TN-S Radial Circuit: Zs Too High for a Type C MCB
A 20 A type C MCB protects a 25 m radial circuit wired in 2.5 mm² line and 1.5 mm² CPC thermoplastic twin-and-earth. The installation is TN-S with a measured external loop impedance Ze of 0.8 Ω. With the circuit isolated, the tester records R1+R2 = 0.90 Ω. The socket outlets are ordinary 13 A double sockets, and the customer reports that the MCB sometimes trips only after several seconds when a 2 kW kettle and a 1.5 kW heater are used together. The consumer unit has no RCD. The electrician proposes replacing the 20 A type C MCB with a 20 A type B MCB and says this will fix the problem. The maximum Zs for a 20 A type C device is 1.09 Ω; for a 20 A type B device it is 2.19 Ω.
1. Calculate Zs and compare it with the maximum permitted value for the 20 A type C MCB. Will the device reliably disconnect within 0.4 s?
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Zs = Ze + R1 + R2 = 0.8 + 0.90 = 1.70 Ω. The maximum Zs for a 20 A type C MCB is 1.09 Ω, so 1.70 Ω is well above the limit. The prospective fault current at the end of the circuit is approximately 230/1.70 = 135 A. A type C device needs between 5 and 10 times its rating for magnetic operation, i.e. 100 A to 200 A. At 135 A it may operate magnetically on a favourable sample, but it is not guaranteed to do so within 0.4 s; it may instead trip thermally after several seconds. That is consistent with the reported delayed tripping, although a loose connection should also be investigated. The circuit does not meet the disconnection requirement and is non-compliant.
2. Evaluate the proposed replacement with a 20 A type B MCB. Does it solve the problem, and what checks must accompany the change?
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A 20 A type B MCB has a maximum Zs of 2.19 Ω, so the measured 1.70 Ω is acceptable. The fault current of 135 A is greater than 5 × 20 = 100 A, so magnetic disconnection is expected well within 0.4 s. The replacement is therefore technically sound, provided the cable is still protected: 2.5 mm² is adequate for 20 A, and the 1.5 mm² CPC must withstand the fault energy. Using the adiabatic equation with a typical let-through time under 0.1 s and k = 115 for thermoplastic copper, S = √(135² × 0.1)/115 = 0.37 mm², so 1.5 mm² passes. The electrician must also confirm that the type B device's short-circuit rating exceeds the prospective fault current at the origin (230/0.8 = 288 A), re-test Zs after the change and investigate the reported tripping, since a 15.2 A load does not explain a 20 A MCB tripping after several seconds.
3. Could a 30 mA RCD be used instead of changing the MCB? Explain the limitations of that approach.
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A 30 mA RCD would operate at a much lower current than the 135 A earth-fault current, so it could provide disconnection and would also give additional protection to the socket outlets. However, using it merely to mask a high Zs is poor practice: the high loop impedance may indicate a poor connection or an undersized CPC, and the RCD does not correct that defect. On a TN-S system the preferred solution is to reduce Zs by changing the device characteristic, increasing the CPC or line conductor size, or shortening the run. An RCD may be acceptable as the protective device for fault protection if it is correctly rated and the circuit arrangement is suitable, but it must not be used to avoid investigating the underlying fault path. The socket circuit still needs 30 mA additional protection regardless.