The problems are on the electrical power worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electrical power is the rate at which an appliance transfers energy, measured in watts. These worked answers correct the 20 problems of the grade 8 science worksheet on electrical power, from watts and kilowatts to motor at three settings: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. Check that you changed minutes to hours or seconds, and watts to kilowatts, before multiplying, because that step causes most lost marks.
Level: Grade 8 science (Year 9 · Class 8) · Curriculum: MS-PS3
Problem 1 Watts and kilowatts
1. The unit of power is the watt (W).2. 3.5 × 1000 = 3500 W.
3. 450 ÷ 1000 = 0.45 kW.
Problem 2 Two power formulas
1. P = V × I, with P in watts, V in volts and I in amperes.2. E = P × t, with E in joules, P in watts and t in seconds.
3. The bill uses the kilowatt-hour (kWh).
Problem 3 Torch bulb
1. P = V × I.2. P = 6.0 V × 0.50 A = 3.0 W.
Problem 4 Hair dryer energy
1. 2 × 60 = 120 s.2. E = P × t = 1500 W × 120 s = 180 000 J.
Problem 5 Oven in kilowatt-hours
1. E = P × t, with P in kW and t in hours.2. E = 3 kW × 2 h = 6 kWh.
Problem 6 Read the rating plate
1. The power is 1200 W.2. 1200 ÷ 1000 = 1.2 kW.
3. E = 1.2 kW × 0.5 h = 0.6 kWh.
Problem 7 Rice cooker current
1. I = P ÷ V.2. I = 700 W ÷ 230 V = 3.04… A, so 3.0 A.
Problem 8 Heater element voltage
1. V = P ÷ I.2. V = 1840 W ÷ 8.0 A = 230 V.
Problem 9 A fan overnight
1. 40 W = 0.040 kW, so E = 0.040 kW × 5 h = 0.20 kWh.2. 5 h = 18 000 s, so E = 40 W × 18 000 s = 720 000 J.
3. 0.20 kWh × 3 600 000 J/kWh = 720 000 J, so the answers agree.
Problem 10 Cost of ironing
1. E = 1.5 kW × 2 h = 3.0 kWh.2. Cost = 3.0 × 1.20 = 3.60 cedis.
Problem 11 Reading the meter
1. 7470 − 7408 = 62 kWh.2. Cost = 62 × 18 = 1116 rupees.
Problem 12 LED against filament lamp
1. Time = 5 × 30 = 150 h. LED: 0.009 kW × 150 h = 1.35 kWh. Filament lamp: 0.060 kW × 150 h = 9.0 kWh.2. 9.0 − 1.35 = 7.65 kWh.
3. Saving = 7.65 × 20 p = 153 p (£1.53).
Problem 13 How long will it run?
1. t = E ÷ P.2. t = 10 kWh ÷ 2.0 kW = 5.0 h.
Problem 14 Weekly energy chart
1. The fridge (10 kWh).2. 4 + 7 + 10 + 3 = 24 kWh.
3. Cost = 24 × $0.14 = $3.36.
4. 7 ÷ 24 × 100 = 29.2…, so 29 %.
Problem 15 Fast kettle, slow kettle
1. Power is energy per second. The 2300 W kettle transfers more joules every second, so it reaches the same total energy sooner.2. Time = E ÷ P, so the ratio is 2300 ÷ 800 = 2.9 times longer.
3. A joule is tiny, so household use would be huge numbers (millions of joules). The kWh is a convenient size for a bill.
Problem 16 Air conditioner season
1. Time = 8 × 30 = 240 h. E = 1.2 × 240 = 288 kWh.2. Cost = 288 × 0.30 = S$86.40.
3. New unit: 0.9 × 240 = 216 kWh. Saved energy = 288 − 216 = 72 kWh, so saving = 72 × 0.30 = S$21.60.
Problem 17 Choosing a fuse
1. Lamp: 60 ÷ 230 = 0.26 A. Toaster: 920 ÷ 230 = 4.0 A. Heater: 2400 ÷ 230 = 10.4 A.2. Lamp: 3 A. Toaster: 5 A. Heater: 13 A.
3. If a fault makes the current rise, a fuse that is too large would not melt in time, so the wires could overheat and start a fire.
Problem 18 Solar panel savings
1. P = 18 V × 5.0 A = 90 W.2. E = 0.090 kW × 6 h = 0.54 kWh.
3. Per day: 0.54 × 60 = 32.4 naira. In 30 days: 32.4 × 30 = 972 naira.
Problem 19 Spot the errors
1. The time must be in hours, not minutes. 90 min = 1.5 h, so E = 2 × 1.5 = 3.0 kWh.2. 30 × 2 is an energy (60 Wh), not a power; the power stays 30 W. E = 0.030 kW × 2 h = 0.060 kWh.
3. Cost = 3.0 × 22 = 66 p.
Problem 20 Motor at three settings
1. 3.0 × 0.20 = 0.60 W; 6.0 × 0.45 = 2.7 W; 9.0 × 0.80 = 7.2 W.2. 2.7 ÷ 0.60 = 4.5 times (the voltage only doubled).
3. E = 7.2 W × 20 s = 144 J.
4. Power is V × I. When the voltage rises, the current also rises, so both factors grow and the product grows faster than either.
