When you are calculating power consumption on inductive items such as flourescent lights (and electric motors) you must calculate the power factor into the equation. Flourescent lights draw more than a simple watts divided by volts will give you.
The amperage information you seek should be on the fixtures. The amps shown on the fixture is a total thats takes into account the power factor. lighting is a continous load so code doesn't allow you to exceed 80% of a breaker (or switch, or wire's) rated capacity. A 20 amp breaker should not be loaded over 16 amps continously. My guess is that about 9 or 10 fixtures on one breaker would be your limit, allowing for a safety factor of 100% on the calculations.
This was just discussed in January, here is a post I made on Jan 14..........
From another board........ This is a discussion pertaining to how many flourescent lights you can put on a timer. A breaker would be the same calculations, just slightly different numbers of lights in the end. The discussion centers on Power Factor and why with flourescents, expecially the older magnetic ballast type, you don't want to come anywhere close to what the simple volts x amps = watts and watts divided by volts = amps formulas, will give you.
The Question......
Could somebody tell me how to figure out how many 40 watt florecent bulbs will work with a timer that says 8.3 amps. I think that 100 watt is close to 1 amp, am I close? If I am I am guessing thst timer would handle 8 100 watt incandesant bulbs and 20 of the 40 watt?
The answer, in two parts...........
Its a lil cornfusing but when I try n explain it sort of helps refresh the whole thing all over again in my old brain. Years ago I could spit it out in my sleep but havent worked with it for a long time. It helps to look at a vector diagram showing the voltage and the lagging inductive current and the in phase resistive current elements. The Power Factor (if I recall correct??) is the cosine of the angle between the Voltage and Current i.e if pure resistive and they are in sync theres no angle (0 degrees) between and the cosine of 0 is one i.e. Unity Power Factor BUTTTTTTT if they are out of phase say an inductive motor load where the current lags the voltage thennnnnn theres an angle between the two and the cosine of that angle has a value so the power factor may be say 80% the bottom line remains that "apparent power" what a straight watts calculation would yield DOES NOT EQUAL 'REAL POWER' what the utility has to generate i.e current to energize the inductors magnetic field PLUS current in the DC resistive part of the inductor (its wire resistance)
John (rusty on this but knew it well a good while back, still interesting to me, however)
and the second part of the answer.........
Again as I posted below and explained, due to the Power Factor problem I NEVER used near as many fixtures/lamps as the straight watts calculations might indicate i.e. I used like 10 or 12 100 watt units (40 + 40 + ballast) on a 20 amp branch circuit.
NOWWWWWWWW lets say you have 4 fixtures each with two 40 watt tubes plus a ballast,,,,,,,,,,,,, lets say thats "about" 100 watts each,,,,,,,,,, then 4 of those would be a rough total of 400 watts,,,,,,,,,,,400 watts divided by 120 volts = 3.3 amps SOOOOOOOOOOOOOOOOOOOOOOOOO Yes a timer rated to carry n switch 8.3 amps ought to work fine
BUTTTTT REMEMBER this doesnt take into account the possibly lousy Power Factor of the ballast as some, especially the older, are poor as its an inductive load (its a transformer)
ANOTHER BUTTTTTTTTTTT still if the timer is rated at 8.3 amps and your load is only 3.6 (but thats if youre using a pure resistive unity PF which the ballast is NOT load) THATS A SAFETY FACTOR OF DOUBLE so you ought to still get by fine. (Those would have to reallyyyyy be poorrrrrrr ballasts if an 8.3 amp rated timer couldnt handle what computes to be a 3.6 amp resistive load) Just be aware if you use watts and treat the ballast as resistive (which its NOT) I wouldnt push things right up to the limit and use so many lamps the straight wattage calculations might say its 7 or 8 amps !!!!!!!!!!!
Watts = Volts x Amps and thats for a pure resistive Power Factor = unity 1 load. Howeverrrrrrrrrrr if the load is inductive (current lags the voltage) the apparent power DOES NOT equal the real power. It takes lagging current n energy to build and maintain the magnetic field surrounding the inductor PLUS it takes in phase (resistive) current to flow throug the straight DC resistance of the ballast. The utility hates poor power factor cuz they have to generate n supply energy for BOTH the lagging current to maintian the magnetic field of a motor/inductor PLUS the in phase resistive current SOOOOOOO their generators have to work harder n theres more inefficiencies.
Hope this helps, DO NOT use so many fixtures the wattage calculations go up near the amps limit of the device TO SOLVE THIS EXACTLY we would need to know the exact loads and power factors but a safety factor of two is sureeeeeeeee plenty. Not knowing the power factor and just guessing I sure wouldnt go much over 5 or 6 amps using straight wattage calculations but 3 to 4.5 or should be fine regardless.
Sorry we cant give any exact answer.
John, A Long Retired EE and a tad rusty on all this butttttttt if you use conservative raitings and ESPECIALLY if you use modern high PF ballats there shouldnt be any problems
I hope they don't mind my reposting, but I have eliminated last name references and don't tell you where I got it, so...........
Charles
That's what I did with my last garage 
What I did was put 3 fixtures on one switch and 6 on the other, scattered all over the ceiling. So, I got three lighting intensities, 3, 6, or 9 lights...