Energy transfers in electrical appliances and the national grid

What you'll learn

This revision guide covers how electrical appliances transfer energy and how the national grid distributes electricity across the UK. You'll learn to calculate power and efficiency, understand energy transfers in common devices, and explain how transformers enable efficient electricity transmission. These topics appear regularly in AQA GCSE Physics exams, particularly in calculation questions and extended response items.

Key terms and definitions

Power — the rate of energy transfer, measured in watts (W) or kilowatts (kW); 1 W = 1 J/s

Efficiency — the proportion of input energy that is usefully transferred, expressed as a percentage or decimal value

National grid — the system of cables and transformers that distributes electricity from power stations to consumers across the country

Transformer — a device that changes the potential difference (voltage) of an alternating current supply

Step-up transformer — increases potential difference while decreasing current for efficient transmission

Step-down transformer — decreases potential difference to safe levels for domestic use (typically 230 V in the UK)

Dissipated energy — energy that is transferred to the surroundings by heating or sound, typically wasted energy

Kilowatt-hour (kWh) — a unit of energy equal to the energy transferred by a 1 kW appliance in 1 hour

Core concepts

Energy transfers in electrical appliances

Electrical appliances transfer energy from the electrical supply to other forms. The type of energy output depends on the appliance's design and purpose.

Useful energy transfers:

• Electric heaters — electrical energy → thermal energy (heat)
• Light bulbs — electrical energy → light (and some thermal energy)
• Electric motors — electrical energy → kinetic energy (and some thermal/sound energy)
• Loudspeakers — electrical energy → sound (and some thermal energy)

No appliance transfers energy with 100% efficiency. Some energy is always dissipated to the surroundings, typically as:

• Thermal energy (heating) in wires and components due to resistance
• Sound energy from vibrations
• Light energy (when unwanted)

Modern appliances like LED bulbs and energy-efficient motors minimise wasted energy transfers, making them more cost-effective and environmentally friendly than older designs.

Calculating power and energy

Power calculations use three key equations you must learn:

P = E ÷ t

Where:

• P = power (W)
• E = energy transferred (J)
• t = time (s)

P = I × V

Where:

• P = power (W)
• I = current (A)
• V = potential difference (V)

P = I² × R

Where:

• P = power (W)
• I = current (A)
• R = resistance (Ω)

Energy transferred can be calculated using:

E = P × t

Where:

• E = energy (J)
• P = power (W)
• t = time (s)

For domestic energy bills, kilowatt-hours are used instead of joules:

E = P × t

Where:

• E = energy (kWh)
• P = power (kW)
• t = time (hours)

To convert: 1 kWh = 3,600,000 J (or 3.6 MJ)

Efficiency calculations

Efficiency measures how much input energy is usefully transferred. The remaining energy is wasted.

Efficiency = (useful energy output ÷ total energy input) × 100%

Or using power:

Efficiency = (useful power output ÷ total power input) × 100%

Efficiency values range from 0% to 100%. Higher efficiency means:

• Less energy wasted
• Lower running costs
• Reduced environmental impact

Typical efficiency values:

• LED bulb: 85-90% (light output)
• Filament bulb: 5-10% (light output)
• Electric motor: 60-90%
• Electric heater: ~100% (all electrical energy becomes thermal energy, which is often the desired output)

Note: Electric heaters appear 100% efficient because their purpose is heating. However, if heat escapes before warming the intended space, they become less effective.

The national grid system

The national grid connects power stations to homes, schools, hospitals and businesses. Electricity travels hundreds of kilometres from generation to consumption.

Key components:

1. Power stations — generate electricity at around 25,000 V
2. Step-up transformers — increase voltage to 400,000 V or 275,000 V
3. Transmission cables — carry high-voltage electricity on pylons
4. Step-down transformers — reduce voltage progressively
5. Local substations — final reduction to 230 V for UK homes

Why use high voltages for transmission?

Higher potential difference reduces current for the same power (P = I × V). Lower current means:

• Reduced energy loss in cables (power loss = I² × R)
• Thinner, cheaper cables can be used
• More efficient transmission over long distances

How transformers work

Transformers only work with alternating current (AC). They consist of:

• Primary coil — connected to input voltage
• Secondary coil — connected to output voltage
• Iron core — links the coils magnetically

The AC in the primary coil creates a changing magnetic field. This induces a potential difference in the secondary coil.

Transformer equation:

(Vp ÷ Vs) = (Np ÷ Ns)

Where:

• Vp = primary potential difference (V)
• Vs = secondary potential difference (V)
• Np = number of turns on primary coil
• Ns = number of turns on secondary coil

For step-up transformers: Ns > Np (more turns on secondary)

For step-down transformers: Ns < Np (fewer turns on secondary)

Transformers are highly efficient (typically >95%), but small amounts of energy are wasted through:

• Heating in the coils (resistance)
• Heating in the iron core
• Sound (vibrations)

Reducing energy waste in transmission

Energy losses in the national grid occur mainly through heating in cables. The power wasted equals I² × R.

Strategies to minimise losses:

1. Use high transmission voltages — reduces current significantly
2. Use thick cables — lower resistance
3. Use materials with low resistivity — typically copper or aluminium

Example: Transmitting 1 MW of power

• At 250 V: current = 4000 A (very high losses)
• At 250,000 V: current = 4 A (minimal losses)

Despite the high efficiency of the grid system (~92%), approximately 8% of generated electricity is lost during transmission and distribution.

Worked examples

Example 1: Power and energy calculation

Question: A 2.0 kW electric kettle operates for 3.5 minutes. Calculate: (a) The energy transferred in joules [3 marks] (b) The current drawn from the 230 V mains supply [2 marks]

Solution:

(a) Convert time to seconds: 3.5 × 60 = 210 s [1 mark]

Convert power to watts: 2.0 kW = 2000 W [1 mark]

E = P × t = 2000 × 210 = 420,000 J (or 420 kJ) [1 mark]

(b) Rearrange P = I × V to find I = P ÷ V [1 mark]

I = 2000 ÷ 230 = 8.7 A [1 mark]

Example 2: Efficiency

Question: An electric motor has a power input of 500 W. It lifts a load, doing 350 J of useful work each second. (a) Calculate the useful power output [1 mark] (b) Calculate the efficiency of the motor [2 marks] (c) State what happens to the wasted energy [1 mark]

Solution:

(a) Useful power output = 350 W [1 mark] (Power = energy per second, so 350 J/s = 350 W)

(b) Efficiency = (useful power output ÷ total power input) × 100% [1 mark]

Efficiency = (350 ÷ 500) × 100% = 70% [1 mark]

(c) The wasted energy is dissipated as thermal energy (heating) in the motor components and some sound energy [1 mark]

Example 3: Transformers

Question: A transformer in the national grid has 200 turns on the primary coil and 8000 turns on the secondary coil. The primary potential difference is 11,000 V. (a) Calculate the secondary potential difference [3 marks] (b) State whether this is a step-up or step-down transformer [1 mark] (c) Explain why transformers are necessary in the national grid [3 marks]

Solution:

(a) Use equation: Vp ÷ Vs = Np ÷ Ns [1 mark]

Rearrange: Vs = Vp × (Ns ÷ Np) [1 mark]

Vs = 11,000 × (8000 ÷ 200) = 11,000 × 40 = 440,000 V [1 mark]

(b) Step-up transformer [1 mark] (Secondary voltage is higher than primary voltage)

(c) Step-up transformers increase voltage for transmission [1 mark], which reduces current and therefore reduces energy losses due to heating in the cables (power loss = I² × R) [1 mark]. Step-down transformers then reduce voltage to safe levels for domestic use (230 V) [1 mark].

Common mistakes and how to avoid them

• Confusing power and energy — Power is the rate of energy transfer (energy per second), not a total amount. Always check units: watts for power, joules or kWh for energy.

• Using wrong time units — When calculating energy from power, ensure time is in seconds for joules or hours for kWh. Don't mix units (e.g., using minutes with kW).

• Efficiency greater than 100% — If your calculation gives efficiency >100%, you've made an error. Check you've used useful output ÷ total input, not the other way round.

• Forgetting that transformers need AC — Transformers only work with alternating current. DC cannot create the changing magnetic field needed for induction.

• Thinking higher voltage means more danger in all contexts — While high voltage is dangerous if contacted, the national grid uses high voltage to reduce current, which actually makes transmission safer by reducing heating.

• Not explaining why high voltage reduces losses — Always link high voltage → low current → reduced power loss in cables using I² × R. The examiner wants the full reasoning chain.

Exam technique for "Energy transfers in electrical appliances and the national grid"

• Command word "calculate" — Show your working clearly. Write the equation, substitute values with units, then give the answer with the correct unit. Marks are available for method even if your final answer is wrong.

• Extended response on the national grid — Structure your answer logically: generation voltage → step-up → transmission → step-down → consumption. Link high voltage to low current to reduced energy losses. Aim for 5-6 well-developed points for 6-mark questions.

• Equation selection — You have three power equations (P = E/t, P = IV, P = I²R). Choose the one that matches the information given. Don't waste time rearranging equations you don't need.

• Efficiency questions — State whether you're using energy or power values and ensure both are in the same units. Express final answers as percentages unless the question specifies otherwise.

Quick revision summary

Electrical appliances transfer energy from electrical supplies to useful forms (light, thermal, kinetic, sound), but no device is 100% efficient. Power measures the rate of energy transfer (P = E/t, P = IV, P = I²R). Efficiency equals useful output divided by total input. The national grid transmits electricity at high voltage (400 kV) using step-up transformers, reducing current to minimise power losses (I²R) in cables. Step-down transformers reduce voltage to safe levels (230 V) for homes. Transformers only work with AC and use electromagnetic induction.