Level 6 — Electronic Skills — Nov/Dec 2025

Level 6 • November/December 2025 • Written assessment

Work through each question before checking the answer. These revision answers are prepared for learners; they are not an official marking scheme.

Question 1

State four factors affecting a conductor’s resistance.

Answer

  • Material resistivity.
  • Length.
  • Cross-sectional area.
  • Temperature.

Question 2

Find equivalent inductance for L1=0.5 H in series with parallel L2=1 H and L3=2 H.

Answer

L1 is in series with L2 and L3 connected in parallel.
L1 is in series with L2 and L3 connected in parallel.
Step-by-step calculation

  1. Assume ideal uncoupled inductors (no mutual inductance).
  2. Lparallel = L2L3/(L2+L3) = 1×2/(1+2) = 2/3 H.
  3. Lequivalent = 0.5 + 2/3 = 7/6 H ≈ 1.167 H.

Question 3

Find current for 27 Ω and 3.3 Ω in series with 4.5 V.

Answer

Step-by-step calculation

  1. Rtotal = 27 + 3.3 = 30.3 Ω.
  2. I = V/R = 4.5/30.3 = 0.1485 A (148.5 mA).

Question 4

Explain two uses of inductors.

Answer

  • Store energy in a magnetic field for switching-converter circuits.
  • Oppose changing current, providing filtering/choke action that reduces ripple or noise.

Question 5

Find the energy used by a 60 W bulb over 5 hours.

Answer

Step-by-step calculation

  1. 60 W = 0.060 kW.
  2. Energy = power × time = 0.060×5 = 0.30 kWh.

Question 6

Draw a PNP transistor symbol and indicate current/voltage direction.

Answer

The PNP emitter arrow points inward toward the base.
The PNP emitter arrow points inward toward the base.

For a forward-active PNP, the emitter is at a higher potential than the base and the collector is below the base. Conventional current enters the emitter and divides into collector and base currents. Thus VEB is positive and IE = IC + IB; the corresponding signed VBE is negative.

Question 7

Describe two Gray-code uses besides Karnaugh maps.

Answer

  • Rotary/position encoders: adjacent positions differ by one bit, reducing ambiguous readings while crossing a boundary.
  • Clock-domain-crossing pointers in asynchronous FIFOs: single-bit transitions reduce the chance of sampling an inconsistent multi-bit pointer.

Question 8

Explain two ROM uses.

Answer

  • Stores firmware/bootstrap instructions needed during startup.
  • Stores stable lookup tables or embedded-device control code that must persist without power.

Question 9

Find the voltage drop for 0.2 A through 15 Ω.

Answer

Step-by-step calculation

  1. Ohm’s law: V = IR.
  2. V = 0.2×15 = 3 V.

Question 10

Give two reasons silicon is preferred over germanium.

Answer

  • Its larger band gap gives lower leakage and better operation at elevated temperatures.
  • Silicon is abundant and forms a useful stable silicon-dioxide insulating layer for integrated-circuit manufacture.

Question 11

Encode 598 and 96.72 in BCD.

Answer

Step-by-step calculation

  1. Encode each decimal digit independently in four bits.
  2. 598 = 5|9|8 = 0101 1001 1000.
  3. 96.72 = 9|6 . 7|2 = 1001 0110 . 0111 0010. Preserve the decimal point; it is not an extra BCD digit.

Question 12

Draw a four-component electronic circuit, explain the components and give four reasons to use binary in an embedded hospital device.

Answer

Battery, switch, current-limiting resistor and LED in a simple low-voltage series circuit.
Battery, switch, current-limiting resistor and LED in a simple low-voltage series circuit.
  • Battery supplies the DC potential difference.
  • Switch opens or closes the current path.
  • Resistor limits LED current to the intended safe operating value.
  • LED converts forward current into visible light; connect with correct polarity.
  • Two voltage states can be represented and distinguished reliably.
  • Logic gates provide simple implementation of control decisions.
  • Binary storage and arithmetic are compatible with digital processors/memory.
  • Defined high/low noise margins reduce susceptibility to small voltage disturbances; a medical device still requires appropriate safety engineering and validation.

Question 13

Describe five safe steps for low-voltage DC lighting, two consequences of doubling R at constant V, and three circuit applications.

Answer

  1. Choose a suitable rated low-voltage DC supply, conductors and components.
  2. Disconnect power before assembly and check the circuit diagram/polarities.
  3. Connect components securely and include appropriate current limiting/protection.
  4. Inspect insulation, continuity and polarity, and check for a short circuit before energising.
  5. Apply power, measure the current/voltage and verify expected operation without overheating.
Step-by-step calculation

  1. Original current I = V/R. With resistance 2R, Inew = V/(2R) = I/2.
  2. At constant V, P = V²/R, so doubling R gives Pnew = P/2.
  • Lighting and household-device control.
  • Power conversion/charging systems.
  • Sensors and industrial monitoring/control systems.

Question 14

Give three IC advantages, three forward-bias applications and four green-electronics examples.

Answer

  • Small size and low weight from integrating many components.
  • Low mass-production cost per function.
  • High reliability due to fewer separate joints/interconnections.
  • LED illumination when a current-limited LED is forward biased.
  • Rectification: a diode conducts on the forward-biased part of an AC waveform.
  • Bipolar-transistor amplification/switching uses an appropriately forward-biased base–emitter junction.
  • Energy-efficient LED lighting.
  • Low-power processors and automatic sleep/standby modes.
  • Solar-powered electronic systems with efficient charging/conversion.
  • Repairable designs, reduced hazardous materials and organised recycling of electronic waste.

Question 15

Relate atomic structure to conduction. For the six-resistor 12 V circuit, find total R, total I, current through 3 Ω and its voltage.

Answer

Atoms contain a nucleus and electrons in allowed energy states. Conductors have mobile electrons/partly filled energy bands. Semiconductor covalent bonding normally binds valence electrons, but thermal energy and doping create mobile electrons and holes that allow controlled conduction.

[(2 Ω + 4 Ω) parallel 8 Ω] in series with 5 Ω and [3 Ω parallel 6 Ω], across 12 V.
[(2 Ω + 4 Ω) parallel 8 Ω] in series with 5 Ω and [3 Ω parallel 6 Ω], across 12 V.
Step-by-step calculation

  1. Left top branch = 2+4 = 6 Ω. Left parallel equivalent = 6×8/(6+8) = 24/7 Ω.
  2. Right parallel equivalent = 3×6/(3+6) = 2 Ω.
  3. Total resistance = 24/7 + 5 + 2 = 73/7 Ω ≈ 10.4286 Ω.
  4. Total current = 12/(73/7) = 84/73 A ≈ 1.1507 A.
  5. Right parallel voltage = I×2 = 168/73 V ≈ 2.3014 V. Both right-hand resistors have this voltage.
  6. Current through 3 Ω = (168/73)/3 = 56/73 A ≈ 0.7671 A.
  7. Check: current through 6 Ω = 28/73 A; the branch currents sum to 84/73 A.
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