K223 / K224 / 2027
Practical electricity overview

Lesson 4 of 4

Hazards and protective devices

  • I can state the hazards of damaged insulation, cable overheating and damp conditions.Syllabus

    Syllabus K223 / K224, 12(d). State the hazards of damaged insulation, cable overheating and damp conditions.

  • I can explain fuses, circuit breakers and fuse ratings.Syllabus

    Syllabus K223 / K224, 12(e). Explain fuses, circuit breakers and fuse ratings.

  • I can explain earthing metal casings and double insulation.Syllabus

    Syllabus K223 / K224, 12(f). Explain earthing metal casings and double insulation.

  • I can explain why switches, fuses and circuit breakers interrupt the live wire.Syllabus

    Syllabus K223 / K224, 12(i). Explain why switches, fuses and circuit breakers interrupt the live wire.

In short

  • A fuse melts and breaks the circuit if the current gets too large. Choose the smallest fuse rating above the normal current.
  • A circuit breaker does the same job and can be reset.
  • Earthing joins a metal case to the ground. If the live wire touches the case, a large current flows to earth and blows the fuse.
  • Double insulation uses two layers of insulation, so no earth wire is needed.

Protection depends on the fault and the current path. Insulation prevents unwanted contact; a protective connection and a suitable device can disconnect a faulty appliance.

Live and neutral form the normal load circuit. Protective earth is a separate connection for faults. For a given potential difference, a lower resistance permits a larger current.

Explain the hazard

Damaged insulation
An exposed live conductor can contact a person or a metal casing. This can create an unintended conducting path. A metal case can become live if a conductor inside touches it.
Overheating cables
Cables have resistance and warm when they carry current. Excessive current can cause enough heating to damage insulation or start a fire. Too many loads on a shared connection, or a cable unsuitable for the load, can exceed the cable's current rating.
Damp conditions
Damp skin can have a lower resistance, while conducting water can create an unwanted path. For the same potential difference, a lower path resistance allows a greater current. Different liquids and wet materials do not all have the same conductivity.

Interrupt excessive current

A fuse contains a conducting link. If a sufficiently excessive current persists, heating melts the link and opens the circuit. Fitting it in live disconnects the appliance from the live supply when it operates.

A fuse's current rating is stated in amperes. It is intended to carry the normal current in the specified conditions while providing protection against excessive current. It does not necessarily melt the instant current rises by a tiny amount above the rating: the size and duration of the excess matter.

An overcurrent circuit breaker automatically opens contacts when the current meets its operating conditions. It can be reset after the fault has been addressed, whereas a melted fuse link must be replaced with the specified fuse. Both must be suitable for the circuit they protect.

Interpret a supplied fuse choice

A 460 W appliance at 230 V

The model has negligible starting surge and a suitable cord. Its permitted fuse choices are 3 A and 13 A.

Normal current I = P/V = 460/230 = 2.0 A.

The 3 A option carries this normal current and gives the lower overcurrent rating. Choosing 13 A simply because it is larger would allow a larger current before protection operates.

This choice follows the stated conditions. Actual equipment uses its specified fuse, which also accounts for its operating behaviour and cord.

Try it step by step

Work it out in steps: Normal current I = P / V in A, then Fuse rating to choose.

  1. Problem 1. A 1150 W heater runs at 230 V. Fuses of 3 A, 5 A and 13 A are available. Which fuse should be fitted?
  2. Problem 2. A 920 W iron runs at 230 V. Fuses of 3 A, 5 A and 13 A are available. Which fuse should be fitted?
  3. Problem 3. A 460 W television runs at 230 V. Fuses of 3 A, 5 A and 13 A are available. Which fuse should be fitted?
Show the answer

Problem 1.

Normal current I = P / V in A: I = P / V = 1150 / 230 = 5.0 A

Fuse rating to choose: A 5 A fuse would blow at the normal current, and 3 A is too small, so the 13 A fuse is fitted: the smallest rating above the normal current.

Problem 2.

Normal current I = P / V in A: I = 920 / 230 = 4.0 A

Fuse rating to choose: The 5 A fuse is the smallest rating above 4.0 A, so it carries the normal current and blows soonest in a fault.

Problem 3.

Normal current I = P / V in A: I = 460 / 230 = 2.0 A

Fuse rating to choose: The 3 A fuse is the smallest rating above 2.0 A.

More detail (optional)

A plug fuse or overcurrent breaker does not guarantee disconnection for every harmful current through a person: such a current can be well below the normal load current. A residual-current device, such as an RCCB, instead compares current supplied and returned and opens the circuit for sufficient imbalance caused by leakage. Its detection principle differs from overcurrent protection.

Check your understanding

A supplied model appliance takes 460 W at 230 V, with negligible starting surge and a suitable cord. Its permitted fuse options are 3 A and 13 A. Which choice follows from these conditions?

  1. 3 A: it carries the 2.0 A normal current and blows sooner if a fault occurs.
  2. Neither, because the fuse rating must equal 230 V.
  3. 13 A, because a larger rating gives better protection and blows less often in normal use.
Show the answer

3 A: it carries the 2.0 A normal current and blows sooner if a fault occurs.

I = P/V = 460/230 = 2.0 A. Under the stated no-surge and suitable-cord conditions, 3 A is the lower permitted rating above the operating current. Real equipment needs its specified fuse.

Give a metal-casing fault a protective return

Trace the protective path all the way back to the source

The simplified supply has a neutral/earth bond at the source. The appliance's neutral and protective-earth wires remain separate. Arrows show one instant; a.c. current reverses during the cycle.

Normal operation: return through neutral

Normal operation: return through neutralA simplified a.c. supply is shown on the left. Source live runs through an intact live-wire fuse to terminal L_a of a resistive load inside a metal case. Terminal N_a returns through a separate blue neutral conductor to the source neutral. A green-and-yellow protective-earth conductor connects the metal case to the source-side neutral and earth bond. The load leads pass through insulating feedthroughs without touching the case. Purple arrows trace the normal load-current loop at one instant: source live, fuse, load, neutral, source. Protective earth carries no normal load current. The source symbol completes the circuit; current does not terminate at an earth symbol.A.c.sourceFuseSupply LLoadL_aN_aMetalcaseSupply N /earth bondProtective earth (PE)

Purple arrows: normal load-current path

The working load has a complete live-and-neutral circuit. Protective earth connects the metal case to the supply bond, but carries no normal load current in this model.

Live touches the case: before the fuse opens

Live touches the case: before the fuse opensThe same circuit now has a live-to-case insulation fault between load terminal L_a and the right metal wall. The fuse is still intact because this panel shows the fault before disconnection. Orange arrows trace a complete additional loop at one instant: source live, live fuse, fault contact, metal case, protective-earth conductor, source-side neutral and earth bond, and back through the source. The load and its neutral branch remain drawn separately. The arrows indicate path and direction, not the relative current magnitudes. The model assumes a sufficiently low fault-loop impedance for the resulting current to operate the appropriate fuse. It does not assign zero voltage to the case or claim that every possible parallel path carries no current.A.c.sourceFuseSupply LLoadL_aN_aMetalcaseSupply N /earth bondProtective earth (PE)Fault

Orange arrows: additional fault-current path

The fault provides an additional path through the case and protective earth back to the source. With the stated low-impedance path, sufficient fault current makes the fuse open the live connection.

The load's terminals L_a and N_a remain separate from the metal case unless the shown fault connects live to it. The small pale blocks mark insulating feedthroughs. Arrow widths do not compare currents, and the fault panel does not imply that the casing is exactly at earth potential.

In normal operation the load current uses live and neutral. The fault panel shows the instant before the fuse opens: a live-to-case fault has a complete low-resistance return through protective earth and the labelled source-side connection. Any current arrows show one instant of the alternating current.

With intact insulation, the metal casing is separate from the live parts. The protective earth connection normally carries negligible current.

If live contacts the casing, the protective connection provides a low-resistance fault path. Follow it from the source: live, protective device, fault, metal casing, protective earth, source-side earth/neutral connection, back to the supply. A sustained current needs that complete return; it does not simply disappear into a ground symbol.

In this model, the path keeps the casing closer to earth potential and allows a sufficiently large fault current for the appropriate fuse or breaker to interrupt live. The low-resistance path and the suitable protective device work together.

More detail (optional)

A broken earth connection can leave the casing at a dangerous potential after a live-to-case fault, even before anyone touches it. A person touching it may then create an additional path. Current can divide between parallel paths; it does not entirely avoid every path with a higher resistance.

Check your understanding

A live wire contacts an earthed metal casing. In the stated fault model, the protective earth gives a low-resistance return to the supply and the live-wire overcurrent protection is suitable. How does this arrangement protect?

  1. A large fault current flows through a complete return path, allowing the protection to interrupt live.
  2. Earth normally carries the load current, so it stops any current from entering the live wire.
  3. The fault current leaves the casing and spreads into the ground without needing a return path.
Show the answer

A large fault current flows through a complete return path, allowing the protection to interrupt live.

The low-resistance protective path keeps the casing closer to earth potential and allows sufficient fault current for the appropriate fuse or breaker to open live. The path and device must both be suitable; earth is not the normal load return.

Separate accessible parts from live parts

Separate live parts from accessible parts with insulation

This schematic section shows two insulating barriers. The nested-square symbol identifies Class II equipment.

Double insulation and the Class II nested-square symbolAt the top is the Class II mark: a small square centred inside a larger square, with no connection between them. Below is an illustrative section rather than a complete appliance drawing. Live parts at the centre are separated from the accessible outer surface by a basic insulating barrier, numbered 1, and an independent supplementary insulating barrier, numbered 2. The outer outline marks the accessible surface. No protective earth connection is used for this protection. Equivalent reinforced insulation is an alternative to two separate barriers; the appearance of a plastic case alone does not establish Class II protection.Class II symbolLive parts12Accessible outer surface
  1. Basic insulation separates the live parts from their surroundings.
  2. Supplementary insulation provides an independent second barrier to accessible parts.

Equivalent reinforced insulation can provide the required protection without two visibly separate layers. A plastic-looking casing alone is not proof of double insulation.

The two labelled insulating barriers illustrate double insulation. The nested-square mark identifies a Class II design; equivalent reinforced insulation can provide the required separation.

Double insulation provides two independent insulating barriers, or equivalent reinforced insulation, between live parts and accessible parts. This design protects against contact without relying on a protective earth connection to those parts.

More detail (optional)

A Class II appliance is designed around that insulation protection. A plastic-looking outer case alone does not establish that an appliance is double insulated. Earthing and double insulation are different ways of addressing a possible live-to-accessible-part fault; identify which design the question describes.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

An electric kettle is rated 2200 W, 230 V. Fuses of 3 A, 5 A and 13 A are available. Calculate the normal current in the kettle and choose the most suitable fuse. Explain why the other two are not suitable. [3]

Saved on this browser only. Anyone using this browser can see it.

Exam-style question

A washing machine has a metal case connected to the earth wire. A fault makes the live wire touch the case. Explain how the earth wire and the fuse together protect a person who then touches the case. [3]

Saved on this browser only. Anyone using this browser can see it.