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O-Level Current of Electricity
What the O-Level syllabus expects for Current of Electricity, and how to practise it.
What the syllabus expects
- State that current is the rate at which charge flows and is measured in amperes.
- Tell the difference between conventional current and electron flow.
- Recall and use charge = current × time in unfamiliar situations or problem solving.
- State that a source's electromotive force (e.m.f.) is the work it does per unit charge to drive charges around a complete circuit, measured in volts.
- Compute the total e.m.f. when several sources are connected in series.
- State that a component's potential difference (p.d.) equals the work done per unit charge to push charges through it, and is measured in volts.
- State that resistance = p.d. / current.
- Apply R = V / I in unfamiliar situations or problem solving.
- Recall and use how a wire's resistance relates proportionally to its length and cross-sectional area in unfamiliar situations or problem solving.
- Describe how raising the temperature affects a metallic conductor's resistance.
- Sketch and interpret I-V characteristic graphs for three cases: an ohmic metallic conductor held at constant temperature, a filament lamp, and a semiconductor diode.
How it's examined
Questions on this topic most often ask you to show. About 3% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (2 marks)
An electric cable is formed from 30 fine copper strands twisted together. Every strand is 0.30 mm in diameter. The resistivity of copper is 1.7 x 10^-8 ohm m. (b) Work out the resistance of a 4.0 m length of this cable.
Show the worked answer
Cross-sectional area of one strand: A1 = pi(d/2)² = pi(0.15 x 10⁻³)² = 7.07 x 10⁻⁸ m². Total area of 30 strands: A = 30 x 7.07 x 10⁻⁸ = 2.12 x 10⁻⁶ m². Resistance: R = rho L / A = (1.7 x 10⁻⁸ x 4.0) / (2.12 x 10⁻⁶) = 0.032 ohm.
Example 2 (2 marks)
An electrical cable consists of 18 fine strands of copper wire, illustrated in the figure (the drawing does not show the exact strand count). The diameter of each strand is 0.30 mm, and the resistivity of copper is 1.7 x 10^-8 ohm m. Work out (a) the resistance of a single strand that is 1.0 m long.
Show the worked answer
R = rho L / A. Radius r = 0.30/2 = 0.15 mm = 1.5 x 10⁻⁴ m. Cross-sectional area A = pi r² = pi x (1.5 x 10⁻⁴)² = 7.07 x 10⁻⁸ m². R = (1.7 x 10⁻⁸ x 1.0) / (7.07 x 10⁻⁸) = 0.24 ohm.
Example 3 (2 marks)
[Section B, Question 15 EITHER] A 10.5 kW, 230 V electric heating unit is joined to the 230 V mains by two copper wires, each 18 m long. The voltage across the heater must not drop below 225 V. Both wires in the cable are copper with resistivity 1.8 ×10⁻⁸ Ω m. Taking the current in the wire as 40 A, work out the largest resistance the cable may have.
Show the worked answer
The heater must receive at least 225 V, so the largest allowed voltage drop across the cable is 230 - 225 = 5 V. With a current of 40 A, the maximum cable resistance is R = V/I = 5/40 = 0.125 ohm.
More O-Level Pure Physics topics
Physical Quantities, Units and Measurement · Kinematics · Dynamics · Turning Effects of Forces · Pressure · Energy · all of O-Level Pure Physics