CIE iGCSE Co-Ordinated Science P4.5.2 The a.c. generator Exam Style Questions Paper 4
Question
On Fig. 3.2, sketch a graph of output voltage against time for the a.c. generator when the wind turbine is turning at a constant speed.
Describe the construction of a basic step-up transformer.
You may include a labelled diagram to aid your description.
Each blade has a surface area of \(90\,\text{m}^2\).
Calculate the force exerted by the wind on each turbine blade.
Describe, in terms of oscillations and energy transfer, what is meant by a longitudinal wave.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic P1.6.3 — Energy resources (Part (a))
• Topic P4.5.2 — The a.c. generator (Part (b))
• Topic P4.5.6 — The transformer (Part (c))
• Topic P1.7 — Pressure (Part (d))
• Topic P3.4 — Sound / Topic P3.1 — General properties of waves (Part (e))
▶️ Answer/Explanation
(a) Does not release \(CO_2\) / renewable / no fuel costs
Wind turbines generate electricity without burning fossil fuels.
This means no greenhouse gases are released and the energy source (wind) is renewable.
(b) Sinusoidal waveform
The trace should be a smooth sine wave, alternating above and below the time axis.
Both the amplitude and the time period of the wave must stay constant, since the turbine turns at constant speed.
(c) Soft-iron core with two coils
A basic transformer has a soft iron core linking two separate coils of wire.
The primary and secondary coils are both wound around the same core.
For a step-up transformer, the number of turns on the secondary coil is greater than the number of turns on the primary coil.
(d) \(648\,000\,\text{N}\) (≈ \(6.5\times10^{5}\,\text{N}\))
Force is calculated using \(F = P \times A\).
\(F = 7200 \times 90 = 648\,000\,\text{N}\), which rounds to \(650\,000\,\text{N}\) (2 s.f.).
(e)(i) \(20\,\text{Hz}\)
The typical range of human hearing is approximately \(20\,\text{Hz}\) to \(20\,000\,\text{Hz}\).
\(20\,\text{Hz}\) is therefore the accepted minimum audible frequency for a healthy human ear.
(e)(ii) Oscillations parallel to energy transfer
In a longitudinal wave, the particles oscillate back and forth in the same direction that the energy is being transferred.
This produces alternating regions of compression and rarefaction along the direction of travel.
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.5.6 — The transformer / Transformer construction (Part (a)(i))
• Topic P4.5.6 — The transformer / Power loss calculations (Part (a)(ii))
• Topic P4.2.2 — Electric current / Direct vs alternating current (Part (b))
• Topic P4.5.2 — The a.c. generator / Slip rings (Part (c)(i))
• Topic P4.5.2 — The a.c. generator / e.m.f. graphs (Part (c)(ii) & (c)(iii))
▶️ Answer/Explanation
(a)(i) A step-up transformer consists of a soft iron core with a primary coil having fewer turns than the secondary coil.
The soft iron core provides a path for the magnetic field, and the turns ratio determines the voltage transformation. In a step-up transformer, the secondary coil has more turns, which increases the voltage.
(a)(ii) \( P = I^2R = 500^2 \times 0.60 = 250,000 \times 0.60 = 150,000 \text{ W} \)
Power loss in a transmission wire is calculated using \( P = I^2R \). The higher the current, the greater the energy lost as heat in the wires. Stepping up the voltage reduces the current and thus reduces power loss.
(b) Direct current (d.c.) flows in one direction only, while alternating current (a.c.) reverses direction periodically.
In d.c., the electric charge flows consistently from positive to negative. In a.c., the direction of charge flow changes repeatedly, typically in a sinusoidal pattern.
(c)(i) Slip rings maintain a continuous electrical connection to each side of the rotating coil, preventing the wires from twisting or tangling.
As the coil rotates, the slip rings rotate with it while the carbon brushes remain stationary, ensuring a constant electrical connection to the external circuit without wires becoming twisted.
(c)(ii) The graph should be a sine wave with positive and negative e.m.f. values, showing constant amplitude and a constant period.
As the coil rotates at constant speed, the e.m.f. varies sinusoidally: maximum positive, zero, maximum negative, and back to zero in one complete rotation. The amplitude and period remain constant.
(c)(iii) The amplitude of the e.m.f. doubles, and the period halves (the frequency doubles).
Doubling the rotation speed means the coil cuts through the magnetic field lines twice as fast, generating double the e.m.f. It also completes each rotation in half the time, so the period is halved.