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Re: Hey do you know your car's alternator only outputs 7-10% while it'srunning?

H

Hammy

You are both wrong.
With AC, Volts x Amps only gives the power with a resistive (heating)
load.
If you have an inductive load this is no longer true. (ie a motor.)
If it was a pure inductor, you could draw as many amps as you liked
and no power would be consumed.
Pure inductors only exist in theory BTW.
If you connect a capacitor across the mains, it will draw current but
no power will be consumed (virtually).
Pure capacitors (as near as dammit) exist.

The pair of you study this:
http://en.wikipedia.org/wiki/Power_factor

I don't know where you Yanks get educated. God help America!
I know you said for "Pure capacitors" and "ideal inductors".But just
in case newb takes you literally caps will dissapete power in the form
of heating due to esr,and they will fail if rms current rateings arent
followed.Well you would think this is obvious a lot of SMPS's still
fail due to insufficent rated capacitors for rms current handling
derated for temp and frequency.

Inductors also have DC losses (DCR),and AC core losses,insulateing
breakdown interwinding capacitance,eddy losses etc.
 
P

Paul Keinanen

And, like the series Xmas lights of yore, when one goes out, they all
go out.

In addition to the normal solenoid (A) in series to automatically
adjust the distance between the electrodes, both the Brush as well as
Thomson-Houston arc-lamp had an additional (high resistance) solenoid
(B) across the electrodes.

If the arc was extinguished, the solenoid B provided loop continuity
and now that the loop current was flowing through it, it also kicked
the electrodes together to attempt a reignition.

There is a brief description in "A history of control engineering,
1800-1930" by Stuart Bennett pages 155-159 (Fig. 5.1c and 5.2)

Some of the pages are available at

http://books.google.com/books?id=1gfKkqB_fTcC&lpg=PA156&pg=PA156#v=onepage&q=&f=false

Paul
 
Y

You

John Fields said:
It's clear to me that if the motor load is resistive and draws 6 amperes
from 115VRMS mains, then the power delivered to a nonreactive mechanical
load cannot be greater than:

Sonny, there is NO SUCH THING as a pure Resistive for a Motor.......
 
H

Hammy

BTW.
RMS means Root of the Mean Squared
It's a sort of "average" that indicates the DC equivalent of the
waveform.
EG For a sine wave the RMS value is 0.636 of the peak value.
For a square wave it is 0.5 of the peak value.

This is not a technically correct explanation BTW but I'm trying to
make it easy for you
Actually its not correct at all.I'll ignore your feeble attempt at an
insult.

I think you may want to take a closure look at capacitors. They do
indeed have a resistive component obviously we aren't talking about a
resistor but it's the sum of all intrinsic extrinsic imperfections
impurities (chemical and packaging imperfections). This is modeled as
a resistor in series with the cap, this is responsible for capacitor
heating from ripple current. Several types of caps have double digit
esr figures go to digikey and look up Tantulm caps they spec the esr
parametrically. Polymer caps have single milli ohms digit or low teens
milli ohms esr but you pay for them in price as well as possibly a
more complicated loop compensation may be required for smps or LDO due
to the esr zero being useless at the HF that a small esr puts it
at.All capacitors have leakage as well; modeled as parallel resistor,
you place an additional resistor for discharging if the load wont
drain it quick enough at power off to comply with safety regulations
or if it's a requirement of the device to insure correct operation.
I'll leave the effect of inductances as a much needed exercise for you
to investigate.

When discussing power typically one uses average waveforms with
exception of say a sine wave with both positive and negative area
under the curve being equal; obviously the average would be zero This
would also apply to bipolar square waves haveing equal area +/-. Hence
rms describes the equivalent heating effect of a resistor.

Oh and by the way the mean of a sine wave is 0.636 the crest factor is
1/sqrt(2) and the rms is sqrt(2).
 
R

Rich Grise

Not if it fails shorted.

You must be very young. I remember, oh, about a half-century ago, laying
the Xmas tree lights out on the floor, and if the string didn't light,
we'd take a known good bulb and go down the string swapping out bulbs,
one at a time; if the string didn't light, we'd take the bulb we just
removed from its socket and swap it out with the next one, and so on.
When the string lights up, you'd throw away the bulb that you just
removed/replaced.

Nowadays, they apparently do have low-V bulbs that are designed to fail
short; I'm danged if I know how they accomplish it. :)

I can guess - one of the filament supports is springy - when the filament
opens, that support springs back, contacting another electrode, shorting
the bulb - but wouldn't that be kinda expensive?

Thanks,
Rich
 
H

Hammy

Actually its not correct at all.I'll ignore your feeble attempt at an
insult.

I think you may want to take a closure look at capacitors. They do
indeed have a resistive component obviously we aren't talking about a
resistor but it's the sum of all intrinsic extrinsic imperfections
impurities (chemical and packaging imperfections). This is modeled as
a resistor in series with the cap, this is responsible for capacitor
heating from ripple current. Several types of caps have double digit
esr figures go to digikey and look up Tantulm caps they spec the esr
parametrically. Polymer caps have single milli ohms digit or low teens
milli ohms esr but you pay for them in price as well as possibly a
more complicated loop compensation may be required for smps or LDO due
to the esr zero being useless at the HF that a small esr puts it
at.All capacitors have leakage as well; modeled as parallel resistor,
you place an additional resistor for discharging if the load wont
drain it quick enough at power off to comply with safety regulations
or if it's a requirement of the device to insure correct operation.
I'll leave the effect of inductances as a much needed exercise for you
to investigate.

When discussing power typically one uses average waveforms with
exception of say a sine wave with both positive and negative area
under the curve being equal; obviously the average would be zero This
would also apply to bipolar square waves haveing equal area +/-. Hence
rms describes the equivalent heating effect of a resistor.

Oh and by the way the mean of a sine wave is 0.636 the crest factor is
1/sqrt(2) and the rms is sqrt(2).

Correction

Oh and by the way the mean of a sine wave is 0.636 the crest factor is
sqrt(2) and the rms is1/ sqrt(2).
 
H

Hammy

BTW.
RMS means Root of the Mean Squared
It's a sort of "average" that indicates the DC equivalent of the
waveform.
EG For a sine wave the RMS value is 0.636 of the peak value.
For a square wave it is 0.5 of the peak value.

Christ your a moron what about duty cycle this has no effect on the
average of your Theoretical square wave? Consider also transient
excursions and rise fall times if you want to get anal about it.

Any idiot can quote formulas some one else derived for well known
waveforms they are in every power electronics book.Mind you you seem
to be haveing trouble interpeting those. It's when youy get to non
periodic waveforms it gets intresting.
This is not a technically correct explanation BTW but I'm trying to
make it easy for you

so right you are there.
 
S

Spehro Pefhany

I can guess - one of the filament supports is springy - when the filament
opens, that support springs back, contacting another electrode, shorting
the bulb - but wouldn't that be kinda expensive?

That's not how they work. It's generally a wire-wound shunt across the
filament supports that has enough oxide that it breaks down well below
line voltage (when the lamp fails) but not at the operating voltage of
the lamp (a few volts). Costs next to nothing to wind a few turns of
wire around the filament supports.
 
R

RoyLFuchs

Abuse complaint forwarded.

All other interested parties please follow with same.
Perhaps this dope will listen when his ISP tells him.

forward the entire message below to

[email protected]
 
A

Archimedes' Lever

Sonny, there is NO SUCH THING as a pure Resistive for a Motor.......

Yes, there is.

A rail gun is a linear motor, and there is ZERO inductance involved.

Mag lev is resistive as well, since it is a DC field on the levitation
coils.

There are other examples. Piezo comes to mind.

Sonny? Jeez, ya retarded ****, get a clue.
 
M

Mycelium

You must be very young. I remember, oh, about a half-century ago, laying
the Xmas tree lights out on the floor, and if the string didn't light,
we'd take a known good bulb and go down the string swapping out bulbs,
one at a time; if the string didn't light, we'd take the bulb we just
removed from its socket and swap it out with the next one, and so on.
When the string lights up, you'd throw away the bulb that you just
removed/replaced.

NO! You said "and like the Xmas lights..." which means that you were
talking about arc lamps in series. THAT is what my comment is about.

Are you drunk, or do you just have a problem remembering what YOU
wrote?

I know how light bulbs fucking work. There is no such thing as such a
bulb "failing shorted" so you fail yet again on common sense as well.


So, while using your BRAIN, re-read what I wrote, which is CLREALY
referencing ARC LAMPS.

Nowadays, they apparently do have low-V bulbs that are designed to fail
short; I'm danged if I know how they accomplish it. :)

They are not incandescents.
I can guess - one of the filament supports is springy - when the filament
opens, that support springs back, contacting another electrode, shorting
the bulb - but wouldn't that be kinda expensive?

Again, use your common sense. and NO, I know of NO light bulbs that
fail shorted.
 
M

Mycelium

Except a fillament virtually can NOT fail shorted.

Except that I was referring to arc lamps.

Also, there is no 'virtually' about it. Short of an impact that jams
both input leads together, an incandescent lamp can NEVER "fail" shorted.

In fact, a good way to 'fix' one is with a swift thud while under power.
If it hits the lead, it will re-weld itself back on, and that usually
(certainly) will not last as long as a properly cinched filament. It
will usually get you a few more hours out of the bulb, but I have seen
them go on for years after a re-attachment.

The bulb has to
besigned to be "failsafe",

It has nothing to do with design. The ONLY failure mode IS an open
filament, and that is the only place such a bulb ever fails short of the
encapsulation being breached. That is just the physics of it by default.
so if the filament does not heat the bulb
shorts

You're nuts.

A light bulb is, by default, a very low resistance device. Any shorting
of its internals or the device which it is mounted in would trip the
branch circuit protection. If not, the mount is suspect of total
non-compliance.

The bulb is specifically designed to couple as little heat as possible
to the base it sits in. That is why you never see a dense epoxy holding
the base on the bulb. It is always "airy" baked material of very low
density and therefore low thermal conductivity.

IF the filament feeder leads were to be shorted (broken bulb under
power)(happens a lot with old style garage shop lights), they would NOT
heat the base, the short would cause a breaker to trip immediately.
 
O

OutsideObserver

You are nothing new, unique, interesting, worthwhile, or anything
good;

Did I ever claim to be such? You are daft!
nothing but a waste of time.

Yet you pay.
Do Usenet a favor and
self-destruct.

Folks like you that spew words like that are the ones that should
succumb to something... like a brain clot that disables you completely
and leaves you speechless so that you cannot tell anyone that you are in
extreme pain. Your family should be cursed with the pain of having to
feed and dress you, etc.

You're crawlin' and you shant get up... boy. (in so many ways)
 
S

Steve Ackman

on Fri said:
...
You are nothing new, unique, interesting, worthwhile, or anything
good; nothing but a waste of time. Do Usenet a favor and
self-destruct.

It's a simple matter.

%BOS
[alt.energy.homepower]
Score: =-9999
NNTP-Posting-Host: 72.197.140.80
%EOS
 
NO! You said "and like the Xmas lights..." which means that you were
talking about arc lamps in series. THAT is what my comment is about.

Are you drunk, or do you just have a problem remembering what YOU
wrote?

I know how light bulbs fucking work. There is no such thing as such a
bulb "failing shorted" so you fail yet again on common sense as well.


So, while using your BRAIN, re-read what I wrote, which is CLREALY
referencing ARC LAMPS.



They are not incandescents.


Again, use your common sense. and NO, I know of NO light bulbs that
fail shorted.
Then you don't know much about decorative series string lighting
technology.
 
Except that I was referring to arc lamps.

Also, there is no 'virtually' about it. Short of an impact that jams
both input leads together, an incandescent lamp can NEVER "fail" shorted.

In fact, a good way to 'fix' one is with a swift thud while under power.
If it hits the lead, it will re-weld itself back on, and that usually
(certainly) will not last as long as a properly cinched filament. It
will usually get you a few more hours out of the bulb, but I have seen
them go on for years after a re-attachment.



It has nothing to do with design. The ONLY failure mode IS an open
filament, and that is the only place such a bulb ever fails short of the
encapsulation being breached. That is just the physics of it by default.


You're nuts.

A light bulb is, by default, a very low resistance device. Any shorting
of its internals or the device which it is mounted in would trip the
branch circuit protection. If not, the mount is suspect of total
non-compliance.

The bulb is specifically designed to couple as little heat as possible
to the base it sits in. That is why you never see a dense epoxy holding
the base on the bulb. It is always "airy" baked material of very low
density and therefore low thermal conductivity.

IF the filament feeder leads were to be shorted (broken bulb under
power)(happens a lot with old style garage shop lights), they would NOT
heat the base, the short would cause a breaker to trip immediately.

We are talking SERIES STRING lighting and they do in fact exist - not
only that they are very common. And they work as described.

It's time you read the thread, and thought it out before you answer.

The answer was in response to this : And, like the series Xmas lights
of yore, when one goes out, they all go out.
 
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