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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.

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Key term

Explain how a photovoltaic cell converts light into electrical energy.

Key term

Give two reasons why the annual mean output of a solar panel in the UK is far below its rated peak output.

Key term

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. 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. 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. 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. 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. 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. 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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