Power is energy transferred per second. Electrical power is P= VI; for a resistor, substituting V= IR gives P=I2R=RV2. Select a formula using quantities for the same component. Electrical energy is E= Pt, with seconds when power is in watts and energy in joules. A kilowatt-hour measures energy: 1kWh=3.6×106J.
Efficiency is useful output energy divided by total input energy, or useful output power divided by input power. State the chosen useful output. Multiply the fraction by 100 for a percentage. For a simple energy tariff, multiply kWh by price per kWh; fixed supply charges are separate. Compare devices on the same delivered service, since a low power rating alone does not guarantee low total energy use.
Example 1
A 12V motor draws 2.0A and delivers 18W mechanically. Find input power and efficiency.
Pin=VI=24W. Efficiency =2418=0.75=75%. The remaining 6W goes to other outputs such as heating.
Example 2
A 1.5 kW heater runs for 2.0h at $0.30 per kWh. Find energy cost, excluding supply charges.
Energy =1.5×2.0=3.0 kWh. Cost =3.0× $0.30= $0.90.
Watch out:Power multiplied by time is energy. Kilowatts measure power; kilowatt-hours measure energy.
Electrical power, energy and efficiency practice | Year 11 Physics | Ascent Arcade