Electrical Energy, Power and the kW·h

Q: Is kW·h a unit of power or energy?

The kW·h is a unit of energy, not power. It equals the energy used by a 1 kW appliance running for 1 hour, and is equivalent to 3.6 × 10⁶ joules.

Q: When can I use the extended power formulae P=I²R and P=V²/R?

These formulae only apply to ohmic components (components that follow Ohm's Law at constant temperature). Use the universal P=IV formula for all other components like filament bulbs.

Physics· 4.2.5· 12 min read

1. Core: Electrical Power Calculations★★☆☆☆⏱ 3 min

📘 Definition

Electrical Power

P = \frac{E}{t} = I \times V

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

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.

2. Core: Electrical Energy Transfer Calculations★★☆☆☆⏱ 3 min

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.

3. Core: The kW·h Unit and Electricity Billing★★★☆☆⏱ 4 min

📘 Definition

Kilowatt-hour (kW·h)

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

The energy transferred by a 1 kW appliance running for 1 hour, used as the standard unit for domestic electricity bills.

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.

4. Extended: Power Calculations with Resistance★★★★☆Extended only⏱ 4 min

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

Goal:

Derive P = I²R from core formulae

Starting from:

P = I \times V

V = I \times R (Ohm's Law)

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

Result:

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

🔬 Derivation

Goal:

Derive P = V²/R from core formulae

Starting from:

P = I \times V

I = \frac{V}{R} (rearranged Ohm's Law)

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

Result:

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.

5. Common Pitfalls

Wrong move:

Using time in hours/minutes when calculating energy in joules

Why:

The joule is defined as a watt-second, so time must be in seconds for J calculations to give a correct result.

Correct move:

Convert all time values to seconds for E=Pt calculations where energy is required in joules; only use hours for kW·h calculations.

Wrong move:

Treating kW·h as a unit of power

Why:

While kW is a power unit, multiplying by time (hours) gives an energy unit, not power.

Correct move:

Always associate kW·h with energy use (for billing) and kW with the power rating of an appliance.

Wrong move:

Forgetting to convert watts to kilowatts for kWh calculations

Why:

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 move:

Divide power in watts by 1000 to get kilowatts before multiplying by time in hours to calculate kWh.

Wrong move:

(Extended) Using P=I²R or P=V²/R for non-ohmic components

Why:

These formulae are derived using Ohm's Law, which only applies to ohmic components at constant temperature.

Correct move:

Only use resistance-based power formulae for ohmic resistors; use the universal P=IV formula for all other components like filament bulbs.

Wrong move:

Mixing up power and energy formulae

Why:

Both quantities use the variables I, V and t, so it is easy to confuse P=IV with E=IVt.

Correct move:

Check the question: if it asks for rate of energy use, calculate power; if it asks for total energy transferred, multiply by time.

6. Quick Reference 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	$3.6 \times 1 0^{6}$ J	$E (k W h) = P (k W) \times t (h)$
Electricity Cost		Currency	$Cost = E (k W h) \times cost per kWh$

7. Frequently Asked

Is kW·h a unit of power or energy?

The kW·h is a unit of energy, not power. It equals the energy used by a 1 kW appliance running for 1 hour, and is equivalent to 3.6 × 10⁶ joules.

Do I need to memorize the kW·h to joule conversion?

Yes, Core and Extended candidates are expected to recall that 1 kW·h = 3.6 × 10⁶ J for exam calculations.

When can I use the extended power formulae P=I²R and P=V²/R?

These formulae only apply to ohmic components (components that follow Ohm's Law at constant temperature). Use the universal P=IV formula for all other components like filament bulbs.

Going deeper

syllabus_documentCIE IGCSE Physics 0625 2026-2028 SyllabusRefer to section 4.2.5 for official requirements

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.