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How do electronic systems work as inputs, processes and outputs, and how are resistors, Ohm's law and potential dividers used?

Electronic systems as input, process and output blocks: sensors and switches as inputs, processing devices, and output transducers, with Ohm's law, series and parallel resistors, and the potential divider used to sense light and temperature, including calculations.

A focused answer to OCR A-Level Product Design on electronic systems: the input, process, output model, sensors and output transducers, Ohm's law, series and parallel resistors, and the potential divider used with an LDR or thermistor, with worked calculations.

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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Quick answer

An electronic system is modelled as three blocks: input (sensors and switches that detect a change, such as a push switch, an LDR or a thermistor), process (a device that makes a decision, such as a transistor, logic gate or microcontroller) and output (a transducer that produces an effect, such as an LED, buzzer or motor). Ohm's law relates voltage, current and resistance: $V = I \times R$ . Resistors in series add ( $R_T = R_1 + R_2$ ); in parallel they combine as $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}$ . A potential divider uses two resistors in series to produce an output voltage $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$ ; replacing one resistor with an LDR or thermistor makes the output respond to light or temperature.

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What this dot point is asking
The input, process, output model
Ohm's law and resistor combinations
The potential divider
Why this matters in products

What this dot point is asking

OCR wants you to model an electronic system as inputs, processes and outputs, identify common components, and use Ohm's law, series and parallel resistor rules, and the potential divider (with an LDR or thermistor) in calculations. Electronics in Product Design is at a systems level but carries real calculation marks.

The input, process, output model

Ohm's law and resistor combinations

The most common Component 01 calculation is rearranging $V = IR$ to find a resistor value, often to limit the current through an LED.

The potential divider

The exam point is to identify which resistor the output is taken across and keep the resistor units consistent.

Why this matters in products

The input, process, output model and the potential divider explain how everyday products sense and respond: a night light switches on in the dark (LDR divider), a fan switches on when hot (thermistor divider), and a doorbell sounds when a switch closes. A designer chooses sensors and sets resistor values so the product responds at the right point.

Calculating a potential-divider output with a thermistor

Write the potential-divider formula

The output across the lower resistor is $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$ . The supply is $6$ V. A thermistor ( $R_1$ , top) is in series with a fixed $4.7$ kilohm resistor ( $R_2$ , bottom, output across it).

Substitute the warm-condition resistances

When warm, the thermistor resistance falls to $1$ kilohm. Convert to consistent units: $R_1 = 1000$ ohms, $R_2 = 4700$ ohms.

Calculate the output

$V_{out} = 6 \times \frac{4700}{1000 + 4700} = 6 \times \frac{4700}{5700} = 6 \times 0.825 = 4.95$ V.

Explain the trend

When cold, the thermistor resistance rises, so $R_1$ grows, the fraction falls and $V_{out}$ drops. So this divider gives a high output when warm and a low output when cold; the rising warm-voltage could switch a transistor and a cooling fan on. Always identify which resistor the output is across and keep units consistent, because that is where the marks are won or lost.

Exam-style practice questions

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

OCR 20193 marksA 12 V supply drives a current of 0.04 A through a resistor in an LED circuit. Calculate the resistance, and state the unit.

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A Component 01 Ohm's law calculation. Marks for the rearrangement, the answer and the unit.

Ohm's law is $V = I \times R$ , so $R = \frac{V}{I} = \frac{12}{0.04} = 300$ ohms. The unit is the ohm (the symbol is the Greek capital omega).

A common dropped mark is forgetting to convert a current given in milliamps to amps (here it is already in amps), or omitting the unit. Always rearrange $V = IR$ correctly: $R = V/I$ .

OCR 20216 marksA potential divider uses a fixed 10 kilohm resistor and an LDR, powered by a 9 V supply, with the output taken across the fixed resistor. In bright light the LDR resistance falls to 2 kilohm. Calculate the output voltage in bright light, and explain how the output changes as it gets darker.

Show worked answer →

A Component 01 potential-divider calculation and explanation. Marks for the formula, the substitution, the answer and the trend.

The output across the lower (fixed) resistor is $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$ , where $R_1$ is the LDR (top) and $R_2$ is the fixed resistor (bottom, output across it). In bright light: $V_{out} = 9 \times \frac{10000}{2000 + 10000} = 9 \times \frac{10000}{12000} = 9 \times 0.833 = 7.5$ V. As it gets darker, the LDR resistance rises, so $R_1$ grows, the fraction $\frac{R_2}{R_1 + R_2}$ falls, and $V_{out}$ across the fixed resistor drops. So this divider gives a high output in light and a low output in dark, which could switch a transistor and a lamp on only in the dark.

A common dropped mark is using the wrong resistor on top, or not converting kilohms consistently; keep units consistent and identify which resistor the output is taken across.

Related dot points

Sources & how we know this

OCR A Level Design and Technology (H404-H406) specification — OCR (2017)