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PeterPan27

Hi all, I've got the 2394 written tomoz. Passed the gola and practical. Just trying to do little brush up.
Got this question in a book..
A ring circuit is 54 metres long and wired in 2.5/1.5 thermoplastic cable 32 amp type c device. Ze is 0.24.
The resistance of 2.5mm copper is 7.41mohms per metre and 1.5mm copper is 12.1m ohms
Question is what's the zs of the circuit
Now I've got the answer of course in back of the book...
Basically 2.5/1.5 has a resistance of 19.51 per metre so it's
Length 54 metres x 19.51 divide by 1000 equals 1.05 and because a ring u divide that by 4 which equals 0.26 ohms making r1 plus r2 0.26 ohm then add that to the ze equals 0.5 which is the zs...
Using the regs type c 32a device is 0.75.
So I got 80 percent of the 0.75 by 0.75x0.8 equal 0.6 so the zs is within regs..
My question is the 19.51 resistance isn't in the question so how do I get that? Is there a page/table in regs I need to learn?
 
Practical guide to inspection testing and certificating by Christopher kitcher......
This is there exact answer

"Zs will be 0.34+0.6=0.94 ohms
So Zs=0.94 ohms
The maximum Zs for a 30A Bs 3036 fuse from table 41.2 inBS 7671 is given at 1.09 ohms.
Using the rule of thumb for temperature correction the maximum permissible value is : 1.09 x 0.8 = 0.87 ohms
This value means the gained value is acceptable"

Ok, mistake in the book it is, then
 
I no the 3 steps off by heart
Steps 1 identify the in going an outgoing
Then measure the line loop , neutral loop and cpc loop an record as r1 rn and r2. The line and N should b within 0.05 ohms of each other. The cpc be 1.67 greater .
Step 2 connect incoming line with outgoing neutral and vice versa. Then calculate the predicted reading I would expect to see at each socket using the formula r1 plus rn divide by 4
Step 3 then do same with line and cpc.. R1 plus r2 divide 4
Any higher reading means loose connections or spur off ring

While doing steps 2 and 3 take reading at every socket making sure within the 0.05 which each other..

Remember that that for other sizes this 1.67 value is different. it comes from L / CPC, in this instance you have 1.67, if you have 4mm / 2.5mm you get 1.6, if 6mm / 2.5mm you get 2.4. It is not like you will see any bigger rings then 2.5/1.5 bet you get my drift
 
i am not being funny or think i know it all but there wont be any calculations as hard as this in the test they are straight forward calculation just know your basics and when to apply corection factors they usually tell u its tested at 20oc so theres not temp correction factor
 
Yeh I know. To be honest I've only really had few days to revise so not expecting to pass just said I would have a go and never know and surly it will give me more insight I I did fail when doing the exam in August..
 
Not sure where you going to use this 0.725 factor at this stage, we all know what it is for and it is for calculating the cable size, but at this moment an from OP the cable size is given, so its too late to use it anywhere.

I wasn't going to , just wondered if it had been considered in the book he had.
 

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