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Cameroon Baccalauréat Série F3 - independent subject practice study materials
Cameroon Baccalauréat Série F3 - independent subject practice study resources

Cameroon Baccalauréat Série F3 - independent subject practice practice questions

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Question 1Electrical circuits, machines and digital logic

Using Boucherot's method (méthode de Boucherot), how is total active power $P_T$ and total reactive power $Q_T$ calculated for an industrial electrical installation with $n$ parallel loads?

  1. P_T = Σ P_i and Q_T = Σ Q_i, and total apparent power is S_T = √(P_T² + Q_T²) Correct answer.
  2. S_T = Σ S_i directly by algebraic summation
  3. P_T = √(Σ P_i²) and Q_T = √(Σ Q_i²)
  4. P_T = Σ P_i / n and Q_T = Σ Q_i / n

Explanation

Boucherot's theorem states that in any AC network, active powers add algebraically ($P_T = \sum P_i$) and reactive powers add algebraically ($Q_T = \sum Q_i$, accounting for inductive $+Q$ and capacitive $-Q$). Apparent powers cannot be added directly algebraically; total apparent power is $S_T = \sqrt{P_T^2 + Q_T^2}$.

Question 2Electrical circuits, machines and digital logic

A 3-phase industrial load consumes an active power $P = 45\text{ kW}$ with a lagging power factor $\cos \varphi = 0.80$ when supplied by a $U = 400\text{ V}$ three-phase line. What is the line current $I$ drawn by the load?

  1. 81.2 A Correct answer.
  2. 65.0 A
  3. 140.6 A
  4. 112.5 A

Explanation

Active power in a three-phase balanced system is $P = \sqrt{3} \cdot U \cdot I \cdot \cos \varphi$. Rearranging for line current: $I = \frac{P}{\sqrt{3} \cdot U \cdot \cos \varphi} = \frac{45000\text{ W}}{\sqrt{3} \times 400\text{ V} \times 0.80} = \frac{45000}{1.73205 \times 400 \times 0.80} = \frac{45000}{554.256} \approx 81.19\text{ A} \approx 81.2\text{ A}$.

Question 3Electrical circuits, machines and digital logic

What is the reactive power $Q_c$ required from a capacitor bank to improve the power factor of a factory drawing $P = 100\text{ kW}$ from $\cos \varphi_1 = 0.70$ ($\tan \varphi_1 = 1.020$) to $\cos \varphi_2 = 0.95$ ($\tan \varphi_2 = 0.329$)?

  1. 69.1 kvar Correct answer.
  2. 102.0 kvar
  3. 32.9 kvar
  4. 134.9 kvar

Explanation

The capacitor bank sizing formula is $Q_c = P \cdot (\tan \varphi_1 - \tan \varphi_2) = 100\text{ kW} \times (1.020 - 0.329) = 100 \times 0.691 = 69.1\text{ kvar}$.

Question 4Electrical circuits, machines and digital logic

A 3-phase distribution sub-panel is fed by a 50-meter 4-core copper cable ($S = 25\text{ mm}^2$, resistivity $\rho = 0.0225\text{ }\Omega\cdot\text{mm}^2/\text{m}$). What is the resistance $R$ of each phase conductor?

  1. 0.045 Ω Correct answer.
  2. 0.090 Ω
  3. 0.0225 Ω
  4. 0.450 Ω

Explanation

Resistance of a single phase conductor is $R = \rho \cdot \frac{L}{S} = 0.0225\text{ }\Omega\cdot\text{mm}^2/\text{m} \times \frac{50\text{ m}}{25\text{ mm}^2} = 0.0225 \times 2 = 0.045\text{ }\Omega$.

Question 5Electrical circuits, machines and digital logic

In a 3-phase balanced distribution line carrying current $I = 60\text{ A}$ at $\cos \varphi = 0.85$ ($\sin \varphi = 0.527$), if the line resistance per phase is $R = 0.05\text{ }\Omega$ and reactance is $X = 0.02\text{ }\Omega$, what is the line-to-line voltage drop $\Delta U$?

  1. 5.51 V Correct answer.
  2. 3.18 V
  3. 9.54 V
  4. 1.84 V

Explanation

The line-to-line voltage drop in a 3-phase system is $\Delta U = \sqrt{3} \cdot I \cdot (R \cos \varphi + X \sin \varphi)$. Substituting: $\Delta U = \sqrt{3} \times 60 \times (0.05 \times 0.85 + 0.02 \times 0.527) = 1.73205 \times 60 \times (0.0425 + 0.01054) = 103.923 \times 0.05304 \approx 5.512\text{ V} \approx 5.51\text{ V}$.

Question 6Electrical circuits, machines and digital logic

A $20\text{ kV} / 400\text{ V}$ three-phase distribution transformer has a rated apparent power $S_n = 630\text{ kVA}$. What is the rated secondary nominal current $I_{n2}$ at $U_0 = 400\text{ V}$?

  1. 909.3 A (~909 A) Correct answer.
  2. 1575 A
  3. 525 A
  4. 31.5 A

Explanation

Secondary nominal current is $I_{n2} = \frac{S_n}{\sqrt{3} \cdot U_0} = \frac{630000\text{ VA}}{\sqrt{3} \times 400\text{ V}} = \frac{630000}{692.82} \approx 909.33\text{ A} \approx 909\text{ A}$.

Question 7Electrical circuits, machines and digital logic

In single-phase AC calculations, a circuit draws a current $I = 10\text{ A}$ under voltage $V = 230\text{ V}$ with a power factor $\cos \varphi = 0.60$ inductive. Calculate the active power $P$, reactive power $Q$, and apparent power $S$.

  1. P = 1380 W, Q = 1840 var, S = 2300 VA Correct answer.
  2. P = 2300 W, Q = 1380 var, S = 1840 VA
  3. P = 1840 W, Q = 1380 var, S = 2300 VA
  4. P = 1380 W, Q = 0 var, S = 1380 VA

Explanation

$S = V \cdot I = 230\text{ V} \times 10\text{ A} = 2300\text{ VA}$. $P = S \cdot \cos \varphi = 2300 \times 0.60 = 1380\text{ W}$. Since $\cos \varphi = 0.60$, $\sin \varphi = \sqrt{1 - 0.6^2} = 0.80$. $Q = S \cdot \sin \varphi = 2300 \times 0.80 = 1840\text{ var}$. Check: $\sqrt{1380^2 + 1840^2} = \sqrt{1904400 + 3385600} = \sqrt{5290000} = 2300\text{ VA}$.

Question 8Electrical circuits, machines and digital logic

What is the purpose of applying the diversity factor (facteur de foisonnement / simultanéité $k_s$) and utilization factor ($k_u$) when sizing an industrial main distribution panel?

  1. To avoid over-sizing transformers and busbars by recognizing that not all connected machines operate simultaneously at full rated power at the exact same instant Correct answer.
  2. To double the cross-sectional area of the neutral conductor for safety
  3. To step down the distribution frequency from 60 Hz to 50 Hz
  4. To compensate for reactive harmonics in LED lighting

Explanation

In real installations, equipment does not run at $100\%$ load continuously ($k_u < 1$), nor do all circuits operate simultaneously ($k_s < 1$). Applying $P_{design} = \sum (P_i \cdot k_u) \cdot k_s$ yields realistic maximum peak demand, preventing wasteful over-sizing of transformers, cables, and switchgear.

Question 9Electrical circuits, machines and digital logic

In a 3-phase asynchronous induction motor nameplate marked `230 V / 400 V`, how must the motor stator stator windings be connected when powered by a $U = 400\text{ V}$ 3-phase mains grid?

  1. Star connection (Couplage Étoile Y) Correct answer.
  2. Delta connection (Couplage Triangle Δ)
  3. Direct single-phase connection with starting capacitor
  4. Zig-zag connection

Explanation

The smaller voltage ($230\text{ V}$) is the maximum voltage permissible across each individual stator winding phase. In a $400\text{ V}$ grid, a Star ($Y$) connection applies $V = U / \sqrt{3} = 400 / 1.732 = 230\text{ V}$ across each phase winding. Connecting in Delta on a 400 V grid would apply 400 V across 230 V windings, burning the insulation.

Question 10Electrical circuits, machines and digital logic

A 4-pole three-phase asynchronous induction motor is connected to a $50\text{ Hz}$ supply. Under full load, the rotor speed is $N = 1440\text{ rpm}$. What is the motor slip $s$ (glissement)?

  1. 4% (s = 0.04) Correct answer.
  2. 6% (s = 0.06)
  3. 2% (s = 0.02)
  4. 10% (s = 0.10)

Explanation

Synchronous speed for a 4-pole ($p = 2$ pole pairs) motor at 50 Hz is $N_s = \frac{60 \cdot f}{p} = \frac{60 \times 50}{2} = 1500\text{ rpm}$. The slip is $s = \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = \frac{60}{1500} = 0.04 = 4\%$.

Question 11Electrical circuits, machines and digital logic

A 3-phase induction motor produces an electromagnetic power $P_{em} = 15\text{ kW}$ transmitted across the air gap. Operating at slip $s = 0.05$, what are the rotor Joule losses ($P_{jr}$) dissipated as heat in the rotor bars?

  1. 750 W (0.75 kW) Correct answer.
  2. 14.25 kW
  3. 150 W
  4. 1500 W

Explanation

Rotor Joule losses are fundamentally linked to electromagnetic air gap power by $P_{jr} = s \cdot P_{em}$. For $s = 0.05$: $P_{jr} = 0.05 \times 15000\text{ W} = 750\text{ W}$. (The remaining total mechanical power developed is $P_{mech} = P_{em}(1 - s) = 15000 \times 0.95 = 14.25\text{ kW}$).

Question 12Electrical circuits, machines and digital logic

A single-phase transformer has a primary winding $N_1 = 1000\text{ turns}$ and secondary winding $N_2 = 100\text{ turns}$. In an open-circuit test with $U_1 = 2300\text{ V}$, it draws $P_{10} = 450\text{ W}$. In a short-circuit test with rated secondary current $I_{2n} = 50\text{ A}$, it consumes $P_{1cc} = 625\text{ W}$. What is the secondary winding resistance $R_s$ referred to the secondary?

  1. 0.25 Ω Correct answer.
  2. 0.125 Ω
  3. 1.25 Ω
  4. 2.50 Ω

Explanation

In a short-circuit test, core losses are negligible and all power consumed is total Joule loss: $P_{1cc} = R_s \cdot I_{2n}^2 \implies R_s = \frac{P_{1cc}}{I_{2n}^2} = \frac{625\text{ W}}{(50\text{ A})^2} = \frac{625}{2500} = 0.25\text{ }\Omega$.

Question 13Electrical circuits, machines and digital logic

Using Kapp's simplified formula for transformer secondary voltage drop $\Delta U_2 = R_s I_2 \cos \varphi_2 + X_s I_2 \sin \varphi_2$, what type of load can produce a negative voltage drop ($\Delta U_2 < 0$), causing the secondary terminal voltage to RISE under load ($U_2 > U_{20}$)?

  1. A leading capacitive load (charge capacitive avec sin φ₂ < 0) Correct answer.
  2. A heavy lagging inductive motor load
  3. A pure resistive heating load (cos φ = 1.0)
  4. An open-circuit zero-load condition

Explanation

For a capacitive load, current leads voltage, so the reactive phase angle is negative ($\sin \varphi_2 < 0$). If $|X_s \sin \varphi_2| > R_s \cos \varphi_2$, the term becomes negative ($\Delta U_2 < 0$), producing the Ferranti effect where output terminal voltage rises above the no-load EMF ($U_2 = U_{20} - \Delta U_2 > U_{20}$).

Question 14Electrical circuits, machines and digital logic

A DC shunt motor has an armature resistance $R_a = 0.5\text{ }\Omega$ and is supplied by $U = 220\text{ V}$. When running at nominal speed, it generates a back-EMF $E = 205\text{ V}$. What is the nominal armature current $I_a$, and what would be the direct starting current $I_{a,start}$ if started without a starting rheostat?

  1. I_a = 30 A; I_a,start = 440 A (nearly 15 times nominal current) Correct answer.
  2. I_a = 15 A; I_a,start = 220 A
  3. I_a = 440 A; I_a,start = 30 A
  4. I_a = 30 A; I_a,start = 30 A

Explanation

Nominal current is $I_a = \frac{U - E}{R_a} = \frac{220\text{ V} - 205\text{ V}}{0.5\text{ }\Omega} = \frac{15}{0.5} = 30\text{ A}$. At starting standstill ($N = 0$), back-EMF is zero ($E = 0$), so $I_{a,start} = \frac{U}{R_a} = \frac{220\text{ V}}{0.5\text{ }\Omega} = 440\text{ A}$, proving why starting resistors or electronic soft starters are mandatory.

Question 15Electrical circuits, machines and digital logic

In a 3-phase synchronous generator (alternateur synchrone), what is the relationship between the mechanical rotational speed $N$ (rpm), electrical grid frequency $f$ (Hz), and the number of pole pairs $p$?

  1. f = \frac{p \cdot N}{60} Correct answer.
  2. f = \frac{N}{60 \cdot p}
  3. f = \frac{60 \cdot p}{N}
  4. f = p \cdot N \cdot 60

Explanation

In synchronous machines, the stator magnetic field rotates at exact synchronism with the rotor: $N = \frac{60 f}{p} \implies f = \frac{p \cdot N}{60}$. For example, a 2-pole ($p = 1$) generator running at 3000 rpm generates $f = \frac{1 \times 3000}{60} = 50\text{ Hz}$.

Question 16Electrical circuits, machines and digital logic

A 3-phase star-connected asynchronous motor rated $15\text{ kW}, 400\text{ V}, 50\text{ Hz}$ is started using a Star-Delta ($Y-\Delta$) starter. How do the line starting current ($I_{start}$) and starting torque ($T_{start}$) in star ($Y$) compare to direct-on-line (DOL) delta starting?

  1. Both starting current and starting torque are reduced by a factor of 3 (divided by 3 / reduced to 33%) Correct answer.
  2. Starting current is divided by 3, but starting torque remains 100% full torque
  3. Starting current is divided by √3 (~58%), and torque is divided by 3
  4. Starting current is halved, and torque is quadrupled

Explanation

In star connection, phase voltage is reduced by $1/\sqrt{3}$ ($230\text{ V}$ instead of $400\text{ V}$). Phase current drops by $1/\sqrt{3}$, and line current drops by an additional $1/\sqrt{3}$, resulting in $I_{startY} = \frac{1}{3} I_{start\Delta}$. Since torque is proportional to the square of voltage ($T \propto V^2$), starting torque drops by $(1/\sqrt{3})^2 = 1/3$ ($T_{startY} = \frac{1}{3} T_{start\Delta}$).

Question 17Electrical circuits, machines and digital logic

In a single-phase controlled thyristor rectifier with firing delay angle $\alpha$, how does the average DC output voltage $V_{dc}(\alpha)$ vary with $\alpha$ for continuous conduction?

  1. V_{dc}(\alpha) = V_{dc0} \cdot \cos(\alpha) Correct answer.
  2. V_{dc}(\alpha) = V_{dc0} \cdot \sin(\alpha)
  3. V_{dc}(\alpha) = V_{dc0} \cdot (1 - \alpha)
  4. V_{dc}(\alpha) = V_{dc0} / \cos(\alpha)

Explanation

For a fully controlled bridge in continuous conduction, integrating the shifted sine wave yields $V_{dc}(\alpha) = \frac{2 V_{max}}{\pi} \cos(\alpha) = V_{dc0} \cos(\alpha)$. When $0^\circ \le \alpha < 90^\circ$, $V_{dc} > 0$ (rectifier mode); when $90^\circ < \alpha < 180^\circ$, $V_{dc} < 0$ (line-commutated inverter mode / onduleur assisté).

Question 18Electrical circuits, machines and digital logic

A DC step-down chopper (hacheur série / buck converter) operates with a DC supply $E = 100\text{ V}$ and a switching duty cycle $\alpha = 0.40$. What is the average output voltage $U_{avg}$ across the load?

  1. 40 V (U_avg = α · E) Correct answer.
  2. 60 V
  3. 250 V
  4. 25 V

Explanation

In a step-down buck chopper, the output voltage during period $T$ is $E$ for time $\alpha T$ and $0$ for time $(1-\alpha)T$. The average output voltage is $U_{avg} = \frac{1}{T} \int_0^{\alpha T} E \, dt = \alpha \cdot E = 0.40 \times 100\text{ V} = 40\text{ V}$.

Question 19Electrical circuits, machines and digital logic

What is the primary function of a Variable Frequency Drive (VFD / variateur de fréquence) in modern AC motor speed control?

  1. It rectifies AC into DC, then uses a PWM inverter to synthesize variable-frequency, variable-voltage AC power while maintaining a constant V/f ratio to preserve rated motor torque Correct answer.
  2. It inserts external resistance into the squirrel-cage rotor bars
  3. It shifts the number of physical pole windings using mechanical relays
  4. It steps up voltage from 400 V to 20 kV for transmission

Explanation

A modern VFD converts fixed $50\text{ Hz}$ AC into DC via a diode bridge, filters it through a DC bus link, and uses an IGBT inverter with sinusoidal PWM to generate variable voltage and frequency ($0-100+\text{ Hz}$). By keeping the ratio $V/f = \text{constant}$, stator magnetic flux is maintained at rated value, delivering full rated torque across the speed range.

Question 20Electrical circuits, machines and digital logic

What is Total Harmonic Distortion (THD / Taux de Distorsion Harmonique) in an electrical power system with fundamental RMS current $I_1$ and harmonic currents $I_2, I_3, I_4, \dots$?

  1. THD = \frac{\sqrt{\sum_{n=2}^\infty I_n^2}}{I_1} Correct answer.
  2. THD = \frac{I_1}{\sum_{n=2}^\infty I_n}
  3. THD = \frac{I_2 + I_3}{I_1}
  4. THD = \sqrt{I_1^2 + I_2^2}

Explanation

Total Harmonic Distortion (THD) measures the harmonic distortion present in a signal, defined as the ratio of the RMS sum of all harmonic components above the fundamental to the RMS amplitude of the fundamental component: $\text{THD} = \frac{\sqrt{I_2^2 + I_3^2 + I_4^2 + \dots}}{I_1}$.

Question 21Electrical circuits, machines and digital logic

In a 3-phase system feeding heavy non-linear loads (such as computers, LED drivers, and single-phase switch-mode power supplies), why does the neutral conductor often carry severe high currents even when phases are balanced?

  1. Triplen harmonics (3rd, 9th, 15th, ...) are in-phase across all three phases and add arithmetically in the neutral conductor: I_N ≈ 3 · I₃ Correct answer.
  2. The neutral conductor acts as an antenna capturing radio waves
  3. The power factor becomes negative and forces current backwards
  4. The earth ground rod develops excessive electrostatic charge

Explanation

Because the three phases are separated by $120^\circ$, third harmonic currents ($3 \times 120^\circ = 360^\circ = 0^\circ$) are zero-sequence and perfectly in-phase with each other. In the neutral wire, instead of canceling out, the third harmonic currents from all three phases sum constructively ($I_N \approx 3 I_3$), often exceeding $100-140\%$ of phase current.

Question 22Electrical circuits, machines and digital logic

A boost converter (hacheur parallèle élévateur) steps up an input voltage $V_{in} = 48\text{ V}$ to an output voltage $V_{out} = 120\text{ V}$. What is the required duty cycle $\alpha$ (assuming ideal lossless continuous conduction)?

  1. 0.60 (60%) Correct answer.
  2. 0.40 (40%)
  3. 0.25 (25%)
  4. 0.75 (75%)

Explanation

The boost converter voltage conversion formula is $V_{out} = \frac{V_{in}}{1 - \alpha}$. Rearranging: $1 - \alpha = \frac{V_{in}}{V_{out}} = \frac{48}{120} = 0.40 \implies \alpha = 1 - 0.40 = 0.60 = 60\%$.

Question 23Electrical circuits, machines and digital logic

What is the function of a snubber circuit (circuit d'aide à la commutation RC) connected across a power thyristor or TRIAC?

  1. To limit the rate of rise of off-state voltage (dv/dt), preventing spurious un-commanded turn-on and protecting the device from transient overvoltage spikes Correct answer.
  2. To provide DC gate firing current continuously
  3. To convert the thyristor into an optocoupler
  4. To cool the heat sink by Peltier effect

Explanation

If the rate of rise of forward voltage ($dv/dt$) across a thyristor is too steep, internal capacitive displacement currents in the junction can trigger regenerative gate turn-on without a gate pulse. A parallel RC snubber absorbs transient $dv/dt$ and dampens voltage spikes during turn-off recovery.

Question 24Electrical circuits, machines and digital logic

What is the synchronization condition required before connecting a 3-phase synchronous alternator in parallel to an energized electrical utility grid?

  1. Equal voltages (U_alt = U_grid), equal frequencies (f_alt = f_grid), identical phase sequence (L1-L2-L3), and phase concordance (in-phase / phase angle Δφ = 0) Correct answer.
  2. Alternator voltage must be exactly double the grid voltage
  3. Alternator frequency must be at least 100 Hz
  4. All three alternator phase lines must be grounded during coupling

Explanation

Paralleling an alternator to a live grid (couplage au réseau) requires four strict synchronization conditions: 1) Equal RMS terminal voltages; 2) Equal frequencies; 3) Identical phase rotation direction; 4) Zero phase angle difference (monitored via synchroscope or dark lamp method). Violating these conditions causes violent electrical short-circuits and mechanical shaft fracture.

Question 25Electrical circuits, machines and digital logic

A $400\text{ V}$ 3-phase induction motor has an efficiency $\eta = 0.88$ and power factor $\cos \varphi = 0.82$ when delivering a rated mechanical shaft power $P_{useful} = 11\text{ kW}$. What is the total active electrical power $P_{elec}$ consumed from the grid?

  1. 12.5 kW Correct answer.
  2. 9.68 kW
  3. 11.0 kW
  4. 15.2 kW

Explanation

Efficiency is defined as $\eta = \frac{P_{useful}}{P_{elec}} \implies P_{elec} = \frac{P_{useful}}{\eta} = \frac{11\text{ kW}}{0.88} = 12.5\text{ kW}$.

Question 26Electrical circuits, machines and digital logic

In transformer vector group designations (indice horaire, e.g. 'Dyn11'), what does 'Dyn11' specify?

  1. Primary connected in Delta (D), Secondary in Star with neutral (yn), with secondary voltage lagging primary by 330° (or leading by 30° / clock position 11 = 11 × 30°) Correct answer.
  2. Direct current transformer with 11 windings
  3. Delta primary, Star secondary, 11 kV insulation rating
  4. Dynamic auto-transformer with 11 tap positions

Explanation

In IEC 60076 transformer notation: 'D' = High-voltage primary in Delta; 'y' = Low-voltage secondary in Star; 'n' = neutral brought out; '11' = clock hour number representing phase displacement angle: Phase displacement $= 11 \times 30^\circ = 330^\circ$ (secondary line voltage leads primary by $+30^\circ$).

Question 27Electrical circuits, machines and digital logic

Why is an open-circuit secondary condition strictly forbidden on an energized Current Transformer (Transformateur de Courant - TC / CT)?

  1. Because without secondary demagnetizing ampere-turns (N₂·I₂), the entire primary current acts as magnetizing current, producing lethal kilovolt secondary voltages and core saturation Correct answer.
  2. Because the primary fuse will immediately blow
  3. Because current transformers only work when connected to DC batteries
  4. Because the secondary winding cools down to freezing temperatures

Explanation

In a CT, primary current is dictated by the main power load. Under normal operation, secondary current $N_2 I_2$ opposes primary ampere-turns $N_1 I_1$, leaving a small net magnetizing flux $\Phi$. If the secondary is opened while primary carries load, the opposing ampere-turns disappear ($I_2 = 0$). All primary current magnetizes the core, driving core saturation and inducing lethal voltages ($>1000-5000\text{ V}$) across open secondary terminals, risking explosion and fatal shock.

Question 28Electrical circuits, machines and digital logic

What is the primary function of an isolator switch (sectionneur) in medium- and low-voltage electrical distribution switchgear?

  1. To provide visible mechanical physical isolation of a de-energized circuit with guaranteed dielectric distance, having NO breaking capacity under load Correct answer.
  2. To automatically interrupt short-circuit currents up to 100 kA
  3. To regulate output voltage automatically under varying load
  4. To measure ground earth loop impedance continuously

Explanation

A sectionneur (isolator / disconnector) is an off-load safety device that provides visible contact separation (coupure pleinement apparente) to guarantee personnel safety during maintenance. It has NO load breaking capacity and must NEVER be opened under load (an arc would form and destroy the switchgear). Load interruption is performed by contactors or circuit breakers.

Question 29Electrical circuits, machines and digital logic

A $400\text{ V}$ 3-phase induction motor has stator winding resistance $R_1 = 0.4\text{ }\Omega$ per phase and rotor resistance referred to stator $R'_2 = 0.3\text{ }\Omega$ per phase, with total leakage reactance $X = 2.0\text{ }\Omega$ per phase. At what slip $s_{max}$ does the motor develop its maximum breakdown torque ($T_{max}$)?

  1. s_max ≈ 0.147 (s_max = R'_2 / √(R_1² + X²)) Correct answer.
  2. s_max = 0.040
  3. s_max = 1.000
  4. s_max = 0.750

Explanation

From the exact equivalent circuit of an induction motor, maximum breakdown torque occurs at slip $s_{max} = \frac{R'_2}{\sqrt{R_1^2 + X^2}}$. Substituting values: $s_{max} = \frac{0.3}{\sqrt{0.4^2 + 2.0^2}} = \frac{0.3}{\sqrt{0.16 + 4.0}} = \frac{0.3}{\sqrt{4.16}} = \frac{0.3}{2.0396} \approx 0.1471 = 14.7\%$.

Question 30Electrical circuits, machines and digital logic

In a 3-phase controlled thyristor rectifier feeding a DC motor, what is the critical condition required to operate the bridge in regenerative inverting mode (onduleur non autonome / régénération sur le réseau)?

  1. The motor back-EMF must be reversed (or armature connections reversed), firing angle must be α > 90°, and an inductive filter must maintain continuous current (Id > 0) Correct answer.
  2. The AC mains supply must be disconnected completely
  3. The firing angle must be set to α = 0°
  4. The diodes must be cooled to liquid nitrogen temperatures

Explanation

Inverter operation converts DC mechanical/braking energy back into the AC grid. Because thyristors can only conduct current in one direction ($I_d > 0$), power flow reversal ($P = V_{dc} \cdot I_d < 0$) requires reversing the polarity of DC voltage: $V_{dc} = V_{dc0} \cos(\alpha) < 0$, which occurs when firing angle $90^\circ < \alpha < 150^\circ-165^\circ$ (maintaining a safety margin $\gamma$ to prevent commutation failure).

Question 31Electrical circuits, machines and digital logic

In a 3-phase asynchronous motor speed control application using a VFD above rated base frequency (e.g. from 50 Hz up to 100 Hz), why does the motor enter the 'Field Weakening / Constant Power' region (zone de défluxage à puissance constante)?

  1. Because the inverter output voltage cannot exceed the maximum available grid DC bus rail voltage (V = V_max = constant), so magnetic flux drops inversely with frequency (Φ ∝ V/f), causing available torque to decrease as 1/f Correct answer.
  2. Because the motor runs backwards above 50 Hz
  3. Because rotor resistance drops to zero at high speeds
  4. Because the supply power factor drops to zero

Explanation

Below 50 Hz, the VFD maintains constant flux $\Phi \propto V/f = \text{constant}$ by increasing voltage with frequency (constant torque region). Above 50 Hz, voltage reaches the maximum line ceiling ($400\text{ V}$) and cannot increase further ($V = V_{max}$). As frequency $f$ continues rising, flux weakens ($\Phi \propto V_{max}/f$), motor torque drops ($T \propto 1/f$), and output power remains constant ($P = T \cdot \omega = \text{constant}$).

Question 32Electrical circuits, machines and digital logic

A factory has a 3-phase transformer supplying $U = 400\text{ V}$. An inductive load draws active power $P_1 = 300\text{ kW}$ at $\cos \varphi_1 = 0.60$ (lagging). An over-excited synchronous motor is added, consuming active power $P_2 = 100\text{ kW}$ at a leading power factor $\cos \varphi_2 = 0.80$ (leading / capacitive). What is the new total overall power factor $\cos \varphi_T$ of the combined installation?

  1. 0.78 lagging (P_T = 400 kW, Q_T = 325 kvar, S_T = 515.4 kVA) Correct answer.
  2. 0.95 leading
  3. 0.60 lagging
  4. 1.00 pure unity

Explanation

For load 1: $P_1 = 300\text{ kW}$, $\tan \varphi_1 = \tan(\arccos 0.60) = 1.3333 \implies Q_1 = 300 \times 1.3333 = +400\text{ kvar}$. For load 2 (over-excited synchronous motor): $P_2 = 100\text{ kW}$, $\tan \varphi_2 = -\tan(\arccos 0.80) = -0.75 \implies Q_2 = 100 \times (-0.75) = -75\text{ kvar}$ (capacitive VAR generation). Total: $P_T = P_1 + P_2 = 400\text{ kW}$; $Q_T = Q_1 + Q_2 = 400 - 75 = +325\text{ kvar}$. Total apparent power: $S_T = \sqrt{400^2 + 325^2} = \sqrt{160000 + 105625} = \sqrt{265625} \approx 515.39\text{ kVA}$. New power factor: $\cos \varphi_T = \frac{P_T}{S_T} = \frac{400}{515.39} \approx 0.776$, i.e. $0.78$ lagging - the over-excited synchronous motor has lifted the plant power factor from 0.60 to 0.78 without any capacitor bank.

Question 33Electrical circuits, machines and digital logic

A single-phase $230\text{ V} / 24\text{ V}$ step-down transformer has a nominal rating $S_n = 240\text{ VA}$. What is the rated full-load secondary current $I_{2n}$?

  1. 10.0 A Correct answer.
  2. 1.04 A
  3. 24.0 A
  4. 5.76 A

Explanation

Secondary nominal current is $I_{2n} = \frac{S_n}{U_{2n}} = \frac{240\text{ VA}}{24\text{ V}} = 10.0\text{ A}$. (Primary current is $I_{1n} = \frac{240\text{ VA}}{230\text{ V}} \approx 1.04\text{ A}$).

Question 34Electrical circuits, machines and digital logic

In electrical machines, what causes the 'Cogging / Detent Torque' phenomenon in permanent magnet brushless DC (BLDC) motors?

  1. Magnetic attraction between the permanent rotor magnets and the stator iron teeth slots when no current is applied Correct answer.
  2. Friction between the carbon brushes and the copper commutator
  3. Harmonic distortion in the PWM inverter switching carrier
  4. Bearing misalignment under heavy radial belt tension

Explanation

Cogging torque is the non-uniform magnetic reluctance torque resulting from the interaction between rotor permanent magnets and stator iron slots. It causes the rotor to prefer discrete angular detent positions at zero current, which is minimized in motor design by skewing stator slots or rotor magnets.

Question 35Electrical circuits, machines and digital logic

A $100\text{ kVA}$ 3-phase diesel generator set operates on a mining site in Cameroon with ambient temperature $T_{amb} = 40^\circ\text{C}$ and altitude $h = 1000\text{ m}$. If the altitude de-rating factor is $k_{alt} = 0.96$ and temperature de-rating factor is $k_{temp} = 0.92$, what is the actual available continuous electrical capacity?

  1. 88.3 kVA (S_actual = S_n · k_alt · k_temp) Correct answer.
  2. 96.0 kVA
  3. 92.0 kVA
  4. 100.0 kVA

Explanation

Actual available generator power is de-rated for ambient conditions: $S_{actual} = S_n \cdot k_{alt} \cdot k_{temp} = 100\text{ kVA} \times 0.96 \times 0.92 = 100 \times 0.8832 = 88.32\text{ kVA} \approx 88.3\text{ kVA}$.

Question 36Measurement, feedback, PLC and automation

In a standard start/stop latching contactor control circuit (auto-maintien), why is a normally open auxiliary contact (KM1: 13-14) wired in parallel across the start pushbutton (NO: S2)?

  1. To latch the contactor coil energized after the momentary start pushbutton is released (seal-in circuit) Correct answer.
  2. To illuminate a warning pilot lamp when the motor stops
  3. To discharge residual static voltage from the stop button
  4. To provide speed feedback to the PLC

Explanation

When momentary push-button $S2$ is pressed, coil $KM1$ energizes, closing power contacts and auxiliary contact `13-14`. When $S2$ is released, current continues flowing to the coil through the closed `13-14` contact (auto-maintien / seal-in latch) until stop button $S1$ is pressed.

Question 37Measurement, feedback, PLC and automation

In a reversing motor starter with two contactors KM1 (Forward) and KM2 (Reverse), why is mechanical and electrical interlocking mandatory?

  1. To prevent both contactors from closing simultaneously, which would short-circuit two supply line phases (L1 and L3) together Correct answer.
  2. To ensure that the motor runs at double speed in reverse
  3. To allow single-phase operation when one fuse blows
  4. To balance the mechanical load on the motor bearings

Explanation

Reversing rotation requires swapping two line phases ($L1$ and $L3$). If $KM1$ and $KM2$ were both energized at the same time, an unrestricted phase-to-phase dead short-circuit would occur. Electrical interlocks wire a normally closed contact of $KM1$ in series with $KM2$'s coil, and vice-versa, supplemented by a physical mechanical lock bar.

Question 38Measurement, feedback, PLC and automation

In an industrial PLC programming standard (IEC 61131-3), which programming language uses graphical rungs with normally open contacts, normally closed contacts, and output coils modeled after relay schematics?

  1. Ladder Diagram (LD / Langage à contacts) Correct answer.
  2. Structured Text (ST / Texte structuré)
  3. Instruction List (IL / Liste d'instructions)
  4. Function Block Diagram (FBD / Schéma blocs)

Explanation

Ladder Diagram (LD) is a graphical language that represents logic as rungs between power rails using relay contact symbols (NO `| |`, NC `|/|`, Coils `( )`), making it intuitive for electrical maintenance technicians.

Question 39Measurement, feedback, PLC and automation

In GRAFCET sequence control, what is a 'Level 1 GRAFCET' (GRAFCET point de vue système / partie opérative) compared to a 'Level 2 GRAFCET' (point de vue partie commande)?

  1. Level 1 describes physical process tasks in plain operational terms without specifying exact sensor/actuator technology, while Level 2 specifies exact technical actuators (KM1, 1M1), sensors (S1, 1S2), and hardware addresses Correct answer.
  2. Level 1 is only for pneumatic circuits, while Level 2 is only for electrical motors
  3. Level 1 can only have 1 step, while Level 2 has 2 steps
  4. Level 1 is written in binary C-code, while Level 2 is written on paper

Explanation

Level 1 GRAFCET (spécification fonctionnelle) describes system actions from the process perspective (e.g. 'Advance carriage', 'Drill hole'). Level 2 GRAFCET (spécification technologique) introduces the specific automation hardware (e.g. action `1M1+`, receptivity `1S2 · a1`).

Question 40Measurement, feedback, PLC and automation

What is the function of a soft-starter (démarreur progressif électronique) for 3-phase induction motors?

  1. It uses back-to-back thyristor phase control to ramp up stator voltage smoothly from an initial pedestal, reducing starting current inrush and eliminating mechanical gear shock Correct answer.
  2. It changes the electrical frequency from 0 to 400 Hz continuously
  3. It operates as a mechanical centrifugal clutch
  4. It converts 3-phase AC into DC for regenerative battery charging

Explanation

A solid-state soft starter uses antiparallel thyristor pairs in each phase. By gradually advancing the firing angle $\alpha$ during starting, the applied AC voltage ramps up smoothly from $\sim 30\%$ to $100\%$, limiting starting current to $2-3 I_n$ and eliminating torque shocks on belts, gearboxes, and pumps. Once at full speed, an internal bypass contactor closes to eliminate thyristor heat dissipation.

Question 41Measurement, feedback, PLC and automation

In a 3-phase induction motor, how is the direction of rotation reversed?

  1. By interchanging any two of the three incoming stator line supply conductors (e.g. swap L1 and L2) Correct answer.
  2. By swapping the neutral and earth connections
  3. By increasing the supply voltage by 15%
  4. By reversing the direction of current in the rotor shorting rings

Explanation

The direction of the stator rotating magnetic field is determined by the phase sequence of the 3-phase supply. Swapping any two line conductors (e.g. $L1$ and $L2$) reverses the spatial phase sequence, causing the stator magnetic field and rotor to rotate in the opposite direction.

Question 42Measurement, feedback, PLC and automation

What is the primary technical reason for using a stepped starting resistor (rhéostat de démarrage) in the rotor circuit of a wound-rotor (slip-ring / moteur à bagues) induction motor?

  1. To increase starting torque up to maximum breakdown torque (T_max) while significantly reducing starting inrush current Correct answer.
  2. To reverse the rotational speed of the motor without changing stator phases
  3. To convert the AC motor into a synchronous generator
  4. To permanently lower the motor operating efficiency to zero

Explanation

In wound-rotor motors, inserting external resistance $R_{ext}$ into the rotor circuit shifts the maximum torque peak ($T_{max}$) directly to zero speed ($s = 1$, at start), while simultaneously limiting the starting current inrush. As the motor accelerates, the external resistors are cut out in steps until the slip rings are short-circuited at full speed.

Question 43Measurement, feedback, PLC and automation

In a 3-phase AC motor circuit, what is the purpose of dynamic DC braking (freinage par injection de courant continu)?

  1. Disconnecting the stator from AC power and injecting DC current into stator windings to create a stationary magnetic field that induces opposing braking torque in the rotating rotor Correct answer.
  2. Reversing two phases while running at full speed without any series resistance
  3. Short-circuiting the power transformer secondary
  4. Applying hydraulic friction pads directly to the motor casing

Explanation

DC injection braking cuts the 3-phase AC supply and immediately applies a low-voltage DC current across two stator terminals. This produces a stationary, non-rotating magnetic field in the air gap. As the rotor continues turning, its conductors cut this stationary field, inducing heavy rotor currents that generate powerful electrodynamic braking torque until the rotor stops.

Question 44Measurement, feedback, PLC and automation

What is the function of an auxiliary control relay (relais auxiliaire / contacteur auxiliaire, e.g. CAD32) compared to a main power contactor (contacteur de puissance, e.g. LC1D)?

  1. Auxiliary relays switch low-current control signals, interlocks, and pilot lamps (typically <10 A) and have no main high-current power poles with arc chutes Correct answer.
  2. Auxiliary relays directly supply 500 kW motors
  3. Auxiliary relays are mechanical fuses that melt under overcurrent
  4. Auxiliary relays step up 230 V AC to 20 kV AC

Explanation

Auxiliary contactors/relays contain only instantaneous and time-delayed auxiliary contacts (NO/NC) designed for low-power automation logic, signaling, and sequencing. Power contactors have heavy-duty main power poles with de-ion arc extinguishing chutes (chambres de coupure) to switch heavy motor starting currents.

Question 45Measurement, feedback, PLC and automation

In a PLC controlling an industrial pump, what is the function of an analog input module configured for a standard $4-20\text{ mA}$ current loop?

  1. To convert continuous industrial sensor signals (e.g. pressure 0–10 bar) into digital numerical values (e.g. 0–27648), where 4 mA represents zero scale and provides live-zero wire-break detection Correct answer.
  2. To power high-pressure hydraulic pumps directly from the sensor wires
  3. To transmit digital Ethernet TCP/IP packets at 1 Gbps
  4. To trip the main circuit breaker when current reaches 20 mA

Explanation

The $4-20\text{ mA}$ standard uses current rather than voltage to eliminate cable resistance losses over long distances. The 'live zero' at $4\text{ mA}$ ($0\%$ measurement) enables immediate detection of a broken wire ($0\text{ mA}$), distinguishing a zero reading from an open-circuit fault.

Question 46Measurement, feedback, PLC and automation

What is the standard utilization category according to IEC 60947-4-1 for contactors switching squirrel-cage induction motors in normal starting, running, and stopping during rotation?

  1. AC-3 Correct answer.
  2. AC-1
  3. DC-1
  4. AC-15

Explanation

Standard contactor utilization categories: AC-1 = Non-inductive or slightly inductive resistive loads (heaters); AC-2 = Slip-ring wound-rotor motors; AC-3 = Squirrel-cage motors with normal starting and breaking during rotation; AC-4 = Severe squirrel-cage motor duties (plugging, jogging, inching / freinage par contre-courant).

Question 47Measurement, feedback, PLC and automation

In a 3-phase autotransformer reduced-voltage motor starter (démarreur par autotransformateur), what is the main advantage compared to a star-delta starter?

  1. It allows selecting intermediate voltage taps (e.g. 50%, 65%, 80% of line voltage) and provides continuous power transition without opening the motor circuit during acceleration Correct answer.
  2. It eliminates the need for any contactors or relays
  3. It allows using 12 V car batteries to start 400 V motors
  4. It operates without drawing any reactive current from the grid

Explanation

Autotransformer starters provide adjustable starting voltage taps (e.g. $65\%$ tap reduces current and torque to $42\%$) and use closed transition (Korndörfer method) so that motor current is never interrupted during transition to full voltage, preventing mechanical and electrical switching surges.

Question 48Electrical protection, earthing and interpretation of supplied models

What is the primary function of the instantaneous magnetic release (déclencheur magnétique) in a low-voltage thermal-magnetic circuit breaker (disjoncteur magnétothermique)?

  1. Instantaneous tripping within milliseconds to protect conductors against destructive high-current short-circuits Correct answer.
  2. Slow time-delayed thermal tripping to protect against prolonged minor overloads
  3. Earth leakage current detection for residual current protection
  4. Automatic power factor regulation

Explanation

A thermal-magnetic breaker has two releases: the bimetallic thermal release operates with an inverse-time delay ($I^2 t$) for moderate overloads; the electromagnetic coil (déclencheur magnétique) trips instantaneously (<20 ms) when current spikes into the short-circuit range.

Question 49Electrical protection, earthing and interpretation of supplied models

According to standard IEC 60898 / 60947-2, what is the magnetic instantaneous trip threshold range for a 'Curve C' (Courbe C) circuit breaker?

  1. 5 to 10 times nominal current (5 In – 10 In) Correct answer.
  2. 3 to 5 times nominal current (3 In – 5 In)
  3. 10 to 20 times nominal current (10 In – 20 In)
  4. 1.13 to 1.45 times nominal current

Explanation

Standard circuit breaker magnetic trip curves: Curve B = $3-5 I_n$ (resistive loads, long cable runs); Curve C = $5-10 I_n$ (standard commercial/domestic general loads, fluorescent lighting, low-surge motors); Curve D = $10-20 I_n$ (heavy inductive starting inrush motors and transformers).

Question 50Electrical protection, earthing and interpretation of supplied models

In the TN earthing system (Régime TN / Mise au neutre), what happens during a direct phase-to-chassis insulation fault?

  1. The fault becomes a direct phase-to-neutral short-circuit with high fault current (I_d ≈ V_0 / Z_loop), allowing standard overcurrent circuit breakers or fuses to trip rapidly within statutory disconnect times Correct answer.
  2. The fault current is zero and the installation continues running indefinitely
  3. The entire building voltage rises to 20 kV
  4. The earth rod melts instantly from overheating

Explanation

In TN systems, exposed conductive parts are bonded directly to the earthed neutral conductor ($PE$ in TN-S or $PEN$ in TN-C). An insulation fault closes a low-impedance metallic fault loop ($Z_s = Z_{AB} + Z_{PE}$), producing a short-circuit current large enough to trigger the magnetic trip of standard circuit breakers within standard times ($t \le 0.4\text{ s}$ at 230 V).