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Sample Senegal Baccalauréat Série T2 Practice Questions
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1A three-phase asynchronous (induction) motor has p = 2 pole pairs (4 poles) and is connected to a 50 Hz industrial power grid. What is its synchronous rotational speed (ns in rpm)?
A.1500 rpm
B.3000 rpm
C.1000 rpm
D.750 rpm
Explanation: The synchronous speed of a rotating magnetic field is given by ns = (60 * f) / p, where f is the grid frequency in Hz and p is the number of pole pairs. For f = 50 Hz and p = 2: ns = (60 * 50) / 2 = 3000 / 2 = 1500 rpm.
2An induction motor operating on a 50 Hz grid has a synchronous speed ns = 1500 rpm and rotates at full load at n = 1440 rpm. What is the rotor slip (s) and the frequency of the induced rotor currents (fr)?
A.s = 0.02 and fr = 1.0 Hz
B.s = 0.04 (4%) and fr = 2.0 Hz
C.s = 0.06 and fr = 3.0 Hz
D.s = 0.08 and fr = 4.0 Hz
Explanation: Rotor slip is s = (ns - n) / ns = (1500 - 1440) / 1500 = 60 / 1500 = 0.04 (4.0%). The frequency of the currents induced in the rotor windings is fr = s * f = 0.04 * 50 Hz = 2.0 Hz.
3A single-phase transformer connected to primary voltage U1 = 230 V delivers an open-circuit secondary voltage U20 = 24 V. During an open-circuit test (essai à vide), the active power drawn is P10 = 18 W. What physical loss component is measured by this test?
A.Total full-load winding copper losses (Pj)
B.Mechanical bearing friction and windage losses
C.Core magnetic losses (iron losses Pfer = hysteresis + eddy currents)
D.Dielectric leakage losses through the transformer oil
Explanation: At rated voltage with the secondary open, the primary carries the relatively small excitation current. When its winding loss is negligible, the input wattmeter reading approximates the core losses from hysteresis and eddy currents. It is an approximation, rather than a measurement of full-load copper loss.
4A single-phase transformer has total secondary-referred resistance Rs = 0.12 Ω and secondary-referred leakage reactance Xs = 0.35 Ω. At rated secondary current I2 = 20 A with an inductive power factor cos(φ2) = 0.80 (sin(φ2) = 0.60), what is the secondary terminal voltage drop (ΔU2) calculated using Kapp's formula?
A.2.4 V
B.4.2 V
C.8.4 V
D.6.12 V
Explanation: According to Kapp's approximation for voltage drop on load: ΔU2 = I2 * (Rs * cos(φ2) + Xs * sin(φ2)). Substituting values: Rs * cos(φ2) = 0.12 * 0.80 = 0.096 Ω. Xs * sin(φ2) = 0.35 * 0.60 = 0.210 Ω. Sum = 0.096 + 0.210 = 0.306 Ω. Therefore ΔU2 = 20 A * 0.306 Ω = 6.12 V.
5A squirrel-cage induction motor is rated to run in delta at the supply line voltage. In the usual approximate starting model, how do its line current and torque in star compare with direct-on-line starting in delta at that same voltage?
A.Line starting current and starting torque are both reduced by a factor of 3 compared to direct-on-line (DOL) start
B.Starting torque is tripled while starting current remains unchanged
C.Starting current is reduced by √3 while starting torque is doubled
D.The motor accelerates at constant torque without any line voltage change
Explanation: During star connection (Y), the voltage applied across each phase winding is V = U / √3 instead of the full delta line voltage U. Since winding current is proportional to voltage and line current in star equals phase current (whereas in delta line current is √3 * I_phase), the starting line current is divided by 3: I_Y = (1/3) * I_Δ. Because electromagnetic torque is proportional to the square of the applied winding voltage (T ∝ V²), starting torque is also reduced by (1/√3)² = 1/3.
6A separately excited DC motor (moteur à courant continu à excitation indépendante) operates with constant excitation magnetic flux Φ. If the armature terminal voltage U is doubled while load torque remains constant, how does the steady-state rotational speed (n) respond?
A.Speed remains strictly unchanged because flux is constant
B.Speed roughly doubles, as electromotive force E = U - Ra*Ia is approximately proportional to speed
C.Speed is halved to maintain constant mechanical power
D.The motor enters runaway (emballement) and destroys itself
Explanation: In a DC motor, the back-EMF is E = U - Ra * Ia = K * Φ * Ω. For constant load torque, armature current Ia = C / (K * Φ) remains constant. Neglecting the small armature resistance drop Ra * Ia, E ≈ U. Therefore, doubling armature voltage U roughly doubles back-EMF E, and since flux Φ is held constant, rotational speed Ω = E / (K * Φ) approximately doubles.
7A three-phase induction motor has an input active power Pa = 12.0 kW when connected to a 400 V, 50 Hz grid. The stator Joule losses are Pjs = 450 W and stator core iron losses are Pfs = 350 W. If the motor operates at slip s = 0.05 (5%), what are the rotor Joule losses (Pjr) and the electromagnetic power (Pem)?
A.Pem = 12.0 kW and Pjr = 600 W
B.Pem = 11.2 kW and Pjr = 1120 W
C.Pem = 11.2 kW and Pjr = 560 W
D.Pem = 10.64 kW and Pjr = 560 W
Explanation: Electromagnetic power transmitted across the air gap from stator to rotor is Pem = Ptr = Pa - Pjs - Pfs = 12,000 W - 450 W - 350 W = 11,200 W = 11.2 kW. The rotor Joule losses are strictly proportional to slip: Pjr = s * Pem = 0.05 * 11,200 W = 560 W.
8A three-phase synchronous generator (alternator) has synchronous reactance Xs = 2.5 Ω per phase (armature resistance Ra is negligible). When delivering rated line current I = 40 A per phase at phase terminal voltage V = 230 V with unity power factor (cos(φ) = 1.0), what is the generated excitation electromotive force per phase (Ev) using the Behn-Eschenburg synchronous impedance model?
A.230 V
B.330 V
C.280 V
D.250.8 V
Explanation: According to the Behn-Eschenburg phasor equation: Ev = V + j * Xs * I. In phasor form with V along the real axis (V = 230 V) and I in phase with V (cos(φ) = 1.0, I = 40 A): j * Xs * I = j * (2.5 * 40) = j * 100 V. The magnitude of the internal EMF is Ev = √(V² + (Xs * I)²) = √(230² + 100²) = √(52,900 + 10,000) = √62,900 ≈ 250.798 V ≈ 250.8 V.
9What is the coupling designation (vector group) of a three-phase transformer whose primary is connected in Star (Y), secondary in Delta (d), with a phase displacement between corresponding primary and secondary line voltages of 330° (lagging by 330°, or leading by 30°)?
A.Yd11
B.Dy5
C.Yy0
D.Yz11
Explanation: In three-phase transformer nomenclature: capital letter represents primary winding (Y = Star), lowercase letter represents secondary winding (d = Delta), and the clock number (indice horaire) indicates phase lag in units of 30° (330° / 30° = 11). Thus, this transformer is designated Yd11.
10A DC shunt motor has armature resistance Ra = 0.5 Ω and is connected to a DC mains supply U = 220 V. At full load, the armature current is Ia = 40 A and rotational speed is N = 1200 rpm. What is the back-electromotive force (E) and the electromagnetic power developed (Pem)?
A.E = 220 V and Pem = 8.8 kW
B.E = 200 V and Pem = 8.0 kW
C.E = 240 V and Pem = 9.6 kW
D.E = 180 V and Pem = 7.2 kW
Explanation: Back-EMF is E = U - Ra * Ia = 220 V - (0.5 Ω * 40 A) = 220 V - 20 V = 200 V. The electromagnetic power converted from electrical to mechanical form is Pem = E * Ia = 200 V * 40 A = 8000 W = 8.0 kW.
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