amp

amp

Probably why Tom Horne specifically stated "as long as the lamps are incandescent..." For calculating purposes, the PF of a resistive load (such as an incandescent light) is PF = 1, thus VA = Watts, for a resistive load. If an inductive load is served (such as fluorescent light fixtures),

LMAO. What do you suppose the power factor of a light bulb is?

Zaf wrote:

Hi, That kind of thinking could cause problems in the field. I said math. As long as one knows what it is. Tony

Hi, That kind of thinking could cause problems in the field. I said math. As long as one knows what it is. Tony

I didn't say it was equal in all cases. What I said was that for incandescent bulbs the value given in watts can be used as the VA value. -- Tom

For a resistive load cosine(phi)=1.

Boden

Tony Hwang wrote:

Boden

Tony Hwang wrote:

Hi,
Which is power factor of 1(phase angle is aero). No loss circuit.
Pure resistive load which does not hardly exist in real world.
Cos0(zer degrees) = 1.
Thanks, Boden.
Tony

Boden wrote:

Boden wrote:

On 28 Oct 2003, Tony Hwang wrote:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ That's the key here. Thank you for saying it, prepare to be belittled and badgered by those too simple minded to understand.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ That's the key here. Thank you for saying it, prepare to be belittled and badgered by those too simple minded to understand.

--

Baisez-les s'ils ne peuvent pas prendre une plaisanterie

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Baisez-les s'ils ne peuvent pas prendre une plaisanterie

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The only simpletons here are the morons like you who insist that the minuscule amount inductance in an incandescent lamp somehow becomes a significant factor when calculating the VA of a resistive load. Tell us again, Tomi Boi, how is it that you came to the conclusion that a toaster is an inductive load and a vacuum motor is a capacitive load? http://tinyurl.com/st7a

Yes, be prepared to be belittled. Incandescent Lights are indeed near perfect resistive loads and as far as I kow they do exist in the real world.

Electric water heaters (and to a lesser degree dryers), near perfect resistors, their job is to dissipate heat. Yes, if you are running an air conditioner or fridge yes, there will be 10-15% reactive portion. High crest factor current waveforms (in electronics) can also impact usable power on a given breaker, but this is well beyond the scope of the question. The question in this case will can ten 65 watt bulbs be safe off a 15 amp circuit and the answer is yes, by more then factor of 2.

Furthermore cos(phi) is an oversimplified for PF. Total power (real + reactive) is the integral of the product of the voltage and current. The current in many electronics devices being nonsinusoidal. Finally, mathamatics is the TOOL quantify the PHYSICS in this problem. Circuit breakers do not do math, they do follow the laws of PHYSICS. Now that you have been belittled by someone simple minded perhaps you will not post that which you are ill informed on.

On 29 Oct 2003, Zaf wrote:

^^^^ ^^^^^^^ So you admit that they are not a pure resistive load. Thank you for the confirmation. I snipped your rant, as it was no help in proving your point. If you had a point, that is.

Thanks for proving that the simple minded are indeed reading the newsgroup, I knew the proof would be offered by someone.

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Baisez-les s'ils ne peuvent pas prendre une plaisanterie

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^^^^ ^^^^^^^ So you admit that they are not a pure resistive load. Thank you for the confirmation. I snipped your rant, as it was no help in proving your point. If you had a point, that is.

Thanks for proving that the simple minded are indeed reading the newsgroup, I knew the proof would be offered by someone.

Baisez-les s'ils ne peuvent pas prendre une plaisanterie

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Click to see the full signature.

On 26 Oct 2003, JosephM wrote:

Absolutely.

15A @ 120V = 1800 watts, so if the fixtures themselves were rated as such (they usually aren't) you could run 6*** 250
watt bulbs without going over 15A (6***25000W.5A)

That's -exactly- the type of circuit that 14 guage wire is made for. Actually, you could put all 12 fixtures on one circuit (and break them up into as many switched branches as you want to.) If you would use a 100W bulb in each, you're still only be talking about 1200W = 10A @ 120V. There's really no need to tie up a 2nd breaker for two circuits.

--

Baisez-les s'ils ne peuvent pas prendre une plaisanterie

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Absolutely.

15A @ 120V = 1800 watts, so if the fixtures themselves were rated as such (they usually aren't) you could run 6

That's -exactly- the type of circuit that 14 guage wire is made for. Actually, you could put all 12 fixtures on one circuit (and break them up into as many switched branches as you want to.) If you would use a 100W bulb in each, you're still only be talking about 1200W = 10A @ 120V. There's really no need to tie up a 2nd breaker for two circuits.

Baisez-les s'ils ne peuvent pas prendre une plaisanterie

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Click to see the full signature.

You forget that, for continuous loads, the circuit is limited to 80% of its rated capacity. This is the calculation you should have used:

15A @ 120V x 80% = 1440 watts.

He's still within the limit, of course, but the limit isn't as high as you claim.

-- Regards, Doug Miller (alphageek-at-milmac-dot-com)

circuit

the

this is turtle.

It seem here your over killing here a good bit. The 12 -- 65 watt lite will pull about 4 to 5 amps and you can combine the 12 lite on 1 -- 20 amp / 1 -- 15 amp circuit with #12/2 and cut down on all the circuits in your home. Less circuits / less trouble down the road. Now you can't have a bunch of other stuff on here too and add the 12 lites together.

Now you could just use the 14 wire and 1 -- 15 amp breaker to run the hold thing.

TURTLE

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