Electrical Machines Questions and Answers – EMF Polygon – 2

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This set of Electrical Machines Multiple Choice Questions & Answers (MCQs) focuses on “EMF Polygon – 2”.

1. The winding’s for a 3-phase alternator are:

(i) 36 slots, 4 poles, span 1 to 8 
(ii) 72 slots, 6 poles, span 1 to 10
(iii) 96 slots, 4 poles, span 1 to 21
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The winding’s having pitch factor of more than 0.97 are __________
a) (i) and (ii) only
b) (ii) and (iii) only
c) (i) and (iii) only
d) (i),(ii) and (iii)
View Answer

Answer: c
Explanation: Kp=cosε/2
(i) Slots per pole = 36/4=9, for a coil span of 8 slots, the coil is short pitched by 1 slot and the chording angle is ε=γ = 20° ⇒ Kp=cos10°=0.985
(ii) Slots per pole = 72/6=12, for a coil span of 10 slots, the coil is short pitched by 2 slots and the chording angle is ε=2γ and γ=180/12 ⇒ ε = 30° ⇒ Kp=cos15°=0.9659
(iii) Slots per pole = 96/4=24, for a coil span of 21 slots, the coil is short pitched by 3 slots and the chording angle is ε=3γ and γ=180/24 ⇒ ε = 24.5° ⇒ Kp=cos12.25°=0.97723.

2. In 48 slot, 4-pole,3 phase alternator, the coil-span is 10. Its distribution and pitch factors are respectively ____________
a) 0.9717,0.966
b) 0.9822, 0.9814
c) 0.9577, 0.9814
d) 0.9577, 0.966
View Answer

Answer: d
Explanation: Slots per pole = 48/4=12, for a coil span of 10 slots, the coil is short pitched by 2 slots and the chording angle is ε=2γ and γ=180/12 ⇒ ε = 30° ⇒ Kp=cosε/2=0.9659.
We know, Kd=(sin(qγ/2))/qsin(γ/2), here q=48/4∗3 = 4 ⇒ Kd=0.957662.
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3. A 3-phase, 4-pole alternator has 48 stator slots carrying a 3-phase distributed winding. Each coil of the winding is short chorded by one slot pitch. The winding factor is given by ______________
a) (cos7.5)/16
b) (cot7.5)/8
c) 1/(8sin7.5)
d) (cot7.5)/16
View Answer

Answer: b
Explanation: Slots per pole = 48/4 = 12 ⇒ γ=180/12=15° and q=48/4∗3=4
coil of the winding is short chorded by one slot pitch ⇒ ε=γ=15°, Kw=Kp∗Kd = cos7.5(sin(4∗15/2))/(4∗sin(15/2)) = (cot7.5)/8.

4. The chording angle for eliminating 5th harmonic should be ____________
a) 30°
b) 34°
c) 36°
d) 35°
View Answer

Answer: c
Explanation: To eliminate 5th harmonic kp5 must be zero, kpn = cosnε/2 ⇒ kp5 = cos5ε/2=0=cos90° ⇒ 5ε/2=90° ⇒ ε = 36°.
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5. Which of the following statements are true?

(i) breadth factor for third harmonic kd3 is more than that for fundamental kd1
(ii) kd3 < kd1
(iii) kd3 may be less or more than kd1 depending upon the number of slots and poles
(iv) coil-span factor for third harmonic kp3>kp1 (coil span factor for fundamental)
(v) kp3 < kp1
(vi) kp3 may be less or more than kp1 depending upon the number of slots and poles

a) (ii), (v)
b) (i), (iv)
c) (iii), (vi)
d) (i), (iii), (iv), (vi)
View Answer

Answer: a
Explanation: Examples will show that kdn < kd1 and has the effect of reducing the nth harmonic EMF in comparison with the fundamental EMF, similarly kpn < kp1.
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6. A 6-pole alternator with 36 slots carries a 2-phase distributed winding. Each coil is short pitched by one slot. The winding factor is given by _____________
a) cot15°/3√2
b) cot15°/4
c) cot15°/2√2
d) cot15°/4
View Answer

Answer: a
Explanation: Slots per pole = 36/6=6, q=6/2=3 ⇒ γ=180°/6=30°, coil is short pitched by one slot ⇒ ε=γ=30° and Kp=cosε/2=cos15° and Kd=sin(γq/2)/q(sinγ/2)=sin45/(3sin15)
Kw=Kp∗Kd=cot15°/3√2.

7. For eliminating nth harmonic from the EMF generated in the phase of a 3-phase alternator, the chording angle should be ______________
a) n∗full pitch
b) (1/n)∗full pitch
c) (2/n)∗full pitch
d) (3/n)∗full pitch
View Answer

Answer: a
Explanation: kpn=cosnε/2=0 ⇒ nε/2=90° ⇒ ε=180°/n=full pitch/n.
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8. Which among the given harmonics are called belt harmonics?
a) 5, 7, 11, 13
b) 3, 6, 9, 12
c) 5, 6, 11, 12
d) 7, 11, 13, 15
View Answer

Answer: a
Explanation: The odd harmonics of the order 5, 7, 11, 13 etc are called belt harmonics.

9. Machine A has 60° phase spread and machine B has 120° phase spread. Both the machines have uniformly distributed winding. The ratio of distribution factors of machine A to machine B is ____________
a) 0.866
b) 1.1
c) 1.55
d) 1.155
View Answer

Answer: c
Explanation: kdA=(sinσ/2)/σ/2, σ=60° ⇒ kdA=0.9556
kdB=(sinσ/2)/σ/2, σ=120° ⇒ kdB=0.827
thus, kdA/kdB=1.155.

Sanfoundry Global Education & Learning Series – Electrical Machines.

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