OCR A Level Physics A: Resistance, resistivity and electrical power (H556) – Revision Notes

Quick recall for OCR A Level Physics A (H556), Version 3.0 (March 2026), sections 4.2.3 (Resistance), 4.2.4 (Resistivity) and 4.2.5 (Power). For full explanations and longer worked examples, read the study guide. Then try the practice questions, the free diagnostics, the OCR A Level Physics hub and the printable checklist.

Definitions and laws

Term	Statement
Resistance (4.2.3(a))	R = V/I at a given p.d.; unit ohm (Ω), one volt per ampere
Ohm’s law (4.2.3(b))	Current is directly proportional to p.d. at constant temperature (and other physical conditions)
Ohmic conductor	Obeys Ohm’s law: straight I-V line through the origin
Resistivity (4.2.4(a))	Material property in R = ρL/A; unit Ω m
NTC thermistor (4.2.4(c))	Semiconductor whose resistance falls as temperature rises
Kilowatt-hour (4.2.5(c))	Energy used by a 1 kW appliance in 1 h; 1 kW h = 3.6 × 10⁶ J

Equations to recall and use

Quantity	Equation	Notes
Resistance	R = V/I	Use V and I at the same point
Resistivity	R = ρL/A, so ρ = RA/L	A = πd²/4 for a round wire
Power	P = VI = I²R = V²/R	Pick the form that matches the data
Energy	W = VIt = Pt	t in seconds for joules
Energy in kW h	E = P (kW) × t (h)	Cost = kW h × price per kW h
Efficiency (3.3.3(c))	useful output energy / total input energy × 100%	You pay for the input

I-V characteristics: recall grid

Component	Graph	Reason
Resistor at constant temperature	Line through origin, straight	R constant
Filament lamp	Flattening curve, symmetrical about origin	Filament heats, ions vibrate more, more collisions, R rises
NTC thermistor	Steepening curve	Heating releases more charge carriers, R falls
Diode	Negligible current in reverse and below about 0.6 V to 0.7 V forward; then steep rise	One-way conduction above a threshold p.d.
LED	Same shape as a diode with higher threshold; emits light	Threshold depends on the LED
LDR	R falls as light intensity rises	Light releases more charge carriers

Reminder: resistance at a point on a curve is V/I, not the slope.

Method in steps: investigating a component (PAG3)

1. Set up the component in series with an ammeter and a variable resistor wired as a potential divider, so the p.d. starts at zero.
2. Voltmeter in parallel with the component only.
3. Increase V in equal steps; record V and I. Add a series protective resistor for a diode or LED.
4. Reverse the polarity for negative values on a diode, lamp or resistor.
5. Plot I against V; calculate V/I in a spreadsheet to show how R varies.
6. For a thermistor, control the temperature separately, with a thermometer in a water bath.

Method in steps: resistivity of a metal wire (PAG3)

1. Read the diameter with a micrometer at spots along the wire, rotating it by 90° each time; take the mean and find A = πd²/4.
2. Measure R for a range of lengths L (equal steps), using small currents so the wire does not heat.
3. Plot R against L. Slope = ρ/A, so ρ = slope × A.
4. Percentage uncertainty in A is twice that in d.

Temperature effects

Material	Resistivity as temperature rises	Explanation
Metal	Increases	Ions vibrate more, more collisions; number density of conduction electrons barely changes
Semiconductor	Decreases	Many more charge carriers released, outweighing extra collisions

Worked reminders

Resistance from a point on a graph. A diode has 0.30 A at 0.80 V. R = 0.80/0.30 = 2.7 Ω. That is its resistance at that point only.

Resistivity. Copper cable of length 25 m and cross-sectional area 1.0 mm² (1.0 × 10⁻⁶ m²), with ρ = 1.7 × 10⁻⁸ Ω m (data supplied): R = 1.7 × 10⁻⁸ × 25 / 1.0 × 10⁻⁶ = 0.425 Ω.

Scaling. If L doubles and d halves, A falls to a quarter, so R rises by 2 × 4 = 8 times.

Power. A 100 W lamp at 230 V: I = 100/230 = 0.435 A; R = 230²/100 = 529 Ω.

Cost. A 3.0 kW heater runs for 2.0 h at 20 p per kW h: energy = 6.0 kW h; cost = 120 p = £1.20.

Conversion. 0.50 kW h = 0.50 × 3.6 × 10⁶ = 1.8 × 10⁶ J.

Reading graphs and data

• A straight line through the origin on an I-V graph means ohmic. Check that the line passes through (0, 0).
• A graph of R against L for a uniform wire is a straight line. Its slope is ρ/A and a positive intercept suggests contact or lead resistance.
• For a lamp, compare V/I at two p.d. values and state that R rose. Then link the rise to temperature.
• For an NTC thermistor, work out the ratio of resistances. If R falls to one fifth, the current at fixed p.d. is five times larger.
• For a diode or LED, quote the approximate p.d. at which the current starts to rise steeply, and state that the current is negligible below it and in reverse.

Power and cost: setting out the working

1. Write the known quantities with units (convert mm to m, minutes to hours or seconds, W to kW).
2. Choose the equation: P = VI if you have V and I, P = I²R if you have I and R, P = V²/R if you have V and R.
3. For energy in joules use W = VIt = Pt with t in seconds. For cost use kW h = P (kW) × t (h).
4. State the answer with its unit and a sensible number of significant figures.

Extra reminder. A 12 V supply drives 3.0 A through a lamp for 90 s. W = 12 × 3.0 × 90 = 3240 J, which is 3.2 kJ to 2 s.f. In kW h this is 3240 / (3.6 × 10⁶) = 9.0 × 10⁻⁴ kW h.

Must-know distinctions

• Resistance depends on shape and material. Resistivity depends only on the material (and temperature).
• Ohm’s law is a statement of proportionality. A component can have a resistance and still be non-ohmic.
• Metals and semiconductors both have more lattice collisions at higher temperature. In a semiconductor the large rise in carrier number wins; in a metal the carrier number barely changes, so collisions win.
• A kilowatt-hour is a unit of energy, not power.

Quick self-test

1. A p.d. of 12 V drives a current of 40 mA through a resistor. Calculate its resistance.
2. State Ohm’s law.
3. Explain why a filament lamp’s I-V graph is not a straight line.
4. A round wire has ρ = 5.5 × 10⁻⁷ Ω m, length 1.0 m and diameter 1.0 mm. Find its resistance.
5. Why is it better to find ρ from the gradient of an R-L graph than from one reading?
6. Compare how the resistivity of a metal and a semiconductor change as temperature increases.
7. A 2.5 A current flows through an 8.0 Ω resistor. Calculate the power.
8. A 60 W heating element has resistance 15 Ω. Calculate the p.d. across it.
9. Convert 0.50 kW h into joules.
10. A 3.0 kW heater runs for 2.0 hours at 20 p per kW h. Calculate the cost.

Answers

1. R = 12 / 0.040 = 300 Ω.
2. The current through a conductor is directly proportional to the p.d. across it, provided the temperature (and other physical conditions) stay constant.
3. The current heats the filament; the metal ions vibrate more, collide more often with electrons, so resistance rises as p.d. increases and the graph flattens.
4. A = π(0.5 × 10⁻³)² = 7.85 × 10⁻⁷ m². R = ρL/A = 5.5 × 10⁻⁷ × 1.0 / 7.85 × 10⁻⁷ = 0.70 Ω.
5. The line averages out random error across several readings and a non-zero intercept (for example contact resistance) does not affect the gradient.
6. The resistivity of a metal increases because of more lattice collisions. The resistivity of a semiconductor decreases because many more charge carriers are released.
7. P = I²R = 2.5² × 8.0 = 50 W.
8. V = √(PR) = √(60 × 15) = 30 V.
9. 0.50 × 3.6 × 10⁶ = 1.8 × 10⁶ J.
10. Energy = 6.0 kW h; cost = 6.0 × 20 = 120 p (£1.20).

Where marks are usually lost

• Writing “resistance is constant” as Ohm’s law and leaving out the temperature condition.
• Taking the slope of a curve as resistance instead of V/I.
• Using diameter as radius: this gives an area four times too big.
• Forgetting to convert mm to m, or mm² to m² (1 mm² = 10⁻⁶ m²).
• Explaining a thermistor with “electrons move faster” instead of “more charge carriers”.
• Using kW with seconds, or W with hours, so the energy unit is wrong.
• Writing a value without a unit, or giving more significant figures than the data supports.
• Not stating a reason for using small currents when finding resistivity: heating changes resistivity.
• Describing a component investigation without the potential divider, so the p.d. cannot start from zero.

Official syllabus

This page is aligned to the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), first assessment 2017, published by OCR, sections 4.2.3 to 4.2.5.