Cambridge IGCSE Physics · topic 4.4
Electrical safety
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A household fault that a fuse may not detect
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]
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.
Show a hint
Begin with part (a): Find normal current, the 46 Ω fault current and the modelled person-path current. Explain why the fuse rating alone is insufficient protection.
Reveal the worked solution
- 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.
Answer (a) Normal current = 5.0 A; earth-fault current = 5.0 A; person-path current = 0.10 A. (b) Earthing provides a fault path; insulation restricts contact; a fuse interrupts excessive current; an RCD detects imbalance to earth.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.