Q1. A synchronous motor always runs at:
A. Slightly less than synchronous speed
B. Slightly more than synchronous speed
C. Exactly synchronous speed
D. Variable speed
Correct Option: 3
Explanation:
In steady-state operation, a synchronous motor runs exactly at synchronous speed because the rotor
magnetic field locks with the rotating magnetic field of the stator. Load variation changes only
the load angle (δ), not the speed.
Q2. Why is a synchronous motor not self-starting?
A. High rotor resistance
B. Zero starting torque
C. Absence of rotor current
D. Low power factor at start
Correct Option: 2
Explanation:
At standstill, the torque produced by the synchronous motor reverses every half cycle, resulting
in zero average torque. Hence, it cannot start by itself.
Q3. Which of the following is used to start a synchronous motor?
A. Auto-transformer starter
B. Damper winding
C. Rotor resistance starter
D. Star-delta starter
Correct Option: 2
Explanation:
Damper windings act like rotor bars of an induction motor during starting. They produce induction
torque and help the motor reach near synchronous speed before DC excitation is applied.
Q4. During starting, a synchronous motor behaves like:
A. DC shunt motor
B. DC series motor
C. Induction motor
D. Universal motor
Correct Option: 3
Explanation:
Due to damper windings, rotor currents are induced and torque is produced based on induction motor
principle during starting.
Q5. With constant excitation, when mechanical load on a synchronous motor increases:
A. Speed decreases
B. Power factor changes
C. Load angle increases
D. Armature voltage increases
Correct Option: 3
Explanation:
For constant excitation, increase in load increases power demand. Since power ∝ sinδ, the load
angle δ increases while speed remains constant.
Q6. In a synchronous motor, air-gap power is equal to:
A. Mechanical output power
B. Rotor copper loss
C. Input electrical power
D. Power developed
Correct Option: 4
Explanation:
Rotor copper loss is negligible because DC excitation is used. Hence, air-gap power is equal to
the power developed by the motor.
Q7. In the equivalent circuit of a synchronous motor, the back EMF is represented as:
A. E
B. V
C. IaRa
D. jXsIa
Correct Option: 1
Explanation:
The back EMF E represents the effect of rotor field excitation and opposes the applied terminal
voltage in the equivalent circuit.
Q8. Power developed by a cylindrical rotor synchronous motor is:
A. EV/Xs cosδ
B. EV/Xs sinδ
C. E²/Xs sinδ
D. V²/Xs sinδ
Correct Option: 2
Explanation:
For cylindrical rotor synchronous motors, power developed per phase is given by
P = (EV/Xs) sinδ.
Q9. A 3-phase synchronous motor has VL = 400 V, Eph = 220 V, Xs = 5 Ω. Maximum power developed per phase is:
A. 9.7 kW
B. 10.16 kW
C. 7.5 kW
D. 12.4 kW
Correct Option: 2
Explanation:
Maximum power occurs at δ = 90°. Using Pmax = (Eph × Vph)/Xs with
Vph = 400/√3 ≈ 231 V gives P ≈ 10.16 kW per phase.
Q10. If excitation of a synchronous motor is increased at constant load:
A. Power factor becomes lagging
B. Power factor becomes leading
C. Speed increases
D. Load angle increases
Correct Option: 2
Explanation:
Increasing excitation increases back EMF, shifting armature current to leading.
Hence, the motor operates at leading power factor.
Q11. The curve between armature current and field current at constant load is called:
A. Regulation curve
B. Torque curve
C. V-curve
D. Inverted V-curve
Correct Option: 3
Explanation:
The armature current initially decreases and then increases with field current,
forming a V-shaped characteristic known as V-curve.
Q12. Stability of a synchronous motor depends mainly on:
A. Speed
B. Field current
C. Load angle
D. Armature resistance
Correct Option: 3
Explanation:
Stability depends on load angle δ. Stable operation is possible only when δ < 90°.
Beyond this, the motor loses synchronism.
Q13. When load increases with constant excitation:
A. δ decreases
B. δ increases
C. Speed decreases
D. PF becomes unity
Correct Option: 2
Explanation:
Torque ∝ sinδ. Hence, an increase in load requires an increase in load angle δ
while speed remains constant.
Q14. Pull-out torque occurs when:
A. δ = 0°
B. δ = 30°
C. δ = 90°
D. δ = 180°
Correct Option: 3
Explanation:
Maximum torque (pull-out torque) occurs at δ = 90°. Beyond this point, the motor
becomes unstable.
Q15. Synchronous motors are preferred for power factor improvement because:
A. High efficiency
B. Constant speed
C. Controllable excitation
D. Low losses
Correct Option: 3
Explanation:
By varying field excitation, synchronous motors can operate at lagging, unity, or
leading power factor, making them suitable for power factor correction.
Q16. A synchronous motor has E = 300 V, V = 400 V, Xs = 10 Ω. Power developed per phase at δ = 30° is:
A. 6 kW
B. 6.93 kW
C. 5.2 kW
D. 8 kW
Correct Option: 1
Explanation:
Using P = (EV/Xs) sinδ = (300×400/10)×sin30° = 6000 W = 6 kW per phase.
Q17. Rotor copper losses in a synchronous motor are:
A. High
B. Zero
C. Equal to stator losses
D. Depends on load
Correct Option: 2
Explanation:
Rotor carries DC current; hence no induced rotor current flows and rotor copper
losses are zero.
Q18. Hunting in synchronous motor is due to:
A. Load fluctuation
B. Supply voltage drop
C. Core loss
D. Armature resistance
Correct Option: 1
Explanation:
Sudden changes in load cause oscillations of rotor about synchronous position,
known as hunting.
Q19. If synchronous reactance is doubled, maximum power developed becomes:
A. Same
B. Double
C. Half
D. Four times
Correct Option: 3
Explanation:
Maximum power is inversely proportional to synchronous reactance. Doubling Xs
reduces maximum power to half.
Q20. Compared to induction motor, synchronous motor has:
A. Lower efficiency
B. Variable speed
C. Higher cost
D. Lower starting torque
Correct Option: 3
Explanation:
Synchronous motors require DC excitation and special starting methods, making
them costlier than induction motors.