Practical Electricity

Resistive elements increase the internal energy of their surroundings

Current through a resistance transfers electrical energy into the internal energy store of the element and its surroundings. Kettles heat water, ovens heat food and space heaters warm the air by this heating effect.

Use P = VI for rate and E = VIt for total transfer

P = VI

E = Pt = VIt

Power is the rate of energy transfer: 920 W means 920 J each second. For steady operation at its rating, a 920 W iron on a 230 V supply takesI = P/V = 920/230 = 4.0 A. If it runs for 300 s, it transfersE = Pt = 920 × 300 = 2.76 × 10⁵ J.

A power rating alone does not give the energy used. A thermostat may switch a heater on and off: at 2.0 kW for a total of 15 min of heating, it uses 0.50 kWh, even if it has been plugged in for an hour. Here we neglect any energy used while the heating element is off.

Guided calculation: separate the rate from the total

A resistor on a steady 12 V d.c. supply draws 2.0 A for 40 s. Write the power first, then the energy. State why the two answers need different units.

Match kilowatts with hours before calculating cost

The kilowatt-hour is a unit of energy:E (kWh) = P (kW) × t (h). A 2.0 kW heater used for 3.5 h consumes 7.0 kWh. At a supplied fictional tariff of 32 cents per kWh, the cost is7.0 × $0.32 = $2.24.

A learner calculates the energy of a 1.2 kW heater used for 20 min as 1.2 × 20 = 24 kWh. Identify the first error before opening the feedback.

State the hazard and the physical reason

Worked pathway: one fault, two possible consequences

In a supplied drawing, a damaged live core is exposed. If a person provides a path from it to earth, current through the body can cause shock. If the exposed live core instead touches neutral, a low-resistance fault can draw a large current through the cable and cause heating. Saying only “electricity escapes” does not identify either complete path.

Predict before checking

A static case study describes intact insulation but too many appliances sharing one cable. No person touches a conductor. Is the immediate concern current through a person or heating in the cable? Explain where the current flows.

Independently explain a different supplied case: damaged insulation permits contact between live and a person's damp skin, with a return path to earth. Why does dampness increase the danger at the same potential difference?

Protection must disconnect an excessive current

A fuse heats and melts under sufficient overcurrent, opening the circuit. An overcurrent circuit breaker opens contacts instead. In the system taught here, they interrupt the live path. A fuse must be replaced after it operates; a breaker can be reset only after the fault has been dealt with by a competent person.

Choose a fuse rating

1. Calculate normal current with I = P/V.
2. For a question that supplies suitable ratings and assumes no starting surge, choose the smallest listed rating above that current.
3. Reject a lower rating because it may melt during normal use; reject an unnecessarily high rating because it gives poor protection.

For the 4.0 A iron above, suppose the question offers 3 A, 5 A and 13 A, with a cable suitable for the selected rating and no starting surge. Choose 5 A. A real installation also depends on cable and device characteristics; this calculation is not an installation design.

A fuse does not prevent every electric shock: a harmful current through a person may be too small to operate an overcurrent device. An RCCB detects leakage current and is a different kind of protection. Do not treat every device called a “circuit breaker” as if it detects the same fault.

Guided choice under supplied conditions

A resistive appliance takes 690 W from 230 V. The classroom choices are 2 A, 5 A and 13 A, all approved for the specified cable, with no starting surge. Calculate normal current, reject the undersized option, then choose the smallest remaining rating above normal current.

Now work independently: a 1380 W resistive load uses the same supply and offers suitable 3 A, 10 A and 13 A ratings with no starting surge. Give its current and your selection, then explain why this does not prove every shock would operate the fuse.

Switches, fuses and circuit breakers belong in the live wire

Opening the live wire isolates the appliance from the high-potential conductor. If a device opens only the neutral wire, current stops but internal parts may remain connected to live and can still cause electric shock. “Off” is not proof that a real circuit is safe to touch. Nor does neutral's normally low potential make it safe to touch: faults or incorrect connections can change its potential.

Find the first incorrect inference

Read this supplied connection account: (1) a switch opens only neutral; (2) the normal operating loop is broken; (3) therefore every internal part is disconnected from live. Identify the first wrong statement and explain it.

Earthing and double insulation control different fault paths

Trace the supplied fault circuit

For the earthed model, follow this sequence: live touches case → current passes through the case and protective-earth conductor → the return connection completes the circuit to the source → sufficient fault current operates the live protective device → that device breaks the live path. The low-resistance path and sufficient fault current are stated features of this example.

Before checking, explain why “current reaches earth and disappears” cannot explain the fuse operation. Name the connection needed to complete the circuit.

Independently compare two supplied designs: A has an accessible metal case bonded to protective earth; B is explicitly identified as Class II with double insulation and the square-inside-square mark. Explain why B does not rely on an earth conductor. Would an unmarked plastic-looking case alone justify that conclusion?

Live supplies; neutral returns; earth protects

The disconnected plug and flexible cable shown here use the modern brown, blue and green/yellow convention. These colours are not a universal guide to historical wiring or every country's installations.

Read a disconnected three-pin plug by role, terminal and colour

Study the inside-view diagram as an explanation of the connections. It is not a rewiring activity.

• Brown live → L, through the fuse.
• Blue neutral → N.
• Green/yellow earth → E, with enough slack to disconnect last if the cable is pulled.

The grip holds the outer sheath so a pull on the cable is taken by the sheath, rather than by the individual conductor connections. Copper carries current because it has low electrical resistance. The insulating material around each core keeps neighbouring conductors separate; the outer sheath also protects the group mechanically.

Justify each material by its job

Use the disconnected diagram to predict what changes if a conductor's copper core is replaced by an insulator, or if the insulation between neighbouring cores is replaced by a conductor. Explain each failure before checking.

Work independently on this account of a disconnected plug drawing: (1) brown connects through the fuse to L; (2) blue connects to N; (3) the grip holds only the individual cores; (4) a pull is therefore kept away from the terminal connections. Identify the first error, its consequence and the feature the corrected drawing must show.

Common mistakes

Check your understanding

Worked example: choose a fuse and calculate running cost

A 1.84 kW resistive, metal-cased appliance runs from a 230 V supply for 15 min each day. Use a fictional tariff of 34 cents per kWh. The question offers suitable 3 A, 5 A and 13 A fuses, assumes no starting surge and specifies a cable suitable for the chosen rating. Choose a fuse, then calculate the cost for 30 days.

1. Normal current = P/V = 1840/230 = 8.0 A.
2. The 13 A fuse is the smallest listed rating above 8.0 A. The 3 A and 5 A fuses could melt during normal use.
3. 15 min = 0.25 h, so daily energy = 1.84 × 0.25 = 0.460 kWh.
4. Energy for 30 days = 0.460 × 30 = 13.8 kWh.
5. Cost = 13.8 × $0.34 = $4.69 to the nearest cent.

Guided practice

Practise this independently

1. A 2.0 kW heater runs for 45 min. Calculate the energy transferred in joules and in kWh. Find the cost at a fictional tariff of 30 cents per kWh.
2. Explain why using an appliance with damaged insulation is dangerous, and why the danger is greater in damp conditions.
3. A live wire touches the metal casing of an earthed kettle. Explain the safety sequence from the fault to disconnection of the supply.
4. State the colour and normal role of the live, neutral and earth conductors. Explain why the switch and fuse are connected in the live wire.
5. A drawing of a disconnected plug shows loose strands of bare copper outside the live terminal and a cable grip holding the individual cores instead of the outer sheath. Identify both errors, their consequences and what the corrected drawing should show.

Try this next

Close the answers and explain question 3 aloud as one unbroken cause-and-effect chain. Then change the power, time and tariff in question 1 and solve it again.

For current public safety guidance, see EMA's electrical-safety guidance. Electrical Safety First also explains the Class II double-insulation mark.

Practise this topic

Start with the six-question topic check. Use the feedback to revisit the matching explanation and worked example, then continue into changed practice and independent assessment.

Check your understanding · Practise what you missed