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Aragon PAU Technology and Engineering II - independent engineering foundations study materials
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Aragon PAU Technology and Engineering II - independent engineering foundations practice questions

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Question 1Materials, testing and selection

A cylindrical steel bar with an initial cross-sectional area of A0 = 100 mm² is subjected to an axial tensile force of F = 50 kN. What is the engineering stress (σ) experienced by the bar?

  1. 500 kPa
  2. 500 MPa Correct answer.
  3. 50 MPa
  4. 5 GPa

Explanation

Engineering stress is defined as force divided by initial area: σ = F / A0. Converting units: F = 50,000 N and A0 = 100 × 10^-6 m². Therefore, σ = 50,000 N / (100 × 10^-6 m²) = 500,000,000 Pa = 500 MPa.

Question 2Materials, testing and selection

A tensile specimen with a gauge length of L0 = 200 mm elongates to a length of L = 200.4 mm under an applied load. What is the engineering strain (ε)?

  1. 0.02 (2.0%)
  2. 0.004 (0.4%)
  3. 0.002 (0.2%) Correct answer.
  4. 0.0002 (0.02%)

Explanation

Engineering strain is the ratio of elongation to original length: ε = ΔL / L0. Elongation ΔL = 200.4 - 200 = 0.4 mm. Thus, ε = 0.4 mm / 200 mm = 0.002, or 0.2%.

Question 3Materials, testing and selection

A structural steel rod of diameter d = 10 mm and original length L0 = 0.5 m is subjected to a tensile force F = 16.5 kN within its elastic limit. Given Young's modulus E = 210 GPa for steel, calculate the total elongation (ΔL) of the rod.

  1. 1.00 mm
  2. 2.00 mm
  3. 0.50 mm Correct answer.
  4. 0.25 mm

Explanation

Cross-sectional area A0 = π d² / 4 = π (0.01 m)² / 4 = 7.854 × 10^-5 m². Stress σ = F / A0 = 16,500 N / (7.854 × 10^-5 m²) = 210.08 MPa. Using Hooke's law (ε = σ / E): strain ε = 210.08 × 10^6 Pa / (210 × 10^9 Pa) = 1.00 × 10^-3. Elongation ΔL = L0 × ε = 0.5 m × 1.00 × 10^-3 = 5.0 × 10^-4 m = 0.50 mm.

Question 4Materials, testing and selection

In a Charpy impact test, a heavy pendulum striker of mass m = 20 kg is released from a height h1 = 1.5 m. After fracturing the notched test specimen, the pendulum swings to a maximum height h2 = 0.6 m on the opposite side. Assuming g = 9.8 m/s², what is the impact energy absorbed by the specimen?

  1. 117.6 J
  2. 176.4 J Correct answer.
  3. 411.6 J
  4. 294.0 J

Explanation

Impact energy absorbed equals the difference in potential energy before and after fracture: ΔE = m g (h1 - h2). ΔE = 20 kg × 9.8 m/s² × (1.5 m - 0.6 m) = 196 × 0.9 = 176.4 J.

Question 5Materials, testing and selection

Which type of indenter is used in the Vickers hardness test?

  1. A cylindrical carbide pin with a flat tip
  2. A square-based diamond pyramid with an angle of 136° between opposite faces Correct answer.
  3. A hardened steel or tungsten carbide ball of 10 mm diameter
  4. A diamond cone with an included angle of 120°

Explanation

The Vickers hardness test utilizes a square-based diamond pyramid indenter with a 136° angle between opposite faces. Indentation diagonals are measured optically.

Question 6Materials, testing and selection

What is the primary microstructural goal and effect of quenching (temple) heat treatment applied to carbon steel?

  1. To transform austenite rapidly into martensite, maximizing hardness and tensile strength Correct answer.
  2. To transform martensite into coarse pearlite to maximize ductility
  3. To remove all carbon from the iron lattice via surface decarburization
  4. To produce a pure ferrite matrix with low yield strength

Explanation

Quenching involves heating steel above its critical upper transformation temperature to form austenite, followed by rapid cooling in water or oil. This prevents diffusion and traps carbon in a supersaturated body-centered tetragonal structure known as martensite, conferring high hardness and strength.

Question 7Materials, testing and selection

Why is tempering (revenido) mandatory immediately following quenching in the heat treatment of tool steel?

  1. To melt grain boundaries and weld microscopic internal microcracks
  2. To increase the hardness even further beyond the quenched state
  3. To relieve internal quenching stresses and increase toughness while reducing extreme brittleness Correct answer.
  4. To convert martensite back into 100% untransformed austenite

Explanation

As-quenched martensite is extremely hard but very brittle and contains severe internal stresses. Tempering (reheating below A1 temperature) allows partial carbon diffusion to form tempered martensite, relieving internal stresses and restoring impact toughness and ductility.

Question 8Materials, testing and selection

Full annealing (recocido de regeneración) of steel involves heating above the critical temperature followed by:

  1. Very slow cooling inside the shut-down furnace to obtain maximum softness and machinability Correct answer.
  2. Rapid quenching in cold brine to freeze the crystal structure
  3. Forced air cooling using high-velocity industrial fans
  4. Immediate mechanical forging while red-hot

Explanation

Full annealing requires heating steel above A3/A1 to form austenite, followed by slow cooling within the furnace (often < 20°C/hour). This produces coarse pearlite and ferrite, ensuring minimum hardness, maximum ductility, and optimal machinability.

Question 9Materials, testing and selection

A binary Cu-Ni phase diagram shows complete liquid and solid solubility. An alloy containing 40 wt% Ni is held at 1200°C in a two-phase (Liquid + α) region. The liquid phase contains wL = 32 wt% Ni and the solid α phase contains wα = 50 wt% Ni. Using the lever rule, determine the mass fraction of the liquid phase (WL).

  1. 36.0%
  2. 44.4%
  3. 64.0%
  4. 55.6% Correct answer.

Explanation

By the lever rule, the fraction of liquid WL = (wα - w0) / (wα - wL). Here w0 = 40, wα = 50, and wL = 32. WL = (50 - 40) / (50 - 32) = 10 / 18 = 0.5556 = 55.6%.

Question 10Materials, testing and selection

Which mechanical property describes a material's capacity to absorb energy during dynamic impact loading prior to fracture?

  1. Creep resistance (resistencia a la fluencia)
  2. Elastic modulus (módulo de elasticidad)
  3. Impact resilience (tenacidad) Correct answer.
  4. Surface hardness (dureza)

Explanation

Impact resilience (tenacidad / resiliencia) measures the amount of energy absorbed by a material during dynamic impact or shock loading before fracture occurs, typically evaluated using Charpy or Izod tests.

Question 11Materials, testing and selection

A tensile specimen with an initial area A0 = 50 mm² experiences a tensile load F = 100 kN. Due to necking, the instantaneous actual cross-sectional area decreases to Ai = 40 mm². Compare the engineering stress (σeng) and true stress (σtrue).

  1. σeng = 2.0 GPa, σtrue = 2.5 GPa Correct answer.
  2. σeng = 2.5 GPa, σtrue = 2.0 GPa
  3. σeng = 200 MPa, σtrue = 250 MPa
  4. σeng = 500 MPa, σtrue = 400 MPa

Explanation

Engineering stress σeng = F / A0 = 100,000 N / (50 × 10^-6 m²) = 2.0 × 10^9 Pa = 2.0 GPa. True stress σtrue = F / Ai = 100,000 N / (40 × 10^-6 m²) = 2.5 × 10^9 Pa = 2.5 GPa.

Question 12Materials, testing and selection

Which combination of indenter geometry and major test load is used in the Rockwell C (HRC) hardness test for hardened steels?

  1. 10 mm tungsten carbide ball with a major load of 3000 kgf
  2. 136° square-based diamond pyramid with a load of 30 kgf
  3. 1/16-inch steel ball indenter with a total major load of 100 kgf
  4. 120° diamond cone indenter with a total major load of 150 kgf Correct answer.

Explanation

Rockwell C (HRC) testing utilizes a 120° diamond spheroconical indenter (Brale indenter) with a preliminary minor load of 10 kgf and a total major load of 150 kgf.

Question 13Materials, testing and selection

In steel heat treatment, normalizing (normalizado) differs from full annealing because the heated steel is cooled in:

  1. An oil bath to achieve full martensitic transformation
  2. Still air at room temperature, producing a finer pearlite microstructure Correct answer.
  3. Agitated ice water to maximize residual thermal stress
  4. A vacuum chamber over several days

Explanation

Normalizing involves austenitizing above Ac3 followed by cooling in still air at ambient room temperature. Air cooling is faster than furnace cooling, producing a finer ferrite-pearlite grain structure with higher yield strength.

Question 14Materials, testing and selection

A cylindrical steel tie rod is designed to carry a maximum static tensile load F = 72 kN. The steel alloy has a yield strength σy = 360 MPa. A safety factor n = 2.5 based on yield strength is specified. Calculate the minimum allowable diameter (d) of the tie rod.

  1. 25.2 mm Correct answer.
  2. 16.0 mm
  3. 35.7 mm
  4. 50.5 mm

Explanation

Allowable working stress σallow = σy / n = 360 MPa / 2.5 = 144 MPa = 144 × 10^6 N/m². Required area A = F / σallow = 72,000 N / (144 × 10^6 N/m²) = 5.0 × 10^-4 m². Minimum diameter d = √(4 A / π) = √(4 × 5.0 × 10^-4 / π) = √(6.366 × 10^-4) = 0.02523 m = 25.2 mm.

Question 15Materials, testing and selection

On the Fe-Fe3C equilibrium phase diagram, the eutectoid reaction occurs at 727°C and 0.77 wt% C. What phase mixture is formed?

  1. Martensite (supersaturated tetragonal phase)
  2. Bainite (feather-like ferrite and fine carbides)
  3. Ledeburite (eutectic matrix of austenite and cementite)
  4. Pearlite (alternating lamellae of α-ferrite and Fe3C cementite) Correct answer.

Explanation

The eutectoid reaction involves solid-state decomposition: γ (austenite, 0.77% C) -> α (ferrite, 0.022% C) + Fe3C (cementite, 6.7% C). The characteristic microstructural mixture formed is called Pearlite.

Question 16Materials, testing and selection

What is mechanical creep (fluencia a alta temperatura) in engineering alloys?

  1. Instantaneous plastic deformation when yield strength is exceeded at room temperature
  2. Time-dependent permanent deformation under constant load at elevated temperatures (T > 0.4 Tm) Correct answer.
  3. Brittle fracture caused by high-frequency acoustic vibrations
  4. Surface wear caused by frictional contact between sliding surfaces

Explanation

Creep is slow, progressive, time-dependent plastic deformation that occurs when a material is subjected to persistent mechanical stress at high temperatures (typically above 40% of absolute melting temperature, T > 0.4 Tm).

Question 17Materials, testing and selection

The endurance limit or fatigue limit (límite de fatiga) of a ferrous metal alloy represents:

  1. The energy required to fracture a specimen in one cycle
  2. The minimum stress required to initiate creep at room temperature
  3. The maximum stress amplitude below which the material can endure an infinite number of cyclic loading cycles without failure Correct answer.
  4. The ultimate tensile stress at which necking begins

Explanation

For ferrous metals (steel), the fatigue limit (endurance limit) is the stress level on an S-N curve below which cyclic loading can be repeated infinitely (e.g., > 10^7 cycles) without causing fatigue failure.

Question 18Materials, testing and selection

An underground steel pipeline (iron, Fe) is protected against electrochemical soil corrosion by connecting it electrically to a sacrificial anode block. Which metal is suitable as a sacrificial anode?

  1. Copper (Cu), because it is more noble and resists corrosion
  2. Gold (Au), because it has a high positive reduction potential
  3. Lead (Pb), because it forms an insoluble oxide skin
  4. Zinc (Zn) or Magnesium (Mg), because they have more negative standard reduction potentials than Fe Correct answer.

Explanation

Sacrificial anode cathodic protection requires a metal with a lower (more negative) reduction potential than iron (e.g., Zn, Mg, Al). The sacrificial metal oxidizes preferentially (anode), donating electrons to the steel pipe and keeping it cathodic (protected).

Question 19Statics, 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 20Statics, 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 21Statics, 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 22Statics, 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 23Statics, 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 24Statics, 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 25Statics, 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 26Statics, 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 27Statics, 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 28Statics, 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 29Statics, 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 30Statics, 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 31Statics, 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 32Electrical circuits, machines and digital logic

According to De Morgan's Laws in Boolean algebra, which expression is logically equivalent to the negated AND operation NOT(A AND B), expressed as (A · B)'?

  1. A ⊕ B
  2. A' + B' Correct answer.
  3. A' · B'
  4. (A + B)'

Explanation

De Morgan's First Law states that the complement of a product (AND) equals the sum (OR) of the complements: (A · B)' = A' + B'.

Question 33Electrical circuits, machines and digital logic

What is the Boolean expression and behavior of a two-input Exclusive-OR (XOR) logic gate with inputs A and B?

  1. Y = A + B; output is HIGH when at least one input is HIGH
  2. Y = (A B)' + (A' B'); output is HIGH when inputs A and B are identical
  3. Y = A ⊕ B = A B' + A' B; output is HIGH when inputs A and B are different Correct answer.
  4. Y = A · B; output is HIGH only when both inputs are HIGH

Explanation

XOR produces a HIGH (1) output if and only if one input is HIGH and the other is LOW (inputs differ). Boolean form: Y = A ⊕ B = A B' + A' B.

Question 34Electrical circuits, machines and digital logic

Simplify the Boolean function F = A B + A B' + A' B to its minimal sum-of-products form.

  1. F = A · B
  2. F = A + B Correct answer.
  3. F = A
  4. F = B

Explanation

Group terms: F = A (B + B') + A' B = A (1) + A' B = A + A' B. By distributive/absorption law (A + A' B = A + B), F simplifies to A + B.

Question 35Electrical circuits, machines and digital logic

A 3-variable logic function F(A, B, C) produces output 1 for minterms m1 (001), m3 (011), m5 (101), and m7 (111). Simplify F(A, B, C).

  1. F = A' C + B C
  2. F = C Correct answer.
  3. F = A
  4. F = B + C

Explanation

Canonical SOP: F = A' B' C + A' B C + A B' C + A B C. Factor out C: F = C (A' B' + A' B + A B' + A B) = C (A' (B' + B) + A (B' + B)) = C (A' + A) = C (1) = C.

Question 36Electrical circuits, machines and digital logic

Simplify the 3-variable logic function F(A, B, C) defined by minterms Σm(0, 2, 4, 6) using a Karnaugh map.

  1. F = C
  2. F = A'
  3. F = C' Correct answer.
  4. F = B'

Explanation

Minterms in binary: m0 = 000, m2 = 010, m4 = 100, m6 = 110. In all 4 minterms, C = 0 (so C' = 1) while A and B take all possible combinations (00, 01, 10, 11). Grouping all 4 cells eliminates A and B, giving F = C'.

Question 37Electrical circuits, machines and digital logic

Simplify the 4-variable Boolean expression F(A, B, C, D) = Σm(2, 3, 6, 7, 10, 11, 14, 15) using a Karnaugh map.

  1. F = B + C
  2. F = D
  3. F = C · D
  4. F = C Correct answer.

Explanation

The minterms correspond to all combinations where variable C = 1 (CD = 10 and 11, for all values of AB). A group of 8 adjacent cells where C = 1 eliminates A, B, and D, simplifying directly to F = C.

Question 38Electrical circuits, machines and digital logic

Simplify the 4-variable Boolean function F(A, B, C, D) = Σm(4, 5, 6, 7, 12, 13, 14, 15) using a Karnaugh map.

  1. F = B + D
  2. F = A
  3. F = B Correct answer.
  4. F = B · C

Explanation

Binary representation: m(4..7) = 0100..0111 (A=0, B=1) and m(12..15) = 1100..1111 (A=1, B=1). This forms a block of 8 adjacent cells where B = 1 for all entries, while A, C, and D vary. The simplified result is F = B.

Question 39Electrical circuits, machines and digital logic

In a 4-variable Karnaugh map with inputs A, B, C, D, minterms m0 (0000), m2 (0010), m8 (1000), and m10 (1010) are all 1s. Grouping these four corner cells yields which simplified expression?

  1. F = B' · D' Correct answer.
  2. F = A' · C'
  3. F = B · D
  4. F = A' · D'

Explanation

The four corner cells of a 4-variable K-map are adjacent due to wrap-around properties. In all four corners (m0, m2, m8, m10), B = 0 (so B' = 1) and D = 0 (so D' = 1), while A and C vary. Grouping the 4 corners gives F = B' · D'.

Question 40Electrical circuits, machines and digital logic

What is the primary function of a 2-to-4 Line Decoder combinational circuit?

  1. Adds two 2-bit binary numbers to produce a 4-bit sum
  2. Decodes a 2-bit binary input code to activate exactly one of 4 output lines Correct answer.
  3. Converts 4-bit BCD input to 2-bit binary code
  4. Multiplexes 4 data input lines onto 2 output lines

Explanation

A 2-to-4 line decoder accepts 2 binary select inputs (A1, A0) and decodes them so that exactly one of the 4 output lines (Y0..Y3) is activated corresponding to the decimal value (0 to 3) of the input.

Question 41Electrical circuits, machines and digital logic

A digital encoder circuit performs the reverse operation of a decoder. A 16-to-4 Line Priority Encoder accepts 16 input lines and outputs:

  1. A 4-bit binary code corresponding to the highest-order active input line Correct answer.
  2. 16 parallel output lines containing single-bit values
  3. A single analog voltage proportional to active inputs
  4. A pulse-width modulated square wave

Explanation

A 16-to-4 priority encoder evaluates 16 input lines (I0..I15) and generates a 4-bit binary code (Y3..Y0) representing the index of the active input line with the highest priority.

Question 42Electrical circuits, machines and digital logic

What fundamental structural characteristic distinguishes sequential logic circuits from combinational logic circuits?

  1. Sequential circuits contain memory elements (flip-flops) and feedback loops, so output depends on present inputs and past history Correct answer.
  2. Combinational circuits depend on clock pulses to operate
  3. Sequential circuits contain only AND/OR logic gates without clock inputs
  4. Combinational circuits store past states in internal RAM registers

Explanation

Combinational circuits have no memory; outputs depend solely on instantaneous present inputs. Sequential circuits incorporate memory elements (latches/flip-flops) and feedback, making output dependent on present inputs and past internal state history.

Question 43Electrical circuits, machines and digital logic

In an active-LOW 3-to-8 line decoder (e.g., 74HC138), input code A2 A1 A0 = 101 (decimal 5) is applied while the enable inputs are valid. Which output pin goes LOW (0)?

  1. Y5 Correct answer.
  2. Y1
  3. Y3
  4. Y7

Explanation

Binary input code 101 represents decimal 5. For an active-LOW decoder, output Y5 is driven LOW (0) while all other outputs (Y0..Y4, Y6, Y7) remain HIGH (1).

Question 44Electrical circuits, machines and digital logic

In a basic NOR-gate SR Flip-Flop (Latch), what occurs if both Set (S) and Reset (R) inputs are held HIGH (S = 1, R = 1) simultaneously?

  1. The output Q goes HIGH permanently
  2. The flip-flop toggles its output state on every clock pulse
  3. An invalid/forbidden state occurs where Q and Q' are forced LOW simultaneously, leading to race conditions Correct answer.
  4. The output remains latched in its previous state Qn

Explanation

S = 1 and R = 1 in a NOR latch forces both outputs Q and Q' to 0 simultaneously, violating the complement property (Q' = NOT Q). When S and R return to 0, unpredictable race conditions determine the final state. This is an invalid/forbidden condition.

Question 45Electrical circuits, machines and digital logic

How does a JK Flip-Flop eliminate the invalid input condition of the SR Flip-Flop when both inputs are HIGH (J = 1, K = 1)?

  1. It operates in transparent data pass-through mode
  2. It operates in toggle mode, complementing the output state (Q_{n+1} = Q'_n) on every active clock edge Correct answer.
  3. It forces output Q to 0 unconditionally
  4. It ignores the clock pulse and freezes

Explanation

The JK flip-flop modifies internal feedback so that J = 1, K = 1 results in toggle mode: on each active clock edge, the output complements (Q_{n+1} = Q'_n).

Question 46Electrical circuits, machines and digital logic

What is the characteristic equation and behavior of an edge-triggered D (Data) Flip-Flop?

  1. Q_{n+1} = D; the output Q assumes the logic state of input D on the active clock transition Correct answer.
  2. Q_{n+1} = D ⊕ Q_n; output toggles whenever D = 1
  3. Q_{n+1} = D'; output assumes inverted D input
  4. Q_{n+1} = 1 when D = 0

Explanation

A D flip-flop transfers the logic state at input D to output Q on the active edge of the clock signal (Q_{n+1} = D), serving as a single-bit storage register.

Question 47Electrical circuits, machines and digital logic

Four T flip-flops are connected in series as a 4-bit binary ripple counter. If a clock signal with frequency fin = 16.0 MHz is applied to the first flip-flop input, what is the output frequency (fout) at the final (fourth) stage Q3?

  1. 1.0 MHz Correct answer.
  2. 4.0 MHz
  3. 2.0 MHz
  4. 8.0 MHz

Explanation

Each T flip-flop operating in toggle mode divides frequency by 2. Cascading 4 flip-flops divides the input clock frequency by 2^4 = 16. Fout = fin / 16 = 16.0 MHz / 16 = 1.0 MHz.

Question 48Electrical circuits, machines and digital logic

What is the relationship between the peak amplitude (Vpeak) of a sinusoidal AC voltage waveform and its effective root-mean-square value (Vrms)?

  1. Vrms = Vpeak / √2 ≈ 0.707 Vpeak Correct answer.
  2. Vrms = Vpeak × √2 ≈ 1.414 Vpeak
  3. Vrms = Vpeak / π ≈ 0.318 Vpeak
  4. Vrms = 2 Vpeak

Explanation

For a pure sinusoidal alternating voltage, Vrms = Vpeak / √2 ≈ 0.707 Vpeak. For standard European mains where Vpeak ≈ 325.3 V, Vrms = 230 V.

Question 49Electrical circuits, machines and digital logic

A single-phase AC series RLC circuit consists of a resistor R = 30 Ω, an inductor with inductive reactance XL = 80 Ω, and a capacitor with capacitive reactance XC = 40 Ω. Calculate the total circuit impedance magnitude (Z).

  1. 150 Ω
  2. 70 Ω
  3. 50 Ω Correct answer.
  4. 110 Ω

Explanation

Impedance magnitude for a series RLC circuit: Z = √[R² + (XL - XC)²]. Net reactance X = XL - XC = 80 - 40 = 40 Ω. Z = √[30² + 40²] = √[900 + 1600] = √2500 = 50 Ω.

Question 50Electrical circuits, machines and digital logic

A single-phase AC load is connected to V = 230 V (rms) at 50 Hz and draws an rms current I = 10 A at a lagging power factor cos φ = 0.80 (sin φ = 0.60). Calculate the apparent power (S), active real power (P), and reactive power (Q).

  1. S = 2.30 kVA, P = 1.84 kW, Q = 1.38 kvar Correct answer.
  2. S = 1.84 kVA, P = 2.30 kW, Q = 1.38 kvar
  3. S = 2.30 kVA, P = 2.30 kW, Q = 1.84 kvar
  4. S = 1.38 kVA, P = 1.84 kW, Q = 2.30 kvar

Explanation

Apparent power S = V × I = 230 V × 10 A = 2300 VA = 2.30 kVA. Active power P = S × cos φ = 2300 × 0.80 = 1840 W = 1.84 kW. Reactive power Q = S × sin φ = 2300 × 0.60 = 1380 var = 1.38 kvar.