DC Motor Drives Questions and Answers

Answer: C

Correct answer: 2.4 Ω

The total circuit resistance required is

and the armature already provides 0.4 ohm, so the resistance to be added is

Answering 2.8 ohm forgets to subtract the armature resistance the circuit already has - a small error here, but a large one in a machine of low armature resistance.

Answer: D

Correct answer: 14.1 kW

The electrical input is

and the shaft output is that multiplied by the efficiency:

Dividing rather than multiplying gives 18.2 kW, which would have the machine producing more than it received.

Answer: A

The first two describe the two working regions of a dc drive correctly. The third is wrong: the added resistance dissipates the voltage it drops, so efficiency falls roughly in proportion to the speed reduction obtained.

Answer: B

Correct answer: 1, 2 and 3

All three are correct and together they set out the converter hierarchy. A semi-converter, having diodes in one half of the bridge, cannot produce a negative mean voltage, so it motors in one direction only. A full converter can invert, giving positive current with either sign of voltage - two quadrants, that is motoring and regenerative braking in one direction. A dual converter adds a second bridge for the reverse current, completing all four.

Answer: C

Correct answer: 1-Q, 2-P, 3-R

Armature voltage control operates below base speed and is efficient, since the converter merely varies the voltage rather than dissipating the difference. Field weakening operates above base speed, where the voltage has reached its ceiling. Armature resistance control also works below base speed, but wastefully - the excess voltage is burnt in the resistor as heat, and it is for that reason obsolete except in the simplest applications.

Answer: D

Correct answer: 1-Q, 2-R, 3-P

Class A has one switch and one diode and can only motor. Class B is its mirror image and can only regenerate. Class E is the full bridge, in which both voltage and current can take either sign, giving all four quadrants. Class C, not listed here, combines A and B for two-quadrant operation in one direction, and two class C units make a class E.

Answer: A

Correct answer: 4, 1, 2, 3

At fixed field the speed follows the armature voltage, so 60 V gives the lowest speed, then 120 V, then 240 V. The fourth condition applies the same 240 V but weakens the field to 80 per cent, and since \( N \propto V/\Phi \) that raises the speed by a further 25 per cent, making it the highest of the four. The sequence 4, 1, 2, 3 traces the whole control strategy: voltage control up to base speed, then field weakening above it.

Answer: B

With the voltage reversed and the current keeping its sign, the power flow reverses, which is exactly inversion. The load must contain an emf of the right polarity to sustain the current, which in a drive is the machine's own back emf.

Answer: C

At no load the current and hence the flux are very small, and the speed must rise enormously to generate the back emf. The reason is precisely the mechanism of the runaway, which is why such machines are geared or coupled rather than belted.

Answer: D

Correct answer: False, since a diode bridge cannot pass power in the reverse direction

A diode conducts in one direction only and its conduction is not controllable, so the bridge can rectify but never invert. When the machine begins to generate, the current has nowhere to go, the dc link voltage rises, and something eventually fails. Regeneration into the mains requires a fully controlled bridge operating beyond 90 degrees, or an active front end; otherwise the energy must be burnt in a braking resistor.