At GCSE
At GCSE you use V = IR, P = IV, and the rules for series and parallel circuits. Household electricity (live/neutral/earth, fuses, the National Grid) and I-V characteristics of fixed resistors, filament lamps and diodes appear every series.
GCSEA-LevelAQAEdexcelOCR
Electricity revision, GCSE & A-Level Physics
Electricity questions reward fluent use of V = IR, P = IV and the rules for series and parallel circuits. At GCSE you also need the household electricity section (mains, fuses, earthing) and the National Grid.
At A-Level the topic extends into emf, internal resistance, potential dividers and Kirchhoff's laws. Required practicals on I-V characteristics and resistivity are reliable mark earners.
Current and charge
Electric current is the rate of flow of charge: I = Q/t, where I is current in amperes (A), Q is charge in coulombs (C), and t is time in seconds. In a metal, current is carried by electrons moving through the lattice of ions.
Conventional current flows from positive to negative, even though in a metal the electrons actually move from negative to positive. One coulomb is the charge carried by a current of 1 ampere flowing for 1 second.
- Current I = Q/t (charge flow per unit time).
- Unit: ampere (A). One ampere = one coulomb per second.
- Conventional current: positive to negative.
- Electron flow: negative to positive (opposite to conventional current).
Potential difference
Potential difference (voltage) is the energy transferred per unit charge: V = E/Q, where V is in volts (V), E is energy in joules (J), and Q is charge in coulombs (C). One volt means one joule of energy is transferred per coulomb of charge.
A battery provides the push that drives electrons around a circuit. The potential difference across a component tells you how much energy is transferred (converted to other forms) by that component for each coulomb that passes through it.
- V = E/Q (energy per unit charge). Unit: volt (V).
- One volt = one joule per coulomb.
- EMF (electromotive force): the energy supplied per coulomb by a source.
- Terminal pd: the voltage actually available across the external circuit.
Resistance and Ohm's law
Resistance is the ratio of potential difference to current: R = V/I. The unit is the ohm. Resistance measures how much a component opposes the flow of current.
Ohm's law states that for an ohmic conductor at constant temperature, current is directly proportional to potential difference. A fixed resistor is ohmic: its I-V graph is a straight line through the origin. A filament lamp is non-ohmic: as current increases, the filament heats up and resistance rises, so the I-V graph curves. A diode is also non-ohmic: it conducts in one direction only.
- R = V/I. Unit: ohm (Ω).
- Ohm's law: V = IR for an ohmic conductor at constant temperature.
- Ohmic: fixed resistor (straight-line I-V graph).
- Non-ohmic: filament lamp (curved I-V), diode (one-way conduction).
Series circuits
In a series circuit, components are connected one after another in a single loop. The same current flows through every component. Potential differences add up: the total pd across the circuit equals the sum of the pds across each component.
Total resistance in series is the sum of individual resistances: R_total = R1 + R2 + R3. Adding a resistor in series increases the total resistance and decreases the current.
- Current: same through every component.
- Potential difference: V_total = V1 + V2 + V3.
- Resistance: R_total = R1 + R2 + R3.
- If one component breaks, the circuit is broken and nothing works.
Parallel circuits
In a parallel circuit, components are connected in separate branches, each connected directly across the supply. The potential difference across each branch is the same. The current splits between the branches: the total current from the supply equals the sum of the currents in each branch.
Total resistance in parallel is less than the smallest individual resistance: 1/R_total = 1/R1 + 1/R2. Adding more resistors in parallel decreases the total resistance because there are more paths for current to flow.
- Potential difference: same across each branch.
- Current: I_total = I1 + I2 (currents add at junctions).
- Resistance: 1/R_total = 1/R1 + 1/R2 (total is less than the smallest resistor).
- If one branch breaks, the others still work. Household circuits are wired in parallel.
Power and energy in circuits
Electrical power is the rate of energy transfer: P = IV. Using V = IR, power can also be written as P = I²R or P = V²/R. The unit is the watt (W), where 1 W = 1 J/s.
Energy transferred is E = Pt = IVt, where t is time in seconds. This is how electricity bills are calculated, though suppliers use kilowatt-hours (kWh): 1 kWh is the energy transferred by a 1 kW appliance running for 1 hour, equal to 3.6 MJ.
- Power: P = IV = I²R = V²/R. Unit: watt (W).
- Energy: E = Pt = IVt. Unit: joule (J).
- 1 kWh = 3.6 MJ. Electricity is sold in kWh.
- A fuse melts if the current exceeds its rating, protecting the appliance and wiring.
Mains electricity and the National Grid
UK mains electricity is alternating current at 230 V and 50 Hz. The live wire (brown) carries the alternating potential, the neutral wire (blue) completes the circuit, and the earth wire (yellow-green) is a safety connection that carries current to the ground if a fault occurs, preventing the casing from becoming live.
The National Grid transmits electricity at high voltage and low current to reduce energy loss in the cables. For a given power, higher voltage means lower current (P = IV), and energy lost as heat is proportional to I²R, so stepping up the voltage drastically reduces transmission losses. Transformers step the voltage up for transmission and down for domestic use.
- Mains: 230 V AC, 50 Hz. Live (brown), neutral (blue), earth (yellow-green).
- Fuses and circuit breakers protect against excessive current.
- Earth wire carries fault current safely to ground.
- National Grid: high voltage, low current, to minimise I²R energy loss.
EMF and internal resistance (A-Level)
A real battery has internal resistance. The electromotive force (emf, ε) is the total energy per coulomb supplied by the source. The terminal potential difference (V) is what is actually available across the external circuit, and it is less than the emf when current flows: V = ε − Ir, where r is the internal resistance.
The lost volts (Ir) represent the energy dissipated inside the battery itself. The equation ε = I(R + r) combines this: the emf drives current through the total resistance (external R plus internal r). A required practical measures internal resistance by plotting V against I and finding the intercept (ε) and gradient (−r).
- ε = I(R + r), where r is internal resistance.
- Terminal pd: V = ε − Ir. Lost volts = Ir.
- Plot V against I: y-intercept = ε, gradient = −r.
- A battery with high internal resistance cannot deliver a large current.
Potential dividers (A-Level)
A potential divider is two resistors in series connected across a supply. The output voltage is taken across one of the resistors. The output voltage is a fraction of the supply voltage: V_out = V_in × R2/(R1 + R2).
Thermistors and LDRs can be used in potential dividers to create sensor circuits. A thermistor's resistance decreases as temperature rises (NTC); an LDR's resistance decreases as light intensity increases. Placing the sensor in the correct position in the divider determines whether V_out rises or falls with the environmental variable.
- V_out = V_in × R2/(R1 + R2).
- Thermistor (NTC): resistance falls as temperature rises.
- LDR: resistance falls as light intensity increases.
- The potentiometer is a variable potential divider.
At A-Level
At A-Level you add emf and internal resistance (ε = I(R + r)), potential dividers, Kirchhoff's laws and the resistivity of a wire (required practical). Questions on power dissipation in real circuits and the behaviour of thermistors and LDRs are common.
Common exam mistakes
Saying current is used up in a circuit.
Current is the same at every point in a series circuit. Energy is transferred, not charge.
Confusing conventional current with electron flow.
Conventional current flows positive to negative; electrons flow negative to positive. Exams use conventional current.
Adding resistances in parallel instead of taking reciprocals.
In parallel, 1/R_total = 1/R1 + 1/R2. The total is always less than the smallest individual resistor.
Saying a filament lamp is ohmic.
A filament lamp is non-ohmic: its resistance increases with current because the filament gets hotter. The I-V graph curves.
Forgetting that high voltage reduces energy loss in the National Grid.
High voltage means low current for the same power (P = IV). Since energy loss is I²R, low current means less waste heat in the cables.
Writing emf = V instead of ε = V + Ir.
Emf is the total energy supplied per coulomb. Terminal pd is less than emf when current flows because some energy is lost inside the cell (Ir).
Electricity exam questions to test yourself
Original exam-style prompts. Answer out loud or in writing first, then open the answer to check your wording.
1. Calculate the total resistance of a 6 Ω and 3 Ω resistor in parallel. [2 marks]+
1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2; so R = 2 Ω.
2. A 12 V battery drives 2 A through a 4 Ω resistor. Calculate the power dissipated. [2 marks]+
P = I²R = 2² × 4 = 4 × 4 = 16 W. (Or P = IV = 2 × 12 = 24 W only if 12 V is across the resistor; check that V = IR = 2 × 4 = 8 V, so P = IV = 2 × 8 = 16 W.)
3. Explain why the National Grid transmits at high voltage. [3 marks]+
For a given power, high voltage means low current (P = IV); energy loss in cables is proportional to I²R; so low current means less energy wasted as heat, making transmission more efficient.
4. Describe the I-V characteristic of a filament lamp and explain its shape. [4 marks]+
The I-V graph is a curve (not a straight line) through the origin; as current increases the filament heats up; higher temperature means more lattice vibrations; electrons collide more often so resistance increases; the gradient decreases at higher voltages.
5. A battery of emf 9 V and internal resistance 1 Ω drives a current of 2 A. Calculate the terminal pd. [3 marks]+
V = ε − Ir = 9 − (2 × 1) = 9 − 2 = 7 V.
6. A potential divider has R1 = 4 kΩ and R2 = 6 kΩ across a 12 V supply. Calculate V_out across R2. [2 marks]+
V_out = V_in × R2/(R1 + R2) = 12 × 6/(4 + 6) = 12 × 6/10 = 7.2 V.
7. Explain how a thermistor can be used in a potential divider to produce a voltage that rises with temperature. [3 marks]+
Use an NTC thermistor as R1 (top of the divider) with a fixed resistor as R2; as temperature rises the thermistor's resistance falls; the voltage across R2 (V_out) rises because a larger share of the total voltage is across R2.
Electricity flashcards and quizzes
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Example flashcards
Q: State Ohm's law as an equation.
A: V = IR (potential difference = current × resistance), valid for an ohmic conductor at constant temperature.
Q: What happens to the total resistance when more resistors are added in parallel?
A: It decreases, more paths are available for current to flow.
Q: Calculate the power dissipated by a 12 Ω resistor carrying 2 A.
A: P = I²R = 4 × 12 = 48 W.
Q: Define emf.
A: The energy transferred per unit charge by a source (e.g. a battery), measured in volts.
Quick summary
Electricity is a high-yield Physics topic for GCSE and A-Level students (AQA, Edexcel, OCR). At A-Level you add emf and internal resistance (ε = I(R + r)), potential dividers, Kirchhoff's laws and the resistivity of a wire (required practical). Examiners reward precise definitions and applied explanations, focus on the core ideas and the small set of terms that come up every series.
Key terms
- Current
- Potential difference
- Resistance
- Ohm's law
- Series circuit
- Parallel circuit
- Power
- EMF
- Internal resistance
Electricity FAQs
Why does the resistance of a filament bulb increase with current?+
More current heats the filament; higher temperature means more lattice vibrations, so electrons collide more often and resistance rises. Filament bulbs are non-ohmic.
Why are domestic appliances connected in parallel?+
Each device gets the full mains voltage, can be switched on or off independently and a fault in one does not break the circuit.
What is the difference between AC and DC?+
Direct current flows in one direction (e.g. from a battery). Alternating current reverses direction periodically; UK mains is 50 Hz AC at 230 V.
Why is high voltage used in the National Grid?+
For a given power, higher voltage means lower current, so less energy is lost as heat in the transmission cables (P_loss = I²R).
Related Physics topics
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