Originally Posted by Curly5759
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
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.
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
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.