Electrodynamics — Science Sage

AC Generator (Alternator)

An AC generator converts mechanical energy into electrical energy using electromagnetic induction. A rectangular coil rotates in a uniform magnetic field between the poles of a magnet.

As the coil rotates, the flux through it changes continuously → induces an EMF (Faraday's law)
The output EMF varies sinusoidally with time
Connected to the external circuit via slip rings (continuous rings) → output is alternating current (AC)

EMF = NBAω sin(θ) = EMF_max sin(ωt)
EMF_max = NBAω

Symbol	Meaning	Unit
N	Number of turns in coil	—
B	Magnetic field strength	T
A	Area of coil	m²
ω	Angular velocity of coil	rad·s⁻¹
θ	Angle between coil plane and B-field	rad

Maximum EMF occurs when the coil plane is parallel to the magnetic field (rate of flux change is greatest). EMF = 0 when the coil is perpendicular to B (flux is at maximum, rate of change = 0).

DC Generator

A DC generator uses a commutator (split-ring) instead of slip rings. The commutator reverses the connections to the external circuit every half-turn, so the output is always in the same direction (rectified/pulsating DC).

AC generator → slip rings → sinusoidal output. DC generator → commutator → rectified (pulsating) DC output.

Electric Motor

An electric motor is the reverse of a generator — it converts electrical energy into mechanical (kinetic) energy. Current-carrying conductors in a magnetic field experience a force (F = BIL), causing rotation.

Uses a commutator to maintain continuous rotation
As the motor spins, it acts like a generator and produces a back-EMF (opposing the supply)
Net voltage across motor = V_supply − EMF_back
At start-up: back-EMF = 0 → large current drawn. At full speed: back-EMF ≈ V_supply → small current drawn

Back-EMF increases as motor speed increases, reducing the current draw and preventing the motor from overheating at full speed.

RMS Values

AC voltage and current are continuously changing. The RMS (root mean square) value is the effective DC equivalent — it delivers the same power as a DC circuit.

V_rms = V_peak / √2 ≈ 0.707 × V_peak
I_rms = I_peak / √2 ≈ 0.707 × I_peak

South African mains supply: V_rms = 230 V → V_peak = 230 × √2 ≈ 325 V.

Power in AC Circuits

P = V_rms × I_rms
P = V²_rms / R
P = I²_rms × R

Worked Example — RMS and Power:
A hairdryer operates at 230 V (rms) and draws 8 A (rms). Calculate: peak voltage, peak current, and power consumption.

V_peak = 230 × √2 = 325.3 V
I_peak = 8 × √2 = 11.3 A
P = V_rms × I_rms = 230 × 8 = 1840 W = 1.84 kW

Transformers

A transformer uses electromagnetic induction to change (transform) AC voltage. It consists of a primary coil and a secondary coil wound on a shared iron core.

V_p / V_s = N_p / N_s
Ideal transformer: V_p × I_p = V_s × I_s (power in = power out)

Step-up transformer: N_s > N_p → V_s > V_p (voltage increases, current decreases)
Step-down transformer: N_s < N_p → V_s < V_p (voltage decreases, current increases)

Transformers only work with AC — DC cannot induce an EMF in the secondary coil because DC produces no changing flux.

Why AC for Power Transmission

Power is transmitted at high voltage and low current to minimise resistive losses (P_loss = I²R). Transformers step voltage up for transmission and step it back down for household use — but transformers only work with AC. This is why AC is the global standard for grid electricity.

P_loss = I²R (power dissipated in transmission lines)
Transmitting at high V → low I → P_loss is minimised

IEB Extension — Transformer Losses and Efficiency

Real transformers are not 100% efficient. Energy losses include:

Eddy currents: induced circular currents in the iron core heat it up. Minimised by using a laminated core (thin layers of iron insulated from each other)
Hysteresis losses: energy lost repeatedly magnetising and demagnetising the core each cycle
Copper losses: resistive heating in the coil windings (I²R losses in the wire)

Efficiency η = (P_out / P_in) × 100% = (V_sI_s) / (V_pI_p) × 100%

Modern power transformers achieve efficiencies of 98–99%. IEB questions often give P_in and P_out and ask for efficiency, or give efficiency and ask for the actual secondary current.

AC Generator — Rotating Coil and EMF Waveform

Generator Controls

Speed (ω) 3 rad·s⁻¹

Turns (N) 5

Live Values

EMF (peak)

EMF (now)

Angle θ

0°

Question 1 of 8

What is the key structural difference between an AC generator and a DC generator?

Question 2 of 8

The peak voltage of South African mains supply is approximately (V_rms = 230 V):

Question 3 of 8

A transformer has 200 primary turns and 50 secondary turns. If the primary voltage is 240 V, the secondary voltage is:

Question 4 of 8

The EMF of an AC generator is maximum when:

Question 5 of 8

An electric heater (pure resistive load) operates at V_rms = 230 V with a resistance of 26.45 Ω. The power it consumes is:

Question 6 of 8

Why are transformers only effective with AC and not DC?

Question 7 of 8

A generator coil (N = 100 turns, B = 0.4 T, A = 0.02 m²) rotates at ω = 50 rad·s⁻¹, so it would produce a certain peak EMF. An identical motor (same coil design, coil resistance 5 Ω) is connected to a 100 V DC supply and spins at a speed where its back-EMF equals 80% of that peak EMF value. What current does the motor draw at this speed?

Question 8 of 8

A step-down transformer connects an 11 kV transmission line (primary, 2200 turns) to a workshop supply at 220 V (secondary). At full load the secondary delivers 50 A, and the transformer's efficiency at this load is 92%. What current is drawn from the 11 kV line (the primary current)?

IEB Extended Questions

IEB Question 1

A transformer has a primary voltage of 11 000 V and a secondary voltage of 230 V. The primary current is 2 A. If the transformer efficiency is 95%, the actual secondary current is approximately:

Show all formulae, substitutions, and units. For transformer problems, state whether it is a step-up or step-down transformer and justify.

Question 1 — AC Generator Analysis (7 marks)

An AC generator has a coil of 80 turns, area 0.05 m², rotating at 50 rev·s⁻¹ in a uniform magnetic field of 0.3 T. (a) Convert the rotational speed to angular velocity ω in rad·s⁻¹. (b) Calculate the peak EMF. (c) Write the equation for EMF as a function of time. (d) Calculate the EMF when θ = 30°. (e) At what angle is the EMF zero? Explain physically why this is so. (f) Calculate V_rms. (g) Sketch the EMF vs time graph for one full revolution, labelling peak values and period.

Question 2 — Transformer Calculation (6 marks)

A step-up transformer is used to transmit power from a 10 kV generator to a 132 kV transmission line. The generator produces 500 kW. (a) Identify this as a step-up or step-down transformer and calculate the turns ratio N_p:N_s. (b) Calculate the current in the primary coil. (c) Calculate the current in the secondary (transmission line) side. (d) The transmission line has a total resistance of 20 Ω. Calculate the power lost as heat in the line. (e) What percentage of the transmitted power is lost? (f) Explain why high voltage transmission reduces energy losses.

Question 3 — RMS and Power (5 marks)

A 60 W light bulb is connected to a 230 V (rms) AC supply. (a) Calculate the rms current through the bulb. (b) Calculate the resistance of the bulb filament when operating. (c) Calculate the peak current. (d) Calculate the peak voltage. (e) If the mains frequency is 50 Hz, how many times per second does the instantaneous power reach its maximum value? Explain.

Question 4 — Motor and Back-EMF (5 marks)

An electric motor is connected to a 120 V supply. At full operating speed it draws 2 A and the back-EMF is 110 V. (a) Calculate the resistance of the motor coil. (b) Calculate the current at start-up (back-EMF = 0). (c) Explain why the motor draws much more current when starting than when running at full speed. (d) Calculate the power converted to mechanical energy at full speed. (e) Calculate the efficiency of the motor at full speed (input power vs mechanical output power).

Question 5 — Transformer Efficiency from Test Data (7 marks)

A transformer with a primary coil of 2000 turns is tested on the workbench. The primary voltage is held constant at 220 V and the secondary voltage remains 11 V throughout. The secondary is connected to three different loads in turn, and the currents are measured:

Trial	I_p (A)	I_s (A)
1	0.50	9.5
2	1.00	19.0
3	1.50	28.4

(a) Calculate the number of secondary turns N_s, using the turns ratio N_p/N_s = V_p/V_s.
(b) For Trial 2, calculate the input power P_in and the output power P_out.
(c) Calculate the efficiency of the transformer for Trial 2.
(d) Calculate the power lost as heat in Trial 2.
(e) Calculate the efficiency for Trial 3 as well, and compare it with your answer for Trial 2. Suggest a physical reason the efficiency is slightly lower at the higher current of Trial 3 (think about which type of loss depends on I²).