Ring Final Testing - Why (r1+r2)/4 is R1+R2.
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- Опубликовано: 5 июн 2024
- Explanation of why the resistance of the line and protective conductors added together and divided by 4 is the same as R1+R2.
The explanation is covered in Guidance Note 3, however not in BS7671 or the On Site Guide.
Ring final circuit tests in more detail: • Tests for Ring Final C...
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Finally!!!! Now I understand why it's division by 4 - the training company didn't explain why, and it bugs me when I don't understand the reason why I'm doing anything. Thanks John, my brain can now stop mithering...
This makes learning so much simpler explained by John - I have paid £****.** to study to become an electrician. I am fairly bright., good on the tools - but you need to know what you are doing. Whilst we are on lockdown, I am on RUclips to try and get my head around everything. The way that everything is clearly explained, John puts our college lecturer to shame. Our lecturer has been doing this for 20 years where we are learning a new skill and he is "expecting" us to to take this in on straight away. Every credit John and your tutorials are brilliant...
You know what John. I've got my inspection next week and the wealth of knowledge I received from your channel was overwhelming.
I thank you for your hard work and I truly from the bottom of my heart hope the best for you. I really hope you continue to teach.
This is a very good explanation. I test regularly and use that calculation all the time, but you find yourself forgetting the theory behind it when you do it so often. This is a great way of refreshing your memory. Thanks.
A video to keep us grounded in these times of madness. Thanks.
I see what you did there. "Grounded" in North America is the term we use for "earthed" 😆😆
What a brilliant well explained video. Never taught it that way at college in the early 80's.
I remember watching your previous videos and could get my head around the /4 . now its all perfectly clear...Thanks
As usual great video John.
Excellent explanation, much clearer than what I was taught at college. Thumbs up to you sir!
Excellent explanation. Hard topic made really easy as always!
As always john a most superb video.very clear methodical and precise.keep the great work and videos.please keep safe sir.
You are a brilliant teacher Sir!
Excellent vid I was trying to explain this to a young fella a week or so ago ..Think I'll get him to watch you're video .... thanks.
No lockdown for me sparks are an essential service .. Have to admit I get nervous now when working with people in close quarters... strange times....I hope you and you're family are well.
Hi John,
I have left comments on "Tests for Ring Final Circuits
" clip and you mentioned you would do another clip soon.
I knew that I must be miss understanding and I just could not get it around then and my guess was wright.
But watching this clip have left me no more words and questions as it was such cristal clear to understand.
Really appreciated for your good work.
Regards
Alan J
Great explanation Thanks John
Brilliant explanation. Thank you
Excellent explanation 👍🏼
Another brilliant video John. I love your style of teaching because as I watch through and my brain starts thinking ahead and generating questions, I'm always safe in the knowledge you will answer them, often in various ways and from different perspectives. Learning is so much more enjoyable when you are on a similar wavelength.
Something was niggling and distracting me throughout this one though... I (we) all know of your understandable dislike of ring final circuits but I was waiting for you to mention it... and waiting.... and waiting... right up until the final moment... but you didn't let us down. Excellent.
Thanks JW, your videos are awesome. I’ve recently completed an installation workshop with one of the largest electrical training providers (I won’t mention any names 4 U at this point). They taught us to carry out this test but not the reason why, not totally happy with the training as there are gaps like this that your excellent videos are filling. Greatly Appreciated..
Another great video with a clear explanation. Perhaps John you could do a video showing the process in practice with real measured values.
At last a decent explanation, this "1/4 issue," keeps coming up on forums and at college. Not many have a modern "Inspection and Testing. guidance Note 3" where there is an excellent text and diagram explanation. I strongly recommend getting it. Furthermore video is the preferred medium for young students. Thanks :-)
Perfect. Thanks John
Great explanation
Excellent explanation
Excellent, thank you.
Brilliant video
Brilliantly explained John, very easy to digest your tutorial. 👍
Well said my suggestion to customers is 2 radial circuits in a kitchen as opposed to a ring. This way if in the unfortunate even a problem occured with one of the radials they wouldn't have complete loss of power for their kitchen utility goods.
Thanks , always find your videos interesting. Some things which can be proven mathematically might also have been mentioned. Where the CPC has a lower Cross Sectional Area (CSA) than the line conductor, R1 + R2 is always less than (r1 + r2)/4, in practise this is not usually not by very much (
Great video
Hi John hope you are well, very nice explaination! It is a weird concept that needs understanding, I got my head round this stuff after reading kirchoff's laws and some study of the whetstone bridge!
Good job!
That Kirchoff guy has a lot to answer for Warren ... 😒😂
The wheatstone bridge is a great circuit... I built a capacitor comparator based on it ! Works for inductance too !! 😎👍☘️🍺
The whetstone bridge is a great way of sharpening your knowledge of electrical circuits.
Nice editing too John
I just passed my 2391 -52 in jan could have done with this then,cracking video.👌
I learned something new. Thanks JW.
Great video Johan, so if I understand this correctly, quarter of the loop essentially is one specific conductor of the ring meaning R1 + R2 means R1 AND R2 as the value read is the value of resistance for each conductor. It is not a case of adding R1 and R2 because that would give the resistances combined
Essentially, there’s two of each conductor line and cpc in each accessory that makes up the ring main meaning there’s always 4 individual r1’s and r2’s in circuit, therefore you divide by 4 to get a unified ZS/ZE (R1+R2)
Its funny but whilst I can get me head around transistor biasing or current-limiting a LED ... I find this stuff strangely obtuse ! 😂😂
Of course... you do a great job of explaining it John 🙄😏
Thanks once again Mr Ward, when I'm in Eire , ring final circuits not allowed & fully insulated cpc in twin & cpc, with all conductors the same size,therefore contractors are often used for shower circuits, much better than trying to ram the cables into a switch box ☘️
Should that have been contactors? Rather than contractors.
I think that if you use contractors for the shower they would have a shocking experience. ;)
brilliant
As soon as you drew it as a circle it made instant sense.. it';s a quadrant of that circle hence the divide by 4
Thank you for your quick response that has helped me I will have to get a small consumer unit because there is no more room in their consumer unit I have watched a lot of your videos and they have help me have you done a video on fire alarms if not will you be doing one in the future The one that I am doing is a dedicated fire alarm panel
JW, have a look in the new GN3. Seems to me there's an error in there (2.6.6 step 3) regards what readings should be expected at each point where cpc is different csa to line
At 4:45 I expected Clifford Stoll to show up and explain Klein Bottles.
I think self-isolation is getting to me...
Great question to ask any potential cowboy Consumer Unit changeout companies ...
A question for you a family member has asked me to put a fire alarm in their enormous house I read the manual it says The mains need to go to the consumer unit with no rcd in it because it will keep tripping how do I go about this when their consumer unit have two rcd in it one for the lights and one for the ring main
Is this a fire alarm system with a dedicated panel, or individual interconnected smoke alarms?
Individual smoke alarms are usually supplied from a local lighting circuit, whether that has a RCD or not doesn't matter - most of the time it will have an RCD due to the cables being concealed in walls/floors.
A fire panel should be supplied on it's own separate circuit, and not via an RCD which also covers other circuits. Mains cabling for the fire panel also needs to be fire rated just as the rest of the installation does. If the mains cabling is surface mounted, then no RCD required, if it has to be concealed then and RCD will be needed, but this should be an RCBO which only supplies the fire panel and nothing else.
What's the point of the r2 test on a radial circuit? I used to think it was good as you don't need to dismantle lights / sockets etc but to measure the zs of the circuit you do need to open / dismantle accessories as you can't add r2 to ze to calculate it.
I like your final comment to use radials rather than ring. I *hate* ring final circuits :)
What sized wire is used for a spur for a socket from 32 amp rated ring main? It's going to be a double socket.
2.5mm², the same size as used for the rest of the circuit.
@@jwflame ok. Thanks.
what is the relationship between r1 and r2 recorded in step 1 and (R1+ R2) recorded in step 3.
Is there an mcb which can check the load on not only the whole ring, but that the load is at least somewhat balanced across both ends to ensure there isn't a break in the circuit?
No.
@@jwflame Well that was succinct :)
How much current will 2.5 mm take before it does melt? Test for future.
It’s actually very very high, but long before it melts the insulation will breakdown and junctions will often have problems. So the 2.5mm melting isn’t usually the problem
Why do we have Ring circuits then john? wasnt it somert to do with a copper shortage ??
Designed in the 1940s so that new homes could be wired using the minimum of resources, one ring circuit for the whole house and a single type of plug for everything, compared to multiple circuits and various different sizes/ratings of plugs used previously.
The rest of the story would make an entire video on it's own.
Look up tom scotts channel and look for his video on british plugs, that explains it a tiny bit more.
@@jwflame
houses built after the 40s were still using radials and at least 3 sizes of the round 3 pin and 2 sizes of the round 2 pin connectors and were still in use at least up to the early 80s so when exactly did every one start using ring circuits and square pin sockets ? , when the house i grew up in was rewired in the early 70s it was ring and square sockets :-)
Hi JW, I am moving my C/U at home, not a straight forward job as it involves getting overhead cable put underground and fitting new external meter cupboard, I am fully qualified BTW, but I want to run something past you before I start. Would it be OK to email at your company email please? Too much info to deal with here but as you are fresh up with latest rules and regs I wonder if you could advise.
My assumption was that its divided by 4 because its basically because you have the resistance measurement from the centre of the ring of the cpc both ways and the line conductor both ways thus dividing by 4,and its only lazy house bashing sparks who want rings done away with because its easier and quicker to run radials,as an industrial spark of 50 years i cant remenver if ever wiring a ring main circuit in 2.5mm cable we wired them in 4mm or sometimes 6mm so overload NEVER became a problem
So basically you are measuring in parallel and not in series which seams to be the more obvious because its one long loop which should increase impedance ie resistance.
“Four divided by four, which in this case, equals one” 😁. I think it’s in all cases 😁
Lets have a Vote for JW to do an Organ Recital for easter........ Bach's fugue in D# minor perhaps ??
Why 4 and not 3 or 5 or 6?
Thanks so much for all your videos! Even though I'm in Canada, I didn't know very much about our 240v "split phase". I got a 120v lathe and blew up the motor, and there were way more options for replacement with 240v motors so I got one, but in my "shop" I only had 120v service. I got an electrician to run 240v service to my shop (60 amps). Thanks again for all your videos, learned lots. Pic of sub panel: i1.lensdump.com/i/jRPibi.jpg (red to black is 240v, red to white or black to white is 120v)
3 dislike peeps wish they had Kyryptonite to defeat the Superhero known as JW !!
JW, you really don't like ring final circuits do you?
They are a decent solution for what they were originally designed for and for their original intended use.
However that was 70+ years ago, and most newer installations with them would be far better using alternative arrangements.
I especially enjoyed the 🙄 when you first mention ring final circuits at a few seconds in! Love it JW!!
I appreciate your videos John, but often have a hard time understanding all that you are saying. I wonder if you would please speak a little slower, and a bit more evenly. You vocal inflection gets lost in some of the transmission. Thanks. Great information.
Closet captains. Not perfect, but they help.
You’re turning 4 cables into one so you divide the result from those 4 cables by 1
first from NZ
John I’m confused.com 🤔
A North American electrician might well ask: what the heck is a ring circuit??
With respect, this explanation is somewhat contrived and does not really explain or prove that R1 + R2 is equal to (r1+r2)/4. And to say that the total resistance is half because you have two resistors in parallel is plain wrong. Any physics textbook will tell you that the total resistance of n resistors in parallel is given by 1/R = 1/r1 + 1/r2 +.............+1/rn. So the total resistance between the two measured points on the circuit, noting your explanation that effectively there are two resistors in parallel, R (i.e.R1 + R2) is given by 1/(R1+R2) = 1/r1 +1/r2, is it not? In the particular case of two resistors in parallel it can be shown that R = r1.r2 / (r1 + r2). The question that my mathematics cannot sort out is how we move from that equation to R = (r1+ r2) / 4. Your explanation is really little more than a mental salve because people then think that they understand it. Which may, of course, be all that they need.
A continuous ring of wire is equivalent to two identical parallel conductors.
If the ring has a resistance of 1 ohm, cutting it into two identical lengths means each length has half the resistance, or 0.5 ohms.
Placing those two 0.5 ohm conductors in parallel will result in half the resistance again or 0.25 ohms.
A quarter of the original 1 ohm.
Another way is to consider that the ring has a length and a cross sectional area.
Cutting the ring into two identical pieces means each piece has half the original length.
Placing those in parallel means the cross sectional area is doubled.
Halving the length of a conductor will halve it's resistance.
Doubling the cross sectional area also halves the resistance.
Half x half = quarter.
John, thank you for taking the time to reply to my comment. My further comments are as follows:
1) Your illustration of the actual situation with the ‘figure of 8’ coupling is a powerful visualisation that I have not seen elsewhere. I had to look hard at the original arrangement and then create a mental animation that unfolded the parallel cross-connected conductors to reveal the circle and thus to see that the two measurement points, although next to one another in the socket, are, in fact, diametrically opposite on the circle. This point is fundamental to our considerations.
2) Your explanation in reply to my comment has an elegant simplicity that works for the particular example of two conductors of equal cross sectional area. Only. It does not work for two conductors of different cross sectional area and this is addressed at sub-section (b) on page 66 of Guidance Note 3. The maximum value of the parallel resistance in the case of two conductors of different csa in figure of 8 configuration will occur at the mid-point of the two conductors and that is because at that point we have a situation where we would have half of each conductor on each side of the test points giving, on each side of each conductor an equal sum of two resistors in series.
3) This is shown in practice, unwittingly, by your friends Gary and Marcus, here: ruclips.net/video/UesYHVsZD-E/видео.html. Their r1 value is 1 ohm (time stamp 1:35) and their r2 value is 1.67 ohms (time stamp 2:50). The value of two resistors in parallel is r1.r2/(r1+r2). This is basic physics. So for Gary and Marcus’ test we have 0.625 ohms. And this is the value measured when they put the test probes at the point where the loop has all the cpc on one side and all the line on the other (time stamp 9:11). But they just brush off the fact that actually this is a completely different value to the 0.67 calculated by Gary! As we move away from the ‘joint’ points, where the conductors are cross-connected, we come closer to the situation of R1+ R2 = (r1+r2)/4, which Gary calculated.
4) So, R1+R2 = (r1 + r2) / 4 is an approximation that is close enough to allow technicians to see that the ring final circuit has no faults. But it is not actually correct for all pairs of conductors, just close enough to be a useful tool. The variance in values is addressed in table 2.8 of Guidance Note 3.
5) Is it of any value, me banging on about this? Hmmm…, well, I just think that this should be presented for what it is, an approximation that gives close enough values. None of the videos that I have seen on You Tube do this and some of the explanations are so contrived as to be nonsense. Obviously your explanation is not nonsense but, as I say, isn’t it worth pointing out that it is only a close approximation that is only completely true when both conductors have the same csa? Otherwise, why bother even trying to explain it at all?
John, thank you for taking the time to reply to my comment.
However, I do think it worth mentioning that your simple and elegant explanation is only completely valid when the csa of the two conductors is the same. In cases where the csa of the two conductors is different, R1+R2 will vary around the created loop, although not by a huge amount. This is addressed in sub-section (b) on page 66 of Guidance Note 3, together with table 2.8. All explanations that I have seen on You Tube omit this important point.
Hi I had the same frustration as you, love this video however it is unfortunate to have picked a value of 1 Ohm for each half section of ring as they do add up to 4ohms in total, which could be misleading to some people.
The way I explained it to myself is if we split the ring into two symmetrical halves then on each side we have r1/2+r2/2
Applying the formula you correctly referred to for resistances in parallel, this translates as 1/R=1/(r1/2+r2/2)+1/(r1/2+r2/2)
>>>This will take you to R=(r1+r2)/4
Of course in practice it is impossible to measure exactly where the two half rings meet symmetrically, but try the maths again with a split such as 3r1/4 + r2/4 and r1/4 + 3r2/4 , the formula also works ;)
Hope I was able to describe the reasoning.
Figure of 8 test.
Sod the clap, £1000 a week to NHS staff, tax free, till this is all over
ring circuits are rubbish