DC Machines Interview Questions and Answers for freshers on “Armature Reaction – 2”.
1. Flux density in the interpolar region drops down because of ______
a) ATa (peak)
b) Large air gap
c) Absence of magnetic poles
d) Depends on other parameters
Answer: b
Clarification: The exact way to find the flux density owing to the simultaneous action of field and armature ampere-turns is to find the resultant ampere-turn distribution ATresultant(∅) = ATf(∅) + ATa(∅). The flux density of ATa(∅) which, because of large air-gap in the interpolar region, has a strong dip along the q-axis even though ATa(peak) is oriented along it.
2. Resultant ampere-turn distribution of a DC machine is given by _________
a) ATresultant(∅) = ATf (∅) – ATa(∅)
b) ATresultant(∅) = – ATf (∅) + ATa(∅)
c) ATresultant(∅) = -ATf (∅) – ATa(∅)
d) ATresultant(∅) = ATf (∅) + ATa(∅)
Answer: d
Clarification: The exact way to find the flux density owing to the simultaneous action of field and armature ampere-turns is to find the resultant ampere-turn distribution ATresultant(∅) = ATf (∅) + ATa(∅), where ∅ is the electrical space angle.
3. Which axis undergo shifting as a result of armature reaction?
a) GNA
b) MNA
c) Both GNA and MNA
d) Remains fixed
Answer: b
Clarification: Apart from distortion of the resultant flux density wave, its MNA also gets shifted from its GNA by a small angle α so that the brushes placed in GNA are no longer in MNA as is the case in the absence of armature current.
4. Armature reaction in a machine is demagnetizing due to _________
a) Machine is designed with iron which is slightly saturated
b) Machine is designed with iron which is unsaturated
c) Depends on the application where machine is being is used
d) Can’t tell
Answer: a
Clarification: The armature reaction in a DC machine is cross-magnetizing causing distortion in the flux density wave shape and a slight shift in MNA. It also causes demagnetization because a machine is normally designed with iron slightly saturated.
5. Which of the following are effects of armature reaction?
a) Increase in iron losses
b) Commutation problems
c) Possibility of commutator sparking
d) Increase in iron losses, commutation problems and commutator sparking
Answer: d
Clarification: Armature reaction in a DC machine is a result of distortion of main field flux distribution by armature current, which produces its own mmf called armature mmf. Directly or indirectly armature reaction is the problem occurring in DC machine as it causes various effects, which reduce machine efficiency.
6. A 250 kW, 400 V, 6-pole dc generator has 720 lap wound conductors. Armature current is ____
a) 625A
b) 6.25A
c) 62.5A
d) 0.625A
Answer: a
Clarification: Armature current multiplied by the armature voltage is called as rating of a DC generator. Thus, 250 kW is the given rating while 400 V is the armature voltage. So, armature current is equal to 250*1000/400 = 625A.
7. What is the total ampere conductors/pole (in SI) if 600 lap wound conductors carry 120A current through conductors (P=4)?
a) 18000
b) 9000
c) 4500
d) 13500
Answer: a
Clarification: Ampere-conductors/pole =ZIc/P= Zla/AP. Ampere conductors per pole is calculated by multiplying total no. of conductors with the current carried by them divided by the total no. of poles.
Ampere-conductors/pole = 600*120/4 =18000.
8. What is the total ampere turns/pole (in AT/pole) if 600 lap wound conductors carry 120A current through conductors (P=4)?
a) 18000
b) 9000
c) 4500
d) 13500
Answer: b
Clarification: Ampere-conductors/pole =ZIc/P= Zla/AP. Ampere turns per pole is calculated by multiplying total no. of conductors with the current carried by them divided by the twice the total no. of poles.
Ampere-turns/pole = 600*120/8 =18000/2= 9000 .
9. If total ampere turns per pole is equal to 6000 A-turns, peak ampere turns for a 4-pole machine is _____
a) 24000
b) 3000
c) 1500
d) 4500
Answer: c
Clarification: Peak flux density in terms of total flux density is given by ATa (peak) = ATa (total) /P. Thus, for a 4-pole machine, ATa (total)= 6000 and P=4. Thus, Peak flux density is equal to 6000/4= 1500.
10. What is the total ampere turns per pole for 720 lap wounded conductors with carrying armature current equal to 625A in a 6-pole machine?
a) 6252 AT/pole
b) 625.2 AT/pole
c) 62.52 AT/pole
d) 8252 AT/pole
Answer: a
Clarification: For a given machine number of parallel paths is equal to 6. So, conductor current will be equal to armature current divide by no. of parallel paths i.e. 625/6. Conductor current = 104.2 A. Total armature ampere-turns, ATa = ½(720*104.2/6)= 6252 AT/pole.
11. For 6252 AT/Poles, if brush shift is of 2.50 mech. Degrees, what will be the demagnetizing ampere-turns per pole for a 6-pole DC machine?
a) 521
b) 5731
c) 5231
d) 571