enter image description here

I am struck in the above problem. The things I was able to figure out are

  1. It is a DC motor

  2. Direction of effective flux [Armature flux + Main Field Flux] due to the brush shift

enter image description here But I am still confused how can we draw voltage waveform out of this information. My workbook says the answer will be option A. Please help me with the problem

  • \$\begingroup\$ Could you find answer? I'm pretty sure it's motor, and I can't differentiate between A & C.. \$\endgroup\$ – Deep Sep 29 at 6:48
  • \$\begingroup\$ Knowing that ans. is A, the only way I can I see to reach there is to consider that armature reaction strengthens field under leading pole tip and weakens it at trading pole tip in the direction of rotation. This changes space B-wave which leads to ans. A (I imagined B wave to be strengthened around leading pole tips). Is this correct explanation., mr. AIR 176😂?? \$\endgroup\$ – Deep Sep 29 at 7:10

We do not know 100% surely how the conductors are connected together and to the commutator lamels, but the sparking is minimized and the output voltage is maximal when every moment as many conductors as possible are connected in series between the brushes and the conductors in use are distributed symmetrically around the magnetic poles.

Voltage diagram (D) would be the ideal, but in practice some fluctuation exists. (B) is a good candidate for the resulted waveform when the brushes are in positions A and B on the commutator. The basic place of the brushes should give full voltage and minimize the sparking.

Just when one lamel leaves the brush, another enters and add the same voltage to the series connection which is dropped off at the same time. The output voltage peak occurs in the middle of the angle position of two contact changes. The contact situation stays stable when the armature rotates angle 0,2*Pi (there are 10 commutator lamels, as many as conductors)

It's also reasonable to assume that the rotation direction isn't reversed between the lamels and conductors.

Still one assumption: We assume no output current, only non-loaded voltage. Thus we avoid all self-induction delay effects. The possiblity to rotate the brushes on the commutator gives a possiblity to compensate them in steady loading situations.

Now we are ready. Rotating the brushes to A'B' distorts the symmetry of the commutation. Commutation becomes late. When new conductor is taken between the brushes, it has higher voltage (=has arrived closer to the pole) than the conductor which is dropped out. Thus The voltage jumps up every time when the connections change.

  • \$\begingroup\$ please take a look at this problem electronics.stackexchange.com/questions/352783/… \$\endgroup\$ – Nikhil Kashyap Jan 29 '18 at 15:24
  • \$\begingroup\$ I am still not able to understand how you differentiated between A and C. Can you please add a diagram. It will help alot in understanding. \$\endgroup\$ – Nikhil Kashyap Jan 29 '18 at 15:39
  • \$\begingroup\$ Look at slide 6 again on my link and consider what the difference in assumption that show agreement with 3 expert answers given that agree and that slide and what the given answer was. such as armature alignment .... We seem to disagree with the given answer. Try to understand what I said. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 3 '18 at 13:27

It is a DC motor

I think it's more likely a DC generator so...

When you had your generator brushes half way between the north and south magnet poles the flux is such that there would be no brush arcing and you would get a smooth voltage waveform as per diagram (B): -

enter image description here

Now, with the brushes moved to the new position, there will be flux problems and arcing hence there is a discontinuity as the brush shorts out coils that have voltage across them. This will mean that diagram (A) or (C) is the resulting voltage waveform.

As this sounds like homework I think you should decide on which of the two is correct - you need to figure out whether the voltage at the new A position is now greater based on current flow shown on the windings.

  • \$\begingroup\$ I think it will be DC motor as dots and cross represent currents. I am preparing for an exam GATE (Graduate Aptitude Test in Engg)and this problem was asked in GATE 2009. I am not able to understand option B itself. Why will the output waveform increase then decrease between 0-0.2pi. How is the graph being drawn? \$\endgroup\$ – Nikhil Kashyap Jan 19 '18 at 20:09
  • 2
    \$\begingroup\$ No, it’s a dc generator. Note that it is asking about the shape of the voltage waveform. If it were a motor they dc supply would be constant irrespective of brush position. \$\endgroup\$ – Andy aka Jan 19 '18 at 20:51
  • \$\begingroup\$ That makes sense. Why is there ripple in the waveform? I can't get how it is being drawn \$\endgroup\$ – Nikhil Kashyap Jan 19 '18 at 21:20
  • \$\begingroup\$ There is ripple because it can never produce perfectly flat dc based on the spacing of poles and the changing directions of flux the windings are cutting. \$\endgroup\$ – Andy aka Jan 19 '18 at 23:47

A 2 Pole Generator with 2 armature conductors would generate a sinusoidal EMF that is now full wave rectified as it commutates at some midpoint. (hopefully equal)

Now with 10 armature conductors the half sine peaks still exist but now there are 10 peaks per cycle that overlap so that the transition from one to the next occurs closer to the peak. (b)

  • \$\begingroup\$ answer in my workbook was A. Can you please explain it in more detail. \$\endgroup\$ – Nikhil Kashyap Feb 2 '18 at 20:25
  • \$\begingroup\$ sure ... I found this slideshare.net/SatyajitPatra2/dc-machines1-47639453 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 2 '18 at 20:48
  • \$\begingroup\$ Its not that helpful. I am confused on why the peak will shift backward as shown in option A? \$\endgroup\$ – Nikhil Kashyap Feb 3 '18 at 10:50
  • \$\begingroup\$ look again at slide 5,6 it agrees with me and others. Perhaps we are overlooking position of brushes is offset with optimal position and phase is shifted. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 3 '18 at 13:40

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