Got a call from a former neighbour the other day... "all my electrics
have gone off!" A quick over the phone diagnosis suggested the socket
circuits RCD was had tripped and would not reset even with all the
downstream MCBs off.

So I went to investigate. Sure enough, it was tripped, and would not
reset. So having enquired if he had been nailing anything to a wall
recently (his attempts at DIY have varying outcomes, the last time he
had a problems like this was when nailing down loft floorboards through
the nice new lighting circuit cables, and then later filling a light
switch back box with wet plaster). He denied all knowledge.

It was protecting three ring circuits, so I disconnected each pair of
neutrals in turn, and tried it again. Still would not reset. Dropped
thee live bus bar out of the bottom and the neutral flylead to the split
load neutral bus bar *still would not reset*. So I decided the RCD was
knackered, and realised I forgot to bring a spare. Went home, came back
with another, wired it in, and the bugger still tripped. This time
however turning off the kitchen sockets circuit cured the problem. First
problem fixed.

So since the only things not unplugged in the kitchen were the cooker
and the washing machine, I turned off the cooker isolator. Problem found.

Combination cooker - gas hob, lekky oven. Figured, bet I know what that
is - knackered element. Stuck an insulation resistance tester between L
& E on the plug and got a reading of 20k ohms. Had the back off the
cooker, and disconnected the element (and like a dope actually removed
it from the oven before testing it), and found that was fine and it was
still leaking like an electric sieve without the element. Disconnected
the fan, no change, The oven light? Nope. Running out of things to
disconnect now... In the end I disconnected the wires connecting the
back of the flex inlet to the rest of the cooker. Still leaky! This was
getting silly. Disconnected the live wire of the flex from the the flex
inlet, still 20k with it dangling in free space. Cut the moulded plug
off the end of the wire, eureka! Sent him up to the shop to procure a
plug while I resembled the cooker[1]!

Then the storey becomes clear... I was explaining to the lady of the
house how it must be that moisture has somehow got into the plug. Ah,
she says, perhaps it happened when I was washing down the wall above the
cooker this morning!

So two and a half hours titting about to find a damp plug! (and an RCD
that had obviously got a bit over sensitive in its old age).

[1] I had to marvel at his brass neck.... He returns with a plug with 8"
of two core flex dangling out of it... I asked where on earth did he
find that? Turns out he went to the local hardware shop (traditional
arkwright style affair), looked at all the nice new plugs on display and
baulked at the £2.50 they wanted for them. Said to the nice girl behind
the counter, have you got any plugs. She pointed him at the display. "Oh
no, not those, they are expensive, haven't you got any cheap ones?". She
obviously took pity on him, vanished into the back of the shop and
returned with a slightly used example and relived him of 50p! (probably
left the boss wondering why his anglepoise is not working!)


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

John Rumm wrote:
Sent him up to the shop to procure a
plug while I resembled the cooker.

In what way did you resemble it? Were you in pieces yourself? Non
compos, so to speak?

Bill

On 19/07/2012 1:19 p.m., Bill Wright wrote:
John Rumm wrote:
Sent him up to the shop to procure a
plug while I resembled the cooker.

In what way did you resemble it? Were you in pieces yourself? Non
compos, so to speak?

To which he replied: "I resemble that remark!"

On 19/07/2012 02:19, Bill Wright wrote:
John Rumm wrote:
Sent him up to the shop to procure a
plug while I resembled the cooker.

In what way did you resemble it? Were you in pieces yourself? Non
compos, so to speak?

My wife thinks I am a hottie, but also prone to gas ;-)


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

On 19/07/2012 01:34, John Rumm wrote:


Hope you charged appropriately!

"John Rumm" wrote in message
o.uk...

8

So two and a half hours titting about to find a damp plug! (and an RCD
that had obviously got a bit over sensitive in its old age).

Didn't the new one trip?
Is that over sensitive too?

On 2012-07-19, John Rumm wrote:

On 19/07/2012 02:19, Bill Wright wrote:
John Rumm wrote:
Sent him up to the shop to procure a
plug while I resembled the cooker.

In what way did you resemble it? Were you in pieces yourself? Non
compos, so to speak?

My wife thinks I am a hottie, but also prone to gas ;-)

applause

On 19/07/2012 09:35, dennis@home wrote:


"John Rumm" wrote in message
o.uk...

8

So two and a half hours titting about to find a damp plug! (and an RCD
that had obviously got a bit over sensitive in its old age).

Didn't the new one trip?

It did when the fault was present... the old one tripped all the time
even with no load connected.

Is that over sensitive too?

No it appeared to be operating within expected parameters... I only had
my old Megger RCD tester with me that has fixed test currents rather
than smooth ramp capabilities, however if did not trip at 15mA and did
at 30mA.


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

On 19/07/2012 01:34, John Rumm wrote:
Got a call from a former neighbour the other day... "all my electrics
have gone off!" A quick over the phone diagnosis suggested the socket
circuits RCD was had tripped and would not reset even with all the
downstream MCBs off.

SNIP

Then the storey becomes clear... I was explaining to the lady of the
house how it must be that moisture has somehow got into the plug. Ah,
she says, perhaps it happened when I was washing down the wall above the
cooker this morning!

It's amazing how different things can be - we have a fair few computers,
printers, etc. and the filter leakage makes our RCD prone to the odd
false trip. However when I forgot that I hadn't put the extension lead
away and the twin sockets filled up to brimming, no trip!

SteveW

On 19/07/2012 15:39, SteveW wrote:
On 19/07/2012 01:34, John Rumm wrote:
Got a call from a former neighbour the other day... "all my electrics
have gone off!" A quick over the phone diagnosis suggested the socket
circuits RCD was had tripped and would not reset even with all the
downstream MCBs off.

SNIP

Then the storey becomes clear... I was explaining to the lady of the
house how it must be that moisture has somehow got into the plug. Ah,
she says, perhaps it happened when I was washing down the wall above the
cooker this morning!

It's amazing how different things can be - we have a fair few computers,
printers, etc. and the filter leakage makes our RCD prone to the odd
false trip. However when I forgot that I hadn't put the extension lead
away and the twin sockets filled up to brimming, no trip!

Indeed, and its the first time I have seen a plug alone showing that
much leakage - especially odd when you consider it was a moulded on
design with not many apparent routes in for water other than the fuse
holder. (it was even more leaky N to E than it was L to E, I found when
testing it in isolation)

--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

"John Rumm" wrote in message
o.uk...

It's amazing how different things can be - we have a fair few computers,
printers, etc. and the filter leakage makes our RCD prone to the odd
false trip. However when I forgot that I hadn't put the extension lead
away and the twin sockets filled up to brimming, no trip!

Indeed, and its the first time I have seen a plug alone showing that much
leakage - especially odd when you consider it was a moulded on design with
not many apparent routes in for water other than the fuse holder. (it was
even more leaky N to E than it was L to E, I found when testing it in
isolation)

Its going to be what it was filled up with.
Water isn't very conductive, you can even run electronics submerged in
water.

Now if the cleaner had a lot of salt in it like some washing up has..

John Rumm wrote:
Got a call from a former neighbour the other day... "all my electrics
have gone off!" A quick over the phone diagnosis suggested the socket
circuits RCD was had tripped and would not reset even with all the
downstream MCBs off.

So I went to investigate. Sure enough, it was tripped, and would not
reset. So having enquired if he had been nailing anything to a wall
recently (his attempts at DIY have varying outcomes, the last time he
had a problems like this was when nailing down loft floorboards
through the nice new lighting circuit cables, and then later filling
a light switch back box with wet plaster). He denied all knowledge.

It was protecting three ring circuits, so I disconnected each pair of
neutrals in turn, and tried it again. Still would not reset. Dropped
thee live bus bar out of the bottom and the neutral flylead to the
split load neutral bus bar *still would not reset*. So I decided the
RCD was knackered, and realised I forgot to bring a spare. Went home,
came back with another, wired it in, and the bugger still tripped.
This time however turning off the kitchen sockets circuit cured the
problem. First problem fixed.

So since the only things not unplugged in the kitchen were the cooker
and the washing machine, I turned off the cooker isolator. Problem
found.
Combination cooker - gas hob, lekky oven. Figured, bet I know what
that is - knackered element. Stuck an insulation resistance tester
between L & E on the plug and got a reading of 20k ohms. Had the back
off the cooker, and disconnected the element (and like a dope
actually removed it from the oven before testing it), and found that
was fine and it was still leaking like an electric sieve without the
element. Disconnected the fan, no change, The oven light? Nope.
Running out of things to disconnect now... In the end I disconnected
the wires connecting the back of the flex inlet to the rest of the
cooker. Still leaky! This was getting silly. Disconnected the live
wire of the flex from the the flex inlet, still 20k with it dangling
in free space. Cut the moulded plug off the end of the wire, eureka!
Sent him up to the shop to procure a plug while I resembled the
cooker[1]!
Then the storey becomes clear... I was explaining to the lady of the
house how it must be that moisture has somehow got into the plug. Ah,
she says, perhaps it happened when I was washing down the wall above
the cooker this morning!

So two and a half hours titting about to find a damp plug! (and an RCD
that had obviously got a bit over sensitive in its old age).


Try this one, you like a puzzle. A new build with a 17th edition dual RCD
set up with the upstairs ring on one RCD and the downstairs ring on the
other RCD. Turn either socket MCB off and both RCDs trip:-)

It's one I had not seen before and it took me 20 minutes to find it (it will
now only take me 3 seconds to find if I ever see the same fault).

--
Adam

On 19/07/2012 17:47, dennis@home wrote:


"John Rumm" wrote in message
o.uk...

It's amazing how different things can be - we have a fair few computers,
printers, etc. and the filter leakage makes our RCD prone to the odd
false trip. However when I forgot that I hadn't put the extension lead
away and the twin sockets filled up to brimming, no trip!

Indeed, and its the first time I have seen a plug alone showing that
much leakage - especially odd when you consider it was a moulded on
design with not many apparent routes in for water other than the fuse
holder. (it was even more leaky N to E than it was L to E, I found
when testing it in isolation)

Its going to be what it was filled up with.
Water isn't very conductive, you can even run electronics submerged in
water.

Now if the cleaner had a lot of salt in it like some washing up has..

I expect it was a mix of a water and proprietary kitchen cleaner - Flash
etc. I did notice a trigger bottle of something sat on the counter.

--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

Try this one, you like a puzzle. A new build with a 17th edition dual RCD
set up with the upstairs ring on one RCD and the downstairs ring on the
other RCD. Turn either socket MCB off and both RCDs trip:-)

It's one I had not seen before and it took me 20 minutes to find it (it will
now only take me 3 seconds to find if I ever see the same fault).


Bet it's something to do with the Neutral line or Earth somewhere...
--
Tony Sayer

"tony sayer" wrote in message
...
Try this one, you like a puzzle. A new build with a 17th edition dual RCD
set up with the upstairs ring on one RCD and the downstairs ring on the
other RCD. Turn either socket MCB off and both RCDs trip:-)

It's one I had not seen before and it took me 20 minutes to find it (it
will
now only take me 3 seconds to find if I ever see the same fault).


Bet it's something to do with the Neutral line or Earth somewhere...

Sharing lives more likely.

On 19/07/2012 20:32, ARWadsworth wrote:

Try this one, you like a puzzle. A new build with a 17th edition dual RCD
set up with the upstairs ring on one RCD and the downstairs ring on the
other RCD. Turn either socket MCB off and both RCDs trip:-)

It's one I had not seen before and it took me 20 minutes to find it (it will
now only take me 3 seconds to find if I ever see the same fault).

Both MCBs feeding one end of each ring rather than both ends of one each?


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

John Rumm wrote:
On 19/07/2012 20:32, ARWadsworth wrote:

Try this one, you like a puzzle. A new build with a 17th edition
dual RCD set up with the upstairs ring on one RCD and the
downstairs ring on the other RCD. Turn either socket MCB off and
both RCDs trip:-) It's one I had not seen before and it took me 20
minutes to find it
(it will now only take me 3 seconds to find if I ever see the same
fault).

Both MCBs feeding one end of each ring rather than both ends of one
each?

Yep. And with the neutals also in the wrong busbar the RCD stays balanced
until one MCB is switched off.

--
Adam

On Saturday, July 21, 2012 7:43:51 AM UTC+1, wrote:
John Rumm wrote:
> On 19/07/2012 20:32, ARWadsworth wrote:
>
> > Try this one, you like a puzzle. A new build with a 17th edition
> > dual RCD set up with the upstairs ring on one RCD and the
> > downstairs ring on the other RCD. Turn either socket MCB off and
> > both RCDs trip:-) It's one I had not seen before and it took me 20
> > minutes to find it
> > (it will now only take me 3 seconds to find if I ever see the same
> > fault).
>
> Both MCBs feeding one end of each ring rather than both ends of one
> each?

Yep. And with the neutals also in the wrong busbar the RCD stays balanced
until one MCB is switched off.

Shouldn't they also both trip if there were unequal loads on the rings?

Onetap wrote:
On Saturday, July 21, 2012 7:43:51 AM UTC+1,
wrote:
John Rumm wrote:
> On 19/07/2012 20:32, ARWadsworth wrote:
>
> > Try this one, you like a puzzle. A new build with a 17th
edition > > dual RCD set up with the upstairs ring on one RCD
and the > > downstairs ring on the other RCD. Turn either
socket MCB off and > > both RCDs trip:-) It's one I had
not seen before and it took me 20 > > minutes to find it
> > (it will now only take me 3 seconds to find if I ever see
the same > > fault).
>
> Both MCBs feeding one end of each ring rather than both ends
of one > each?

Yep. And with the neutals also in the wrong busbar the RCD stays
balanced
until one MCB is switched off.

Shouldn't they also both trip if there were unequal loads on the
rings?

Both did trip when either MCB was turned off. With both MCBs on the circuit
is balanced.
--
Adam

On 21/07/2012 09:11, Onetap wrote:
On Saturday, July 21, 2012 7:43:51 AM UTC+1, wrote:
John Rumm wrote:
> On 19/07/2012 20:32, ARWadsworth wrote:
>
> > Try this one, you like a puzzle. A new build with a 17th edition
> > dual RCD set up with the upstairs ring on one RCD and the
> > downstairs ring on the other RCD. Turn either socket MCB off and
> > both RCDs trip:-) It's one I had not seen before and it took me 20
> > minutes to find it
> > (it will now only take me 3 seconds to find if I ever see the same
> > fault).
>
> Both MCBs feeding one end of each ring rather than both ends of one
> each?

Yep. And with the neutals also in the wrong busbar the RCD stays balanced
until one MCB is switched off.

Shouldn't they also both trip if there were unequal loads on the rings?

No, you have in effect made one hybrid circuit out of the two. So any
load on either ring ends up drawing power through both MCBs and hence
both RCDs, and also returning it though both. So the total sum adds up
to zero for each RCD even if some of the balancing current is being fed
from the other (and vice versa)


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal loads on the rings?

No, you have in effect made one hybrid circuit out of the two. So any
load on either ring ends up drawing power through both MCBs and hence
both RCDs, and also returning it though both. So the total sum adds up
to zero for each RCD even if some of the balancing current is being fed
from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge my knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance between two conductors using a differential current transformer. This measures the difference between the current flowing through the live conductor and that returning through the neutral conductor. If these do not sum to zero, there is a leakage of current to somewhere else (to earth/ground, or to another circuit), and the device will open its contacts." Much as I thought.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2, you'd have 5A through L and 3A through N on Ring 1 RCD, which should trip. Similarly 3A & 5A on L & N respectively on Ring 2 RCD which should trip simultaneously.

What am I missing here?

On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:
> On Saturday, July 21, 2012 7:43:51 AM UTC+1, wrote:
>> John Rumm wrote:
>> > On 19/07/2012 20:32, ARWadsworth wrote:
>> >
>> > > Try this one, you like a puzzle. A new build with a 17th edition
>> > > dual RCD set up with the upstairs ring on one RCD and the
>> > > downstairs ring on the other RCD. Turn either socket MCB off and
>> > > both RCDs trip:-) It's one I had not seen before and it took me 20
>> > > minutes to find it
>> > > (it will now only take me 3 seconds to find if I ever see the same
>> > > fault).
>> >
>> > Both MCBs feeding one end of each ring rather than both ends of one
>> > each?
>>
>> Yep. And with the neutals also in the wrong busbar the RCD stays balanced
>> until one MCB is switched off.
>
> Shouldn't they also both trip if there were unequal loads on the rings?

No, you have in effect made one hybrid circuit out of the two. So any
load on either ring ends up drawing power through both MCBs and hence
both RCDs, and also returning it though both. So the total sum adds up
to zero for each RCD even if some of the balancing current is being fed
from the other (and vice versa)


Bugger; I just wrote a post querying that and the penny dropped 40 milliseconds after I hit the post button.

I had assumed both rings were supplied through RCDs, so didn't get it. They are supplied through MCBs with one main RCD on the distribution board, which works as stated. Apologies for being dense.

On 21/07/2012 11:54, Onetap wrote:
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote: > On Saturday, July 21, 2012
7:43:51 AM UTC+1, wrote: >> John
Rumm wrote: >> > On 19/07/2012 20:32, ARWadsworth
wrote: >> > >> > > Try this one,
you like a puzzle. A new build with a 17th edition >>
> > dual RCD set up with the upstairs ring on one RCD
and the >> > > downstairs ring on the other
RCD. Turn either socket MCB off and >> > > both
RCDs trip:-) It's one I had not seen before and it took me
20 >> > > minutes to find it >> >
> (it will now only take me 3 seconds to find if I ever see
the same >> > > fault). >> >
>> > Both MCBs feeding one end of each ring rather
than both ends of one >> > each? >> >>
Yep. And with the neutals also in the wrong busbar the RCD stays
balanced >> until one MCB is switched off. > >
Shouldn't they also both trip if there were unequal loads on
the rings?

No, you have in effect made one hybrid circuit out of the two. So
any load on either ring ends up drawing power through both MCBs and
hence both RCDs, and also returning it though both. So the total
sum adds up to zero for each RCD even if some of the balancing
current is being fed from the other (and vice versa)


Bugger; I just wrote a post querying that and the penny dropped 40
milliseconds after I hit the post button.

I had assumed both rings were supplied through RCDs, so didn't get
it. They are supplied through MCBs with one main RCD on the
distribution board, which works as stated. Apologies for being
dense.

In fact you were closer with your first post! ;-)

17th Edition CUs typically have (at least) two RCDs:

http://wiki.diyfaq.org.uk/index.php?...Consumer_Units

So in this case there were two MCBs but also two RCDs as well. When
either MCB was turned off, both RCDs tripped.




--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

On 21/07/2012 11:50, Onetap wrote:
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal
loads on the rings?

No, you have in effect made one hybrid circuit out of the two. So
any load on either ring ends up drawing power through both MCBs and
hence both RCDs, and also returning it though both. So the total
sum adds up to zero for each RCD even if some of the balancing
current is being fed from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge my
knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance between two
conductors using a differential current transformer. This measures
the difference between the current flowing through the live conductor
and that returning through the neutral conductor. If these do not sum
to zero, there is a leakage of current to somewhere else (to
earth/ground, or to another circuit), and the device will open its
contacts." Much as I thought.

Indeed.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2, you'd
have 5A through L and 3A through N on Ring 1 RCD, which should trip.

Ah well if you have mangled the circuits together, then just because
ring 1 is drawing 5A it does not mean all of that current is being drawn
through RCD 1. Some will be coming through the other device as well. So
while the whole lot is mashed up in parallel, then it hangs together.

Similarly 3A & 5A on L & N respectively on Ring 2 RCD which should
trip simultaneously.

What am I missing here?



--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

Onetap wrote:
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal
loads on the rings?

No, you have in effect made one hybrid circuit out of the two. So
any
load on either ring ends up drawing power through both MCBs and
hence
both RCDs, and also returning it though both. So the total sum adds
up
to zero for each RCD even if some of the balancing current is being
fed
from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge my
knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance between
two conductors using a differential current transformer. This
measures the difference between the current flowing through the live
conductor and that returning through the neutral conductor. If these
do not sum to zero, there is a leakage of current to somewhere else
(to earth/ground, or to another circuit), and the device will open
its contacts." Much as I thought.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2, you'd
have 5A through L and 3A through N on Ring 1 RCD, which should trip.
Similarly 3A & 5A on L & N respectively on Ring 2 RCD which should
trip simultaneously.

What am I missing here?

I'll give it a go.

http://i428.photobucket.com/albums/q...o2020/ring.jpg


Now put a load on that circuit.

If the load is at the halfwaypoint then both RCDs share the same current.

Move the load closer to the left RCD and more current passes through the
left RCD than the right RCD. That will not create an imbalance and trip the
RCD.

The resistance of the live and neutral cables to the left RCD are the same
as each other. It will be a lower resistance than the cables to the right
RCD however the resistance of the live and neutral cables to the right RCD
are the same as each other.


--

HTH

Cheers

Adam

On Saturday, July 21, 2012 12:41:13 PM UTC+1, wrote:

Ah, right, got it.

The essential bit of information ( that I had read but my brain didn't register) was RING circuit. Hence 2 lives and 2 neutrals for each RING, one of each connected to the two RCDs. I had been thinking of this as 2 radial circuits.

On Thu, 19 Jul 2012 02:19:12 +0100, Bill Wright
wrote:

John Rumm wrote:
Sent him up to the shop to procure a
plug while I resembled the cooker.

In what way did you resemble it? Were you in pieces yourself? Non
compos, so to speak?

Hot and bothered, iwt.

Onetap wrote:
On Saturday, July 21, 2012 12:41:13 PM UTC+1,
wrote:

Ah, right, got it.

The essential bit of information ( that I had read but my brain
didn't register) was RING circuit. Hence 2 lives and 2 neutrals for
each RING, one of each connected to the two RCDs. I had been thinking
of this as 2 radial circuits.

And of course if the neutrals had been left in the correct busbar then the
RCDs would have tripped:-)

--
Adam

In article , ARWadsworth adamwadsworth@blue
yonder.co.uk scribeth thus
Onetap wrote:
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal
loads on the rings?

No, you have in effect made one hybrid circuit out of the two. So
any
load on either ring ends up drawing power through both MCBs and
hence
both RCDs, and also returning it though both. So the total sum adds
up
to zero for each RCD even if some of the balancing current is being
fed
from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge my
knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance between
two conductors using a differential current transformer. This
measures the difference between the current flowing through the live
conductor and that returning through the neutral conductor. If these
do not sum to zero, there is a leakage of current to somewhere else
(to earth/ground, or to another circuit), and the device will open
its contacts." Much as I thought.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2, you'd
have 5A through L and 3A through N on Ring 1 RCD, which should trip.
Similarly 3A & 5A on L & N respectively on Ring 2 RCD which should
trip simultaneously.

What am I missing here?

I'll give it a go.

http://i428.photobucket.com/albums/q...o2020/ring.jpg


Now put a load on that circuit.

If the load is at the halfwaypoint then both RCDs share the same current.

Move the load closer to the left RCD and more current passes through the
left RCD than the right RCD. That will not create an imbalance and trip the
RCD.

The resistance of the live and neutral cables to the left RCD are the same
as each other. It will be a lower resistance than the cables to the right
RCD however the resistance of the live and neutral cables to the right RCD
are the same as each other.



As long as the current flowing in and out of each RCD is within its
rated tripping current thats to say balance, then its no matter how its
connected. It won't know any different.

All its looking for is same in .. and same out..


Now in Adam's example that would as it stands be true but what if you
connected a load say live to one end of the ring and neutral to the
other in that example. Thats at opposite ends of the ring.

Given sufficient current then it would be possible to get an unbalance
but that would depend on the current flowing and the loop resistance of
the cabling, and if there was an open circuit in that ring somewhere;?..
--
Tony Sayer

tony sayer wrote:
In article , ARWadsworth
adamwadsworth@blue yonder.co.uk scribeth thus
Onetap wrote:
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal
loads on the rings?

No, you have in effect made one hybrid circuit out of the two.
So any
load on either ring ends up drawing power through both MCBs and
hence
both RCDs, and also returning it though both. So the total sum
adds up
to zero for each RCD even if some of the balancing current is
being fed
from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge
my knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance
between two conductors using a differential current transformer.
This measures the difference between the current flowing through
the live conductor and that returning through the neutral
conductor. If these do not sum to zero, there is a leakage of
current to somewhere else (to earth/ground, or to another
circuit), and the device will open its contacts." Much as I
thought.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2,
you'd have 5A through L and 3A through N on Ring 1 RCD, which
should trip. Similarly 3A & 5A on L & N respectively on Ring 2
RCD which should trip simultaneously.

What am I missing here?

I'll give it a go.

http://i428.photobucket.com/albums/q...o2020/ring.jpg


Now put a load on that circuit.

If the load is at the halfwaypoint then both RCDs share the same
current.

Move the load closer to the left RCD and more current passes
through the left RCD than the right RCD. That will not create an
imbalance and trip the RCD.

The resistance of the live and neutral cables to the left RCD are
the same as each other. It will be a lower resistance than the
cables to the right RCD however the resistance of the live and
neutral cables to the right RCD are the same as each other.



As long as the current flowing in and out of each RCD is within its
rated tripping current thats to say balance, then its no matter how
its connected. It won't know any different.

All its looking for is same in .. and same out..


Now in Adam's example that would as it stands be true but what if you
connected a load say live to one end of the ring and neutral to the
other in that example. Thats at opposite ends of the ring.

:-) BS 1363 plugs usually would stop this thing happening on a ring circuit.

Assuming that you did find a way put a load with a live at one end and a
neutral at the other end then I believe both RCDs should trip. That's one I
have never considered.

Given sufficient current then it would be possible to get an unbalance
but that would depend on the current flowing and the loop resistance
of the cabling, and if there was an open circuit in that ring
somewhere;?..

It would have to be a big current. The end to end resistance reading between
r1 and rn should be wiithin 0.05 ohms of each other (in most cases)

An open circuit would cause the RCD to trip as the resistance of the cables
would not be equal.

--
Adam

In article , ARWadsworth adamwadsworth@blue
yonder.co.uk scribeth thus
tony sayer wrote:
In article , ARWadsworth
adamwadsworth@blue yonder.co.uk scribeth thus
Onetap wrote:
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal
loads on the rings?

No, you have in effect made one hybrid circuit out of the two.
So any
load on either ring ends up drawing power through both MCBs and
hence
both RCDs, and also returning it though both. So the total sum
adds up
to zero for each RCD even if some of the balancing current is
being fed
from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge
my knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance
between two conductors using a differential current transformer.
This measures the difference between the current flowing through
the live conductor and that returning through the neutral
conductor. If these do not sum to zero, there is a leakage of
current to somewhere else (to earth/ground, or to another
circuit), and the device will open its contacts." Much as I
thought.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2,
you'd have 5A through L and 3A through N on Ring 1 RCD, which
should trip. Similarly 3A & 5A on L & N respectively on Ring 2
RCD which should trip simultaneously.

What am I missing here?

I'll give it a go.

http://i428.photobucket.com/albums/q...o2020/ring.jpg


Now put a load on that circuit.

If the load is at the halfwaypoint then both RCDs share the same
current.

Move the load closer to the left RCD and more current passes
through the left RCD than the right RCD. That will not create an
imbalance and trip the RCD.

The resistance of the live and neutral cables to the left RCD are
the same as each other. It will be a lower resistance than the
cables to the right RCD however the resistance of the live and
neutral cables to the right RCD are the same as each other.



As long as the current flowing in and out of each RCD is within its
rated tripping current thats to say balance, then its no matter how
its connected. It won't know any different.

All its looking for is same in .. and same out..


Now in Adam's example that would as it stands be true but what if you
connected a load say live to one end of the ring and neutral to the
other in that example. Thats at opposite ends of the ring.

:-) BS 1363 plugs usually would stop this thing happening on a ring circuit.

Indeed it would .. normally..


Assuming that you did find a way put a load with a live at one end and a
neutral at the other end then I believe both RCDs should trip. That's one I
have never considered.

What if the whole circuit is reversed by accident?. But we are under an
assumption this is somehow abnormal ;?...

Given sufficient current then it would be possible to get an unbalance
but that would depend on the current flowing and the loop resistance
of the cabling, and if there was an open circuit in that ring
somewhere;?..

It would have to be a big current. The end to end resistance reading between
r1 and rn should be wiithin 0.05 ohms of each other (in most cases)

Yes it would. Normally.

An open circuit would cause the RCD to trip as the resistance of the cables
would not be equal.

Yes..


--
Tony Sayer

"John Rumm" wrote in message
...

Shouldn't they also both trip if there were unequal loads on the rings?

No, you have in effect made one hybrid circuit out of the two. So any load
on either ring ends up drawing power through both MCBs and hence both
RCDs, and also returning it though both. So the total sum adds up to zero
for each RCD even if some of the balancing current is being fed from the
other (and vice versa)

He was just lucky.
Try putting some load on it and it will soon unbalance and trip.
Some induction motors will do it nicely if you have one or two.

"Onetap" wrote in message
...
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal loads on
the rings?

No, you have in effect made one hybrid circuit out of the two. So any
load on either ring ends up drawing power through both MCBs and hence
both RCDs, and also returning it though both. So the total sum adds up
to zero for each RCD even if some of the balancing current is being fed
from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge my
knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance between two
conductors using a differential current transformer. This measures the
difference between the current flowing through the live conductor and that
returning through the neutral conductor. If these do not sum to zero,
there is a leakage of current to somewhere else (to earth/ground, or to
another circuit), and the device will open its contacts." Much as I
thought.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2, you'd have 5A
through L and 3A through N on Ring 1 RCD, which should trip. Similarly 3A
& 5A on L & N respectively on Ring 2 RCD which should trip simultaneously.

What am I missing here?

You aren't.
There is nothing in the circuit that would keep the currents balanced, if
they are its pure luck.

"Onetap" wrote in message
...

Bugger; I just wrote a post querying that and the penny dropped 40
milliseconds after I hit the post button.

I had assumed both rings were supplied through RCDs, so didn't get it.
They are supplied through MCBs with one main RCD on the distribution
board, which works as stated. Apologies for being dense.


Its a split unit, he didn't say it was on the same RCD.

On 21/07/2012 20:22, dennis@home wrote:


"Onetap" wrote in message
...
On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm wrote:
On 21/07/2012 09:11, Onetap wrote:

> Shouldn't they also both trip if there were unequal loads
on the rings?

No, you have in effect made one hybrid circuit out of the two. So any
load on either ring ends up drawing power through both MCBs and hence
both RCDs, and also returning it though both. So the total sum adds up
to zero for each RCD even if some of the balancing current is being fed
from the other (and vice versa)



Sorry, I don't get that. I've been to wikipedia to try to bridge my
knowledge gap and still don't get it.

Wiki says; "RCDs operate by measuring the current balance between two
conductors using a differential current transformer. This measures the
difference between the current flowing through the live conductor and
that returning through the neutral conductor. If these do not sum to
zero, there is a leakage of current to somewhere else (to
earth/ground, or to another circuit), and the device will open its
contacts." Much as I thought.

So, if you have 5A load on Ring 1 sockets and 3A on Ring 2, you'd
have 5A through L and 3A through N on Ring 1 RCD, which should trip.
Similarly 3A & 5A on L & N respectively on Ring 2 RCD which should
trip simultaneously.

What am I missing here?

You aren't.
There is nothing in the circuit that would keep the currents balanced,
if they are its pure luck.

In the situation Adam described, luck had nothing to do with it.

--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

"John Rumm" wrote in message
o.uk...

In the situation Adam described, luck had nothing to do with it.

You have a split consumer unit with the two neutral bus bars connected
together by several metres of 2.5 mm2 cable.
There is nothing there to balance the current through the RCDs.

dennis@home wrote:
"John Rumm" wrote in message
o.uk...

In the situation Adam described, luck had nothing to do with it.

You have a split consumer unit with the two neutral bus bars connected
together by several metres of 2.5 mm2 cable.
There is nothing there to balance the current through the RCDs.

Best that you just shut up you now and stop showing yourself up.

--
Adam

On 21/07/2012 23:51, dennis@home wrote:


"John Rumm" wrote in message
o.uk...

In the situation Adam described, luck had nothing to do with it.

You have a split consumer unit with the two neutral bus bars connected
together by several metres of 2.5 mm2 cable.

and hence the neutral terminals of two RCDs

and an equal length of 2.5mm^2 cable connecting the live terminals of
those same two RCDs...

in effect, two potentiometer wires with the tapping positions ganged
together.

You now place a load between them at some distance from the first RCD

RR1
R1L --\/\/\/\/\---- R2L
^
|
|
_|_
| |
| L |
| |
|___|
|
|
|
v
R1N --\/\/\/\/\----R2N
RR2

So if we treat the socket position (i.e. tap position) as a proportion
of the distance where 0.5 would be half way, and 1 the far end, we can
do a bit of network analysis.

So basically we are interested in the currents at the live terminals of
the two RCDs R1L, and R2L, and also at their neutral terminals R1N and R2N

So, using I = V/R we can say:

The current at R1L is dictated by the resistive path through a portion
of RR1 in series with L in series with the parallel load the through the
two haves of RR2

R1L = V / (RR1*TAP + L + 1/ (1/(RR2 * TAP ) + 1/(RR2 * (1-TAP))))

R2L is very similar but for the first TAP term:

R2L = V / (RR1*(1-TAP) + L + 1/ (1/(RR2 * TAP ) + 1/(RR2 * (1-TAP))))

So as the tap position moves from left to right, IR1L falls, and IR2L
rises.

The neutral currents are similar except its the RR1 resistance that is
treated as the parallel pair. If you do the sums you will see that IR1L
will always equal IR1N regardless of the tap position. Same for the
other RCD.

Adding the other circuit to the mix is just adding a copy of the above
with a different tap position and value of L, so makes no overall
difference - its load is superimposed on that of the first circuit, but
the current sharing is fixed by the same rules.

The result is a near perfect balance save for any very small differences
in total conductor length between L and N due to the way the
terminations in the sockets are made.

There is nothing there to balance the current through the RCDs.

Other than the laws of physics, no.



--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

dennis@home wrote:
"John Rumm" wrote in message
...

Shouldn't they also both trip if there were unequal loads on
the rings?

No, you have in effect made one hybrid circuit out of the two. So
any load on either ring ends up drawing power through both MCBs and
hence both RCDs, and also returning it though both. So the total
sum adds up to zero for each RCD even if some of the balancing
current is being fed from the other (and vice versa)

He was just lucky.
Try putting some load on it and it will soon unbalance and trip.
Some induction motors will do it nicely if you have one or two.

It had a load on it. Your post only goes to prove that in dennis world the
laws of physics are once again different to the rest of the world.

--
Adam

"John Rumm" wrote in message
...
On 21/07/2012 23:51, dennis@home wrote:


"John Rumm" wrote in message
o.uk...

In the situation Adam described, luck had nothing to do with it.

You have a split consumer unit with the two neutral bus bars connected
together by several metres of 2.5 mm2 cable.

and hence the neutral terminals of two RCDs

and an equal length of 2.5mm^2 cable connecting the live terminals of
those same two RCDs...

in effect, two potentiometer wires with the tapping positions ganged
together.

You now place a load between them at some distance from the first RCD

RR1
R1L --\/\/\/\/\---- R2L
^
|
|
_|_
| |
| L |
| |
|___|
|
|
|
v
R1N --\/\/\/\/\----R2N
RR2

So if we treat the socket position (i.e. tap position) as a proportion of
the distance where 0.5 would be half way, and 1 the far end, we can do a
bit of network analysis.

So basically we are interested in the currents at the live terminals of
the two RCDs R1L, and R2L, and also at their neutral terminals R1N and R2N

So, using I = V/R we can say:

The current at R1L is dictated by the resistive path through a portion of
RR1 in series with L in series with the parallel load the through the two
haves of RR2

R1L = V / (RR1*TAP + L + 1/ (1/(RR2 * TAP ) + 1/(RR2 * (1-TAP))))

R2L is very similar but for the first TAP term:

R2L = V / (RR1*(1-TAP) + L + 1/ (1/(RR2 * TAP ) + 1/(RR2 * (1-TAP))))

So as the tap position moves from left to right, IR1L falls, and IR2L
rises.

The neutral currents are similar except its the RR1 resistance that is
treated as the parallel pair. If you do the sums you will see that IR1L
will always equal IR1N regardless of the tap position. Same for the other
RCD.

Adding the other circuit to the mix is just adding a copy of the above
with a different tap position and value of L, so makes no overall
difference - its load is superimposed on that of the first circuit, but
the current sharing is fixed by the same rules.

The result is a near perfect balance save for any very small differences
in total conductor length between L and N due to the way the terminations
in the sockets are made.

There is nothing there to balance the current through the RCDs.

Other than the laws of physics, no.

Other than the fact you are ignoring all the other circuits which have live
feed from one side and return to both neutral bus bars due to the neutral
bus bars being connected.

Like I said just luck that it didn't keep tripping.