Define resistance and apply Ohm’s law
Relate resistance to voltage and current and state the conditions for Ohm’s law.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 9. Circuit calculations and component sketches are original study practice; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use charge, time, energy, power, area and consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How is resistance defined at an operating point?
What you will learn
- Define resistance as potential difference divided by current
- Recall and use V = IR
- State Ohm's law and its constant-temperature condition
Resistance relates potential difference to current
For a component at a stated operating condition, resistance R = V/I, the ratio of potential difference across it to current through it. Its unit is the ohm, Ω, where 1 Ω = 1 V A⁻¹.
For an ohmic conductor at constant temperature, current is directly proportional to potential difference. This is Ohm's law. In that range V/I stays constant and V = IR; changing temperature may change the resistance and invalidate the same constant-R model.
Check which quantity is held constant
A straight I–V line through the origin has a constant ratio V/I if voltage is on the horizontal axis and current on the vertical axis. The gradient is I/V = 1/R in that graph convention; if axes are reversed, the gradient is R.
When calculating a resistance from a point on a non-ohmic I–V curve, use the ratio V/I at that operating point. Do not call it a constant resistance over the whole curve.
Pause and explain
What condition belongs in the statement of Ohm’s law?
Worked example
A resistor has 5.0 V across it and current 0.25 A. Find resistance.
Show the worked solution
- Use R = V/I.
- R = 5.0/0.25.
- State ohms.
Answer R = 20 Ω.
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Test this topic →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.
A 9.0 V potential difference drives 0.30 A through an ohmic resistor. Find R.
Give me a hint
Divide voltage by current.
Compare my reasoning
- R = V/I.
- R = 9.0/0.30.
- Give the unit Ω.
30 Ω.
Look for these in your work
- Correct ratio used
- Value calculated
- Ohm unit stated
A resistor of 18 Ω carries 0.20 A. Find its potential difference.
Give me a hint
Apply Ohm's law by multiplying the resistance by the current.
Compare my reasoning
- V = 18 × 0.20.
- V = 3.6 V.
- State the potential difference.
3.6 V.
Look for these in your work
- Ohm's law applied
- Product calculated
- Volt unit stated
A learner measures 2.0 V and 0.50 A for a filament lamp, then claims the lamp has a constant resistance of 4.0 Ω at every current. Explain the error.
Give me a hint
Distinguish an operating-point ratio from a constant resistance over a range.
Compare my reasoning
- At that point R = V/I = 4.0 Ω.
- The filament heats as current increases, so its resistance changes.
- One V/I measurement does not establish a constant resistance across all operating points.
4.0 Ω is the resistance ratio at the stated operating point, but a filament lamp is non-ohmic over a changing-current range because its temperature and resistance change.
Look for these in your work
- Point resistance calculated
- Temperature effect noted
- Overgeneralisation corrected
Common mistakes
- Inverting V/I when finding resistance
- Claiming every component obeys Ohm's law at every operating point
- Ignoring temperature when assessing a conductor's I–V relationship
What can you explain now?
How is resistance defined at an operating point?
Compare with the explanation
R = V/I
It is potential difference divided by current.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How can a component have a resistance at one operating point without obeying Ohm's law over its full range?
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.