The sine wave, the DC generator and the transformer
Week 10 · Chapter 3 (C): the sine wave of a generator and its frequency (50 Hz at home), the effective value (0.707 of the maximum), rectification and the DC generator (split-ring commutator), the transformer (turns, voltage and current ratios, step-up and step-down) · 3 lessons
2
From AC to DC: the commutator
Physics Sec 3 Term 1, Chapter 3: pp. 73–79
GoalExplain how a split-ring commutator makes the external current flow in one direction, and why it still pulsates.
New words
commutator (split ring) · one ring cut into two halves, each joined to one end of the coil; the brushes touch it and slide from one half to the otherpulsating direct current · a current that always flows the same way but rises and falls, reaching zero twice in every turnDC generator · a generator whose slip rings are replaced by a commutator
Explanation
In the AC generator each end of the coil is joined to its own slip ring, so the current in the outside circuit reverses every half turn. In the DC generator the two slip rings are replaced by one ring cut into two halves, the commutator, and each half is joined to one end of the coil. The two brushes stay still. When the plane of the coil passes the position perpendicular to the field, the emf in the coil is zero and is about to reverse; at that very moment each brush slides from one half to the other half. So the connection to the outside also reverses, and the current in the external circuit always flows the same way. It is still not steady: it rises to a peak and falls to zero twice in every turn, so the graph is a row of humps. Using many coils turned at angles to one another, with many segments in the commutator, makes the humps overlap and the current nearly steady.
Worked examples
Turn
Plane of the coil
emf of the coil
Outside current with slip rings
Outside current with commutator
0°
perpendicular to the field
0
0
0
90°
parallel to the field
maximum, one way
one way
direction A
180°
perpendicular to the field
0
0
0
270°
parallel to the field
maximum, the other way
the other way
direction A again
The table follows one coil through a turn. The first and third rows are the moments when the brushes change halves.
A coil turns at 20 turns per second. With slip rings the current has f = 20 Hz. With a commutator the current never reverses, but it has 2 humps per turn, so 20 × 2 = 40 humps per second.
A simple model of two coils at right angles, with the output always taken from the coil with the larger emf: the output varies between 0.707 of the peak and the peak. With a peak of 10 V it stays between 10 × 0.707 = 7.07 V and 10 V, never reaching zero.
Class activity
Paper practice. Draw the commutator as a circle cut into two halves with the two brushes touching it on opposite sides. Draw it again after a half turn. Mark which end of the coil each brush touches in both drawings, and check that the outside current has not changed direction.
Practice
1
A coil turns at 30 turns per second. Find the frequency of the emf with slip rings, and the number of humps per second with a commutator.
Slip rings: f = 30 Hz. Commutator: 30 × 2 = 60 humps per second.
2
A coil with a commutator turns at 600 turns per minute. How many humps are there per second?
600 ÷ 60 = 10 turns per second; 10 × 2 = 20 humps per second.
3
In what position of the coil do the brushes change from one half of the commutator to the other, and what is the emf then?
When the coil’s plane is at right angles to the field; the coil’s emf is zero at that moment.
4
In the simple two-coil model of the lesson the peak is 12 V. Between what two values does the output vary?
