Electric circuits: Ohm’s law, resistors and the cost of electricity
الأسبوع 8 · الدرس الثاني «الدوائر الكهربية»: قانون أوم وتوصيل المقاومات على التوالي وعلى التوازي والقدرة والطاقة الكهربية وتكلفة الاستهلاك المنزلي والسلامة · 3 دروس
3
Power, energy and the cost of electricity
Science Prep 3 Term 1, Unit 2: pp. 46–53
GoalCalculate electric power, energy in joules and in kW·h, and the cost of electricity with an assumed price, and name basic safety rules.
New words
electric power (P) · the electric energy used each second, P = E/t; unit watt (W) = joule per secondkilowatt-hour (kW·h) · the energy used by a 1000 W device in one hour: 3.6 × 10⁶ Jenergy efficiency label · a label on an appliance that shows its monthly energy use and a rating, the top class being the most efficient
Explanation
Electric power is the energy a device uses each second: P = E ÷ t. Its unit is the watt, 1 W = 1 J/s, so a 9 W bulb uses 9 J every second. The energy used is E = P × t in joules, with t in seconds. Because the joule is very small, electricity bills use the kilowatt-hour: 1 kW·h = 1000 W × 3600 s = 3.6 × 10⁶ J. To get kW·h, multiply the power in kW by the time in hours; the cost is then the kW·h times the price per kW·h. Careful: kW is power, kW·h is energy, and the price here is always an assumption; real prices vary, so we write “assume a price of …”. LED bulbs use less energy than CFLs for the same light, so they cost less to run. Heaters, irons and air conditioners use the most energy. An energy efficiency label shows the monthly consumption and a rating. Safety: do not overload a socket, keep wires insulated, keep hands and tools dry, switch the source off before changing anything, and never use household sources in school experiments. Never open an appliance.
Quantity
Formula
Unit
power
P = E ÷ t
W = J/s
energy
E = P × t
J
energy for bills
kW × h
kW·h
cost
kW·h × price
pounds
Worked examples
A 100 W lamp is on for 30 s. E = P × t = 100 W × 30 s = 3000 J. A 9 W bulb in 1 minute uses 9 W × 60 s = 540 J.
An 800 W iron is used for 2 h. Power = 0.8 kW. Energy = 0.8 kW × 2 h = 1.6 kW·h. Assume a price of 2 pounds per kW·h: cost = 1.6 × 2 = 3.20 pounds.
Compare an LED lamp and an old filament lamp that give the same light, both on 5 h a day for 20 days (100 h). Assume 9 W for the LED and 60 W for the filament lamp (invented values). LED: 0.009 kW × 100 h = 0.9 kW·h. Filament: 0.060 kW × 100 h = 6 kW·h. At an assumed 2 pounds per kW·h: LED 1.80 pounds, filament 12.00 pounds. The LED saves 10.20 pounds.
1 kW·h in joules: 1000 W × 3600 s = 3 600 000 J = 3.6 × 10⁶ J.
Class activity
Bill game, with a grown-up. Read the label of one appliance together, without opening it, and copy its power in watts. Invent how many hours it is used in a month and calculate the energy in kW·h. Assume a price of 2 pounds per kW·h and calculate the cost. Do the same for two other appliances and add a made-up monthly bill for one family. Say that the label values and the price are only examples. Give one safety rule for each appliance.
Practice
1
What does a 12 W bulb use in 5 s?
E = 12 W × 5 s = 60 J.
2
A 2 kW heater runs for 3 h. Find the energy and, assuming 1.50 pounds per kW·h, the cost.
E = 2 kW × 3 h = 6 kW·h. Cost = 6 × 1.50 = 9.00 pounds.
3
A 60 W fan is on for 10 h. Find the energy in kW·h and the cost at an assumed 2 pounds per kW·h.