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This lesson covers the three fundamental quantities in electric circuits — current, voltage (potential difference) and resistance — as required by the Edexcel GCSE Physics specification (1PH0). You need to understand what each quantity means, how they are measured, the relationship between them (Ohm's law), and how to use the key equations confidently.
Electric current is the rate of flow of electric charge. In a metal conductor, the charge carriers are electrons. When a circuit is complete, electrons flow through the wires and components.
I=tQ
Where:
This equation tells you that 1 ampere is a flow of 1 coulomb of charge per second.
| Quantity | Symbol | Unit | Unit Symbol |
|---|---|---|---|
| Current | I | ampere | A |
| Charge | Q | coulomb | C |
| Time | t | second | s |
Exam Tip: The equation I = Q/t can be rearranged to Q = It or t = Q/I. Make sure you can rearrange confidently — the exam may give you any two of the three quantities and ask you to calculate the third.
There are two ways to describe the direction of flow in a circuit:
When electricity was first studied, scientists assumed that positive charges moved through the circuit. By the time electrons were discovered and found to flow from negative to positive, the convention was already established. We still use conventional current direction today.
Exam Tip: If an exam question asks about "current direction," use conventional current (positive to negative) unless the question specifically asks about electron flow.
In metallic conductors (wires, resistors), the charge carriers are free electrons (also called delocalised electrons). These electrons are not bound to individual atoms and can move freely through the metal lattice.
In electrolytes (ionic solutions), the charge carriers are ions — both positive ions (cations) moving towards the negative electrode and negative ions (anions) moving towards the positive electrode.
Voltage — more correctly called potential difference (p.d.) — is the energy transferred per unit charge that passes between two points in a circuit.
V=QW
Where:
This equation tells you that 1 volt means 1 joule of energy is transferred per coulomb of charge.
| Quantity | Symbol | Unit | Unit Symbol |
|---|---|---|---|
| Potential difference | V | volt | V |
| Energy / work done | W | joule | J |
| Charge | Q | coulomb | C |
Exam Tip: A very common exam mistake is connecting the ammeter in parallel or the voltmeter in series. Remember: Ammeter = Along the wire (series); Voltmeter = aVoid the component (parallel, across it).
Resistance is a measure of how much a component opposes (resists) the flow of electric current through it. A component with high resistance requires a greater potential difference across it to drive the same current.
V=IR
Where:
This can be rearranged:
| Quantity | Symbol | Unit | Unit Symbol |
|---|---|---|---|
| Potential difference | V | volt | V |
| Current | I | ampere | A |
| Resistance | R | ohm | Ω |
Exam Tip: Ohm's law (V = IR) is one of the most important equations in GCSE physics. You will use it in almost every electricity question. Practise rearranging it until it becomes second nature.
A current of 3 A flows through a lamp for 2 minutes. How much charge passes through the lamp?
A charge of 450 C flows through a resistor in 90 seconds. What is the current?
A 12 V battery drives a current of 0.5 A through a resistor. What is the resistance?
A current of 2 A flows through a 15 Ω resistor. What is the potential difference across it?
A charge of 200 C passes through a heater with a potential difference of 6 V across it. How much energy is transferred?
graph TD
A["Ohm’s Law: V = IR"] --> B["To find V:<br/>V = I × R"]
A --> C["To find I:<br/>I = V / R"]
A --> D["To find R:<br/>R = V / I"]
B --> E["If I doubles and R stays the same,<br/>V doubles"]
C --> F["If V stays the same and R doubles,<br/>I halves"]
D --> G["If V = 12 V and I = 3 A,<br/>R = 4 Ω"]
style A fill:#2c3e50,color:#fff
style B fill:#2980b9,color:#fff
style C fill:#2980b9,color:#fff
style D fill:#2980b9,color:#fff
style E fill:#27ae60,color:#fff
style F fill:#27ae60,color:#fff
style G fill:#27ae60,color:#fff
Being fluent with the three headline equations — I = Q/t, V = W/Q and V = IR — is the single most reliable way to pick up marks on an Edexcel electricity paper. Work through the following examples carefully and make sure you understand every substitution, unit conversion and rearrangement.
A current of 0.25 A flows through a torch bulb for 4 minutes. Calculate the charge that has flowed.
Step 1 — identify the equation. We know current and time, and we want charge, so rearrange I = Q/t to Q = It.
Step 2 — convert units. The equation uses seconds, so 4 minutes = 4 × 60 = 240 s.
Step 3 — substitute. Q = 0.25 × 240 = 60 C.
Step 4 — state the answer with units. Charge Q = 60 C.
Exam Tip: Edexcel markschemes reward unit conversion as a separate step. Show the "× 60" explicitly. Students who jump straight to
Q = 0.25 × 4lose both the working mark and the final answer.
A cell transfers 12 J of energy to every 4 C of charge that passes through it. Calculate the potential difference across the cell.
Using V = W/Q: V = 12 / 4 = 3 V. The cell has a potential difference of 3 V.
A 6 V battery drives a current of 2 A through a resistor for 30 s. Calculate: (a) the resistance, (b) the charge that flowed.
(a) R = V/I = 6 / 2 = 3 Ω.
(b) Q = It = 2 × 30 = 60 C.
| Quantity | What it is | Equation | Unit | Measuring instrument | Connected |
|---|---|---|---|---|---|
| Current (I) | Rate of flow of charge | I = Q/t | ampere (A) | ammeter | in series |
| Potential difference (V) | Energy transferred per unit charge | V = W/Q | volt (V) | voltmeter | in parallel |
| Resistance (R) | Opposition to current flow | R = V/I | ohm (Ω) | ohmmeter / calculated | n/a |
Common Mistake: Confusing current and voltage. Current is what flows through a component; voltage is what pushes current through and is measured across two points. If a student writes "the voltage flowing through the resistor", an examiner will deduct marks even if the numerical answer is correct.
Common Mistake: Forgetting that an ohmic conductor only obeys Ohm's law at constant temperature. A filament lamp is not an ohmic conductor because its resistance increases as it heats up.
Common Mistake: Quoting charge in amperes. Charge is measured in coulombs (C), not amperes. Amperes are for current (coulombs per second).
flowchart LR
A[Cell/Battery] -->|gives energy to charge| B[Charge carriers]
B -->|flow as current| C[Component e.g. bulb]
C -->|energy transferred to surroundings| D[Light + Heat]
D -.->|charge returns to cell| A
style A fill:#2c3e50,color:#fff
style C fill:#e67e22,color:#fff
The diagram reminds you that the same charge loops around a simple circuit. What changes is the energy each coulomb carries — the cell lifts it "up" in energy; the component transfers this energy to the surroundings.
A grade 4–5 response to "Explain the difference between current and potential difference" might say: "Current is the flow of charge measured in amps. Potential difference is the push from the cell measured in volts." This is accurate but imprecise.
A grade 7–9 response uses the precise terms: "Current is the rate of flow of charge (I = Q/t), measured in amperes, with the ammeter placed in series. Potential difference is the energy transferred per unit charge (V = W/Q), measured in volts, with the voltmeter placed in parallel. In an ohmic conductor at constant temperature, V and I are directly proportional, obeying Ohm's law (V = IR)." Notice the use of the formal terms charge, rate, per unit charge, and Ohm's law — this precision distinguishes higher grades. Always distinguish clearly between series vs parallel connections when describing measurement.
Edexcel alignment: This content is aligned with Edexcel GCSE Physics (1PH0) specification Topic 7 Electricity and circuits — specifically 7.2 Circuits and current, 7.3 Resistance, and parts of 7.5 Energy transfer. Assessed on Paper 2.