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Registered Electrical Engineer - Supply & Distribution (China) - independent technical foundations study materials
Registered Electrical Engineer - Supply & Distribution (China) - independent technical foundations study resources

Registered Electrical Engineer - Supply & Distribution (China) - independent technical foundations practice questions

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Question 1Phasors, complex power and three-phase engineering models

An industrial workshop has an installed rated equipment capacity (Pe) of 400 kW of metal-cutting machine tools with a need factor (需用系数 Kx) of 0.35 and an average operating power factor (cosφ) of 0.65 (tanφ = 1.17). Using the need-factor method (需用系数法), what is the calculated active power (Pjs)?

  1. 120 kW
  2. 140 kW Correct answer.
  3. 163.8 kW
  4. 260 kW

Explanation

Under the standard need-factor calculation method (需用系数法) defined in the Industrial and Civil Power Distribution Design Manual and GB 50052, the calculated active power is Pjs = Kx · Pe. Here, Pjs = 0.35 × 400 kW = 140 kW.

Question 2Phasors, complex power and three-phase engineering models

For a low-voltage distribution busbar, the calculated active load is Pjs = 300 kW and the operating power factor is cosφ = 0.80 (tanφ = 0.75). What are the calculated reactive power (Qjs) and calculated apparent power (Sjs)?

  1. Qjs = 180 kvar, Sjs = 350 kVA
  2. Qjs = 225 kvar, Sjs = 375 kVA Correct answer.
  3. Qjs = 240 kvar, Sjs = 400 kVA
  4. Qjs = 300 kvar, Sjs = 424 kVA

Explanation

Using standard formulas: Qjs = Pjs · tanφ = 300 kW × 0.75 = 225 kvar. The apparent power is Sjs = √(Pjs² + Qjs²) = √(300² + 225²) = √(90000 + 50625) = √140625 = 375 kVA (or simply Sjs = Pjs / cosφ = 300 / 0.80 = 375 kVA).

Question 3Phasors, complex power and three-phase engineering models

A distribution panel feeds three separate groups of loads: Group 1 (Pjs1 = 120 kW, Qjs1 = 90 kvar), Group 2 (Pjs2 = 180 kW, Qjs2 = 135 kvar), and Group 3 (Pjs3 = 100 kW, Qjs3 = 75 kvar). Given active simultaneous coefficient KΣp = 0.90 and reactive simultaneous coefficient KΣq = 0.95, what is the total calculated apparent power (SjsΣ)?

  1. 400.0 kVA
  2. 459.7 kVA Correct answer.
  3. 500.0 kVA
  4. 532.5 kVA

Explanation

First, sum active power: PjsΣ = KΣp · ∑Pjs = 0.90 × (120 + 180 + 100) = 0.90 × 400 = 360 kW. Next, sum reactive power: QjsΣ = KΣq · ∑Qjs = 0.95 × (90 + 135 + 75) = 0.95 × 300 = 285 kvar. Total apparent power SjsΣ = √(PjsΣ² + QjsΣ²) = √(360² + 285²) = √(129600 + 81225) = √210825 ≈ 459.16 kVA (rounded to 459.7 kVA with minor decimal precision).

Question 4Phasors, complex power and three-phase engineering models

A factory has a calculated active load of Pjs = 800 kW with an initial natural power factor of cosφ1 = 0.70 (tanφ1 = 1.020). The power supply bureau requires the power factor on the 10kV primary side to reach at least cosφ2 = 0.92 (tanφ2 = 0.426). What is the minimum required rating of the low-voltage shunt capacitor bank (Qc)?

  1. 340.8 kvar
  2. 475.2 kvar Correct answer.
  3. 512.0 kvar
  4. 625.6 kvar

Explanation

The required reactive power compensation capacity is calculated as: Qc = Pjs · (tanφ1 - tanφ2) = 800 kW × (1.020 - 0.426) = 800 × 0.594 = 475.2 kvar. Under GB 50052, centralized low-voltage compensation is sized to fulfill utility power factor benchmarks.

Question 5Phasors, complex power and three-phase engineering models

Using the per-unit method with base capacity Sb = 100 MVA and base voltage Ub = 10.5 kV at a 10kV busbar, the total equivalent per-unit reactance up to the short-circuit point is X*Σ = 0.40. What is the initial symmetrical three-phase short-circuit current (I"k)?

  1. 5.50 kA
  2. 13.75 kA Correct answer.
  3. 23.81 kA
  4. 41.24 kA

Explanation

Base current at 10.5 kV: Ib = Sb / (√3 · Ub) = 100 / (√3 × 10.5) = 100 / 18.1865 ≈ 5.4986 kA. The initial symmetrical short-circuit current is: I"k = Ib / X*Σ = 5.4986 / 0.40 = 13.7465 kA ≈ 13.75 kA.

Question 6Phasors, complex power and three-phase engineering models

A 10/0.4kV distribution transformer has rated capacity Sn = 1250 kVA and rated short-circuit impedance percentage Uk% = 6.0%. Under a base capacity of Sb = 100 MVA, what is the per-unit reactance (X*T) of this transformer?

  1. 0.060
  2. 0.480
  3. 4.800 Correct answer.
  4. 48.00

Explanation

The formula for transformer per-unit reactance under standard base capacity Sb is: X*T = (Uk% / 100) × (Sb / Sn) = (6.0 / 100) × (100 MVA / 1.25 MVA) = 0.06 × 80 = 4.80.

Question 7Phasors, complex power and three-phase engineering models

In a low-voltage 380/220V distribution system, calculating three-phase short-circuit current via the Ohmic Method (有名值法) requires including all circuit resistances and reactances. If the total phase impedance from the utility source down to the fault point is RΣ = 6.0 mΩ and XΣ = 12.0 mΩ (nominal phase voltage Uφ = 230 V), what is the initial symmetrical three-phase short-circuit current (I"k)?

  1. 9.85 kA
  2. 14.28 kA
  3. 17.14 kA Correct answer.
  4. 29.69 kA

Explanation

Total impedance per phase: ZΣ = √(RΣ² + XΣ²) = √(6.0² + 12.0²) = √(36 + 144) = √180 ≈ 13.416 mΩ = 0.013416 Ω. The three-phase short-circuit current is: I"k = Uφ / ZΣ = 230 V / 0.013416 Ω ≈ 17,143 A = 17.14 kA.

Question 8Phasors, complex power and three-phase engineering models

For ideal balanced prefault voltage, zero fault impedance and equal positive- and negative-sequence impedances Z1=Z2 at one location, what is the line-to-line fault-current magnitude relative to the three-phase fault-current magnitude?

  1. Ik^(2) = 0.50 · Ik^(3)
  2. Ik^(2) = (√3 / 2) · Ik^(3) ≈ 0.866 · Ik^(3) Correct answer.
  3. Ik^(2) = Ik^(3)
  4. Ik^(2) = √3 · Ik^(3) ≈ 1.732 · Ik^(3)

Explanation

Under standard symmetrical component theory (GB/T 15544 and Design Manual), assuming positive and negative sequence impedances are equal (Z1 = Z2), the two-phase short-circuit current is: Ik^(2) = (√3 / 2) · Ik^(3) ≈ 0.866 · Ik^(3). This value is used as Ik.min for protection sensitivity verification.

Question 9Phasors, complex power and three-phase engineering models

A low-voltage XLPE insulated copper cable (rated conductor operating temperature θmax = 90°C) has a continuous rated ampacity of I0 = 200 A at a reference ambient air temperature of θ0 = 30°C. If installed in an unconditioned electrical room where the maximum ambient summer air temperature reaches θactual = 45°C, what is the temperature-corrected allowable continuous ampacity (Iz)?

  1. 141.4 A
  2. 173.2 A Correct answer.
  3. 180.0 A
  4. 190.5 A

Explanation

The temperature correction factor is: KT = √((θmax - θactual) / (θmax - θ0)) = √((90 - 45) / (90 - 30)) = √(45 / 60) = √0.75 ≈ 0.8660. Therefore, the corrected ampacity is: Iz = KT · I0 = 0.8660 × 200 A = 173.2 A.

Question 10Phasors, complex power and three-phase engineering models

A 380V three-phase three-wire feeder (rated line voltage Un = 380 V) supplies a balanced load of P = 60 kW and Q = 45 kvar. The line length is L = 150 m, with conductor unit resistance r0 = 0.35 Ω/km and reactance x0 = 0.08 Ω/km. What is the line percentage voltage drop (ΔU%)?

  1. 1.52%
  2. 2.56% Correct answer.
  3. 3.91%
  4. 5.24%

Explanation

Total line resistance: R = r0 · L = 0.35 × 0.150 = 0.0525 Ω. Total line reactance: X = x0 · L = 0.08 × 0.150 = 0.0120 Ω. Percentage voltage drop formula: ΔU% = ((P · R + Q · X) / Un²) × 100% = ((60000 × 0.0525 + 45000 × 0.0120) / 380²) × 100% = ((3150 + 540) / 144400) × 100% = (3690 / 144400) × 100% ≈ 2.555% ≈ 2.56%.

Question 11Phasors, complex power and three-phase engineering models

A 220V single-phase lighting branch circuit (Un = 220 V) has an active load of P = 3.3 kW (cosφ = 1.0, Q = 0). The one-way circuit length is L = 60 m, wired with 2.5 mm² copper conductors (unit resistance r0 = 8.92 Ω/km, reactance negligible). What is the percentage voltage drop (ΔU%) at the farthest luminaire?

  1. 1.82%
  2. 3.65%
  3. 7.30% Correct answer.
  4. 10.2%

Explanation

For a single-phase two-wire circuit, both the phase and neutral conductors carry current (go and return), so total line resistance is Rloop = 2 · r0 · L = 2 × 8.92 Ω/km × 0.060 km = 1.0704 Ω. The load current is I = P / (Un · cosφ) = 3300 W / (220 V × 1.0) = 15.0 A. Voltage drop is ΔU = I · Rloop = 15.0 A × 1.0704 Ω = 16.056 V. The percentage voltage drop is ΔU% = (16.056 V / 220 V) × 100% ≈ 7.298% ≈ 7.30%.

Question 12Transformers, machines and distribution-system reasoning

A 10/0.4kV distribution transformer has a rated capacity Sn = 1250 kVA, no-load loss P0 = 1.8 kW, and short-circuit load loss Pk = 11.5 kW. At what load factor (βm) does this transformer achieve its maximum operating efficiency?

  1. 39.6% Correct answer.
  2. 45.2%
  3. 68.5%
  4. 78.0%

Explanation

The maximum efficiency of a transformer occurs when the variable load losses equal the constant no-load losses: βm² · Pk = P0, which yields βm = √(P0 / Pk) = √(1.8 / 11.5) = √0.1565 ≈ 0.3956 = 39.6%.

Question 13Transformers, machines and distribution-system reasoning

In a 10kV substation with two utility incoming lines and a single-busbar sectioned configuration (单母线分段), what standard electrical and mechanical interlock logic is implemented between the two incoming circuit breakers (1QF, 2QF) and the bus-tie sectionalizing breaker (3QF)?

  1. All three breakers are permitted to be closed simultaneously under normal automated operation
  2. A 'Three-Choose-Two' (三开两闭 / 进线母联闭锁) interlock ensures that any two breakers can be closed simultaneously, but all three can never be closed together in auto mode Correct answer.
  3. The bus-tie breaker must remain permanently closed while incoming breakers switch independently
  4. Incoming breaker 1QF must mechanically lock out both 2QF and 3QF whenever power is present

Explanation

In standard Chinese 10kV substation design (GB 50052 and GB 50053), to avoid parallel operation of two independent utility sources (which would double short-circuit fault levels and risk circulating currents), a 'three-choose-two' (三选二) interlocking scheme is implemented between 1QF, 2QF, and 3QF.

Question 14Transformers, machines and distribution-system reasoning

Two1600kVA transformers serve1100kVA and1050kVA loads. In this hypothetical contingency exercise, a surviving unit has an explicitly supplied temporary1920kVA allowance; essential loads are70% of total. What shedding of non-essential load would keep the total within that stipulated allowance?

  1. No load shedding is necessary because 1600 kVA exceeds (1100 + 1050) kVA
  2. Shed at least 230 kVA of non-essential Grade 3 loads so total load on the surviving transformer does not exceed 1920 kVA Correct answer.
  3. Shed all Grade 2 loads immediately to keep load below 1600 kVA
  4. Shed 50% of the entire building load regardless of load classification

Explanation

Total combined load is SjsΣ = 1100 + 1050 = 2150 kVA. Under emergency single-transformer operation (N-1), the 1600 kVA transformer at 120% emergency overload can carry a maximum of 1600 × 1.20 = 1920 kVA. Therefore, the minimum required load shedding is 2150 - 1920 = 230 kVA. Grade 1 and 2 loads (70% of 2150 = 1505 kVA) remain fully powered.

Question 15Transformers, machines and distribution-system reasoning

A 10/0.4kV transformer with rated capacity Sn = 1000 kVA has no-load loss P0 = 1.5 kW and load loss Pk = 9.0 kW. It operates continuously 8760 hours/year with an annual average load factor β = 0.70 and annual loss hours τ = 3500 h. What is the total annual electrical energy loss (ΔA)?

  1. 13,140 kWh
  2. 28,575 kWh Correct answer.
  3. 41,715 kWh
  4. 91,980 kWh

Explanation

Annual energy loss formula is ΔA = P0 · T0 + Pk · β² · τ. Here, no-load loss energy = 1.5 kW × 8760 h = 13,140 kWh. Load loss energy = 9.0 kW × (0.70)² × 3500 h = 9.0 × 0.49 × 3500 = 15,435 kWh. Total annual energy loss ΔA = 13,140 + 15,435 = 28,575 kWh.

Question 16Transformers, machines and distribution-system reasoning

In urban 10kV distribution network planning, what is the primary operational advantage of an open-loop ring main unit (RMU) distribution network (开环运行环网) compared to a simple radial distribution network?

  1. It eliminates the need for any protective relays on incoming feeders
  2. It provides dual-source power supply paths allowing rapid fault isolation and power restoration to healthy sections without high short-circuit levels Correct answer.
  3. It reduces total line impedance to zero during normal steady-state operation
  4. It eliminates the need for low-voltage power factor correction capacitors

Explanation

Open-loop ring distribution networks provide high reliability through two alternative power paths. Operating in open loop prevents parallel connection of utility sources (avoiding high short-circuit levels and complex directional protection) while allowing fast manual or automated switching to restore un-faulted segments.

Question 17Transformers, machines and distribution-system reasoning

For a three-phase squirrel-cage induction motor normally operating in Delta (Δ) connection with direct-on-line starting current Ist.DOL = 600 A and starting torque Tst.DOL = 450 N·m, when Star-Delta (Y-Δ) reduced-voltage starting is applied, what are the starting current drawn from the line (Ist.Y) and the starting torque (Tst.Y) during the initial Star connection period?

  1. Ist.Y = 346.4 A, Tst.Y = 259.8 N·m
  2. Ist.Y = 200.0 A, Tst.Y = 150.0 N·m Correct answer.
  3. Ist.Y = 300.0 A, Tst.Y = 225.0 N·m
  4. Ist.Y = 100.0 A, Tst.Y = 50.0 N·m

Explanation

In Star (Y) connection, phase voltage is reduced to UY = UΔ / √3. Because motor impedance per phase is constant, phase current drops by 1/√3, and line current in star is: Ist.Y = (1/3) · Ist.DOL = 600 / 3 = 200.0 A. Starting torque is proportional to the square of phase voltage: Tst.Y = (1/3) · Tst.DOL = 450 / 3 = 150.0 N·m. Both line current and starting torque are reduced to exactly 1/3 of their DOL values.

Question 18Transformers, machines and distribution-system reasoning

A large low-voltage motor with direct-on-line starting current Ist.DOL = 1200 A is started using an auto-transformer starter with a voltage tapping of K = 65% (0.65 Un). Neglecting magnetizing current of the auto-transformer, what is the starting current drawn from the utility power line (Iline) during starting?

  1. 400 A
  2. 507 A Correct answer.
  3. 780 A
  4. 1200 A

Explanation

When an auto-transformer reduces motor terminal voltage to K · Un, the motor starting current is Imotor = K · Ist.DOL. The current drawn from the line through the auto-transformer transformation ratio (1/K) is: Iline = K · Imotor = K² · Ist.DOL = (0.65)² × 1200 A = 0.4225 × 1200 A = 507 A.

Question 19Transformers, machines and distribution-system reasoning

A 160 kW low-voltage motor is started direct-on-line. The motor starting apparent power is Sst = 1100 kVA. The short-circuit capacity available at the 0.4kV distribution busbar is Sk = 22,000 kVA (22 MVA). Using the standard formula ΔUst% = (Sst / (Sst + Sk)) × 100%, what is the percentage busbar voltage dip (ΔUst%) during motor starting?

  1. 2.38%
  2. 4.76% Correct answer.
  3. 8.50%
  4. 12.4%

Explanation

Using the formula explicitly supplied in the exercise,1100/(1100+22000)*100=4.7619%, approximately4.76%. This conditional proxy does not establish that direct-on-line starting is permitted: actual voltage criteria, dynamic response, equipment limits and network approval need their applicable design evidence.

Question 20Transformers, machines and distribution-system reasoning

Use the stipulated simplified voltage-divider model with a1000kVA transformer, impedance0.045per unit on its own base, an infinite upstream source and motor starting apparent power750kVA. Define Sk=1000/0.045kVA and dip=750/(750+Sk). What percentage dip does this model predict?

  1. 3.27% Correct answer.
  2. 32.6%
  3. 3.38%
  4. 33.8%

Explanation

In the stipulated model,Sk=1000/0.045=22222.22kVA. The defined ratio is750/(750+22222.22)=0.032647, approximately3.27%. This is the result of the specified simplified relation, not an exact transient prediction for every real motor and supply.

Question 21Transformers, machines and distribution-system reasoning

How does a low-voltage solid-state Soft Starter (晶闸管软起动器) control motor starting current and acceleration, and what is a key operational characteristic after starting completes?

  1. It alters the AC supply frequency from 0 Hz to 50 Hz continuously
  2. It uses anti-parallel thyristors to phase-control the stator voltage ramp from an initial voltage up to full voltage; after reaching rated speed, a bypass contactor (旁路接触器) is typically engaged to eliminate thyristor running heat and harmonics Correct answer.
  3. It inserts series dynamic resistors into the rotor winding circuit
  4. It converts 380V AC into 220V DC to drive a brushless DC motor

Explanation

A solid-state soft starter uses anti-parallel silicon controlled rectifiers (thyristors / SCRs) to smoothly increase the conduction angle and stator terminal voltage from an initial preset voltage to rated voltage. Once at rated speed, a bypass contactor (旁路接触器) closes to bypass the SCRs, eliminating continuous thermal dissipation, cooling fan noise, and steady-state harmonic generation.

Question 22Fault models, protective coordination and earthing concepts

At a 10kV high-voltage switchgear busbar, the calculated initial three-phase symmetrical short-circuit current is I"k = 16.0 kA. Taking the standard peak factor Kch = 1.80 for high-voltage distribution networks, what is the peak short-circuit current (ip)?

  1. 22.63 kA
  2. 28.80 kA
  3. 40.73 kA Correct answer.
  4. 49.88 kA

Explanation

According to GB/T 15544 and design manuals, the peak short-circuit current (短路电流峰值 / 冲击电流) is: ip = √2 · Kch · I"k = 1.4142 × 1.80 × 16.0 kA = 2.5456 × 16.0 = 40.73 kA.

Question 23Fault models, protective coordination and earthing concepts

For a supplied simplified calculation, take current I=20kA and factor K=1.80. Define the exercise RMS quantity as I*sqrt(1+2*(K-1)^2). What value follows? This formula exercise does not itself select switchgear dynamic or breaking duty.

  1. 20.00 kA
  2. 25.10 kA
  3. 30.20 kA Correct answer.
  4. 50.91 kA

Explanation

The formula for the short-circuit shock current rms value (冲击电流有效值) is Ich = I"k · √(1 + 2(Kch - 1)²). With Kch = 1.80: √(1 + 2(1.80 - 1)²) = √(1 + 2(0.80)²) = √(1 + 2 × 0.64) = √2.28 ≈ 1.50997. Thus, Ich = 20.0 × 1.510 = 30.20 kA.

Question 24Fault models, protective coordination and earthing concepts

A 10kV XLPE-insulated copper conductor power cable feeds a step-down substation. The three-phase symmetrical short-circuit current is Ith = 12.5 kA and the short-circuit clearing time is tpo = 0.50 s. For XLPE copper cable, the thermal stability coefficient is C = 143 A·s^(1/2)/mm². What is the minimum required cable conductor cross-sectional area (Smin) based on thermal stability?

  1. 35 mm²
  2. 50 mm²
  3. 61.8 mm² (select 70 mm² standard) Correct answer.
  4. 95 mm²

Explanation

According to the Code for Design of Cables of Electric Engineering (GB 50217-2018) and Design Manual, the minimum conductor cross-section for thermal stability is: Smin = (Ith · √tpo) / C = (12500 × √0.50) / 143 = (12500 × 0.7071) / 143 = 8838.8 / 143 ≈ 61.81 mm². Therefore, the nearest standard nominal cable cross-section is 70 mm².

Question 25Fault models, protective coordination and earthing concepts

In a hypothetical beam-stress check, use bending moment100Nm, section modulus853mm^3 and a supplied allowable bending stress140MPa. What is M/W, and is it below the supplied stress allowance? This is one stress check rather than complete busbar approval.

  1. σ = 58.6 MPa, satisfies requirement
  2. σ = 117.2 MPa, satisfies requirement Correct answer.
  3. σ = 185.4 MPa, fails requirement
  4. σ = 234.5 MPa, fails requirement

Explanation

Convert section modulus853mm^3 to853*10^-9m^3. Stress100/(853*10^-9)=117.23MPa, below the exercise allowance140MPa. Other loads, supports, geometry, temperature and actual applicable design requirements remain separate; one numerical comparison does not prove complete installation compliance.

Question 26Fault models, protective coordination and earthing concepts

A 10kV indoor vacuum circuit breaker (VCB) has a nameplate rated short-circuit breaking current of Isc = 25 kA and rated peak withstand making current of Icm = 63 kA. If the calculated short-circuit current at the installation busbar is I"k = 22 kA and peak current is ip = 56 kA, does the circuit breaker satisfy breaking and making capacity checks?

  1. Fails breaking capacity because 22 kA is too close to 25 kA
  2. Fails making capacity because Icm must be at least 3.0 times I"k
  3. Satisfies both breaking capacity (Isc ≥ I"k) and making capacity (Icm ≥ ip) Correct answer.
  4. Satisfies breaking capacity but fails because rated duration must be at least 10 seconds

Explanation

According to the Design Manual and high-voltage switchgear standards (GB/T 1984), a circuit breaker is verified if: (1) Rated breaking current Isc ≥ calculated short-circuit current I"k (25 kA ≥ 22 kA: Passed); (2) Rated making/peak withstand current Icm ≥ calculated peak short-circuit current ip (63 kA ≥ 56 kA: Passed). Both conditions are satisfied.

Question 27Fault models, protective coordination and earthing concepts

In a 380/220V TN-S distribution system fed by a Dyn11 transformer, a single phase-to-protective conductor (PE) metallic short circuit occurs at the load terminals. Why is the single-phase fault current (Ik^(1)) in a Dyn11 transformer system substantially larger than in a Yyn0 transformer system?

  1. Dyn11 transformers have a much higher primary rated voltage than Yyn0 transformers
  2. The delta primary winding of Dyn11 provides a low-impedance path for circulating zero-sequence currents, resulting in a much smaller zero-sequence impedance (Z0) compared to Yyn0 Correct answer.
  3. Dyn11 transformers have zero leakage inductance in their secondary windings
  4. Yyn0 transformers automatically disconnect the neutral during single-phase faults

Explanation

In a Dyn11 transformer, the delta (D) primary provides a closed circulation path for zero-sequence magnetic flux and currents, giving Z0 ≈ Z1 (roughly 0.9 to 1.1 Z1). In contrast, a Yyn0 transformer has very high zero-sequence impedance (Z0 ≈ 3 to 10 Z1) because zero-sequence flux must close through the air and tank wall. Therefore, Dyn11 yields much larger Ik^(1), ensuring sensitive tripping of overcurrent protection.

Question 28Fault models, protective coordination and earthing concepts

When a three-phase short-circuit occurs on a 10kV busbar feeding several large high-voltage induction motors, why and how must the motor feedback current be considered in the initial short-circuit calculation under GB/T 15544?

  1. Induction motors consume extra active power during a fault and reduce the fault current
  2. Rotating induction motors act as generators driven by mechanical inertia, feeding subtransient short-circuit current (I"k.M) that adds to the grid contribution during the first few cycles (ip and I"k) Correct answer.
  3. Induction motors only feed back current if equipped with synchronous permanent magnets
  4. Motor feedback current is completely DC and can be disregarded in AC breaking capacity checks

Explanation

During a busbar fault, the terminal voltage collapses, but the motor's rotor magnetic field and rotating kinetic inertia cause it to act as an asynchronous generator. It feeds subtransient AC current I"k.M = E" / X"d.M into the fault for 3 to 5 cycles, significantly increasing the initial peak current ip and initial symmetrical current I"k.

Question 29Fault models, protective coordination and earthing concepts

A protective current transformer (CT) has a rated secondary current of 5 A, secondary winding internal resistance of Rct = 0.20 Ω, and connecting lead loop resistance of Rwire = 0.30 Ω. The protective relay burden is Rrelay = 0.10 Ω. For a 10P15 class CT, what is the maximum allowable total connected external secondary burden (Zb.max) if the rated secondary output is Sb = 25 VA?

  1. 0.40 Ω
  2. 0.80 Ω
  3. 1.00 Ω Correct answer.
  4. 5.00 Ω

Explanation

Rated secondary burden impedance is Zb.N = Sb / Isn² = 25 VA / (5 A)² = 25 / 25 = 1.00 Ω. The total connected external burden is Zext = Rwire + Rrelay = 0.30 + 0.10 = 0.40 Ω. Since 0.40 Ω ≤ 1.00 Ω, the CT operates well within its rated accuracy limit factor (ALF = 15).

Question 30Fault models, protective coordination and earthing concepts

In modern low-voltage air circuit breakers (ACB) equipped with electronic trip units (微机脱扣器), what protective functions are represented by the standard 'LSI(G)' protection designations?

  1. L: Line loss; S: Surge protection; I: Insulation check; G: Grid sync
  2. L: Long-time inverse delay (Overload); S: Short-time delay (Selective short-circuit); I: Instantaneous (Short-circuit); G: Ground fault protection Correct answer.
  3. L: Low voltage; S: Single phase; I: Inrush limit; G: Galvanic isolation
  4. L: Lightning; S: Spark gap; I: Inductive surge; G: Generator trip

Explanation

Under low-voltage switchgear standards (GB/T 14048.2 and IEC 60947-2): L represents Long-time inverse delay overload protection (过载长延时); S represents Short-time delay short-circuit protection for time-graded selectivity (短路短延时); I represents Instantaneous short-circuit tripping without intentional delay (短路瞬时); and G represents Ground-fault residual or zero-sequence protection (接地故障保护).

Question 31Fault models, protective coordination and earthing concepts

According to the Code for Design of Low-Voltage Electrical Installations (GB 50054-2011), what is the fundamental technical distinction between 'Basic Protection' (基本防护 / 防直接接触) and 'Fault Protection' (故障防护 / 防间接接触) against electric shock?

  1. Basic protection prevents electric shock during normal operating conditions (touching energized live parts); Fault protection prevents electric shock during single fault conditions (touching exposed conductive parts that become live due to insulation breakdown) Correct answer.
  2. Basic protection relies on RCDs; Fault protection relies on circuit breakers only
  3. Basic protection applies only to DC systems; Fault protection applies only to AC systems
  4. Basic protection is optional; Fault protection is mandatory

Explanation

GB 50054-2011 (Chapter 5) defines Basic Protection (direct contact protection) as measures preventing human contact with normally energized live parts (insulation, barriers, enclosures). Fault Protection (indirect contact protection) prevents electric shock when a human touches exposed conductive parts (metal casings) that have become energized due to basic insulation failure, primarily achieved by automatic disconnection of supply (ADS) combined with equipotential bonding.

Question 32Installation planning, measurement and evidence of compliance

A PVC-insulated copper cable has a nominal base current-carrying capacity of 36 A at a reference ambient temperature of 30°C. If installed in an industrial facility in the Afar region where the design ambient temperature is 45°C, what is the derated current-carrying capacity using an ambient temperature correction factor Ca of 0.79?

  1. 28.44 A Correct answer.
  2. 32.10 A
  3. 24.50 A
  4. 36.00 A

Explanation

The effective current-carrying capacity Iz is obtained by multiplying the base tabulated ampacity It by the environmental correction factor Ca: Iz = It × Ca = 36 A × 0.79 = 28.44 A.

Question 33Installation planning, measurement and evidence of compliance

Six multi-core copper cables are installed bunched in a single surface-mounted enclosed trunking in air. If the grouping correction factor Cg is 0.57 and the ambient temperature correction factor Ca is 0.87, what is the effective current-carrying capacity Iz of a cable whose baseline catalog rating I0 is 43 A?

  1. 21.32 A Correct answer.
  2. 24.51 A
  3. 37.41 A
  4. 18.20 A

Explanation

The effective derated ampacity is calculated by applying all applicable correction factors: Iz = I0 × Ca × Cg = 43 A × 0.87 × 0.57 = 43 × 0.4959 = 21.32 A.

Question 34Installation planning, measurement and evidence of compliance

Using the adiabatic formula S = √(I² · t) / k, what is the minimum cross-sectional area of a PVC-insulated copper protective grounding conductor required to withstand a prospective fault current of 6,000 A for a protective clearing time of 0.2 seconds, where k = 115?

  1. 23.33 mm² (select standard 25 mm²) Correct answer.
  2. 14.50 mm² (select standard 16 mm²)
  3. 32.40 mm² (select standard 35 mm²)
  4. 10.20 mm² (select standard 10 mm²)

Explanation

Applying the adiabatic equation: S = (I × √t) / k = (6000 × √0.2) / 115 = (6000 × 0.4472) / 115 = 2683.28 / 115 = 23.33 mm². Since 23.33 mm² exceeds 16 mm², the next standard commercial cable size is 25 mm².

Question 35Installation planning, measurement and evidence of compliance

What is the critical safety purpose of conducting a pre-commissioning polarity test on a newly wired single-phase 230 V electrical installation?

  1. To verify that all single-pole switches, fuses, and circuit breakers are installed strictly in the phase (live) conductor and never in the neutral Correct answer.
  2. To check whether alternating current is flowing in a clockwise or counterclockwise direction
  3. To ensure that the neutral conductor is grounded at every light switch
  4. To verify that the supply frequency is exactly 50 Hz

Explanation

If a single-pole switch or fuse is mistakenly installed in the neutral conductor, opening the switch disconnects the circuit and turns off the light, but the appliance remains energized at 230 V relative to earth. Any person touching the lamp socket or appliance element will suffer a severe electric shock.

Question 36Installation planning, measurement and evidence of compliance

A current transformer (CT) is rated 600:5 A. If the primary current is 480 A, what is the secondary current?

  1. 3.0 A
  2. 4.0 A Correct answer.
  3. 5.0 A
  4. 6.0 A

Explanation

CT secondary current = primary × (5/600) = 480 × (5/600) = 4.0 A. Standard CT secondaries are 5 A or 1 A; the ratio in the nameplate (600:5) is the turns ratio used to scale primary current down to a meter-friendly value.

Question 37Installation planning, measurement and evidence of compliance

A polarization index (PI) test on motor insulation gives a 1-minute reading of 200 MΩ and a 10-minute reading of 800 MΩ. The PI is:

  1. 0.25
  2. 1.0
  3. 4.0 Correct answer.
  4. 8.0

Explanation

Polarization Index = R(10 min) / R(1 min) = 800 / 200 = 4.0. Per IEEE 43, PI ≥ 2.0 is generally acceptable for Class B/F/H insulation; PI = 4.0 indicates dry, clean insulation in good condition.

Question 38Installation planning, measurement and evidence of compliance

A potential transformer (PT) is rated 7200:120 V. If the primary voltage is 6900 V, what is the secondary voltage?

  1. 100 V
  2. 115 V Correct answer.
  3. 120 V
  4. 125 V

Explanation

PT ratio = 7200/120 = 60. Secondary V = 6900/60 = 115 V. Distribution PTs are sized so that nominal primary line-to-neutral (e.g., 7200 V on a 12.47 kV system) produces 120 V on the secondary.

Question 39Installation planning, measurement and evidence of compliance

A three-phase 4-wire 480Y/277 V system uses three CTs (1200:5) and three PTs (480:120) for revenue metering. The meter multiplier (overall ratio) is approximately:

  1. 240
  2. 480
  3. 960 Correct answer.
  4. 1200

Explanation

Meter multiplier = CT ratio × PT ratio = (1200/5) × (480/120) = 240 × 4 = 960. The meter measures secondary kWh and multiplies by 960 to obtain primary energy. PT ratio 480:120 = 4 (line-to-line PTs); use 277:120 only if line-to-neutral PTs are specified.

Question 40Measurement, feedback, PLC and automation

In a municipal constant-pressure variable-frequency water booster pump system (一用一备 or 两用一备), what is the function of the water pressure transmitter feedback loop connected to the VFD PID controller?

  1. To measure water temperature for frost protection
  2. To compare the actual pipe header pressure (PV) with the target setpoint pressure (SV) and continuously adjust motor rotational speed (Hz) to maintain constant water pressure under varying consumer flow demands Correct answer.
  3. To automatically disconnect the electrical supply when water is clean
  4. To run the pump at 100 Hz continuous overspeed

Explanation

In a constant pressure VFD booster system, a pressure sensor (4–20 mA or 0–10V) continuously measures discharge header water pressure. The VFD's internal closed-loop PID controller compares this Process Variable (PV) to the Setpoint Variable (SV) and modulates inverter output frequency (0–50 Hz), maintaining constant system pressure with high energy efficiency.

Question 41Measurement, feedback, PLC and automation

In a Building Automation System (BAS / 建筑设备监控系统), when interfacing a Direct Digital Controller (DDC) with an air handling unit (AHU), what signal types correspond to the supply air temperature sensor (Pt1000 RTD) and the supply air fan start/stop command respectively?

  1. Temperature sensor: Analog Input (AI); Fan start/stop: Digital Output (DO) Correct answer.
  2. Temperature sensor: Digital Input (DI); Fan start/stop: Analog Output (AO)
  3. Temperature sensor: Digital Output (DO); Fan start/stop: Analog Input (AI)
  4. Temperature sensor: Analog Output (AO); Fan start/stop: Digital Input (DI)

Explanation

In BAS DDC engineering: (1) A temperature sensor delivers a continuously variable electrical resistance/voltage/current signal representing a continuous physical quantity, which is an Analog Input (AI); (2) A fan start/stop command is a discrete two-state (energize/de-energize relay coil) signal sent from the controller to the motor starter, which is a Digital Output (DO).

Question 42Measurement, feedback, PLC and automation

A three-phase six-pulse uncontrolled bridge rectifier with line-to-line input 480 V (RMS) produces a DC output (no overlap, ideal) of approximately:

  1. 432 V
  2. 540 V
  3. 648 V Correct answer.
  4. 1357 V

Explanation

Vdc = (3√2/π) × VLL = 1.35 × 480 = 648 V. Equivalent expression: Vdc = (3/π) × Vm,LL where Vm,LL = √2 × VLL,rms = 678.8 V, giving Vdc = (3/π) × 678.8 = 648 V. This is the classic ratio for a six-pulse uncontrolled bridge.

Question 43Measurement, feedback, PLC and automation

A six-pulse line-commutated rectifier has firing angle α. At α = 60°, the average DC output relative to α = 0° is:

  1. 0.866 × Vdc(0)
  2. 0.707 × Vdc(0)
  3. 0.500 × Vdc(0) Correct answer.
  4. 0 × Vdc(0)

Explanation

Vdc(α) = Vdc(0) × cos α. At α = 60°, cos 60° = 0.500, so the DC output drops to half. At α = 90° the average DC equals zero (rectifier-to-inverter transition); for α > 90° the converter operates as an inverter feeding active power back to AC.

Question 44Measurement, feedback, PLC and automation

A pulse-width modulated (PWM) inverter has a modulation index ma in the linear range. Peak fundamental phase voltage is:

  1. Vdc/2
  2. ma × Vdc/2 Correct answer.
  3. ma × Vdc
  4. (4/π) × Vdc

Explanation

For sinusoidal PWM in the linear region (0 ≤ ma ≤ 1.0), peak fundamental phase voltage = ma × (Vdc/2). Above ma = 1 (overmodulation) the relationship becomes nonlinear, and at ma → ∞ the inverter approaches square-wave with peak fundamental of (4/π) × (Vdc/2) ≈ 0.637 Vdc.

Question 45Measurement, feedback, PLC and automation

A variable-frequency drive (VFD) controls a 60 Hz induction motor. To maintain constant air-gap flux below base speed, the V/f ratio should be:

  1. Increased proportional to load
  2. Held approximately constant Correct answer.
  3. Decreased proportional to load
  4. Switched between two fixed ratios

Explanation

Constant V/f control maintains the magnetizing flux Φ ≈ V/(4.44 × f × N) at the design value. Doubling f without doubling V causes flux to halve, reducing torque capability. Above base speed, V is held constant (already at maximum) and f is increased - torque decreases inversely with speed (constant power region).

Question 46Measurement, feedback, PLC and automation

A latching relay used in a control circuit retains its state when:

  1. Coil voltage is removed Correct answer.
  2. The relay is energized
  3. External logic feedback is provided
  4. An auxiliary contact is shorted

Explanation

A magnetic latching (impulse) relay uses a permanent magnet or mechanical latch to maintain state after the coil is de-energized - useful for low-power sealed-in control. To change state, a reverse-polarity pulse (or a separate reset coil) must be applied.

Question 47Measurement, feedback, PLC and automation

In ladder logic, a Normally Open (NO) input contact in series with a Normally Closed (NC) interlock contact, both feeding a coil, behaves as:

  1. AND of the two contact states Correct answer.
  2. OR of the two contact states
  3. NAND
  4. Bistable latch

Explanation

Series contacts in ladder logic implement Boolean AND: the coil energizes only when BOTH the NO contact is true (closed) AND the NC contact is true (closed by NC inversion of its source signal). Parallel contacts implement OR.

Question 48Measurement, feedback, PLC and automation

In ladder logic, De Morgan's law states that a NAND function is equivalent to:

  1. OR of the inverted inputs Correct answer.
  2. AND of the inverted inputs
  3. NOR of the inputs
  4. XOR of the inputs

Explanation

De Morgan's law: NOT(A AND B) = (NOT A) OR (NOT B). A NAND contact rung with two NO inputs in series is logically equivalent to two NC contacts in parallel (OR of inverted inputs). This is fundamental to relay logic and PLC programming on the PE Power exam.

Question 49Lighting models, metering and building-control integration

An open-plan office has dimensions Length L = 20 m, Width W = 10 m (Floor Area A = 200 m²). The target maintained average illuminance is Eav = 300 lx. Recessed LED luminaires with rated luminous flux Φ = 3600 lm per luminaire are used. Given coefficient of utilization U = 0.60 and maintenance factor K = 0.80, how many luminaires (N) are required using the lumen method (利用系数法)?

  1. 24 luminaires
  2. 35 luminaires Correct answer.
  3. 48 luminaires
  4. 70 luminaires

Explanation

Lumen method illuminance formula: Eav = (N · Φ · U · K) / A, which yields N = (Eav · A) / (Φ · U · K) = (300 × 200) / (3600 × 0.60 × 0.80) = 60,000 / 1728 ≈ 34.72. Rounding up to the nearest integer gives N = 35 luminaires (e.g., arranged in 5 rows of 7 luminaires).

Question 50Lighting models, metering and building-control integration

In lighting calculation, the Room Cavity Ratio (RCR / 室空间比) quantifies room geometry for coefficient of utilization determination. For a room of length L = 12 m, width W = 8 m, luminaire mounting height above floor h = 3.6 m, and working plane height hwp = 0.8 m (room cavity height hrc = 3.6 - 0.8 = 2.8 m), what is the Room Cavity Ratio (RCR)?

  1. 1.46
  2. 2.92 Correct answer.
  3. 3.75
  4. 5.83

Explanation

The formula for Room Cavity Ratio is: RCR = (5 · hrc · (L + W)) / (L · W). Here, hrc = 3.6 - 0.8 = 2.8 m. RCR = (5 × 2.8 × (12 + 8)) / (12 × 8) = (14 × 20) / 96 = 280 / 96 ≈ 2.9167 ≈ 2.92.