Electrical power, P = VI and P = I²R

The power of an appliance is the energy it transfers each second.

One watt (W) is one joule per second. For anything in a circuit:

power = potential difference × current
P = V × I

Power in watts (W), p.d. in volts (V), current in amps (A). Recall this one (AQA, Edexcel and OCR).

1. A kettle on the 230 V mains takes a current of 10 A
2. P = 230 × 10
3. P = 2300 W = 2.3 kW

A power rating in kW has to become watts before you use it: 2.3 kW = 2300 W.

Put V = I R into P = V I and you get a second power equation.

It is useful when you know the resistance instead of the p.d.:

power = (current)² × resistance
P = I² × R

Recall this one too. Only the current is squared.

1. A car rear-window heater has a resistance of 1.5 Ω and a current of 8 A
2. P = 8² × 1.5 = 64 × 1.5
3. P = 96 W

Rearranging: R = P ÷ I², and (Higher) I = √(P ÷ R).

Currents in mA must be changed to amps before squaring: 50 mA = 0.050 A, so I² = 0.0025, not 2500.

A plug fuse melts if the current gets too big.

It must be rated just above the normal current: too low and it melts every time you switch on, too high and it may not melt in a fault. UK plug fuses come in 3 A, 5 A and 13 A.

1. A vacuum cleaner is rated 920 W on the 230 V mains
2. I = P ÷ V = 920 ÷ 230 = 4 A
3. 3 A is too small, so fit the 5 A fuse

Open the Ohm's law tab. Type 230 in Voltage and 2.3 in Power with the unit set to kW. The calculator gives the current (10 A) and the resistance (23 Ω). Now change the power to 60 W. The current drops to about 0.26 A and the resistance rises to about 880 Ω. On the same supply, which has the bigger resistance: a high-power appliance or a low-power one?