Q1. The electromagnetic torque developed in a synchronous motor is proportional to:
A. sin δ
B. cos δ
C. tan δ
D. sin² δ
Correct Option: 1
Explanation:
For a cylindrical rotor synchronous motor, power developed is
P = (3EV/Xs) sin δ. Since torque T = P/ωs, electromagnetic torque
is directly proportional to sin δ.
Q2. Which torque is responsible for maintaining synchronism in a synchronous motor?
A. Starting torque
B. Reluctance torque
C. Pull-out torque
D. Synchronizing torque
Correct Option: 4
Explanation:
Synchronizing torque is the restoring torque that keeps the rotor
magnetic field locked with the stator rotating magnetic field when
load changes.
Q3. The maximum power developed by a cylindrical rotor synchronous motor occurs at load angle δ equal to:
A. 30°
B. 45°
C. 60°
D. 90°
Correct Option: 4
Explanation:
From the power equation P = Pmax sin δ, maximum power occurs when
sin δ = 1, i.e., δ = 90°. This corresponds to the pull-out condition.
Q4. A synchronous motor running at no-load draws armature current mainly to:
A. Produce torque
B. Overcome friction losses
C. Establish magnetic field
D. Supply mechanical output
Correct Option: 3
Explanation:
At no-load, a synchronous motor does not deliver useful mechanical
output. Armature current is mainly drawn to establish the rotating
magnetic field.
Q5. The reluctance torque in a synchronous motor exists mainly due to:
A. Field current
B. Armature reaction
C. Non-uniform air gap
D. Damper winding
Correct Option: 3
Explanation:
Reluctance torque exists due to non-uniform air gap and difference
between direct-axis and quadrature-axis reactances, mainly in salient
pole machines.
Q6. A 3-phase, 400 V, star-connected synchronous motor has Xs = 10 Ω per phase. If Eph = 440 V and δ = 30°, power developed is:
A. 8.1 kW
B. 15.24 kW
C. 16.3 kW
D. 24.4 kW
Correct Option: 2
Explanation:
Vph = 400/√3 = 231 V.
Power developed,
P = (3EV/Xs) sin δ = (3×440×231/10)×0.5 ≈ 15.24 kW.
Q7. A synchronous motor develops maximum power of 50 kW. Pull-out torque at 1500 rpm is:
A. 159 N·m
B. 318 N·m
C. 636 N·m
D. 955 N·m
Correct Option: 2
Explanation:
ωs = 2πN/60 = 157 rad/s.
Torque T = P/ωs = 50000/157 ≈ 318 N·m.
Q8. Which of the following does NOT affect the power developed by a synchronous motor?
A. Supply voltage
B. Synchronous reactance
C. Load angle
D. Armature resistance (neglected)
Correct Option: 4
Explanation:
When armature resistance is neglected, the power equation
P = (EV/Xs) sin δ does not include armature resistance.
Q9. The alternative expression for power developed in a synchronous motor is based on:
A. Armature current
B. Field current
C. Synchronizing power coefficient
D. Slip
Correct Option: 3
Explanation:
Power can also be expressed as P = Psyn sin δ, where Psyn is the
synchronizing power coefficient used in stability analysis.
Q10. A 3-phase synchronous motor has V = 6.6 kV, E = 7 kV, Xs = 20 Ω/phase and δ = 25°. Power developed is:
A. 0.85 MW
B. 0.98 MW
C. 1.48 MW
D. 2.10 MW
Correct Option: 2
Explanation:
Vph = 3810 V, Eph = 4041 V.
P = (3EV/Xs) sin δ ≈ 0.98 MW.
Q11. In a salient pole synchronous motor, power developed depends on:
A. sin δ only
B. cos δ only
C. sin δ and sin 2δ
D. cos δ and cos 2δ
Correct Option: 3
Explanation:
Power equation of salient pole motor includes electromagnetic power
(sin δ) and reluctance power component (sin 2δ).
Q12. The reluctance power component in a salient pole synchronous motor is proportional to:
A. sin δ
B. cos δ
C. sin 2δ
D. cos 2δ
Correct Option: 3
Explanation:
Reluctance power arises due to saliency and is proportional to sin 2δ.
Q13. A salient pole synchronous motor has Xd = 1.2 Ω and Xq = 0.8 Ω. Reluctance torque exists because:
A. Xd = Xq
B. Xd ≠ Xq
C. Depends on load
D. Depends on speed
Correct Option: 2
Explanation:
Reluctance torque exists only when direct-axis and quadrature-axis
reactances are unequal (Xd ≠ Xq).
Q14. Which torque is absent in a cylindrical rotor synchronous motor?
A. Electromagnetic torque
B. Synchronizing torque
C. Reluctance torque
D. Shaft torque
Correct Option: 3
Explanation:
Cylindrical rotor machines have uniform air gap, so Xd = Xq and
reluctance torque does not exist.
Q15. The pull-out torque of a synchronous motor corresponds to:
A. Minimum power
B. Zero load
C. Maximum power
D. Unity power factor
Correct Option: 3
Explanation:
Pull-out torque occurs at δ = 90°, where power developed is maximum.
Beyond this point, synchronism is lost.
Q16. A synchronous motor develops 30 kW at δ = 30°. Maximum power developed is:
A. 34.6 kW
B. 40 kW
C. 60 kW
D. 90 kW
Correct Option: 3
Explanation:
Using P = Pmax sin δ → Pmax = 30/sin30° = 60 kW.
Q17. In steady-state operation, increase in load on a synchronous motor results in:
A. Decrease in speed
B. Increase in slip
C. Increase in load angle
D. Increase in excitation
Correct Option: 3
Explanation:
A synchronous motor runs at constant speed. Increase in load increases
torque angle δ while speed remains unchanged.
Q18. A synchronous motor has Ns = 1000 rpm and developed power 20 kW. Electromagnetic torque is:
A. 95.5 N·m
B. 191 N·m
C. 286 N·m
D. 382 N·m
Correct Option: 2
Explanation:
ωs = 2Ï€×1000/60 = 104.7 rad/s.
T = P/ωs = 20000/104.7 ≈ 191 N·m.
Q19. Which expression represents power developed in a cylindrical rotor synchronous motor?
A. P = EV/Xs sin δ
B. P = V²/Xs cos δ
C. P = E²/Xs sin δ
D. P = V/E sin δ
Correct Option: 1
Explanation:
Standard power equation of cylindrical rotor synchronous motor is
P = (EV/Xs) sin δ.
Q20. A synchronous motor operating at δ = 90° indicates:
A. Stable operation
B. No-load
C. Pull-out condition
D. Maximum efficiency
Correct Option: 3
Explanation:
At δ = 90°, power is maximum and synchronizing torque becomes zero.
Motor is at pull-out condition.