76) The O.C.C of a DC generator is always called its ………… characteristic

1. internal
2. external
4. none of the above
EXPLANATION
No answer description required for this question.

77) In a cumulatively compounded generator, the shunt and series fields ……………… each other

1. aid
2. oppose
3. does not effect
4. none of the above
EXPLANATION
No answer description required for this question.

78) Which of the following will not prevent a self excited shunt generator from building upto its full load voltage

1. wrong direction of rotation
2. open field
3. no residual magnetism
4. speed too high
EXPLANATION
No answer description required for this question.

79) The O.C.C of a DC generator is

1. similar for all type of generator
2. different for all type of generator
3. none of the above
EXPLANATION
No answer description required for this question.

80) A 4 pole DC shunt generator with a wave wound armature having 390 conductors has to supply a load of 500 lamps each of 100 W, 250 V. Allowing 10 V of the voltage drop in the connecting leads between the generator and the load and a contact drop of 1 V per brush. What is the generated e.m.f ? The flux/pole = 30 mWb ; Ra = 0.05 Ω and Rsh = 65 Ω.

1. 282.8 V
2. 272.2 V
3. 257.5 V
4. 279.6 V
EXPLANATION
Load current IL = 500×100/250 = 200 A

Voltage across shunt = 250+10 = 260 V

Shunt current Ish = 260/65 = 4 A

Armature current Ia = 200+4 = 204 A

Generated Voltage Eg = V + Drop in leads + Brush drop + IaRa

= 250+10+(2×1)+(204×0.05) = 272.2 V

81) A shunt generator has a full load current of 196 A at 220 V. The stray losses are 720 W and shunt field coil resistance is 55 Ω. It has a full load efficiency of 88 % . What is the value of armature resistance ?

1. 0.107 Ω
2. 0.258 Ω
3. 0.358 Ω
4. 0.5 Ω
EXPLANATION
Output = 220×196 = 43120 W

η = 88% = 0.88

Electrical input = 43120/0.88 = 49000 W

Total losses = 49000-43120 = 5880 W

Ish = 220/55 = 4 A

Ia = 196+4 = 200 A

Shunt Cu loss = 220× =880 W

Stray losses, Wc = 880+720 = 1600 W

Armature Cu loss = 5880 – 1600 = 4280 W

I2aRa = 4280 or Ra = 4280/(200)2 = 0.107 Ω

82) The ideal commutation in a coil is not possible because each coil has

1. resistance
2. inductance
3. capacitance
4. none of the above
EXPLANATION
No answer description required for this question.

83) In a multi polar DC machine, for the same value of Φ, Z and N,  the armature e.m.f

1. is more for wave than lap winding
2. is less for wave than lap winding
3. depends whether running as motor or generator
4. none of the above
EXPLANATION
Eg = PΦZN/60A

For the same Φ, Z and N

Eg  is directly proportional to  P/A

84) The dummy coil in a DC machines are useful to

1. increase efficiency
2. improve communication
3. mechanically balance armature
4. reduce machine cost
EXPLANATION
No answer description required for this question.

85) The armature resistance of a 6 pole lap wound DC machine is 0.05 Ω. If the armature is rewound using a wave winding, then armature resistance will be

1. 0.30 Ω
2. 0.15 Ω
3. 0.10 Ω
4. 0.45 Ω
EXPLANATION
Let r ohm be the resistance per parallel path. Then r/6 = 0.05 or r = 6 × 0.05 = 0.03 Ω. When the armature is rewound for wave connection, then there will be two parallel paths, each of resistance = 3r = 3×0.3 = 0.9 Ω. Therefore, armature resistance for wave wound machine = 0.9/2 = 0.45 Ω

86) In a 4 pole, 25 kW, 200 V wave wound DC shunt generator, the current in each parallel path will be

1. 125 A
2. 62.5 A
3. 31.25 A
4. 250 A
EXPLANATION
Load current IL = 25×103/200 = 125 A

For a shunt generator Ia = IL + Ish

Since, Ish is small, it can be neglected so that Ia = IL = 125 A. As the machine is wave wound, the number of parallel paths is A = 2. Therefore, current per parallel path = 125/2 = 62.5 A

87) A DC generator having a shunt field of 50 Ω was generating normally at 1000 r.p.m. The critical resistance of the machine was 80 Ω. Due to some reasons, the speed of the prime mover become such that the generator just failed to generate. The speed at that time must have been

1. 800 r.p.m
2. 1600 r.p.m
3. 625 r.p.m
4. 500 r.p.m
EXPLANATION
N1×50 = N2×80

or

1000×50 = N2×80

N2 = 625 r.p.m

88) An 8 pole lap wound DC shunt generator is supplying 295 A to a load. The field current of the generator is 5 A. The current per parallel path is

1. 37.5 A
2. 300 A
3. 150 A
4. 75 A
EXPLANATION
Ia = IL + Ish = 295+5 = 300 A.  Now the number of parallel paths is A = P = 8. Therefore, current per parallel path = Ia/A = Ia/P = 300/8 = 37.5 A

89) Fleming’s right hand rule is applicable to

1. DC motor
2. alternator
3. DC generator
4. transformer
EXPLANATION
No answer description required for this question.

90) The commutator in a DC machine can convert

1. AC to DC
2. DC to AC
3. both 1 and 2
4. none of the above