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CAP Électricien - independent technical foundations study materials
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CAP Électricien - independent technical foundations practice questions

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Question 1Electrical circuits, machines and digital logic

An electric convector heater with a pure resistance of 23 Ω is connected to a 230 V single-phase AC circuit. What is the current flowing through the heater?

  1. 10 A Correct answer.
  2. 5 A
  3. 12 A
  4. 23 A

Explanation

According to Ohm's law, I = U / R. For a resistive load connected to 230 V with R = 23 Ω, the current is I = 230 V / 23 Ω = 10 A.

Question 2Electrical circuits, machines and digital logic

Which physical law states that the thermal power dissipated in a conductor is directly proportional to its resistance and the square of the electric current (P = R × I²)?

  1. Joule's law (Effet Joule) Correct answer.
  2. Lenz's law
  3. Faraday's law
  4. Coulomb's law

Explanation

Joule's law (l'effet Joule) defines the thermal energy released by an electric current passing through a conductor with resistance R, expressed as P = R × I² in watts (or Q = R × I² × t in joules).

Question 3Electrical circuits, machines and digital logic

A steady current of 16 A flows through a circuit conductor with a total resistance of 0.5 Ω. What is the thermal power dissipated along the cable by the Joule effect?

  1. 64 W
  2. 128 W Correct answer.
  3. 256 W
  4. 8 W

Explanation

Using Joule's law, P = R × I². Here, P = 0.5 Ω × (16 A)² = 0.5 × 256 = 128 W of dissipated thermal power.

Question 4Electrical circuits, machines and digital logic

The public electrical distribution network in France delivers alternating current at a standard nominal frequency of 50 Hz. What is the period (T) of this sinusoidal voltage wave?

  1. 10 ms
  2. 20 ms Correct answer.
  3. 50 ms
  4. 5 ms

Explanation

The period T is the inverse of the frequency: T = 1 / f. For f = 50 Hz, T = 1 / 50 s = 0.020 s = 20 ms.

Question 5Electrical circuits, machines and digital logic

The electrical resistance of a conductor is given by Pouillet's formula R = ρ × L / S. If the length (L) of a wire is doubled and its cross-sectional area (S) is halved, how does its resistance change?

  1. It remains unchanged
  2. It doubles (increases by 2 times)
  3. It quadruples (increases by 4 times) Correct answer.
  4. It is reduced by half

Explanation

Resistance is directly proportional to length and inversely proportional to cross-section: R' = ρ × (2L) / (S / 2) = 4 × (ρ × L / S) = 4R. Therefore, the resistance quadruples.

Question 6Electrical circuits, machines and digital logic

A single copper conductor has a length of 50 m and a cross-section of 1.5 mm². Taking copper resistivity at ambient temperature as ρ = 0.018 Ω·mm²/m, what is the electrical resistance of this conductor?

  1. 0.60 Ω Correct answer.
  2. 1.35 Ω
  3. 0.18 Ω
  4. 0.06 Ω

Explanation

Using R = ρ × L / S: R = (0.018 Ω·mm²/m × 50 m) / 1.5 mm² = 0.9 / 1.5 = 0.60 Ω.

Question 7Electrical circuits, machines and digital logic

A single-phase 230 V electric motor draws an RMS current of 8 A with a power factor (cos φ) of 0.85. What is the active power (P) consumed by the motor?

  1. 1840 W
  2. 1564 W Correct answer.
  3. 2165 W
  4. 1320 W

Explanation

Active power in single-phase AC is calculated as P = U × I × cos φ. Here, P = 230 V × 8 A × 0.85 = 1564 W.

Question 8Electrical circuits, machines and digital logic

What is the unit of apparent power (S = U × I) in alternating current circuits?

  1. Watt (W)
  2. Volt-ampere reactive (var)
  3. Volt-ampere (VA) Correct answer.
  4. Joule (J)

Explanation

Apparent power (puissance apparente S) is measured in volt-amperes (VA). Active power (P) is measured in watts (W), and reactive power (Q) is measured in volt-amperes reactive (var).

Question 9Electrical circuits, machines and digital logic

An inductive AC load consumes an active power P = 1200 W and has an apparent power S = 1500 VA. What is the reactive power (Q) absorbed by this load?

  1. 300 var
  2. 900 var Correct answer.
  3. 2700 var
  4. 1000 var

Explanation

According to the power triangle relation S² = P² + Q², the reactive power is Q = √(S² - P²). Here, Q = √(1500² - 1200²) = √(2,250,000 - 1,440,000) = √810,000 = 900 var.

Question 10Electrical circuits, machines and digital logic

Two heating elements with resistances R1 = 60 Ω and R2 = 40 Ω are connected in parallel across a 230 V supply. What is the equivalent resistance (Req) of this circuit?

  1. 100 Ω
  2. 24 Ω Correct answer.
  3. 50 Ω
  4. 20 Ω

Explanation

For two resistors in parallel, Req = (R1 × R2) / (R1 + R2). Here, Req = (60 × 40) / (60 + 40) = 2400 / 100 = 24 Ω. In parallel circuits, Req is always lower than the smallest branch resistance.

Question 11Electrical circuits, machines and digital logic

In a series circuit consisting of three different resistors connected across a DC voltage source, which electrical quantity remains identical through all three components?

  1. The electric current (I) Correct answer.
  2. The voltage drop (U)
  3. The dissipated power (P)
  4. The electrical resistance (R)

Explanation

In any series circuit without branch junctions, the electric current is identical at every point in the loop. The total voltage divides across each resistor proportionally to its resistance.

Question 12Electrical circuits, machines and digital logic

Two resistors R1 = 30 Ω and R2 = 70 Ω are connected in series across a 230 V AC source. What is the voltage drop across resistor R1?

  1. 69 V Correct answer.
  2. 161 V
  3. 115 V
  4. 76.7 V

Explanation

Using the voltage divider formula: U1 = U × [R1 / (R1 + R2)] = 230 V × [30 / (30 + 70)] = 230 × (30 / 100) = 69 V.

Question 13Electrical circuits, machines and digital logic

In a residential single-phase installation with an effective (RMS) voltage of 230 V, what is the approximate peak maximum voltage (Umax)?

  1. 230 V
  2. 325 V Correct answer.
  3. 400 V
  4. 460 V

Explanation

For a sinusoidal alternating waveform, peak voltage is Umax = Ueff × √2. With Ueff = 230 V, Umax = 230 × 1.414 ≈ 325 V.

Question 14Electrical circuits, machines and digital logic

How must an ammeter be connected to measure the electric current flowing through a circuit load?

  1. In parallel across the terminals of the load
  2. In series with the load by opening the circuit Correct answer.
  3. Between the phase conductor and the protective earth conductor
  4. Between the neutral conductor and the protective earth conductor

Explanation

An ammeter has near-zero internal resistance and must always be inserted in series into the circuit loop so that all current passes through it. Connecting an ammeter in parallel creates a dangerous dead short circuit.

Question 15Electrical circuits, machines and digital logic

In a balanced three-phase 400 V star-connected network, what is the mathematical relationship between the line-to-line voltage (U = 400 V) and line-to-neutral voltage (V = 230 V)?

  1. U = V × 2
  2. U = V × √3 Correct answer.
  3. U = V / √3
  4. U = V × √2

Explanation

In a three-phase system with star (étoile) neutral, the compound line-to-line voltage U is related to the simple phase-to-neutral voltage V by U = V × √3. With V = 230 V, U = 230 × 1.732 ≈ 400 V.

Question 16Electrical circuits, machines and digital logic

A balanced three-phase heating unit is connected to a 400 V supply (line-to-line) and draws a line current of 10 A with cos φ = 1.0. What is the total active power dissipated by the heater?

  1. 4000 W
  2. 6928 W Correct answer.
  3. 12000 W
  4. 2300 W

Explanation

Total active power in a three-phase balanced system is calculated as P = √3 × U × I × cos φ. Here, P = 1.732 × 400 V × 10 A × 1.0 = 6928 W (approximately 6.93 kW).

Question 17Electrical circuits, machines and digital logic

How is the power factor (cos φ) of an alternating current installation defined mathematically?

  1. The ratio of active power to apparent power (P / S) Correct answer.
  2. The ratio of apparent power to active power (S / P)
  3. The ratio of reactive power to active power (Q / P)
  4. The product of active power and reactive power (P × Q)

Explanation

Power factor is defined as the ratio of active power (P in watts) to apparent power (S in volt-amperes): cos φ = P / S. It ranges from 0 to 1 and indicates how effectively electrical power is converted into useful work.

Question 18Electrical circuits, machines and digital logic

An electric radiant panel with a nominal power rating of 2000 W operates at full capacity for 4.5 hours. How much electrical energy does it consume?

  1. 4.5 kWh
  2. 9.0 kWh Correct answer.
  3. 900 kWh
  4. 90 kWh

Explanation

Electrical energy is the product of power and time: E = P × t. Converting 2000 W to 2 kW: E = 2 kW × 4.5 h = 9.0 kWh.

Question 19Electrical circuits, machines and digital logic

A 230 V single-phase circuit supplies a resistive load drawing 10 A. The total loop resistance of the feed conductors (phase + neutral) is 0.46 Ω. What is the percentage voltage drop (ΔU%) at the load terminals?

  1. 1.0%
  2. 2.0% Correct answer.
  3. 4.6%
  4. 0.5%

Explanation

The absolute voltage drop is ΔU = R × I = 0.46 Ω × 10 A = 4.6 V. The percentage drop relative to 230 V is (4.6 V / 230 V) × 100% = 2.0%, which easily complies with the NF C 15-100 limit (3% for lighting, 5% for other uses).

Question 20Transformers, 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 21Transformers, 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 22Transformers, 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 23Transformers, 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 24Transformers, 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 25Transformers, 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 26Transformers, 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 27Transformers, 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 28Transformers, 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 29Fault models, protective coordination and earthing concepts

Calculate the maximum prospective initial symmetrical three-phase short-circuit current (Ik'') at the low-voltage busbars directly fed by a 1000 kVA, 400 V transformer with a relative short-circuit impedance of uk = 6.0% (assuming an ideal rigid upstream medium-voltage grid, c_max = 1.05).

  1. Approximately 3.5 kA
  2. Approximately 12.1 kA
  3. Approximately 25.3 kA Correct answer.
  4. Approximately 65.0 kA

Explanation

Calculations: 1) Rated secondary current I_r = S_r / (sqrt(3) * U_n) = 1000 kVA / (sqrt(3) * 400 V) = 1443.4 A. 2) Transformer internal short-circuit current I_ktrafo = I_r / u_k = 1443.4 A / 0.06 = 24.06 kA. 3) Factoring in the standard voltage coefficient c_max = 1.05 (IEC 60909): Ik'' = 1.05 * 24.06 kA = 25.26 kA ≈ 25.3 kA. Switchgear busbars and main circuit breakers must be rated for at least I_cu ≥ 26 kA / 36 kA.

Question 30Fault models, protective coordination and earthing concepts

An electrical engineer designs a selective protection scheme. A feeder circuit is protected by an upstream NH00 160 A gG fuse. Downstream in the sub-distribution panel, branch circuits are protected by MCBs (Type C 16 A) and an MCCB (100 A). How is total selectivity (Vollselektivität) verified between the protective devices across the entire short-circuit current range?

  1. By ensuring that both upstream and downstream devices have identical nominal current ratings.
  2. By verifying that the downstream circuit breaker trips faster than 10 seconds under standard overload conditions.
  3. By placing all circuit breakers on the same DIN rail without separating partitions.
  4. By comparing time-current tripping curves for overloads, and verifying for short circuits that the total let-through energy of the downstream device (I²t_operating) is strictly less than the pre-arcing melting energy (I²t_melt) of the upstream fuse. Correct answer.

Explanation

Selectivity between series protective devices requires: 1) Non-overlapping time-current characteristics in the thermal overload region; 2) For short-circuits (instantaneous magnetic / arcing region), energy selectivity requires the downstream device's total clearing let-through energy (I²t_total / I²t_operating) to remain lower than the melting let-through energy (I²t_pre-arcing / I²t_Schmelz) of the upstream fuse (IEC 60947-2 / NIN 5.3.6).

Question 31Fault models, protective coordination and earthing concepts

For a stipulated PV model, each module has open-circuit voltage41.5V at25C and linear voltage coefficient minus0.28% per degree C. A22-module series string is evaluated at minus10C, a supplied exercise temperature. What is its approximate voltage compared with a1000V inverter input ceiling?

  1. Uoc(max) = 1002.5 V; exceeds 1000 V DC and violates design limits (requires reducing string length to 21 modules). Correct answer.
  2. Uoc(max) = 913.0 V; completely safe with standard margin.
  3. Uoc(max) = 750.0 V; because voltage drops as temperature decreases.
  4. Uoc(max) = 1500.0 V; because solar panels double their output in winter.

Explanation

The temperature change is minus35C. The linear model predicts41.5*(1+0.0028*35)=45.567V per module and22*45.567=1002.474V. This exceeds the stated1000V ceiling. A21-module version predicts956.907V under these assumptions, but actual design also needs the real module/inverter specifications and all applicable margins.

Question 32Fault models, protective coordination and earthing concepts

In a mission-critical financial data center, Residual Current Monitors (RCM / Differenzstrom-Überwachungsgeräte according to EN 62020) are installed on all main server power distribution units instead of standard tripping RCDs. Why is RCM monitoring preferred in critical continuous facilities?

  1. Because RCM devices convert all residual currents into bitcoins.
  2. Because RCM devices continuously measure differential leakage currents in real time and trigger early warning alarms to the building management system without tripping or shutting down power to critical servers, allowing maintenance teams to locate insulation degradation before a catastrophic blackout occurs. Correct answer.
  3. Because RCM devices eliminate all electrical grounding requirements.
  4. Because RCM devices allow servers to run without electricity.

Explanation

According to EN 62020 and NIN 5.3.8, Residual Current Monitors (RCMs) continuously measure differential fault currents without automatic disconnection. In critical data centers and industrial continuous processes where unexpected power shutdowns cause immense financial or operational damage, RCMs provide early alerts when insulation deteriorates, allowing planned maintenance before faults trigger breakers.

Question 33Fault models, protective coordination and earthing concepts

Why is an overcurrent protective device (such as a standard MCB or gG fuse) rarely able to provide automatic disconnection of supply (ADS) for earth faults in a TT system?

  1. The high resistance of the local earth electrode (RA) and source earth restricts fault current to values far below the instantaneous magnetic trip threshold of the MCB Correct answer.
  2. TT systems operate without an alternating frequency
  3. Circuit breakers do not conduct fault currents when connected to earth rods
  4. The neutral conductor in TT systems carries twice the line voltage during a fault

Explanation

In a TT system, the earth fault current path passes through the earth mass between the consumer earth electrode (RA) and the transformer earth electrode (RB). Because typical earth resistances are tens or hundreds of ohms, the fault current If = U0 / (RA + RB) is limited to a few amperes (e.g., 230 V / 25 ohms = 9.2 A), which cannot trip a 16 A or 32 A MCB quickly. Hence, an RCD is virtually always required for ADS in TT installations.

Question 34Fault models, protective coordination and earthing concepts

Why may a three-phase variable-frequency drive require a Type B RCD when an RCD is part of the protective design?

  1. Its power electronics can produce smooth DC residual current that a Type A RCD is not designed to detect; the required RCD type must follow the drive manufacturer's data and installation design Correct answer.
  2. Type A RCDs trip excessively fast during normal motor acceleration due to mechanical vibrations
  3. Type B RCDs generate active high-frequency carrier signals to extinguish arcs
  4. Three-phase VFDs reverse line voltage polarity, requiring a bi-directional mechanical latch

Explanation

A rectifier and DC link can produce residual-current waveforms, including smooth DC, outside a Type A RCD's detection capability. A Type B device detects smooth DC as well as the waveforms covered by Type A. This is not a blanket rule for every drive: the circuit design, protective measure, drive characteristics, upstream devices, and manufacturer's instructions determine the required RCD type.

Question 35Installation 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 36Installation 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 37Installation 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 38Installation 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 39Installation 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 40Installation 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 41Installation 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 42Installation 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 43Installation 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 44Installation 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 45Measurement, 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 46Measurement, 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 47Measurement, 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 48Measurement, 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 49Measurement, 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 50Measurement, 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.