DC Motor Drives Questions and Answers

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.

Answer: A

Writing \( E = k\phi\omega \) makes the two dependences explicit, and it is the relation on which every method of speed control rests. Reducing the flux at fixed voltage therefore raises the speed, which is field weakening.

Answer: B

Correct answer: the product of flux and armature current

The torque is

so it depends on the flux and the armature current together, and on nothing else. With the flux held constant - as it is below base speed on a separately excited machine - torque and current are simply proportional, which is why a dc drive controls torque by controlling current. In a series machine the flux itself is produced by the armature current, and only then does the torque become proportional to the square of the current.

Answer: C

Correct answer: \( V = E + I_aR_a \)

The applied voltage must supply the back emf and drive the current through the armature resistance:

The back emf takes almost all of it in normal running, the resistive term being a few per cent. At standstill \( E = 0 \), so the whole of \( V \) appears across \( R_a \) and the current would be enormous - which is why a starter is needed. The form with a minus sign describes the machine acting as a generator.

Answer: D

Rearranging the circuit equation isolates the speed, and the expression shows the three handles available: the applied voltage, the armature resistance and the flux. Each gives its own method of control with its own drawbacks.

Answer: A

With the current nearly zero, the resistive drop vanishes and the speed is set by voltage and flux alone. The drop from that value to the loaded speed is the machine's speed regulation, which is small in a shunt or separately excited machine.

Answer: B

Armature voltage control gives constant torque capability down to standstill without wasting energy, since the flux and hence the torque per ampere are unchanged. Field weakening is reserved for speeds above the base value.

Answer: C

Once the armature voltage has reached its ceiling, the only way to go faster is to reduce the flux, which raises the speed for the same back emf. Torque falls in proportion, so the region above base speed is one of constant power.

Answer: D

Correct answer: constant torque capability

Below base speed the flux is held at its rated value and the speed is raised by increasing the armature voltage. Since \( T = k\Phi I_a \) and the current is limited by heating rather than by speed, the machine can develop rated torque at any speed in this range, and the power rises in proportion to the speed. Above base speed the voltage has reached its ceiling and the flux must be weakened, giving constant power with falling torque instead.

Answer: A

The resistance drops voltage and dissipates it as heat, so efficiency falls in proportion to the speed reduction, and the speed also becomes very sensitive to load. It survives only in simple traction and crane starters.