Full chapter
Practical electricity
All 4 topics and the revision summary on one page.
01
Electrical heating and power
An appliance transfers electrical energy. Its power tells you how quickly it does so; its operating time determines how much energy it uses.
Current is charge passing per second. Potential difference is energy transferred per charge. Power is the rate of energy transfer: one watt means one joule per second.
Use the heating effect
In a resistive heating element, electrical work transfers energy into the material. Its internal energy increases as it warms. Energy then transfers from the hot element to cooler objects and surroundings.
- Electric kettle
- The heating element transfers energy to the water. Heating the water is the useful purpose; some energy also warms the kettle and surroundings.
- Electric oven
- Hot elements transfer energy to the oven and food, including by radiation and through the warmed air.
- Electric heater
- The element heats its surroundings by radiation and by heating air that can circulate through the room.
At a steady operating temperature, an element can keep receiving electrical energy while transferring energy away at the same rate. Its temperature need not keep rising. Charge continues through the circuit; it is not used up or changed into heat.
Connect voltage, current and power
If charge Q passes through a component with p.d. V, the electrical energy transferred is E = VQ. For steady current I over time t, Q = It. Combining these gives:
Use the p.d. across the component and current through that same component. Convert small units first: 1 mA = 0.001 A and 1 mV = 0.001 V.
These expressions give electrical input. They do not by themselves tell you how much becomes a particular useful output. A motor's mechanical output, for example, would need further information about the motor or its load.
Power, then energy
A heater operating for 12 minutes
A model resistive heater operates at a stated 230 V and 5.0 A, with constant power for 12 minutes.
P = VI = 230 x 5.0 = 1150 W = 1.15 kW.
The duration is 12 x 60 = 720 s, so E = Pt = 1150 x 720 = 828 000 J.
This is the electrical input during the stated operating time. The calculation uses the supplied operating values for this resistive appliance.
Determine electrical input using low voltage
Pair the component's voltage with its current
A is in series; V connects across the component's endpoints P and Q. These supplied readings are for a low-voltage d.c. circuit.
For a steady 180 s interval:
P = VI = 6.0 × 0.40 = 2.4 W.
E = Pt = 2.4 × 180 = 432 J.
The ideal voltmeter takes negligible current, so the ammeter gives the component current. These calculations find electrical input; a particular useful output may be smaller.
For this component, P = 6.0 x 0.40 = 2.4 W. Over 180 s, E = 2.4 x 180 = 432 J.
- Use a suitable low-voltage supply and component. Connect the ammeter in series and voltmeter across the component whose input you want.
- Choose meter ranges that include the expected readings and give useful resolution. Record the actual readings with their units and measure the operating time.
- Check the readings during the interval. A component that warms can change its current, so one starting reading may not represent the whole run.
- Calculate power from matching voltage and current readings, then account for the time for which that power applies.
E = Pt uses a constant power or the average power over the interval. If voltage stays fixed while current changes, the interval's average current can be used in E = VIt. If both voltage and current change, find the power for each interval and add its energy; multiplying two separate average readings is not generally enough.
Optional 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?
02
Energy use and cost
An electricity charge depends on energy used. Work out the power and operating time in matching units before applying the price per unit of energy.
Power is energy transferred per second. At constant power, E = Pt; use the interval's average power if it varies. A larger power means a faster energy transfer, but the total also depends on time.
A kilowatt-hour is an energy unit
One kilowatt-hour, written kW h or kWh, is the energy transferred by a power of 1 kW maintained for 1 hour.
A kilowatt is a unit of power; a kilowatt-hour is a unit of energy. The hour multiplies the power. Writing kW/h would mean a rate of change of power and does not describe the energy used here.
Use hours with kilowatts
The same heater, with a supplied tariff
The heater uses 1.15 kW for 12 minutes. First, t = 12/60 = 0.20 h.
Energy = 1.15 x 0.20 = 0.230 kW h. This is also 0.230 x 3 600 000 = 828 000 J.
Use an example tariff of $0.30 per kW h. Cost = 0.230 x $0.30 = $0.069, or 6.9 cents.
If the answer is required to the nearest cent, this is $0.07. Keep the unrounded energy and cost until the final step. The tariff here is a supplied example price.
Compare the complete use pattern
Appliance A
0.10 kW for 5.0 h
Energy = 0.10 x 5.0 = 0.50 kW h.
Appliance B
1.0 kW for 0.25 h
Energy = 1.0 x 0.25 = 0.25 kW h.
B has ten times the power but runs for much less time. It uses half as much energy as A in these stated patterns. At the same tariff, its energy cost is also half as much.
A thermostat can turn a heater on and off. If a 1 kW heater is powered for only part of an hour, its energy use is not automatically 1 kW h. Use its total powered-on time at that power, or a supplied average power over the whole interval. Include any other energy use if the question specifies it.
Check the time unit. Twelve minutes is 720 s or 0.20 h. It is neither 12 h nor 0.12 h. Choose seconds for a calculation in joules and hours for a calculation using kilowatts and kilowatt-hours.
Optional check A heater uses a constant 1.15 kW for 12 minutes. Using a supplied example tariff of $0.30 per kW h, what is the energy cost before rounding to the nearest cent?
03
Live, neutral, earth and the plug
The three wires have different roles. Understanding where each connects explains both normal operation and why a switch must break the live connection.
A sustained current needs a complete conducting path. A break can stop current while a potential difference still exists. Review circuit connections if you need help following the paths.
Identify each conductor
- Live: L, brown
- Live is at a substantial alternating potential difference relative to earth. Together with neutral, it supplies the normal load circuit.
- Neutral: N, blue
- Neutral is the normal return conductor, maintained near earth potential in the simplified supply model. It carries normal load current. Its name does not make it safe to touch in every real circuit.
- Earth: E, green and yellow
- Protective earth connects exposed conductive parts, such as a metal casing, to the protective return path where required. It normally carries negligible current in a correctly operating appliance.
Mains is an alternating supply: current direction reverses. Calling neutral the return conductor describes its circuit role; it does not mean charge always travels in one direction through the appliance.
Read the plug connections
Identify terminals in an inside wiring view
The cable enters at the bottom. This labelled schematic shows connections inside the plug; it is not the pin-facing view or a complete circuit.
- L: liveBrown
- N: neutralBlue
- E: earthGreen/yellow
The fuse is in the live connection. The grip holds the outer sheath so a pull on the cable is not taken directly by the terminal connections.
The brown core connects to L through the fuse, the blue core to N, and the green/yellow core to E. Use the terminal letters and the stated viewpoint: looking at the pins from the opposite side reverses left and right in the drawing.
The grip secures the cable's outer sheath so that a pull on the cable is not transferred directly to the core connections. The insulating covering around each core keeps the conductors apart. The plug view shows these connections; the fault circuit explains how the earth connection works with protection.
Open the live connection
An open circuit can still contain parts connected to live
Both circuits have the same idealised a.c. supply and resistive load. An amber band marks connection to live potential; it is not a current arrow. Brown and blue still identify the cores.
Open a single-pole switch in live
The open gap disconnects the load from live. The load remains connected to neutral in this simplified circuit.
Open only the neutral wire
Current is zero, but the load and the blue segment between the load and the open gap remain connected to live. Stopping operation has not disconnected live.
Live alternates in potential relative to earth. In this model neutral is maintained near earth potential; that does not make neutral universally safe to touch.
A single-pole switch opens one conductor. Placing it in live breaks the connection between the live supply and the appliance when the switch is open.
If the switch opens only neutral, the normal circuit is incomplete and the appliance stops operating. However, the load remains connected to live. In the ideal no-current model, there is no voltage drop across the load, so even the load-side segment of the opened neutral connection can be at live potential. Its blue insulation still identifies its intended wiring role.
For the same reason, a fuse or circuit breaker must interrupt live when it operates. Interrupting only neutral can stop operation while leaving the appliance connected to live. Some devices open more than one conductor; the essential protection here includes breaking the live connection.
Optional check In a simple mains appliance circuit, a single-pole switch opens only the neutral wire. The appliance stops operating. Which explanation is correct?
04
Hazards and protective devices
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.
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.
Optional check 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?
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
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
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.
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.
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.
Optional check 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?
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.
- Basic insulation separates the live parts from their surroundings.
- 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.
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.
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.
Revision summary
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.
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