Q1. In an alternator, the armature winding is placed on the stator mainly because:
A. Rotor speed is low
B. High voltage can be easily insulated
C. Rotor inertia is reduced
D. Copper loss is minimized
Correct Option: B
Explanation:
The armature winding carries high-voltage AC output. Placing it on the stator makes insulation easier, improves cooling, and simplifies maintenance. Insulating high voltage on a rotating rotor is mechanically difficult due to centrifugal forces.
Q2. Concentric (chain) winding in alternators is characterized by:
A. Equal coil span for all coils
B. Coils of different spans placed concentrically
C. One coil per phase only
D. Lap winding arrangement
Correct Option: B
Explanation:
In concentric winding, coils of different spans are placed one inside another with a common center. This reduces end connections, saves copper, and is commonly used in alternator stator windings.
Q3. In a two-layer winding of an alternator:
A. One conductor per slot is used
B. Two conductors of same phase occupy one slot
C. Upper and lower layers belong to different coils
D. It is used only for DC machines
Correct Option: C
Explanation:
In two-layer winding, each slot contains two conductors—one in the upper layer and one in the lower layer—belonging to different coils. This improves emf waveform and distribution.
Q4. For the same line voltage, which connection gives higher phase voltage?
A. Star
B. Delta
C. Both same
D. Depends on load
Correct Option: B
Explanation:
In star connection, phase voltage = VL/√3, whereas in delta connection, phase voltage equals line voltage. Hence delta gives higher phase voltage for the same line voltage.
Q5. The ratio of line voltage to phase voltage in a star-connected alternator is:
A. 1
B. √2
C. √3
D. 3
Correct Option: C
Explanation:
In star connection, line voltage is √3 times the phase voltage. Therefore, VL/Vph = √3.
Q6. Short-pitching of armature winding in alternator is done mainly to:
A. Increase emf
B. Reduce copper loss
C. Eliminate harmonics
D. Increase speed
Correct Option: C
Explanation:
Short-pitching (chording) is mainly used to eliminate lower-order harmonics, improve waveform, and reduce noise. Although emf reduces slightly, harmonic suppression is the main objective.
Q7. A coil in an alternator is short-pitched by 30°. Find the pitch factor (kp).
A. 0.866
B. 0.966
C. 0.707
D. 0.933
Correct Option: B
Explanation:
Pitch factor kp = cos(α/2). For α = 30°, kp = cos(15°) = 0.966.
Q8. Distribution factor (kd) of an alternator winding is always:
A. Greater than 1
B. Equal to 1
C. Less than 1
D. Zero
Correct Option: C
Explanation:
Due to phase difference between emfs induced in distributed conductors, the resultant emf is less than the arithmetic sum. Hence distribution factor is always less than 1.
Q9. An alternator has 9 slots per pole and 3 slots per pole per phase. Find distribution factor.
A. 0.96
B. 0.92
C. 0.955
D. 1
Correct Option: C
Explanation:
Using the standard distribution factor formula, the calculated value comes out approximately equal to 0.955.
Q10. Winding factor (kw) is the product of:
A. kp × kd
B. kp + kd
C. kd / kp
D. kp − kd
Correct Option: A
Explanation:
Winding factor represents the combined effect of short-pitching and distribution of winding. Hence, kw = kp × kd.
Q11. If pitch factor = 0.96 and distribution factor = 0.95, find winding factor.
A. 0.912
B. 0.90
C. 0.95
D. 0.99
Correct Option: A
Explanation:
Winding factor kw = kp × kd = 0.96 × 0.95 = 0.912.
Q12. Which harmonic is completely eliminated by short-pitching of 120° electrical?
A. 3rd
B. 5th
C. 7th
D. Fundamental
Correct Option: A
Explanation:
Short-pitching by 120° electrical completely eliminates the 3rd harmonic emf, which is especially important in star-connected alternators.
Q13. Effect of increasing harmonics in alternator emf is:
A. Increased efficiency
B. Improved voltage regulation
C. Distorted waveform
D. Higher speed
Correct Option: C
Explanation:
Harmonics distort the emf waveform, increase losses, cause heating, and reduce power quality.
Q14. An alternator has 4 poles, runs at 1500 rpm. Find frequency.
A. 25 Hz
B. 50 Hz
C. 60 Hz
D. 75 Hz
Correct Option: B
Explanation:
f = (P × N) / 120 = (4 × 1500) / 120 = 50 Hz.
Q15. An alternator has 8 poles and frequency 50 Hz. Find speed.
A. 375 rpm
B. 750 rpm
C. 1500 rpm
D. 3000 rpm
Correct Option: B
Explanation:
N = (120 × f) / P = (120 × 50) / 8 = 750 rpm.
Q16. EMF equation of an alternator is:
A. E = 4.44 f Φ T kw
B. E = 2.22 f Φ T
C. E = ΦN
D. E = V / I
Correct Option: A
Explanation:
The standard EMF equation of an alternator per phase is E = 4.44 f Φ T kw, where kw is the winding factor.
Q17. An alternator has flux per pole = 0.05 Wb, frequency = 50 Hz, turns/phase = 100, kw = 0.9. Find induced emf/phase.
A. 900 V
B. 1000 V
C. 1110 V
D. 999 V
Correct Option: D
Explanation:
E = 4.44 × 50 × 0.05 × 100 × 0.9 = 999 V.
Q18. Alternator size mainly depends upon:
A. Voltage rating
B. Frequency
C. Speed
D. Output power & speed
Correct Option: D
Explanation:
Alternator size depends on output power and speed. Higher power increases size, while higher speed reduces size.
Q19. For the same output power, alternator size will be smaller if:
A. Speed is low
B. Speed is high
C. Frequency is low
D. Voltage is low
Correct Option: B
Explanation:
At higher speeds, less magnetic material is required and flux per pole reduces, making the alternator more compact.
Q20. Why large alternators operate at higher voltages?
A. Reduce iron loss
B. Reduce copper loss
C. Reduce insulation
D. Increase frequency
Correct Option: B
Explanation:
Operating at higher voltage reduces current for the same power output, which significantly reduces copper losses (I²R) and improves efficiency.