Password avoidance


C

Char Jackson

Actually, it is and IIRC it has to do with wires ability to carry power more
efficiently at higher voltages. As I understand it, that's also one of the
reasons utilities transmit power at kilovolts instead of 120 volts.
But, since I can't quote any statistics and am too lazy to google any more,
I'm going to say you win. I think the difference is minor anyway so let's
revert to theory and say you're right.
Dave
You're confusing Ohm's Law with transmission theory, two totally
different things. You correctly stated one form of Ohm's Law above,
P=IE. That darn equals sign gets in the way of your current position.
:)
 
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C

Char Jackson

My bad, you are correct. The acronym I should have use is KVA
(KiloVoltAmps). IIRC KWH is true power and KVA is apparent power. I don't
know that much about the difference, I know I used to rent generators rated
in KWH and now they are in KVA and there is a slight difference in the
output ability. 2) There is a formula for this one. :-D
Dave
Oy. :) Glad we got to the bottom of it.
 
D

Dave

Char Jackson said:
You're confusing Ohm's Law with transmission theory, two totally
different things. You correctly stated one form of Ohm's Law above,
P=IE. That darn equals sign gets in the way of your current position.
:)
So, when you transmit power it acts one way, but when you apply it to an
appliance it acts another? I was never taught that and never found it in
application either. YRMV.
I actually did not state Ohm's law, it is I=E/R. What I stated was Watt's
law, which is P=I*E. You do realize one is relevant to power and one is
relevant to resistance?
I would agree, except you are assuming the power is the same in both
instances, or more correctly, the current draw, which I am saying is not.
So, in my case, you can't solve for P in both 120 and 240 by assuming I is
automatically going to be half if E is doubled.
I hope you're not going to have us Thevonize a circuit, I hated that more
than anything. Plus, I'd have to dig out some reference books just to get
all the steps correct. :-D
Dave
 
C

Chris Sidener

TOM said:
I got to thinking about the lamps used on theater marques. I think they
last as long as the do because the switching cycle is short enough the
filament doesn't have to go through the expansion/contraction of a
household-type on/of cycle.

Probably wrong, but it sounds sort of logical to me... :>))
Long ago, in the 50s and before I got my BSEE from Purdue, I
worked part-time at a Westinghouse Sales facility. We stocked
"ruggedized" light bulbs for traffic lights and such. They were
marked "130v" and had a heavy duty filament structure. They
cost 2 times as much as standard bulbs, but I was told
that they lasted 5 times as long!

Chris
 
C

Char Jackson

So, when you transmit power it acts one way, but when you apply it to an
appliance it acts another? I was never taught that and never found it in
application either. YRMV.
Yes, transmission lines have known losses which need to be accounted
for and factored into their designs, but that has nothing to do with
the current discussion. There are no long distance transmission lines
in a residential or even commercial property, so the losses from such
lines are irrelevant to the current discussion. I suspect the reason
you were never taught that is because this is far outside your area of
expertise, which I believe is what you said in a previous post.

Take your example of the electric dryer. If you have two dryers, one
designed for 120v and the other designed for 240v, the 120v model will
draw twice as many amps as the 240v model. Both dryers will use the
same number of Watts, (because P=IE), assuming the only differences
between the two models are the voltages they are designed to operate
with.
I actually did not state Ohm's law, it is I=E/R. What I stated was Watt's
law, which is P=I*E. You do realize one is relevant to power and one is
relevant to resistance?
I would agree, except you are assuming the power is the same in both
instances, or more correctly, the current draw, which I am saying is not.
If we're having a technical discussion, then P (power) is expressed in
Watts, not amps. Current draw refers to amps, but power refers to
Watts. Watts are the product of amps times voltage. (P=IE)
So, in my case, you can't solve for P in both 120 and 240 by assuming I is
automatically going to be half if E is doubled.
If you're saying that "P equals IE" is wrong, and should be restated
as "P is approximately equal to IE", well we better get the textbooks
updated with this new formula. In the meantime, yes, if E is doubled
then I will be halved and P will remain unchanged. It doesn't just
work that way in the textbooks, it works that way in the field, too.
I hope you're not going to have us Thevonize a circuit, I hated that more
than anything. Plus, I'd have to dig out some reference books just to get
all the steps correct. :-D
Not a bad idea to dust off the books now and then. :)
 
G

Gene E. Bloch

Gene


Actually Gordon is 100% correct with that statement. I agree . Same scenario if you
open and close a 100w light in the room on a 110v 60hz circuit it will take up twice
that amount of power than if you would leave the light on for a 24hr period without
shutting it off. Same exact principle for the computer
It could hardly be the same exact principle for the computer, since the
light bulb is a hot filament - a resistive load - with its resistance
changing as it heats up, whereas a computer is a reactive load.

I think you guys need to plug a Kill A Watt in and look what (watt?)
happens. You should also measure the cold resistance of a filament bulb to
get some idea how much current it might draw before it reaches its
operating temperature - and note that it takes well under a second to get
to that temperature.

Here's a hint: 100W bulb = 90 ohms. At 120V, that's 1.33A, or 160W at
startup. For under 1 second...

So tell me - I *really* want to know - how can this bulb use up twice as
much power when I turn it on, compared to its steady state consumption for
a day?
 
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G

Gene E. Bloch

I was told, by the journeyman electrician I worked for, that the terms
"open" and "close" originated "way back when." He assumed that when
candles were used to provide light, they had shutters; you open the
shutters to light the room and close the shutters to darken the room.

Another idea was the "barn door" shutters on Klieg lights:

http://en.wikipedia.org/wiki/Klieg_light

Barn doors:
http://en.wikipedia.org/wiki/Barn_doors

Closer to the original off topic:
http://michaelbluejay.com/electricity/myths.html
Have you ever seen an old-fashioned knife switch? One glance at that (or
for that matter at the interior of a modern wall switch) would tell you
where those terms come from. They are quite literal...
 
G

Gene E. Bloch

Yes, transmission lines have known losses which need to be accounted
for and factored into their designs, but that has nothing to do with
the current discussion. There are no long distance transmission lines
in a residential or even commercial property, so the losses from such
lines are irrelevant to the current discussion. I suspect the reason
you were never taught that is because this is far outside your area of
expertise, which I believe is what you said in a previous post.

Take your example of the electric dryer. If you have two dryers, one
designed for 120v and the other designed for 240v, the 120v model will
draw twice as many amps as the 240v model. Both dryers will use the
same number of Watts, (because P=IE), assuming the only differences
between the two models are the voltages they are designed to operate
with.


If we're having a technical discussion, then P (power) is expressed in
Watts, not amps. Current draw refers to amps, but power refers to
Watts. Watts are the product of amps times voltage. (P=IE)


If you're saying that "P equals IE" is wrong, and should be restated
as "P is approximately equal to IE", well we better get the textbooks
updated with this new formula. In the meantime, yes, if E is doubled
then I will be halved and P will remain unchanged. It doesn't just
work that way in the textbooks, it works that way in the field, too.


Not a bad idea to dust off the books now and then. :)
It seems that people are ignoring Ohm's law.

If you double the voltage into a fixed device, you will not halve the
current but in fact double it. Ohm's law says E = IR (that's for DC; it's
more complicated in AC with capacitance and inductance to consider - i.e.,
reactive components, but that wouldn't affect this argument).

The result is that the putative dryer will consume four times the power (in
the short amount of time before it bursts into flame).

The case where the power is the same at 120 vs 140 volts is where we have
two entirely separate devices, each one properly designed for its voltage.
 
G

Gene E. Bloch

Oy. :) Glad we got to the bottom of it.
We didn't. KVA is a measure of power (when the power factor is included),
KWH is a measure of energy consumed.

Energy consumed is power times time, or equivalently, power is energy per
unit time.
 
C

Char Jackson

It seems that people are ignoring Ohm's law.

If you double the voltage into a fixed device, you will not halve the
current but in fact double it. Ohm's law says E = IR (that's for DC; it's
more complicated in AC with capacitance and inductance to consider - i.e.,
reactive components, but that wouldn't affect this argument).

The result is that the putative dryer will consume four times the power (in
the short amount of time before it bursts into flame).

The case where the power is the same at 120 vs 140 volts is where we have
two entirely separate devices, each one properly designed for its voltage.
I totally agree with your assessment. I may have written poorly or
clumsily above, but at least my dryer example had two dryers. ;-)
 
G

Gene E. Bloch

I totally agree with your assessment. I may have written poorly or
clumsily above, but at least my dryer example had two dryers. ;-)
Yes, you have been accurate & clear, IMO.

BTW, I failed to notice my error above:
"120 vs 140 volts" should have been "120 vs 240 volts"
 
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D

Dave

Char Jackson said:
Yes, transmission lines have known losses which need to be accounted
for and factored into their designs, but that has nothing to do with
the current discussion. There are no long distance transmission lines
in a residential or even commercial property, so the losses from such
lines are irrelevant to the current discussion. I suspect the reason
you were never taught that is because this is far outside your area of
expertise, which I believe is what you said in a previous post.
One final post and I'm done with this discussion. Now you've changed from
conductors ability to transmit, stated as MHO, to resistance or impedance, a
conductor's inability to transmit. It looks like you have trouble staying on
target and shift from one side to the other just to be right. I suspect your
errors in your postings is due to you trying to be right no matter the cost
so you don't stay on topic. Yes, I'm willing to admit my limitations, but
now realize you aren't and are limited, but won't admit it.
Take your example of the electric dryer. If you have two dryers, one
designed for 120v and the other designed for 240v, the 120v model will
draw twice as many amps as the 240v model. Both dryers will use the
same number of Watts, (because P=IE), assuming the only differences
between the two models are the voltages they are designed to operate
with.
Again, you are assuming power (P) will remain the same and current (I) will
halve when you double voltage (E or V), which I'm saying you don't know, but
are assuming.
If we're having a technical discussion, then P (power) is expressed in
Watts, not amps. Current draw refers to amps, but power refers to
Watts. Watts are the product of amps times voltage. (P=IE)
I'm sorry, but we were in fact having a technical discussion due to your
incessant postings trying to make it look like you know more than I do, or
something to that affect. I really don't care if you do, but will not let
you make me look bad when you so obviously don't know as much as you think
you do. For instance, I don't know why you'd take the time to reiterate
Watt's Law (which you erroneously called Ohm's Law) except trying to avoid
what you posted. See, you didn't recognize the difference when it was first
posted and now you're some kind of expert on Watt's Law.
If you're saying that "P equals IE" is wrong, and should be restated
as "P is approximately equal to IE", well we better get the textbooks
updated with this new formula. In the meantime, yes, if E is doubled
then I will be halved and P will remain unchanged. It doesn't just
work that way in the textbooks, it works that way in the field, too.
Now you're trying to save face again by twisting what I wrote. It was clear
and still is clear, I did not say or suggest P=I*E is anything but exact. As
far as textbooks and formulas, when you know the difference between Watt's
Law, Ohm's Law, MHO, Impedance and etc. we can discuss again. Well, that's
probably not going to happen as you don't want to discuss, you want to
embarrass and so far I'd say it's you that needs the textbooks.
Not a bad idea to dust off the books now and then. :)
Again, when you decide you're ready to read a textbook let me know, I'll
ship you some for free. You'll LOVE the one with Thevenin's Theorum.
 
D

Dave

snip
It seems that people are ignoring Ohm's law.

If you double the voltage into a fixed device, you will not halve the
current but in fact double it. Ohm's law says E = IR (that's for DC; it's
more complicated in AC with capacitance and inductance to consider - i.e.,
reactive components, but that wouldn't affect this argument).

The result is that the putative dryer will consume four times the power
(in
the short amount of time before it bursts into flame).

The case where the power is the same at 120 vs 140 volts is where we have
two entirely separate devices, each one properly designed for its voltage.
Someone correct me here if I'm wrong, but it's my understanding when you
wire a dryer that's dual voltage capable, with 120V you use maybe half the
heating elements. When you wire it with 240V you use all the heating
elements. So everything changes by an unknown except Voltage which we know
is 240V. We can now start to solve the problem by finding the resistance
(Ohms) with an Ohmmeter, which is now a known and solve for current (Amps)
or find current with an Ammeter and solve for resistance (Ohms). We can now
solve for power (Watts) now as well. We can cross-check with Ohm's Law to
verify. We can solve for Watts before Ohms, but we have to have two known's
before we can solve. We cannot assume any factor and come up with a factual
result.
To go back to your post, if you wire a dryer that's not dual voltage
capable, which I think is what you meant judging by the data in your post,
then you are correct, you will let the smoke out of the dryer. One final
note, while your assessment of Ohm's law is true, you are assuming all the
people in this discussion don't know the difference between devices that are
multi-voltage capable and that's not so. Judging by Char's reply to your
post he doesn't know the difference either, just like he didn't know the
difference between Watt's Law and Ohm's Law.
Dave
 
S

Sam E

Sam said:
[snip]
You may be using a computer for 60 years or more. 60 years, and ONE
logon per day. Now you've got over 21,914 logons. At 5 seconds each,
that's more than 24 hours wasted on repeatedly entering your password.
That is fine
What's "this"? Are you continuing to imagine that I was LIMITING
myself to ONE DAY? How many times do you log on during your LIFE? You
spend a lot more than 5 seconds logging on.
if it is your own computer.
Do it at work that demands security, and you'll find yourself out of a job.
That is totally ridiculous. All it takes is an outsider to get by
password security and it is all over with. What the OP wants is not
worth it.
You replied to the wrong person. I asked questions.
 
G

Gene E. Bloch

snip


Someone correct me here if I'm wrong, but it's my understanding when you
wire a dryer that's dual voltage capable, with 120V you use maybe half the
heating elements. When you wire it with 240V you use all the heating
elements. So everything changes by an unknown except Voltage which we know
is 240V. We can now start to solve the problem by finding the resistance
(Ohms) with an Ohmmeter, which is now a known and solve for current (Amps)
or find current with an Ammeter and solve for resistance (Ohms). We can now
solve for power (Watts) now as well. We can cross-check with Ohm's Law to
verify. We can solve for Watts before Ohms, but we have to have two known's
before we can solve. We cannot assume any factor and come up with a factual
result.
To go back to your post, if you wire a dryer that's not dual voltage
capable, which I think is what you meant judging by the data in your post,
then you are correct, you will let the smoke out of the dryer. One final
note, while your assessment of Ohm's law is true, you are assuming all the
people in this discussion don't know the difference between devices that are
multi-voltage capable and that's not so. Judging by Char's reply to your
post he doesn't know the difference either, just like he didn't know the
difference between Watt's Law and Ohm's Law.
Dave
You are right in your supposition. I was talking about plugging a dryer
(really any device, but dryer is the example that was chosen) that is wired
for 120V into 240V power. I don't think I was aware that a 120V machine
could normally be rejumpered for 240V operation. Thanks for that
information.

The multi-voltage capable dryer certainly could be thought of as an
implementation of what we were describing as two dryer models, one for 120V
and one for 240V. Clearly it's a more elegant idea than having two distinct
models, especially in an place where both voltages are available. It might
be as simple as wiring a pair of motor coils in parallel for 120V and
serial for 240V, the same for heating coils, and having transformer taps
for the power supply for the digital circuits.

Probably most of us are aware that the switching power supplies commonly
used as USB charger wall warts can be plugged into a range of voltages at
different frequencies. I don't know how they are wired, but clearly they
are designed by the right engineers (not designed by me, for sure!).
 
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G

Gene E. Bloch

Have you ever seen an old-fashioned knife switch? One glance at that (or
for that matter at the interior of a modern wall switch) would tell you
where those terms come from. They are quite literal...
I was careless in the above. I forgot to point out that when you *open* a
switch you get darkness and when you *close* a switch you get light.
However, my grandparents and other relatives of their generation (they were
immigrants born around 1880) used to say "close the light" to mean turn the
light off.

Also, to engineers at least, an open circuit is a circuit that is not
powered or otherwise not complete.

BTW, if I were a poor person in the days of candles, I don't think I'd want
to close a shutter on them to cut off their light. I'd be burning candles
without getting the benefit of their light, which is poor economics.
 
G

GreyCloud

Sam said:
Sam said:
[snip]

You may be using a computer for 60 years or more. 60 years, and ONE
logon per day. Now you've got over 21,914 logons. At 5 seconds each,
that's more than 24 hours wasted on repeatedly entering your password.
That is fine
What's "this"? Are you continuing to imagine that I was LIMITING
myself to ONE DAY? How many times do you log on during your LIFE? You
spend a lot more than 5 seconds logging on.

if it is your own computer.
Do it at work that demands security, and you'll find yourself out of a job.
That is totally ridiculous. All it takes is an outsider to get by
password security and it is all over with. What the OP wants is not
worth it.
You replied to the wrong person. I asked questions.
I remember a DOD gal that managed a VMS cluster and she got pretty lazy
and wrote down her password, only because VMS does not allow
passwordless systems, and the system got compromised. She spent the
next year going from government facitly to facility teaching security
after that snafu.
It isn't about how many times a year you have to login... why do you
think MS provided a password mechanism in the first place?
It helps to keep outsiders on the net from just waltzing in and snooping
thru your files.
 
G

Gene E. Bloch

snip


Someone correct me here if I'm wrong, but it's my understanding when you
wire a dryer that's dual voltage capable, with 120V you use maybe half the
heating elements. When you wire it with 240V you use all the heating
elements. So everything changes by an unknown except Voltage which we know
is 240V. We can now start to solve the problem by finding the resistance
(Ohms) with an Ohmmeter, which is now a known and solve for current (Amps)
or find current with an Ammeter and solve for resistance (Ohms). We can now
solve for power (Watts) now as well. We can cross-check with Ohm's Law to
verify. We can solve for Watts before Ohms, but we have to have two known's
before we can solve. We cannot assume any factor and come up with a factual
result.
To go back to your post, if you wire a dryer that's not dual voltage
capable, which I think is what you meant judging by the data in your post,
then you are correct, you will let the smoke out of the dryer. One final
note, while your assessment of Ohm's law is true, you are assuming all the
people in this discussion don't know the difference between devices that are
multi-voltage capable and that's not so. Judging by Char's reply to your
post he doesn't know the difference either, just like he didn't know the
difference between Watt's Law and Ohm's Law.
Dave
Do *you* know what Watt's law is?

http://encyclopedia2.thefreedictionary.com/Watt's+law

Recall that Watt died in 1819, long before the unit of power was named
after him (1882), and before the formula for power in an electric circuit
was worked out.
 
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