Quote:
Originally Posted by MEXANDME
I understand how to figure voltage drop, i.e., 475 x 2 = 950 x 140w =133000 / 18960 (#8 wire) =7.02 vd. Then 7.02 + 10.5 (minimum voltage at light) = 17.52, so use the 19v tap.
My question was how does using a 22v tap for a 140w load effect the TF versus using the 12v tap for a 140w load.
I was told that on the 22v tap you can have 180w max, the 19v tap 200w max, the 17v tap 220w max, etc.
I was told that if you were to put 240w on the 22v tap, the coil would get extremely hot and burn up the TF in a few weeks.
If this is true, is there a formula/equation to figure the various senarios?
Thanks,
ROS

There's a few things going on here.
All transformer common taps have a 25 amp limit (per UL listed restriction). At 12 volts, this translates to 300 watts (12 x 25 = 300).
Voltage taps, on the other hand, have no limit imposed by UL  except that they are limited based on restrictions of the ratings of the terminal block itself and the wiring that connects the tap to other transformer components.
Some manufacturers spec these components so the full load of the transformer can be carried on any one voltage tap  others restrict this load.
So, it's likely that you need not be concerned about the 22volt tap (since it's likely the manufacturer would rate its load fairly high  but check to make sure) but instead you need to focus on the common tap for that wire pair.
As I said above, there's a 25 amp limit on the common  300 watts at 12 volts. You might think that the equation changes for the 22 volt tap, but it doesn't. As long as you have 12 volts at the fixture, it doesn't matter what voltage tap you use  the equation is the same.
The big concern here is the wattage represented by voltage loss in your wire. If you are starting at 22 volts and ending up with 12 volts at the fixture, then you are losing 10 volts  or consuming nearly double the wattage. That would bring you to about 260 watts on the common. (NEC limits you to 80% max. load on a circuit  240 watts).
The forumula you site is correct, you just need to calculate the wattage contribution of the voltage loss to ensure you don't overload the common. You can work that into your existing formula or just use our calculator to get the total watts (lamps plus wire).
Here's the link to the calculator.