Ring final circuit question

[Peter Monfort]

When you do the continuity test on a ring final circuit and you have to connect L1 to N2 and vice versa, this assumes you know which L cable is the incoming or outgoing.

I understand that if you connect L1 to N1 and vice versa the ohm readings rise then fall as you go round the ring circuit. Why is this? I can't explain this to a friend who asked.

If the connections are done correctly the readings stay the same as you go round.

Thanks

 

[Andy78]

Hi Peter, are you a trainee ?

The reason is that in cross connecting the line and neutral conductors you are creating a closed loop through all the live conductors. When you measure at the socket you are measuring the same loop at each point and so should get roughly the same reading.

 

[Peter Monfort]

Hi
Yes...in training.

I understand why they should be the same reading.

but what happens if you connect L1 to N1 and cross connect L2 to N2?

the readings change then don't they? Why is this?

 

[Andy78]

Hi
Yes...in training.

I understand why they should be the same reading.

but what happens if you connect L1 to N1 and cross connect L2 to N2?

the readings change then don't they? Why is this?

Have a think about what you are actually measuring with the cables connected in both situations. Sorry but I'm away back to work now.

If you ask a moderator there is a great training sub forum here you can apply for entry to.

 

[Peter Monfort]

I Sussed it.

if you do it the correct way L1 to N2 and L2 to N1 you effectively create two identical loops. Taking a reading at any socket effectively measures the resistance across two loops of equal length in parallel.
So using the equation for resistances in parallel the resistance will be the same each time.

if you connect L1 to N1 and L2 to N2 you lose the equal loop sizes at each socket. Thus when measuring the resistance across two unequal length loops and the total resistance will change becoming higher the further you go round the ring and then falling again.

Makes sense now I have drawn it out more simply and thought of each leg of the ring as a resistor.

 

[telectrix]

give that man a cigar.