Electrical power is P = VI, and electrical energy is E = Pt, or E = VIt. Keep one consistent unit set: watts with seconds for joules, or kilowatts with hours for kilowatt-hours.
This lesson belongs to the SPM Physics electricity section. It builds on applying potential difference, resistance and Ohm’s law.
Which unit set should I use?
Choose the set before you substitute, then keep it throughout.
| Set | Power | Time | Energy |
|---|---|---|---|
| Physics (SI) | watt (W) | second (s) | joule (J) |
| Household | kilowatt (kW) | hour (h) | kilowatt-hour (kWh) |
Worked example: a kettle
A 2.0 kW kettle runs for 3 minutes. Find the energy used in joules and in kWh.
In joules. P = 2.0 kW = 2 000 W. t = 3 × 60 = 180 s. E = Pt = 2 000 × 180 = 360 000 J, which is 360 kJ.
In kilowatt-hours. t = 3 ÷ 60 = 0.05 h. E = 2.0 × 0.05 = 0.10 kWh.
The two answers agree: 0.10 kWh × 3 600 000 = 360 000 J. If the tariff in a question is RM0.40 per kWh, the cost is 0.10 × 0.40 = RM0.04.
The mistake that costs marks
The common slip is to mix unit sets: multiplying 2 000 W by 3 minutes and writing 6 000 J.
| Step | Wrong | Right |
|---|---|---|
| Time | 3 (minutes) | 180 s |
| Power | 2 000 W | 2 000 W |
| Energy | 2 000 × 3 = 6 000 J | 2 000 × 180 = 360 000 J |
The wrong answer is 60 times too small. A watt is a joule per second, so the time must be in seconds.
Check yourself
A 240 V appliance draws a current of 5.0 A for 2 hours. Find its power in kW and the energy used in kWh.
Answer
P = VI = 240 × 5.0 = 1 200 W = 1.2 kW.
E = Pt = 1.2 × 2 = 2.4 kWh.
Check: 1 200 W for 7 200 s gives 8 640 000 J, and 2.4 × 3 600 000 = 8 640 000 J. The two agree.
What to study next
Energy and power also appear in explanation questions with tables. The electrical reasoning from diagrams and measurements section covers those. Then work the electricity practice set. The units and significant figure checker checks units and rounding on your answers.
For a teacher to go through unit pairing with you, see online one-to-one Physics tuition.