GCSE physics electricity and circuits: Ohm's law explained
Electricity is one of the highest-value topics in GCSE Physics. It comes up every year, it carries a lot of marks, and the good news is that most of it follows a small set of rules you can learn and practise. The single most important rule is Ohm's law.
This guide covers everything you need to know about Ohm's law, how to use it in series and parallel circuits, and how to interpret V-I characteristic graphs for different components. There are worked examples throughout so you can see exactly how to set out your calculations.
Paper
One
of the AQA GCSE Physics papers puts electricity front and centre – Paper 1 includes the full electricity topic alongside energy, particle model, and atomic structure
What is Ohm's law?
Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided the temperature remains constant. In equation form it looks like this: V = IR.
V is the potential difference measured in volts (V). I is the current measured in amps (A). R is the resistance measured in ohms (Ω).
The equation tells you three things depending on how you rearrange it. If you know any two of the three quantities, you can always find the third. To find current, use I = V ÷ R. To find resistance, use R = V ÷ I.
A useful trick is the V-I-R triangle. Write V at the top and I and R side by side at the bottom. Cover the quantity you want to find and the remaining two show you the formula. Cover V and you see I × R. Cover I and you see V ÷ R. Cover R and you see V ÷ I.
Worked example: Using V = IR
A resistor has a resistance of 15 Ω. The current flowing through it is 0.4 A. What is the potential difference across the resistor?
Start by writing the equation: V = IR. Substitute the values: V = 0.4 × 15. That gives V = 6 V. The potential difference across the resistor is 6 V.
Now try it the other way round. A lamp has a potential difference of 12 V across it and a current of 2 A flowing through it. What is its resistance? Rearrange to R = V ÷ I. Substitute: R = 12 ÷ 2 = 6 Ω.
| Quantity | Symbol | Unit | Unit symbol | Equation |
|---|---|---|---|---|
| Potential difference | V | Volts | V | V = IR |
| Current | I | Amps | A | I = V ÷ R |
| Resistance | R | Ohms | Ω | R = V ÷ I |
| Total resistance (series) | R_total | Ohms | Ω | R_total = R₁ + R₂ + R₃ |
Series circuits and their rules
In a series circuit, components are connected one after the other in a single loop. There is only one path for the current to follow.
The current is the same at every point in a series circuit. If 0.5 A flows out of the battery, then 0.5 A flows through every component in the loop.
The potential difference is shared between the components. The individual potential differences across each component add up to the total potential difference of the supply. So if a battery supplies 9 V and there are two resistors in series, the potential difference across the first plus the potential difference across the second will equal 9 V.
The total resistance in a series circuit is the sum of the individual resistances. If you connect a 4 Ω resistor and a 6 Ω resistor in series, the total resistance is 4 + 6 = 10 Ω.
Worked example: Series circuit
Two resistors of 8 Ω and 12 Ω are connected in series to a 6 V battery. What is the current in the circuit?
First find the total resistance: R_total = 8 + 12 = 20 Ω. Then use I = V ÷ R. Substitute: I = 6 ÷ 20 = 0.3 A. The current flowing through every part of this circuit is 0.3 A.
Parallel circuits and their rules
In a parallel circuit, the current has more than one path to follow. Components are connected across each other, forming separate branches.
The potential difference across each branch is the same. Every branch receives the full supply voltage. If the battery is 12 V, every component connected in parallel has 12 V across it.
The current splits at junctions. The total current from the supply equals the sum of the currents through each branch. If one branch draws 0.2 A and another draws 0.3 A, the total current is 0.5 A.
At GCSE level you are expected to understand parallel circuits qualitatively rather than calculate their total resistance with a formula. The key idea you need is that adding more resistors in parallel actually decreases the total resistance of the circuit – which can feel counter-intuitive at first. The quantitative reciprocal formula (1/R_total = 1/R₁ + 1/R₂) is A-level content, so you are not expected to use it in a GCSE exam.
Why does adding resistors in parallel reduce total resistance? Think of it like opening extra lanes on a motorway. Every new branch gives the current another route to flow through, so the overall opposition to current drops. You do not need a formula to explain this in a GCSE exam – just the reasoning.
Worked example: Current in parallel
Two resistors of 6 Ω and 3 Ω are connected in parallel across a 12 V supply. What is the current through each branch?
Because the potential difference across each branch is the same as the supply, each resistor has 12 V across it.
Branch 1: I = V ÷ R = 12 ÷ 6 = 2 A.
Branch 2: I = V ÷ R = 12 ÷ 3 = 4 A.
The total current leaving the supply is the sum of the branch currents: 2 + 4 = 6 A. Notice that the 3 Ω resistor draws twice as much current as the 6 Ω resistor, because it has half the resistance for the same voltage. This is a typical GCSE-level calculation – you only need Ohm's law applied to each branch separately.
V-I characteristics for different components
A V-I characteristic graph plots voltage (x-axis) against current (y-axis) for a component. The shape of the line tells you how the component behaves. You need to know the graphs for three components: An ohmic resistor, a filament lamp, and a diode.
Ohmic resistor (fixed resistor)
The V-I graph for an ohmic resistor is a straight line through the origin. This means the current is directly proportional to the voltage – exactly what Ohm's law predicts. The resistance stays constant regardless of the current flowing through it. A steeper line means a lower resistance because more current flows for the same voltage.
The filament lamp IV characteristic
The V-I graph for a filament lamp is a curve that starts steep and gradually flattens. At low voltages the graph looks almost linear, but as the current increases, the filament heats up. The higher temperature causes the metal atoms to vibrate more, which increases the resistance. This is why the curve flattens – the same increase in voltage produces a smaller increase in current. A filament lamp is a non-ohmic conductor because its resistance changes with temperature.
The diode IV characteristic curve
A diode only allows current to flow in one direction. In the forward direction, no current flows until the voltage reaches about 0.7 V (the threshold voltage). After that, current increases rapidly. In the reverse direction, the resistance is extremely high and essentially no current flows, so the line sits flat along the x-axis. The V-I graph is therefore not symmetrical – it has a sharp upward curve on one side and a flat line on the other.
In the exam, you may be asked to describe or sketch V-I characteristics. Remember: Straight line through the origin = ohmic resistor, curve that flattens = filament lamp, current in one direction only = diode.
Required practical: Investigating resistance
One of the required practicals for GCSE Physics involves investigating how the length of a wire affects its resistance. You connect a piece of wire into a circuit with a battery, an ammeter (in series), and a voltmeter (in parallel across the wire). You then measure the current and voltage for different lengths of wire and calculate the resistance using R = V ÷ I.
You should find that resistance is directly proportional to length. If you double the length of the wire, the resistance doubles. This is because the electrons have to travel through more material and collide with more ions along the way.
To get reliable results, keep the wire at a constant temperature by using a low current and switching off between readings. Use at least five different lengths and repeat each measurement to calculate a mean. Plot a graph of resistance against length – it should be a straight line through the origin.
Electricity revision checklist
Make sure you can confidently do all of these before your exam.
- State Ohm's law and use V = IR in all three rearrangements
- Calculate total resistance in a series circuit by adding individual resistances
- Explain qualitatively that total resistance in parallel is less than the smallest branch resistance
- Apply Ohm's law to each branch of a parallel circuit to find branch currents
- Explain how current and voltage behave in series vs parallel circuits
- Sketch and interpret V-I graphs for a resistor, filament lamp, and diode
- Describe the required practical for investigating resistance and wire length
- Identify ohmic and non-ohmic conductors from their V-I characteristics
Common mistakes to avoid
Mixing up series and parallel rules is the most common error. Remember: Current is the same everywhere in series, voltage is the same everywhere in parallel. If you get these the wrong way round, every calculation that follows will be wrong.
Another frequent mistake is trying to use the A-level reciprocal formula for parallel resistance in a GCSE answer. It is not on the GCSE specification, so examiners are not looking for it. Stick to applying Ohm's law to each branch separately and reasoning about how total resistance changes qualitatively.
When reading V-I graphs, pay attention to the axes. Some exam boards plot current on the x-axis and voltage on the y-axis, which reverses the shape of the curves. Always check the axis labels before interpreting the graph.
Finally, always include units in your answers. An answer of 6 without a unit could mean 6 V, 6 A, or 6 Ω – the examiner cannot award the mark without the correct unit.
