On 12/26/2016 01:19 PM, FromTheRafters wrote:

[snip]

Yet you still seem to think that Ohm's Law applies to open circuits.
What is the 'voltage drop' across the open contacts of such a switch?

It's the supply voltage, unless there's another open somewhere. Current
is 0 (so power is 0 too).

P = IE

0 = 0 * 120V

With an ON switch, E=0 and (for example 1A current)

0 = 1A * 0

0W for the switch, whether on or off. It's different if it could be part
way (as in a rheostat for a dimmer).

On 12/26/2016 02:23 PM, James Wilkinson Sword wrote:

[snip]

Unlikely for that to happen, and in any case, they're very low current
devices.


Its not low current if there is a short circuit. Then current could be
very high.

--
Mark Lloyd
http://notstupid.us/

"Nothing could be more anti-Biblical than letting women vote."
[Editorial, Harper's Magazine, November 1853]

James Wilkinson Sword presented the following explanation :
On Mon, 26 Dec 2016 21:19:05 -0000, Tekkie® wrote:

posted for all of us...



On Sat, 24 Dec 2016 22:09:00 -0000, "James Wilkinson Sword"
wrote:

They do want disconnects to be installed in a uniform way, just to
avoid confusion when someone may need to operate it in a hurry

You push it the way is isn't to turn it off, you don't have options. And
I doubt most people would remember which way in a panic.

--
If your developed muscle memory is "off is down" you don't have to
think in an emergency

+50 he is making assumptions.

What?

Muscle forgetfulness.

On 12/26/2016 05:24 PM, Uncle Monster wrote:

[snip]


I'm over 60 and I can still hear a battery operated analogue wall clock ticking across the room when it's quiet enough. When I was working on CRT TV's, monitors and terminals, I could hear the horizontal oscillators. Back when ultrasonic motion detectors were common in the security industry, I could hear the harmonics from those things. I've always liked peace and quiet so I protected my hearing. When I went to a nightclub to repair some equipment, I wore hearing protection. I had the earmuffs and some of the in the ear hearing protectors which had a little disk valve inside that closed when there was loud sound. Those earplugs were pretty slick because I could hear someone talking on a construction site but when a worker dropped a pipe or cranked up a hammer drill next to us, the valve closed and blocked the loud sounds. The new ones are even better. I've had to tell nurses to lower their voice because I'm not deaf like most of the older folks at the center. My friend Stinky is 53 and quite hard of hearing because as a young man, he listened to Def Leppard at full volume on his full ear covering head phones. ヽ(ヅ)ノ

[8~{} Uncle Earie Monster



One I had no choice but to be around very loud music. I found it helped
to open my mouth to equalize pressure.

--
Mark Lloyd
http://notstupid.us/

"Nothing could be more anti-Biblical than letting women vote."
[Editorial, Harper's Magazine, November 1853]

On 12/26/2016 05:55 PM, James Wilkinson Sword wrote:

[snip]

I can often hear ultrasonic mole scarers, but it might be they're faulty
as they don't all do it. Although whenever I've said "isn't that
annoying" to someone, they have no idea what I'm talking about. I
tested myself at some science fair when I was 12, and I had better range
than most. And a doctor about 5 years ago did a hearing test and said I
have better hearing than a kid (I'm 41).


My father used to spend a tot of time listening to a certain classical
music radio station. There was a constant annoying high-frequency hiss
on it. HE couldn't hear the hiss.

BTW, he had the radio on all night, and just a loud as day.

--
Mark Lloyd
http://notstupid.us/

"Nothing could be more anti-Biblical than letting women vote."
[Editorial, Harper's Magazine, November 1853]

Sam E brought next idea :
On 12/26/2016 01:19 PM, FromTheRafters wrote:

[snip]

Yet you still seem to think that Ohm's Law applies to open circuits.
What is the 'voltage drop' across the open contacts of such a switch?

It's the supply voltage, unless there's another open somewhere. Current is 0
(so power is 0 too).

P = IE

0 = 0 * 120V

With an ON switch, E=0 and (for example 1A current)

0 = 1A * 0

0W for the switch, whether on or off. It's different if it could be part way
(as in a rheostat for a dimmer).

True about the voltage being the supply voltage, but that is not
'voltage drop' because, as you mention, there is no power (energy)
being dissipated in an open circuit.

On Mon, 26 Dec 2016 22:20:37 -0000, "James Wilkinson Sword"
wrote:

On Mon, 26 Dec 2016 22:12:19 -0000, wrote:

On Mon, 26 Dec 2016 16:48:10 -0500, Ralph Mowery
wrote:

Depends on what you call low current.

As little as 10 ma can be painful, 20 ma sevear pain and maybe
difficulty breathing. At 50 ma you start to get to the danger zone.
Depending on the path of the current you may be able to take more or
less.

Even less than 10 ma is enough to make you hirt yourself.
Most people can feel around 1 to 2 ma.


We used to always say the 5ma GFCI will keep you from being
electrocuted but it doesn't keep you from falling off the ladder.

And serves no purpose at all if you touch live and neutral, which is just as likely unless you have stuff earthed everywhere. I tend not to earth things, as all it does is increase the chance of shock.
You already said you have a wood floor so there is not much ground
around you. The rules change pretty quickly when you are slab on
grade. Concrete is a pretty good conductor. If you are standing on a
tile floor and touch something hot, you get it the worst way, right
through your core.

On Mon, 26 Dec 2016 23:53:37 -0000, "James Wilkinson Sword"
wrote:

Why don't the PIRs only run at night?

They look at the IR from your body and the lens on the front chops it
as you walk across the field. If you crack one open you will see bars
in there that block the IR. The detector sees the signal change as
your hot spot moved past those bars. If you actually want to defeat
one, put a hot IR emitter in front of it and bring up the power slowly
(it only detects sharp changes) until you drive the detector into
saturation, then it is blind. If you have enough sunlight on them they
are blind too. It will also kill the detector. The ones I have outside
go bad pretty fast if the sun hits them. I try to keep them shaded if
they look east, south or west (in the arc of the sun)

On Monday, December 26, 2016 at 9:48:22 AM UTC-5, trader_4 wrote:
On Sunday, December 25, 2016 at 4:46:09 PM UTC-5, wrote:
On Sat, 24 Dec 2016 18:00:52 -0800 (PST), DerbyDad03
wrote:

Snip

If an EE knows how to wire a split switched receptacle, it is something (s)he
learned from a source unrelated to the degree they earned - unless perhaps it
was an elective.

I doubt you have an EE degree, but if you do, are you telling us that
where you got it, they had to instruct you on every single possible thing
you will ever see? Where I got my degree, they taught us to think like
an engineer, use electrical principles from Ohms Law to Maxwell's equations
and apply them to the real world. It takes nothing more than the most
basic understanding of electricity to be able to figure out how to wire
a split-receptacle, something I already knew before I entered high school,
let alone college. And no, no one had to specifically instruct me in how
it worked. I saw one, figured it out in maybe a minute.



It sure wasn't an elective at RIT back when I attended.

If you graduated from there, they should be embarrassed.

You can doubt all you want, but you would be wrong.

You totally missed the the point of my post and I'm not going to waste any
effort trying to explain it because you'll probably argue with that too.

BTW...how do you go from saying "DerbyDad, as usual, is 100% right" to
doubting my claim that I have a BSEE? Praise followed by condemnation?
Flip-flop much?

On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.
Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

On Mon, 26 Dec 2016 18:49:10 -0600, Mark Lloyd
wrote:

On 12/26/2016 11:54 AM, James Wilkinson Sword wrote:

[snip]

Called 3-way in the US. I think I understand why, but 2-way sounds more
reasonable.

It can be 3-way (or more) if you want, you add a switch in the middle
that swaps the poles. But if there's two switches, calling it 3-way is
illogical.

IIRC, what I heard is that its called 3-way because there's wiring going
to 3 places (each switch, and the light). It does sound strange. I'd
like to use 3-way for THREE switches.

[snip]

That is just an old electrician tradition. When you have the cross
connected DPDT switch, used as an intermediate switch in a pair of
travelers it is a "4 way" even though it functions exactly the same.
It just has 4 ungrounded conductors.

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

On Monday, December 26, 2016 at 2:19:09 PM UTC-5, FromTheRafters wrote:
After serious thinking trader_4 wrote :
On Monday, December 26, 2016 at 10:41:22 AM UTC-5, FromTheRafters wrote:
trader_4 wrote on 12/26/2016 :
On Sunday, December 25, 2016 at 4:46:09 PM UTC-5, wrote:
On Sat, 24 Dec 2016 18:00:52 -0800 (PST), DerbyDad03
wrote:

Snip

If an EE knows how to wire a split switched receptacle, it is something
(s)he learned from a source unrelated to the degree they earned - unless
perhaps it was an elective.

I doubt you have an EE degree, but if you do, are you telling us that
where you got it, they had to instruct you on every single possible thing
you will ever see? Where I got my degree, they taught us to think like
an engineer, use electrical principles from Ohms Law to Maxwell's equations
and apply them to the real world. It takes nothing more than the most
basic understanding of electricity to be able to figure out how to wire
a split-receptacle, something I already knew before I entered high school,
let alone college. And no, no one had to specifically instruct me in how
it worked. I saw one, figured it out in maybe a minute.



It sure wasn't an elective at RIT back when I attended.

If you graduated from there, they should be embarrassed.

I disagree. When I first learned about electricity and Ohm's Law, my
instructor used the 'plumbing analogy' and mentioned that it is not
exactly correct but 'good enough for most electrical work'. If you
wanted to go into physics or electronics it has many failures.

Where does Ohms Law fail with regard to a switched outlet? It doesn't.
Neither do Kirchoff's Laws, which are all that's needed to understand
a simple switched circuit.



For one thing, current (measured in amps) does *not* flow through a
conductor even though almost all electrical people will insist that it
does. Current is the *rate* at which *charge* flows and the rate
doesn't *go* anywhere. The number of charge *carriers* (usually, but
not always electrons) past a certain point (or cross sectional area)
can be counted and multiplied by the amount of charge each carries, to
get the current.


A silly nit which has nothing to do with anything. Again, what does that
have to do with wiring a switched outlet?



Understanding Ohm's Law and/or Maxwell's equations may be a
prerequisite for your chosen profession, but that doesn't mean everyone
who knows them must also know how to wire household switches.

If you know them and really understand them, then there has to be
something wrong with you or how you were taught that you can have
a degree in EE and not be able to apply Ohms Law and the most
elementary of circuit principles to solve the problem of how a
switched outlet works. Good grief, has all of America gone dumb?
Some of you seem to think that the purpose of college and an EE degree
is to teach you how to do specific tasks. It's not. It's to teach
science, physics and engineering that you can then apply to all the
new things you encounter. That is what engineering is, the application
of science to actual, real world problems. What you are suggesting is
that colleges are there to train essentially robots, that can only deal
with that which they've explicitly been shown.




There are
other endeavors within the electricity fields than construction work.
You would be completely bamboozled by ECM gear and someone who learned
about it from the ground up (basic E & E) might assume, like you do,
that the things they learned apply to all other fields, but they don't.

I never said electricity, Ohms Law, Maxwell;s equations, etc apply to
anything other than electricity. The switched outlet here is a very
basic circuit that anyone who has an EE should be able to figure out.
It's irrelevant that the circuit is in a house.
And if you can't, then you or your educational system have failed.

Yet you still seem to think that Ohm's Law applies to open circuits.
What is the 'voltage drop' across the open contacts of such a switch?

Diversion into rat hole detected. Diversion into rat hole rejected.

On Monday, December 26, 2016 at 4:44:38 PM UTC-5, Tommy Silva wrote:
On 12/26/2016 12:01 PM, trader_4 wrote:
On Monday, December 26, 2016 at 10:41:22 AM UTC-5, FromTheRafters wrote:

Understanding Ohm's Law and/or Maxwell's equations may be a
prerequisite for your chosen profession, but that doesn't mean everyone
who knows them must also know how to wire household switches.
If you know them and really understand them, then there has to be
something wrong with you or how you were taught that you can have
a degree in EE and not be able to apply Ohms Law and the most
elementary of circuit principles to solve the problem of how a
switched outlet works. Good grief, has all of America gone dumb?
Some of you seem to think that the purpose of college and an EE degree
is to teach you how to do specific tasks. It's not. It's to teach
science, physics and engineering that you can then apply to all the
new things you encounter. That is what engineering is, the application
of science to actual, real world problems. What you are suggesting is
that colleges are there to train essentially robots, that can only deal
with that which they've explicitly been shown.

My neighbor is a EE. EEs are a special kind of stupid.

Maybe your neighbor got his degree in one of those colleges that some
here have apparently been to. Where they only educate you on how to
solve certain things they give you, instead of instructing on basic
principles so you can solve almost anything.


For example, two years ago a county snow plow knocked over about a dozen neighborhood mailboxes.
I replaced mine by pulling my old 4x4 post out with a bumper jack and dropping a new $5 post into the existing hole. Complete job took less than 30 minutes.

OTOH, EE neighbor bought a splint-style post repair kit from a big box store. The repaired post lasted about 6 months until a summer storm blew it over.
Next repair attempt was another post repair kit but this time he added some guy wires fastened to tent stakes.
As you might imagine, the tent stakes pulled out of the ground and the whole mess fell over again.

EE currently has added 3 concrete blocks to the tent stakes to keep them from pulling out of the ground. It seems to be holding at the moment. ;-)

I'm sure everyone can come up with similar, regardless of any degrees.
It proves nothing.

On Monday, December 26, 2016 at 7:41:19 PM UTC-5, Mark Lloyd wrote:
On 12/26/2016 09:00 AM, trader_4 wrote:

All that is solved with a neutral there and it's no
big deal to run when you're installing a new circuit.


Considering having neutral in switch boxes, I have a few places where
I've installed receptacles next to switches. Normal wall receptacles can
be (and often are) behind things where they're hard to get to when you
need to plug something in temporarily. Those switch-height receptacles
are really used a lot.

--
Mark Lloyd
http://notstupid.us/

"Nothing could be more anti-Biblical than letting women vote."
[Editorial, Harper's Magazine, November 1853]

What any of that has to do with requiring a neutral at a switch, IDK.

On Monday, December 26, 2016 at 9:04:29 PM UTC-5, DerbyDad03 wrote:
On Monday, December 26, 2016 at 9:48:22 AM UTC-5, trader_4 wrote:
On Sunday, December 25, 2016 at 4:46:09 PM UTC-5, wrote:
On Sat, 24 Dec 2016 18:00:52 -0800 (PST), DerbyDad03
wrote:

Snip

If an EE knows how to wire a split switched receptacle, it is something (s)he
learned from a source unrelated to the degree they earned - unless perhaps it
was an elective.

I doubt you have an EE degree, but if you do, are you telling us that
where you got it, they had to instruct you on every single possible thing
you will ever see? Where I got my degree, they taught us to think like
an engineer, use electrical principles from Ohms Law to Maxwell's equations
and apply them to the real world. It takes nothing more than the most
basic understanding of electricity to be able to figure out how to wire
a split-receptacle, something I already knew before I entered high school,
let alone college. And no, no one had to specifically instruct me in how
it worked. I saw one, figured it out in maybe a minute.



It sure wasn't an elective at RIT back when I attended.

If you graduated from there, they should be embarrassed.

You can doubt all you want, but you would be wrong.

You totally missed the the point of my post and I'm not going to waste any
effort trying to explain it because you'll probably argue with that too.

BTW...how do you go from saying "DerbyDad, as usual, is 100% right" to
doubting my claim that I have a BSEE? Praise followed by condemnation?
Flip-flop much?

Because the way Clare presented it, I thought it was Clare speaking,
not you. But it doesn't change the core of the issue. If I had known
it was you, I would have been a bit kinder in my words.

But sorry. If you think an EE can't figure out how a switched outlet
works with what they learned to get their degree, then there is something
wrong with you or where you got you degree. RIT didn't teach you
the most basic circuit theory?

On Tuesday, December 27, 2016 at 9:16:13 AM UTC-5, trader_4 wrote:
On Monday, December 26, 2016 at 4:44:38 PM UTC-5, Tommy Silva wrote:
On 12/26/2016 12:01 PM, trader_4 wrote:
On Monday, December 26, 2016 at 10:41:22 AM UTC-5, FromTheRafters wrote:

Understanding Ohm's Law and/or Maxwell's equations may be a
prerequisite for your chosen profession, but that doesn't mean everyone
who knows them must also know how to wire household switches.
If you know them and really understand them, then there has to be
something wrong with you or how you were taught that you can have
a degree in EE and not be able to apply Ohms Law and the most
elementary of circuit principles to solve the problem of how a
switched outlet works. Good grief, has all of America gone dumb?
Some of you seem to think that the purpose of college and an EE degree
is to teach you how to do specific tasks. It's not. It's to teach
science, physics and engineering that you can then apply to all the
new things you encounter. That is what engineering is, the application
of science to actual, real world problems. What you are suggesting is
that colleges are there to train essentially robots, that can only deal
with that which they've explicitly been shown.

My neighbor is a EE. EEs are a special kind of stupid.

Maybe your neighbor got his degree in one of those colleges that some
here have apparently been to. Where they only educate you on how to
solve certain things they give you, instead of instructing on basic
principles so you can solve almost anything.

Another indication that you are really good at missing points.

On Monday, December 26, 2016 at 9:06:49 PM UTC-5, wrote:
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.
Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

+1

Not that it matters, this is an attempt at diversion by Rafters.
If you got a degree in EE and can't figure out a simple switch circuit,
claim that you need a course to cover that, then something is either
wrong with you or wrong with where you got an EE degree.

On Tuesday, December 27, 2016 at 9:21:41 AM UTC-5, DerbyDad03 wrote:
On Tuesday, December 27, 2016 at 9:16:13 AM UTC-5, trader_4 wrote:
On Monday, December 26, 2016 at 4:44:38 PM UTC-5, Tommy Silva wrote:
On 12/26/2016 12:01 PM, trader_4 wrote:
On Monday, December 26, 2016 at 10:41:22 AM UTC-5, FromTheRafters wrote:

Understanding Ohm's Law and/or Maxwell's equations may be a
prerequisite for your chosen profession, but that doesn't mean everyone
who knows them must also know how to wire household switches.
If you know them and really understand them, then there has to be
something wrong with you or how you were taught that you can have
a degree in EE and not be able to apply Ohms Law and the most
elementary of circuit principles to solve the problem of how a
switched outlet works. Good grief, has all of America gone dumb?
Some of you seem to think that the purpose of college and an EE degree
is to teach you how to do specific tasks. It's not. It's to teach
science, physics and engineering that you can then apply to all the
new things you encounter. That is what engineering is, the application
of science to actual, real world problems. What you are suggesting is
that colleges are there to train essentially robots, that can only deal
with that which they've explicitly been shown.

My neighbor is a EE. EEs are a special kind of stupid.

Maybe your neighbor got his degree in one of those colleges that some
here have apparently been to. Where they only educate you on how to
solve certain things they give you, instead of instructing on basic
principles so you can solve almost anything.

Another indication that you are really good at missing points.

If you got a degree in EE and claim that you need more instruction,
a special course in how to understand how a switched outlet works,
then you should ask for your money back. Sorry, but it is that simple.

On Tue, 27 Dec 2016 00:12:04 -0000, Uncle Monster wrote:

On Monday, December 26, 2016 at 5:53:53 PM UTC-6, James Wilkinson Sword wrote:
On Mon, 26 Dec 2016 23:44:34 -0000, Uncle Monster wrote:

On Monday, December 26, 2016 at 12:01:49 PM UTC-6, James Wilkinson Sword wrote:
On Mon, 26 Dec 2016 14:22:02 -0000, FromTheRafters wrote:

Mark Lloyd explained on 12/25/2016 :
On 12/25/2016 03:26 PM, James Wilkinson Sword wrote:

[snip]

Hearing a 50Hz mains hum isn't that bad, but imagine hearing a higher
frequency.

Once I had a TV where the horizontal frequency (almost 16KHz) was audible. It
was unpleasant to be around for very long. IIRC, I sold it to someone with
hearing loss who couldn't hear it.

I could hear most of them.

What year?

I could also hear the detectors in stores
back when they were sonic. Not anymore though, and not because the
technology has changed, but because my hearing has changed.

Detectors?
--
Back in the 1970's, ultrasonic motion detectors for security systems were quite common in businesses. The last of the old Polaroid instant cameras used an ultrasonic transducer to measure distance to the subject of a picture and focus the camera. There were some ultrasonic motion detectors at one time that detected a shopper in the isle at a retailer so an electronic advertising display could start playing music and the voice of a pitchman. I haven't seen one for a while but I could hear the harmonics and clicks. ヽ(€¢€¿€¢)ノ

What I think's strange is motion detectors stay on when the alarm is deactivated. Why don't the PIRs only run at night?
--


The PIR motion detectors have and indicator light for a "walk test" to indicate that the unit is operating correctly and covering the area it's meant to. There is a jumper on the circuit board inside the PIR that can be moved to disable the walk test light but most security system installers just leave the jumper in place. ヽ(ヅ)ノ

[8~{} Uncle Infrared Monster

I know, but surely the main control unit should switch all the detectors off unless you either have the alarm activated or you select test mode. What a waste of electricity, wearing out the PIRs, and annoying people with lights and clicks.

--
A great way to lose weight is to eat naked in front of a mirror. Restaurants will almost always throw you out before you can eat too much.

On Tue, 27 Dec 2016 00:45:30 -0000, Mark Lloyd wrote:

On 12/26/2016 11:52 AM, James Wilkinson Sword wrote:

[snip]


I have all the rooms bright, and the bedroom dimmer so it's not dazzling
when I wake up.


If you're watching TV, a ceiling light is likely to be in the wrong
place. That's one reason for using a lamp in the bedroom. The other is
to have a switch I can reach from bed (I need it DARK, and "night light"
is essentially a dirty word).

I don't see the problem with a ceiling light and a TV.

--
"I have left orders to be awakened at any time in case of national emergency, even if I'm in a cabinet meeting." - Ronald Reagan

On Tue, 27 Dec 2016 00:53:29 -0000, Mark Lloyd wrote:

On 12/26/2016 12:01 PM, James Wilkinson Sword wrote:

[snip]

Once I had a TV where the horizontal frequency (almost 16KHz) was
audible. It
was unpleasant to be around for very long. IIRC, I sold it to someone
with
hearing loss who couldn't hear it.

I could hear most of them.

What year?


IIRC, around 1995, 20-inch RCA.

Your hearing must be better than mine then. I might have heard it once or twice, but only on ones that were otherwise playing up.

--
Sit on my lap, and we'll talk about the first thing that comes up.

On Tue, 27 Dec 2016 01:07:56 -0000, Mark Lloyd wrote:

On 12/26/2016 02:23 PM, James Wilkinson Sword wrote:

[snip]

Unlikely for that to happen, and in any case, they're very low current
devices.


Its not low current if there is a short circuit. Then current could be
very high.

Now you're asking for two failures at once.

--
The record of having had intercourse the most frequently goes to a boy who was recorded to have had intercourse about 52,000 times over a period of 30 years. This means he had intercourse on average 33.3 times a week.

On Tue, 27 Dec 2016 09:07:14 -0500, FromTheRafters
wrote:

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

If you look at a load being 0 ohms to infinity in the theoretical
world of mathematics the analogy works just fine and so does the math.
In reality on the micro scale there probably is some current flow
unless you are doing this in a vacuum and there is no capacitive
coupling. You do not change the rules just because the switch is open
or the wire is broken. That is particularly true if you are a real
world trouble shooter.

On 12/26/2016 12:01 PM, trader_4 wrote:

My neighbor is a EE. EEs are a special kind of stupid.

For example, two years ago a county snow plow knocked over about a dozen neighborhood mailboxes.
I replaced mine by pulling my old 4x4 post out with a bumper jack and dropping a new $5 post into the existing hole. Complete job took less than 30 minutes.

OTOH, EE neighbor bought a splint-style post repair kit from a big box store. The repaired post lasted about 6 months until a summer storm blew it over.
Next repair attempt was another post repair kit but this time he added some guy wires fastened to tent stakes.
As you might imagine, the tent stakes pulled out of the ground and the whole mess fell over again.

EE currently has added 3 concrete blocks to the tent stakes to keep them from pulling out of the ground. It seems to be holding at the moment. ;-)


Question: What do electrical engineers use for birth control?
Answer: Their personality.

explained on 12/27/2016 :
On Tue, 27 Dec 2016 09:07:14 -0500, FromTheRafters
wrote:

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

If you look at a load being 0 ohms to infinity in the theoretical
world of mathematics the analogy works just fine and so does the math.
In reality on the micro scale there probably is some current flow
unless you are doing this in a vacuum and there is no capacitive
coupling. You do not change the rules just because the switch is open
or the wire is broken. That is particularly true if you are a real
world trouble shooter.

The first time this discussion came about, it was about a statement
about Ohm's law that I said IMO was wrong. Sure, I *could* have used
the argument that reality and theory were not the same thing and the
statement about zero amps was already wrong on the face of it. This
*is* about circuit theory, and *not* about reality.

Okay, consider a voltage source charging a capacitor. Consider
everything to be ideal except for the conductors. Only the conductors
can dissipate power, and only in the form of heat. After first applying
the voltage across the cap, a lot of current flows through the
conductors (we can, by convention, say that current flows through a
conductor even though it is not strictly true) and the resistance of
the conductors causes power to be dissipated as heat. The load device
(the cap) has less than the source voltage across it at this time
because the total of the source voltage has been lessened by the amount
of voltage drop across the conductors.

This lost power can be deduced from the total 'voltage drop' across the
conductors' resistance or as the difference between the source voltage
and the voltage at the device being driven (the cap in this case). This
is the voltage drop due to the resistance of the conductors and the
current through them. When the current is zero, that is the cap is
fully charged (in theory it can be, but as you say in practice it can
only be very very close) the loss through dissipation is also zero (or
very very close) -- *not* suddenly as high as the supply voltage as it
seems you are suggesting.

At this time (theoretically) there is no longer any current anywhere in
the circuit, and no 'voltage drop' anywhere, only the supplied voltage
from the source across the cap. No dissipation, no voltage drop. It is
equivalent to two ideal voltage sources connected to each other, since
there is no current, the conductors might as well be ideal.

Basically, this is the same scenario as before, except I have
substituted a charging cap for an opening switch, the difference being
that now we can see the trend of current and voltage drop as they both
approach zero rather than just a sudden change.

On 12/27/2016 08:18 AM, trader_4 wrote:

[snip]

What any of that has to do with requiring a neutral at a switch, IDK.


Neutral was needed for a receptacle. Maybe you didn't notice the
receptacle was installed into what was formerly a SWITCH BOX. That is, I
got a benefit from neutral being present at a switch.

--
Mark Lloyd
http://notstupid.us/

"God not only plays dice. He sometimes throws the dice where they cannot
be seen." [Stephen Hawking]

On 12/27/2016 08:27 AM, James Wilkinson Sword wrote:

[snip]

I don't see the problem with a ceiling light and a TV.


Consider that people's eyes adjust to the amount of light. Its hard to
see something when a MUCH BRIGHTER thing is nearby (like a ceiling light
in almost the same direction as the TV screen). The same thing can
happen when reading (a bright light BEHIND the book makes the book darker).

--
Mark Lloyd
http://notstupid.us/

"God not only plays dice. He sometimes throws the dice where they cannot
be seen." [Stephen Hawking]

On 12/27/2016 08:29 AM, James Wilkinson Sword wrote:

[snip]

Those will power the 20W CFLs, but have difficulty with less power than
that.


That sounds like something to try. Its in a room where it wouldn't hurt
to have more light.

--
Mark Lloyd
http://notstupid.us/

"God not only plays dice. He sometimes throws the dice where they cannot
be seen." [Stephen Hawking]

On 12/27/2016 08:30 AM, James Wilkinson Sword wrote:

[snip]


Its not low current if there is a short circuit. Then current could be
very high.

Now you're asking for two failures at once.



When you get near a switch, there's a big red flashing light and a loud
warning "DEFECTIVE GROUND! -- NO NOT TOUCH SCREWS!!!!". :-)

--
Mark Lloyd
http://notstupid.us/

"God not only plays dice. He sometimes throws the dice where they cannot
be seen." [Stephen Hawking]

On Tue, 27 Dec 2016 01:52:40 -0000, wrote:

On Mon, 26 Dec 2016 22:20:37 -0000, "James Wilkinson Sword"
wrote:

On Mon, 26 Dec 2016 22:12:19 -0000, wrote:

On Mon, 26 Dec 2016 16:48:10 -0500, Ralph Mowery
wrote:

Depends on what you call low current.

As little as 10 ma can be painful, 20 ma sevear pain and maybe
difficulty breathing. At 50 ma you start to get to the danger zone.
Depending on the path of the current you may be able to take more or
less.

Even less than 10 ma is enough to make you hirt yourself.
Most people can feel around 1 to 2 ma.


We used to always say the 5ma GFCI will keep you from being
electrocuted but it doesn't keep you from falling off the ladder.

And serves no purpose at all if you touch live and neutral, which is just as likely unless you have stuff earthed everywhere. I tend not to earth things, as all it does is increase the chance of shock.
You already said you have a wood floor so there is not much ground
around you. The rules change pretty quickly when you are slab on
grade. Concrete is a pretty good conductor. If you are standing on a
tile floor and touch something hot, you get it the worst way, right
through your core.

Rules are for the obedience of fools.

--
Using Opera's mail client: http://www.opera.com/mail/

On Tue, 27 Dec 2016 01:52:40 -0000, wrote:

On Mon, 26 Dec 2016 22:20:37 -0000, "James Wilkinson Sword"
wrote:

On Mon, 26 Dec 2016 22:12:19 -0000, wrote:

On Mon, 26 Dec 2016 16:48:10 -0500, Ralph Mowery
wrote:

Depends on what you call low current.

As little as 10 ma can be painful, 20 ma sevear pain and maybe
difficulty breathing. At 50 ma you start to get to the danger zone.
Depending on the path of the current you may be able to take more or
less.

Even less than 10 ma is enough to make you hirt yourself.
Most people can feel around 1 to 2 ma.


We used to always say the 5ma GFCI will keep you from being
electrocuted but it doesn't keep you from falling off the ladder.

And serves no purpose at all if you touch live and neutral, which is just as likely unless you have stuff earthed everywhere. I tend not to earth things, as all it does is increase the chance of shock.
You already said you have a wood floor so there is not much ground
around you. The rules change pretty quickly when you are slab on
grade. Concrete is a pretty good conductor. If you are standing on a
tile floor and touch something hot, you get it the worst way, right
through your core.

Rules are for the obedience of fools.

--
Using Opera's mail client: http://www.opera.com/mail/

On Tue, 27 Dec 2016 20:18:18 -0000, Mark Lloyd wrote:

On 12/27/2016 08:30 AM, James Wilkinson Sword wrote:

[snip]


Its not low current if there is a short circuit. Then current could be
very high.

Now you're asking for two failures at once.



When you get near a switch, there's a big red flashing light and a loud
warning "DEFECTIVE GROUND! -- NO NOT TOUCH SCREWS!!!!". :-)

I don't bother earthing those.


--
Our records indicate that you were once felt up by Jimmy Savile and could be entitled to £2147 in compensation.
Just reply "Hows about that then" to register, or to opt out just reply "Stop Jimmy Stop".
Register before the end of the month and get a free "Lawyers 4 U".

On Tue, 27 Dec 2016 20:43:29 -0000, "James Wilkinson Sword"
wrote:

Rules are for the obedience of fools.

428 disobeying fools were killed or seriously injured in UK last year
in electrical incidents.

On Tuesday, December 27, 2016 at 2:37:17 PM UTC-5, FromTheRafters wrote:
explained on 12/27/2016 :
On Tue, 27 Dec 2016 09:07:14 -0500, FromTheRafters
wrote:

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

If you look at a load being 0 ohms to infinity in the theoretical
world of mathematics the analogy works just fine and so does the math.
In reality on the micro scale there probably is some current flow
unless you are doing this in a vacuum and there is no capacitive
coupling. You do not change the rules just because the switch is open
or the wire is broken. That is particularly true if you are a real
world trouble shooter.

The first time this discussion came about, it was about a statement
about Ohm's law that I said IMO was wrong. Sure, I *could* have used
the argument that reality and theory were not the same thing and the
statement about zero amps was already wrong on the face of it. This
*is* about circuit theory, and *not* about reality.

Okay, consider a voltage source charging a capacitor. Consider
everything to be ideal except for the conductors. Only the conductors
can dissipate power, and only in the form of heat. After first applying
the voltage across the cap, a lot of current flows through the
conductors (we can, by convention, say that current flows through a
conductor even though it is not strictly true) and the resistance of
the conductors causes power to be dissipated as heat. The load device
(the cap) has less than the source voltage across it at this time
because the total of the source voltage has been lessened by the amount
of voltage drop across the conductors.

This lost power can be deduced from the total 'voltage drop' across the
conductors' resistance or as the difference between the source voltage
and the voltage at the device being driven (the cap in this case). This
is the voltage drop due to the resistance of the conductors and the
current through them. When the current is zero, that is the cap is
fully charged (in theory it can be, but as you say in practice it can
only be very very close) the loss through dissipation is also zero (or
very very close) -- *not* suddenly as high as the supply voltage as it
seems you are suggesting.

At this time (theoretically) there is no longer any current anywhere in
the circuit, and no 'voltage drop' anywhere, only the supplied voltage
from the source across the cap. No dissipation, no voltage drop. It is
equivalent to two ideal voltage sources connected to each other, since
there is no current, the conductors might as well be ideal.

Basically, this is the same scenario as before, except I have
substituted a charging cap for an opening switch, the difference being
that now we can see the trend of current and voltage drop as they both
approach zero rather than just a sudden change.

And Ohms Law still applies and works. V = IR. I=0, gives V =0,
the voltage drop across the conductors is zero. Nor was there any
division by zero, so don't start in with that again, please.

On Tue, 27 Dec 2016 14:07:14 -0000, FromTheRafters wrote:

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

Bull****. An open switch drops all the voltage and no current flows.

--
They say Confucius does his crosswords with a pen.

On Tuesday, December 27, 2016 at 9:07:20 AM UTC-5, FromTheRafters wrote:
was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

Baloney. Hook a capacitor in series with a resistor and battery.
When connected, the voltage drop across the cap will start to
build up from zero and after a sufficient time, the voltage drop
across the cap will equal the voltage of the battery.
Kirchoff's Voltage Law works in circuits with no current flowing.

On Tue, 27 Dec 2016 14:37:11 -0500, FromTheRafters
wrote:

explained on 12/27/2016 :
On Tue, 27 Dec 2016 09:07:14 -0500, FromTheRafters
wrote:

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

I have noticed that "Gordo" has dropped out of the on going responses,
when it became technical. As Gordoseemingly graduated from Health Kit
U - he did not get educated in the nuances of Ohms's law . He relies
on a ~$4 Lowes circuit tester -his mommy gave him



You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

If you look at a load being 0 ohms to infinity in the theoretical
world of mathematics the analogy works just fine and so does the math.
In reality on the micro scale there probably is some current flow
unless you are doing this in a vacuum and there is no capacitive
coupling. You do not change the rules just because the switch is open
or the wire is broken. That is particularly true if you are a real
world trouble shooter.

The first time this discussion came about, it was about a statement
about Ohm's law that I said IMO was wrong. Sure, I *could* have used
the argument that reality and theory were not the same thing and the
statement about zero amps was already wrong on the face of it. This
*is* about circuit theory, and *not* about reality.

Okay, consider a voltage source charging a capacitor. Consider
everything to be ideal except for the conductors. Only the conductors
can dissipate power, and only in the form of heat. After first applying
the voltage across the cap, a lot of current flows through the
conductors (we can, by convention, say that current flows through a
conductor even though it is not strictly true) and the resistance of
the conductors causes power to be dissipated as heat. The load device
(the cap) has less than the source voltage across it at this time
because the total of the source voltage has been lessened by the amount
of voltage drop across the conductors.

This lost power can be deduced from the total 'voltage drop' across the
conductors' resistance or as the difference between the source voltage
and the voltage at the device being driven (the cap in this case). This
is the voltage drop due to the resistance of the conductors and the
current through them. When the current is zero, that is the cap is
fully charged (in theory it can be, but as you say in practice it can
only be very very close) the loss through dissipation is also zero (or
very very close) -- *not* suddenly as high as the supply voltage as it
seems you are suggesting.

At this time (theoretically) there is no longer any current anywhere in
the circuit, and no 'voltage drop' anywhere, only the supplied voltage
from the source across the cap. No dissipation, no voltage drop. It is
equivalent to two ideal voltage sources connected to each other, since
there is no current, the conductors might as well be ideal.

Basically, this is the same scenario as before, except I have
substituted a charging cap for an opening switch, the difference being
that now we can see the trend of current and voltage drop as they both
approach zero rather than just a sudden change.

trader_4 explained :
On Tuesday, December 27, 2016 at 2:37:17 PM UTC-5, FromTheRafters wrote:
explained on 12/27/2016 :
On Tue, 27 Dec 2016 09:07:14 -0500, FromTheRafters
wrote:

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

If you look at a load being 0 ohms to infinity in the theoretical
world of mathematics the analogy works just fine and so does the math.
In reality on the micro scale there probably is some current flow
unless you are doing this in a vacuum and there is no capacitive
coupling. You do not change the rules just because the switch is open
or the wire is broken. That is particularly true if you are a real
world trouble shooter.

The first time this discussion came about, it was about a statement
about Ohm's law that I said IMO was wrong. Sure, I *could* have used
the argument that reality and theory were not the same thing and the
statement about zero amps was already wrong on the face of it. This
*is* about circuit theory, and *not* about reality.

Okay, consider a voltage source charging a capacitor. Consider
everything to be ideal except for the conductors. Only the conductors
can dissipate power, and only in the form of heat. After first applying
the voltage across the cap, a lot of current flows through the
conductors (we can, by convention, say that current flows through a
conductor even though it is not strictly true) and the resistance of
the conductors causes power to be dissipated as heat. The load device
(the cap) has less than the source voltage across it at this time
because the total of the source voltage has been lessened by the amount
of voltage drop across the conductors.

This lost power can be deduced from the total 'voltage drop' across the
conductors' resistance or as the difference between the source voltage
and the voltage at the device being driven (the cap in this case). This
is the voltage drop due to the resistance of the conductors and the
current through them. When the current is zero, that is the cap is
fully charged (in theory it can be, but as you say in practice it can
only be very very close) the loss through dissipation is also zero (or
very very close) -- *not* suddenly as high as the supply voltage as it
seems you are suggesting.

At this time (theoretically) there is no longer any current anywhere in
the circuit, and no 'voltage drop' anywhere, only the supplied voltage
from the source across the cap. No dissipation, no voltage drop. It is
equivalent to two ideal voltage sources connected to each other, since
there is no current, the conductors might as well be ideal.

Basically, this is the same scenario as before, except I have
substituted a charging cap for an opening switch, the difference being
that now we can see the trend of current and voltage drop as they both
approach zero rather than just a sudden change.

And Ohms Law still applies and works. V = IR. I=0, gives V =0,
the voltage drop across the conductors is zero. Nor was there any
division by zero, so don't start in with that again, please.

I wouldn't dream of it, but given I=0 and nothing else, you can't nail
down V nor can you nail down R by using Ohm's Law. Ohm's Law does *not*
tell you that if the current is zero, the voltage must also be zero as
that Al Gebra poster suggested.

Ohm's Law is not only about the V=IR equation, it is about a formula
describing the relationships between these values in a circuit and
requires the other equation forms too. They are derived with the
stipulation that I and R cannot be . . . well, you told me not to go
there, so I won't.

On Tue, 27 Dec 2016 20:43:29 -0000, "James Wilkinson Sword"
wrote:

On Tue, 27 Dec 2016 01:52:40 -0000, wrote:

On Mon, 26 Dec 2016 22:20:37 -0000, "James Wilkinson Sword"
wrote:

On Mon, 26 Dec 2016 22:12:19 -0000, wrote:

On Mon, 26 Dec 2016 16:48:10 -0500, Ralph Mowery
wrote:

Depends on what you call low current.

As little as 10 ma can be painful, 20 ma sevear pain and maybe
difficulty breathing. At 50 ma you start to get to the danger zone.
Depending on the path of the current you may be able to take more or
less.

Even less than 10 ma is enough to make you hirt yourself.
Most people can feel around 1 to 2 ma.


We used to always say the 5ma GFCI will keep you from being
electrocuted but it doesn't keep you from falling off the ladder.

And serves no purpose at all if you touch live and neutral, which is just as likely unless you have stuff earthed everywhere. I tend not to earth things, as all it does is increase the chance of shock.
You already said you have a wood floor so there is not much ground
around you. The rules change pretty quickly when you are slab on
grade. Concrete is a pretty good conductor. If you are standing on a
tile floor and touch something hot, you get it the worst way, right
through your core.

Rules are for the obedience of fools.
No, rules are for the PROTECTION of fools.