DC Motor Drives Questions and Answers

Answer: C

Correct answer: Changing the supply frequency

A dc machine has no supply frequency to change - its speed depends on armature voltage, flux and armature resistance, and nothing else. Frequency control belongs to ac machines, whose synchronous speed is \( 120f/P \). The other three are the three classical dc speed control methods: armature voltage below base speed, field weakening above it, and series resistance, which works but wastes the energy as heat.

Answer: D

Correct answer: extra heating without extra useful torque

The mean torque follows the mean current, so the ripple adds nothing useful. But heating follows the root mean square current, which the ripple raises, so the machine runs hotter for the same output and may have to be derated. The ripple also causes torque pulsations and extra commutator sparking. A smoothing inductor, or a higher pulse number, is fitted to reduce it.

Answer: A

Correct answer: reduce the harmonics drawn from the supply

Two six-pulse bridges fed from transformer windings 30 degrees apart give twelve pulses per cycle. On the supply side the fifth and seventh harmonics cancel, leaving the eleventh and thirteenth as the lowest present and much reducing the distortion the installation injects into the network. On the dc side the ripple frequency doubles and its amplitude falls, so less smoothing is needed - though not none. The mean voltage and the quadrant capability are unchanged.

Answer: B

Raising, holding and controlled lowering between them require torque and speed of both signs, which is four-quadrant duty. A dual converter or a four-quadrant chopper provides it, with a mechanical brake for holding at rest.

Answer: C

Independent control of flux and armature current gives a predictable and adjustable characteristic, which the series machine's load-dependent flux does not. The series machine's advantage is its starting torque, which suits traction.

Answer: A

The average output of a single phase semi converter is \(V_{dc} = \dfrac{V_m}{\pi}(1 + \cos\alpha)\). It is greatest for \(\alpha = 0\), when it equals \(2V_m/\pi\).

\[ V_m = \sqrt{2} \times 208 = 294.16\ \text{V} \]

\[ \begin{aligned} V_{dc,max} = \frac{2 \times 294.16}{\pi} \\ &= 187.36\ \text{V} \end{aligned} \]

The values 18.736 V, 1.8736 V and 1873.6 V are 10, 100 and 10 times too small or large. The answer is 187.36 V, option A.