# Resistance

> CIE IGCSE Physics · 0625 2026-2028
> Source: https://www.owlsprep.com/study/cie-0625-u4-resistance/

This guide covers all Core and Extended resistance content for CIE IGCSE Physics 0625 (2026-2028 syllabus), including Ohm’s Law calculations, component resistance behaviours, and series/parallel resistance rules for Extended candidates.

**Prerequisites:** [Electric current and potential difference](https://www.owlsprep.com/study/cie-0625-u4-current-potential-difference/); [Basic circuit components and diagrams](https://www.owlsprep.com/study/cie-0625-u4-circuit-diagrams/)

## Learning objectives

- Define resistance using the formula R = V/I and recall the unit of resistance (ohm, Ω)
- Calculate resistance, current or potential difference using Ohm's Law for ohmic and non-ohmic components
- Describe how resistance changes for fixed resistors, filament lamps and diodes
- (Extended only) Calculate total resistance for series and parallel resistor combinations
- Describe the practical method to measure resistance of a component

## Core: Definition of Resistance and Ohm's Law

**Resistance** — Resistance is a measure of how much a component opposes the flow of electric current through it, defined as the ratio of potential difference across the component to the current flowing through it.

*Notation:* R

*Example:* A 10 Ω resistor allows 0.2 A of current to flow when 2 V is applied across it.

The formula relating resistance, potential difference and current is known as Ohm's Law, which applies to ohmic conductors (components where resistance remains constant at constant temperature, regardless of the current flowing through them).

$$R = \frac{V}{I}$$

Where: $R$ = resistance in ohms (Ω), $V$ = potential difference in volts (V), $I$ = current in amperes (A).

**Worked example:** Calculate the resistance of a fixed resistor if a potential difference of 12 V causes a current of 0.3 A to flow through it.

1. Write down known values and formula: $V = 12$ V, $I = 0.3$ A, $R = V/I$
2. $$R = \frac{12}{0.3}$$
3. Calculate the result: $R = 40$ Ω

> **Exam tip:** Always convert units to base units before substituting into the formula: divide current in mA by 1000 to get A, and convert mV to V by dividing by 1000.

## Extended Only: Resistance Characteristics of Common Components

Different components have distinct resistance behaviours that are frequently tested in exams. You will be expected to link current-voltage (I-V) graph shapes to these behaviours.

- Fixed resistor (ohmic conductor): Resistance is constant at constant temperature, so its I-V graph is a straight line through the origin.
- Filament lamp: As current increases, the metal filament heats up, increasing resistance, so the I-V graph curves as current rises.
- Diode: Resistance is very high in the reverse direction, so current only flows in the forward direction above a small threshold voltage (~0.7 V for silicon diodes).

**Worked example:** A student measures a current of 0.1 A through a filament lamp when the potential difference is 2 V. When the potential difference is increased to 6 V, the current is 0.2 A. Explain why the resistance changes.

1. Calculate resistance at both readings using $R = V/I$
2. $$R_1 = \frac{2}{0.1} = 20 \text{ } Ω$$
3. $$R_2 = \frac{6}{0.2} = 30 \text{ } Ω$$
4. The higher current increases the temperature of the filament, which increases the resistance of the metal wire as temperature rises.

> **tip**
>
> If a question asks why a filament lamp's resistance increases with current, always link the change to rising temperature of the filament, not just the increase in current.

## Extended Only: Series and Parallel Resistance Rules

**Total Circuit Resistance** — The equivalent single resistance that would draw the same current from the power supply as the full combination of components in the circuit.

For resistors connected in series, the total resistance is the sum of the individual resistances:

$$R_{total} = R_1 + R_2 + R_3 + ...$$

For resistors connected in parallel, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances:

$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$$

**Worked example:** Calculate the total resistance of a circuit with two resistors of 4 Ω and 6 Ω connected (a) in series, (b) in parallel.

1. (a) Series: sum the individual resistances
2. $$R_{total} = 4 + 6 = 10 \text{ } Ω$$
3. (b) Parallel: use the reciprocal formula
4. $$\frac{1}{R_{total}} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}$$
5. $$R_{total} = \frac{12}{5} = 2.4 \text{ } Ω$$

> **Exam tip:** For two resistors in parallel, you can use the shortcut formula $R_{total} = (R_1 \times R_2) / (R_1 + R_2)$ to save time, but always show your working to earn full marks.

## Core + Extended: Practical Measurement of Resistance

You will be asked to describe experiments to measure the resistance of a component in both written and practical assessments. Follow this standard method:

1. Set up a series circuit with the test component, an ammeter (in series), a voltmeter (in parallel across the component), a variable resistor, and a power supply.
2. Adjust the variable resistor to change the current through the component, recording at least 3 pairs of current and potential difference readings.
3. Calculate resistance for each pair using $R = V/I$: average results for ohmic components, or plot an I-V graph for non-ohmic components to analyse resistance changes.

**Check your understanding**

1. Why is the voltmeter connected in parallel across the test component?

   *Why:* Ammeters are connected in series to measure current through the circuit, while voltmeters are connected in parallel to measure the voltage drop across a specific component.

## Common pitfalls

- **Wrong:** Forgetting to convert mA to A before calculating resistance
  - Why it fails: This produces a resistance value 1000x higher than the correct answer, leading to lost calculation marks.
  - Correct: Always divide current in mA by 1000 to get base units (A) before substituting into $R = V/I$.
- **Wrong:** Assuming all components follow Ohm's Law (constant resistance)
  - Why it fails: Filament lamps, diodes, thermistors and LDRs are non-ohmic, so their resistance changes with current, temperature or light level.
  - Correct: Only assume constant resistance for fixed resistors at constant temperature; explicitly link resistance changes to the relevant variable (e.g. temperature for lamps) for non-ohmic components.
- **Wrong:** Adding parallel resistors directly instead of using the reciprocal formula (Extended only)
  - Why it fails: Parallel resistance is always lower than the smallest individual resistor, so direct addition produces an incorrectly high total resistance.
  - Correct: Use the reciprocal formula for parallel resistance, and check your result is smaller than the lowest individual resistor in the combination.
- **Wrong:** Stating that diodes have zero resistance in the forward direction
  - Why it fails: Diodes have a small threshold voltage (~0.7 V) so their resistance is low but not zero once this voltage is reached.
  - Correct: State that diodes have very high resistance in the reverse direction, and low resistance in the forward direction above the threshold voltage.

## Cheatsheet

| Concept | Core Rule/Formula | Extended Rule/Formula | Unit |
| --- | --- | --- | --- |
| Resistance definition | $R = V/I$ | Same as Core | Ω (ohm) |
| Fixed resistor I-V graph | Out of scope | Straight line through the origin (constant resistance at constant temperature) | - |
| Filament lamp I-V graph | Out of scope | Curve of decreasing gradient (resistance rises as temperature rises) | - |
| Series resistance | Out of scope | $R_{total} = R_1 + R_2 + ...$ | Ω |
| Parallel resistance | Out of scope | $1/R_{total} = 1/R_1 + 1/R_2 + ...$ | Ω |

## What's next

Now that you have mastered resistance concepts, you can apply these rules to more complex circuit problems, including those involving variable components like thermistors and LDRs that are frequently tested in both Core and Extended papers. You will also use resistance calculations when learning about electrical power and energy transfer, a high-weight topic for Paper 2 and Paper 4 exams. Extended candidates should practice combining series and parallel resistance calculations with current and potential difference rules to solve multi-step circuit problems, which are often worth 3-4 marks in extended theory papers.

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