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Electricity and magnetism · Subtopic 3 of 8

Resistance

Resistance tells you how hard it is for current to flow: R=VIR=\dfrac VI, in ohms (Ω). A fixed resistor at constant temperature has a straight-line II–VV graph; a filament lamp curves because its resistance rises as it heats; a diode lets current through in one direction only. Longer, thinner wires have more resistance.

Cambridge 5054 Paper 1Cambridge 5054 Paper 2Cambridge 0625 Papers 2, 4Cambridge 0625 Papers 1, 3Edexcel 4PH1 Paper 1PEdexcel 4PH1 Paper 2P≈ 30 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use R=VIR=\dfrac VI, rearranged as needed.
  • Describe how to measure resistance with an ammeter and voltmeter.
  • Sketch and explain II–VV graphs for a resistor, filament lamp and diode.
  • Use R∝LR\propto L and R∝1AR\propto\dfrac1A for wires.
Pause and think

The same battery is connected first to a short, thick copper wire, then to a long, thin one. Which carries more current? Why?

Section 2 of 14

Core idea

Start hereWhy a lamp glows but the wires don't

The same current passes through the thin filament and the thick copper leads, yet only the filament glows. The filament resists the current far more, so that's where the energy is transferred.

The big idea

R=VIR=\frac VI (Ω). Ohmic conductor: I∝VI\propto V. R∝R\propto length, R∝1areaR\propto\frac1{\text{area}}.

1. I–V graphs

Current–voltage graphs
−6−4−2246−1−0.50.510potential difference (V)current (A)resistor (straight)filament lampdiode
Resistor: straight line through the origin, constant RR. Lamp: curves over, because a hotter filament resists more. Diode: almost no current backwards.
V=IRV=IR
where
RR
resistance (Ω = V/A).
  • Ohmic resistor: straight line through the origin.
  • Filament lamp: curve flattens; resistance rises as it heats.
  • Diode: current one way only.
Worked exampleAn ohmic resistor

6 V gives 0.3 A.

Goal
RR; current at 9 V.
Strategy
R=V/IR=V/I; constant RR.
  1. R=60.3=20 ΩR=\frac6{0.3}=20\ \Omega
    Definition.
  2. I=920=0.45 AI=\frac9{20}=0.45\text{ A}
    Same RR.
Answer20 Ω20\ \Omega; 0.450.45 A.
✓ CheckVV ×1.5, II ×1.5 ✓.

2. Wire dimensions

Resistance of a wire is proportional to its length
204060801001234560length (cm)resistance (Ω)R ∝ L
Measured points scatter slightly about a straight line through the origin.
Worked exampleA 4 Ω wire

Double the length; separately, double the diameter.

Goal
New resistances.
Strategy
R∝LAR\propto\frac LA; area ∝d2\propto d^2.
  1. 2×4=8 Ω2\times4=8\ \Omega
    Longer path.
  2. A×4⇒R=44=1 ΩA\times4\Rightarrow R=\frac44=1\ \Omega
    Double diameter: four times the area.
Answer8 Ω8\ \Omega; 1 Ω1\ \Omega.
✓ CheckThick wires for high currents ✓.
Common misconceptionDiameter vs area Doubling the diameter quarters the resistance, not halves it.
Pause and think

Why does a thermistor's resistance fall when hot?

Summary

  • V=IRV=IR.
  • R∝LR\propto L, R∝1/AR\propto1/A.
Key words
Ohm
V/A.
HistoryOhm's rejected discovery. Georg Ohm, a schoolteacher, published his law that current is proportional to voltage in 18271827. Critics called it "a web of naked fancies" and he lost his job. Fifteen years later the Royal Society gave him its top medal, and the unit of resistance is named after him.
Fun factSuperconductors: zero resistance. Some materials, cooled to extremely low temperatures, lose all resistance. A current started in a superconducting loop has kept flowing for years with no battery. MRI scanners use superconducting coils to make their strong magnetic fields.
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