This set of DC Machines Multiple Choice Questions & Answers (MCQs) focuses on “Commutation Process -1”.
1. Commutation is delayed due to __________
a) Leakage reactance
b) Effect of armature reaction
c) Leakage reactance and armature reaction
d) Because of other factors
View Answer
Explanation: The leakage inductance Lc of the coil undergoing commutation has induced in it reactance voltage Lc (dic/dt) which opposes the change in current thereby delaying commutation. The effect of armature reaction causes shift in MNA, delaying the whole commutation process ultimately.
2. Why brushes shifting method is not employed in practical commutation?
a) Expensive
b) Construction problems
c) Causes demagnetization
d) Used practically
View Answer
Explanation: Brushes are located at GNA’s, a small voltage is induced in the commutating coil. It opposes current commutation as the commutating coil is cutting the flux which has the same sign as that of the pole being left behind. It could be partially remedied by shifting the brushes towards MNA but that causes direct demagnetization and is therefore not employed in practice.
3. Which are the methods for getting an ideal commutation?
a) Resistance commutation
b) Voltage commutation
c) Current commutation
d) Resistance and voltage commutation
View Answer
Explanation: Adding resistance between commutator segments and brushes, thus, reducing L/R and consequently getting faster commutation is one of the method. In voltage commutation we, introduce narrow poles called as interpoles to fasten the process.
4. In resistance commutation method we add resistance between __________________
a) Brushes and external circuit
b) Commutator and armature winding
c) At field winding
d) Commutator and brush
View Answer
Explanation: High contact resistance between commutator segments and brushes, achieved by using carbon brushes, adds resistance to the circuit of the commutating coil thereby reducing the time constant (L/R) of the current transient (ic(t)), helping it to change faster in the desired direction.
5. What is the effect on time constant of transient current, as a result of resistance commutation?
a) Remains same
b) Decreases
c) Increases
d) Increases then decreases
View Answer
Explanation: We add high contact resistance between commutator segments and brushes thus, it adds resistance to the circuit of the commutating coil thereby reducing the time constant (L/R) of the current transient (ic(t)), helping it to change faster in the desired direction.
6. Which voltage is neutralized in voltage commutation process?
a) Armature
b) Reactance
c) Field
d) Cannot be determined
View Answer
Explanation: To speed up the commutation process, the reactance voltage must be neutralized by injecting a suitable polarity dynamical (speed) voltage into the commutating coil. In order that this injection is restricted to commutating coils, narrow interpoles are provided in the interpolar region.
7. Interpoles are excited with ____________
a) Armature current
b) Field current
c) Separate supply
d) Mains current
View Answer
Explanation: These interpoles apply a local correction to the air-gap flux density wave such that a pip of appropriate flux density exists over the commutating coil to induce in it a voltage of the same sign as that of coil current after commutation. Hence, they are excited with armature current.
8. Interpoles are excited by keeping them in ____________ with armature.
a) Series
b) Parallel
c) Anywhere
d) Not kept with armature
View Answer
Explanation: In voltage commutation method we use interpoles to speed up the commutation process. For neutralization of reactance voltage at all loads, the interpoles must be excited by armature current by connecting them in series with armature.
9. Polarity of interpoles is one pole ahead in the direction of armature rotation in _____________
a) Motor
b) Generator
c) Always ahead
d) Always behind
View Answer
Explanation: polarity of an interpole is that of the main pole ahead in the direction of armature rotation for the generating mode and that of the main pole left behind with respect to the direction of rotation for motoring mode.
10. What is the size of interpolar air gap compare to main pole air gap?
a) Small
b) Same
c) More
d) Can be of any size
View Answer
Explanation: The interpolar air-gap is kept larger than that of the main pole so that their magnetic circuit is linear resulting in cancellation of the reactance voltage (a linear derivative term) at all loads. Large air-gap results in greater amount of leakage flux which is accommodated by tapering the interpoles with a wider base.
11. Formula for cancellation of reactance voltage on an average basis using interpoles ________
a) 2[Bi(av)liva] Nc = Lc (dic/dt) = Lc (2lc/tc)
b) [Bi(av)liva] Nc = Lc (dic/dt) = Lc (2lc/tc)
c) 2[Bi(av)liva] Nc = Lc (dic/dt) = Lc (lc/tc)
d) Cannot be determined
View Answer
Explanation: 2[Bi(av)liva] Nc = Lc (dic/dt) = Lc (2lc/tc).
With Bi determined by above equation, the ampere-turns needed to cancel the armature reaction ampere-turns and then to create the necessary flux density are given by, ATi = ATa (peak)+(Bi/µ0 )lgi
12. For a given 4-pole machine, carrying armature current 56.82 A, with 846 conductors. The mean flux density in the air gap under the interpoles is 0.5 Wb/m2 on full load and radial gap length is 0.3 cm. Ampere-turns required for an interpole is ____________
a) 3198
b) 2099
c) 4198
d) 6297
View Answer
Explanation: Required ampere-turns are given by ATi = ATa (peak) +(Bi/µ0 )lgi . Substituting the
ATi =[(56.82*846)/ (2*2*4 )] +(0.5/4π*10-7 )* 0.3*10-2 =4198. Turns can be found by dividing with armature current.
13. A 440 V, 4-pole, 25 kW, dc generator has a wave-connected armature winding with 846 conductors. The mean flux density in the air-gap under the interpoles is 0.5 Wb/m2 on full load and the radial gap length is 0.3 cm. Number of turns required on each interpole is _________
a) 74
b) 84
c) 64
d) 54
View Answer
Explanation: Corresponding value of ampere-turns for an interpole
ATi = ATa (peak) +(Bi/µ0) lgi.
= Ia* Z/2*A*P + (Bi/µ0) lgi.
Assuming Ia = Iline Ia = 25*103/440= 56.82 A.
ATi = [(56.82*846)/ (2*2*4)] +(0.5/4π*10-7 )* 0.3*10-2 =4198.
Ni= 4198/56.82= 74.
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