AndyD please e-mail me the torrent to download the SANS10142-1 from Demonoid. And could some1 please send me the 2012 installation rules syllabus, i wish 2 write both papers b4 year end. My e-mail address is ilangasolsol@gmail.com or collin_dube@yahoo.com
Installation Rules-past papers.
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AndyD please e-mail me the torrent to download the SANS10142-1 from Demonoid. And could some1 please send me the 2012 installation rules syllabus, i wish 2 write both papers b4 year end. My e-mail address is ilangasolsol@gmail.com or collin_dube@yahoo.com_______________________________________________
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If u around east of gauteng try ekurhuleni college brakpan campus.there I know 4 fact coz I jst started on monday,its from 18h00 to 20h00Comment
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good morning guys to you all can any one assist me with past intallation rules papers, 2011nov,aprl ,aug,2007,2006,2010;2009 all paper 1&2 ,your assistance will be highly from appreciated from vuvu by the way thanks in advanced.im currently writting this try- mester.Comment
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Thanx cvmostert! Papers received. I now have most of the papers with memos from the last 6 years. Anyone with memo's for 2012 and 2011? Cant find them anywhere...We are not human beings undergoing a spiritual phase.....we are spiritual beings undergoing a human phase....
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Hey guys, i am seriously stuck on probably what is a rather simple question, but i cannot fathom how the memo gets the answer for this one. I suppose this is why you need to attend a college course to pass this exam right??
2nd Paper 2010 August, Question 1.
DB 230V AC supplying: 10m run then splits to 10m run to 20A load, and 20m run to 40A load.
This is all you are given.
Calculate the minimum cable size.
Now i'm assuming this is all 1ph and copper cables as it doesn't say.
The memo gives me an equation to use: VD = (I x pL) / A; therfore, A = (I x pL) / VD
Ok, now first of all, i know that A is the area. I know that I is the amperage. I am assuming that VD is the volt drop?? I have no idea what pL is???
The totals the memo inserts are: A = (I x pL) / VD --- A = (40 x 0.022 x 30) / 5.5
= 4.8mm therefore a 6mm cable should be used.
And further states that a 6mm cable = 0.0073 -- this i understand from Table 6.2(b)
But where do the other totals come from in getting to the 4.8mm cable size??
How does pL = 0.022 x 30??
How does VD = 5.5???
Please could someone explain this to me as i dont know who to ask to clarify this? I have looked back over past exam papers as well, as well as thru my Regs book extensively, but am failing to find any explanation??
The Regs book does not seem to explain the working out behind these answers properly! Or maybe I am just clueless??
Please, any help MUCH appreciated!We are not human beings undergoing a spiritual phase.....we are spiritual beings undergoing a human phase....
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Another thing...
Just working my way through past papers, and am realising that the number of times the memo sheet is incorrect in its calculations is scary! Are the guys who are marking these papers electrically minded, or are they just paid to sit and mark ONLY according to their answer sheet?? Over the last 4 papers i have worked through, about 10 or 15 marks could have been lost due to an INCORRECT answer sheet!! I am guessing here that if your answers dont match the ones on the memo, u'll be losing your marks?! And, not being allowed to challenge the marking, you'd be doing a rewrite at another R500! Seriously hoping this isn't the case...!!We are not human beings undergoing a spiritual phase.....we are spiritual beings undergoing a human phase....
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