OxfordAQA International AS and A-level Physics: Current, p.d., resistance, I-V characteristics and resistivity (9630) – Revision Notes

These notes condense sections 3.4.1 to 3.4.3 of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4: basics of electricity, current-voltage characteristics and resistivity. The content is shared International AS and A-level content. Longer explanations and worked examples are in the study guide, then test yourself with the practice questions.

Navigation: course hub, printable checklist and the free diagnostics. Unit conversions and uncertainty are in measurements and their errors.

Definitions to learn

• Electric current: rate of flow of charge.
• Potential difference: the work done (energy transferred) per unit charge moving between two points.
• Resistance: the ratio of the p.d. across a component to the current through it.
• Ohm’s law: for a conductor, I is proportional to V as long as physical conditions such as temperature do not change.
• Resistivity: the property of a material, defined through ρ = RA/L, that links the resistance of a sample to its length and cross-sectional area.
• Superconductor: a material with zero resistivity at and below its critical temperature.

Equations and units

Quantity	Equation	Unit	Note
Current	I = ΔQ/Δt	A = C s⁻¹	Time in seconds
Charge from electrons	Q = Ne	C	e = 1.60 × 10⁻¹⁹ C
P.d.	V = W/Q	V = J C⁻¹	W is energy transferred
Resistance	R = V/I	Ω = V A⁻¹	Value at a point
Resistivity	ρ = RA/L	Ω m	A = πd²/4 for a round wire
Rearranged	R = ρL/A; L = RA/ρ		Check powers of ten

Useful conversions: 1 mA = 10⁻³ A, 1 kΩ = 10³ Ω, 1 mm² = 10⁻⁶ m², 1 mm = 10⁻³ m, 1 min = 60 s.

Method boxes

Charge or current questions

1. Convert time to seconds and current to amps.
2. Q = It for constant current. For N electrons, Q = Ne.
3. State the unit with the answer.

Resistance at a point on a curve

1. Read the coordinates (V, I) of the point. Check which axis is which.
2. Calculate R = V/I.
3. Do not use the gradient or its reciprocal.

Resistivity from a graph of R against L

1. Gradient = ρ/A for a straight line through the origin.
2. Find A from the diameter: A = πd²/4, using d in metres.
3. ρ = gradient × A.

I-V characteristics at a glance

Component	Shape (I against V)	Why
Ohmic conductor (metal at constant temperature)	Straight line through the origin	R constant
Filament lamp	S-shaped, symmetrical about the origin; flattens at high	V
Semiconductor diode	Very little current in reverse, or forward below roughly 0.6 V; steep rise beyond that	Conducts one way only

If V is plotted vertically and I horizontally, each curve is reflected in the line y = x, so the lamp curve bends upwards: V rises faster than I. Ohmic gradient is R on a V-I graph and 1/R on an I-V graph.

Ideal meters: ammeter zero resistance, voltmeter infinite resistance, unless the question says otherwise.

Temperature effects

Material	Resistance as T rises	Reason
Metal	Increases	Ions vibrate more, so more collisions with electrons
ntc thermistor	Decreases	More charge carriers released, which outweighs the extra vibration
Superconductor	Zero below the critical temperature	No energy lost to the lattice

Only ntc thermistors are on the specification. A thermistor with a fixed resistor and a cell can act as a temperature sensor: as temperature rises, current rises and so does the p.d. across the fixed resistor.

Superconductor applications

• Strong magnetic fields from large currents in coils with no heating, as in scanners and accelerators.
• Cables for transmitting power with no energy lost as heat.
• The material must be kept below its critical temperature, which usually means expensive cooling.
• Critical field is not assessed.

Measuring a characteristic: checklist

1. Ammeter in series with the component, voltmeter in parallel across it.
2. A variable supply (or a circuit that lets you vary the p.d. from near zero upwards) to give a wide range of readings.
3. Reverse the connections to cover negative p.d. for a lamp or diode.
4. For a diode, increase the forward p.d. in small steps near the threshold, and take care not to exceed the rated current.
5. Allow the filament to settle at each reading.
6. Tabulate V, I and R = V/I, then plot.

Resistance-temperature graphs

• An ntc thermistor graph falls steeply at low temperature and flattens at high temperature.
• To use it as a sensor, read R at the measured temperature, or find the temperature from a measured R.
• A metal wire graph rises with temperature. A superconductor graph is a horizontal line along R = 0 below the critical temperature, and rises sharply to a non-zero value above it.
• Quote values read from the graph with units, and show the reading lines or points you used.

Small worked reminders

A. A charge of 120 C passes in 2.5 min. I = 120/150 = 0.80 A.

B. A wire has R = 6.0 Ω, L = 1.5 m and d = 0.50 mm. A = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m². ρ = RA/L = 6.0 × 1.96 × 10⁻⁷ / 1.5 = 7.9 × 10⁻⁷ Ω m.

C. A thermistor of resistance 2.5 kΩ is connected to a 3.0 V supply. I = 3.0/2500 = 1.2 mA. If it warms and its resistance falls to 1.5 kΩ, I = 2.0 mA.

D. A filament lamp has V = 6.0 V when I = 0.30 A. R = 6.0/0.30 = 20 Ω. This is the resistance at that point only.

Must-know distinctions

• Resistance and resistivity: resistance belongs to a particular sample, in ohms. Resistivity belongs to the material at a given temperature, in ohm metres.
• Ohm’s law and R = V/I: the definition R = V/I always applies. Ohm’s law says R is constant, which is true only under certain conditions.
• Metal and ntc thermistor: opposite trends, with different causes.
• Low resistance and zero resistance: a good conductor still has resistivity. A superconductor has none below its critical temperature.

Quick self-test

1. A current of 0.80 A flows for 150 s. What charge passes?
2. How much energy is transferred when 24 C passes through a p.d. of 9.0 V?
3. What is the resistance of a component with 3.6 V across it and 45 mA through it?
4. Calculate the resistance of a 5.0 m copper wire of cross-sectional area 0.50 mm², given ρ = 1.7 × 10⁻⁸ Ω m.
5. A wire has its length and its diameter both doubled. By what factor does R change?
6. Why does a filament lamp not obey Ohm’s law?
7. Describe the behaviour of a diode in reverse bias.
8. Explain why the resistance of a metal rises with temperature.
9. How many electrons pass a point each second when the current is 1.0 mA?
10. State what is meant by critical temperature.
11. Give two applications of superconductors.
12. On a V-I graph (I on the horizontal axis), a lamp curve passes through (0.30 A, 6.0 V). Why is it wrong to find R from the gradient at that point?

Answers

1. Q = It = 0.80 × 150 = 120 C.
2. W = VQ = 9.0 × 24 = 216 J.
3. R = 3.6 / 0.045 = 80 Ω.
4. R = ρL/A = 1.7 × 10⁻⁸ × 5.0 / (0.50 × 10⁻⁶) = 0.17 Ω.
5. L doubles (R × 2); A increases by 4 (R ÷ 4). Net R halves.
6. Its temperature rises with current, so its resistance is not constant. Current is not proportional to p.d.
7. Almost no current flows. The resistance is very large.
8. Ions vibrate with greater amplitude, so electrons collide with them more often. The number of free electrons barely changes, so resistivity rises.
9. N = I/e = 1.0 × 10⁻³ / 1.60 × 10⁻¹⁹ = 6.3 × 10¹⁵ per second.
10. The temperature at and below which the material has zero resistivity.
11. For example, strong magnetic fields and low-loss power transmission cables.
12. R = V/I is the ratio of the coordinates (6.0/0.30 = 20 Ω). The gradient of a curve is not that ratio.

Where marks are usually lost

• Using the gradient of a curve to find resistance.
• Reading the axes the wrong way round when V is on the horizontal axis.
• Giving the power of ten wrongly in A = πd²/4 when d is in mm.
• Using the radius in place of the diameter, or the reverse.
• Writing “the lamp gets hotter so the current goes up”, when current rises more slowly because resistance increases.
• Describing a diode as “not conducting” in forward bias, without mentioning the threshold.
• Saying thermistor resistance “goes down because there are fewer electrons”. More carriers are available.
• Leaving out units, or giving a resistivity in ohms.
• Stating “resistance is directly proportional to temperature” for a metal. The specification asks only for the qualitative effect.

Official syllabus

Source: OxfordAQA International AS and A-level Physics (9630), Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.4.1 to 3.4.3, shared International AS and A-level content.