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 (Electricity) of OCR Gateway Science A, defines potential difference and resistance, works through $V = IR$V=IR in every direction, introduces the standard circuit symbols, and shows how an ammeter and a voltmeter are connected to measure current and voltage.

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, recognise and draw the common circuit symbols, explain how to connect an ammeter (in series) and a voltmeter (in parallel), and carry out worked calculations.

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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). As you saw in the last lesson, one volt means one joule of energy transferred per coulomb of charge ($V = E/Q$V=E/Q). 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 — this 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. This is why long, thin wires have more resistance than short, thick ones, and why a hot filament has more resistance than a cold one (a point explored fully in the I–V characteristics lesson).

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 there is less room for the electrons to flow through — 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 to make heating elements, which is deliberately chosen for its higher resistance so that it heats up. Temperature matters too: for most metals, raising the temperature raises the resistance, because the hotter atoms vibrate more and obstruct the electrons more often.

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

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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 — A current given in milliamps

A current of $250 \text{ mA}$250 mA flows through a component of resistance $8\ \Omega$8 Ω. Calculate the potential difference across it.

Step 1 — convert the current to amperes: $250 \text{ mA} = 0.25 \text{ A}$250 mA=0.25 A.

Step 2 — write the equation: $V = IR$V=IR.

Step 3 — substitute: $V = 0.25 \times 8$V=0.25×8.

Step 4 — calculate: $V = 2 \text{ V}$V=2 V.

Answer: the potential difference is $2 \text{ V}$2 V. (Had the current been left as $250$250 instead of $0.25$0.25, the answer would have come out $1000$1000 times too big — always convert milliamps to amps first.)

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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Circuit Symbols

Circuits are drawn using standard symbols so that any physicist can read them. You must be able to recognise and draw the common ones. The diagram below shows the symbols you need.