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S3 Physics: Household Electricity, Power and Cost (10 Advanced Items)
S3 Physics: Household Electricity, Power and Cost (10 Advanced Items)
Ten advanced items: several appliances over many days, fixed charges on a bill, current and fuse choice, savings from LED replacement.
Focus Combine the power first, then multiply by hours and days; add fixed charges separately.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 30 |
| Difficulty mix | Intermediate 10 |
| Chapter | Chapter: S3 Physics: Household Electricity, Power and Cost |
| Level | S3 |
02 Questions (10 items)
- IntermediateFour 60 W lamps run 5 hours a day for 30 days. Find the total energy in kWh.
- IntermediateAppliance A is 2,000 W for 0.5 hour; B is 500 W for 1.5 hours. Which uses more energy and by how much?
- IntermediateA month uses 320 units at $1.2 per unit plus a $20 fixed charge. Find the total bill.
- IntermediateA 3 kW water heater runs 1.5 hours a day for 30 days at $1.2 per unit. Find the energy and the cost.
- IntermediateA 2,000 W appliance runs on 220 V. Find the current and say whether a 10 A fuse is suitable.
- IntermediateReplacing a 60 W bulb with a 9 W LED for 5 hours a day over 30 days: how much energy and money is saved at $1.2 per unit?
- IntermediateHow long does 1.5 kWh take with a 1,200 W appliance? And with 600 W?
- IntermediateConvert 2.5 kWh into joules.
- IntermediateA 5 W standby light runs 24 hours a day for 30 days at $1.5 per unit. Find the energy and cost.
- IntermediateA 13 A, 220 V socket already supplies 800 W. Can a 2,200 W appliance be added?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 4 × 60 W = 240 W; 240 × 5 = 1,200 Wh = 1.2 kWh a day; 1.2 × 30 = 36 kWh. | Combine the power first, then multiply by hours and by days. |
| 2 | A: 2,000 × 0.5 = 1 kWh; B: 500 × 1.5 = 0.75 kWh; A uses 0.25 kWh more. | High power does not always mean more energy; compare power × time. |
| 3 | 320 × $1.2 = $384, plus $20 gives $404. | The fixed charge does not depend on usage, so add it separately. |
| 4 | 3 × 1.5 = 4.5 kWh a day; × 30 = 135 kWh; 135 × $1.2 = $162. | kW × hours gives kWh directly; multiply by the tariff for the cost. |
| 5 | I = 2,000 ÷ 220 ≈ 9.09 A; 10 A is too close to the limit, so a 13 A fuse should be used. | A fuse should sit a little above the normal current; 9.09 A on a 10 A fuse will keep blowing. |
| 6 | Saves 51 W; 51 × 5 = 255 Wh a day = 0.255 kWh; × 30 = 7.65 kWh; × $1.2 = $9.18. | What is saved is the difference in power, not the whole 60 W. |
| 7 | 1,500 Wh ÷ 1,200 W = 1.25 h; 1,500 ÷ 600 = 2.5 h. | For the same energy, halving the power doubles the time. |
| 8 | 2.5 × 3.6 × 10⁶ = 9,000,000 J. | One kWh is 3.6 × 10⁶ J, so multiply by the number of units. |
| 9 | 5 × 24 = 120 Wh a day = 0.12 kWh; × 30 = 3.6 kWh; × $1.5 = $5.4. | Standby power is small but adds up over a long time. |
| 10 | Limit = 220 × 13 = 2,860 W; 800 + 2,200 = 3,000 W, which exceeds it, so no. | Compare the total power with the limit; exceeding it trips the circuit or blows the fuse. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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