AC generator: principle and parts
An AC generator (alternator) changes mechanical energy into electrical energy. It works on electromagnetic induction: turning a coil in a magnetic field changes the flux through it, so an emf is induced.
Main parts
- Armature: a coil of many turns wound on a soft iron core.
- Field magnet: strong poles N and S that give a steady field B.
- Slip rings: two metal rings that turn with the coil, one for each end.
- Brushes: carbon pieces that press on the slip rings and take current out to the circuit.
Emf of an AC generator
Let the coil (N turns, area A) turn with angular speed ω. At time t the normal makes angle θ = ωt with B. Flux: Φ = N B A cos ωt.
By Faraday's law, e = −dΦ/dt = N B A ω sin ωt. The peak emf is e₀ = NBAω.
- Coil face ⟂ to field (θ = 0°): flux is maximum, emf = 0.
- Coil plane parallel to field (θ = 90°): flux is 0, emf is maximum. Here the sides cut the field lines fastest.
In one full turn the emf changes direction twice. Frequency f = ω/2π.
Transformer: principle and construction
A transformer changes an AC voltage to a higher or lower AC voltage. It works on mutual induction.
Two coils are wound on one laminated soft iron core: the primary (Np turns, connected to the input) and the secondary (Ns turns, connected to the load). AC in the primary makes a changing flux in the core. The same flux passes through the secondary and induces an emf there.
A transformer does not work on steady DC, because steady DC makes no change in flux.
Turns ratio: step-up and step-down
Each turn of both coils gets the same emf, dΦ/dt. So Vs/Vp = Ns/Np.
For an ideal transformer (no loss) input power = output power: Vp Ip = Vs Is, so Is/Ip = Np/Ns.
- Step-up: Ns > Np. Voltage goes up, current goes down.
- Step-down: Ns < Np. Voltage goes down, current goes up.
A transformer does not make energy. When it raises voltage, it lowers current by the same factor.
Energy losses and efficiency
Efficiency η = output power / input power × 100%. Real transformers reach 95–99%. Losses:
- Copper loss: heating I²R in the windings. Reduced with thick copper wire.
- Eddy current loss: swirling currents in the iron core. Reduced with a laminated core (thin insulated sheets).
- Hysteresis loss: energy lost as the core is magnetised back and forth. Reduced with soft iron.
- Flux leakage: some flux of the primary misses the secondary. Reduced by winding coils one over the other.
Why transmit at high voltage?
For the same power P = VI, a high V means a small I. Line loss I²R becomes much smaller. That is why power is sent at 220 kV or 400 kV and stepped down near homes.
Try it: a practical
Try it: In the 3D, keep Vp = 220 V and Np = 5. Predict Vs for Ns = 10, then for Ns = 2. Check your answers in the readout (440 V and 88 V). For the generator, pause, then drag the "Coil angle" slider from 0° to 90°. Note where e is 0 and where it is largest. At home: look at the label on a phone charger. It says input 100–240 V AC and output 5 V: that is a step-down.
Key formulas and definitions
- Flux through a turning coil: Φ = N B A cos ωt
- Generator emf: e = N B A ω sin ωt; peak e₀ = NBAω
- ω = 2πf
- Transformer: Vs/Vp = Ns/Np
- Ideal transformer: Vp Ip = Vs Is ⇒ Is/Ip = Np/Ns
- Efficiency: η = (Vs Is)/(Vp Ip) × 100%
- Line loss: P_loss = I² R_line
Worked examples
1. A 100-turn coil of area 0.05 m² turns at 50 rev/s in a 0.2 T field. Find the peak emf.
ω = 2π × 50 = 314 rad/s. e₀ = NBAω = 100 × 0.2 × 0.05 × 314 = 314 V.
2. For the generator above, find the emf when the coil plane has turned 30° from the position where its face is ⟂ to the field.
e = e₀ sin 30° = 314 × 0.5 = 157 V.
3. A generator gives peak emf 200 V at 25 rev/s. What is the peak emf at 50 rev/s?
e₀ ∝ ω. Doubling the speed doubles it: e₀ = 400 V. The frequency also doubles to 50 Hz.
4. A transformer has 200 primary and 1000 secondary turns. The input is 230 V. Find the output voltage. Is it step-up or step-down?
Vs = Vp × Ns/Np = 230 × 1000/200 = 1150 V. Step-up (Ns > Np).
5. A step-down transformer changes 220 V to 11 V. The primary has 2000 turns. Find the secondary turns and the ratio of currents.
Ns = Np × Vs/Vp = 2000 × 11/220 = 100 turns. Is/Ip = Np/Ns = 20, so the secondary current is 20 times the primary current.
6. An ideal transformer gives 11 V, 4 A to a load from 220 V mains. Find the primary current.
Vp Ip = Vs Is ⇒ Ip = 11 × 4 / 220 = 0.2 A.
7. A transformer takes 5 A at 220 V and gives 2 A at 500 V. Find its efficiency.
Input = 220 × 5 = 1100 W. Output = 500 × 2 = 1000 W. η = 1000/1100 × 100 ≈ 90.9%.
8. 100 kW is sent through lines of resistance 5 Ω. Find the line loss when sent at (a) 1000 V, (b) 20000 V.
(a) I = 100000/1000 = 100 A, loss = 100² × 5 = 50 kW (half the power!). (b) I = 5 A, loss = 25 × 5 = 125 W. High voltage cuts the loss 400 times.
Common mistakes
- Saying emf is maximum when flux is maximum. Emf is maximum when flux is ZERO (coil plane parallel to B), because flux is changing fastest there.
- Thinking a transformer works with a battery. Steady DC gives no change in flux, so no emf is induced in the secondary.
- Thinking a step-up transformer increases power. It raises voltage but lowers current; power stays the same or a little less.
- Using Is/Ip = Ns/Np. Current goes the other way: Is/Ip = Np/Ns.