AC generator
A coil rotating in a magnetic field has a changing flux, so an emf ε = NBAω sin ωt is induced. Slip rings and brushes take the AC out.
Principle and parts
An AC generator changes mechanical energy into electrical energy. It works on electromagnetic induction. A coil, called the armature, is turned in a uniform magnetic field. As it turns, the angle between the field and the coil's area vector changes. So the flux through the coil changes, and an emf is induced. The main parts are: the armature coil of N turns; a strong field magnet; two slip rings fixed to the axle, one joined to each end of the coil; and two carbon brushes pressing on the slip rings. The brushes stay still and connect the rotating coil to the outside circuit. The turning force comes from a turbine driven by falling water or steam, or from an engine.
Deriving ε = NBAω sin ωt
- 1A coil of N turns, each of area A, rotates at a constant angular speed ω in a uniform field B.
- 2At time t, the angle between B and the area vector is θ = ωt (taking θ = 0 at t = 0).
- 3Flux linkage: NΦ = NBA cos ωt.
- 4Induced emf: ε = −N dΦ/dt = −NBA d(cos ωt)/dt.
- 5Since d(cos ωt)/dt = −ω sin ωt, ε = NBAω sin ωt.
- 6★So, ε = ε₀ sin ωt, with peak value ε₀ = NBAω. The peak occurs when the plane of the coil is parallel to B.
★ key step
Flux and emf as the coil turns
| Angle θ = ωt | Plane of the coil | Flux linkage NΦ | Induced emf ε |
|---|---|---|---|
| 0° | Perpendicular to B | Greatest, NBA | Zero |
| 90° | Parallel to B | Zero | Greatest, +ε₀ |
| 180° | Perpendicular to B | Greatest, −NBA (reversed) | Zero |
| 270° | Parallel to B | Zero | Greatest, −ε₀ |
| 360° | Perpendicular to B | Greatest, NBA | Zero |
Energy chain in an AC generator
How to draw the AC generator
- Draw the curved pole pieces of a strong magnet, marked N on the left and S on the right.
- Between them, draw a rectangular coil ABCD on an axle, with a curved arrow showing its rotation.
- At the end of the axle, draw two separate rings, the slip rings R₁ and R₂, each joined to one end of the coil.
- Draw two carbon brushes B₁ and B₂ pressing on the rings.
- Join the brushes to an external load (a bulb or resistor) or a galvanometer.
- Beside it, sketch the output: a sine curve of ε against t, with the peak ε₀ marked.
Worked example: peak and rms emf
A generator coil has 100 turns, each of area 0.1 m². It rotates at 50 revolutions per second in a uniform field of 0.1 T. Find the peak emf and the rms emf. (π = 3.14)
- ω = 2πν = 2 × 3.14 × 50 = 314 rad/s.
- ε₀ = NBAω = 100 × 0.1 T × 0.1 m² × 314 rad/s.
- 100 × 0.1 × 0.1 = 1, so ε₀ = 314 V.
- rms emf = ε₀/√2 = 314 ÷ 1.414 ≈ 222 V.
Peak emf ≈ 314 V; rms emf ≈ 222 V
Turbines in hydro and thermal power plants across India turn generators to give 50 Hz AC. A diesel generator set, used in many shops and flats during power cuts, is a smaller AC generator turned by an engine.
5 marks: “State the principle of an AC generator. Draw a labelled diagram and derive the expression for the emf induced in its coil.” Electromagnetic induction; labelled diagram with magnet, coil, slip rings and brushes; NΦ = NBA cos ωt; ε = −N dΦ/dt = NBAω sin ωt; peak ε₀ = NBAω.
2 marks: “At what position of the rotating coil is the induced emf (a) greatest and (b) zero? Explain.” (a) Plane parallel to B: the flux is zero but changing fastest. (b) Plane perpendicular to B: the flux is greatest but momentarily not changing.
1 marks: “Name the parts of an AC generator that connect the rotating coil to the external circuit.” Slip rings and carbon brushes.
Generator emf ε = NBAω sin ωt: say 'NBA omega', for Number of turns, B-field, Area and omega.