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Armature Reaction and Commutation Questions and Answers

Answer: C

Correct answer: the resultant flux density in the air gap is zero

A coil crossing this axis has no emf induced by the main field, which is where its current can be reversed most easily. Under load the axis shifts - forward in a generator, backward in a motor - so the brushes, which are fixed, are no longer at the right place unless something restores it.

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12.

Under load, the flux density under one edge of a main pole is increased and under the other decreased. The net flux per pole

Answer: D

Equal ampere-turns are added at one tip and subtracted at the other, but the added ones act on iron already near saturation and produce less flux than the removed ones took away. That is the origin of the demagnetising effect in a machine with no brush shift.

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Answer: A

In a lightly saturated machine the distortion is large because the iron responds readily, and the flux redistributes freely. A heavily saturated circuit resists the redistribution, though it then shows a larger net demagnetising effect.

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Answer: B

Correct answer: a coil is short-circuited by a brush

The brush spans more than one segment, so for a brief interval it connects both ends of one coil together. During that interval the coil’s current must reverse completely, and the whole difficulty of commutation lies in getting it to do so before the short circuit is broken.

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Answer: C

Correct answer: reverses uniformly through the commutation period

Linear commutation is the ideal: the current changes at a constant rate and reaches its full reversed value exactly as contact is broken, so no energy remains in the coil’s inductance and no arc is drawn. Real commutation lags behind this because of the reactance voltage the reversal itself induces.

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Answer: D

Correct answer: the trailing edge of the brush

When commutation is delayed, current is still flowing in the coil at the instant the segment leaves the brush, and the remaining energy in the coil’s inductance is released as an arc. It happens at the trailing edge because that is where contact is finally broken.

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Answer: A

Correct answer: the leading edge of the brush

Too strong an interpole reverses the current early, so a reverse current is already established while the brush is still making contact. The excess must then discharge at the leading edge. It is the less common fault, since interpoles are usually set weak rather than strong, but it points to the opposite adjustment.

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Answer: B

Correct answer: the rising contact resistance as the segment leaves the brush

As the segment slides out from under the brush the contact area shrinks, so its resistance rises steeply, and that rising resistance forces the current out of the leaving path and into the entering one. It suffices in small machines but not in large ones, where the reactance voltage is too great for resistance alone to overcome.

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Answer: C

Correct answer: the reactance voltage is too large for contact resistance alone

The reactance voltage is proportional to the rate of change of a substantial current in a substantial inductance, and in a large machine it far exceeds what a rising contact resistance can force against. An interpole instead induces an equal and opposite emf in the coil, cancelling the reactance voltage directly.

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Answer: D

Grades vary between about eight and twelve for ordinary carbon and rather more for metal-graphite, which is what fixes the brush area for a given current. Exceeding it overheats the contact and burns the commutator surface.

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