Q1. An A.C. series motor is mainly modified from which motor?
A. DC shunt motor
B. DC series motor
C. Induction motor
D. Synchronous motor
Correct Option: 2
Explanation:
An A.C. series motor is derived from a DC series motor. Both have armature and field windings connected in series. When DC series motor is operated on AC supply, issues like poor power factor and commutation arise, so modifications such as laminated core and compensating winding are introduced.
Q2. Why does a DC series motor run unsatisfactorily on AC supply?
A. High copper loss
B. Poor commutation and low power factor
C. Low torque
D. Excessive speed
Correct Option: 2
Explanation:
On AC supply, alternating flux produces high reactance voltage in armature, resulting in poor power factor and difficult commutation. This causes sparking and reduced efficiency.
Q3. In an A.C. series motor, the laminated core is used to:
A. Reduce copper loss
B. Reduce eddy current loss
C. Increase torque
D. Improve cooling
Correct Option: 2
Explanation:
Alternating flux induces eddy currents in the core. Laminated core increases resistance to circulating currents, thereby reducing eddy current loss and improving efficiency.
Q4. Which winding is added in an A.C. series motor to improve power factor?
A. Shunt field winding
B. Compensating winding
C. Interpole winding
D. Damper winding
Correct Option: 2
Explanation:
Compensating winding neutralizes armature reaction and reduces reactance voltage, which improves the power factor of the A.C. series motor.
Q5. The direction of rotation of an A.C. series motor can be reversed by:
A. Reversing armature current only
B. Reversing field current only
C. Reversing both armature and field currents
D. It cannot be reversed
Correct Option: 3
Explanation:
Torque T ∝ ΦIa. Reversing only one current reverses torque. To reverse rotation, both armature and field current directions must be reversed simultaneously.
Q6. Universal motor operates satisfactorily on:
A. Only AC supply
B. Only DC supply
C. Both AC and DC supply
D. Only single-phase AC
Correct Option: 3
Explanation:
Universal motor is specially designed to operate on both AC and DC supply with acceptable commutation, laminated core, and compensating winding.
Q7. Why is a universal motor capable of very high speed?
A. Low armature resistance
B. High torque-to-weight ratio
C. Weak magnetic field at no load
D. Large air-gap
Correct Option: 3
Explanation:
At no load, armature current is small, causing weak field flux. Since speed is inversely proportional to flux, the motor attains very high speed.
Q8. In a universal motor, if supply voltage is reduced by 10%, the speed will approximately:
A. Reduce by 10%
B. Reduce by 20%
C. Remain constant
D. Increase
Correct Option: 1
Explanation:
Speed of universal motor is approximately proportional to voltage. Hence, a 10% reduction in voltage causes nearly 10% reduction in speed.
Q9. The flux in a universal motor is proportional to armature current. If armature current is increased by 20%, torque will increase by approximately:
A. 20%
B. 44%
C. 40%
D. 60%
Correct Option: 2
Explanation:
Since Φ ∝ I and T ∝ ΦI, torque becomes proportional to I². Increasing current by 20% gives T = (1.2)² = 1.44, i.e., 44% increase.
Q10. The speed of a universal motor is controlled by:
A. Field weakening
B. Armature resistance control
C. Voltage control
D. All of the above
Correct Option: 4
Explanation:
Speed of universal motor can be controlled by voltage variation (TRIAC), field control, and armature resistance control.
Q11. A universal motor draws 4 A at 230 V AC. If power factor is 0.8, input power is:
A. 736 W
B. 800 W
C. 920 W
D. 1000 W
Correct Option: 1
Explanation:
Input power P = VIcosφ = 230 × 4 × 0.8 = 736 W.
Q12. In an A.C. series motor, the compensating winding is connected:
A. In parallel with armature
B. In series with armature
C. Across supply
D. In parallel with field
Correct Option: 2
Explanation:
Compensating winding must carry the same current as armature to neutralize armature reaction, hence connected in series.
Q13. Which loss is comparatively higher in an A.C. series motor?
A. Mechanical loss
B. Copper loss
C. Core loss
D. Stray loss
Correct Option: 3
Explanation:
Due to alternating flux, hysteresis and eddy current losses increase significantly, making core loss higher in A.C. series motor.
Q14. At no-load, the speed of a universal motor is:
A. Zero
B. Rated speed
C. Very high (dangerous)
D. Slightly above rated
Correct Option: 3
Explanation:
At no load, flux is weak and speed becomes excessively high, which can be dangerous for the motor.
Q15. A universal motor develops 0.5 N·m torque at 5000 rpm. Output power is:
A. 130 W
B. 200 W
C. 262 W
D. 400 W
Correct Option: 3
Explanation:
Output power P = (2Ï€NT)/60 = (2Ï€ × 5000 × 0.5)/60 ≈ 262 W.
Q16. Why are universal motors not used for constant speed applications?
A. Poor efficiency
B. High noise
C. Speed varies widely with load
D. High cost
Correct Option: 3
Explanation:
Speed of universal motor changes significantly with load, hence it is unsuitable for constant-speed applications.
Q17. If flux in a universal motor is halved while armature current remains constant, torque becomes:
A. Same
B. Half
C. Double
D. One-fourth
Correct Option: 2
Explanation:
Torque T ∝ ΦI. If flux becomes half and current remains constant, torque also becomes half.
Q18. Which method is most commonly used for speed control of universal motors in household appliances?
A. Field control
B. Armature control
C. Voltage control using TRIAC
D. Pole changing
Correct Option: 3
Explanation:
TRIAC-based voltage control is compact, efficient, and widely used in fans, mixers, and vacuum cleaners.
Q19. A universal motor runs at 6000 rpm on no-load and 4000 rpm on full-load. Speed regulation is:
A. 25%
B. 33.3%
C. 40%
D. 50%
Correct Option: 4
Explanation:
Speed regulation = (N0 − Nf) / Nf × 100 = (6000 − 4000) / 4000 × 100 = 50%.
Q20. If a universal motor has torque proportional to current squared, doubling current will increase torque by:
A. 2 times
B. 3 times
C. 4 times
D. 8 times
Correct Option: 3
Explanation:
Since T ∝ I², doubling current gives T = (2I)² = 4I², i.e., torque becomes four times.