Resistance, resistivity and circuits

Resistance and Ohm's law, series and parallel combinations, electrical power, I-V characteristics, resistivity, and how temperature and light affect resistance.

Understand that resistance is defined by $R = \frac{V}{I}$ and that Ohm's law is a special case where for constant temperature

Be able to derive the equations for combining resistances in series and parallel using the principles of charge and energy conservation, and be able to use these equations

Be able to use the equations $P = VI$ , $W = VIt$ and be able to derive and use related equations, e.g. $P = I^2R$ and $P = \frac{V^2}{R}$

Understand how to sketch, recognise and interpret current-potential difference graphs for components, including ohmic conductors, filament bulbs, thermistors and diodes

Be able to use the equation $R = \frac{\rho l}{A}$

CORE PRACTICAL 7: Determine the electrical resistivity of a material

Understand how changes of resistance with temperature may be modelled in terms of lattice vibrations and number of conduction electrons, and apply this model to metallic conductors and negative temperature coefficient thermistors

Understand how changes of resistance with illumination may be modelled in terms of the number of conduction electrons, and apply this model to LDRs

A $6.0\,\Omega$ resistor is connected in series with a parallel combination of a $4.0\,\Omega$ resistor and a $12\,\Omega$ resistor. The combination is connected to a $12\,\text{V}$ supply of negligible internal resistance.

Parallel combination: $\left(\tfrac{1}{4.0} + \tfrac{1}{12}\right)^{-1} = 3.0\,\Omega$ (1)
Total $R = 6.0 + 3.0 = 9.0\,\Omega$ (1)
Supply current $= 12/9.0 = 1.33\,\text{A}$ , so pd across parallel section $= 1.33 \times 3.0 = 4.0\,\text{V}$ (1)
Current in $12\,\Omega$ resistor $= 4.0/12 = 0.33\,\text{A}$ (1)

A nichrome wire has length $0.75\,\text{m}$ and diameter $0.28\,\text{mm}$ . The resistivity of nichrome is $1.10 \times 10^{-6}\,\Omega\,\text{m}$ .

Calculate the resistance of the wire.

$A = \pi (0.14 \times 10^{-3})^2 = 6.16 \times 10^{-8}\,\text{m}^2$ (1)

The current–potential difference ( $I$ – $V$ ) graph for a filament lamp is a curve whose gradient decreases as $V$ increases.

Explain why the resistance of the filament increases as the current through it increases.

As current increases, the filament's temperature increases (1)
The metal ions/lattice vibrate with greater amplitude (1)
So conduction electrons collide with the ions more frequently, increasing resistance (and the drift velocity per volt is lower) (1)

Specification points

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Sample questions