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IGCSE · Cambridge (CIE) · 0652

Explain filament-lamp current–voltage curvature

Connect a filament's increasing temperature and resistance to a less-steep I–V curve as voltage magnitude rises.

Section P4.2.4 · Resistance

Resistance · Cambridge IGCSE Physical Science 0652 · P4.2.4

Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.

Before you begin

  • Interpret an I–V graph with current vertical.
  • Calculate V/I at a non-zero point.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What does V/I at a non-zero point calculate?

What you will learn

  • Sketch the filament-lamp I–V pattern.
  • Explain curvature using heating rather than a constant-resistance extrapolation.
Original illustrative lamp I–V curve passes through the supplied points (2 V, 0.40 A) and (6 V, 0.75 A), with symmetric reversed polarity. Heating increases operating-point resistance and reduces the gradient. This curve is not a universal filament-lamp law; tangent reciprocal and V/I are different on a curve.Filament warms as input risesCurrent / APotential difference / V-6601-1Less steep at larger magnitudes
Original illustrative lamp I–V curve passes through the supplied points (2 V, 0.40 A) and (6 V, 0.75 A), with symmetric reversed polarity. Heating increases operating-point resistance and reduces the gradient. This curve is not a universal filament-lamp law; tangent reciprocal and V/I are different on a curve.

Trace the heating effect

As the magnitude of p.d. across a filament lamp increases, current increases and the filament heats. Its metallic resistance rises with temperature. On I-vertical, V-horizontal axes, the curve becomes less steep at larger voltage magnitudes than a constant-resistance origin line.

An ordinary filament lamp responds similarly to the two polarities in this model. The negative-voltage branch has negative current, with corresponding curvature as the magnitude rises. It does not have the diode's strong one-direction preference. The sketch describes a qualitative lamp response, not every device called a lamp.

Read each operating point

Use R = V/I at a stated non-zero point to calculate its operating resistance. Supplied points (2.0 V, 0.40 A) and (6.0 V, 0.75 A) give 5.0 Ω and 8.0 Ω. The higher quotient supports changing resistance rather than a constant-R extension from the cool state.

For a curved graph, do not automatically treat a local tangent's reciprocal as the same V/I operating resistance. The ratio uses a line from the origin to the chosen point. Detailed thermal properties and calibrated lamp predictions need measured data; the illustrative diagram is a schematic curve rather than a device law.

Sources for this lesson

Pause and explain

A filament heats as voltage magnitude rises. Which I–V description fits?

Put the idea to work

Worked example

A filament lamp has supplied points (2.0 V, 0.40 A) and (6.0 V, 0.75 A). Compare operating resistances.

Show the worked solution
  1. First R = 2.0/0.40 = 5.0 Ω.
  2. Second R = 6.0/0.75 = 8.0 Ω.
  3. Resistance is larger in the higher-voltage state, consistent with heating.

Answer 5.0 Ω then 8.0 Ω; resistance increases.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Explain the curvature of a filament-lamp I–V graph on I-vertical, V-horizontal axes.

Give me a hint

Trace voltage magnitude, heating and resistance changes in order.

Compare my reasoning
  1. Larger voltage magnitude increases current and heating.
  2. The hotter filament has greater resistance.
  3. Current grows less rapidly, so the curve becomes less steep.

Filament heating raises resistance and makes the I–V curve less steep at larger magnitudes.

Look for these in your work

  • Name the heating mechanism.
  • Specify the axes and magnitude trend.
2 · Independent

Calculate R for lamp points (2.0 V, 0.40 A) and (6.0 V, 0.75 A).

Give me a hint

Use the ratio at each stated point rather than one fixed resistance.

Compare my reasoning
  1. 2.0/0.40 = 5.0 Ω.
  2. 6.0/0.75 = 8.0 Ω.
  3. The two operating resistances differ.

5.0 Ω; 8.0 Ω.

Look for these in your work

  • Calculate each state.
  • Avoid a constant-resistance claim.
3 · Transfer

A learner predicts 1.2 A at 6 V from a lamp's earlier 5 Ω cool-state resistance, despite a measured 0.75 A. Assess the prediction.

Give me a hint

Compare the state-dependent quotient with the assumed fixed R.

Compare my reasoning
  1. A fixed 5 Ω would predict 6/5 = 1.2 A.
  2. The measured state gives 6/0.75 = 8 Ω.
  3. Heating makes the cool-state extrapolation unsuitable.

The fixed 5 Ω extrapolation fails; the measured hotter-state resistance is 8 Ω.

Look for these in your work

  • Identify the invalid constancy assumption.
  • Use the given higher-state evidence.

Common mistakes

  • Keep I and V axis order explicit.
  • Do not keep a cool-state R for a heated state.
  • A schematic lamp curve is not a calibrated device law.
Recall without your notes

What can you explain now?

Why does the lamp's I–V curve become less steep as voltage magnitude increases?

Compare with the explanation

Filament heating increases resistance.

The current then grows less rapidly than a constant-resistance extrapolation would predict.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Explain lamp I–V curvature and compare operating resistances while retaining thermal and schematic limits.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physical Science

Course scope and assessment

This page maps the 2025–2026 Cambridge IGCSE Physical Science 0652 syllabus: 38 Chemistry sections and 46 Physics sections. Every section links to original 0620/0625 teaching or one of five new 0652 lessons, with guided, independent and transfer practice and explained answers. Cambridge limits entry to Administrative Zone 3 and identifies November 2026 as the final exam series. The section crosswalk is complete as a first coverage map; a numbered-objective and Core/Supplement audit, independent subject-teacher review and cumulative assessment are still needed. Written practice does not award a Cambridge grade. Online activities cannot demonstrate supervised laboratory competence or complete Paper 5/6.

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