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Mecklenburg-Vorpommern Abitur Electrical Engineering - independent engineering foundations study materials
Mecklenburg-Vorpommern Abitur Electrical Engineering - independent engineering foundations study resources

Mecklenburg-Vorpommern Abitur Electrical Engineering - independent engineering foundations practice questions

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Question 1Electrical circuits, machines and digital logic

Two resistors R1 = 10 Ω and R2 = 20 Ω are connected in series to a 30 V DC supply. What is the current in the circuit?

  1. 1.0 A Correct answer.
  2. 1.5 A
  3. 3.0 A
  4. 0.67 A

Explanation

Series Req = 10 + 20 = 30 Ω. I = 30 V / 30 Ω = 1.0 A.

Question 2Electrical circuits, machines and digital logic

Resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance?

  1. 2.0 Ω Correct answer.
  2. 9.0 Ω
  3. 4.5 Ω
  4. 1.5 Ω

Explanation

1/Req = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 ⇒ Req = 2.0 Ω. Equivalently Req = (6×3)/(6+3) = 18/9 = 2 Ω.

Question 3Electrical circuits, machines and digital logic

In a voltage divider, R1 = 4.0 kΩ (top) and R2 = 6.0 kΩ (bottom) share a 10 V DC supply. What is the voltage across R2?

  1. 6.0 V Correct answer.
  2. 4.0 V
  3. 10 V
  4. 2.5 V

Explanation

U2 = U · R2/(R1+R2) = 10 × 6/(4+6) = 10 × 0.6 = 6.0 V.

Question 4Electrical circuits, machines and digital logic

Two series resistors drop 8.0 V and 12 V respectively. If the current is 0.40 A, what is the total resistance?

  1. 50 Ω Correct answer.
  2. 20 Ω
  3. 5.0 Ω
  4. 30 Ω

Explanation

Total voltage U = 8 + 12 = 20 V. R = U/I = 20/0.40 = 50 Ω. (Also R = R1+R2 = 8/0.4 + 12/0.4 = 20+30 = 50 Ω.)

Question 5Electrical circuits, machines and digital logic

A 100 Ω resistor has 50 V across it. What power does it dissipate?

  1. 25 W Correct answer.
  2. 2.0 W
  3. 5000 W
  4. 0.5 W

Explanation

P = U²/R = 50²/100 = 2500/100 = 25 W. (Also I = 0.5 A, P = UI = 25 W.)

Question 6Electrical circuits, machines and digital logic

In a parallel DC circuit, branch currents are 1.2 A and 0.80 A. What is the total supply current?

  1. 2.0 A Correct answer.
  2. 0.40 A
  3. 1.5 A
  4. 0.96 A

Explanation

Kirchhoff's current law: I_total = I1 + I2 = 1.2 + 0.80 = 2.0 A for two parallel branches.

Question 7Electrical circuits, machines and digital logic

A wire has resistance 0.50 Ω per meter. What is the resistance of a 40 m length of the same wire (same cross-section and temperature)?

  1. 20 Ω Correct answer.
  2. 0.0125 Ω
  3. 40.5 Ω
  4. 80 Ω

Explanation

R ∝ length at constant ρ and A, so R = 0.50 Ω/m × 40 m = 20 Ω.

Question 8Electrical circuits, machines and digital logic

A battery with emf 12.0 V and internal resistance r = 0.50 Ω supplies a 5.5 Ω load. What is the terminal voltage across the load?

  1. 11.0 V Correct answer.
  2. 12.0 V
  3. 6.0 V
  4. 0.50 V

Explanation

I = E/(R+r) = 12/(5.5+0.5) = 12/6 = 2.0 A. Terminal voltage U = I·R = 2.0×5.5 = 11.0 V (or U = E − I·r = 12 − 1 = 11 V).

Question 9Electrical circuits, machines and digital logic

Four 8.0 Ω resistors are connected: two series pairs in parallel with each other. What is Req?

  1. 8.0 Ω Correct answer.
  2. 2.0 Ω
  3. 16 Ω
  4. 32 Ω

Explanation

Each series pair: 8+8 = 16 Ω. Two 16 Ω in parallel: Req = 16/2 = 8.0 Ω.

Question 10Electrical circuits, machines and digital logic

A potentiometer as a rheostat uses 0–10 kΩ. At a dial setting that inserts 2.5 kΩ in series with a 7.5 kΩ fixed resistor across 20 V, what current flows?

  1. 2.0 mA Correct answer.
  2. 2.0 A
  3. 8.0 mA
  4. 0.50 mA

Explanation

Rtot = 2.5 + 7.5 = 10 kΩ = 10 000 Ω. I = 20 / 10 000 = 0.002 A = 2.0 mA.

Question 11Electrical circuits, machines and digital logic

Ideal ammeter resistance is ≈0 Ω and ideal voltmeter resistance is ≈∞. A real ammeter of 0.10 Ω is placed in series with a 4.9 Ω load on 5.0 V. What current does the ammeter read?

  1. 1.0 A Correct answer.
  2. 1.02 A
  3. 50 A
  4. 0.10 A

Explanation

Rtot = 4.9 + 0.10 = 5.0 Ω. I = 5.0/5.0 = 1.0 A. (Without the ammeter burden, I would be 5/4.9 ≈ 1.02 A.)

Question 12Electrical circuits, machines and digital logic

An inductor L = 100 mH is driven at f = 50 Hz. What is its inductive reactance XL?

  1. 31.4 Ω Correct answer.
  2. 5.0 Ω
  3. 314 Ω
  4. 0.0314 Ω

Explanation

XL = 2πfL = 2π × 50 × 0.100 ≈ 6.283 × 5 ≈ 31.4 Ω.

Question 13Electrical circuits, machines and digital logic

A capacitor C = 100 µF is used at f = 50 Hz. What is XC?

  1. 31.8 Ω Correct answer.
  2. 318 Ω
  3. 5.0 Ω
  4. 0.0318 Ω

Explanation

XC = 1/(2πfC) = 1/(2π × 50 × 100×10⁻⁶) = 1/(2π × 0.005) ≈ 1/0.0314 ≈ 31.8 Ω.

Question 14Electrical circuits, machines and digital logic

A series RC circuit has R = 40 Ω and XC = 30 Ω on a 50 V RMS source. What is the RMS current?

  1. 1.0 A Correct answer.
  2. 1.25 A
  3. 0.71 A
  4. 2.0 A

Explanation

Z = √(40²+30²) = √(1600+900) = √2500 = 50 Ω. I = U/Z = 50/50 = 1.0 A RMS.

Question 15Electrical circuits, machines and digital logic

An AC voltage has peak value û = 325 V. What is the RMS value (sinusoidal)?

  1. 230 V Correct answer.
  2. 325 V
  3. 460 V
  4. 163 V

Explanation

For a sine wave, U_RMS = û/√2 ≈ 325/1.414 ≈ 230 V (European mains peak ≈ √2 × 230).

Question 16Electrical circuits, machines and digital logic

A series RLC circuit has R = 20 Ω, XL = 50 Ω, XC = 30 Ω. What is Z?

  1. 28.3 Ω Correct answer.
  2. 100 Ω
  3. 40 Ω
  4. 20 Ω

Explanation

X = XL − XC = 50 − 30 = 20 Ω. Z = √(R² + X²) = √(400+400) = √800 ≈ 28.3 Ω.

Question 17Electrical circuits, machines and digital logic

At series resonance for RLC, XL = XC. If R = 10 Ω and U = 20 V RMS, what is the current?

  1. 2.0 A Correct answer.
  2. 0 A
  3. depends on L only
  4. ∞

Explanation

At series resonance Z = R (X cancels), so I = U/R = 20/10 = 2.0 A.

Question 18Electrical circuits, machines and digital logic

Find the resonant frequency of L = 40 mH and C = 100 µF (series).

  1. 80 Hz Correct answer.
  2. 50 Hz
  3. 159 Hz
  4. 8.0 Hz

Explanation

f0 = 1/(2π√(LC)) = 1/(2π√(0.04 × 100×10⁻⁶)) = 1/(2π√(4×10⁻⁶)) = 1/(2π × 2×10⁻³) = 1/(0.01257) ≈ 80 Hz.

Question 19Electrical circuits, machines and digital logic

A load draws 400 W real power at 230 V RMS with current 2.5 A RMS. What is the power factor?

  1. 0.70 Correct answer.
  2. 1.00
  3. 0.50
  4. 1.74

Explanation

S = U·I = 230 × 2.5 = 575 VA. pf = P/S = 400/575 ≈ 0.70.

Question 20Electrical circuits, machines and digital logic

A purely capacitive load has XC = 46 Ω on 230 V RMS. What is the reactive power magnitude |Q|?

  1. 1150 var Correct answer.
  2. 230 var
  3. 46 var
  4. 5290 var

Explanation

I = U/XC = 230/46 = 5.0 A. |Q| = U·I = 230×5 = 1150 var (or U²/XC = 52900/46 = 1150 var).

Question 21Electrical circuits, machines and digital logic

Phase angle φ of a series RL load with R = 30 Ω and XL = 30 Ω is:

  1. 45° Correct answer.
  2. 0°
  3. 90°
  4. 30°

Explanation

tan φ = XL/R = 30/30 = 1 ⇒ φ = 45° (current lags voltage by 45°).

Question 22Electrical circuits, machines and digital logic

Two inductors 4.0 mH and 12 mH (no mutual coupling) are in parallel. What is Leq?

  1. 3.0 mH Correct answer.
  2. 16 mH
  3. 8.0 mH
  4. 48 mH

Explanation

1/Leq = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 ⇒ Leq = 3.0 mH.

Question 23Electrical circuits, machines and digital logic

Ideal transformer 230 V / 23 V supplies a 2.3 Ω resistive load on the secondary. What primary current is drawn (ideal, ignore magnetizing current)?

  1. 1.0 A Correct answer.
  2. 10 A
  3. 0.10 A
  4. 100 A

Explanation

Is = Us/R = 23/2.3 = 10 A. Power Pin = Pout ⇒ Up·Ip = Us·Is ⇒ Ip = (23×10)/230 = 1.0 A. Also Ip/Is = Ns/Np = 1/10.

Question 24Electrical circuits, machines and digital logic

For R = 5 Ω, XL = 100 Ω, XC = 40 Ω at a given frequency, what is the impedance magnitude?

  1. 60.2 Ω Correct answer.
  2. 145 Ω
  3. 55 Ω
  4. 5 Ω

Explanation

X = 100 − 40 = 60 Ω. Z = √(5² + 60²) = √(25+3600) = √3625 ≈ 60.2 Ω.

Question 25Electrical circuits, machines and digital logic

Balanced three-phase real power with UL = 400 V, IL = 10 A, pf = 0.80 is:

  1. 5.54 kW Correct answer.
  2. 3.20 kW
  3. 4.00 kW
  4. 6.93 kW

Explanation

P = √3 · UL · IL · pf = 1.732 × 400 × 10 × 0.80 ≈ 5540 W ≈ 5.54 kW.

Question 26Electrical circuits, machines and digital logic

A coil has R = 20 Ω and L = 0.10 H at 50 Hz. What is its impedance magnitude?

  1. 37.4 Ω Correct answer.
  2. 20 Ω
  3. 31.4 Ω
  4. 51.4 Ω

Explanation

XL = 2π×50×0.10 ≈ 31.4 Ω. Z = √(20²+31.4²) ≈ √(400+986) ≈ √1386 ≈ 37.2–37.4 Ω.

Question 27Electrical circuits, machines and digital logic

A series RC has R = XC = 25 Ω. Relative to the voltage, the current phase is:

  1. 45° leading Correct answer.
  2. 45° lagging
  3. 0°
  4. 90° leading

Explanation

φ = arctan(−XC/R) = arctan(−1) = −45° for voltage as reference means current leads by 45° in the RC convention (current leads voltage in a capacitive circuit).

Question 28Electrical circuits, machines and digital logic

Energy stored in C = 470 µF charged to 100 V is:

  1. 2.35 J Correct answer.
  2. 47 J
  3. 0.047 J
  4. 4.7 J

Explanation

W = ½CV² = 0.5 × 470×10⁻⁶ × 100² = 0.5 × 470×10⁻⁶ × 10000 = 2.35 J.

Question 29Electrical circuits, machines and digital logic

Energy stored in L = 200 mH carrying 5.0 A is:

  1. 2.5 J Correct answer.
  2. 1.0 J
  3. 5.0 J
  4. 0.5 J

Explanation

W = ½ L I² = 0.5 × 0.20 × 25 = 0.10 × 25 = 2.5 J.

Question 30Electrical circuits, machines and digital logic

An ideal silicon diode in forward bias is often modeled with VF ≈ 0.70 V. With a 5.0 V source and series R = 430 Ω, what is the forward current?

  1. 10 mA Correct answer.
  2. 11.6 mA
  3. 0.70 mA
  4. 5.0 mA

Explanation

I = (U − VF)/R = (5.0 − 0.70)/430 = 4.3/430 = 0.010 A = 10 mA.

Question 31Electrical circuits, machines and digital logic

Zener diode VZ = 5.1 V regulates with series Rs = 220 Ω from 12 V, load RL = 510 Ω. What is the current through Rs?

  1. 31.4 mA Correct answer.
  2. 23.5 mA
  3. 54.5 mA
  4. 5.1 mA

Explanation

Voltage across Rs is 12 − 5.1 = 6.9 V. I_Rs = 6.9/220 ≈ 0.0314 A = 31.4 mA.

Question 32Electrical circuits, machines and digital logic

Silicon NPN switch: VBB = 5.0 V, RB = 43 kΩ, VBE(on) ≈ 0.70 V. What is IB?

  1. 100 µA Correct answer.
  2. 116 µA
  3. 5.0 µA
  4. 70 µA

Explanation

IB = (VBB − VBE)/RB = (5.0 − 0.70)/43000 = 4.3/43000 = 100×10⁻⁶ A = 100 µA.

Question 33Electrical circuits, machines and digital logic

Ideal non-inverting op-amp with Rg = 2.0 kΩ (to ground) and Rf = 8.0 kΩ. What is Av?

  1. 5.0 Correct answer.
  2. 4.0
  3. 0.25
  4. −4.0

Explanation

Av = 1 + Rf/Rg = 1 + 8/2 = 1 + 4 = 5.0.

Question 34Electrical circuits, machines and digital logic

An ideal integrator op-amp has R = 10 kΩ and C = 100 nF. The time constant τ = RC is:

  1. 1.0 ms Correct answer.
  2. 10 ms
  3. 1.0 µs
  4. 100 ms

Explanation

τ = RC = 10×10³ × 100×10⁻⁹ = 10×10³ × 10⁻⁷ = 1.0×10⁻³ s = 1.0 ms.

Question 35Electrical circuits, machines and digital logic

First-order RC low-pass with R = 1.59 kΩ and C = 100 nF has cutoff fc = 1/(2πRC) closest to:

  1. 1.0 kHz Correct answer.
  2. 100 Hz
  3. 10 kHz
  4. 159 Hz

Explanation

fc = 1/(2π × 1590 × 100×10⁻⁹) ≈ 1/(2π × 1.59×10⁻⁴) ≈ 1/(1.0×10⁻³) ≈ 1000 Hz = 1.0 kHz.

Question 36Electrical circuits, machines and digital logic

Peak detector / envelope: capacitor charged to û = 15 V through an ideal diode. With no load, the held DC level is:

  1. 15 V Correct answer.
  2. 10.6 V
  3. 7.5 V
  4. 0 V

Explanation

Ideal peak detector holds the peak û = 15 V on C (until discharge paths exist).

Question 37Electrical circuits, machines and digital logic

Op-amp comparator (open-loop) with ±12 V rails, Vin+ = 1.0 V, Vin− = 0 V. Saturation output is closest to:

  1. +12 V Correct answer.
  2. −12 V
  3. 0 V
  4. +1.0 V

Explanation

Vin+ > Vin− ⇒ output saturates toward the positive rail ≈ +12 V (ideal comparator).

Question 38Electrical circuits, machines and digital logic

Inverting summer: two inputs 1.0 V and 2.0 V through 10 kΩ each, Rf = 10 kΩ, ideal op-amp. What is Uout?

  1. −3.0 V Correct answer.
  2. +3.0 V
  3. −1.5 V
  4. −0.3 V

Explanation

Uout = −Rf(U1/R1 + U2/R2) = −10k(1/10k + 2/10k) = −(0.1+0.2)×10? Wait: −(1/10k·1 + 2/10k·1)×10k = −(0.1+0.2) = −0.3? No: −Rf/R · (U1+U2) = −1·(1+2) = −3.0 V.

Question 39Measurement, feedback, PLC and automation

Gray code for decimal 3 (binary 11) is commonly:

  1. 10 Correct answer.
  2. 11
  3. 01
  4. 00

Explanation

Binary 3 is 11₂. Binary-to-Gray: keep MSB, then each Gray bit is XOR of adjacent binary bits → 10₂. So decimal 3 ↔ Gray 10.

Question 40Measurement, feedback, PLC and automation

A 555 timer astable with RA = 4.7 kΩ, RB = 4.7 kΩ, C = 100 nF. Period T ≈ 0.693(RA+2RB)C is closest to:

  1. 1.0 ms Correct answer.
  2. 0.33 ms
  3. 10 ms
  4. 100 µs

Explanation

RA+2RB = 4.7k + 9.4k = 14.1k. T ≈ 0.693 × 14100 × 100e−9 ≈ 0.693 × 1.41e−3 ≈ 9.8e−4 s ≈ 1.0 ms.

Question 41Measurement, 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 42Measurement, 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 43Measurement, 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 44Measurement, 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 45Measurement, 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 46Measurement, 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 47Measurement, 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 48Engineering models, drawings, evidence and project decisions

A voltmeter with 10 MΩ input resistance measures a 5.0 V Thevenin source that has 10 MΩ internal resistance. What voltage does the meter read?

  1. 2.5 V Correct answer.
  2. 5.0 V
  3. 10 V
  4. 0 V

Explanation

Voltage divider: U_m = 5 × (10M)/(10M+10M) = 2.5 V. Meter loading halves the reading.

Question 49Engineering models, drawings, evidence and project decisions

In the European first-angle projection method used in German technical drawings, where is the view obtained by looking from the LEFT placed relative to the front view?

  1. To the right of the front view Correct answer.
  2. To the left of the front view
  3. Above the front view
  4. Below the front view

Explanation

In first-angle projection (Projektionsmethode 1) the object lies between the observer and the projection plane, so each view is thrown onto the plane BEYOND the object and therefore appears on the side opposite the viewing direction. Looking from the left places that view on the right; looking from above places the plan below.

Question 50Engineering models, drawings, evidence and project decisions

A shaft is dimensioned 40 ± 0.05 mm. What are the limits of size and the tolerance?

  1. 40.05 / 39.95 mm; tolerance 0.10 mm Correct answer.
  2. 40.05 / 39.95 mm; tolerance 0.05 mm
  3. 40.5 / 39.5 mm; tolerance 1.0 mm
  4. 40.05 / 40.00 mm; tolerance 0.05 mm

Explanation

Upper limit of size = 40 + 0.05 = 40.05 mm, lower limit = 40 − 0.05 = 39.95 mm. The tolerance (Maßtoleranz) is the DIFFERENCE between the two limits: 40.05 − 39.95 = 0.10 mm, i.e. twice the individual deviation.