K323 / 2027
Practical electricity overview

Chapter revision

Revision summary

Key ideas, equations and common mistakes. Open any topic below for the full explanation.

Electrical input, power and energy

Kettles, ovens and heaters use the heating effect of current in a resistive element. Electrical work increases the material's internal energy; energy then transfers to water, food, air and surroundings. Input energy need not all become the named useful output.

Power: P = VI
P is energy transferred per second, in W. Use V in volts across the chosen component and I in amperes through it. One watt is one joule per second.
Energy: E = VIt = Pt
With steady operating values, use t in seconds for E in joules. E = Pt also works with the average power over the interval. At fixed V, an average I can be used; if both change, account for the changing power.
Conversions
1 kW = 1000 W; 1 mA = 0.001 A; 1 mV = 0.001 V; 1 min = 60 s; 1 h = 3600 s.

A low-voltage determination needs an ammeter in series, a voltmeter across the same component and a measured time. Choose suitable ranges and check whether the readings remain steady. Electrical readings alone do not determine useful output energy.

Kilowatt-hours and cost

Energy in kW h = power in kW x time in h. One kW h is 3 600 000 J. Cost = energy in kW h x the stated tariff per kW h. A rating gives power; operating time or average power is also needed for energy use.

For the supplied heater: 1.15 kW x 0.20 h = 0.230 kW h. At the example $0.30 per kW h, cost is $0.069 = 6.9 cents. Keep intermediate values before rounding.

Connections and live-wire interruption

  • Live: brown, substantial alternating p.d. relative to earth; connects through the plug fuse.
  • Neutral: blue, normal load return, near earth potential in the supply model.
  • Earth: green/yellow, protective connection to exposed conductive parts where required; negligible normal current.
  • Plug: identify L/N/E by terminal labels and viewpoint. The cable grip secures the outer sheath.
  • Switches, fuses and breakers: must interrupt live. Opening only neutral can stop current while leaving the appliance connected to live.

Match protection to the fault

  • Damaged insulation: exposed live parts or a live-to-case contact can create an unwanted current path.
  • Overheated cable: excessive current can damage insulation or cause fire.
  • Damp conditions: lower resistance or new conducting paths can increase current through a person.
  • Fuse: excess current heats and melts its link. Its rating is in amperes; actual operating time depends on the current.
  • Overcurrent breaker: opens contacts automatically under its operating conditions and can be reset after the fault is addressed.
  • Earthing: a complete low-resistance protective return allows appropriate protection to disconnect live after a casing fault.
  • Double insulation: two independent barriers or equivalent reinforced insulation protect accessible parts without relying on protective earth.

Normal load current returns through neutral. Protective earth has a different role. An RCCB detects an imbalance between supplied and returned current; a fuse or overload breaker responds to excessive current.

Back to electrical heating and power

Practise exam questions

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

Exam-style question

A 2.0 kW heater is used for 3.0 hours each day for 30 days. Electricity costs 30 cents per kW h. Calculate the energy used in kW h and the total cost. [3]

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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]

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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]

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Quick recall

Card 1 of 5. Answer in your head, then check.

A low-voltage component has a steady 6.0 V across it and 0.40 A through it for 180 s. Which result describes its electrical input?

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