Resistance and Ohm's Law

When charge flows through a component, it encounters opposition to its flow. This opposition is called resistance, and understanding it is essential for analysing every circuit you will meet at A-Level.

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What Is Resistance?

Resistance is a measure of how much a component opposes the flow of current. It is defined as the ratio of the potential difference across a component to the current flowing through it:

R = V / I

where:

• R = resistance in ohms (Ω)
• V = potential difference in volts (V)
• I = current in amperes (A)

One ohm means that a potential difference of one volt drives a current of one ampere through the component.

This equation can be rearranged to give two other useful forms:

• V = IR (to find the p.d. across a component)
• I = V / R (to find the current through a component)

Worked Example 1

A current of 0.25 A flows through a 48 Ω resistor. What is the potential difference across it?

Solution:

• V = IR = 0.25 × 48 = 12 V

Worked Example 2: Finding Current Through Multiple Resistors

A 9.0 V battery is connected to a circuit containing a single 180 Ω resistor. Calculate the current. The resistor is then replaced with a 360 Ω resistor. What happens to the current?

Solution:

• With 180 Ω: I = V/R = 9.0/180 = 0.050 A (50 mA)
• With 360 Ω: I = V/R = 9.0/360 = 0.025 A (25 mA)
• Doubling the resistance halves the current, as expected from I = V/R with constant V.

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Ohm's Law

Ohm's law states that for a conductor at constant temperature, the current through it is directly proportional to the potential difference across it.

Mathematically: V ∝ I (at constant temperature)

This means that if you double the voltage, the current doubles. A graph of V against I is a straight line through the origin. The gradient of this line equals the resistance.

It is important to understand that Ohm's law is not the same as V = IR. The equation V = IR is a definition of resistance — it applies to every component, whether or not it obeys Ohm's law. Ohm's law is an additional statement about the behaviour of certain components: that their resistance remains constant as the p.d. changes.

A component that obeys Ohm's law is called an ohmic conductor. A component that does not is called a non-ohmic conductor.

Critical exam distinction: V = IR is always true as a definition. Ohm's law is a special case where R is constant. Examiners love testing whether students confuse these two ideas.

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I-V Characteristics

The best way to understand how a component behaves is to look at its I-V characteristic — a graph showing how current varies with potential difference.

flowchart TD
    A[Given an I-V graph] --> B{Is it a straight line through the origin?}
    B -->|Yes| C[Ohmic conductor - constant R]
    B -->|No| D{Does the curve flatten at higher V?}
    D -->|Yes| E{Is it symmetrical?}
    E -->|Yes| F[Filament lamp - R increases with temperature]
    E -->|No| G{Current only in one direction?}
    G -->|Yes| H[Diode - threshold ~0.6 V for silicon]
    D -->|No| I[Check for other non-ohmic behaviour]

Ohmic Resistor (at constant temperature)

The I-V graph is a straight line through the origin. The resistance is constant at all values of V and I. The gradient of the line equals 1/R.

Filament Lamp

The graph curves away from the current axis. As the voltage increases, the filament heats up, causing the metal atoms to vibrate more. This increases the resistance. A filament lamp is therefore a non-ohmic conductor — its resistance increases with temperature.

At low voltages, the lamp is approximately ohmic because it has not yet heated up significantly.

Semiconductor Diode

A diode only allows significant current flow in one direction — the forward direction. In forward bias (positive voltage applied to the anode), no significant current flows until the voltage reaches a threshold of approximately 0.6 V for silicon. Beyond this, current increases rapidly.

In reverse bias (negative voltage), only a tiny leakage current flows (typically microamps). If the reverse voltage becomes too large, the diode can break down.

Light-Dependent Resistor (LDR)

An LDR is made from a semiconductor material (typically cadmium sulphide). In the dark, few charge carriers are available and the resistance is very high (often hundreds of kΩ or more). As light intensity increases, more electron-hole pairs are liberated, increasing the number of charge carriers and reducing the resistance.

Summary: more light → more charge carriers → lower resistance

Thermistor (NTC type)

A negative temperature coefficient (NTC) thermistor is made from a semiconductor material. At low temperatures, the resistance is high. As temperature increases, more electrons gain enough energy to escape from their atoms, increasing the number of charge carriers. The resistance decreases.

Summary: higher temperature → more charge carriers → lower resistance

Complete Component Summary Table

Component	I-V Shape	How R Changes	Ohmic?	Key Physics
Ohmic resistor	Straight line through origin	Constant	Yes	Fixed number of carriers, constant collision rate
Filament lamp	Curve flattening at high V	Increases with V	No	Heating increases lattice vibrations
Diode	Near-zero below 0.6 V, then steep	Very high → low	No	Threshold voltage for forward conduction
LDR	Depends on light level	Decreases with light	Varies	Photons liberate electron-hole pairs
NTC thermistor	Depends on temperature	Decreases with temperature	Varies	Thermal energy frees charge carriers

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Factors Affecting the Resistance of a Wire

For a uniform conductor, the resistance depends on three factors:

1. Length (L): Longer wires have more resistance. R ∝ L
2. Cross-sectional area (A): Thicker wires have less resistance. R ∝ 1/A
3. Material: Different materials have different abilities to conduct. This is characterised by the resistivity (ρ), which you will study in the next lesson.

These factors combine to give:

R = ρL / A

Worked Example 3

A copper wire has a length of 2.0 m and a cross-sectional area of 1.0 × 10⁻⁶ m². The resistivity of copper is 1.7 × 10⁻⁸ Ω m. Calculate the resistance.

Solution:

• R = ρL / A = (1.7 × 10⁻⁸ × 2.0) / (1.0 × 10⁻⁶)
• R = 3.4 × 10⁻⁸ / 1.0 × 10⁻⁶
• R = 0.034 Ω (or 34 mΩ)

Worked Example 4: Comparing Two Wires

Wire A is 1.0 m long with diameter 0.50 mm. Wire B is 2.0 m long with diameter 1.0 mm. Both are made of the same material. What is the ratio R_A / R_B?

Solution:

• R = ρL/A and A = π(d/2)² = πd²/4
• R_A = ρ(1.0) / [π(0.50 × 10⁻³)²/4]
• R_B = ρ(2.0) / [π(1.0 × 10⁻³)²/4]
• R_A/R_B = (1.0/0.50²) / (2.0/1.0²) = (1.0/0.25) / (2.0/1.0) = 4.0/2.0 = 2.0
• Wire A has twice the resistance despite being half the length, because it is much thinner.

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Microscopic Model of Resistance

At the microscopic level, resistance arises because the moving charge carriers (electrons in a metal) collide with the vibrating lattice ions. These collisions transfer kinetic energy from the electrons to the ions, which heats the conductor.

When temperature increases, the lattice ions vibrate with greater amplitude, making collisions more frequent. This is why the resistance of metals increases with temperature — there are more obstacles for the electrons to navigate.

In semiconductors, the opposite effect dominates: higher temperatures liberate more charge carriers, and this increase in the number of carriers outweighs the increased collision rate. This is why the resistance of thermistors and LDRs decreases as temperature or light intensity increases.

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Common Exam Mistakes

Mistake	Why It Is Wrong	Correct Approach
"V = IR is Ohm's law"	V = IR is a definition of R that applies to all components	Ohm's law states V ∝ I at constant temperature (R is constant)
Reading R from the gradient of an I-V graph	Gradient of I vs V is 1/R, not R	R = V/I at a point, or gradient = 1/R for a straight-line I-V graph
Assuming a filament lamp has constant R	Its resistance increases as it heats	State that R changes with temperature
Forgetting units for area	Using mm² instead of m² in R = ρL/A	Always convert to SI: 1 mm² = 1 × 10⁻⁶ m²

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A-Level Deep Dive: Resistance and Ohm's Law

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic 3.2 (resistance and Ohm's law) establishes the definition of resistance $R = V/I$R=V/I, the statement of Ohm's law as the special case in which $R$R is constant at constant temperature, and the qualitative I–V characteristics of an ohmic conductor, a filament lamp, a semiconductor diode, an LDR and an NTC thermistor (refer to the official Pearson Edexcel specification document for exact wording). It threads through every subsequent strand: 3.3 (resistivity, $R = \rho L/A$R=ρL/A), 3.4 (series/parallel networks) and 3.5 ($V = \varepsilon - Ir$V=ε−Ir). The Edexcel formula booklet lists $V = IR$V=IR but does not list Ohm's law as an equation — examiners expect it to be quoted in words and applied as a condition, not as an algebraic identity.

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Worked example with full mark scheme

Question (8 marks):

A student investigates the I–V characteristic of a small filament lamp rated $6.0\ \text{V}, 0.50\ \text{A}$6.0 V,0.50 A and records:

$V/\text{V}$V/V	$0.50$0.50	$1.0$1.0	$2.0$2.0	$4.0$4.0	$6.0$6.0
$I/\text{A}$I/A	$0.18$0.18	$0.27$0.27	$0.36$0.36	$0.45$0.45	$0.50$0.50

(a) Calculate the resistance of the lamp at $V = 0.50\ \text{V}$V=0.50 V and at $V = 6.0\ \text{V}$V=6.0 V. (2)

(b) Use your answers to (a) to explain whether the lamp is an ohmic conductor. (2)

(c) Sketch the I–V characteristic on labelled axes. (2)

(d) Explain, in terms of the microscopic behaviour of the metal, why the resistance changes as the lamp heats up. (2)

Solution with mark scheme:

(a) M1 — apply $R = V/I$R=V/I. A1 — $R \approx 2.8\ \Omega$R≈2.8 Ω at $0.50\ \text{V}$0.50 V and $R = 12\ \Omega$R=12 Ω at $6.0\ \text{V}$6.0 V. Common error: chord gradient instead of the point ratio.

(b) B1 — the lamp is not ohmic: $R$R has changed by a factor of about 4 as the p.d. rose twelvefold. B1 — Ohm's law requires $V \propto I$V∝I at constant temperature, and the filament heats with applied p.d., so the condition fails.

(c) B1 — axes $I/\text{A}$I/A vertical, $V/\text{V}$V/V horizontal. B1 — approximately linear through origin at low $V$V, curving away from the current axis at higher $V$V, passing through $(6.0\ \text{V}, 0.50\ \text{A})$(6.0 V,0.50 A).

(d) B1 — current heats the filament; lattice ions vibrate more, electron–lattice collisions become more frequent. B1 — by $I = nAve$I=nAve (with $n$n, $A$A, $e$e fixed), more scattering reduces drift velocity at a given field, so smaller current at the same p.d. — larger resistance. Naming the mechanism is essential for the second mark.

Total: 8 marks (M1 A1 + B1 B1 + B1 B1 + B1 B1).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A student is given an unmarked component and connects it to a variable d.c. supply with an ammeter in series and a voltmeter in parallel.

(a) Explain how the ammeter must be connected and why it should have a very low resistance. (2)

(b) The current is essentially zero for $V$V between $-5.0\ \text{V}$−5.0 V and $+0.55\ \text{V}$+0.55 V, and rises sharply for $V > +0.6\ \text{V}$V>+0.6 V. Identify the component and justify. (2)

(c) Suggest one modification to protect the component from damage at high forward currents. (2)

Mark scheme by AO: