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Maximum Amperage load calculations based on Light wattage

Maximum Amperage load calculations based on Light wattage

#1
05-27-03, 06:19 PM
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Join Date: Mar 2003
Location: Louisiana
Posts: 21
Maximum Amperage load calculations based on Light wattage

Could you please give me a formula to calculate the maximum amperage that fluorescent lights will draw, based on the wattage of the fluorescent bulbs? In this situation, three each, dual bulb, 4 foot fluorescent light fixtures. The bulbs installed are 40 watt bulbs. There will be a total of 6 bulbs on this single circuit. Am I correct in the assumption that this is only an approximate 5 amperge draw?

#2
05-27-03, 06:28 PM
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Join Date: Feb 2003
Location: Indiana
Posts: 317
In a flourescent fixture the lamps are nominal the real load is the ballast which should be labled as to the load. electronic ballasts use the least ammount of power.

#3
05-27-03, 06:30 PM
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Join Date: Dec 2001
Location: New York
Posts: 1,287
The general 'basic' formula is Watts=volts x amps
So if there are (6) 40W lamps, ignoring the losses of the ballast, you have 2A.
This ballast
http://www.universalballast.com/prod.../M132R120C.pdf
has 6Watt loss on a 32 watt lamp.

#4
05-27-03, 06:52 PM
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Join Date: Mar 2003
Location: Louisiana
Posts: 21
Under the previous description , a 15 amp breaker should be more than sufficient for the described lighting, with nothing else on the circuit?

#5
05-27-03, 07:12 PM
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Join Date: Feb 2003
Location: Indiana
Posts: 317
yes

#6
05-27-03, 08:00 PM
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Way, way more than sufficient.