DC Motors Questions and Answers

Answer: A

With the flux reduced to residual magnetism, the speed must rise enormously for the back emf to balance the supply, and a lightly loaded machine can reach a speed at which it flies apart. Field failure protection exists for exactly that case.

Answer: B

Correct answer: the flux is at its full value when armature current flows

Torque is \( k\Phi I_a \), so the machine needs full flux at the moment the armature is energised, or it will produce little torque for a very large current and may fail to accelerate. There is a second reason: with a weak field the machine would run away as it accelerated. The field circuit is highly inductive and takes time to establish its current, which is why the field switch closes first.

Answer: C

Correct answer: develop less torque for the same armature current

Since \( T = k\Phi I_a \), a weakened field gives less torque for every ampere drawn, so the machine accelerates poorly while taking the same heavy current - the worst combination. It also risks running away once it does accelerate, since speed varies inversely with flux. Both are reasons the field is established at full strength before the armature is connected.

Answer: D

Because the armature current is in the opposite sense to that in a generator at the same terminal polarity, the shift is backward rather than forward. A machine that must run both as motor and as generator therefore cannot use a fixed brush shift.

Answer: A

Cutting the supply and recording the speed against time gives the rate of retardation, and with the inertia known the rotational loss follows. Repeating the test with a known extra loss lets the inertia itself be found.

Answer: B

A series machine cannot be run light for a no-load test, so two are coupled with their fields in series, one motoring and one generating, and their combined losses measured. It is the series machine's equivalent of the back-to-back test.

Answer: C

Correct answer: it cannot be run safely on no load

Swinburne’s test measures the losses with the machine running light. A series motor on no load draws almost no current, so its field almost vanishes and its speed rises without limit until the armature is destroyed by centrifugal force. The test therefore cannot be performed, and a series machine is tested instead by a direct load test or by the Hopkinson method with two identical machines.

Answer: D

Correct answer: iron loss together with friction and windage

The rotational losses are those that depend on speed and flux rather than on load current: hysteresis and eddy current loss in the armature iron, bearing and brush friction, and windage. They are measured together in a no-load test and treated as constant, in contrast with the armature copper loss, which varies with the square of the current.

Answer: A

Correct answer: 200 V

From the armature circuit equation,

The back emf takes 91 per cent of the applied voltage. Note that this same machine at standstill would draw \( 220/0.5 = 440 \) A, eleven times its running current, which is why it needs a starter.

Answer: B

Correct answer: 800 W

The armature copper loss is

This is the variable loss, rising with the square of the current, and it is also the difference between the 8.8 kW drawn by the armature and the 8.0 kW converted to mechanical form.