Upstream RCD Tripping?

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davetheglitz

Electrician
Joined
Mar 18, 2008
Messages
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Location
Saltash, Cornwall
Sorry Guys - this goes on a bit!

Background

Called out today after Western Power had disconnected my elderly customers ring main after they lost power. Apparantly the 30mA RCD in the house had gone - but so had the supply to the meter. After a lot of playing around they disconnected the ring main and told the customer it was down to them. No paperwork was left - so I'm only going on hearsay concerning the fault.

On checking - no fault on the ring main. >3M insulation. Connected up - Zs at sockets around 60 ohms (TT) and no trip at 15mA - correct at 30mA.

Overall Zs all circuits in > 2M.

No outside feeds from the ring main, no signs of damp.

The house is remote - it didn't have mains electricity until 1994 and it had to come in around a mile to a local transformer.

I'm at a loss to explain the trip - unless it's something like the fridge going into defrost causing the trip. However this doesn't explain the loss of supply to the meter.

Here is the Question (at last!)

I've been told that it is feasible for a tripping fault to be caused by a fault 'upstream' from the RCD. However I can't get my head around how this could work. Has anyone found anything similar - and if so has anyone got any explanations!

Thanks

 
voltage drop mate, if the trannie aint giving you enough voltage then the current rises (as we all know)

but inherently so will the leakage that already exists, or so I would believe. ?:|

or could we have dropped a N somewhere in the supply?

anyone else with any thoughts on this?

 
Hi Albert,

I don't follow this as the resistance is the constant not power- so voltage drop would cause current (and leakage) to drop.

If the dropped neutral was between the transformer supply (I presume this would have some RCD device) and the household RCD would it be possible for a transient due to the inductance of the transformer to provide a voltage spike sufficient to allow 30mA to go to earththrough 2M?

Damn - just missed Tukey going one up against Germany!

Thanks fot the thoughts.

See you all later!

Dave

 
Hi Martyn,

I = V/R so if R is constant I is proportional to V - hence decrease in V gives a decrease in I - not an increase!

Thanks for the pdf - can't bring myself to read it yet - but if I find the salient bit I will post it here!

Cheers

Dave

PS - did you get that compressor starter job?

 
Hi Martyn,I = V/R so if R is constant I is proportional to V - hence decrease in V gives a decrease in R - not an increase!

Thanks for the pdf - can't bring myself to read it yet - but if I find the salient bit I will post it here!

Cheers

Dave

PS - did you get that compressor starter job?
dont think you quite grasped this yet,

decrease in V=increase in I,

R is the constant,(by your own admission)

look at the 1st year triangle

voltage drop WILL ALWAYS cause current to rise,

regardless of the circumstances.

 
Hi Martyn,I = V/R so if R is constant I is proportional to V - hence decrease in V gives a decrease in R - not an increase!

Thanks for the pdf - can't bring myself to read it yet - but if I find the salient bit I will post it here!

Cheers

Dave

PS - did you get that compressor starter job?
Last point first - yes, but not until next month! Thanks for info:x

Now: as regard I=V/R.

I`ve drunk too much beer. It`s my wife`s fault (she`s gone to bed, so I can type freely:^ O)

My previous post was incorrect. I typed withoput thinking - sorry :8}

There are a couple of sections in there r.e. faults upstream on RCD`s.

Wait 1........

http://www.talk.electricianforum.co.uk/showpost.php?p=5629&postcount=9

That was the first of a few mind-numbing posts on the general subject.

(There`s more if you want it????)

 
dont think you quite grasped this yet,decrease in V=increase in I,

R is the constant,(by your own admission)

look at the 1st year triangle

voltage drop WILL ALWAYS cause current to rise,

regardless of the circumstances.
Hi Albert - think it through again. By your arguement if the voltage dropped to zero, the current would go to infinity. Don't sound right when you think of it like that!

I = V/R. if R is constant at 1 ohm I = V ->so I is proportional to V. They go up and down together.

Cheers

Dave

 
Hi Albert - think it through again. By your arguement if the voltage dropped to zero, the current would go to infinity. Don't sound right when you think of it like that!I = V/R. if R is constant at 1 ohm I = V ->so I is proportional to V. They go up and down together.

Cheers

Dave
inversely,

obviously I can only go up if R is constant, if there is no V then R is neglible.

why do I come on here to teach 1st month basics to someone that already knows everything?????

Im only trying to help and if you cant grasp the basics then perhaps you are doing something that is beyond your capabilities and knowledge.....

 
the equasion is actually V=I/Rwork the rest out for yourself.
Sorry Albert much as you protest - it isn't. V = I x R. Often you see this as a triangle with V at the top. Use the cover up rule to get the unknown.

Consider a resistor with current flowing through it. The voltage is what drives the current through the resistor. If the voltage increases - then so does the current.

Do the water analogy. A resistor is like a restriction in a pipe. Current is like the flow rate and pressure is like the voltage. Increase the pressure(voltage) and the flow (current) will increase! Directly proportional.

Perhaps you should review your first month basics - I don't know everything which is why I ask questions - but I've got ONC, HNC and an Hons Degree in Electrical and Electronic Engineering and am an ex Science Teacher so I feel qualified to deliberate on ohms law. More importantly the internet backs me up! See the link below.

http://www.kpsec.freeuk.com/ohmslaw.htm

You've got a lot of experience, a lot of know how and a lot of good sense - and I've learnt a lot from some of your posts. However if you are going to come on here to teach first month basics to someone who doesn't have the ability to comprehend please have the courtesy to check that you are right in the first place! :)

Remember - this ain't the SF forum! :)

I suspect you are thinking in terms of transformer secondary windings. In this case power is the constant - so for a lower number of turns you get lower voltage and higher current capability.

Now I've put this s****y post on I discovered a mistake in a previous post where I typed an R instead of an I. May have confused matters!! Sorry! X( (thats an embarrassed smiley!)

 
Sorry Albert much as you protest - it isn't. V = I x R. Often you see this as a triangle with V at the top. Use the cover up rule to get the unknown. Consider a resistor with current flowing through it. The voltage is what drives the current through the resistor. If the voltage increases - then so does the current.

Do the water analogy. A resistor is like a restriction in a pipe. Current is like the flow rate and pressure is like the voltage. Increase the pressure(voltage) and the flow (current) will increase! Directly proportional.

Perhaps you should review your first month basics - I don't know everything which is why I ask questions - but I've got ONC, HNC and an Hons Degree in Electrical and Electronic Engineering and am an ex Science Teacher so I feel qualified to deliberate on ohms law. More importantly the internet backs me up! See the link below.

http://www.kpsec.freeuk.com/ohmslaw.htm

You've got a lot of experience, a lot of know how and a lot of good sense - and I've learnt a lot from some of your posts. However if you are going to come on here to teach first month basics to someone who doesn't have the ability to comprehend please have the courtesy to check that you are right in the first place! :)

Remember - this ain't the SF forum! :)

I suspect you are thinking in terms of transformer secondary windings. In this case power is the constant - so for a lower number of turns you get lower voltage and higher current capability.

Now I've put this s****y post on I discovered a mistake in a previous post where I typed an R instead of an I. May have confused matters!! Sorry! X( (thats an embarrassed smiley!)
it is, typo TBH, sorry tho

I suspect you may be right, (I'll come to that in a mo)

tho on that note Im still not convinced, cos if V drops, then I must increase to provide the same flow to R. considering R to be the load, ie, the installation.

if one variable is constant, then for each increase in one the other must decrease?

normal conditions

V = 230

so in a dimmer if R increases then I decreases... ?

Im lost now.....

back to my other off subject point, seeing as Ive got this one all f ooked up, :eek:

why wont a normal generator run a welder?

is it due to the fact a genny makes a flat wave and not a sine wave?

or because when you strike the welder the current draw rises so fast the genny cant keep up(almost stalling) and thus the voltage drops to near 0.

therefore in trying to figure this out I may have misunderstood some variables.

 
STEPTOE

If the POWER REMAINS THE SAME, then yes current will increase for a lower voltage

eg a 110v 600w drill will draw twice as much current as a 240v 600w drill. If the voltage on the mains drops, then so does the power and the 240v drill will draw less current and not have the same power. Can you grasp that?

 
Just spotted the last bit STEPTOE

My dad is a shipbuilder/welder and he uses jennys all the time when welding.

 
STEPTOEIf the POWER REMAINS THE SAME, then yes current will increase for a lower voltage

eg a 110v 600w drill will draw twice as much current as a 240v 600w drill. If the voltage on the mains drops, then so does the power and the 240v drill will draw less current and not have the same power. Can you grasp that?
yes, that is why during the '80s&'90s E7 heaters made in England wouldnt work properly in N,Ireland, NI was already on 230v, (remember the Dimplex strike?)

so if power draw remains the same and Voltage drops, then Current rises, is that correct.?

tho surely your comparison to a 110v drill is dependant on the transformer windings being the source of the greater current draw?

but I understand where your getting at.

 
Some nice info I've gleaned from stuff sent to me by KME - Thanks mate! :D So it seems lightning or a fuse blowing upstream can cause an RCD trip. My interpretation is that the high transient voltage will cause greater leakage current due to cable capacitance - hence RCD trips.
Hi DaveTG... :x

haven't seen you postin on here much over past few days.... ?:|

Hope you haven't been locked away readin all the stuff KME sent you??? :eek: :^O:^O:^O:^O

 
Hi SL,

Very perceptive of you - nothing to do with the Forum - but I've been agonising whether to start off on my own as rather than carry on as an employee! Am I mad considering the current economic climate - or have I been watching too much of the Apprentice!!!

Cash flow forcasts - business plans - or is it all just gut feelings!! Interesting times!

Cheers

Dave

 
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