Solve problems involving electrical power and energy in DC circuits, applying $P = VI = I^2 R = V^2 / R$ and electrical energy $W = P t$

What this dot point is asking

QCAA wants you to compute electrical power and energy in DC circuits, switch fluently between the three equivalent power formulas, and apply the results to household-cost problems (which use kilowatt-hours, not joules).

Power in a resistor

The rate at which electrical energy is converted to other forms (heat in a resistor, light in a bulb, mechanical work in a motor) is:

SI unit: watt (W $=$ J s$^{-1}$). For an ohmic component, $V = IR$, which gives two equivalent forms:

Use whichever has the two quantities you already know.

Energy

Energy is power multiplied by time:

SI unit: joule. In domestic context the unit is the kilowatt-hour (kWh):

A $2$ kW heater run for $1$ hour consumes $2$ kWh of energy.

Resistive heating ($I^2 R$ losses)

When current flows through a resistor, electrical energy converts to heat. The power dissipated is $P = I^2 R$. This is why transmission lines use high voltage and low current: at the same power $P = VI$, halving the current quarters the resistive losses ($I^2 R$).

How this appears in IA1 and EA

IA1

Often an oscilloscope or data-logger trace with $V$ and $I$ measured, asking for instantaneous and average power.

EA Paper 1

Multiple choice on which power formula is appropriate, and conversions between watts and kilowatt-hours.

EA Paper 2

Combined with the series-and-parallel dot point to find the power dissipated in each individual resistor of a small circuit, then total energy delivered by the battery over a stated time.

Examples in context

Example 1. A Gladstone alumina refinery worker runs a $1.8 \text{ kW}$ portable angle grinder for $30$ minutes on a $230 \text{ V}$ outlet. The current is $I = P/V = 1800/230 = 7.83 \text{ A}$, and the energy used is $W = Pt = 1800 \times 1800 = 3.24 \text{ MJ}$ or $0.90 \text{ kWh}$. At an industrial tariff of about $0.18$ dollars per kWh, the run costs around sixteen cents. The same calculation, scaled to a $250 \text{ A}$ pot-line rectifier, lets process engineers verify electricity-bill line items against operating logs to within one per cent.

Example 2. Bundaberg cane farmers use $7.5 \text{ kW}$ three-phase irrigation pumps. Over a five-hour set, energy drawn is $W = 7500 \times 18000 = 1.35 \times 10^8 \text{ J} = 37.5 \text{ kWh}$. At $0.28$ dollars per kWh and ten sets a week, the seasonal bill is significant. Because $P = I^2 R$, an aged motor whose effective resistance has crept up by $10$ per cent dissipates an extra $750 \text{ W}$ as heat at the same current draw, an example QCAA EA Unit 1 thematic items often use as the stimulus for an economic-physics calculation.

Try this

Q1. State the unit of electrical energy used by utilities and convert $1 \text{ kWh}$ to joules. [2 marks]

• Cue. $1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$.

Q2. A $2.0 \text{ kW}$ kettle is plugged into a $240 \text{ V}$ supply for $4.0$ minutes. Calculate the current, the resistance, and the energy delivered. [4 marks]

• Cue. $I = 8.33 \text{ A}$; $R = V/I = 28.8 \text{ ohms}$; $W = Pt = 4.8 \times 10^5 \text{ J}$.

Q3. A Bundaberg shed runs a $3.0 \text{ kW}$ motor with $85$ per cent electrical efficiency. (a) Calculate the input power and the heat dissipated. (b) Calculate the energy bill for $6.0 \text{ h per day, 30 days}$ at $0.30$ dollars per kWh. (c) Discuss one design change that would reduce $I^2 R$ losses in the supply cable. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $P_{in} = 3.53 \text{ kW}$, lost $0.53 \text{ kW}$; (b) $635 \text{ kWh}$ at $190.50$ dollars; (c) thicker cross-section copper to lower $R$.