having a brainfart......

[gavin.sibley]

hello all,

having a complete meltdown on a job this afternoon and i need an answer to what i think may be a very very stupid question.

if on an energised circuit you disconnect the neutral at the board (lighting circuit) will that removed neutral rise in voltage too to potentially 240v? reason i ask is I've been chasing what i thought was a live switching fault all afternoon and suddenly realised it may be a neutral issue somewhere?

thanks all,
gavin

 

[ian.settle1]

What do you think?

 

[gavin.sibley]

i think it would as the "neutral" doesn't have a reference point to keep it at 0v, so would rise with the voltage of the rest of the circuit, but, as i said may just be having a brainfart.....

 

[micknew]

Yes for sure on the neutral you disconnected.

 

[Marvo]

Depends if there's lamps (bulbs) in the fitting and whether the switch is on or off.

Draw the circuit on paper and label it with the voltages you'd expect to see with respect to earth at all the different points.

 

[gavin.sibley]

thanks for all the replies, really got myself in a knot with this one!

the voltage showing on the unknown cable (which i mistakinly presumed as live as it was red) varied depending on switch position and which switch, when i measured continuity between this and the known live source this measured around 20-25 ohm, now i have come home and started thinking clearly, im thinking that this changing voltage is actually a floating neutral cable and that the 25 ohm resistance is just the resitstance through the lamps in the circuit which is why it is so high. any objections?

 

[micknew]

No objections to that lol.

 

[micknew]

Brain fart gone, you'll feel better now.

 

[Marvo]

A 25 ohm light circuit would be a low resistance. You'd need a lot of lamps connected to get 25 ohms.

 

[gavin.sibley]

by quite alot is there anyway to put a figure on that? the whole house is all on one circuit, would that potentially get it down that low or would we be looking at alot more?

 

[micknew]

A 25 ohm light circuit would be a low resistance. You'd need a lot of lamps connected to get 25 ohms.

But if its a tungsten lamp/lamps the resistance increases dramatically as the temperature of the lamp element increases.

 

[micknew]

Of course there could be other problems, I don't know the resistance of a cold tungsten lamp offhand.

 

[gnuuser]

Of course there could be other problems, I don't know the resistance of a cold tungsten lamp offhand.

take a cold lamp and measure its resistance and make a note of it
if all the lamps are the same use the proper formula to calculate the resistance total
even if they are different values you can still calculate the total resistance and compare it to your readings.
preparing for a job is more than getting your materials around
set yourself up with a testing regime that covers all bases
this will prevent possible brain farts and make you look good in the eyes of your customers
(impressing them with professionalism is a sure way to get some free advertising)

 

[tazz]

You Know the Wattage, you Know the voltage......so you can calculate the current & hence the resistance, without a meter....!!

 

[Richard Burns]

If you have a circuit with the neutral disconnected but line connected then there is no current flowing and so there is no volt drop occurring on the circuit.
Therefore the potential difference to earth at the end of the disconnected neutral will be equal to the supply voltage to earth.
If the neutral were connected then voltage would be dropped over the load and the voltage to earth at the connected neutral would be (very close to) zero.

If you have a resistance of 25 ohms then using V=IR and P=VI you can say that P = V²/R so your load would be about 2.1 kW or 9 A, quite high for lighting but possible.