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5.4 Electrical Currents and Energy Transfer

GCSE Physics · AQA · Chapter 5: Electricity in the Home

A free GCSE Physics lesson for AQA (8463) from OBA School: charge flow Q = I t, energy transfer E = Q V, energy in a circuit, appliances and kilowatt-hours. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Calculate the charge that flows using Q = I t, and rearrange it
  • Explain why a resistor heats up when charge flows through it
  • Calculate the energy transferred using E = Q V and E = V I t
  • Show that the energy supplied by a battery equals the total energy transferred to the components in the circuit
  • Describe the energy transfers in household appliances and calculate energy in kilowatt-hours and its cost

Key words

  • Charge, Q: The quantity of electricity carried by electrons, measured in coulombs (C).
  • Current, I: The rate of flow of charge: the charge passing a point each second. 1 A = 1 C/s.
  • Potential difference: The energy transferred by each coulomb of charge that passes through a component. 1 V = 1 J/C.
  • Energy store: Where energy is held, e.g. the chemical store of a battery or the thermal store of a resistor.
  • Kilowatt-hour (kWh): The energy transferred by a 1 kW appliance in 1 hour. 1 kWh = 3 600 000 J.

Charge flow

A current in a metal wire is a flow of electrons, each carrying a tiny negative charge. The charge that flows past a point depends on the current and how long it flows for:

charge flow, Q (C) = current, I (A) × time, t (s)

Rearranged: I = Q ÷ t t = Q ÷ I

Worked example: charge through a lamp

A current of 0.40 A passes through a lamp for 5.0 minutes. Calculate the charge that flows.
t = 5.0 × 60 = 300 s
Q = I t = 0.40 A × 300 s = 120 C

Answer: 120 C

Exam tip: Time must be in seconds. Hours → seconds: multiply by 3600. Small currents may be in milliamps: 1 mA = 0.001 A.

Why a resistor heats up

The battery does work pushing electrons through the resistor. As they move, the electrons keep colliding with the vibrating metal ions of the resistor. Each collision passes energy to the ions, so they vibrate more. The resistor gets hotter and heats its surroundings.

Energy is transferred from the chemical energy store of the battery, by the electric current, to the thermal energy store of the resistor, and then by heating to the surroundings. The electrical work done by the battery equals the energy transferred to the resistor.

Energy and potential difference

The pd across a component tells you how much energy each coulomb transfers to it. A pd of 1 volt means 1 joule per coulomb. So the energy transferred is:

energy transferred, E (J) = charge flow, Q (C) × potential difference, V (V)

Since Q = I t, this can also be written E = V I t, which is the same as E = P t because P = V I.

Worked example: energy from charge

45 C of charge flows through a motor with a pd of 9.0 V across it.
E = Q V = 45 C × 9.0 V = 405 J

Answer: 405 J

Worked example: energy from current and time

A 6.0 V bulb takes a current of 0.25 A for 2.0 minutes.
t = 2.0 × 60 = 120 s
E = V I t = 6.0 × 0.25 × 120 = 180 J

Answer: 180 J

Energy in a series circuit

In the circuit below, each coulomb leaving the battery carries 9 J. It gives 6 J to the lamp and 3 J to the resistor, so no energy is left when it returns. Energy is conserved: the energy supplied by the battery equals the total energy transferred to the components. The charge is not used up – only its energy is transferred.

Worked example: checking energy is conserved

The circuit runs for 40 s with a current of 0.50 A.
Q = I t = 0.50 × 40 = 20 C
Energy from battery = Q V = 20 × 9.0 = 180 J
Energy to lamp = 20 × 6.0 = 120 J; energy to resistor = 20 × 3.0 = 60 J
120 J + 60 J = 180 J

Answer: energy supplied (180 J) = energy transferred to the components (180 J)

Appliances and kilowatt-hours

Appliances are designed to transfer energy from the mains or a battery into a useful store. Some of the energy is always wasted, usually by heating the surroundings.

  • Electric drill; Supply battery; Main useful energy transfer kinetic energy store of the spinning motor and drill bit
  • Kettle; Supply mains; Main useful energy transfer thermal energy store of the water
  • Electric fan; Supply mains; Main useful energy transfer kinetic energy store of the blades and air
  • Phone charger; Supply mains; Main useful energy transfer chemical energy store of the phone battery

Joules are very small units for home energy use, so electricity companies measure energy in kilowatt-hours (kWh). An electricity meter records the number of kWh supplied.

energy (kWh) = power (kW) × time (h) cost = energy (kWh) × price per kWh

1 kWh = 1000 W × 3600 s = 3 600 000 J (3.6 MJ)

Worked example: cost of using a tumble dryer

A 2.4 kW tumble dryer runs for 1.5 hours. Electricity costs 28p per kWh.
E = 2.4 kW × 1.5 h = 3.6 kWh
cost = 3.6 × 28p = 100.8p

Answer: about £1.01

Common mistakes

  • Leaving time in minutes or hours in Q = I t or E = Q V. Use seconds (except when working in kWh).
  • Mixing up charge and current. Charge is measured in coulombs (C); current is charge per second, in amps (A).
  • Saying the current or charge is “used up” in a lamp. The same current flows all round a series circuit; it is energy that is transferred.
  • Using watts and seconds to get kWh. For kilowatt-hours use kilowatts and hours.

Revision questions

Try these, then check your answers below each one.

Question 1. A current of 2.0 A flows through a motor for 3.0 minutes. Calculate the charge that flows.

Answer: 360 C

Working: t = 3.0 × 60 = 180 s. Q = I t = 2.0 × 180 = 360 C.

Question 2. A charge of 25 C passes through a component with a pd of 6.0 V across it. Calculate the energy transferred.

Answer: 150 J

Working: E = Q V = 25 C × 6.0 V = 150 J.

Question 3. Which statement describes one coulomb of charge? (Choose from: The charge that passes a point when a current of 1 A flows for 1 s, The energy carried by 1 A in 1 s, The charge on one electron, The current that flows when the pd is 1 V.)

Answer: The charge that passes a point when a current of 1 A flows for 1 s

Working: Q = I t = 1 A × 1 s = 1 C. (One electron has a far smaller charge, 1.6 × 10−¹⁹ C.)

Related lessons

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