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 ($I^2 R$).

Worked example

A household has the following appliances running for the stated times: a $200$ W television for $4$ h, a $3000$ W heater for $2$ h, and a $100$ W fridge running $24$ h continuously. The electricity tariff is $0.30$ dollars per kWh. Calculate the daily cost.

Television: $0.200$ kW $\times 4$ h $= 0.80$ kWh.

Heater: $3.0$ kW $\times 2$ h $= 6.0$ kWh.

Fridge: $0.100$ kW $\times 24$ h $= 2.4$ kWh.

Total: $0.80 + 6.0 + 2.4 = 9.2$ kWh.

Cost: $9.2 \times 0.30 = 2.76$ dollars per day.

Common traps

Confusing power with energy. A $100$ W bulb has a power rating. To get energy used, multiply by the time. The same $100$ W bulb running $1$ h uses $360\,000$ J; running $10$ h uses $3.6 \times 10^6$ J.

Mixing watts and kilowatts. A $2400$ W kettle is $2.4$ kW, not $2400$ kW. Convert at the start.

Using the wrong power formula for non-ohmic devices. $P = VI$ always works. $P = I^2 R$ and $P = V^2 / R$ assume Ohm's law holds. For a diode or a lamp at high current, use $P = VI$ from measured operating values.

Forgetting the factor of $3600$ in kWh to joule conversion. $1$ kWh $= 3.6 \times 10^6$ J, not $3600$ J.

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.

In one sentence

Electrical power is $P = VI$ in any circuit and equivalently $P = I^2 R = V^2 / R$ in an ohmic resistor, with SI units of watts, and electrical energy is $W = P t$ measured in joules in physics problems but in kilowatt-hours on household bills ($1$ kWh $= 3.6 \times 10^6$ J).