Free practice · 2365 Level 3
2365 Level 3 practice questions
Twenty free practice questions from the City & Guilds 2365 Level 3 units: environmental technology, electrical science, fault diagnosis, inspection and testing, and electrical design. Try each one, then open the answer to read why it's right.
These come from the same question bank as the full course, which has over a thousand Level 3 questions, study notes for every topic and a mock exam for every unit.
Question 1 · Unit 301
A PV string inverter develops a fault and shows no output on a sunny day. Before any d.c. work, the installer must remember that the array is:
- AStill live whenever light falls on it
- BEarthed and therefore dead
- COnly live at night
- DSafe because the inverter has failed
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Answer: A. PV modules produce a d.c. voltage whenever light falls on them, regardless of inverter state, so the d.c. side must be treated as live and isolated at the d.c. isolator before work.
Question 2 · Unit 301
Upgrading loft insulation before installing a heat pump is recommended so that the:
- AHeat pump can be smaller and run more efficiently
- BFlow temperature must be increased
- CRadiators can be made smaller
- DElectrical supply can be reduced to 110 V
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Answer: A. Reducing heat loss first lowers the building's demand, so a smaller heat pump running at efficient low flow temperatures can meet it — the 'fabric first' principle, improving running costs.
Question 3 · Unit 301
A 6 kWp PV array generates an estimated 5,400 kWh in a year. Its specific yield is:
- A32,400 kWh/kWp
- B900 kWh/kWp
- C1.1 kWh/kWp
- D450 kWh/kWp
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Answer: B. Specific yield = 5400 ÷ 6 = 900 kWh/kWp, a typical figure for a well-sited UK array, allowing comparison between systems of different sizes.
Question 4 · Unit 301
A PV array of 16 modules each rated 410 W has a peak output of:
- A25.6 kW
- B6.56 kW
- C0.41 kW
- D4.10 kW
Show the answer
Answer: B. Peak output = 16 × 410 = 6560 W = 6.56 kWp at standard test conditions. This figure sizes the inverter and informs the network connection application.
Question 5 · Unit 302
Three 60 µF capacitors connected in parallel give a total capacitance of:
- A60 µF
- B20 µF
- C180 µF
- D540 µF
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Answer: C. Capacitors in parallel add: 60 + 60 + 60 = 180 µF. (In series they would combine like resistors in parallel, giving 20 µF.)
Question 6 · Unit 302
A load draws 12 kW at 0.6 power factor. After correction to 0.95, the apparent power changes from 20 kVA to about:
- A7.2 kVA
- B20 kVA
- C31.6 kVA
- D12.6 kVA
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Answer: D. At 0.6 PF, kVA = 12/0.6 = 20 kVA; corrected to 0.95, kVA = 12/0.95 = 12.6 kVA. The lower kVA means less current for the same useful power.
Question 7 · Unit 302
A transformer is 96% efficient and delivers 4.8 kW to its load. Its input power is approximately:
- A5.0 kW
- B9.6 kW
- C4.8 kW
- D4.6 kW
Show the answer
Answer: A. Input = 4.8 / 0.96 = 5.0 kW. The 0.2 kW difference is lost as core (iron) and winding (copper) losses.
Question 8 · Unit 302
A lamp provides 800 lux at 1 m. By the inverse-square law, the illuminance at 2 m is:
- A1600 lux
- B400 lux
- C200 lux
- D800 lux
Show the answer
Answer: C. E ∝ 1/d²; doubling distance from 1 m to 2 m divides illuminance by 2² = 4, giving 800 ÷ 4 = 200 lux.
Question 9 · Unit 303
A two-person job involves fault finding near exposed busbars in a distribution board. The safest approach is to:
- ARely on rubber-soled work boots alone
- BRemove the main earth connection first
- CWork it live in order to save time
- DIsolate, lock off and prove dead before access
Show the answer
Answer: D. Access near exposed live busbars requires safe isolation: isolate, secure with a lock-off, post a notice and prove dead. Working live or relying on footwear is not an acceptable control.
Question 10 · Unit 303
A cooker circuit reads near-zero resistance between line and neutral when isolated and disconnected from the appliance. This indicates:
- AA high-resistance joint
- BA short circuit in the fixed wiring
- CCorrect insulation resistance
- DA healthy open circuit
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Answer: B. With the appliance disconnected, the fixed wiring should read high resistance between line and neutral; near-zero indicates a short circuit (damaged cable or trapped conductor) in the wiring that must be found and repaired.
Question 11 · Unit 303
On a radial circuit with a suspected break, testing continuity from the board to the midpoint accessory tells the electrician whether the fault is:
- AIn the first or the second half of the circuit
- BWithin the protective device
- CAn overload condition
- DCaused entirely by a simple reversed polarity fault
Show the answer
Answer: A. A continuity test to the midpoint shows whether the conductor is intact up to that point, locating the break in the first or second half — the basis of the half-split method on a radial circuit.
Question 12 · Unit 303
After rectifying a fault that involved a damaged section of cable, best practice is to:
- AIgnore it if the circuit works
- BTape over the damage and re-energise
- CLeave the damaged cable in place and simply increase the device rating
- DReplace the damaged length rather than rely on a temporary repair
Show the answer
Answer: D. A damaged cable section should be replaced (or properly jointed in an accessible enclosure) so the installation is restored to a safe, compliant condition, rather than a temporary fix that may fail.
Question 13 · Unit 304
For the R1+R2 test, the line and cpc are linked at the board and resistance is measured at each point. A reading that climbs steadily toward the circuit's far end indicates:
- AA dead short circuit right at the board end of the run
- BSimple reversed polarity
- CA faulty test instrument always
- DNormal behaviour as conductor length increases
Show the answer
Answer: D. R1+R2 naturally increases with distance from the board as conductor length grows; a steady rise to the furthest point is expected. Sudden jumps would indicate a loose joint.
Question 14 · Unit 304
During an EICR, a circuit with no earth fault protection where it is now required (e.g. cables in walls without RCD) is typically coded:
- AC3 improvement only
- BSatisfactory, no action
- CC2 (potentially dangerous)
- DNo code required
Show the answer
Answer: C. Absence of RCD additional protection where now required (e.g. concealed cables) is generally coded C2 (potentially dangerous), making the report unsatisfactory and requiring remedial action.
Question 15 · Unit 304
At handover, demonstrating the operation of the RCD test button to the client is good practice because it:
- AShows them how to perform the recommended periodic user test
- BIncreases the circuit rating
- CAutomatically calibrates the RCD's residual trip current setting
- DReplaces formal testing
Show the answer
Answer: A. Showing the client how and how often to use the RCD test button supports ongoing safety, as users are advised to test RCDs periodically (commonly quarterly). It does not calibrate the device.
Question 16 · Unit 304
The part of BS 7671 that sets out the requirements for inspection and testing is:
- APart 7
- BPart 1
- CPart 6
- DPart 4
Show the answer
Answer: C. Part 6 of BS 7671 covers inspection and testing, including initial verification and periodic inspection. Part 4 covers protection, Part 7 special locations.
Question 17 · Unit 305
A designer is told the supply is 230 V TN-C-S with a Ze of 0.35 Ω and a PFC of 1.6 kA. These figures are used to:
- AChoose the internal wall paint colour scheme
- BSet the required lighting lux level
- CVerify disconnection times and breaking capacity
- DSize the incoming water pipe run
Show the answer
Answer: C. Ze feeds the Zs calculation (disconnection times) and PFC sets the minimum device breaking capacity, so these supply characteristics are central to protection design.
Question 18 · Unit 305
A 230 V circuit has a design current of 18 A and a cable with tabulated capacity 24 A after derating. A suitable device satisfying Ib ≤ In ≤ Iz is:
- A32 A
- B16 A
- C6 A
- D20 A
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Answer: D. In must be ≥ Ib (18 A) and ≤ Iz (24 A): a 20 A device fits. 16 A is below Ib (nuisance tripping); 32 A exceeds Iz (cable unprotected).
Question 19 · Unit 305
Selectivity between a 100 A upstream fuse and a 32 A downstream MCB is desirable so that a fault on the final circuit:
- AAlways blows the 100 A upstream main fuse first
- BHas no effect on either of the devices
- CDisconnects the whole building supply
- DTrips only the 32 A device, keeping others supplied
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Answer: D. With correct selectivity the 32 A MCB clears a final-circuit fault while the 100 A fuse stays intact, so only the affected circuit loses supply and the rest of the installation keeps running.
Question 20 · Unit 305
A design that is fully documented with calculations and assumptions benefits later work because it:
- ALets future changes be checked against the design
- BPrevents any future modification at all
- CRemoves the need for any certification at all
- DReplaces the need for any testing
Show the answer
Answer: A. Documented calculations and assumptions allow later additions or alterations to be assessed against the original design (e.g. spare capacity, diversity), supporting safe and compliant modifications.
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