# Electrical Energy, Power and the kW·h

> Physics · CIE IGCSE 0625
> Source: https://www.owlsprep.com/study/cie-0625-u4-electrical-energy-power-and-the/

This guide covers all core and extended content for electrical energy, power, and the kW·h unit as specified for CIE IGCSE Physics 0625, including domestic billing calculations.

**Prerequisites:** [Understanding of current, voltage and resistance](https://www.owlsprep.com/study/cie-0625-u4-current-voltage-resistance/); [Basic energy transfer concepts](https://www.owlsprep.com/study/cie-0625-u1-energy-transfers/)

## Learning objectives

- Recall and apply core formulae for electrical power (P=IV, P=E/t)
- Calculate electrical energy transferred using E=Pt or E=IVt
- Use the kW·h unit to calculate domestic electricity costs
- (Extended only) Derive and apply P=I²R and P=V²/R for ohmic components

## Core: Electrical Power Calculations

**Electrical Power** — The rate of electrical energy conversion by a component, measured in watts (W).

*Notation:* P = \frac{E}{t} = I \times V

*Example:* A 50 W phone charger converts 50 joules of electrical energy every second.

Power is calculated using two core formulae: dividing total energy transferred by time taken, or multiplying current by potential difference across the component. All units must be SI units for these calculations: power in watts, energy in joules, time in seconds, current in amps, voltage in volts.

**Worked example:** A portable fan draws a current of 1.5 A from a 12 V battery. Calculate the power of the fan.

1. Recall the core power formula relating current and voltage: P = I × V
2. $$P = 1.5 \, \text{A} \times 12 \, \text{V} = 18 \, \text{W}$$

> **Exam tip:** Remember 1 watt = 1 joule per second, so you can cross-check power answers using energy and time values if you are unsure of your calculation.

## Core: Electrical Energy Transfer Calculations

Total electrical energy transferred by a component is calculated by multiplying power by the time the component is active, or directly by multiplying current, voltage and time. The SI unit for energy is the joule (J).

$$E = P \times t = I \times V \times t$$

**Worked example:** A 40 W LED bulb is left switched on for 3 hours. Calculate the total electrical energy transferred to the bulb in joules.

1. Convert time from hours to seconds: 3 hours = 3 × 60 × 60 = 10800 s
2. Use the energy formula E = P × t
3. $$E = 40 \, \text{W} \times 10800 \, \text{s} = 432000 \, \text{J} = 4.32 \times 10^5 \, \text{J}$$

> **Exam tip:** Always convert time to seconds for joule calculations, unless you are explicitly calculating energy in kilowatt-hours.

## Core: The kW·h Unit and Electricity Billing

**Kilowatt-hour (kW·h)** — The energy transferred by a 1 kW appliance running for 1 hour, used as the standard unit for domestic electricity bills.

*Notation:* 1 \, \text{kW·h} = 3.6 \times 10^6 \, \text{J}

*Example:* A 2 kW heater running for 0.5 hours uses 1 kW·h of energy.

To calculate electricity costs, first find the total energy used in kW·h, then multiply by the cost per kW·h set by the energy provider.

**Worked example:** A 1.2 kW electric oven is used for 2.5 hours. If electricity costs \$0.14 per kW·h, calculate the total cost of using the oven.

1. Calculate energy used in kW·h: E = P(kW) × t(h)
2. $$E = 1.2 \, \text{kW} \times 2.5 \, \text{h} = 3 \, \text{kW·h}$$
3. Calculate total cost: Cost = Energy × Cost per kW·h
4. $$\text{Cost} = 3 \times 0.14 = \$0.42$$

> **Exam tip:** Double-check units for kWh calculations: power must be in kilowatts (divide watts by 1000) and time must be in hours, not seconds or minutes.

## Extended: Power Calculations with Resistance

Extended candidates can combine the core power formula P = I×V with Ohm's Law (V = I×R) to derive two additional power formulae for ohmic components:

**Derivation:** Derive P = I²R from core formulae

1. Substitute V = I×R into P = I×V
2. $$P = I \times (I \times R) = I^2 R$$

*Conclusion:* Power dissipated by an ohmic resistor equals the square of the current multiplied by resistance.

**Derivation:** Derive P = V²/R from core formulae

1. Substitute I = V/R into P = I×V
2. $$P = \frac{V}{R} \times V = \frac{V^2}{R}$$

*Conclusion:* Power dissipated by an ohmic resistor equals the square of the potential difference divided by resistance.

**Worked example:** A 20 Ω ohmic resistor has a potential difference of 10 V across it. Calculate the power dissipated by the resistor.

1. Use the extended power formula P = V²/R
2. $$P = \frac{(10 \, \text{V})^2}{20 \, \Omega} = \frac{100}{20} = 5 \, \text{W}$$

> **Exam tip:** Only use the resistance-based power formulae for ohmic components. For non-ohmic components like filament bulbs, always use P = IV.

## Common pitfalls

- **Wrong:** Using time in hours/minutes when calculating energy in joules
  - Why it fails: The joule is defined as a watt-second, so time must be in seconds for J calculations to give a correct result.
  - Correct: Convert all time values to seconds for E=Pt calculations where energy is required in joules; only use hours for kW·h calculations.
- **Wrong:** Treating kW·h as a unit of power
  - Why it fails: While kW is a power unit, multiplying by time (hours) gives an energy unit, not power.
  - Correct: Always associate kW·h with energy use (for billing) and kW with the power rating of an appliance.
- **Wrong:** Forgetting to convert watts to kilowatts for kWh calculations
  - Why it fails: The kW·h unit uses kilowatts as the power unit, so watt values will give results 1000x larger than the correct value if not converted.
  - Correct: Divide power in watts by 1000 to get kilowatts before multiplying by time in hours to calculate kWh.
- **Wrong:** (Extended) Using P=I²R or P=V²/R for non-ohmic components
  - Why it fails: These formulae are derived using Ohm's Law, which only applies to ohmic components at constant temperature.
  - Correct: Only use resistance-based power formulae for ohmic resistors; use the universal P=IV formula for all other components like filament bulbs.
- **Wrong:** Mixing up power and energy formulae
  - Why it fails: Both quantities use the variables I, V and t, so it is easy to confuse P=IV with E=IVt.
  - Correct: Check the question: if it asks for *rate* of energy use, calculate power; if it asks for total energy transferred, multiply by time.

## Cheatsheet

| Quantity | Symbol | Units | Core Formulae | Extended Formulae |
| --- | --- | --- | --- | --- |
| Electrical Power | P | Watts (W) | $P = E/t$, $P = I \times V$ | $P = I^2 R$, $P = V^2/R$ |
| Electrical Energy | E | Joules (J) | $E = P \times t$, $E = I \times V \times t$ |  |
| Kilowatt-hour | kW·h | $3.6 \times 10^6$ J | $E(kWh) = P(kW) \times t(h)$ |  |
| Electricity Cost |  | Currency | $\text{Cost} = E(kWh) \times \text{cost per kWh}$ |  |

## What's next

Now you have mastered electrical energy and power calculations, you are ready to move on to series and parallel circuit rules, which are frequently tested alongside power questions in both core and extended papers. You will also use these power calculation skills when studying domestic electricity safety features, where appliance power ratings determine the correct fuse size for a circuit. Regular practice of structured calculation questions will help you avoid common unit conversion errors and ensure you score full marks on this high-frequency topic.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-0625-u4-electrical-energy-power-and-the/