Choose protective ratings from supplied circuit limits
Use stated normal current, available ratings and cable limits in a simplified selection problem.
Section P4.5 · Dangers of electricity
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Calculate I = P/V.
- Compare currents and understand a protected-current limit.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How do you find normal current for a supplied 460 W load at 230 V?
What you will learn
- Use stated normal current, available ratings and cable limits in a simplified selection problem.
- Choose only within both supplied limits, and recognise when no listed rating is suitable.
Selection needs more than appliance power
A suitable supplied overcurrent rating permits normal operation and protects the stated cable or circuit limit. In an explicitly simplified no-starting-surge exercise, choose the lowest available rating above the normal current that is no greater than the supplied protected-current limit. If none fits, the given arrangement has no acceptable option.
For a supplied 460 W, 230 V load, normal current is 2 A. With available 3 A and 13 A fuses and a stated 6 A cable limit, 3 A fits while 13 A is too high. This explains a paper model, not a replacement specification for an actual appliance.
Apply the same constraints to trip settings
A supplied adjustable overcurrent trip setting must likewise allow stated normal operation and protect the given limit under the exercise assumptions. For a 6 A normal load, an 8 A limit and options 5 A, 7 A and 10 A, the supplied answer is 7 A. Five is too low and ten fails the limit.
Real selection also uses manufacturer instructions, starting current, device time–current characteristics, cable conditions and installation design. Do not change household fuse or trip ratings from a classroom calculation. Where a real specification exists, it takes priority over a generic smallest-above-current rule.
Sources for this lesson
Pause and explain
Normal current is 6 A, protected limit 8 A and supplied trip options are 5 A, 7 A and 10 A. Ignore startup. Which fits?
Worked example
Select from 3 A and 13 A for a 460 W, 230 V supplied load with a 6 A protected limit; ignore startup surge.
Show the worked solution
- I = 460/230 = 2 A.
- 3 A is above 2 A and no greater than 6 A.
- 13 A exceeds the protected limit.
Answer 3 A in the explicitly supplied simplified model.
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.
Choose between 3 A and 13 A for the supplied 460 W, 230 V load and 6 A cable limit, ignoring startup.
Give me a hint
Calculate normal current and apply both inequalities.
Compare my reasoning
- Normal current is 2 A.
- 3 A permits it and is within 6 A.
- 13 A is above the protected limit.
3 A in the supplied model.
Look for these in your work
- Calculate the load current.
- Check the cable limit.
Select a supplied trip setting of 5 A, 7 A or 10 A for normal 6 A and protected limit 8 A, with no startup surge.
Give me a hint
Reject settings failing either normal operation or protection.
Compare my reasoning
- 5 A is below normal 6 A.
- 10 A exceeds the limit 8 A.
- 7 A satisfies both stated constraints.
7 A in the supplied model.
Look for these in your work
- Explain both rejected settings.
- Keep the simplified assumptions explicit.
Normal current is 7 A, protected limit 8 A and available fuses are 3 A and 13 A, ignoring startup. Should the higher fuse simply be used?
Give me a hint
Check whether any available rating satisfies both constraints.
Compare my reasoning
- 3 A is below normal 7 A.
- 13 A is above the protected 8 A limit.
- Neither is suitable in the supplied arrangement.
No suitable supplied option; do not select 13 A merely to make it operate.
Look for these in your work
- Reject both for stated reasons.
- Do not raise protection to hide incompatibility.
Common mistakes
- Use the protected limit too.
- No suitable option means an incompatible arrangement.
- Real specifications and operating characteristics matter.
What can you explain now?
What two current conditions must a supplied protective choice satisfy?
Compare with the explanation
Permit normal operation and protect the stated limit
Both the ordinary operating current and the protected circuit rating constrain the selection.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Choose only within both supplied limits, and recognise when no listed rating is suitable.
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.