DC Generators Questions and Answers

Answer: C

One pole tip is strengthened and the other weakened by equal ampere-turns, but the iron of the strengthened tip is already near saturation, so the net flux falls. That is why the effect is small in a lightly saturated machine and appreciable in a heavily loaded one.

Answer: D

Reversing tens or hundreds of amperes in a fraction of a millisecond induces a substantial emf of self-induction, which tries to keep the current flowing and produces sparking at the trailing brush edge. Interpoles generate an equal and opposite emf.

Answer: A

Correct answer: using carbon brushes whose contact resistance forces the current to transfer

As the brush passes from one segment to the next, the area of contact with the leaving segment shrinks and that with the arriving segment grows. With a resistive brush material the resistance of the leaving path rises accordingly, and the current is progressively forced to transfer to the new segment rather than lingering and arcing at the break. Carbon is used precisely for this contact resistance, which is why it is preferred to copper despite its higher drop.

Answer: B

Interpoles, or a deliberate shift of the brushes, give the coil an emf that drives the current reversal instead of resisting it. Interpoles are preferred because their correction tracks the load and needs no adjustment when the direction changes.

Answer: C

Correct answer: iron loss together with friction and windage

The constant losses are those that do not vary with load: hysteresis and eddy current loss in the armature iron, which depend on flux and speed, and the mechanical losses of bearing friction, brush friction and windage. The shunt field copper loss is usually counted with them, since the field current is held constant. The armature copper loss is the variable one, rising with the square of the current.

Answer: D

Correct answer: armature copper loss

The armature copper loss is \( I_a^2 R_a \), so it rises with the square of the load current and is the only significant loss that varies with load. Iron, friction and windage losses depend on flux and speed, both held constant in normal operation, and the shunt field loss depends on the terminal voltage. The division into constant and variable losses is what makes the maximum efficiency condition so simple.

Answer: A

Correct answer: the variable 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 where

that is where the variable loss has grown to equal the constant loss. Machines are usually designed so that this occurs somewhat below rated load, near the load at which they will actually spend most of their working life.

Answer: B

Running the machine light gives the constant losses, and the copper loss at any load is then computed from the measured resistances. It is quick and needs almost no power, but it cannot reveal how the machine behaves hot.

Answer: C

Correct answer: does not check the machine’s behaviour under load

Swinburne’s test measures the no-load losses and the resistances and computes the efficiency at any load from them, so it is cheap and needs very little power. But the machine never carries load during the test, so nothing is learnt about commutation at load, about temperature rise, or about the effect of armature reaction. A regenerative or Hopkinson test, which loads two identical machines against each other, answers those questions at the cost of needing a matched pair.

Answer: D

One machine drives the other as a generator and the generated power is returned electrically, so only the losses of the pair have to be supplied. It tests both machines at full load with a small power input, which is its whole point.