
21. The pantograph type isolator is preferred where
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Think about the direction in which its arms move.
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Answer: A. Floor space is limited, since it connects vertically to an overhead busbar
The pantograph isolator opens and closes by a vertical scissor movement to the busbar above, so it needs very little horizontal space.
22. Compared with an isolator, a load-break switch is able to
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Its name tells you what kind of current it can break.
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Answer: B. Make and break normal load current but not interrupt fault current
A load-break switch has simple arc-control to switch rated load currents; fault interruption is left to a fuse or circuit breaker.
23. The main purpose of series current-limiting reactors in a power system is to
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They are connected in series with the circuit, not across it.
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Answer: C. Limit short-circuit current to within the breaking capacity of circuit breakers
Series reactors add reactance in the fault path, reducing fault current so that breakers of moderate rating can be used and equipment is protected from fault stresses.
24. Current-limiting reactors are usually of the air-cored type because
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What happens to an iron core at very high currents?
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Answer: A. Their reactance must remain constant even at high fault currents
An iron core would saturate under fault current and its reactance would fall exactly when it is needed; air-cored reactors do not saturate.
25. A disadvantage of connecting reactors in series with each generator (generator reactors) is
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These reactors carry the full generator current at all times.
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Answer: D. A constant voltage drop and power loss during normal operation
Generator reactors carry full load current all the time, causing continuous voltage drop and I²R loss; modern generators have enough leakage reactance, so they are rarely used.
26. The main advantage of bus-bar (tie-bar) reactors over feeder reactors is that
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Consider how much current normally flows between bus-bar sections.
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Answer: C. They cause little voltage drop and loss in normal operation since little current flows through them
Bus-bar reactors connect sections of the bus; under normal conditions little power flows between sections, so the reactor drop and loss are small, while fault feed from other sections is limited.
27. A 20 MVA, 11 kV generator has a reactance of 10%. What series reactor (in ohms per phase) is needed to limit the three-phase short-circuit level at its terminals to 100 MVA?
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Find the total per-unit reactance needed, subtract the generator's share, then convert to ohms.
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Answer: B. 0.605 Ω
Total reactance needed = 20/100 = 0.2 pu, so reactor = 0.1 pu on 20 MVA. Base impedance = 11²/20 = 6.05 Ω, so X = 0.1 × 6.05 = 0.605 Ω.
28. A three-phase 11 kV reactor carries a rated current of 500 A and has a reactance of 3%. Its reactance per phase is about
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Use the phase voltage, not the line voltage.
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Answer: B. 0.381 Ω
%X = I·X / V_phase × 100, so X = 0.03 × (11000/√3) / 500 = 0.381 Ω.
29. Two identical 10 MVA generators, each of 10% reactance, run in parallel on a common bus. Neglecting other impedances, the three-phase fault level at the bus is
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Parallel generators halve the reactance on the same base.
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Answer: C. 200 MVA
The two 10% reactances in parallel give 5% on a 10 MVA base. Fault MVA = 10 × 100/5 = 200 MVA.
30. A reactor has 10% reactance on a 10 MVA base. Its reactance on a 50 MVA base (same voltage) is
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Per-unit impedance is proportional to the base MVA.
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Answer: D. 50%
Per-unit reactance changes in proportion to the base MVA: 10% × 50/10 = 50%.
31. Shunt reactors are connected at the ends of long EHV lines mainly to
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Think of what happens to the receiving-end voltage of a lightly loaded long line.
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Answer: B. Absorb the line's charging reactive power and limit over-voltage at light load
At light load the line's charging current causes a voltage rise (Ferranti effect); shunt reactors absorb this reactive power and keep the voltage within limits.
32. A three-phase, star-connected shunt reactor is rated 20 Mvar at 132 kV. Its reactance per phase is about
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For a balanced three-phase reactor, Q = V_L²/X.
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Answer: C. 871 Ω
X per phase = V_L² / Q = 132² / 20 = 871.2 Ω.
6.2 Circuit breakers
32 questions · AElE0602
33. SF6 gas is an excellent arc-quenching medium in circuit breakers mainly because it is
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The key property is about what the gas does to free electrons.
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Answer: A. Strongly electronegative, capturing free electrons to form heavy negative ions
SF6 molecules attach free electrons to form heavy, slow ions, so the arc space regains dielectric strength very quickly after current zero.
34. At the same pressure, the dielectric strength of SF6 compared with air is about
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It is better than air, but not by orders of magnitude.
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Answer: D. 2 to 3 times higher
SF6 has roughly 2.5 times the dielectric strength of air at equal pressure, and the value rises further when it is pressurised.
35. Vacuum circuit breakers are most widely used for
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Think of the typical distribution voltages used by utilities.
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Answer: A. Medium voltages, typically 3.3 kV to 33 kV
Vacuum interrupters are compact and need little maintenance; they dominate the medium-voltage range, while SF6 breakers are used at higher transmission voltages.
36. In a vacuum circuit breaker, the arc after contact separation is sustained by
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There is no gas or liquid inside the interrupter.
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Answer: D. Metal vapour released from the contact surfaces
There is no gas in a vacuum interrupter; the arc burns in metal vapour evaporated from the contacts, which condenses rapidly at current zero.
37. The contacts of modern vacuum interrupters are commonly made of
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An alloy of copper with a refractory metal.
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Answer: A. Copper-chromium alloy
Cu-Cr contacts give low chopping current, good anti-welding properties and high dielectric recovery in vacuum.
38. Current chopping in a circuit breaker is most likely to produce dangerous over-voltages when it interrupts
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The over-voltage comes from energy stored in an inductance.
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Answer: D. Small inductive currents, such as the magnetising current of an unloaded transformer
When a small inductive current is forced to zero before its natural zero, the energy ½Li² stored in the inductance transfers to the stray capacitance, giving high over-voltage.
39. The gas produced in largest proportion when an arc burns in transformer oil inside an oil circuit breaker is
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It is the lightest gas, with excellent cooling property.
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Answer: D. Hydrogen
The arc decomposes the oil, releasing mainly hydrogen, which has high thermal conductivity and helps to cool and de-ionise the arc.
40. In a minimum oil circuit breaker, the oil is used mainly
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The 'minimum' quantity of oil is enough only for one job.
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Answer: A. As the arc-quenching medium, with solid insulation used for insulation to earth
Unlike the bulk oil breaker, the minimum oil breaker keeps the oil in a small insulated chamber for arc extinction; porcelain and other solid materials insulate it from earth.
41. A major disadvantage of bulk oil circuit breakers is
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Consider what can happen to a large volume of hydrocarbon near an arc.
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Answer: D. The risk of fire and the large quantity of oil needing maintenance
Bulk oil breakers use a large volume of flammable oil, which carries fire/explosion risk and requires regular testing and replacement as it carbonises.
42. A notable drawback of air-blast circuit breakers is
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The arc is extinguished by a high-pressure jet released into the atmosphere.
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Answer: A. High noise during operation and the need for a compressed-air plant
Air-blast breakers are fast and free of fire risk, but the high-pressure blast is very noisy and an air compressor system must be maintained.
43. In a low-voltage air circuit breaker, the arc is extinguished mainly by
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Think of what arc runners do to the length of the arc.
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Answer: B. Lengthening and cooling it with arc runners and arc chutes (high-resistance method)
The arc rises along arc runners into arc chutes, where it is stretched, split and cooled until its resistance becomes so high that it cannot be maintained.
44. The high-resistance method of arc interruption is mainly suitable for
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Which kind of current has no natural zero to help interruption?
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Answer: D. DC circuits and low-power AC circuit breakers
Raising arc resistance dissipates large arc energy, so it is practical only at low power and in DC circuits, which have no natural current zero.
45. The transient voltage that appears across the breaker contacts immediately after the arc is extinguished at current zero is called
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It is a transient, not the steady power-frequency voltage that follows.
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Answer: A. Restriking voltage
The high-frequency transient that follows current zero is the restriking (transient recovery) voltage; when it dies out, the power-frequency voltage that remains is the recovery voltage.
46. Resistance switching (a resistor connected across the breaker contacts) is used to
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The resistor is placed across the contacts, during the arcing period.
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Answer: B. Reduce the rate of rise of restriking voltage and damp switching transients
A shunt resistor damps the LC oscillation after current zero, lowering the RRRV and the peak restriking voltage so the arc is less likely to restrike.
47. In the circuit-breaker operating sequence O – t – CO – t′ – CO, the symbol 'CO' means
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Read the letters in order as two separate operations.
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Answer: A. A closing operation followed immediately (without intentional delay) by an opening operation
The rated operating sequence uses O for open and CO for close followed at once by open, as happens when a breaker recloses onto a persisting fault.
48. Which of the following is a type test rather than a routine test of a circuit breaker?
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A type test proves the design and is not repeated on every unit.
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Answer: C. Short-circuit making and breaking capacity test
Short-circuit tests require a large testing station and are done on a representative sample to prove the design; the others are done on every breaker produced.
49. Synthetic testing of circuit breakers is used because it
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Very large breakers exceed the capacity of direct testing stations.
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Answer: D. Uses separate current and voltage sources to simulate large short-circuit duty with smaller plant
A high-current low-voltage source supplies the fault current and a separate high-voltage source applies the recovery voltage, reproducing the full duty at much lower test power.
50. For proper selection, the symmetrical breaking capacity of a circuit breaker must be
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Consider the worst current the breaker may have to interrupt.
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Answer: C. At least equal to the maximum prospective fault level at its location
A breaker must safely interrupt the largest fault current that can flow at its point of installation, so its breaking capacity must not be lower than that fault level.
51. In a miniature circuit breaker (MCB), overload and short-circuit protection are provided respectively by
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Slow heating versus a sudden magnetic pull.
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Answer: B. A bimetallic (thermal) element and an electromagnetic trip coil
The bimetal bends slowly with sustained overload current (inverse time), while the magnetic coil trips instantly on high short-circuit current.
52. Which MCB tripping characteristic is most suitable for loads with very high inrush current, such as transformers or X-ray machines?
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Pick the curve with the highest instantaneous trip multiple.
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Answer: A. Type D
Type D MCBs trip magnetically at 10 to 20 times rated current, so they ride through large inrush currents; B (3–5 In) suits resistive loads and C (5–10 In) suits motors and general loads.
53. Compared with an MCB, a moulded case circuit breaker (MCCB) typically offers
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MCCBs are used on main and sub-main distribution boards.
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Answer: D. Higher current ratings and adjustable trip settings
MCCBs are available in ratings up to a few thousand amperes, with higher breaking capacity and often adjustable thermal and magnetic settings, unlike typical fixed-setting MCBs.
54. A circuit breaker on an 11 kV three-phase system has a symmetrical breaking current of 20 kA. Its breaking capacity is about
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Use the three-phase power formula with line values.
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Answer: C. 381 MVA
Breaking capacity = √3 × V × I = √3 × 11 kV × 20 kA ≈ 381 MVA.
55. A circuit breaker has a rated symmetrical breaking current of 20 kA (r.m.s.). Its rated making current (peak) is about
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The standard multiplying factor is 2.55.
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Answer: C. 51 kA
Rated making current = 2.55 × symmetrical breaking current = 2.55 × 20 ≈ 51 kA peak, allowing for √2 and the maximum DC offset.
56. At contact separation, the AC component of a fault current is 10 kA r.m.s. and the DC component is 7 kA. The asymmetrical breaking current is about
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Combine the two components as r.m.s. values, not by simple addition.
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Answer: C. 12.2 kA
I_asym = √(I_ac² + I_dc²) = √(10² + 7²) ≈ 12.2 kA.
57. A three-phase circuit breaker is rated 2500 MVA at 132 kV. Its rated symmetrical breaking current is about
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Rearrange S = √3·V·I for I.
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Answer: B. 10.9 kA
I = MVA / (√3 × kV) = 2500 / (√3 × 132) ≈ 10.9 kA.
58. In a circuit breaker test circuit, the inductance is 6 mH and the stray capacitance is 0.01 µF. The natural frequency of the restriking voltage is about
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Use f = 1/(2π√LC) with SI units.
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Answer: B. 20.5 kHz
f = 1 / (2π√(LC)) = 1 / (2π√(6×10⁻³ × 10⁻⁸)) ≈ 20.5 kHz.
59. For a breaker circuit with L = 4 mH and C = 0.01 µF, the value of resistance across the contacts that just gives critical damping of the restriking transient is about
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The critical value is half of √(L/C).
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Answer: B. 316 Ω
Critical resistance R = ½√(L/C) = 0.5 × √(4×10⁻³ / 10⁻⁸) = 0.5 × 632 ≈ 316 Ω.
60. A breaker chops a magnetising current of 4 A in a transformer of inductance 10 H with stray capacitance 1000 pF. Neglecting losses, the prospective over-voltage is about
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Equate magnetic energy in L to electric energy in C.
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Answer: B. 400 kV
Equating ½Li² = ½Cv², v = i√(L/C) = 4 × √(10 / 10⁻⁹) = 4 × 10⁵ V = 400 kV.
61. A circuit breaker has a 1-second short-time current rating of 40 kA. Assuming the same I²t, its 3-second rating would be about
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Thermal withstand depends on I²t.
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Answer: B. 23.1 kA
For equal I²t, I₃ = 40 × √(1/3) ≈ 23.1 kA.
62. A 1 MVA, 11/0.415 kV transformer has 5% impedance. Assuming an infinite source, the three-phase fault current at its LV terminals, which the LV breaker must interrupt, is about
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Divide the full-load current by the per-unit impedance.
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Answer: C. 27.8 kA
Full-load current = 1000 / (√3 × 0.415) ≈ 1391 A; fault current = 1391 / 0.05 ≈ 27.8 kA.
63. A Type C MCB is rated 16 A. Its instantaneous (magnetic) trip operates for currents in the range of about
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Type C corresponds to 5 to 10 times In.
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Answer: C. 80 A to 160 A
Type C MCBs trip magnetically between 5 and 10 times rated current: 5 × 16 = 80 A to 10 × 16 = 160 A.
64. For a circuit breaker interrupting a fault on an 11 kV system, the peak value of restriking voltage in an undamped circuit is about
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Start from the peak phase voltage and double it.
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Answer: C. 18.0 kV
Peak phase voltage = 11√2/√3 ≈ 8.98 kV; for an undamped LC circuit the restriking voltage can reach twice this, about 18.0 kV.
6.3 Protective relays
32 questions · AElE0603
65. The ability of a protection system to disconnect only the faulty section while leaving the healthy parts in service is called
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It is about which section gets tripped, not how fast or how small a fault is seen.
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Answer: C. Selectivity
Selectivity (discrimination) means only the breakers nearest the fault trip, so the minimum part of the system is disconnected.
66. An attracted-armature relay is mainly used where the required operation is
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There is no rotating disc, so there is no built-in delay.
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Answer: D. Instantaneous
Attracted-armature relays close their contacts as soon as the magnetic pull exceeds the restraint, giving fast instantaneous operation.
67. Induction disc type relays can be used
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The operating torque needs time-varying, phase-shifted fluxes.
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Answer: A. Only on AC circuits
The disc is driven by eddy currents from two phase-displaced alternating fluxes; a steady DC flux produces no driving torque.
68. An IDMT overcurrent relay shows a definite minimum operating time at very high currents mainly because
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What limits the flux in an iron core?
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Answer: A. Its magnetic circuit saturates, so torque stops increasing with current
At high multiples of setting the relay's electromagnet saturates, so further increase in current hardly increases the torque, giving a near-constant minimum time.
69. In an induction disc overcurrent relay, the time multiplier setting (TMS) adjusts the operating time by
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The current setting is changed by the plug; the time setting by something mechanical.
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Answer: D. Changing the distance the disc travels before the contacts close
The TMS moves the backstop, altering the angle the disc must rotate, so the operating time changes in proportion to TMS for the same current.
70. A directional overcurrent relay needs, in addition to current,
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Direction is decided by comparing a current with something else.
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Answer: C. A voltage (polarising) input to determine the direction of power flow
Direction of fault current is judged by its phase angle relative to a reference voltage, so a VT input is needed for polarisation.
71. For earth-fault protection of a three-phase feeder, the earth-fault relay is commonly connected
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Earth-fault current shows up as an imbalance in the sum of phase currents.
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Answer: C. In the residual circuit of three line CTs, or to a core-balance CT surrounding all phase conductors
The phasor sum of three phase currents is zero under normal or phase-fault conditions; any residual current indicates an earth fault.
72. Earth-fault relays can usually be given much lower current settings than phase overcurrent relays because
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What does the residual circuit see during normal full-load operation?
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Answer: A. Normal balanced load current produces no residual current
In a balanced system the residual current is near zero even at full load, so the earth-fault relay can be set at a small fraction of rated current for high sensitivity.
73. An undervoltage relay is commonly used with induction motors to
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Think of what should happen when the supply disappears and later returns.
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Answer: D. Disconnect the motor when supply voltage fails or falls too low, preventing uncontrolled restart
On loss or severe dip of voltage the motor would draw high current and stall, and could restart dangerously when supply returns; the undervoltage relay trips it instead.
74. A contactor differs from a circuit breaker mainly in that a contactor is designed
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A motor starter may switch the same motor many times a day.
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Answer: D. For frequent making and breaking of normal load currents, not for interrupting fault currents
Contactors are electrically held switches with long mechanical life for repeated load switching, e.g. motor starting; fault interruption is left to fuses or breakers.
75. The thermal overload relay in a motor starter mainly protects the motor against
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Bimetals respond slowly to heat.
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Answer: B. Sustained overload currents that would overheat its windings
A bimetallic overload relay responds to I²t heating with an inverse time characteristic, matching the motor's thermal capability; short circuits need fuses or MCCB.
76. In a typical motor circuit, protection against short circuits is usually provided by
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Short circuits must be cleared very fast by a device with high breaking capacity.
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Answer: A. Back-up HRC fuses or an MCCB with instantaneous trip
Thermal overload relays are too slow and have low breaking capacity, so short circuits are cleared by fuses or a breaker ahead of the contactor.
77. Differential protection operates on the principle of
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Kirchhoff's current law applied to a zone.
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Answer: C. Comparing the currents entering and leaving the protected zone
Under normal conditions and external faults the two currents balance; an internal fault creates a difference that operates the relay.
78. Percentage (biased) differential relays are used instead of simple differential relays mainly to
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Spill current grows with through-fault current.
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Answer: A. Prevent mal-operation on heavy external faults due to CT mismatch and saturation
The restraint winding raises the required differential current in proportion to through current, so spill currents from unequal CTs during through-faults do not cause tripping.
79. Transformer differential relays use second-harmonic restraint to avoid tripping during
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Which current is rich in the 2nd harmonic and appears at switch-on?
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Answer: D. Magnetising inrush when the transformer is energised
Inrush current flows only in one winding and looks like an internal fault, but it is rich in second harmonic, which is used to block the relay.
80. For differential protection of a delta-star power transformer, the CTs are connected
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The CT connection is opposite to the winding it serves.
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Answer: C. In delta on the star side and in star on the delta side
This corrects the 30° phase shift of the transformer and removes zero-sequence currents from the star side, so the relay currents balance on through faults.
81. The Buchholz relay is installed
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It responds to gas from decomposing oil.
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Answer: B. In the pipe between the main tank and the conservator of an oil-filled transformer
Gases produced by incipient faults rise toward the conservator and collect in the Buchholz chamber, operating the alarm and trip floats.
82. A distance relay basically measures
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Voltage divided by current at the relaying point.
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Answer: C. The impedance (V/I) between the relay location and the fault
Since line impedance is proportional to length, the measured V/I indicates the distance to the fault; the relay operates if it is below the zone setting.
83. Which distance relay is inherently directional?
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Look for the characteristic that passes through the origin.
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Answer: B. Mho (admittance) relay
The mho characteristic on the R-X diagram is a circle passing through the origin, so it operates only for faults in the forward direction.
84. A reactance relay is preferred for short transmission lines because
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On a short line, the arc's resistance can be comparable with line impedance.
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Answer: A. Its operation is not affected by fault arc resistance
Arc resistance is large compared with the impedance of a short line; a reactance relay measures only X, so arc resistance does not cause under-reach.
85. A plain impedance relay needs an additional directional unit because
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Where is the centre of its characteristic on the R-X diagram?
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Answer: D. Its circular characteristic centred at the origin makes it non-directional
The impedance relay operates for any impedance inside the circle around the origin, including faults behind it, so a directional element supervises it.
86. Zone 1 of a distance relay is normally set to about 80–90% of the protected line length in order to
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Zone 1 has no time delay.
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Answer: B. Avoid over-reaching into the next line due to CT, VT and line-data errors
Zone 1 trips instantaneously, so it must not reach beyond the remote bus; the margin covers measurement errors. The rest of the line is covered by zone 2.
87. Restricted earth-fault (REF) protection of a transformer is used to
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'Restricted' refers to the zone covered.
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Answer: C. Detect earth faults within the transformer winding zone, especially near the star point
REF compares the neutral CT current with the residual of the line CTs, giving sensitive protection for earth faults inside the protected winding only.
88. An overcurrent relay is connected to a 400/5 A CT and has a plug setting of 125%. For a fault current of 6000 A, the plug setting multiplier (PSM) is
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PSM = relay current ÷ (plug setting × CT secondary rating).
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Answer: B. 12
Relay current = 6000 × 5/400 = 75 A; pick-up = 1.25 × 5 = 6.25 A; PSM = 75 / 6.25 = 12.
89. A standard inverse IDMT relay follows t = 0.14 × TMS / (PSM^0.02 − 1). For PSM = 10 and TMS = 0.5, the operating time is about
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Compute 10^0.02 carefully; it is just above 1.
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Answer: B. 1.49 s
t = 0.14 × 0.5 / (10^0.02 − 1) = 0.07 / 0.0471 ≈ 1.49 s.
90. An IDMT relay operates in 3.0 s at TMS = 1.0 for a certain fault current. With TMS changed to 0.3 and the same current, the operating time is
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TMS scales the curve vertically.
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Answer: B. 0.9 s
Operating time is proportional to TMS: 3.0 × 0.3 = 0.9 s.
91. A 100 km, 132 kV line has an impedance of 0.5 Ω/km. Zone 1 is set at 80% of the line. With CT 400/1 A and VT 132 kV/110 V, the zone 1 setting in secondary ohms is about
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Z_sec = Z_pri × (CT ratio ÷ VT ratio).
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Answer: A. 13.3 Ω
Primary zone 1 = 0.8 × 50 = 40 Ω. Secondary Z = Z_primary × CT ratio / VT ratio = 40 × 400 / 1200 ≈ 13.3 Ω.
92. A biased differential relay operates when the differential current exceeds 20% of the average of the two CT secondary currents. If these are 5.0 A and 4.2 A, will the relay operate?
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Compute the difference and 20% of the average separately, then compare.
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Answer: D. No; the threshold is 0.92 A, more than the 0.8 A difference
Differential current = 5.0 − 4.2 = 0.8 A; average through current = 4.6 A; threshold = 0.2 × 4.6 = 0.92 A > 0.8 A, so it does not operate.
93. An earth-fault relay with a 20% current setting is fed from 200/1 A CTs. The minimum primary earth-fault current that will operate it is
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Convert the relay pick-up back through the CT ratio.
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Answer: B. 40 A
Relay pick-up = 0.2 × 1 A = 0.2 A secondary, which corresponds to 0.2 × 200 = 40 A primary.
94. In a radial system, the downstream relay clears a fault in 0.6 s. With a grading (discrimination) margin of 0.4 s, the upstream relay should operate for the same fault in at least
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Back-up must wait for the primary relay plus a margin.
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Answer: D. 1.0 s
The upstream relay is backup; its time must exceed the downstream time by the grading margin: 0.6 + 0.4 = 1.0 s.
95. The star winding of a transformer is earthed through a resistor that limits a full-voltage earth fault to rated full-load current. If the REF relay is set at 15% of full-load current, the percentage of the winding protected is
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Fault current is proportional to distance of the fault from the neutral.
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Answer: C. 85%
Earth-fault current is proportional to the fraction of winding between the neutral and fault. Faults within 15% of the winding from the neutral give less than the setting, so 85% is protected.
96. A 50 km line has zone 1 set at 80% and zone 2 at 120% of its length. A solid fault occurs 45 km from the relaying point on this line. It will be cleared by
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Compare the fault distance with each zone reach in kilometres.
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Answer: A. Zone 2, after its time delay
Zone 1 reaches 40 km and zone 2 reaches 60 km, so a fault at 45 km lies in zone 2 and is cleared with the zone 2 time delay (unless a carrier scheme is used).
6.4 Lightning protection
31 questions · AElE0604
97. In most cloud-to-ground lightning flashes, the charge lowered to earth from the base of the thundercloud is
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The upper part of the cloud carries the opposite charge to the base.
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Answer: A. Negative
The lower part of a thundercloud usually carries negative charge, and about 90% of cloud-to-ground flashes lower negative charge to earth.
98. The part of a lightning flash that carries the highest current and produces the brightest luminosity is the
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It begins after the channel to ground is complete.
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Answer: A. Return stroke
After the stepped leader bridges the gap to earth, the return stroke travels up the ionised channel carrying tens of kiloamperes; it is the visible flash.
99. The peak current of a typical cloud-to-ground lightning stroke lies in the range of
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It is measured in kiloamperes.
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Answer: A. About 10 kA to 100 kA
Measured stroke currents typically lie between about 10 kA and 100 kA, with a median near 30 kA.
100. A standard 1.2/50 µs lightning impulse voltage has
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The first number is short because the rise is steep.
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Answer: D. A front time of 1.2 µs and a time to half-value of 50 µs
The standard lightning impulse rises to peak in 1.2 µs (front) and decays to half its peak at 50 µs (tail).
101. Indirect (induced) lightning over-voltages on an overhead line are produced when
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No direct contact between lightning and the line is involved.
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Answer: A. A charged cloud discharges elsewhere, releasing the bound charge on the line as travelling waves
A charged cloud induces opposite bound charge on the line; when the cloud discharges nearby, this charge is suddenly freed and travels along the line as surges.
102. Induced lightning over-voltages are usually of most concern for
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Compare the typical size of induced surges with the insulation level of different lines.
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Answer: B. Distribution lines with relatively low insulation levels
Induced surges rarely exceed a few hundred kV, which EHV insulation can withstand, but they can flash over medium- and low-voltage distribution insulation.
103. The isokeraunic (keraunic) level of a place refers to
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It is counted in days per year.
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Answer: D. The number of thunderstorm days in a year
It is the number of days per year on which thunder is heard, and is used to estimate the lightning flash density for line design.
104. Using Ng = 0.04 Td^1.25, the ground flash density for a region with 50 thunderstorm days per year is about
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Raise 50 to the power 1.25 first.
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Answer: B. 5.3 flashes/km²/year
Ng = 0.04 × 50^1.25 = 0.04 × 133 ≈ 5.3 flashes per km² per year.
105. The main function of a lightning arrester is to
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It acts like a safety valve for voltage.
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Answer: A. Divert surge current to earth, limit the voltage across equipment and then restore insulation
An arrester acts as a near-open circuit at normal voltage, conducts during an over-voltage to clamp it, and then stops the power-follow current so normal service continues.
106. A lightning arrester protecting a transformer is connected
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It must offer a path from the line to ground.
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Answer: B. Between each line conductor and earth, as close as possible to the transformer terminals
The arrester is a shunt device from line to earth; placing it near the equipment minimises the extra voltage due to travelling-wave reflections over the separation distance.
107. An ideal lightning arrester should
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Think of a switch that closes only for over-voltage and reopens by itself.
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Answer: C. Draw no current at normal voltage, conduct freely on over-voltage, and stop power-follow current afterwards
These three features ensure the arrester does not disturb normal operation, limits over-voltages effectively, and restores normal service by itself.
108. A major disadvantage of the simple rod-gap arrester is that
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What happens to the arc once the surge has passed?
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Answer: D. It cannot interrupt the power-follow current after the surge has passed
Once the gap sparks over, the arc is maintained by the power-frequency voltage, causing a short circuit that must be cleared by a breaker; its spark-over also varies with weather and polarity.
109. In a horn-gap arrester, the arc is extinguished because
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Look at the shape of the horns.
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Answer: B. It rises along the diverging horns, lengthens and cools until it breaks
Heated air and electromagnetic forces push the arc upward along the diverging horns, increasing its length and resistance until it is extinguished.
110. In an expulsion type lightning arrester, the arc is extinguished by
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The arrester's name describes what happens to the gases.
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Answer: C. Gases produced from the fibre tube lining, which expel the arc products
The arc inside a fibre tube vaporises some of the material; the high-pressure gas is expelled through the open end, de-ionising the gap and quenching the power-follow arc.
111. A conventional valve type (silicon carbide) lightning arrester consists of
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SiC alone would pass too much current at normal voltage.
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Answer: D. Non-linear SiC resistor blocks in series with spark gaps
SiC blocks are not non-linear enough to withstand system voltage continuously, so series gaps are needed to isolate them at normal voltage.
112. The main advantage of metal-oxide (ZnO) arresters over silicon carbide arresters is that
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Consider whether series gaps are still needed.
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Answer: A. Their extreme non-linearity allows gapless construction with negligible current at normal voltage
ZnO has a far more non-linear V-I characteristic, so the leakage current at operating voltage is tiny and series gaps are unnecessary, giving faster and more reliable protection.
113. The residual (discharge) voltage of a lightning arrester is
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It is the voltage that the protected equipment actually sees.
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Answer: B. The voltage across its terminals while the discharge current is flowing
Residual voltage is the I·R drop across the arrester during discharge; it sets the protective level that the equipment insulation must withstand with margin.
114. Lightning arresters are located as close as possible to the transformer they protect because
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Think about what happens to a surge at a high-impedance terminal.
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Answer: A. Travelling-wave reflections raise the voltage at the transformer as the separation increases
A transformer terminal behaves almost like an open end; the reflected wave adds to the incident wave, and the over-voltage at the transformer rises with distance from the arrester.
115. A surge absorber (such as the Ferranti type) protects equipment mainly by
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Steep fronts stress the first few turns of a winding.
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Answer: D. Reducing the steepness of the surge front and absorbing part of its energy
A surge absorber, made of inductance and loss elements, flattens the wavefront and dissipates energy, protecting windings from steep-fronted surges.
116. The main purpose of the overhead earth (ground) wire on a transmission line is to
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It sits above the phase conductors.
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Answer: C. Shield the phase conductors from direct lightning strokes
The earth wire at the top of the towers intercepts lightning strokes and conducts them to earth through the towers, so the phase conductors are shielded.
117. Traditionally, the shielding (protective) angle between the earth wire and the outer phase conductor of a transmission line is kept at about
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It is less than half a right angle.
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Answer: A. 30°
A protective angle of about 30° (smaller for EHV lines) gives good shielding against direct strokes to the phase conductors.
118. Back-flashover on a transmission line occurs when
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The flashover goes from the earthed side to the line, the reverse of normal.
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Answer: B. Lightning strikes the tower or earth wire and the tower-top potential becomes high enough to flash over the insulator to a phase conductor
Stroke current through the tower and footing resistance raises the tower potential; if the insulator string's withstand is exceeded, it flashes over from tower to conductor.
119. Back-flashovers on a line can be reduced most effectively by
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Tower-top voltage depends on the footing resistance.
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Answer: C. Lowering the tower footing resistance, for example with counterpoise wires
The tower-top voltage during a stroke is roughly I × R_footing; buried counterpoise conductors or rods lower R and hence the voltage across the insulators.
120. A 20 kA lightning stroke hits a phase conductor of surge impedance 400 Ω midway along the line. The voltage surge travelling in each direction is about
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Half of the current travels each way.
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Answer: C. 4000 kV
The stroke current divides equally into two directions, so V = (I/2) × Z = 10 kA × 400 Ω = 4000 kV.
121. A 900 kV surge travels on an overhead line of surge impedance 400 Ω into a cable of surge impedance 50 Ω. The voltage transmitted into the cable is
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Use the transmission coefficient 2Z₂/(Z₁+Z₂).
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Answer: C. 200 kV
Transmitted voltage = 2Z₂/(Z₁ + Z₂) × V = (100/450) × 900 = 200 kV.
122. A 300 kV surge reaches the open-circuited end of a transmission line. The voltage at the open end rises to
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At an open end, current becomes zero and the voltage wave reflects fully.
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Answer: D. 600 kV
At an open end the reflection coefficient is +1, so the reflected wave adds to the incident wave: 300 + 300 = 600 kV.
123. A lightning surge travels on an overhead line at about the speed of light. The time it takes to travel 30 km is about
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Speed of light is 3 × 10⁸ m/s.
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Answer: D. 100 µs
t = d / v = 30 000 m / (3 × 10⁸ m/s) = 1 × 10⁻⁴ s = 100 µs.
124. An earth wire is 10 m vertically above the level of a phase conductor. For a shielding angle of 30°, the maximum horizontal offset of the conductor from the earth wire for it to be shielded is about
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Use tan of the shielding angle.
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Answer: B. 5.8 m
Horizontal offset = 10 × tan 30° = 10 × 0.577 ≈ 5.8 m.
125. On an effectively earthed system with a highest system voltage of 145 kV, using an 80% coefficient of earthing, the minimum voltage rating of the lightning arrester is about
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Multiply the highest line voltage by the coefficient of earthing.
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Answer: B. 116 kV
Arrester rating = coefficient of earthing × highest system voltage = 0.8 × 145 = 116 kV.
126. A lightning arrester has a residual voltage of 350 kV. If a protective margin of 20% above the arrester protective level is required, the minimum BIL of the protected transformer is
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Add 20% to the arrester's protective level.
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Answer: C. 420 kV
Minimum BIL = 1.2 × 350 = 420 kV.
127. A 30 kA lightning stroke hits a tower whose footing resistance is 10 Ω. Neglecting tower inductance and earth-wire current sharing, the tower-top potential is about
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Ohm's law with kiloamperes.
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Answer: C. 300 kV
V = I × R = 30 kA × 10 Ω = 300 kV.
6.5 Earthing
31 questions · AElE0605
128. Equipment earthing means
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It concerns parts that do not carry current normally.
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Answer: D. Connecting non-current-carrying metal parts such as enclosures and frames to earth
Equipment (body) earthing bonds exposed metal parts to earth; system earthing refers to earthing the neutral of the supply system.
129. Besides keeping the enclosure near earth potential, an important purpose of equipment earthing is to
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A fault must be cleared, not just tolerated.
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Answer: A. Provide a low-impedance path so that enough fault current flows to operate the fuse or breaker quickly
A good earth path lets a phase-to-frame fault produce a large current, which makes the protective device disconnect the supply before a dangerous touch voltage persists.
130. In an 11 kV system with an isolated (unearthed) neutral, a solid earth fault occurs on one phase. The voltage of each healthy phase to earth becomes about
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The faulted phase is now at earth potential; what voltage remains between it and the others?
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Answer: C. 11 kV
With the faulted phase at earth potential, the healthy phases rise from phase voltage (6.35 kV) to line voltage, √3 × 6.35 = 11 kV.
131. The 'arcing ground' phenomenon, which can cause severe over-voltages, occurs mainly in
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The fault current there is purely capacitive.
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Answer: D. Systems with an isolated (ungrounded) neutral
In an ungrounded system the earth-fault current is capacitive; the arc repeatedly extinguishes and restrikes, charging the line capacitance to high voltages.
132. A Peterson coil (arc suppression coil) earths the neutral through
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The coil's current and the line capacitance current are 180° apart.
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Answer: A. An inductance tuned so that its current cancels the capacitive earth-fault current
Resonant earthing chooses the coil's inductive current equal and opposite to the capacitive fault current, so the fault arc self-extinguishes.
133. Solid (effective) earthing of EHV systems is preferred mainly because it
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Insulation is very expensive at high voltages.
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Answer: D. Keeps healthy-phase voltages low during earth faults, allowing reduced insulation levels and arrester ratings
With the neutral solidly earthed, healthy phases stay close to phase voltage during an earth fault, so insulation and arresters can be designed for lower voltages, a large saving at EHV.
134. The main purpose of earthing a system neutral through a resistor is to
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It is a compromise between solid and isolated neutral.
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Answer: A. Limit earth-fault current to reduce damage while still allowing earth-fault relays to operate
A neutral resistor limits the fault current (reducing burning and mechanical stress) and damps arcing grounds, while leaving enough current for reliable relay operation.
135. A zig-zag (earthing) transformer is used in a power system to
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A delta winding has no star point.
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Answer: C. Provide an artificial neutral point for earthing a delta-connected system
A delta system has no neutral; the zig-zag transformer offers high impedance to positive-sequence but low impedance to zero-sequence current, creating a neutral for earthing.
136. In pipe earthing, alternate layers of charcoal and salt are placed around the electrode mainly to
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The goal is a lower earth resistance.
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Answer: A. Retain moisture and reduce the resistivity of the surrounding soil
Salt dissolves to form an electrolyte and charcoal holds moisture, so the effective soil resistivity around the electrode, and hence earth resistance, is reduced.
137. The resistance of an earth electrode is commonly measured by the
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It uses auxiliary electrodes driven into the ground.
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Answer: D. Fall-of-potential method using an earth tester with current and potential probes
A current is passed between the electrode and a remote current probe; the potential probe placed at about 62% of that distance gives the true electrode resistance.
138. Soil resistivity at an earthing site decreases when
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Conduction in soil is electrolytic.
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Answer: C. Its moisture and dissolved-salt content increase
Soil conducts mainly through electrolytes in its moisture, so more water and salts lower its resistivity; drying or freezing raises it sharply.
139. In a TT earthing system,
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Both letters refer to 'Terre' (earth), directly.
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Answer: D. The consumer's exposed metal parts are earthed through a local electrode independent of the supply earth
The first T means the source is directly earthed; the second T means installation metalwork is earthed through its own separate electrode.
140. Touch voltage is defined as the potential difference between
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It involves the hand.
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Answer: B. An earthed metal structure a person is touching and the ground where the person stands
Touch voltage acts from hand to feet; step voltage is between the two feet about 1 m apart; ground potential rise is the electrode-to-remote-earth voltage.
141. The main purpose of equipotential bonding in an installation is to
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Current needs a potential difference to flow through a body.
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Answer: A. Minimise voltage differences between metal parts that can be touched at the same time
Bonding water pipes, gas pipes, structural steel and equipment together keeps them at nearly the same potential during a fault, so dangerous touch voltages cannot appear between them.
142. A 30 mA residual current device (RCD/RCCB) is installed mainly to
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30 mA is about the body current, not the load current.
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Answer: A. Protect people against electric shock due to earth leakage
A 30 mA RCD trips quickly when leakage to earth exceeds about 30 mA, a level chosen to prevent fatal shock; it does not detect overloads or phase-to-neutral faults.
143. A residual current device detects a fault by sensing
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Kirchhoff's current law between phase and neutral.
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Answer: C. An imbalance between the phase and neutral currents passing through its core
Phase and neutral pass through a toroidal core; normally their fluxes cancel, but leakage to earth leaves a net flux that induces a trip signal.
144. Ventricular fibrillation of the heart, the main cause of death by electric shock, can be caused by a power-frequency body current as small as about
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It is well above the threshold of perception but still in milliamperes.
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Answer: B. 50 mA
About 1 mA is the threshold of perception and about 10 mA the let-go limit; currents of the order of 50 mA and above through the chest can cause fibrillation.
145. Before working on an isolated high-voltage cable or line, a worker must
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Never trust that isolated equipment is dead.
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Answer: D. Test it for absence of voltage, then discharge it and connect it to earth
Safe practice is isolate, prove dead with an approved tester, then discharge and earth the conductors (under a permit-to-work) to remove trapped and induced charge.
146. If a person is stuck to a live low-voltage conductor, the first correct action of a rescuer is to
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The rescuer must not become a second victim.
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Answer: B. Switch off the supply, or separate the person using dry insulating material
Touching the victim directly would make the rescuer part of the circuit; the supply must be disconnected or the victim freed with a dry insulating object.
147. The earth continuity (protective) conductor in a building installation connects
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It is the green/yellow conductor.
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Answer: C. Exposed metal parts of equipment and sockets to the main earthing terminal
The protective conductor runs with the circuit and ties all exposed conductive parts back to the main earthing terminal and electrode.
148. In a three-pin plug, the earth pin is made longer than the other two pins so that
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Think of the order in which the pins engage.
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Answer: D. It makes contact first and breaks contact last
The longer earth pin ensures the appliance body is earthed before the live pin connects and stays earthed until the live pin disconnects.
149. A hemispherical earth electrode of radius 0.5 m is buried flush with the surface in soil of resistivity 100 Ω·m. Its earth resistance (R = ρ/2πr) is about
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Substitute directly in R = ρ/(2πr).
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Answer: B. 31.8 Ω
R = ρ / (2πr) = 100 / (2π × 0.5) ≈ 31.8 Ω.
150. Four identical earth rods of 20 Ω each are connected in parallel, far enough apart that mutual effects can be neglected. The combined earth resistance is
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Treat them as parallel resistors.
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Answer: A. 5 Ω
Equal resistances in parallel: R = 20 / 4 = 5 Ω. In practice, closely spaced rods give somewhat more because of mutual interference.
151. A substation earth grid has a resistance of 0.5 Ω. During an earth fault, 1000 A flows into the grid. The rise of grid potential is
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Use Ohm's law.
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Answer: B. 500 V
Ground potential rise = I × R = 1000 × 0.5 = 500 V.
152. A person touching faulty equipment experiences a touch voltage of 100 V. Taking the body resistance as 1000 Ω, the body current is
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Use Ohm's law and convert to milliamperes.
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Answer: C. 100 mA
I = V / R = 100 / 1000 = 0.1 A = 100 mA, enough to cause ventricular fibrillation.
153. Using Dalziel's relation I = 0.116/√t (for a 50 kg person), the tolerable body current for a shock duration of 0.5 s is about
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Divide by the square root of the time, not the time itself.
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Answer: C. 164 mA
I = 0.116 / √0.5 = 0.116 / 0.707 ≈ 0.164 A = 164 mA.
154. An 11 kV system neutral is to be earthed through a resistor that limits the earth-fault current to 500 A. The required resistance is about
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The neutral resistor sees phase voltage during an earth fault.
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Answer: B. 12.7 Ω
R = V_phase / I = (11000/√3) / 500 = 6351 / 500 ≈ 12.7 Ω.
155. Each phase of a 50 Hz system has a capacitance to earth of 1 µF. The inductance of a Peterson coil that exactly compensates the earth-fault capacitive current, L = 1/(3ω²C), is about
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ω = 2πf ≈ 314 rad/s.
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Answer: B. 3.38 H
L = 1 / (3 × (2π × 50)² × 1 × 10⁻⁶) = 1 / (3 × 98 696 × 10⁻⁶) ≈ 3.38 H.
156. An 11 kV, 50 Hz isolated-neutral system has a capacitance of 1 µF per phase to earth. The capacitive current at a single line-to-earth fault, 3·V_ph·ω·C, is about
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Use the phase voltage and multiply by 3.
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Answer: C. 6.0 A
I = 3 × (11000/√3) × 314.16 × 1 × 10⁻⁶ ≈ 3 × 6351 × 314.16 × 10⁻⁶ ≈ 6.0 A.
157. In a TT installation protected by a 30 mA RCD, the touch voltage must not exceed 50 V (R_A × IΔn ≤ 50 V). The maximum allowable earth electrode resistance R_A is about
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Divide the voltage limit by the RCD's rated residual current in amperes.
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Answer: D. 1667 Ω
R_A ≤ 50 / 0.030 ≈ 1667 Ω.
158. In a TN system with U₀ = 230 V, the earth fault loop impedance at a socket is 0.8 Ω. The prospective earth-fault current is about
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Divide the phase-to-earth voltage by the loop impedance.
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Answer: B. 288 A
I = U₀ / Zs = 230 / 0.8 = 287.5 A ≈ 288 A.
6.6 Substations
30 questions · AElE0606
159. A switching substation is one in which
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Its name suggests what it does and what it does not do.
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Answer: D. Lines at the same voltage are connected and switched without any voltage transformation
Switching substations only connect or disconnect circuits at one voltage level, improving flexibility and sectionalising of the network.
160. Gas-insulated substations (GIS) are preferred mainly where
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SF6 has much higher dielectric strength than air.
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Answer: A. Space is limited, such as in cities or underground and hydro power houses
Because SF6 insulation allows very small clearances, GIS occupies a small fraction of the area of an air-insulated substation and is well protected from pollution.
161. The main drawback of the single bus-bar arrangement is that
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Everything depends on one common conductor.
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Answer: D. A bus fault or bus maintenance requires shutting down the whole substation
The single bus is simple and cheap, but all circuits depend on one bus, so any bus outage interrupts every circuit.
162. Sectionalising a single bus with a circuit breaker or isolator has the advantage that
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Think of dividing one bus into parts.
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Answer: D. A fault or maintenance on one section affects only the circuits on that section
Dividing the bus into sections limits the outage to the affected section while the rest of the substation continues in service.
163. The main and transfer bus arrangement allows
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The extra bus is used temporarily during breaker maintenance.
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Answer: A. Any circuit breaker to be taken out for maintenance without interrupting its circuit
When a feeder breaker needs maintenance, the circuit is transferred to the transfer bus and fed through the bus-coupler (transfer) breaker.
164. Which bus-bar arrangement gives the highest reliability and flexibility, but at the highest cost?
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Look for the scheme with two breakers per circuit.
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Answer: D. Double bus with double breaker
Each circuit has two breakers and can be connected to either bus, so loss of any one bus or breaker does not interrupt any circuit, but the breaker count is twice the number of circuits.
165. In a ring bus arrangement,
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Visualise breakers arranged in a closed loop.
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Answer: C. Each circuit is connected between two breakers, and the number of breakers equals the number of circuits
The breakers form a closed ring with circuits connected at the nodes, so each circuit is fed from two sides; opening the ring for maintenance reduces security.
166. In a breaker-and-a-half arrangement,
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The name refers to breakers per circuit.
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Answer: A. Three circuit breakers serve two circuits between two main buses
Each diameter has three breakers in series between the two buses with two circuits connected, so each circuit effectively uses 1.5 breakers.
167. The earthing system of an outdoor substation usually consists of
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It must control potential gradients over the whole yard.
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Answer: D. A buried mesh (earth mat) of conductors connected to rods, to which all equipment and structures are bonded
A grid of buried conductors gives low resistance and limits step and touch voltages over the whole yard; all metal structures, tanks, neutrals and arresters are connected to it.
168. A layer of crushed rock (gravel) is spread over the surface of a substation yard mainly to
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It is in series with the person's feet.
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Answer: A. Increase surface resistivity, reducing body current from step and touch voltages
The high-resistivity gravel adds resistance in series with a person's feet, so for a given step or touch voltage the body current is smaller; it also keeps the surface dry and free of weeds.
169. The secondary winding of a CT or VT is earthed at one point mainly to
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Think of what happens if the primary voltage leaks onto the secondary.
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Answer: C. Protect personnel and instruments if insulation between primary and secondary fails
Earthing the secondary circuit prevents it from rising to a dangerous potential through capacitive coupling or insulation breakdown from the high-voltage primary.
170. The secondary of a current transformer in service must never be left open-circuited because
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The primary current is fixed by the line, not by the CT.
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Answer: B. A dangerously high voltage is induced and the core may saturate and overheat
With no secondary ampere-turns to oppose it, the whole primary current magnetises the core, producing very high flux, high peak secondary voltage and core heating.
171. The station battery in a substation is provided to
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Protection must work exactly when AC voltage collapses.
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Answer: C. Supply DC for tripping, protection and control even when AC supply is lost
Breaker trip coils, relays and control circuits must work during faults when the AC may collapse, so a battery with charger provides a reliable DC supply.
172. A capacitive voltage transformer (CVT) at an EHV substation also serves as
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Capacitors pass high-frequency signals easily.
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Answer: A. A coupling capacitor for power line carrier communication (PLCC)
The capacitor stack of a CVT couples high-frequency carrier signals onto the line, so the same unit provides metering voltage and the PLCC coupling.
173. A wave trap (line trap) installed in series with a transmission line at a substation
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It is a tuned circuit used with PLCC.
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Answer: D. Blocks carrier-frequency signals while passing power-frequency current
The wave trap is a parallel LC circuit tuned to the carrier frequency; it presents high impedance to carrier signals, keeping them on the line, and negligible impedance at 50 Hz.
174. In an on-load tap changer, a transition resistor or reactor is used so that
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Two conditions must be avoided: open circuit and short circuit of turns.
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Answer: B. Load current is not interrupted and the tapped turns are not short-circuited during a tap change
During a change, both taps are briefly bridged through an impedance that limits the circulating current while maintaining continuity of the load current.
175. Tap changers are usually placed on the high-voltage winding of a transformer because
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Compare current and number of turns on each side.
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Answer: C. It carries less current and has more turns, allowing finer voltage steps
Lower current reduces switching duty on the tap-changer contacts, and the larger number of turns gives smaller percentage change per tap.
176. A booster transformer is used in a distribution system to
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'Boost' refers to adding voltage along the feeder.
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Answer: B. Inject a series voltage to raise the voltage at an intermediate point of a feeder
Its secondary is connected in series with the line and its primary is excited from the same supply, so it adds a controlled voltage to compensate the feeder drop.
177. A synchronous condenser is
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It is a rotating machine.
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Answer: A. A synchronous motor running without mechanical load whose excitation controls reactive power
Over-excited, it supplies lagging reactive power (acts like a capacitor); under-excited, it absorbs reactive power, giving smooth control of voltage.
178. Compared with static capacitor banks, a synchronous condenser has the advantage that it
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Excitation can be varied smoothly in both directions.
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Answer: D. Gives continuous control and can both supply and absorb reactive power
Excitation control gives smooth adjustment over the lagging and leading range; capacitor output falls with V² and is changed only in steps.
179. A static VAR compensator (SVC) commonly uses
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It is 'static' because it has no rotating parts.
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Answer: C. A thyristor-controlled reactor with fixed or thyristor-switched capacitors
Varying the firing angle of the thyristor-controlled reactor, together with switched capacitors, gives fast, continuously adjustable reactive power without moving parts.
180. An advantage of a STATCOM over an SVC is that the STATCOM
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Compare a controlled current source with a fixed admittance.
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Answer: B. Can supply nearly its rated reactive current even when the system voltage is low
A STATCOM is a voltage-source converter whose current is controlled independently of the AC voltage, whereas an SVC behaves like a fixed admittance at its limit, so its output falls as V².
181. An on-load tap changer provides a range of ±10% in 16 equal steps. The voltage change per step is
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The total range from −10% to +10% is 20%.
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Answer: B. 1.25%
Total range = 20%; 20% / 16 steps = 1.25% per step.
182. An 11000/400 V transformer must keep 400 V on the secondary when the primary voltage falls to 10.5 kV. The required reduction in HV turns (using an off-load tap) is about
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Keep volts per turn in mind: fewer primary turns raise the secondary voltage.
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Answer: A. 4.55%
Required ratio = 10500/400, so HV turns must be 10500/11000 = 0.9545 of nominal, a reduction of about 4.55% (not 4.76%, which wrongly divides by 10.5 kV).
183. A load of 10 MW at 0.8 power factor lagging is to be raised to 0.95 lagging using a synchronous condenser. The reactive power the condenser must supply is about
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Find tan φ for each power factor.
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Answer: B. 4.2 MVAr
Q = P (tan φ₁ − tan φ₂) = 10 × (0.750 − 0.329) ≈ 4.2 MVAr.
184. A three-phase, delta-connected capacitor bank must supply 3 MVAr at 11 kV, 50 Hz. The capacitance per phase is about
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In delta, each capacitor sees the line voltage.
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Answer: B. 26.3 µF
Each phase supplies 1 MVAr at 11 kV (line voltage across each delta branch): C = Q / (V²ω) = 10⁶ / (11000² × 314.16) ≈ 26.3 µF.
185. A voltmeter connected to an 11000/110 V potential transformer reads 104 V. The primary voltage is
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Multiply by the PT ratio.
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Answer: C. 10.4 kV
Primary voltage = 104 × (11000/110) = 104 × 100 = 10 400 V.
186. An ammeter connected to a 200/5 A current transformer reads 3.5 A. The line current is
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Multiply by the CT ratio.
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Answer: C. 140 A
Line current = 3.5 × (200/5) = 3.5 × 40 = 140 A.
187. A substation with 6 circuits uses the breaker-and-a-half arrangement. The number of circuit breakers required is
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Count breakers per circuit from the arrangement's name.
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Answer: B. 9
Each pair of circuits uses 3 breakers, so 6 circuits need 6 × 1.5 = 9 breakers.
188. A substation is fed from a 10 MVA transformer of 8% reactance connected to an infinite source. The three-phase fault level on its secondary bus is
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Divide the rating by the per-unit reactance.
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Answer: C. 125 MVA
Fault MVA = rated MVA × 100 / %X = 10 × 100 / 8 = 125 MVA.