DC Generators Questions and Answers

Answer: A

Correct answer: \( P\Phi ZN/60A \)

Each conductor cuts \( \Phi P \) webers per revolution and there are \( N/60 \) revolutions per second, so a single conductor generates \( P\Phi N/60 \) volts. With \( Z \) conductors arranged in \( A \) parallel paths, \( Z/A \) of them are in series in each path, giving

Everything about the machine’s voltage follows from this: the emf is proportional to flux and to speed, and it depends on the winding only through \( P/A \), which is \( 1 \) for lap and \( P/2 \) for wave.

Answer: B

Correct answer: the number of poles

In a lap winding each coil is connected back to the adjacent commutator segment, so the winding closes on itself after passing under one pair of poles and \( A = P \). Every pole therefore has its own parallel path. Since the current divides between as many paths as there are poles, each path carries only \( I_a/P \), which is why lap winding suits heavy-current machines.

Answer: C

Correct answer: two

In a wave winding the coil ends are carried forward to segments nearly two pole pitches apart, so the winding travels right round the armature before closing, and there are always exactly two parallel paths whatever the pole number: \( A = 2 \). The whole armature current divides between just those two, so the current per path is large but the number of conductors in series is high, which gives the high voltage.

Answer: D

Correct answer: low voltage and high current

Lap winding gives \( A = P \) parallel paths, so the armature current divides many ways and each path carries little. That is exactly what a heavy-current machine needs. The price is that only \( Z/P \) conductors are in series, so the emf is comparatively low. Wave winding does the reverse, putting \( Z/2 \) conductors in series for a high voltage at modest current.

Answer: A

Small differences in the flux under each pole give the paths slightly different emfs, and without a low-resistance connection between corresponding points the difference drives current through the brushes. A wave winding needs no such arrangement.

Answer: B

Correct answer: equal to the number of poles

A lap winding has as many parallel paths as poles, and each path must be brought out at a brush, so the machine carries \( P \) brush arms - alternately positive and negative around the commutator. A wave winding, having only two paths, needs in principle only two brushes, though more are often fitted in practice to share the current and reduce the current density at each contact.

Answer: C

Correct answer: 167 V

For a lap winding \( A = P \), so the two cancel in the emf equation:

Note how the pole number drops out entirely for a lap winding: the emf depends only on flux per pole, conductors and speed.

Answer: D

Correct answer: 334 V

Rewound as a wave winding, \( A = 2 \) instead of \( 4 \), so with everything else unchanged the emf doubles:

The ratio is \( P/2 = 2 \) for a four-pole machine, so the wave-wound emf is exactly twice the lap-wound figure of 167 V - at half the current-carrying capability, since there are now two paths instead of four.

Answer: A

Each conductor's emf reverses as it passes from one pole to the next, and the commutator reverses the external connection at the same instant so the terminal polarity stays fixed. It is the defining component of the dc machine.

Answer: B

Correct answer: neutralise the reactance voltage and improve commutation

Sparking that worsens with load points to the reactance voltage induced in the coil undergoing commutation, which grows with the current being reversed. Interpoles - narrow poles set on the magnetic neutral axis and wound in series with the armature - produce a flux proportional to that current, inducing an emf that opposes the reactance voltage and forces the current to reverse in the time available. Being series-connected, they compensate automatically at every load.