Legacy Systems & Control: electronics and switching
Cover the older option's circuit calculations, measurement, switching, relays and control-system concepts alongside 2028 common electronics.
Exam scope and assessment notes
StudyForA provides original teaching and scenario-based practice, not Cambridge past papers, official marks or an authenticated practical course. It cannot assess a candidate's actual workshop skills, finished product or Component 2 portfolio. Component 2 is school-based, internally assessed and externally moderated; the teacher authenticates candidate work. Any real making requires centre facilities, risk controls and direct qualified supervision. Candidate project work must remain their own. Follow the syllabus for the exam year.
Before you begin
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which instrument is used to measure potential difference across a component?
What you will learn
- Use voltage, current and resistance relationships and simple electrical power calculations in syllabus contexts.
- Describe circuit-building safety, switches, relays and control behaviour in the legacy Systems & Control option.
Use circuit quantities consistently
The older Systems & Control syllabus uses Ohm's law, V = IR, and electrical power, P = VI, for basic circuit reasoning. A current-limiting resistor can be selected from a specified supply, component voltage and target current using the appropriate voltage difference divided by current.
Ammeters, voltmeters and multimeters measure different quantities and must be used as directed. School tasks may ask candidates to interpret a reading or select a circuit component. Real circuit construction, soldering and mains work are outside this online lesson and require teacher-approved low-risk equipment and supervision.
Explain control components
Switches may be normally open or normally closed; pole/throw descriptions explain which contacts are connected. A relay allows a low-power control circuit to switch another circuit, but the ratings and isolation must suit the application.
Legacy assessment includes constructing and reading circuit diagrams, relays and control circuits, as well as components such as resistors, capacitors, transistors, diodes and sensors. State the input, processing, output and feedback role, and consider a failure-safe state. Component 2 real candidate work must remain the candidate's own and be supervised by the centre.
Sources for this lesson
Pause and explain
A 5 V supply provides 0.1 A to a simple load. What is the electrical power under those stated conditions?
Worked example
A 6 V supply is used with a component that drops 2 V and needs 0.02 A. Find the ideal series resistor and identify the assumptions.
Show the worked solution
- The resistor must drop the remaining voltage: 6 V − 2 V = 4 V.
- Use R = V ÷ I = 4 ÷ 0.02.
- R = 200 Ω in the ideal calculation.
- Check current, power rating, tolerances and the component datasheet before a supervised circuit test.
Answer The ideal resistor is 200 Ω. This calculation assumes the stated supply and component voltage/current are correct and steady. A real selection also checks power dissipation, tolerance, available standard values, temperature, component datasheet and safe low-voltage setup.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 9 V source and a 3 V component operate at 0.03 A. Calculate the ideal series resistor and give two checks before a real test.
Give me a hint
Find the voltage left for the resistor, then use R = V/I.
Compare my reasoning
- Calculate resistor voltage: 9 − 3 = 6 V.
- Calculate R = 6 ÷ 0.03.
- Check resistor power rating and component/supply data before teacher-approved testing.
R = 200 Ω ideally. Also check the resistor's power dissipation and tolerance, and verify component current and supply behaviour from reliable specifications. Any physical test needs a teacher-approved safe circuit.
Look for these in your work
- Uses voltage difference
- Calculates resistance
- Names component and safety checks
A normally closed switch is used in a simple warning circuit. Explain what happens when the switch is activated and why the chosen default state matters.
Give me a hint
Describe the contact state before activation, then the change and its safety purpose.
Compare my reasoning
- State the switch's normal connection.
- Describe how activation changes the circuit path.
- Explain whether the default condition is suitable for the warning task.
A normally closed contact conducts in its unactivated state and opens when activated. Whether that is suitable depends on whether activation should stop or trigger the output and what happens on a broken wire. Define the intended fail-safe behaviour before selecting the switch.
Look for these in your work
- Explains NC state
- Traces circuit behaviour
- Considers fail-safe intent
A control circuit uses a relay to switch a motor. What information is needed before deciding that the relay is appropriate?
Give me a hint
Check coil/control requirements, contact ratings, motor startup current, isolation and safe state.
Compare my reasoning
- Identify the control and switched supply conditions.
- Check relay coil, contact and motor ratings from datasheets.
- Plan isolation, protection and failure behaviour with a qualified teacher.
Check the relay coil voltage/current, contact voltage and current ratings including motor starting current, isolation, switching frequency and protection. Decide what happens on control failure. A real motor circuit requires qualified design and supervision; do not wire it from this example.
Look for these in your work
- Checks both sides of the relay
- Uses component ratings
- Recognises real-world safety boundary
Common mistakes
- Using the full supply voltage when the component already drops voltage
- Confusing current and voltage meter connections
- Treating a classroom value calculation as a safe circuit build instruction
What can you explain now?
A 5 V supply provides 0.1 A to a simple load. What is the electrical power under those stated conditions?
Compare with the explanation
0.5 W.
P = VI = 5 × 0.1 = 0.5 W.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Recall the main decision in “Legacy Systems & Control: electronics and switching”, name the evidence that supports it, and identify one limitation that still needs checking.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.