Level 6 — Electronic Skills — Mar/Apr 2026

Level 6 • March/April 2026 • 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

List two factors affecting conductivity beyond atomic structure.

Answer

  • Temperature.
  • Impurity/dopant concentration.

Question 2

Name the two BJT types.

Answer

  • NPN.
  • PNP.

Question 3

Label A, B and C in the electrical-symbol diagram.

Answer

Source symbols A, B and C.
Source symbols A, B and C.
Label Component
A Open switch
B AC voltage source (circle with sine-wave mark)
C Fuse (small rectangle containing a continuous line)

Question 4

Name two n-type dopants.

Answer

  • Phosphorus.
  • Arsenic.

Question 5

Indicate electron flow in the three-resistor 5 V circuit.

Answer

Three 4.7 kΩ resistors in series with a 5 V supply; the battery’s long line is positive.
Three 4.7 kΩ resistors in series with a 5 V supply; the battery’s long line is positive.
Electron-flow arrows run from the negative terminal through the external resistors toward the positive terminal.
Electron-flow arrows run from the negative terminal through the external resistors toward the positive terminal.

Electrons leave the negative terminal and move through the lower resistor, up the right-hand resistor and back through the upper resistor to the positive terminal. This is opposite to conventional current.

Question 6

Find current when 0.24 C passes in 15 ms.

Answer

Step-by-step calculation

  1. Convert time: 15 ms = 0.015 s.
  2. I = Q/t = 0.24/0.015 = 16 A.

Question 7

Find uncompressed size for a 3840 × 2160 image at 32 bits/pixel.

Answer

Step-by-step calculation

  1. Pixels = 3840×2160 = 8,294,400.
  2. Bits = 8,294,400×32 = 265,420,800.
  3. Bytes = 265,420,800/8 = 33,177,600.
  4. Decimal size = 33.1776 MB. Binary size = 33,177,600/1,048,576 = 31.640625 MiB, excluding file overhead.

Question 8

Explain silicon valence/conduction bands through covalent bonding.

Answer

Silicon atoms share four valence electrons in covalent bonds. In a large crystal, interacting atomic energy levels split into closely spaced allowed levels, forming bands. The valence band contains bonding-electron states; the conduction band contains higher-energy states in which electrons can move through the crystal. A forbidden energy gap separates the bands.

Question 9

Explain thermal energy and the band gap at 0 K and room temperature.

Answer

At 0 K, ideal intrinsic silicon has a full valence band and an empty conduction band, so there are no thermally generated mobile carriers. At room temperature, some electrons gain sufficient thermal energy to cross the band gap. Each promoted electron leaves a hole in the valence band; both the electrons and holes contribute to conduction. The relatively small semiconductor band gap permits this limited carrier generation.

Question 10

A device uses 1.8 MJ in 30 minutes at 250 V. Find its power.

Answer

The question calls the device both a fan and a heater; the energy/time calculation is unchanged.

Step-by-step calculation

  1. Energy = 1.8 MJ = 1,800,000 J.
  2. Time = 30×60 = 1800 s.
  3. P = E/t = 1,800,000/1800 = 1000 W = 1 kW.

Question 11

Interpret the resistor marking 4R7K.

Answer

R acts as the decimal point for resistance in ohms: 4.7 Ω. K is the tolerance letter: ±10%. This tolerance K should not be mistaken for the kilo-ohm prefix.

Question 12

Draw a 240 V circuit with series resistors R1, R2 and R3.

Answer

R1, R2 and R3 connected end-to-end in one loop with a 240 V source.
R1, R2 and R3 connected end-to-end in one loop with a 240 V source.

The same current flows through all three resistors. The voltage drops sum to 240 V; Rtotal = R1 + R2 + R3. This is a circuit-theory drawing, not a practical mains-wiring instruction.

Question 13

Explain why cache is above RAM in the hierarchy using speed, cost and capacity.

Answer

Registers, cache, RAM and secondary storage in descending speed/cost per byte and increasing typical capacity.
Registers, cache, RAM and secondary storage in descending speed/cost per byte and increasing typical capacity.
  • Speed: cache has lower latency and is closer to the CPU.
  • Cost: it is more expensive per byte, typically using SRAM.
  • Capacity: only a relatively small cache is economical, while main RAM supplies a larger working-memory capacity.

Question 14

Recommend temporary working memory for multimedia editing.

Answer

RAM holds the active files/data while the application works because it supports fast random access and adequate working capacity. SSD storage is useful for persistent files and scratch storage but does not replace RAM.

Question 15

Draw germanium’s atomic structure, compare PN forward/reverse bias, and explain gallium-doped p-type germanium with a diagram.

Answer

Germanium has atomic number 32 and shell populations 2, 8, 18, 4.
Germanium has atomic number 32 and shell populations 2, 8, 18, 4.

Neutral germanium has 32 protons and 32 electrons, with four valence electrons. Neutron count depends on the isotope.

Forward bias connects P to positive and N to negative; reverse bias connects P to negative and N to positive.
Forward bias connects P to positive and N to negative; reverse bias connects P to negative and N to positive.
Forward bias Reverse bias
Reduces the junction barrier/depletion width Increases the barrier/depletion width
Permits substantial current once appropriate operating conditions are reached Only small leakage flows until breakdown; current limiting remains necessary
Trivalent gallium replacing a germanium atom leaves a hole in the four-bond lattice.
Trivalent gallium replacing a germanium atom leaves a hole in the four-bond lattice.

Gallium has atomic number 31 and three valence electrons (shell distribution 2,8,18,3). Replacing a tetravalent Ge atom with Ga provides only three bonding electrons, leaving an electron vacancy. The acceptor can take an electron from a neighbouring bond, leaving a mobile hole. Holes are the majority carriers and electrons the minority carriers. The bulk material remains electrically neutral; p-type does not mean the whole crystal has a net positive charge.

Question 16

Find copper resistance at 80°C from R₀=100 Ω and α₀=0.0043/°C; give four IC advantages; calculate supply current and parallel voltage in Figure 3.

Answer

Step-by-step calculation

  1. R = R₀[1+α₀(T−0)] = 100[1+0.0043×80].
  2. 0.0043×80 = 0.344; R = 100×1.344 = 134.4 Ω.
  • Small size and low weight.
  • Low mass-production cost.
  • High reliability from fewer interconnections.
  • Lower power requirements for many equivalent functions, depending on the design.
33 Ω in series with [15 Ω parallel 27 Ω] and 47 Ω across 240 V.
33 Ω in series with [15 Ω parallel 27 Ω] and 47 Ω across 240 V.
Step-by-step calculation

  1. Parallel resistance = 15×27/(15+27) = 405/42 = 9.642857 Ω.
  2. Total = 33+9.642857+47 = 89.642857 Ω.
  3. Supply current = 240/89.642857 = 2.6773 A.
  4. Parallel voltage = I×Rparallel = 2.6773×9.642857 = 25.8167 V. Both parallel resistors have this potential difference.

Question 17

Convert 10.110₂, 1A4₁₆ and 125₈ to decimal; add 454+234 in BCD; and encode the sum of X=24₈ and Y=88₁₀ in Excess-3.

Answer

Step-by-step calculation

  1. 10.110₂ = 1×2 + 0×1 + 1/2 + 1/4 + 0/8 = 2.75₁₀.
  2. 1A4₁₆ = 1×256 + 10×16 + 4 = 420₁₀.
  3. 125₈ = 1×64 + 2×8 + 5 = 85₁₀.
Step-by-step calculation

  1. Write each digit separately: 454 = 0100 0101 0100; 234 = 0010 0011 0100.
  2. Add units: 0100 + 0100 = 1000 (8), a valid BCD digit.
  3. Add tens: 0101 + 0011 = 1000 (8), a valid BCD digit.
  4. Add hundreds: 0100 + 0010 = 0110 (6), a valid BCD digit.
  5. Result = 0110 1000 1000 (BCD) = 688₁₀. No digit needs the +0110 BCD correction in this addition.
Step-by-step calculation

  1. X = 24₈ = 2×8+4 = 20₁₀.
  2. Sum = 20+88 = 108₁₀.
  3. Excess-3 is per digit: 1+3=4→0100; 0+3=3→0011; 8+3=11→1011.
  4. Result = 0100 0011 1011.

Question 18

Give three SSD advantages; calculate a 1450 W iron’s 45-minute cost at KSh 30/kWh; explain three ROM types; and find simultaneous 1200 MB application instances on 16 GB RAM with 2 GB reserved.

Answer

  • Faster random access than typical mechanical HDDs.
  • No moving parts, improving shock resistance.
  • Quiet operation and often lower power use.
Step-by-step calculation

  1. 1450 W = 1.45 kW; 45 minutes = 0.75 hour.
  2. Energy = 1.45×0.75 = 1.0875 kWh.
  3. Cost = 1.0875×30 = KSh 32.625 ≈ KSh 32.63.
  • PROM: programmed once to hold permanent instructions/data.
  • EPROM: can be erased with ultraviolet light and reprogrammed.
  • EEPROM/flash: non-volatile memory that can be electrically erased and rewritten; flash normally erases blocks.
Step-by-step calculation

  1. Available RAM = 16−2 = 14 GB.
  2. Using the exam convention 1 GB = 1024 MB: 14×1024 = 14,336 MB.
  3. 14336/1200 = 11.9467; maximum = 11 whole instances. 12 need 14,400 MB.
  4. If decimal GB were intended instead, 14,000/1200 = 11.6667, which also allows 11 instances.
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