Under the Ministerial Ordinance on Technical Standards for Electrical Equipment (電気設備に関する技術基準を定める省令), which is the range of "high voltage" (高圧) for AC?
Over 300 V up to 6,600 V
Over 600 V up to 7,000 V
Over 750 V up to 7,000 V
Over 600 V up to 10,000 V
AnswerB. Over 600 V up to 7,000 V
Article 2 of the Ordinance defines AC high voltage as over 600 V up to 7,000 V. "Over 300 V up to 6,600 V" mixes up the 300 V used in indoor wiring rules with the 6,600 V nominal distribution voltage. "Over 750 V" confuses it with the DC low-voltage limit (750 V). "Up to 10,000 V" is wrong because anything above 7,000 V is extra-high voltage (特別高圧).
Q2 | Fences around HV equipment
When high-voltage equipment is installed with a fence around it (outside factory premises), the sum of the fence height and the distance from the fence to live parts must be at least how many metres?
3 m
4 m
6 m
5 m
AnswerD. 5 m
Article 21(2) of the Interpretation requires the sum of fence height and distance to live parts to be at least 5 m. 3 m and 4 m are too short. 6 m is the value for extra-high-voltage equipment over 35,000 V up to 160,000 V (Article 22).
Q3 | Mounting HV equipment high up
In an urban area, high-voltage equipment (with cable connections) is mounted high up so that people cannot touch it. At least how high above the ground must it be?
4.5 m
3.5 m
4 m
5 m
AnswerA. 4.5 m
Article 21(3) of the Interpretation requires at least 4.5 m above ground (4 m outside urban areas). 4 m is the value outside urban areas, so it is not enough here. 3.5 m does not appear in the rule. 5 m confuses it with the fence-height-plus-distance value or the rule for extra-high-voltage equipment (Article 22).
Q4 | Cubicle enclosure
When high-voltage equipment is housed in a metal box so that no live parts are exposed, which earthing work does Article 21 of the Interpretation of Technical Standards require for that metal box?
Class A earthing (A種接地工事)
Class B earthing (B種接地工事)
Class D earthing (D種接地工事)
Class C earthing (C種接地工事)
AnswerC. Class D earthing (D種接地工事)
Article 21(4) of the Interpretation says the equipment may be housed in "a concrete box or a metal box with Class D earthing". Class A applies to the metal frames of high-voltage equipment itself (Article 29) and to surge arresters, not to this box rule. Class B is applied on the low-voltage side of a transformer linking high and low voltage. Class C is for frames of low-voltage equipment over 300 V.
Q5 | Devices that produce arcs
A high-voltage circuit breaker that produces an arc when it operates is installed without a fire-resistant enclosure. What is the minimum distance from combustible materials such as a wooden ceiling?
0.5 m
1.5 m
2 m
1 m
AnswerD. 1 m
Under Article 23 of the Interpretation, high-voltage switchgear that produces arcs must be at least 1 m from combustibles. 2 m is the value for extra-high voltage. 0.5 m and 1.5 m do not appear in the rule.
Q6 | Where surge arresters are required
At the service entrance of a customer supplied from a high-voltage overhead line, from what received power is a surge arrester required?
50 kW or more
500 kW or more
300 kW or more
2,000 kW or more
AnswerB. 500 kW or more
Article 37(1)(iii) of the Interpretation requires an arrester at the service entrance of customers receiving 500 kW or more from a high-voltage overhead line. 50 kW, 300 kW and 2,000 kW are not the threshold in this rule.
Q7 | Earthing of surge arresters
Which class of earthing is, in principle, required for a surge arrester installed on a high-voltage circuit?
Class A earthing
Class B earthing
Class C earthing
Class D earthing
AnswerA. Class A earthing
Article 37(3) of the Interpretation requires Class A earthing for arresters on high- and extra-high-voltage circuits. Class B is for the low-voltage side of transformers, and Classes C and D are for frames of low-voltage equipment; none of these is the rule for arresters that discharge lightning current to earth.
Q8 | Earthing of HV equipment frames
Which pair correctly gives the earthing work for the metal case of a high-voltage transformer and its earth resistance (when no exception such as a fence or insulating stand applies)?
Class A, 100 Ω or less
Class C, 10 Ω or less
Class A, 10 Ω or less
Class D, 100 Ω or less
AnswerC. Class A, 10 Ω or less
Article 29 of the Interpretation requires Class A earthing for metal cases of high-voltage equipment, and Article 17 sets Class A resistance at 10 Ω or less. "Class A, 100 Ω" mixes in the Class D value. "Class C, 10 Ω" is for low-voltage equipment over 300 V. "Class D, 100 Ω" is for low-voltage equipment of 300 V or less.
Q9 | CB type and PF-S type
What is the main protective switching device of a "PF-S type" cubicle high-voltage receiving unit?
A vacuum circuit breaker combined with an overcurrent relay
A current-limiting fuse combined with a load-break switch
A disconnector combined with a voltage transformer
A surge arrester combined with a zero-phase current transformer
AnswerB. A current-limiting fuse combined with a load-break switch
In the PF-S type, a current-limiting fuse (PF) and a load-break switch (S) together form the main protective device. A breaker with an overcurrent relay is the CB type. A disconnector cannot switch load current and a voltage transformer only measures, so they are not a main breaker. An arrester and a ZCT are for overvoltage and earth-fault detection.
Q10 | Purpose of the sectionalizing switch
What is the main purpose of installing a load-break switch with an earth-fault relay at the point of demarcation of responsibility (責任分界点)?
To improve the power factor on the premises and lower the bill
To discharge lightning overvoltage to earth
To transform the received voltage down to low voltage
To keep an earth fault on the premises from spreading to the distribution line
AnswerD. To keep an earth fault on the premises from spreading to the distribution line
A sectionalizing switch with an earth-fault relay opens when an earth fault occurs on the premises, so the fault does not black out the whole distribution line. Power factor is improved by capacitors, overvoltage is handled by arresters, and voltage is reduced by transformers.
Q11 | Handling a disconnector
Which statement about a disconnector (断路器, DS) is correct?
It must not be opened while load current is flowing
It can interrupt short-circuit current
It can switch load current but cannot interrupt short-circuit current
It opens automatically on a signal from an overcurrent relay
AnswerA. It must not be opened while load current is flowing
A disconnector has no arc-quenching ability, so it must not be opened while load current flows. Interrupting short-circuit current and opening automatically on a relay signal describe a circuit breaker. "Switches load current but not short-circuit current" describes a load-break switch (LBS).
Q12 | Order of switching off
A high-voltage circuit has a circuit breaker (CB) and a disconnector (DS) in series. Which is the correct order to de-energize it for inspection?
Open the disconnector first, then the breaker
Open the breaker and disconnector at the same time
Open the breaker first, then the disconnector
Open only the disconnector and leave the breaker closed
AnswerC. Open the breaker first, then the disconnector
A disconnector cannot break load current, so open the breaker first to stop the current, then open the disconnector to isolate the circuit. Opening the disconnector first, or both together, or only the disconnector, would draw an arc at the disconnector while current flows. To re-energize, close the disconnector first and then the breaker.
Q13 | Load-break switch
When a high-voltage load-break switch (LBS) is used on the primary side of a transformer, why is it combined with a current-limiting fuse?
Because the switch alone cannot switch load current
Because the fuse improves the power factor
Because the fuse detects earth-fault current
Because the switch alone cannot interrupt short-circuit current
AnswerD. Because the switch alone cannot interrupt short-circuit current
A load-break switch can switch load current but cannot interrupt short-circuit current, so the current-limiting fuse clears short circuits. "Cannot switch load current" describes a disconnector. A fuse does not improve power factor, and earth faults are detected by a ZCT and earth-fault relay.
Q14 | Vacuum circuit breaker
Which statement about a vacuum circuit breaker (VCB) is correct?
It blows compressed air at the arc to interrupt current
It opens its contacts inside a vacuum vessel, extinguishing the arc to interrupt current
It can operate only when no load current is flowing
Once it operates, it melts and must be replaced
AnswerB. It opens its contacts inside a vacuum vessel, extinguishing the arc to interrupt current
A vacuum circuit breaker opens its contacts in a vacuum vessel to quench the arc. Blowing compressed air is how an air-blast breaker works. Operating only without load current describes a disconnector, and melting and needing replacement describes a fuse.
Q15 | Enclosed fuse performance
Which performance does the Interpretation of Technical Standards require of an enclosed fuse (other than a current-limiting fuse) used as an overcurrent breaker on a high-voltage circuit?
Withstands 1.3 times rated current and melts within 120 minutes at 2 times
Withstands 1.25 times rated current and melts within 2 minutes at 2 times
Withstands 1.1 times rated current and melts within 60 minutes at 1.6 times
Withstands 1.5 times rated current and melts within 10 minutes at 3 times
AnswerA. Withstands 1.3 times rated current and melts within 120 minutes at 2 times
Article 34(2) of the Interpretation requires an enclosed fuse to withstand 1.3 times rated current and melt within 120 minutes at 2 times (or be a compliant current-limiting fuse). "1.25 times / 2 minutes" is the rule for non-enclosed fuses (Article 34(3)). "1.1 times / 1.6 times / 60 minutes" is the low-voltage fuse rule for 30 A or less (Article 33), not the high-voltage rule. "1.5 times / 3 times / 10 minutes" does not appear.
Q16 | CT secondary circuit
An ammeter is to be removed from the secondary of an energized current transformer (CT). What is the correct procedure?
Remove it, leaving the secondary open
Disconnect the Class D earth wire of the secondary first, then remove it
Short-circuit the secondary terminals first, then remove it
Increase the primary current first, then remove it
AnswerC. Short-circuit the secondary terminals first, then remove it
Opening the secondary of an energized CT produces a high voltage across it and can cause insulation breakdown or burning, so short the secondary first, then remove the meter. Leaving it open is exactly what must be avoided. Removing the Class D earth required by Article 28 increases danger if insulation between primary and secondary fails. Raising the primary current does not reduce the danger.
Q17 | CT ratio calculation
A current transformer with a ratio of 150/5 A has 3.2 A flowing in its secondary. What is the primary current?
480 A
30 A
0.11 A
96 A
AnswerD. 96 A
The ratio is 150 ÷ 5 = 30, so the primary current is 3.2 × 30 = 96 A. 480 A is 3.2 × 150, forgetting to divide by 5. 30 A is just the ratio. 0.11 A is 3.2 × 5 ÷ 150, dividing the wrong way.
Q18 | VT ratio calculation
A voltage transformer (VT) with a ratio of 6,600/110 V shows 105 V on its secondary voltmeter. What is the primary voltage?
6,600 V
6,300 V
6,930 V
11,550 V
AnswerB. 6,300 V
The ratio is 6,600 ÷ 110 = 60, so the primary voltage is 105 × 60 = 6,300 V. 6,600 V is the rated value, for a secondary reading of 110 V. 6,930 V is 6,600 × 105 ÷ 100, dividing by 100 instead of 110. 11,550 V is 105 × 110, multiplying without using the ratio.
Q19 | Capacity of an open-delta (V) connection
Three 50 kVA single-phase transformers were used in delta. One failed, so the remaining two are connected in open delta (V connection) to supply a three-phase load. About how many kVA can be supplied at most?
100 kVA
150 kVA
86.6 kVA
75 kVA
AnswerC. 86.6 kVA
An open-delta bank can deliver √3 times the rating of one unit: √3 × 50 ≈ 86.6 kVA. 100 kVA is simply the sum of two units, which a V connection cannot deliver. 150 kVA is the capacity of the full delta bank. 75 kVA is half of 150 kVA.
Q20 | Short-circuit current
A three-phase short circuit occurs at the secondary terminals of a 300 kVA, 210 V secondary, 5 % impedance three-phase transformer. About how many kA is the short-circuit current (ignore source-side impedance)?
16.5 kA
28.6 kA
4.1 kA
0.82 kA
AnswerA. 16.5 kA
Rated current = 300,000 ÷ (√3 × 210) ≈ 824.8 A. Short-circuit current = rated current × 100 ÷ %Z = 824.8 × 100 ÷ 5 ≈ 16,496 A ≈ 16.5 kA. 28.6 kA forgets the √3 (300,000 ÷ 210 × 20). 4.1 kA multiplies by %Z (824.8 × 5) instead of 100 ÷ %Z. 0.82 kA is just the rated current.
Q21 | Zero-phase current transformer
In high-voltage receiving equipment, which device is combined with an earth-fault relay to detect earth faults?
Voltage transformer (VT)
Zero-phase current transformer (ZCT)
Series reactor (SR)
Disconnector (DS)
AnswerB. Zero-phase current transformer (ZCT)
A ZCT passes all three conductors through its core and detects the difference in current (zero-phase current) caused by an earth fault, sending it to the earth-fault relay. A VT measures voltage, a series reactor protects capacitors from harmonics and inrush, and a disconnector isolates circuits.
Q22 | Overcurrent relay characteristic
Which statement correctly describes the "inverse-time characteristic" of an overcurrent relay in receiving equipment?
The larger the current, the longer the operating time
It operates after a fixed time regardless of current
It operates when the current falls below the setting
The larger the current, the shorter the operating time
AnswerD. The larger the current, the shorter the operating time
An inverse-time characteristic means the larger the current, the faster the relay operates: quickly for short circuits, slowly for slight overloads. "Larger current, longer time" is the reverse. A fixed time regardless of current is a definite-time characteristic. A relay that operates when current falls below the setting is not an overcurrent relay.
Q23 | Directional earth-fault relay
In a receiving installation with long high-voltage cables on the premises, why is a directional earth-fault relay used instead of a plain earth-fault relay?
To prevent unwanted operation, during an external earth fault, caused by current through the cable's capacitance to earth
To interrupt larger short-circuit currents on the premises
To switch off capacitors automatically when the power factor becomes leading
To discharge lightning overvoltage to earth quickly
AnswerA. To prevent unwanted operation, during an external earth fault, caused by current through the cable's capacitance to earth
Long cables have large capacitance to earth, so during an earth fault outside the premises, zero-phase current can flow through them and trip a plain earth-fault relay unnecessarily. A directional relay also uses zero-phase voltage to judge the direction and operates only for faults on the premises. Interrupting capacity depends on the breaker, not the relay type. Switching capacitors and handling lightning are not its role.
Q24 | Protection coordination
Which statement correctly describes protection coordination (保護協調)?
When a fault occurs, all breakers open at the same time for safety
The utility substation breaker operates before the customer's breaker
Only the protective device nearest the fault on the supply side operates, keeping the outage as small as possible
Only one of the overcurrent relay or the earth-fault relay is installed
AnswerC. Only the protective device nearest the fault on the supply side operates, keeping the outage as small as possible
Protection coordination adjusts pickup currents and times so that only the device nearest the fault on the supply side trips. Opening all breakers at once widens the outage. If the utility breaker tripped first, the whole distribution line would go dark, so the customer side is set to act first. Overcurrent and earth faults are different, so one relay type alone is not enough.
Q25 | Effects of power-factor correction
A power-factor-correction capacitor is connected in parallel with a load. Which effect is correct (the load's active power does not change)?
The load's active power increases, raising motor output
Line current rises, so voltage drop increases
The reactive power the load itself needs disappears
Line current falls, so power loss decreases
AnswerD. Line current falls, so power loss decreases
The capacitor's leading reactive power cancels the load's lagging reactive power, so the apparent power and current from the source fall, reducing losses (proportional to current squared) and voltage drop. Active power and motor output do not change. Current rising is the reverse. The load still needs the same reactive power; the capacitor simply supplies it locally.
Q26 | Capacitor size for PF correction
A load of 300 kW at a power factor of 0.6 (lagging) is to be improved to 0.8 (lagging). What capacitor size in kvar is needed?
175 kvar
225 kvar
400 kvar
60 kvar
AnswerA. 175 kvar
tanθ1 = 0.8 ÷ 0.6 = 4/3 and tanθ2 = 0.6 ÷ 0.8 = 0.75. Q = 300 × (4/3 − 0.75) = 300 × 0.583… = 175 kvar. 225 kvar is the reactive power remaining after correction (300 × 0.75). 400 kvar is the reactive power before correction (300 × 4/3); cancelling all of it would give a power factor of 1. 60 kvar wrongly multiplies by the difference of power factors, 300 × (0.8 − 0.6).
Q27 | Correcting to unity power factor
What capacitor size in kvar is needed to correct a 240 kW load at a power factor of 0.8 (lagging) to a power factor of 1.0?
144 kvar
180 kvar
300 kvar
320 kvar
AnswerB. 180 kvar
At PF 0.8, tanθ = 0.6 ÷ 0.8 = 0.75, so reactive power is 240 × 0.75 = 180 kvar. Correcting to 1.0 cancels all of it: 180 kvar. 144 kvar uses sinθ instead of tanθ (240 × 0.6). 300 kvar confuses it with the apparent power (240 ÷ 0.8 = 300 kVA). 320 kvar uses the tanθ for PF 0.6 (240 × 4/3).
Q28 | Series reactor
What is the main purpose of combining a series reactor with a high-voltage power-factor-correction capacitor?
To increase active power and improve motor efficiency
To detect zero-phase current and operate the earth-fault relay
To reduce voltage distortion from harmonics and limit inrush current at switch-on
To discharge the residual charge of a disconnected capacitor
AnswerC. To reduce voltage distortion from harmonics and limit inrush current at switch-on
The JEMA guide explains that a series reactor reduces harmonic voltage distortion and limits capacitor inrush current (in HV receiving equipment, typically 6 % of the capacitor reactance). It does not increase active power. Detecting zero-phase current is the ZCT's job, and discharging residual charge is done by a discharge coil or resistor.
Q29 | Capacitors at light load
If a power-factor-correction capacitor stays connected at light load, such as at night, what is likely to happen?
The lagging power factor worsens and voltage drops sharply
Active power increases and the energy meter reads higher
The transformer core saturates and an earth fault occurs
The power factor becomes leading and the receiving-end voltage rises
AnswerD. The power factor becomes leading and the receiving-end voltage rises
At light load the lagging reactive power is small, so the capacitor's leading reactive power exceeds it, the power factor becomes leading, and the receiving-end voltage can rise. A worse lagging power factor is the reverse. Capacitors do not add active power. Core saturation and earth faults are unrelated.
Q30 | Earthing of instrument transformer secondaries
Which class of earthing is applied to the secondary circuit of a high-voltage instrument transformer (VT or CT)?
Class A earthing
Class D earthing
Class B earthing
Class C earthing
AnswerB. Class D earthing
Article 28 of the Interpretation requires Class D earthing on the secondary circuit of high-voltage instrument transformers. Class A applies to extra-high-voltage instrument transformer secondaries (Article 28(2)) and to frames of HV equipment. Class B is for the low-voltage side of HV/LV transformers, and Class C for frames of low-voltage equipment over 300 V.
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