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Scott Lurndal wrote:

Actually those number are ohms per 1000 feet.

Yes. assuming the same length, material, etc.

No. The resistance should be 169 ohms at 130 volts. You seem to have gotten 156 ohms by dividing 120 volts by 0.768231 amps. There is no justification for this. The resistance of the filament will vary with its temperature. It will be lower than 169 ohms at 120 volts. but there is no reason to assume that it will be 156 ohms.

See previous comment. You seem to have derived a result from your assumptions (you assumed the current at 120 volts is 0.769231 amps in your previous calculation).

This result is based upon an incorrect assumption that the current is 0.7629231 when the voltage is 120 volts.

The power at 120 volts will be between the 85 watts (the value that would be calculated if the resistance is constant with temperature) and the 100 watts at 130 volts. Thus 92.3 watts is a reasonable guess for the power at 120 volts but you have not presented anything to prove it.

Dan

"IanM" wrote:

When it comes to an incandescent lamp filament, there is more than just the geometry (length and cross section)of the wire at work to determine the filament resistance.

Coatings on the wire, shape of the winding, the metal alloy are just a few of things that come into play to arrive at the final design.

For a given wattage of lamp, the total lamp resistance of the lamp increases in direct proportion to the rated voltage.

Lew

With light bulbs, the explanation is__ _simple_.__ Your assumption is in
error -- "everything else" does__ _not_ __remain unchanged. <grin>

(Note: Lew had the 'results' right, albeit with a somewhat incorrect description of the 'causation')

'Rough service' bulbs use a different composition in the filament element than normal service bulbs. It has a***much*** higher resistance per unit of
cross-section area, and thus ***does*** require a larger cross-section to get
the 'appropriate' resistance for the slightly lower amperage needed for the
same number of watts as the lower-voltage bulb. Note: the 'wire length' of
the actual 'element' in a rough service bulb is usually much shorter (not as
much 'coil') -- necessitating a higher resistance 'per unit length'; over and
above the compensation for the larger cross-section.

BTW, the relationship between bulb life and applied voltage is an__ _eleventh_
___order_ equation -- i.e., a 5% decrease in applied voltage will result in
an over 70% increase in bulb life. Note: decreasing the voltage by 5%__ _will_
__result in a _more_than_5%_ decrease in the light output, so you__ _do_ __end up
paying 'more ***per***lumen***of***light***output***'.

NOTE: the cost of the electricity to operate a 'typical' household light-bulb (25-150 watt) is generally***several***times* the cost of the bulb itself.

Hint: imagine a square copper wire of a fixed length, now double its thickness and width, _in_steel_, and explain how that is equivalent to four copper wires of the original thickness in parallel for 1/4 the resistance! <grin>

Actually it isn't, it's temperature dependant. The resistance goes up with increasing temperature so that as the voltage increases the dissipated power doesn't increase in a linear way.

If you have a light fixture that's difficult to reach, Radio Shack is your friend. Our great room had a couple track lights well over 30' from the floor and a ceiling fan's 4 lights were around 25' high. Even with my straight ladder braced against the ceiling, it was awkward as the dickens getting the ladder inside, extended and within reach of either track OR the fan in particular. Eventually, some of the track bulbs departed and a couple of the fan bulbs as well. It was about time to paint the room, so I had the painter replace all bulbs with new.

I then visited Radio Shack and bought some 5a diodes for about a dime each. It was simple to install the diodes in series with the light switch, working on the floor at the switch and not up on a ladder. Each diode was well within the range of ampacity of the fixture/lights, of course.

The resulting lamps had a slightly yellower glow to them, but I never had to worry about replacing another bulb. (No diode on the fan's fan circuit, of course).

Another comment, about two lamps in a pump house wired in series, reminded me of our porch lights. There, we had a carriage light on either side of our entry doors on the front porch. The glass in the carriage lights was bubbled and cross hatched by design. I never really liked the way it looked when lighted, so I got a couple bulbs with spiral "flame type" indentations in them. They're quite common at any hardware, light or big box store. I then wired the two lights in series, rather than parallel, dropping the voltage across the two matched bulbs by 50%. The resulting light was a beautiful yellowish glow and the combined effect of the cross hatched glass in the fixtures, coupled with the spiral indentations in the bulb gave a flickering effect to the lights when someone walked or drove in front of them. It really looked swell and I soon began just leaving them on 24/7/365. When we moved from the house to another, we took the porch carriage lights with us; the bulbs were 13 years old by then, and they lasted another 10 years at the second location in the same fixtures.

"Yeahbut" applies. <grin>

A light bulb is__ _not_ __a 'constant resistance' device. It takes__ _more_ __than
a__ _simple_ __high-school physics application of Ohm's Law to get an accurate
answer.

Proof: measure the 'cold' resistance of a, say, (nominal) 100 watt (@120v) lightbulb. It only a þw= -- as in _less_than_ten_ -- ohms. Which is why a light bulb draws__ _lots_ __of power (for a very short time) when it is turned
on. {Google for 'light bulb inrush current' for the gory details, if you're
interested.} Incidentally, this also explains while the vast majority of
bulbs burn out__ _when_ __you turn them on.

Incandescent Lightbulbs run***HOT*** -- 'white hot' (grin), in fact. This
-greatly- increases the filament resistance over what it is at 'room
temperature'.

Reducing the voltage of a nominal 130v bulb to 120v will result in a__
_decrease_ __(albeit relatively minor) in the resistance of the bulb.
As a result of that change, the current flow at 120V will be__ _more_
__than 12/13 the current flow at 130v.

Thus, a '100 watt @ 130v' bulb has a nominal operating current draw of 0.7692A at 130v. a 'hot' resistance of roughly 169 ohms.

__
_if_ __one makes the mistake of assuming constant resistance, then at 120v the
consumption would be about .7100A (120/130 * .7692A, or about 85.2watt @ 120v).

BUT, this calculation is in***error***. Because the resistance of the bulb will__
_decrease_ __-- due to the fact that it is not running as 'hot' at 120V as it
does at 130V.

The exact figures depend on__ _exactly_ __how the bulb is constructed, but for
'typical' 130V bulbs, the power consumption will be around 88-90 watts, at
120V. (Roughly 92.3 watts is the 'absolute maximum')

#### Site Timeline

- posted on December 11, 2009, 8:45 pm

Actually those number are ohms per 1000 feet.

Yes. assuming the same length, material, etc.

No. The resistance should be 169 ohms at 130 volts. You seem to have gotten 156 ohms by dividing 120 volts by 0.768231 amps. There is no justification for this. The resistance of the filament will vary with its temperature. It will be lower than 169 ohms at 120 volts. but there is no reason to assume that it will be 156 ohms.

See previous comment. You seem to have derived a result from your assumptions (you assumed the current at 120 volts is 0.769231 amps in your previous calculation).

This result is based upon an incorrect assumption that the current is 0.7629231 when the voltage is 120 volts.

The power at 120 volts will be between the 85 watts (the value that would be calculated if the resistance is constant with temperature) and the 100 watts at 130 volts. Thus 92.3 watts is a reasonable guess for the power at 120 volts but you have not presented anything to prove it.

Dan

- posted on December 11, 2009, 7:24 pm

When it comes to an incandescent lamp filament, there is more than just the geometry (length and cross section)of the wire at work to determine the filament resistance.

Coatings on the wire, shape of the winding, the metal alloy are just a few of things that come into play to arrive at the final design.

For a given wattage of lamp, the total lamp resistance of the lamp increases in direct proportion to the rated voltage.

Lew

- posted on December 11, 2009, 8:18 pm

On 12/11/2009 01:24 PM, Lew Hodgett wrote:

I don't think this is correct.

Power = V^2/R

Based on that, to keep wattage constant resistance must increase as the square of the voltage. Double the voltage and the resistance has to increase by a factor of four.

Chris

I don't think this is correct.

Power = V^2/R

Based on that, to keep wattage constant resistance must increase as the square of the voltage. Double the voltage and the resistance has to increase by a factor of four.

Chris

- posted on December 11, 2009, 9:32 pm

"Chris Friesen" wrote:

You are correct.

Yep.

Lew

You are correct.

Yep.

Lew

- posted on December 11, 2009, 8:51 pm

With light bulbs, the explanation is

(Note: Lew had the 'results' right, albeit with a somewhat incorrect description of the 'causation')

'Rough service' bulbs use a different composition in the filament element than normal service bulbs. It has a

BTW, the relationship between bulb life and applied voltage is an

NOTE: the cost of the electricity to operate a 'typical' household light-bulb (25-150 watt) is generally

Hint: imagine a square copper wire of a fixed length, now double its thickness and width, _in_steel_, and explain how that is equivalent to four copper wires of the original thickness in parallel for 1/4 the resistance! <grin>

- posted on December 11, 2009, 9:28 am

Actually it isn't, it's temperature dependant. The resistance goes up with increasing temperature so that as the voltage increases the dissipated power doesn't increase in a linear way.

- posted on December 11, 2009, 12:10 am

If you have a light fixture that's difficult to reach, Radio Shack is your friend. Our great room had a couple track lights well over 30' from the floor and a ceiling fan's 4 lights were around 25' high. Even with my straight ladder braced against the ceiling, it was awkward as the dickens getting the ladder inside, extended and within reach of either track OR the fan in particular. Eventually, some of the track bulbs departed and a couple of the fan bulbs as well. It was about time to paint the room, so I had the painter replace all bulbs with new.

I then visited Radio Shack and bought some 5a diodes for about a dime each. It was simple to install the diodes in series with the light switch, working on the floor at the switch and not up on a ladder. Each diode was well within the range of ampacity of the fixture/lights, of course.

The resulting lamps had a slightly yellower glow to them, but I never had to worry about replacing another bulb. (No diode on the fan's fan circuit, of course).

Another comment, about two lamps in a pump house wired in series, reminded me of our porch lights. There, we had a carriage light on either side of our entry doors on the front porch. The glass in the carriage lights was bubbled and cross hatched by design. I never really liked the way it looked when lighted, so I got a couple bulbs with spiral "flame type" indentations in them. They're quite common at any hardware, light or big box store. I then wired the two lights in series, rather than parallel, dropping the voltage across the two matched bulbs by 50%. The resulting light was a beautiful yellowish glow and the combined effect of the cross hatched glass in the fixtures, coupled with the spiral indentations in the bulb gave a flickering effect to the lights when someone walked or drove in front of them. It really looked swell and I soon began just leaving them on 24/7/365. When we moved from the house to another, we took the porch carriage lights with us; the bulbs were 13 years old by then, and they lasted another 10 years at the second location in the same fixtures.

--

Nonny

ELOQUIDIOT (n) A highly educated, sophisticated,

Nonny

ELOQUIDIOT (n) A highly educated, sophisticated,

Click to see the full signature.

- posted on December 11, 2009, 12:10 am

"Stuart" wrote:
---------------------------------------------------

Actually the load, in this case 100 watts, remains the same, but the lumen out decreases with increase in voltage rating of the lamp.

I used to make a very good living designing and selling industrial lighting systems.

As power costs increase the effiency of lamps becomes more and more important.

It was a straight forward process, based on total cost of ownership, to justify $30.00 lamps with 20,000 hour lamp life when power costs were less than $0.03/KWH.

With today's power costs, it's a slam dunk.

Lew

Actually the load, in this case 100 watts, remains the same, but the lumen out decreases with increase in voltage rating of the lamp.

I used to make a very good living designing and selling industrial lighting systems.

As power costs increase the effiency of lamps becomes more and more important.

It was a straight forward process, based on total cost of ownership, to justify $30.00 lamps with 20,000 hour lamp life when power costs were less than $0.03/KWH.

With today's power costs, it's a slam dunk.

Lew

- posted on December 11, 2009, 1:18 am

Lew Hodgett wrote:

And yet you don't know how to rerate an incandescent bulb. Just goes to show that he who has the best line of bullshit wins.

And yet you don't know how to rerate an incandescent bulb. Just goes to show that he who has the best line of bullshit wins.

- posted on December 11, 2009, 3:33 am

"J. Clarke" wrote:

After you have had a chance to review a lamp catalog, get back to me.

Lew

After you have had a chance to review a lamp catalog, get back to me.

Lew

- posted on December 11, 2009, 4:07 am

Lew Hodgett wrote:

And what do you believe that a lamp catalog will tell me? If you think that it will tell me tht a lamp rated for 100 watts at 130 volts will draw 100 watts at 120 volts then you need to take remedial reading.

And what do you believe that a lamp catalog will tell me? If you think that it will tell me tht a lamp rated for 100 watts at 130 volts will draw 100 watts at 120 volts then you need to take remedial reading.

- posted on December 11, 2009, 1:19 am

scrawled the following:

Newp. I get 100W worth of light out of $2 (delivered) 23W CFLs. No replacements necessary for 2 years now, but all my Feit CFLs have failed in under a year, including the 4 they replaced at their cost.

-- To know what you prefer instead of humbly saying Amen to what the world tells you you ought to prefer, is to have kept your soul alive. -- Robert Louis Stevenson

Newp. I get 100W worth of light out of $2 (delivered) 23W CFLs. No replacements necessary for 2 years now, but all my Feit CFLs have failed in under a year, including the 4 they replaced at their cost.

-- To know what you prefer instead of humbly saying Amen to what the world tells you you ought to prefer, is to have kept your soul alive. -- Robert Louis Stevenson

- posted on December 11, 2009, 12:09 am

Lew Hodgett wrote:

If he's using 130v bulbs on 120v then he's using about 15% less KWH than he would be using 120v of the same nominal wattage. If that gives him enough light then he's got no problem. If he has to add bulbs or go up a wattage level to get the illumination level he needs then things get more complicated.

If he's using 130v bulbs on 120v then he's using about 15% less KWH than he would be using 120v of the same nominal wattage. If that gives him enough light then he's got no problem. If he has to add bulbs or go up a wattage level to get the illumination level he needs then things get more complicated.

- posted on December 11, 2009, 12:41 am

"J. Clarke" wrote:

NOT!

100 watts is 100 watts regardless the voltage rating of the lamp.

The current flowing thru the lamp is reduced which reduces lumen output when the voltage rating of the lamp is increased.

(E = I*R for a resistance load.)

Lew

NOT!

100 watts is 100 watts regardless the voltage rating of the lamp.

The current flowing thru the lamp is reduced which reduces lumen output when the voltage rating of the lamp is increased.

(E = I*R for a resistance load.)

Lew

- posted on December 11, 2009, 1:15 am

Lew Hodgett wrote:

If you have visions of becoming an electrical engineer, don't quit your day job.

For a resistive load P=E^2/R. If the lamp is rated for 100 watts at 130 volts then it will dissipate (120^2/130^2)*100 watts at 120v or about 85 watts if its resistance remains constant.

If you have visions of becoming an electrical engineer, don't quit your day job.

For a resistive load P=E^2/R. If the lamp is rated for 100 watts at 130 volts then it will dissipate (120^2/130^2)*100 watts at 120v or about 85 watts if its resistance remains constant.

- posted on December 11, 2009, 3:42 am

"J. Clarke" wrote:

Review lamp data found in any lamp catalog.

The proof is left to the student.

Lew

Review lamp data found in any lamp catalog.

The proof is left to the student.

Lew

- posted on December 11, 2009, 4:09 am

Lew Hodgett wrote:

Proof? This is high school physics.

If you disagree please be kind enough to show us some support instead of some vague "review a lamp catalog".

Proof? This is high school physics.

If you disagree please be kind enough to show us some support instead of some vague "review a lamp catalog".

- posted on December 11, 2009, 10:09 pm

"Yeahbut" applies. <grin>

A light bulb is

Proof: measure the 'cold' resistance of a, say, (nominal) 100 watt (@120v) lightbulb. It only a þw= -- as in _less_than_ten_ -- ohms. Which is why a light bulb draws

Incandescent Lightbulbs run

Reducing the voltage of a nominal 130v bulb to 120v will result in a

Thus, a '100 watt @ 130v' bulb has a nominal operating current draw of 0.7692A at 130v. a 'hot' resistance of roughly 169 ohms.

BUT, this calculation is in

The exact figures depend on

- posted on December 11, 2009, 5:12 am

J. Clarke wrote:

The assumption that the resistance will remain constant is a bad one. As has already been pointed out elsewhere in this thread, the resistance of a light bulb varies with the temperature of the filament. A colder filament will have a lower resistance. A lower resistance will result in a higher current and a higher power. The actual power at 120 volts will be somewhere between the 85 watts that you calculated and the 100 watts that it would dissipate at 130 volts.

The assumption that the resistance will remain constant is a bad one. As has already been pointed out elsewhere in this thread, the resistance of a light bulb varies with the temperature of the filament. A colder filament will have a lower resistance. A lower resistance will result in a higher current and a higher power. The actual power at 120 volts will be somewhere between the 85 watts that you calculated and the 100 watts that it would dissipate at 130 volts.

- posted on December 11, 2009, 5:50 am

Dan Coby wrote:

The standard number given is V^1.6. But the point is that a 100 watt bulb only draws 100 watts at the design voltage, it's not 100 watts at all voltages as Brainiac claims.

The standard number given is V^1.6. But the point is that a 100 watt bulb only draws 100 watts at the design voltage, it's not 100 watts at all voltages as Brainiac claims.

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