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Resistance, resistivity and component behaviour

Key idea: Resistance is the p.d.-to-current ratio at an operating point. Material, geometry, temperature and carrier behaviour determine how that ratio changes.

  • GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 2 of 3

Check your understanding

By the end of this lesson, you should be able to

  • Define resistance and use V = IR and R = ρl/A.
  • Sketch and interpret I–V curves for an ohmic resistor, filament lamp, diode and NTC thermistor.
  • Explain temperature effects and analyse terminal p.d. and output power for a source with internal resistance.

Learn the idea

Big question: What can the shape of an I–V graph reveal that one resistance calculation cannot?

Interpret four distinct I–V behaviours

Resistance at an operating point is R = V/I. An ohmic resistor at constant temperature has a straight I–V line through the origin. A filament lamp's curve becomes less steep on an I-vertical graph as heating raises metal resistivity.

A diode carries very little reverse current and rises sharply after forward threshold. An NTC thermistor becomes more conducting as temperature rises because many more mobile carriers become available. Always label axes before using the word gradient.

Check your understanding: Does being able to calculate V/I prove a component is ohmic?

No. Ohmic behaviour requires V proportional to I, so resistance is constant over the range at constant temperature.

Connect material and source behaviour

For a uniform wire, R = ρl/A. Longer wires give carriers more collisions; a larger area provides more parallel paths. Resistivity ρ is a material property at a stated temperature. In a metal, heating increases lattice vibration, so carriers have a smaller drift velocity for the same electric field and resistivity rises.

A real source has terminal p.d. V = ε − Ir while delivering current. The lost volts Ir represent energy per charge transferred internally, and output power is IV = I(ε − Ir).

Check your understanding: Why is terminal p.d. equal to e.m.f. when no current flows?

Ir is zero, so there is no internal voltage drop.

A-Level current–voltage characteristicsFour qualitative current against potential difference graphs. An ohmic conductor is a straight line through the origin. A filament lamp becomes less steep as voltage magnitude rises. A diode carries almost no reverse current and rises steeply in forward bias. An NTC thermistor becomes steeper as self-heating reduces resistance.Ohmic conductorVIFilament lampVISemiconductor diodeVINTC thermistorVIforward risewarmer, lower R
Scroll diagram horizontally to read all labels.
These are I-against-V graphs, so the local gradient represents conductance. The lamp heats and becomes less conducting; the NTC thermistor heats and becomes more conducting.
Real source with internal resistance and external loadA circuit model with ideal electromotive force epsilon and internal resistance r inside the source boundary, connected in series to an external load R. Current leaves the positive terminal and voltage labels show epsilon equals V plus Ir during discharge.real source+−ideal e.m.f. εinternal rlost p.d. = IrRterminal p.d. VIdischarging: ε = V + Irsource energy = load transfer + internal heating
Discharging-source model: the ideal e.m.f. and internal resistance are in series. The terminal p.d. across the load is V = ε − Ir.

Key ideas

  • Gradient meaning depends on which quantity is on each axis.
  • State ‘constant temperature’ for ohmic behaviour.
  • Lost volts Ir represent energy per charge transferred inside the source.

Relationships to know

  • R = V/I
  • R = ρl/A
  • Vterminal = ε − Ir
  • Poutput = IV

Follow the reasoning

Worked example

Analyse a real source at one operating point

Question: A cell has e.m.f. 9.0 V and internal resistance 1.2 Ω. It supplies a 4.8 Ω resistor. Find current, terminal p.d., external output power and internal heating power.

  1. Step 1: Combine the complete loop resistance

    Why: Internal resistance carries the same current as the external resistor.

    Working: Rtotal = 4.8 + 1.2 = 6.0 Ω; I = ε/Rtotal = 9.0/6.0 = 1.5 A.

  2. Step 2: Find terminal p.d.

    Why: It is either the external drop or e.m.f. minus lost volts.

    Working: V = I(4.8) = 7.2 V; also 9.0 − 1.5(1.2) = 7.2 V.

  3. Step 3: Separate output and internal powers

    Why: The source supplies energy to both paths.

    Working: Pout = IV = 1.5(7.2) = 10.8 W; Pinternal = I²r = 1.5²(1.2) = 2.7 W.

Answer: I = 1.5 A, terminal p.d. = 7.2 V, output power = 10.8 W and internal heating = 2.7 W.

Check: Source power εI = 13.5 W equals 10.8 W + 2.7 W.

Now try it with support

Practise with support

A cell has e.m.f. 6.0 V and internal resistance 0.50 Ω while delivering 2.0 A. Find terminal p.d. and output power.

Hints

  1. Use V = ε − Ir.
  2. Output power uses terminal p.d., not e.m.f.
View the guided answer

V = 6.0 − 2.0(0.50) = 5.0 V. Poutput = IV = 2.0(5.0) = 10 W.

Your turn

Practise independently

A wire has length 1.5 m, area 0.20 mm² and resistivity 1.7 × 10⁻⁸ Ω m. Find its resistance and predict the effect of doubling its length.

Check your answer

A = 0.20 mm² = 2.0 × 10⁻⁷ m². R = ρl/A = (1.7 × 10⁻⁸)(1.5)/(2.0 × 10⁻⁷) = 0.128 Ω. Doubling l at fixed A and ρ doubles R to 0.255 Ω.

Common mistakes and exam guidance

Watch out for

  • Calling every component ohmic because R = V/I can be calculated.
  • Explaining an NTC thermistor as if it were a metal filament; its carrier number changes strongly with temperature.

In an exam

  • Label axes before describing whether an I–V graph becomes steeper or shallower.
  • For internal resistance, distinguish e.m.f., terminal p.d., lost volts and output power explicitly.

Put the ideas together

Exam-style practice [8 marks]

A wire of length 2.0 m and area 0.30 mm² has resistivity 1.7 × 10⁻⁸ Ω m. Find its resistance. Then explain and sketch how the I–V graphs of this wire and an NTC thermistor differ as they heat during measurement.

Plan before you answer

  • Convert area before using R = ρl/A.
  • State the physical carrier explanation for each material.
  • Label I and V axes and describe gradient changes.
View the marking points and model answer

Marking points

  1. Uses A = 3.0 × 10⁻⁷ m².
  2. Obtains R ≈ 0.113 Ω.
  3. Wire graph approximately straight only if temperature is controlled.
  4. Heating metal increases lattice vibration/resistivity.
  5. Metal I–V curve becomes less steep with self-heating.
  6. Heating NTC increases mobile carrier number.
  7. NTC resistance falls.
  8. NTC I–V curve becomes steeper with self-heating; sketches and axes consistent.

Model answer

R = (1.7 × 10⁻⁸)(2.0)/(3.0 × 10⁻⁷) = 0.113 Ω. At controlled temperature the metal wire is approximately ohmic. If it self-heats, stronger lattice vibrations increase resistivity, so an I-against-V curve bends to a smaller gradient. In an NTC thermistor, heating releases many more mobile carriers, resistance falls and the curve bends to a larger gradient. Both sketches should pass through the origin and show the stated symmetry for reversed polarity.

Finish from memory

Three-question recap

  1. State the condition in the definition of an ohmic conductor.

    Check

    V is proportional to I at constant temperature.

  2. How does an NTC thermistor's resistance change when heated?

    Check

    It decreases.

  3. What are lost volts?

    Check

    Ir, the energy per charge transferred inside a source.

Try this next

Place these components inside series, parallel and potential-divider networks.

Continue to Series, parallel and potential-divider networksReturn to CircuitsReview Circuit symbols and diagrams

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H1 Physics
Edition
GCE A-Level H1 Physics 2027