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Electronics & Analogue Control

How electronic systems sense and respond — circuit symbols, Ohm's law, power, input and output transducers, voltage dividers, and the transistor as a switch.

Topic 3 booklet — PDF coming soon Data booklet (formulae) Go to self-check

Exam = tested in the written paper most years. Assignment = also used in assignment Task 1 (input sensing circuit & Vout). About a third of the whole paper is calculations.

Sections:
Concept 1

Electricity and circuits Exam

Engineers design circuits to control how energy flows. The three quantities that matter in every circuit are voltage (V, volts — the energy given to each unit of charge), current (I, amperes — the rate of flow of charge) and resistance (R, ohms — how strongly a component opposes current).

Circuit symbols — from the data booklet

Engineers use the standard British Standard (BS) symbols listed in the N5 Engineering Science data booklet. Learn to recognise and draw every one.

Cell / battery +
Cell / battery
Supplies electrical energy. Long line is +.
Push/toggle switch
Switch
Opens and closes the circuit.
Lamp
Lamp
Produces light when current flows.
Light-emitting diode
LED
Light when current flows the correct way round.
Resistor (rectangle)
Resistor
Limits the current in a circuit.
Variable resistor
Variable resistor
Resistance adjusted by hand.
Light-dependent resistor
LDR
Resistance changes with light level.
Thermistor
Thermistor (NTC)
Resistance changes with temperature.
Diode
Diode
Allows current one way only.
Buzzer
Buzzer
Produces sound when current flows.
Motor M
Motor
Rotates when current flows.
Voltmeter and ammeter V A
Voltmeter / ammeter
Measure voltage across / current through.
Relay
Relay
Small current switches a larger circuit.
NPN bipolar transistor
NPN transistor
Fast electronic switch — see Concept 6.

Series and parallel circuits

SeriesParallel
Connectionone after the other in a single loopon separate branches
Currentthe same at every point (It everywhere)splits between branches and adds back together (It = I1 + I2 + …)
Voltagesupply voltage shared between components (Vt = V1 + V2 + …)supply voltage is the same across each branch
Total resistanceRt = R1 + R2 + …1/Rt = 1/R1 + 1/R2 + …

Measuring voltage and current

MeterMeasuresHow to connectResistance of meter
Voltmetervoltage (V)in parallel with the componentvery high
Ammetercurrent (A)in series with the componentvery low
Pre-power-up check — always check a circuit before switching on the supply:

Series or parallel?

Which type of circuit does each rule describe? Choose, then check. (Task 2 in your booklet.)

Concept 2

Ohm's law and resistance Exam

Ohm's law links voltage, current and resistance — it is the most important formula in electronics. Your marks come from showing the working: substitute the numbers, rearrange if needed, then give the answer with a unit.

Ohm's law
V = I R
voltage = current × resistance · V, A, Ω
Resistors in series
Rt = R1 + R2 + …
add the resistances · Ω
Resistors in parallel
1Rt = 1R1 + 1R2 + …
reciprocals — answer is less than the smallest R
Prefixes the markers expect you to convert: 1 kΩ = 1000 Ω · 1 MΩ = 1 000 000 Ω · 1 mA = 0.001 A. Convert first, then substitute.

Ohm's law — V = I R

Worked example — substitute & solve
Substitute & solve
A resistor has a current of 0.2 A flowing through it when the voltage across it is 6 V. Calculate the resistance.
V=I R
6=0.2 × Rnumbers in first
R=6 ÷ 0.2now rearrange
R=30 Ω
Worked example — rearrange with a prefix
Rearrange — numbers in first
A 2.2 kΩ resistor has a voltage of 11 V across it. Calculate the current.
Convert first
R = 2.2 kΩ = 2200 Ω kΩ → Ω
V=I R
11=I × 2200numbers in first
I=11 ÷ 2200now rearrange
I=0.005 A (5 mA)

Resistors in series — Rt = R1 + R2 + …

Worked example — substitute & solve
Substitute & solve
Three resistors of 100 Ω, 220 Ω and 470 Ω are connected in series. Calculate the total resistance.
Rt=R1 + R2 + R3
Rt=100 + 220 + 470
Rt=790 Ω

Resistors in parallel — the reciprocal formula

Worked example — substitute & solve
Substitute & solve
Two resistors of 100 Ω and 200 Ω are connected in parallel. Calculate the total resistance.
1/Rt=1/R1 + 1/R2
1/Rt=1/100 + 1/200
1/Rt=0.015
Rt=1 ÷ 0.015 = 66.7 Ωflip at the end

Sense check: 66.7 Ω is less than 100 Ω — the parallel total is always less than the smallest resistor.

🔌 Ohm's law explorer — live

Drag the voltage and resistance. The current updates instantly from I = V ÷ R.

current (mA)
power (W) — P = IV

Challenge: set the circuit so exactly 20 mA flows.

Concept 3

Electrical power Exam

Electrical power can be calculated three ways, depending on which quantities you know. All three are on the data sheet — choose the one that matches the question's givens.

Know V and I
P = I V
power = current × voltage · W
Know I and R
P = I2 R
square the current first · W
Know V and R
P = V2R
square the voltage first · W
Worked example — P = IV (substitute & solve)
Substitute & solve
A lamp has 12 V across it and a current of 0.5 A. Calculate the power.
P=I V
P=0.5 × 12
P=6 W
Worked example — P = I²R with a prefix conversion
Substitute & solve — convert first
A 1 kΩ resistor has 50 mA flowing through it. Calculate the power dissipated.
Convert first
I = 50 mA = 0.05 A mA → A
R = 1 kΩ = 1000 Ω kΩ → Ω
P=I2 R
P=0.052 × 1000square the current first
P=0.0025 × 1000
P=2.5 W
Worked example — P = V²/R (substitute & solve)
Substitute & solve
A 1 kΩ resistor has 5 V across it. Calculate the power dissipated.
P=V2 ÷ R
P=52 ÷ 1000square the voltage first
P=25 ÷ 1000
P=0.025 W (25 mW)
Worked example — rearrange (find the current)
Rearrange — numbers in first
A lamp uses 18 W from a 12 V supply. Calculate the current.
P=I V
18=I × 12numbers in first
I=18 ÷ 12now rearrange
I=1.5 A

Pick the power formula

For each question, choose the formula you would reach for first, then check.

Concept 4

Input and output devices Exam

Engineered systems follow input → process → output. An input transducer changes a real-world input into an electrical signal; an output transducer changes an electrical signal into a real-world output.

Input transducerReacts to…Engineering example
Switchbeing pressed or movedstart button on a machine
Variable resistorbeing turned by handvolume control
LDRthe level of light falling on itautomatic street lighting
Thermistor (NTC)temperaturetemperature sensor in an oven
Output transducerOutputEngineering example
Lamp / LEDlightwarning indicator
Buzzersoundalarm system
Motormovement (kinetic energy)fan, pump, robot arm
Relayswitches a separate circuitswitches a high-power motor

How the two sensors behave

SensorConditionResistance
LDRbright lightlow (a few hundred ohms)
darknesshigh (often hundreds of kilohms)
NTC thermistorhigh temperaturelow
low temperaturehigh

Match the sensor to the job

Choose the most suitable input transducer for each engineering job, then check. (Task 3 in your booklet.)

Concept 5

Voltage dividers and sensors ExamAssignment

A voltage divider is two resistors in series that split the supply voltage. The output V2 is taken across the lower resistor. Because the resistors are in series, the larger resistance always has the larger voltage across it — the voltages are in the same ratio as the resistances.

An analogue signal can take any value within a range and changes smoothly — like the output of an LDR divider as the light slowly fades at sunset. Replace one resistor with an LDR or thermistor and the divider becomes a sensor circuit: as the sensor's resistance changes, V2 changes.

Voltage divider — R1 over R2, output across R2 +Vs R1 V2 (output) R2 0 V
The output V2 is taken across the lower resistor R2.
Voltage divider
V1V2 = R1R2
voltages in the same ratio as the resistances
Worked example — substitute & solve
Substitute & solve
In a voltage divider, R₁ = 1 kΩ and R₂ = 2 kΩ. The voltage across R₁ (V₁) is 3 V. Calculate the voltage across R₂ (V₂).
V1 ÷ V2=R1 ÷ R2
3 ÷ V2=1000 ÷ 2000numbers in first
3 ÷ V2=0.5
V2=3 ÷ 0.5 = 6 Vbigger R = bigger V
Worked example — sensor divider
Substitute & solve
A sensor voltage divider has an LDR as R₁ and a fixed resistor as R₂. In normal light R₁ = 4 kΩ and R₂ = 1 kΩ. The voltage across the LDR (V₁) is 4 V. Calculate V₂.
V1 ÷ V2=R1 ÷ R2
4 ÷ V2=4000 ÷ 1000 = 4
V2=4 ÷ 4 = 1 V

Where you put the sensor decides what the circuit does

Sensor positionIn bright light / when hot…In darkness / when cold…
LDR as R₁ (top)LDR resistance low → V2 highLDR resistance high → V2 low
LDR as R₂ (bottom)LDR resistance low → V2 lowLDR resistance high → V2 high
How to describe a sensor circuit (the most-missed marks): say what happens to the sensor's resistance, then to Vout, then to the transistor and output. "As it gets darker the LDR's resistance increases, so Vout across the LDR increases; when Vout is high enough the transistor switches on and the lamp lights."

🎚️ Voltage-divider explorer — live

The supply is fixed at 12 V. Drag R₁ and R₂ and watch how the 12 V is shared — the bigger resistance always takes the bigger share.

V₁ across R₁ (V)
V₂ across R₂ — the output (V)

Challenge: make the output V₂ = 9 V (within ±0.1 V). Hint: what ratio of R₂ to R₁ puts three-quarters of the supply across R₂?

Concept 6

The transistor, relay and protection diode Exam

A transistor is a fast, automatic electronic switch: a small input voltage at the base controls a much larger output current. An NPN transistor switches on when the base voltage reaches about 0.7 V.

ConnectionLetterFunction
BaseBthe input — a small signal here switches the transistor on or off
CollectorCconnected to the output device, through the supply
EmitterEthe path back to 0 V

Relays and the protection diode

A relay is an electromagnetic switch: a small current through its coil produces a magnetic field that closes a separate set of contacts — so a small electronic circuit can switch a much larger one. When the coil switches off it produces a voltage spike that can destroy the transistor, so a protection diode is connected across the coil, in reverse to the supply.

A full control circuit

Sensor voltage divider → transistor → relay with protection diode → output +V 0 V LDR (R₁) R₂ V₂ → base CEB relay coil protection diode M separate supply relay contacts
Sensor voltage divider → transistor → relay (with protection diode) → high-power output.

Every control circuit in this topic is built from the same four blocks: an input transducer in a voltage divider, a transistor that switches when V2 reaches about 0.7 V, and an output device — or a relay to switch a separate, larger circuit.

Choosing the sensor position

CircuitPosition of sensorOutput switches on when…
Light-sensingLDR as R₁ (top)the light level is high
Dark-sensingLDR as R₂ (bottom)the light level is low
High-temperaturethermistor as R₁ (top)the temperature is high
Low-temperaturethermistor as R₂ (bottom)the temperature is low

Build the control circuit

A dark-sensing security lamp needs its blocks in the right order. Tap a block to place it; tap a placed block to remove it. One block is not needed.

Senses the input Switches at ≈0.7 V Switches the big circuit The output
Assignment link. When you build and test a sensor control circuit in your assignment (Task 1: input sensing circuit and Vout), see Booklet 8 — Assignment Skills: wire the divider as the right kind of sensor (dark / cold), describe the direction Vout changes, and record results that name each component's function.

Common mistakes — watch out for these in the exam

  • Giving an answer with no unit, or the wrong unit (V, A, Ω, W).
  • Connecting meters wrongly — an ammeter goes in series, a voltmeter in parallel.
  • Drawing the battery the wrong way round — the long line is +.
  • Describing a sensor circuit without using Vout — say what happens to the resistance, then Vout, then the transistor/output.
  • Forgetting the transistor is the switch, and the relay lets it switch a larger current.
Booklet check

Check your booklet work

Try each task in your booklet first, then open the matching answer. Calculations show the final value with the key steps.

Try This — final answers

Try This — Ohm's law Booklet p.9
  1. R = 12 ÷ 0.5 = 24 Ω
  2. I = 5 ÷ 220 = 0.023 A (23 mA)
  3. R = 24 ÷ 2 = 12 Ω
Try This — Series resistance Booklet p.10
  1. 220 + 330 = 550 Ω
  2. 1000 + 2000 + 4700 = 7700 Ω (7.7 kΩ)
  3. 4 × 100 = 400 Ω
Try This — Parallel resistance Booklet p.11
  1. Two equal resistors halve: 470 ÷ 2 = 235 Ω
  2. 1/Rt = 1/100 + 1/1000 = 0.011 → 90.9 Ω ≈ 91 Ω
  3. 1/Rt = 1/100 + 1/200 + 1/400 = 0.0175 → 57.1 Ω
Try This — Power (P = VI, P = I²R, P = V²/R) Booklet p.12–14

P = VI:

  1. 230 × 5 = 1150 W
  2. 24 × 0.8 = 19.2 W
  3. I = 18 ÷ 12 = 1.5 A

P = I²R:

  1. 0.2² × 100 = 4 W
  2. 0.04² × 470 = 0.75 W
  3. 50 mA = 0.05 A; 0.05² × 1000 = 2.5 W

P = V²/R:

  1. 12² ÷ 220 = 0.65 W
  2. 9² ÷ 4700 = 0.017 W (17 mW)
  3. V = √(P × R) = √(1 × 100) = 10 V
Try This — Voltage divider & sensor divider Booklet p.21 & p.23

Voltage divider:

  1. V₁ = 6 V, R₁ = 2 kΩ, R₂ = 1 kΩ → V₂ = 6 × 1/2 = 3 V
  2. V₁ = 4 V, R₁ = 470 Ω, R₂ = 940 Ω → V₂ = 4 × 940/470 = 8 V
  3. V₂ = 4 V, R₁ = 1 kΩ, R₂ = 2 kΩ → V₁ = 4 × 1/2 = 2 V

Sensor divider:

  1. LDR R₁ = 2 kΩ (V₁ = 4 V), R₂ = 3 kΩ → V₂ = 4 × 3/2 = 6 V
  2. R₁ = 10 kΩ (V₁ = 2 V), thermistor R₂ = 20 kΩ → V₂ = 2 × 20/10 = 4 V
  3. R₁ = 2 kΩ (V₁ = 3 V), LDR R₂ = 6 kΩ → V₂ = 3 × 6/2 = 9 V

Section practice — final answers

Practice — Electricity and Circuits Booklet p.7
  1. Volt (V).
  2. Ampere (A).
  3. Ohm (Ω).
  4. In parallel with (across) the component.
  5. In series with the component.
  6. The current is the same at every point.
  7. The voltage is the same across each branch (equal to the supply).
  8. 6 V — the 12 V supply is shared equally between two identical lamps.
  9. Any two: supply voltage correct; wires/components secure; resistor values correct; polarity of LEDs/diodes/supply correct.
  10. The ammeter is in series, so its resistance adds to the circuit — a high resistance would reduce the very current it is trying to measure.
Practice — Electrical Calculations Booklet p.9–14
  1. V = I R.
  2. I = 9 ÷ 470 = 0.019 A (19 mA)
  3. 220 + 470 + 1000 = 1690 Ω (1.69 kΩ)
  4. Two equal in parallel halve: 220 ÷ 2 = 110 Ω
  5. P = 24 × 1.5 = 36 W
  6. P = I²R → I = √(1 ÷ 100) = 0.1 A
  7. P = 12² ÷ 1000 = 0.144 W (144 mW)
  8. 1000 ÷ 2 = 500 Ω
  9. 1000 + 2200 + 4700 = 7900 Ω (7.9 kΩ)
  10. V = 0.05 × 220 = 11 V; P = 11 × 0.05 = 0.55 W
Task 3 & Practice — Input and Output Devices Booklet p.17–20

Task 3 — match the sensor: dark → LDR · warm room → thermistor · oven door opened → switch · dimming knob → variable resistor · floor sensor → switch (pressure switch).

  1. A component that changes a real-world input into an electrical signal.
  2. A component that changes an electrical signal into a real-world output.
  3. The LDR's resistance increases as the light level decreases.
  4. The NTC thermistor's resistance increases as the temperature decreases.
  5. Input: thermistor. Output: motor (fan).
  6. Read the value your class recorded for normal room light in your data table.
  7. Read your table — interpolate between the recorded temperatures if needed.
  8. Lamp and LED.
  9. A relay lets a small electronic circuit switch a high-power circuit — e.g. switching a mains-powered motor from a low-voltage sensor circuit.
  10. A switch (limit / guard switch).
Practice — Voltage Dividers and Sensors Booklet p.25–26
  1. A signal that can take any value within a range and changes smoothly.
  2. To produce a smaller voltage from the supply voltage.
  3. V₁ = 3 V, R₁ = 220 Ω, R₂ = 440 Ω → V₂ = 3 × 440/220 = 6 V
  4. The LDR's resistance increases, so more voltage is dropped across the LDR (R₁) — V₂ decreases.
  5. The thermistor's (R₂'s) resistance decreases, so V₂ decreases.
  6. LDR R₁ = 2 kΩ (V₁ = 2 V), R₂ = 3 kΩ → V₂ = 2 × 3/2 = 3 V
  7. The output is across only one of two series resistors — the two voltages add up to the supply, so each part must be less than the supply.
  8. R₁ = 1 kΩ, R₂ = 4 kΩ, V₂ = 8 V → V₁ = 8 × 1/4 = 2 V
  9. e.g. a volume control or lamp dimmer — turning the knob changes the output voltage smoothly.
  10. Bottom position (R₂) — in darkness the LDR's resistance is high, so V₂ is high and the output switches on.
Practice — Transistor Switching, Relays and Output Control Booklet p.29–30
  1. It acts as a fast electronic switch — a small base voltage switches a larger output current.
  2. Base, collector, emitter.
  3. About 0.7 V.
  4. An electromagnetic switch: a small current through the coil closes separate contacts, switching a larger circuit.
  5. It protects the transistor from the voltage spike produced when the relay coil switches off.
  6. Bottom (R₂) — so V₂ rises as it gets dark.
  7. As it gets darker the LDR's resistance increases (1), so the voltage across the LDR / V₂ increases (1); when it reaches about 0.7 V the transistor switches on and the lamp lights (1).
  8. The transistor circuit is low-voltage; the relay's contacts are electrically separate, so its small coil current can safely switch the 230 V mains circuit.
  9. Thermistor as R₁ (top). When the machine gets hot the thermistor's resistance falls, so V₂ (across R₂) rises; at about 0.7 V the transistor switches on the fan.
  10. Any two with fixes, e.g.: LED/battery the wrong way round — check polarity before power-up; missing protection diode across the relay coil — always fit one, reversed to the supply; meters connected wrongly — voltmeter in parallel, ammeter in series; loose breadboard wires — push components fully home and test step by step.
Check yourself

Check yourself

Mixed multiple choice

One question from every part of the topic. Choose an answer for each, then mark them.

Rate your confidence — the booklet's success criteria

These are the success criteria from your booklet. Red = not yet, Amber = getting there, Green = confident. Saved on this device.

Ready for exam questions? Open the Past Paper Finder and look for the electronics and analogue-control questions.

Sources & credits: The Topic 3 booklet © R Stewart, 2026. The Past Paper Finder is compiled by Mr McDonald, 2024; past-paper questions © Qualifications Scotland (SQA). The N4/N5 data booklet is reproduced for educational use, © Qualifications Scotland (SQA).