Electric Power & Energy
Key idea: O Level practical electricity: heating effect, power P = VI, energy E = VIt, kWh, and electricity cost calculations.
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The core idea
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Learning objectives
- Explain electrical heating in common appliances
- Apply P = VI
- Apply E = VIt
- Calculate electrical energy and cost in kWh
- Identify the hazard from damaged insulation
- Identify the hazard from overheating cables
- Identify the hazard from damp conditions
- Explain how fuses and circuit breakers protect circuits
- Choose and justify an appropriate fuse rating
- Explain why metal casings are earthed
- Explain why double-insulated appliances do not need an earth wire
- State the meanings and roles of live, neutral and earth
- Describe the wiring of a mains plug
- Explain live-wire placement of switches, fuses and circuit breakers
1. Definition
A. Electric power
Electric power, P (W), is the rate of electrical energy transfer:
P = VI
B. Electrical energy transferred
Electrical energy, E (J), transferred in time t (s) is:
E = VIt
You should be able to describe the heating effect of electricity, use P = VI and E = VIt, and calculate electricity cost from kWh.
2. Key Ideas
- Heating effect: energy is transferred electrically and increases the internal-energy stores of the appliance and its contents.
- Power: P = VI (W), where V is in V and I is in A.
- Energy: E = VIt (J), where t is in s.
- Household energy unit: kilowatt-hour (kWh):
E(kWh) = P(kW) × t(h)
- Conversion: 1 kWh = 3.6 MJ.
- Cost:
cost = energy (kWh) × tariff ($/kWh)
3. Detailed Explanations
A. Heating effect of electricity
When current flows through a heating element, energy is transferred electrically and increases the internal-energy stores of the element and its surroundings. This is why kettles, ovens, heaters and irons get hot.
B. Power vs energy (don’t mix them up)
- Power tells you how fast energy is transferred (J per second).
- Energy depends on both power and time:
E = Pt and since P = VI, E = VIt
C. kWh and electricity bills
Electricity bills use kWh because it matches how appliances are rated (kW) and how long they run (hours).
Convert P to kW, convert time to hours, find energy in kWh, then multiply by the tariff.
4. Common Mistakes
- Using kW without converting to W (or vice versa).
- Using E = VIt but leaving t in hours instead of seconds.
- Mixing up J and kWh (they are different units of energy).
- Writing the tariff as
$/kWinstead of$/kWh.
5. Exam Tips
- Write the formula first: P = VI or E = VIt.
- Track units carefully (V, A, W, s; or kW, h, kWh).
- For cost questions: compute energy in kWh, then multiply by the tariff.
6. Worked Examples
Modelled example 1
Power from V and I
Problem
Study the worked solution
Choose the power relationship
Method
Use P = VI.Reason
Voltage and current are both given.Working
P = (230)(8.0) = 1840 WExpress in kilowatts
Method
Divide by 1000.Reason
Household ratings and billing calculations commonly use kW.Working
P = 1.84 kW
Guided practice 2
Energy used (kWh)
Problem
Match kW with hours
Hints
Hint 1: unit route
Hint 2: operation
View solution step by step
Use the billing-energy route
Method
Multiply kW by hours.Reason
The product has the household energy unit kWh.Working
E = (1.84)(2.0) = 3.68 kWh
Common misconception 3
Cost of electricity
Learner response
Find the billed energy before the cost
View solution step by step
Find energy used
Method
Multiply power by operating time.Reason
The tariff charges for energy in kWh, not power in kW.Working
E = (1.84)(2.0) = 3.68 kWhApply the tariff
Method
Multiply energy by $0.28 per kWh.Reason
The kWh units cancel, leaving dollars.Working
cost = (3.68)(0.28) = $1.03
Examiner practice 4
Energy in joules
Examination question
Use seconds with E = VIt
View solution step by step
Convert time
1 markMethod
Convert minutes to seconds.Reason
E = VIt gives joules when time is in seconds.Working
t = (5.0)(60) = 300 sCalculate energy
2 marksMethod
Substitute in E = VIt.Reason
The voltage, current and SI time are now consistent.Working
E = (12)(2.0)(300) = 7200 J
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark time conversion, substitution and result.
Challenge 5
Cost of using a heater
Billing transfer
Choose the kW–hour route
Hints
Hint 1: time unit
Hint 2: two stages
View solution step by step
Find energy in kWh
Method
Multiply the kW rating by time in hours.Reason
Tariffs are quoted per kWh.Working
E = (1.20)(0.75) = 0.90 kWhFind cost
Method
Multiply by the tariff.Reason
Each kWh costs $0.30.Working
cost = (0.90)(0.30) = $0.27
7. Mind Stretchers
Mind stretcher 1: Convert kWh to JExtension
An appliance uses 0.80 kWh of energy. Convert this to joules.
Show Answer
0.80 kWh = 0.80 × 3.6 MJ = 2.88 MJ = 2.88 × 10⁶ J
Mind stretcher 2: Comparing billsExtension
Appliance A is 2.0 kW and runs for 0.50 h. Appliance B is 0.80 kW and runs for 2.0 h. Which uses more energy?
Show Answer
Energy A: 2.0 × 0.50 = 1.0 kWh.
Energy B: 0.80 × 2.0 = 1.6 kWh.
Appliance B uses more energy.
8. Practice and next step
- For every calculation, choose either the joule route (W and s) or the billing route (kW and h) before substituting values.
- Use the Practical Electricity Structured Practice to combine P = VI, E = VIt and kWh cost questions.
- Next, learn how excess current becomes a hazard in Dangers Of Electricity.
Continue with the next resource in this course.
Course and syllabus information
- Course
- SEC G3 Physics
- Edition
- SEC G3 Physics 2027