What does the Eduqas resistive components and sensing module cover?

It covers fixed and variable resistors (the four-band colour code, preferred E-series values, tolerance, and using a variable resistor as a potentiometer or rheostat), driving LEDs with a current-limiting series resistor, potential dividers (the divider equation, choosing resistor values for a target output, and the loading effect), sensing circuits with light-dependent resistors and NTC thermistors in potential dividers, and capacitors with the charge equation and the RC time constant that creates time delays.

Which equations recur most across this module?

The LED current-limiting resistor $R = \frac{V_S - V_L}{I}$, the potential-divider equation $V_\text{out} = V_\text{in}\frac{R_2}{R_1 + R_2}$, the capacitor charge $Q = CV$, and the time constant $\tau = RC$ with the 63% (charge) and 37% (discharge) figures after one time constant. Ohm's law underlies all of them.

What is the most common mistake in this module?

Unit slips dominate: leaving the LED current in milliamps, the resistors in mixed units, or (worst) the capacitance in microfarads instead of farads when using Q = CV or the time constant, which throws the answer out by a million. Other frequent errors are using the whole supply voltage across the LED resistor instead of subtracting the forward voltage, putting the wrong resistor on top in a divider, and getting the LDR or thermistor resistance direction backwards.

How do LDRs and thermistors make sensing circuits?

A light-dependent resistor's resistance falls as light rises, and an NTC thermistor's resistance falls as temperature rises. Putting either into a potential divider turns the resistance change into a voltage change, so the output voltage tracks light or temperature. Where you place the sensor (top or bottom of the divider) decides whether the output rises or falls with the input, and a variable partner resistor sets the switching threshold. The output then feeds a comparator that switches when the voltage crosses the threshold.

How should I revise resistive components and sensing?

Make the LED resistor calculation and the divider equation automatic, because they appear in almost every paper. Practise reading and writing the colour code, rounding to preferred values, and computing tolerance ranges. Drill choosing divider resistors for a target voltage and reasoning about the loading effect. Rehearse the LDR and thermistor resistance directions and how to place a sensor for a rising or falling output. Finally, master Q = CV and the time constant with the capacitance always in farads.

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Eduqas GCSE Electronics: resistive components and sensing (colour code, LED resistors, dividers, sensors, capacitors)

A deep-dive Eduqas GCSE Electronics guide to the resistive components and sensing module within Component 1. Covers fixed and variable resistors with the colour code, preferred values and tolerance, driving LEDs with a current-limiting resistor, potential dividers and choosing resistor values, sensing with LDRs and thermistors, and capacitors with the RC time constant and time delays.

Generated by Claude Opus 4.814 min readC490 Component 1Updated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this module actually demands
Components and the LED resistor
Dividers, sensing and timing
How this module is examined
Check your knowledge

What this module actually demands

Resistive components and sensing turns the circuit theory of the first module into real building blocks. It is where the two most-examined calculations in the course live: sizing the current-limiting resistor for an LED, and finding the output of a potential divider. It then uses the divider to build sensing subsystems with LDRs and thermistors, and introduces the capacitor and the RC time constant that every later timing circuit relies on. The examiners reward fluent calculation in base units, correct component choices to preferred values, and clear reasoning about how a sensing output changes with light or temperature.

This guide walks through the topics in order and sets out the exam patterns Eduqas repeats. Each topic has a matching dot-point page with practice; this overview ties them together.

Components and the LED resistor

Fixed and variable resistors read the four-band colour code (two digits, a multiplier, a tolerance), explain preferred E-series values and tolerance ranges, and use a variable resistor as a potentiometer (three terminals, variable voltage) or a rheostat (two terminals, variable resistance). Driving LEDs explains why an LED needs a series resistor, uses the forward voltage drop, and sizes the resistor with $R = \frac{V_S - V_L}{I}$ .

Dividers, sensing and timing

Potential dividers split the supply with $V_\text{out} = V_\text{in}\frac{R_2}{R_1 + R_2}$ , choose resistor values for a target output, and account for the loading effect. Sensing with LDRs and thermistors puts a sensor into a divider so the output voltage tracks light or temperature, choosing which way round to get a rising output. Capacitors and time delays use $Q = CV$ and the time constant $\tau = RC$ to make predictable delays, the basis of the 555 timer.

How this module is examined

A typical Eduqas profile for this content:

Calculations. The LED current-limiting resistor, the divider output and choosing divider resistors, the loaded output, the charge $Q = CV$ , and the time constant $\tau = RC$ .
Component questions. Reading and writing the colour code, choosing a preferred value, and potentiometer versus rheostat wiring.
Sensing questions. Describing and explaining how a divider output changes with light or temperature, and designing a sensor for a rising output.
Explanation. Why an LED needs a resistor, why loading lowers a divider output, and why a larger RC gives a longer delay.

Check your knowledge

A mix of recall and calculation questions covering the module. Attempt them under timed conditions, then check against the solutions.

A resistor is yellow, violet, brown, gold. State its value and tolerance. (2 marks)
An LED ( $V_L = 2.0\ \text{V}$ ) runs at $20\ \text{mA}$ from a $9.0\ \text{V}$ supply. Find the series resistor. (2 marks)
A divider has $R_1 = 3.0\ \text{k}\Omega$ (top) and $R_2 = 1.0\ \text{k}\Omega$ (bottom) across $8.0\ \text{V}$ . Find the output across $R_2$ . (2 marks)
State what happens to an NTC thermistor's resistance as it gets hotter. (1 mark)
Find the charge stored on a $470\ \mu\text{F}$ capacitor charged to $5.0\ \text{V}$ . (2 marks)
A $100\ \mu\text{F}$ capacitor charges through a $22\ \text{k}\Omega$ resistor. Find the time constant. (2 marks)

Solutions

Step 1: Q1 - reading the resistor colour code

The four-band code gives two digits, a multiplier, and a tolerance. Read yellow (4), violet (7), brown (multiplier $\times 10$ ), gold ( $\pm 5\%$ ):

$4, 7, \times 10 = 470\ \Omega$ , $\pm 5\%$ .

Step 2: Q2 - LED current-limiting resistor

The resistor carries the voltage that the LED does not take. Subtract the LED forward voltage from the supply to find the resistor voltage, then apply Ohm's law. Convert the current to amperes first:

$V_R = 9.0 - 2.0 = 7.0\ \text{V}$ ; $R = \dfrac{7.0}{0.020} = 350\ \Omega$ (nearest preferred value $390\ \Omega$ ).

Step 3: Q3 - potential divider output

The output voltage is the fraction of the total resistance that $R_2$ contributes, multiplied by the supply voltage:

V_\text{out} = 8.0 \times \frac{1.0}{3.0 + 1.0} = 8.0 \times \frac{1}{4} = 2.0\ \text{V}

Step 4: Q4 - NTC thermistor behaviour

NTC stands for Negative Temperature Coefficient. As temperature rises, the resistance of an NTC thermistor falls, which is the opposite of a normal metal resistor.

Answer: it decreases (NTC: resistance falls as temperature rises).

Step 5: Q5 - charge stored on a capacitor

Use $Q = CV$ , but the capacitance must be in farads, not microfarads. Convert $470\ \mu\text{F}$ to $470 \times 10^{-6}\ \text{F}$ :

Q = CV = (470 \times 10^{-6}) \times 5.0 = 2.35 \times 10^{-3}\ \text{C} = 2.35\ \text{mC}

Step 6: Q6 - RC time constant

The time constant $\tau = RC$ gives the time to charge to about 63 percent of the supply voltage. Work in base units (ohms and farads give seconds):

\tau = RC = 22\,000 \times 100 \times 10^{-6} = 2.2\ \text{s}

Sources & how we know this

WJEC Eduqas GCSE (9-1) Electronics specification (C490) — WJEC Eduqas (2017)