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100+ Free Befähigungsprüfung Elektrotechnik Practice Questions
Befähigungsprüfung für das reglementierte Gewerbe Elektrotechnik (Österreich) practice questions are available now; exam metadata is being verified.
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Key Facts: Befähigungsprüfung Elektrotechnik Exam
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The Austrian Befähigungsprüfung Elektrotechnik (NQR Level 6) is a multi-part modular examination administered by the WKO Meisterprüfungsstellen. Passing requires at least Grade 4 (Genügend) on a 1–5 scale across all modules. Since Jan 1, 2024, exam fees are 100% federally funded for the first two attempts. The curriculum spans Electrical Engineering Fundamentals & Machines (20%), Electrical Installations & Safety Engineering per ÖVE/ÖNORM E 8101 (25%), Lightning, Surge & Explosion Protection (15%), Building Technology, Lighting, KNX/DALI, PV & EV Charging (20%), and Statutory Standards, Safety Rules, Verification & Costing (20%). This bank offers 100 English-language practice MCQs preserving Austrian technical and statutory standards.
Sample Befähigungsprüfung Elektrotechnik Practice Questions
Try these sample questions to test your Befähigungsprüfung Elektrotechnik exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1In a symmetrical three-phase AC system (Drehstromnetz) with a line-to-neutral phase voltage of 230 V (Strangspannung U_ph), what is the nominal line-to-line voltage (Leiterspannung U) between any two phase conductors?
A.230 V
B.325 V
C.400 V
D.500 V
Explanation: In a symmetrical three-phase system, the line-to-line voltage (Leiterspannung U) is related to the phase voltage (Strangspannung U_ph) by the square root of 3 factor (Verkettungsfaktor √3 ≈ 1.732): U = √3 × U_ph = √3 × 230 V ≈ 400 V.
2Which formula correctly calculates the total apparent power (Scheinleistung S) in a symmetrical three-phase AC network?
A.S = 3 × U × I
B.S = √3 × U × I
C.S = √3 × U × I × cos φ
D.S = U × I / √3
Explanation: The apparent power (Scheinleistung S) in a symmetrical three-phase network with line-to-line voltage U and line current I is given by S = √3 × U × I, measured in volt-amperes (VA) or kilovolt-amperes (kVA).
3What is the physical and economic consequence of operating an industrial facility with an uncompensated, low power factor (e.g., cos φ = 0.65 inductive)?
A.Active power consumption increases while line currents decrease.
B.Line currents and I²R distribution losses increase, causing voltage drops and utility reactive energy penalty charges.
C.Transformer core saturation occurs, reducing the system operating frequency.
D.Insulation resistance across protective earth conductors breaks down proportionally to inductive reactance.
Explanation: A low power factor (cos φ) means a high reactive current (Blindstrom) oscillates between inductive loads and the source without doing useful work. This increases total line current (I = P / (√3 × U × cos φ)), causing higher ohmic line losses (P_v = 3 × I² × R), greater voltage drops along cables and transformers, and punitive tariff surcharges by grid operators for excessive reactive energy (Blindstromtarife).
4An electrical installation draws an active power P = 50 kW at cos φ_1 = 0.70 (tan φ_1 = 1.020). What reactive power compensation capacity (Q_c) is required to improve the power factor to cos φ_2 = 0.95 (tan φ_2 = 0.329)?
A.16.5 kvar
B.34.6 kvar
C.51.0 kvar
D.67.5 kvar
Explanation: The required capacitor bank reactive power is calculated using Q_c = P × (tan φ_1 - tan φ_2) = 50 kW × (1.020 - 0.329) = 50 kW × 0.691 = 34.55 kvar ≈ 34.6 kvar.
5In an ideal single-phase transformer with primary voltage U_1, primary turns N_1, secondary turns N_2, and secondary voltage U_2, which equation defines the transformation ratio (Übersetzungsverhältnis ü)?
A.ü = U_1 / U_2 = N_1 / N_2 = I_2 / I_1
B.ü = U_1 / U_2 = N_2 / N_1 = I_1 / I_2
C.ü = U_2 / U_1 = N_1 / N_2 = I_1 / I_2
D.ü = U_1 × U_2 = N_1 × N_2 = I_1 × I_2
Explanation: For an ideal transformer, the voltage ratio is directly proportional to the turns ratio (U_1 / U_2 = N_1 / N_2), while the current ratio is inversely proportional (I_2 / I_1) because apparent power is conserved (U_1 × I_1 = U_2 × I_2). Thus, ü = U_1 / U_2 = N_1 / N_2 = I_2 / I_1.
6A three-phase distribution transformer has a rated power S_r = 630 kVA, rated secondary voltage U_2 = 400 V, and a relative short-circuit voltage u_k = 4.0%. What is the prospective symmetrical initial short-circuit current (I_k'') directly at the secondary terminals assuming an infinite primary grid?
A.9.09 kA
B.22.7 kA
C.36.4 kA
D.90.9 kA
Explanation: First, calculate the rated secondary current: I_r = S_r / (√3 × U_2) = 630,000 VA / (1.732 × 400 V) = 909.3 A. The short-circuit current is then determined by the impedance voltage percentage u_k%: I_k'' = I_r × (100% / u_k%) = 909.3 A × (100 / 4.0) = 909.3 A × 25 = 22,733 A ≈ 22.7 kA.
7What does the vector group designation 'Dyn5' on a three-phase transformer nameplate indicate according to ÖVE/ÖNORM standards?
A.Primary delta, secondary star with accessible neutral, secondary voltage lags primary by 150° (5 × 30°).
B.Primary star, secondary delta with neutral, secondary voltage leads primary by 150°.
C.Primary delta, secondary zigzag with neutral, secondary voltage lags primary by 50°.
D.Primary double delta, secondary star, rated for 5 kV operation.
Explanation: In transformer vector group designations: 'D' = high-voltage winding in Delta (Dreieck), 'y' = low-voltage winding in Star (Stern), 'n' = neutral conductor brought out to a terminal, '5' = hour number (Uhrenzahl), meaning the secondary low-voltage vector lags the primary high-voltage vector by 5 × 30° = 150°.
8Under what load condition does an electrical power transformer achieve its maximum operating efficiency (Wirkungsgrad η_max)?
A.At no-load (Leerlauf) when secondary current is zero.
B.At exactly 100% rated full load (Vollast) under all conditions.
C.When load-dependent copper losses (Kupferverluste P_Cu) equal constant iron core losses (Eisenverluste P_Fe).
D.When short-circuit voltage u_k equals total load impedance.
Explanation: Maximum transformer efficiency (η_max) occurs when the variable load-dependent copper losses (ohmic winding losses P_Cu = I² × R) equal the constant, voltage-dependent iron losses (hysteresis and eddy current losses P_Fe / P_0). In distribution transformers, this maximum typically occurs around 50% to 70% of rated load.
9What is the synchronous speed (Synchrondrehzahl n_s) of a 4-pole three-phase asynchronous motor connected to a standard 50 Hz power supply?
A.750 rpm
B.1000 rpm
C.1500 rpm
D.3000 rpm
Explanation: Synchronous speed is calculated as n_s = (f × 60) / p, where f = 50 Hz and p is the number of pole pairs. A 4-pole motor has p = 2 pole pairs. Thus: n_s = (50 × 60) / 2 = 3000 / 2 = 1500 rpm (min⁻¹).
10A 4-pole three-phase induction motor operates on a 50 Hz supply with a measured full-load rotor speed n = 1440 rpm. What is the operational slip (Schlupf s) of this motor?
A.2.0%
B.4.0%
C.6.0%
D.8.0%
Explanation: Synchronous speed for a 4-pole motor (p = 2) at 50 Hz is n_s = 1500 rpm. Slip is calculated as s = (n_s - n) / n_s = (1500 - 1440) / 1500 = 60 / 1500 = 0.04 = 4.0%.
About the Befähigungsprüfung Elektrotechnik Practice Questions
Verified exam format metadata for Befähigungsprüfung für das reglementierte Gewerbe Elektrotechnik (Österreich) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.