What is covered in Grade 12 Physical Sciences Electrodynamics?

Electrodynamics connects electricity and magnetism through electromagnetic induction. This chapter covers AC and DC generators, AC and DC motors, and transformers. Faraday's law and Lenz's law are applied to explain how these devices work.

What is Faraday's Law in Physical Sciences?

The emf, E, induced around a single loop of conductor is proportional to the rate of change of the magnetic flux, φ, through the area, A, of the loop. This can be stated mathematically as: E = −N(Δφ/Δt), where φ = B·A cosθ and B is the strength of the magnetic field.

A generator is a device that converts mechanical energy into electrical energy.

What is the formula for Faraday's law (induced emf)?

The formula for Faraday's law (induced emf) is: E = −N(Δφ/Δt). Where: E = induced emf (V); N = number of turns; Δφ = change in flux (Wb); Δt = time (s) The SI unit is V.

Is Grade 12 Physical Sciences Electrodynamics on the CAPS curriculum?

Yes. Electrodynamics is part of the official CAPS Physical Sciences curriculum for Grade 12, Term 3 in South Africa. The content on MathSciBuddy follows the Annual Teaching Plan (ATP) set by the Department of Basic Education.

Electrodynamics Grade 12 Physical Sciences CAPS Notes

Electrodynamics explains how mechanical energy is converted into electrical energy (generators) and how electrical energy is transmitted efficiently over long distances using transformers. Understanding AC generators and the rms values of alternating current is essential for real-world electrical engineering.

Week 4

10.1 AC Generators

Explain how an AC generator works and why it produces a sinusoidal outputCalculate peak voltage, rms voltage: Vrms = Vpeak/√2Calculate peak current, rms current: Irms = Ipeak/√2

An AC generator (alternator) converts mechanical energy into electrical energy using electromagnetic induction. A rectangular coil is rotated in a magnetic field. As the coil rotates, the rate of change of magnetic flux through the coil changes continuously, inducing a sinusoidally varying emf. The result is an alternating current (AC) that reverses direction twice per revolution.

Definition

Faraday's Law of Electromagnetic Induction

The emf induced in a conductor is directly proportional to the rate of change of magnetic flux linkage. The faster the flux changes, the greater the induced emf. Mathematically: emf = −N(ΔΦ/Δt), where N is the number of turns and ΔΦ/Δt is the rate of change of magnetic flux.

An AC generator: a rectangular coil rotates in a uniform magnetic field between the poles of a magnet. Slip rings and brushes allow current to flow to the external circuit. The output is a sinusoidal AC voltage.

Key parts of an AC generator

Armature coil: the rotating rectangular coil that cuts magnetic field lines
Permanent magnet: provides the external magnetic field
Slip rings: rotating metal rings attached to the coil ends
Carbon brushes: stationary contacts that press against the slip rings to carry current out
Output: sinusoidal AC voltage

The output voltage is sinusoidal: v(t) = Vpeak·sin(θ). The peak voltage Vpeak is the maximum voltage reached. Because AC constantly changes, we use the root-mean-square (rms) value to represent its equivalent DC effect. The rms voltage produces the same heating effect in a resistor as an equal DC voltage.

Formula

RMS voltage

V_{rms} = \frac{V_{peak}}{\sqrt{2}}

Vrms = root-mean-square voltage (V), Vpeak = maximum (peak) voltage (V)

SI unit: V

Formula

RMS current

I_{rms} = \frac{I_{peak}}{\sqrt{2}}

Irms = root-mean-square current (A), Ipeak = maximum (peak) current (A)

SI unit: A

Sinusoidal AC output from a generator: voltage varies between +Vpeak and −Vpeak. The horizontal dashed line shows Vrms = Vpeak/√2, the equivalent DC heating value.

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Exam Tip

Exam tip: South African household supply is 220 V rms at 50 Hz. The peak voltage is Vpeak = 220 × √2 ≈ 311 V. Appliance ratings always use rms values. The division by √2 ≈ 1.414 is exact only for a pure sinusoidal waveform.

Worked Example

An AC generator produces a peak voltage of 340 V. Calculate (a) the rms voltage, and (b) the rms current when connected to a 100 Ω resistor.

Given

Vpeak = 340 V
R = 100 Ω

Find

Vrms and Irms

Solution

1Vrms = Vpeak / √2 = 340 / 1.414 = 240.5 V ≈ 240 V
2Irms = Vrms / R = 240 / 100 = 2.40 A
3Alternatively: Ipeak = Vpeak / R = 340 / 100 = 3.40 A; Irms = 3.40 / √2 = 2.40 A ✓

Answer: Vrms ≈ 240 V; Irms ≈ 2.40 A

Practice Question

An AC generator has a peak voltage of 180 V. (a) Calculate the rms voltage. (b) If connected to a 60 Ω load, calculate the rms current and the average power dissipated.

(6 marks)

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Watch Out

Common mistake: using Vpeak in power calculations instead of Vrms. Power in AC circuits is P = VrmsIrms = Irms²R = Vrms²/R. Never use peak values in power formulas.

Week 4

10.2 Transformers

Explain how a transformer works: Vs/Vp = Ns/NpExplain how transformers are used in power transmission to reduce energy losses

A transformer uses mutual electromagnetic induction to change AC voltage from one value to another. It consists of two coils (primary and secondary) wound on a shared iron core. When AC flows in the primary coil, it creates a changing magnetic flux in the core. This changing flux induces an emf in the secondary coil. The ratio of voltages equals the ratio of turns.

A step-up transformer: the secondary coil has more turns than the primary. The iron core channels the magnetic flux from the primary to the secondary coil.

Formula

Transformer equation

\frac{V_p}{V_s} = \frac{N_p}{N_s}

Vs = secondary voltage (V), Vp = primary voltage (V), Ns = secondary turns, Np = primary turns

For an ideal transformer (no energy losses), power is conserved: Pp = Ps, so VpIp = VsIs. If voltage is stepped up, current is stepped down proportionally. A step-up transformer increases voltage (Ns > Np). A step-down transformer decreases voltage (Ns < Np). Transformers only work with AC — they do not work with DC because DC creates no changing flux.

Step-Up vs Step-Down Transformers

Property	Step-Up Transformer	Step-Down Transformer
Turns ratio	Ns > Np	Ns < Np
Voltage	Vs > Vp (increased)	Vs < Vp (decreased)
Current	Is < Ip (decreased)	Is > Ip (increased)
Use in grid	Power station → transmission lines	Transmission lines → homes
Power	Conserved (ideal): VpIp = VsIs	Conserved (ideal): VpIp = VsIs

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Real World

Power transmission: Eskom generates electricity at ~22 000 V, then steps it up to 765 000 V (765 kV) for long-distance transmission. High voltage means low current, so P_loss = I²R is minimised. Near homes, step-down transformers reduce voltage to 220 V.

Worked Example

A transformer has 500 turns on the primary and 2 000 turns on the secondary. The primary is connected to a 220 V AC supply. (a) Calculate the secondary voltage. (b) If the secondary current is 0.5 A, calculate the primary current (assume ideal transformer).

Given

Np = 500
Ns = 2 000
Vp = 220 V
Is = 0.5 A

Find

Vs and Ip

Solution

1Vs / Vp = Ns / Np → Vs = Vp × (Ns/Np) = 220 × (2000/500) = 220 × 4 = 880 V
2Ideal transformer: VpIp = VsIs → Ip = VsIs / Vp = (880 × 0.5) / 220 = 440 / 220 = 2 A

Answer: Vs = 880 V (step-up); Ip = 2 A

Practice Question

A step-down transformer converts 11 000 V to 220 V. If the primary has 5 000 turns, how many turns does the secondary have?

(3 marks)

Practice Question

Explain why it is more efficient to transmit electrical power at high voltage and low current rather than low voltage and high current.

(4 marks)

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Watch Out

Transformers only work with AC. A DC source produces a constant (non-changing) magnetic flux in the core, so no emf is induced in the secondary. This is a common exam question.

Week 5

10.3 Power in AC Circuits and Electrical Safety

Calculate peak voltage, rms voltage: Vrms = Vpeak/√2Calculate peak current, rms current: Irms = Ipeak/√2

Formula

AC average power

P_{avg} = V_{rms}I_{rms}

P = average power (W), Vrms = rms voltage (V), Irms = rms current (A)

SI unit: W

The average power dissipated in an AC resistive circuit equals the product of rms voltage and rms current. This is equivalent to the power that would be dissipated by a DC circuit at the same voltage and current values. The instantaneous power alternates between zero and a maximum of VpeakIpeak, but the average is half of this maximum.

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Note

Average power in AC: P_avg = Vpeak·Ipeak / 2 = Vrms·Irms. Both expressions are equivalent because Vrms = Vpeak/√2 and Irms = Ipeak/√2, so Vrms·Irms = Vpeak·Ipeak / 2.

Worked Example

A kettle rated at 2 000 W is connected to a 220 V rms AC supply. Calculate (a) the rms current, (b) the peak current, and (c) the peak voltage.

Given

P = 2 000 W
Vrms = 220 V

Find

Irms, Ipeak, Vpeak

Solution

1Irms = P / Vrms = 2 000 / 220 ≈ 9.09 A
2Ipeak = Irms × √2 = 9.09 × 1.414 ≈ 12.85 A
3Vpeak = Vrms × √2 = 220 × 1.414 ≈ 311 V

Answer: Irms ≈ 9.09 A; Ipeak ≈ 12.85 A; Vpeak ≈ 311 V

Practice Question

An AC circuit has a peak current of 14.14 A and a peak voltage of 311 V. Calculate (a) the rms current, (b) the rms voltage, and (c) the average power dissipated.

(6 marks)

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Exam Tip

Quick check: if Vpeak ≈ 311 V and Ipeak ≈ 14.14 A, you are almost certainly working with the standard 220 V / 50 Hz South African household supply. Recognising these numbers saves calculation time in exams.

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Watch Out

Do not confuse the transformer equation (Vs/Vp = Ns/Np) with the rms formula. They are used in completely different contexts. The transformer equation relates the two coils; rms relates peak to average for a single sinusoidal quantity.