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Power and energy resources: original mixed practice with explanations

Practice only helps if you can see exactly why an answer is right, so every question here is worked in full.

These eleven questions cover power, energy in kWh, efficiency, generation data, availability and reliability, and trade-off reasoning from power and energy resources. All data here is invented for practice and is not taken from any exam paper. They go from easier to harder.

Write your answer with its unit on paper first, then open the answer. Use g = 10 N/kg where gravity is needed. Conversions: 1 kWh = 3.6 MJ, 1 min = 60 s, 1 h = 3600 s.

Questions

1. A heater transfers 90 kJ of energy in 45 s. Calculate its power.

Show answer

90 kJ = 90 000 J. P = E ÷ t = 90 000 ÷ 45 = 2000 W (2.0 kW).

2. A 100 W lamp is on for 2.0 hours. Find the energy transferred in kWh and in joules.

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100 W = 0.100 kW. E = 0.100 × 2.0 = 0.20 kWh. In joules: 100 × 7200 s = 720 000 J (0.72 MJ). Check: 0.20 × 3.6 MJ = 0.72 MJ.

3. A 1.2 kW hairdryer must transfer 540 kJ. How long does this take, in seconds and minutes?

Show answer

1.2 kW = 1200 W. t = E ÷ P = 540 000 ÷ 1200 = 450 s. 450 ÷ 60 = 7.5 minutes.

4. A 2.0 kW heater runs for 3.0 hours. Electricity costs RM0.30 per kWh (an invented price for this question). Find the cost.

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Energy = 2.0 × 3.0 = 6.0 kWh. Cost = 6.0 × 0.30 = RM1.80.

5. A crane lifts an 800 N load through 15 m in 20 s. Its input power is 1000 W. Find (a) the useful power and (b) the efficiency.

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(a) Useful work = 800 × 15 = 12 000 J. Useful power = 12 000 ÷ 20 = 600 W.

(b) Input energy = 1000 × 20 = 20 000 J. Efficiency = 12 000 ÷ 20 000 = 0.60 = 60%.

6. An LED lamp (10 W) and a filament lamp (50 W) give the same light for 5 hours a day over 30 days. Find the difference in energy used, in kWh.

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Time = 5 × 30 = 150 h. LED: 10 × 150 = 1500 Wh = 1.5 kWh. Filament: 50 × 150 = 7500 Wh = 7.5 kWh. Difference = 7.5 − 1.5 = 6.0 kWh.

7. A wind farm with an installed capacity of 2.5 MW has these average outputs in four 3-hour blocks: 1.0 MW, 2.0 MW, 0.5 MW and 1.5 MW. Find (a) the total energy in MWh, (b) the average power over the 12 hours and (c) that average as a percentage of capacity.

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(a) Energy = 1.0×3 + 2.0×3 + 0.5×3 + 1.5×3 = 3 + 6 + 1.5 + 4.5 = 15 MWh.

(b) Average power = 15 ÷ 12 = 1.25 MW.

(c) 1.25 ÷ 2.5 = 0.50 = 50%.

8. For each case, say whether it is mainly a question of availability of the resource or reliability of the source, and explain briefly. (a) A reservoir is low after a long dry spell. (b) A gas-fired station with plenty of fuel is offline after a fault. (c) A solar farm produces no output at night when demand is high.

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(a) Availability: less water is stored, so less of the resource is present.

(b) Reliability: the fuel is available, but the source cannot deliver power.

(c) Sunlight is not available at night, which means the solar farm is not reliable for night demand unless energy is stored or another source supplies it. Both words apply, each to a different part of the statement.

9. Water flows through a hydroelectric station at 1000 kg per second and falls through 50 m. Calculate (a) the power of the falling water and (b) the electrical power output if the station is 80% efficient.

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(a) Energy per second = m g h = 1000 × 10 × 50 = 500 000 J each second, so the power is 500 kW.

(b) 0.80 × 500 = 400 kW.

10. A town needs 600 kWh each day. Solar panels give 0.8 kWh per m² per day on average. (a) Find the area needed. (b) In a cloudy week they give 0.4 kWh per m² per day. Find the new area. (c) State one assumption that your answer depends on.

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(a) 600 ÷ 0.8 = 750 m².

(b) 600 ÷ 0.4 = 1500 m², double the area because the yield is halved.

(c) For example: “the demand stays at 600 kWh each day”, or “the panel yield is the stated average”. A conclusion using this should begin “If …”.

11. An electric shower is rated at 3.0 kW and runs for 8.0 minutes. Find the energy transferred in kJ, in MJ and in kWh.

Show answer

8.0 min = 480 s. E = 3000 × 480 = 1 440 000 J = 1440 kJ = 1.44 MJ. In kWh: 3.0 × (8.0 ÷ 60) = 3.0 × 0.1333 = 0.40 kWh. Check: 0.40 × 3.6 = 1.44 MJ.

If you got these wrong

What went wrongGo to
Answer was a factor of 60 or 1000 out (Q1, Q2, Q3, Q11)Calculate power from energy and time
Compared power ratings instead of energy, or mixed useful and input power (Q4, Q5, Q6)Compare two devices over an equal task
Added powers instead of multiplying by time (Q7, Q9)Read supplied generation data
Mixed up availability, reliability or renewable (Q8)Distinguish resource availability from reliability
Gave an opinion instead of a conditional conclusion (Q10)Explain a trade-off using stated assumptions
Energy and efficiency steps unclear (Q5, Q9)Work, energy and efficiency

Record each repeated slip in the mistake log and retest queue and try a fresh question on it a few days later.

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