Amps = Watts/Volts, Right?

[SteveJ34]

Somewhat Off Topic but an electrical question....

I've got to get an "IguanaWorld" habitat project completed so I can get back to my xmas bartop projects.

I'm down to the final stages of completing the electrical for some recessed light fixtures and  one flourescent fixture.

I have (4) Max 100 Watt fixtures and (1) 40 Watt  flourescent fixture. I'm planning to wire this to a single plug (with multiple switches inline for the various fixtures) which would go to what is a standard 15amp household circuit.

I'm no expert....I know enough to be dangerous......but from my basic calculations, there shouldn't be any problem with the amperage required to run these (5) lights of said circuit, as...

440/110 = 4.4 amps

Is my thinking correct?

Granted, there are multiple outlets on that single circuit but there is nothing else in that room except a small TV, radio, and bedside lamp.

I think I should be fine, eh?

[Samstag]

Yeah you've got that right (assuming the voltage is 110).  That's fine on a 15A circuit if it's the only thing(s) using it.

[Granger]

You should be fine, but just for information sake, you are correct on your formula in a DC circuit, but in an AC circuit watts = volts x amps x the coefficient of power(relationship of how capacitive/reactive circuit is)

[SteveJ34]

You should be fine, but just for information sake, you are correct on your formula in a DC circuit, but in an AC circuit watts = volts x amps x the coefficient of power(relationship of how capacitive/reactive circuit is)

Thanks for the insight.

I figured the true AC circuit formula was not quite as simple as what I was operating from which is why I posed the question in the hopes that someone more knowledgable than I would respond.

For future reference, I wonder how one determines the "coeffiicient of power" for a given electrical device, ie: a 100 watt max recessed lighting fixture for example....obviously what kind of bulb you stick in it probably is the determining factor, eh?

[lcddream]

the coefficient of power basically would mean the "power factor"  or efficiency of the load in question. for instance since a fluorescent light uses an electronic ballast it is given a power factor of say .9 so if the light had a consumtion of 32 watts, you would divide 32 by .9 to figure what the potential total reactive power for that lamp would be. For regular resistive loads (not electronic based) such as incandescent light bulbs power factor would be a non-issue, and even for fluorescent lights it doesn't become a big issue. Reactive power means that the light/other equipment can pull more power than it actually needs because the electronic component is less efficient. Basically the lower the pf the worse off you are, a pf of 1.0 is ideal.

Just remember the NEC only allows for 80% of a circuit breakers  rated capacity to be used.  Which means 15A breaker X  120v X .8 is the amount of watts allowed on this circuit by code. (1440 Watts).

[Shape D.]

4 Amps
440 / 110 = 4

[SteveJ34]

4 Amps
440 / 110 = 4

Ah, yes, me quick math not so good before my first cup of joe in the a.m.