DC Motors Questions and Answers

Answer: D

Correct answer: A slip ring assembly on the rotor

Slip rings provide a continuous connection to a rotating winding and belong to synchronous machines and wound rotor induction machines. A dc machine uses a commutator, which is a rotary switch reversing the connection to each coil as it passes the poles - a quite different device, and the essential one. The other three items are all parts of a dc motor.

Answer: A

Correct answer: inspection of brushes and commutator

The sliding contact is the one part of a dc machine that wears in normal service, so brush length, spring pressure, bedding and commutator surface are checked routinely and brushes replaced as they wear down. Windings and bearings need attention on the same schedule as any other machine. It is this recurring brush maintenance that has driven dc drives out of most new installations.

Answer: B

Correct answer: its speed at rated voltage, rated flux and rated load

A nameplate speed is the speed under the full set of rated conditions, and it is the datum against which regulation and field weakening range are quoted. The machine runs somewhat faster than this on no load, and faster again with the field weakened; and the highest safe speed is a separate figure, set by centrifugal stress and commutation and usually stated on the plate as a maximum.

Answer: C

Correct answer: the copper loss equals the constant loss

Writing the efficiency with constant loss \( W_c \) and variable loss \( I^2R \) and differentiating with respect to current gives the maximum at

Machines are designed so that this occurs somewhat below rated load, near where they will spend most of their working life. It is not at full output, and certainly not at no load, where there is no output at all to be efficient with.

Answer: D

Correct answer: appreciably lower than at rated load

At light load the constant losses - iron, friction, windage and the shunt field - are unchanged, but the output has fallen to a quarter, so they form a much larger fraction of the input. The efficiency curve rises steeply from zero at no load, peaks somewhat below rated load, and falls away slowly beyond it, so quarter load lies well down the rising limb.

Answer: A

Stator copper and iron losses come off the input, the remainder crosses the air gap as the product of back emf and armature current, and mechanical losses come off before the shaft. Reversing that flow is generator operation.

Answer: B

Correct answer: the product of the back emf and the armature current

\( EI_a \) is the power crossing from the electrical side to the mechanical side, which is what the air gap power means. It is less than the armature input \( VI_a \) by the copper loss \( I_a^2R_a \), and greater than the shaft output by the rotational losses. Using the terminal voltage instead would include a loss that never crosses the gap.

Answer: C

That proportionality is what makes an armature current loop equivalent to a torque loop, which is the foundation of every dc drive control scheme. The square law belongs to the series machine, where the flux rises with the current as well.

Answer: D

Correct answer: doubles the torque developed

With the flux held constant, \( T = k\Phi I_a \) is simply proportional to current, so doubling the current doubles the torque. The square law belongs to the series machine below saturation, where the flux itself rises with the current - and confusing the two is exactly what the quadrupling distractor tests.

Answer: A

Correct answer: rise to about 125 per cent of its former value

The torque required has not changed, and \( T = k\Phi I_a \), so with the flux at 80 per cent the current must rise to compensate:

Field weakening therefore raises the speed by 25 per cent and the armature current by 25 per cent at the same torque, which is why the constant-power region above base speed is limited by armature heating as much as by anything else.