how do you test a lighting circuit with an ohm meter? i know the 1st step is to switch off the power supply feeding the lighting circuit, then what do i do? i heard that tere should be a reading of 0.5 m ohms
Not sure why you want to test it but 612.3.2 tells us we require 1Mohm resistance between all cables when measured at 500 volt where 500 volt may produce damage we can reduce to 250 volt. This is called insulation resistance it will need the circuit making dead and that includes both live conductors Line and Neutral which means the main switch will need to be off not just a MCB since all neutrals are connected together it may need access into the consumer unit to do this test.
There is also a test for what is referred to as R1 + R2 here we use a lot lower 612.2.1 between 4 and 24 volt at not less than 200ma this also requires access into the consumer unit to fit test links.
In both cases with these currents and voltages a special tester is required. For the R1 + R2 there is another method using a different tester called an earth loop impedance meter which unlike the other tests is done on a live circuit this test is preferred with a circuit already in use as the wires are not disturbed so you are unlikely to introduce a fault while testing the low ohm meter test is normally done with new installations to prove it is safe before energising.
All use expensive meters unlikely to be owned by DIY people and also can produce danger while testing is being done so I would not recommend hire so maybe you can say why you want to test the lighting circuit with an ohm meter?
...say why you want to test the lighting circuit with an ohm meter? [/quote]
i done an experiment and managed to get the lighting circuit working perfectly.
the MAIN CU is located in the front room.
16 AMP WIRE (MAX 24 AMPS) connected to 16 amp MCB IN MAIN CU.
FRom main CU , 16 amp wire leads to UPSTAIRS HALL WAY(IT USED to power up water tank that is no longer in use).Bought 5 mtrs of 16 amp wire to connect with the wire that used to power up water tank (used junction box 16 amps).New wire powers up new SECONDARY UNIT.
ALtogether 15 mts of 16 amp wire used.
18 (mv) * 15 mtr / 1000 = .27
new SU has 40 amp RCBO
16 Amps mcb
6 amp mcb
6 amp lighting wire connected to 6 amp mcb
the wire goes out to the garden (distance 6 mtr).
used 2 switches, one for switching light on from downstairs and the other to switch on from upstars.the dis tace between the switches is 3 mtrs.
2 coms connected
l1 from upstairs switch connected to l1 switch in conservatory (pull on)
same method for L2
switch in conservatory recieves LIVE FEED in L1 AND SWITCH live in L2
i will assume that the 6 amp wire used altogether is 14 or 15 mtrs.
27 (mv) * 15 mtrs / 1000 = .0405
Powering up 18 watt energy saving bulb so
18/240 = 0.075
15 mtrs * .075 * 29 (mv) / 1000 = 0.032625
+ .0405 = 0.073125
i Don't know if the maths is correct bu t the circuit works perfectly .
I am careful on calling any cable but flex by amp rating as it is so dependent on route but since you state 18mV/A/m I realise it is 2.5mm² as to yes 0.018*15 = 0.27 Ohms or 0.018*15*16 = 4.32 Volts drop with no correction for temperature.
“new SU has 40 amp RCBO
16 Amps mcb
6 amp mcb”
Are you sure 40A RCBO too big for cable a RCD would make more sense? And since feed from a 16A MCB the second 16A MCB is not required. But using a RCBO as an RCD is OK and using a 16A MCB as a connector block is OK also so not breaking any rules.
Not sure what 6A lighting wire is but since you say 27mV/A/m I would assume somewhere around 1.5mm² which as 29mV/A/m but think you have misplaced the decimal point very easy with slide rule I know I make 0.029*16 = 0.435 MS Excel or OO Math are a lot easier to use but at 2.61 Volts drop at 6A well within requirements.
As to if one should use rating of protection or rating of item depends on why you are working it out. To check volt drop is within 3% then ratting of bulb but to ensure the trip will open on short circuit then MCB rating.
A B6 MCB will require 5(B) * 6 = 30A to trip on the magnetic overload. 230/30 = 7.666667 ohms max book says 7.67 ohms if it was C6 then 3.8 and D6 = 1.9 the calculations you have made are for R1 + Rn which is correct with RCD protection but for R1 + R2 then since the earth is thinner (0.029 + 0.044) / 2 gives 0.0365 for 1.5mm² twin and earth and (0.018 + 0.029) / 2 gives 0.235 for 2.5mm² twin and earth.
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