Current, resistance and potential difference — AQA GCSE Physics

SPEC 8463 • MODULE 2 • LESSON 2.02

V = I R

The most-used equation in Paper 1 electricity. Two definitions, one equation, one specific phrase that earns or loses the mark — and the IV line that tells you when you're allowed to use it.

Building on Lesson 2.01

Last lesson: charge moves, and current is the rate at which it moves. But that doesn't tell you why charge flows in the first place — or what controls how much.

Three Quantities, One Relationship

Three quantities run every electricity calculation in Paper 1:

Potential difference (V) — what drives the charge.

Current (I) — the response: how much charge flows.

Resistance (R) — what opposes the flow.

They're locked together by a single equation: V = I R. Get the definitions exact and the calculations follow.

Spec Note — Either Term Is Accepted

The AQA spec is explicit: "Questions will be set using the term potential difference. Students will gain credit for the correct use of either potential difference or voltage." Use whichever you find clearer — both score.

Defining Potential Difference

Potential difference (pd, or voltage) is the quantity that makes charges move. The mark-scheme definition links it directly to energy.

Potential Difference

The energy transferred per unit charge passing between two points in a circuit. Symbol: V. Unit: the volt (V).

What This Definition Means in Practice

If you put 1 coulomb of charge between two points with a 6 V pd between them, 6 joules of energy are transferred as the charge moves between those two points. Double the pd → double the energy per charge. Half the pd → half the energy per charge. pd and energy are proportional.

The Volt — 1 J Per Coulomb

The volt is defined directly from energy and charge. It's worth memorising the unit identity — it shows up later in lesson 2.08 (the energy-charge equation E = Q V).

1 V = 1 J / C

one volt = one joule of energy per coulomb of charge

Higher pd

Each coulomb of charge transfers more energy as it passes through. A 230 V mains lamp transfers 230 J per coulomb — far more than a 1.5 V cell's 1.5 J per coulomb.

Lower pd

Each coulomb transfers less energy. Lower-voltage cells (e.g. a button cell) deliver less energy per coulomb of charge — that's why a button cell can't power a lamp brightly.

Defining Resistance

Resistance is the third quantity in V = I R — and the one that limits current. Mark schemes accept the short version below.

Electrical Resistance

The opposition to the flow of electric charge. Calculated as R = V / I. Symbol: R. Unit: the ohm (Ω).

The Spec Sentence to Memorise

The AQA spec states it plainly: "The greater the resistance of the component, the smaller the current for a given potential difference across the component."

For the same pd: more resistance → less current. That's the qualitative version of V = I R, and it's a 1-mark recall point on its own.

The Ohm — 1 V Per Amp

Like the volt, the ohm has a clean unit identity that links it back to V and I.

1 Ω = 1 V / A

one ohm = one volt of pd per ampere of current

Reading That Backwards

If a component has a resistance of 1 Ω, then a 1 V pd across it drives a current of 1 A through it. A 10 Ω component? Same 1 V pd drives only 0.1 A — ten times the resistance, one tenth of the current.

Symbol Notes

The symbol for ohm is the Greek capital letter omega: Ω. In handwritten work, examiners accept "ohms" or the omega symbol. Don't write "O" (zero or letter O) — it's ambiguous and risks losing the unit mark.

The Equation: V = I R

The equation that links all three. Recall and apply — it's on the recall list, not the equation sheet.

V = I × R

potential difference (V) = current (A) × resistance (Ω)

QuantitySymbolUnit

Potential differenceVvolts (V)

CurrentIamperes (A) — "amp" is also accepted

ResistanceRohms (Ω)

Three Forms — Same Equation

Depending on what you're asked for, rearrange:

V = I × R (find pd) · I = V / R (find current) · R = V / I (find resistance)

The V = I R Triangle

If rearranging is what slows you down, the triangle is your friend. Cover the quantity you want — what's left tells you the calculation.

Cover V → V = I × R

Cover I → I = V / R

Cover R → R = V / I

V on top → division when you find I or R; multiplication when you find V.

Show the Equation You're Using

Even if you use the triangle in your head, always write the equation form you're applying as the first line of your working. AQA mark schemes routinely award the first mark for stating V = IR (or its rearrangement) — don't skip past it.

Worked Example — Find V

"A current of 0.60 A flows through a 20 Ω resistor. Calculate the potential difference across the resistor."3 MARKS

E

Equation: V = I × R

S

Substitute: V = 0.60 × 20

U

Unit: volts (V)

A

Answer: V = 12 V

✓

Sense check: Higher current or higher resistance gives higher pd — and 12 V is a sensible everyday voltage. Answer looks right.

Worked Example — Rearrange For R

"A 12 V cell drives a current of 0.40 A through a resistor. Calculate the resistance of the resistor."3 MARKS

E

Equation: V = I × R, rearranged → R = V / I

S

Substitute: R = 12 / 0.40

U

Unit: ohms (Ω)

A

Answer: R = 30 Ω

✓

Mark-scheme tip: Always write the rearrangement before substituting. The line "R = V / I" can score the equation mark even if your arithmetic goes wrong later.

The Slip That Loses Marks

One symbol misread. Three marks gone. This is the most common single error in Ohm's law calculations.

✗ Loses all 3 marks

V = I + R

V = 0.60 + 20

V = 20.6 V

The "+" instead of "×" gives a numerically plausible-looking answer (V around 20). It's wrong arithmetically and wrong physics. Mark scheme awards 0 — the equation itself is incorrect.

✓ Full marks

V = I × R

V = 0.60 × 20

V = 12 V

Multiplication, not addition. One symbol away from a full-mark answer.

Why This Happens

Under exam pressure, the equation triangle gets misread or misremembered. The mitigation is simple: write "V = I × R" with the multiplication sign visible — not "V = IR" where the symbol is implicit. It takes one extra second and removes the slip entirely.

Examiner Alert — Mixed Units in V = I R

Just like Q = I t needs everything in SI units, V = I R demands the same discipline. Mix prefixes inside the equation and the answer is wrong by a factor of 1000.

Convert Before Substituting

If a question gives current in mA, pd in mV, or resistance in kΩ, convert before you put numbers in. The conversion line goes between writing the equation and substituting:

I = 250 mA ÷ 1000 = 0.250 A

Then substitute the converted value. This is the same habit you built in lesson 2.01 — it pays off here too.

The Three Conversions to Memorise

mA → A (÷ 1000) · mV → V (÷ 1000) · kΩ → Ω (× 1000)

Mark schemes for V = I R questions follow the pattern from Q = I t: a follow-through clause may save the substitution mark, but the final answer mark goes to students whose value is numerically correct in SI units.

Ohmic Conductors

For most components R is constant — they obey V = I R cleanly. These are ohmic conductors. The definition has a phrase examiners specifically watch for.

Ohmic Conductor

A conductor in which the current is directly proportional to the potential difference across it, at constant temperature. Equivalently: the resistance remains constant as the current changes.

Two Equivalent Statements (Both Spec-Accepted)

The AQA spec gives the relationship two ways — both are mark-scheme accurate:

1. "Current is directly proportional to potential difference (at constant temperature)" — verbatim spec 4.2.1.4.

2. "Resistance remains constant as the current changes" — also verbatim spec 4.2.1.4.

Use whichever you find clearer. Doubling V doubles I; trebling V trebles I — and R never changes.

Examples

A fixed resistor at constant temperature is the textbook ohmic conductor. So is a metal wire, provided its temperature doesn't change. The constant-temperature condition is critical — if the component heats up, its resistance can change, and it stops being ohmic.

What's NOT Ohmic (Preview of Lesson 2.04)

Lamps, diodes, thermistors, and LDRs all have changing resistance. Their I–V graphs are not straight lines. We'll cover each in detail in 2.04 (RP4 — IV characteristics) — for now, just remember that "ohmic" is a special case, not the default.

Examiner Alert — Two Words That Decide The Mark

The phrase "at constant temperature" is part of the definition of an ohmic conductor — not an optional extra, not a footnote. Drop it and the mark goes.

✗ Risks 0

"An ohmic conductor is one where current is directly proportional to potential difference."

Almost right — but mark schemes have rejected this because it's incomplete . Without "constant temperature", the statement isn't physically true: a filament lamp at low pd is briefly proportional too.

✓ Mark-scheme accurate

"An ohmic conductor is one where current is directly proportional to potential difference, at constant temperature."

The condition is what makes it specific to ohmic conductors. Three small words. One mark earned.

Why Temperature Matters

When a conductor's temperature rises, the metal ions vibrate more vigorously, and electrons collide with them more often. More collisions = more opposition = higher resistance. So for current to stay strictly proportional to pd, the conductor must not heat up — hence the "constant temperature" clause.

The I–V Graph for an Ohmic Conductor

An I–V graph plots current (I, vertical axis) against potential difference (V, horizontal axis). For an ohmic conductor it gives one specific shape: a straight line through the origin.

I–V graph: ohmic conductor at constant temperature

Reading the Graph

Straight line → I and V are proportional. Through the origin → no current can flow without a driving pd; remove the source and the current is zero. Gradient = 1 / R → a steeper line means a smaller resistance (more current per volt). The graph for a 10 Ω resistor would be steeper than the graph for a 100 Ω resistor.

Practice Question

A 12 V battery is connected across a fixed resistor of resistance 3.0 kΩ.

Calculate the current flowing through the resistor.

Give your answer in amperes.3 MARKS

E

Equation: V = I × R, rearranged → I = V / R

Convert

Resistance to ohms: R = 3.0 kΩ × 1000 = 3000 Ω

S

Substitute: I = 12 / 3000

U

Unit: amperes (A)

A

Answer: I = 0.0040 A (or 4.0 × 10⁻³ A, or 4.0 mA)

⚠️

Mark loss check: Substituting 3.0 (kΩ) directly gives I = 4.0 A — wrong by a factor of 1000. The kΩ → Ω conversion line is the difference between full marks and a follow-through mark.

Three Things to Remember

The discipline that separates Level 3 answers from Level 1 in every Ohm's law question.

1. Definitions Are Worth Marks On Their Own

Potential difference = energy transferred per unit charge. Resistance = opposition to the flow of charge. Ohmic conductor = I directly proportional to V at constant temperature. Each is a 1-mark recall question and the wording is the mark — don't paraphrase loosely.

2. Write the Equation Form Before Substituting

"V = I × R" with the multiplication sign visible. Or its rearrangement: "R = V / I", "I = V / R". Always before you put numbers in. M1 is yours regardless of what happens next.

3. Convert Prefix Units Before Substituting

mA → A (÷ 1000), mV → V (÷ 1000), kΩ → Ω (× 1000). Show the conversion as a separate line. The follow-through clause in mark schemes only saves the substitution mark — the answer mark goes to whoever puts SI units in.

Summary

Definitions (Mark-Scheme Verbatim)

Potential difference = energy transferred per unit charge · symbol V · unit volt (V)

1 V = 1 J/C (one joule per coulomb)

Resistance = opposition to the flow of (electric) charge · calculated as R = V / I · unit ohm (Ω)

1 Ω = 1 V/A (one volt per ampere)

Ohmic conductor : current directly proportional to pd at constant temperature — equivalently, resistance remains constant as the current changes

The Equation (Recall)

Spec form: potential difference = current × resistance · V = I × R

Three forms: V = I R · I = V / R · R = V / I

Convert mA → A (÷ 1000), mV → V (÷ 1000), kΩ → Ω (× 1000) before substituting

Worked example: 12 V across 3.0 kΩ → I = 12 / 3000 = 4.0 × 10⁻³ A

The I–V Graph for an Ohmic Conductor

Straight line through the origin at constant temperature

Gradient = 1 / R — steeper line, smaller resistance

Doubling V doubles I → resistance is constant

Lamps, diodes, thermistors, LDRs do not obey this — that's lesson 2.04

Spec Quotes — Both Terms / Both Units Accepted

The AQA spec gives two explicit acceptances: "voltage" earns full credit alongside "potential difference", and "amp" is acceptable for "ampere". Use whichever you find clearer; both score on the mark scheme.

Checkpoint

You should now be able to:

✅ Define potential difference as the energy transferred per unit charge, with unit volts (1 V = 1 J/C), and use it interchangeably with "voltage".

✅ Define resistance as the opposition to the flow of charge, calculated as V / I, with unit ohms (1 Ω = 1 V/A).

✅ Apply V = I × R in any of its three forms, with confident unit conversion (mA → A, mV → V) shown as separate lines of working.

✅ State the ohmic-conductor definition correctly — including the "at constant temperature" clause without prompting.

✅ Recognise the I–V graph of an ohmic conductor as a straight line through the origin, with gradient equal to 1/R.

Next Lesson

2.03 — Resistance of a wire & Required Practical 3. We'll investigate how resistance depends on the length of a wire — one of the most reliably-examined practicals on Paper 1 — and meet the equipment, method, results, and the typical errors examiners flag.