# DC Machines Questions and Answers – Ways to Reduce Effects of Armature Reaction – 2

This set of DC Machines Questions and Answers for Experienced people focuses on “Ways to Reduce Effects of Armature Reaction – 2”.

1. Axis undergo shifting as a result of armature reaction, can be balanced by ______
a) Increase in armature current
b) Decrease in armature current
c) Introducing interpoles
d) Removing interpoles

Explanation: 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, due to armature reaction. This effect is countered by the interpoles placed in GNA.

2. The choice of average coil voltage determines the minimum number of commutator segments for its design.
a) True
b) False

Explanation: The maximum allowable voltage between adjacent segments is 30–40 V, limiting the average voltage between them to much less than this figure. The choice of the average coil voltage determines the minimum number of commutator segments for its design, to avoid any flashover and ultimately short circuit.

3. Compensating winding will provide incomplete neutralization ____________
a) Under pole region
b) In interpolar region
c) Everywhere
d) Complete neutralization

Explanation: The compensating winding neutralizes the armature mmf directly under the pole while in the interpolar region, there is incomplete neutralization. Further, the effect of the resultant armature mmf in interpolar region is rendered insignificant because of large interpolar gap.

4. Cross-magnetizing effect of armature reaction can be reduced by __________
a) Removing saturation in teeth and pole-shoe
b) Making smooth pole shoes
c) Introducing saturation in teeth and pole-shoe
d) Cannot be determined

Explanation: The cross-magnetizing effect of the armature reaction can be reduced by making the main field ampere-turns larger compared to the armature ampere-turns such that the main field mmf exerts predominant control over the air-gap flux. This is achieved by introducing saturation in the teeth and pole-shoe.

5. Cross-magnetizing effect of armature reaction can be reduced by __________
a) Removing saturation in teeth and pole-shoe
b) Making smooth pole shoes
c) Chamfering the pole shoes
d) Cannot be determined

Explanation: By chamfering the pole-shoes which increases the air-gap at the pole tips. This method increases the reluctance to the path of main flux in a DC machine but its influence on the cross-flux is much greater. This is because the cross flux has to cross the air-gap twice.
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6. To counter the effect of shift in MNA due to armature reaction, which of the following component can be shifted?
a) Poles
b) Commutator
c) Brushes
d) Cannot be determined

Explanation: To counter the effect of shift in MNA due to armature reaction, the brushes could be shifted. A small brush shift in appropriate direction, in the direction of rotation for generator and in opposite direction for motor, also helps in commutation.

7. Calculate the number of conductors on each pole piece required in a compensating winding for a 6-pole lap-wound dc armature containing 286 conductors. The compensating winding carries full armature current. Assume ratio of pole arc/ pole pitch = 0.7.
a) 6
b) 8
c) 9
d) 7

Explanation: The number of ampere-turns required for compensating winding is ATcw /pole = ATa (peak) *(pole arc/pole pitch) = [IaZ/(AP2)] * (pole arc/pole pitch).
Ncw/pole = (Z/2AP) * (pole arc/pole pitch) = [286 / (2*6*6)] 0.7 = 2.78.
Compensating conductors/pole = 2 * 2.78 = 6 (nearest integer).

8. A compensating winding with ampere-turns greater than peak ampere turns is required in order to neutralize the effect of armature reaction because _____________
a) Pole arc = Pole pitch
b) Pole arc > Pole pitch
c) Pole arc < Pole pitch
d) Can’t be determined using pole arc, pole pitch

Explanation: The number of ampere-turns required for compensating winding in a DC machine is ATcw /pole = ATa (peak) *(pole arc/pole pitch) = [IaZ/(AP2)] * (pole arc/pole pitch). Thus, if compensating winding ampere turns are more then, pole arc is definitely greater than pole pitch.

9. If pole arc is less than pole pitch, a compensating winding will have ampere-turns _________ (compare to peak ampere turns).
a) Less
b) Equal
c) More
d) Can’t be specified

Explanation: The ampere-turns required for compensating winding in a DC machine is ATcw /pole = ATa (peak) *(pole arc/pole pitch) = [IaZ/(AP2)] * (pole arc/pole pitch). Thus, if compensating winding ampere turns are less then, pole arc is smaller than pole pitch and vice-versa.

10. What is the pole arc/pitch ratio, if 360 AT compensating winding is used where 1960AT is peak value?
a) 0.7
b) 0.8
c) 0.9
d) 0.6

Explanation: The ampere-turns required for compensating winding in a DC machine is ATcw /pole = ATa (peak) *(pole arc/pole pitch). If compensating winding of 360AT is used the, 360/1960 will give ratio of pole pitch /pole arc, equal to 0.6.

11. Only drawback of compensating winding is _______
a) Cost
b) Unavailability of material
c) Construction
d) Not a single drawback

Explanation: Compensating winding is the best method in order to prevent the effect of armature reaction and its consequences. Only problem is compensating winding is expensive, but it is must to use them in machines with heavy overloads occur.

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