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This lesson brings together all the topics from the AQA GCSE Physics Electricity unit (4.2) and provides structured exam practice. You will review the key equations, common question types, mark scheme expectations and strategies for maximising your marks on electricity questions.
You must know and be able to use all of the following equations. Some are given on the AQA equation sheet; others must be memorised.
| Equation | Meaning |
|---|---|
| Q = I x t | Charge = current x time |
| V = I x R | Potential difference = current x resistance |
| P = I x V | Power = current x potential difference |
| P = I^2 x R | Power = current squared x resistance |
| E = P x t | Energy transferred = power x time |
| E = Q x V | Energy transferred = charge x potential difference |
| Equation | Meaning |
|---|---|
| V(p)/V(s) = n(p)/n(s) | Transformer equation (ratio of voltages equals ratio of turns) |
Exam Tip: Even though some equations are on the equation sheet, you should have them memorised so you can apply them instantly. In the exam, time is precious — do not waste it searching the equation sheet for basic formulae like V = IR or P = IV.
These questions give you two values and ask you to calculate the third.
Example: A resistor has a current of 2.5 A flowing through it. The potential difference across it is 10 V. Calculate the resistance.
R = V / I = 10 / 2.5 = 4 ohms
These questions require two or more calculations, often using different equations.
Example: A 1.5 kW heater is used for 45 minutes. Electricity costs 30 pence per kWh. Calculate the cost.
Step 1: Convert time: 45 minutes = 45/60 = 0.75 hours
Step 2: Calculate energy: E = P x t = 1.5 x 0.75 = 1.125 kWh
Step 3: Calculate cost: Cost = 1.125 x 30 = 33.75 pence
These questions involve series and/or parallel circuits.
Example: Two resistors (8 ohms and 12 ohms) are connected in series to a 10 V battery.
a) Total resistance: R = 8 + 12 = 20 ohms
b) Current: I = V/R = 10/20 = 0.5 A
c) p.d. across the 8-ohm resistor: V = IR = 0.5 x 8 = 4 V
d) p.d. across the 12-ohm resistor: V = IR = 0.5 x 12 = 6 V
Check: 4 + 6 = 10 V (matches the battery — correct)
Exam Tip: In circuit calculation questions, always check your answers are consistent. In a series circuit, the p.d. values must add up to the supply voltage. In a parallel circuit, the branch currents must add up to the total current. If they do not, you have made an error somewhere.
Model Answer:
Electricity is transmitted at high voltage to reduce energy losses in the transmission cables.
Power is related to current and voltage by P = I x V. For a given power to be transmitted, increasing the voltage means the current can be reduced.
The power dissipated (wasted as heat) in the cables is given by P(lost) = I^2 x R. Since the resistance of the cables is fixed, reducing the current significantly reduces the power wasted as heat. Because power loss depends on current squared, even a small reduction in current produces a large reduction in power loss.
Step-up transformers are used at the power station to increase the voltage (and decrease the current) before the electricity enters the transmission lines. Step-down transformers at local substations then reduce the voltage to 230 V for safe domestic use.
This makes the transmission of electricity much more efficient, meaning less fuel is needed and costs are reduced.
Model Answer:
The earth wire and fuse work together to protect the user from electric shock if a fault occurs.
If the live wire comes loose inside an appliance and touches the metal casing, the casing becomes live at 230 V. This is dangerous because anyone touching the casing would receive an electric shock — a large current would flow through their body to earth.
However, with an earth wire connected to the metal casing, the current flows through the earth wire to the ground instead of through the user. The earth wire provides a very low resistance path to earth, so a very large current flows.
This large current causes the fuse to heat up, melt and break. The fuse is connected in the live wire, so when it blows, it disconnects the appliance from the live supply. The appliance is now safe to touch.
The fuse must be rated just above the normal working current of the appliance. If it were rated too high, it would not blow during a fault. If rated too low, it would blow during normal use.
Model Answer:
A filament lamp does not obey Ohm's Law because its resistance changes as the current through it changes.
When current first flows through the filament, the filament is cool and has a relatively low resistance. As the current increases, the filament heats up significantly (filament lamps operate at very high temperatures — the tungsten filament glows white-hot).
At higher temperatures, the metal ions in the tungsten filament vibrate with greater amplitude. This means the electrons (charge carriers) collide more frequently with the vibrating ions as they pass through the filament.
More frequent collisions mean the electrons lose more energy and find it harder to flow, which is what we observe as increased resistance.
Because the resistance increases as current increases, the current is no longer directly proportional to the potential difference. The I–V graph is a curve rather than a straight line, showing that Ohm's Law is not obeyed.
Exam Tip: For 6-mark questions, plan your answer before writing. Use a logical structure: cause -> effect -> consequence. Include relevant equations where appropriate. Write in full sentences and use correct scientific terminology throughout. Aim for at least six distinct physics points.
The two required practicals in this unit are heavily examined. Here is a summary of what examiners look for:
| Examiner Expectation | What to Include |
|---|---|
| Circuit description | Power supply, ammeter in series, voltmeter in parallel, wire attached to ruler |
| Independent variable | Length of wire (e.g., 10 cm to 100 cm) |
| Dependent variable | Resistance (calculated from R = V/I) |
| Control variables | Same wire (material, diameter), temperature (low current, switch off between readings) |
| Repeats | At least 3 repeats; calculate mean |
| Graph | R vs. length; expect straight line through origin |
| Safety | Low voltage; switch off between readings |
| Examiner Expectation | What to Include |
|---|---|
| Circuit description | Power supply, variable resistor in series, ammeter in series, voltmeter in parallel |
| How to vary p.d. | Adjust variable resistor; reverse connections for negative values |
| Components tested | Resistor, filament lamp, diode |
| Graph shapes | Resistor: straight line; lamp: S-curve; diode: one-direction only |
| Explanation of shapes | Resistor: constant R; lamp: R increases with temperature; diode: high R in reverse, low R in forward above threshold |
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