DC Motors Questions and Answers

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.

Answer: C

Correct answer: 2.25 Ω

The total resistance required is

and the armature already provides 0.25, so the external resistance is

Answering 2.5 forgets that the armature is part of the circuit - a small error here, but a serious one in a machine of very low armature resistance.

Answer: D

Correct answer: the growing back emf takes over the limiting of the current

As the machine accelerates its back emf rises, so \( V - E \) falls and with it the current. Each step of resistance can then be removed to bring the current back up to the accelerating value without exceeding the limit, and by the time the machine is at speed the back emf alone is doing the limiting and no external resistance is needed. Armature resistance and flux change hardly at all during the start.

Answer: A

The armature resistive drop grows with current, leaving less of the applied voltage for the back emf, so the speed droops by a few per cent from no load to full load. Armature reaction weakens the flux slightly, which offsets part of that droop.