Define the kilowatt-hour (kWh) and calculate the cost of using electrical appliances where the energy unit is the kWh

4.2.5 Electrical Energy and Electrical Power

Objective (Cambridge IGCSE 0625)

• State and use the core equations P = IV and E = IVt.
• Show that for a constant‑voltage circuit E = Pt and relate power to resistance (P = I²R, P = V²/R).
• Define the kilowatt‑hour (kWh) as the practical unit of electrical energy.
• Convert between kWh and joules (extension).
• Calculate the energy used by an appliance and the cost when the price is given in £/kWh (or any other currency).

1. Core equations – power and energy

When a source supplies a current I through a potential difference V, electrical energy is transferred to the circuit elements.

• Power (rate of energy transfer)
  P = IV  [W]

• Energy transferred in a time t
  E = IVt  [J]

• If the voltage is constant (the usual case for household appliances) we can substitute P = IV into the energy expression:
  E = Pt  [J]

  This is the form most often used for cost calculations.

2. Power expressed in terms of resistance

Using Ohm’s law V = IR we can rewrite the power formula in two additional useful ways.

These forms are handy when the power rating is not printed on the appliance but the resistance of a heating element, for example, is known.

3. The kilowatt‑hour (kWh)

The electricity industry uses the kilowatt‑hour as the billing unit of energy.

In SI units:

4. Units checklist (AO2 reminder)

• Power: W (watts) → kilowatts: divide by 1 000.
• Time: s (seconds) → hours: divide by 3 600.
• Energy: J (joules) → kilowatt‑hours: divide by 3.6 × 10⁶.
• When using the formula E = Pt, ensure P is in kW and t in hours to obtain E directly in kWh.

5. Cost of electricity

If the supplier charges C (currency per kWh), the cost K of running an appliance is

where E_kWh is the energy consumed expressed in kilowatt‑hours.

6. Step‑by‑step calculation method

1. Determine the power.

   • If a power rating is printed, use it directly.
   • If not, calculate it with P = IV or with the resistance forms P = I²R / P = V²/R.

2. Convert power to kilowatts. P_kW = P_W ÷ 1000
3. Record the operating time in hours. t_h
4. Calculate energy used. E_kWh = P_kW × t_h
5. Find the cost. K = E_kWh × C

7. Worked examples

Example 1 – Appliance with a printed power rating

A 1500 W electric heater is switched on for 3 h each day. Electricity price = £0.20 /kWh.

1. P_kW = 1500 W ÷ 1000 = 1.5 kW
2. E_kWh = 1.5 kW × 3 h = 4.5 kWh
3. K = 4.5 kWh × £0.20/kWh = £0.90 per day
4. Monthly cost (30 days) = £0.90 × 30 = £27.0

Example 2 – Using measured current and voltage (AO1 – applying P = IV)

A resistive heater draws I = 2 A from a 230 V mains supply and runs for 1 hour. Electricity price = £0.20 /kWh.

1. Step 0: P = IV = 2 A × 230 V = 460 W
2. P_kW = 460 W ÷ 1000 = 0.460 kW
3. E_kWh = 0.460 kW × 1 h = 0.460 kWh
4. K = 0.460 kWh × £0.20/kWh = £0.092 (≈ 9 p)

Example 3 – Only the resistance is known (using P = V²/R)

A heating element has a resistance R = 50 Ω** and is connected to the UK mains (230 V). It operates for 2 h. Electricity price = £0.20 /kWh.

1. Step 0: P = V²/R = (230 V)² ÷ 50 Ω = 52900 ÷ 50 ≈ 1058 W
2. P_kW = 1058 W ÷ 1000 ≈ 1.058 kW
3. E_kWh = 1.058 kW × 2 h ≈ 2.12 kWh
4. K = 2.12 kWh × £0.20/kWh ≈ £0.424 (≈ 42 p)

8. Extension – converting between kWh and joules

Although not required for basic cost calculations, the ability to move between the SI unit (joule) and the billing unit (kWh) is useful.

• To convert kWh → J: multiply by 3.6 × 10⁶.
• To convert J → kWh: divide by 3.6 × 10⁶.

Practice question

“A battery stores 7.2 MJ of energy. Express this amount in kWh, and state how many 60 W LED lamps could be run for 5 hours each using this energy.”

Solution sketch:

1. 7.2 MJ ÷ 3.6 × 10⁶ J kWh⁻¹ = 2.0 kWh.
2. Energy per lamp for 5 h: 0.060 kW × 5 h = 0.30 kWh.
3. Number of lamps = 2.0 kWh ÷ 0.30 kWh ≈ 6 lamps.

9. Typical household appliances (230 V supply)

10. Key points to remember

• P = IV – power is the product of current and voltage.
• E = IVt – energy transferred in a time t.
• For a constant voltage source, E = Pt (useful for appliances with a fixed power rating).
• Alternative power forms: P = I²R and P = V²/R.
• 1 kWh = 3.6 MJ = 3.6 × 10⁶ J (conversion for extension work).
• Always convert watts to kilowatts before using the kWh formula.
• Cost = (energy in kWh) × (price per kWh).
• Reducing either the power rating (e.g., using LED lights) or the operating time lowers the electricity bill.