Thank you Dave for helping us to remain technically minded as we appraoch each job. It's always a good reminder to give attention to how we design each circuit to make sure that it complies. As always much appreciated and keep up the good work. 👏🏾
One of the best Elec tech sites on the web helped me understand how to wire sockets & lights into my son's ring circuit house which I had not come across before although I was competent to carry out all the work myself
A handy tip that's stuck with me since I was told it about the < & > symbols is that the < looks a bit like a squashed L (for less).I'd never had difficulty with them but this is just saves that bit of thinking. Also, when using a calculator, the = button is your friend. So 20÷.88= (then) ÷.725= gives the answer without the additional step or using M+ and MR. Thanks again for your incredible content.
Hi Dave Can we please have a video about installing single cables for lighting circuits in steel conduit and how we prevent Eddie currents, thanks for everything mate
It is all in the question, the data/clues are all there to work out the reference method. Have another look at the video and follow my working through.
@@learnelectrics4402 Thanks for the quick reply, I understand Iz is current carrying capacity but where i struggle to understand is when working out the formula sequence e.g Should it be; Iz ≥ In = It Ca x Cg Or It ≥ In = Iz Ca x Cg
Thanks, another great video. Does this apply also to a ring final, wired in 2.5mm TC+E? I’m bashing my brain against this because on the one hand OSG part 7 says we can use 2.5mm cable for an I n of 32 A, and an I t 32/0.8 = 40 A; but the current carrying capacity of 2.5mm cable is nowhere near that? I can’t help thinking I’ve missed something that is written in the books somewhere that explains the case for ring finals … ?
With a ring circuit, you have two 2.5mm wires - making 5mm of conductor. That is why the figures seem higher. Take a look at this video. Dave. th-cam.com/video/6GNb1SSEjQg/w-d-xo.html
Hi great video. I am currently doing design project through college and i have the exact issue that you’ve stated in 14:00. my I n is greater than my I t due to the 25c ambient temperature (which is required in the spec). my understanding is that I t should be bigger than I n, under all circumstances. so how can i get around this issue to change the values and make the calculation work? really hope you can get back to me on this 😅 many thanks Daniel
Great vid! Quick q.. for question 1 why is the ref method C and not 100? ref method 100 mentions insulation doesn't exceed 100mm for which it doesn't as it's 50mm? thanks!
Ref method C is the starting point for the reference methods. This has the highest permissible current for a size of conductor. All the others are reductions on the Ref C figures. Hope this helps.
Can I ask a question on using T&E in trunking in a ring cicuit with a 32A breaker. Trunking is reference method B which is not in table F6, its in table F5i. This table is for multicore cables. With no other factors to derate, 32A would need a 6mm CSA cable. This can't be right can it? Also, using table F6, we could never use 2.5mm CSA cable on a 32A breaker, but most ring circuits are wired like this and BS7671 says we can use a cable capable of carrying 20A in a ring circuit, on a 32a breaker.
The regs don't like to see twin and earth in trunking, You have a conductor with a grey sheath on it and then you put it in a box. That is two layers of plastic that are preventing heat loss.Take a look at the video below, there are several on ring circuits, and yes, 2.5 on a 32A breaker is ok. th-cam.com/video/-PAruseUf04/w-d-xo.html
@@learnelectrics4402 In my case, this is a C&G exam question where they specify T&E running in dado conduit. They say the right answer is 4mm2 but I have no idea why.
@@learnelectrics4402 Further to this question, I think I am right in saying that for Ring Final circuits, there is no calculation for cable size. The regs say that 2.5mm2 (good for 20A) is ok with a 32A breaker with a floor area of less than 100m2. However C&G teaching says to use the design current for the ring, not the tabulated current, to look up the appropriate cable size. This is why they say 4mm2. I just just don't understand why they teach this when its not in the Regs.
give this man a medal. no one is doing it like dave. thank you dave appreciate all your hard work
Wow, thanks, that is really appreciated. Dave.
Brilliant
Thanks for watching.
Thank you Dave for helping us to remain technically minded as we appraoch each job. It's always a good reminder to give attention to how we design each circuit to make sure that it complies. As always much appreciated and keep up the good work. 👏🏾
Very welcome and thanks for the positive comments. Dave.
One of the best Elec tech sites on the web helped me understand how to wire sockets & lights into my son's ring circuit house which I had not come across before although I was competent to carry out all the work myself
p😮😮
Glad it helped, thanks for watching.
Thanks for watching.
Wow! This is very informative, thanks for sharing it with us. It worth it and the best tutorial on the topic for me.
Glad it was helpful Daniel. Keep spreading the word and lots more to come. Dave.
Thanks Dave-informative video, and very understandable graphics
Glad you enjoyed it and thanks for the positive feedback. Dave.
This video is amazing! Keep up the good work because I’m learning a lot.
Will do, thanks for watching. Dave.
Another great simple elegant presentation.
Thanks again, thanks for the support.
A handy tip that's stuck with me since I was told it about the < & > symbols is that the < looks a bit like a squashed L (for less).I'd never had difficulty with them but this is just saves that bit of thinking.
Also, when using a calculator, the = button is your friend. So 20÷.88= (then) ÷.725= gives the answer without the additional step or using M+ and MR.
Thanks again for your incredible content.
Excellent support. Thanks for the input, it will help many. Dave.
Another great simple elegant presentation. Super @:)
Thanks a lot, appreciated.
great video to re-cap. thank you.
Glad it was useful. Dave.
Excellent, correct design is always very important 👍
Absolutely. Thanks for your supportive comments Brian, appreciated.
Very useful and informative video thanks
Thank you, appreciated and thanks for watching.
Very very useful. Thanks.
You are welcome, thanks for watching. Dave.
Fantastic tutorials keep up the good work 👍
thanks for the comments. Yes, lots more to come. Dave.
Hi Dave
Can we please have a video about installing single cables for lighting circuits in steel conduit and how we prevent Eddie currents, thanks for everything mate
Its on the list. Thanks.
Great video, watched it twice now 😂
Great to know. Thanks for the feedback, appreciated. Dave.
Thanks for the video...
My pleasure. Dave.
This is perfect
Glad it helped, thank you. Dave.
Thank you so much!
You're welcome. Dave.
Brilliant explanation .but just wondering where about in question said what reference method is that.
Thanks
It is all in the question, the data/clues are all there to work out the reference method. Have another look at the video and follow my working through.
@@learnelectrics4402 thanks for your reply .boss page .I just have looked again you are right 👍
Great work 👍, I have a question what about Iz as i am getting it mixed up with It.
Iz is the amount of current a conductor can carry 24/7 under the installation conditions that affect it.
@@learnelectrics4402 Thanks for the quick reply, I understand Iz is current carrying capacity but where i struggle to understand is when working out the formula sequence e.g
Should it be;
Iz ≥ In = It
Ca x Cg
Or
It ≥ In = Iz
Ca x Cg
brilliant thank you so much
Glad it helped. Dave.
Thanks for the video, where can i find the guide ?
Amazon is a good price for the on site guide. make sure it is the latest version, amendment two, brown cover.
Thanks, another great video. Does this apply also to a ring final, wired in 2.5mm TC+E? I’m bashing my brain against this because on the one hand OSG part 7 says we can use 2.5mm cable for an I n of 32 A, and an I t 32/0.8 = 40 A; but the current carrying capacity of 2.5mm cable is nowhere near that? I can’t help thinking I’ve missed something that is written in the books somewhere that explains the case for ring finals … ?
With a ring circuit, you have two 2.5mm wires - making 5mm of conductor. That is why the figures seem higher. Take a look at this video. Dave.
th-cam.com/video/6GNb1SSEjQg/w-d-xo.html
Hi great video. I am currently doing design project through college and i have the exact issue that you’ve stated in 14:00. my I n is greater than my I t due to the 25c ambient temperature (which is required in the spec).
my understanding is that I t should be bigger than I n, under all circumstances. so how can i get around this issue to change the values and make the calculation work? really hope you can get back to me on this 😅
many thanks Daniel
Increase the cable size, that is the best way, this will increase It. Or decrease the breaker rating to reduce In, but not always practical.
The best !
Excellent, that's good to know. Thanks for watching. Dave.
Great vid! Quick q.. for question 1 why is the ref method C and not 100? ref method 100 mentions insulation doesn't exceed 100mm for which it doesn't as it's 50mm? thanks!
Ref method C is the starting point for the reference methods. This has the highest permissible current for a size of conductor. All the others are reductions on the Ref C figures. Hope this helps.
@@learnelectrics4402 so do we always use ref method c for the exam questions on the 2365 203 exam?
Amazing
Thank you, great feedback.
Can I ask a question on using T&E in trunking in a ring cicuit with a 32A breaker. Trunking is reference method B which is not in table F6, its in table F5i. This table is for multicore cables. With no other factors to derate, 32A would need a 6mm CSA cable. This can't be right can it?
Also, using table F6, we could never use 2.5mm CSA cable on a 32A breaker, but most ring circuits are wired like this and BS7671 says we can use a cable capable of carrying 20A in a ring circuit, on a 32a breaker.
The regs don't like to see twin and earth in trunking, You have a conductor with a grey sheath on it and then you put it in a box. That is two layers of plastic that are preventing heat loss.Take a look at the video below, there are several on ring circuits, and yes, 2.5 on a 32A breaker is ok.
th-cam.com/video/-PAruseUf04/w-d-xo.html
@@learnelectrics4402 In my case, this is a C&G exam question where they specify T&E running in dado conduit. They say the right answer is 4mm2 but I have no idea why.
@@learnelectrics4402 Further to this question, I think I am right in saying that for Ring Final circuits, there is no calculation for cable size. The regs say that 2.5mm2 (good for 20A) is ok with a 32A breaker with a floor area of less than 100m2.
However C&G teaching says to use the design current for the ring, not the tabulated current, to look up the appropriate cable size. This is why they say 4mm2. I just just don't understand why they teach this when its not in the Regs.
Smashing 💪🏻
Thank you, appreciated
thank you easy
Thank you, appreciated.
Please what of the cable calculation for single and three phase motors
Take a look at this video. If you know the load, you can calculate the amps and cable size.
th-cam.com/video/bXiE9bdFLsU/w-d-xo.html
Thanks
You're very welcome. Thanks for watching. Dave.
great
Thank you, apprteciated.
How do I make a donation
The best donation is to tell others about the LearnElectrics channel. Spread the word. Thanks, Dave.
I'm here again
You're very welcome.
hi mark niels
hi ryan
@@harveylennox1128 hey babe
Thanks for watching.
Thank you for watching.
Thank you.