Electrical power: power as energy transferred per second, the relationships linking power to current, voltage and resistance, and using power to find the energy and cost of running an appliance.

What this key area is asking

The SQA wants you to define electrical power as energy transferred per second, select and use the power relationships ($P = IV$, $P = I^2R$, $P = \frac{V^2}{R}$ and $P = \frac{E}{t}$), and use them to find the energy used and the cost of running an appliance.

Power as energy per second

The three power relationships

All three give the electrical power, but each uses a different pair of quantities. Use $P = IV$ when you know the current and voltage; use $P = I^2R$ when you know the current and resistance; use $P = \frac{V^2}{R}$ when you know the voltage and resistance. They are linked by Ohm's law ($V = IR$): substituting $V = IR$ into $P = IV$ gives $P = I^2R$, and substituting $I = \frac{V}{R}$ gives $P = \frac{V^2}{R}$.

The cost of electricity

Electricity bills are charged per kilowatt-hour (kWh), the energy used by a $1 \text{ kW}$ appliance in $1$ hour. To find the cost: work out the energy used in kilowatt-hours (power in kW multiplied by time in hours), then multiply by the price of one unit. For example, a $2 \text{ kW}$ heater run for $3$ hours uses $2 \times 3 = 6 \text{ kWh}$; at $30 \text{p}$ per unit that costs $6 \times 30 = 180 \text{p}$, or £$1.80$.

The kilowatt-hour is a unit of energy, not power, even though it contains the word "watt". It is used on bills because the joule is far too small for household amounts of energy: one kWh is $3\,600\,000 \text{ J}$. Appliances that transfer a lot of energy each second (high power) and run for a long time cost the most, which is why heaters, kettles and tumble dryers dominate an electricity bill while a phone charger costs almost nothing.

Power ratings and fuses

Appliances carry a power rating (for example "$2300 \text{ W}$, $230 \text{ V}$") that tells you how much energy they transfer each second at the stated voltage. From the rating you can work out the normal operating current using $P = IV$, rearranged to $I = \frac{P}{V}$. This current is what decides the correct fuse: the fuse must be rated just above the normal current so that it melts and breaks the circuit if a fault makes the current rise dangerously.

Try this

Q1. A lamp operates at $12 \text{ V}$ and draws $2.0 \text{ A}$. Calculate its power. [2 marks]

• Cue. $P = IV = 12 \times 2.0 = 24 \text{ W}$.

Q2. A $1500 \text{ W}$ heater runs for $40 \text{ s}$. Calculate the energy used. [2 marks]

• Cue. $E = Pt = 1500 \times 40 = 60\,000 \text{ J}$.

Q3. A current of $3.0 \text{ A}$ flows through a $10 \text{ }\Omega$ resistor. Calculate the power dissipated. [2 marks]

• Cue. $P = I^2R = (3.0)^2 \times 10 = 90 \text{ W}$.