Okay i believe I understand so the load neutral would then receive an extra 5 volts am I understand that right? 

In a way, yes. A diagram might help:

If you like the relative voltages, in red, show the voltage differences across each part (cable and load), whereas the blue figures show the voltages relative to Earth.

But none of that directly explains why the lamp glows. If it's an incandescent, a substantial current must be available (much more than capacitive coupling would normally provide) and if the light were conventionally switched and the switch in the off position, there shouldn't be a path for such a current. There must be something odd with the setup or a  fault somewhere.

   - Andy.

Okay i believe I understand so the load neutral would then receive an extra 5 volts am I understand that right? 

In a way, yes. A diagram might help:

If you like the relative voltages, in red, show the voltage differences across each part (cable and load), whereas the blue figures show the voltages relative to Earth.

But none of that directly explains why the lamp glows. If it's an incandescent, a substantial current must be available (much more than capacitive coupling would normally provide) and if the light were conventionally switched and the switch in the off position, there shouldn't be a path for such a current. There must be something odd with the setup or a  fault somewhere.

   - Andy.

Ah okay so across the line conductor where lossing 5v and then again across the load we lose a further 5 volts which goes down to earth and the neutral conductor. 

Currently I am only a 3rd year electrical apprentice I was wondering if there was any materials I could read into to understand this all better 

Thank you. 

AJJewsbury said:

There must be something odd with the setup or a  fault somewhere.

Given that it was on YouTube, it might have been faked.

Any chance of a link, Sam?

https://youtube.com/shorts/y08JoizQOgw?feature=share4

Heres a link to the video it's a YouTube short without much information but if you look at the first comment on the video it provides a little more detail personally I think it's wrong and there's something else causing the issue. 

Cant see any comments at all just a picture of a lamp in a holder accompanied by annoying music. Could be almost anything - lamp in series with fan or another light fitting used to be a classic with decorators making a mess of loop through rose (aka 3 plate) ceiling fittings. I suspect that the use of phones to take a before and after pic have reduced this a bit

Equally could be an LED 'filament' lamp and as above capacitance.

I agree, shower  induced voltage drop is likely spurious.

Mike

Thank you. That is an LED lamp.

That is an LED lamp.

agreed - e.g. https://www.toolstation.com/sylvania-led-rt-a60-filament-clear-gls-lamp/p85671

Real filament (incandescent) GLS lamps (at least of the ordinary domestic type) generally had the filament as a flat semicircle at towards the top held up on a number of fine wires from a central glass post (or as an obvious coil in much older ones) - like this one https://www.easy-lightbulbs.com/products/gls-100w-b22d-bc-240v-philips-clear-light-bulb . The 'sticks' are LEDs trying to look like filaments.

We're likely back to the capacitive coupling theory.

   - Andy.

Okay so you reckon that the light is gaining an induced voltage from the shower circuit caused by capacitive coupling. Would this effect be increased with a higher voltage drop? 

we lose a further 5 volts which goes down to earth and the neutral conductor. 

Erm, voltage is probably best thought of as a pressure difference between two points - volts don't travel as such. It's the current (amps) that flows - and flowing along something with resistance (which everything has) results in the pressure drop (voltage difference). Text book wise it's all just Ohm's Law. V=IR. You've got the same current flowing through all parts of this simple circuit - the overall current is regulated by  the overall resistance of the circuit (load and all the connecting wires) and the supply  voltage as a result is divided up across each component of the circuit - so the voltage of across each part in series adds up to the original supply  voltage. There's nothing special about the N conductor in this regard - it carries current and has a voltage difference along its length in exactly the same way as the line conductor.

   - Andy.