Understanding Electrical Power: The Complete GCSE & A-Level Guide
Electrical power is the rate at which electrical energy is transferred in a circuit. It is one of the most fundamental concepts in GCSE and A-Level physics, and understanding all three power formulae — along with Ohm's Law — allows you to solve virtually any resistive circuit problem.
The Three Power Formulae
All three formulae are equivalent and can be derived from each other using Ohm's Law (V = IR):
- P = IV — Use when you know current and voltage.
- P = I²R — Use when you know current and resistance.
- P = V²/R — Use when you know voltage and resistance.
The unit of power is the Watt (W). One watt equals one joule per second (1 W = 1 J/s). For larger values, 1 kilowatt (kW) = 1,000 W, and 1 megawatt (MW) = 1,000,000 W.
Ohm's Law Connection: V = IR
Ohm's Law states that the voltage across a resistor is proportional to the current through it: V = IR. This is the bridge between the three power formulae. If you know any two of P, I, V, R, you can calculate the other two.
Power of Common Household Appliances
| Appliance | Typical Power | Current at 230 V |
|---|
| Electric shower | 7,000–10,000 W | 30.4–43.5 A |
| Electric kettle | 2,000–3,000 W | 8.7–13.0 A |
| Microwave oven | 800–1,200 W | 3.5–5.2 A |
| Hair dryer | 1,200–2,400 W | 5.2–10.4 A |
| Laptop computer | 45–65 W | 0.2–0.28 A |
| LED light bulb | 9 W | 0.04 A |
| Desktop PC | 200–400 W | 0.87–1.74 A |
| Flat screen TV (50") | 70–150 W | 0.3–0.65 A |
Energy Calculation: E = Pt
Energy transferred = Power × Time. For electricity bills, energy is measured in kilowatt-hours (kWh). To convert: Energy (kWh) = Power (kW) × Time (hours). The cost is then: Cost (p) = Energy (kWh) × unit rate (p/kWh).
Three-Phase Power
In commercial and industrial settings, three-phase AC power is used. The formula is: P = √3 × VL × IL × cos φ, where VL is line voltage, IL is line current, and cos φ is the power factor. The √3 factor (≈ 1.732) arises from the 120° phase difference between the three supply phases.
Power Factor in AC Circuits
In AC circuits containing inductors or capacitors, the current and voltage are not in phase. The power factor (PF = cos φ) accounts for this. Real power (W) = Apparent power (VA) × Power factor. A low power factor means much of the apparent power is wasted as reactive power. Industrial consumers are charged for low power factor, which is why power factor correction capacitors are widely used.
Worked GCSE Example 1
Q: A hairdryer operates at 230 V and draws a current of 8 A. Calculate the power.
- Formula: P = IV
- P = 8 A × 230 V
- P = 1,840 W = 1.84 kW
Worked GCSE Example 2
Q: A resistor of 47 Ω has a current of 200 mA flowing through it. Calculate the power dissipated.
- Convert: 200 mA = 0.2 A
- Formula: P = I²R
- P = (0.2)² × 47 = 0.04 × 47
- P = 1.88 W
Worked A-Level Example
Q: A 60 W light bulb is rated at 230 V. Calculate its resistance when operating at full power.
- Formula: P = V²/R → R = V²/P
- R = (230)² / 60 = 52,900 / 60
- R = 881.7 Ω
- Current: I = P/V = 60/230 = 0.261 A
Frequently Asked Questions
What is the formula for electrical power?
The primary formula is P = IV, where P is power in watts (W), I is current in amperes (A), and V is voltage in volts (V). Two alternative forms derived via Ohm's Law are P = I²R (when resistance is known instead of voltage) and P = V²/R (when current is not known). All three give identical results for the same circuit conditions.
How do I calculate power if I only know voltage and resistance?
Use the formula P = V²/R. For example, a 12 V supply connected across a 6 Ω resistor: P = 12² ÷ 6 = 144 ÷ 6 = 24 W. You can verify this using Ohm's Law first — I = V/R = 12/6 = 2 A — then P = IV = 2 × 12 = 24 W. Both methods confirm the same answer.
What is the difference between watts and kilowatts?
1 kilowatt (kW) = 1,000 watts (W), and 1 megawatt (MW) = 1,000,000 W. On electricity bills, energy is measured in kilowatt-hours (kWh). A 1 kW appliance running for 1 hour uses 1 kWh. The average UK electricity unit rate is approximately 24–27p per kWh (2025 figures), so 1 kWh costs roughly 24–27 pence.
How do I calculate electricity cost from power?
Follow three steps: (1) Convert power to kW: divide watts by 1,000. (2) Calculate energy: Energy (kWh) = Power (kW) × Time (hours). (3) Cost = Energy (kWh) × unit rate (p/kWh). Example: A 2,500 W kettle used for 5 minutes (= 5/60 hours ≈ 0.0833 h) at 25p/kWh: Energy = 2.5 × 0.0833 = 0.208 kWh. Cost = 0.208 × 25 = 5.2p.
What is power factor and why does it matter?
Power factor (PF) is only relevant in AC circuits and equals cos φ, where φ is the phase angle between voltage and current. A PF of 1.0 means all supplied power does useful work (purely resistive load). Inductive loads (motors, transformers) and capacitive loads reduce the PF below 1.0, meaning more current must be supplied for the same useful power output. This increases energy losses in cables. Industrial premises pay penalties for low power factor, which is why correction capacitors are installed.
How is three-phase power different from single-phase?
Single-phase power uses one live wire and a neutral: P = VI cos φ. Three-phase uses three live wires at 120° phase separation: P = √3 × VL × IL × cos φ, where VL is the line-to-line voltage (typically 400 V in the UK) and IL is the line current. Three-phase delivers more power for the same conductor size, so it is used for industrial machinery, large buildings, and the National Grid. The UK domestic supply is single-phase 230 V.
Why does a hairdryer on the same circuit as a kettle trip the fuse?
Each appliance draws significant current. A 2 kW kettle at 230 V draws I = P/V = 2000/230 ≈ 8.7 A. A 2 kW hairdryer draws another ≈ 8.7 A. Total ≈ 17.4 A, which exceeds the 13 A rating of a standard UK plug fuse, or approaches the 20 A limit of a ring-main circuit. Fuses and circuit breakers protect the wiring from overheating (P = I²R losses in cables). Always check the combined current load on any circuit.