THREE PHASE SYNCHRONOUS GENERATORS
Introduction:
An electrical machine, which converts mechanical energy into electrical energy of
alternating current in nature, is called an ALTERNATOR or AC GENERATOR. It is also
called as SYNCHRONOUS GENERATOR as its operation is in synchronous with other
generators or other AC sources when it is operated along with them.
Principle of operation:
Synchronous Generator (Alternator) operates on the fundamental principle of Faraday’s
Laws of electromagnetic induction, i.e. whenever the magnetic flux linking the armature
conductor changes, an Electro Motive Force (EMF) is induced in the conductor. The
direction of induced E.M.F. is given by Fleming’s Right-hand rule. When there is a
relative motion between the conductor and magnetic field, the induced EMF is called as
dynamically induced EMF.
In alternators, there are two types of construction
i) Stationary field winding and rotating armature winding-like DC generators
(only for small capacity alternators of few KW rating)
ii) Stationary armature winding and rotating field winding (Suitable for MW size
Synchronous Generators)
The field windings of alternators require direct current for excitation. Excitation is
supplied by a DC generator called an exciter which is mechanically coupled with the
rotor shaft. The DC exciter supplies required power for the rotating field winding to
produce magnetic field. As the prime mover rotates, the rotor of an alternator also rotates
and the stator conductors being stationary are cut by magnetic flux of rotor poles, hence
an E.M.F. is induced in the stator conductors. As the rotor magnetic poles are
alternatively North and South, they induce an alternating E.M.F in the stator conductors.
The frequency of alternating E.M.F. depends on the number of North and South poles
moving past the conductors in one second.
Constructional features and types of 3 phase Synchronous generators:
Compared to the DC generators, the major differences in construction of Synchronous
Generators of MW size are:
a) In a DC generator the armature winding rotates and the field system is stationary
whereas in an Alternator of MW size, the armature winding is mounted on a
stationary frame and the field winding on a rotating frame.
b) The Stationary armature of an alternator is connected directly to load which
receives AC supply and the rotating field winding is connected to DC supply
using two low capacity slip rings. This makes the construction of alternator
simpler.
Advantages of having the stationary armature and rotating field:
a) Insulation of stationary armature conductors working at high voltage is easier.
b) Tapping of electrical energy from a stationary armature is simpler.
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c) The use of slip rings and brushes are eliminated as the load is directly connected
to the alternator terminals
d) The machine can operate at higher speeds enabling a larger output from the
alternator.
Construction of Alternator (Synchronous generator) :
Basically an alternator consist of two parts: a) Stator b) Rotor as shown in Fig1a.
Fig 1a. Construction of Alternator
Fig 1b Stator slots
a) Stator: It consists of stator frame, stator core and windings.
i) Stator Frame: It is a cast iron or welded steel protective frame and gives
support to the entire machine assembly. In small machines it is made of a single
piece of cast iron. In large sized machines, the frame is fabricated by sections of
cast iron sheet steel welded together to form a cylindrical drum.
ii) Stator Core: It is made of special magnetic iron or steel alloy laminations.
They are laminated to minimize the core losses. These laminations are insulated from one
another and pressed together to form the core. Slots are provided on its inner periphery to
house the stator conductors. Slots provided on the stator are of three types: Wide Open,
Semi-Closed and closed are shown in Fig1b.
The wide open type slots are more commonly used because the defective coils can be
easily removed and replaced. The laminations also have openings which make axial and
radial ventilations for the purpose of cooling.
iii) Stator Windings (armature windings): These are insulated copper conductors
housed in stator slots in some specific manner of inter connections.
b) Rotor: It is the rotating part, with North and South poles attached to it. Poles carry
field windings which are supplied with direct current through two slip rings and brushes.
Rotors are of two types:
i)
Salient Pole Type
ii)
Smooth Cylindrical or Non-salient Pole Type
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i) Salient Pole Type: this type of rotor is like a magnetic fly wheel made of cast
iron or steel and a number of alternate North and South poles are bolted to it as
shown in
Fig 2a. The Salient or projecting poles are made of thick steel
laminations, riveted together and fixed to the rotor poles. The ends of the field
windings are connected to the DC supply through slip rings carrying brushes.
Such rotors have large diameter and small axial length.
Fig.2a. Salient-pole rotor
Advantages:
a) Less expensive
b) It provides sufficient space for field coils
Disadvantages:
a) If the salient poles are driven at high speed, it will cause excessive
windage losses and would tend to produce noise.
b) As the rotor is not robust in construction, it cannot withstand mechanical
stress if driven at high speed.
Application: Used for low and medium speed alternators
Examples: Hydro Electric Power Plants, Diesel Power Plants and Gas Turbine
Power Plants.
ii) Smooth Cylindrical or Non-salient Pole Type: These rotors are cylindrical in
construction and are made from solid forged steel alloy having a number of slots
on its outer periphery at regular intervals for accommodating field coils. Field
coils are connected to a DC supply by means of slip rings and brushes for
excitation purpose. The regions forming the central polar areas are left unslotted.
The field coils are so arranged around these polar areas that flux density is
maximum on the central polar areas. These types of rotors are characterized by
small diameter and very long axial length.
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Fig. 2b Phase-wound (Non-salient) Rotor
Advantages:
a) Gives better balance
b) Noiseless operation
c) Less windage loss
d) Better E.M.F. waveform
Application: Used in very high speed turbo alternators.
Example: Steam turbines
Frequency of the induced E.M.F. or Relationship between Speed and Frequency:
Consider an alternator whose rotor is being driven at a constant speed of N r.p.m.
Let P = Number of poles
f = frequency of induced E.M.F.
In one complete revolution of the rotor, each of the North and South poles
move past all the stator conductors. When one pair of North and South poles moves past
the armature conductor, the induced E.M.F. undergoes one full cycle. Therefore in a Ppole machine, in one complete revolution of the field system, the induced E.M.F. in the
armature conductors will complete
P
2
cycles of waveform.
Number of cycles of EMF induced/second= no. of cycles of emf per revolution x No. of
revolutuions per sec
i.e. Frequency in cycles/second =
i.e.
f
P N

2 60
PN
Hz
120
or
N
120 f
r.p.m
P
Note: In order to keep the frequency constant, the speed N must remain unchanged.
Synchronous generator connected to grid runs at a constant speed known as synchronous
speed in steady state.
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E.M.F. Equation:
Consider a 3-Phase alternator with P-Poles driven at a constant speed of N r.p.m.
Let Z = Number of conductors or coil sides in series per phase
= 2T ----where T is the number of turns per phase.
 = useful flux/pole in Webers
f = frequency in Hz
Kd = distribution factor or winding factor or breadth factor
Kp = pitch factor
Consider an armature conductor on the stator of an alternator. Let the alternator
rotor move through one revolution in t=60/N seconds. In one revolution of the rotor, all
the P poles on the rotor move past each of the stator conductors.
Flux cut by the conductor in one rev olution P Wb
According to Faraday’s second law of electromagnetic induction,
Av erageE.M.F.induced in the conductor  E av 
d
dt
where d = flux cut/stator conductor in one revolution of rotor = P
dt = time taken for one revolution of rotor =
So the average emf induced per conductor Eav=
Wb
60
seconds
N
=
The emf induced in Z number of conductors, Eav=
But N= 120f/P ,
Eav=
RMS value of emf/phase =1.11 x
=
==
=
=
volt
The above equation of induced E.M.F./phase is true only if the winding is concentrated
in one slot. But practically the winding for each phase under each pole is distributed and
for such cases Kp and Kd are considered.
Thus, E.M.F./phase will be, E r.m.s./phase = 4.44fTKpKd Volts.
If the alternator is star connected, then the line voltage is
El =
x 4.44fTKpKd Volts.
Winding factors:(Kw=KpKd)
For the following advantages windings always short pitched by 1 or 2 slots
•
The primary advantage of short-pitch coils is we achieve saving of copper.
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•
Short pitched winding reduces the MMF harmonics produced by the armature
winding .
•
Short pitched winding reduces the EMF harmonics induced in the windings,
without reducing the magnitude of the fundamental EMF wave to a great extent.
Pitch factor Kp= emf induced in a short pitched coil/ emf induced in a full pitched coil
Kp = cos α/2
where α is called short pitched angle.
Note: For full pitched windings Kp = 1
Fig. 3a Resultant emf for full pitched coil
Fig. 3b Resultant emf for short pitched coil
Concentrated windings in which all conductors of a given phase per pole are concentrated
in a single slot, are not commercially used because they have the following
disadvantages,
1. They fail to use the entire inner periphery of the stator iron efficiently.
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2. They make it necessary to use extremely deep slots where the windings are
concentrated. This causes an increase in the mmf required to setup the desired airgap
flux.
3. Deep slots also increase the armature leakage flux and the armature reactance.
4. They result in low copper-to-iron ratios by not using the armature iron completely.
5. They fail to reduce harmonics as effectively as distributed windings.
Distribution factor Kd = (emf induced in a distributed winding)/ (emf induced in a
concentrated winding)
= vector sum of the emf/ arithmetic sum of the emf
Distribution factor Kd = ( sin mβ/2) / (m sin β/2)
where m = number of slots per pole per phase, β = slot angle
Example Problems:
1. A 3Φ, 50 Hz, star connected salient pole alternator has 216 slots with 5
conductors per slot. The winding is distributed with Kd= 0.959 and full pitched.
The flux per pole is 30 mwb and the alternator runs at 250 rpm. Determine the
phase and line voltages of emf induced.
Solution: Ns = 250 rpm, f = 50 Hz, P = 120 x f/Ns = 120 x 50/250 = 24 poles
Kd= 0.9597 , Kp = 1 for full pitched winding.
Total no of conductor Z = conductor/ slot x number of slots
= 216 x 5 = 1080
Conductor per phase Zph= 1080/3=360
Turns per phase Tph= Zph/2 = 360/2=180
Therefore Eph = 4.44 KpKd f Ф Tph = 4.44 x 1 x 0.9597 x 50 x 30 m x 180
= 1150.488 volts
Hence the line Voltage EL = √3 Eph = √3 x1150.488 = 1992.65 volts
2. A three phase 16-pole alternator has a star connected winding with 144slots and
10 conductors/slot. The flux/pole is 0.03Wb and the speed is 375 r.p.m. Find the
frequency, the phase E.M.F. and line E.M.F. Assume full pitched coils.
f=PN/120=(16*375)/120=50 Hz
n=slots/pole
=144/16=9
slot angle= β
β=180/n=180/9=20 degree electrical
m=(slots/pole)/phase
=144/(16x3)=3
Kd = ( sin mβ/2) / (m sin β/2)
= sin(3*20/2)/(3*sin(20/2))=0.9598
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Total no of conductor Z = number of slots x
conductor/ slot
= 144 x 10 = 1440
Conductor per phase Zph= 1440/3=480
Turns per phase Tph= Zph/2 = 480/2=240
Therefore Eph = 4.44 Kp Kd f Ф Tph
= 4.44 x 1 x 0.9598 x 50 x 0.030 x 240
= 1534.144 volts
The line Voltage EL = √3 Eph
= √3 x1534.144
= 2657.215 volts
3. A 8 pole, 750 rpm three phase star connected alternator has 72 slots, each of
which having 10 conductors. Calculate the RMS value of the emf per phase if the
flux per pole is 0.1 wb and the winding factor is 0.96.
Solution: Winding factor Kw=Kp*Kd= 0.96(given)
P=8, Speed N=750 rpm
f=PN/120=(8*750/120)=50 Hz
Total no of conductor Z = number of slots x conductor/ slot
= 72 x 10 = 720
Conductor per phase Zph= 720/3=240
Turns per phase Tph= Zph/2 = 240/2=120
Eph = 4.44 Kw f Ф Tph
= 4.44 x 0.96 x 50 x 0.10 x 120
= 2557.44 volts
The line Voltage EL = √3 Eph
= √3 x2557.44
= 4429.616 volts
4. A 8 pole, three phase, 50 Hz, star connected a.c. generator has 24 stator slots.
Find the number of conductors per slot if the flux per pole is 60mWb and the
terminal voltage is 1100 V. Assume full pitch coils
Solution: Given P=8, f=50, S=24
N=120f/P=(120*50)/8=750 rpm
n=slots/pole
=24/8=3
slot angle= β
β=180/3=60 degree electrical
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m=(slots/pole)/phase
=24/(8x3)=1
Kd = ( sin(mβ/2)) / (m sin (β/2))
= sin(1*60/2)/(1*sin(60/2))=1
Kp=1; full pitched coils
Flux = Ф=0.06 wb
Terminal voltage=1100 V(line voltage given)
Eph=E1/sqrt(3)
= 1100/sqrt(3)=635.085 V
Total no of conductors Z = to be determined
Eph = 4.44 Kp Kd f Ф Tph
Therefore 635.085 = 4.44 Kp Kd f Ф Tph
= 4.44 x 1 x 1 x 50 x 0.06 x Tph
Tph=47.68 ≈48
Conductors per phase= Zph=48*2 =96
Z=Zph*number of phases=96*3=288
Conductors/slot=288/24=12
5. A 4 pole, three phase, 50 Hz, star connected a.c. generator has 24 stator slots.
Find the number of conductors per slot if the flux per pole is 61.7mWb and the
terminal voltage is 1100 V. Assume full pitch coils.
Given P=4, f=50, S=24
N=120f/P=(120*50)/4=1500 rpm
n=slots/pole
=24/4=6
slot angle= β
β=180/6=30 degree electrical
m=(slots/pole)/phase
=24/(4x3)=2
Kd = ( sin(mβ/2)) / (m sin (β/2))
= sin(2*30/2))/(2*sin(30/2))=0.9659
Kp=1; full pitched coils
Flux = Ф=0.06 wb
Terminal voltage=1100 V(line voltage given)
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Eph=E1/sqrt(3)
= 1100/sqrt(3)=635.085 V
Total no of conductors Z = to be determined
Eph = 4.44 Kp Kd f Ф Tph
Therefore 635.085 = 4.44 Kp Kd f Ф Tph
= 4.44 x 1 x 0.9659 x 50 x 0.0617 x Tph
Tph=48.002 ≈48
Conductors per phase= Zph=48*2 =96
Z=Zph*number of phases=96*3=288
Conductors/slot=288/24=12
6. A 8 pole, 750 rpm three phase star connected alternator has 72 slots, each of
which is having 10 conductors. Calculate the RMS value of the emf per phase and
line voltage. The flux per pole is 0.1 wb. The coils are short pitched by one slot
and the windings are distributed
P=8, Speed N=750 rpm
f=PN/120=(8*750/120)=50 Hz
Total no of conductor Z = number of slots x conductor/ slot
= 72 x 10 = 720
Conductor per phase Zph= 720/3=240
Turns per phase Tph= Zph/2 = 240/2=120
n=slots/pole
=72/8=9
slot angle= β =180/9=20 degree electrical
m=(slots/pole)/phase
=72/(8x3)=3
Coils are short pitched by one slot
hence short pitch angle α= β =20 degree
Kp= cos (α/2)= cos(20/2)=0.9848
Kd = ( sin(mβ/2)) / (m sin (β/2))
= sin(3*20/2))/(3*sin(20/2))=0.9598
Kw=Kp*kd =0.9848*0.9598 =0.9452
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Therefore Eph = 4.44 Kw f Ф Tph
= 4.44 x 0.9452 x 50 x 0.10 x 120
= 2518.0128 volts
The line Voltage EL = √3 Eph
= √3 x2518.0128 =4361.326 volts
7. A three phase 16-pole alternator has a star connected winding with 144slots and
10 conductors/slot. The flux/pole is 0.03Wb and the speed is 375 r.p.m. Find the
frequency, the phase E.M.F. and line E.M.F. The coils are short chorded by two
slots.
f=PN/120=(16*375)/120=50 Hz
n=slots/pole =144/16=9
slot angle=β=180/n=180/9=20 degree electrical
m=(slots/pole)/phase
=144/(16x3)=3
Short pitch angle α= 2*β =2*20=40 degree
Kp= cos (40/2)= cos(40/2)=0.9397
Kd = ( sin mβ/2) / (m sin β/2)
= sin(3*20/2)/(3*sin(20/2))=0.9598
Total no of conductor Z = number of slots x conductor/ slot
= 144 x 10 = 1440
Conductor per phase Zph= 1440/3=480
Turns per phase Tph= Zph/2 = 480/2=240
Therefore Eph = 4.44 Kp Kd f Ф Tph
= 4.44 x 0.9397 x 0.9598 x 50 x 0.030 x 240
= 1441.635 volts
The line Voltage EL = √3 Eph
= √3 x 1441.635 = 2496.985 volts
8. A three phase star connected alternator with 12 poles generates 1100V on open
circuit at a speed of 500 rpm. Assume 180 turns per phase, a distribution factor of
0.9598 and coils are short pitched by one slot. Find useful flux per pole,
frequency, winding factor and estimate the number of stator slots and
conductors/slot if slots per pole per phase is 3.
f=PN/120=(12*500)/120=50 Hz
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Kd = ( sin(mβ/2)) / (m sin (β/2))=0.9598(given)
Slots/pole/phase=m=3
Kd= sin(3*β/2))/(3*sin(β/2))=0.9598
Solving, we get, slot angle =β=20 degree ele
•
n=slots/pole = 180/β =180/20=9
Total number of slots= S=(slots/pole)*poles
=9*12=108
Short pitch angle α= 1*β =1*20=20 deg
Kp = cos (20/2)= 0.9848
Kw=Kp*kd =0.9848*0.9598 =0.9452
Eph=1100/√3=635.085
•
Therefore Eph = 4.44 Kw f Ф Tph
635.085= 4.44 x 0.9452 x 50 x Ф x 180
Ф = 0.0168144 wb
Total conductors Z=180*3*2=1080
Conductors/slot=1080/108=10
9. A three phase 60 Hz, 16-pole alternator has a star connected winding with 144
slots and 5 conductors/layer. The winding is double layer. The flux/pole is 24.8
mwb is sinusoidally distributed. Find the speed, the phase E.M.F. and line E.M.F.
The coils are short chorded by one slot.
N=120f/P=(120*60)/16=450 rpm
n=slots/pole
=144/16=9
slot angle=β=180/n=180/9=20 degree electrical
m=(slots/pole)/phase
=144/(16x3)=3
Short pitch angle α= 1*β =1*20=20 degree
Kp= cos (20/2)= cos(20/2)=0.9848
Kd = ( sin mβ/2) / (m sin β/2)
= sin(3*20/2)/(3*sin(20/2))=0.9598
Total no of conductors Z
12
Z= number of slots *(conductors/ layer) *number of layers
= 144 x 5x2 = 1440
Conductor per phase Zph= 1440/3=480
Turns per phase Tph= Zph/2 = 480/2=240
Therefore Eph = 4.44 Kp Kd f Ф Tph
= 4.44 x 0.9848 x 0.9598 x 60 x 0.0248 x 240
= 1498.739 volts
The line Voltage EL = √3 Eph
= √3 x 1498.739 = 2595.892 volts
10. A 4 pole, three phase, 50 Hz, star connected alternator has 60 stator slots with 4
conductors/slot. Coils are short pitched by 3 slots. If the Phase spread is 60
degree, find the line voltage induced for a flux of 0.943 wb sinusoidally
distributed in space All the turns per phase are in series. Calculate the speed of the
alternator
N=120f/P=(120*50)/4=1500 rpm
n=slots/pole =60/4=15
Phase spread=m β=60 deg
m=(slots/pole)/phase =60/(4x3)=5
slot angle=β=60/5=12 degree electrical
Also, β=180/n=180/15=12 deg
Short pitch angle α= 3*β =3*12=36 degree
Kp= cos (α /2)= cos(36/2)=0.9511
Kd = ( sin mβ/2) / (m sin β/2)
= sin(5*12/2)/(5*sin(12/2))=0.9567
Total no of conductors Z
Z= number of slots *(conductors/ slot)
= 60 x 4 = 240
Conductor per phase Zph= 240/3=80
Turns per phase Tph= Zph/2 = 80/2=40
Therefore Eph = 4.44 Kp Kd f Ф Tph
= 4.44 x 0.9511 x 0.9567 x 50 x 0.943 x 40
= 7619.5025 volts
The line Voltage EL = √3 Eph
= √3 x 7619.5025 = 13197.366 volts
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11. A 4 pole, three phase, 50 Hz, star connected alternator slots/pole/phase=4 with 4
conductors/slot arranged in two layer winding. Coil span=150 deg electrical. Find
the no load line voltage induced emf for a flux of 0.12 wb sinusoidally distributed
in space. Also calculate the winding factor
m=(slots/pole)/phase = 4(given)
Total number of slots=4*3*4=48
n=slots/pole=4*3=12
slot angle=β=180/n=180/12=15 degree ele
Coil span=150 deg electrical
Short pitch angle=180-coil span(in degree ele)
Short pitch angle α= 180-150 =30 degree
Kp= cos (α /2)= cos(30/2)=0.9659
Kd = ( sin mβ/2) / (m sin β/2)
= sin(4*15/2)/(4*sin(15/2))=0.9577
Winding factor=Kp*Kd=0.9659*0.9577=0.9250
Total no of conductors Z
Z= number of slots *(conductors/ slot)
= 48 x 4 = 192
Conductor per phase Zph= 192/3=64
Turns per phase Tph= Zph/2 = 64/2=32
Therefore Eph = 4.44 Kp Kd f Ф Tph
= 4.44 x 0.9659 x 0.9577 x 50 x 0.12 x 32
= 788.5802 volts
The line Voltage EL = √3 Eph
= √3 x 788.5802 = 1365.861 volts
12. Calculate the no-load terminal voltage of a three-phase, 8 pole, star connected
alternator running at 750 rpm having following data:
flux/pole= 50mwb , Stator slots=72 , Number of conductors/slot=10. Calculate the
percentage rise in open circuit terminal voltage if the windings were concentrated
and accommodated in 24 slots with depth increased to house 30 conductors per
slot
14
Solution: case-1 : with72 slots
m=(slots/pole)/phase = (72/8)/3 =3
n=slots/pole=72*8=9
slot angle=β=180/n=180/9=20 degree ele
Coil span=180 deg electrical (Assuming full pitched coils)
Short pitch angle α= 180-180 =0 degree
Kp= cos (α /2)= cos(0/2)=1
Kd = ( sin mβ/2) / (m sin β/2)
= sin(3*20/2)/(3*sin(20/2))=0.9598
Winding factor=Kp*Kd=1*0.9598=0.9598
Total no of conductors Z
Z= number of slots *(conductors/ slot)
= 72 x 10 = 720
Conductor per phase Zph= 720/3=240
Turns per phase Tph= Zph/2 = 240/2=120
Therefore Eph = 4.44 Kp Kd f Ф Tph
= 4.44 x 1 x 0.9598 x 50 x 0.05 x 120
= 1278.4536 volts
The line Voltage EL = √3 Eph
= √3 x 1278.4536 = 2214.3466 volts
Solution: case-2 : with 24 slots
m=(slots/pole)/phase = (24/8)/3 =1
n=slots/pole=24*8=3
slot angle=β=180/3=180/3=60 degree ele
Coil span=180 deg electrical (Assuming full pitched coils)
Short pitch angle α= 180-180 =0 degree
Kp= cos (α /2)= cos(0/2)=1
Kd = ( sin mβ/2) / (m sin β/2)
= sin(1*60/2)/(1*sin(60/2))=1
Winding factor=Kp*Kd=1*1=1
Total no of conductors Z
Z= number of slots *(conductors/ slot)
= 24 x 30 = 720
15
Conductor per phase Zph= 720/3=240
Turns per phase Tph= Zph/2 = 240/2=120
Therefore Eph = 4.44 Kp Kd f Ф Tph
= 4.44 x 1 x 1 x 50 x 0.05 x 120
= 1332 volts
The line Voltage EL = √3 Eph
= √3 x 1332 = 2307.0917 volts
Percentage increase in terminal voltage
=((2307.0917- 2214.3466)/ 2214.3466)*100
=4.19%
Tutorial Problems:
1. A three phase 16-pole alternator has a star connected winding with 144slots and
10conductors/slot. The flux/pole is 0.03Wb and the speed is 375r.p.m. Find the
frequency, the phase E.M.F. and line E.M.F. assume pitch factor Kp = 1 and
distribution factor Kd = 0.96 [50Hz, 1534.464, 2657.77V]
2. A 6 pole, 3 phase, 50Hz alternator has an armature with 90 slots and 8 conductors
per slot and revolves at 1000rpm. The flux per pole being 0.05Wb. Calculate the
emf generated if the winding factor is 0.97 and all the conductors in each phase
are in series.(1292.04V/phase)
3. A 3 phase 16 pole alternator has a star connected winding with 144 slots and 10
conductors per slot. The flux per pole is 0.03Wb and the speed is 375 rpm. Find
the frequency and the phase and line emf if the distribution factor Kd=0.96 and
the pitch factor =1.[50Hz,2657.76V]
4. A 3 phase , 50Hz star connected alternator has 200 conductors per phase and flux
per pole of 0.06Wb. Find i) emf generated per phase ii) emf between terminals.
Assume the winding to be full pitched and distribution factor to be 0.966.
[1286.7,2228.65V]
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