Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Option D: Energy and the Environment
Photovoltaic cells and wind power
- 3 things to know
- Component 3
Overview
Recall questions on how a photovoltaic cell converts light to electricity and what limits its efficiency, the power available from a wind turbine and its dependence on the wind speed and the blade area, worked output calculations, and both sides of the case for each.
What you need to know
Answer each one before you reveal it.
Give two reasons why the annual mean output of a solar panel in the UK is far below its rated peak output.
Give the equation for the power available in a flowing fluid passing through an area A.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Explain how a photovoltaic cell converts light into electrical energy.
Model answer
A photon absorbed in the semiconductor gives an electron enough energy to cross the band gap, leaving a hole behind. The internal electric field at the p-n junction sweeps the electron and the hole in opposite directions, so a potential difference appears across the cell.
- 2
Why does a photon of energy below the band gap produce no current in a photovoltaic cell?
Model answer
It cannot lift an electron across the gap into the conduction band, so no free charge carrier is created. Such a photon passes through the cell or is absorbed as heat.
- 3
Explain why a photon of energy well above the band gap still only supplies a fixed amount of useful energy.
Model answer
Only the band gap energy is needed to free the electron, and the excess is quickly lost to lattice vibrations as heat. That loss is one of the main reasons the efficiency of a single-junction cell is limited.
- 4
A photovoltaic panel of area 1.6 m² operates at 20% efficiency in an insolation of 800 W m⁻². Calculate its electrical power output.
Model answer
The incident power is 800 × 1.6 = 1280 W, so the output is 1280 × 0.20 = 256 W. Working back, 256/1280 = 0.20, the efficiency.
- 5
A domestic array must deliver a mean 3.0 kW in an insolation of 750 W m⁻² at an efficiency of 18%. Calculate the area required.
Model answer
The useful power per square metre is 750 × 0.18 = 135 W m⁻², so the area is 3000/135 = 22 m². Working back, 22.2 × 135 = 3.0 × 10³ W.
- 6
Give two reasons why the annual mean output of a solar panel in the UK is far below its rated peak output.
Model answer
The Sun is below the horizon for half the time and low in the sky for much of the rest, so the insolation on a fixed panel is usually a small fraction of the peak value. Cloud cover scatters and absorbs much of the remaining radiation.
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