A charge of 120\,C flows through a wire in 40\,s. Calculate the current. [2 marks]

WJEC WJEC 2019-style practice: A current of 1.5\,A flows through a heater for 40\,s. Calculate the charge that flows.

A topic 1.1 Calculate question on Q = It. State the values: current I = 1.5\,A and time t = 40\,s (1 mark for selecting the correct equation from the formula list). Substitute: Q = It = 1.5 × 40 = 60\,C (2 marks for the calculation and the unit coulombs). Markers reward the equation, the substitution and the answer in coulombs. A common error is to divide rather than multiply, or to omit the unit.

WJEC WJEC 2021-style practice: A resistor has a potential difference of 9.0\,V across it and a current of 0.30\,A through it. Calculate its resistance, and state what happens to the current if the pd is halved at constant resistance.

A topic 1.1 Calculate and Explain question. Use V = IR rearranged (Higher candidates rearrange; the formula list gives V = IR): R = VI = 9.00.30 = 30\, (2 marks for the rearrangement and the answer with units). If the pd is **halved** to 4.5\,V at constant resistance, then I = VR = 4.530 = 0.15\,A, so the **current also halves** (2 marks for the reasoning and the new current). Markers reward the resistance with units and the proportional change. A common error is to think halving the voltage leaves the current unchanged.

WalesPhysicsSyllabus dot point

What are current, potential difference and resistance, and how are they linked?

Electric current as the rate of flow of charge, the charge equation, potential difference as energy transferred per unit charge, resistance, and the equation linking potential difference, current and resistance.

A focused answer to WJEC GCSE Physics topic 1.1 on current, potential difference and resistance, covering current as the rate of flow of charge, the charge equation, potential difference as energy per unit charge, resistance, and the equation linking them.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Electric current and charge
Potential difference
Resistance
What affects resistance
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What this topic is asking

WJEC wants you to define electric current as the rate of flow of charge, use the charge equation, define potential difference as energy per unit charge, and link potential difference, current and resistance. This is part of topic 1.1 Electric circuits in Unit 1 of WJEC GCSE Physics (3420).

Electric current and charge

A larger current means more charge passes each second. One ampere is one coulomb per second. Conventional current is taken to flow from the positive to the negative terminal (opposite to the electron flow), and this is the direction used in circuit diagrams. Current is measured with an ammeter connected in series with the component.

Potential difference

A higher potential difference gives each coulomb of charge more energy to deliver to the components. Across a component such as a lamp, the potential difference tells you how much energy each coulomb gives up there, transferred as heat and light. Potential difference is measured with a voltmeter connected in parallel across the component.

Resistance

Rearranging gives $R = \dfrac{V}{I}$ (to calculate resistance) and $I = \dfrac{V}{R}$ (to calculate current). Higher tier candidates are expected to rearrange the equation; Foundation candidates are given it in the form they need. A high resistance lets only a small current flow for a given voltage. Resistance arises because the moving electrons collide with the fixed ions in the conductor, transferring energy and heating it.

What affects resistance

The resistance of a wire depends on its length (longer means more resistance), its cross-sectional area (thicker means less resistance), the material it is made of, and its temperature. For most metal conductors, heating increases the resistance, because the ions vibrate more strongly and obstruct the flow of electrons. Measuring how the resistance of a wire varies, or finding the resistance of a component, is a specified practical that can appear in Unit 3 and the written papers.

Finding the current and charge for a resistor

A $15\,\Omega$ resistor is connected across a $6.0\,\text{V}$ supply. Calculate the current through the resistor and the charge that flows in $2$ minutes.

step 1: Find the current

Rearrange $V = IR$ for current: $I = \dfrac{V}{R} = \dfrac{6.0}{15} = 0.40\,\text{A}$ .

step 2: Write down the time in seconds

Two minutes is $t = 2 \times 60 = 120\,\text{s}$ .

step 3: Find the charge

Use $Q = It = 0.40 \times 120 = 48\,\text{C}$ .

step 4: State the answers

The current is $0.40\,\text{A}$ and the charge that flows in two minutes is $48\,\text{C}$ . Both $V = IR$ and $Q = It$ are on the WJEC formula list, but you must select them and substitute correctly with the time in seconds.

Try this

Q1. A charge of $120\,\text{C}$ flows through a wire in $40\,\text{s}$ . Calculate the current. [2 marks]

Cue. Rearrange $Q = It$ : $I = \dfrac{Q}{t} = \dfrac{120}{40} = 3.0\,\text{A}$ .

Q2. State what is meant by a potential difference of one volt. [1 mark]

Cue. One joule of energy is transferred per coulomb of charge.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20193 marksA current of

1.5\,\text{A}

flows through a heater for

40\,\text{s}

. Calculate the charge that flows.

Show worked answer →

A topic 1.1 Calculate question on $Q = It$ . State the values: current $I = 1.5\,\text{A}$ and time $t = 40\,\text{s}$ (1 mark for selecting the correct equation from the formula list). Substitute: $Q = It = 1.5 \times 40 = 60\,\text{C}$ (2 marks for the calculation and the unit coulombs). Markers reward the equation, the substitution and the answer in coulombs. A common error is to divide rather than multiply, or to omit the unit.

WJEC 20214 marksA resistor has a potential difference of

9.0\,\text{V}

across it and a current of

0.30\,\text{A}

through it. Calculate its resistance, and state what happens to the current if the pd is halved at constant resistance.

Show worked answer →

A topic 1.1 Calculate and Explain question. Use $V = IR$ rearranged (Higher candidates rearrange; the formula list gives $V = IR$ ): $R = \dfrac{V}{I} = \dfrac{9.0}{0.30} = 30\,\Omega$ (2 marks for the rearrangement and the answer with units). If the pd is halved to $4.5\,\text{V}$ at constant resistance, then $I = \dfrac{V}{R} = \dfrac{4.5}{30} = 0.15\,\text{A}$ , so the current also halves (2 marks for the reasoning and the new current). Markers reward the resistance with units and the proportional change. A common error is to think halving the voltage leaves the current unchanged.

Related dot points

Sources & how we know this

WJEC GCSE Physics specification (3420) from 2016 — WJEC (2016)