Potential Difference and Resistance

If current is the flow of charge, what makes that charge flow, and what holds it back? The "push" that drives the current is the potential difference (voltage) supplied by the cell, and the "opposition" to the flow is the resistance of the components. Together with current, these two quantities are tied by one of the most important relationships in all of electricity: $V = IR$V=IR. This lesson, part of Topic P3 of OCR Gateway Combined Science A, defines potential difference and resistance, works through $V = IR$V=IR in every direction, explains how to connect an ammeter and a voltmeter, and sets out the required practical for finding how the resistance of a wire depends on its length.

By the end of this lesson you should be able to define potential difference and resistance with their units, recall and rearrange $V = IR$V=IR, explain how to connect an ammeter (in series) and a voltmeter (in parallel), describe the required practical on the resistance of a wire, and carry out worked calculations.

This lesson builds AO1 understanding of potential difference and resistance, AO2 application when you rearrange and use $V = IR$V=IR and set up the required practical on the resistance of a wire, and AO3 analysis when you interpret the results of that experiment.

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Potential Difference (Voltage)

The potential difference (p.d.) across a component, also called the voltage, is the energy transferred per unit charge as charge passes through it. It is, in effect, the "push" that drives the charge round the circuit: a cell with a larger voltage gives each coulomb of charge more energy and so drives a larger current through a given component.

Potential difference is given the symbol $V$V and measured in volts ($\text{V}$V). One volt means one joule of energy transferred per coulomb of charge. Across the cell, the p.d. measures the energy given to each coulomb; across a component such as a lamp, the p.d. measures the energy transferred by each coulomb as it passes through.

Exam Tip: Potential difference is energy transferred per coulomb (joules per coulomb). It is measured across a component (between two points), not "through" it — which is why a voltmeter goes across the component, in parallel.

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Resistance

Resistance is a measure of how difficult it is for charge to flow through a component. A component with a high resistance opposes the current strongly, so only a small current flows for a given voltage; a low resistance lets a larger current through for the same voltage. Resistance is given the symbol $R$R and measured in ohms, symbol $\Omega$Ω (the Greek letter omega).

Resistance arises because, as the electrons drift through a component, they collide with the vibrating atoms of the material, which slows their flow and transfers energy to the atoms (heating the component). The more collisions, the greater the resistance. The resistance of a piece of wire depends on three things: its length, its thickness (cross-sectional area), and the material it is made from. A longer wire has more resistance, because the electrons must travel further and undergo more collisions on the way. A thinner wire has more resistance, because the charge is squeezed through a smaller area. And some materials are naturally better conductors than others: copper, for example, has a much lower resistance than the nichrome alloy used for heating elements, which is chosen for its higher resistance so that it heats up.

This is not just theory — it explains real engineering choices you see every day. The wires that carry electricity around your home are made of copper and are kept relatively thick, precisely so that they have a low resistance: you do not want the connecting wires to waste energy as heat or to get warm, because that would be inefficient and could even be a fire risk. By contrast, the heating element inside a toaster, kettle or hairdryer is a long, thin coil of nichrome — a material deliberately chosen for its high resistance — so that it does the opposite and gets hot enough to glow or to warm the air. The same physics, applied in reverse, gives two completely different jobs. A useful way to summarise it: low resistance for carrying electricity, high resistance for producing heat.

Watch the vocabulary: it is easy to muddle "conductor" and "resistance". A good conductor (like copper) has a low resistance; a poor conductor has a high resistance. An insulator (like plastic) has such a high resistance that virtually no current flows through it at all, which is why the copper wires in a plug are wrapped in plastic.

Exam Tip: Resistance ($\Omega$Ω) measures the opposition to the flow of charge. For a fixed voltage, more resistance means less current. A longer or thinner wire has more resistance. The unit is the ohm, symbol $\Omega$Ω.

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The Key Equation: $V = IR$V=IR

Potential difference, current and resistance are linked by:

$V = IR$V=IR

where $V$V is the potential difference in volts ($\text{V}$V), $I$I is the current in amperes ($\text{A}$A), and $R$R is the resistance in ohms ($\Omega$Ω). In words: the voltage across a component equals the current through it multiplied by its resistance. This single equation underlies almost every circuit calculation in the topic.

It rearranges to make any quantity the subject:

$V = IR \qquad I = \dfrac{V}{R} \qquad R = \dfrac{V}{I}$V=IRI=RV​R=IV​

Worked Example 1 — Finding voltage

A current of $0.5 \text{ A}$0.5 A flows through a resistor of resistance $20\ \Omega$20 Ω. Calculate the potential difference across it.

Step 1 — write the equation: $V = IR$V=IR.

Step 2 — substitute: $V = 0.5 \times 20$V=0.5×20.

Step 3 — calculate: $V = 10 \text{ V}$V=10 V.

Answer: the potential difference is $10 \text{ V}$10 V.

Worked Example 2 — Rearranging for resistance

A lamp has a potential difference of $6 \text{ V}$6 V across it and a current of $3 \text{ A}$3 A flowing through it. Calculate its resistance.

Step 1 — rearrange for resistance: $R = \dfrac{V}{I}$R=IV​.

Step 2 — substitute: $R = \dfrac{6}{3}$R=36​.

Step 3 — calculate: $R = 2\ \Omega$R=2 Ω.

Answer: the resistance of the lamp is $2\ \Omega$2 Ω.

Worked Example 3 — Rearranging for current

A resistor of $1500\ \Omega$1500 Ω is connected to a $3 \text{ V}$3 V supply. Calculate the current flowing.

Step 1 — rearrange for current: $I = \dfrac{V}{R}$I=RV​.

Step 2 — substitute: $I = \dfrac{3}{1500}$I=15003​.

Step 3 — calculate: $I = 0.002 \text{ A}$I=0.002 A (which is $2 \text{ mA}$2 mA).

Answer: the current is $0.002 \text{ A}$0.002 A, or $2 \text{ mA}$2 mA.

Worked Example 4 — Reading off a resistor from a pair of measurements

In an experiment, a component has a potential difference of $4.5 \text{ V}$4.5 V across it and a current of $150 \text{ mA}$150 mA through it. Calculate its resistance.

Step 1 — convert the current to amperes first: $150 \text{ mA} = 0.15 \text{ A}$150 mA=0.15 A.

Step 2 — rearrange for resistance: $R = \dfrac{V}{I}$R=IV​.

Step 3 — substitute: $R = \dfrac{4.5}{0.15}$R=0.154.5​.

Step 4 — calculate: $R = 30\ \Omega$R=30 Ω.

Answer: the resistance is $30\ \Omega$30 Ω. (Had the current been left as $150$150, the answer would have come out as a tiny fraction of an ohm — a thousand times too small — which is why the conversion matters.)

Exam Tip: Lay out every $V = IR$V=IR calculation as equation, substitution, answer with unit. Keep currents in amperes and resistances in ohms; a current given in milliamps ($\text{mA}$mA) must be converted to amps first ($1 \text{ mA} = 0.001 \text{ A}$1 mA=0.001 A).

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Measuring Current and Voltage

To take the readings needed for $V = IR$V=IR, you use two meters, and how you connect them is a frequent exam point — the two rules are opposite to each other: