DC Motors Questions and Answers

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.

Answer: C

If the field opens, the speed of a DC motor will increase, dangerously.

The speed of a DC motor is

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

When the field circuit of a shunt motor opens, the flux falls to the small residual value. The back emf collapses, the armature current rises sharply, and to restore the back emf the motor races to a very high speed. Unloaded, it can run away and damage itself, which is why field-failure protection is fitted.

The speed does not decrease, stop or become zero. Those would follow only if no torque were produced at all, but the residual flux and the very large armature current still produce enough torque to accelerate the motor.

Answer: A

The key gives speed and torque as what the output of a dc motor depends on. The output of a motor is its mechanical power, which is the product of the torque developed on the shaft and the speed, and the output power in watts is torque times angular speed. The back emf is proportional to the speed and the flux, and it governs how much current the armature draws, but it is an internal quantity of the motor and is not itself the output. The applied voltage and the load resistance are electrical factors on the input side, and a motor load is not described by a load resistance. The answer is speed and torque.

Answer: C

The assertion is true and the reason is false (option C). The back e.m.f. of a d.c. motor follows from the supply voltage minus the drops across the armature (and series field) resistance, \(E_b = V - I_a(R_a + R_{se})\), so it depends on those resistances.

At no load the armature current is small, so the drop is small and the back e.m.f. is nearly equal to the supply voltage, which is its maximum, not its minimum. The reason is therefore false.

Answer: A

Both the flux and the armature current appear to the first power, and everything else is fixed by the machine's construction. In practice the relation is used in the shortened form that torque is proportional to the product of flux and current.

Answer: B

Correct answer: opposes the applied voltage and limits the current

A rotating armature generates an emf of its own, and by Lenz’s law it opposes the applied voltage:

Since \( E \) takes most of \( V \) in normal running, the current is small - and at standstill, where \( E \) is zero, it is enormous. The back emf is present whenever the machine turns, not merely while accelerating.

Answer: C

Correct answer: reduces the back emf and admits more current

Load the machine and it slows a little; the back emf falls with the speed, the difference \( V - E \) grows, and more current flows through the armature resistance. More current means more torque, which is exactly what the added load demanded, and the machine settles at a slightly lower speed. No controller is involved - the regulation is inherent in the equations, and it is why a shunt motor holds its speed so well.

Answer: D

Correct answer: the back emf multiplied by the armature current

Of the power \( VI_a \) drawn by the armature, \( I_a^2R_a \) is dissipated as heat in its resistance and the remainder is converted:

so \( EI_a \) is the electromagnetic power crossing to the mechanical side. The shaft output is less again by the iron, friction and windage losses.

Answer: A

With nothing to oppose the applied voltage but a small armature resistance, the current can be twenty times rated or more. A starting resistance or a converter with a current limit is therefore essential on any machine of size.

Answer: B

Correct answer: 1,000 A

At rest there is no back emf, so the armature resistance alone limits the current:

That is many times the rated current of such a machine, and it would destroy the armature and the commutator in seconds. It is the whole reason a dc motor cannot be switched directly onto the supply and needs a starter.