Simple loop generator:

Consider a single turn loop ABCD rotating clock wise in a uniform magnetic
field with a constant speed as shown in figure. As the loop rotates, the flux
linking the coil sides AB and CD changes continuously. Hence the emf
induced in these coil sides also changes but the emf induced in one coil side
adds to induced in the other.

(i)
(ii)
(iii)
(iv)
(v)
(vi)

When the loop is in position 1, the generated emf is zero because

the coil sides (AB and CD) are cutting no flux but are moving
parallel to it.
When the loop is in position 2, the coil side is moving at an angle to
the flux and there fore a low emf is generated as indicated by point
2.
When the loop is in position 3, the coil sides are at right angle to the
flux and are, therefore cutting the at a maximum rate. Hence at this
instant, the generated emf is maximum.
At position 4, the generated emf is less because the coil sides are
cutting flux at an angle.
At position 5, no magnetic lines are cut and hence induced emf is
zero.
At position 6, the coil sides move under a pole of opposite polarity
and hence the direction of generated emf is reversed. The
maximum emf in this direction will be when the loop is at position 7
and zero at position 1.

Action of a commutator:

(i)

(ii)

In figure the coil sides AB and CD are under N pole and S pole
respectively. Note that segment c1connects the coil side AB to point
P of the load resistance R and the segment c2 connects the coil side
CD to point Q of the load . Also note the direction of current through
load. It is from Q to P.
After a half revolution of the loop (i.e. 180 rotation), the coil side AB
under S pole and the coil side CD under N pole. The current in the
coil side now flow in the reverse direction but the segment C1 and C2
have also moved through 180 i.e. segment C1 is now in constant
with (+ve) brush and segment C2 in contact with (-ve) brush. Note
that commutator has reversed the coil connections to the load i.e.
coil side AB is now connected to point Q of the load and coil side CD
to the point P of the load but direction of current remain same as
before.

Construction of DC Generator:

Field system: The function of field system is to produce uniform magnetic

field within which armature rotates.
Armature: The armature core is keyed to machine shaft. It is consisting of
slotted soft iron lamination. The purpose of laminating the core is to reduce
eddy current loss.
Armature Winding: The slots of the armature core hole insulated
conductors that are connected in a switch manner is known as armature
winding . This is the winding in which working emf induces. The armature
conductors are connected in series-parallel.The conductors being connected
in series so as to increase voltage and in parallel so as to increase current.
Commutator: A commutator is a mechanical rectifier which converts the
alternating voltage generated in the armature winding into direct voltage
across the brushes. Commutators are made of copper and are insulated from
each other by mica sheet. Depending upon the manner in which armature
conductor are connected to commutator segment three are two types of
armature winding :
(a) Lap winding (Z/P)
(b)Wave winding

Simple lap winding:

(a) There are as many as parallel path as the no. of pole (p).
(b)Each parallel path has Z/p conductors in series where z and p are the
total no. of armature conductors and poles respectively.
(c) Total armature current ,Ia = p*current / no, of parallel path.
(d)Emf generated =emf / parallel path.

Simple wave winding:

(a) There are tow parallel paths irrespective of no. of poles of machine
(b)Each parallel path has Z / 2 conductors in series .
(c) Emf generated = emf / parallel path.
(d)Total armature current ,Ia = 2*current / no. of parallel path.

EMF equation a d.c generator:

Let,

=Flux/pole in webers(wb)

Z=Total number of armature conductor.

P=Number of poles.
A=Number of parallel paths=2 for wave winding

=p for lap winding.

N=Speed of armature in r.p.m.
Eg=emf of the generator
Flux cut by one conductor in one revolution of the armature,
D=p webers
Since the armature rotates N times in 60sec
armature rotates 1 times in 60/N=dt sec
Emf gerated/conductor=d/dt
= p/(60/N)
= pN/60 volts.
EMf of generator Eg =Emf per parallel path
=Emf per conductor*no. of conductors in series per
parallel path.
= pN/60 *Z/A
= pZN/60A
For wave winding= pZN/60*2
For lap winding = pZN/60*p

Types of DC generator:
i.
ii.

Separately excited dc generators

Self excited dc generators

Separately excited dc generators:

A dc generator whose field magnet winding is supplied from an
independent external dc source is called a separately excited generator.

Armature current, Ia=IL

Terminal voltage,V= Eg-IaRa
Electric power developed= EgIa
Power delivered tothe load= (Eg-IaRa)Ia=VIa

Self excited generated:

(i)
(ii)
(iii)

Series generator
Shunt generator
Compound generator

Series generator:
In the series wound generator the field winding is connected in series with
armature winding so that whole armature current flows through the field
winding as well as the load.
Armature current, Ia=Ise=IL=I
Terminal voltage,V=Eg I(Ra+Rse)
Power developed in armature= EgIa
Power delivered to load= [Eg I(Ra+Rse)]Ia=VIa

Shunt generator:
In a shunt generator[Fig.1.35],the field winding is connected in parallel with
the armature winding so that terminal voltage of the generator is applied
across it. The shunt winding has many turns of fine wire having high
resistance.
Shunt field current, Ish =V/ Rsh
Armature current Ia = IL + Ish
6

Terminal voltage,V= Eg IaRa

Power developed in armature= EgIa
Power delivered to load=VIL

Compound generator:

(a)

Short shunt
Series field current, Ise = IL
Shunt field current, Ish=V+ IseRse/ Rsh
Terminal voltage,V= Eg IaRa- IseRse
Power develop in armature,P= EgIa
Power delivered to load=VIL

(b)Long shunt:
Series field current Ise = Ia= IL+ Ish
Shunt field current Ish=V/ Rsh
Terminal voltage,V= Eg-Ia(Ra+ Rsh)-brush drop
Power developed in armature = EgIa
Power delivered to load=VIL
Problem 1: A shunt generator delivers 450 A of 230 V and the resistor of
the shunt field and armeture are 50 ohm and 0.03 ohm respectively.
Calculate the generated emf.
Solve:

Terminal Voltage,V= Eg - Ia.Ra

Or,

Eg=V + Ia.Ra

Shunt field current , Ish=v/Rsh = 230/50

=4.6 A.
Armature current , Ia = IL + Ish
=454.6
Generated emf, Eg= 230 +(454.6*0.03)
= 243.64 V
Problem 2: A long shunt compound generator delivers a load current of at
500 V and has armeture series field and shunt field resistance of 0.05 ohm,
0.03 ohm and 250 ohm respectively. Calculate generator voltage and ac
current. Allow 1V perbrush for contact drop.
Solve:

Shunt field current, Ish=V/Rsh=500/250

8

=2A
series field current,

Ise=Ia=Il+Ish=50+2
=52 A

Armature voltage drop, IaRa=52x0.05

=2.6 V
Series winding voltage drop, IseRse=52x0.03
=1.56 V
Brush drop=2x1=2 V
Generated emf, Eg=V+IaRa+IseRse+Brush drop
=500+2.6+1.56+2
=506.16 V
Problem-3: A short shunt compound generator deliver a load current of 30 A
at 220 V and has armature,series field and shunt field resistances of 0.05
ohm, 0.30 ohm and 200 ohm respectively. Calculate the induced emf and the
armature current. Allow 1.0 V per brush for contact drop.
Solve:

Shunt field current, Ish=V+IseRse/Rsh

=220+30x0.30/200
=1.145 A
Armature current, Ia=Il+Ish
=30+1.145=31.145 A
Armature voltage drop, IaRa=31.145X0.05
9

=1.56 V
Series winding voltage drop, IseRse=30x0.30
=9 V
Brush drop=2x1=2 V
Generated emf, Eg=V+IaRa+IseRse+Brush drop
=220+1.56+9+2
=232.56 V
Problem-4: In a long shunt compound generator,the terminal voltage is 230 V
when generator delivers 150 A. Determine
i. Induced emf
ii. Total power generated and
iii. Distribution of this power.
Given that shunt field,series field,diverter and armature resistance are 92
ohm,0.015 ohm,0.03 ohm and 0.032 ohm respectively.
Solve:

Shunt field current, Ish=V/Rsh=230/92

=2.5 A
Armature current, Ia=Il+Ish
=150+2.5=152.5 A
10

Since series field resistance and diverter are at parallel, so their combined
RT=Rse.Rd/Rse+Rd
=0.015x0.032/0.015+0.032

Total armature circuit resistance is,

=0.032+0.01=0.042 ohm
i. Induced emf, Eg=V+Ia(Ra+RT)
=230+152.5x0.042= 236.405 V
ii. Total power generated, EgIa=236.405x152.5
=36051.76 W
iii. In armature, I2aRa=(152.5)2x0.032=744.2 W
In series field and diverter, I2seRT=(152.5)2x0.01
=232.56 W
In shunt field, I2shRsh=(2.5)2x92=575 W
Power delivered to load, VIL=230x150
=34500 W

P-4:- The following information is given for a 300-kw,600-v,long-shunt

compound generator: Shunt field resistance=75,armature resistance
including brush resistance =0.03,commutating field winding resistance
=0.011,series field resistance =0.012,divertor resistance =0.036. When
the machine is delivering full load, calculate the voltage and power
generated by the armature.
Solution:- Power output= 300,000w
Output current=300000/600
=500A
Ish =600/75=8A

11

Ia =500+8=508A
Since the series field resistance and divertor resistance are in parallel their
combined resistance is=(0.012*0.036)/0.048=0.009
Total armature circuit resistance =0.03 -0.011+0.009=0.05
Voltage drop=508*0.05=25.4v
Voltage generated by armature=600+25.4=625.4V
Power generated=625.4*508=317,700w=317.7kw
P-5:- A four-pole generator, having wave-wound armature winding has 51
slots, each slot containing 20 conductors. What will be the voltage generated
in the machine when driven at 1500 rpm assuming the flux per pole to be
7.0mWb?
Solution:Here

Now Eg =ZNP/60A volt

= 7*10^-3Wb,

Z = 51*20=1020,
A= P= 4, N= 1500 rpm.
Eg = ( 7*10^-3 * 1020*1500*4)/(60*2)
=178.5V
P -6:- A 10Kw, 250V, d.c., 6-pole shunt generator runs at 1000 rpm, when
delivering full-load. The armature has 534 lap-connected conductors. Fullload Cu loss is 0.64 Kw. The total brush drop 1 volt. Determine the flux per
pole. Neglect shunt current.
Solution:- Since shunt current is negligible, there is no shunt Cu loss. The
copper loss occurs in armature only.
I=Ia=10,000/250=40A; (Ia)^2Ra=Arm.Cu loss or 40^2*Ra=0.64*10^3;
Ra=0.4
IaRa drop = 0.4*40=16V; Brush drop= 2*1 =2V

12

Generated emf. Eg=250+16+1=267V

Now, Eg=(ZNP

)/60A volt

267=(*534*1000*6)/60*6

=30*10^-3Wb =30 mWb.

P-7: A shunt generator delivers195A at terminal p.d. of 250V. The armature

resistance and shunt field resistance are 0.02 and 50 respectively. The
iron and friction losses equal 950W. Find a. E.M.F. generated b. cu losses c.
output of the prime mover d. commercial, mechanical and electrical
efficiencies.
Soln: a. Ish =250/50=5A; Ia 195+5=200A
Armature voltage drop=Ia Ra=200*0.02=4V
Generated e.m.f=250+4=254V
b.Armature cu loss= Ia2 Ra=(200)2*0.02=800W
shunt cu loss=V Ish =250*5=1250W
Total cu loss = 1250+800=2050W
c.Stray loss=950W;Total losses=2050+950=3000W
output=250*195=48750W;Input=48750+3000=51750W
Output of prime mover=51750W
d.Generated input=51750W;Stray losses=950W
Electrical power produced in armature=51750-750=50800
m =(50800/51750)*100=98.2%
Electrical or cu loss=2050W
e=[48750*100/948750+2050)]=95.9%
and c=(48750/51750)*100=94.2%

13

P-8: A 500V d.c shunt motor draws a line current of 5A on light load . If
armature resistance is 0.15 and field resistance is 200 , determine the
effiency of the machine running as a ganarator, delivering a load current
of 40A.
Soln: i. As a motor , on light load out of 5A of line current 2.5A are required
for field circuit and 2.5A are required for armature. Neglecting cu loss in
armature at no load,the armature power goes towards armature coreloss
and mechanical loss at the rated speed , This amounts to
(500*2.5)=1250W
ii. As a generator,for a line current of 40A , total current for the armature
is 42.5A . output of generator =500*40=20kw
Total losses as a generator =1250+field cu loss+arm. Cu loss
=(1250+1250+42.52*0.15)w=2.771kw
Efficiency=[20*100/(20+2.771)]=87.83%
P-9:The armature of a four pole d.c. shunt generator is lap wound and
generates 216v when running at600rpm . Armature has 144 slots, with 6
conductors per slot.If this armature is rewound, wave connected, find the
e.m.f generated with the same flux per pole but running at 500rpm.
Soln: Total no. of armature conductors=Z=144*6=864
For a lap winding no. of parallel paths in armature=no. of poles.In the
e.m.f. equation, E=(ZN/60)(p/a)
P=a
E= ZN/60
=216/8640=25 milli-webers
If the armature is rewound with wave connection, no. of parallel paths=2
Hence at 500r.p.m. with 25mwb as the flux per pole
The armature emf=(25*103*864*500/60)*2
=360V

Losses of d.c machine

14

The losses in a dc machine may be defined into three phases.

1. Copper losses
2. Iron or core losses
3. Mechanical losses
1. Copper losses_
a. Armatur copper loss = Ia2Ra
b. Shunt field copper loss =Ish2Rsh
c. Series field copper loss = Ise2Rse
2. Iron or core losses_
a. Hysteresis losses: Hysteresis loss occurs in the armature of the
dc machine since any given part of the armatureis subject to
magnatic riversals as it passes under successive poles. It is
given by_
Ph=B1.6max f v watt
Bmax =maximum flux density in the armature
f=frequency of the magnatic revarsal=NP/120
N= rpm, p= no. of poles , V = volume of armature
in m3
=steinetz hystersis co efficient.
b. Eddy current loss_
When armature rotates in the magnatic field of the poles, an
emf is induced in it which circulates eddy current loss. In order
to reduce this lossthe armature core is build up of thin
laminations insulated from each other by a thin layer of
varnish.
Pe= KeB2maxf2t2v watt
Ke= constant,
Bmax =maximum flux density in the core,
f= frequency of the magnatic reversals,
t= thickness of lamination,
v= volume of core in m3
3. Mechanical losses:

15

a. Frictional losses: for example( bearing friction, brush friction

etc)
b. Windage loss: (air friction of rotating armature)
Iron loss and mechanical loss are togther called stray losses.
Constant and variable loss.
Constant loss

1.
2.
3.

Variable loss
1. copper loss in armature winding
2. copper lossin series field winding

Iron loss
Mechanical loss
Shunt field loss
Total loss= constant loss+variable loss

Power Stages:
A) Mechanical power input > Iron and friction losses
B) Electrical power developed in armature EgIa > Cu losses
C) Electrical power output VIL

i)

Mechanical Efficiency:
m =

B
A

EgIa
Mec h anical Power Input

ii)

Electrical Efficiency:
C
VIl
e = B = Eg . Ia

iii)

Commercial or Overall Efficiency:

B
C
C
(VI )l
A
c =
= A = BMec h anical Power Input
B

Condition for Maximum Efficiency:

The efficiency of a d.c. generator is not constant but varies with load.
Consider a shunt generator delivering a load current IL at a terminal voltage
V.
Generator Output = VIL
Generator Input = Output + Losses

16

= VIL + I2Ra + Wc
= VIL + (IL + ISh)2 Ra+ Wc
The shunt field current ISh is generally small as compared to IL and therefore
can be neglected.
Generator Input = VIL + I2Ra + Wc
=

Output
Input
VIl
VIl+ I 2 R a+W c

1
= 1+( I 2 R a + W c )
V
VIl

----------(1)

Now,
d
1
(
)
dIl I 2 R a W c
+
V
VIl
Ra W c

V VIl 2

=0

=0

Ra W c
=
V VIl 2
IL2Ra = Wc
Variable loss = Constant loss.

Open Circuit Characteristic of a D.C. Generator

The O.C.C. for a d.c. generator is determined as follows. The field winding of
the d.c. generator (series or shunt) is disconnected from the machine and is
separately excited from an external d.c. source as shown in Fig. (3.1) (ii). The
generator is run at fixed speed (i.e., normal speed). The field current (If) is

17

increased from zero in steps and the corresponding values of generated

e.m.f.
(E0) read off on a voltmeter connected across the armature terminals. On
plotting
the relation between E0 and If, we get the open circuit characteristic as
shown in
Fig. (3.1) (i).

Fig. 3.1
The following points may be noted from O.C.C.:
(i) When the field current is zero, there is some generated e.m.f. OA. This is
due to the residual magnetism in the field poles.
(ii) Over a fairly wide range of field current (upto point B in the curve), the
curve is linear. It is because in this range, reluctance of iron is negligible as
compared with that of air gap. The air gap reluctance is constant and hence
linear relationship.
(iii) After point B on the curve, the reluctance of iron also comes into picture.
It
is because a t higher flux densities, r for iron decreases and reluctance of
iron is no longer negligible. Consequently, the curve deviates from linear
relationship.
(iv) After point C on the curve, the magnetic saturation of poles begins and
E0
tends to level off.
The reader may note that the O.C.C. of even self-excited generator is
obtained
by running it as a separately excited generator.

Critical field resistance of a shunt-generator:

18

Fig.17.29
The voltage build up in a shunt generator depends upon field circuit
resistance. If the field circuit resistance is r1 (line OA), then generator will
build up a voltage OM as shown fig 17.29. If the field circuit resistance is
increased to r2 (line OB), the generator will build up a voltage OL, slightly
less than OM, as the field circuit resistance is increased, the slop of
resistance line also increase. When the field resistance line becomes tangent
(line OC) to O.C.C. the generator would just excite. If the field circuit
resistance is increased beyond this point (say line OD), the generator will fail
to excite. The field circuit resistance is represented by line OC (tangent to
O.C.C.) is called critical field resistance Rc for the shunt generator. If may be
defined as under:
The maximum field circuit resistance (for a given
speed) with which the shunt generator would just excite is known as its
critical circuit.

Characteristics of Series Generator

Fig. 17.30, shows the connections of a series wound generator. Since there is
only one current (that which flows through the whole machine), the load
current
is the same as the exciting current.

Fig 17.30
19

(i)

(ii)

(iii)

O.C.C. Curve 1 shows the open circuit characteristics (O.C.C) of a

series generator. It can be obtained experimentally by disconnecting
the field winding from the machine and exciting it from a separate
d.c source as discussed in section 22.
Internal characteristics: curve 2 shows the total or internal
characteristics of a series generator. It gives the relation between
the generated e.m.f. E on load and armature current. Due to
armature reaction, the flux in the machine will be less than the
e.m.f.E0 generated under no load conditions. Consequently, internal
characteristics curve lies below the O.C.C. curve; the difference
between them representing the effect of a armature reaction (see
fig.17.30).
External characteristics: curve 3 shows the external
characteristics of a series generator. It gives the relation between
terminal voltage v and load current IL..
V

= E Ia (Ra + Rse)

Fig.17.31
Therefore; external characteristics curve will lie below internal
characteristics curve by an amount equal to ohmic drop [i.e.; Ia (Ra+ Rse ) ] in
the machine as shown in fig.17.30(ii). The internal and external
characteristics of a dc series generator can be plotted from one another as
shown in fig. 17.31. Suppose we are given the internal characteristics of the
generator. Let the line OC represent the resistance of the whole machine i.e;
Ra +Rse. if the load current is OB, drop in the machine is AB i.e.
AB = ohmic drop in the machine
=OB ( Ra + Rse )

20

Now raise a perpendicular from point B and make a point b on this line such
that ab = AB. Then point b will lie on external characteristics of the
generator. Following similar procedure, other points of external
characteristics can be located. It is easy to see that we can also plot internal
characteristics from the external characteristics.

CHARACTERISTICS OF A SHUNT GENERATOR:

Fig. 17.32 (i) shows the connections of a shunt wound generator. The
armature current Ia splits up into two parts small fraction Ish following through
shunt field winding while the major part IL goes to the external load.

Fig. 17.32
(i)

(ii)

(iii)

O.C.C. The O.C.C. of a shunt generator is similar in shape to that of

a series generator shown in fig 17.32 (ii). The line OA represents the
shunt field circuit resistance. When the generator is run at normal
speed, it will up a voltage OM. At no load, the terminal voltage of
the generator will be constant (=OM ) represented by the horizontal
dotted line MC
Internal characteristic: When the generator is loaded, flux per
pole is reducing due to armature reaction. Therefore e.m.f. E
generated on load is less than the e.m.f. generated at no load. As a
result, the terminal characteristics (E/Ia) drop down slightly as shown
in fig. 17.32 (ii).
External characteristic: Curve 2 shows the external
characteristics of a shunt generator. It gives the relation between
the terminal voltage V and load current IL.
V = E - Ia Ra
=E ( IL + LSh ) Ra

Therefore, external characteristics curve will lie below the internal

characteristics curve by an amount equal to drop in the armature circuit [i.e.;
(IL + Ish ) Ra] as shown in fig.17.32 (ii).
21

Note: It may be seen from the external characteristics that change in

terminal voltage from no-load to full load is small. The terminal voltage can
always be maintained constant by adjusting the field rheostat R.

CHARACTERISTICS OF COMPOUND GENERATORS:

In a compound generator, both series and shunt excitation are combined as
shown in fig. 17.33. The shunt winding can be connected either across the
armature only or across armature plus series field (long-shunt connection G).
T he compound generator can be cumulatively compounded or differentially
compounded generator. The latter is rarely used in partice. Therefore we
shall discuss the characteristics of cumulatively compounded generator. It
may be noted that external characteristics of long and short shunt
compounded generators are almost identical.

External characteristic:

Fig. 17.34 shows the external characteristics of

a cumulatively compound generator. The series excitation aids the shunt
excitation. The degree of compounding depends upon the increase in series
excitation with the increase in load current.
(i)

(ii)

The series winding turns are so adjusted that with the increase in
load current the terminal voltage increases, it is called overcompounded generator. In such a case as the load current
increases, the series m.m.f. increases and tends to increase the flux
and hence the generated voltage. The the increase in generated
voltage is
greater than the Ia Ra drop so that instead of decreasing, the
terminal voltage increases as shown by curve A in fig. 17.34.
If series winding turns are so adjusted that with the increase in load
current, the terminal voltage substantially remains constant, it is
called flat-compounded generator. The series winding of such a
22

(iii)

machine has lesser number of turns than the one in overcompounded machine and, therefore, does not increase the flux as
much for a given load current. Consequently, the full-load voltage is
nearly equal to the no-load voltage indicated by curve B is in fig.
17.34.
If series field winding has lesser number of turns than for a flat
compounded machine, the terminal voltage falls with increase in
load current as indicate by curve C in fig. 17.34. Such a machine is
called under-compounded generator.

Theory of an Ideal transformer

An ideal transformer is one that has,
(i)
(ii)

no winding resistance
no leakage flux i.e the same flux links both the windings

(iii) no iron losses(i.e eddy current and hysteresis losses) in the core.
Consider an ideal transformer on no load i.e. secondary is open-circuited as
shown in Fig.19.2(i). Under such conditions, the primary is simply a coil of
pure inductance. When an alternating voltage V1 is applied to the primary, it
draws a small magnetising current Im which lags behind the applied voltage
by 900. This alternating current Im produces an alternating flux which is
proportional to and in phase with it. The alternating flux links both the
windings and induces e.m.f. E1 in the primary and e.m.f. E2 in the secondary.
The primary e.m.f E1 is, at every instant, equal to and in opposition to V1.
Both e.m.f E1 and E2 lag behind flux by 900 which is shown in the phasor
diagram in Fig.19.2(ii). However, their magnitudes depend upon the number
of primary and secondary turns.

23

Transformer
A transformer is a static piece of equipment used either for raising or
lowering the voltage of an a.c. supply with a corresponding decrease or
increase in current. It essentially consists of two windings, the primary and
secondary, wound on a common laminated magnetic core as shown in
Fig.19.1. The winding connected to the a.c. source is called primary winding
and the one connected to load is called secondary winding.

24

Working: When an alternating voltage V1 is applied to the primary, an

alternating flux is set up in the core. This alternating flux links both the
windings and induces e.m.f.s E1 and E2 in them according to Faradays laws of
electromagnetic induction. The e.m.f. E1 is termed as primary e.m.f. and
e.m.f E2 is termed as secondary e.m.f.
d
E1 = - N1 dt

Clearly,

And

E2 = - N2

d
dt

E2/E1 = N2/N1

E.M.F EQUATION OF A TRANSFORMER

Consider that an alternating v 1 of frequency f is applied to the
primary as shown in fi g.(1). The sinusoidal fl ux produced by the
primary can be represented as:
= m sint
The instantaneous e.m.f e1 induced in the primarty,

25

d
e1 = -N1 dt

d
= -N1 dt

( m sint)

= N 1 m cost = -2fN 1 m cost

= 2fN 1 m sin(t-90 0 )..(i)
It is clear from the above equation that maximum value of induced
e.m.f in the primary is:
E m1 = 2fN 1 m
The r.m.s value E 1 of the primary is:

E1 =

Em 1
2

2 f N 1 m
2

= 4.44f N 1 m
Similarly
E2 = 4.44f N 2 m
In an ideal transformer, E1 = V1 and E2 = V2

Voltage transformation ratio (k)

From the above equations of induced e.m.f , we have
E2/E1 = N2/N1 = K
The constant k is called voltang transformation ratio. Thus if k=5(i.e.
N2/N1=5),then,E2 =5E1
For an ideal transformer:

26

(i)
(ii)

E1= V1 and E2 = V2 as there is no voltage drop in the windings.

E2/E1 = V2/V1 = N2/N1 = K
There are no losses. Therefore , vot-amperes input To the primary
are edual to the output volt-amperes i,e
V1 I1 = V2 I2

I1/I2 = V1/V2 = 1/k

Hence,currents are in the inverse ratio of voltage transformarmation ratio.

This simply means that if we raise the voltage,there is a corresponding
decrease of current.
PRACTICAL TRANSFORMER
A practical transformer differs from the ideal transformer in many respects.
The practical transformer has (i) iron losses (ii)winding resistances (iii)
magnetic leakage,giving rise to leakage reactances.
(i)

Iron losses: Since the iron core is subjected to alternating flux,

(ii)

there occurs eddy current and hysteresis losses in it. These two
losses together are known as iron losses or core losses.the iron
losses depend upon the supply frequency, it may be noted that
magnitude of iron losses is quite small in a pratical transformer.
winding resistances: since the windings consist of copper
conductor ,it immediately follows that both primary and secondary
will have winding resistance.the primary resistance R1 and
secondary resistance R2 act in series with the respective windings
as shown in fig.19.3.

27

Leakage reactances: Both primary and secondary currens produce

flux. The flux which links both the windings is the useful flux and is called
mutual flux .however,primary current would produce some flux 1 which
would produce some flux 1 that would not link the primary winding. The flux
such as 1 or 2 which links only one winding is called leakage flux . the
leakage flux paths are mainly through the air . The effect of these leakage
fluxes would be the same as though inductive reactance were connected in
series with each winding of transformer that had no ieakage flux as shown in
fig 19.3 . in other words ,the effect of primary leakage flux 1 is to introduce
an inductive reactance X1 in series with the primary winding as shown in fig
19.3. Similarly,the secondary leakage flux 2 introduces an inductive
reactance X2 in series with the secondary winding.

PRACTICAL TRANSFORMER ON ON LOAD

Consider a partical transformer on on load,I,e. secondary on open-circuit as
shown in fig.19.5(i).the primary will draw a small current I0 to supply(i)the
iron losses and (ii) a very small amount of copper loss in the primary.hence
the primary no load current I0 is not 900 behind the applied voltage V1 but
lags it by an angle 0900 as shown in phasor diagram in fig.19.5(ii). No
load input power, W0 =V1I0cos0

28

(i) The component I w in phase with the applied voltage V 1 . This is

known as active component and supplies the iron and a very small
primary copper loss.
Iw=I 0 cos 0
(ii) The component I m lagging behind V 1 by 90 0 and is known as
magnetising component. It is this component which produces the
mutual flux in the core.
I m = I 0 sin 0

I 0 I m2 I W2
Therefore,
No load p.f.,

cos 0 = I W /I 0

Ideal Transformer on Load:- Let connect a load ZL across the

secondary of an ideal transformer as shown in figure. The secondary e.m.f E2
will causes a current I2 to follow through the load.

I2

E 2 V2

ZL ZL

The angle at which I2 lead or lags V2 depends upon the resistance of the load.
In the present case we have considered inductive load so that current I2 lags
behind V2 by 2
29

The secondary current I2 sets up an m.m.f N2I2 which produce a flux in the
opposite direction to the flux originally set up in the primary by
magnetizing current. This will change the flux in the core from the original
value. However the flux in the core should not the change in the original
value. In ordered to fulfill this condition the primary must developed an
m.m.f which exactly counterbalances the secondary m.m.f N2I2 . Hence a
primary current I1 must follow such that.
N1I1 = N2I2
I1 = (N2 / N1) I2 = K I2
Phasor diagram:- Shows the Phasor diagram of an ideal transformer on load.
Note that in drawing the Phasor diagram, the valu of K has been assumed
unity so that primary Phasor are equal to secondary Phasors. The secondary
current I2 lags behind V2 by 2. It causes a primary current I1 = K I2 = I I2 =
I2 which is in antiphase with it.
1 = 2
Cos 1 = cos 2
Thus, power factor on the primary side is equal to the p.f on the secondary
side.

Transformer with resistance and leakage

resistance:- Fig. shows a practical transformer having winding
resistance and leakage resistance. This is the actual condition that exit in a
transformer. This is voltage drop in R1 and X1 so that prima n y e.m.f E1 is less
than the applied voltage V1. Similarly there is voltage drop in R1 and X1 so that
30

secondary terminal voltage V2 is less than the secondary e.m.f E2. Let us take
the usual case of inductive load which causes the secondary current I 2 to lag
behind the secondary voltage V2 by 2. The total primary current I1 must two
requirements viz.
a

It must supply the no load current I0 to meet the iron losses in the
transformer and to provide flux in the core.

It must supply a current I2 to counteract the demagnetizing effect of

secondary current I2. The magnitude of I2 will be such that:

N1 I 2' N 2 I 2
I 2'

N2
I 2 KI 2
N1

Or

The total primary current I1

will be the Phasor sum of

I 2'

I0
and

I1 = I2+I0

V1 E1 I 1 ( R1 jX 1 )
= - E1 + I1 Z1

V 2 E 2 I 2 ( R2 jX 2 )

E2 I 2 Z 2

31

I 1 I 0 ( KI 2 )
Where

Phasor diagram:- Fig. shows the Phasor diagram of a practical transformer

for the usual case of inductive load. Both E1 and E2 lag the mutual flux by
900. Note that current e.m.f that opposes that applied voltage V1 is -E1.
Therefore of we add I1R1 and I1X1 to -E1. We get the applied primary voltage
V1.

Load power factor

Primary power factor

Input power to transformer

Output power of transformer

cos 2
cos 1

P1 V1 I1 cos 1
P2 V2 I 2 cos 2

Impedance Ratio:- Consider a transformer having impedance Z2 in

the secondary as shown in figure.

32

Z2

Z1

V2
I2

Z1
I1

Z2
V
I
( 2 )( 1 )
Z1
V1
I2

Z2
K2
Z1
Or
Note the importance of above relation. We can transfer the parameters from
one winding to the others, thus;
i

A resistance R1 in the primary becomes K2R1 when transferred to the

secondary.

ii

A resistance R2 in the secondary becomes R2/ K2 when transferred to

the primary.

iii

A resistance X1 in the primary becomes K2 X1 when transferred to the

secondary.

iv

A resistance X2 in the secondary becomes X2/K2 when transferred to

the primary.

33

Shifting Impedance In a Transformer:

Fig. shows a transformer

where resistance and reactance are shown external to the winding. The
resistance and reactance of one winding can be transferred to the other by
appropriately using the factor K2. This makes the analysis of the transformer
a simple affair because then we have to work in one winding only.

Referred to primary:- when secondary resistance or reactance is

transferred to primary it is divided by K2 . it is then called equivalent
secondary resistance or reactance referred to primary and is denoted

by

R2' orX 2'

Equivalent resistance of transformer referred to primary,

R01 R1 R2' R1 R2 / K 2
Equivalent reactance of transformer referred to primary,

X 01 X 1 X 2' X 1 X 2 / K 2
Equivalent impedance of transformer referred to primary,

Z 01 R012 X 012

34

ii

Referred to secondary:- when primary resistance or reactance is

transferred to the secondary it is multiplied by K2 . it is then called
equivalent primary resistance or reactance referred to the secondary

and its denoted by

R1' orX 1'

Equivalent resistance of transformer referred to secondary

R02 R2 R1' R2 R1 K 2
Equivalent reactance of transformer referred to secondary

X 02 X 2 X 1' X 2 X 1 K 2
Equivalent impedance of transformer referred to secondary,

Z 02 R022 X 022

Approximate Equivalent Circuit of A Transformer:

35

The no load current I0 is only 1 3% of the rated primary current and may be
neglected without any serious error. The transformer can then show as in
figure.
This is an approximate representation because no load current had been
neglected. Note that all the circuit elements have been shown external so
that the transformer is the ideal one.

Equivalent Circuit Of Transformer To Primary:- All the secondary

quantities are referred to primary ,we get the equivalent circuit of
the transformer referred to primary as shown in figure.

The equivalent circuit in fig. is an electrical circuit and can be solved for
various current voltages. Thus if we find V2 and I2 , then actual secondary
values can be determind as under:

Actual secondary voltage,

Actual secondary current,

ii

V2 KV2'

I 2 I 2' / K

Equivalent Circuit Of Transformer To Secondary:- All the primary

quantities are referred to secondary ,we get the equivalent circuit
of the transformer referred to secondary as shown in figure.
36

The equivacircuit circuit shown in fig . is an electrical circuit and can be

solved for various voltages and currents . thus is we find V1 and I1 , then
actual primary values can be determind as under:

Actual secondary voltage,

V1 V1' / K

I1 I1' K
Actual secondary current,

Voltage regulation :
The voltage regulation of a transformer is a arithmatic difference between
the no load secondary voltage (0V2) and the secondary voltage V2 on load
expresed as percetage of no load voltage i.e
% age voltage regulation =

100

Where,

V2 = No-load secondery voltage

=KV1:
V2 = Secondary voltage on load

We know ,

V2 V2 = I2 R02 cos 2

37

I2X02 sin 2

The +ve sign is for lagging p.f and ve sign is for leading p.f .

Transformer test:
The circuit constants, eficiency and voltage regulation of a transformer can
be determined by two simple test 1. Open circuit test and 2. Shot circuit
test. These tests are very convenient as they provide the required
information without actually loading the transformer .

1. Open circuit or no load test:

In this test, the rated voltage is applied to the primary (usually low voltage
winding ) while the secondary is left open circuited. The applied primary
voltage V1 is measured by the voltmeter, the no load current I0 by ammeter
and no-load input power W0 by wattmeter as shown in fig.

As the normal rated voltage is applied to the primary, therefor normal iron
losses will occur in the transfomer core. Hence wattmeter will record the iron
losses and small copper loss in the primary. Since no load current I0 is very
small , Cu losses in the primary under no load condition are negligible as
compared with iron losses. Hence wattmeter reading practically gives the
iron losses in the transformer. It is reminded that iron losses are the same at
all load .
Iron losses,

P1 = Wattmeter reading = W0

No load current

= ammeter reading = I0

Applied voltage

= voltmeter reading = V1

Input power ,

W0 = V1I0 cos 0

No load p.f.

cos 0 = W0/V1I0
38

Iw = I0 cos 0 ; Im = I0 sin 0
Thus open circuit test gives Pi , I0, cos 0 , Iw , and Im.
2. Short-Circuit or impedance test.
In this test, the secondary in short-circuited by a tick conductor and variable
low voltage is applied to the primary as shown in fig. The low input voltage is
gradually raised till at voltage Vsc, full-load current I1 flows in the primary.
Then I2 in the secondary also has full-load value since I1/I2 = N2/N1 . Under
such conditions, the copper loss in the windings is the same as that on full
load.

There is no output from the transformer under short circuit conditions.

Therefore , input power is all lost and this loss is almost since the voltage Vsc
is very small. Hence the wattmeter will practically register the full-load
copper losses in the transformer windings. Fig 2shows the equivalent circuit
of a transformer on short circuit as referred to primary; the no load current
being neglected due to its smallness.
Full load Cu loss, Pc = Watt meter reading = Ws
Applied voltage,

= Voltmeter reading = Vsc

F.L primary current = Armature reading

= I1

P I 2 R I 2R' I 2R
c
1 1 1 2
1 01
R
P / I2
01
c
1

39

Where

R
01

is the total resistance of transformer referred to primary. Total

Z
impedance referred to primary.

01

V / I
sc 1

X 01

2
2
Z 01
R01

Total leakage reactance referred to primary,

cos s Pc / Vsc I1
Short circuit power factor

R01andX 01
The short circuit test gives full load Cu loss,

LOSSES IN A TRANSFORMER
The power losses in a transformer are of two types, namely ;
1. Core or Iron losses.
2. Copper losses .
These losses appear in the form of heat and produce 1. An increase in
temperature and 2. A drop in efficiency
1. Core or Iron losses (Pi) : These consist of hysteresis and eddy current
losses and occur in the transformer core due to the alternating flux. These
can be determined by open-circuit test .
Hysteresis loss = Khf Bm1.6 watts/m3
Eddy current loss = Ke f2Bm2 t2 watts/m3
Both hysteresis and eddy current losses depend upon (i) maximum flux
density Bm in the core and (ii) supply frequency f. Since transformer are
connected to constant-frecuency, constant voltage supply, both f and Bm are
constant . Hence, core or iron losses are practically the same at all loads.
Iron or core losses, Pi = Hysteresis loss + Eddy current loss
= Constant losses .
The hysteresis loss can be minimised by using steel of high silicon content
whereas eddy current loss can be reduced by using core of thin laminations .
40

2. Copper losses : These losses occur in both the primary and secondary
windings due to their ohmicresistence. These can be determined by shortcircuit test.
Total Cu losses PC = I12 R1+ I22 R2
= I12 R01 Or I22 R02
It is clear that copper losses vary as the square of load current . Thus if
copper losses are 400W at a load current of 10A, Then they will be ()
400 =100 w at a load current of 5A .
Total losses in a transformer = Pi +PC
= Constant losses + Variable losses
It may be noted that in a transformer, copper losses account for about 90%
of the total losses .
EFFICIENCY FROM TRANSFORMER TEST.
F.L Iron loss = Pi
F.L Cu loss = PC
Total F.L losses = Pi + PC
We can now find the full-load efficiency of the transformer at any p.f without
actually loading the transformer.

F.L efficiency F.L =

( Fullload VA X p . f ) +
Fullload VA X p . f

Also for any load equal to x full-load

Corresponding total losses = Pi + x2 PC

Corresponding x =

( x Fullload VA X p . f )+
x Fullload VA X p . f

Note that iron losses remains the same at all loads.

41

Condition for maximum efficiency :

Output Power = V2 I2 cos 2
If R02 is the total residence of the transformer referred to secondary, then ,
Total Cu loss PC = I22 R02
Total losses

= Pi + PC

So , Transformer

..(i)

For a normal transformer, V2 is approximately constant. Hence for a load of

given p.f efficiency depends upon load current I2. It is clear from exp(i) above
that numerator is constant and for the efficiency to be maximum, the
denominator should be minimum i.e
d
d

Or,

Or,
Or,

d
d

0-

=0

( V2 cos2 + Pi/I2 + I2 R02 ) = 0

+ R02

=0

Pi = I22 R02 (ii)

i.e Iron losses = Copper losses

Hence efficiency of a transformer will be maximum when copper losses are
equal to constant or iron losses .
From equ,,..(ii) above ,the load current I2 corresponding to maximum is given
by ;

42

Pi / R02
I2 =

Output kVA Corresponding To Maximum Efficiency .

Let,

PC = copper losses at full-load kVA

Pi = Iron losses
x = Fraction of full-load kVA at which efficiency is

maximum
Total Cu losses = x2 PC
x2 PC = Pi .for maximum efficiency

so,

Pi / PC
Or,

x=

(Iron loss)
( F . LCU LOSS)

So, output kVA corresponding to maximum efficiency

(Iron loss)
= x full load kVA = full load kVA ( F . LCU LOSS)
It may be noted that the value of kVA at which the efficiency is maximum is
independent of p.f of the load.
Example 32.1. The maximum flux density in the core of a 250/3000 volts,
50-Hz single phase transformer is 1.2 Wb/m2. If the e.m.f. per turn is 8 volt,
determine
(1)primary and secondary turns (2) area of the core.

Solution. (1)

E1 = N1e.m.f. induced/ turn

N1=250/8=32; N2=3000/8=375

(2)

E2=4.44 f N2 Bm A
3000=4.44503751.2A; A=0.03 m2

43

Example 32.2.The core of a 100-Kva, 11000/550 v, 50-Hz, 1-ph, core type

transformer has a cross section of 20 cm 20 cm. Find (1) the number of
H.V. and L.V. turns per phase and (2) the e.m.f. per turn if the maximum core
density is not to exceed 1.3 tesla. Assume a stacking factor of o.9. What will
happen if its primary voltage is increased by 10% on no- load?
Solution.
(1)

A=(0.20.2)0.9=0.036 m2

Bm=1.3 T,

11000=4.4450N11.30.036, N1=1060
550=4.4450N21.30.036, N2=53
Or,

N2=KN1=53

(2)e.m.f./ turn=11000/1060=10.4 V
Example 32.3. A single phase transformer has 400 primary and 1000
secondary turns The net cross sectional area of the core is 60 cm2.If the
primary winding be connected to a50 hz supply at 520 v, calculate(1)the
peak value of flux density in the core (2)the voltage induced in the
secondary winding.
Solution.
(1) K=N2/N1=1000/400==2.5
E2/E1=K ; E2=KE1=2.5520=1300 v
(2)

E1=4.44 f N1 Bm A
500=4.4450400Bm(6010 -4) ;Bm =0.976

Wb/m2

Example 32.8.A single phase, 50 hz, core type transformer has square cores
of 20 cn side. Permissible maximum flux density is 1Wb/m2.calculate the
number of turns per limb on the High and Low voltage sides for a 3000/220 V
ratio.
Solution. Starting with calculation for L.V.turns, T2,

44

4.4450[(202010-4)1]T2=220
T2=220/8.88=24.77
T2=26
T1/T2=V1/V2; T1=263000/220=354
Number of turns on eacg Limb =177
Number of turns on each Limb=13

Example 32.7. A 25 KVA single phase transformer has 250 turns on the
primary and 40 turns on the secondary winding .The primary is connected
to 1500 volt,50 Hz mains.(1)primary and secondary currents on full load,
(2)secondary e.m.f.,(3)maximum flux in the core.
Solution.

(1)

If V2= secondary voltage rating=secondary e.m.f.

V2/1500=40/250; V2 =240 volts

(2)

primary currents=25000/1500=16.67 amp

Secondary currents =25000/240=104.2 amp

(3)

If m is the maximum core-flux in Wb,

1500=4.4450m250, m=0.o27 Wb or 27 mWb

Example 32.6.A single phase transformer has 500 turns in the primary and
1200 turns in the secondary. The cross- sectional area of the core is 80 sq.
cm. If the primary winding is connected to a 50Hz supply at 500 v, calculate
(1)Peal flux density, and (2) Voltage induced in the secondary.
Soolution. From the e.m.f. equation for transformer ,
500=4.4450m500 ; m=1/222 Wb
(1) peak flux density, Bm=m/(8010-4)=0.563 wb/m2
(2)

V2/V1=N2/N1; V2=5001200/500=1200 volts

Example 32.4. A 25 KVA transformer has 500 turns on the primary and 50
turns on the secondary winding. The primary is connected to 3000 V,50 Hz
supply. Find the full load primary and secondary currents, the secondary
45

e.m.f. and the maximum fiux in the core. Neglect leakage drops and no load
primary current.
Solution.

K=N2/N1=50/500=1/10

Now, full load=I1=25000/3000=8.33 A,I2=I1/K=108.33=83.3A

e.m.f. per turn on primary side=3000/500=6V
Secondary e.m.f.=650=300 V
E1=4.44 f N1 m
3000=4.4450500m; m= 27mWb

Dc motor principle:
A machine that converts dc power into mechanical power is known as a dc
motor. The principle of dc motor can be stated as When a current carrying
conductor is placed in a magnetic field, the conductor experiences a
mechanical force. The direction of this force is given by fleming left hand
rule and its magnitudes is given by
F = BIL

Voltage equation of a dc motor:

Let in a d.c. motor(Fig.)

V = applied voltage
Eb = back e.m.f
46

Ra = armature resistance
Ia = armature current
Since back e.m.f Eb acts in opposition to the applied voltage V, the net
voltage across the armature circuit is V-Eb. The armature current Ia is given by
:
I

V = E b+ I a Ra.(i)
Power equation of d.c. motor:
Multiplying both sides of equ. (i),
V I a = EbIa+I2aRa
Eb Ia = V I a - I2aRa
Condition for maximum power:
The mechanical power developed by the motor is, Pm= Eb Ia
For maximum power,
d
d

=0

V -2 IaRa = 0

V = 2IaRa
IaRa = V/2

From equation (i)

V = E b+ I a Ra
= Eb+ V/2
Eb = V/2

Armature torque of a dc motor:

Let in a dc motor ,
47

r = average radius of armature in (m)

l = effective length of each con doctor in
(m)
Z = total no. of armature conductors
A = number of parallel path
i = current in each conductor = Ia/A
B = average flux density in Wb/m3
= flux per pole in Wb
P = no. of poles
Torque in one conductor, T = Fr
=B i l r
Total torque in Z no. of conductors
Ta = Z F r
=ZBil r
= Z (/a)(Ia/A)lr
=

ZI a P
2 A

(N-m)

= 1.59 ZIa (P/A)

Since Z, P and A are fixed
Ta Ia
(i)

For shunt motor,

Ta Ia [ is constant ]

(ii)

For series motor,

Ta Ia2

Prob-1: A 220 V d.c. machine has an armature resistance of 0.5 ohm. If

the full-load armature current is 20 A, find the induced e.m.f. when the
machine acts as (i) generator (ii) motor.
Solve:

48

Generator

Motor

From the Fig.1, shunt current is considered negligible because its value is not
given.
(i) As generator[Fig.i] Eg = V+IaRa =220+(0.5x20)=230 V
(ii) As motor [Fig.ii]
Eb = V-IaRa = 220-(0.5x20)=210 V
Prob-2: A separately excited D.C. generator has armature circuit 1000 r.p.m,
it delivers a current of 100 A at 250 V to a load of constant resistance. If the
generator speed drop to 700 r.p.m, with field current unaltered, find the
current delivered to load.
Solve:
RL = 250/100= 2.5 ohms.
Eg1 = 250+(100x0.1)+2 = 262 V
At 700 r.p.m., Eg2 = 262x(700/1000) = 183.4 V
If Ia is the new current, Eg2-2-(Iax0.10)= 2.5 Ia
This gives Ia = 96.77 amp.
Prob-3: A 220 V, shunt motor has armature resistance of 0.8 ohm and field
resistance of 200 ohm. Determine the back e.m.f. when giving an output of
7.46 kw at 85 percent efficiency.
Solve:
Motor input power = (7.46X103)/0.85 W
Motor input current = (7460/0.85)/440 = 19.95 A
Ish = 440/200 = 2.2 A
Ia = IL-Ish = 19.95-2.25 =17.75 A
Now
Eb = V-IaRa
= 440-(17.75x0.8) =425.8 V
Prob-4: A 25 kw, 250 V, d.c. shunt motor has armature resistance of 0.06
ohm and field resistance of 100 ohm. Determine the total armature power
49

developed when working(i) as a generator delivering 25 kw output and

(ii) as a motor taking 25 kw input.
Solve:

Generator

Motor

As Generator[Fig.i]
Output current = 25000/250 =100A
Ish = 250/100 = 2.5 A;
Ia = 102.5 A
Generated e.m.f. = V+IaRa
= 250+(102.5x0.06) = 256.15 V
Power developed in armature = EbIa
= (256.15x102.5)/1000 = 26.25 kw
As motor [Fig.ii],
Motor input current = 100 A ; Ish = 2.5 A ; Ia = 97.5 A
Eb = 250-(97.5x0.06) = 244.15 V
Power developed in armature = EbIa
= (244.15x97.5)/1000 = 23.8 kw
Prob-5: A d.c. motor takes an armature current of 110 A at 480 V. The
armature circuit resistance is 0.2 ohm. The machine has 6-poles and the
armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb.
Calculate- (i) the speed and
(ii) the gross torque developed by the armature.
Solve:
Here Eb = 480-(110x0.2)= 458 V,
=0.05 W,
Now
Eb = (ZNP)/60A
Or
458=(0.05x864xNx6)/(60x6)
N = 636 r.p.m.
Ta = 0.159xZIa x(P/A)
= 0.159x0.05x864x110x(6x6)
50

Z = 864

= 756.3 N-m
Prob-6: A 250 V, 4-pole, wave-wound d.c. series motor has 782 conductors
on its armature. It has armature and series field resistance of 0.75 ohm. The
motor takes a current of 40 A. Estimate its speed and gross torque developed
if it has a flux per pole of 25 mWb.
Solve:
Here Eb = (ZNP)/60A .................(i)
Now, Eb = V-IaRa
= 250-(40x0.75) = 220 V
From equation (i),
220 = (25x10-3x782xNx4)/60x2
N = 337 r.p.m
Ta = 0.159xZIa x(P/A)
= 0.159x25x10-3x782x40x2
= 249 N-m
Prob-7: A d.c. shunt machine develops an a.c. e.m.f. of 250 V at 1500 r.p.m.
Find its torque and mechanical power developed for an armature current of
50 A.
Solve:
Mechanical power developed in the armature = EbIa
= 250x50=12,500
W
Now,

Ta = 9.55(EbIa/N)
= 9.55(12500/1500) = 79.6 N-m

Prob-8: Determine developed torque and shaft torque of 220 V, 4-pole

series motor with 800 conductors wave-connected supplying a load of 8.2 kw
by taking 45 A from the mains. The flux per pole is 25 mWb and its armature
circuit resistance is 0.6 ohm.
Solve:
Developed torque, Ta = 0.159xZIa x(P/A)
= 0.159x25x10-3x800x45(4/2)
= 286.2 N-m
Eb = V-IaRa = 220-(45x0.6) = 193 V
51

Now,
Or
Also,
Or

Eb = (ZNP)/60A
193= 25x10-3x800xN(4/2)
N = 289 r.p.m
2NTsh = Output
Tsh(2 x 289) = 8200
Tsh = 4.52 N -m

Prob-9: A 220-V, d.c. shunt motor runs at 500 r.p.m. when the armature
circuit is 50 A. Calculate the speed if the torque is doubled. Given that Ra =
0.2 ohm.
Solve:
We know, Ta Ia
Since is constant, Ta Ia
Now, Ta1 Ia1
and Ta2 Ia2
Ta2/Ta1 = Ia2/Ia1
Ia2 = 100 A
Now, N2/N1 = Eb2/Eb1.................(i)
Eb1 = 220- (50x0.2)= 210 V
From equation (i),
N2/500 = 200/210
N2 = 476 r.p.m.

Eb2 = 220-(100x0.2) = 200 V

Prob-10: A 500 V, 37.3 KW, 1000 r.p.m. d.c. shunt motor has on full-load an
efficiency of 90 percent. The armature circuit resistance is 0.24 ohm and
there is total voltage drop of 2V at the brushes. The field current is 1.8 A.
Determine - (i) full-load line current
(ii) full load shaft torque in N-m and
(iii) total resistance in motor starter to limit the starting
current to 1.5 times the full load current.
Solve:
(i) Motor input = 37300/0.9 = 41444 W
F.L. line current = 41444/500 = 82.9 A
(ii) Tsh = 9.55 ( Output/N)
= 9.55(37300/1000) = 356 N-m
(iii) Starting line current = 1.5 x 82.9 = 124.3 A
Arm. current at starting = 124.3-1.8 = 122.5 A
If R is the starter resistance( which is in series with armature),
then- 122.5(R+0.24) +2 = 500
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R = 3.825 ohm.
Prob- 11: A 4-pole, 220 V shunt motor has 540 lap-wound conductor. It takes
32 A from the supply mains and develops output power of 5.595 KW. The
field winding takes 1 A. The armature resistance is 0.09 ohm and the flux per
pole is 30 mWb.
Calculate- (i) the speed and
(ii) the torque developed in N-m.
Solve:
Ia = IL - Ish = 32-1 =31 A ; Eb = V-IaRa = 220-(31x0.09) = 217.2 V
Now,
Eb = (ZNP)/60A
217.2 = ( 30x10-3x540xNx4)x(60x4)
217.2 = 0.27 N
(i)
N = 804. 444 r.p.m.
(ii)
Tsh = 9.55 ( Output/N)
= 9.55(5595/804.444)
= 66.42 N-m
Prob- 12: Determine the torque established by the armature of a four-pole
D.C. motor having 774 conductors, two paths in parallel, 24 mWb of poleflux and the armature current is 50 Amps.
Solve:
We know,
Torque, T = 0.159xZIa x(P/A) Nw-m
Two paths in parallel for a 4-pole case means a wave winding.
T = 0.159x(24x10-3)x774x50x(4/2)
= 295.36 Nw-m
Prob- 13: A 500 V D.C. shunt motor draws a line-current of 5 A on light-load.
If armature resistance is 0.15 ohm and field resistance is 200 ohm,
determine the efficiency of the machine running as a generator delivering a
load current of 40 Amps.
Solve:
(i) No load, running as a motor:
Input power = 500x5 = 2500 W
Field copper-loss = 500x2.5 = 1250 W
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(ii) As a generator, delivering 40 A to load:

Output delivered = 500x40x10-3 = 20 KW
Losses: (a) Field copper-loss = 1250 W
(b) Arm. copper-loss = 42.52x0.15 = 271 W
(c) No load losses = 1250 W
Total losses = 2.771 KW
Generator efficiency = (Output power/Input power)x100 %
= (20/22.771)x100%
= 87.83 %
Prob-14: A d.c. series motor takes 40 A at 220 V
and runs at 800 r.p.m.
If the armature and field resistance are 0.2 ohm and 0.1 ohm respectively
and the iron and friction losses are 0.5 KW, find the torque developed in the
armature. What will be the output of the motor ?
Solve:
Armature torque is given by, Ta = 9.55 ( EbIa/N) N-m
Now,
Eb = V-Ia(Ra+Rse)
= 220-40(0.2+0.1)
= 208 V
Ta = 9.55x208x(40/800) = 99.3 N-m
Cu loss in armature and series field resistance = 402x0.3 = 480 W
Iron and friction losses = 500 W
Total losses = 480+500 = 980 W
Motor power input = 220x40 = 8800 W
Motor output = 8800-980 = 7820 W = 7.82 KW
Prob-15: A cutting tool exerts a tangential force of 400 N on a steel bar of
diameter 10 cm which is being turned in a simple lathe. The lathe is driven
by a chain at 840 r.p.m. from a 220 V d.c. Motor size is the motor pulley if the
lathe pulley has a diameter of 24 cm ?
Solve:
Torque Tsh = Tangential force x radius = 400x0.05 = 20 N-m
Output power = Tshx2N Watt
= 20x2(840/60) Watt = 1760 W
Motor input = Output power/efficiency
= 1760/0.8 = 2200 W
Current drawn by motor = 2200/220 = 10 A
Let N1 and D1 be the speed and diameter of the driver pulley respectively and
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N2 and D2 the respective speed and diameter of the lathe pulley.

Then
N1xD1 = N2xD2
Or
1800xD1 =840x0.24
D1 = 201.6/1800 m
= 0.112 m

55