Power transmission, transformer efficiency and the DC motor
Week 11 · Chapter 3 (D): transformers in power transmission (I²R loss) and their efficiency, the two transformer examples, the DC motor (construction, Fleming’s left-hand rule, the commutator, one revolution, several coils, motor against galvanometer); the summary and exercises (pp. 86–96) are not used · 3 lessons
3
One revolution of a motor: why the current reverses
Physics Sec 3 Term 1, Chapter 3: pp. 80–85
GoalFollow a coil through one revolution, explain why the current must be reversed, and find the torque at different angles.
New words
torque · the turning effect of the couple on the coil, τ = BIAN sin θangle θ · the angle between the normal to the coil and the fieldinertia · the tendency of the turning coil to keep turning through the position where the torque is zero
Explanation
Start with the plane of the coil parallel to the field. Then θ = 90° between the normal and the field, and the torque τ = BIAN sin θ has its largest value. As the coil turns, θ gets smaller and the torque falls. When the plane is perpendicular to the field, θ = 0° and the torque is zero: the two forces are now along one line and pull the coil apart rather than turn it. The coil does not stop, because its inertia carries it past this position. At this moment each half of the split ring slides under the other brush, so the coil’s current changes sign. Now the force on each side is reversed too, and the torque keeps the same sense of rotation: it grows to its maximum when the plane is parallel again. Without the commutator (with two plain slip rings) the torque would change sense after the perpendicular position and pull the coil back, so it would only rock to and fro. A motor with several coils at equal angles, and a commutator with twice as many segments as coils, has a larger and smoother torque, because while one coil is passing through zero the others are near their maximum.
Turn
Plane of the coil
Torque
What happens at the commutator
start
parallel to the field
maximum
each brush on its own half
¼ turn
perpendicular to the field
zero
brushes cross the gap; the halves swap
½ turn
parallel again
maximum, same sense
current in the coil is now reversed
¾ turn
perpendicular
zero
the halves swap again
Worked examples
θ
0°
30°
60°
90°
τ (N·m)
0
0.072
0.125
0.144
The coil of the last lesson has a largest torque BIAN = 0.144 N·m. When the normal makes θ = 30° with the field, τ = 0.144 × sin 30° = 0.144 × 0.5 = 0.072 N·m. At θ = 60° it is 0.144 × 0.866 = 0.125 N·m. At θ = 90° (plane parallel) it is 0.144 N·m, and at θ = 0° (plane perpendicular) it is zero.
Why the commutator matters. With the commutator, the torque of one coil is always in one sense: its size follows |sin θ|, which rises from zero to a maximum and falls to zero in every half turn. Without it the sign of sin θ would be allowed to go negative after the half turn, the torque would reverse, and the coil would rock. The graph shows both.
θ
0°
30°
45°
60°
90°
one coil (N·m)
0
0.072
0.102
0.125
0.144
two coils (N·m)
0.144
0.197
0.204
0.197
0.144
Two coils at 90° to each other. The torque of each follows |sin θ| and |cos θ|, with 0.144 N·m as the largest value of one coil. Total = 0.144 × (|sin θ| + |cos θ|). At θ = 0°: 0 + 0.144 = 0.144 N·m. At 30°: 0.072 + 0.125 = 0.197 N·m. At 45°: 0.102 + 0.102 = 0.204 N·m. The total never falls to zero (it stays between 0.144 and 0.204 N·m), so the motor runs more smoothly than with one coil.
Class activity
Paper only. Make a four-row table like the one above for a motor with one coil, and mark in a different colour the moments when (a) the torque is zero, (b) the commutator swaps the halves. Then sketch the torque curve for one coil with and without a commutator over one full turn, and mark the position where the two curves separate.
Practice
1
The largest torque of a coil (BIAN) is 0.50 N·m. Find the torque when the normal to the coil makes 30° with the field.
τ = 0.50 × sin 30° = 0.50 × 0.5 = 0.25 N·m.
2
Find the torque on that coil when its plane is perpendicular to the field.
The normal is along the field, θ = 0° and sin 0° = 0, so the torque is 0.
3
The torque is zero at the perpendicular position. Why does the coil not stop there?
The coil is already turning, so its inertia carries it through the position. The commutator then reverses the current and the torque pushes it on in the same sense.
4
A student replaces the split ring by two plain slip rings and uses the same DC source. Describe what the coil does.
Past the perpendicular position the torque changes sense and pulls the coil back, so it rocks to and fro near that position instead of turning round.
