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/m2 & 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

0.64Pole pitch

Pole arc = 0.64×pole pitch = 0.64×4

D = 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 =60

NZ P

A

400=0.010803 1540 8

60 8

N

N =

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 = 120

tVA and IL=

12

tVA

Ia =Ish+IL=11

120 12 120

t tV V V

Eg= 0.02 1000 350 6

35060 60 2

ZN PV

A

(A=2 wave wound)

Eg = Vt+IaRa

350 = Vt+11 128.8

0.8120 120

t tV V or Vt=

350 120

128.8

326.08V

load current (IL)= 326.08

27.1712

t

L

VA

R

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%

8859.6W Ans.

Sol.

Terminal Voltage (Vt) = 250V

Ra = 0.015

Total contact drop at brushes = 2V

Eg1= 1

60

ZN P

A

Eg1 1N

Ia=IL

For 200KW load

IL1= 3200 10

800250

A

Eg= Vt+ILRa+2

= 250+800×0.015+2=264V

For 125KW load

IL2=3125 10

500250

A

Eg2= V+IL2Ra+2 = 250+500×0.015+2

Eg2= 259.5V

Eg2 N2

1 1

2 2

g

g

E N

E N

1

2

264

259.5

N

N

Percentage reduction speed 1 2

1

264 259.5100 100

264

N N

N

=

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

2 22 1

1 1 2 1

b b

b b

E EN

N E E

1 2

1 2

a aI I

600 212.5 15

700 212.5

R

R = 212.5 – 0.857 212.5

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= 1

60

ZN P

A

Ebo= 60

oZN P

A

1 1b

bo o

E N

E N

No = N1×1

220859

200

bo

b

E

E

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

1250

A

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 oN & Eb1 1 1N

1 1 1

bo o o

b

E N

E N

armature reaction weakens the field flux by 3%

1 (1 0.03) o = 0.97

o

No = 1000 rpm

N1= 1

1

b o o

bo

E N

E

= 240.2 1

1000249.2 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 a

PZI

A For wave connection A=2

=0.159×25×10–3×800×45×4

2

=

Armature circuit resistance R = Rsc+Ra = 0.6

Eb = Vt–IaR = 220–45×0.6 = 193V

Eb= ZNP

A

or, 193 = 25×10-3×800×4

2N

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 =

286.2 N-m Ans.

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=6 0.018 840 1200

60 60 6

P ZN

A

Q. 9. 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=4 0.02 810 1000

60 60 2

P ZN

A

302.04V Ans.

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 A =P= 8 for lap wound Eg=

8 0.07 500 1000

60 60 8

P ZN

A

Now for wave wound Generator, Eg=583.33, A = 2

N = 60 583.33 60 2

8 0.07 500

gE A

P Z

Q.11 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, A = 8, Eg=500V, N = 400rpm, Z = 144×6=864

Eg=60

P ZN

A

Or, 60gE A

P Z N

=

500 60 8

8 864 400

583.33V Ans.

250rpm Ans.

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, Ia = 150A

Generated Emf, Eg=312 60 10 1152 300

60 60 12

P ZN

A

345.6V

Power Output = Eg×Ia=345.6×150 =

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=440

220

t

sh

V

R 2A

Ia= IL+Ish= 50+2=52A

Eg = Vt+IaRa= 440+(52×0.25) =

51.84KW Ans.

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=220

1.1200

t

sh

VA

R

P = Vt×IL

IL=20 1000

220

90.909A

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 =

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 Rsh=15

Eg = Vt+IaRa

Ia = 127 120

0.02

g t

a

E V

R

350A

Ish=120

15sh

V

R 8A

IL=Ia–Ish=350–8=

20834.57W Ans.

342A Ans.

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=220

110sh

V

R 2A

Eg=Vt+IaRa

Ia=g

a

E V

R

=

230 220

0.025

=400A

Load current= IL=Ia–Ish= 400–2 =

Load Power=Vt×IL=220×398=

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 Ans. (i) 1A (ii) 50A (iii) 225V

Sol.

(i) Ish=230

230

t

sh

V

R

Field current =

(ii) Armature current,

Ia=IL–Ish=51–1=

(iii)

Eb=Vt–IaRa= 230–(50×0.1)=

398A Ans.

87.56KW ANS.

1A Ans.

50A Ans.

225V Ans.

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= 250

125

t

sh

V

R 2A

Ia=IL+Ish= 80+2=82A

Eg=Vt+IaRa=250+82×0.1=258.2V

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

Power developed = Eb×Ia = 242.2×78 =

21.172KW Ans.

18.89KW Ans.

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=230

0.92250sh

VA

R

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 (sh α I )

1 1 1

2 2 2

b sh

b sh

E N I

E N I

2

222.5 800 0.92

218 1000 shI

Ish2 =800 0.92 218

1000 222.5

=0.8211A

Ish2= 0.7211( )

t

sh

V

R R

230

0.7211250R

0.7211R+(0.7211×250) = 230

R = 230 (0.7211 250)

0.7211

=

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)

1 1

2 2

b

b

E N

E N =

193 1000

193 17.5 600R

1158 = 1930–175 R

R = 1930 1158

175

68.95 Ans.

4.41 Ans.