Volt drop calculation

[pcasey]

Hi, Thanks in advance for any advice!

I was called out to a client’s swimming pool, as there was a fault on the pool lights. I found that someone had installed 2 x 12 volt 300 watt lamps, recessed underwater into the pool walls, using 10mm cable in parallel so that there are 2 neutrals and 2 lives at the transformer. Neither light was working. I found that the transformer situated in the pool plant room was 40 meters away from the furthest of the 2 pool lights, which seemed an excessive distance to me. I also found that a 50 Amp fuse had been fitted in between the secondary 12-volt side of the transformer and the lights and this had burnt out, so causing the problem. My question is this, is the calculation to measure voltage drop on a 230-volt AC circuit the same that I must use to calculate the voltage drop on a 12-volt AC circuit? The reason I ask is because when I use the normal formula i.e.



4.4 (volt drop per A per mtr for 10mm cable) x 25A (load of 1 x 300w lamps) x 40 (cable run in meters) - Divided by 1000 = 4.4 volts

I can't imagine that this can be right as even using 16mm cable with a Vd/A/mtr of 2.8 the volt drop would still be 2.8 volts.

in fact using this formula I would need to fit the transformer no more than 5 meters away to achieve a volt drop of no more than 3%

(4.4 x 25 x 5) / 1000 = 0.55 volts = load voltage of 11.45 volts

Although I've not worked on pool lights before, I have seen many pools where the plant room is 10 or 15 meters away so I’m guessing that I need to alter the equation for 12 volt systems but not sure how.

Again, many thanks for any help, Phil.

 

[telectrix]

you calculate the VD on 12V the same as with 240V. obviously, as the current is 20x @12V, VD is much higher in proportion.

 

[Flanders]

Yes the voltage drop will be high when the old 12 volt down lights lamps were all the rage we had to use 2.5 tri ratted cable to feed each 50W lamp about 5 meters from a central transformer.

 

[pcasey]

Cheers, I thought I was going mad! That said, the client tells me that the lights were working for years which makes no sense to me if the load voltage at the lights was less than 8 volts, according to the calculation. Could the lights have worked at this voltage and if so, does a voltage drop damage a transformer? Cheers again for any advice!

 

[Flanders]

When he says years does he mean he's has them on for 1/2 an hour 3 time a year or on all the time it sounds like due to the voltage drop excessive load was put on the transformer and burnt them out

 

[pcasey]

I think he's had them on for a few hours every now and then, for the odd pool party, lucky sod lol. I think that he may have left them on overnight after a few too many beers and that caused the problem. Could you explain why the voltage drop increases the load? I'm guessing it's because of a change in resistance but not sure, cheers, Phil.

 

[Flanders]

Yes more resistance more load V/IR

 

[telectrix]

resistance is futile .

 

[Flanders]

Beam me up scotty

 

[telectrix]

alas, scotty is no more. he's gone to the great starship engine room in the sky.

 

[telectrix]

alas, scotty is no more. he's gone to the great starship engine room in the sky.

 

[pcasey]

without sounding too thick, and bearing in mind I've been in France for 10 years lol.....if I increase the resistance of a circuit while keeping the voltage constant or even lowering it, won't the current be lower?

 

[telectrix]

yes, but in this case why is the resistance increasing? only temperature rise will cause that. the load will try and pull wattage, to achieve this ( P=VI) as the voltage drops, the current will rise.

 

[Flanders]

The resistance isn't increasing is it not just high in the first place ?

 

[pcasey]

it makes perfect sense when I look at P=VI as lowering the voltage due to increased resistance increases the load however I'm confused when trying to understand how that's compatible with V=IR as if the resistance is higher surely the current flow decreases. I guess it's getting late and I need a good nights sleep! Thanks for your help, I really do appreciate it!

 

[telectrix]

if the load is purely resistive, then that resistance will only increase with temperature rise untill the lamp reaches is normal operating temp.

 

[Flanders]

Sorry I'm wrong About V=IR ignore this I think i'm only confusing the issue.

 

[Geoffsd]

Sorry I'm wrong About V=IR ignore this I think i'm only confusing the issue.

No you aren't. You can't be. Ohms law is constant.

I=V/R so if the resistance goes up the current will go down.
If the voltage goes down the current will go down.

P=V²/R so if the resistance goes up the wattage will go down.
If the voltage goes down the wattage will go down.

The confusion I think is because we are talking about the increase in resistance of the cable before the load which results in lower voltage at the load which means the load has lower wattage and so lower current.
Also when the temperature and resistance of the load itself increases this will result in further reduction in wattage and current although the stated values will be for the operating temperature.

 

[pcasey]

Thanks for continuing the thread as I'm still at a loss as to why the 50A fuse burnt out if higher resistance, in the cable or load, lowers the current that flows. I thought that the only consequence of volt drop in this circuit would be that the lamps would be dim, or is it me that's being dim? lol. There was 25 A flowing down each 10mm cable so I guess that may have generated some heat which could have burnt out the terminal on the fuse....does that sound plausible?

 

[DPG]

At that sort of current, any loose/poor connections are going to get hot. Daz