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The tale of the tripping RCD
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John Rumm
Jul 18, 2012, 8:34:31 PM7/18/12
to
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 Ł2.50 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 |
\=================================================================/
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 Ł2.50 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 |
\=================================================================/
Bill Wright
Jul 18, 2012, 9:19:12 PM7/18/12
to
John Rumm wrote:
Sent him up to the shop to procure a
> plug while I resembled the cooker.
Sent him up to the shop to procure a
In what way did you resemble it? Were you in pieces yourself? Non
compos, so to speak?
Bill
Gib Bogle
Jul 19, 2012, 12:29:53 AM7/19/12
to
John Rumm
Jul 19, 2012, 1:55:57 AM7/19/12
to
newshound
Jul 19, 2012, 4:20:41 AM7/19/12
to
On 19/07/2012 01:34, John Rumm wrote:
>
Hope you charged appropriately!
>
Hope you charged appropriately!
dennis@home
Jul 19, 2012, 4:35:07 AM7/19/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
news:6eOdnR3cqdSKy5rN...@brightview.co.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).
Is that over sensitive too?
Adam Funk
Jul 19, 2012, 6:52:22 AM7/19/12
to
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 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 ;-)
John Rumm
Jul 19, 2012, 10:34:47 AM7/19/12
to
On 19/07/2012 09:35, dennis@home wrote:
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:6eOdnR3cqdSKy5rN...@brightview.co.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
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:6eOdnR3cqdSKy5rN...@brightview.co.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?
even with no load connected.
> Is that over sensitive too?
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.
SteveW
Jul 19, 2012, 10:39:25 AM7/19/12
to
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>
> 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.
> 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!
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
John Rumm
Jul 19, 2012, 11:43:01 AM7/19/12
to
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)
dennis@home
Jul 19, 2012, 12:47:58 PM7/19/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
>> 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)
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
Jul 19, 2012, 3:44:54 PM7/19/12
to
etc. I did notice a trigger bottle of something sat on the counter.
ARWadsworth
Jul 19, 2012, 3:32:17 PM7/19/12
to
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
tony sayer
Jul 19, 2012, 4:08:48 PM7/19/12
to
>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...
>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).
>
--
Tony Sayer
dennis@home
Jul 19, 2012, 6:46:29 PM7/19/12
to
"tony sayer" <to...@bancom.co.uk> wrote in message
news:i7YLAhPQ...@bancom.co.uk...
John Rumm
Jul 19, 2012, 7:54:44 PM7/19/12
to
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?
> 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).
ARWadsworth
Jul 21, 2012, 2:43:51 AM7/21/12
to
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
> 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?
until one MCB is switched off.
--
Adam
Onetap
Jul 21, 2012, 4:11:47 AM7/21/12
to
On Saturday, July 21, 2012 7:43:51 AM UTC+1, adamwa...@blueyonder.co.uk 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?
> 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?
> Yep. And with the neutals also in the wrong busbar the RCD stays balanced
> until one MCB is switched off.
ARWadsworth
Jul 21, 2012, 4:25:37 AM7/21/12
to
is balanced.
--
Adam
John Rumm
Jul 21, 2012, 5:28:18 AM7/21/12
to
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)
Message has been deleted
Onetap
Jul 21, 2012, 6:54:41 AM7/21/12
to
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, adamwa...@blueyonder.co.uk wrote:
> On 21/07/2012 09:11, Onetap wrote:
> >> John Rumm wrote:
> >> &gt; On 19/07/2012 20:32, ARWadsworth wrote:
> >> &gt;
> >> &gt; &gt; Try this one, you like a puzzle. A new build with a 17th edition
> >> &gt; &gt; dual RCD set up with the upstairs ring on one RCD and the
> >> &gt; &gt; downstairs ring on the other RCD. Turn either socket MCB off and
> >> &gt; &gt; both RCDs trip:-) It&#39;s one I had not seen before and it took me 20
> >> &gt; &gt; minutes to find it
> >> &gt; &gt; (it will now only take me 3 seconds to find if I ever see the same
> >> &gt; &gt; fault).
> >> &gt;
> >> &gt; Both MCBs feeding one end of each ring rather than both ends of one
> >> &gt; 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.
> 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)
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.
John Rumm
Jul 21, 2012, 7:27:32 AM7/21/12
to
17th Edition CUs typically have (at least) two RCDs:
http://wiki.diyfaq.org.uk/index.php?title=17th_Edition_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.
John Rumm
Jul 21, 2012, 7:35:41 AM7/21/12
to
On 21/07/2012 11:50, Onetap wrote:
> On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm 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
>>
>> 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)
>>
>
>
> 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?
ARWadsworth
Jul 21, 2012, 7:41:13 AM7/21/12
to
Onetap wrote:
> On Saturday, July 21, 2012 10:28:18 AM UTC+1, John Rumm 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)
> >
>
>
> >
> > 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.
>
> 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.
> 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.
> trip simultaneously.
>
> What am I missing here?
http://i428.photobucket.com/albums/qq6/adamko2020/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
Onetap
Jul 21, 2012, 8:00:33 AM7/21/12
to
On Saturday, July 21, 2012 12:41:13 PM UTC+1, adamwa...@blueyonder.co.uk 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.
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.
Grimly Curmudgeon
Jul 21, 2012, 8:15:05 AM7/21/12
to
ARWadsworth
Jul 21, 2012, 8:24:33 AM7/21/12
to
RCDs would have tripped:-)
--
Adam
tony sayer
Jul 21, 2012, 8:53:14 AM7/21/12
to
In article <jue4gr$p7b$1...@dont-email.me>, ARWadsworth <adamwadsworth@blue
yonder.co.uk> scribeth thus
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
yonder.co.uk> scribeth thus
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
ARWadsworth
Jul 21, 2012, 9:35:49 AM7/21/12
to
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;?..
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
tony sayer
Jul 21, 2012, 11:23:47 AM7/21/12
to
In article <jueb7p$ssm$1...@dont-email.me>, ARWadsworth <adamwadsworth@blue
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 ;?...
>
--
Tony Sayer
>
>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.
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.
>> 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.
Yes..
>An open circuit would cause the RCD to trip as the resistance of the cables
>would not be equal.
>
--
Tony Sayer
dennis@home
Jul 21, 2012, 3:20:15 PM7/21/12
to
"John Rumm" <see.my.s...@nowhere.null> 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)
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.
dennis@home
Jul 21, 2012, 3:22:16 PM7/21/12
to
"Onetap" <one...@talk21.com> wrote in message
news:99244111-573a-4353...@googlegroups.com...
> 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)
>>
>
>
>>
>> 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.
> 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?
There is nothing in the circuit that would keep the currents balanced, if
they are its pure luck.
dennis@home
Jul 21, 2012, 3:26:44 PM7/21/12
to
news:48f9b9d6-6251-48ec...@googlegroups.com...
> 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.
> 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.
>
John Rumm
Jul 21, 2012, 6:37:49 PM7/21/12
to
dennis@home
Jul 21, 2012, 6:51:14 PM7/21/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
> In the situation Adam described, luck had nothing to do with it.
together by several metres of 2.5 mm2 cable.
There is nothing there to balance the current through the RCDs.
ARWadsworth
Jul 21, 2012, 7:05:21 PM7/21/12
to
--
Adam
John Rumm
Jul 21, 2012, 11:37:29 PM7/21/12
to
On 21/07/2012 23:51, dennis@home wrote:
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:hK2dnQEqwoegspbN...@brightview.co.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
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:hK2dnQEqwoegspbN...@brightview.co.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 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.
ARWadsworth
Jul 22, 2012, 4:43:29 AM7/22/12
to
laws of physics are once again different to the rest of the world.
--
Adam
dennis@home
Jul 22, 2012, 1:13:39 PM7/22/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
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.
ARWadsworth
Jul 22, 2012, 1:58:18 PM7/22/12
to
--
Adam
John Rumm
Jul 22, 2012, 3:36:01 PM7/22/12
to
yes, that's right, connected together!
So the feeds to all the other circuits are also being drawn from the
pair of RCDs.
You see that bit above where I point out that other circuits are just
duplicates of the same arrangement and make no difference to the balance
condition?
> Like I said just luck that it didn't keep tripping.
dennis@home
Jul 23, 2012, 3:10:06 AM7/23/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
With a ten amp load 30 mA is 0.035% so the cables and joints would have to
be within 0.35% or about 3 mm per meter of run.
At 30A its about 1 mm per meter of run.
Add in the starting current for an induction motor and/or a cooker and/or a
shower and its even less.
Now while I agree it is possible to match the lengths reasonably well many
consumer units look very untidy if you cut the cables and then put them into
terminations with all that slack lying about.
So as I said its luck that it didn't keep tripping.
As an aside.
I tend to recall some discussion about using twin RCDs as protection against
incorrect wiring of rings (but I don't recall where I heard it).
Something like one MCB - feeding two RCDs one for each leg.
They trip whenever the ring is broken.
However the tolerance issue makes it a none starter for most people.
>
>> Like I said just luck that it didn't keep tripping.
>
> Yes you said it...
Just to be even more clear..
the guy that installed it was unlucky,
the gut that used it was lucky in that it didn't trip all the time.
That is as far as it goes, I don't want another argument about a petty
issue.
Mike Tomlinson
Jul 23, 2012, 3:09:00 AM7/23/12
to
En el art�culo <jueb7p$ssm$1...@dont-email.me>, ARWadsworth <adamwadsworth@
blueyonder.co.uk> escribi�:
>:-) BS 1363 plugs usually would stop this thing happening on a ring circuit.
Containing the fuse that Woddles says is not needed...
--
(\_/)
(='.'=)
(")_(")
blueyonder.co.uk> escribi�:
>:-) BS 1363 plugs usually would stop this thing happening on a ring circuit.
--
(\_/)
(='.'=)
(")_(")
John Rumm
Jul 23, 2012, 12:42:35 PM7/23/12
to
> With a ten amp load 30 mA is 0.035% so the cables and joints would have
> to be within 0.35% or about 3 mm per meter of run.
cable resistance alone as the only resistive elements in the whole circuit.
Have a peak back at the circuit diagram, and ponder the significance of
the load.
Lets say the load is passing 10A, that would make the total circuit
resistance around 23 ohms...
So to add 30mA of imbalance, we would need a change in total circuit
resistance through the load of 23 - 230 / 10.03 = 0.07 ohms.
At 7.41 mOhms/m that would need 9.4 extra metres of 2.5mm^2 wire in one
leg. Even allowing for needing an extra few inches of wire in the CU to
keep things tidy, that sounds pretty implausible.
> So as I said its luck that it didn't keep tripping.
> As an aside.
> I tend to recall some discussion about using twin RCDs as protection
> against incorrect wiring of rings (but I don't recall where I heard it).
> Something like one MCB - feeding two RCDs one for each leg.
> They trip whenever the ring is broken.
> However the tolerance issue makes it a none starter for most people.
>
>>
>>> Like I said just luck that it didn't keep tripping.
>>
>> Yes you said it...
>
> I have said it again.
>
> Just to be even more clear..
> the guy that installed it was unlucky,
> the gut that used it was lucky in that it didn't trip all the time.
> That is as far as it goes, I don't want another argument about a petty
> issue.
dennis@home
Jul 23, 2012, 3:43:29 PM7/23/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
>> Have you thought about the tolerances in that?
>
> Indeed I have, although I get the feeling you probably have not...
>
>> With a ten amp load 30 mA is 0.035% so the cables and joints would have
>> to be within 0.35% or about 3 mm per meter of run.
>
> Yes, see what I mean? I get the impression that you are treating the cable
> resistance alone as the only resistive elements in the whole circuit.
I am treating the return path as two resistors in parallel.
>> With a ten amp load 30 mA is 0.035% so the cables and joints would have
>> to be within 0.35% or about 3 mm per meter of run.
>
> Yes, see what I mean? I get the impression that you are treating the cable
> resistance alone as the only resistive elements in the whole circuit.
there is no need to do anything more.
>
> Have a peak back at the circuit diagram, and ponder the significance of
> the load.
the current through the parallel resistances.
>
> Lets say the load is passing 10A, that would make the total circuit
> resistance around 23 ohms...
>
> So to add 30mA of imbalance, we would need a change in total circuit
> resistance through the load of 23 - 230 / 10.03 = 0.07 ohms.
you just need to split the current through the two paths.
this is just the ratio of their resistance, their resistance being
proportional to their length.
>
> At 7.41 mOhms/m that would need 9.4 extra metres of 2.5mm^2 wire in one
> leg. Even allowing for needing an extra few inches of wire in the CU to
> keep things tidy, that sounds pretty implausible.
thing.
Try working out the difference in current flowing down two parallel bits of
wire when one is 0.03% longer.
ARWadsworth
Jul 23, 2012, 4:38:54 PM7/23/12
to
That's true.
--
Adam
John Rumm
Jul 23, 2012, 5:15:09 PM7/23/12
to
On 23/07/2012 20:43, dennis@home wrote:
> I am treating the return path as two resistors in parallel.
> there is no need to do anything more.
Yup that's probably what is misdirecting you. Its the position of the
> I am treating the return path as two resistors in parallel.
> there is no need to do anything more.
socket on the circuit which dictates that ratio and its not particularly
relevant so long as the proportional distance on the line conductor is
the same. Its the feed to return ratio that matters, not the return /
return one.
> Try working out the difference in current flowing down two parallel bits
> of wire when one is 0.03% longer.
dennis@home
Jul 23, 2012, 5:50:01 PM7/23/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
> On 23/07/2012 20:43, dennis@home wrote:
>
>> I am treating the return path as two resistors in parallel.
>> there is no need to do anything more.
>
> Yup that's probably what is misdirecting you. Its the position of the
> socket on the circuit which dictates that ratio and its not particularly
> relevant so long as the proportional distance on the line conductor is the
> same. Its the feed to return ratio that matters, not the return / return
> one.
The socket position doesn't matter.
>
>> I am treating the return path as two resistors in parallel.
>> there is no need to do anything more.
>
> Yup that's probably what is misdirecting you. Its the position of the
> socket on the circuit which dictates that ratio and its not particularly
> relevant so long as the proportional distance on the line conductor is the
> same. Its the feed to return ratio that matters, not the return / return
> one.
You are confusing the return paths that need to be balanced.
Its the line and neutral that have to be balanced not the two lines or the
two neutrals.
If the neutral and the line aren't exactly the same length some of the
current may flow down the other pair if they don't have exactly the same
difference.
basically if the line and neutral don't match each other on each side there
will be an imbalance.
>
>> Try working out the difference in current flowing down two parallel bits
>> of wire when one is 0.03% longer.
>
> Given one might already be 5 times the length of the other, bugger all.
Try it again with the correct wires.
>> Try working out the difference in current flowing down two parallel bits
>> of wire when one is 0.03% longer.
>
> Given one might already be 5 times the length of the other, bugger all.
John Rumm
Jul 23, 2012, 6:20:22 PM7/23/12
to
On 23/07/2012 22:50, dennis@home wrote:
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:-_WdnZk668BDI5DN...@brightview.co.uk...
>> On 23/07/2012 20:43, dennis@home wrote:
>>
>>> I am treating the return path as two resistors in parallel.
>>> there is no need to do anything more.
>>
>> Yup that's probably what is misdirecting you. Its the position of the
>> socket on the circuit which dictates that ratio and its not
>> particularly relevant so long as the proportional distance on the line
>> conductor is the same. Its the feed to return ratio that matters, not
>> the return / return one.
>
> The socket position doesn't matter.
Make your mind up... a moment ago your "treating the return path as two
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:-_WdnZk668BDI5DN...@brightview.co.uk...
>> On 23/07/2012 20:43, dennis@home wrote:
>>
>>> I am treating the return path as two resistors in parallel.
>>> there is no need to do anything more.
>>
>> Yup that's probably what is misdirecting you. Its the position of the
>> socket on the circuit which dictates that ratio and its not
>> particularly relevant so long as the proportional distance on the line
>> conductor is the same. Its the feed to return ratio that matters, not
>> the return / return one.
>
> The socket position doesn't matter.
resistors in parallel.". Its the socket position that dictates the ratio
of those two resistors (and the ratio of the feed paths)
> You are confusing the return paths that need to be balanced.
> Its the line and neutral that have to be balanced not the two lines or
> the two neutrals.
Its like watching a multimaster!
getting confused yet?
> If the neutral and the line aren't exactly the same length some of the
> current may flow down the other pair if they don't have exactly the same
> difference.
> basically if the line and neutral don't match each other on each side
> there will be an imbalance.
>>
>>> Try working out the difference in current flowing down two parallel bits
>>> of wire when one is 0.03% longer.
>>
>> Given one might already be 5 times the length of the other, bugger all.
>
> Try it again with the correct wires.
dennis@home
Jul 23, 2012, 6:39:38 PM7/23/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
You have stated they vary a bit due to how they are cut when you install
stuff, I have pointed out that they are almost always cut to different
lengths in the consumer unit.
I have also pointed out that minor variations have a big effect on the
current balance and told you how to work it out, yet you come up with some
weird 9.5 m which is just silly.
You only need a variation of about 0.3% in the length to imbalance it by 30
mA with a 10A load (kettle to you).
That's ~3 mm per meter of run, Do you keep all your connections so the line
and neutral never vary by that much?
>
>> basically if the line and neutral don't match each other on each side
>> there will be an imbalance.
>>>
>>>> Try working out the difference in current flowing down two parallel
>>>> bits
>>>> of wire when one is 0.03% longer.
>>>
>>> Given one might already be 5 times the length of the other, bugger all.
>>
>> Try it again with the correct wires.
>
> never mind den. Adam fixed it, its all gone away.
However the fact it existed at all is interesting, you have made frequent
claims that faults don't exist in ring mains quite a few times.
tony sayer
Jul 24, 2012, 1:45:14 PM7/24/12
to
>> And where would that difference come from?
>
>Are you deliberately acting dumb?
>
>You have stated they vary a bit due to how they are cut when you install
>stuff, I have pointed out that they are almost always cut to different
>lengths in the consumer unit.
>
>I have also pointed out that minor variations have a big effect on the
>current balance and told you how to work it out, yet you come up with some
>weird 9.5 m which is just silly.
>
>You only need a variation of about 0.3% in the length to imbalance it by 30
>mA with a 10A load (kettle to you).
>That's ~3 mm per meter of run, Do you keep all your connections so the line
>and neutral never vary by that much?
Can we see a worked example to support this statement?.....
>
>Are you deliberately acting dumb?
>
>You have stated they vary a bit due to how they are cut when you install
>stuff, I have pointed out that they are almost always cut to different
>lengths in the consumer unit.
>
>I have also pointed out that minor variations have a big effect on the
>current balance and told you how to work it out, yet you come up with some
>weird 9.5 m which is just silly.
>
>You only need a variation of about 0.3% in the length to imbalance it by 30
>mA with a 10A load (kettle to you).
>That's ~3 mm per meter of run, Do you keep all your connections so the line
>and neutral never vary by that much?
--
Tony Sayer
dennis@home
Jul 24, 2012, 3:32:13 PM7/24/12
to
"tony sayer" <to...@bancom.co.uk> wrote in message
news:WyXWyWFq...@bancom.co.uk...
>>You only need a variation of about 0.3% in the length to imbalance it by
>>30
>>mA with a 10A load (kettle to you).
>>That's ~3 mm per meter of run, Do you keep all your connections so the
>>line
>>and neutral never vary by that much?
>
> Can we see a worked example to support this statement?.....
10A current * 0.3% = 30mA (tripping current)
Now take two 10 m long conductors in parallel say R1 and R2.
Same length = same resistance = same current (10/2 = 5 Amps flows down each
leg)
Now make one conductor 30 mm longer and work out the current down each leg.
I will leave that as an exercise to the reader as they won't believe me if I
do it.
To help out the effective circuit is
line----R1----
| |
--R2------------------------------Load-------------------R'1-------
| |
-----R'2-------------neutral
You can ignore the load it only sets the current flowing through the
circuit.
You can also ignore the actual voltage drop across the resistors as its only
the ratio of the current that matters.
R1 is the resistance (length of the circuit to the load) + the RCD
resistance.
R2 is the resistance of the other leg + the other RCD resistance.
We will make it easy and assume the RCDs are identical so they can be
ignored.
R'1 and R'2 are the neutral returns and the other side of the RCDs, we will
ignore the RCD as before.
If you still need help try
http://www.electronics-tutorials.ws/resistor/res_4.html
tony sayer
Jul 24, 2012, 4:36:26 PM7/24/12
to
In article <jumtes$o4r$4...@news.albasani.net>, dennis@home <den...@killspam.kicks-
ass.net> scribeth thus
>
>I will leave that as an exercise to the reader as they won't believe me if I
>do it.
>
Well you could be using bell wire or 16 mm as stated we need to know the cable
resistance you are specifying before we can go any further..
>
>To help out the effective circuit is
>
>line----R1----
> | |
> --R2------------------------------Load-------------------R'1-------
>
> | |
>
> -----R'2-------------neutral
That isn't rendering itself too well here. Perhaps this might..
L in L out -----1-----------------------------------------out L L out
RCD 1 RCD 2
N in N out ------------------------------------------2----out N N out
Do you now imply that the load is attached say to points 1 and 2 in each line?..
Now consider the Live line from the Left RCD to the point 1 thats "x" ohms from
point 1 to the Right RCD thats say "y" ohms
Neutral line from LH RCD to point 2 thats z ohms and from point 2 to the RH RCD
thats "w" ohms is this what you imply?..
>
--
Tony Sayer
ass.net> scribeth thus
>
>
>"tony sayer" <to...@bancom.co.uk> wrote in message
>news:WyXWyWFq...@bancom.co.uk...
>
>>>You only need a variation of about 0.3% in the length to imbalance it by
>>>30
>>>mA with a 10A load (kettle to you).
>>>That's ~3 mm per meter of run, Do you keep all your connections so the
>>>line
>>>and neutral never vary by that much?
>>
>> Can we see a worked example to support this statement?.....
>
>Sure..
>
>10A current * 0.3% = 30mA (tripping current)
>
>Now take two 10 m long conductors in parallel say R1 and R2.
>Same length = same resistance = same current (10/2 = 5 Amps flows down each
>leg)
>
>Now make one conductor 30 mm longer and work out the current down each leg.
>
Right now what is the resistance per metre of those conductors?..
>
>"tony sayer" <to...@bancom.co.uk> wrote in message
>news:WyXWyWFq...@bancom.co.uk...
>
>>>You only need a variation of about 0.3% in the length to imbalance it by
>>>30
>>>mA with a 10A load (kettle to you).
>>>That's ~3 mm per meter of run, Do you keep all your connections so the
>>>line
>>>and neutral never vary by that much?
>>
>> Can we see a worked example to support this statement?.....
>
>Sure..
>
>10A current * 0.3% = 30mA (tripping current)
>
>Now take two 10 m long conductors in parallel say R1 and R2.
>Same length = same resistance = same current (10/2 = 5 Amps flows down each
>leg)
>
>Now make one conductor 30 mm longer and work out the current down each leg.
>
>
>I will leave that as an exercise to the reader as they won't believe me if I
>do it.
>
resistance you are specifying before we can go any further..
>
>To help out the effective circuit is
>
>line----R1----
> | |
> --R2------------------------------Load-------------------R'1-------
>
> | |
>
> -----R'2-------------neutral
L in L out -----1-----------------------------------------out L L out
RCD 1 RCD 2
N in N out ------------------------------------------2----out N N out
Do you now imply that the load is attached say to points 1 and 2 in each line?..
Now consider the Live line from the Left RCD to the point 1 thats "x" ohms from
point 1 to the Right RCD thats say "y" ohms
Neutral line from LH RCD to point 2 thats z ohms and from point 2 to the RH RCD
thats "w" ohms is this what you imply?..
>
--
Tony Sayer
dennis@home
Jul 24, 2012, 5:04:26 PM7/24/12
to
news:FvvC1cGK...@bancom.co.uk...
results.
>
>>
>>I will leave that as an exercise to the reader as they won't believe me if
>>I
>>do it.
>>
>
> Well you could be using bell wire or 16 mm as stated we need to know the
> cable
> resistance you are specifying before we can go any further..
>
>>
>>To help out the effective circuit is
>>
>>line----R1----
>> | |
>> --R2------------------------------Load-------------------R'1-------
>>
>> | |
>>
>> -----R'2-------------neutral
>
> That isn't rendering itself too well here. Perhaps this might..
>
>
> L in L out -----1-----------------------------------------out L
> L out
> RCD 1 RCD
> 2
> N in N out ------------------------------------------2----out N
> N out
>
>
> Do you now imply that the load is attached say to points 1 and 2 in each
> line?..
>
>
> Now consider the Live line from the Left RCD to the point 1 thats "x" ohms
> from
> point 1 to the Right RCD thats say "y" ohms
I have changed yours to match what i said, the load is across 1 and 2.
L in and N in of the two rcds are connected together in the consumer unit.
>
L in L out ----R1--------------1--------R2-----out L L in
RCD 1 RCD 2
N in N out ----R'1------------2-----R'2-------out N N in
> In article <jumtes$o4r$4...@news.albasani.net>, dennis@home
> <den...@killspam.kicks-
> ass.net> scribeth thus
>>
>>
>>"tony sayer" <to...@bancom.co.uk> wrote in message
>>news:WyXWyWFq...@bancom.co.uk...
>>
>>>>You only need a variation of about 0.3% in the length to imbalance it by
>>>>30
>>>>mA with a 10A load (kettle to you).
>>>>That's ~3 mm per meter of run, Do you keep all your connections so the
>>>>line
>>>>and neutral never vary by that much?
>>>
>>> Can we see a worked example to support this statement?.....
>>
>>Sure..
>>
>>10A current * 0.3% = 30mA (tripping current)
>>
>>Now take two 10 m long conductors in parallel say R1 and R2.
>>Same length = same resistance = same current (10/2 = 5 Amps flows down
>>each
>>leg)
>>
>>Now make one conductor 30 mm longer and work out the current down each
>>leg.
>>
>
> Right now what is the resistance per metre of those conductors?..
It doesn't matter you can choose any value you like, it doesn't affect the
> <den...@killspam.kicks-
> ass.net> scribeth thus
>>
>>
>>"tony sayer" <to...@bancom.co.uk> wrote in message
>>news:WyXWyWFq...@bancom.co.uk...
>>
>>>>You only need a variation of about 0.3% in the length to imbalance it by
>>>>30
>>>>mA with a 10A load (kettle to you).
>>>>That's ~3 mm per meter of run, Do you keep all your connections so the
>>>>line
>>>>and neutral never vary by that much?
>>>
>>> Can we see a worked example to support this statement?.....
>>
>>Sure..
>>
>>10A current * 0.3% = 30mA (tripping current)
>>
>>Now take two 10 m long conductors in parallel say R1 and R2.
>>Same length = same resistance = same current (10/2 = 5 Amps flows down
>>each
>>leg)
>>
>>Now make one conductor 30 mm longer and work out the current down each
>>leg.
>>
>
> Right now what is the resistance per metre of those conductors?..
results.
>
>>
>>I will leave that as an exercise to the reader as they won't believe me if
>>I
>>do it.
>>
>
> Well you could be using bell wire or 16 mm as stated we need to know the
> cable
> resistance you are specifying before we can go any further..
>
>>
>>To help out the effective circuit is
>>
>>line----R1----
>> | |
>> --R2------------------------------Load-------------------R'1-------
>>
>> | |
>>
>> -----R'2-------------neutral
>
> That isn't rendering itself too well here. Perhaps this might..
>
>
> L in L out -----1-----------------------------------------out L
> L out
> RCD 1 RCD
> 2
> N in N out ------------------------------------------2----out N
> N out
>
>
> Do you now imply that the load is attached say to points 1 and 2 in each
> line?..
>
>
> Now consider the Live line from the Left RCD to the point 1 thats "x" ohms
> from
> point 1 to the Right RCD thats say "y" ohms
L in and N in of the two rcds are connected together in the consumer unit.
>
L in L out ----R1--------------1--------R2-----out L L in
RCD 1 RCD 2
N in N out ----R'1------------2-----R'2-------out N N in
tony sayer
Jul 24, 2012, 7:07:43 PM7/24/12
to
In article <jun2kp$3hn$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
Well if the resistance's are equal lets say R1 is the same as R'1 And R2
is the same as R'2 then the currents will be balanced flowing thru Both
RCD's..
Course if that point along the line is say moved then ,more or less
current will flow thru each RCD and long as the current flows are equal
through Each RCD i.e. same in and same out then they are in balance and
no tripping will occur whatever the absolute current actually is..
But in each instance R1 must be equal to R'1 and R2 must be equal to R'2
Agreed?..
Now were you suggesting that R1 could be slightly different from R'1 or
R2 differs from R'2 and that was caused by differing lengths of the
cable in use?..
And by 30 mm?..
--
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
is the same as R'2 then the currents will be balanced flowing thru Both
RCD's..
Course if that point along the line is say moved then ,more or less
current will flow thru each RCD and long as the current flows are equal
through Each RCD i.e. same in and same out then they are in balance and
no tripping will occur whatever the absolute current actually is..
But in each instance R1 must be equal to R'1 and R2 must be equal to R'2
Agreed?..
Now were you suggesting that R1 could be slightly different from R'1 or
R2 differs from R'2 and that was caused by differing lengths of the
cable in use?..
And by 30 mm?..
--
Tony Sayer
dennis@home
Jul 25, 2012, 3:03:37 AM7/25/12
to
news:W+SWhPI$qyDQ...@bancom.co.uk...
> Agreed?..
Sounds about right.
>
> Now were you suggesting that R1 could be slightly different from R'1 or
> R2 differs from R'2 and that was caused by differing lengths of the
> cable in use?..
Well if you have a fixed voltage across them, which you must have as they
are tied together at each end.
Then the current flows through them in the ration of their resistance.
eg. 1 ohm and 2 ohm gives one third of the current through the 2 ohm and two
thirds through the 1 ohm.
that applies if you have 10 ohm and 20 ohm or 1 milliohm and 2 milliohm.
>
> And by 30 mm?..
That was chosen because its 0.3% of 10 m. It is also happens to be 30 mA of
10A, 30 mA being about the amount needed to trip an RCD.
So to take an example if the loop is identical R1=R2=R'1=R'2 then the
current will split equally as R1=R'1 and R2=R'2.
Now if you do it where R1 == R1+0.3% you get an imbalance and the ratios
split
so the current through R1 becomes I*0.5 * 1 * 1/1.003 and through R'1
becomes I*0.5 * 1.003
If its a bigger load the current flow difference is more.
The only way around it is using superconductors or putting balancing
circuits in each leg.
BTW I ignored the construction of the RCD as that only makes it worse as
they will also split the current differently if they add any significant
resistance to the paths. As it stands there is no need to balance the path
resistance in an RCD only the need to register zero when the current flows.
The paths could be different if the current transformer isn't wound
symmetrically (say one winding on top of the other. They are designed to be
not used in parallel after all.
> Agreed?..
Sounds about right.
>
> Now were you suggesting that R1 could be slightly different from R'1 or
> R2 differs from R'2 and that was caused by differing lengths of the
> cable in use?..
are tied together at each end.
Then the current flows through them in the ration of their resistance.
eg. 1 ohm and 2 ohm gives one third of the current through the 2 ohm and two
thirds through the 1 ohm.
that applies if you have 10 ohm and 20 ohm or 1 milliohm and 2 milliohm.
>
> And by 30 mm?..
That was chosen because its 0.3% of 10 m. It is also happens to be 30 mA of
10A, 30 mA being about the amount needed to trip an RCD.
So to take an example if the loop is identical R1=R2=R'1=R'2 then the
current will split equally as R1=R'1 and R2=R'2.
Now if you do it where R1 == R1+0.3% you get an imbalance and the ratios
split
so the current through R1 becomes I*0.5 * 1 * 1/1.003 and through R'1
becomes I*0.5 * 1.003
If its a bigger load the current flow difference is more.
The only way around it is using superconductors or putting balancing
circuits in each leg.
BTW I ignored the construction of the RCD as that only makes it worse as
they will also split the current differently if they add any significant
resistance to the paths. As it stands there is no need to balance the path
resistance in an RCD only the need to register zero when the current flows.
The paths could be different if the current transformer isn't wound
symmetrically (say one winding on top of the other. They are designed to be
not used in parallel after all.
tony sayer
Jul 25, 2012, 8:58:07 AM7/25/12
to
In article <juo5ob$1kh$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
>
>
the current off in this circuit?.
Moving the take off points, say there're at the centre and now your
going to move them apart by say 30 mm around 1 inch? Thats to say A
might be moved to the Left hand RCD whilst B stays the same. Agreed ?.
There is a difference in length of the points your taking the current
off and thats now moving apart.
That being so you'll see that the resistance and size of the cable does
now come into the equation.
You could use say copper bell wire or you could be using say copper buss
bar’s certainly a difference in resistivity with length there!.
You did say that the resistance of the cable doesn't matter.
Do you still believe that to be true?. .
>
>So to take an example if the loop is identical R1=R2=R'1=R'2 then the
>current will split equally as R1=R'1 and R2=R'2.
>
>Now if you do it where R1 == R1+0.3% you get an imbalance and the ratios
>split
>so the current through R1 becomes I*0.5 * 1 * 1/1.003 and through R'1
>becomes I*0.5 * 1.003
>
>If its a bigger load the current flow difference is more.
>The only way around it is using superconductors or putting balancing
>circuits in each leg.
>
>BTW I ignored the construction of the RCD as that only makes it worse as
>they will also split the current differently if they add any significant
>resistance to the paths. As it stands there is no need to balance the path
>resistance in an RCD only the need to register zero when the current flows.
>The paths could be different if the current transformer isn't wound
>symmetrically (say one winding on top of the other. They are designed to be
>not used in parallel after all.
>
Matters not as long as the see equal currents thru them they don't
care;!..
--
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
>
>
>"tony sayer" <to...@bancom.co.uk> wrote in message
>news:W+SWhPI$qyDQ...@bancom.co.uk...
>
>> Agreed?..
>
>Sounds about right.
>
>>
>> Now were you suggesting that R1 could be slightly different from R'1 or
>> R2 differs from R'2 and that was caused by differing lengths of the
>> cable in use?..
>
>Well if you have a fixed voltage across them, which you must have as they
>are tied together at each end.
>Then the current flows through them in the ration of their resistance.
>eg. 1 ohm and 2 ohm gives one third of the current through the 2 ohm and two
>thirds through the 1 ohm.
>that applies if you have 10 ohm and 20 ohm or 1 milliohm and 2 milliohm.
>
>>
>> And by 30 mm?..
>
>That was chosen because its 0.3% of 10 m. It is also happens to be 30 mA of
>10A, 30 mA being about the amount needed to trip an RCD.
Now OK thus far now your talking of moving the points where your taking
>news:W+SWhPI$qyDQ...@bancom.co.uk...
>
>> Agreed?..
>
>Sounds about right.
>
>>
>> Now were you suggesting that R1 could be slightly different from R'1 or
>> R2 differs from R'2 and that was caused by differing lengths of the
>> cable in use?..
>
>Well if you have a fixed voltage across them, which you must have as they
>are tied together at each end.
>Then the current flows through them in the ration of their resistance.
>eg. 1 ohm and 2 ohm gives one third of the current through the 2 ohm and two
>thirds through the 1 ohm.
>that applies if you have 10 ohm and 20 ohm or 1 milliohm and 2 milliohm.
>
>>
>> And by 30 mm?..
>
>That was chosen because its 0.3% of 10 m. It is also happens to be 30 mA of
>10A, 30 mA being about the amount needed to trip an RCD.
the current off in this circuit?.
Moving the take off points, say there're at the centre and now your
going to move them apart by say 30 mm around 1 inch? Thats to say A
might be moved to the Left hand RCD whilst B stays the same. Agreed ?.
There is a difference in length of the points your taking the current
off and thats now moving apart.
That being so you'll see that the resistance and size of the cable does
now come into the equation.
You could use say copper bell wire or you could be using say copper buss
bar’s certainly a difference in resistivity with length there!.
You did say that the resistance of the cable doesn't matter.
Do you still believe that to be true?. .
>
>So to take an example if the loop is identical R1=R2=R'1=R'2 then the
>current will split equally as R1=R'1 and R2=R'2.
>
>Now if you do it where R1 == R1+0.3% you get an imbalance and the ratios
>split
>so the current through R1 becomes I*0.5 * 1 * 1/1.003 and through R'1
>becomes I*0.5 * 1.003
>
>If its a bigger load the current flow difference is more.
>The only way around it is using superconductors or putting balancing
>circuits in each leg.
>
>BTW I ignored the construction of the RCD as that only makes it worse as
>they will also split the current differently if they add any significant
>resistance to the paths. As it stands there is no need to balance the path
>resistance in an RCD only the need to register zero when the current flows.
>The paths could be different if the current transformer isn't wound
>symmetrically (say one winding on top of the other. They are designed to be
>not used in parallel after all.
>
care;!..
--
Tony Sayer
dennis@home
Jul 25, 2012, 2:04:30 PM7/25/12
to
news:dyqUidJf...@bancom.co.uk...
>>
>>That was chosen because its 0.3% of 10 m. It is also happens to be 30 mA
>>of
>>10A, 30 mA being about the amount needed to trip an RCD.
>
> Now OK thus far now your talking of moving the points where your taking
> the current off in this circuit?.
>
> Moving the take off points, say there're at the centre and now your
> going to move them apart by say 30 mm around 1 inch? Thats to say A
> might be moved to the Left hand RCD whilst B stays the same. Agreed ?.
> There is a difference in length of the points your taking the current
> off and thats now moving apart.
>
> That being so you'll see that the resistance and size of the cable does
> now come into the equation.
No it doesn't.
The resistance is proportional to length, the actual resistance doesn't
matter.
For example
If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
amps through them then 1 amp will go through the 2 ohm and 2 amps will go
through the 1 ohm resistor.
the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
10A flowing through.
>
> You could use say copper bell wire or you could be using say copper buss
> bar’s certainly a difference in resistivity with length there!.
You sure could, it wouldn't make any difference to the balance (excluding
the effects of the wires glowing).
>
> You did say that the resistance of the cable doesn't matter.
>
> Do you still believe that to be true?. .
I know it doesn't matter.
We are talking about current balances through parallel resistances and its
all about ratios.
>>So to take an example if the loop is identical R1=R2=R'1=R'2 then the
>>current will split equally as R1=R'1 and R2=R'2.
>>
>>Now if you do it where R1 == R1+0.3% you get an imbalance and the ratios
>>split
>>so the current through R1 becomes I*0.5 * 1 * 1/1.003 and through R'1
>>becomes I*0.5 * 1.003
>>
>>If its a bigger load the current flow difference is more.
>>The only way around it is using superconductors or putting balancing
>>circuits in each leg.
>>
>>BTW I ignored the construction of the RCD as that only makes it worse as
>>they will also split the current differently if they add any significant
>>resistance to the paths. As it stands there is no need to balance the path
>>resistance in an RCD only the need to register zero when the current
>>flows.
>>The paths could be different if the current transformer isn't wound
>>symmetrically (say one winding on top of the other. They are designed to
>>be
>>not used in parallel after all.
>>
>
> Matters not as long as the see equal currents thru them they don't
> care;!..
I know, but the internal construction could be different for the line and
neutral.
They are designed to be used in a single line in and neutral in circuit so
the current always passes through whatever their internal construction is.
It would be possible to design one where there were say 10 turns of wire for
the line and 10 turns of wire for the neutral on concentric formers. The
wire length would be different and hence the resistance but the current
would still balance.
I expect they do whatever is cheap.
>>
>>That was chosen because its 0.3% of 10 m. It is also happens to be 30 mA
>>of
>>10A, 30 mA being about the amount needed to trip an RCD.
>
> Now OK thus far now your talking of moving the points where your taking
> the current off in this circuit?.
>
> Moving the take off points, say there're at the centre and now your
> going to move them apart by say 30 mm around 1 inch? Thats to say A
> might be moved to the Left hand RCD whilst B stays the same. Agreed ?.
> There is a difference in length of the points your taking the current
> off and thats now moving apart.
>
> That being so you'll see that the resistance and size of the cable does
> now come into the equation.
The resistance is proportional to length, the actual resistance doesn't
matter.
For example
If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
amps through them then 1 amp will go through the 2 ohm and 2 amps will go
through the 1 ohm resistor.
the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
10A flowing through.
>
> You could use say copper bell wire or you could be using say copper buss
> bar’s certainly a difference in resistivity with length there!.
the effects of the wires glowing).
>
> You did say that the resistance of the cable doesn't matter.
>
> Do you still believe that to be true?. .
We are talking about current balances through parallel resistances and its
all about ratios.
>>So to take an example if the loop is identical R1=R2=R'1=R'2 then the
>>current will split equally as R1=R'1 and R2=R'2.
>>
>>Now if you do it where R1 == R1+0.3% you get an imbalance and the ratios
>>split
>>so the current through R1 becomes I*0.5 * 1 * 1/1.003 and through R'1
>>becomes I*0.5 * 1.003
>>
>>If its a bigger load the current flow difference is more.
>>The only way around it is using superconductors or putting balancing
>>circuits in each leg.
>>
>>BTW I ignored the construction of the RCD as that only makes it worse as
>>they will also split the current differently if they add any significant
>>resistance to the paths. As it stands there is no need to balance the path
>>resistance in an RCD only the need to register zero when the current
>>flows.
>>The paths could be different if the current transformer isn't wound
>>symmetrically (say one winding on top of the other. They are designed to
>>be
>>not used in parallel after all.
>>
>
> Matters not as long as the see equal currents thru them they don't
> care;!..
neutral.
They are designed to be used in a single line in and neutral in circuit so
the current always passes through whatever their internal construction is.
It would be possible to design one where there were say 10 turns of wire for
the line and 10 turns of wire for the neutral on concentric formers. The
wire length would be different and hence the resistance but the current
would still balance.
I expect they do whatever is cheap.
tony sayer
Jul 25, 2012, 6:05:29 PM7/25/12
to
>
>If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
>amps through them then 1 amp will go through the 2 ohm and 2 amps will go
>through the 1 ohm resistor.
>the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
>10A flowing through.
>
>>
>> You could use say copper bell wire or you could be using say copper buss
>> bar’s certainly a difference in resistivity with length there!.
>
>You sure could, it wouldn't make any difference to the balance (excluding
>the effects of the wires glowing).
>
Lets draw this out again...
>If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
>amps through them then 1 amp will go through the 2 ohm and 2 amps will go
>through the 1 ohm resistor.
>the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
>10A flowing through.
>
>>
>> You could use say copper bell wire or you could be using say copper buss
>> bar’s certainly a difference in resistivity with length there!.
>
>You sure could, it wouldn't make any difference to the balance (excluding
>the effects of the wires glowing).
>
here we have the Two RCD's and the Two conductors..
R1 R2
L ----------\/\/\/\/----Lo---\/\/\/\/\------- L
RCD RCD
N ----------\/\/\/\/----Lo---\/\/\/\/\---------N
R1a R2a
Lets assume that R1 is say 10 ohms and R2 is 20 ohms
Assume also that R1a is 20 ohms and R2a is 10 ohms
The Load is represented as Lo and is taken from the shown junction.
The feed Live and Neutral, the supply side of the RCD's are connected
together and have No resistance at all they are simply connected
together OK from the same incoming supply?.
Now take any suitable voltage supply you like and assume the load is say
10 amps now what is the current flowing thru each RCD on the L and N
lines thru that RCD thats to say in each coil in each RCD?.
What do you work those currents to be?. Please show your working.
>
>I know, but the internal construction could be different for the line and
>neutral.
>They are designed to be used in a single line in and neutral in circuit so
>the current always passes through whatever their internal construction is.
>It would be possible to design one where there were say 10 turns of wire for
>the line and 10 turns of wire for the neutral on concentric formers. The
>wire length would be different and hence the resistance but the current
>would still balance.
remember we're looking for a very small -difference- in current flow in
each conductor..
>I expect they do whatever is cheap.
>
as reliably as they can;!..
--
Tony Sayer
tony sayer
Jul 25, 2012, 6:36:52 PM7/25/12
to
In article <FogyEzAp...@bancom.co.uk>, tony sayer
<to...@bancom.co.uk> scribeth thus
Whoops! mean to demo this;!..
Are you looking at it like this..
|----\/\/-----|
L | |---Lo
|----\/\/\/---| |
|
|
|----\/\/\/---| |
N | |---Lo
|----|/\/\/---|
Here I think your assuming that the resists are in parallel and the
circuit flow is from Live thru the upper rissys then to the load then
back thru the bottom rissys to the Neutral line..
Is that what your getting at?..
--
Tony Sayer
<to...@bancom.co.uk> scribeth thus
Are you looking at it like this..
|----\/\/-----|
L | |---Lo
|----\/\/\/---| |
|
|
|----\/\/\/---| |
N | |---Lo
|----|/\/\/---|
Here I think your assuming that the resists are in parallel and the
circuit flow is from Live thru the upper rissys then to the load then
back thru the bottom rissys to the Neutral line..
Is that what your getting at?..
--
Tony Sayer
dennis@home
Jul 25, 2012, 6:46:52 PM7/25/12
to
news:FogyEzAp...@bancom.co.uk...
You have the neutrals swapped and it is totally out of balance.
Its like having 10 meters connected to one leg and 20 to the other and the
opposite for the line.
>
> The Load is represented as Lo and is taken from the shown junction.
>
> The feed Live and Neutral, the supply side of the RCD's are connected
> together and have No resistance at all they are simply connected
> together OK from the same incoming supply?.
>
> Now take any suitable voltage supply you like and assume the load is say
> 10 amps now what is the current flowing thru each RCD on the L and N
> lines thru that RCD thats to say in each coil in each RCD?.
>
> What do you work those currents to be?. Please show your working.
What workings, that's do it in the head stuff.
R1 has 6.666 amps, R2 has 3.333 amps
R1a has 3.333 amps, R2a has 6.666 amps.
The question you need to ask is what's the current if R1a = R1 * 1.003
and/or R2 = R2a * 1.003.
>
>
>>
>>I know, but the internal construction could be different for the line and
>>neutral.
>
> Why should they be?. They need to be as symmetric as possible..
>
>>They are designed to be used in a single line in and neutral in circuit so
>>the current always passes through whatever their internal construction is.
>>It would be possible to design one where there were say 10 turns of wire
>>for
>>the line and 10 turns of wire for the neutral on concentric formers. The
>>wire length would be different and hence the resistance but the current
>>would still balance.
>
> Well you'd need in practice an arrangement that is very finely balanced
> remember we're looking for a very small -difference- in current flow in
> each conductor..
And 10 turns doesn't = 10 turns?
>>
>>If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
>>amps through them then 1 amp will go through the 2 ohm and 2 amps will go
>>through the 1 ohm resistor.
>>the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
>>10A flowing through.
>>
>>>
>>> You could use say copper bell wire or you could be using say copper buss
>>> bar’s certainly a difference in resistivity with length there!.
>>
>>You sure could, it wouldn't make any difference to the balance (excluding
>>the effects of the wires glowing).
>>
>
>
> Lets draw this out again...
>
>
> here we have the Two RCD's and the Two conductors..
>
>
> R1 R2
> L ----------\/\/\/\/----Lo---\/\/\/\/\------- L
> RCD RCD
> N ----------\/\/\/\/----Lo---\/\/\/\/\---------N
> R1a R2a
>
> Lets assume that R1 is say 10 ohms and R2 is 20 ohms
>
> Assume also that R1a is 20 ohms and R2a is 10 ohms
That is silly,
>>If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
>>amps through them then 1 amp will go through the 2 ohm and 2 amps will go
>>through the 1 ohm resistor.
>>the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
>>10A flowing through.
>>
>>>
>>> You could use say copper bell wire or you could be using say copper buss
>>> bar’s certainly a difference in resistivity with length there!.
>>
>>You sure could, it wouldn't make any difference to the balance (excluding
>>the effects of the wires glowing).
>>
>
>
> Lets draw this out again...
>
>
> here we have the Two RCD's and the Two conductors..
>
>
> R1 R2
> L ----------\/\/\/\/----Lo---\/\/\/\/\------- L
> RCD RCD
> N ----------\/\/\/\/----Lo---\/\/\/\/\---------N
> R1a R2a
>
> Lets assume that R1 is say 10 ohms and R2 is 20 ohms
>
> Assume also that R1a is 20 ohms and R2a is 10 ohms
You have the neutrals swapped and it is totally out of balance.
Its like having 10 meters connected to one leg and 20 to the other and the
opposite for the line.
>
> The Load is represented as Lo and is taken from the shown junction.
>
> The feed Live and Neutral, the supply side of the RCD's are connected
> together and have No resistance at all they are simply connected
> together OK from the same incoming supply?.
>
> Now take any suitable voltage supply you like and assume the load is say
> 10 amps now what is the current flowing thru each RCD on the L and N
> lines thru that RCD thats to say in each coil in each RCD?.
>
> What do you work those currents to be?. Please show your working.
R1 has 6.666 amps, R2 has 3.333 amps
R1a has 3.333 amps, R2a has 6.666 amps.
The question you need to ask is what's the current if R1a = R1 * 1.003
and/or R2 = R2a * 1.003.
>
>
>>
>>I know, but the internal construction could be different for the line and
>>neutral.
>
> Why should they be?. They need to be as symmetric as possible..
>
>>They are designed to be used in a single line in and neutral in circuit so
>>the current always passes through whatever their internal construction is.
>>It would be possible to design one where there were say 10 turns of wire
>>for
>>the line and 10 turns of wire for the neutral on concentric formers. The
>>wire length would be different and hence the resistance but the current
>>would still balance.
>
> Well you'd need in practice an arrangement that is very finely balanced
> remember we're looking for a very small -difference- in current flow in
> each conductor..
dennis@home
Jul 25, 2012, 7:05:00 PM7/25/12
to
news:+5bswVBE...@bancom.co.uk...
>
> Whoops! mean to demo this;!..
>
>
> Are you looking at it like this..
>
>
> |----\/\/-----|
> L | |---Lo
> |----\/\/\/---| |
> |
> |
> |----\/\/\/---| |
> N | |---Lo
> |----|/\/\/---|
>
>
> Here I think your assuming that the resists are in parallel and the
> circuit flow is from Live thru the upper rissys then to the load then
> back thru the bottom rissys to the Neutral line..
Its not an assumption its what happens when you swap the two halves of the
ring.
Its all very simple once you actually throw away all the cr@p.
>
> Whoops! mean to demo this;!..
>
>
> Are you looking at it like this..
>
>
> |----\/\/-----|
> L | |---Lo
> |----\/\/\/---| |
> |
> |
> |----\/\/\/---| |
> N | |---Lo
> |----|/\/\/---|
>
>
> Here I think your assuming that the resists are in parallel and the
> circuit flow is from Live thru the upper rissys then to the load then
> back thru the bottom rissys to the Neutral line..
ring.
Its all very simple once you actually throw away all the cr@p.
tony sayer
Jul 25, 2012, 7:23:55 PM7/25/12
to
In article <jupt0t$nou$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
>
>
>You have the neutrals swapped and it is totally out of balance.
Yes, so it is. But the neutrals aren't swapped at all. The resistance's
in circuit have changed...
>Its like having 10 meters connected to one leg and 20 to the other and the
>opposite for the line.
Yes it is but you can now see that the difference in Resistance is the
deciding factor yes or no?.
This is the same as taking the Load off the lines between the two RCD's
at different points albeit an exaggeration in that example.
Agree?..
.
>
>>
>> The Load is represented as Lo and is taken from the shown junction.
>>
>> The feed Live and Neutral, the supply side of the RCD's are connected
>> together and have No resistance at all they are simply connected
>> together OK from the same incoming supply?.
>>
>> Now take any suitable voltage supply you like and assume the load is say
>> 10 amps now what is the current flowing thru each RCD on the L and N
>> lines thru that RCD thats to say in each coil in each RCD?.
>>
>> What do you work those currents to be?. Please show your working.
>
>What workings, that's do it in the head stuff.
>
>R1 has 6.666 amps, R2 has 3.333 amps
>R1a has 3.333 amps, R2a has 6.666 amps.
>
>The question you need to ask is what's the current if R1a = R1 * 1.003
>and/or R2 = R2a * 1.003.
Now how would you get that to alter?.
What would have to alter to achieve that?.
A slightly different current flowing?..
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
>
>
>"tony sayer" <to...@bancom.co.uk> wrote in message
>news:FogyEzAp...@bancom.co.uk...
>>>
>>>If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
>>>amps through them then 1 amp will go through the 2 ohm and 2 amps will go
>>>through the 1 ohm resistor.
>>>the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
>>>10A flowing through.
>>>
>>>>
>>>> You could use say copper bell wire or you could be using say copper buss
>>>> bar’s certainly a difference in resistivity with length there!.
>>>
>>>You sure could, it wouldn't make any difference to the balance (excluding
>>>the effects of the wires glowing).
>>>
>>
>>
>> Lets draw this out again...
>>
>>
>> here we have the Two RCD's and the Two conductors..
>>
>>
>> R1 R2
>> L ----------\/\/\/\/----Lo---\/\/\/\/\------- L
>> RCD RCD
>> N ----------\/\/\/\/----Lo---\/\/\/\/\---------N
>> R1a R2a
>>
>> Lets assume that R1 is say 10 ohms and R2 is 20 ohms
>>
>> Assume also that R1a is 20 ohms and R2a is 10 ohms
>
>That is silly,
No, don't think so..
>news:FogyEzAp...@bancom.co.uk...
>>>
>>>If you have a 2 ohm resistor in parallel with a 1 ohm resistor and shove 3
>>>amps through them then 1 amp will go through the 2 ohm and 2 amps will go
>>>through the 1 ohm resistor.
>>>the same is true if you have a 20 ohm resistor and a 10 ohm resistor with
>>>10A flowing through.
>>>
>>>>
>>>> You could use say copper bell wire or you could be using say copper buss
>>>> bar’s certainly a difference in resistivity with length there!.
>>>
>>>You sure could, it wouldn't make any difference to the balance (excluding
>>>the effects of the wires glowing).
>>>
>>
>>
>> Lets draw this out again...
>>
>>
>> here we have the Two RCD's and the Two conductors..
>>
>>
>> R1 R2
>> L ----------\/\/\/\/----Lo---\/\/\/\/\------- L
>> RCD RCD
>> N ----------\/\/\/\/----Lo---\/\/\/\/\---------N
>> R1a R2a
>>
>> Lets assume that R1 is say 10 ohms and R2 is 20 ohms
>>
>> Assume also that R1a is 20 ohms and R2a is 10 ohms
>
>That is silly,
>You have the neutrals swapped and it is totally out of balance.
in circuit have changed...
>Its like having 10 meters connected to one leg and 20 to the other and the
>opposite for the line.
deciding factor yes or no?.
This is the same as taking the Load off the lines between the two RCD's
at different points albeit an exaggeration in that example.
Agree?..
.
>
>>
>> The Load is represented as Lo and is taken from the shown junction.
>>
>> The feed Live and Neutral, the supply side of the RCD's are connected
>> together and have No resistance at all they are simply connected
>> together OK from the same incoming supply?.
>>
>> Now take any suitable voltage supply you like and assume the load is say
>> 10 amps now what is the current flowing thru each RCD on the L and N
>> lines thru that RCD thats to say in each coil in each RCD?.
>>
>> What do you work those currents to be?. Please show your working.
>
>What workings, that's do it in the head stuff.
>
>R1 has 6.666 amps, R2 has 3.333 amps
>R1a has 3.333 amps, R2a has 6.666 amps.
>
>The question you need to ask is what's the current if R1a = R1 * 1.003
>and/or R2 = R2a * 1.003.
What would have to alter to achieve that?.
A slightly different current flowing?..
>
>>
>>
>>>
>>>I know, but the internal construction could be different for the line and
>>>neutral.
>>
>> Why should they be?. They need to be as symmetric as possible..
>>
>>>They are designed to be used in a single line in and neutral in circuit so
>>>the current always passes through whatever their internal construction is.
>>>It would be possible to design one where there were say 10 turns of wire
>>>for
>>>the line and 10 turns of wire for the neutral on concentric formers. The
>>>wire length would be different and hence the resistance but the current
>>>would still balance.
>>
>> Well you'd need in practice an arrangement that is very finely balanced
>> remember we're looking for a very small -difference- in current flow in
>> each conductor..
>
>And 10 turns doesn't = 10 turns?
Yes thats precisely the need to be the same..
>>
>>
>>>
>>>I know, but the internal construction could be different for the line and
>>>neutral.
>>
>> Why should they be?. They need to be as symmetric as possible..
>>
>>>They are designed to be used in a single line in and neutral in circuit so
>>>the current always passes through whatever their internal construction is.
>>>It would be possible to design one where there were say 10 turns of wire
>>>for
>>>the line and 10 turns of wire for the neutral on concentric formers. The
>>>wire length would be different and hence the resistance but the current
>>>would still balance.
>>
>> Well you'd need in practice an arrangement that is very finely balanced
>> remember we're looking for a very small -difference- in current flow in
>> each conductor..
>
>And 10 turns doesn't = 10 turns?
>
>>
>>>I expect they do whatever is cheap.
>>>
>> I expect they do whatever's necessary to make sure it works as designed
>> as reliably as they can;!..
>>
>>
>> --
>> Tony Sayer
>>
--
>>
>>>I expect they do whatever is cheap.
>>>
>> I expect they do whatever's necessary to make sure it works as designed
>> as reliably as they can;!..
>>
>>
>> --
>> Tony Sayer
>>
Tony Sayer
tony sayer
Jul 25, 2012, 7:36:01 PM7/25/12
to
In article <jupu2u$plj$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
>
>
Ok now that above is the same as below..
Or is it?.
What is different?.
R1 R2
L ----------\/\/\/\/----Lo---\/\/\/\/\------- L
RCD RCD
N ----------\/\/\/\/----Lo---\/\/\/\/\---------N
R1a R2a
--
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
>
>
Or is it?.
What is different?.
R1 R2
L ----------\/\/\/\/----Lo---\/\/\/\/\------- L
RCD RCD
N ----------\/\/\/\/----Lo---\/\/\/\/\---------N
R1a R2a
Tony Sayer
A.Lee
Jul 26, 2012, 1:34:17 AM7/26/12
to
dennis@home <den...@killspam.kicks-ass.net> wrote:
> You only need a variation of about 0.3% in the length to imbalance it by 30
> mA with a 10A load (kettle to you).
> That's ~3 mm per meter of run, Do you keep all your connections so the line
> and neutral never vary by that much
No it isnt.
> You only need a variation of about 0.3% in the length to imbalance it by 30
> mA with a 10A load (kettle to you).
> That's ~3 mm per meter of run, Do you keep all your connections so the line
> and neutral never vary by that much
That is your fundamental error. The resistance of the cable is so small
for most domestic installs that it makes no difference at all if the
neutral cable is a metre or more longer than the Line cable, or vice
versa.
It is the Load that causes the biggest resistance, by a very long way,
with the cable having such a small influence, it is not worth even
thinking about the cables effect on RCD imbalance.
In your case, if the cable had such an effect, every shower in the
Country would trip the RCD when it was used, as the cable always has a
difference in length between L & N, typically 100 - 150mm, add the ~35A
load, and your idea would cause such an imbalance that it would not be
usable.
--
To reply by e-mail, change the ' + ' to 'plus'.
dennis@home
Jul 26, 2012, 4:19:37 AM7/26/12
to
news:kG4aPtBL...@bancom.co.uk...
It is so exaggerated that even a small load would trip it and the fault
would have never got past testing in the first place.
Its not going to happen either as the two conductors usually run in the same
sheath.
>>> The Load is represented as Lo and is taken from the shown junction.
>>>
>>> The feed Live and Neutral, the supply side of the RCD's are connected
>>> together and have No resistance at all they are simply connected
>>> together OK from the same incoming supply?.
>>>
>>> Now take any suitable voltage supply you like and assume the load is say
>>> 10 amps now what is the current flowing thru each RCD on the L and N
>>> lines thru that RCD thats to say in each coil in each RCD?.
>>>
>>> What do you work those currents to be?. Please show your working.
>>
>>What workings, that's do it in the head stuff.
>>
>>R1 has 6.666 amps, R2 has 3.333 amps
>>R1a has 3.333 amps, R2a has 6.666 amps.
>>
>
>
>>The question you need to ask is what's the current if R1a = R1 * 1.003
>>and/or R2 = R2a * 1.003.
>
> Now how would you get that to alter?.
The extra 30 mm of cable in either of the conductors!!!!!!!!
> What would have to alter to achieve that?.
>
> A slightly different current flowing?..
>>
>>>
>>>
>>>>
>>>>I know, but the internal construction could be different for the line
>>>>and
>>>>neutral.
>>>
>>> Why should they be?. They need to be as symmetric as possible..
>>>
>>>>They are designed to be used in a single line in and neutral in circuit
>>>>so
>>>>the current always passes through whatever their internal construction
>>>>is.
>>>>It would be possible to design one where there were say 10 turns of wire
>>>>for
>>>>the line and 10 turns of wire for the neutral on concentric formers. The
>>>>wire length would be different and hence the resistance but the current
>>>>would still balance.
>>>
>>> Well you'd need in practice an arrangement that is very finely balanced
>>> remember we're looking for a very small -difference- in current flow in
>>> each conductor..
>>
>>And 10 turns doesn't = 10 turns?
>
> Yes thats precisely the need to be the same..
And if one is wound around the other it has more wire and more resistance
(if its the same wire like it is in a ring circuit).
So its possible to produce an RCD that can add extra resistance to one leg
that still works as an RCD.
You need to take them all apart to see if they do that.
would have never got past testing in the first place.
Its not going to happen either as the two conductors usually run in the same
sheath.
>>> The Load is represented as Lo and is taken from the shown junction.
>>>
>>> The feed Live and Neutral, the supply side of the RCD's are connected
>>> together and have No resistance at all they are simply connected
>>> together OK from the same incoming supply?.
>>>
>>> Now take any suitable voltage supply you like and assume the load is say
>>> 10 amps now what is the current flowing thru each RCD on the L and N
>>> lines thru that RCD thats to say in each coil in each RCD?.
>>>
>>> What do you work those currents to be?. Please show your working.
>>
>>What workings, that's do it in the head stuff.
>>
>>R1 has 6.666 amps, R2 has 3.333 amps
>>R1a has 3.333 amps, R2a has 6.666 amps.
>>
>
>
>>The question you need to ask is what's the current if R1a = R1 * 1.003
>>and/or R2 = R2a * 1.003.
>
> Now how would you get that to alter?.
> What would have to alter to achieve that?.
>
> A slightly different current flowing?..
>>
>>>
>>>
>>>>
>>>>I know, but the internal construction could be different for the line
>>>>and
>>>>neutral.
>>>
>>> Why should they be?. They need to be as symmetric as possible..
>>>
>>>>They are designed to be used in a single line in and neutral in circuit
>>>>so
>>>>the current always passes through whatever their internal construction
>>>>is.
>>>>It would be possible to design one where there were say 10 turns of wire
>>>>for
>>>>the line and 10 turns of wire for the neutral on concentric formers. The
>>>>wire length would be different and hence the resistance but the current
>>>>would still balance.
>>>
>>> Well you'd need in practice an arrangement that is very finely balanced
>>> remember we're looking for a very small -difference- in current flow in
>>> each conductor..
>>
>>And 10 turns doesn't = 10 turns?
>
> Yes thats precisely the need to be the same..
(if its the same wire like it is in a ring circuit).
So its possible to produce an RCD that can add extra resistance to one leg
that still works as an RCD.
You need to take them all apart to see if they do that.
tony sayer
Jul 26, 2012, 5:29:25 AM7/26/12
to
>> Yes it is but you can now see that the difference in Resistance is the
>> deciding factor yes or no?.
>>
>> This is the same as taking the Load off the lines between the two RCD's
>> at different points albeit an exaggeration in that example.
>>
>>
>> Agree?..
>
>It is so exaggerated that even a small load would trip it and the fault
>would have never got past testing in the first place.
Yes it is exaggerated .. and thats done for example and clarity of the
>> deciding factor yes or no?.
>>
>> This is the same as taking the Load off the lines between the two RCD's
>> at different points albeit an exaggeration in that example.
>>
>>
>> Agree?..
>
>It is so exaggerated that even a small load would trip it and the fault
>would have never got past testing in the first place.
principles involved..
>
>Its not going to happen either as the two conductors usually run in the same
>sheath.
conductors we're concerned with not anything to do with capacity of them
at all..
>
>>>> The Load is represented as Lo and is taken from the shown junction.
>>>>
>>>> The feed Live and Neutral, the supply side of the RCD's are connected
>>>> together and have No resistance at all they are simply connected
>>>> together OK from the same incoming supply?.
>>>>
>>>> Now take any suitable voltage supply you like and assume the load is say
>>>> 10 amps now what is the current flowing thru each RCD on the L and N
>>>> lines thru that RCD thats to say in each coil in each RCD?.
>>>>
>>>> What do you work those currents to be?. Please show your working.
>>>
>>>What workings, that's do it in the head stuff.
>>>
>>>R1 has 6.666 amps, R2 has 3.333 amps
>>>R1a has 3.333 amps, R2a has 6.666 amps.
>>>
>>
>>
>>>The question you need to ask is what's the current if R1a = R1 * 1.003
>>>and/or R2 = R2a * 1.003.
>>
>> Now how would you get that to alter?.
>
>The extra 30 mm of cable in either of the conductors!!!!!!!!
OK?..
But does not the resistance of the conductors depend on the amount of
conductor, thats to say a bigger cross section conductor is going to
have less resistance per unit volume than a smaller one?..
What's got more resistance a conductor of say 1 mm diameter or one of
say 20 mm diameter?..
>
>> What would have to alter to achieve that?.
>>
>> A slightly different current flowing?..
>>>
>>>>
>>>>
>>>>>
>>>>>I know, but the internal construction could be different for the line
>>>>>and
>>>>>neutral.
>>>>
>>>> Why should they be?. They need to be as symmetric as possible..
>>>>
>>>>>They are designed to be used in a single line in and neutral in circuit
>>>>>so
>>>>>the current always passes through whatever their internal construction
>>>>>is.
>>>>>It would be possible to design one where there were say 10 turns of wire
>>>>>for
>>>>>the line and 10 turns of wire for the neutral on concentric formers. The
>>>>>wire length would be different and hence the resistance but the current
>>>>>would still balance.
>>>>
>>>> Well you'd need in practice an arrangement that is very finely balanced
>>>> remember we're looking for a very small -difference- in current flow in
>>>> each conductor..
>>>
>>>And 10 turns doesn't = 10 turns?
>>
>> Yes thats precisely the need to be the same..
>
>And if one is wound around the other it has more wire and more resistance
>(if its the same wire like it is in a ring circuit).
>So its possible to produce an RCD that can add extra resistance to one leg
>that still works as an RCD.
Your looking for a device that will pass up to say 63 odd amps.
It needs tho to detect a difference of around 30 ma, in practice less,
of current -differential- in those Two conductors going thru it..
>You need to take them all apart to see if they do that.
Did you get around to looking at the other circuit arrangement yet?..
The one with the supposed parallel resistance's therein?..
>
>
>
--
Tony Sayer
dennis@home
Jul 26, 2012, 7:45:52 AM7/26/12
to
"A.Lee" <alan@darkroom.+.com> wrote in message
news:1knttsz.1v926jp1iucgxuN%alan@darkroom.+.com...
> dennis@home <den...@killspam.kicks-ass.net> wrote:
>
>
>> You only need a variation of about 0.3% in the length to imbalance it by
>> 30
>> mA with a 10A load (kettle to you).
>> That's ~3 mm per meter of run, Do you keep all your connections so the
>> line
>> and neutral never vary by that much
>
> No it isnt.
> That is your fundamental error.
Its not an error on my part.
>
>
>> You only need a variation of about 0.3% in the length to imbalance it by
>> 30
>> mA with a 10A load (kettle to you).
>> That's ~3 mm per meter of run, Do you keep all your connections so the
>> line
>> and neutral never vary by that much
>
> No it isnt.
> That is your fundamental error.
> The resistance of the cable is so small
> for most domestic installs that it makes no difference at all if the
> neutral cable is a metre or more longer than the Line cable, or vice
> versa.
Even the IIE regs knows that can't be true.
If it were there would be no need for rules like not putting loads near one
end of a ring.
It doesn't matter how low the resistance of the cable is it will still cause
the flow to split where the cables are in parallel as in this case and in
the case of a normal ring.
>
> It is the Load that causes the biggest resistance, by a very long way,
> with the cable having such a small influence, it is not worth even
> thinking about the cables effect on RCD imbalance.
currents and not the way they split.
>
> In your case, if the cable had such an effect, every shower in the
> Country would trip the RCD when it was used, as the cable always has a
> difference in length between L & N, typically 100 - 150mm, add the ~35A
> load, and your idea would cause such an imbalance that it would not be
> usable.
this case.
There is only one path for the current down the line, through the load,
return through the neutral.
You may well have problems if you ran two different cables from the consumer
unit to the shower but that would be in overloading one cable not in
tripping anything.
This is a specific case where there are two paths for the line to the load
and two paths back for the neutral.
It only occurs where cables are paralleled like in rings.
And in this case the circuit was also wired *incorrectly* in having two RCDs
paralleled too.
dennis@home
Jul 26, 2012, 8:36:53 AM7/26/12
to
news:$+HcxeD1...@bancom.co.uk...
circuit, like running the line and neutral differently.
It doesn't help with the case in question if you change things so much they
no longer resemble the original.
>
>>
>>>>> The Load is represented as Lo and is taken from the shown junction.
>>>>>
>>>>> The feed Live and Neutral, the supply side of the RCD's are connected
>>>>> together and have No resistance at all they are simply connected
>>>>> together OK from the same incoming supply?.
>>>>>
>>>>> Now take any suitable voltage supply you like and assume the load is
>>>>> say
>>>>> 10 amps now what is the current flowing thru each RCD on the L and N
>>>>> lines thru that RCD thats to say in each coil in each RCD?.
>>>>>
>>>>> What do you work those currents to be?. Please show your working.
>>>>
>>>>What workings, that's do it in the head stuff.
>>>>
>>>>R1 has 6.666 amps, R2 has 3.333 amps
>>>>R1a has 3.333 amps, R2a has 6.666 amps.
>>>>
>>>
>>>
>>>>The question you need to ask is what's the current if R1a = R1 * 1.003
>>>>and/or R2 = R2a * 1.003.
>>>
>>> Now how would you get that to alter?.
>>
>>The extra 30 mm of cable in either of the conductors!!!!!!!!
>
> Yes. therefore the resistance of the conductors varies with the length
> OK?..
>
> But does not the resistance of the conductors depend on the amount of
> conductor, thats to say a bigger cross section conductor is going to
> have less resistance per unit volume than a smaller one?..
It doesn't matter what the actual conductance is.
Look at the equations the balance only depends on the ratio and not the
resistance.
>
>
> What's got more resistance a conductor of say 1 mm diameter or one of
> say 20 mm diameter?..
We have been through this, the ratio is the same for 10:20 micro ohms as it
is for 10:20 mega ohms.
the current will split in the same proportions whichever you have.
You can wire one leg in 1 mm and one in 20mm if you want, however we are
talking about a ring main (final if you must) and it is wired in normal T&E.
As above, one winding needs to be bigger than the other if they are
concentric (wound around each other) this takes more wire.
>
> Your looking for a device that will pass up to say 63 odd amps.
>
> It needs tho to detect a difference of around 30 ma, in practice less,
> of current -differential- in those Two conductors going thru it..
>
>>You need to take them all apart to see if they do that.
>
> No you don't the principles of the device are sufficient..
>
>
> Did you get around to looking at the other circuit arrangement yet?..
>
> The one with the supposed parallel resistance's therein?..
The wires are in parallel, they have resistance, there are resistances in
parallel, where is the problem?
Are you trying to change the circuit to something else?
Do you not understand that if you have two equal lengths of 2.5 mm T&E
feeding a load the current will split equally down them because of their
resistance?
Nothing to do with the load at all, just because they are wires with
resistance?
Do you not understand that if one wire is twice as long the current will
split in proportion to their resistance which is proportional to their
length? One wire will carry 1/3 of the current and the other will carry 2/3
of the current.
Do you not see there are two such wire combinations in the circuit, one in
the line and one in the neutral?
Do you fail to see that if one of the wires in one of those combinations
(say the line) is slightly different (the 30mm in the example) then the
current will split differently to the two wires on the other combination
(say the neutral)?
Does this not make an imbalance in the RCDs?
Does this imbalance not trip the RCDs if the load draws enough current for
the imbalance to exceed 30mA?
Do all electrical circuits installed in domestic installations keep the
wires balanced to 30 mm or do electricians lay the wires in and cut them to
be neat?
Do you not see that 30 mA is a very small current in a circuit that can take
tens of amps and that even small tolerances on cables in a ring can produce
such small imbalances? 30 mA is only 0.3% of 10 amps and only 0.1% of 30
amps.
>>> Yes it is but you can now see that the difference in Resistance is the
>>> deciding factor yes or no?.
>>>
>>> This is the same as taking the Load off the lines between the two RCD's
>>> at different points albeit an exaggeration in that example.
>>>
>>>
>>> Agree?..
>>
>>It is so exaggerated that even a small load would trip it and the fault
>>would have never got past testing in the first place.
>
> Yes it is exaggerated .. and thats done for example and clarity of the
> principles involved..
>
>>
>>Its not going to happen either as the two conductors usually run in the
>>same
>>sheath.
>
> What has the got to do with this ?.. Its just the resistance of the
> conductors we're concerned with not anything to do with capacity of them
> at all..
Well it means the installer has to go to great lengths to produce such a
>>> deciding factor yes or no?.
>>>
>>> This is the same as taking the Load off the lines between the two RCD's
>>> at different points albeit an exaggeration in that example.
>>>
>>>
>>> Agree?..
>>
>>It is so exaggerated that even a small load would trip it and the fault
>>would have never got past testing in the first place.
>
> Yes it is exaggerated .. and thats done for example and clarity of the
> principles involved..
>
>>
>>Its not going to happen either as the two conductors usually run in the
>>same
>>sheath.
>
> What has the got to do with this ?.. Its just the resistance of the
> conductors we're concerned with not anything to do with capacity of them
> at all..
circuit, like running the line and neutral differently.
It doesn't help with the case in question if you change things so much they
no longer resemble the original.
>
>>
>>>>> The Load is represented as Lo and is taken from the shown junction.
>>>>>
>>>>> The feed Live and Neutral, the supply side of the RCD's are connected
>>>>> together and have No resistance at all they are simply connected
>>>>> together OK from the same incoming supply?.
>>>>>
>>>>> Now take any suitable voltage supply you like and assume the load is
>>>>> say
>>>>> 10 amps now what is the current flowing thru each RCD on the L and N
>>>>> lines thru that RCD thats to say in each coil in each RCD?.
>>>>>
>>>>> What do you work those currents to be?. Please show your working.
>>>>
>>>>What workings, that's do it in the head stuff.
>>>>
>>>>R1 has 6.666 amps, R2 has 3.333 amps
>>>>R1a has 3.333 amps, R2a has 6.666 amps.
>>>>
>>>
>>>
>>>>The question you need to ask is what's the current if R1a = R1 * 1.003
>>>>and/or R2 = R2a * 1.003.
>>>
>>> Now how would you get that to alter?.
>>
>>The extra 30 mm of cable in either of the conductors!!!!!!!!
>
> Yes. therefore the resistance of the conductors varies with the length
> OK?..
>
> But does not the resistance of the conductors depend on the amount of
> conductor, thats to say a bigger cross section conductor is going to
> have less resistance per unit volume than a smaller one?..
Look at the equations the balance only depends on the ratio and not the
resistance.
>
>
> What's got more resistance a conductor of say 1 mm diameter or one of
> say 20 mm diameter?..
is for 10:20 mega ohms.
the current will split in the same proportions whichever you have.
You can wire one leg in 1 mm and one in 20mm if you want, however we are
talking about a ring main (final if you must) and it is wired in normal T&E.
concentric (wound around each other) this takes more wire.
>
> Your looking for a device that will pass up to say 63 odd amps.
>
> It needs tho to detect a difference of around 30 ma, in practice less,
> of current -differential- in those Two conductors going thru it..
>
>>You need to take them all apart to see if they do that.
>
> No you don't the principles of the device are sufficient..
>
>
> Did you get around to looking at the other circuit arrangement yet?..
>
> The one with the supposed parallel resistance's therein?..
parallel, where is the problem?
Are you trying to change the circuit to something else?
Do you not understand that if you have two equal lengths of 2.5 mm T&E
feeding a load the current will split equally down them because of their
resistance?
Nothing to do with the load at all, just because they are wires with
resistance?
Do you not understand that if one wire is twice as long the current will
split in proportion to their resistance which is proportional to their
length? One wire will carry 1/3 of the current and the other will carry 2/3
of the current.
Do you not see there are two such wire combinations in the circuit, one in
the line and one in the neutral?
Do you fail to see that if one of the wires in one of those combinations
(say the line) is slightly different (the 30mm in the example) then the
current will split differently to the two wires on the other combination
(say the neutral)?
Does this not make an imbalance in the RCDs?
Does this imbalance not trip the RCDs if the load draws enough current for
the imbalance to exceed 30mA?
Do all electrical circuits installed in domestic installations keep the
wires balanced to 30 mm or do electricians lay the wires in and cut them to
be neat?
Do you not see that 30 mA is a very small current in a circuit that can take
tens of amps and that even small tolerances on cables in a ring can produce
such small imbalances? 30 mA is only 0.3% of 10 amps and only 0.1% of 30
amps.
tony sayer
Jul 26, 2012, 10:57:15 AM7/26/12
to
In article <jurdl6$a05$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
>
>
OK I can see where your going wrong with this now.
Consider the diagram below. This is the oddball circuit that was
concocted. Two RCD's linked together by a cable. I've shown the incoming
side of it for clarity.
Now as long as the cable resistance's are the same thats to say R1 is
the same as R1a and R2 is the same as R2a then all will be well as the
load current shown as connected across the Two lines will pull equal
amounts Thru the RCD's. Well thats to say from the overall supply.
Now when they are getting different thats to say R1 is now different to
R1a and R2 is different to R2 a then the currents flowing THROUGH the
RCD's become unbalanced.
R1 R2
L in -- L ----------\/\/\/\/----Lo---\/\/\/\/\------- L --- L in
RCD RCD
N in -- N ----------\/\/\/\/-----Lo---\/\/\/\/\- -------N --- N in
R1a R2a
It appears to be the same as below where the resistance's in the circuit
are shown as paralleled together. However if they are differing then at
first glance they appear as simply Two resistors in paralled and the
current across Both of them may seem to be equal just the one resist
taking more current cos of its lower ohms.
If this were as just that, then what you are saying would be true.
Assume that now exaggerated, but for clarity, R1 is 10 ohms R2 is 20
ohms with their ends connected as shown then the OVERALL resistance
would not matter it would be the same as if you reversed R1 and R2
Same as R1a and R2a call them 10 and 20 ohms respectively.
So now we have R1 is 10 ohms R2 is 20
R1a is 10 ohms R2a is 20.
However reverse one leg lets make R1 20 ohms and R2 10 ohms still the
same current will flow overall thru the load.
The current flowing across the whole circuit would be the same.
If it were that simple
R1
|-r---\/\/-----|
L | R2 |---Lo
|-r---\/\/\/---| |
|
R1a |
|-r---\/\/\/---| |
N | R2a |---Lo
|-r---|/\/\/---|
Now all would be well in the above which looks like the circuit upper
above except we have those awkward RCD's in circuit and they are
detecting the current flows thru the individual resistors..
The small "r" denotes the RCD coils in each leg of the respective
RCD's..
See what will now happen if you alter the current thru the circuit by
varying the resistance there will be a current unbalance flowing thru
the RCD's in circuit.
This again .. lets take it to extremis and make Two of the resistors
open circuit shall we infinite ohms?..
R1 R2
L in -- L Lo---\/\/\/\/\------- L --- L in
RCD RCD
N in -- N ----------\/\/\/\/-----Lo- --N --- N in
R1a R2a
Now the circuit would still be there current would flow but what's now
happening to the RCD's?.
One is seeing all its live current going out the other is seeing all its
neutral current but no live to balance it hence a rather rapid trip;!..
The cable has resistance due to its length and cross section all along
its length.
That is the other factor. By altering the points in the circuit where
you have the load connected to them the resistance's will become
different.
Now can you not see if the cable lengths and diameter alter then the
resistance's and currents flowing will vary?.
And that is dependant on the size of the cables used in the example.
Remember this also alters with length.
An interesting excersise for you to do is take a typical cable say a
lump of 2.5 T&E and tell us if its 10 metres long how much we'd have the
alter the load taken point to achieve a current unbalance of greater
than 25 ma.
Now do that with say .1 mm cable and then say 16 mm cross section and
now do the lengths change from the centre tapping point?..
Lets see your results...
--
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
>
>
OK I can see where your going wrong with this now.
Consider the diagram below. This is the oddball circuit that was
concocted. Two RCD's linked together by a cable. I've shown the incoming
side of it for clarity.
Now as long as the cable resistance's are the same thats to say R1 is
the same as R1a and R2 is the same as R2a then all will be well as the
load current shown as connected across the Two lines will pull equal
amounts Thru the RCD's. Well thats to say from the overall supply.
Now when they are getting different thats to say R1 is now different to
R1a and R2 is different to R2 a then the currents flowing THROUGH the
RCD's become unbalanced.
R1 R2
L in -- L ----------\/\/\/\/----Lo---\/\/\/\/\------- L --- L in
RCD RCD
N in -- N ----------\/\/\/\/-----Lo---\/\/\/\/\- -------N --- N in
R1a R2a
It appears to be the same as below where the resistance's in the circuit
are shown as paralleled together. However if they are differing then at
first glance they appear as simply Two resistors in paralled and the
current across Both of them may seem to be equal just the one resist
taking more current cos of its lower ohms.
If this were as just that, then what you are saying would be true.
Assume that now exaggerated, but for clarity, R1 is 10 ohms R2 is 20
ohms with their ends connected as shown then the OVERALL resistance
would not matter it would be the same as if you reversed R1 and R2
Same as R1a and R2a call them 10 and 20 ohms respectively.
So now we have R1 is 10 ohms R2 is 20
R1a is 10 ohms R2a is 20.
However reverse one leg lets make R1 20 ohms and R2 10 ohms still the
same current will flow overall thru the load.
The current flowing across the whole circuit would be the same.
If it were that simple
R1
|-r---\/\/-----|
L | R2 |---Lo
|-r---\/\/\/---| |
|
R1a |
|-r---\/\/\/---| |
N | R2a |---Lo
|-r---|/\/\/---|
Now all would be well in the above which looks like the circuit upper
above except we have those awkward RCD's in circuit and they are
detecting the current flows thru the individual resistors..
The small "r" denotes the RCD coils in each leg of the respective
RCD's..
See what will now happen if you alter the current thru the circuit by
varying the resistance there will be a current unbalance flowing thru
the RCD's in circuit.
This again .. lets take it to extremis and make Two of the resistors
open circuit shall we infinite ohms?..
R1 R2
L in -- L Lo---\/\/\/\/\------- L --- L in
RCD RCD
N in -- N ----------\/\/\/\/-----Lo- --N --- N in
R1a R2a
Now the circuit would still be there current would flow but what's now
happening to the RCD's?.
One is seeing all its live current going out the other is seeing all its
neutral current but no live to balance it hence a rather rapid trip;!..
The cable has resistance due to its length and cross section all along
its length.
That is the other factor. By altering the points in the circuit where
you have the load connected to them the resistance's will become
different.
Now can you not see if the cable lengths and diameter alter then the
resistance's and currents flowing will vary?.
And that is dependant on the size of the cables used in the example.
Remember this also alters with length.
An interesting excersise for you to do is take a typical cable say a
lump of 2.5 T&E and tell us if its 10 metres long how much we'd have the
alter the load taken point to achieve a current unbalance of greater
than 25 ma.
Now do that with say .1 mm cable and then say 16 mm cross section and
now do the lengths change from the centre tapping point?..
Lets see your results...
--
Tony Sayer
Adam Funk
Jul 26, 2012, 11:23:04 AM7/26/12
to
On 2012-07-26, A.Lee wrote:
> dennis@home <den...@killspam.kicks-ass.net> wrote:
>
>
>> You only need a variation of about 0.3% in the length to imbalance it by 30
>> mA with a 10A load (kettle to you).
>> That's ~3 mm per meter of run, Do you keep all your connections so the line
>> and neutral never vary by that much
>
> No it isnt.
> That is your fundamental error. The resistance of the cable is so small
> for most domestic installs that it makes no difference at all if the
> neutral cable is a metre or more longer than the Line cable, or vice
> versa.
>
> It is the Load that causes the biggest resistance, by a very long way,
> with the cable having such a small influence, it is not worth even
> thinking about the cables effect on RCD imbalance.
(Just for the avoidance of doubt, I agree with you & John & Adam W
> dennis@home <den...@killspam.kicks-ass.net> wrote:
>
>
>> You only need a variation of about 0.3% in the length to imbalance it by 30
>> mA with a 10A load (kettle to you).
>> That's ~3 mm per meter of run, Do you keep all your connections so the line
>> and neutral never vary by that much
>
> No it isnt.
> That is your fundamental error. The resistance of the cable is so small
> for most domestic installs that it makes no difference at all if the
> neutral cable is a metre or more longer than the Line cable, or vice
> versa.
>
> It is the Load that causes the biggest resistance, by a very long way,
> with the cable having such a small influence, it is not worth even
> thinking about the cables effect on RCD imbalance.
about this.)
> In your case, if the cable had such an effect, every shower in the
> Country would trip the RCD when it was used, as the cable always has a
> difference in length between L & N, typically 100 - 150mm, add the ~35A
> load, and your idea would cause such an imbalance that it would not be
> usable.
electric shower some years ago & don't recall that kind of
difference, but maybe I'm forgetting something.)
A.Lee
Jul 26, 2012, 11:39:02 AM7/26/12
to
Adam Funk <a24...@ducksburg.com> wrote:
> > In your case, if the cable had such an effect, every shower in the
> > Country would trip the RCD when it was used, as the cable always has a
> > difference in length between L & N, typically 100 - 150mm, add the ~35A
> > load, and your idea would cause such an imbalance that it would not be
> > usable.
>
> Where is the extra 100 to 150 mm of one wire? (I installed an
> electric shower some years ago & don't recall that kind of
> difference, but maybe I'm forgetting something.)
At the CU end. If the cable enters the CU at one end, the neutral bar
> > In your case, if the cable had such an effect, every shower in the
> > Country would trip the RCD when it was used, as the cable always has a
> > difference in length between L & N, typically 100 - 150mm, add the ~35A
> > load, and your idea would cause such an imbalance that it would not be
> > usable.
>
> Where is the extra 100 to 150 mm of one wire? (I installed an
> electric shower some years ago & don't recall that kind of
> difference, but maybe I'm forgetting something.)
may be next to the entry, the line terminal at the other end. The only
sure way is to have the cable enter the CU at mid point between the line
terminal and neutral bar, which in real life, is not normally possible.
dennis@home
Jul 26, 2012, 12:24:46 PM7/26/12
to
"Adam Funk" <a24...@ducksburg.com> wrote in message
news:ov38e9x...@news.ducksburg.com...
> On 2012-07-26, A.Lee wrote:
>
>> dennis@home <den...@killspam.kicks-ass.net> wrote:
>>
>>
>>> You only need a variation of about 0.3% in the length to imbalance it by
>>> 30
>>> mA with a 10A load (kettle to you).
>>> That's ~3 mm per meter of run, Do you keep all your connections so the
>>> line
>>> and neutral never vary by that much
>>
>> No it isnt.
>> That is your fundamental error. The resistance of the cable is so small
>> for most domestic installs that it makes no difference at all if the
>> neutral cable is a metre or more longer than the Line cable, or vice
>> versa.
>>
>> It is the Load that causes the biggest resistance, by a very long way,
>> with the cable having such a small influence, it is not worth even
>> thinking about the cables effect on RCD imbalance.
>
> (Just for the avoidance of doubt, I agree with you & John & Adam W
> about this.)
If you read the first few posts about the *actual* circuit you may well
>
>> dennis@home <den...@killspam.kicks-ass.net> wrote:
>>
>>
>>> You only need a variation of about 0.3% in the length to imbalance it by
>>> 30
>>> mA with a 10A load (kettle to you).
>>> That's ~3 mm per meter of run, Do you keep all your connections so the
>>> line
>>> and neutral never vary by that much
>>
>> No it isnt.
>> That is your fundamental error. The resistance of the cable is so small
>> for most domestic installs that it makes no difference at all if the
>> neutral cable is a metre or more longer than the Line cable, or vice
>> versa.
>>
>> It is the Load that causes the biggest resistance, by a very long way,
>> with the cable having such a small influence, it is not worth even
>> thinking about the cables effect on RCD imbalance.
>
> (Just for the avoidance of doubt, I agree with you & John & Adam W
> about this.)
change your mind.
>
>
>> In your case, if the cable had such an effect, every shower in the
>> Country would trip the RCD when it was used, as the cable always has a
>> difference in length between L & N, typically 100 - 150mm, add the ~35A
>> load, and your idea would cause such an imbalance that it would not be
>> usable.
>
> Where is the extra 100 to 150 mm of one wire? (I installed an
> electric shower some years ago & don't recall that kind of
> difference, but maybe I'm forgetting something.)
question as described by adam.
the shower is a red herring and has very little to do with what is being
discussed.
The extra bit of wire is a sum of the tolerances, including those that
electricians make in the consumer unit to make things neat.
If you think about it are the distances between the neutral bars and the MCB
outputs exactly the same for each cable terminated or do you cut them so
they are neat?
John claims that the circuit will always balance while I claim its luck that
it did.
Unless you put some very tight (IMO impossibly tight) tolerances on cable
lengths it does not balance.
You may notice that John probably understands this now as he has gone very
quite.
But just for you I will go through it again..
First the basics as I can't get certain posters to understand them.
Wires have resistance proportional to their length (assuming they are normal
wires)
A ring circuit has two line wires in parallel feeding the load.
A ring circuit has two neutral wires in parallel returning from the load.
All the current has to pass through the load so its effects are a net zero
on the balance of currents.
The current through the two lines will be split in proportion to the
resistance (length)
The same applies to the neutral
Normally the two lines match the two neutrals as they are in the same pair
of T&E cables
This means you will get exactly the same current through the line and
neutral in the one cable and the same applies to the other cable.
Now on the faulty circuit these are tied at the ends to two RCDs and if they
are exactly the same nothing happens, they don't trip.
Now if you make one core slightly longer anywhere in the circuit the current
will be imbalanced on one of the cables (the resistances are different so
the split is different).
Now as you only need an imbalance of 30 mA to trip the RCD it doesn't take
much difference in the length.
In the *example* I gave with a 10A load you need 0.3% imbalance to get 30
mA.
Now 0.3% of a cable is not much, its 3mm in a metre or in the example I gave
its 30mm in 10 metres.
A 30A load will only need a difference of 0.1%.
Now do you understand why I said it was luck it didn't trip. Getting it to
within 0.3% without trying isn't the norm.
The Natural Philosopher
Jul 26, 2012, 1:05:17 PM7/26/12
to
are for a single RCD.
If there are two in parallel with neutrals connected downstream of the
RCDs it makes a huge difference depending on where they are connected.
--
To people who know nothing, anything is possible.
To people who know too much, it is a sad fact
that they know how little is really possible -
and how hard it is to achieve it.
John Rumm
Jul 26, 2012, 1:47:59 PM7/26/12
to
cut it at the furthest point away from the entry to where it ever may
need to be. That way you can move circuits and MCBs about at will. It
also leave plenty of wire in the CU to dress things round neatly. If
you cut everything "just right" you inevitably end up with a birds nest
before long.
--
Cheers,
John.
/=================================================================\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\=================================================================/
John Rumm
Jul 26, 2012, 3:37:00 PM7/26/12
to
On 26/07/2012 17:24, dennis@home wrote:
> John claims that the circuit will always balance while I claim its luck
> that it did.
No, if you recall I said "In the situation Adam described, luck had
> John claims that the circuit will always balance while I claim its luck
> that it did.
nothing to do with it."
A statement I still maintain.
> Unless you put some very tight (IMO impossibly tight) tolerances on
> cable lengths it does not balance.
> You may notice that John probably understands this now as he has gone
> very quite.
minutes as usual until you got bored. Since I figure no one would take
much notice anyway, it hardly seemed to matter. However since there
seems to be ongoing interest and no end in site, lets see if we can put
it to bed.
> But just for you I will go through it again..
>
> First the basics as I can't get certain posters to understand them.
>
> Wires have resistance proportional to their length (assuming they are
> normal wires)
>
> A ring circuit has two line wires in parallel feeding the load.
> A ring circuit has two neutral wires in parallel returning from the load.
> All the current has to pass through the load so its effects are a net
> zero on the balance of currents.
magnitude of any imbalance. Unlike "normal" circumstances where an RCDs
imbalance will be typically caused by insulation failure and hence the
leakage very much dependant on circuit voltage, this situation is more
affected by total load current. So the load is very significant factor.
> The current through the two lines will be split in proportion to the
> resistance (length)
> The same applies to the neutral
> Normally the two lines match the two neutrals as they are in the same
> pair of T&E cables
>
> This means you will get exactly the same current through the line and
> neutral in the one cable and the same applies to the other cable.
>
> Now on the faulty circuit these are tied at the ends to two RCDs and if
> they are exactly the same nothing happens, they don't trip.
> Now if you make one core slightly longer anywhere in the circuit the
> current will be imbalanced on one of the cables (the resistances are
> different so the split is different).
>
> Now as you only need an imbalance of 30 mA to trip the RCD it doesn't
> take much difference in the length.
unlikely to happen.
> In the *example* I gave with a 10A load you need 0.3% imbalance to get
> 30 mA.
> Now 0.3% of a cable is not much, its 3mm in a metre or in the example I
> gave its 30mm in 10 metres.
one being that 10m is rather short for a ring circuit[1]. You also seem
to be assuming that the extra length is at one end of the cable. If you
are maintaining that one might normally cut the neutral longer at the CU
(which I suspect is untrue for most electricians) that really ought to
apply to both ends of the wire, since they both terminate at the same CU.
[1] The shortest ring I recall wiring was for a kitchen approx 7' by
11'. That took around 22m of cable (wired in the ceiling void, with each
socket drop taking 2 - 3m of cable).
> A 30A load will only need a difference of 0.1%.
>
> Now do you understand why I said it was luck it didn't trip.
Let's explore why...
If we take the following model of the circuit in question:
http://wiki.diyfaq.org.uk/index.php?title=File:RCDsCrosswired.gif
We have got a 230V supply at 50Hz with a Zs of 0.1 ohms (the supply
impedance has no effect on the result, but may as well be included for
completeness, along with a "book" TN-S value of Ze of 0.8 Ohms).
We have two ring circuits as you might find in a typical medium sized
semi, a downstairs one with a total of 40m of 2.5mm^2 T&E, and a
slightly longer upstairs one with 53m.
If we take the resistance of 2.5mm^2 wire as 7.41 mOhm/m, it means we
have a round trip resistance of about 0.4 ohm for the longer circuit,
and about 0.3 ohms for the shorter.
The situation as described was a new build, second fix wiring complete,
with a different electrician (Adam) commissioned to test the system. The
system is in use by other trades completing the building work. So I have
used in this example a total load of 10A on one circuit, and 2.5A on the
other, representing a mix of work lights, power tools, chargers etc. At
230V these correspond to load impedances of 23 and 92 ohms.
To demonstrate the position of the loads makes no difference, I have
placed the first a quarter of the way along the circuit, and the second
at about a third.
Now Dennis has been maintaining that small variations in conductor
length amounting to just a handful of mm will result in a trip. So for
this example I have included half a metre of extra neutral conductor on
each circuit, which I hope most will agree is excessive (if I were
reading that kind of difference on a loop round trip test, I would be
expecting a fault somewhere!)
At 7.41 mOhm/m that amounts to a total of 3.7 mOhms extra resistance,
which I have split equally between the ends. I have also introduced the
same mismatch on the second circuit.
Now for analysis, I have summed the currents in all of the cable ends
connected to the first RCD. In a perfectly balanced situation these
would obviously be zero. However in this case with an extra metre of
wire floating about in there, they are not.
(I have also summed current flow in the other ends as well however since
these are identical to the first, they line on the graph would be
obscured - so I have included an inversion in there just to make it visible)
So if you have a look at :
http://wiki.diyfaq.org.uk/index.php?title=File:RCDsCrosswiredCurrentBalanc.gif
You can see we only get around 19.5mA RMS of imbalance. Unlikely to trip
the RCD. In reality you would need to add another 10A or so of load to
be sure of doing it with this configuration. You can play with the
parameters to move in and out of trip scenarios, but the basic upshot is
that you are unlikely to trip the circuit with a typical (much smaller)
mismatch of conductor lengths, and with normal loads on it.
Now assuming that I have not made some glaring error in calculation,
this is why I disagree with Dennis' assessment of it just being "luck".
(if anyone wants the LT Spice model to play with, then let me know and I
will stick it on the FAQ web site)
dennis@home
Jul 26, 2012, 4:19:18 PM7/26/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
news:tK2dnQQZFaPHAYzN...@brightview.co.uk...
than screwing a mess of wire into the holes.
Some lines will be short and some long, the same goes for the neutrals.
>
> [1] The shortest ring I recall wiring was for a kitchen approx 7' by 11'.
> That took around 22m of cable (wired in the ceiling void, with each socket
> drop taking 2 - 3m of cable).
>
>> A 30A load will only need a difference of 0.1%.
>>
>> Now do you understand why I said it was luck it didn't trip.
>
> I understand why you said it, I just don't agree with the assessment.
>
> Let's explore why...
>
> If we take the following model of the circuit in question:
>
> http://wiki.diyfaq.org.uk/index.php?title=File:RCDsCrosswired.gif
each wire.
Now do the same with one addition as might be expected rather than the case
where you cut all the wires too long.
A tolerance is where you take the worst case and you have deliberately
chosen the best case.
You have made the odd assumption that when the CU is wired all the neutrals
will be too long and not some distribution based on where the bus bars are
fitted in the CU.
The same goes for the live too.
If you are still convinced that all electricians just lop the cable, strip
it and push it into a termination then I suspect some electricians will
complain.
dennis@home
Jul 26, 2012, 4:22:06 PM7/26/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
news:tK2dnQQZFaPHAYzN...@brightview.co.uk...
>
> No, I had intended to just let you wiggle and change arguments every 5
> minutes as usual until you got bored. Since I figure no one would take
> much notice anyway, it hardly seemed to matter. However since there seems
> to be ongoing interest and no end in site, lets see if we can put it to
> bed.
I haven't changed what I said at all.
> minutes as usual until you got bored. Since I figure no one would take
> much notice anyway, it hardly seemed to matter. However since there seems
> to be ongoing interest and no end in site, lets see if we can put it to
> bed.
I was correct when I started and I am correct now, despite your optimistic
spice models.
Mike Tomlinson
Jul 26, 2012, 4:18:43 PM7/26/12
to
En el art�culo <tK2dnQQZFaPHAYzN...@brightview.co.uk>, John
Rumm <see.my.s...@nowhere.null> escribi�:
>> Now as you only need an imbalance of 30 mA to trip the RCD it doesn't
>> take much difference in the length.
>
>In a real world situation it actually takes quite a bit. Enough that its
>unlikely to happen.
It would also depend on the response curve of the RCD - whether it's
type B, C, D etc.
--
(\_/)
(='.'=)
(")_(")
Rumm <see.my.s...@nowhere.null> escribi�:
>> Now as you only need an imbalance of 30 mA to trip the RCD it doesn't
>> take much difference in the length.
>
>In a real world situation it actually takes quite a bit. Enough that its
>unlikely to happen.
type B, C, D etc.
--
(\_/)
(='.'=)
(")_(")
tony sayer
Jul 26, 2012, 4:33:43 PM7/26/12
to
>Now the circuit would still be there current would flow but what's now
>happening to the RCD's?.
>
>One is seeing all its live current going out the other is seeing all its
>neutral current but no live to balance it hence a rather rapid trip;!..
>
>
>
>
>
>The cable has resistance due to its length and cross section all along
>its length.
>
>That is the other factor. By altering the points in the circuit where
>you have the load connected to them the resistance's will become
>different.
>
>Now can you not see if the cable lengths and diameter alter then the
>resistance's and currents flowing will vary?.
>
>And that is dependant on the size of the cables used in the example.
>Remember this also alters with length.
>
>An interesting excersise for you to do is take a typical cable say a
>lump of 2.5 T&E and tell us if its 10 metres long how much we'd have the
>alter the load taken point to achieve a current unbalance of greater
>than 25 ma.
>
>Now do that with say .1 mm cable and then say 16 mm cross section and
>now do the lengths change from the centre tapping point?..
>
>Lets see your results...
>
Got any yet Dennis?...
>happening to the RCD's?.
>
>One is seeing all its live current going out the other is seeing all its
>neutral current but no live to balance it hence a rather rapid trip;!..
>
>
>
>
>
>The cable has resistance due to its length and cross section all along
>its length.
>
>That is the other factor. By altering the points in the circuit where
>you have the load connected to them the resistance's will become
>different.
>
>Now can you not see if the cable lengths and diameter alter then the
>resistance's and currents flowing will vary?.
>
>And that is dependant on the size of the cables used in the example.
>Remember this also alters with length.
>
>An interesting excersise for you to do is take a typical cable say a
>lump of 2.5 T&E and tell us if its 10 metres long how much we'd have the
>alter the load taken point to achieve a current unbalance of greater
>than 25 ma.
>
>Now do that with say .1 mm cable and then say 16 mm cross section and
>now do the lengths change from the centre tapping point?..
>
>Lets see your results...
>
--
Tony Sayer
dennis@home
Jul 26, 2012, 5:06:05 PM7/26/12
to
news:3UD+77Cn...@bancom.co.uk...
You have had all the results you need.
The 0.1 and 16 mm cable are irrelevant to the discussion.
All they do is change the resistance per metre and its already been shown
that makes no difference to the imbalance.
The 0.1 and 16 mm cable are irrelevant to the discussion.
All they do is change the resistance per metre and its already been shown
that makes no difference to the imbalance.
John Rumm
Jul 26, 2012, 6:01:03 PM7/26/12
to
Generally the only two parameters that vary on RCDs[1] are the trip
threshold, and the delay time (if any). B, C, & D types are the fault
current (i.e. magnetic) trip response on MCBs.
[1] RCBOs Obviously have a MCB style trip curve - that that only comes
into play in fault current scenarios.
tony sayer
Jul 26, 2012, 6:38:37 PM7/26/12
to
In article <jusbft$b2e$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
>
>
Which is an excellent demonstration, if one was needed, that you still
do not understand the principles of the operation of the RCD units in
the circuit configuration under discussion....
--
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
>
>
do not understand the principles of the operation of the RCD units in
the circuit configuration under discussion....
--
Tony Sayer
dennis@home
Jul 26, 2012, 7:00:35 PM7/26/12
to
news:eyRUSrDt...@bancom.co.uk...
I fully understand them, it appears you do not.
take moving the load around the ring..
If you put the load close to an end, say 1m from the CU and 30m from the CU.
Now 97% of the current flows down the short bit.
So any tolerance in the length of that bit of wire has a bigger effect on
the current imbalance than on the longer length.
It just makes it more likely to trip than with it in the centre.
As for the .1 and 16 mm cables that is just a red herring and has nothing to
do with this at all.
Are you trying to cover up something like John appeared to do with his
carefully crafted spice model?
It took all of ten seconds to see what he had done BTW.
take moving the load around the ring..
If you put the load close to an end, say 1m from the CU and 30m from the CU.
Now 97% of the current flows down the short bit.
So any tolerance in the length of that bit of wire has a bigger effect on
the current imbalance than on the longer length.
It just makes it more likely to trip than with it in the centre.
As for the .1 and 16 mm cables that is just a red herring and has nothing to
do with this at all.
Are you trying to cover up something like John appeared to do with his
carefully crafted spice model?
It took all of ten seconds to see what he had done BTW.
John Rumm
Jul 26, 2012, 7:30:49 PM7/26/12
to
On 26/07/2012 21:19, dennis@home wrote:
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
>> There are a number of aspects you are not considering here. The
>> obvious one being that 10m is rather short for a ring circuit[1]. You
>> also seem to be assuming that the extra length is at one end of the
>> cable. If you are maintaining that one might normally cut the neutral
>> longer at the CU (which I suspect is untrue for most electricians)
>> that really ought to apply to both ends of the wire, since they both
>> terminate at the same CU.
>
> I am saying they are likely to cut the cores to give a neat finish
> rather than screwing a mess of wire into the holes.
Not sure quite what holes you are screwing wire into...
>> obvious one being that 10m is rather short for a ring circuit[1]. You
>> also seem to be assuming that the extra length is at one end of the
>> cable. If you are maintaining that one might normally cut the neutral
>> longer at the CU (which I suspect is untrue for most electricians)
>> that really ought to apply to both ends of the wire, since they both
>> terminate at the same CU.
>
> I am saying they are likely to cut the cores to give a neat finish
> rather than screwing a mess of wire into the holes.
> Some lines will be short and some long, the same goes for the neutrals.
nicer to cut all cables equal so they can reach the furthest way
(without stretching wires across the box in a straight line) so you can
dress everything consistently and neatly. If you then need to change a
circuit from one side of the CU to the other, move MCB, or switch out to
a RCBO you have the wire to do it all neatly.
>> [1] The shortest ring I recall wiring was for a kitchen approx 7' by
>> 11'. That took around 22m of cable (wired in the ceiling void, with
>> each socket drop taking 2 - 3m of cable).
>>
>>> A 30A load will only need a difference of 0.1%.
>>>
>>> Now do you understand why I said it was luck it didn't trip.
>>
>> I understand why you said it, I just don't agree with the assessment.
>>
>> Let's explore why...
>>
>> If we take the following model of the circuit in question:
>>
>> http://wiki.diyfaq.org.uk/index.php?title=File:RCDsCrosswired.gif
>
> Well I am not surprised that balances, you have added the same
> resistance to each wire.
realistic) but it will still not trip.
You were the one suggesting that the neutrals would be longer. Is there
any particular reason that you would only do this for one of them now?
> Now do the same with one addition as might be expected rather than the
> case where you cut all the wires too long.
> A tolerance is where you take the worst case and you have deliberately
> chosen the best case.
length.
I also chose typical real world round trip circuit impedances (most
rings in domestic situations being between 0.4 and 0.6 ohms - sometimes
a bit less on single room circuits like kitchens, but even then, 0.2 or
lower would normally be a fault).
> You have made the odd assumption that when the CU is wired all the
> neutrals will be too long and not some distribution based on where the
> bus bars are fitted in the CU.
no more than about 50 - 75 mm long. Even if you trim every wire to
length you are talking about resistance differences of under 0.0005 ohms
which will make naff all difference.
Also consider that we are talking about a ring circuit here, where good
practice would suggest both neutrals go into the same bus bar terminal,
and hence are likely to be cut identically even if the installer has
mixed up the cables.
I have allowed you half a metre of extra wire!
> The same goes for the live too.
tolerances even further.
> If you are still convinced that all electricians just lop the cable,
> strip it and push it into a termination then I suspect some electricians
> will complain.
Why not ask some of them how to wire a CU neatly? They may even have
photo's for you.
tony sayer
Jul 26, 2012, 7:32:06 PM7/26/12
to
>>
>> Which is an excellent demonstration, if one was needed, that you still
>> do not understand the principles of the operation of the RCD units in
>> the circuit configuration under discussion....
>
>I fully understand them, it appears you do not.
>
>take moving the load around the ring..
>
>If you put the load close to an end, say 1m from the CU and 30m from the CU.
>Now 97% of the current flows down the short bit.
>So any tolerance in the length of that bit of wire has a bigger effect on
>the current imbalance than on the longer length.
>It just makes it more likely to trip than with it in the centre.
Are you understand what we were discussing at all?..
>> Which is an excellent demonstration, if one was needed, that you still
>> do not understand the principles of the operation of the RCD units in
>> the circuit configuration under discussion....
>
>I fully understand them, it appears you do not.
>
>take moving the load around the ring..
>
>If you put the load close to an end, say 1m from the CU and 30m from the CU.
>Now 97% of the current flows down the short bit.
>So any tolerance in the length of that bit of wire has a bigger effect on
>the current imbalance than on the longer length.
>It just makes it more likely to trip than with it in the centre.
I am discussing this rather unusual configuration with TWO RCD's..
R1 R2
L in -- L ----------\/\/\/\/----Lo---\/\/\/\/\------- L --- L in
RCD RCD
N in -- N ----------\/\/\/\/-----Lo---\/\/\/\/\- -------N --- N in
R1a R2a
holds true...
resistors represents "total" circuit conductor resistance
R1
L in -- L ----------\/\/\/\/---------->
RCD Load across these two points
N in -- N ----------\/\/\/\/----------->
R2
The load can be -anywhere- in that circuit it matters not where its
taken from R1 and R2 can be whatever they are within current capacity
and voltage drop i.e. as long or as short a lump of wire you want..
The current flowing, whatever that is, flows in and out of the single
RCD it can only do that, it has nowhere else to go!..
You do see and understand the difference don't you?..
>
>As for the .1 and 16 mm cables that is just a red herring and has nothing to
>do with this at all.
>Are you trying to cover up something like John appeared to do with his
>carefully crafted spice model?
>It took all of ten seconds to see what he had done BTW.
>
--
Tony Sayer
dennis@home
Jul 27, 2012, 3:04:49 AM7/27/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
They will vary because of the CU, they will vary if they aren't cut at
exactly right angles. they will vary if they are not inserted exactly the
same in each socket, they will vary for many reasons. Electrical
installation is not precision engineering and the tolerances on lengths is
not zero as you claim.
>
>> Now do the same with one addition as might be expected rather than the
>> case where you cut all the wires too long.
>
> Sorry, don't understand that.
Well that's why you keep getting the wrong answer.
>
>> A tolerance is where you take the worst case and you have deliberately
>> chosen the best case.
>
> I chose a relatively pessimistic case - with a significant imbalance in
> length.
>
> I also chose typical real world round trip circuit impedances (most rings
> in domestic situations being between 0.4 and 0.6 ohms - sometimes a bit
> less on single room circuits like kitchens, but even then, 0.2 or lower
> would normally be a fault).
doesn't apply are zero and infinity.
>
>> You have made the odd assumption that when the CU is wired all the
>> neutrals will be too long and not some distribution based on where the
>> bus bars are fitted in the CU.
>
> The neutral bus bars are almost always adjacent anyway, and in many CUs no
> more than about 50 - 75 mm long. Even if you trim every wire to length you
> are talking about resistance differences of under 0.0005 ohms which will
> make naff all difference.
how about the live bars?
do the cables all enter the same hole?
BTW if one of the bars is 0.0005 ohms and one is 0.00025 ohms and they are
in parallel they will split current in the same proportions as a 10 ohm and
20 ohm one. Not that it matters in this case. Just because something is of
low ohms doesn't mean it doesn't have a significant effect.
>
> Also consider that we are talking about a ring circuit here, where good
> practice would suggest both neutrals go into the same bus bar terminal,
> and hence are likely to be cut identically even if the installer has mixed
> up the cables.
>
> I have allowed you half a metre of extra wire!
well will not have an effect and has nothing to do with wiring tolerances.
>
>> The same goes for the live too.
>
> If you start making some of those longer then you actually relax the
> tolerances even further.
A tolerances isn't something you get to choose in real life, its what
happens when real people do things.
>
>> If you are still convinced that all electricians just lop the cable,
>> strip it and push it into a termination then I suspect some electricians
>> will complain.
>
> Not sure how you drew that conclusion.
>
> Why not ask some of them how to wire a CU neatly? They may even have
> photo's for you.
dennis@home
Jul 27, 2012, 3:06:42 AM7/27/12
to
news:6CXd+iE2...@bancom.co.uk...
I have said applies to that and not to the imaginary circuit you have just
posted.
>>>
>>> Which is an excellent demonstration, if one was needed, that you still
>>> do not understand the principles of the operation of the RCD units in
>>> the circuit configuration under discussion....
>>
>>I fully understand them, it appears you do not.
>>
>>take moving the load around the ring..
>>
>>If you put the load close to an end, say 1m from the CU and 30m from the
>>CU.
>>Now 97% of the current flows down the short bit.
>>So any tolerance in the length of that bit of wire has a bigger effect on
>>the current imbalance than on the longer length.
>>It just makes it more likely to trip than with it in the centre.
>
>
> Are you understand what we were discussing at all?..
>
> I am discussing this rather unusual configuration with TWO RCD's..
>
> R1 R2
> L in -- L ----------\/\/\/\/----Lo---\/\/\/\/\------- L --- L in
> RCD RCD
> N in -- N ----------\/\/\/\/-----Lo---\/\/\/\/\- -------N --- N in
> R1a R2a
>
>
>
> You it seems are thinking of this with ONE where what you are stating
> holds true...
You are being stupid, I drew the circuit with the two RCDs in and everything
>>> Which is an excellent demonstration, if one was needed, that you still
>>> do not understand the principles of the operation of the RCD units in
>>> the circuit configuration under discussion....
>>
>>I fully understand them, it appears you do not.
>>
>>take moving the load around the ring..
>>
>>If you put the load close to an end, say 1m from the CU and 30m from the
>>CU.
>>Now 97% of the current flows down the short bit.
>>So any tolerance in the length of that bit of wire has a bigger effect on
>>the current imbalance than on the longer length.
>>It just makes it more likely to trip than with it in the centre.
>
>
> Are you understand what we were discussing at all?..
>
> I am discussing this rather unusual configuration with TWO RCD's..
>
> R1 R2
> L in -- L ----------\/\/\/\/----Lo---\/\/\/\/\------- L --- L in
> RCD RCD
> N in -- N ----------\/\/\/\/-----Lo---\/\/\/\/\- -------N --- N in
> R1a R2a
>
>
>
> You it seems are thinking of this with ONE where what you are stating
> holds true...
I have said applies to that and not to the imaginary circuit you have just
posted.
Mike Tomlinson
Jul 27, 2012, 3:07:36 AM7/27/12
to
En el art�culo <p86dndfyjN2GI4zN...@brightview.co.uk>, John
Rumm <see.my.s...@nowhere.null> escribi�:
>Are you sure you are not confusing MCB response curves with RCD types?
Yes, I am. Thanks for th4e correction.
Rumm <see.my.s...@nowhere.null> escribi�:
>Are you sure you are not confusing MCB response curves with RCD types?
tony sayer
Jul 27, 2012, 3:38:22 AM7/27/12
to
In article <jutem3$3ud$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
>
>
Well make your mind up!
I am not being stupid. I am trying to take this argument seriously.
Now as I've drawn that quite clearly can you draw out what you are on
about again so that it renders properly, clearly, and theres _no_
ambiguity.
Or describe it as best you can and I'll draw what It seems you mean and
imply as from all you have said in recent posts implies that you have
more than the one circuit into a single RCD not more RCD's than the one.
Now is this the case or not?...
--
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
>
>
I am not being stupid. I am trying to take this argument seriously.
Now as I've drawn that quite clearly can you draw out what you are on
about again so that it renders properly, clearly, and theres _no_
ambiguity.
Or describe it as best you can and I'll draw what It seems you mean and
imply as from all you have said in recent posts implies that you have
more than the one circuit into a single RCD not more RCD's than the one.
Now is this the case or not?...
--
Tony Sayer
dennis@home
Jul 27, 2012, 7:59:24 AM7/27/12
to
news:k7xKQ7Eu...@bancom.co.uk...
Well as the OP said two RCDs,
the drawing I posted said two RCDs,
and its impossible to get an imbalance with one RCD no matter how many two
wire circuits you hang off it you can draw your own conclusions!
the drawing I posted said two RCDs,
and its impossible to get an imbalance with one RCD no matter how many two
wire circuits you hang off it you can draw your own conclusions!
John Rumm
Jul 27, 2012, 8:07:50 AM7/27/12
to
On 27/07/2012 00:32, tony sayer wrote:
>> As for the .1 and 16 mm cables that is just a red herring and has nothing to
>> do with this at all.
>> Are you trying to cover up something like John appeared to do with his
>> carefully crafted spice model?
>
> No, see above.
>
>> It took all of ten seconds to see what he had done BTW.
>
> Which was what then?..
I was being devious you see...
>> As for the .1 and 16 mm cables that is just a red herring and has nothing to
>> do with this at all.
>> Are you trying to cover up something like John appeared to do with his
>> carefully crafted spice model?
>
> No, see above.
>
>> It took all of ten seconds to see what he had done BTW.
>
> Which was what then?..
I made it too close to real life with round trip resistances like those
you find on real circuits, and added extra wire length to *both* ends of
one of the wires to create the imbalance. Apparently one is supposed to
cut either the return or the feed longer in the CU, but only on the left
hand end of the wire to get the "normal" result. So alas my model just
behaved the same way the real life example did, which is obviously
undesirable.
Apparently if you cut all the cables in a cack handed way, and squint a
bit, you can make a case for the setup unbalancing enough to trip the
RCDs with only 0.3% extra wire... I must be doing that wrong as well
though, because adding the pre-requisite 16cm of wire to my longer
circuit, and the extra 12cm to the short one, then tweaking the loads
down to 7 ohms to put both circuits into slight overload on their 32A
MCBs, we still only get 27.6mA RMS of imbalance which might not trip it.
Perhaps my circuits were too long. Ideally they should only be long
enough for a lap of a shoe box - then all those extra inches you need on
one end of one of the wires in the CU will make a much bigger
proportional difference.
Perhaps its non representative since I placed my averaged loads at the
25% and 33% positions on the rings. Obviously with all those workmen
having their individual loads spread out all over the place, it would
make more sense to represent these as an averaged load at the 3% point
(this is also better apparently, since it makes the ring circuit behave
much more like a radial, and those are Dennis approved, so must be better).
Adam Funk
Jul 27, 2012, 8:05:27 AM7/27/12
to
On 2012-07-26, John Rumm wrote:
> On 26/07/2012 16:39, A.Lee wrote:
>> Adam Funk <a24...@ducksburg.com> wrote:
>>
>>>> In your case, if the cable had such an effect, every shower in the
>>>> Country would trip the RCD when it was used, as the cable always has a
>>>> difference in length between L & N, typically 100 - 150mm, add the ~35A
>>>> load, and your idea would cause such an imbalance that it would not be
>>>> usable.
>>>
>>> Where is the extra 100 to 150 mm of one wire? (I installed an
>>> electric shower some years ago & don't recall that kind of
>>> difference, but maybe I'm forgetting something.)
>>
>> At the CU end. If the cable enters the CU at one end, the neutral bar
>> may be next to the entry, the line terminal at the other end. The only
>> sure way is to have the cable enter the CU at mid point between the line
>> terminal and neutral bar, which in real life, is not normally possible.
OK, but that's not *every* shower in the country, just the ones
> On 26/07/2012 16:39, A.Lee wrote:
>> Adam Funk <a24...@ducksburg.com> wrote:
>>
>>>> In your case, if the cable had such an effect, every shower in the
>>>> Country would trip the RCD when it was used, as the cable always has a
>>>> difference in length between L & N, typically 100 - 150mm, add the ~35A
>>>> load, and your idea would cause such an imbalance that it would not be
>>>> usable.
>>>
>>> Where is the extra 100 to 150 mm of one wire? (I installed an
>>> electric shower some years ago & don't recall that kind of
>>> difference, but maybe I'm forgetting something.)
>>
>> At the CU end. If the cable enters the CU at one end, the neutral bar
>> may be next to the entry, the line terminal at the other end. The only
>> sure way is to have the cable enter the CU at mid point between the line
>> terminal and neutral bar, which in real life, is not normally possible.
installed by people who like to keep the wires short.
> It makes more sense to take the cable from wherever it enters and then
> cut it at the furthest point away from the entry to where it ever may
> need to be. That way you can move circuits and MCBs about at will. It
> also leave plenty of wire in the CU to dress things round neatly. If
> you cut everything "just right" you inevitably end up with a birds nest
> before long.
places, of course) for that reason. It's been a while, but I'm pretty
sure there's some slack on the shower wires in my CU.
dennis@home
Jul 27, 2012, 8:24:40 AM7/27/12
to
"John Rumm" <see.my.s...@nowhere.null> wrote in message
> I was being devious you see...
>
> I made it too close to real life with round trip resistances like those
> you find on real circuits, and added extra wire length to *both* ends of
> one of the wires to create the imbalance. Apparently one is supposed to
> cut either the return or the feed longer in the CU, but only on the left
> hand end of the wire to get the "normal" result. So alas my model just
> behaved the same way the real life example did, which is obviously
> undesirable.
>
> Apparently if you cut all the cables in a cack handed way, and squint a
> bit, you can make a case for the setup unbalancing enough to trip the RCDs
> with only 0.3% extra wire... I must be doing that wrong as well though,
> because adding the pre-requisite 16cm of wire to my longer circuit, and
> the extra 12cm to the short one, then tweaking the loads down to 7 ohms to
> put both circuits into slight overload on their 32A MCBs, we still only
> get 27.6mA RMS of imbalance which might not trip it. Perhaps my circuits
> were too long. Ideally they should only be long enough for a lap of a shoe
> box - then all those extra inches you need on one end of one of the wires
> in the CU will make a much bigger proportional difference.
Tolerances work both ways, some wires will be short some long.
The worst case in this being one short and one long, not both long.
Try adding 16 cm to the short one and nothing to the long one and report
back.
>
> Perhaps its non representative since I placed my averaged loads at the 25%
> and 33% positions on the rings. Obviously with all those workmen having
> their individual loads spread out all over the place, it would make more
> sense to represent these as an averaged load at the 3% point (this is also
> better apparently, since it makes the ring circuit behave much more like a
> radial, and those are Dennis approved, so must be better).
tony sayer
Jul 28, 2012, 7:12:55 AM7/28/12
to
In article <65Kdne2HEKQOGY_N...@brightview.co.uk>, John
Rumm <see.my.s...@nowhere.null> scribeth thus
I still think that Dennis is seeing this as something different;!...
--
Tony Sayer
Rumm <see.my.s...@nowhere.null> scribeth thus
--
Tony Sayer
dennis@home
Jul 28, 2012, 10:49:56 AM7/28/12
to
news:Mw3wyeJ3...@bancom.co.uk...
> I still think that Dennis is seeing this as something different;!...
I think you are trying to avoid the truth.
You may notice that the figures now backup what I said in the first place
and not the perfect balance John suggested.
All that now remains is just how much tolerance is there on the circuit,
which I have no doubt will be a another source of contention.
> I still think that Dennis is seeing this as something different;!...
You may notice that the figures now backup what I said in the first place
and not the perfect balance John suggested.
All that now remains is just how much tolerance is there on the circuit,
which I have no doubt will be a another source of contention.
tony sayer
Jul 28, 2012, 1:33:36 PM7/28/12
to
In article <jv0u6l$hiu$1...@news.albasani.net>, dennis@home
<den...@killspam.kicks-ass.net> scribeth thus
>
>
>
>You may notice that the figures now backup what I said in the first place
>and not the perfect balance John suggested.
>All that now remains is just how much tolerance is there on the circuit,
>which I have no doubt will be a another source of contention.
>
I simply asked you to draw it out _Clearly_ as the original didn't
render that well. It certainly appeared that the RCD's weren't
designated as they ought to have been.
Will you have another go then we can see it clearly for the avoidance of
any doubt.
Not that much to ask surely ?..
--
Tony Sayer
<den...@killspam.kicks-ass.net> scribeth thus
>
>
>"tony sayer" <to...@bancom.co.uk> wrote in message
>news:Mw3wyeJ3...@bancom.co.uk...
>
>
>> I still think that Dennis is seeing this as something different;!...
>
>I think you are trying to avoid the truth.
Not at all..
>news:Mw3wyeJ3...@bancom.co.uk...
>
>
>> I still think that Dennis is seeing this as something different;!...
>
>I think you are trying to avoid the truth.
>
>You may notice that the figures now backup what I said in the first place
>and not the perfect balance John suggested.
>All that now remains is just how much tolerance is there on the circuit,
>which I have no doubt will be a another source of contention.
>
render that well. It certainly appeared that the RCD's weren't
designated as they ought to have been.
Will you have another go then we can see it clearly for the avoidance of
any doubt.
Not that much to ask surely ?..
--
Tony Sayer
Mike Tomlinson
Jul 28, 2012, 3:39:23 PM7/28/12
to
En el art�culo <x$nPxeMwJ...@bancom.co.uk>, tony sayer
<to...@bancom.co.uk> escribi�:
>Not that much to ask surely ?..
What you're experiencing now is classic DenWriggle (c) (r) (tm).
Mr Fisher, when cornered in an argument, is like a fish out of water -
slippery and hard to pin down.
<to...@bancom.co.uk> escribi�:
>Not that much to ask surely ?..
Mr Fisher, when cornered in an argument, is like a fish out of water -
slippery and hard to pin down.
John Rumm
Jul 28, 2012, 8:45:08 PM7/28/12
to
On 27/07/2012 13:24, dennis@home wrote:
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:65Kdne2HEKQOGY_N...@brightview.co.uk...
>
>> I was being devious you see...
>
> You still are.
>
>>
>> I made it too close to real life with round trip resistances like
>> those you find on real circuits, and added extra wire length to *both*
>> ends of one of the wires to create the imbalance. Apparently one is
>> supposed to cut either the return or the feed longer in the CU, but
>> only on the left hand end of the wire to get the "normal" result. So
>> alas my model just behaved the same way the real life example did,
>> which is obviously undesirable.
>>
>> Apparently if you cut all the cables in a cack handed way, and squint
>> a bit, you can make a case for the setup unbalancing enough to trip
>> the RCDs with only 0.3% extra wire... I must be doing that wrong as
>> well though, because adding the pre-requisite 16cm of wire to my
>> longer circuit, and the extra 12cm to the short one, then tweaking the
>> loads down to 7 ohms to put both circuits into slight overload on
>> their 32A MCBs, we still only get 27.6mA RMS of imbalance which might
>> not trip it. Perhaps my circuits were too long. Ideally they should
>> only be long enough for a lap of a shoe box - then all those extra
>> inches you need on one end of one of the wires in the CU will make a
>> much bigger proportional difference.
>
> Why add it to both wires?
The suggestion seemed to be that one would need extra wire in the CU to
>
>
> "John Rumm" <see.my.s...@nowhere.null> wrote in message
> news:65Kdne2HEKQOGY_N...@brightview.co.uk...
>
>> I was being devious you see...
>
> You still are.
>
>>
>> I made it too close to real life with round trip resistances like
>> those you find on real circuits, and added extra wire length to *both*
>> ends of one of the wires to create the imbalance. Apparently one is
>> supposed to cut either the return or the feed longer in the CU, but
>> only on the left hand end of the wire to get the "normal" result. So
>> alas my model just behaved the same way the real life example did,
>> which is obviously undesirable.
>>
>> Apparently if you cut all the cables in a cack handed way, and squint
>> a bit, you can make a case for the setup unbalancing enough to trip
>> the RCDs with only 0.3% extra wire... I must be doing that wrong as
>> well though, because adding the pre-requisite 16cm of wire to my
>> longer circuit, and the extra 12cm to the short one, then tweaking the
>> loads down to 7 ohms to put both circuits into slight overload on
>> their 32A MCBs, we still only get 27.6mA RMS of imbalance which might
>> not trip it. Perhaps my circuits were too long. Ideally they should
>> only be long enough for a lap of a shoe box - then all those extra
>> inches you need on one end of one of the wires in the CU will make a
>> much bigger proportional difference.
>
> Why add it to both wires?
reach the neutral bus bars. If so, it seems reasonable to assume that if
this is the case, then it must apply to both ends of the cable.
> Tolerances work both ways, some wires will be short some long.
> The worst case in this being one short and one long, not both long.
> Try adding 16 cm to the short one and nothing to the long one and report
> back.
neutral of the most highly loaded circuit, and add it all at one end to
the shortest leg of the circuit so that it has the greatest effect.
Imbalance is about 15.4mA RMS
If I sprinkle little random additions around the other wires, I tend to
see less imbalance rather than more. Now that is not to say you can't
get trip conditions with the right combination of extra length in the
"right/wrong" places, and high enough circuit currents, but you have to
contrive rather untypical installations to do it. For the situation that
was originally described, its far more probable that you won't get
enough imbalance to cause a trip.
Still, common sense and basic physics should tell you that, however
since you seem relish doing neither, you can stick to calling it luck.
>> Perhaps its non representative since I placed my averaged loads at the
>> 25% and 33% positions on the rings. Obviously with all those workmen
>> having their individual loads spread out all over the place, it would
>> make more sense to represent these as an averaged load at the 3% point
>> (this is also better apparently, since it makes the ring circuit
>> behave much more like a radial, and those are Dennis approved, so must
>> be better).
>
> What have radials got to do with this?
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