Electric Current, Potential Difference, and Resistance Revision Notes | A Level AQA Physics

1. The Big Picture: What Are We Actually Measuring?

Picture a copper wire connected to a battery. Inside that wire, billions of electrons are drifting in one direction — not randomly, but driven by the battery. Three quantities describe everything happening in that wire: how many charges move per second, how much energy each charge carries, and how much the wire resists that movement. Those three quantities are current, potential difference, and resistance.

At A Level, you need more than a working definition — you need to understand what each quantity physically represents, use the correct equations with confidence, and apply them to unfamiliar problems. This lesson builds all three skills from the ground up.

2. Electric Current: Rate of Flow of Charge

Think of charge carriers (electrons in a metal) as water molecules flowing through a pipe. Current is the measure of how much charge flows past a fixed point every second — the electrical equivalent of flow rate. The more charges that pass per second, the higher the current.

The equation that captures this definition precisely is $I = \frac{\Delta Q}{\Delta t}$. The delta notation ($\Delta$) signals that we are dealing with a change in charge over a change in time — this matters in circuits where charge accumulates or is released over a finite interval.

A current of 1 A means that 1 coulomb of charge passes a point every second. To put that in perspective, 1 coulomb is approximately $6.24 \times 10^{18}$ electron charges — so even a modest current involves an enormous number of charge carriers moving together.

Worked example: Calculating charge transferred from current and time

A current of 0.40 A flows through a resistor for 3 minutes. Calculate the total charge transferred. [3 marks]

1. $\Delta t = 3 \times 60 = 180 \text{ s}$ — [M1] Unit conversion is essential — the equation requires seconds, not minutes. This is a common mark-losing error.
2. $I = \frac{\Delta Q}{\Delta t} \Rightarrow \Delta Q = I \times \Delta t$ — [M1] Rearranging correctly for $\Delta Q$ before substituting values.
3. $\Delta Q = 0.40 \times 180 = 72 \text{ C}$ — [A1] Arithmetic check: $0.40 \times 180 = 0.40 \times 100 + 0.40 \times 80 = 40 + 32 = 72$. Units: A × s = C. Correct.

Answer: $\Delta Q = 72 \text{ C}$

Current is the foundation of everything else in this lesson. Once you are confident with $I = \frac{\Delta Q}{\Delta t}$, the next question is: what drives that current in the first place? That is where potential difference comes in.

3. Potential Difference: Energy Per Unit Charge

A battery does not push electrons directly — it gives them energy. Potential difference (p.d.) measures how much energy is transferred per coulomb of charge that moves between two points. Think of it like the height difference between two points on a water slide: the greater the height difference, the more energy each water molecule gains as it falls.

A potential difference of 6 V across a component means that 6 joules of energy are transferred for every coulomb of charge that passes through it. The energy is transferred from the charge carriers to the component — for example, as heat in a resistor or light in a lamp.

Worked example: Calculating work done using potential difference

A 12 V battery drives a current of 2.0 A through a lamp for 5.0 s. Calculate the work done by the battery. [4 marks]

1. $\Delta Q = I \times \Delta t = 2.0 \times 5.0 = 10 \text{ C}$ — [M1] Using $I = \frac{\Delta Q}{\Delta t}$ rearranged. This links back to the current equation from the previous section.
2. $V = \frac{W}{Q} \Rightarrow W = V \times Q$ — [M1] Rearranging for work done $W$.
3. $W = 12 \times 10 = 120 \text{ J}$ — [A1] Arithmetic: $12 \times 10 = 120$. Units: V × C = J. Correct.
4. Power $= IV = 2.0 \times 12 = 24$ W; Energy $= P \times t = 24 \times 5.0 = 120$ J — [A1] Cross-check using $P = IV$ and $E = Pt$ confirms the answer. Both routes give 120 J.

Answer: $W = 120 \text{ J}$

Notice how the two equations $I = \frac{\Delta Q}{\Delta t}$ and $V = \frac{W}{Q}$ work together in multi-step problems. Examiners frequently set questions that require both — always identify which quantities you have before choosing which equation to apply first.

4. Resistance: Opposing the Flow

You now know that current is charge flow per second and that potential difference drives that flow by supplying energy. The missing piece is: why does the same voltage produce different currents in different components? The answer is resistance.

Resistance quantifies how strongly a component opposes the flow of charge. A component with high resistance requires a large potential difference to drive even a small current through it. A component with low resistance allows a large current for the same potential difference.

This definition of resistance is general — it applies to any component at any instant, regardless of whether the component obeys Ohm's law. Ohm's law is a special case that states $R$ is constant for a given conductor at constant temperature. The definition $R = \frac{V}{I}$ holds universally, even for non-ohmic components such as filament lamps and diodes where $R$ changes with current.

Worked example: Calculating resistance from potential difference and current

A component has a potential difference of 9.0 V across it and a current of 0.30 A through it. Calculate its resistance. [2 marks]

1. $R = \frac{V}{I}$ — [M1] State the equation before substituting.
2. $R = \frac{9.0}{0.30} = 30 \text{ } \Omega$ — [A1] Arithmetic: $9.0 \div 0.30 = 30$. Units: V ÷ A = Ω. Correct.

Answer: $R = 30 \text{ } \Omega$

The three equations $I = \frac{\Delta Q}{\Delta t}$, $V = \frac{W}{Q}$, and $R = \frac{V}{I}$ form a tightly connected set. Mastering their rearrangements and knowing when to chain them together is the core skill this topic tests.

5. Bringing It Together: The Required Practical

The equations for current, potential difference, and resistance are not abstract — they are measured directly in the laboratory. The standard practical for this topic uses an ammeter and voltmeter to determine the resistance of a component, and understanding the method is itself an exam requirement.

The straight line through the origin on the $V$–$I$ graph is the visual signature of an ohmic conductor. The gradient of this line gives resistance directly: $\text{gradient} = \frac{\Delta V}{\Delta I} = R$. When drawing the gradient triangle in an exam, always show the triangle on the graph and include units in $\frac{\Delta V}{\Delta I}$.

6. Multi-Step Problems: Chaining the Equations

Exam questions rarely ask you to apply a single equation in isolation. More often, you need to chain two or three equations together, carrying a result from one step into the next. The worked example below models the approach examiners expect.

Worked example: Multi-step problem: current, charge, and resistance

A resistor of resistance 15 Ω has a potential difference of 6.0 V across it. Calculate (a) the current through the resistor, and (b) the charge that flows through it in 4.0 minutes. [4 marks]

1. $R = \frac{V}{I} \Rightarrow I = \frac{V}{R} = \frac{6.0}{15}$ — [M1] Rearranging $R = \frac{V}{I}$ for current.
2. $I = \frac{6.0}{15} = 0.40 \text{ A}$ — [A1] Arithmetic: $6.0 \div 15 = 0.40$. Units: V ÷ Ω = A. Correct.
3. $\Delta t = 4.0 \times 60 = 240 \text{ s}$ — [M1] Always convert minutes to seconds before using $I = \frac{\Delta Q}{\Delta t}$.
4. $\Delta Q = I \times \Delta t = 0.40 \times 240 = 96 \text{ C}$ — [A1] Arithmetic: $0.40 \times 240 = 0.40 \times 200 + 0.40 \times 40 = 80 + 16 = 96$. Units: A × s = C. Correct.

Answer: (a) $I = 0.40 \text{ A}$; (b) $\Delta Q = 96 \text{ C}$

The three equations in this topic are among the most frequently examined in the entire A Level course. Getting full marks is about more than knowing the formulae — it is about presenting your working in a way that earns every available method mark, even if you make an arithmetic slip.

7. Exam Technique and Mark-Winning Strategies

For extended-response questions that ask you to compare or evaluate, structure your answer as: opening statement → first scientific point with reasoning → second point (or counter-argument) → conclusion. For example, comparing two components on a $V$–$I$ graph requires you to describe the shape of each curve, explain what the shape tells you about resistance, and then reach a comparative judgement.

These three equations — $I = \frac{\Delta Q}{\Delta t}$, $V = \frac{W}{Q}$, and $R = \frac{V}{I}$ — together define the language of circuit analysis at A Level. Current quantifies charge flow rate, potential difference quantifies energy per unit charge, and resistance quantifies opposition to that flow. Every calculation in this topic is a variation on these three relationships.