Quiz: Electrical Engineering 13 May 2021

Quiz: Electrical Engineering
Exam: DFCCIL
Topic: DC Generator

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. Which of the following generator has zero percentage regulation?
(a) Flat compound
(b) Over compound
(c) Under compound
(d) Both flat compound and under compound

Q2. A dc generator efficiency is maximum when:
(a) Variable loss + constant loss=1
(b) Variable loss – constant loss =1
(c) Constant loss – variable loss =1
(d) Variable loss – constant loss =0

Q3. If once destroyed, the residual magnetism of a shunt generator can be restored by connecting its shunt field.
(a) to a battery
(b) in reverse
(c) to earth
(d) to transformer

Q4. Full-load terminal voltage in an over-compounded generator with respect to no-load terminal voltage is–
(a) equal
(b) zero
(c) less
(d) more

Q5. Pole shoe of a D.C. machine is laminated for the purpose of
(a) Decreasing hysteresis loss
(b) Decreasing eddy current loss
(c) Decreasing both hysteresis and eddy current loss
(d) Manufacturing ease

Q6. A lap wound D.C. machine has 400 conductors and 8 poles. The voltage induced per conductor is 2V. the machine generates a voltage of
(a) 100V
(b) 200V
(c) 400V
(d) 800V

Q7. In a 4-pole, 20KW, 200V wave wound DC shunt generator, the current in each parallel path will be
(a) 25A
(b) 10A
(c) 50A
(d) 100A

Q8. Condition for maximum power delivered in D.C. Generator is_______:
(a) Back EMF = (2 × Supply voltage)
(b) Back EMF = Supply voltage
(c) Back EMF = (Supply voltage/4)
(d) Back EMF = (Supply voltage/9)

Q9. Under-commutation gives rise to …….
(a) sparking at the leading edge of the brush
(b) sparking at the trailing edge of the brush
(c) no sparking at all
(d) sparking at the middle of the brush

Q10. The e.m.f of a DC generator depends on…………
(a) Frequency
(b) Speed
(c) Commutation
(d) Brush contact drop

SOLUTIONS

S5. Ans.(b)
Sol. Pole shoe provides support to the field coils and spread out the flux over armature periphery. Pole shoe is laminated to avoid heating and eddy current losses.

S6. Ans.(a)
Sol. for Lap winding, A = P = 8
so, there will be 8 parallel paths and conductor per parallel path = 400/8=50
Total emf induced = 50 × 2 = 100V

S7. Ans.(c)
Sol. Total current(I_T )=P/V=(20×1000)/200=100 A
For, Wave wound: number of parallel paths=2
∴current in each parallel path=I_T/2=100/2=50 A.

S8. Ans.(a)
Sol.

S9. Ans.(b)
Sol. Due to under – commutation, flux at trailing edges gets crowded, so there is sparking.

S10. Ans.(b)
Sol. E=ZϕNP/60A
∴E α N
So, the e.m.f of a DC generator depends on speed.

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