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On-Site Guide BS 7671:2018 - 7.2.2 Socket-outlet circuits

Former Community Member
Former Community Member
Hello, I am studying the On-Site Guide and have reached Section 7 Final Circuits.


7.2.2 Socket-outlet circuits

...

As a rule of thumb for rings, unfused spur lengths should not exceed 1/8 the cable length from the spur to the furthest point of the ring.

...


I should like to ask, what does this mean?


The length of the cable used for the non-fused spur is the length of the cable from where it connects to ring final circuit at the terminals of a socket-outlet (or junction box...), to the terminals at the other end of the cable where it connects to a single or twin socket-outlet.


What is the length ‘from the spur to the furthest point of the ring’? Does ‘from the spur’ mean from the the point where the spur connects to the circuit? Does ‘to the furthest point of the ring’ mean to the furthest point away from where the spur connects to the circuit i.e. half the length of the ring final circuit? If so, the maximum lengths would be the same for all spurs on that circuit.


Or, does it mean from where the spur connects to the circuit to the point which is the furthest away from the protection device in the consumer unit i.e. a point half way around the circuit beginning from the consumer unit? If so, the maximum length of the spur cable would be longest nearer to the consumer unit and shorter as the furthest point away is reached... Surely this cannot be correct, because at the furthest point the maximum length would be zero - !


So, in conclusion: 

1/ establish the full length of the ring final circuit;

2/ find the furthest point of the ring by dividing the full length by two;

3/ divide that length by 8 to find the maximum length of a spur cable for that particular ring final circuit.


Please would someone confirm my understanding or explain what the sentence within 7.2.2 actually means?


Many thanks!


Parents
  • 0
    Look at it this way, if you have a socket-outlet at the furthest point on a ring final circuit, the electrical power runs half way around the first half of the circuit and half way around the second half of the circuit (i.e. one full circuit); if you have a socket outlet at, or very close to, the consumer unit, the electrical power, effectively, runs all the way around the whole circuit. My conclusion being that, the electrical power on a ring-final circuit will, indeed, run the same distance around that circuit regardless of the location of the socket-outlet.

    Not quite. The current drawn though a socket will be divided between the two sides (legs) of the ring in proportion to the conductivity (1/resistance) of each leg. So a socket at the exact mid point drawing 13A will see 6.5A flowing through both sides. A socket near to the CU will see almost all its current flow through the shortest leg, and very little go the long way around.


    You can verify that simply using Ohm's Law - noting that the voltage at both legs at the CU must be the same (as they're connected together) and likewise the voltage on both legs where they meet at a socket must also be the same (again as they're solidly connected together) - but the longer leg will have a higher resistance so Ohm's law say that it must therefore carry a proportionally lower current if V=IR is to hold true for both legs at the same time.


       - Andy.
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  • 0
    Look at it this way, if you have a socket-outlet at the furthest point on a ring final circuit, the electrical power runs half way around the first half of the circuit and half way around the second half of the circuit (i.e. one full circuit); if you have a socket outlet at, or very close to, the consumer unit, the electrical power, effectively, runs all the way around the whole circuit. My conclusion being that, the electrical power on a ring-final circuit will, indeed, run the same distance around that circuit regardless of the location of the socket-outlet.

    Not quite. The current drawn though a socket will be divided between the two sides (legs) of the ring in proportion to the conductivity (1/resistance) of each leg. So a socket at the exact mid point drawing 13A will see 6.5A flowing through both sides. A socket near to the CU will see almost all its current flow through the shortest leg, and very little go the long way around.


    You can verify that simply using Ohm's Law - noting that the voltage at both legs at the CU must be the same (as they're connected together) and likewise the voltage on both legs where they meet at a socket must also be the same (again as they're solidly connected together) - but the longer leg will have a higher resistance so Ohm's law say that it must therefore carry a proportionally lower current if V=IR is to hold true for both legs at the same time.


       - Andy.
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