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Question 1Electrical circuits, machines and digital logic

A single-phase 230 V sub-circuit supplies a fixed resistive heating load of 3.68 kW. What is the calculated design current (Ib) for this installation?

  1. 16.0 A Correct answer.
  2. 10.0 A
  3. 8.0 A
  4. 20.5 A

Explanation

For a single-phase resistive load where cos phi = 1.0, the design current is calculated as Ib = P / U = 3680 W / 230 V = 16.0 A. This calculation represents the baseline current used to select the upstream protective device (In) and determine the required cable cross-section under DS/HD 60364.

Question 2Electrical circuits, machines and digital logic

A three-phase 400 V symmetrical resistive load has a total active power rating of 20.78 kW. What is the line current (Ib) drawn by this load?

  1. 30.0 A Correct answer.
  2. 17.3 A
  3. 52.0 A
  4. 10.0 A

Explanation

For a balanced three-phase resistive load (cos phi = 1.0), the line current is calculated using Ib = P / (sqrt(3) x U) = 20780 W / (1.732 x 400 V) = 20780 / 692.8 = 30.0 A. This current defines the required rating for the upstream three-pole circuit breaker and cable conductors.

Question 3Electrical circuits, machines and digital logic

Calculate the percentage voltage drop (Delta U %) in a 230 V single-phase circuit wired with 2.5 mm² copper conductors (conductivity gamma = 56 m/(ohm x mm²)) over a one-way length of 35 m, supplying a 10 A load at unity power factor (cos phi = 1.0).

  1. 2.17% Correct answer.
  2. 4.35%
  3. 1.09%
  4. 3.50%

Explanation

The single-phase voltage drop formula is Delta U = (2 x L x I x cos phi) / (gamma x S). Inserting the parameters: Delta U = (2 x 35 m x 10 A x 1.0) / (56 x 2.5 mm²) = 700 / 140 = 5.0 V. Expressed as a percentage of nominal 230 V: (5.0 V / 230 V) x 100% = 2.17%. Because 2.17% <= 3.0% (and <= 5.0%), this cable cross-section complies with DS/HD 60364.

Question 4Electrical circuits, machines and digital logic

A 400 V three-phase balanced distribution circuit carries a continuous load of 25 A (cos phi = 0.90) over a cable route of 40 m using 6 mm² copper conductors (gamma = 56 m/(ohm x mm²)). What is the line-to-line voltage drop (Delta U) in volts?

  1. 4.64 V Correct answer.
  2. 8.03 V
  3. 2.68 V
  4. 9.28 V

Explanation

For a three-phase circuit, the line-to-line voltage drop is Delta U = (sqrt(3) x L x I x cos phi) / (gamma x S). Inserting the values: Delta U = (1.732 x 40 x 25 x 0.90) / (56 x 6) = 1558.8 / 336 = 4.64 V. Expressed as a percentage of 400 V, this represents 1.16%, which is well within standard limits.

Question 5Electrical circuits, machines and digital logic

A 230 V single-phase cable run of 45 m delivers 16 A to an outdoor water heater. The allowable maximum voltage drop is 3% (6.9 V). Using copper (gamma = 56 m/(ohm x mm²)) and cos phi = 1.0, calculate the minimum theoretical cross-sectional area (S) needed to satisfy the voltage drop constraint.

  1. 3.73 mm² (requiring selection of standard 4.0 mm² cable) Correct answer.
  2. 2.35 mm² (permitting selection of standard 2.5 mm² cable)
  3. 1.48 mm² (permitting selection of standard 1.5 mm² cable)
  4. 5.85 mm² (requiring selection of standard 6.0 mm² cable)

Explanation

From Delta U = (2 x L x I x cos phi) / (gamma x S), rearranging for cross-section yields S = (2 x L x I x cos phi) / (gamma x Delta U). Substituting values: S = (2 x 45 m x 16 A x 1.0) / (56 x 6.9 V) = 1440 / 386.4 = 3.727 mm². Because 3.73 mm² exceeds standard 2.5 mm², the designer must specify standard 4.0 mm² cable to satisfy the 3% voltage drop limit.

Question 6Electrical circuits, machines and digital logic

In a commercial installation, an electrician must calculate the line-to-line voltage drop for a 400 V three-phase sub-main supplying a 50 A balanced inductive load (cos phi = 0.80, sin phi = 0.60) over a distance of 50 m. The cable is 10 mm² copper (R = 2.25 mohm/m at operating temperature) with inductive reactance X = 0.08 mohm/m. What is the line-to-line voltage drop (Delta U)?

  1. 8.00 V (2.00%) Correct answer.
  2. 13.86 V (3.47%)
  3. 4.62 V (1.16%)
  4. 16.00 V (4.00%)

Explanation

Total phase resistance R = 2.25 mohm/m x 50 m = 0.1125 ohm. Total reactance X = 0.08 mohm/m x 50 m = 0.004 ohm. For a balanced three-phase system: Delta U = sqrt(3) x I x (R x cos phi + X x sin phi). Substituting values: Delta U = 1.732 x 50 A x (0.1125 x 0.80 + 0.004 x 0.60) = 86.60 x (0.0900 + 0.0024) = 86.60 x 0.0924 = 8.00 V line-to-line. As a percentage of 400 V, (8.00 / 400) x 100% = 2.00%, well within the 5% limit.

Question 7Transformers, machines and distribution-system reasoning

What happens to single-phase 230 V consumer loads in a three-phase building installation if a mechanical break occurs in the main upstream neutral conductor (rupture du neutre)?

  1. The phase conductors automatically disconnect via bimetallic expansion
  2. All single-phase voltages drop immediately to exactly 0 V across all circuits
  3. The line voltage rises to 20,000 V medium-voltage levels
  4. Single-phase circuits on lightly loaded phases experience dangerous overvoltages up to 400 V, destroying electronic appliances and creating fire risks Correct answer.

Explanation

In a three-phase four-wire system with an open neutral, the artificial neutral point shifts towards the phase carrying the heaviest load. Consequently, the voltage across lightly loaded single-phase appliances can surge up to phase-to-phase voltage (approaching 400 V AC), while the heavily loaded phase experiences severe undervoltage. This overvoltage destroys switch-mode power supplies, LED drivers, and household electronics, and represents a major cause of electrical fires.

Question 8Transformers, machines and distribution-system reasoning

An electrical project manager designs an indoor transformer substation for an industrial plant. The distribution transformer has a rating of 1000 kVA, 16 kV / 400 V, and a vector group of Dyn11. Why is the Dyn11 vector group standardly specified in Swiss low-voltage distribution networks instead of a Yyn0 configuration?

  1. Dyn11 eliminates the need for any protective earthing conductors in the entire low-voltage network.
  2. The delta primary winding blocks third-harmonic (triplen) zero-sequence currents from propagating into the medium-voltage grid, and the star secondary with neutral allows 100% neutral loading with superior phase unbalance handling. Correct answer.
  3. Dyn11 operates with zero reactive power losses under all load conditions, resulting in an automatic power factor of cos phi = 1.0.
  4. Dyn11 transformers do not generate any magnetic flux or heat during full-load operation.

Explanation

The Dyn11 vector group (Delta primary, Star secondary with 30-degree lead) provides major advantages for distribution grids: the delta high-voltage winding allows zero-sequence harmonic currents (such as 3rd harmonics from single-phase rectifiers) to circulate internally without entering the HV grid, while the grounded neutral star on the low-voltage side accommodates heavy single-phase unbalanced neutral currents without severe neutral point displacement.

Question 9Transformers, machines and distribution-system reasoning

What is the synchronous speed (synkront omdrejningstal ns) of a 4-pole three-phase asynchronous motor operating on a standard 50 Hz Danish mains supply?

  1. 1500 rpm Correct answer.
  2. 3000 rpm
  3. 1000 rpm
  4. 750 rpm

Explanation

Synchronous speed is calculated from the grid frequency (f) and the number of poles (p) using ns = (120 x f) / p (or (60 x f) / pole pairs). For f = 50 Hz and p = 4 poles, ns = (120 x 50) / 4 = 6000 / 4 = 1500 rpm. The actual full-load rotor speed is slightly lower (e.g. 1430 to 1460 rpm) due to rotor slip.

Question 10Transformers, machines and distribution-system reasoning

In a three-phase motor reversing circuit (højre/venstre-omkobling), how is the direction of rotation reversed?

  1. By interchanging any two of the incoming phase conductors feeding the stator windings (e.g. swapping L1 and L2) Correct answer.
  2. By swapping the neutral and protective earth conductors
  3. By inverting the DC polarity of the overload relay coil
  4. By connecting all three stator phase windings in series with an electrolytic capacitor

Explanation

The rotational direction of a three-phase induction motor depends directly on the phase rotation of the stator magnetic field. Swapping any two active phase lines (such as L1 and L2, or L2 and L3) reverses the spatial rotation of the magnetic field from clockwise to counter-clockwise, causing the rotor to spin in the opposite direction.

Question 11Transformers, machines and distribution-system reasoning

What is the primary function of a thermal overload relay (termisk overstrømsrelæ / bimetalrelæ) in a motor starter?

  1. To protect motor windings against sustained thermal overheating caused by mechanical overload or phase loss by deflecting bimetallic strips and tripping the control circuit Correct answer.
  2. To clear instantaneous high-energy short-circuit faults within 5 milliseconds
  3. To regulate motor speed based on mechanical load variations
  4. To provide harmonic filtering for the upstream power grid

Explanation

A thermal overload relay contains resistive heating elements wrapped around bimetallic strips that bend when carrying overcurrent. Under prolonged moderate overcurrent (such as a jammed conveyor or loss of one phase), the bimetals deflect, tripping an auxiliary contact (95-96 NC) to de-energize the main contactor coil before stator winding insulation degrades. Short-circuit protection must be provided by upstream fuses or magnetic circuit breakers.

Question 12Transformers, machines and distribution-system reasoning

A 400 V three-phase induction motor has a rated mechanical shaft output power of P = 7.5 kW, an efficiency eta = 0.88, and an operational power factor cos phi = 0.82. What is the rated full-load line current (In) drawn from the 400 V supply?

  1. 15.0 A Correct answer.
  2. 10.8 A
  3. 26.0 A
  4. 8.7 A

Explanation

Mechanical output power is related to electrical input current by P = sqrt(3) x U x I x cos phi x eta. Solving for line current: I = P / (sqrt(3) x U x cos phi x eta) = 7500 W / (1.732 x 400 V x 0.82 x 0.88) = 7500 / 500.0 = 15.00 A. This rated full-load current (15 A) determines the selection and calibration of the motor protective device and feed conductors.

Question 13Transformers, machines and distribution-system reasoning

During star-delta (stjerne-trekant) starting of a three-phase induction motor, how do the starting line current (Istart) and starting torque (Tstart) in star connection compare to direct-on-line (DOL) values in delta connection?

  1. Both starting line current and starting torque are reduced to 1/3 (approx. 33%) of their direct-on-line values Correct answer.
  2. Starting current is reduced to 1/3, but starting torque remains 100% of nominal
  3. Starting current is reduced by 58% (1/sqrt(3)), while torque is reduced by 33%
  4. Starting torque is tripled while starting current is halved

Explanation

In star connection, the voltage across each phase winding is reduced by a factor of 1/sqrt(3) (to 230 V instead of 400 V). The winding current is therefore reduced by 1/sqrt(3). Because line current in star equals phase current (whereas line current in delta is sqrt(3) times phase current), the total line current drops by (1/sqrt(3)) x (1/sqrt(3)) = 1/3 (33%). Because induction motor torque is proportional to the square of winding voltage (T proportional to U²), starting torque also drops to (1/sqrt(3))² = 1/3 (33%).

Question 14Transformers, machines and distribution-system reasoning

In a star-delta starter where the thermal overload relay (termisk bimetalrelæ) is wired inside the delta loop in series with the individual phase windings, to what current value must the relay dial be calibrated?

  1. The motor phase winding current: Isetting = In / sqrt(3) = approx. 0.58 x In Correct answer.
  2. The full motor rated line current: Isetting = In
  3. The star starting current: Isetting = 0.33 x In
  4. The DOL locked-rotor current: Isetting = 6.0 x In

Explanation

When the overload relay is located inside the delta loop (e.g., directly under the main contactor or delta contactor), it carries only the current flowing through one motor winding, which in delta is Iphase = Iline / sqrt(3) = In / 1.732 approx. 0.58 x In. Setting the relay to 0.58 x In ensures full thermal protection during continuous delta running. If set to In, the motor would be unprotected against severe overloads.

Question 15Transformers, machines and distribution-system reasoning

What is the function of a dynamic braking resistor (bremsemodstand) connected to a variable frequency drive (frekvensomformer)?

  1. To dissipate kinetic energy regenerated by the motor as heat, preventing the drive's intermediate DC bus voltage from exceeding overvoltage trip thresholds during rapid deceleration Correct answer.
  2. To limit inrush charging current into the DC link smoothing capacitors upon power-up
  3. To reduce radio frequency emissions conducted onto the 400 V supply grid
  4. To provide auxiliary heating inside the control cabinet in cold outdoor environments

Explanation

When an AC motor decelerates a heavy inertial load faster than natural coasting (or during downhill crane lowering), rotor speed exceeds stator synchronous speed, causing the motor to operate as an induction generator. The regenerated energy is fed back through the inverter diodes into the DC bus capacitors, causing DC link voltage to soar. A brake chopper transistor switches the excess energy into an external dynamic braking resistor to be dissipated safely as heat.

Question 16Transformers, machines and distribution-system reasoning

A 4-pole three-phase asynchronous motor operating on a 50 Hz supply rotates at a measured full-load shaft speed of n = 1440 rpm. What is the motor's rotor slip (slip s) under these conditions?

  1. 4.0% (s = 0.04) Correct answer.
  2. 2.0% (s = 0.02)
  3. 6.0% (s = 0.06)
  4. 14.4% (s = 0.144)

Explanation

Synchronous speed for a 4-pole motor at 50 Hz is ns = (120 x 50) / 4 = 1500 rpm. Rotor slip is calculated using s = (ns - n) / ns = (1500 - 1440) / 1500 = 60 / 1500 = 0.04. Expressed as a percentage: 0.04 x 100% = 4.0%. Slip is the fundamental mechanism whereby the rotating magnetic field cuts the rotor bars to induce rotor current and develop torque.

Question 17Fault models, protective coordination and earthing concepts

What criteria must be verified between the prospective maximum short-circuit current (Ik max) at the breaker terminal and the breaking capacity (Pouvoir de Coupure - Pdc) of the circuit breaker?

  1. Pdc (Icu or Icn) ≤ Ik max
  2. Pdc (Icu or Icn) ≥ Ik max Correct answer.
  3. Pdc = 0.5 × Ik max
  4. Pdc must be equal to nominal rating In

Explanation

To safely interrupt an electrical fault without exploding or welding contacts, the circuit breaker's rated ultimate breaking capacity (Icu for industrial EN 60947-2 or Icn for domestic EN 60898) must be greater than or equal to the maximum prospective short-circuit current (Ik max) calculated at that point in the installation (Pdc ≥ Ik max).

Question 18Fault models, protective coordination and earthing concepts

What is the primary characteristic that distinguishes a TN-S system from a TN-C system?

  1. In TN-S, the neutral and protective functions are combined into a single PEN conductor
  2. In TN-S, the neutral (N) and protective conductor (PE) are completely separated throughout the entire installation Correct answer.
  3. In TN-S, residual current devices (DDR) are strictly prohibited
  4. In TN-S, no earthing electrode is required at the source transformer

Explanation

In a TN-S earthing scheme, the letter 'S' stands for 'Separated'. The neutral conductor (N) and the protective earth conductor (PE) are separate physical conductors throughout the entire distribution network. In contrast, TN-C ('Combined') utilizes a single common conductor (PEN) for both protective earthing and neutral current return.

Question 19Fault models, protective coordination and earthing concepts

What is the fundamental functional difference between a Type 1 and a Type 2 surge protective device (SPD / parafoudre)?

  1. Type 1 SPDs protect against continuous overvoltages; Type 2 SPDs protect against short-circuits
  2. Type 1 SPDs are tested with a 10/350 µs current waveform to withstand direct lightning strikes; Type 2 SPDs are tested with an 8/20 µs waveform to absorb indirect induced switching and lightning surges Correct answer.
  3. Type 1 SPDs are installed exclusively on terminal computers; Type 2 SPDs are installed at high-voltage transmission towers
  4. Type 1 SPDs contain only electronic diodes; Type 2 SPDs contain only quartz sand fuses

Explanation

Type 1 SPDs are heavy-duty lightning arresters characterized by an impulse discharge current (Iimp) tested with a 10/350 µs current wave, representing the high energy content of a direct lightning strike; they are mandatory at the service entrance of buildings equipped with external lightning rods (paratonnerres). Type 2 SPDs are tested with an 8/20 µs wave (In and Imax) to protect against indirect lightning surges and grid switching transients.

Question 20Fault models, protective coordination and earthing concepts

On an agricultural site or construction site classified as damp (UL = 25 V), the main service distribution is protected by a 500 mA RCD. What is the maximum allowable resistance of the earth electrode (RA)?

  1. 167 Ω
  2. 100 Ω
  3. 25 Ω
  4. 50 Ω Correct answer.

Explanation

Applying the formula RA ≤ UL / IΔn under the damp condition where UL = 25 V: RA ≤ 25 V / 0.5 A = 50 Ω. In agricultural buildings (NF C 15-100 Part 7-705) where animals are particularly vulnerable to step and touch voltages, earth resistance must not exceed 50 Ω unless lower sensitivity RCDs (e.g. 300 mA or 30 mA) are deployed.

Question 21Fault models, protective coordination and earthing concepts

During an infrared thermographic survey (thermographie infrarouge) of a main LV switchboard under load, a severe thermal anomaly is detected on one terminal connection. What is the physical mechanism causing this hot spot?

  1. The breaker is about to trip magnetically due to reverse phase sequence
  2. The circuit breaker is experiencing high-frequency skin effect
  3. The copper conductor has absorbed excessive moisture from the ambient air
  4. A loose or improperly torqued terminal screw increases electrical contact resistance (Rc), causing concentrated Joule heating (P = Rc · I²) Correct answer.

Explanation

When an electrical terminal screw is improperly torqued, mechanical clamping pressure drops, reducing microscopic contact surface area and elevating contact resistance (Rc). In accordance with Joule's law (P = Rc × I²), the high load current traversing this localized resistance generates intense thermal dissipation, leading to insulation scorching, terminal oxidation runaway, and potential electrical fire.

Question 22Fault models, protective coordination and earthing concepts

How does ambient temperature affect the open-circuit voltage (Uoc) of a silicon photovoltaic solar module?

  1. Uoc increases proportionally as cell temperature rises
  2. Uoc decreases as cell temperature rises (negative temperature coefficient, typically -0.3%/°C to -0.35%/°C) Correct answer.
  3. Uoc remains completely invariant regardless of temperature changes
  4. Uoc drops to zero whenever temperature rises above 25°C

Explanation

Semiconductor physics dictates that the bandgap of crystalline silicon narrows as temperature rises, which directly reduces the module's open-circuit voltage (Uoc). Consequently, PV modules exhibit a negative temperature voltage coefficient (typically -0.28% to -0.35% per °C). Crucially for safety, the highest open-circuit voltage (Uoc max) occurs on cold sunny winter mornings, which must be calculated at -10°C or -15°C to avoid exceeding the inverter's maximum input rating.

Question 23Fault models, protective coordination and earthing concepts

When working in an industrial stationary battery room (accumulateurs stationnaires pour onduleur / TGBT), what dual physical hazards are specifically highlighted by NF C 18-510?

  1. High-frequency radio emissions and sonic vibration
  2. Severe direct-current short-circuit arc flash (which cannot be interrupted by normal zero-crossing) and explosive hydrogen gas release during charging Correct answer.
  3. Nuclear alpha radiation and magnetic levitation of tools
  4. Biological mold contamination and legionella

Explanation

Industrial battery banks store massive chemical energy. During charging, lead-acid and Ni-Cd cells release explosive hydrogen gas (H2), requiring ATEX-compliant ventilation. Electrically, batteries maintain constant DC potential that cannot be switched off at the cell terminals; a dropped wrench or short-circuit produces an instantaneous multi-thousand ampere DC arc flash that vaporizes metal, requiring insulated tools (NF EN 60900) and full face shield protection.

Question 24Fault models, protective coordination and earthing concepts

At a main low-voltage switchboard (TGBT) located immediately downstream of a 1600 kVA transformer, prospective short-circuit current reaches 38 kA. In conducting an electrical risk assessment under NF C 18-510, what danger is critical even though the voltage is only 400 V?

  1. Electromagnetic gravity distortion that pulls metal tools toward the busbars
  2. Catastrophic arc flash incident energy (thermal energy in cal/cm² and acoustic blast wave) capable of causing fatal 3rd-degree burns and blast trauma several meters away Correct answer.
  3. Direct conversion of air nitrogen into deadly liquid ammonia
  4. Instantaneous cryogenic freezing of copper conductors

Explanation

Arc flash thermal energy is proportional to the prospective fault current and fault clearing time (E ∝ Isc × t), not voltage alone. At the main TGBT close to a 1600 kVA transformer, a 38 kA arc flash releases plasma temperatures exceeding 10,000°C in milliseconds. The massive radiant heat (incident energy exceeding 40 cal/cm²), concussive pressure wave, and vaporized copper shrapnel can be lethal without arc-rated flash suits and face shields complying with NF EN 166 and IEEE 1584.

Question 25Installation planning, measurement and evidence of compliance

What fundamental relationship must be satisfied between the circuit operating current (Ib), the protective device nominal rating (In), and the permissible cable current (Iz)?

  1. Ib ≤ Iz ≤ In
  2. In ≤ Ib ≤ Iz
  3. Iz ≤ In ≤ Ib
  4. Ib ≤ In ≤ Iz Correct answer.

Explanation

NF C 15-100 clause 433.1 defines the fundamental overcurrent coordination rule: Ib ≤ In ≤ Iz. The nominal rating of the protective device (In) must be greater than or equal to the calculated design current of the load (Ib) to prevent nuisance tripping, while not exceeding the continuous current-carrying capacity of the cable (Iz) to prevent thermal degradation of conductor insulation.

Question 26Installation planning, measurement and evidence of compliance

A 3-phase cable (5x16 mm² Cu, XLPE insulation, 90°C) is routed in an ambient temperature of 40°C on an unperforated cable tray grouped touching alongside 3 other multi-core circuits (total 4 circuits, bundling factor f2 = 0.70). Base current-carrying capacity for Verlegeart C at 30°C is I0 = 98 A. The temperature derating factor for XLPE at 40°C is f1 = 0.91. What is the permissible continuous current-carrying capacity (Iz) of the cable?

  1. 98.0 A
  2. 89.2 A
  3. 68.6 A
  4. 62.4 A Correct answer.

Explanation

Pursuant to NIN 5.2.3, the continuous permissible current-carrying capacity is calculated by applying correction factors: Iz = I0 * f1 * f2 = 98 A * 0.91 * 0.70 = 62.42 A ≈ 62.4 A. Therefore, the upstream overcurrent protective device rating In must satisfy In ≤ Iz (e.g. In ≤ 50 A or 63 A depending on load characteristics).

Question 27Installation planning, measurement and evidence of compliance

What is the technical and safety purpose of measuring Line Impedance (ZL / L-N and L-L) in addition to Fault Loop Impedance (Zs / L-PE) during electrical installation testing?

  1. To check whether the building Wi-Fi network has sufficient bandwidth.
  2. To determine the prospective line short-circuit currents (Ik(L-N) and Ik(L-L)) to verify that overcurrent protective devices can clear phase-to-neutral shorts, and to calculate actual operating voltage drop (ΔU) under full load. Correct answer.
  3. To adjust the water pressure inside heat pump circulation pipes.
  4. Line impedance testing is prohibited under Swiss NIN standards.

Explanation

Measuring line impedance ZL (between phase and neutral, or between two phases) verifies the prospective short-circuit current for line-to-neutral or phase-to-phase faults. This confirms that circuit breakers trip magnetically for L-N faults and allows calculation of actual circuit voltage drop (ΔU = Ib * ZL) pursuant to NIN 5.2.5 and EN 61557-3.

Question 28Installation planning, measurement and evidence of compliance

In what specific grounding scenario is the dual-clamp earth resistance measurement method (Erdungszangenmessung / schlagfreie Zangenmethode) technically valid without disconnecting the main earthing strap?

  1. On an isolated single earthing spike of a remote mountain telecommunications hut with zero interconnected grid connections.
  2. In a meshed, multi-grounded system (such as an urban interconnected TN network or lightning protection system with multiple parallel down-conductors connected to a common ring earth), where the parallel return path resistance is negligibly low compared to the individual electrode under test. Correct answer.
  3. On an isolated plastic rainwater downpipe.
  4. Only when the entire electrical installation is submerged in distilled water.

Explanation

The clamp-on earth testing method (Erdungszange) uses an injection coil to induce a high-frequency voltage into the loop and a detection coil to measure the resulting current. It strictly requires a closed loop through multiple parallel grounding paths (R_parallel ≈ 0). The instrument measures Rx + R_parallel ≈ Rx. If applied to an isolated single ground electrode without parallel return paths, it cannot measure earth resistance.

Question 29Installation planning, measurement and evidence of compliance

A single-core PVC-insulated cable is routed through an industrial boiler room where the ambient air temperature is measured at 45 °C. If the temperature correction factor k1 is 0.79 and no grouping applies, what is the effective current-carrying capacity (Iz) of a conductor with a tabulated rating (I0) of 34 A?

  1. 26.86 A Correct answer.
  2. 34.00 A
  3. 43.04 A
  4. 22.10 A

Explanation

The effective current-carrying capacity is Iz = I0 x k1 x k2. With no grouping (k2 = 1.0), Iz = 34 A x 0.79 = 26.86 A. An overcurrent protective device selected for this circuit must not exceed In = 25 A to guarantee that the conductor does not exceed 70 °C under operating conditions.

Question 30Installation planning, measurement and evidence of compliance

When installing cables directly embedded in thermal insulation inside an external cavity wall (Reference Method A1), what thermal effect dictates the severe derating of current capacity?

  1. Thermal insulation drastically reduces heat dissipation away from the conductor, causing conductor core temperatures to escalate rapidly Correct answer.
  2. Dielectric breakdown of the air inside mineral wool generates circulating capacitive eddy currents
  3. Thermal insulation causes inductive coupling between the phase and neutral conductors
  4. The chemical blowing agents in insulation degrade copper conductivity over time

Explanation

Building insulation materials (such as mineral wool, fiberglass, or EPS foam) have very high thermal resistivity (low thermal conductivity). When cables run within insulation, heat generated by ohmic I²R losses cannot escape into the surrounding building structure, causing the core temperature to rise dramatically unless current capacity is severely restricted under Method A1/A2.

Question 31Installation planning, measurement and evidence of compliance

An electrician is designing a sub-distribution feed with an anticipated design current Ib = 42 A. The installation uses three loaded conductors in conduit inside an insulated wall (Method A2). Tabulated ratings for 70 °C PVC are: 6 mm² = 34 A; 10 mm² = 46 A; 16 mm² = 61 A. Which nominal MCB rating (In) and standard cable cross-section must be selected assuming no correction factors?

  1. In = 50 A MCB and 16 mm² cable Correct answer.
  2. In = 50 A MCB and 10 mm² cable
  3. In = 40 A MCB and 6 mm² cable
  4. In = 63 A MCB and 10 mm² cable

Explanation

The basic rule is Ib <= In <= Iz. Here Ib = 42 A, so the protective device must be rated at least 50 A (since 40 A is less than Ib = 42 A, which would cause nuisance tripping). Next, Iz must be greater than or equal to In = 50 A. For 10 mm², Iz is 46 A, which is less than In = 50 A (violating In <= Iz). Therefore, the next standard size of 16 mm² (Iz = 61 A) must be selected.

Question 32Installation planning, measurement and evidence of compliance

An industrial circuit has design current Ib = 28 A and is protected by a 32 A device. Four three-phase circuits share closed trunking (k2 = 0.65), and the 40 °C temperature factor for the selected PVC cable is k1 = 0.87. What minimum tabulated current-carrying capacity I0 coordinates the cable with the 32 A device?

  1. 56.58 A Correct answer.
  2. 32.18 A
  3. 15.83 A
  4. 62.40 A

Explanation

The selected 32 A protective device must not exceed the corrected cable capacity: In <= Iz = I0 x k1 x k2. Therefore I0 >= 32 / (0.87 x 0.65) = 56.58 A. Using 28 A alone would check only the design load and would not coordinate the cable with the stated 32 A device.

Question 33Installation planning, measurement and evidence of compliance

A luminaire is marked with the Class II double-square symbol. How is protection against electric shock provided, and what should the installer do with the circuit protective conductor?

  1. Double or reinforced insulation provides protection; do not connect PE to the Class II enclosure, but terminate any circuit PE safely for continuity where required Correct answer.
  2. Connect PE to any accessible screw because Class II requires two earth paths
  3. Rely only on the upstream fuse because Class II means overcurrent protection
  4. Supply it only from 12 V SELV because Class II and SELV are identical

Explanation

Class II equipment uses double or reinforced insulation and has no protective-earth connection to its enclosure. Do not improvise a bond to the product. If a protective conductor is present in the fixed wiring, maintain and safely park its continuity in a suitable terminal as the installation design requires; Class II does not mean SELV and does not remove circuit protective-device requirements.

Question 34Installation planning, measurement and evidence of compliance

An electrician tests continuity on a 2.5 mm² copper protective conductor over a route length of 25 m. Given copper resistivity rho = 0.0175 ohm x mm²/m at 20 °C, what is the theoretical expected resistance of the conductor?

  1. 0.175 ohms Correct answer.
  2. 0.350 ohms
  3. 1.750 ohms
  4. 0.088 ohms

Explanation

The resistance of a single conductor is calculated using R = (rho x L) / S = (0.0175 ohm x mm²/m x 25 m) / 2.5 mm² = 0.4375 / 2.5 = 0.175 ohms. When performing the test with test leads nulled (zeroed), the measured resistance should closely match 0.175 ohms. A substantially higher reading would indicate a loose terminal screw or corroded splice.

Question 35Installation planning, measurement and evidence of compliance

What does pressing the integral test button (T-knap) on a residual current device (RCD) actually verify during an inspection?

  1. It verifies that the internal mechanical latch, spring mechanism, and sensor coil operate properly by creating an artificial internal leakage current across the toroid Correct answer.
  2. It measures the earth electrode resistance of the building
  3. It measures the exact disconnection time in milliseconds
  4. It checks the continuity of the downstream protective conductors in socket outlets

Explanation

The test button connects an internal resistor between a phase on the load side and the neutral on the supply side (bypassing the toroidal transformer core), producing an artificial unbalance that triggers the mechanical tripping mechanism. While essential for regular user testing to prevent mechanical sticking, it does not measure tripping time, tripping current, or the integrity of the external PE earthing network.

Question 36Measurement, feedback, PLC and automation

In modern commercial building automation, KNX IP Secure and BACnet/SC (Secure Connect) protocols are implemented. What is the primary network engineering objective when connecting operational technology (OT / Gebäudeautomation) to an enterprise IT backbone?

  1. Merging all heating, ventilation, CCTV, access control, and public guest Wi-Fi networks into a single unencrypted flat broadcast domain.
  2. Eliminating all firewalls to speed up KNX telegram delivery.
  3. Establishing strict physical and logical network segmentation (VLANs, DMZs, firewall access control lists) and end-to-end cryptographic authentication/encryption to prevent unauthorized cyber access or malware propagation from the office network into critical building control systems. Correct answer.
  4. Assigning static public IPv4 addresses with default manufacturer passwords to every individual room sensor.

Explanation

Building Automation Systems (BAS / OT) control critical infrastructure (HVAC, fire dampers, access control, power distribution). Connecting OT to IT requires strict segmentation (VLANs, firewalls, zero-trust architecture), disabling unnecessary ports, and deploying secure protocols (KNX IP Secure with AES-128 encryption, BACnet/SC with TLS 1.3) to safeguard against unauthorized access and lateral malware attacks.

Question 37Measurement, feedback, PLC and automation

In European industrial PLC automation, what is the fundamental electrical difference between a PNP (sourcing) inductive proximity sensor and an NPN (sinking) sensor connected to a digital input module?

  1. A PNP sensor switches the positive (+24 V DC) supply to the PLC input terminal, whereas an NPN sensor switches the negative (0 V DC) ground to the PLC input terminal Correct answer.
  2. A PNP sensor operates on AC voltage, while an NPN sensor operates on DC voltage
  3. A PNP sensor is normally closed, while an NPN sensor is always normally open
  4. A PNP sensor generates an analog 4-20 mA current loop, while NPN generates a pulsed output

Explanation

PNP sensors have an internal PNP transistor with the emitter tied to +24 V DC and the collector connected to the output (sourcing output). When triggered, the sensor switches +24 V DC into the PLC input, which is the standard configuration in European industrial machinery. NPN sensors have an open-collector transistor tied to 0 V DC (sinking output), switching the input line to ground (standard in Japanese/Asian machinery).

Question 38Measurement, feedback, PLC and automation

In KNX system engineering, what is the operational difference between an Individual Address (fysisk adresse) and a Group Address (gruppeadresse)?

  1. An Individual Address (e.g. 1.1.20) uniquely identifies the physical location of a hardware device during commissioning, whereas a Group Address (e.g. 2/1/15) links communication objects to execute collective control functions like switching or dimming Correct answer.
  2. An Individual Address controls high voltage, while a Group Address controls extra-low voltage
  3. Individual Addresses are assigned to actuators, while Group Addresses are assigned only to power supplies
  4. Individual Addresses change dynamically during normal operation, while Group Addresses are permanent hardware MAC addresses

Explanation

In KNX, every device has a unique Individual Address formatted as Area.Line.Device (e.g. 1.2.10) programmed via the physical programming button in ETS software. It serves to identify and configure that specific piece of hardware. Group Addresses (formatted as Main/Middle/Sub, e.g. 1/1/5) represent functional actions (such as 'Living Room Ceiling Lights ON/OFF') and are linked to communication objects of pushbuttons, motion sensors, and relay actuators.

Question 39Measurement, 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 40Measurement, 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 41Measurement, 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 42Measurement, 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 43Lighting models, metering and building-control integration

Why can a DALI lighting-control pair be installed in the same multicore cable as mains conductors when the selected cable and equipment are approved for that arrangement?

  1. The DALI conductors and connected equipment use insulation suitable for the highest voltage present, so the separation rule can be met within one approved cable Correct answer.
  2. DALI carries high-voltage direct current that repels AC electric fields
  3. The DALI protocol uses fiber-optic light pulses inside the copper wire
  4. DALI ballast electronics are magnetically shielded by lead insulation

Explanation

DALI is not treated as SELV, and a shared multicore arrangement is possible when every conductor and connected interface has insulation suitable for the highest voltage present and the equipment manufacturer's instructions permit it. The two-wire DALI bus is polarity-insensitive, so DA/DA is more accurate than implying fixed DA+ and DA− conductors.

Question 44Lighting models, metering and building-control integration

A commercial office floor has dimensions of 20 meters by 15 meters. If the design lighting power density is specified as 10 W/m², what is the total lighting load for this floor?

  1. 300 W
  2. 1.5 kW
  3. 3.0 kW Correct answer.
  4. 30 kW

Explanation

The total floor area is calculated as: Area = 20 m * 15 m = 300 m². The total lighting load is: Load = 300 m² * 10 W/m² = 3,000 W, which is equivalent to 3.0 kW.

Question 45Lighting models, metering and building-control integration

Illumination level for a workplace task is commonly specified in which unit?

  1. Lux (lm/m²) Correct answer.
  2. Henry
  3. Weber only
  4. Siemens

Explanation

Illuminance is measured in lux (lumens per square metre). Lighting design selects lamp type, luminaire distribution, mounting height and maintenance factors to achieve required lux levels for the task.

Question 46Lighting models, metering and building-control integration

Stroboscopic effect with discharge lighting near rotating machinery is hazardous because:

  1. It always increases lux levels safely
  2. Moving parts may appear stationary or slow, misleading operators Correct answer.
  3. It eliminates all glare
  4. It converts a.c. motors to d.c. motors

Explanation

Discharge lamps on a.c. can flicker at supply frequency; under this light, rotating machinery may appear stopped or slow (stroboscopic effect), causing serious safety risk. Multi-phase supply to luminaires or high-frequency electronic ballasts mitigate the effect.

Question 47Lighting models, metering and building-control integration

Electronic ballasts / LED drivers for lighting can introduce which power-quality concern?

  1. Only improved earth electrode resistance
  2. Harmonic currents and non-linear load effects on the supply Correct answer.
  3. Automatic removal of all bonding
  4. Conversion of the building earthing to unearthed IT only

Explanation

Non-linear lighting electronic loads draw distorted current, injecting harmonics that overload neutrals, heat transformers and disturb sensitive equipment. The Grade C syllabus explicitly covers non-linear load effects and improvement methods (filtering, oversizing neutrals, active filters, etc.).

Question 48Lighting models, metering and building-control integration

One method of mitigating harmonic problems from non-linear loads is to:

  1. Ignore neutral conductor sizing because harmonics cancel perfectly always
  2. Use appropriately sized neutrals, filters or other engineered mitigation where needed Correct answer.
  3. Remove all earthing connections
  4. Replace all RCDs with larger fuses only

Explanation

Triplen harmonics can add in the neutral. Mitigation includes oversized neutrals, K-rated transformers, passive/active filters, phase diversity and limiting non-linear load penetration. Removing earthing or swapping RCDs for fuses is unsafe and ineffective.

Question 49Engineering models, drawings, evidence and project decisions

An electrical contracting enterprise calculates its annual productive billable hours per full-time installer. The gross contractual annual working time is 2,080 hours (52 weeks x 40 hours). Deductions include: 5 weeks paid vacation (200 h), 9 public holidays (72 h), average sick/accident leave (64 h), and continuing education/internal meetings (94 h). What is the realistic annual productive billable capacity per installer?

  1. 2,080 hours
  2. 1,650 hours Correct answer.
  3. 1,200 hours
  4. 1,950 hours

Explanation

Calculating productive billable hours: Gross hours (2,080 h) minus Vacation (200 h) minus Holidays (72 h) minus Sick leave (64 h) minus Non-productive training/meetings (94 h) = 2,080 - 430 = 1,650 productive hours. In Swiss electrical calculation practice (EIT.swiss), an average of 1,600 to 1,680 billable hours per full-time employee is used to determine the hourly cost rate.

Question 50Engineering models, drawings, evidence and project decisions

When authoring a formal judicial expert opinion (Gerichtsgutachten) for a Swiss civil or criminal court regarding an electrical accident, what fundamental legal and ethical standards must a Dipl. Elektroinstallations- und Sicherheitsexperte uphold?

  1. The expert must always manipulate facts to ensure the contractor who hired them wins the lawsuit.
  2. The expert can invent technical norms that do not exist in official legislation.
  3. The expert must deliver a personal verdict declaring who should go to prison, superseding the judge.
  4. The expert must maintain strict impartiality and personal independence (Unparteilichkeit und Unabhängigkeit), evaluate all facts based on the specific technical standards (NIN / ESTI / EleG) in force at the time the installation was constructed, clearly separate verified factual findings from technical deductions, and answer only the specific technical questions formulated by the court without deciding legal matters reserved for the judge. Correct answer.

Explanation

Under the Swiss Civil Procedure Code (ZPO Art. 183 ff.) and Criminal Procedure Code (StPO Art. 182 ff.), a court-appointed expert witness (Gerichtsexperte) is an impartial assistant to justice. The expert must remain strictly objective, declare any conflicts of interest, evaluate technical compliance against the specific standards applicable at the construction date (keine Rückwirkung neuerer Normen), provide clear evidence-based reasoning, and avoid usurping the judicial evaluation of law.