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

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

Explanation: K

_{p}=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° ⇒ K

_{p}=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° ⇒ K

_{p}=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° ⇒ K

_{p}=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

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° ⇒ K

_{p}=cosε/2=0.9659.

We know, K

_{d}=(sin(qγ/2))/qsin(γ/2), here q=48/4∗3 = 4 ⇒ K

_{d}=0.957662.

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

Explanation: Slots per pole= 48/4 =12 ⇒ γ=180/12=15° and q=48/4&lowest;3=4

coil of the winding is short chorded by one slot pitch ⇒ ε=γ=15°, K

_{w}=K

_{p}&lowest;K

_{d}= cos7.5(sin(4∗15/2))/(4∗sin(15/2)) = (cot7.5)/8.

4. The chording angle for eliminating 5^{th} harmonic should be ____________

a) 30°

b) 34°

c) 36°

d) 35°

View Answer

Explanation: To eliminate 5

^{th}harmonic k

_{p5}must be zero, k

_{pn}= cosnε/2 ⇒ k

_{p5}= cos5ε/2=0=cos90° ⇒ 5ε/2=90° ⇒ ε =36°.

5. Consider the following statements:

(i) breadth factor for third harmonic k_{d3} is more than that for fundamental k_{d1}

(ii) k_{d3} < k_{d1}

(iii) k_{d3} may be less or more than k_{d1} depending upon the number of slots and poles

(iv) coil-span factor for third harmonic k_{p3}>k_{p1} (coil span factor for fundamental)

(v) k_{p3} < k_{p1}

(vi) k_{p3} may be less or more than k_{p1} depending upon the number of slots and poles

From these, the correct answer is

a) (ii),(v)

b) (i),(iv)

c) (iii),(vi)

d) (i),(iii),(iv),(vi)

View Answer

Explanation: Examples will show that k

_{dn}< k

_{d1}and has the effect of reducing the n

^{th}harmonic EMF in comparison with the fundamental EMF, similarly k

_{pn}< k

_{p1}.

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

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 K

_{p}=cosε/2=cos15° and K

_{d}=sin(γq/2)/q(sinγ/2)=sin45/(3sin15)

K

_{w}=K

_{p}∗K

_{d}=cot15°/3√2

7. For eliminating n^{th} 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

Explanation: k

_{pn}=cosnε/2=0 ⇒ nε/2=90° ⇒ ε=180°/n= full pitch/n.

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

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

Explanation: k

_{dA}=(sinσ/2)/σ/2, σ=60° ⇒ k

_{dA}=0.9556

k

_{dB}=(sinσ/2)/σ/2, σ=120° ⇒ k

_{dB}=0.827

thus, k

_{dA}/ k

_{dB}=1.155.

**Sanfoundry Global Education & Learning Series – Electrical Machines.**

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