Adiabatic Lecture.

Dad my head hurts. Is this guy any good?

Nice video ... I think he has made one fundamental error though - he's only considered faults at the very end of the circuit.

Even with Ze at 0.35Ω a fault nearer the start of the circuit would give rise to a considerably larger current (say around 657A rather than the 151.3A he considered) - so using the same 0.1s disconnection time, the S=√(I²t)/k would yield around 1.8mm² (not so reassurring for a 1.5mm² c.p.c.)

There's also the issue that assuming 0.35Ω for Ze isn't really considering the worst case, neither is the 1.2 factor for the conductor resistance which presumes that the conductors will have been heated to 70ºC prior to the fault - which they may well not have been in practice. On the plus side his assumption of a 0.1s disconnection time is rather unrealistic the other way (even if BS 7671 fails to provide any better information),

There is a tradition in many textbooks of considering only faults at the far end - and that was justifiable when the protective device is a fuse since they tend to have increasing energy let-through with reducing fault currents so far end faults were the worst case for both disconnection times (ADS) and conductor protection. I suspect a lot of people have taken the old explanations and just substituted fuses with MCBs without realizing that MCB charactreristics are quite different.

On the RCD element of an RCBO, we can only assume (for larger earth fault currents) that it'll disconnect within 40ms (as the standard specifies - it may be faster under particular circumstances under tests that we do, but that's not a guarantee that it'll always be that fast under all allowable conditions). Even 40ms might not be quick enough for conductor protection if the fault currents are high - a 1.0mm² would only be good up to about 575A and a 1.5mm² up to 862.5A (presuming k=115).

   - Andy.

Dad my head hurts. Is this guy any good?

Nice video ... I think he has made one fundamental error though - he's only considered faults at the very end of the circuit.

Even with Ze at 0.35Ω a fault nearer the start of the circuit would give rise to a considerably larger current (say around 657A rather than the 151.3A he considered) - so using the same 0.1s disconnection time, the S=√(I²t)/k would yield around 1.8mm² (not so reassurring for a 1.5mm² c.p.c.)

There's also the issue that assuming 0.35Ω for Ze isn't really considering the worst case, neither is the 1.2 factor for the conductor resistance which presumes that the conductors will have been heated to 70ºC prior to the fault - which they may well not have been in practice. On the plus side his assumption of a 0.1s disconnection time is rather unrealistic the other way (even if BS 7671 fails to provide any better information),

There is a tradition in many textbooks of considering only faults at the far end - and that was justifiable when the protective device is a fuse since they tend to have increasing energy let-through with reducing fault currents so far end faults were the worst case for both disconnection times (ADS) and conductor protection. I suspect a lot of people have taken the old explanations and just substituted fuses with MCBs without realizing that MCB charactreristics are quite different.

On the RCD element of an RCBO, we can only assume (for larger earth fault currents) that it'll disconnect within 40ms (as the standard specifies - it may be faster under particular circumstances under tests that we do, but that's not a guarantee that it'll always be that fast under all allowable conditions). Even 40ms might not be quick enough for conductor protection if the fault currents are high - a 1.0mm² would only be good up to about 575A and a 1.5mm² up to 862.5A (presuming k=115).

   - Andy.