Synchronous Motors Questions and Answers

Answer: A

Correct answer: lags the terminal voltage by the load angle

A motor takes power in, so its rotor field is dragged behind the stator field and the excitation emf lags the terminal voltage. In a generator the rotor is driven ahead and the emf leads. The sign of the load angle is therefore what distinguishes the two modes of the same machine.

Answer: B

The net voltage acting round the armature circuit is the phasor difference between the applied voltage and the back emf, and the synchronous impedance limits the current it drives. Both the magnitude and the phase of that difference matter.

Answer: C

The rotor falls further behind the field until the developed torque matches the load, which lengthens the phasor difference driving the current. The speed cannot change at all, which is the machine's defining property.

Answer: D

Normal excitation is defined as the field current giving unity power factor at that particular load, so it is not a fixed value but changes as the load does. More excitation than that gives a leading power factor and less gives a lagging one.

Answer: A

As the load grows the demagnetising component of armature reaction grows with it, so the field must be raised to keep the reactive exchange at zero. Normal excitation therefore rises along a locus as the load increases.

Answer: B

Correct answer: power factor against field current at constant load

The power factor rises to unity as the excitation is increased to normal and falls again as the machine becomes over-excited, so the curve peaks in the middle and slopes away on both sides. That inverted V is the mirror of the ordinary V-curve of armature current, which reaches its minimum at the same point.

Answer: C

Correct answer: falls to a minimum and then rises again

This is the V-curve. Under-excited, the machine draws a lagging magnetising current from the supply; over-excited, it draws a leading one. Between them lies normal excitation, where the machine supplies its own magnetising requirement and the armature current is at its minimum, all of it real.

Answer: D

Correct answer: a leading current of similar magnitude

Both machines carry the same load, so both draw the same real current; the difference is in the reactive component. The under-excited machine takes lagging reactive current from the supply and the over-excited machine supplies it, so their currents are of similar size but on opposite sides of unity power factor.

Answer: A

Correct answer: at which it can be brought into synchronism from near synchronous speed

Pull-in torque governs the last step of starting: the machine has been brought close to synchronous speed on its damper winding, and the field is then applied to lock it into step. If the load torque exceeds the pull-in value the rotor cannot be captured, and the machine continues to slip.

Answer: B

It corresponds to the peak of the power-angle curve, and beyond that point no load angle exists at which the developed torque can match the demand. Raising the excitation raises it in direct proportion, which is why an important drive is run somewhat over-excited.