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EEA Electrical Engineer Exam - independent technical foundations study materials
EEA Electrical Engineer Exam - independent technical foundations study resources

EEA Electrical Engineer Exam - independent technical foundations practice questions

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

A residential building consists of 10 identical apartments, each having an individual connected maximum demand of 8 kW. If the electrical design engineer applies an overall diversity factor of 0.65 for the main building distribution board, what is the diversified design load?

  1. 52 kW Correct answer.
  2. 80 kW
  3. 64 kW
  4. 45 kW

Explanation

The total connected load of the 10 apartments is 10 × 8 kW = 80 kW. Applying the diversity factor of 0.65, the diversified design load is 80 kW × 0.65 = 52 kW.

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

An industrial manufacturing facility operates with an active real load of 400 kW at an uncorrected power factor of 0.72 lagging. What rating of shunt capacitor bank in kvar is required to improve the facility's power factor to 0.95 lagging?

  1. 254 kvar Correct answer.
  2. 180 kvar
  3. 320 kvar
  4. 125 kvar

Explanation

Original angle φ1 = arccos(0.72) = 43.95° → tan φ1 = 0.9639. Target angle φ2 = arccos(0.95) = 18.19° → tan φ2 = 0.3287. Required capacitor kvar Qc = P × (tan φ1 - tan φ2) = 400 kW × (0.9639 - 0.3287) = 400 × 0.6352 = 254.08 kvar ≈ 254 kvar.

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

Using the standard formula R = [ρ / (2π · L)] · ln(4L / d), calculate the theoretical earth resistance of a single vertical copper-bonded steel rod of length L = 3.0 meters and diameter d = 0.016 meters driven in soil with a uniform resistivity ρ = 60 Ω·m.

  1. 21.07 ohms Correct answer.
  2. 10.54 ohms
  3. 35.20 ohms
  4. 42.14 ohms

Explanation

Given L = 3.0 m, d = 0.016 m, and ρ = 60 Ω·m: 4L / d = (4 × 3.0) / 0.016 = 12 / 0.016 = 750. Then ln(750) = 6.620. Now evaluate the coefficient: ρ / (2π × L) = 60 / (2 × 3.14159 × 3.0) = 60 / 18.8496 = 3.183 Ω. Total resistance R = 3.183 × 6.620 = 21.07 Ω.

Question 4Phasors, 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 5Phasors, 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 6Phasors, 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 7Phasors, 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 8Phasors, 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 9Phasors, 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 10Phasors, 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 11Phasors, 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 12Phasors, 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 13Phasors, 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 14Phasors, 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 15Transformers, machines and distribution-system reasoning

What is the rated full-load secondary phase-to-phase current of a three-phase 630 kVA, 15 kV / 400 V distribution transformer at rated voltage?

  1. 909.3 A Correct answer.
  2. 525.0 A
  3. 1,575.0 A
  4. 1,200.0 A

Explanation

Full-load secondary current is calculated as I_FL = S / (√3 × V_L) = 630,000 VA / (√3 × 400 V) = 630,000 / 692.82 = 909.33 A ≈ 909.3 A.

Question 16Transformers, machines and distribution-system reasoning

On oil-immersed power transformers equipped with an oil conservator tank, what are the distinct operational functions of the two contact stages in a Buchholz relay?

  1. The upper float triggers an alarm upon slow gas accumulation, while the lower float/vane trips the circuit breaker upon rapid oil surge Correct answer.
  2. The upper float trips the low-voltage breaker, while the lower float trips the high-voltage breaker
  3. Both contacts trip simultaneously upon any change in oil temperature
  4. The upper contact starts the cooling fans, while the lower contact disconnects the neutral

Explanation

A Buchholz relay is mounted in the pipe connecting the main transformer tank to the conservator. Stage 1 (upper float) detects slow gas generation caused by minor incipient faults (e.g. core insulation degradation) and triggers an alarm. Stage 2 (lower float/vane) detects violent oil rushes caused by severe internal short circuits and immediately trips the isolating circuit breakers.

Question 17Transformers, machines and distribution-system reasoning

What is the primary operational advantage of implementing a Ring Main Unit (RMU) configuration in urban 15 kV medium-voltage underground distribution networks?

  1. It provides two alternative supply paths to distribution substations, allowing rapid isolation of cable faults and power restoration from the healthy feeder Correct answer.
  2. It automatically converts medium voltage to low voltage without requiring a distribution transformer
  3. It eliminates the need for underground cable trenches and conduit systems
  4. It eliminates all harmonic currents generated by non-linear consumer loads

Explanation

A Ring Main Unit (RMU) connects distribution substations in a closed ring that is normally operated open at a designated midpoint. If a cable section fails, the faulted section can be isolated via load-break switches, and power restored to all consumers by closing the ring tie switch, ensuring high service reliability.

Question 18Transformers, machines and distribution-system reasoning

How does an On-Line Double Conversion Uninterruptible Power Supply (UPS) operate under normal conditions to ensure zero transfer time during a power outage?

  1. The inverter continuously powers the critical load from the internal DC bus, which is simultaneously fed by the rectifier and floating battery Correct answer.
  2. The load is fed directly from the utility grid via a static switch while the inverter remains completely powered off
  3. The load is powered by a fly-wheel generator that takes 5 seconds to transfer during a blackout
  4. The battery is connected in series with the AC mains line to absorb voltage spikes

Explanation

In an online double-conversion UPS, incoming AC is rectified to DC, which continuously charges the battery and powers the inverter. The inverter continuously generates clean AC for the critical load. When utility power fails, battery power seamlessly sustains the DC bus with absolutely zero transfer time or phase discontinuity.

Question 19Transformers, machines and distribution-system reasoning

A three-phase, 400 V, 75 kW induction motor has a full-load current of 135 A and a direct-on-line (DOL) starting current equal to 6.5 times full-load current (877.5 A). If the motor is reconfigured to start using a Star-Delta starter, what will be the starting line current drawn from the supply?

  1. 292.5 A Correct answer.
  2. 506.6 A
  3. 438.8 A
  4. 135.0 A

Explanation

In a Star-Delta starter, the motor windings are connected in star during starting, reducing the winding voltage by 1/√3. Consequently, the starting current drawn from the line is reduced to exactly one-third (1/3) of the DOL starting current: I_start(star) = I_start(DOL) / 3 = 877.5 A / 3 = 292.5 A.

Question 20Transformers, machines and distribution-system reasoning

What primary parameters of a power transformer are determined by performing an open-circuit (no-load) test with rated voltage applied to the low-voltage winding?

  1. Core (iron) losses and magnetizing branch impedance (Rc and Xm) Correct answer.
  2. Full-load copper (I²R) losses and series leakage impedance
  3. Dielectric breakdown strength of the winding insulation
  4. Mechanical short-circuit withstand rating of the tank

Explanation

In an open-circuit test at rated voltage, the current drawn is only the small no-load magnetizing current (1–3% of rated). Copper losses are negligible, so the wattmeter reading directly measures the core (iron) hysteresis and eddy-current losses, from which shunt parameters Rc and Xm are derived.

Question 21Transformers, machines and distribution-system reasoning

An industrial manufacturing facility connected to the EEU 15 kV distribution network operates with an uncorrected average power factor of 0.75. When the plant installs automatic power factor correction capacitors to achieve 0.95 lagging, which significant internal technical benefit is realized within the factory's own distribution system?

  1. Reduction of internal I²R thermal losses in cables and transformers, decreased internal voltage drops, and released distribution transformer capacity for additional machinery Correct answer.
  2. Complete elimination of active real power (kW) consumption from the grid
  3. Increase in incoming line frequency from 50 Hz to 60 Hz
  4. Elimination of the requirement for overcurrent circuit breakers

Explanation

Improving power factor from 0.75 to 0.95 reduces total line current (apparent current I = P / [√3 · V · cos φ]) by over 21%. Because conductor thermal losses are proportional to I², internal distribution heating losses decrease by approximately 38%. Furthermore, internal voltage drop is reduced, and transformer kVA capacity is freed up to power additional production equipment.

Question 22Fault models, protective coordination and earthing concepts

What is the primary technical function of a main switch-disconnector installed at the service entrance of a building distribution board?

  1. To provide simultaneous isolation of all live poles from the supply under no-load or normal load conditions for safe maintenance Correct answer.
  2. To detect low-magnitude earth leakage currents and automatically disconnect within 30 milliseconds
  3. To regulate incoming utility voltage variations and maintain constant 230 V output
  4. To step down medium-voltage distribution to low-voltage consumer levels

Explanation

A switch-disconnector is an isolating device capable of making, carrying, and breaking currents under normal circuit conditions, providing an air gap isolation across all live conductors (and neutral where required) to guarantee worker safety during maintenance.

Question 23Fault models, protective coordination and earthing concepts

A 1,000 kVA, 400 V three-phase distribution transformer has a percentage impedance (%Z) of 5.0%. Assuming an infinite primary bus, what is the prospective symmetrical three-phase short-circuit fault current at the secondary terminals?

  1. 28.87 kA Correct answer.
  2. 14.43 kA
  3. 57.74 kA
  4. 20.00 kA

Explanation

First, calculate the transformer full-load secondary current: I_FL = 1,000,000 / (√3 × 400) = 1,443.38 A. The symmetrical short-circuit current is I_sc = I_FL / Z_pu = 1,443.38 A / 0.05 = 28,867.5 A ≈ 28.87 kA.

Question 24Fault models, protective coordination and earthing concepts

A 15 kV distribution bus is supplied by a 20 MVA substation power transformer having a per-unit impedance of 0.08 (8.0%) on its own rating base. Neglecting upstream utility source impedance, what is the three-phase symmetrical fault level on the 15 kV bus?

  1. 250 MVA Correct answer.
  2. 160 MVA
  3. 200 MVA
  4. 320 MVA

Explanation

The symmetrical short-circuit fault MVA is calculated as: Fault MVA = MVA_base / Z_pu = 20 MVA / 0.08 = 250 MVA.

Question 25Fault models, protective coordination and earthing concepts

Why is it strictly prohibited to leave the secondary terminals of an energized Current Transformer (CT) open-circuited while primary load current is flowing?

  1. Without secondary demagnetizing current, all primary current acts as magnetizing current, producing lethal peak voltages across the secondary and severe core overheating Correct answer.
  2. The primary winding will instantly melt due to excessive secondary current feedback
  3. The upstream circuit breaker will fail to trip during an overload
  4. The transformer oil will decompose and generate hydrogen gas

Explanation

Under normal closed-loop conditions, secondary ampere-turns oppose primary ampere-turns, keeping core flux small. If the secondary is open-circuited, the secondary current becomes zero and the entire primary ampere-turns act as magnetizing force. This drives the core into extreme saturation, producing sharp flux changes that induce thousands of volts across the open secondary terminals, presenting a fatal electric shock hazard and causing core destruction.

Question 26Fault models, protective coordination and earthing concepts

What is the standard rated secondary current specified for instrument current transformers (CTs) used in medium- and low-voltage distribution switchgear protection and metering?

  1. 1 A or 5 A Correct answer.
  2. 10 A or 20 A
  3. 0.1 A or 0.5 A
  4. 100 mA or 500 mA

Explanation

IEC 61869-2 establishes standard secondary current ratings for current transformers as 1 A or 5 A. In switchgear assemblies with short secondary lead runs, 5 A is standard, while 1 A is preferred for long cable runs to minimize secondary burden.

Question 27Fault models, protective coordination and earthing concepts

A 15 kV medium-voltage underground cable with copper conductors and XLPE insulation (k = 143) must withstand a prospective symmetrical short-circuit current of 18 kA for a clearing time of 0.5 seconds. What is the minimum required conductor cross-sectional area?

  1. 89.0 mm² (select standard 95 mm²) Correct answer.
  2. 62.9 mm² (select standard 70 mm²)
  3. 112.5 mm² (select standard 120 mm²)
  4. 48.2 mm² (select standard 50 mm²)

Explanation

Using the thermal short-circuit adiabatic formula: S = (I × √t) / k = (18,000 A × √0.5 s) / 143 = (18,000 × 0.7071) / 143 = 12,727.9 / 143 = 89.0 mm². The next available standard commercial cross-section is 95 mm².

Question 28Fault models, protective coordination and earthing concepts

How is a TN-S earthing system characterized from the distribution transformer to the final point of utilization?

  1. Separate neutral (N) and protective earth (PE) conductors are maintained throughout the entire distribution network Correct answer.
  2. Neutral and protective earth functions are combined in a single conductor throughout the entire system
  3. The installation has no connection to the transformer neutral and uses only a local earth rod
  4. The system neutral is isolated from ground and monitored by an insulation monitor

Explanation

In a TN-S system (Terre-Neutre Séparé), the source neutral is connected directly to earth, and separate Neutral (N) and Protective Earth (PE) conductors run independently from the source all the way to every point of utilization.

Question 29Fault models, protective coordination and earthing concepts

In a TN-C-S earthing arrangement (also known as Protective Multiple Earthing - PME), what are the distinct paths of the neutral and protective conductors?

  1. Neutral and protective earth are combined in a single PEN conductor from the utility supply to the consumer intake, then separated into independent N and PE conductors inside the building Correct answer.
  2. Neutral and protective earth are separated outside the building and combined inside final branch circuits
  3. The protective conductor is connected directly to water pipes with no connection to the utility neutral
  4. Neutral and protective earth conductors are connected through a high-impedance reactor at the consumer switchboard

Explanation

In TN-C-S systems, the utility distribution network uses a combined Protective Earth and Neutral (PEN) conductor (TN-C). At the consumer intake service head, the PEN conductor is bonded to the main earthing terminal and split into distinct, separate Neutral (N) and Protective Earth (PE) conductors (TN-S) for internal distribution.

Question 30Fault models, protective coordination and earthing concepts

How does flameproof (explosion-proof, designation 'Ex d') electrical equipment achieve safety when operated in a hazardous industrial environment?

  1. Its robust enclosure withstands an internal gas explosion without structural damage and cools exiting gases so they cannot ignite the surrounding atmosphere Correct answer.
  2. It completely prevents all explosive gases from ever entering the enclosure
  3. It operates at an energy level so low that electrical sparks have insufficient thermal energy to ignite gas
  4. It purges the interior of the enclosure continuously with flammable gas

Explanation

Flameproof (Ex d) protection assumes flammable gas will enter the enclosure and ignite. The heavy enclosure is engineered with flame paths (machined joints) to withstand the internal explosion pressure and cool the escaping exhaust gases below the ignition temperature of the external atmosphere.

Question 31Fault models, protective coordination and earthing concepts

What is the primary safety objective of applying a physical padlock and warning tag (Lockout / Tagout - LOTO) to an isolated electrical circuit breaker before beginning maintenance work?

  1. To physically prevent unauthorized or accidental re-energization of the circuit while personnel are working on it Correct answer.
  2. To measure the leakage current flowing to ground through the padlock
  3. To notify the utility company of a billing meter disconnection
  4. To prevent theft of the circuit breaker from the distribution panel

Explanation

Lockout/Tagout (LOTO) is an essential electrical safety procedure that physically locks the circuit breaker or disconnector in the OFF position using a technician's personal padlock, accompanied by a tag identifying the worker. This guarantees that nobody can inadvertently restore power while work is underway.

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 office room with dimensions 12 m by 8 m (area = 96 m²) requires a maintained illuminance of 500 lux. The selected LED luminaire produces 4,000 lumens. If the utilization factor (UF) is 0.60 and the maintenance factor (MF) is 0.80, how many luminaires are required using the Lumen Method?

  1. 25 luminaires Correct answer.
  2. 20 luminaires
  3. 30 luminaires
  4. 16 luminaires

Explanation

Using the Lumen Method formula: Total Required Flux (Φ) = (E × A) / (UF × MF) = (500 lux × 96 m²) / (0.60 × 0.80) = 48,000 / 0.48 = 100,000 lumens. The number of luminaires N = Total Flux / Luminaire Flux = 100,000 / 4,000 = 25 luminaires.

Question 50Lighting 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).