Sketch and explain a constant-resistance current–voltage graph
Put current on the vertical axis, p.d. on the horizontal axis, and connect a straight origin line to constant V/I.
Section 4.2.4 · Resistance
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Use R = V/I.
- Read signed graph axes.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What current follows at 4 V through a constant 4 Ω resistor?
What you will learn
- Sketch the I–V pattern of a constant-resistance resistor.
- Relate its gradient to 1/R and retain the constant-temperature condition.
Define the axis convention
For the current–voltage graph here, horizontal position is p.d. V and vertical position is current I. A resistor with constant R obeys I = V/R, giving a straight line through the origin. Reversing the p.d. reverses current in the same model, so the line continues into the negative-voltage, negative-current quadrant.
The gradient is I/V = 1/R on these axes. A steeper line therefore represents lower resistance, not higher resistance. If a different graph swaps the axes, re-derive the gradient meaning rather than transferring the same rule blindly.
Keep the material conditions
Constant resistance is the model condition, usually supported by nearly constant temperature and suitable operating limits. If a resistor heats substantially, its resistance can change and the straight-line assumption may no longer fit. An origin point alone does not prove linearity over the whole range.
For a non-zero point on the constant line, R = V/I. On an illustrative 4 Ω line, V = 4 V gives I = 1 A and V = −4 V gives I = −1 A. At the origin 0/0 is not a usable quotient; infer the constant resistance from the non-zero slope or points.
Pause and explain
On I-vertical, V-horizontal axes, which straight origin line has lower R?
Worked example
A constant-resistance I–V line passes through (4 V, 1 A). Find R and its I-versus-V gradient.
Show the worked solution
- R = V/I = 4/1 = 4 Ω.
- Gradient = I/V = 1/4 A/V.
- The straight origin line has constant resistance under the stated conditions.
Answer R = 4 Ω; gradient 0.25 A/V.
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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.
Describe the I–V graph of a constant-resistance resistor with current vertical and p.d. horizontal.
Give me a hint
Use I = V/R under fixed-resistance conditions and include polarity reversal.
Compare my reasoning
- The relation is proportional.
- Draw a straight line through the origin.
- Continue into the negative V, negative I quadrant for reversed polarity.
A straight origin line with matching positive and negative quadrants.
Look for these in your work
- Label both axes.
- Keep resistance and temperature conditions explicit.
An origin line passes through (6 V, 0.50 A). Find R and I/V gradient.
Give me a hint
Use different quotient directions for resistance and graph slope.
Compare my reasoning
- R = 6/0.50 = 12 Ω.
- Gradient = 0.50/6 = 1/12 A/V.
- The quantities are reciprocals on these axes.
12 Ω; gradient 1/12 A/V ≈ 0.0833 A/V.
Look for these in your work
- Distinguish the quotients.
- State the graph axes.
A learner labels the I–V gradient as R and calls the steeper line more resistive. Correct both statements.
Give me a hint
Re-derive vertical change divided by horizontal change.
Compare my reasoning
- The gradient is ΔI/ΔV.
- For a straight origin line this is 1/R.
- A steeper gradient means lower R.
Gradient is 1/R; the steeper line has lower resistance on these axes.
Look for these in your work
- Use the stated axes.
- Do not confuse a reciprocal relation.
Common mistakes
- Specify the I-vertical, V-horizontal axes.
- A steeper I–V line means lower R.
- Use non-zero points rather than 0/0 at the origin.
What can you explain now?
What shape does a constant-resistance I–V graph have?
Compare with the explanation
A straight line through the origin.
I = V/R gives current proportional to p.d. for fixed R on the specified axes.
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Make it stick
Sketch a constant-resistance I–V line and interpret its reciprocal-resistance gradient correctly.
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.