Part P Circuit Design and Protection 2 — Questions and Answers
Question 1: An electrician is designing an electric shower circuit. The shower is rated at 9.5kW at 230V. What is the design current (Ib)?
- 32A
- 38A
- 41.3A (Correct answer)
- 45A
Correct answer: 41.3A
The design current is calculated using Ib = P/V = 9500/230 = 41.3A. This means a 45A MCB and appropriately rated cable (typically 10mm² twin and earth) would be required. The cable must be rated to carry at least the MCB rating (In) after applying all relevant correction factors for the installation conditions.
Question 2: What is an RCBO and what advantage does it offer over separate MCB and RCD protection?
- A remote-controlled circuit breaker operated by smartphone
- A combined MCB and RCD in a single device providing individual circuit RCD protection (Correct answer)
- A type of surge protection device
- A replacement for the main switch
Correct answer: A combined MCB and RCD in a single device providing individual circuit RCD protection
An RCBO (Residual Current Circuit Breaker with Overcurrent protection) combines the functions of an MCB and RCD in a single module-width device. Each circuit has its own independent RCD protection, so a fault on one circuit does not trip other circuits. This eliminates the nuisance of a shared RCD tripping and affecting multiple circuits simultaneously.
Question 3: When selecting a cable for a circuit, the relationship Ib ≤ In ≤ Iz must be satisfied. What do these symbols represent?
- Ib = breaking current, In = nominal voltage, Iz = impedance
- Ib = design current, In = rated current of protective device, Iz = current-carrying capacity of cable (Correct answer)
- Ib = insulation resistance, In = neutral current, Iz = zone rating
- Ib = bonding current, In = nominal frequency, Iz = zonal impedance
Correct answer: Ib = design current, In = rated current of protective device, Iz = current-carrying capacity of cable
This fundamental inequality ensures that: the design current (Ib, the actual current the circuit will carry) does not exceed the rating of the protective device (In, the MCB/fuse rating), and the protective device rating does not exceed the current-carrying capacity of the cable (Iz, the maximum safe continuous current for the cable in its installation conditions).
Question 4: A 6mm² twin and earth cable is installed in a 20m run from the consumer unit to a cooker switch. The design current is 28A. What formula is used to calculate voltage drop?
- Vd = I²R
- Vd = (mV/A/m × Ib × L) / 1000 (Correct answer)
- Vd = P × I
- Vd = Zs × In
Correct answer: Vd = (mV/A/m × Ib × L) / 1000
Voltage drop is calculated using Vd = (mV/A/m × Ib × L) / 1000, where mV/A/m is the voltage drop per ampere per metre from the cable tables (BS 7671 Appendix 4), Ib is the design current in amps, and L is the cable length in metres. The result in millivolts is divided by 1000 to convert to volts for comparison with the permitted limits.
Question 5: What is the purpose of discrimination (selectivity) between protective devices in a domestic installation?
- To ensure all circuit breakers trip simultaneously during a fault
- To ensure only the protective device closest to the fault operates, leaving other circuits unaffected (Correct answer)
- To reduce the cost of the installation
- To increase the earth fault loop impedance
Correct answer: To ensure only the protective device closest to the fault operates, leaving other circuits unaffected
Discrimination ensures that during a fault, only the device nearest to the fault trips, while upstream devices remain closed. In a domestic consumer unit, this means a fault on one circuit should trip only that circuit's MCB, not the main switch or RCD protecting other circuits. This is achieved through appropriate device selection and time-current coordination.
Question 6: What is the adiabatic equation used for in electrical installation design?
- Calculating the voltage drop in long cable runs
- Calculating the minimum cross-sectional area of a protective conductor to withstand fault current without damage (Correct answer)
- Calculating the diversity of a circuit
- Determining the number of socket outlets required
Correct answer: Calculating the minimum cross-sectional area of a protective conductor to withstand fault current without damage
The adiabatic equation S = √(I²t)/k calculates the minimum cross-sectional area (S) of a protective conductor that can carry the fault current (I) for the disconnection time (t) without the conductor temperature exceeding its limit. The factor k accounts for the conductor material and insulation type. This ensures the CPC can safely carry the fault current.
Part P Circuit Design and Protection 2
An electrician is designing an electric shower circuit.
The shower is rated at 9.5kW at 230V.
What is the design current (Ib)?