Determine resistance with separate current and voltage measurements
Describe a supervised low-voltage voltmeter–ammeter experiment and calculate resistance while controlling heating and meter loading.
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Connect an ammeter in series and a voltmeter in parallel.
- Calculate R = V/I.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How should the voltage meter be connected to the test component?
What you will learn
- Describe the resistance-measurement apparatus and procedure.
- Calculate a matched V/I and identify heating and meter limitations.
Build the measurement path
Use an isolated low-voltage source, switch, suitable current control and the test component under teacher supervision. Place an ammeter in the current path and a voltmeter across the component's terminals. Choose suitable ranges and polarity. Close the switch briefly, record a stable simultaneous V and I pair, and calculate R = V/I.
For repeated readings, vary the source or current control in suitable small steps and record matching pairs rather than combining one trial's voltage with another's current. Keep contacts secure and identify whether the component temperature stays nearly constant. Switch off between readings where needed to limit heating.
Interpret the measured resistance
A 3.0 V reading and 0.60 A matching component current give 5.0 Ω. If later quotients differ beyond the stated measurement variation, resistance may be changing, for example through heating; the data do not automatically prove a unique cause without controlling other conditions.
A real voltmeter takes a small current, and a real ammeter can have a small p.d. across it. In the illustrated arrangement, an upstream ammeter can include the voltmeter's branch current. The simple V/I estimate treats that current as negligible or uses ideal meters; retain this approximation rather than claiming perfectly exact sample resistance. Written analysis prepares a practical task but does not demonstrate competent apparatus use.
Sources for this lesson
Pause and explain
An upstream ammeter includes a small voltmeter current. What supports the simple sample V/I estimate?
Worked example
A matched component trial gives 3.0 V and 0.60 A, with negligible voltmeter current. Find the resistance estimate.
Show the worked solution
- Use the matching component V and I.
- Apply R = V/I.
- 3.0/0.60 = 5.0 Ω.
Answer 5.0 Ω under the negligible-meter-loading approximation.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Outline a resistance-measurement experiment using an ammeter and voltmeter.
Give me a hint
Specify connections, ranges, matching readings and the resistance calculation.
Compare my reasoning
- Use supervised isolated low-voltage apparatus with A in series and V across the component.
- Select suitable ranges, close briefly and record matching stable V and I.
- Calculate R = V/I while stating heating and meter approximations.
Record matched V and I with series A and parallel V, then calculate R = V/I.
Look for these in your work
- Include both meter connections.
- State practical controls and limitations.
Trials at nearly constant temperature give (2.0 V, 0.40 A) and (3.0 V, 0.60 A). Compare the resistance estimates.
Give me a hint
Calculate each quotient independently before judging constancy.
Compare my reasoning
- 2.0/0.40 = 5.0 Ω.
- 3.0/0.60 = 5.0 Ω.
- The two stated estimates agree.
Both 5.0 Ω; they agree for these stated trials.
Look for these in your work
- Use matching pairs.
- Do not prove every possible operating state from two readings.
Current is read before heating and voltage after heating. Can their quotient establish the component's resistance at either state?
Give me a hint
Check whether the measurements share one operating condition.
Compare my reasoning
- The component conditions changed.
- The voltage and current are not a matched pair.
- Record simultaneous or controlled same-state values.
No; use a matched pair at one controlled operating state.
Look for these in your work
- Identify the state mismatch.
- Describe a valid measurement pair.
Common mistakes
- Use series A and parallel V.
- Record matching states and control heating.
- Real meter effects limit the estimate.
What can you explain now?
Why record voltage and current at the same operating state?
Compare with the explanation
Because both must describe the same component conditions.
Mixing trials can create a quotient unrelated to any actual measured resistance state.
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Make it stick
Describe and analyse resistance measurements with matched states, correct meter topology and explicit practical 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.