2.54PH1 · Section 2: Electricity

Electrical power and energy

Key learnings

Power: P = IV, P = I²R, P = V²/R
Power = rate of energy transfer; $P = E/t$ ; unit watt (W), where 1 W = 1 J/s. For a component the primary form is $P = IV$ ( $P$ in W, $I$ in A, $V$ in V). Substituting $V = IR$ gives $P = I^2R$ ; substituting $I = V/R$ gives $P = V^2/R$ . Pick the form matching what you are given: know $I$ and $R$ → $P = I^2R$ ; know $V$ and $R$ → $P = V^2/R$ . All three are on the formula sheet. Mini-example: 0.5 A at 12 V → $P = IV = 6$ W.
Energy E = IVt = Pt; the kWh
Energy transferred: $E = IVt = Pt$ , with $E$ in joules (J), $I$ in A, $V$ in V and time $t$ in seconds. The kilowatt-hour (kWh) is the domestic energy unit: $1$ kWh $= 1000$ W $× 3600$ s $= 3\,600\,000$ J $= 3.6$ MJ. In kWh, $E = P_{\text{kW}} × t_{\text{hours}}$ , and cost $=$ energy (kWh) $×$ price per unit. A "unit" on a bill means 1 kWh, not 1 joule. Mini-example: a 2 kW appliance for 3 h uses $2 × 3 = 6$ kWh.
Power definition; choosing a fuse
Power = rate of energy transfer; unit watt (W). Two elements: (1) rate of energy transfer, (2) unit W — "energy" alone is wrong, power is energy per unit time. Fuse choice uses $I = P/V$ : a 2 kW (2000 W) kettle on 230 V draws $I = 2000/230 ≈ 8.7$ A, so pick a 13 A fuse (the next standard rating above the working current). A fuse below the working current blows needlessly; one far above gives no protection.

Common mistakes

Drawn from real examiner reports.

Power written as energy (no ×t)

Many use $P = IV$ to get a power in watts, then write that as the energy in joules — forgetting to multiply by time. Energy and power differ: $E = Pt$ . A 50 W heater for 300 s gives $E = 50 × 300 = 15\,000$ J, not 50 J. Examiners give 1 mark for the power but withhold the energy mark unless time appears; $E = IVt$ needs current, voltage AND time in the working.

November 2024 Paper 2P Q5(b)–(c): candidates who used the power of the heater without multiplying by any time scored a maximum of 1 mark. This was highlighted as a significant source of lost marks across the cohort.

Time not converted to seconds

$E = IVt$ needs time in seconds to give joules. Substituting minutes (e.g. 45 instead of 2700 s) makes the answer 60× too small. The June 2024 report states plainly "time should be in seconds". For kWh instead, time is in hours: a 5-minute run is $5/60 = 0.083$ h, not 5 h. Convert first: minutes → s ×60, hours → s ×3600.

June 2024 Paper 2P (general examiner advice, p. 46): "time should be in seconds; current should be in amps rather than milli-amps" — flagged as a recurring unit error across multiple questions.

Costing in joules, not kWh

Tariffs are in pence per kWh, so cost $= E_{\text{kWh}} × p$ . Using joules directly (e.g. $3\,600\,000$ J × 0.28 p) gives a nonsensical number instead of $1$ kWh × 28 p $= 28$ p. Convert joules to kWh (÷ $3\,600\,000$ ) before multiplying by the rate. Also: defining power as "energy" loses the mark — it is the RATE of energy transfer.

Power vs energy wording

Power (W) is the RATE of energy transfer; energy (J or kWh) is the TOTAL transferred. Saying "the kettle uses 2 kW of energy" is wrong — it uses 2 kW of POWER. A 2 kW kettle for 3 min transfers $E = 2000 × 180 = 360\,000$ J, but its power stays 2 kW. Match the unit to the quantity: W for power, J or kWh for energy.

Fuse rating is not the power

A 13 A fuse does not mean the appliance uses 13 W — the rating is the maximum safe CURRENT, not power. Find the working current from $I = P/V$ : a 3000 W hairdryer on 230 V draws $I = 3000/230 ≈ 13$ A, so a 13 A fuse fits. Choose the next standard fuse above the working current (commonly 3 A, 5 A or 13 A).

Rounding too early skews the cost

Rounding an intermediate value too soon can push the final cost outside the mark-scheme tolerance — e.g. using $I ≈ 11$ A instead of 10.9 A. Keep extra figures through the working and round only the final answer. In "show that" questions, carry at least one more significant figure than the target value to avoid circular reasoning.

Exam tips

Energy calc scaffold; state the unit
For domestic energy questions: (1) find power — $P = IV$ , else use the given $P$ ; (2) energy — $E = Pt$ ( $t$ in s) or $E = P_{\text{kW}} × t_{\text{h}}$ in kWh; (3) cost — $E_{\text{kWh}} ×$ tariff; (4) write the unit on the answer line — the examiner flags this.
Pick the power form for your data
Choose the power equation that fits the quantities given: $P = IV$ (have I and V), $P = I^2R$ (have I and R), $P = V^2/R$ (have V and R). Picking the right form avoids an unnecessary extra step and a rounding error. All three are printed on the formula sheet.
Keep kWh and J units consistent
For joules, put power in watts and time in seconds; for kWh, put power in kW and time in hours. Do not mix — watts × hours, or kW × seconds, gives an answer off by a big factor. Decide which energy unit the question wants first, then convert both inputs to match.
"Show that" — carry an extra sig fig
In "show that" or "verify" questions, work to at least one more significant figure than the quoted answer before rounding. This proves the result independently and avoids circular reasoning. Then round to match the question's stated precision and add the unit.

Revision notes