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post #1 of 9 Old 12-17-2013, 08:12 PM Thread Starter
Curly5759
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Garage electrical Question

All,
I want to rewire my garage (actually a barn) and add some flouescent lights.
Currently, the main drive area has 3 individual 60 watt bulbs. I want to changed them to this style of light
http://www.homedepot.com/p/Lithonia-...specifications
and it uses 32 watts. (110 volt circuit)

I looked at a few online calculators, but I don't read greek.

My question is: how many amps does each light fixture pull? I don't want to overload the circuit.

Thanks
Curly

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post #2 of 9 Old 12-17-2013, 08:22 PM
jnicewan
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A 60w bulb uses about .5 amps each
32w is going to be about .3 amps
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post #3 of 9 Old 12-17-2013, 08:45 PM
laybackman
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Quote:
Originally Posted by jnicewan View Post
A 60w bulb uses about .5 amps each
32w is going to be about .3 amps
Agreed!

watts/volts= amps

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post #4 of 9 Old 12-17-2013, 08:56 PM
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That fixture is actually 64 watts. It uses two T8 32 watt bulbs. With three light fixtures, there is an additional 12 watts or ~.1 amps@ 110VAC.

0.1A should not make a difference in a circuit unless that circuit is already on the edge of max. The vast majority of house lighting circuits are only running about 1/3-1/2 capacity of the wire used. But, each install varies, so check your wire before adding additional load to any circuit. (Typical CYA statement there)

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post #5 of 9 Old 12-18-2013, 10:45 AM Thread Starter
Curly5759
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Thanks All!

Curly

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post #6 of 9 Old 12-18-2013, 12:27 PM
vadslram
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Garage
How many hundred lights are you looking to put up?

If you are doing a wiring job I would put all your lights on the same dedicated breaker. It makes it simpler. You could run a dozen of those fixtures on one circuit wired with cheap (relatively) 14/2 on a 15 amp breaker. I'd run it on 12/2 myself but everything in my house is over kill.
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post #7 of 9 Old 12-18-2013, 12:32 PM
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i've had great luck with 100w equivalent cfl bulbs and they pull even less power. fixtures are about a buck a each and the bulbs are dirt cheap compared to fluorescent

just another idea
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post #8 of 9 Old 12-18-2013, 01:08 PM Thread Starter
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I actually have CFL bulbs in the fixtures now. They don't provide enough light for my taste. As I get older I find I am needing brighter light to work with. I installed 2 of the fixtures I linked over my work bench and I like the way they look, so I figured I'd add some more to match.

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post #9 of 9 Old 12-18-2013, 02:43 PM
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Quote:
Originally Posted by Curly5759 View Post
All,
I want to rewire my garage (actually a barn) and add some flouescent lights.
Currently, the main drive area has 3 individual 60 watt bulbs. I want to changed them to this style of light
http://www.homedepot.com/p/Lithonia-...specifications
and it uses 32 watts. (110 volt circuit)

I looked at a few online calculators, but I don't read greek.

My question is: how many amps does each light fixture pull? I don't want to overload the circuit.

Thanks
Curly
Ohm's Law: E = I x R
Watt's Law: P = I x E

Equations may be rearranged and/or combined using basic algebra to isolate unknown quantities. AC or DC values don't matter - just the numbers are what's important.

Use a nominal 110V for household wiring, 220V for full-voltage household wiring (like your oven or dryer circuit - or a decent welder,) and a nominal 12V for automotive calculations.

E is voltage (electromotive force,) expressed in VAC or VDC.
I is current, expressed in Amperes
R is resistance, expressed on Ohms
P is power, expressed in Watts

Example:
For a 100W household bulb, how much current is drawn? How much resistance in the filament? Assume zero conductor loss and a nominal applied voltage of 110VAC.

RESISTANCE OF BULB: Using Ohm's Law, E = I x R. E is 110VAC, so 110 = I x R. Both values are unknown, so sort out the current required.

CURRENT OF CIRCUIT: Using Watt's Law, P = I x E. Replacing values gives 100 = I x 110. Rearrangement gives us 100/110 = I. Therefore, 0.(90) (the "90" factor repeats) Amperes, or just short of 1A. Design and fuse for 1A on that branch.

Returning to resistance, we know know I. So, 110 = 0.(90)... x R. 110/0.(90)... = R. 121 ohms = R.

As you can see, Ohm's Law and Watt's Law are very simple equations, and that actually took a good deal less time to type than it took to solve. Using the nominal circuit voltages tends toward an inbuilt safety factor (since the actual operating voltage is usually a bit higher,) and conductor loses are negligible, once you figure out what your requirements are and choose accordingly.

Ampacity charts are available online (choose a reputable source) for wire gage vs. current vs. conductor length, and I would suggest picking up a copy of Ugly's Electrical Reference if you do a fair bit of electrical work (it's a small pocket-size electrical design reference, written for the apprentice & journey-level tradesman, with yellow covers and a red comb binding. Well worth the money you'll spend for it - about $10, usually.)

"Ampacity" refers to the current-carrying capacity of a wire, which is usually a function of wire gage (cross-sectional area of the conductor) and length.

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