DC Motors Questions and Answers

Answer: B

The opposition of the back emf to the supply is explained by Lenz's law.

When the armature of a DC motor rotates in the field, its conductors cut the flux and an emf is induced in them. Lenz's law says that an induced emf always acts to oppose the change that produces it; here it opposes the current driven by the supply, so it acts against the supply voltage, which is why it is called the back emf.

Faraday's laws give the magnitude of the induced emf but not its direction. Fleming's left hand rule gives the direction of the force on a current-carrying conductor (motor action), and the right hand rule gives the direction of an induced current, not the reason it opposes the supply.

Answer: A

The speed may be increased by increasing the field circuit resistance.

The speed of a DC motor is \(N \propto \frac{V - I_{a}R_{a}}{\phi}\). Adding resistance in the shunt field circuit reduces the field current and so the flux, and with less flux the motor must run faster to generate the back emf the supply requires. This field weakening is the standard way to raise a shunt motor above its normal speed.

Decreasing the field resistance strengthens the field and lowers the speed. Adding armature resistance increases the \(I_{a}R_{a}\) drop and also lowers the speed. Reducing the load raises the speed only slightly in a shunt motor and is not a method of speed control.

Answer: D

Both Statement I and Statement II are true.

Statement I is true: a dc series motor's speed rises sharply as its load falls, and on no load it can race to a dangerous speed. Line shaft drives are belt drives, where the load can be lost if a belt slips or breaks, so series motors are not used for them; shunt motors are.

Statement II is true: in a series motor the field flux is produced by the armature current, so \(\phi \propto I_{a}\) and the torque \(T \propto \phi I_{a} \propto I_{a}^{2}\) before saturation. The torque against armature current curve is therefore a parabola (becoming nearly straight once the field saturates).

Answer: C

All three controls, (a), (b) and (c), play a significant role.

The speed of a dc motor is

\[ N \propto \frac{V - I_{a}R_{a}}{\phi} \]

so it can be changed through each quantity on the right. Flux control (a field rheostat) weakens \(\phi\) and gives speeds above normal. Armature resistance control adds resistance in series with the armature and gives speeds below normal. Supply voltage control, as in the Ward-Leonard system or a controlled rectifier, varies \(V\) smoothly over a wide range.

Every option leaving one of them out is incomplete.

Answer: A

The speed of the motor decreases.

For a dc series motor, \(N \propto \frac{V - I_{a}(R_{a} + R_{se} + R_{ext})}{\phi}\). An extra resistance in series increases the voltage drop, which reduces the back emf available at the same load, so the speed falls. This is the armature resistance (rheostatic) method of speed control, which gives speeds below the rated speed.

The speed does not stay the same or rise, because nothing in the change weakens the field or raises the voltage. The motor settles at a new lower speed rather than oscillating.

Answer: C

Plugging is done by reversing the armature polarity.

In plugging (reverse current braking) the supply connections to the armature are reversed while the field is left as it is. The back emf, which keeps its original direction while the motor is still turning, now adds to the supply voltage, so a large reverse current flows and produces a strong braking torque. A resistance is inserted to limit this current, and the supply must be cut at zero speed or the motor will start in reverse.

Connecting a resistance across the armature (with the supply removed) is dynamic or rheostatic braking. Reversing both armature and field leaves the direction of torque unchanged, and reversing the field of a shunt motor is not the standard plugging method.

Answer: A

The speed becomes slightly less than doubled.

Motor torque is \(T = k\phi I_{a}\). With a constant load torque, halving the field (and so, assuming a linear magnetic circuit, the flux) forces the armature current to double. The speed is \(N \propto \frac{V - I_{a}R_{a}}{\phi}\).

\[ \begin{aligned} \frac{N_{2}}{N_{1}} &= \frac{V - 2I_{a}R_{a}}{V - I_{a}R_{a}} \times \frac{\phi}{\phi/2} \\ &= 2 \times \frac{V - 2I_{a}R_{a}}{V - I_{a}R_{a}} \end{aligned} \]

The second factor is a little less than 1, because the doubled current causes a larger armature drop, so the speed rises to slightly less than double. Exactly double would ignore the armature drop, and the 'half' options reverse the effect: weakening the field raises a dc motor's speed.