5000 watt inverter schematics circuit
If you have (90) 50w lamps = 4500 watts total. Assuming a 120/240 panel, if you put 1/2 on one "side" of the panel and the other 1/2 on the other "side" of the panel, that would be 2250 watts on each half. The generator should be rated *at least* 125% of the load; 4500w x 1.25 = 5625W. Using a 4500W generator on this load will cause it to overheat and shorten its life as it is running at 100% of capacity all the time..
One half of the panel is 120V to neutral, and the other is 120V to neutral - or 240V between both circuit breaker terminals. Ohms law for DC circuits and purely resistive AC circuits says Volts x Amps = Watts; or Watts / Volts = Amps. So, 2250W / 120V = 18.75A on each pole of a 2 pole circuit breaker that feeds the sub panel. A #12 copper wire is rated for 20 amps; but as per National Electrical Code - must be de-rated to 80% of rating which means it is good up to 16 amps maximum. A #10 copper wire is rated for 30 amps, but it too must be derated to 80%, making it good for 24 amps maximum. So, if you are going to feed a sub panel supplying (90) 50watt lamps, you will need to run a #10/3 copper cable from a two pole 30 amp circuit breaker at the generator to a 120/240 volt "main lug only" sub panel rated for at least 30 amps.
Divide your load evenly across the sub panel - (4) 15 amps circuits via (2) two pole 15 amp circuit breakers on each "side" of the panel if you run (2) 14/3 cables out to the lights - or (4) single pole 15 amp circuit breakers if you run (4) 14/2 cables out to the lights. No circuit breaker terminal should have more than 23 lamps that means you have (2) w/ 22 lamps and (2) with 23 lamps. The circuit w/ 23 lamps will draw 23 lamps x 50w = 1150W. 1150W / 120V = 9.6A. The 22 lamp load will be 22 x 50w = 1100W. 110W / 120V = 9.2A. Which is well within the 12A maximum allowed (after derating as required by code) by a #14 copper wire rated for 15A.
Mar 10, 2014 |