Voltage Drop in Three Phase Lighting Circuits

Let's start with basics and assume you have done no research for your homework assignment.

When does your homework have to be handed in, it could take quite a while to arrive at an answer.

Just to add, why would you use an autotransformers or any transformer for that matter. A neutral core in the cable makes economic sense to me.

I don't see a point in any of your questions but OK let's say I have plenty of time so time is not an issue...

Secondly, the setup is as I described it in my first post, there's no point in reinventing the wheel and "imaganing" that there's no autotransformer. The light fixture comes with a column that has an autotransformer at the base of the column which output is then connected to the lights ballast.

I assume advantages are that you don't need a neutral, smaller current and hence smaller voltage drop so longer circuits are achievable but I really don't want to go into a discussion of comparing this circuit arrangement with some other arrangement.

I liked your response better than the OPs

Voltage drop?

What will that tell you?

I'd use a meter if you really want to know. You should have a fair idea what the supply voltage is. Measure it at the last light in the string.

But, what do I know?

I need to find the voltage drop to check if the value is compliant (less than 5% of the supply voltage)

I don't have a luxury of using a meter but I would love to do a small scale experiment in the lab if I could. My main problem is understanding how the currents are interacting at each "branch out".

Look up the wire data (ohms per 1000 ft). Determine how much current in each wire and apply Ohm's Law. I don't know if the autotransformers are set up that way, but they might have taps for adjusting to the modestly different primary voltages.

Hey,

I have cable data, that's easy but I don't know how to determine current in each wire that's where I need some assistance. There's no taps on the AT, only two actives are connected to it and that's it. Have a look at

http://www.gelighting.com/na/grainger/downloads/ballast/10-0102GraingerSS-480V-277VAutotransformers.pdf

When all else fails :1-888-GEBALLAST (432-2552)

"the return path" on 3 phase?? most interesting concept

The luminaire is connected to the two actives. Imagine a 800W luminaire rated for 400V connected to the two of the three phases. What would be the running current in one of the phases? So, if it's conneced to the Red and White phase with 6mm PVC single core cable 50m distance, what is the voltage drop that I need to compare to the standard so I see that the installation is compliant, that's the question. The third phase (blue) is not loaded...

Of course there would never be just one light on the circuit and usually the connecting phases would alternate so the blue phase would be loaded as well but I guess you can see the point?

Regards,

So you know the loading, the voltage and the length.

Why can't you calculate the volt drop?

I don't know the regulations you work to. In the UK we use BS7671 which sets a maximum volt drop of 3% for lighting circuits. Power is allowed 6%.

A few years ago I wrote a spread sheet for electrical calculations (it's still growing) which includes cumulative volt drop. It's now in wide use by members of three UK electrical web sites.

As you're an anonymous poster I can't help.

Hey,

I was posting as anonymous because I don't post much to this forum (even though I follow it a lot) and I'm slightly intimidated by all the smartasses that come along and try to ruin almost every topic with their smartass comments without actually providing any value/help or anything that even resembles an answer to the question OP asks... I admit a lot of questions asked here are silly and repeated but sharing the knowledge is supposed to be something purer than just acting smart in front of other people who ask for help.

Now that I've said that, I wouldn't mind having a look at your spreadsheet even though I'm in Australia as the standards are I believe somewhat similar... :D

Cheers,

When all else fails :1-888-GEBALLAST (432-2552)

The practice does make tracking response threads rather difficult, and abstruse because of it.

slightly intimidated? at what? most of us here are grownups. I can't speak for everyone but I know most of the regulars around here don't come here out of obligation to perform work for free. most come for a bit of commradery and a bit of levity.Had you began your post with, "I have a lab/homework assignment and I'm a bit stumped and could use a little help", you would have gotten a different set of replies, seeing as how you know lengths and voltages already your cables drop should have already been clear. you didn't supply resistance values on your t-formers and since they probably aren't superconductors we are just left to guess, or joke about the formation of your question. I wont do your assignment for you but I don't mind aiding you in understanding the concepts involved. btw, if I were you I'd make an effort to befriend Mr Slack, that can only benefit you.

GA!

Hi Jay! Bad habit, posting as anonomous! Even I have been known to snipe at anons. I thought this question was quite reasonable though.

I can't understand why simple ohms law should not apply either to each line separately, so I am waiting for a good answer too! Even though it was my trade for so many years, I never "designed" a multi phase circuit, and many three phase mysteries remain a dark art to me.

(Pure laziness on my part...I admit it. More fun to study history than electrics.)

There are really knowledgable people like RedFred and others who get it though. We both await the answers!

Cheers mate! Hoist a lager for me!

Why British Standard 7671 is such an unknown and its application so uncommon is a complete mystery here <sigh>.

So what percentage volt drop did you calculate?

I don't know British Standard 7671 ... I assume because I am in the USA.

Would that be why?

If not, alternatively, it is: because I am lazy.

Did you use the pf and loss in control gear from the manufacturer in your calculation?.