46) If the flux of a DC generator is halved and its speed is doubled, the generated e.m.f will be

  1. halved
  2. doubled
  3. remians same
  4. quadrupled
ANSWER
Answer: ( 3 )
EXPLANATION
No answer description required for this question.

47) Stray losses consist of ………

  1. magnetic and mechanical losses
  2. magnetic and copper losses
  3. mechanical and copper losses
  4. none of the above
ANSWER
Answer: ( 1 )
EXPLANATION
Usually magnetic and mechanical losses are collectively called stray losses.

48) A 30 kW, 300 V DC shunt generator has armature and field resistance of 0.05 Ω and 100 Ω respectively. What is the generated e.m.f

  1. 292.25 V
  2. 311.25 V
  3. 350.15 V
  4. none of the above
ANSWER
Answer: ( 1 )
EXPLANATION
IL= 30×103/300 = 100 A

Ish = V/Rsh = 300/100 = 3 A

Ia = IL + Ish = 100 + 3 = 103 A

Generated e.m.f Eg = V + IaRa = 300 + 103×0.05 = 300+5.15 = 305.15 V

49) The main function of the compensating winding in a DC machine is

  1. assist in commutation
  2. reducing demagnetization effect of armature reaction
  3. reduce distortion effect of armature reaction
  4. eliminate reactance voltage
ANSWER
Answer: ( 3 )
EXPLANATION
No answer description required for this question.

50) The efficiency of a DC generator means its

  1. electrical efficiency
  2. overall efficiency
  3. mechanical efficiency
  4. none of the above
ANSWER
Answer: ( 2 )
EXPLANATION
No answer description required for this question.

51) The overall efficiency of a DC generator is maximum when its variable loss is equal to

  1. constant loss
  2. stray loss
  3. iron loss
  4. mechanical loss
ANSWER
Answer: ( 1 )
EXPLANATION
No answer description required for this question.

52) In a DC shunt generator, the constant losses are equal to

  1. stray losses
  2. mechanical losses
  3. stray losses and shunt Cu losses
  4. none of the above
ANSWER
Answer: ( 3 )
EXPLANATION
No answer description required for this question.

53) A 4 pole, lap wound DC shunt generator has an armature winding consisting of 220 turns each of 0.04 Ω. The armature resistance is

  1. 0.025 Ω
  2. 0.05 Ω
  3. 0.055 Ω
  4. 1 Ω
ANSWER
Answer: ( 4 )
EXPLANATION
Total resistance of 220 turns = 220 × 0.04 = 0.88 Ω

Since there are 4 parallel paths in armature

Resistance of each path = 0.88/4 = 0.22 Ω

Now, there are four such resistance in parallel each of value 0.22 Ω

Here, Armature resistance Ra = 0.22/4 = 0.055 Ω 

54) In a long shunt compound parallel generator, the shunt field is connected in parallel with

  1. armature
  2. series field
  3. parallel combination of armature and series field
  4. series combination of armature and series field
ANSWER
Answer: ( 4 )
EXPLANATION
No answer description required for this question.

55) Wave wound generator provides

  1. less current but more voltage
  2. more current but less voltage
  3. more current and more voltage
  4. none of the above
ANSWER
Answer: ( 1 )
EXPLANATION
No answer description required for this question.

56) In DC machine , the commutator acts as …….

  1. a rectifirer
  2. an amplifier
  3. a load
  4. none of the above
ANSWER
Answer: ( 1 )
EXPLANATION
No answer description required for this question.

57) The eddy current power loss in a DC generator is……

  1. directly proportional to thickness of each lamination
  2. inversely proportional to thickness of each lamination
  3. directly proportional to square of thickness of each lamination
  4. none of the above
ANSWER
Answer: ( 1 )
EXPLANATION
The eddy current power loss Pe in a DC generator is given by:

Pe = KeB2maxf2t2V watts

Where Bmax = maximum flux density in the core

f = frequency of the magnetic reversals

t = thickness of each laminations

V = Volume of armature core

58) A DC shunt generator supplies a load of 7.5 kW at 200 V. The armature resistance is 0.6 Ω and field resistance is 80 Ω. The generated e.m.f is

  1. 224 V
  2. 448 V
  3. 123.5 V
  4. 202.5 V
ANSWER
Answer: ( 1 )
EXPLANATION
Ish = 200/80 = 2.5 A

IL = 7.5 × 1000 / 200= 37.5 A

Armature current Ia = IL + Ish  = 37.5 + 2.5 = 40 A

Generated Voltage Eg = V+ IaRa = 200+40×0.6 = 224 V

59) The field circuit resistance of a DC shunt generator is 200 Ω. When the generator output is 100 kW, the terminal voltage is 500 V and the generated e.m.f is  525 V. The armature resistance is

  1. 1 Ω
  2. 0.5 Ω
  3. 0.25 Ω
  4. 0.123 Ω
ANSWER
Answer: ( 4 )
EXPLANATION
Armature voltage drop, IaRa  = Eg – V = 525 – 500 = 25 V

Ish = 500/200 = 2.5 A

IL = 100×103 / 500 = 200 A

Ia = IL + Ish  = 200+2.5 = 202.5 A

Ra = IaRa/Ia = 25/202.5 = 0.123 Ω

60) In a short shunt compound wound generator, the shunt field is connected in parallel with

  1. series field
  2. armature
  3. parallel combination of series field and armature
  4. series combination of series field and armature
ANSWER
Answer: ( 2 )
EXPLANATION
No answer description required for this question.

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