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What is electric current, and what determines a conductor's resistance?

Electric charge and current, potential difference, Ohm's law and resistance, resistivity, and the I-V characteristics of components.

A CCEA A-Level Physics answer on electric charge and current, potential difference, Ohm's law and resistance, the resistivity of a material, and the current-voltage characteristics of a resistor, a filament lamp and a diode.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

CCEA wants you to define charge and current, define potential difference, state and apply Ohm's law and the idea of resistance, use resistivity, and recognise and explain the current-voltage characteristics of common components. A resistivity calculation and a characteristic-graph explanation are exam staples.

The answer

Charge, current and potential difference

Charge is carried by electrons in a metal, each carrying 1.6×10−19 C1.6 \times 10^{-19}\ \text{C}. Conventional current is taken to flow from positive to negative, opposite to the electron drift. The drift speed is given by I=nAveI = nAve, where nn is the number density of charge carriers.

Resistance, Ohm's law and resistivity

A longer or thinner wire has more resistance; resistivity is a property of the material itself, measured in Ω m\Omega\,\text{m}, and it rises with temperature for a metal.

Current-voltage characteristics

  • A fixed resistor (or metal wire at constant temperature) gives a straight line through the origin, obeying Ohm's law.
  • A filament lamp gives an S-shaped curve bending towards the voltage axis, because the filament heats up and its resistance rises with temperature.
  • A diode conducts only in the forward direction above a threshold of about 0.6 V0.6\ \text{V}, and has very high resistance in reverse.

Worked example: drift speed in a copper wire

Examples in context

Example 1. Choosing house wiring. A high-current circuit (a cooker or shower) uses thick cable to keep resistance low, R=ρL/AR = \rho L / A, so that the heat dissipated I2RI^2 R in the cable stays small and the wire does not overheat. Lighting circuits, carrying little current, can use thinner, cheaper cable.

Example 2. A rectifier diode in a charger. A diode's one-way characteristic lets mains AC pass in only one direction, the first stage in converting it to the DC a phone needs. Below about 0.6 V0.6\ \text{V} forward, or under any reverse voltage, almost no current flows, which is exactly the behaviour seen on its I-V graph.

Try this

Q1. A wire of length 2.0 m2.0\ \text{m} and cross-sectional area 0.50 mm20.50\ \text{mm}^2 has resistivity 5.0×10−7 Ω m5.0 \times 10^{-7}\ \Omega\,\text{m}. Find its resistance. [3 marks]

  • Cue. R=ρLA=5.0×10−7×2.00.50×10−6=2.0 ΩR = \frac{\rho L}{A} = \frac{5.0 \times 10^{-7} \times 2.0}{0.50 \times 10^{-6}} = 2.0\ \Omega.

Q2. Define potential difference. [1 mark]

  • Cue. The energy transferred per unit charge between two points.

Q3. A charge of 90 C90\ \text{C} passes a point in 30 s30\ \text{s}. Find the current. [2 marks]

  • Cue. I=Qt=9030=3.0 AI = \frac{Q}{t} = \frac{90}{30} = 3.0\ \text{A}.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20196 marksA copper wire has a length of 1.5 m and a diameter of 0.40 mm. The resistivity of copper is 1.7 times 10 to the minus 8 ohm metre. Calculate the cross-sectional area and the resistance of the wire. The wire carries a current of 2.5 A; calculate the charge passing a point in 1.0 minute and the number of electrons this represents.
Show worked answer →

The cross-sectional area is

A=πr2=π(0.20×10−3)2=π×4.0×10−8=1.26×10−7A = \pi r^2 = \pi (0.20 \times 10^{-3})^2 = \pi \times 4.0 \times 10^{-8} = 1.26 \times 10^{-7} m squared.

The resistance is

R=ρLA=1.7×10−8×1.51.26×10−7=2.55×10−81.26×10−7=0.20R = \frac{\rho L}{A} = \frac{1.7 \times 10^{-8} \times 1.5}{1.26 \times 10^{-7}} = \frac{2.55 \times 10^{-8}}{1.26 \times 10^{-7}} = 0.20 ohm.

The charge passing in 60 s is

Q=It=2.5×60=150Q = It = 2.5 \times 60 = 150 C.

The number of electrons is

N=Qe=1501.6×10−19=9.4×1020N = \frac{Q}{e} = \frac{150}{1.6 \times 10^{-19}} = 9.4 \times 10^{20} electrons.

Markers reward the area from the diameter (halving it first), the resistivity equation, Q=ItQ = It, and dividing by the electron charge.

CCEA 20214 marksSketch the current-voltage characteristic of a filament lamp and explain its shape. State how the resistance of the filament changes as the current increases, and explain why in terms of the lattice and the charge carriers.
Show worked answer →

The characteristic is an S-shaped curve through the origin that bends towards the voltage axis as the voltage increases (the gradient of current against voltage decreases), and it is symmetric for reversed polarity.

As the current increases the filament dissipates more power and heats up. The hotter metal lattice ions vibrate more strongly, so the free electrons collide with them more often, impeding the flow. This means the resistance of the filament increases as the current (and temperature) increases.

Markers reward a correctly shaped symmetric curve, the statement that resistance increases, and the lattice-vibration explanation linking temperature to more frequent collisions.

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