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Resistance

I–V characteristics of a metal wire and a filament lamp

  • 3 things to know
  • Component 2

Overview

Shape of the I–V characteristic of a filament lamp and curvature of the I–V characteristic of a filament lamp.

What you need to know

Answer each one before you reveal it.

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Key term

For a graph of current against potential difference, what does the gradient at a point represent?

Key term

Why does the I–V characteristic of a filament lamp still pass through the origin?

Key term

Outline how the I–V characteristic of a metal wire at constant temperature is obtained experimentally.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Describe the shape of the I–V characteristic of a metal wire kept at constant temperature.

    Model answer

    It is a straight line through the origin, with the same gradient for both directions of current.

  2. 2

    For a graph of current against potential difference, what does the gradient at a point represent?

    Model answer

    The gradient of an I–V graph is 1/R, the reciprocal of the resistance at that point.

  3. 3

    Describe the shape of the I–V characteristic of a filament lamp.

    Model answer

    It passes through the origin and curves so that the gradient becomes progressively smaller as the current increases, giving an S shape when both polarities are plotted.

  4. 4

    Explain the curvature of the I–V characteristic of a filament lamp.

    Model answer

    As the current increases the filament gets hotter, its resistance rises, and so equal increases in potential difference produce steadily smaller increases in current.

  5. 5

    Why does the I–V characteristic of a filament lamp still pass through the origin?

    Model answer

    With no potential difference applied there is no current, and this holds however the resistance varies.

  6. 6

    How is the resistance of a component found from a point on its I–V graph?

    Model answer

    The resistance is the value of V at that point divided by the value of I at that point, not the gradient of the curve.

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