Unit-5
Question Bank
Q. 1.
A 8 pole lap wound DC generator has 70 slots in its armature with 22 conductors per slot. The
ratio of pole are to pole pitch is 0.64. The diameter of the bore of the pole shoe is 0.48m. The
length of the pole shoe is 0.28m. The air gap flux density is 0.32Wb/m 2 & the generated emf in
the armature is 400V. Find the speed of generator.
Ans. 1442
Sol.
P = 8, No. of slots =70, conductors/slot=22

Total No. of conductors = 70×22 = 1540
The ratio of pole arc to pole pitch = 0.64
Diameter of bore of pole shoe=0.48m & length of pole shoe = 0.28m
Air gap flux density = 0.32Wb/m2
emf generated (Eg) =400V
A = P = 8 (Lap wound)
Pole arc
 0.64
Pole pitch
Pole arc = 0.64×pole pitch = 0.64×

D
4
= 0.1205m
Area of pole shoe (A) = pole arc ×length of pole shoe
= 0.1205×0.28 = 0.03376m2
  0.3376  0.32  0.010803Wb
Eg =
 NZ
400=
N=
60

P
A
0.010803  N 1540 8

60
8
1442rpm Ans.
Q. 2.
A 6-pole DC shunt generator has the following data
Field resistance = 120  , armature resistance = 0.8 
Number of conductors = 350 (Wave connected)
Flux per pole= 0.02Wb
Load resistance across the terminals = 12  , armature rotates at 1000 rpm. Calculate power
absorbed by load
Ans.
8859.6W
Sol.
Rsh = 120  ,
Ra = 0.8  ,
Z = 350,  = 0.02Wb
Let the terminal voltage be Vt Volts.
Ish =
Vt
A
120
Ia =Ish+IL=
Eg=
 ZN
60
and IL=
Vt
A
12
Vt Vt 11V
 
120 12 120

P 0.02  1000  350 6

  350V (A=2 wave wound)
A
60
2
Eg = Vt+IaRa
350 = Vt+

11Vt
128.8Vt
350  120
 0.8 
 326.08V
or Vt=
120
120
128.8
load current (IL)=
Vt 326.08

 27.17 A
RL
12
8859.6W Ans.
And power absorbed by load = VtIL=326.08×27.17 =
Q. 3.
A separately excited DC generator has terminal voltage 250V with constant field excitation. If
the load changes from 200KW to 125KW, find percentage reduction in speed. The armature
resistance is 0.015  and total contact drop at brushes is 2V. Neglect the armature reaction.
The flux & total number of armature conductors remain constant.
Ans.
1.704%
Sol.
Terminal Voltage (Vt) = 250V
Ra = 0.015 
Total contact drop at brushes = 2V
Eg1=
 ZN1 P
60 A
Eg1  N1
Ia=IL
For 200KW load
200  103
 800 A
250
IL1=
Eg= Vt+ILRa+2
= 250+800×0.015+2=264V
For 125KW load
IL2=
125 103
 500 A
250
Eg2= V+IL2Ra+2 = 250+500×0.015+2
Eg2= 259.5V
Eg2  N2
Eg1
Eg 2

N1
N2
N1
264

N 2 259.5
Percentage reduction speed
N1  N 2
264  259.5
 100 
 100 =
N1
264
1.704% Ans.
Q. 4.
A 220V DC series motor has an armature resistance of 0.3  and field resistance of 0.2  . It
runs at a speed of 700 rpm taking a current of 15A. Calculate the resistance to be inserted in
series with the armature to reduce the speed to 600rpm. The input current remains constant.
Assume that the magnetization characteristics is st. line.
Ans. 2.02 
Sol.
Vt = 220V,
Ra=0.3  ,
Rsc=0.2 
I=Ia1=Ia2=15A
N1=700rpm,
N2=600rpm
Eb1= VtIa1(Ra+Rsc) = 220–15×(0.2+0.3)
Eb1=212.5V
Eb2=Vt–Ia1(Ra+Rsc+R)
=220–15×(0.2+0.3+R)
Eb2= (212.5–15R)V
N 2 Eb 2 1 Eb 2

 
N1 Eb1 2 Eb1
 I a1  I a 2 


1  2 
600 212.5  15 R

700
212.5
212.5 – 0.857  212.5
R=
=
15
2.02  Ans.
Q. 5.
A 220V, 1.5KW, 859rpm, separately excited DC motor has armature resistance of 2.5  and it
draws a current of 8A at rated load condition. If the field current & the armature current are
fixed at the value of rated speed at rated load, what will be the no-load speed of the motor?
Assume losses remain constant between no-load full-load operation.
Ans.
944.9rpm
Sol.
Let Eb1 be back emf at rated load and Eb2 at no load
Iao=0 (no load)
Ia=8A (at rated load)
Armature resistance (Ra) = 2.5 
Supply Voltage (Vt) = 220V
Rated speed (N) = 859 rpm = N1 (say)
Eb1= Vt–IaRa = 220–8×2.5=200V
Eb2= Vt–IaoRa= 220–0 = 220V
Eb1=
Ebo=
 ZN1 P
60 A
 ZN o P
60 A
Eb1 N1

Ebo N o
 Ebo 
220

  859 
200
 Eb1 
No = N1× 
944.9rpm Ans.
Q. 6.
A 250V, DC shunt motor on no-load, runs at a speed of 1000 rpm and takes a current of 5A the
armature and shunt field resistances are 0.2  and 250  respectively. Calculate the speed
when the motor is on-load, and is taking current of 50A. Assume that the armature reaction
weakens the field by 3%.
Ans. 993.7rpm
Sol.
Motor on no-load
line current, IL=5A
Field current Ish=
250
 1A
250
Iao=IL–Ish=5–1 = 4A

Ebo=Vt-IaoRa= 250–4×0.2 = 249.2V
On load
IL=50A
Ia1=50–1 = 49A

Eb1=Vt–Ia1Ra=250–49×0.2 = 240.2V
Ebo  o No
& Eb1  1 N1
Ebo o N o

Eb1 N11

armature reaction weakens the field flux by 3%
1  (1  0.03)  o = 0.97 o
No = 1000 rpm

 Eb1  o N o

 Ebo  1
N1= 
 240.2   1 

  1000 
 249.2   0.97 
=
993.7rpm Ans.
Q. 7.
Determine developed torque and shaft torque of 220V, 4-pole series motor with 800
conductors wave connected & supplying a load of 8.2KW by taking 45A from the mains. The
flux per pole is 25mWb and its armature circuit resistance is 0.6  .
Ans. 286.2N-m
270.5N-m
Sol
Developed Torque Ta=0.159  ZI a 
P
For wave connection A=2
A
=0.159×25×10–3×800×45×
4
2
286.2 N-m Ans.
=
Armature circuit resistance R = Rsc+Ra = 0.6 
Eb = Vt–IaR
Eb=
=
220–45×0.6 = 193V
 ZNP
A
4
2
or,
193 = 25×10-3×800×  N
or,
N = 4.825 rpm
Tsh = Shaft torque ,
Out put Power= 8.2KW = 8200W

2  N Tsh= Out Put Power
Or,
2π×4.825 Tsh = 8200
Tsh =
8200
=
2  4.825
270.5N-m Ans.
Q. 8.
A 6-Pole, lap wound d.c. generator has 840 armature conductors and flux per pole of
0.018Wb. The generator is run at 1200 rpm. Calculate the emf generated.
Ans. 302.4V
Sol.
P = 6,
Z = 840,  =0.018Wb,
N = 1200rpm
In lap wound generator, A = P=6
Generated emf, Eg=
Q. 9.
P ZN 6  0.018  840 1200


60 A
60  6
302.04V Ans.
Calculate the emf generated by a 4-pole wave wound armature with 45 slots, with 18
conductors per slot when driven at 1000 rpm. The flux per pole is 0.02Wb.
Ans. 540V
Sol.
P = 4,
Z=18×45 = 810,  =0.02Wb
N = 1000rpm, For wave wound generator, A = 2
Eg=
P ZN 4  0.02  810 1000


60 A
60  2
540V Ans.
Q. 10
A lap-connected 8-pole generator has 500 armature conductors and useful flux of 0.07Wb.
Determine the induced emf when it runs at 1000rpm. What must be the speed at which it is to
be driven to produce the same emf if it is wave wound?
Ans.
583.3V,
250rpm
Sol.
P = 8,
Z =500,  =0.07Wb,
N = 1000 rpm
P ZN 8  0.07  500 1000


60 A
60  8
583.33V Ans.
Now for wave wound Generator, Eg=583.33,
Eg  60 A
N=
Q.11
P Z

583.33  60  2

8  0.07  500
A =P= 8 for lap wound Eg=
A=2
250rpm Ans.
A lap wound DC generator having 8-poles develops emf of 500V at 400rpm. The armature has
144 slots and each slot contains 6 conductors. Calculate the flux per pole.
Ans. 0.0868Wb
Sol.
P = 8,
Eg=
Or,
A = 8, Eg=500V,
N = 400rpm, Z = 144×6=864
P ZN
60 A

Eg  60  A
PZ  N
=
500  60  8

8  864  400
0.0868Wb Ans.
Q.12.
Determine the power output of a dc motor armature having 1152 lap-connected conductors
carrying 150A and rotating at 300rpm in a 12-pole. The flux/pole is 60mWb.
Ans.
51.84KW
Sol.
Z=1152,P = 12,
A = 12, N = 300rpm,
 =60mWb=60×10-3Wb,
Generated Emf, Eg=
Power Output
Ia = 150A
P ZN 12  60 103 1152  300

 345.6V
60 A
60 12
= Eg×Ia=345.6×150 =
51.84KW Ans.
Q. 13. A dc shunt generator has an armature resistance of 0.25  and the resistance of shunt
field is 220  . While delivering a load current of 50A, it has terminal voltage of 440V.
Determine the generated emf.
Ans. 453V
Sol.
Ish=
Vt
440

 2A
Rsh 220
Ia= IL+Ish= 50+2=52A
Eg = Vt+IaRa= 440+(52×0.25) =
453V Ans.
Q. 14. A 20KW, 220V dc shunt generator has an armature resistance of 0.07  and a shunt field
resistance of 200  . Find power developed in the armature when it delivers rated output.
Ans.
20834.57W
Sol.
Ish=
Vt
220

 1.1A
Rsh 200
P = Vt×IL
IL=
20  1000
 90.909A
220
Ia= IL+Ish= 90.909+1.1 = 92.009A
Eg=Vt+IaRa= 220+(92.009×0.07) = 226.44V
Power developed = EgIa = 226.44×92.009 =
20834.57W Ans.
Q. 15. A shunt Generator has an induced voltage on open circuit of 127V. When the machine is
loaded, terminal voltage is 120V. Find the load current if the field resistance is 15  and
armature resistance is 0.02  . Ignore armature reaction.
Ans.
342A
Sol.
Eg =127,
Vt = 120V,
Ra=0.02 
Eg = Vt+IaRa
Ia =
Ish=
E g  Vt
Ra

127  120
 350A
0.02
V
120

 8A
Rsh 15
342A Ans.
IL=Ia–Ish=350–8=
Rsh=15 
Q. 16. A d.c. shunt generator is supplying load connected bus bar voltage of 220V. It has an armature
resistance of 0.025  and field resistance of 110  . Calculate the value of load current and
load power when it generates an emf of 230V.
Ans.
398A, 87.56KW
Sol.
Eg =230V
Ish=
V
220

 2A
Rsh 110
Eg=Vt+IaRa
Ia=
Eg  V
Ra
=
230  220
=400A
0.025
398A Ans.
Load current= IL=Ia–Ish= 400–2 =
Load Power=Vt×IL=220×398=
87.56KW ANS.
Q. 17. A 230V d.c. shunt Motor takes 51A at full load. Resistance of armature and field windings are
0.1  and 230  . Determine (i) field current (ii) armature current
(iii) back emf developed at full load
Sol.
(i)
Ish=
Vt
230

Rsh 230
Field current =
(ii)
1A Ans.
Armature current,
Ia=IL–Ish=51–1=
50A Ans.
(iii)
Eb=Vt–IaRa= 230–(50×0.1)=
225V Ans.
Ans. (i) 1A (ii) 50A (iii) 225V
Q. 18. A 250V d.c. shunt Machine has line current of 80A. It has armature & field resistance of 0.1 
and 125  respectively. Calculate Power developed in armature when running as (a)
Generator (b) Motor.
Ans. (a) 21.172KW (b)
1889KW
Sol.
(a)
As a generator
Ish=
Vt
250

 2A
Rsh 125
Ia=IL+Ish= 80+2=82A
Eg=Vt+IaRa=250+82×0.1=258.2V
21.172KW Ans.
Power Developed = EgIa= 258.2×82=
(b)
As a Motor
Ia=IL–Ish=80–2 = 78A
Eb=Vt–IaRa
= 250–(78×0.1) = 242.2V
18.89KW Ans.
Power developed = Eb×Ia = 242.2×78 =
Q. 19. A 230V dc shunt Motor runs at 800rpm and takes armature current of 50A. Find the resistance
to be added to the field circuit to increase speed to 1000 rpm at an armature current of 80A.
Assume armature resistance 0.15  and field winding resistance = 250 
Ans. 68.95

Sol.
Ia = 50A,
Ish=
V
230

 0.92 A
Rsh 250
IL=Ia+Ish=50.92A
Eb1=Vt–IaRa
= 230–50×0.15 = 222.5V
Now R is connected to field arnding
N2 = 1000rpm
Ia= 80A, Eb2=Vt–IaRa
= 230–80×0.15 = 218V
Since Eb   N
Eb  NIsh (  α Ish )
Eb1 N1 I sh1

Eb 2 N 2 I sh 2
222.5 800  0.92

218 1000  I sh 2
Ish2 =
800  0.92 218

=0.8211A
1000
222.5
Ish2=
Vt
 0.7211
( R  Rsh )
230
 0.7211
R  250
0.7211R+(0.7211×250) = 230
R=
230  (0.7211 250)
=
0.7211
68.95  Ans.
Q. 20. A 200V dc shunt Motor running at 1000 rpm takes armature current of 17.5A. It is required to
reduce the speed to 600 rpm. What must be the value of resistance to be inserted in the
armature circuit if the original armature resistance is 0.4  ? Take armature current to be
constant during this process.
Ans. 4.41 
Sol.
Ia=17.5
Eb1=Vt–IaRa
= 200–(17.5×0.4) = 193V
N1=1000rpm
Eb2=Vt–Ia(Ra+R)
= 200–17.5(0.4+R) = 193–17.5R
N2 = 600rpm
Eb  N (  =Constant)
Eb1 N1
193
1000


=
Eb 2 N 2 193  17.5R 600
1158 = 1930–175 R
R=
1930  1158

175
4.41  Ans.