Electrical Safety Worked ExamplesIGCSE Physics 0625

(a) Damaged insulation. The exposed conductor may be the live wire. Anyone touching it becomes part of a path from the live wire to earth, so the hazard is an electric shock. (A bare live wire touching the neutral wire could also cause a short circuit.)

(b) Overheating cable. A cable carrying a large current heats up. If it gets hot enough the insulation can melt or catch alight, so the hazard is fire (and, once the insulation has melted, bare wires and a shock).

(c) Damp conditions. Water (especially with dissolved impurities) conducts electricity and wet skin has a much lower resistance than dry skin. Water can bridge the gap between a live part and the person, so the hazard is an electric shock.

(d) Overloading a socket. Every appliance takes its current through the single wall socket and the wiring behind it. The total current can exceed what the socket and cables are designed to carry, so they overheat, with a risk of fire.

Positions $1$ and $2$ are on the live wire; positions $3$ and $4$ are on the neutral wire. The live wire is the one at a high voltage, so both safety devices must be able to disconnect it.

The fuse must be in the live wire. If the fuse were in the neutral wire and melted, the lamp would stop working but its wiring would still be connected to the live wire — a person touching it could still get a shock.

The switch must also be in the live wire, for the same reason: when it is open, the lamp must be completely disconnected from the high voltage.

So both devices go on the live wire, at positions $1$ and $2$. Option A puts the switch in the neutral wire, C puts both in the neutral wire and D puts the fuse in the neutral wire — each leaves the lamp connected to the live wire at some point.

(a) Follow the wire from X back to the supply. It goes straight to the live wire without passing through the switch, so X is still at the high voltage of the live wire ($230\text{ V}$) when the switch is open.

The lamp is off only because the circuit is broken after the lamp, in the neutral wire. If the owner touches X, a current can pass from the live wire through their body to earth — they receive an electric shock.

(b) Move the switch into the live wire, between the supply and X. Opening the switch then disconnects the lamp from the live wire, so no part of the lamp is at a high voltage.

(a) When the live wire touches the case, the case would be at a high voltage. The earth wire connects the case to earth through a path of very low resistance.

So a very large current flows from the live wire, through the case and down the earth wire — much larger than the normal current of the heater.

This large current passes through the fuse, which is in the live wire. The fuse wire heats up and melts, breaking the circuit.

Now the heater (and its case) is disconnected from the live wire, so the case is no longer at a high voltage and a person touching it does not get an electric shock.

(b) Without the earth wire, the case stays at a high voltage but the fuse does not melt (the heater still works normally). A person touching the case becomes the path to earth, so a current passes through them — an electric shock.

(a) Each element in parallel has the full $230\text{ V}$ across it. Use $I = \dfrac{V}{R}$ for each.

The fuse is in the main (unbranched) part of the circuit, so it carries the sum of the branch currents.

(b) A fuse must carry the normal current without melting, so its rating must be above $4.5\text{ A}$. The $3\text{ A}$ fuse would melt as soon as the heater was switched on.

The rating should also be as close as possible above the normal current, so that the fuse melts quickly if a fault makes the current rise. A $13\text{ A}$ fuse would let a fault current nearly three times the normal value flow without melting, so the cable could overheat. The $5\text{ A}$ fuse is the correct choice.

In a double-insulated appliance, the user is protected from shock by the insulation itself: the case is non-conducting, so even if a live wire touched it, the case could not give a shock. That is why no earth wire is needed. So A describes the job of the insulation, not the fuse.

The fuse is still needed. If a fault (for example a short circuit inside the motor) makes the current too large, the cable and wiring would overheat and could cause a fire. The fuse melts and breaks the circuit first. This is B.

C is wrong: when the mixer is switched off there is no current in either wire anyway. D is wrong: a fuse has negligible resistance and does not reduce the current in normal use — it only melts when the current exceeds its rating.

(a) The appliances are in parallel, each with $230\text{ V}$ across it. Use $P = IV$, so $I = \dfrac{P}{V}$, for each (converting kW to W).

Add the currents in the parallel branches.

(b) The setting must be above $19\text{ A}$, otherwise the switch would trip during normal use ($6\text{ A}$ and $16\text{ A}$ are too low). It should be as close as possible above it, so $20\text{ A}$ — a $32\text{ A}$ setting would allow a much larger current, and the cables could overheat, before it tripped.

(c) When the current rises above the set value (because of a fault or an overload), the trip switch opens automatically and breaks the circuit in the live wire. This stops the current before the cables overheat. (Inside, the current passes through an electromagnet; a large enough current makes the magnetic force strong enough to pull the contacts apart.)

(d) A trip switch can simply be reset (switched back on) once the fault is cleared, whereas a melted fuse must be replaced. It also acts faster than a fuse.

(a) The appliances are in parallel, so the total power is the sum of the individual powers (in watts).

Use $P = IV$ with the total power.

(b) $18.8\text{ A}$ is much more than the $13\text{ A}$ the lead is designed for. The large current makes the cable and the socket connections overheat; the insulation may melt, giving a risk of fire (and of electric shock from exposed wires).

(c) Remove the kettle's $2200\text{ W}$ and recalculate.

$9.26\text{ A}$ is less than $13\text{ A}$, so the lead is no longer overloaded.

(a) Use $I = \dfrac{V}{R}$, converting kilohms to ohms.

Repeat with the wet-hand resistance.

(b) Water conducts electricity, and wet skin has a much lower resistance than dry skin. With the same voltage, the current through the person is $40$ times larger. Currents of a few tens of milliamperes through the body can be fatal, so damp conditions greatly increase the risk of a serious electric shock.

(c) The fuse melts only when the current in the live wire exceeds $3\text{ A}$. The current through the person, $0.20\text{ A}$, is far below $3\text{ A}$, so the fuse does not melt — yet this current is easily large enough to kill.

(a) The power dissipated in a resistance is $P = I^2R$.

Repeat for the larger current.

(b) Compare the two results. Because $P$ depends on $I^2$, multiplying the current by $4$ multiplies the heating by $4^2 = 16$.

(c) Use the same equation with the smaller resistance.

A thicker wire has a smaller resistance, so for the same current much less thermal energy is produced per second (one quarter here). A thin cable carrying a large current can overheat, melt its insulation and start a fire; a thicker cable stays cool enough to be safe.

Fault 1: the fuse is in the neutral wire. If a fault makes the current too large, the fuse melts but the lamp stays connected to the live wire, so parts of it remain at a high voltage. The fuse should be in the live wire.

Fault 2: the switch is in the neutral wire. With the switch open the lamp is off, but it is still connected to the live wire; a person changing the bulb could get an electric shock. The switch should be in the live wire.

Fault 3: the earth wire is not connected to the metal case. If the live wire touched the case, the case would become live. With no low-resistance path to earth, no large current would flow to melt the fuse, so anyone touching the case would get an electric shock. The earth wire must be connected to the metal case.

(a) As the diameter increases, the fusing current increases. The graph is a curve that gets steeper, so the fusing current increases more and more rapidly — it is not directly proportional to the diameter (the line is not straight).

(b) Read across from $5.0\text{ A}$ on the vertical axis to the curve, then down to the horizontal axis. $5.0\text{ A}$ lies between the points at $0.20\text{ mm}$ ($4.5\text{ A}$) and $0.25\text{ mm}$ ($6.3\text{ A}$), nearer the first.

(c) The fuse must not melt at the normal current of $3.6\text{ A}$: the $0.15\text{ mm}$ wire melts at $2.9\text{ A}$, so it would melt in normal use.

The fuse should melt at a current only slightly above $3.6\text{ A}$. The $0.20\text{ mm}$ wire melts at $4.5\text{ A}$, which is just above normal; the $0.25\text{ mm}$ wire needs $6.3\text{ A}$, letting a much larger fault current flow. So use the $0.20\text{ mm}$ wire.

(d) Any one sensible precaution: do not touch the wire while the current is flowing (it becomes very hot); keep the wire on the heatproof mat; wear eye protection in case of molten metal; use a low-voltage supply, never the mains.