P = E / t, so E = P × t

• Unit of power watt (W) = 1 joule per second. "A 9 W LED lamp" uses 9 joules of electrical energy every second.
• The joule is a very small unit of energy, so in household billing the larger unit kilowatt-hour (kW·h) is used:
• 1 kW·h = 1000 W × 3600 s = 3.6 × 10⁶ J.

1. Power in kW = power in W ÷ 1000.
2. Find the number of operating hours.
3. Energy in kW·h = power in kW × hours.
4. Cost = energy in kW·h × price per kW·h.

• (The price per kW·h used in the book’s example is 1.55 Egyptian pounds.
• For your own calculations use the price given in the question.)

Worked example 7: comparing two lamps

• A CFL of 20 W and an LED of 12 W give about the same light.
• Compare the monthly (30-day) costs if each is used 12 h per day at 1.55 pounds per kW·h.
• Hours = 30 × 12 = 360 h.
• CFL 0.020 kW × 360 = 7.2 kW·h; cost = 7.2 × 1.55 = 11.16 pounds.
• LED 0.012 × 360 = 4.32 kW·h; cost = 4.32 × 1.55 = 6.70 pounds.
• Saving = about 4.46 pounds per month per lamp. The LED is more efficient.

Worked example 8: a kettle

• A 1500 W kettle is used 0.5 h per day. What is the weekly energy and cost at 1.55 pounds per kW·h?
• P = 1.5 kW. Hours per week = 0.5 × 7 = 3.5 h. Energy = 1.5 × 3.5 = 5.25 kW·h. Cost = 5.25 × 1.55 = 8.1375, about 8.14 pounds.

Worked example 9: equal energy

• Appliance X has power 2000 W, appliance Y has 1000 W.
• X runs for 3 hours.
• How long must Y run to use the same energy?
• E = 2 kW × 3 h = 6 kW·h; time for Y = 6 / 1 = 6 hours.
• The lower the power, the longer it takes to use the same energy.

• Typical powers (for reference) electric heater 1–2 kW; iron 1–2.5 kW; air conditioner 1.5–2.5 kW; laptop about 50 W; fan 50–100 W; television about 100 W.
• Heating appliances and air conditioners consume the most energy.
• Increasing the operating hours (cause) increases the energy and the cost (effect).