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Saxony-Anhalt Abitur Engineering Sciences - independent engineering foundations study materials
Saxony-Anhalt Abitur Engineering Sciences - independent engineering foundations study resources

Saxony-Anhalt Abitur Engineering Sciences - independent engineering foundations practice questions

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Question 1Materials, testing and selection

Which value is normally used as the modulus of elasticity (Elastizitätsmodul) of structural steel S235 in engineering calculations?

  1. 70 000 N/mm²
  2. 210 000 N/mm² Correct answer.
  3. 21 000 N/mm²
  4. 2 100 N/mm²

Explanation

E ≈ 210 000 N/mm² (210 GPa) for all common structural steels. It links stress and strain in Hooke's law σ = E · ε and is essentially independent of the steel grade - S235 and S355 have the same stiffness but different yield strengths.

Question 2Materials, testing and selection

A round steel tie bar with a diameter of 20 mm carries an axial tensile force of 62.8 kN. What is the normal stress in the bar?

  1. ≈ 100 N/mm²
  2. ≈ 157 N/mm²
  3. ≈ 200 N/mm² Correct answer.
  4. ≈ 3 140 N/mm²

Explanation

S = π/4 · d² = π/4 · 20² = 314.2 mm². σ = F/S = 62 800 N / 314.2 mm² ≈ 200 N/mm². For S235 (Re = 235 N/mm²) this bar is already close to the yield point.

Question 3Materials, testing and selection

A steel rod of length 2 500 mm and cross-section 200 mm² is loaded with 42 kN. With E = 210 000 N/mm², how much does it elongate elastically?

  1. 2.5 mm Correct answer.
  2. 0.25 mm
  3. 25 mm
  4. 1.0 mm

Explanation

σ = F/S = 42 000/200 = 210 N/mm². ε = σ/E = 210/210 000 = 0.001. ΔL = ε · L = 0.001 · 2 500 mm = 2.5 mm. Equivalently ΔL = F·L/(E·S).

Question 4Materials, testing and selection

A cube of an unknown metal measuring 40 mm × 40 mm × 40 mm has a mass of 172.8 g. Which metal is it most likely to be?

  1. Aluminium (ρ ≈ 2.7 g/cm³) Correct answer.
  2. Steel (ρ ≈ 7.85 g/cm³)
  3. Copper (ρ ≈ 8.9 g/cm³)
  4. Magnesium (ρ ≈ 1.74 g/cm³)

Explanation

V = 4 cm × 4 cm × 4 cm = 64 cm³. ρ = m/V = 172.8 g / 64 cm³ = 2.70 g/cm³, which matches aluminium.

Question 5Materials, testing and selection

A galvanised (zinc-coated) steel sheet is scratched down to the bare steel. Why does the steel still not rust at the scratch?

  1. The zinc layer flows into the scratch and seals it mechanically
  2. Zinc is less noble than iron, so it becomes the anode and corrodes preferentially, protecting the steel cathodically Correct answer.
  3. Zinc is more noble than iron and therefore protects it as a cathode
  4. The zinc raises the carbon content of the steel locally, making it corrosion resistant

Explanation

In the electrochemical series zinc (≈ −0.76 V) is less noble than iron (≈ −0.44 V). In the presence of an electrolyte the zinc dissolves as the sacrificial anode and the exposed steel becomes the cathode, where no metal dissolution occurs. This active (cathodic) protection is why galvanising still works at cut edges and scratches.

Question 6Materials, testing and selection

A square softwood post with a cross-section of 120 mm × 120 mm is loaded in compression parallel to the grain. If the permissible compressive stress is 8.5 N/mm², what is the maximum permissible axial load?

  1. ≈ 1.0 kN
  2. ≈ 61.2 kN
  3. ≈ 122.4 kN Correct answer.
  4. ≈ 1 224 kN

Explanation

A = 120 mm × 120 mm = 14 400 mm². F = σ · A = 8.5 N/mm² × 14 400 mm² = 122 400 N = 122.4 kN. (For a slender post a buckling check would also be required, which can reduce this value considerably.)

Question 7Materials, testing and selection

A tie rod of S235 steel (yield strength Re = 235 N/mm²) must carry a static tensile force of 90 kN with a safety factor of 1.5 against yielding. What is the minimum required cross-sectional area?

  1. ≈ 383 mm²
  2. ≈ 575 mm² Correct answer.
  3. ≈ 862 mm²
  4. ≈ 255 mm²

Explanation

σ_perm = Re / ν = 235 / 1.5 = 156.7 N/mm². A = F / σ_perm = 90 000 N / 156.7 N/mm² ≈ 574.5 mm². A round bar would need d = √(4A/π) ≈ 27.0 mm, so Ø28 mm would be selected.

Question 8Materials, testing and selection

A proportional tensile specimen has an original gauge length L0 = 100 mm and a diameter of 10 mm. After fracture the gauge length measures 117 mm, and the maximum force recorded during the test was 42.4 kN. What are the percentage elongation after fracture A and the tensile strength Rm?

  1. A = 17 %, Rm ≈ 540 N/mm² Correct answer.
  2. A = 17 %, Rm ≈ 424 N/mm²
  3. A = 117 %, Rm ≈ 540 N/mm²
  4. A = 14.5 %, Rm ≈ 540 N/mm²

Explanation

A = (Lu − L0)/L0 × 100 % = (117 − 100)/100 × 100 % = 17 %. S0 = π/4 · 10² = 78.54 mm², so Rm = 42 400 N / 78.54 mm² ≈ 540 N/mm². An elongation of 17 % indicates a ductile steel with substantial plastic reserve.

Question 9Statics, strength and mechanical transmission

The moment of a force about a point is calculated as:

  1. Force × the perpendicular distance from the point to the line of action of the force Correct answer.
  2. Force × the total length of the member
  3. Force ÷ the distance to the point
  4. Force × the distance measured along the line of action of the force

Explanation

M = F · a, where a is the perpendicular (shortest) distance from the reference point to the line of action. This perpendicular distance is called the lever arm (Hebelarm), and the unit is Nm or kNm.

Question 10Statics, strength and mechanical transmission

How many independent equilibrium conditions are available for a rigid body in a plane (2D) force system?

  1. Two: ΣFx = 0 and ΣFy = 0
  2. Three: ΣFx = 0, ΣFy = 0 and ΣM = 0 Correct answer.
  3. Four, one for each quadrant
  4. Six: three force and three moment equations

Explanation

A rigid body in a plane has three degrees of freedom (two translations and one rotation), so equilibrium requires exactly three equations: ΣFx = 0, ΣFy = 0 and ΣM = 0 about any chosen point.

Question 11Statics, strength and mechanical transmission

A movable (roller) support - verschiebliches Auflager - on a beam provides how many unknown reaction components?

  1. One, acting perpendicular to the sliding surface Correct answer.
  2. Two, one horizontal and one vertical
  3. Three, including a restraining moment
  4. None; it carries no load

Explanation

A roller support blocks only the displacement perpendicular to its sliding surface, so it contributes exactly one unknown reaction force in that direction. It allows both movement along the surface and rotation.

Question 12Statics, strength and mechanical transmission

A simply supported beam has one pinned support and one roller support. How many reaction components does it have, and is it statically determinate?

  1. Two components; statically indeterminate
  2. Three components; statically determinate Correct answer.
  3. Four components; statically indeterminate to the first degree
  4. Three components; kinematically unstable

Explanation

Pinned support = 2 components, roller support = 1 component, total 3 unknowns. In a plane there are exactly 3 equilibrium equations, so the reactions can be found from statics alone: the system is statically determinate (statisch bestimmt).

Question 13Statics, strength and mechanical transmission

Two forces act at the same point: F1 = 300 N horizontally and F2 = 400 N vertically. What is the magnitude and direction of the resultant?

  1. 500 N at 53.1° to the horizontal Correct answer.
  2. 700 N at 45°
  3. 500 N at 36.9° to the horizontal
  4. 100 N at 53.1° to the horizontal

Explanation

For perpendicular forces R = √(300² + 400²) = √250 000 = 500 N. The angle to the horizontal is α = arctan(F2/F1) = arctan(400/300) = 53.13°.

Question 14Statics, strength and mechanical transmission

A simply supported beam AB of span 6.0 m carries a single point load F = 24 kN at 2.0 m from support A. What are the support reactions A and B?

  1. A = 8 kN, B = 16 kN
  2. A = 16 kN, B = 8 kN Correct answer.
  3. A = 12 kN, B = 12 kN
  4. A = 18 kN, B = 6 kN

Explanation

Taking moments about B: A · 6.0 = 24 · 4.0 → A = 16 kN. Vertical equilibrium: B = 24 − 16 = 8 kN. The reaction is always larger at the support NEARER the load, because that support has the shorter lever arm to resist with.

Question 15Statics, strength and mechanical transmission

A simply supported beam of span L = 8.0 m carries a uniformly distributed load q = 5 kN/m over its whole length. What is the maximum bending moment?

  1. 20 kNm
  2. 40 kNm Correct answer.
  3. 80 kNm
  4. 160 kNm

Explanation

For a full-span UDL on a simply supported beam, Mmax = q·L²/8 = 5 × 8²/8 = 5 × 64/8 = 40 kNm, occurring at midspan where the shear force passes through zero. The reactions are q·L/2 = 20 kN each.

Question 16Statics, strength and mechanical transmission

For a simply supported beam of span 6.0 m with a single point load of 24 kN at 2.0 m from A (reactions A = 16 kN, B = 8 kN), what is the maximum bending moment and where does it occur?

  1. 32 kNm at the load point, 2.0 m from A Correct answer.
  2. 48 kNm at midspan
  3. 32 kNm at midspan
  4. 16 kNm at the load point

Explanation

For a single point load the shear force is constant on each side and changes sign exactly at the load, so the bending moment peaks there. M = A · a = 16 kN × 2.0 m = 32 kNm. Checking from the other side gives the same value: 8 kN × 4.0 m = 32 kNm.

Question 17Statics, strength and mechanical transmission

In a beam, at which location does the bending moment reach a local maximum or minimum?

  1. Where the shear force is greatest
  2. Where the shear force passes through zero or changes sign Correct answer.
  3. Always at midspan
  4. Always directly at the supports

Explanation

The shear force is the derivative of the bending moment: dM/dx = V. Extreme values of M therefore occur where V = 0. In a shear-force diagram every crossing of the zero line marks a peak in the moment diagram.

Question 18Statics, strength and mechanical transmission

A cantilever of length 3.0 m carries a uniformly distributed load q = 4 kN/m over its full length plus a point load F = 10 kN at the free end. What is the bending moment at the fixed support?

  1. 30 kNm
  2. 48 kNm Correct answer.
  3. 39 kNm
  4. 66 kNm

Explanation

The UDL has resultant q·L = 12 kN acting at the centroid, 1.5 m from the support, giving q·L²/2 = 4 × 9/2 = 18 kNm. The end load gives F·L = 10 × 3.0 = 30 kNm. Superposition: M = 18 + 30 = 48 kNm.

Question 19Statics, strength and mechanical transmission

A simply supported beam AB of span 10.0 m carries a UDL q = 6 kN/m over the whole span plus a point load F = 20 kN at 2.5 m from A. What is the reaction at A?

  1. 40 kN
  2. 45 kN Correct answer.
  3. 35 kN
  4. 50 kN

Explanation

By superposition. UDL: total 6 × 10 = 60 kN, symmetric, so 30 kN at each support. Point load: A receives 20 × 7.5/10 = 15 kN (the lever arm to the FAR support divided by the span). Total A = 30 + 15 = 45 kN. Check: B = 30 + 5 = 35 kN, and 45 + 35 = 80 kN = 60 + 20.

Question 20Statics, strength and mechanical transmission

A simply supported beam of span 6.0 m carries a UDL of 8 kN/m over the left half only (from 0 to 3.0 m). Determine the reaction at A and the position and value of the maximum bending moment.

  1. A = 18 kN; Mmax = 20.25 kNm at x = 2.25 m from A Correct answer.
  2. A = 12 kN; Mmax = 36 kNm at midspan
  3. A = 18 kN; Mmax = 18 kNm at x = 3.0 m, the end of the loaded length
  4. A = 6 kN; Mmax = 20.25 kNm at x = 2.25 m from A

Explanation

The load resultant is 8 × 3 = 24 kN acting at 1.5 m from A. ΣM about B: A × 6.0 = 24 × 4.5 → A = 18 kN, so B = 6 kN. In the loaded region V(x) = 18 − 8x, which is zero at x = 2.25 m. There M = 18 × 2.25 − 8 × 2.25²/2 = 40.5 − 20.25 = 20.25 kNm.

Question 21Statics, strength and mechanical transmission

Two parallel downward forces act on a rigid beam: F1 = 200 N at x = 0 and F2 = 300 N at x = 2.0 m. Where does their resultant act?

  1. 1.0 m from F1
  2. 1.2 m from F1 Correct answer.
  3. 0.8 m from F1
  4. 1.5 m from F1

Explanation

R = 200 + 300 = 500 N. By the moment theorem (Momentensatz), the moment of the resultant about a point equals the sum of the moments of the individual forces. About the position of F1: 500 · x = 200 · 0 + 300 · 2.0 → x = 600/500 = 1.2 m. The resultant always lies nearer the larger force.

Question 22Electrical circuits, machines and digital logic

Three resistors of 100 Ω, 220 Ω and 680 Ω are connected in series. What is the total resistance?

  1. 1 000 Ω Correct answer.
  2. 62.4 Ω
  3. 333 Ω
  4. 220 Ω

Explanation

In series the resistances simply add: Rges = 100 + 220 + 680 = 1 000 Ω. The same current flows through all three, and the supply voltage divides between them in proportion to their resistances.

Question 23Electrical circuits, machines and digital logic

Two resistors of 100 Ω each are connected in parallel. What is the total resistance?

  1. 200 Ω
  2. 100 Ω
  3. 50 Ω Correct answer.
  4. 0.02 Ω

Explanation

For two equal resistors in parallel R = R1/2 = 100/2 = 50 Ω. Generally R = (R1·R2)/(R1+R2) = 10 000/200 = 50 Ω. The current divides equally between the two identical branches.

Question 24Electrical circuits, machines and digital logic

An electric heater draws 9.0 A at 230 V. What is its power consumption?

  1. 2 070 W Correct answer.
  2. 25.6 W
  3. 239 W
  4. 207 W

Explanation

P = U · I = 230 V × 9.0 A = 2 070 W ≈ 2.07 kW. Its equivalent resistance is R = U/I = 25.6 Ω.

Question 25Electrical circuits, machines and digital logic

An unloaded voltage divider consists of R1 = 4.7 kΩ and R2 = 10 kΩ in series across 15 V. What voltage appears across R2?

  1. 4.80 V
  2. 10.20 V Correct answer.
  3. 7.50 V
  4. 15.0 V

Explanation

U2 = U · R2/(R1 + R2) = 15 × 10/(4.7 + 10) = 15 × 10/14.7 = 10.20 V. The same current flows through both resistors, so the voltage splits in the ratio of the resistances.

Question 26Electrical circuits, machines and digital logic

The divider above (R1 = 4.7 kΩ, R2 = 10 kΩ, 15 V) is now loaded with RL = 10 kΩ connected across R2. What happens to the output voltage?

  1. It rises to about 12.0 V
  2. It stays at 10.20 V, because the divider ratio is fixed by the resistors
  3. It falls to about 7.7 V Correct answer.
  4. It falls to about 5.0 V

Explanation

The load sits in parallel with R2: 10 kΩ ∥ 10 kΩ = 5 kΩ. The new output is U2 = 15 × 5/(4.7 + 5) = 15 × 5/9.7 ≈ 7.73 V. This is exactly why a divider is designed with a cross-current (Querstrom) several times the load current when the output must stay stable.

Question 27Electrical circuits, machines and digital logic

A red LED with a forward voltage of 2.0 V is to be operated at 20 mA from a 9 V supply. What series resistor is required?

  1. 350 Ω Correct answer.
  2. 450 Ω
  3. 100 Ω
  4. 550 Ω

Explanation

The resistor must drop the difference between supply and forward voltage: UR = 9 − 2.0 = 7.0 V. R = UR/I = 7.0 V / 0.020 A = 350 Ω. In practice the next higher standard value (390 Ω in the E12 series) is chosen so the current stays below 20 mA.

Question 28Electrical circuits, machines and digital logic

For the LED series resistor above (350 Ω carrying 20 mA), what is the power dissipated in the resistor and the minimum sensible power rating?

  1. 0.04 W, so a 0.125 W resistor
  2. 0.14 W, so a 0.25 W resistor Correct answer.
  3. 0.18 W, so a 0.25 W resistor
  4. 1.40 W, so a 2 W resistor

Explanation

PR = UR · I = 7.0 V × 0.020 A = 0.14 W, or equivalently I²·R = 0.020² × 350 = 0.14 W. A standard 0.25 W resistor gives a comfortable margin; a 0.125 W type would run at over 100 % of its rating.

Question 29Electrical circuits, machines and digital logic

At a node, currents of 3.0 A and 2.5 A flow in, and 1.8 A flows out through one branch. What current flows out through the only remaining branch?

  1. 3.7 A Correct answer.
  2. 7.3 A
  3. 0.7 A
  4. 5.5 A

Explanation

Kirchhoff's current law: the sum of the currents entering a node equals the sum leaving it. 3.0 + 2.5 = 5.5 A in, of which 1.8 A already leaves, so the remaining branch carries 5.5 − 1.8 = 3.7 A.

Question 30Electrical circuits, machines and digital logic

A DC circuit supplies 12 A through a two-core copper cable 25 m long with a conductor cross-section of 4 mm². With ρ = 0.0178 Ω·mm²/m, what is the voltage drop along the cable?

  1. 1.33 V
  2. 2.67 V Correct answer.
  3. 5.34 V
  4. 0.21 V

Explanation

The current must flow out and back, so the conductor length is 2 × 25 = 50 m. R = ρ·l/A = 0.0178 × 50/4 = 0.2225 Ω. ΔU = R · I = 0.2225 × 12 = 2.67 V, i.e. about 1.2 % of a 230 V supply.

Question 31Electrical circuits, machines and digital logic

A 24 V source feeds R1 = 20 Ω in series with the parallel combination of R2 = 60 Ω and R3 = 30 Ω. What current flows through R3?

  1. 0.20 A
  2. 0.40 A Correct answer.
  3. 0.60 A
  4. 0.80 A

Explanation

R2 ∥ R3 = (60 × 30)/(60 + 30) = 20 Ω. Total resistance = 20 + 20 = 40 Ω, so I_total = 24/40 = 0.60 A. The voltage across the parallel block is 0.60 × 20 = 12 V, therefore I_R3 = 12/30 = 0.40 A.

Question 32Electrical circuits, machines and digital logic

An electric motor draws 2.3 kW from a 230 V supply and delivers 1.84 kW of mechanical power. What is its efficiency and what current does it draw?

  1. η = 80 %, I = 10.0 A Correct answer.
  2. η = 80 %, I = 8.0 A
  3. η = 125 %, I = 10.0 A
  4. η = 20 %, I = 10.0 A

Explanation

η = Pout/Pin = 1.84/2.3 = 0.80 = 80 %. The current is determined by the power the motor TAKES FROM the supply: I = Pin/U = 2 300 W / 230 V = 10.0 A. The 460 W difference is lost as heat, friction and iron/copper losses.

Question 33Electrical circuits, machines and digital logic

In a purely inductive AC circuit, what is the phase relationship between current and voltage?

  1. The current lags the voltage by 90° Correct answer.
  2. The current leads the voltage by 90°
  3. Current and voltage are in phase
  4. The current lags the voltage by 180°

Explanation

In an ideal coil u = L · di/dt, so the voltage is greatest when the current changes fastest, i.e. at the current's zero crossing. The current therefore lags the voltage by a quarter period (90°). The classic mnemonic is that in an inductance the current 'comes late'.

Question 34Electrical circuits, machines and digital logic

How many diodes does a B2 bridge (Graetz) rectifier contain, and how many output pulses occur per mains period?

  1. 2 diodes, 1 pulse
  2. 4 diodes, 2 pulses Correct answer.
  3. 4 diodes, 4 pulses
  4. 6 diodes, 3 pulses

Explanation

The B2 bridge uses four diodes conducting in diagonal pairs, so both half-waves of the input reach the load with the same polarity. That gives two output humps per mains period - hence 'zweipulsig'. On a 50 Hz supply the ripple frequency is 100 Hz.

Question 35Electrical circuits, machines and digital logic

A B2 bridge rectifier is fed with 230 V AC (RMS). Ignoring diode forward voltages, what is the ideal mean DC output voltage without a smoothing capacitor?

  1. 230 V
  2. 207 V Correct answer.
  3. 325 V
  4. 115 V

Explanation

For the two-pulse bridge Udi = (2√2/π) · U ≈ 0.9 · U = 0.9 × 230 V ≈ 207 V. The factor 0.9 is the ratio of the arithmetic mean of a full-wave rectified sine to its RMS value.

Question 36Electrical circuits, machines and digital logic

What happens to the DC output of a B2 rectifier when a sufficiently large smoothing capacitor is connected in parallel with the load?

  1. The mean voltage rises towards the peak value and the ripple falls Correct answer.
  2. The mean voltage falls and the ripple rises
  3. The mean voltage is unchanged but the ripple frequency doubles
  4. The output becomes alternating again

Explanation

The capacitor charges to near the peak value (about 325 V for a 230 V input) and then supplies the load between conduction peaks, so it only discharges partially. The mean voltage therefore rises above the unsmoothed 207 V towards the peak, and the residual ripple falls as the capacitance or the load resistance increases.

Question 37Electrical circuits, machines and digital logic

Why is a capacitor connected in parallel with the inductive ballast of a fluorescent lamp (Kompensation)?

  1. To increase the lamp's active power output
  2. To supply the inductive reactive power locally, raising cos φ and reducing the current in the supply cable Correct answer.
  3. To convert the AC into DC for the lamp
  4. To increase the voltage across the lamp

Explanation

The ballast choke draws lagging reactive power; the parallel capacitor supplies leading reactive power that largely cancels it. The active power P is unchanged, but the apparent power S = P/cos φ and hence the line current fall, which reduces cable losses (I²R) and lets the same installation carry more load.

Question 38Electrical circuits, machines and digital logic

What is the reactance of a 10 µF capacitor at 50 Hz?

  1. 318 Ω Correct answer.
  2. 3 183 Ω
  3. 31.8 Ω
  4. 0.00314 Ω

Explanation

XC = 1/(2πf·C) = 1/(2π × 50 Hz × 10 × 10⁻⁶ F) = 1/0.003142 ≈ 318 Ω. Capacitive reactance falls as either frequency or capacitance rises.

Question 39Electrical circuits, machines and digital logic

A single-phase motor takes P = 1.5 kW at 230 V / 50 Hz with cos φ1 = 0.60. It is to be compensated to cos φ2 = 0.95. What capacitance is required?

  1. ≈ 30 µF
  2. ≈ 91 µF Correct answer.
  3. ≈ 120 µF
  4. ≈ 9 µF

Explanation

tan φ1 = 1.333 and tan φ2 = 0.329. The reactive power to be compensated is Qc = P(tan φ1 − tan φ2) = 1 500 × (1.333 − 0.329) ≈ 1 507 var. Then C = Qc/(2πf·U²) = 1 507/(2π × 50 × 230²) ≈ 9.1 × 10⁻⁵ F ≈ 91 µF.

Question 40Electrical circuits, machines and digital logic

A photovoltaic module has an open-circuit voltage of 45 V, a short-circuit current of 9.5 A, and a maximum power point at 37 V / 8.6 A. What is its fill factor?

  1. 0.74 Correct answer.
  2. 0.82
  3. 0.90
  4. 1.34

Explanation

FF = (Umpp · Impp)/(Uoc · Isc) = (37 × 8.6)/(45 × 9.5) = 318.2/427.5 ≈ 0.74. The fill factor measures how 'square' the I-V curve is; good crystalline silicon modules reach roughly 0.7 to 0.8.

Question 41Electrical circuits, machines and digital logic

What is the task of a charge controller (Laderegler) in a stand-alone photovoltaic system with a battery?

  1. It converts the battery's DC into 230 V AC for household appliances
  2. It limits charging voltage and current and prevents deep discharge of the battery Correct answer.
  3. It increases the modules' short-circuit current
  4. It stores the surplus energy in its own capacitors

Explanation

The charge controller protects the battery at both ends of its operating range: it caps the charging voltage so the cells are not overcharged and gassed, and it disconnects the load at the deep-discharge threshold. For lithium systems it also works together with the battery management system on cell balancing and temperature limits.

Question 42Measurement, feedback, PLC and automation

For which input combinations does a two-input XOR (exclusive OR) gate produce a 1 at its output?

  1. Only when both inputs are 1
  2. Only when the two inputs are different Correct answer.
  3. Only when both inputs are 0
  4. For every combination except 0/0

Explanation

XOR outputs 1 exactly when the inputs are unequal: 0/1 and 1/0 give 1, while 0/0 and 1/1 give 0. It is therefore also called an inequality or antivalence gate and is the core of binary adders and parity checks.

Question 43Measurement, feedback, PLC and automation

What distinguishes a closed-loop control (Regelung) from an open-loop control (Steuerung)?

  1. A Regelung continuously measures the actual value and feeds it back for comparison with the setpoint, forming a closed loop Correct answer.
  2. A Regelung always responds faster than a Steuerung
  3. A Steuerung uses electricity while a Regelung uses pneumatics
  4. A Regelung works without any sensors

Explanation

The defining criterion is the structure of the action path. A Steuerung has an open action chain (offener Wirkungsablauf) - the effect of the output is never checked. A Regelung closes the loop (geschlossener Wirkungskreis): the actual value is measured, compared with the setpoint, and the deviation drives the correction, so disturbances are automatically counteracted.

Question 44Measurement, feedback, PLC and automation

Applying De Morgan's theorem, which expression is equivalent to NOT (A AND B)?

  1. (NOT A) AND (NOT B)
  2. (NOT A) OR (NOT B) Correct answer.
  3. A OR B
  4. NOT (A OR B)

Explanation

De Morgan's rule states that negating a conjunction turns it into a disjunction of the negated terms: NOT(A AND B) = (NOT A) OR (NOT B). This is why a NAND function can be realised by an OR gate with inverted inputs, a substitution used constantly when converting between circuit forms in a Funktionsplan.

Question 45Measurement, feedback, PLC and automation

In a basic RS flip-flop built from two cross-coupled NOR gates, which input combination is not permitted?

  1. S = 0, R = 0
  2. S = 1, R = 0
  3. S = 0, R = 1
  4. S = 1, R = 1 Correct answer.

Explanation

With both inputs active the flip-flop is told to set and reset at the same time. Both outputs are forced to 0, so Q and NOT-Q are no longer complementary. Worse, when the inputs are released together, the state that results depends on tiny differences in gate delays - an unpredictable race condition.

Question 46Measurement, feedback, PLC and automation

How does a PLC (speicherprogrammierbare Steuerung, SPS) execute its program?

  1. Once only, at power-up
  2. Cyclically: read the inputs into the process image, execute the program, then write the output image to the outputs Correct answer.
  3. In parallel, with all instructions executed simultaneously
  4. Only when an operator presses a start button

Explanation

The PLC runs a fixed cycle: it samples all inputs into the process input image, works through the program using those frozen values, and finally transfers the process output image to the physical outputs. The cycle time, typically a few milliseconds, sets the worst-case reaction time of the whole controller.

Question 47Measurement, feedback, PLC and automation

In an automation system, which component converts a physical quantity such as pressure or temperature into an electrical signal?

  1. The actuator (Aktor)
  2. The sensor Correct answer.
  3. The CPU of the PLC
  4. The bus coupler

Explanation

The sensor sits at the interface between the process and the control system, converting a non-electrical quantity into an electrical signal the controller can read. Networked sensors that deliver their data directly onto a fieldbus are a key building block of the Internet of Things in production.

Question 48Measurement, feedback, PLC and automation

Simplify the Boolean function Y = A·B + A·(NOT B).

  1. Y = A Correct answer.
  2. Y = B
  3. Y = A + B
  4. Y = 0

Explanation

Factor out A: Y = A · (B + NOT B). Since B + NOT B = 1 always, Y = A · 1 = A. The output depends solely on A, so B is a redundant input and two gates can be deleted from the Funktionsplan.

Question 49Manufacturing, machining models and metrology

A gear drive has a driving pinion with 20 teeth running at 1 500 min⁻¹ and a driven gear with 75 teeth. What is the gear ratio i and the output speed?

  1. i = 3.75; n2 = 400 min⁻¹ Correct answer.
  2. i = 3.75; n2 = 5 625 min⁻¹
  3. i = 0.27; n2 = 5 625 min⁻¹
  4. i = 3.75; n2 = 1 500 min⁻¹

Explanation

i = z2/z1 = 75/20 = 3.75, and i = n1/n2, so n2 = n1/i = 1 500/3.75 = 400 min⁻¹. A ratio greater than 1 is a reduction: the output turns more slowly than the input, but with correspondingly higher torque.

Question 50Manufacturing, machining models and metrology

A motor delivers 4.0 kW at 1 500 min⁻¹. What torque does it produce?

  1. 2.67 Nm
  2. 25.5 Nm Correct answer.
  3. 160 Nm
  4. 255 Nm

Explanation

P = M · ω with ω = 2πn. Here ω = 2π × 1 500/60 = 157.1 rad/s, so M = P/ω = 4 000 W / 157.1 rad/s ≈ 25.5 Nm.