Define electrical power and energy, apply the power equations, and calculate energy transferred and the cost of running a device

What this dot point is asking

SEAB wants you to define electrical power and electrical energy, to use the three forms of the power equation, and to calculate the energy a device transfers and the cost of running it. The central insight is that power is the rate of transferring energy, so once you know the power you find the energy simply by multiplying by time. The power equation has three forms because Ohm's law lets you swap voltage and current for each other.

The answer

Power is the rate of energy transfer

Electrical power is the rate at which a component converts electrical energy into other forms such as heat, light or motion. The basic equation is:

where $P$ is power in watts ($\text{W}$), $V$ is voltage in volts and $I$ is current in amperes. One watt is one joule per second.

The three forms of the power equation

Because Ohm's law gives $V = IR$, you can substitute to get two more useful forms:

Choose the form that uses the quantities you are given. If you know the current and resistance, use $I^2 R$; if you know the voltage and resistance, use $V^2/R$; if you know voltage and current, use $VI$.

Energy is power times time

Electrical energy transferred is the power multiplied by the time for which it flows:

With power in watts and time in seconds, the energy comes out in joules. This is the same energy you would get from $E = VIt$, since $P = VI$.

The kilowatt-hour

The joule is a small unit, so household energy is measured in kilowatt-hours ($\text{kWh}$). One kilowatt-hour is the energy used by a $1\ \text{kW}$ device running for one hour. To find the cost of running a device:

Keep the power in kilowatts and the time in hours, and you get the energy directly in kilowatt-hours ready to multiply by the tariff.

Examples in context

Example 1. Choosing a resistor's power rating. A resistor is not only specified by its resistance but by how much power it can safely dissipate. If a circuit forces $0.5\ \text{A}$ through a $20\ \Omega$ resistor, it dissipates $I^2 R = 0.25 \times 20 = 5\ \text{W}$, so a quarter-watt resistor would overheat and a $5\ \text{W}$ or larger rating is needed. Power calculations protect components from burning out.

Example 2. Comparing two kettles. A $3\ \text{kW}$ kettle boils water faster than a $2\ \text{kW}$ kettle because it transfers energy at a higher rate. For the same final energy delivered to the water, the more powerful kettle simply takes less time, since $E = Pt$ means a larger $P$ needs a smaller $t$. Power decides speed; energy decides the bill.

Try this

• Cue. A lamp is rated $12\ \text{V}$, $24\ \text{W}$. Find the current it draws. Rearrange $P = VI$ to $I = P/V = 24/12 = 2.0\ \text{A}$.

• Cue. A $1500\ \text{W}$ iron runs for $30\ \text{minutes}$. Find the energy used in kilowatt-hours. Convert: power $= 1.5\ \text{kW}$, time $= 0.5\ \text{h}$, so $E = 1.5 \times 0.5 = 0.75\ \text{kWh}$.

• Cue. Explain why a thick supply cable is used for a high-power appliance. High power means high current, and a thick cable has low resistance so it dissipates little $I^2 R$ heat and does not overheat.