SCHME OF EVALUTION

II/IV B.Tech(Regular) DEGREE EXAMINATION

April,2017 Electronics & Instrumentation Engineering

14EI 403 Electrical Technology Max.Marks:60 marks

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1.Answer all questions. (1×12=12Marks)

Answer one question from each unit. (4×12=48Marks)

a) The function of commutator in DC-Machines is convert alternating emf into

unidirectional emf and vice varsa.

b) The voltage balance equation for DC-Generator,

c) short shunt compound generator:

d) Slip corresponding to backward torque is (2-S)

e) Core with high permeability (or)high grade silicon steel core

f) Poles=4,Frequency=50 Hz,Speed(N)=1000 rpm,Slip=Ns-Nr/Ns

Ns=120×f/p=120×50/4=1500 rpm.Slip=1500-1000/1500=0.3336.

g) Applications of stepper motors convers,tape readers,floppys,disk drives,cnc lathes,paper

feeder on printers

h) or synchronous speed.

i) Star delta starter or DOL starter or Auto-transformer starter.

j) Application of Synchronous Motor:-i) Synchronous motor having no load connected to its

shaft is used for power factor improvement.

ii) Synchronous motor finds application where operating speed is less (around 500 rpm)

and high power is required. For power requirement from 35 kW to 2500 KW, the size,

weight and cost of the corresponding three phase induction motor is very high. Hence

these motors are preferably used. Ex- Reciprocating pump, compressor, rolling mills.

k) This is the angle subtended by the single slot. β=1800/No. of slots/pole

l) i) Heat dissipation is less ii) space required for the armature winding less iii) Size of the

machine is less.

UNIT-I

2 EMF EQUATION OF DC GENERATOR = 3M & DERIVATION =3M, Total=6M

(a) EMF EQUATION OF DC GENERATOR [6M]

DC Generator EMF Equation is

Simple wave wound generator Numbers of parallel paths are only 2 = A Therefore, Induced

emf

for wave type of winding generator is

Simple lap-wound generator Here, number of parallel paths is equal to number of

conductors in one path i.e. P = A Therefore, Induced emf for lap-wound generator is

2 b) CALCULATED EMF = 3M, NEW SPEED = 3M TOTAL=6M

(OR)

3 EXPLANATION=2M,OPERATION=2M,DIAGRAM OF 3-POINT STARTER=2M (6M)

a) Three Point Starter:

Construction:

Working Of Three Point Starter:

1) Initially when a DC supply is switched ON with handle in the OFF position.

2) The handle is now moved clockwise to the first stud, the shunt field winding is directly

connected across the supply while the whole starting resistance is inserted in series with the

armature circuit.

3) As the handle is gradually moved over to the final stud, the starting resistance is cut out

of the armature circuit in steps. The handle is now held magnetically by the no volt release

coil which is energized by shunt field current.

4) If the supply voltage is suddenly interrupted or if the field excitation is accidently cut, the

no volt release coil is demagnetized and the handle goes back to the OFF position under the

pull of the spring. If no volt coil were not used, then in case of failure of supply. The handle

would remain on the final stud. If then supply is restored, the motor will be directly

connected across the supply, resulting in an excessive armature current.

5) If the motor is overloaded (or any fault occurs) it will draw excessive current from the

supply. This current will increase the ampere turns of the overload release coil and pull the

armature, thus short circuited the no volt release coil. The no volt coil is demagnetized and

the handle is pulled to the OFF position by the spring .thus the motor is automatically

disconnected from the supply.

3 b) CALCULATED SPEED =4M, VOLTAGE EQUATION=2M TOTAL=6M

UNIT-II

4 DERIVE THE PARAMETERS OF SINGLE PHASE TRANSFORMER=3M

O.C & S.C TEST DATA=3 M

TOTAL=6M

a) 1. O.C. and S.C. Tests on Single Phase Transformer

The efficiency and regulation of a transformer on any load condition and at any

power factor condition can be predetermined by indirect loading method. In this

method, the actual load is not used on transformer. But the equivalent circuit

parameters of a transformer are determined by conducting two tests on a transformer

which are,

2. Short circuit test (S.C.Test)

The parameters calculated from these test results are effective in determining the

regulation and efficiency of a transformer at any load and power factor condition, without

actually loading the transformer. The advantage of this method is that without much power

loss the tests can be performed and results can be obtained. Let us discuss in detail how to

perform these tests and how to use the results to calculate equivalent circuit parameters.

1.1 Open Circuit Test (O.C. Test)

The experimental circuit to conduct O.C test is shown in the Fig. 1.

Fig 1. Experimental circuit for O.C. test

The transformer primary is connected to a.c. supply through ammeter, wattmeter and

variac. The secondary of transformer is kept open. Usually low voltage side is used as

primary and high voltage side as secondary to conduct O.C test.

The primary is excited by rated voltage, which is adjusted precisely with the help of a

variac. The wattmeter measures input power. The ammeter measures input current. The

voltemeter gives the value of rated primary voltage applied at rated frequency.

Sometimes a voltmeter may be connected across secondary to measure secondary

voltage which is V2 = E2 when primary is supplied with rated voltage. As voltmeter

resistance is very high, though voltmeter is connected, secondary is treated to be open

circuit as voltmeter current is always negligibly small.

When the primary voltage is adjusted to its rated value with the help of variac, readings

of ammeter and wattmeter are to be recorded.

The observation table is as follows

Vo = Rated voltage

Wo = Input power

Io = Input current = no load current

As transformer secondary is open, it is on no load. So current drawn by the primary is

no load current Io. The two components of this no load current are,

Im = Io sin Φo

Ic = Io cos Φo

where cos Φo = No load power factor

And hence power input can be written as,

Wo = Vo Io cos Φo

The phasor diagram is shown in the Fig. 2.

Fig. 2

As secondary is open, I2 = 0. Thus its reflected current on primary is also zero. So we

have primary current I1 =Io. The transformer no load current is always very small, hardly 2

to 4 % of its full load value. As I2 = 0, secondary copper losses are zero. And I1 = Io is very

low hence copper losses on primary are also very very low. Thus the total copper losses in

O.C. test are negligibly small. As against this the input voltage is rated at rated frequency

hence flux density in the core is at its maximum value. Hence iron losses are at rated

voltage. As output power is zero and copper losses are very low, the total input power is

used to supply iron losses. This power is measured by the wattmeter i.e. Wo. Hence the

wattmeter in O.C. test gives iron losses which remain constant for all the loads.

... Wo = Pi = Iron losses

Calculations : We know that,

Wo = Vo Io cos Φ

cos Φo = Wo /(Vo Io ) = no load power factor

Once cos Φo is known we can obtain,

Ic = Io cos Φo

and Im = Io sin Φo

Once Ic and Im are known we can determine exciting circuit parameters as,

Ro = Vo /Ic Ω

and Xo = Vo /Im Ω

Note : The no load power factor cos Φo is very low hence wattmeter used must be low

power factor type otherwise there might be error in the results. If the meters are connected

on secondary and primary is kept open then from O.C. test we get Ro'and Xo' with which we

can obtain Ro and Xo knowing the transformation ratio K.

1.2 Short Circuit Test (S.C. Test)

In this test, primary is connected to a.c. supply through variac, ammeter and voltmeter

as shown in the Fig. 3.

Fig. 3 Fig 1. Experimental circuit for O.C. test

The secondary is short circuited with the help of thick copper wire or solid link. As high

voltage side is always low current side, it is convenient to connect high voltage side to

supply and shorting the low voltage side.

As secondary is shorted, its resistance is very very small and on rated voltage it may

draw very large current. Such large current can cause overheating and burning of the

transformer. To limit this short circuit current, primary is supplied with low voltage which is

just enough to cause rated current to flow through primary which can be observed on an

ammeter. The low voltage can be adjusted with the help of variac. Hence this test is also

called low voltage test or reduced voltage test. The wattmeter reading as well as voltmeter,

ammeter readings are recorded. The observation table is as follows,

Now the current flowing through the windings are rated current hence the total copper

loss is full load copper loss. Now the voltage supplied is low which is a small fraction of the

rated voltage. The iron losses are function of applied voltage. So the iron losses in reduced

voltage test are very small. Hence the wattmeter reading is the power loss which is equal to

full load copper losses as iron losses are very low.

... Wsc = (Pcu) F.L. = Full load copper loss

Calculations : From S.C. test readings we can write,

Wsc = Vsc Isc cos Φsc

... cos Φsc = Vsc Isc /Wsc = short circuit power factor

Wsc = Isc2 R1e = copper loss

... R1e =Wsc /Isc2

while Z1e =Vsc /Isc = √(R1e2

+ X1e2)

... X1e = √(Z1e

2 - R1e2)

Thus we get the equivalent circuit parameters R1e, X1e and Z1e. Knowing the

transformation ratio K, the equivalent circuit parameters referred to secondary also can be

obtained.

Note : In short, if meters are connected to primary of transformer in S.C. test, calculations

give us R1e and Z1e if meters are connected to secondary of transformer in S.C. test

calculations give us R2e and Z2e.

1.3 Calculation of Efficiency from O.C. and S.C. Tests

We know that,

From O.C. test, Wo = Pi

From S.C. test, Wsc = (Pcu) F.L.

Thus for any p.f. cos Φ2 the efficiency can be predetermined. Similarly at any load which is

fraction of full load then also efficiency can be predetermined as,

where n = fraction of full load

4 b) CALCULATION OF EFFICIENCY = 3M, FULL LOAD EFFICIENCY = 3M

TOATAL=6M

(OR)

5 DRAW=2M,EXPLANATION=2M,EQUIVALENT OF TRANSFORMER=2M (6M)

a) Equivalent Circuit of Transformer

Equivalent impedance of transformer is essential to be calculated because the electrical

power transformer is an electrical power system equipment for estimating different

parameters of electrical power system which may be required to calculate total internal

impedance of an electrical power transformer, viewing from primary side or secondary side

as per requirement. This calculation requires equivalent circuit of transformer referred to

primary or equivalent circuit of transformer referred to secondary sides respectively.

Percentage impedance is also very essential parameter of transformer. Special attention is to

be given to this parameter during installing a transformer in an existing electrical power

system. Percentage impedance of different power transformers should be properly matched

during parallel operation of power transformers. The percentage impedance can be derived

from equivalent impedance of transformer so, it can be said that equivalent circuit of

transformer is also required during calculation of % impedance.

Equivalent Circuit of Transformer Referred to Primary

For drawing equivalent circuit of transformer referred to primary, first we have to establish

general equivalent circuit of transformer then, we will modify it for referring from primary

side. For doing this, first we need to recall the complete vector diagram of a transformer

which is

shown the above figure.

Let us consider the transformation ratio be,

In the figure above, the applied voltage to the primary is V1 and voltage across the primary

winding is E1. Total current supplied to primary is I1. So the voltage V1 applied to the

primary is partly dropped by I1Z1 or I1R1 + j.I1X1 before it appears across primary winding.

The voltage appeared across winding is countered by primary induced emf E1. So voltage

equation of this portion of the transformer can be written as,

The equivalent circuit for that equation can be drawn as below,

From the vector diagram above, it is found that the total primary current I1 has two

components, one is no - load component Io and the other is load component I2′. As this

primary current has two components or branches, so there must be a parallel path with

primary winding of transformer. This parallel path of current is known as excitation branch

of equivalent circuit of transformer. The resistive and reactive branches of the excitation

circuit can be represented as

The load component I2′ flows through the primary winding of transformer and induced

voltage across the winding is E1 as shown in the figure right. This induced voltage E1

transforms to secondary and it is E2 and load component of primary current I2′ is

transformed to secondary as secondary current I2. Current of secondary is I2. So the voltage

E2 across secondary winding is partly dropped by I2Z2 or I2R2 + j.I2X2 before it appears

across load. The load voltage is V2.

The complete equivalent circuit of transformer is shown below.

Now if we see the voltage drop in secondary from primary side, then it would be ′K′ times

greater and would be written as K.Z2.I2. Again I2′.N1 = I2.N2

Therefore,

From above equation, secondary impedance of transformer referred to primary is,

So, the complete equivalent circuit of transformer referred to primary is shown in the figure

below,

5 b) H.V SIDE PARAMETERS = 3M, L.V SIDE PARAMETERS = 3M

TOTAL=6M

UNIT-III

6 DERIVATION OF TORQUE=3M, TOTAL=6M

3-PHASE INDUCTION MOTOR RUNNING CONDITION OF TORQUE=3M

a) Torque Equation of Three Phase Induction Motor

The torque produced by three phase induction motor depends upon the following three

factors: Firstly the magnitude of rotor current, secondly the flux which interact with the

rotor of three phase induction motor and is responsible for producing emf in the rotor part of

induction motor, lastly the power factor of rotor of the three phase induction motor.

Combining all these factors together we get the equation of torque as-

Where, T is the torque produced by induction motor, φ is flux responsible for producing

induced emf, I2 is rotor current, cosθ2 is the power factor of rotor circuit.The flux φ

produced by the stator is proportional to stator emf E1. i.e φ ∝ E1 We know that

transformation ratio K is defined as the ratio of secondary voltage (rotor voltage) to that of

primary voltage (stator voltage).

Rotor current I2 is defined as the ratio of rotor induced emf under running condition , sE2 to

total impedance, Z2 of rotor side, and total impedance Z2 on rotor side is

given by , Putting this value in above equation we get,

s= slip of Induction motor

We know that power factor is defined as ratio of resistance to that of impedance. The power

factor of the rotor circuit is

Putting the value of flux φ, rotor current I2, power factor cosθ2 in the equation of torque we

get, Combining similar term we get,

Removing proportionality constant we get,

Where ns is synchronous speed in r. p. s, ns = Ns / 60. So, finally the equation of torque

becomes, Derivation of K in torque

equation.

In case of three phase induction motor, there occur copper losses in rotor. These rotor

copper losses are expressed as Pc = 3I22R2 We know that rotor current,

Substitute this value of I2 in the equation of rotor copper losses, Pc. So, we get

The ratio of P2 : Pc : Pm = 1 : s : (1 - s) Where, P2 is the

rotor input, Pc is the rotor copper losses, Pm is the mechanical power developed.

Substitute the value of Pc in above equation we get,

On simplifying we get, The

mechanical power developed Pm = Tω, Substituting the value of Pm

We know that the rotor speed N = Ns(1 - s) Substituting this value of rotor speed in above

equation we get,

Ns is speed in revolution per minute (rpm) and ns is speed in revolution per sec (rps) and the

relation between the two is Substitute this value of Ns in above equation and

simplifying it we get

Comparing both the equations, we get, constant K = 3 / 2πns

Equation of Starting Torque of Three Phase Induction Motor

Starting torque is the torque produced by induction motor when it is started. We know that

at start the rotor speed, N is zero.

So, the equation of starting torque is easily obtained by simply putting the value of s = 1 in

the equation of torque of the three phase induction motor,

The starting torque is also known as standstill torque.

6 b) CALCULATE Ns & Nr = 2M, SPEED AT FULL LOAD = 2M, RATIO OF

MAX.TORQUE =2M. TOTAL =6M

(OR)

7 EXPLANATION OF SINGLE PHASE INDUCTION MOTOR=3M

SELF STARTING METHODS=3M TOTAL = 6M

a) The reason behind this is very interesting. Single phase induction motor has distributed

stator winding and a squirrel-cage rotor. When fed from a single phasesupply, its stator

winding produces a flux ( or field ) which is only alternating i.e. onewhich alternates

along one space axis only.

1. Single phase induction motors or asynchronous motors.

2. Single phase synchronous motors.

3. Commutator motors.

This article will provide fundamentals, description and working principle of single phase

induction motor.

Construction of Single Phase Induction Motor

Stator: As its name indicates stator is a stationary part of induction motor. A single phase ac

supply is given to the stator of single phase induction motor. Rotor: The rotor is a rotating

part of induction motor. The rotor is connected to the mechanical load through the shaft.

The rotor in single phase induction motor is of squirrel cage rotor type. The construction of

single phase induction motor is almost similar to the squirrel cage three phase motor except

that in case of asynchronous motor the stator have two windings instead of one as compare

to the single stator winding in three phase induction motor.

Stator of Single Phase Induction Motor

The stator of the single phase induction motor has laminated stamping to reduce eddy

current losses on its periphery. The slots are provided on its stamping to carry stator or main

winding. In order to reduce the hysteresis losses, stamping are made up of silicon steel.

When the stator winding is given a single phase ac supply, the magnetic field is produced

and the motor rotates at a speed slightly less than the synchronous speed Ns which is given

by

Where, f = supply voltage frequency, P = No. of poles of the motor. The construction of the

stator of asynchronous motor is similar to that of three phase induction motor except there

are two dissimilarity in the winding part of the single phase induction motor.

Firstly the single phase induction motors are mostly provided with concentric coils. As the

number of turns per coil can be easily adjusted with the help of concentric coils, the mmf

distribution is almost sinusoidal.

1. Except for shaded pole motor, the asynchronous motor has two stator windings namely

the main winding and the auxiliary winding. These two windings are placed in space

quadrature with respect to each other.

2. Rotor of Single Phase Induction Motor

The construction of the rotor of the single phase induction motor is similar to

the squirrel cage three phase induction motor. The rotor is cylindrical in shape and has slots

all over its periphery. The slots are not made parallel to each other but are bit skewed as the

skewing prevents magnetic locking of stator and rotor teeth and makes the working of

induction motor more smooth and quieter i.e less noise. The squirrel cage rotor consists of

aluminum, brass or copper bars. These aluminum or copper bars are called rotor conductors

and are placed in the slots on the periphery of the rotor. The rotor conductors are

permanently shorted by the copper or aluminum rings called the end rings. In order to

provide mechanical strength these rotor conductor are braced to the end ring and hence form

a complete closed circuit resembling like a cage and hence got its name as squirrel cage

induction motor.

Working Principle of Single Phase Induction Motor

This alternating current produces an alternating flux called main flux. This main flux

also links with the rotor conductors and hence cut the rotor conductors. According to the

Faraday’s law of electromagnetic induction, emf gets induced in the rotor. As the rotor

circuit is closed one so, the current starts flowing in the rotor. This current is called the rotor

current. This rotor current produces its own flux called rotor flux. Since this flux is

produced due to induction principle so, the motor working on this principle got its name as

induction motor. According to double field revolving theory, any alternating quantity can be

resolved into two components, each component have magnitude equal to the half of the

maximum magnitude of the alternating quantity and both these component rotates in

opposite direction to each other. For example - a flux, φ can be resolved into two

components

Each of these components rotates in opposite direction i. e if one φm / 2 is rotating in

clockwise direction then the other φm / 2 rotates in anticlockwise direction. When a single

phase ac supply is given to the stator winding of single phase induction motor, it produces

its flux of magnitude, φm. According to the double field revolving theory, this alternating

flux, φm is divided into two components of magnitude φm /2. Each of these components will

rotate in opposite direction, with the synchronous speed, Ns.

Now at starting, both the forward and backward components of flux are exactly opposite to

each other. Also both of these components of flux are equal in magnitude.

Methods for Making Single Phase Induction as Self Starting Motor

From the above topic we can easily conclude that the single phase induction motors are not

self starting because the produced stator flux is alternating in nature and at the starting the

two components of this flux cancel each other and hence there is no net torque. The solution

to this problem is that if the stator flux is made rotating type, rather than alternating type,

which rotates in one particular direction only. Then the induction motor will become self

starting. Now for producing this rotating magnetic field we require two alternating flux,

having some phase difference angle between them. Split phase induction motor,

1. Capacitor start inductor motor,

2. Capacitor start capacitor run induction motor,

3. Shaded pole induction motor.

4. Permanent split capacitor motor or single value capacitor motor.

7 b) DIFFERENCE BETWEEN SLIP RING INDUCTION MOTOR AND SLIP RING

INDUCTION MOTOR =6M

Basis For Comparison Slip Ring Induction Motor Induction Squirrel

Defintion The rotor of the motor is

Constructed as a slip ring type.

The rotor of the motor is a squirrel

cage type

Rotor Cylindrical laminated core with

parallel slots and each slot consist

one bar.

The slots of the rotor are not

parallel, but are skewed.

Construction Complicated Simple

Resistance Added external to the rotor The rotor bar is permanently

shorted at the end of the ring, thus

it is not possible to add any

external resistance.

Starter The rotor resistance starter can

be used.

Rotor resistance starter cannot be

used.

Starting Torque High Low

Brushes Present Absent

Maintenance Frequent maintenance required Less maintenance required

Copper Loss High Low

Efficieny Low High

Speed Control Possible Not Possible

Power Factor Low High

Cost Costly Cheap

Starting Current Low High

Uses Use in hoist,cranes,elevator where

high Torque is required.

Use in lathe machines, fan,

blower,Profiting machines,etc.

UNIT=IV

8 a) SLOT ANGLE=2M, CALCULATION OF m =1M, DISTRIBUTION

FACTORS=3M TOTAL=6M

8. WORKING PRINCIPLE OF ALTERNATOR=3M TOTAL=6M

IMPORTANCE OF THE STATOR WINDING STATIONARY ON

ALTERNATOR=3M

b) The working principle of alternator is very simple. It is just like basic principle of DC

generator. It also depends upon Faraday's law of electromagnetic induction which says the

current is induced in the conductor inside a magnetic field when there is a relative motion

between that conductor and the magnetic field.

For understanding working of alternator let's think about a single rectangular turn placed in

between two opposite magnetic pole as shown below.

Say this single turn loop ABCD can rotate against axis a-b. Suppose this loop starts rotating

clockwise. After 90° rotation the side AB or conductor AB of the loop comes in front of S-

pole and conductor CD comes in front of N-pole. At this position the tangential motion of

the conductor AB is just perpendicular to the magnetic flux lines from N to S pole. Hence

rate of flux cutting by the conductor AB is maximum here and for that flux cutting there will

be an induced current in the conductor AB and direction of the induced current can be

determined by Flemming's right hand rule. As per this rule the direction of this current will

be from A to B. At the same time conductor CD comes under N pole and here also if we

apply Fleming right hand rule we will get the direction of induced current and it will be

from C to D.

Now after clockwise rotation of another 90o the turn ABCD comes at vertical position as

shown below. At this position tangential motion of conductor AB and CD is just parallel to

the magnetic flux lines, hence there will be no flux cutting that is no current in the

conductor. While the turn ABCD comes from horizontal position to vertical position, angle

between flux lines and direction of motion of conductor, reduces from 90° to 0° and

consequently the induced current in the turn is reduced to zero from its maximum value.

After another clockwise rotation of 90° the turn again come to horizontal position and here

conductor AB comes under N-pole and CD comes under S-pole, and here if we again apply

Flemming's right hand rule, we will see that induced current in conductor AB, is from point

B to A and induced current in the conductor CD is from D to C.

As at this position the turn comes at horizontal position from its vertical position, the current

in the conductors comes to its maximum value from zero. That means current is circulating

in the close turn from point B to A, from A to D, from D to C and from C to B. Just reverse

of the previous horizontal position when the current was circulating as A → B → C → D →

A.

While the turn further proceeds to its vertical position the current is again reduced to zero.

So if the turn continues to rotate the current in the turn continually alternate its direction.

During every full revolution of the turn, the current in the turn gradually reaches to its

maximum value then reduces to zero and then again it comes to its maximum value but in

opposite direction and again it comes to zero. In this way the current completes one full sine

wave form during each 360o revolution of the turn. So we have seen how an alternating

current is produced in a turn is rotated inside a magnetic field. From this, we will now come

to the actual working principle of alternator.

Now we cut the loop and connect its two ends with two slip rings and stationary brush is

placed on each slip ring. If we connect two terminals of an external load with these two

brushes, we will get an alternating current in the load. This is our elementary model

of alternator.

Having understood the very basic principle of alternator, let us now have an insight into its

basic operational principal of a practical alternator. During discussion of basic working of o,

we have considered that the magnetic field is stationary and conductors (armature) is

rotating. But generally in practical construction of alternator, armature conductors are

stationary and field magnets rotate between them. The rotor of an alternator or

a synchronous generator is mechanically coupled to the shaft or the turbine blades, which on

being made to rotate at synchronous speed Ns under some mechanical force results in

magnetic flux cutting of the stationary armature conductors housed on the stator. As a direct

consequence of this flux cutting an induced emf and current starts to flow through the

armature conductors which first flow in one direction for the first half cycle and then in the

other direction for the second half cycle for each winding with a definite time lag of

120o due to the space displaced arrangement of 120o between them as shown in the figure

below.

This particular phenomena results in 3φ power flow out of the alternator which is then

transmitted to the distribution stations for domestic and industrial uses.

Importance of the stator winding stationary on alternator

1) The output current can be collected directly from fixed terminals on the stator without

going through brushes.

2) It is easier to insulate stationary windings for high a.c voltages.

3) Only two slip-rings are require for d.c supply to excite the field winding on the rotor.

4) It requires less maintenance and less damage to the winding

5) The stationary armature construction causes less leakage flux, hence improves voltage

regulation.6) Armature can have more number of conductors, hence e.m.f. induced will be

more. (OR)

9 a) PHASE AND LINE VOLTAGE OF ALTERNATOR=4M,

CALCULATE Kd & Kp =2M TOTAL = 6M