An LED (V_L = 2.0\ V) runs at 10\ mA from a 5.0\ V supply. Find the resistor. [2 marks]

WJEC Eduqas Eduqas 2020-style practice: An LED with a forward voltage of 2.0\ V is to operate at 15\ mA from a 9.0\ V supply. Calculate the value of the series resistor required and explain why the resistor is needed.

The resistor must drop the voltage that is not across the LED. Voltage across the resistor: V_R = V_S - V_L = 9.0 - 2.0 = 7.0\ V. Apply Ohm's law with the chosen current: R = V_RI = 7.00.015 = 467\. The nearest preferred value above this is 470\ (choosing slightly higher keeps the current just under 15\ mA, which is safe). Why it is needed (1 mark): an LED has very low resistance once it conducts, so without a series resistor the current would be very large and the LED would be destroyed. The resistor limits the current to a safe value. Markers reward V_R = 7.0\ V, R ≈ 470\, and the limiting explanation.

WJEC Eduqas Eduqas 2022-style practice: An LED has a forward voltage of 1.8\ V and is run from a 5.0\ V supply through a 220\ resistor. Calculate the current through the LED.

Voltage across the resistor: V_R = V_S - V_L = 5.0 - 1.8 = 3.2\ V. Current (the same through the resistor and LED, as they are in series): I = V_RR = 3.2220 = 0.0145\ A = 14.5\ mA. Markers reward subtracting the forward voltage to get V_R = 3.2\ V and the current 14.5\ mA. The common error is using the full 5.0\ V across the resistor instead of 3.2\ V.

EnglandElectronicsSyllabus dot point

Why does an LED need a series resistor, and how do you calculate its value?

Driving LEDs: the need for a current-limiting series resistor, the forward voltage drop of an LED, and calculating the resistor value for a chosen current.

An Eduqas GCSE Electronics answer on driving LEDs: why an LED needs a series current-limiting resistor, the LED forward voltage drop, and calculating the resistor value from the supply voltage, forward voltage and forward current.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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What this dot point is asking

Eduqas wants you to explain why a light-emitting diode (LED) needs a series current-limiting resistor, use the LED's forward voltage drop, and calculate the resistor value for a chosen operating current. This is the single most common calculation in the whole course and appears in almost every paper.

The answer

Why an LED needs a series resistor

The forward voltage drop

Calculating the resistor value

Choosing a current-limiting resistor

A green LED ( $V_L = 2.2\ \text{V}$ ) is to run at $12\ \text{mA}$ from a $6.0\ \text{V}$ supply. Find the resistor value and a suitable preferred value.

step 1: Find the voltage across the resistor

The LED takes $2.2\ \text{V}$ , so the resistor takes the rest: $V_R = V_S - V_L = 6.0 - 2.2 = 3.8\ \text{V}$ .

step 2: Apply Ohm's law with the chosen current

Convert the current: $12\ \text{mA} = 0.012\ \text{A}$ . Then $R = \dfrac{V_R}{I} = \dfrac{3.8}{0.012} = 317\ \Omega$ .

step 3: Choose a preferred value

The nearest E12 preferred value at or above $317\ \Omega$ is $330\ \Omega$ . Using $330\ \Omega$ gives a current of $\frac{3.8}{330} = 11.5\ \text{mA}$ , safely just under $12\ \text{mA}$ .

Examples in context

The current-limiting LED resistor is the most frequently examined calculation in the course and a staple of practical work. Every indicator LED on a project board, every status light driven by a logic gate or microcontroller pin, and every seven-segment display segment needs this calculation. In the non-exam assessment you specify the exact preferred resistor for each indicator. The same "subtract the device voltage, then apply Ohm's law" method reappears when sizing base resistors for transistors.

Try this

Q1. Why must an LED have a series resistor? [1 mark]

Cue. It limits the current to a safe value; without it the LED's low resistance would let too much current flow and destroy it.

Q2. An LED ( $V_L = 2.0\ \text{V}$ ) runs at $10\ \text{mA}$ from a $5.0\ \text{V}$ supply. Find the resistor. [2 marks]

Cue. $V_R = 5.0 - 2.0 = 3.0\ \text{V}$ ; $R = \frac{3.0}{0.010} = 300\ \Omega$ .

Q3. State a typical forward voltage for a red LED. [1 mark]

Cue. About $2\ \text{V}$ (roughly $1.8$ to $2.2\ \text{V}$ ).

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20204 marksAn LED with a forward voltage of

2.0\ \text{V}

is to operate at

15\ \text{mA}

from a

9.0\ \text{V}

supply. Calculate the value of the series resistor required and explain why the resistor is needed.

Show worked answer →

The resistor must drop the voltage that is not across the LED. Voltage across the resistor: $V_R = V_S - V_L = 9.0 - 2.0 = 7.0\ \text{V}$ .

Apply Ohm's law with the chosen current: $R = \dfrac{V_R}{I} = \dfrac{7.0}{0.015} = 467\ \Omega$ . The nearest preferred value above this is $470\ \Omega$ (choosing slightly higher keeps the current just under $15\ \text{mA}$ , which is safe).

Why it is needed (1 mark): an LED has very low resistance once it conducts, so without a series resistor the current would be very large and the LED would be destroyed. The resistor limits the current to a safe value.

Markers reward $V_R = 7.0\ \text{V}$ , $R \approx 470\ \Omega$ , and the limiting explanation.

Eduqas 20222 marksAn LED has a forward voltage of

1.8\ \text{V}

and is run from a

5.0\ \text{V}

supply through a

220\ \Omega

resistor. Calculate the current through the LED.

Show worked answer →

Voltage across the resistor: $V_R = V_S - V_L = 5.0 - 1.8 = 3.2\ \text{V}$ .

Current (the same through the resistor and LED, as they are in series): $I = \dfrac{V_R}{R} = \dfrac{3.2}{220} = 0.0145\ \text{A} = 14.5\ \text{mA}$ .

Markers reward subtracting the forward voltage to get $V_R = 3.2\ \text{V}$ and the current $14.5\ \text{mA}$ . The common error is using the full $5.0\ \text{V}$ across the resistor instead of $3.2\ \text{V}$ .

Related dot points

Sources & how we know this

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