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
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
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.
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?
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
- First R = 2.0/0.40 = 5.0 Ω.
- Second R = 6.0/0.75 = 8.0 Ω.
- 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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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.
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
- Larger voltage magnitude increases current and heating.
- The hotter filament has greater resistance.
- 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.
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
- 2.0/0.40 = 5.0 Ω.
- 6.0/0.75 = 8.0 Ω.
- 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.
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
- A fixed 5 Ω would predict 6/5 = 1.2 A.
- The measured state gives 6/0.75 = 8 Ω.
- 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.
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.
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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.