Coaching Centre
Electrical power, energy and safety
Power as the rate of energy transfer (P = VI = I²R = V²/R), energy and the kilowatt-hour, why globes are dimmer in series, the 230 V household supply, active, neutral and earth wires, circuit breakers, RCDs, earth faults and the effects of current on the body.
On a cold winter evening, millions of Australian homes switch on heaters, ovens and kettles within the same half hour, and the National Electricity Market has to find several thousand extra megawatts almost on the spot. Your bill counts that energy in kilowatt-hours, your switchboard protects you from it with breakers and RCDs, and every number involved comes from one short equation: P = VI.
Power: the rate of energy transfer
Power is energy transferred per second, $P = W/t$, in watts (1 W = 1 J s−1). In a circuit, each coulomb transfers V joules (V = W/q), and I coulombs pass each second, so
$$P = VI$$
Substituting V = IR or I = V/R gives two more forms, useful when you know the resistance:
$$P = I^2R = \frac{V^2}{R}$$
Use $I^2R$ when the current is fixed (series components, transmission lines) and $V^2/R$ when the voltage is fixed (appliances in parallel on the mains). In a resistor all of this power becomes heat; in a motor or globe some becomes motion or light.
An electric kettle is rated 2.4 kW at 230 V. Find the current it draws, its resistance, and the energy it uses in a 3.0 minute boil.
$I = \dfrac{P}{V} = \dfrac{2400}{230} = 10.4$ A.
$R = \dfrac{V^2}{P} = \dfrac{230^2}{2400} = 22$ Ω.
$E = Pt = (2400)(180) = 4.3 \times 10^{5}$ J.
Energy and the kilowatt-hour
Electricity retailers bill energy in kilowatt-hours: the energy used by a 1 kW appliance running for 1 hour. With P in kW and t in hours, $E = Pt$ gives kWh directly.
$$1\ \text{kWh} = (1000\ \text{J s}^{-1})(3600\ \text{s}) = 3.6 \times 10^{6}\ \text{J}$$
A 2.0 kW heater runs for 5.0 h a day for a 90-day winter. Electricity costs 30 c per kWh. Find the energy used and the cost.
$E = Pt = (2.0)(5.0 \times 90) = 900$ kWh $= 3.2 \times 10^{9}$ J.
900 × 30 c = 27 000 c, or 270 dollars for the winter.
Series or parallel: which is brighter?
A globe's brightness depends on the power it dissipates, not the current alone. Identical globes in parallel each get the full supply voltage and full power. In series they share the voltage, the current is smaller, and each glows far less brightly.
Two 60 W, 230 V globes are connected in series across 230 V. Assuming their resistance stays the same, find the power of each.
$R = \dfrac{V^2}{P} = \dfrac{230^2}{60} = 882$ Ω.
$I = \dfrac{230}{2 \times 882} = 0.130$ A.
$P = I^2R = (0.130)^2(882) = 15$ W: a quarter of the rated power, because each globe has half the voltage and $P \propto V^2$. (In reality the cooler filaments have lower resistance, so each would run at a little more than 15 W.)
Household circuits
Year 9 (topic 6F) introduced power boards and why overloading them is dangerous. Here is the full picture. The mains supply in Australia is 230 V AC at 50 Hz. (Many older books, including some current textbooks, still quote 240 V; Australia adopted the 230 V standard in 2000, and the actual voltage at a power point is usually somewhere between about 225 V and 250 V.)
- Active (brown, older red): at 230 V relative to earth. Switches, fuses and circuit breakers go in the active so that turning them off disconnects the dangerous wire.
- Neutral (blue, older black): the return path, held close to 0 V at the switchboard.
- Earth (green/yellow): connected to the metal case of appliances and to a stake in the ground. It normally carries no current.
- Appliances are in parallel, so each gets 230 V and can be switched on and off independently. Their currents add, which is why too many appliances on one circuit overloads it.
- Circuit breakers (and older fuses) protect the wiring: if the current exceeds the rating (often 16 A or 20 A for power circuits), the breaker trips before the cables overheat and start a fire.
- Residual current devices (RCDs) protect people. An RCD compares the current in the active with the current in the neutral. If they differ by more than 30 mA, current must be leaking to earth (perhaps through someone), and it cuts the supply in a few hundredths of a second. RCDs have been required on power circuits in new homes since 1991, and the 2018 wiring rules require them on all final circuits in new work.
- Double-insulated appliances (marked with a square inside a square) have no exposed metal that could become live, so they need no earth wire.
Earth faults and electric shock
The danger in a shock is the current through the body, especially across the chest. Dry skin gives a hand-to-foot resistance of perhaps 1000–100 000 Ω; wet skin far less. With 230 V across about 1000 Ω, $I = V/R \approx 0.23$ A. Roughly: 1 mA can be felt, 10–20 mA makes muscles clamp so you can't let go, and 50 mA or more through the chest for a fraction of a second can send the heart into fibrillation. That is why the RCD threshold is 30 mA.
- With just the kettle on for 1.0 h: 2,400 W, 10.4 A, 2.4 kWh and 72 c. Check each with P = VI and E = Pt.
- Turn the heater on too: 4,400 W would need 19.1 A, so the 16 A breaker trips. Turn the heater off and reset.
- Turn on "Loose active wire touches kettle case" with the earth connected: the fault current runs to earth and the breaker trips at once.
- Reset, turn off "Earth wire connected", and turn the fault on again. Nothing trips and the kettle still works, but the case is live. Now turn on "Someone touches the kettle": the RCD trips.
- Reset and turn off "RCD fitted", then repeat. Read what happens and explain why both earth wires and RCDs are needed.
Test yourself
A 230 V hair dryer draws 8.0 A. Its power is
P = VI = 230 × 8.0 = 1840 W ≈ 1.8 kW.
The current through a resistor is doubled. The power it dissipates
P = I²R, so doubling I multiplies P by 4.
How much energy does a 150 W television use in 4.0 hours?
E = Pt = 0.150 kW × 4.0 h = 0.60 kWh (which is 2.2 × 106 J). 600 kWh forgets to convert W to kW; 600 J forgets to convert hours to seconds.
The main purpose of a residual current device (RCD) is to
An RCD detects a difference of more than 30 mA between active and neutral currents and disconnects within milliseconds.
Switches and circuit breakers are always placed in the active wire because
With the switch in the active, an "off" appliance has no part at 230 V. A switch in the neutral would leave the appliance live.
Label the household circuit.
Sort the safety features.
Power is the rate of transfer and equals potential difference multiplied by . Electricity is billed in . Household appliances are connected in across the V supply. An RCD trips when more than 30 mA leaks to .
A power circuit protected by a 16 A circuit breaker supplies a 2.4 kW kettle and a 2.0 kW heater at 230 V. (a) Calculate the total current and explain what happens when both are switched on. (b) The heater has a metal case. Its active wire comes loose and touches the case. Analyse what happens if (i) the earth wire is intact and (ii) the earth wire is broken but the circuit has an RCD.
(a) The appliances are in parallel, so their powers add: P = 2400 + 2000 = 4400 W. I = P/V = 4400/230 = 19.1 A. This exceeds the 16 A rating, so the circuit breaker trips, cutting the supply before the cables can overheat.
(b)(i) With the earth wire intact, the case is connected to ground through a very low resistance. When the active touches the case, a very large fault current flows through the earth wire, far above 16 A, so the circuit breaker trips almost instantly. The case is never left at a dangerous voltage.
(ii) With the earth broken, no current flows through the fault, so the breaker does not trip and the heater may appear to work normally, but its case is now at 230 V. If someone touches it, current flows through their body to earth. This current leaves through the active but does not return through the neutral, so the RCD detects an imbalance. Once it exceeds 30 mA the RCD disconnects the supply within a few hundredths of a second. The person receives a brief shock, but the current is cut off before it can cause ventricular fibrillation, which is likely above about 50 mA for longer exposures.