installing Malibu landscape lighting
Freezout
I bought a Malibu kit. It comes with 75 feet of cable. The instructions
say that, as a rule of thumb, 16 gauge cable can be used for uns up to100 feet,
14-gauge for up to 150 feet. However, it doesn't say what gauge
is included in the kit. A look at the cable and I see it has "18 AWG 2/C
underground low energy circuit cable" stamped into it. Does this
mean it is 18 gauge? (If so, how nice of them to include something even
cheaper than what they describe).
The instructions go on to say, "For added efficiency , you can mix heavier
gauge
cable with existing 16 or 14-gauge, but heavier cable should be installed
closest
to the transformer. Does this mean I can add to the 75 foot length with a
higher
gauge if I like? If so, I take it then the most I could add to it would be 25
feet
of 16 gauge. Although, it seems like it would be even less than 25 feet, since
it's limit is at 100 feet even if it were *all* 16 guage, much less only one
quarter
of the it. How does this work?
Also, can I fork off the cable in two different directions, or does it have to
be in one continuous line? If I can do this, does it effect the total
length of cable I can use. I've seen these cable couplers that allow you
to attach more cable. They couple the two pieces of cable side-by-side in
a thing you clamp over them with penetrates the cable to create a connection
between them. I'm wondering if I could use one of these to feed a branch
of cable off in another direction. If I can, it would be more efficient for me
because I wouldn't have to double back the cable where it's already been
laid to then head in the opposite direction (ie, I have to lay it in a sort of
"T" shaped configuration.)
Thanks,
Free...@aol.com
JCoggins
Seems like a lot of Malibu 12-volt stuff comes with transformers rated about
300 watts -- maybe less. It's not so much a hazard as it is an
energy-waster, which is one reason why they encourage you to use a larger
gauge more "efficient" wire. The other is voltage drop on a long run.
Suggest you try a 14 AWG (and, since this is outdoors, get it rated for
direct burial). If you want to use the 18 AWG, put it farther out on the
last half of the run where its higher resistance will cause less trouble.
If you're lighting each side of a walkway, you don't have to go out and back
in one long run. Two separate parallel runs that both connect to the
transformer will work better.
Freezout wrote in message <19981125170956...@ng113.aol.com>...
Freezout
>If it says 18 AWG then it's 18 gauge -- and that does seem awfully small.
>Seems like a lot of Malibu 12-volt stuff comes with transformers rated about
>300 watts -- maybe less.
This comes with an 88 watt power pack and
eight 7 watt bulbs, which I believe will be
more than enough. All of the lights and cable
will be above ground, either mounted on the
deck railing or side of the house.
> It's not so much a hazard as it is an
>energy-waster, which is one reason why they encourage you to use a larger
>gauge more "efficient" wire. The other is voltage drop on a long run.
Energy waster in the sense that it will cost
me more to run the lights using this cable,
or in the sense that the lights won't shine
as brightly?
>Suggest you try a 14 AWG (and, since this is outdoors, get it rated for
>direct burial). If you want to use the 18 AWG, put it farther out on the
>last half of the run where its higher resistance will cause less trouble.
>
>If you're lighting each side of a walkway, you don't have to go out and back
>in one long run. Two separate parallel runs that both connect to the
>transformer will work better.
Okay to connect two cables to the power pack
then. That's one of the things I was wondering.
Still wondering about the use of the cable
JCoggins
AWG cable they gave you is probably rated for 88W/12V = 7.3 amps. But the
tables I have nearby say #18 has a resistance of 6.39 ohms per 1000 feet,
compared to 2.52 ohms per 1000 for #14. So you're going to waste a little
more energy in the smaller cable; it's lost as heat.
Don't see why you can't "T" the cable (more than once even), especially at
this low voltage. At higher voltages it wouldn't be quite so easy. But make
the connections water-tight. Usually, that means a junction box of some
kind.
Freezout wrote in message <19981126110609...@ng-cf1.aol.com>...
JCoggins
If your 88W transformer approaches full load, say 7 amps @ 12V, the power
loss in 75 feet of #18 AWG will be huge! Power loss depends on the square of
the current, which would be (7*7 = 49). Multiplying that times wire
resistance is 49 * [6 ohms *(75ft / 1000 ft)] = 22 watts. But this is only
one way; you have that much resistance back again for double the loss -- or
44 watts.
Amazing that Malibu would sell wire that wastes half the available power.
But if that's what you say you got .... guess I'd suggest the #14 AWG, which
would have only 1/3 the loss.
What did this kit cost, BTW?
JCoggins wrote in message <73k7he$jhd$1...@ash.prod.itd.earthlink.net>...
Freezout
>Didn't really look at this too closely last time but ... hold on.
>
>If your 88W transformer approaches full load, say 7 amps @ 12V, the power
>loss in 75 feet of #18 AWG will be huge!
Okay. I know nothing about wiring, electricity, etc. Does this loss manifest
itself in my monthly bill, or merely by dimmer lights?
I can take the dimmer lights,
because I'm actually concerned about them being too
bright anyway, and shining in the neighbors windows
at night. (There's only about 8-ft between houses
along the walkway.) I just want enough light so
as not to stumble. I may not even use all of the
eight 7 watt lamps provided.
>Power loss depends on the square of
>the current, which would be (7*7 = 49). Multiplying that times wire
>resistance is 49 * [6 ohms *(75ft / 1000 ft)] = 22 watts. But this is only
>one way; you have that much resistance back again for double the loss -- or
>44 watts.
>
>Amazing that Malibu would sell wire that wastes half the available power.
Surprised me because intially I was looking at buying
the components separately and the Malibu cable they
were selling were lower guages. I didn't expect
Malibu to spring this kind of thing on my when
I bought the kit.
>But if that's what you say you got .... guess I'd suggest the #14 AWG, which
>would have only 1/3 the loss.
>
>What did this kit cost, BTW?
This cost $50 at places like Lowes and Home Depot.
They have another kit that has deck lights a step
down from these ( I think maybe 6 watt lamps and
without the covers that you can snap on to direct
the light down ward) for $40.
Kit prices are considerably cheaper than if I were
to buy separately. Kind of makes me wonder now
if perhaps the 88w transformers in the kits are of
a lower grade than the 88w transformers they
sell separately, given what they do with their cable.
JCoggins
watts to keep the cable warm. The wasted energy is roughly equivalent to the
load on a VCR.
Freezout wrote in message <19981127135128...@ng146.aol.com>...
Bob Kaplow
> A look at the cable and I see it has "18 AWG 2/C
> underground low energy circuit cable" stamped into it. Does this
> mean it is 18 gauge? (If so, how nice of them to include something even
Yes!
> Also, can I fork off the cable in two different directions, or does it have to
> be in one continuous line? If I can do this, does it effect the total
> length of cable I can use. I've seen these cable couplers that allow you
> to attach more cable. They couple the two pieces of cable side-by-side in
> a thing you clamp over them with penetrates the cable to create a connection
> between them. I'm wondering if I could use one of these to feed a branch
> of cable off in another direction. If I can, it would be more efficient for me
> because I wouldn't have to double back the cable where it's already been
> laid to then head in the opposite direction (ie, I have to lay it in a sort of
> "T" shaped configuration.)
The clamps are used to connect fixtures along the cable without cutting the
cable. But they can be used to "T" as well. Theoretically the distance you
need to control is from the transformer to the farthest single point, but in
reality I'd suggest limiting the whole run to whatever they resay for a
single run, especially if you use the wimpy 18ga stuff.
BTW, you can actually wire a closed loop, where the wire connects back to
itself. Essentially, you're feeding both ends of the wire, reducing the loss
to the far end considerably. That's how most of mine (NOT malibu...) are
connected. Be sure you match the polarity on the wire if you do this, or
you'll blow lots of fuses. The wire usually has one side with little ridges,
while the other side is smooth.
Bob Kaplow
danh...@infonet.isl.net
>In article <19981125170956...@ng113.aol.com>, free...@aol.com (Freezout) writes:
>> A look at the cable and I see it has "18 AWG 2/C
>> underground low energy circuit cable" stamped into it. Does this
>> mean it is 18 gauge? (If so, how nice of them to include something even
>
>Yes!
>
>> Also, can I fork off the cable in two different directions, or does it have to
>> be in one continuous line? If I can do this, does it effect the total
>> length of cable I can use. I've seen these cable couplers that allow you
>> to attach more cable. They couple the two pieces of cable side-by-side in
>> a thing you clamp over them with penetrates the cable to create a connection
>> between them. I'm wondering if I could use one of these to feed a branch
>> of cable off in another direction. If I can, it would be more efficient for me
>> because I wouldn't have to double back the cable where it's already been
>> laid to then head in the opposite direction (ie, I have to lay it in a sort of
>> "T" shaped configuration.)
>
>The clamps are used to connect fixtures along the cable without cutting the
>cable. But they can be used to "T" as well. Theoretically the distance you
>need to control is from the transformer to the farthest single point, but in
>reality I'd suggest limiting the whole run to whatever they resay for a
>single run, especially if you use the wimpy 18ga stuff.
Also, when you split off another feed line like this, use TWO of the clamp
on connectors. The connections made by these things are sufficiently
unreliable that the extra conductivity of the second one will help assure
trouble-free operation.
Dan Hicks
Hey!! My advice is free -- take it for what it's worth!
http://www.millcomm.com/~danhicks
---== http://www.newsfeeds.com - Largest Usenet Server In The World! ==---
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danh...@infonet.isl.net
less power as the resistance of the circuit increases (V**2/R is the power
formula).
In <19981128210103...@ng98.aol.com>, free...@aol.com (Freezout) writes:
>In other words, are you saying that if I use an
>88 watt transformer, I will use 88 watts of energy
>regardless of the number of lamps (equal to or
>less than the max allowed) or the gauge
>of the cable?
>
>"JCoggins" <cog...@earthlink.net> wrote:
>
>>You'll pay for the whole 88 watts; you'll use 44 watts for light, plus 44
>>watts to keep the cable warm. The wasted energy is roughly equivalent to the
>>load on a VCR.
>>
>>Freezout wrote in message <19981127135128...@ng146.aol.com>...
>>>"JCoggins" <cog...@earthlink.net> wrote:
>>>>If your 88W transformer approaches full load, say 7 amps @ 12V, the power
>>>>loss in 75 feet of #18 AWG will be huge!
>>>
>>>Okay. I know nothing about wiring, electricity, etc. Does this loss
>>manifest
>>>itself in my monthly bill, or merely by dimmer lights?
Freezout
Freezout
Freezout
>The clamps are used to connect fixtures along the cable without cutting the
>cable. But they can be used to "T" as well. Theoretically the distance you
>need to control is from the transformer to the farthest single point, but in
>reality I'd suggest limiting the whole run to whatever they resay for a
>single run, especially if you use the wimpy 18ga stuff.
After looking it over, I think I've found
a way to do it efficiently in a continuous run,
so I won't need to "T." The total
run will be about 150 feet. The instructions
say as a rule of thumb, 16 gauge can be used
for up to 100 ft, and 14 gauge for up to 150.
If I'm understanding this right, buying a 100
feet of 16 guage and then tacking on 50 ft of the
18 guage that came with the set wouldn't cut it.
But how about buying 100 ft of 14 gauge and
tacking on 50 ft of the 18 at the end? Is
this proper. As far as dimmer lights go on
that last 50ft, this would actually be good, since
at this point if I only got 4 watts out of the
7 watt lamps, that may be all the light I want
anyway.
Topher Eliot
>After looking it over, I think I've found
>a way to do it efficiently in a continuous run,
>so I won't need to "T." The total
>run will be about 150 feet. The instructions
>say as a rule of thumb, 16 gauge can be used
>for up to 100 ft, and 14 gauge for up to 150.
>
>If I'm understanding this right, buying a 100
>feet of 16 guage and then tacking on 50 ft of the
>18 guage that came with the set wouldn't cut it.
Right, that's not good enough.
>But how about buying 100 ft of 14 gauge and
>tacking on 50 ft of the 18 at the end? Is
>this proper.
Not really, and I wouldn't do it. The cost
of the wire is small compared to your labor
ins installing all this stuff. It would be a
real pisser to get it all put in and decide
you had to redo it with thicker wire.
Topher Eliot
el...@alum.mit.edu
Visit the house maintenance archive at http://www.geocities.com/heartland/7400
JCoggins
offered (as noted) refer to a gross load that draws full load current from
the transformer. I figured it would be more or less intuitive that if you
use fewer lights, you also use less energy. Sorry if it wasn't.
You can also use V^2/R to calculate power (as suggested in another post
here) but, like many freshmen studying electricity, you'll find it's a
specialized forumula that gives bad results if you don't understand exactly
what voltage and resistance you're dealing with.
Freezout wrote in message <19981128210103...@ng98.aol.com>...
Freezout
>I have no idea how many lights you're actually going to use. The numbers I
>offered (as noted) refer to a gross load that draws full load current from
>the transformer. I figured it would be more or less intuitive that if you
>use fewer lights, you also use less energy. Sorry if it wasn't.
I'll repeat what I said in my initial post that you
quoted (below). I know nothing about wiring, electricity, etc. :)
You said "You'll pay for the whole 88 watts; you'll use 44 watts for light,
plus 44
watts to keep the cable warm." That makes
the assumption that 44 watts of lamps are used
if I understand correctly. So the point I
figured you might be trying to make then is
that if I lowered the number of lamps to say
36 watts, then it would be the remaining 62 watts
that would keep the cable warm-- that I 88 watts
of energy was going to be expended regardless of
the gauge of cable or number of lights I used.
I'm not sure how the waste of 44watts in your initial
sentence comes about. That's what I'm trying to understand.
And I'm only interested in waste in terms of money.
Waste in terms of energy lost in a wire that doesn't
effect my bill is not something I'm worried about,
unless it turns out the lamps are too dim for my purposes.
>You can also use V^2/R to calculate power (as suggested in another post
>here) but, like many freshmen studying electricity, you'll find it's a
>specialized forumula that gives bad results if you don't understand exactly
>what voltage and resistance you're dealing with.
Thanks. I don't think there's any need for a formula
here. :) I am merely trying to find out if by "waste",
we are talking about cost of operation. And, knowing
nothing about electricity, I haven't a clue how to make
use of that formula.
JCoggins
watt-hours. That's watts used multiplied times hours of use. (The actual
units work out to be energy: a joule per unit time times actual time.)
How your particular outdoor lighting system actually performs depends on
much cable you string, what size it is, and how many lights you install;
also on losses that take place in the transformer, since it's not perfectly
efficient (less efficient at lower loads). But since you haven't installed
anything yet, nobody knows these numbers.
Let's guess that, due to the small gauge cable, you do waste an average of
44 watts an hour for several hours or 132 watt-hours per day. Over a 30-day
billing cycle, that's 3,960 watt-hours or almost 4 KWH (kilowatt hours). I
don't know what your utility charges you per KWH; call and find out. But if
you use 1000 KWH total, the energy wasted in the cable alone would increase
your bill by about 0.4%. The lights will add about that much more.
Add that on top of the initial $50 capital investment and I don't think
you're getting a good return. But enjoy the lights.
Freezout wrote in message <19981129125526...@ng94.aol.com>...
Freezout
>
>Let's guess that, due to the small gauge cable, you do waste an average of
>44 watts an hour for several hours or 132 watt-hours per day. Over a 30-day
>billing cycle, that's 3,960 watt-hours or almost 4 KWH (kilowatt hours). I
>don't know what your utility charges you per KWH; call and find out. But if
>you use 1000 KWH total, the energy wasted in the cable alone would increase
>your bill by about 0.4%. The lights will add about that much more.
>
>Add that on top of the initial $50 capital investment and I don't think
>you're getting a good return. But enjoy the lights.
Well, I have to have the lights, so whatever
it costs I'll to pay. Tell me if this sums up
what you are saying correctly:
All other things being equal, it will cost me
more money each month to operate these
lights if I wire with 18 gauge cable than if
I wire with 14 gauge cable.
Thanks,
Free...@aol.com
Bill Neisius
JCoggins wrote in message <73s45d$4gh$1...@oak.prod.itd.earthlink.net>...
>Let's guess that, due to the small gauge cable, you do waste an average of
>44 watts an hour for several hours or 132 watt-hours per day. Over a 30-day
>billing cycle, that's 3,960 watt-hours or almost 4 KWH (kilowatt hours). I
>don't know what your utility charges you per KWH; call and find out. But if
>you use 1000 KWH total, the energy wasted in the cable alone would increase
>your bill by about 0.4%. The lights will add about that much more.
>
>Add that on top of the initial $50 capital investment and I don't think
>you're getting a good return. But enjoy the lights.
Now wait a second...
If you put "88 watts" of lights on a 12V transformer, that would be a load
of
approximately 1.6 ohms. In a previous message, it was revealed that the
wire has 6ohms/1000ft of resistance, so 75ft of wire is 0.45 ohms.
Since the wire resistance is in series with the lamp resistance, the total
load
on the transformer would be the sum of the two, or approximately 2 ohms.
That means that the "88 watts" of lights would only be consuming 72 watts
of power from the transformer... so actually you would be SAVING 18%
energy!
JCoggins
180-kilowatt hours on a 1000 KWH bill? Remarkable. But when you think about
it, there IS a full moon somewhere -- all the time.
Bill Neisius wrote in message <73ufga$t15$1...@fir.prod.itd.earthlink.net>...
Freezout
>Tell me if this sums up
>what you are saying correctly:
>
>All other things being equal, it will cost me
>more money each month to operate these
>lights if I wire with 18 gauge cable than if
>I wire with 14 gauge cable.
>
>Thanks,
>Free...@aol.com
In response,
"JCoggins" <cog...@earthlink.net> wrote:
>If you consider power lost in the cable as wasted energy (it didn't go into
>the lights) then, yes: for a fixed load of X watts, the smaller gauge will
>cost you more to operate than the larger.
What difference does it make whether it goes to the
lights or is lost in the cable have on the answer to
the question? You've lost me, again. I'm not sure
I get your meaning. It seems as if you are saying that
both the 18 and 14 gauge will consume the
same amount of energy during operation, but
there will be a difference in the amount of energy that
reaches the lights. Fine. But it also seems that
you are saying that the 18 gauge will cost more to
operate than the 14 gauge. But this is counter
to the previous point, for if both wires consume
the same amount of energy (regardless of how
much goes to the lights), then they will cost the same.
I keep trying to make this simple, but don't seem to
be getting anywhere. :)
>Now answer me a question. In a business environment, would you approve
>installation of a motor or some outdoor lights or other capital equipment in
>which half the power was dissipated in the wiring?
Forget about any other situation other than what
I'm dealing with. It will keep things simple.
Freezout
Freezout
>That may be what he's trying to say, but it's not correct. You'll use
>less power as the resistance of the circuit increases (V**2/R is the power
>formula).
>
> free...@aol.com (Freezout) writes:
>>In other words, are you saying that if I use an
>>88 watt transformer, I will use 88 watts of energy
>>regardless of the number of lamps (equal to or
>>less than the max allowed) or the gauge
>>of the cable?
Dan, help me out here. What I want to know
is all other things being equal, will I spend more
money each year to operate the lights if I
use a 18 gauge cable as opposed to a 14 gauge?
Thanks,
Free...@aol.com
danh...@infonet.isl.net
The difference is negligible, but you'll spend more money with the larger
14 gauge cable. However, with the 14 gauge you will waste less money
heating the cable (ie, you will get more light output).
Freezout
>free...@aol.com (Freezout) writes:
>>Dan, help me out here. What I want to know
>>is all other things being equal, will I spend more
>>money each year to operate the lights if I
>>use a 18 gauge cable as opposed to a 14 gauge?
>
>The difference is negligible, but you'll spend more money with the larger
>14 gauge cable. However, with the 14 gauge you will waste less money
>heating the cable (ie, you will get more light output).
>
>Dan Hicks
Thank you! As I said in my prior posts, a decrease in
light output at the end of the run could actually be good
because I don't want it glaring in the neighbors
windows. So I will go ahead and use this 18 gauge
they gave me for the last 50ft rather than throw it
away.
Michel Gagnon
> All other things being equal, it will cost me
> more money each month to operate these
> lights if I wire with 18 gauge cable than if
> I wire with 14 gauge cable.
>
> Thanks,
> Free...@aol.com
Yes.
And to answer clearly a previous question, if you have an 88 W
transformer, it is like having 2 faucets on a bathtub.
If you install only 1 10-W lightbulb, you will take 10 W (plus 1 or 2 W
in the transformer). So the remaining 78 W is a reserve.
If you install 10 10-W lightbulbs, you will need 100 W, but your
transformer is only able to give 88 W. So it will overheat and will burn
shortly (too much "exercise" for the transformer).
In a nutshell, if you install a 88 W transformer, make sure the total
load (total watts) of lamps is under 60 or 65 W.
As for cable size, don't install more than 40 W (i.e. 3,2 A at 12 V) on
a single run (maybe 50 W is the total cable lenght is short). In all
cases, you will be better with 14 gauge, but this is especially true if
you have a long distance to cover and a high wattage. (Think of it as
watering your flowers with a small-diametre garden hose).
--
Michel Gagnon -- Michel...@videotron.ca
Montréal (Québec, Canada)
JCoggins
It would be very hard to destroy it from an overload.
Michel Gagnon wrote in message
<1djb0qq.mu...@ppp099.133.mmtl.videotron.net>...
Bill Neisius
Freezout wrote in message <19981203201336...@ng32.aol.com>...
>Michel...@videotron.ca (Michel Gagnon) wrote:
>
>>Freezout <free...@aol.com> wrote:
>>
>>> All other things being equal, it will cost me
>>> more money each month to operate these
>>> lights if I wire with 18 gauge cable than if
>>> I wire with 14 gauge cable.
>>>
>>> Thanks,
>>> Free...@aol.com
>>
>>
>>Yes.
>
>Hmm. This is the exact opposite of the response
>I got from Dan Hicks, who said I would spend more money each month (ie,
higher
>electric bill)
>with the 14 gauge-- although he did say the difference
>was negligible.
If you use smaller diameter wire, it will have more resistance... more
resistance
implies that LESS current will be flowing through it. Your electricty bill
is based
on watts (well, actually volt-amps, but whatever...) which is the product
of
voltage and current.
The smaller diameter wire will use less energy.
The smaller diameter wire is cheaper.
The bulbs will last longer.
If you put 88watts of lamps on a "zero loss" cable, 12 hours/day for 30
days = 31680 watthr
If you use 75 feet of cable with a loss of 6 ohms/1000 feet, that's 0.45
ohms
added in series with the lamp load... the total load of the lossy wire plus
the lamps
would be 72watts.
72watts run 12 hours/day for 30 days = 25920 watthr
The savings is 5760 watthr/month (18%)
Using $0.14157/kWh (Southern California Edison), that's $0.82 saved each
month...
guess you don't need to worry about how long to "payback", since it cost
less
in the first place...
Freezout
>Freezout <free...@aol.com> wrote:
>
>> All other things being equal, it will cost me
>> more money each month to operate these
>> lights if I wire with 18 gauge cable than if
>> I wire with 14 gauge cable.
>>
>> Thanks,
>> Free...@aol.com
>
>
>Yes.
Hmm. This is the exact opposite of the response
I got from Dan Hicks, who said I would spend more money each month (ie, higher
electric bill)
with the 14 gauge-- although he did say the difference
was negligible.
(snip)
>In a nutshell, if you install a 88 W transformer, make sure the total
>load (total watts) of lamps is under 60 or 65 W.
I don't plan on using more than 64 watts, and
no less than 56.
>As for cable size, don't install more than 40 W (i.e. 3,2 A at 12 V) on
>a single run (maybe 50 W is the total cable lenght is short). In all
>cases, you will be better with 14 gauge, but this is especially true if
>you have a long distance to cover and a high wattage. (Think of it as
>watering your flowers with a small-diametre garden hose).
I assume you give this advice, not for safety reasons,
but so I won't experience a drop in light output
at the end of the run-- correct?
If so, I actually have a situation where I am concerned
about having too much light output in the area
where the end of the run is, so it actually works to my
advantage if the light output dims.
Freezout
Freezout
>Freezout wrote in message <19981203201336...@ng32.aol.com>...
>>Michel...@videotron.ca (Michel Gagnon) wrote:
>>
>>>Freezout <free...@aol.com> wrote:
>>>
>>>> All other things being equal, it will cost me
>>>> more money each month to operate these
>>>> lights if I wire with 18 gauge cable than if
>>>> I wire with 14 gauge cable.
>>>>
>>>> Thanks,
>>>> Free...@aol.com
>>>
>>>
>>>Yes.
>>
>>Hmm. This is the exact opposite of the response
>>I got from Dan Hicks, who said I would spend more money each month (ie,
>higher
>>electric bill)
>>with the 14 gauge-- although he did say the difference
>>was negligible.
>
>
>If you use smaller diameter wire, it will have more resistance... more
>resistance
>implies that LESS current will be flowing through it. Your electricty bill
>is based
>on watts (well, actually volt-amps, but whatever...) which is the product
>of
>voltage and current.
>
>The smaller diameter wire will use less energy.
>The smaller diameter wire is cheaper.
>The bulbs will last longer.
>
>If you put 88watts of lamps on a "zero loss" cable, 12 hours/day for 30
>days = 31680 watthr
>
>If you use 75 feet of cable with a loss of 6 ohms/1000 feet, that's 0.45
>ohms
>added in series with the lamp load... the total load of the lossy wire plus
>the lamps
>would be 72watts.
>
>72watts run 12 hours/day for 30 days = 25920 watthr
>
>The savings is 5760 watthr/month (18%)
>
>Using $0.14157/kWh (Southern California Edison), that's $0.82 saved each
>month...
>guess you don't need to worry about how long to "payback", since it cost
>less
>in the first place...
Where have *you* been? Wish you were
around to answer my question the first time.
Would have only needed one response. :)
Seemed like I was having to jump through
hoops to get a straightforward answer.
I had set aside last night to get the job done
anyway, and I used 100ft of 16 followed
by the 75 of 18 that came with it.
Anyway, you and Dan, agree, and both
gave clear, straight forward responses,
so I have to believe you guys. Sorry, J. :)
Thanks for the help from all of you.
JCoggins
is fair-to-middling' close to the 0.4% cited somewhere else in this ...
well, remarkable thread.
MBAs? Executive for sure.
Freezout wrote in message <19981204181637...@ng-cf1.aol.com>...
Freezout
I was able to get all the lights situated
last night and they turned out fine.
This again was with 100ft of 16 gauge connected
to the 88watt transformer, and 75 ft of 18 gauge
connected to the 16 gauge. I have four 7 watt
lamps connected to the first 100 ft (16 gauge)
and four 7 watt lamps, plus one 4 watt,
connected to the final 75 ft of 18 gauge. Again,
I didn't want the last five lamps to be too
bright anyway, so whatever dimming occured
has worked out fine. They light the walkway
without beaming bright light at the neighbor's
windows.
Freezout
JCoggins
hope it works well for many years to come), you have increased your electric
bill by about 0.5% based on 3 hours of usage per day.
The 32 watts of lights on 75 feet of #16 wire represent about 4.98 ohms
(wire plus lights). Likewise, the 28 watts on 100 feet of #16 wire
represent about 5.5 ohms. Since these resistances are in parallel, reducing
to one equivalent circuit resistance gives ~2.6 ohms. At 12 volts, this
draws about 4.6 amps and uses 55W (using I^2R), which is about 5 KWHr more
per month. That's 0.5% more on a 1000 KwHr account.
You are saving nothing. But, like I said, enjoy the lights.
Freezout wrote in message <19981125170956...@ng113.aol.com>...
>Needed some deck lighting along a walkway and along the stairway of a deck.
>I bought a Malibu kit. It comes with 75 feet of cable. The instructions
>say that, as a rule of thumb, 16 gauge cable can be used for uns up to100
feet,
>14-gauge for up to 150 feet. However, it doesn't say what gauge
>is included in the kit. A look at the cable and I see it has "18 AWG 2/C
>underground low energy circuit cable" stamped into it. Does this
>mean it is 18 gauge? (If so, how nice of them to include something even
>cheaper than what they describe).
>
>The instructions go on to say, "For added efficiency , you can mix heavier
>gauge
>cable with existing 16 or 14-gauge, but heavier cable should be installed
>closest
>to the transformer. Does this mean I can add to the 75 foot length with a
>higher
>gauge if I like? If so, I take it then the most I could add to it would be
25
>feet
>of 16 gauge. Although, it seems like it would be even less than 25 feet,
since
>
>it's limit is at 100 feet even if it were *all* 16 guage, much less only
one
>quarter
>of the it. How does this work?
>
>
>Also, can I fork off the cable in two different directions, or does it have
to
>be in one continuous line? If I can do this, does it effect the total
>length of cable I can use. I've seen these cable couplers that allow
you
>to attach more cable. They couple the two pieces of cable side-by-side in
>a thing you clamp over them with penetrates the cable to create a
connection
>between them. I'm wondering if I could use one of these to feed a branch
>of cable off in another direction. If I can, it would be more efficient
for me
>because I wouldn't have to double back the cable where it's already been
>laid to then head in the opposite direction (ie, I have to lay it in a sort
of
>"T" shaped configuration.)
>
>Thanks,
>Free...@aol.com
Freezout
>Now that you've got it all installed (and congratulations ... I sincerely
>hope it works well for many years to come), you have increased your electric
>bill by about 0.5% based on 3 hours of usage per day.
Well, I knew it was going to cost money to run
the lights. 'lectricty aint free. But since
the 16 gauge and 18 gauge will consume
less electricity than if I had used 14 guage
(see the other two responses I got),
at least I know there was no value in going
ahead in buying all 14 gauge and tossing the 18
that came with the kit. That's what my initial
question was trying to find out. I figured I could
get the answer to that question simply and clearly
in one response. Didn't figure it would draw out this long. But curiousity
was what kept me in pursuit of
an answer.
>The 32 watts of lights on 75 feet of #16 wire represent about 4.98 ohms
>(wire plus lights).
I used 18 gauge, not 16, here.
> Likewise, the 28 watts on 100 feet of #16 wire
>represent about 5.5 ohms. Since these resistances are in parallel,
Don't know what parallel is, all the cable was wired
in one strand, 18 attaching to the 16, and 16 attaching
to the transformer.
>reducing
>to one equivalent circuit resistance gives ~2.6 ohms. At 12 volts, this
>draws about 4.6 amps and uses 55W (using I^2R), which is about 5 KWHr more
>per month. That's 0.5% more on a 1000 KwHr account.
5% more than what??? than if I had no lights?
>You are saving nothing. But, like I said, enjoy the lights.
I'm using less electricity than if I had wired with all 14
gauge, though it's negligible.
Sylvan Butler
>I'm using less electricity than if I had wired with all 14
>gauge, though it's negligible.
Not only is it negligible, I question its very existance.
The voltage applied to your circuit is fixed (supplied by the
transformer). Therefore the current flow hence power usage depends on
the resistance in the circuit.
The voltage "used" by your wiring is more with smaller wire (bigger
guage number) because there is more resistance in smaller wire.
The voltage across the lights determines the amount of light you get,
higher volts means more light. Also, current flow thru the lights is
not constant. Lower voltage will result in more current (if the
current is available) because the lamp does not run as hot and
therefore has slightly lower resistance.
So, smaller wire means more wire resistance and less lamp resistance.
Increased wire resistance means more power lost in the wiring, and
somewhat less light. Depending on the specific type and number of
lamps, you may or may not have more or less total circuit resistance.
You'd have to measure current flow with both wiring options to be sure.
sdb
--
Do NOT send me unsolicited commercial e-mail (UCE)!
Watch out for munged e-mail address.
User should be sylvan and host is cyberhighway.net.
JCoggins
grasp.
Sylvan Butler wrote in message <74ct4f$eoe$2...@news.cyberhighway.net>...
>Freezout (free...@aol.com) on 6 Dec 1998 02:39:01 GMT wrote:
>>I'm using less electricity than if I had wired with all 14
>>gauge, though it's negligible.
>
>Not only is it negligible, I question its very existance.
>
Freezout
>Thanks for helping to a clarify an issue that has been difficult
>for some to grasp.
>Sylvan Butler wrote in message <74ct4f$eoe$2...@news.cyberhighway.net>...
>>Freezout (free...@aol.com) wrote:
>>>I'm using less electricity than if I had wired with all 14
>>>gauge, though it's negligible.
>>
>>Not only is it negligible, I question its very existance.
>>
>>You'd have to measure current flow with both wiring options to be sure.
>>
>>sdb
Well, I'm definitely one of the "some."
But then again, I wonder if xsylvan...@cyberhighway.net (Sylvan Butler)
is aware of the whole situation. Primarily, does he understand that
I would be using the same number and types of lamps regardless of what
wiring I used. I know JCoggins knows that, as I made this point
clearly many times in my previous posts. But I'm not sure if Sylvan
read all the other posts.
I would like to know if Sylvan would agree or disagree with this statement:
"All other things being equal, it will cost me
more money each month to operate these
lights if I wire with 18 gauge cable than if
I wire with 14 gauge cable."
Now, danh...@infonet.isl.net, Michel...@videotron.ca (Michel Gagnon),
and "Bill Neisius" <neisi...@netcom.com>, are all in straight forward
agreement with the above statement [they would be the other
of the "some", I suppose. :) ]
JCoggins disagreed with the above statement.
So given I know nothing about electricity, I went with the three that agreed.
But I wouldn't mind learning something and "grasping" all this :),
so let me know Sylvan.
Again, that is ***all other things being equal***
I emphasis this, Sylvan, because you made a reference to less light output
from the lamps using a the higher numbered gauge, and I can't see
the relevance of this on energy use unless I were to take advantage
of the higher light output by using fewer lamps, or lamps rated for
lower watts. And with **all other things being equal** this is quite
obviously not a consideration.
Thanks,
Freezout
.
Sylvan Butler
>is aware of the whole situation. Primarily, does he understand that
>I would be using the same number and types of lamps regardless of what
Yes.
>I would like to know if Sylvan would agree or disagree with this statement:
>"All other things being equal, it will cost me
>more money each month to operate these
>lights if I wire with 18 gauge cable than if
>I wire with 14 gauge cable."
Disagree. Wiring with 18ga (smaller) wire instead of wiring with 14ga
(larger) wire will reduce the light output and increase the electricity
lost in the wiring. This will result in a minor _increase_ in power
usage if it results in anything measurable.
Overall change in electricity usage depends on how your particular
bulbs react to the reduced voltage (due to more being lost in the
wire), as well as how many bulbs you are powering. The more bulbs you
power, the more power INCREASE you will see when you drop the voltage
(due to longer cable runs).
However, since the 18ga was "free" and you would need to spend extra
for the 14ga, the amortized cost of that 14ga was eliminated by using
18ga and you will likely save money overall.
>I emphasis this, Sylvan, because you made a reference to less light output
>from the lamps using a the higher numbered gauge, and I can't see
Correct.
>the relevance of this on energy use unless I were to take advantage
>of the higher light output by using fewer lamps, or lamps rated for
The issue is that these are incandescent lamps. They make light by
getting hot. When cold they draw almost infinite power, and their
demand decreases the hotter they get.
When fed a reduced voltage they will not get as hot. Therefore each
bulb will draw more current than "normal." More current means more
power lost in the wiring and more power used.
For (an extreme) example, try running two 60watt 120v bulbs on a 12v
power supply (eg a car battery). Normally these bulbs will draw a
total of 1amp at 120v. Proportionally they should draw 10amps at 12v
to provide the same total of 120watts. But they won't. They will draw
far far far more because they won't heat up enough to limit their power
demand.
Freezout
>The issue is that these are incandescent lamps. They make light by
>getting hot. When cold they draw almost infinite power, and their
>demand decreases the hotter they get.
>
>When fed a reduced voltage they will not get as hot. Therefore each
>bulb will draw more current than "normal." More current means more
>power lost in the wiring and more power used.
Thank you. Given (as I've said many times)
that I know nothing about electricity, don't
you think this point about how incandescent
lights work might have been important in
explaining your point? :) Well, it was. Things
are actually starting to make sense now. Well,
almost. I wonder if I can actually learn something here.
Will these lights cost more to run in the winter
than the summer?
Okay, so I now understand this idea that
the amount of energy these lamps draw
depends on how hot they get. Now lets see
if I understand your argument. The lights
use more energy with the 14 gauge setup-- call this
additional energy amount "Z." But the
cable uses more energy with 18/16 setup I have--
call this additional energy amount "X." You
are saying that amount Z will be less than
amount X. Correct?
Hell, if it is, I may have actually learned something. :)
JCoggins
question but thought it would probably focus on why you have 60W of lights
installed but (with the wires and distances cited) appear to be using only
about 55W.
It would be pointless to wonder about that now, since Mr. Butler has already
explained how dynamic this simple problem really is.
Freezout wrote in message <19981206192518...@ng97.aol.com>...
>JCoggins" <cog...@earthlink.net> wrote:
>
>>Thanks for helping to a clarify an issue that has been difficult
>>for some to grasp.
>
>>Sylvan Butler wrote in message <74ct4f$eoe$2...@news.cyberhighway.net>...
>>>Freezout (free...@aol.com) wrote:
>>>>I'm using less electricity than if I had wired with all 14
>>>>gauge, though it's negligible.
>>>
>>>Not only is it negligible, I question its very existance.
>>>
>>>You'd have to measure current flow with both wiring options to be sure.
>>>
>>>sdb
>
>Well, I'm definitely one of the "some."
>But then again, I wonder if xsylvan...@cyberhighway.net (Sylvan
Butler)
>is aware of the whole situation. Primarily, does he understand that
>I would be using the same number and types of lamps regardless of what
>wiring I used. I know JCoggins knows that, as I made this point
>clearly many times in my previous posts. But I'm not sure if Sylvan
>read all the other posts.
>
>I would like to know if Sylvan would agree or disagree with this statement:
>
>"All other things being equal, it will cost me
>more money each month to operate these
>lights if I wire with 18 gauge cable than if
>I wire with 14 gauge cable."
>
>Now, danh...@infonet.isl.net, Michel...@videotron.ca (Michel Gagnon),
>and "Bill Neisius" <neisi...@netcom.com>, are all in straight forward
>agreement with the above statement [they would be the other
>of the "some", I suppose. :) ]
>
>JCoggins disagreed with the above statement.
>
>So given I know nothing about electricity, I went with the three that
agreed.
>
>But I wouldn't mind learning something and "grasping" all this :),
>so let me know Sylvan.
>
>Again, that is ***all other things being equal***
>
>I emphasis this, Sylvan, because you made a reference to less light output
>from the lamps using a the higher numbered gauge, and I can't see
>the relevance of this on energy use unless I were to take advantage
>of the higher light output by using fewer lamps, or lamps rated for
Bill Neisius
Sylvan Butler wrote in message <74f8ge$4eu$2...@news.cyberhighway.net>...
>Disagree. Wiring with 18ga (smaller) wire instead of wiring with 14ga
>(larger) wire will reduce the light output and increase the electricity
>lost in the wiring. This will result in a minor _increase_ in power
>usage if it results in anything measurable.
>
>Overall change in electricity usage depends on how your particular
>bulbs react to the reduced voltage (due to more being lost in the
>wire), as well as how many bulbs you are powering. The more bulbs you
>power, the more power INCREASE you will see when you drop the voltage
>(due to longer cable runs).
So... the power used by the circuit INCREASES when the resistance
in the wire increases? One wonders what will happen when the wire
resistance becomes 'infinite', such as when the timer opens the circuit;
in effect adding a piece of wire with 'infinite' resistance...
The bulbs resistance does depend on the voltage. Using a 7watt
intermatic bulb gave the following:
12.08V: 0.54A == 22.37 ohms
9.74V: 0.48A == 20.29 ohms
5.06V: 0.35A == 14.46 ohms
3.18V: 0.28A == 11.36 ohms
This comes pretty close to R = sqrt(V*41.4). Plugging this value into
a spread sheet, with 10 lamps spaced 10 feet apart on a wire with
6 ohms/1000ft resistance:
Total wattage of bulbs: 70W
Total power consumed by circuit: 66W
Total power consumed by lamps: 59W
Total power consumed by wire: 7W
The voltage across the bulbs varies from about 11.6V at the first
bulb, to 10.1 at the last bulb; wattage varies from 6.7W down to 5.4W.
Doubling the wire loss to 12 ohms/1000ft gives the following:
Total wattage of bulbs: 70W
Total power consumed by circuit: 62W
Total power consumed by lamps: 50W
Total power consumed by wire: 12W
In no case does increasing wire loss cause more power to be consumed
by the circuit. Makes sense really: if reducing the voltage caused more
more current to flow, the increased current flow would cause an increased
voltage drop along the wire, which in turn would cause more current to
flow and increase the voltage drop etc...
JCoggins
on voltage. Resistance varies with temperature which mostly varies with
current (and some with ambient). This is why cable ampacity is derated at
higher temps -- not voltages.
I've seen many tables relating metal resistance versus temperature. I'd love
to see one that shows change with voltage. Heck, I'll even take a
spreadsheet.
Bill Neisius wrote in message <74h2pv$rup$1...@birch.prod.itd.earthlink.net>...
Freezout
>Just what we need -- one more response. Actually, I did expect a further
>question but thought it would probably focus on why you have 60W of lights
>installed but (with the wires and distances cited) appear to be using only
>about 55W.
I'm not sure what you are talking about. Are you asking
why I do not want the lights I installed to shine to
their capacity? If so, I'm pretty sure I explained that
I actually prefered dimmer lights in the area at the
end of the run so as not to beam excess light to my
neighbors windows. I only want enough to illuminate
the walk.
Now I could have bought a 4 watt bulb set, but that
set did not come with covers you snap over the lights
to direct it downward. I would have had to buy
covers, and the additional cost would have ended up
costing me far more than if I had just bought a 7 watt
set to begin with.
Or perhaps you question refers to the fact that I bought
an eight lamp set, which at 7 watts adds up to 56, and
not 60. But I had nine areas I wanted to light up, so
I bought an additional 4 watt.
At any rate, at least I understand your argument a bit
more (enough to satisfy my initial curiosity), even
if the debate was not resolved amongst those in the
thread that actually know something about electricity
(unlike myself. )
Of course if Butler is correct, then even if I do spend
a tiny bit more on electricity with my setup, it is offset
by the $40 savings in cable had by not buying all 14 gauge . Never dreamed it
would take so many posts
to understand this, but as I said, there came a point
where my curiousity was more motivation than any savings.
But hey, I'm enjoying the lights.
JCoggins
that. And a lot of capacitance and resistance parameters to look up in
tables depending on what stage you're talking about.
Besides, from a user's perspective, they have receptacles but no
"traditional" neutral and no ground. Maybe just a hot only?
Harrison Hall wrote in message <366d1f3...@netnews.worldnet.att.net>...
>On Mon, 7 Dec 1998 06:52:05 -0500, "JCoggins" <cog...@earthlink.net>
>wrote:
>
>>Just what we need -- one more response. Actually, I did expect a further
>>question but thought it would probably focus on why you have 60W of lights
>>installed but (with the wires and distances cited) appear to be using only
>>about 55W.
>>
>>It would be pointless to wonder about that now, since Mr. Butler has
already
>>explained how dynamic this simple problem really is.
>>
>>===
>
>
>>>Freezout wrote in message <19981206192518...@ng97.aol.com>...
>>>
>>>Thanks for helping to a clarify an issue that has been difficult
>>>for some to grasp.
>>>
>>>===
>
>
> But what type of second order surge protector are you going to
>use? <snort>
>
> Harry
Sylvan Butler
>xsylvan...@cyberhighway.net (Sylvan Butler) wrote:
>>The issue is that these are incandescent lamps. They make light by
>>getting hot. When cold they draw almost infinite power, and their
>>demand decreases the hotter they get.
>>
>>When fed a reduced voltage they will not get as hot. Therefore each
>>bulb will draw more current than "normal." More current means more
>>power lost in the wiring and more power used.
>Thank you. Given (as I've said many times)
>that I know nothing about electricity, don't
>you think this point about how incandescent
>lights work might have been important in
>explaining your point? :) Well, it was. Things
That's is why I originally wrote:
>>The voltage across the lights determines the amount of light you get,
>>higher volts means more light. Also, current flow thru the lights is
>>not constant. Lower voltage will result in more current (if the
>>current is available) because the lamp does not run as hot and
>>therefore has slightly lower resistance.
>Will these lights cost more to run in the winter
>than the summer?
Not measurably. The electricity thru them heats them up far more than
the change in temp from winter to summer.
>Okay, so I now understand this idea that
>the amount of energy these lamps draw
>depends on how hot they get. Now lets see
Right.
>if I understand your argument. The lights
>use more energy with the 14 gauge setup-- call this
No, less energy with the 14ga, because they get hotter because of the
greater voltage and so allow less amps. (power=volts * amps)
>additional energy amount "Z." But the
>cable uses more energy with 18/16 setup I have--
Right.
>call this additional energy amount "X." You
>are saying that amount Z will be less than
>amount X. Correct?
Essentially. X will be >0 and Z will be <=0
>Hell, if it is, I may have actually learned something. :)
sdb
Sylvan Butler
>So... the power used by the circuit INCREASES when the resistance
>in the wire increases? One wonders what will happen when the wire
Not as simple as that. :)
>The bulbs resistance does depend on the voltage. Using a 7watt
Yup.
>intermatic bulb gave the following:
> 12.08V: 0.54A == 22.37 ohms
> 9.74V: 0.48A == 20.29 ohms
> 5.06V: 0.35A == 14.46 ohms
> 3.18V: 0.28A == 11.36 ohms
Good! Some actual numbers. :)
>In no case does increasing wire loss cause more power to be consumed
>by the circuit. Makes sense really: if reducing the voltage caused more
Hmm. I think your spreadsheet assumed some linear relationships where
in reality are non-linear. (Primarily the bulb resistance... I don't
believe summarizing it to a linear equation is a valid assumption.)
>more current to flow, the increased current flow would cause an increased
>voltage drop along the wire, which in turn would cause more current to
>flow and increase the voltage drop etc...
Did you take into account a mix of different wire sizes? Or what
happens if you double (or more) the wattage of the bulb and the effect
of V on R?
As I pointed out in my initial post (or maybe it was the second...),
the magnitude of this effect depends on the number and type of bulbs in
the circuit.
But in any case, I don't think any electrical cost savings or not will
be measurable. The difference first noticed is typically light intensity.
Sylvan Butler
>Of course if Butler is correct, then even if I do spend
>a tiny bit more on electricity with my setup, it is offset
I don't think your setup would produce a measurable difference in
electricity usage changing only the gauge of wiring. Not with less
than 100watts...
Now if you compare electricity used to lumens output... You _MAY_ be
able to measure that. But again, so few watts to start with and so
little light...
>by the $40 savings in cable had by not buying all 14 gauge . Never dreamed it
Exactly. Let's see, at $0.10/kwh, you have to save 400,000watts for
one hour, or 4watts for 100,000 hours, or 20watts for 2000hours
(typical bulb life?) to break even.
And guess what!!! Operating a bulb at a slightly lower voltage (caused
perhaps by more loss in the wiring) will often increase its operating
life (there are some exceptions). So you have to add in the cost of
avoided bulb replacement to the cost of avoided 14ga cable.
I really think you made the correct decision. Just don't expect any
monetary savings just from lower electricity usage. :)
JCoggins
determine a useful working (hot) resistance. But you can apply a constant
voltage source, vary bulb temperature by other physical means and measure
current change. In addition to establishing that voltage and resistance are
independent, it is also the basis for many instrument temperature
measurements.
The relationship between resistance and temperature is not linear -- as you
noted. This is one of the ideas on which superconductors, RTDs and
thermocouples are based. In the case of the latter two, resistance and
temperature are typically considered linear over certain defined ranges in
which the devices are used. Outside of those ranges, accuracy falls
appreciably.
Sylvan Butler wrote in message <74hluu$tua$1...@news.cyberhighway.net>...
>Bill Neisius (neisi...@netcom.com) on Mon, 7 Dec 1998 09:19:06 -0800
wrote:
>
>>The bulbs resistance does depend on the voltage.
>
>Yup.
danh...@infonet.isl.net
>So... the power used by the circuit INCREASES when the resistance
>in the wire increases? One wonders what will happen when the wire
>in effect adding a piece of wire with 'infinite' resistance...
>
>intermatic bulb gave the following:
>
> 12.08V: 0.54A == 22.37 ohms
> 9.74V: 0.48A == 20.29 ohms
> 5.06V: 0.35A == 14.46 ohms
> 3.18V: 0.28A == 11.36 ohms
>
>a spread sheet, with 10 lamps spaced 10 feet apart on a wire with
>6 ohms/1000ft resistance:
>
> Total wattage of bulbs: 70W
> Total power consumed by circuit: 66W
> Total power consumed by lamps: 59W
> Total power consumed by wire: 7W
>
>The voltage across the bulbs varies from about 11.6V at the first
>bulb, to 10.1 at the last bulb; wattage varies from 6.7W down to 5.4W.
>
>
>Doubling the wire loss to 12 ohms/1000ft gives the following:
>
> Total wattage of bulbs: 70W
> Total power consumed by circuit: 62W
> Total power consumed by lamps: 50W
> Total power consumed by wire: 12W
>
>by the circuit. Makes sense really: if reducing the voltage caused more
>voltage drop along the wire, which in turn would cause more current to
>flow and increase the voltage drop etc...
Bill, I have to agree. I haven't done the equations, but it seems to me
that you can't get an increasing power with increasing cable resistance
unless there's a significant "negative" resistance somewhere, and I don't
thing a common incandescent bulb can be made to exhibit negative
resistance. (A gas discharge bulb is of course a different matter.)
Bill Neisius
JCoggins wrote in message <74h6sg$t2u$1...@holly.prod.itd.earthlink.net>...
>Great lookin' numbers. But bulb resistance never has (and never will)
depend
>on voltage. Resistance varies with temperature which mostly varies with
>current (and some with ambient). This is why cable ampacity is derated at
>higher temps -- not voltages.
Ok... here's how the resistance changed for different currents:
0.54A: 12.08V == 22.37 ohms
0.48A: 9.74V == 20.29 ohms
0.35A: 5.06V == 14.46 ohms
0.28A: 3.18V == 11.36 ohms
Temperature depends on power, which is the product of voltage and current:
power = voltage * current
Temperature is equally dependant on voltage and current...
A cable operating at a high temperature can be derated in 'ampacity'
or voltage (voltacity?)... but the reality is that the cable is usually
connected
to a fixed 110 volt source, so all you can control is the current...
>I've seen many tables relating metal resistance versus temperature. I'd
love
>to see one that shows change with voltage. Heck, I'll even take a
>spreadsheet.
see below.
>Bill Neisius wrote in message
<74h2pv$rup$1...@birch.prod.itd.earthlink.net>...
John Coggins
resurrect this and nail me with it so I'm gonna clarify it now, just for a
record.
No, a thermocouple is not a good example of a device whose resistance varies
with temperature (though it's emf or voltage does). The others are ... but I
left this incorrect one in the list on the chance that an individual would
say: (1) by silence, they didn't know the difference; (2) by speaking up,
they did; even if it contradicted a previous statement. Regards.
JCoggins wrote in message <74hnvf$psf$1...@ash.prod.itd.earthlink.net>...
>An incandescent(!) bulb's nominal voltage and rated power output do
>determine a useful working (hot) resistance. But you can apply a constant
>voltage source, vary bulb temperature by other physical means and measure
>current change. In addition to establishing that voltage and resistance
are
>independent, it is also the basis for many instrument temperature
>measurements.
>
>The relationship between resistance and temperature is not linear -- as you
>noted. This is one of the ideas on which superconductors, RTDs and
>thermocouples are based. In the case of the latter two, resistance and
>temperature are typically considered linear over certain defined ranges in
>which the devices are used. Outside of those ranges, accuracy falls
>appreciably.
>
>
>Sylvan Butler wrote in message <74hluu$tua$1...@news.cyberhighway.net>...
>>Bill Neisius (neisi...@netcom.com) on Mon, 7 Dec 1998 09:19:06 -0800
>wrote:
>>
>>>The bulbs resistance does depend on the voltage.
>>
>>Yup.
>>
>>>In no case does increasing wire loss cause more power to be consumed
>>>by the circuit. Makes sense really: if reducing the voltage caused more
>>
Bill Neisius
Sylvan Butler wrote in message <74f8ge$4eu$2...@news.cyberhighway.net>...
>For (an extreme) example, try running two 60watt 120v bulbs on a 12v
>power supply (eg a car battery). Normally these bulbs will draw a
>total of 1amp at 120v. Proportionally they should draw 10amps at 12v
>to provide the same total of 120watts. But they won't. They will draw
>far far far more because they won't heat up enough to limit their power
>demand.
Ah HA! Caught ya... You've never actually connected a 60watt bulb to
a 12v power supply, have you? Sylvan, you really shouldn't make-up
facts just to prove a point. I think you need to spend some time in the
garage
and run a few experiments...
First problem is that the cold resistance of a typical 'Sylvania' 60watt
bulb is 17 ohms. Even if you connect 2 of them in parallel, that would
only be 8.5 ohms. At 12V, the MAXIMUM current that could flow
through each bulb is 0.7A or a total of 1.4A for both. Naturally, as the
filament heats up, the resistance increases and drops the current
further... I measure 0.16A for a single 60watt bulb at 12V, or 0.32A for
two
in parallel... not 'far far far more' than 10A!
The other problem is that 'watts' is a measure of the work done by the
electricity. '120 watts' of work done on the filament of a bulb causes it
to heat up to a certain temperature... it doesn't matter if that's
120v*1A or 12V*10A: the electricity is doing the same work in heating
the filament. If the bulbs somehow pull 'much more' than 10A (lets'
say...20A?), then there's 240watts of work being done on the filaments...
if it doesn't show up as heat, then Where Does It Go?
[unless the energy is being converted into mass, causing the wire to get
heavier?
"Beam me up Captain" :) ]
Sylvan Butler
>Sylvan Butler wrote in message <74f8ge$4eu$2...@news.cyberhighway.net>...
>>For (an extreme) example, try running two 60watt 120v bulbs on a 12v
>>power supply (eg a car battery). Normally these bulbs will draw a
>>total of 1amp at 120v. Proportionally they should draw 10amps at 12v
>>to provide the same total of 120watts. But they won't. They will draw
>>far far far more because they won't heat up enough to limit their power
>Ah HA! Caught ya... You've never actually connected a 60watt bulb to
>a 12v power supply, have you? Sylvan, you really shouldn't make-up
No, haven't. Just to a 12v starting battery, more than 20 years ago.
(I was just a kid then, no nice power supply, but there was a car in
the driveway...)
>First problem is that the cold resistance of a typical 'Sylvania' 60watt
>bulb is 17 ohms. Even if you connect 2 of them in parallel, that would
Hmm, I'll have to measure what I've got around here... Last time I
measured a 120v light bulb I swear it read 0ohms (no, not infinite)!
>only be 8.5 ohms. At 12V, the MAXIMUM current that could flow
>through each bulb is 0.7A or a total of 1.4A for both. Naturally, as the
>filament heats up, the resistance increases and drops the current
>further... I measure 0.16A for a single 60watt bulb at 12V, or 0.32A for
>two
>in parallel... not 'far far far more' than 10A!
Sure isn't.
Did you see any inrush current? Probably not with 17ohms cold...
What I remember from my experiment, is that it blew my 15amp fuse
(glass 1.25x.25", in the cig lighter plug) without any visible light
from the bulb.
>The other problem is that 'watts' is a measure of the work done by the
>electricity. '120 watts' of work done on the filament of a bulb causes it
>to heat up to a certain temperature... it doesn't matter if that's
>120v*1A or 12V*10A: the electricity is doing the same work in heating
Right...
>the filament. If the bulbs somehow pull 'much more' than 10A (lets'
>say...20A?), then there's 240watts of work being done on the filaments...
>if it doesn't show up as heat, then Where Does It Go?
Good point. I sure drew some wrong conclusions from that experiment!
How about...
It goes... away because 1) your supply cannot provide it, or 2) the
fuse blows (like mine :) .
So the resistance is non-linear, but not that non-linear. I wonder if
the reactance of a coiled filament on A.C. vs D.C. makes a measurable
difference... Probably not at 60hz. Any thoughts?