Compare conductor p d when current is held constant
Apply the resistance trend under the specific constant-current condition and distinguish it from fixed supply voltage.
Section 4.3.3 · Action and use of circuit components
Exam scope and assessment notes
Core and Extended · 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 V = IR.
- Distinguish a controlled quantity and a changing quantity.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which relation connects p.d., current and resistance?
What you will learn
- State the constant-current resistance–p.d. trend.
- Identify when the condition is not met.
Keep current fixed
For a conductor, p.d. increases as resistance increases when current is held constant. V = IR makes the condition explicit: with the same I, a larger R requires a larger V. At 0.20 A, resistances of 10 Ω and 20 Ω require 2 V and 4 V respectively.
A constant-current comparison is a stated operating condition, not an automatic property of every cell circuit. The source or experimental control must provide the required p.d. while maintaining that current within its range. A diagram label cannot guarantee unlimited output voltage.
Distinguish a fixed-p.d. comparison
If a conductor instead remains across a fixed 4 V supply and its resistance rises from 10 Ω to 20 Ω, its p.d. stays 4 V while current falls from 0.40 A to 0.20 A. The same resistance increase therefore gives different quantity changes under different controls.
A thermistor, LDR or adjustable resistor can change resistance, but the resulting voltage trend depends on its surrounding circuit and control conditions. Identify the conductor and what is held constant before predicting an output. This lesson establishes the Core relation; divider redistribution is separately labelled Supplement.
Pause and explain
A conductor remains across a fixed 4 V supply while its resistance doubles. What is its p.d.?
Worked example
Current is held at 0.20 A while a conductor resistance rises from 10 Ω to 20 Ω. Find its p.d. in each state.
Show the worked solution
- Use the same current for both states.
- First V = 0.20 × 10 = 2 V.
- Second V = 0.20 × 20 = 4 V.
Answer P.d. increases from 2 V to 4 V at constant 0.20 A.
Try a topic test
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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.
At constant 0.20 A, compare p.d.s for 10 Ω and 20 Ω.
Give me a hint
Use the same current in both V = IR calculations.
Compare my reasoning
- Vfirst = 0.20 × 10 = 2 V.
- Vsecond = 0.20 × 20 = 4 V.
- The p.d. doubles when resistance doubles at fixed current.
2 V and 4 V.
Look for these in your work
- Keep current fixed.
- State both operating states.
At constant 0.10 A, a conductor changes from 30 Ω to 50 Ω. Find the p.d.s and increase.
Give me a hint
Calculate each state before taking the difference.
Compare my reasoning
- Vfirst = 0.10 × 30 = 3 V.
- Vsecond = 0.10 × 50 = 5 V.
- The increase is 5 − 3 = 2 V.
3 V to 5 V; increase 2 V.
Look for these in your work
- Use the stated current.
- Distinguish value and change.
A 10 Ω conductor changes to 20 Ω across fixed 4 V. Find current in both states and assess a claim that its p.d. must rise.
Give me a hint
Identify which quantity this comparison actually holds fixed.
Compare my reasoning
- Current initially is 4/10 = 0.40 A.
- Current finally is 4/20 = 0.20 A.
- P.d. stays 4 V because the source condition fixes it.
0.40 A to 0.20 A; p.d. remains 4 V, so the claimed rise is incorrect.
Look for these in your work
- Use the fixed-p.d. condition.
- Contrast it with constant current.
Common mistakes
- The trend requires constant current.
- Fixed voltage gives a different comparison.
- A real source has an operating range.
What can you explain now?
At constant current, what happens to p.d. when resistance increases?
Compare with the explanation
P.d. increases.
V = IR links a larger resistance to a larger p.d. when the current stays fixed.
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Make it stick
State the constant-current condition before applying the increasing-resistance increasing-p.d. trend.
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