We just bought a cabin. It has a 100 amp service breaker, and a service
panel with 5 20 amp breakers.

I know on appliances, all I have to do to find out the amperage is RTFM, but
for lights and such, how do I calculate just how much wattage I can put on
any breaker circuit?

The lighting is inadequate. I will need more. I have a licensed
electrician friend who will come and wire everything, so it will be done
right and safe. I just need to get an idea how many lights we are talking
about so I can do some shopping. I don't want to max everything out and put
up a ton of lights. I just want to balance them, and not put so many that I
am approaching critical mass.

TIA

Steve

SteveB wrote:
We just bought a cabin. It has a 100 amp service breaker, and a service
panel with 5 20 amp breakers.

I know on appliances, all I have to do to find out the amperage is RTFM, but
for lights and such, how do I calculate just how much wattage I can put on
any breaker circuit?

The lighting is inadequate. I will need more. I have a licensed
electrician friend who will come and wire everything, so it will be done
right and safe. I just need to get an idea how many lights we are talking
about so I can do some shopping. I don't want to max everything out and put
up a ton of lights. I just want to balance them, and not put so many that I
am approaching critical mass.

TIA

Steve



Wattage devided by the voltage (120) is amps. Or
just add up the wattage. 20 amps x 120v = 2400
watts. Lights are easy since the wattage is
written on each bulb.

You don't want to exceed 80% of the capacity of the circuit. If you are
using a 15 amp circuit don't exceed 1200 watts on the circuit
"SteveB" wrote in message
news:j1aKe.84965$4o.53276@fed1read06...
We just bought a cabin. It has a 100 amp service breaker, and a service
panel with 5 20 amp breakers.

I know on appliances, all I have to do to find out the amperage is RTFM,
but for lights and such, how do I calculate just how much wattage I can
put on any breaker circuit?

The lighting is inadequate. I will need more. I have a licensed
electrician friend who will come and wire everything, so it will be done
right and safe. I just need to get an idea how many lights we are talking
about so I can do some shopping. I don't want to max everything out and
put up a ton of lights. I just want to balance them, and not put so many
that I am approaching critical mass.

TIA

Steve

"SteveB" wrote in message
news:j1aKe.84965$4o.53276@fed1read06...
We just bought a cabin. It has a 100 amp service breaker, and a service panel
with 5 20 amp breakers.

I know on appliances, all I have to do to find out the amperage is RTFM, but
for lights and such, how do I calculate just how much wattage I can put on any
breaker circuit?

The lighting is inadequate. I will need more. I have a licensed electrician
friend who will come and wire everything, so it will be done right and safe.
I just need to get an idea how many lights we are talking about so I can do
some shopping. I don't want to max everything out and put up a ton of lights.
I just want to balance them, and not put so many that I am approaching
critical mass.

TIA

Steve

this is Turtle.

In normal lighting circuits it is hard to over load normal lighting wattages to
just light up a cabin. Just take 1 -- 20 amp breaker and circuit for lights only
and that gives you 20 -- 100 watt light bulbs to light up the cabin. One Circuit
is what I thinik you need for all lighting to the cabin.

Now really 19.6 -- 100 watt light bulbs to be exact.

TURTLE

"TURTLE" wrote in message
...

"SteveB" wrote in
message news:j1aKe.84965$4o.53276@fed1read06...
We just bought a cabin. It has a 100 amp service
breaker, and a service panel with 5 20 amp breakers.

I know on appliances, all I have to do to find out
the amperage is RTFM, but for lights and such, how
do I calculate just how much wattage I can put on
any breaker circuit?

The lighting is inadequate. I will need more. I
have a licensed electrician friend who will come and
wire everything, so it will be done right and safe.
I just need to get an idea how many lights we are
talking about so I can do some shopping. I don't
want to max everything out and put up a ton of
lights. I just want to balance them, and not put so
many that I am approaching critical mass.

TIA

Steve

this is Turtle.

In normal lighting circuits it is hard to over load
normal lighting wattages to just light up a cabin.
Just take 1 -- 20 amp breaker and circuit for lights
only and that gives you 20 -- 100 watt light bulbs to
light up the cabin. One Circuit is what I thinik you
need for all lighting to the cabin.

Now really 19.6 -- 100 watt light bulbs to be exact.

TURTLE

Not the best idea: Pop the breaker an dthe whole place
goes dark. They should be mixed on at least two
breakers, and instead of 20 bulbs, that would be 16
bulbs. Over 80% usage will allow normal variations in
the grid, cabin, over time, breakers, etc. to begin to
heat the breaker, thus degrading it over time and
leaving no safety overhead. So with a min two lines
you've got 32 bulbs now, lots more than you'll need.
Not sure where the 80% figure comes from, nec, ul, mfg,
whatever, but it's reality.

HTH,
Pop

"Pop" wrote in message
...

"TURTLE" wrote in message
...

"SteveB" wrote in message
news:j1aKe.84965$4o.53276@fed1read06...
We just bought a cabin. It has a 100 amp service breaker, and a service
panel with 5 20 amp breakers.

I know on appliances, all I have to do to find out the amperage is RTFM, but
for lights and such, how do I calculate just how much wattage I can put on
any breaker circuit?

The lighting is inadequate. I will need more. I have a licensed
electrician friend who will come and wire everything, so it will be done
right and safe. I just need to get an idea how many lights we are talking
about so I can do some shopping. I don't want to max everything out and put
up a ton of lights. I just want to balance them, and not put so many that I
am approaching critical mass.

TIA

Steve

this is Turtle.

In normal lighting circuits it is hard to over load normal lighting wattages
to just light up a cabin. Just take 1 -- 20 amp breaker and circuit for
lights only and that gives you 20 -- 100 watt light bulbs to light up the
cabin. One Circuit is what I thinik you need for all lighting to the cabin.

Now really 19.6 -- 100 watt light bulbs to be exact.

TURTLE

Not the best idea: Pop the breaker an dthe whole place goes dark. They should
be mixed on at least two breakers, and instead of 20 bulbs, that would be 16
bulbs. Over 80% usage will allow normal variations in the grid, cabin, over
time, breakers, etc. to begin to heat the breaker, thus degrading it over time
and leaving no safety overhead. So with a min two lines you've got 32 bulbs
now, lots more than you'll need. Not sure where the 80% figure comes from,
nec, ul, mfg, whatever, but it's reality.

HTH,
Pop

This is Turtle.

He was asking what number of breakers would be need for lighting for the cabin.
I said 19.6 for a 20 amp breaker will support the 19.6 light bulbs and will be
the 80% of the amperate of the 20 amp breaker. He can split up the bulb in all
area , but all lighting would not need more than 1 --- 20 amp circuit, no matter
how he run it.

With two circiuts of 20 amps he could put not 32 -- 100 watt light bulbs but
39.2 --- 100 light bulbs. This would still be compliant of the 80% rule of the
two 20 amp breaker circuits.

Now the degrading of the breaker to trip at lower amperages is just a effect you
have to deal with 20 to 30 years from now and just wait 20 years or so and think
about them.

TURTLE

wrote in message
...
On Wed, 10 Aug 2005 21:28:49 -0500, "TURTLE"
wrote:

With two circiuts of 20 amps he could put not 32 -- 100 watt light bulbs but
39.2 --- 100 light bulbs. This would still be compliant of the 80% rule of the
two 20 amp breaker circuits.


What 80% rule?

This urban legend has taken on a life of it's own. There are lots of
80% rules in the NEC (even some 50% rules) but none of them apply to
lighting fixtures in a dwelling.


This is Turtle.

When Speaking of a Electrical breaker in a breaker box suppling electricity and
the 80% rule does not matter if the elctricity is supplied to a whole home or to
a one light bulb. If you have a 20 amp breaker in the box. You better not put
more that a 80% of 20 amp load on it. So You can have a 16 amp load to be
supplied to the 20 amp breaker.

i know now that your not a electrician for you would know exactly what I was
speaking about and you would be telling me about the 80% rules. So 80% Applies
to everything that will pull amps from a breaker and if you want to change
anything. Please be welcome to do as you please.

TURTLE

Dear Steve, the formula is

volts times amps = watts

So, you can say that the 110 volt circuit, times 20 amps = 2200 watts.

Hint: Try using flourescent, or compact fluorescent bulbs. They use a lot
less power.

--

Christopher A. Young
Learn more about Jesus
www.lds.org
www.mormons.com


"SteveB" wrote in message
news:j1aKe.84965$4o.53276@fed1read06...
We just bought a cabin. It has a 100 amp service breaker, and a service
panel with 5 20 amp breakers.

I know on appliances, all I have to do to find out the amperage is RTFM, but
for lights and such, how do I calculate just how much wattage I can put on
any breaker circuit?

The lighting is inadequate. I will need more. I have a licensed
electrician friend who will come and wire everything, so it will be done
right and safe. I just need to get an idea how many lights we are talking
about so I can do some shopping. I don't want to max everything out and put
up a ton of lights. I just want to balance them, and not put so many that I
am approaching critical mass.

TIA

Steve

Stormin Mormon wrote:

volts times amps = watts

So, you can say that the 110 volt circuit, times 20 amps = 2200 watts.

Hint: Try using flourescent, or compact fluorescent bulbs. They use a lot
less power.

At a lower power factor... watts = 0.4 x volts x amps,
for some undercounter fluorescents.

Nick

In article , "TURTLE" wrote:


When Speaking of a Electrical breaker in a breaker box suppling electricity and
the 80% rule does not matter if the elctricity is supplied to a whole home or to
a one light bulb. If you have a 20 amp breaker in the box. You better not put
more that a 80% of 20 amp load on it. So You can have a 16 amp load to be
supplied to the 20 amp breaker.

Wrong.

i know now that your not a electrician for you would know exactly what I was
speaking about and you would be telling me about the 80% rules. So 80% Applies
to everything that will pull amps from a breaker and if you want to change
anything. Please be welcome to do as you please.

What you say here is completely wrong.

It's quite obvious that *you* are not an electrician, or you would know that
the 80% rule refers to continuous loads, defined by the NEC as "a load where
the maximum current is expected to continue for three hours or more." This
does *not* apply to residential lighting circuits, or indeed to most other
loads.

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

Get a copy of my NEW AND IMPROVED TrollFilter for NewsProxy/Nfilter
by sending email to autoresponder at filterinfo-at-milmac-dot-com
You must use your REAL email address to get a response.

"Doug Miller" wrote in message
m...
In article , "TURTLE"
wrote:


When Speaking of a Electrical breaker in a breaker box suppling electricity
and
the 80% rule does not matter if the elctricity is supplied to a whole home or
to
a one light bulb. If you have a 20 amp breaker in the box. You better not put
more that a 80% of 20 amp load on it. So You can have a 16 amp load to be
supplied to the 20 amp breaker.

Wrong.

i know now that your not a electrician for you would know exactly what I was
speaking about and you would be telling me about the 80% rules. So 80% Applies
to everything that will pull amps from a breaker and if you want to change
anything. Please be welcome to do as you please.

What you say here is completely wrong.

It's quite obvious that *you* are not an electrician, or you would know that
the 80% rule refers to continuous loads, defined by the NEC as "a load where
the maximum current is expected to continue for three hours or more." This
does *not* apply to residential lighting circuits, or indeed to most other
loads.

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

Get a copy of my NEW AND IMPROVED TrollFilter for NewsProxy/Nfilter
by sending email to autoresponder at filterinfo-at-milmac-dot-com
You must use your REAL email address to get a response.

"Doug Miller" wrote in message
m...
In article , "TURTLE"
wrote:


When Speaking of a Electrical breaker in a breaker box suppling electricity
and
the 80% rule does not matter if the elctricity is supplied to a whole home or
to
a one light bulb. If you have a 20 amp breaker in the box. You better not put
more that a 80% of 20 amp load on it. So You can have a 16 amp load to be
supplied to the 20 amp breaker.

Wrong.

i know now that your not a electrician for you would know exactly what I was
speaking about and you would be telling me about the 80% rules. So 80% Applies
to everything that will pull amps from a breaker and if you want to change
anything. Please be welcome to do as you please.

What you say here is completely wrong.

It's quite obvious that *you* are not an electrician, or you would know that
the 80% rule refers to continuous loads, defined by the NEC as "a load where
the maximum current is expected to continue for three hours or more." This
does *not* apply to residential lighting circuits, or indeed to most other
loads.

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


This is Turtle.

first i don't do residentiual electric work and only commercial HVAC which makes
me do the electric for them.

Second if you was running power to lights and the lighting wattage was 2,400
watts / 24 -- 100 watt light bulbs and then would you say you would hook all
them up to a 20 amp breaker which the load would be matching a 2,400 watt load.
Then or would you only connect 19.6 light bulb on the 20 amp breaker to not over
load the breaker and trip if all the lights were turned on at one time. Awwwww
that would be the 80% rule. When every you pull over 80% of the breakers rating.
Your asking it to trip and you have to come back to rebalance the loads.

Tell me what you would do with this problem !

TURTLE

wrote in message
...
On Thu, 11 Aug 2005 11:04:06 -0500, "TURTLE"
wrote:

This is Turtle.

first i don't do residentiual electric work and only commercial HVAC which
makes
me do the electric for them.


Turtle, I specifically addressed residential lighting loads. Do you
have every light in your house on at the same time?

I bet Turtle doesn't, but if he has kids, probably. Like at my house.

Steve

In article , "TURTLE" wrote:

"Doug Miller" wrote in message
om...
In article , "TURTLE"
wrote:


When Speaking of a Electrical breaker in a breaker box suppling electricity
and
the 80% rule does not matter if the elctricity is supplied to a whole home or

to
a one light bulb. If you have a 20 amp breaker in the box. You better not put
more that a 80% of 20 amp load on it. So You can have a 16 amp load to be
supplied to the 20 amp breaker.

Wrong.

i know now that your not a electrician for you would know exactly what I was
speaking about and you would be telling me about the 80% rules. So 80%
Applies
to everything that will pull amps from a breaker and if you want to change
anything. Please be welcome to do as you please.

What you say here is completely wrong.

It's quite obvious that *you* are not an electrician, or you would know that
the 80% rule refers to continuous loads, defined by the NEC as "a load where
the maximum current is expected to continue for three hours or more." This
does *not* apply to residential lighting circuits, or indeed to most other
loads.


This is Turtle.

first i don't do residentiual electric work and only commercial HVAC which
makes me do the electric for them.

Then perhaps you shouldn't be trying to answer questions about residential
electrical work...

Second if you was running power to lights and the lighting wattage was 2,400
watts / 24 -- 100 watt light bulbs and then would you say you would hook all
them up to a 20 amp breaker which the load would be matching a 2,400 watt
load.
Then or would you only connect 19.6 light bulb on the 20 amp breaker to not
over
load the breaker and trip if all the lights were turned on at one time. Awwwww
that would be the 80% rule. When every you pull over 80% of the breakers
rating.

No, that's just plain wrong. The 80% rule DOES NOT APPLY in this situation
because it is NOT a "continuous load" as defined in the NEC.
Your asking it to trip and you have to come back to rebalance the loads.

Nonsense. Twenty 100W light bulbs is NOT going to trip a 20A breaker.

Tell me what you would do with this problem !

I wouldn't do anything about it -- because it is NOT a problem.

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

It's time to throw all their damned tea in the harbor again.

wrote:
Stormin Mormon wrote:


volts times amps = watts

So, you can say that the 110 volt circuit, times 20 amps = 2200 watts.

Hint: Try using flourescent, or compact fluorescent bulbs. They use a lot
less power.


At a lower power factor... watts = 0.4 x volts x amps,
for some undercounter fluorescents.

Nick

Do fluorescent lights have a power factor? would
like to know. But, that said, if the undercounter
light fixture say 15W, then that is how much is
used regardless of power factor and w=va still
applies. or am I wrong and the labels are incorrect.

"Doug Miller" wrote in message
m...
In article , "TURTLE"
wrote:

"Doug Miller" wrote in message
. com...
In article , "TURTLE"
wrote:


When Speaking of a Electrical breaker in a breaker box suppling electricity
and
the 80% rule does not matter if the elctricity is supplied to a whole home
or

to
a one light bulb. If you have a 20 amp breaker in the box. You better not
put
more that a 80% of 20 amp load on it. So You can have a 16 amp load to be
supplied to the 20 amp breaker.

Wrong.

i know now that your not a electrician for you would know exactly what I was
speaking about and you would be telling me about the 80% rules. So 80%
Applies
to everything that will pull amps from a breaker and if you want to change
anything. Please be welcome to do as you please.

What you say here is completely wrong.

It's quite obvious that *you* are not an electrician, or you would know that
the 80% rule refers to continuous loads, defined by the NEC as "a load where
the maximum current is expected to continue for three hours or more." This
does *not* apply to residential lighting circuits, or indeed to most other
loads.


This is Turtle.

first i don't do residentiual electric work and only commercial HVAC which
makes me do the electric for them.

Then perhaps you shouldn't be trying to answer questions about residential
electrical work...

Second if you was running power to lights and the lighting wattage was 2,400
watts / 24 -- 100 watt light bulbs and then would you say you would hook all
them up to a 20 amp breaker which the load would be matching a 2,400 watt
load.
Then or would you only connect 19.6 light bulb on the 20 amp breaker to not
over
load the breaker and trip if all the lights were turned on at one time. Awwwww
that would be the 80% rule. When every you pull over 80% of the breakers
rating.

No, that's just plain wrong. The 80% rule DOES NOT APPLY in this situation
because it is NOT a "continuous load" as defined in the NEC.
Your asking it to trip and you have to come back to rebalance the loads.

Nonsense. Twenty 100W light bulbs is NOT going to trip a 20A breaker.

Tell me what you would do with this problem !

I wouldn't do anything about it -- because it is NOT a problem.

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

It's time to throw all their damned tea in the harbor again.

This is Turtle.

You missed the question of what you would do in this case. I stated that I would
not put but 20 -- 100 watt light bulbs on a 20 amp circuit / breaker. The 20
light bulbs would be the max i would put on the 20 amp breaker because of the
80% rule.

what is the max number of 100 watt light bulbs that you can put on a 20 amp
breaker ?

So , What is the answer to this question ?

TURTLE

Now, I am really confused. I never thought this would be such a complex
issue.

It seemed like a simple question, but now I am not so sure. I am thinking
of cancelling the sale because of all of this.

I mean, I lie awake nights now, thinking of amperage, wattage, voltage, and
things I didn't even know existed last week.

The current debate has caused me to think of a lot of related issues.

Will the atmospheric pressure change affect these bulbs? What if I buy them
here in Nevada and take them to Utah? Will the change in elevation or time
zone affect their performance?

Will the "long life" bulbs violate the religious beliefs of the area where
there are strong afterlife philosophies? Would I be in violation of local
traditions by bringing in these "long life" bulbs?

What happens if I violate this 80% rule? Say, by 2%? Was this one of the
commandments that was dropped by Mel Brooks playing Moses in "The History of
the World, Part One"?

If I use bulbs made outside of the US, can I, in good conscience, ask my
union electrician buddy to work on this project?

There are five 20 amp breakers. That will give me up to 20,000 watts in a
1200 square foot cabin. Or, a mere 16,000 if I listen to the 80%
philosophy. My question is, IF the Space Shuttle were flying, will it be
visible from space? That's a lot of light.

Should I just use Halogens?

Do you think I will need eye protection?

Do you think it will cause sunburn or carpet discoloration?

In the meantime, I believe I will just stick to flashlights, candles, and
kerosene lights.

All this other stuff is just so complicated and confusing.

Right now, I'm going to take six valiums and try to wind down.

Steve

George E. Cawthon wrote:

...watts = 0.4 x volts x amps, for some undercounter fluorescents.

Do fluorescent lights have a power factor?

Sure... 0.4 in the case above.

But, that said, if the undercounter light fixture say 15W, then that
is how much is used regardless of power factor

Yes.

and w=va still applies. or am I wrong...

You am wrong. W = PFxVxA.

Nick

In article , "TURTLE" wrote:

"Doug Miller" wrote in message
om...


I wouldn't do anything about it -- because it is NOT a problem.

This is Turtle.

You missed the question of what you would do in this case.

No, you missed my answer - right above.

I stated that I would not put but 20 -- 100 watt light bulbs on a 20 amp
circuit / breaker. The 20 light bulbs would be the max i would put on the 20
amp breaker because of the 80% rule.

Yes, I know that's what you stated. But you're wrong. You're having a very
hard time grasping this simple concept: the 80% rule DOES NOT APPLY in this
situation.

what is the max number of 100 watt light bulbs that you can put on a 20 amp
breaker ?

So , What is the answer to this question ?

20A x 120V / 100W = 24 bulbs.

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

It's time to throw all their damned tea in the harbor again.

wrote:
George E. Cawthon wrote:


...watts = 0.4 x volts x amps, for some undercounter fluorescents.



Do fluorescent lights have a power factor?


Sure... 0.4 in the case above.


But, that said, if the undercounter light fixture say 15W, then that
is how much is used regardless of power factor


Yes.


and w=va still applies. or am I wrong...


You am wrong. W = PFxVxA.

Nick


Did you go to embassy doublespeak school? If you
did, congratulations in your studies. You
certainly confused me. You answered the 2nd
question yes and the third saying I am wrong.
Those answers are inconsistent. The subject was
totaling up the wattage or amps. If the fixtures
says 15 watts and it has a power factor, the
powerfactor is already applied. If you total it
up with other appliances you don't apply the power
factor again.

I think I'll just stick with totaling up the
wattage of fixtures/appliances.

George E. Cawthon wrote:

wrote:

...watts = 0.4 x volts x amps, for some undercounter fluorescents.

Do fluorescent lights have a power factor?

Sure... 0.4 in the case above.

But, that said, if the undercounter light fixture say 15W, then that
is how much is used regardless of power factor

Yes.

and w=va still applies. or am I wrong...

You am wrong. W = PFxVxA.

Did you go to embassy doublespeak school?

Ignorance and belligerance are an unfortunate combination.

...You answered the 2nd question yes and the third saying I am wrong.
Those answers are inconsistent.

No... In the case above, the real power is 15W = 0.4x120Vx0.3125A, but
"w=va" = 120Vx0.3125A = 37.5 VA overestimates it by a factor of 2.5.

If the fixtures says 15 watts and it has a power factor, the
powerfactor is already applied.

Yes.

I think I'll just stick with totaling up the
wattage of fixtures/appliances.

That's a good idea, in most cases.

Nick

"Doug Miller" wrote in message
. ..
In article , "TURTLE"
wrote:

"Doug Miller" wrote in message
. com...


I wouldn't do anything about it -- because it is NOT a problem.

This is Turtle.

You missed the question of what you would do in this case.

No, you missed my answer - right above.

I stated that I would not put but 20 -- 100 watt light bulbs on a 20 amp
circuit / breaker. The 20 light bulbs would be the max i would put on the 20
amp breaker because of the 80% rule.

Yes, I know that's what you stated. But you're wrong. You're having a very
hard time grasping this simple concept: the 80% rule DOES NOT APPLY in this
situation.

what is the max number of 100 watt light bulbs that you can put on a 20 amp
breaker ?

So , What is the answer to this question ?

20A x 120V / 100W = 24 bulbs.

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

It's time to throw all their damned tea in the harbor again.

This is Turtle.

I see you would load up a break to it's max. amps and if the people in the house
happen to turn on all 24 light bulbs. It will blow the breaker in about 2 or 3
hours and then call you to ask why did my break throw. You maybe able to do this
by NEC but I will not load anything up to the max. what so ever.

Now what I did up above by just putting 20 light bulbs on a 20 amp breaker is
Nothing wrong with doing it. You had said it was wrong well i tell you it is
nothing wrong with it if i wanted to do it. Now you can explain any wrong as NEC
goes please explain it to me. Let me explain to you what right and wrong is.
Right is you can do it. Wrong is you Can't do it. Now take these two words and
explain wrong as you say above here.

TURTLE

"SteveB" wrote in message
news:saWKe.697$DW1.169@fed1read06...
Now, I am really confused. I never thought this would be such a complex
issue.

It seemed like a simple question, but now I am not so sure. I am thinking of
cancelling the sale because of all of this.

I mean, I lie awake nights now, thinking of amperage, wattage, voltage, and
things I didn't even know existed last week.

The current debate has caused me to think of a lot of related issues.

Will the atmospheric pressure change affect these bulbs? What if I buy them
here in Nevada and take them to Utah? Will the change in elevation or time
zone affect their performance?

Will the "long life" bulbs violate the religious beliefs of the area where
there are strong afterlife philosophies? Would I be in violation of local
traditions by bringing in these "long life" bulbs?

What happens if I violate this 80% rule? Say, by 2%? Was this one of the
commandments that was dropped by Mel Brooks playing Moses in "The History of
the World, Part One"?

If I use bulbs made outside of the US, can I, in good conscience, ask my union
electrician buddy to work on this project?

There are five 20 amp breakers. That will give me up to 20,000 watts in a
1200 square foot cabin. Or, a mere 16,000 if I listen to the 80% philosophy.
My question is, IF the Space Shuttle were flying, will it be visible from
space? That's a lot of light.

Should I just use Halogens?

Do you think I will need eye protection?

Do you think it will cause sunburn or carpet discoloration?

In the meantime, I believe I will just stick to flashlights, candles, and
kerosene lights.

All this other stuff is just so complicated and confusing.

Right now, I'm going to take six valiums and try to wind down.

Steve

This is Turtle.

I would just go ahead and Punt on First 10 .

TURTLE

"TURTLE" wrote

I would just go ahead and Punt on First 10 .

TURTLE

I would, but I am afraid that I might hit a light bulb.

Steve

In article , "TURTLE" wrote:

Now what I did up above by just putting 20 light bulbs on a 20 amp breaker is
Nothing wrong with doing it. You had said it was wrong well i tell you it is
nothing wrong with it if i wanted to do it.

Please read more carefully, Turtle. I never said that putting only 20 100W
bulbs on a 20A breaker was wrong -- I said you were wrong to claim that 19
bulbs was the maximum permitted because of the 80% rule.

Now you can explain any wrong as
NEC goes please explain it to me.

I've explained it several times already, but you're not paying attention.
Let's try again. The 80% rule applies to continuous loads. Residential
lighting is not a "continuous load" as defined in the NEC. Therefore the 80%
rule does not apply to residential lighting circuits.

Let me explain to you what right and wrong is.
Right is you can do it. Wrong is you Can't do it.

Yep, and by that definition putting twenty-four 100W light bulbs on a 20A 120V
circuit is right. So is twenty bulbs. Or five bulbs. Or one.

Now take these two words and
explain wrong as you say above here.

There's nothing wrong with loading a circuit to less than its capacity. What's
wrong is your understanding of the capacity of a 20A circuit when used for
residential lighting, and your understanding of the 80% rule.



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

It's time to throw all their damned tea in the harbor again.

In article ,
wrote:
George E. Cawthon wrote:

...watts = 0.4 x volts x amps, for some undercounter fluorescents.

Do fluorescent lights have a power factor?

Sure... 0.4 in the case above.

But, that said, if the undercounter light fixture say 15W, then that
is how much is used regardless of power factor

Yes.

and w=va still applies. or am I wrong...

You am wrong. W = PFxVxA.

Nick


QUESTION about this "power factor":

(it has been *so long* since I understood any
of that stuff, that I've forgotten all but a
few words describing it.)

With DC, the pf is 1.0?

With AC, I'm not sure what it is.

Of course there's the "rms" stuff, trying to
get an average value of a sine-wave.

The pf, I recall from *ages* (decades) ago, had something to
do with the voltage and current "waves" getting out of sync
with each other, due to a coil or a capacitor (one shifting
in one direction, one in the other).

I recall something about having to use trig to get
the pf, maybe it was the sine or cosine of the degrees
of lead or lag?

If so, then since those functions range between plus and
minus one, then maybe the pf I dimly recall is the
reciprocal of yours?

Anyway -- here's my question:

How do you get a substantial pf for a fluorscent (sp?)
light? (Huge electric motor, I understand how.)

And a pf for an incandescent light, that would involve
no phase shift at all?

Obvously, I could use some mental fill-in!

Thanks!

David

David Combs wrote:

With DC, the pf is 1.0?

Yes.

With AC, I'm not sure what it is.

With AC, it's the sum of instantaneous products of current and voltage
over a cycle, divided by the product of their average values.
It's 1.0 if current is proportional to voltage.

It's also the number you see when you push the PF button on a Kill-a-Watt
meter. I did that with a "0.1A 120V" Little Giant fountain pump the size
of a golfball and saw 0.42. The meter said the pump used 5 watts.

Nick

"With AC, it's the sum of instantaneous products of current and voltage

over a cycle, divided by the product of their average values.
It's 1.0 if current is proportional to voltage. "

David's memory is correct. For AC, its the cosine of the phase angle
between the voltage and current. When there is zero phase shift, the
power factor is 1.0

wrote:

"With AC, it's the sum of instantaneous products of current and voltage
over a cycle, divided by the product of their average values.
It's 1.0 if current is proportional to voltage. "

David's memory is correct. For AC, its the cosine of the phase angle
between the voltage and current. When there is zero phase shift, the
power factor is 1.0

What's the phase shift of a switching power supply? :-)

Nick

On Sun, 11 Sep 2005 03:58:33 -0700, trader4 wrote:

"With AC, it's the sum of instantaneous products of current and voltage

over a cycle, divided by the product of their average values.
It's 1.0 if current is proportional to voltage. "

David's memory is correct. For AC, its the cosine of the phase angle
between the voltage and current. When there is zero phase shift, the
power factor is 1.0

As Nick Pine eludes to, this isn't a good definition because it doesn't
take into account the harmonic content of the waveforms. It works for
purely sinusoidal voltage abd current though. Power factor is more
appropriately the Power divided by RMS Volts * RMS Amps, or PF == P/VA.
The power can always be found by averaging the instantaneous V*A over the
cycle, as nicksanspam wrote above.

--
Keith

On Sun, 11 Sep 2005 02:08:43 +0000, David Combs wrote:

In article ,
wrote:
George E. Cawthon wrote:

...watts = 0.4 x volts x amps, for some undercounter fluorescents.

Do fluorescent lights have a power factor?

Sure... 0.4 in the case above.

But, that said, if the undercounter light fixture say 15W, then that
is how much is used regardless of power factor

Yes.

and w=va still applies. or am I wrong...

You am wrong. W = PFxVxA.

Nick


QUESTION about this "power factor":

(it has been *so long* since I understood any
of that stuff, that I've forgotten all but a
few words describing it.)

With DC, the pf is 1.0?

PF is kinda meangless with DC, since P=VA.

With AC, I'm not sure what it is.

Of course there's the "rms" stuff, trying to
get an average value of a sine-wave.

Not average. When Volts is multiplied by amps (power) there is a squared
term in there. RMS == Root Means Square (root of the mean square), so
when you calculate power, the squared term is taken into account. Average
voltage isn't all that useful; the average for sine wave is zero. ;-)

The pf, I recall from *ages* (decades) ago, had something to do with the
voltage and current "waves" getting out of sync with each other, due to
a coil or a capacitor (one shifting in one direction, one in the other).

Or a non-linear load adding harmonic content to the waveforms.

I recall something about having to use trig to get the pf, maybe it was
the sine or cosine of the degrees of lead or lag?

Cosine of the angle between them, but that really only works for
pure sine waves.

If so, then since those functions range between plus and minus one, then
maybe the pf I dimly recall is the reciprocal of yours?

Anyway -- here's my question:

How do you get a substantial pf for a fluorscent (sp?) light?

The fluorescent is a non-linear load. It chops the current waveform,
adding harmonics to it. The above formula isn't very useful in this case.

(Huge electric motor, I understand how.)

A "huge electric motor" should have a PF very close to one if it's heavily
loaded. If it's creating no mechanical work, then it will have a worse
PF.

And a pf for an incandescent light, that would involve no phase shift at
all?

The phase shift will be small, but it will only conduct current during
part of the cycle, chpping the current waveform. Since the current isn't
being used during the entire waveform, the power uring this part of the
cycle isn't useful.

Obvously, I could use some mental fill-in!

Thanks!

--
Keith

In article ,
keith wrote:
On Sun, 11 Sep 2005 03:58:33 -0700, trader4 wrote:

"With AC, it's the sum of instantaneous products of current and voltage

over a cycle, divided by the product of their average values.
It's 1.0 if current is proportional to voltage. "

David's memory is correct. For AC, its the cosine of the phase angle
between the voltage and current. When there is zero phase shift, the
power factor is 1.0

As Nick Pine eludes to, this isn't a good definition because it doesn't
take into account the harmonic content of the waveforms. It works for
purely sinusoidal voltage abd current though. Power factor is more
appropriately the Power divided by RMS Volts * RMS Amps, or PF == P/VA.
The power can always be found by averaging the instantaneous V*A over the
cycle, as nicksanspam wrote above.

--
Keith

Thanks, all!

Now, where (on net) can I learn more, esp about
that paragraph above?

(Yeah, I can google, but I'm not quite sure for
exactly what, and *maybe* one of you already have
a good site in mind.)

David

In article ,
keith wrote:
On Sun, 11 Sep 2005 02:08:43 +0000, David Combs wrote:

....

Of course there's the "rms" stuff, trying to
get an average value of a sine-wave.

Not average. When Volts is multiplied by amps (power) there is a squared
term in there. RMS == Root Means Square (root of the mean square), so
when you calculate power, the squared term is taken into account.
Yes, I now recall that.

Average
voltage isn't all that useful; the average for sine wave is zero. ;-)
Indeed!

The pf, I recall from *ages* (decades) ago, had something to do with the
voltage and current "waves" getting out of sync with each other, due to
a coil or a capacitor (one shifting in one direction, one in the other).

Or a non-linear load adding harmonic content to the waveforms.

....
....
How do you get a substantial pf for a fluorscent (sp?) light?

The fluorescent is a non-linear load. It chops the current waveform,
adding harmonics to it.

The harmonics being like the terms in a (*very* dimly recalled) fourier series
or transform that sum up to approximate the chopping-caused square or
impulse or whatever waveform it is?

As I asked in my other reply, do you know offhand any
good sites that cover this stuff (with drawings, too)?

The above formula isn't very useful in this case.

So, what is?

(Huge electric motor, I understand how.)

A "huge electric motor" should have a PF very close to one if it's heavily
loaded. If it's creating no mechanical work, then it will have a worse
PF.

Jeez, on the surface, that sounds like the *opposite*, maybe,
of what one would expect???

Maybe you could say some more?

(I knew way back then that going for a double-E would be
way over my ability!)


And a pf for an incandescent light, that would involve no phase shift at
all?

The phase shift will be small, but it will only conduct current during
part of the cycle, chpping the current waveform. Since the current isn't
being used during the entire waveform, the power uring this part of the
cycle isn't useful.

Wow... Maybe you could say a bit more about that!

(This stuff is NOT simple!)


Thanks to all for a fascinating look into the dark.

David