This set of DC Machines Multiple Choice Questions & Answers (MCQs) focuses on “Wave Winding”.
1. What will be the value of “Yf + Yb” for a wave winding?
a) Equal to Yc
b) Half of the Yc value
c) Double of the Yc value
d) Four times Yc value
Explanation: In the wave winding, as the number of coil-sides is double the number of segments, the top coil-side of the second coil will be numbered as (1+2*Yc). After numbering other coil sides,
1 + 2*Yc – Yf = 1+ Yb
So Yf + Yb = 2Yc.
2. For a progressive wave winding Yc = ______
Explanation: Starting at segment 1 and after going through P/2 coils or Yc (P/2) segments, the winding should end in segment 2 for progressive winding or segment (C) for retrogressive winding. That is mathematically,
Yc (P/2) = (C+1)
Yc = 2(C+1)/P
3. Number of parallel paths in wave winding are ______
a) Equal to P
b) Equal to P/2
d) Depends on other parameters
Explanation: In wave winding all coils are divided into 2 groups- all coils carrying clockwise current are series connected and so are all coils with counter-clockwise current- and these 2 groups are in parallel because the winding is closed. Thus, a wave winding has 2 parallel paths irrespective of number of poles.
4. What is the spacing between the brushes for a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coils.
a) 4 segments
b) 8 segments
c) 16 segments
d) 12 segments
Explanation: Only 2 brushes are required in this case as the number of poles in wave winding is equal to 2. So, spacing between the brushes is equal to total number of segments i.e. total slots divided by 2. Spacing between brushes = C/A = 16/2 = 8 segments.
5. What is the relation between conductor current and armature current in wave winding?
a) Ic = Ia
b) Ic = 2Ia
c) Ic = 4Ia
d) Ic = Ia/2
Explanation: the number of parallel paths in the in a wave winding is equal to 2. So, armature current will get divided equally into total number of conductors/paths. Conductor current in a wave wounded machine is half of the Ia.
6. For a conductor current equal to 4mA, Current carried by a particular brush in a 2-pole machine will be _____
Explanation: Conductor current in a wave wounded machine is half of the Ia. So, Ia= 8mA. All positive and all negative brushes are respectively connected in parallel to feed the external circuit. Thus, IBRUSH = Ia /(P/2). Solving we get Brush current = 8mA.
7. Equalizer rings are needed in the wave winding.
Explanation: The armature coil forms 2 parallel paths under the influence of all pole-pairs so that the effect of the magnetic circuit asymmetry is equally present in both the parallel paths resulting in equal parallel-path voltages. Thus, equalizer rings are not needed in wave winding.
8. For a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coil, Yf value is ____
Explanation: Ycs = 16/6 = 2 slots (nearest lower integral value)
Yb= 2*2+1 = 5
Yc= 2(16-1)/6 = 5 segments
Yf = 2Yc – Yb = 5.
9. Wave winding machines are used in ______ currents applications.
d) Can be used anywhere
Explanation: Lap winding machine has the advantage of large number of parallel paths and lower conductor current and is therefore used in low voltage and high current applications. Wave winding has fixed number of parallel paths so, wave wounded machine is used in low currents application.
10. For a wave wounded machine number of brushes for small, large machines respectively is ________ _________
a) 2, 2
b) 4, 2
c) 2, P
d) Both values depend on the given conditions
Explanation: For a small wave wounded machine number of parallel paths are 2, thus 2 brushes are used. For a large machine total number of brushes is equal to the total number of poles. The spacing between adjacent brushes is C/P commutator segments.
Sanfoundry Global Education & Learning Series – DC Machines.
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1000+ Multiple Choice Questions and Answers.