MYP · 19 · extended response
A household fault that a fuse may not detect
Physics in society · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Hypothetical fault current versus path resistance
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Current / A
Horizontal axis (x): Path resistance / Ω
Read the exact data as a table
| Path resistance / Ω | Current / A |
|---|---|
| 46 | 5 |
| 230 | 1 |
| 2300 | 0.1 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 230 V appliance normally uses 1150 W and has a metal case connected to earth. In a hypothetical insulation fault, the effective live-to-earth path resistance becomes 46 Ω. In a separate scenario with direct contact with live or a failed protective-earth connection, model the person-to-earth path as 2300 Ω with 230 V across it. This is not the voltage across a person touching a correctly earthed case. A fitted fuse is rated 13 A, and an RCD trips when live–neutral current imbalance exceeds 30 mA. These are simplified fault calculations, not instructions for live testing. Supplied data (Path resistance / Ω; Current / A): (46, 5), (230, 1), (2300, 0.1).
- (a)
Find normal current, the 46 Ω fault current and the modelled person-path current. Explain why the fuse rating alone is insufficient protection.
3 marks - (b)
Explain how earthing, double insulation, a fuse and an RCD address different failure conditions.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find normal current, the 46 Ω fault current and the modelled person-path current. Explain why the fuse rating alone is insufficient protection.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Normal operating current is P/V = 1150/230 = 5.0 A.
- 2
The fault path would carry 230/46 = 5.0 A, while the person-path model gives 230/2300 = 0.10 A = 100 mA.
- 3
Neither additional path alone exceeds a 13 A fuse rating, and fuses have time–current characteristics. A potentially harmful earth leakage can therefore exist without prompt fuse operation.
(b)
Explain how earthing, double insulation, a fuse and an RCD address different failure conditions.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
A protective earth conductor links accessible metal to earth and can carry fault current. Double-insulated equipment uses separate insulation barriers and suitable construction instead of an exposed earthed case.
- 2
A fuse or overcurrent breaker protects the circuit against excessive current, with an operating time that depends on current; it does not directly measure a small body current.
- 3
An RCD compares live and neutral currents. A 100 mA earth leakage exceeds the stated 30 mA threshold, but trip timing and the complete installation still matter. Classroom calculations cannot establish that contact is safe.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Normal operating current is P/V = 1150/230 = 5.0 A.
- The fault path would carry 230/46 = 5.0 A, while the person-path model gives 230/2300 = 0.10 A = 100 mA.
- Neither additional path alone exceeds a 13 A fuse rating, and fuses have time–current characteristics. A potentially harmful earth leakage can therefore exist without prompt fuse operation.
- A protective earth conductor links accessible metal to earth and can carry fault current. Double-insulated equipment uses separate insulation barriers and suitable construction instead of an exposed earthed case.
- A fuse or overcurrent breaker protects the circuit against excessive current, with an operating time that depends on current; it does not directly measure a small body current.
- An RCD compares live and neutral currents. A 100 mA earth leakage exceeds the stated 30 mA threshold, but trip timing and the complete installation still matter. Classroom calculations cannot establish that contact is safe.
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