Overloading one side of ring circuit

I've seen several references on different threads to the effect that one should avoid overloading one side of a ring circuit. I've also read that it's acceptable to route the return leg of a ring circuit by taking the 'return' cable back the same way as the 'outward' cable.

So, if this installation method is followed, is it necessary (or advisable), in order to avoid overloading one side of the circuit, to feed alternate power sockets from the 'outward' cable and the remaining sockets from the 'return' cable?

It's not so much the sockets as the appliances. There is a tendency for the washing machine, tumble drier and dishwasher to cluster together with the kettle and toaster in the kitchen ,and if you have a dual-fuel oven or a combination microwave, that may be very close too. All these appliances have heating elements so can draw high currents (though not continually and, with luckj, not all on at the same time).

You can have as many sockets as you like on one leg with radios, table lamps, TVs and hifi plugged into them.

The whole point of a ring is that there is no "side", the current flows down both sides to the equipment. You can overload a ring by plugging in too much equipment but the MCBs in the CU will stop the cable overheating. If you are planning on putting large loads on a ring then the electrician might design in multiple rings, or even seperate radials, for specific items depending on their power rating, but not socket order.

As for the how the cable is routed, it will have to come together at some point to meet up at the CU! Where they follow the same route can be anywhere.

If you did do the alternate socket procedure you would need to document the cable route since it would be very unusual and not something an electrician would think to check out when testing the system. That's a hint that you don't need to, or even possibly shouldn't, carry out the procedure.

It is a valid concern, and you shouldn't have a length of cable of 10m feeding a number of socket outlets and the return leg of 10m without any socket outlets, such a configuration would be an unbalenced ring, you should aim to have the loads balenced around the ring, if its actually ring shaped and goes round the walls of a room for example, it should take care of its self to a certain extent, but in the case of a number of socket outlets along a single wall, yes alternate which leg of the ring different outlets are on to balence them, or install a radial circuit

HandyJon said:

The whole point of a ring is that there is no "side", the current flows down both sides to the equipment.

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Yes, but if the load is not exactly in the middle of the ring, then the current flow on each leg of the ring will not be equal, and in the case of a very unbalenced ring, you can overload one side of the ring

You can overload a ring by plugging in too much equipment but the MCBs in the CU will stop the cable overheating.

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Its possible for a ring cable to be overloaded, even if the circuit in general isn't, if the loads are unbalencec around the ring

If you are planning on putting large loads on a ring then the electrician might design in multiple rings, or even seperate radials, for specific items depending on their power rating, but not socket order.

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au contraire, while number and type of circuits is something considered in the design, so is balence of ring circuits, because as above, its a good idea to do your best to design out the possibily of an unnbalenced ring

As for the how the cable is routed, it will have to come together at some point to meet up at the CU! Where they follow the same route can be anywhere.

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Obviously they come together at the CU, its a ring, and I'm not quite sure what exactly the second part of this statement is saying

If you did do the alternate socket procedure you would need to document the cable route since it would be very unusual and not something an electrician would think to check out when testing the system.

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It wouldn't be that unusual, its often done for example where a conduit runs along a wall with many outlets on it, and the circuit used is a ring, and both legs return to the orgin through the same conduit, to do otherwise would lead to a ring thats unbalenced by design. And not sure what test you are talking about, but the only thing that I can think it would cause a problem with, is if the tests indicate a fault and your are trying to trace this fault, it would be usful to know which order the ring is in then, but I would have thought that most electricians would not be thrown by it, its not like it isn't good practice to design ring circuits so they are balenced...

That's a hint that you don't need to, or even possibly shouldn't, carry out the procedure.

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I beg to differ, as explained above

I've heard the concept of the unbalanced ring, but I'd be interested to know how it manifests itself. An increase in cable temperature on one side, perhaps? Or voltage drop? Used to be more of a prob in those long-gone days before rings, when a heavy load near the start of a radial would drop the noticably the volts at the end.

JohnD said:

I've heard the concept of the unbalanced ring, but I'd be interested to know how it manifests itself. An increase in cable temperature on one side, perhaps? Or voltage drop? Used to be more of a prob in those long-gone days before rings, when a heavy load near the start of a radial would drop the noticably the volts at the end.

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A load on any circuit will create volt drop on the circuit, all cable has a resistance, and if you apply ohms law V=IR you can use that to work out the voltage dropped over a piece of cable. You can see this in practice if you switch on an electric shower, the lights will dim a little.

On a ring circuit, things are no different, exact that because of the nature of the ring, the resistances are less. If we take a point a third of the way around the ring and we plug in an electric heater, there will be volt drop over the circuit, and the point the heater is plugged in will be at a slightly lower voltage than the terminals in the distrubution board, the volt drop over the two different lengths of cable will be the same (because they are connected together at the point we are measureing voltage at), but the resistance of the longer leg will be more than that of the shorter leg (twice as much, because its twice as long), if we use ohms law to work out the current in the ring legs we get voltdrop over resistance, and because the resistance in one leg is half of that in the other leg, it follows that current will be double, so we have two thirds flowing through one leg, and one third flowing through the other leg. usually in a properly designed ring this doesn't matter too much, the cable is rated to carry at least two thrids of the breaker rating, but if you have a few ring sockets very close to the orgin on one leg of the ring, and have quite a few high power appliances in the ring, its possible that you'll be overloading one of the legs.

I hope this explanation is helpful

I don't think Ohms Law should be used to calculate how much current flows through which leg of a ring circuit. Kirchoffs Law should be used in this case. If you had a ring circuit and plugged a 3kW load into a socket 1mtr for the CU and the other leg totalled 100mts back to the CU, although the volt drop/resistance on the first leg is lower, as the current increase, the resistance increases and the current flows through the other leg (Kirchoffs Law).

If the ring circuits are designed in accordance with BS-7671 there is no problem and any contractor will do this for you.

Pensdown said:

I don't think Ohms Law should be used to calculate how much current flows through which leg of a ring circuit.

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Why ever not?! Does ohms law suddenly stop applying if the circuit is a ring?....

Kirchoffs Law should be used in this case.

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While kirchoff's law is undoubtedly true for a ring as it is any other circuit, I fail to see how it helps you calculate the current flowing in each ring leg

If you had a ring circuit and plugged a 3kW load into a socket 1mtr for the CU and the other leg totalled 100mts back to the CU, although the volt drop/resistance on the first leg is lower

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The resistance is lower, the volt drop over each leg of the ring will however be the same (they go into the same terminals, how can you have two different voltage readings?)

as the current increase, the resistance increases and the current flows through the other leg (Kirchoffs Law).

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By what process does an increase in current cause a increase in resistance? (temperature increases aside - and don't try to tell me that temperature increases in conductors ensures the ring stays in balence)


lets work your example of your 3kw electric fire through, a 2.5mm² copper conductor has a resistance of 7.41 milliohms per metre, so combined resistance of the phase and neutral conductors on the 1m leg of your ring will be 0.01482 ohms, and on the 100m leg, 1.482 ohms, putting these two resisters in parallel will result in a resistance of 0.01467 ohms, we also have the current, 13A, we can apply ohms law and find that the volt drop to that point on the ring is 0.19071, and now we have this, coupled with the resistance of each leg, we can apply ohms law again for each leg of the ring and find the current in each leg, and we find that the legs carry 12.8684 amps on the 1m leg, and 0.128684 amps in the 100m leg (100th of the current in the shorter leg... coincidence...I think not )

Same principle applies to any ring, not just ones that are out of volt drop spec (aforementioned ring drops more than 4% of supply voltage assuming full load at mid point)

So you're saying that there's a lot more current carried in the short leg, so it would warm up a tiny bit, but are there any other undesirable effects?

the current in the ring will be spread out between both legs. altho if everything is to 1 side, 1 leg will probably carry slightly more poer than the other, but not by much (and not enough to overheat the cable)

JohnD said:

So you're saying that there's a lot more current carried in the short leg, so it would warm up a tiny bit, but are there any other undesirable effects?

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No thats it , but if we use the example of the ring with 1m and 100m legs, if we instead of one elctric fire, plugged two in, and a 1.5kw hair dryer as well, we'll be pulling 32A through one leg and all of 300ma through the other, if we have 3 fires and one hairdryier we then have 45A and because of the way breakers work it'll allow it a bit ( cables like breakers are designed to take a 1.45 overload for a while) and that current for a cable designed to carry 32A constantly won't pose a problem, but here we are putting 45A through a cable that is designed to take 20A constantly, and 20A x 1.45 = 29A on short overloads

As it happens, the cable is under overload current when the breaker is at design current, and when the breaker is at short duration overload current, the cable will be setting fire to the wallpaper

EDIT: just to clarify, same dangers as a broken ring, over a ring thats intact, but just cronically unbalenced, pretty unlikely to occur this extremely in the real world, but 26A/6A wouldn't be too unlikely in a badly designed ring, and if we are assuming 20A current capacity like before (minimum allowed for ring, and prefectly possible depedning on derating factors) 26A won't burn down the house, but will accelerate ageing of the cable

Ooerrr. What have I started? What is it they say: "Ask two lawyers for advice and you'll three different opinions". Must be the same with sparks.

Adam_151 said:

we'll be pulling 32A through one leg and all of 300ma through the other,

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dare i ask how the **** did you come up with that. more will go thru 1 leg than the other, BUT not by that much. power will spread out, with slightly more current using the less resistive route. but not 32A and 0.3A...

andy said:

dare i ask how the **** did you come up with that. more will go thru 1 leg than the other, BUT not by that much. power will spread out, with slightly more current using the less resistive route. but not 32A and 0.3A...

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As proved above, directly in relation to the resistances, 100x more current will through through the path with 100th of the resistance of the other