On Wed, 28 Dec 2016 05:11:38 -0000, wrote:

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:

snip
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.

No, we don't give a **** if a fool kills himself., the rest of us can make our own decisions.

So, are you a fool or can you make your own decisions?

--
Why is a person who plays the piano called a pianist but a person who drives a racing car not called a racist?

"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core and
it is use up too 400 Hrz. Presently we have ferret iron cores transformers
that will work into very high Frequency depend on density of iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit unless
you are running mile and miles of line. In normal use on open line if there
or no load/current there is no Voltage drop, you must have current present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can be
an open circuit there are current and voltage drop present when transmitter
is on!
My question is why would I go into electrical theory to some one that wants
to know how or can he change duplex receptacle from single source to double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them..

You do know that North American wiring standards, voltages and frequency of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core and
it is use up too 400 Hrz. Presently we have ferret iron cores transformers
that will work into very high Frequency depend on density of iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit unless
you are running mile and miles of line. In normal use on open line if there
or no load/current there is no Voltage drop, you must have current present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can be
an open circuit there are current and voltage drop present when transmitter
is on!
My question is why would I go into electrical theory to some one that wants
to know how or can he change duplex receptacle from single source to double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

--
If you eat a judge's uniform you might contract a lawsuit.

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core and
it is use up too 400 Hrz. Presently we have ferret iron cores transformers
that will work into very high Frequency depend on density of iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit unless
you are running mile and miles of line. In normal use on open line if there
or no load/current there is no Voltage drop, you must have current present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when transmitter
is on!
My question is why would I go into electrical theory to some one that wants
to know how or can he change duplex receptacle from single source to double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core and
it is use up too 400 Hrz. Presently we have ferret iron cores transformers
that will work into very high Frequency depend on density of iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit unless
you are running mile and miles of line. In normal use on open line if there
or no load/current there is no Voltage drop, you must have current present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when transmitter
is on!
My question is why would I go into electrical theory to some one that wants
to know how or can he change duplex receptacle from single source to double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument. As I see it, you have say 240 volts at the supply, and that voltage is dropped as you go round the circuit. It could be 10 volts across some wires, and 230 across a heating element. When it's switched off, all the drop is across the switch..

--
A group of white South Africans recently killed a black lawyer because he was black. That was wrong. They should have killed him because he was a lawyer.

On Tuesday, December 27, 2016 at 9:15:53 PM UTC-5, FromTheRafters wrote:
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.


Everyone that passed high school algebra can nail it down:

V = IR When I=0, V = 0

I suppose on a first week algebra exam you're answer would be
to argue, bitch, moan and insist that V can't be zero, that it's
unsolvable. And once again, no division by zero was involved.

The case where there is no potential difference and hence no current
flow, is one of the defining limits of Ohm's Law and it still works.
As I told you a hundred times, do a graph of V = IR and it goes right
through the origin. There is no undefined singularity, like you claim.

I suppose on a first week algebra exam you're answer would be
to argue, bitch, moan and insist that it can't be zero.




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.

No it doesn't require the other forms too. You only need to pretend
it does, because you want to force division by zero.


They are derived with the
stipulation that I and R cannot be . . . well, you told me not to go
there, so I won't.

Better not, because the rest of us went there decades ago and remember
what we learned. You seem to like to try to redefine everything from
basic circuit theory to the definition of current, pick some silly nit
as if you alone have found something new. Your most recent
silliness is to claim that referring to current "flowing" is incorrect.
We all know that current is composed of electrons and when we say
current is flowing in a conductor, we mean it's electrons that are
flowing. This is like bitching about someone referring to a tidal flow
as that and not saying instead "water molecules". Boy, I feel sorry
for the teachers who tried to educate you.

On Wednesday, December 28, 2016 at 4:33:37 PM UTC-5, James Wilkinson Sword wrote:
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core and
it is use up too 400 Hrz. Presently we have ferret iron cores transformers
that will work into very high Frequency depend on density of iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit unless
you are running mile and miles of line. In normal use on open line if there
or no load/current there is no Voltage drop, you must have current present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when transmitter
is on!
My question is why would I go into electrical theory to some one that wants
to know how or can he change duplex receptacle from single source to double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument. As I see it, you have say 240 volts at the supply, and that voltage is dropped as you go round the circuit. It could be 10 volts across some wires, and 230 across a heating element. When it's switched off, all the drop is across the switch.

--
A group of white South Africans recently killed a black lawyer because he was black. That was wrong. They should have killed him because he was a lawyer.

Don't forget that Rafters came up with this gem:

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.

Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

In article ,
says...


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*


There is always some current flowing. It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

All of this is picking the nits off the nits.

So ther is never a 0 in the equation.
You just have to look very hard to find the components.

On Wednesday, December 28, 2016 at 4:47:10 PM UTC-6, Ralph Mowery wrote:
In article ,
says...


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*


There is always some current flowing. It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

All of this is picking the nits off the nits.

So ther is never a 0 in the equation.
You just have to look very hard to find the components.

I don't think the contrarians will be happy until you get down to the quantum level. (・_・ヾ

[8~{} Uncle Atomic Monster

In article ,
says...



Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

The capacitor will never fully charge. There is a time constant where
the capacitor charges to about 63.2 % . Usually 5 time constants are
concidered fully charged, but it never reaches a true full charge.

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

On Wed, 28 Dec 2016 22:54:43 -0000, Ralph Mowery wrote:

In article ,
says...



Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

The capacitor will never fully charge. There is a time constant where
the capacitor charges to about 63.2 % . Usually 5 time constants are
concidered fully charged, but it never reaches a true full charge.

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

What if someone shorts the resistor for a bit?

--
Some people are like slinkies, not really good for anything, but they bring a smile to your face when pushed down the stairs.

James Wilkinson Sword submitted this idea :
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency
of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I
suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that
in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable
putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core
and
it is use up too 400 Hrz. Presently we have ferret iron cores
transformers
that will work into very high Frequency depend on density of
iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless
you are running mile and miles of line. In normal use on open line if
there
or no load/current there is no Voltage drop, you must have current
present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when
transmitter
is on!
My question is why would I go into electrical theory to some one that
wants
to know how or can he change duplex receptacle from single source to
double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument.

Not philosophical so much as unimportant to the jobs done by electrical
engineers. People can use computers just fine without knowing how they
work, and the same goes for electrical circuit theory and electrical
work and those who use calculus and the mathematics behind it.

As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the circuit.
It could be 10 volts across some wires, and 230 across a heating element.
When it's switched off, all the drop is across the switch.

I'm not suggesting that there is no difference of potential across the
switch, I am saying that the difference of potential across the switch
is not there because of dissipation. Heat is not the only way that
energy can be dissipated, but it *is* the usual one.

Emphasis asterisks are added by me.

"Voltage drop describes how energy is supplied of a voltage source that
is reduced *as electric current moves through the passive elements
(elements that do not supply voltage) of an electrical circuit*.
Voltage drops across internal resistances of the source, across
conductors, across contacts, and across connectors are undesired as the
supplied energy is lost (dissipated). Voltage drops across loads and
across other active circuit elements are desired as the supplied energy
performs useful work."

From: https://en.wikipedia.org/wiki/Voltage_drop

"Dissipation

Energy
"The conversion of mechanical energy into heat is called energy
dissipation." €“ François Roddier[1] *The term is also applied to the
loss of energy due to generation of unwanted heat in electric and
electronic circuits*."

From: https://en.wikipedia.org/wiki/Dissipation

and

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces *through which electric current
can flow*."

From: https://en.wikipedia.org/wiki/Electronic_circuit

Electric current can't flow when the switch is open, and it is no
longer a *circuit* by the definition used in circuit theory. Even when
the switch replaced by a capacitor and the capacitor is charging, no
current flows through the capacitor. There is no dissipation in the
capacitor due to the no current condition, and hence no voltage drop
across the capacitor.

It is easy to misunderstand concepts when one throws out word meanings
as 'merely semantics'. Not that you have done so to the best of my
recollection, but others sure have. If one doesn't 'pick nits' they can
often lose subtle underpinnings of complex concepts.

In article , says...



The capacitor will never fully charge. There is a time constant where
the capacitor charges to about 63.2 % . Usually 5 time constants are
concidered fully charged, but it never reaches a true full charge.

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

What if someone shorts the resistor for a bit?

The wire doing the shorting has some very low resistance, not counting
the wires from the source fo the voltage.
Just as there is no perfect insulator, there is no perfect conductor.

On Wed, 28 Dec 2016 23:17:41 -0000, FromTheRafters wrote:

James Wilkinson Sword submitted this idea :
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
snip

snip

You do know that North American wiring standards, voltages and frequency
of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I
suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that
in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable
putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core
and
it is use up too 400 Hrz. Presently we have ferret iron cores
transformers
that will work into very high Frequency depend on density of
iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless
you are running mile and miles of line. In normal use on open line if
there
or no load/current there is no Voltage drop, you must have current
present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when
transmitter
is on!
My question is why would I go into electrical theory to some one that
wants
to know how or can he change duplex receptacle from single source to
double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument.

Not philosophical so much as unimportant to the jobs done by electrical
engineers. People can use computers just fine without knowing how they
work, and the same goes for electrical circuit theory and electrical
work and those who use calculus and the mathematics behind it.

As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the circuit.
It could be 10 volts across some wires, and 230 across a heating element.
When it's switched off, all the drop is across the switch.

I'm not suggesting that there is no difference of potential across the
switch, I am saying that the difference of potential across the switch
is not there because of dissipation. Heat is not the only way that
energy can be dissipated, but it *is* the usual one.

Emphasis asterisks are added by me.

"Voltage drop describes how energy is supplied of a voltage source that
is reduced *as electric current moves through the passive elements
(elements that do not supply voltage) of an electrical circuit*.
Voltage drops across internal resistances of the source, across
conductors, across contacts, and across connectors are undesired as the
supplied energy is lost (dissipated). Voltage drops across loads and
across other active circuit elements are desired as the supplied energy
performs useful work."

From: https://en.wikipedia.org/wiki/Voltage_drop

"Dissipation

Energy
"The conversion of mechanical energy into heat is called energy
dissipation." €“ François Roddier[1] *The term is also applied to the
loss of energy due to generation of unwanted heat in electric and
electronic circuits*."

From: https://en.wikipedia.org/wiki/Dissipation

and

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces *through which electric current
can flow*."

From: https://en.wikipedia.org/wiki/Electronic_circuit

Electric current can't flow when the switch is open, and it is no
longer a *circuit* by the definition used in circuit theory. Even when
the switch replaced by a capacitor and the capacitor is charging, no
current flows through the capacitor. There is no dissipation in the
capacitor due to the no current condition, and hence no voltage drop
across the capacitor.

It is easy to misunderstand concepts when one throws out word meanings
as 'merely semantics'. Not that you have done so to the best of my
recollection, but others sure have. If one doesn't 'pick nits' they can
often lose subtle underpinnings of complex concepts.

Take a very thin wire used to supply your load. Big voltage drop, right? As the wire gets thinner and thinner, the voltage drop increases, right? As the wire become infinitely thin, the voltage drop increases to the load voltage, an open circuit.

--
"I'm prescribing these pills for you," said the doctor to the overweight patient, who tipped the scales at about three hundred pounds.
"I don't want you to swallow them. Just spill them on the floor twice a day and pick them up, one at a time."

On Wed, 28 Dec 2016 23:21:52 -0000, Ralph Mowery wrote:

In article , says...



The capacitor will never fully charge. There is a time constant where
the capacitor charges to about 63.2 % . Usually 5 time constants are
concidered fully charged, but it never reaches a true full charge.

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

What if someone shorts the resistor for a bit?

The wire doing the shorting has some very low resistance, not counting
the wires from the source fo the voltage.
Just as there is no perfect insulator, there is no perfect conductor.

Then pre charge the capacitor with a bigger load. It is possible to make the capacitor and the battery have precisely the same voltage.

--
Wedding rings: the world's smallest handcuffs.

In article , says...

On Wed, 28 Dec 2016 23:21:52 -0000, Ralph Mowery wrote:

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

What if someone shorts the resistor for a bit?

The wire doing the shorting has some very low resistance, not counting
the wires from the source fo the voltage.
Just as there is no perfect insulator, there is no perfect conductor.

Then pre charge the capacitor with a bigger load. It is possible to make the capacitor and the battery have precisely the same voltage.

Try to keep up. Even if it was possible to count the number of
electrons in the capacitor and make them equal to the voltage source,
the dielectric in the capacitor will let a few electrons pass, creating
some current when it is hooked up.

On Wed, 28 Dec 2016 23:48:58 -0000, Ralph Mowery wrote:

In article , says...

On Wed, 28 Dec 2016 23:21:52 -0000, Ralph Mowery wrote:

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

What if someone shorts the resistor for a bit?

The wire doing the shorting has some very low resistance, not counting
the wires from the source fo the voltage.
Just as there is no perfect insulator, there is no perfect conductor.

Then pre charge the capacitor with a bigger load. It is possible to make the capacitor and the battery have precisely the same voltage.

Try to keep up. Even if it was possible to count the number of
electrons in the capacitor and make them equal to the voltage source,
the dielectric in the capacitor will let a few electrons pass, creating
some current when it is hooked up.

Ever heard of the word negligible?

--
California lawmakers are now proposing an amendment that would allow 14 year olds a quarter vote and 16 year olds a half a vote in all state elections.
How stupid is this? Don't they have enough trouble counting WHOLE votes? How are they going to figure out fractions?!

On Wednesday, December 28, 2016 at 5:47:10 PM UTC-5, Ralph Mowery wrote:
In article ,
says...


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*


There is always some current flowing. It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

All of this is picking the nits off the nits.


I agree, but it's Rafter's specialty.



So ther is never a 0 in the equation.

Even in the ideal case, where there is zero current flow,
Ohm's Law, Kirchoff's Voltage Law, still work. For some bizarre
reason, Rafters claims that when we have V = IR, if I is zero,
some law of mathematics involving division by zero is violated,
so we can't solve for V. Everyone else here agrees we can and
the answer is zero.

What any of this has to do with wiring a switch, IDK, but here we are.

On Wednesday, December 28, 2016 at 5:54:11 PM UTC-5, Ralph Mowery wrote:
In article ,
says...



Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

The capacitor will never fully charge.

Theoretically, in the ideal capacitor case, that's true.
But it doesn't have to fully charge for my example to be correct.
Rafters claimed voltage drop is about energy loss, dissipation.
There is voltage drop across that cap, starting from 0 and
asymptotically approaching the value of the voltage source
over time and it does not involve energy loss or dissipation.

On Wednesday, December 28, 2016 at 6:17:46 PM UTC-5, FromTheRafters wrote:
James Wilkinson Sword submitted this idea :
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency
of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I
suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that
in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable
putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core
and
it is use up too 400 Hrz. Presently we have ferret iron cores
transformers
that will work into very high Frequency depend on density of
iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless
you are running mile and miles of line. In normal use on open line if
there
or no load/current there is no Voltage drop, you must have current
present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when
transmitter
is on!
My question is why would I go into electrical theory to some one that
wants
to know how or can he change duplex receptacle from single source to
double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument.

Not philosophical so much as unimportant to the jobs done by electrical
engineers.

Say what? Basic electrical principles are not relevant to the work
done by EEs?


People can use computers just fine without knowing how they
work, and the same goes for electrical circuit theory and electrical
work and those who use calculus and the mathematics behind it.

As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the circuit.
It could be 10 volts across some wires, and 230 across a heating element.
When it's switched off, all the drop is across the switch.

I'm not suggesting that there is no difference of potential across the
switch, I am saying that the difference of potential across the switch
is not there because of dissipation. Heat is not the only way that
energy can be dissipated, but it *is* the usual one.


Wandering in the wilderness again.



Emphasis asterisks are added by me.

"Voltage drop describes how energy is supplied of a voltage source that
is reduced *as electric current moves through the passive elements
(elements that do not supply voltage) of an electrical circuit*.
Voltage drops across internal resistances of the source, across
conductors, across contacts, and across connectors are undesired as the
supplied energy is lost (dissipated). Voltage drops across loads and
across other active circuit elements are desired as the supplied energy
performs useful work.



From: https://en.wikipedia.org/wiki/Voltage_drop

"Dissipation

Energy
"The conversion of mechanical energy into heat is called energy
dissipation." €“ François Roddier[1] *The term is also applied to the
loss of energy due to generation of unwanted heat in electric and
electronic circuits*."

From: https://en.wikipedia.org/wiki/Dissipation

and

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces *through which electric current
can flow*."

From: https://en.wikipedia.org/wiki/Electronic_circuit

Electric current can't flow when the switch is open, and it is no
longer a *circuit* by the definition used in circuit theory. Even when
the switch replaced by a capacitor and the capacitor is charging, no
current flows through the capacitor.

That's technically true, but the result is the same as if the current
was flowing through the capacitor.


There is no dissipation in the
capacitor due to the no current condition, and hence no voltage drop
across the capacitor.


Wrong again. From your own reference, wiki, on voltage drop:

"Analogous to Ohm's law for direct-current circuits, electrical impedance may be expressed by the formula {\displaystyle E=IZ} E=IZ. So, the voltage drop in an AC circuit is the product of the current and the impedance of the circuit."

Bingo!

Hence, there is voltage drop across the capacitor, it follows Ohm's
LAw applied to it's impedance. There has to be voltage drop
otherwise Kirchoff's Law would not work. Put a volt meter across
a resistor in a circuit, you can see a voltage. It's the voltage drop
across the resistor. Put a volt meter across a capacitor or inductor
in a circuit and you will also see it's voltage drop. The fact that
in one element it's simple resistance, the other it;s impedance,
doesn't change the fact that all those voltage drops around the circuit
have to add up to zero.

And if the no current condition is what gives us the no "dissipation"
and then no voltage drop, then how do you explain an inductor?
Current flows through an inductor, but an inductor, like a cap,
doesn't dissipate energy either. The simple answer is that the
voltage drop across both is defined by their impedance and the
current flowing through them. Ohm's Law, Kirchoff's Law work,
you treat them as impedances.




It is easy to misunderstand concepts when one throws out word meanings
as 'merely semantics'. Not that you have done so to the best of my
recollection, but others sure have. If one doesn't 'pick nits' they can
often lose subtle underpinnings of complex concepts.

You're the one who totally misunderstands concepts. You're even
here lecturing us about using the term "current flow", claiming
that it's actually electrons that are flowing. As if we don't
all know that. Good grief, please just stop already.

On Thu, 29 Dec 2016 00:55:38 -0000, trader_4 wrote:

On Wednesday, December 28, 2016 at 6:17:46 PM UTC-5, FromTheRafters wrote:
James Wilkinson Sword submitted this idea :
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency
of
normal AC power for homes. business and industry are quite different from
those of Europe? Your inductive hum is lower and more masculine. I
suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that
in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable
putting
education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core
and
it is use up too 400 Hrz. Presently we have ferret iron cores
transformers
that will work into very high Frequency depend on density of
iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless
you are running mile and miles of line. In normal use on open line if
there
or no load/current there is no Voltage drop, you must have current
present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when
transmitter
is on!
My question is why would I go into electrical theory to some one that
wants
to know how or can he change duplex receptacle from single source to
double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument.

Not philosophical so much as unimportant to the jobs done by electrical
engineers.

Say what? Basic electrical principles are not relevant to the work
done by EEs?


People can use computers just fine without knowing how they
work, and the same goes for electrical circuit theory and electrical
work and those who use calculus and the mathematics behind it.

As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the circuit.
It could be 10 volts across some wires, and 230 across a heating element.
When it's switched off, all the drop is across the switch.

I'm not suggesting that there is no difference of potential across the
switch, I am saying that the difference of potential across the switch
is not there because of dissipation. Heat is not the only way that
energy can be dissipated, but it *is* the usual one.


Wandering in the wilderness again.



Emphasis asterisks are added by me.

"Voltage drop describes how energy is supplied of a voltage source that
is reduced *as electric current moves through the passive elements
(elements that do not supply voltage) of an electrical circuit*.
Voltage drops across internal resistances of the source, across
conductors, across contacts, and across connectors are undesired as the
supplied energy is lost (dissipated). Voltage drops across loads and
across other active circuit elements are desired as the supplied energy
performs useful work.



From: https://en.wikipedia.org/wiki/Voltage_drop

"Dissipation

Energy
"The conversion of mechanical energy into heat is called energy
dissipation." €“ François Roddier[1] *The term is also applied to the
loss of energy due to generation of unwanted heat in electric and
electronic circuits*."

From: https://en.wikipedia.org/wiki/Dissipation

and

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces *through which electric current
can flow*."

From: https://en.wikipedia.org/wiki/Electronic_circuit

Electric current can't flow when the switch is open, and it is no
longer a *circuit* by the definition used in circuit theory. Even when
the switch replaced by a capacitor and the capacitor is charging, no
current flows through the capacitor.

That's technically true, but the result is the same as if the current
was flowing through the capacitor.


There is no dissipation in the
capacitor due to the no current condition, and hence no voltage drop
across the capacitor.


Wrong again. From your own reference, wiki, on voltage drop:

"Analogous to Ohm's law for direct-current circuits, electrical impedance may be expressed by the formula {\displaystyle E=IZ} E=IZ. So, the voltage drop in an AC circuit is the product of the current and the impedance of the circuit."

Bingo!

Hence, there is voltage drop across the capacitor, it follows Ohm's
LAw applied to it's impedance. There has to be voltage drop
otherwise Kirchoff's Law would not work. Put a volt meter across
a resistor in a circuit, you can see a voltage. It's the voltage drop
across the resistor. Put a volt meter across a capacitor or inductor
in a circuit and you will also see it's voltage drop. The fact that
in one element it's simple resistance, the other it;s impedance,
doesn't change the fact that all those voltage drops around the circuit
have to add up to zero.

And if the no current condition is what gives us the no "dissipation"
and then no voltage drop, then how do you explain an inductor?
Current flows through an inductor, but an inductor, like a cap,
doesn't dissipate energy either. The simple answer is that the
voltage drop across both is defined by their impedance and the
current flowing through them. Ohm's Law, Kirchoff's Law work,
you treat them as impedances.




It is easy to misunderstand concepts when one throws out word meanings
as 'merely semantics'. Not that you have done so to the best of my
recollection, but others sure have. If one doesn't 'pick nits' they can
often lose subtle underpinnings of complex concepts.

You're the one who totally misunderstands concepts. You're even
here lecturing us about using the term "current flow", claiming
that it's actually electrons that are flowing. As if we don't
all know that. Good grief, please just stop already.

I think "FromTheRafters"'s head is up in the rafters in a daydream.

--
23% of all photocopier faults worldwide are caused by people sitting on them and photocopying their buttocks.

Ralph Mowery formulated the question :
In article ,
says...


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*


There is always some current flowing.

Yes, gfretwell mentioned this already and I knew this already. This is
why Ohm's Law works as well as it does. However, when someone
stipulates that the current *is* zero, it means he is talking theory if
you go by what you just mentioned about reality. I objected to the
assumed correctness of the statement, and the argument ensued.


It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

That argument 'floats'.

You don't even need to use limits when you state a current (or
resistance) is close to zero and apply Ohm's Law. The algebra used to
derive the three equations which make up the Ohm's Law formula requires
that I and R be non-zero.

All of this is picking the nits off the nits.

Perhaps, but it is 'on topic' nitpicking.

And I didn't start it, I only mentioned that a certain statement was
incorrect IMO. There was really no need to argue the point, but they
apparently wanted to. I don't mind defending my point of view, and
don't back down just because they decide to call me names either.

So ther is never a 0 in the equation.
You just have to look very hard to find the components.

I agree. Mr. Ohm himself wasn't using open circuits when he made his
empirical observations. He had a meter and a source and he put
differing lengths of assorted materials in a 'circuit' and observed the
results. When scientists wanted to see if Ohm's Law worked at very
small scales, they didn't construct a four atom wide conductor and
leave it isolated and say "Zero amps, and zero volts, yep - it still
works" either.

trader_4 used his keyboard to write :
On Wednesday, December 28, 2016 at 5:47:10 PM UTC-5, Ralph Mowery wrote:
In article ,
says...


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*


There is always some current flowing. It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

All of this is picking the nits off the nits.


I agree, but it's Rafter's specialty.



So ther is never a 0 in the equation.

Even in the ideal case, where there is zero current flow,
Ohm's Law, Kirchoff's Voltage Law, still work. For some bizarre
reason, Rafters claims that when we have V = IR, if I is zero,
some law of mathematics involving division by zero is violated,
so we can't solve for V. Everyone else here agrees we can and
the answer is zero.

You can't know that *everyone* agrees until *everyone* says so, and I
don't mean someone using "everyone" as a nym either.

What any of this has to do with wiring a switch, IDK, but here we are.

You should be able to stop at any time, but you can't can you? The
reason this switch thread devolved into this is because of your remark
about electrical engineers not being worth their salt if they don't
know about some minutia about cutting a plate on a duplex receptacle.

https://dengarden.com/home-improveme...alf-hot-outlet

Instead of helping the poster, you decided to attempt to make yourself
look better than him (as you almost always do) by denigrating him and
the place he got his degree from. IMO he learned a good deal more about
the subject than you did, and put it to use in a much more complicated
field.

trader_4 wrote on 12/28/2016 :
On Wednesday, December 28, 2016 at 4:33:37 PM UTC-5, James Wilkinson Sword
wrote:
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks
What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency
of normal AC power for homes. business and industry are quite different
from those of Europe? Your inductive hum is lower and more masculine. I
suppose that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line of work and when you are ask question Simplicity is the right answer
not what degree you have or what the Ohms law is, these days you learn
that in third grade of school. Regarding school it is not always how much
of education have but do you know how to use it. There are many people
that have College Degrees but are sweeping floors because are; incapable
putting education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core
and it is use up too 400 Hrz. Presently we have ferret iron cores
transformers that will work into very high Frequency depend on density of
iron/material that core is made of, it has became very popular. Then we
have Air core which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless you are running mile and miles of line. In normal use on open line
if there or no load/current there is no Voltage drop, you must have
current present to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when
transmitter is on!
My question is why would I go into electrical theory to some one that
wants to know how or can he change duplex receptacle from single source
to double supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument. As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the
circuit. It could be 10 volts across some wires, and 230 across a heating
element. When it's switched off, all the drop is across the switch.

--
A group of white South Africans recently killed a black lawyer because he
was black. That was wrong. They should have killed him because he was a
lawyer.

Don't forget that Rafters came up with this gem:

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.

Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

Hell, you can't even spell it right.

Show me the capacitors.

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Ralph Mowery pretended :
In article ,
says...



Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

The capacitor will never fully charge. There is a time constant where
the capacitor charges to about 63.2 % . Usually 5 time constants are
concidered fully charged, but it never reaches a true full charge.

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

As long as we are back in the real world, my stipulation about ideal
voltage source and ideal capacitor is gone. I never argued about the
real world circuits.

http://www.allaboutcircuits.com/text...acitor-quirks/

https://en.wikipedia.org/wiki/Voltage_source

trader_4 formulated on Wednesday :
On Wednesday, December 28, 2016 at 6:17:46 PM UTC-5, FromTheRafters wrote:
James Wilkinson Sword submitted this idea :
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson
Sword wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks
What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency
of
normal AC power for homes. business and industry are quite different
from those of Europe? Your inductive hum is lower and more masculine. I
suppose
that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line
of work and when you are ask question Simplicity is the right answer not
what degree you have or what the Ohms law is, these days you learn that
in
third grade of school. Regarding school it is not always how much of
education have but do you know how to use it. There are many people that
have College Degrees but are sweeping floors because are; incapable
putting
education in use!
Another subject there is many different Transformers, Most common is
iron core, that could be made of row iron stocked pieces that makes the
core and
it is use up too 400 Hrz. Presently we have ferret iron cores
transformers
that will work into very high Frequency depend on density of
iron/material
that core is made of, it has became very popular. Then we have Air core
which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless
you are running mile and miles of line. In normal use on open line if
there
or no load/current there is no Voltage drop, you must have current
present
to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna
can be
an open circuit there are current and voltage drop present when
transmitter
is on!
My question is why would I go into electrical theory to some one that
wants
to know how or can he change duplex receptacle from single source to
double
supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the
gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument.

Not philosophical so much as unimportant to the jobs done by electrical
engineers.

Say what? Basic electrical principles are not relevant to the work
done by EEs?


People can use computers just fine without knowing how they
work, and the same goes for electrical circuit theory and electrical
work and those who use calculus and the mathematics behind it.

As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the
circuit. It could be 10 volts across some wires, and 230 across a heating
element. When it's switched off, all the drop is across the switch.

I'm not suggesting that there is no difference of potential across the
switch, I am saying that the difference of potential across the switch
is not there because of dissipation. Heat is not the only way that
energy can be dissipated, but it *is* the usual one.


Wandering in the wilderness again.



Emphasis asterisks are added by me.

"Voltage drop describes how energy is supplied of a voltage source that
is reduced *as electric current moves through the passive elements
(elements that do not supply voltage) of an electrical circuit*.
Voltage drops across internal resistances of the source, across
conductors, across contacts, and across connectors are undesired as the
supplied energy is lost (dissipated). Voltage drops across loads and
across other active circuit elements are desired as the supplied energy
performs useful work.



From: https://en.wikipedia.org/wiki/Voltage_drop

"Dissipation

Energy
"The conversion of mechanical energy into heat is called energy
dissipation." €“ François Roddier[1] *The term is also applied to the
loss of energy due to generation of unwanted heat in electric and
electronic circuits*."

From: https://en.wikipedia.org/wiki/Dissipation

and

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces *through which electric current
can flow*."

From: https://en.wikipedia.org/wiki/Electronic_circuit

Electric current can't flow when the switch is open, and it is no
longer a *circuit* by the definition used in circuit theory. Even when
the switch replaced by a capacitor and the capacitor is charging, no
current flows through the capacitor.

That's technically true, but the result is the same as if the current
was flowing through the capacitor.


There is no dissipation in the
capacitor due to the no current condition, and hence no voltage drop
across the capacitor.


Wrong again. From your own reference, wiki, on voltage drop:

"Analogous to Ohm's law for direct-current circuits, electrical impedance may
be expressed by the formula {\displaystyle E=IZ} E=IZ. So, the voltage drop
in an AC circuit is the product of the current and the impedance of the
circuit."

Bingo!

Hence, there is voltage drop across the capacitor, it follows Ohm's
LAw applied to it's impedance. There has to be voltage drop
otherwise Kirchoff's Law would not work. Put a volt meter across
a resistor in a circuit, you can see a voltage. It's the voltage drop
across the resistor. Put a volt meter across a capacitor or inductor
in a circuit and you will also see it's voltage drop. The fact that
in one element it's simple resistance, the other it;s impedance,
doesn't change the fact that all those voltage drops around the circuit
have to add up to zero.

And if the no current condition is what gives us the no "dissipation"
and then no voltage drop, then how do you explain an inductor?
Current flows through an inductor, but an inductor, like a cap,
doesn't dissipate energy either. The simple answer is that the
voltage drop across both is defined by their impedance and the
current flowing through them. Ohm's Law, Kirchoff's Law work,
you treat them as impedances.




It is easy to misunderstand concepts when one throws out word meanings
as 'merely semantics'. Not that you have done so to the best of my
recollection, but others sure have. If one doesn't 'pick nits' they can
often lose subtle underpinnings of complex concepts.

You're the one who totally misunderstands concepts. You're even
here lecturing us about using the term "current flow", claiming
that it's actually electrons that are flowing.

It's a common misconception, and just by convention the word current is
used with flow to simplify things. I mentioned it in passing. It just
seemed to you like a lecture because you probably had to look it up.

As if we don't
all know that. Good grief, please just stop already.

You seem to be having trouble with the word analogous now while trying
to move the goalposts. IMO, they should refer to that 'voltage drop' as
'reactive voltage drop' like they do with all of the other AC terms
that are not related to purely resistive devices. Anyway, we weren't
discussing an analog to Ohm's Law and reactance. I only used the
capacitor as a way to slow down the act of opening the switch and
having the current start high and approach zero instead of suddenly
just being zero for demonstration purposes. It made it easier to see
the trend and the mathematical discontinuity.

I'll stop now as you seem to be getting overly emotional.

On 12/22/2016 11:57 AM, Dave C wrote:
I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks

Call an electrician.

On Thu, 29 Dec 2016 02:18:26 -0000, FromTheRafters wrote:

trader_4 wrote on 12/28/2016 :
On Wednesday, December 28, 2016 at 4:33:37 PM UTC-5, James Wilkinson Sword
wrote:
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch.. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks
What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency
of normal AC power for homes. business and industry are quite different
from those of Europe? Your inductive hum is lower and more masculine. I
suppose that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line of work and when you are ask question Simplicity is the right answer
not what degree you have or what the Ohms law is, these days you learn
that in third grade of school. Regarding school it is not always how much
of education have but do you know how to use it. There are many people
that have College Degrees but are sweeping floors because are; incapable
putting education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core
and it is use up too 400 Hrz. Presently we have ferret iron cores
transformers that will work into very high Frequency depend on density of
iron/material that core is made of, it has became very popular. Then we
have Air core which is use in High frequency including Micro wave..
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless you are running mile and miles of line. In normal use on open line
if there or no load/current there is no Voltage drop, you must have
current present to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when
transmitter is on!
My question is why would I go into electrical theory to some one that
wants to know how or can he change duplex receptacle from single source
to double supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument. As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the
circuit. It could be 10 volts across some wires, and 230 across a heating
element. When it's switched off, all the drop is across the switch..

--
A group of white South Africans recently killed a black lawyer because he
was black. That was wrong. They should have killed him because he was a
lawyer.

Don't forget that Rafters came up with this gem:

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.

Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

Hell, you can't even spell it right.

Show me the capacitors.

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Where does it say you can't use them?

--
Why isn't there mouse-flavored cat food?

On Wednesday, December 28, 2016 at 8:51:01 PM UTC-5, FromTheRafters wrote:
Ralph Mowery formulated the question :
In article ,
says...


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*


There is always some current flowing.

Yes, gfretwell mentioned this already and I knew this already. This is
why Ohm's Law works as well as it does.

Ohm's law works with zero current flowing:

V = IR

I = 0, what is V?

Answer from everyone else here, V = 0




However, when someone
stipulates that the current *is* zero, it means he is talking theory if
you go by what you just mentioned about reality. I objected to the
assumed correctness of the statement, and the argument ensued.


A current of zero in a circuit is not "theory".



It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

That argument 'floats'.

You don't even need to use limits when you state a current (or
resistance) is close to zero and apply Ohm's Law. The algebra used to
derive the three equations which make up the Ohm's Law formula requires
that I and R be non-zero.


Complete bull ****.


All of this is picking the nits off the nits.

Perhaps, but it is 'on topic' nitpicking.

And I didn't start it, I only mentioned that a certain statement was
incorrect IMO. There was really no need to argue the point, but they
apparently wanted to. I don't mind defending my point of view, and
don't back down just because they decide to call me names either.

So ther is never a 0 in the equation.
You just have to look very hard to find the components.

I agree. Mr. Ohm himself wasn't using open circuits when he made his
empirical observations.

Where you there?


He had a meter and a source and he put
differing lengths of assorted materials in a 'circuit' and observed the
results. When scientists wanted to see if Ohm's Law worked at very
small scales, they didn't construct a four atom wide conductor and
leave it isolated and say "Zero amps, and zero volts, yep - it still
works" either.

And Ohm's Law works for zero current. And it works for impedances too,
another concept you completely fail to grasp.

On Wednesday, December 28, 2016 at 9:13:50 PM UTC-5, FromTheRafters wrote:
trader_4 used his keyboard to write :
On Wednesday, December 28, 2016 at 5:47:10 PM UTC-5, Ralph Mowery wrote:
In article ,
says...


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*


There is always some current flowing. It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

All of this is picking the nits off the nits.


I agree, but it's Rafter's specialty.



So ther is never a 0 in the equation.

Even in the ideal case, where there is zero current flow,
Ohm's Law, Kirchoff's Voltage Law, still work. For some bizarre
reason, Rafters claims that when we have V = IR, if I is zero,
some law of mathematics involving division by zero is violated,
so we can't solve for V. Everyone else here agrees we can and
the answer is zero.

You can't know that *everyone* agrees until *everyone* says so, and I
don't mean someone using "everyone" as a nym either.

Cite for us the posters that agree with you. Every one I've seen says
you're wrong, not that it matters that much, because it doesn't change
the laws of physics.



What any of this has to do with wiring a switch, IDK, but here we are.

You should be able to stop at any time, but you can't can you? The
reason this switch thread devolved into this is because of your remark
about electrical engineers not being worth their salt if they don't
know about some minutia about cutting a plate on a duplex receptacle.

Go figure. I simply stated that if you have a degree in electrical
engineering, you should be able to understand and solve a simple
switch circuit. That's all. Go figure. You started with all the
stupid, BS, nit nonsense. Like chastising us for saying "current flow",
because, according to you, current does not flow.




https://dengarden.com/home-improveme...alf-hot-outlet

Instead of helping the poster, you decided to attempt to make yourself
look better than him (as you almost always do) by denigrating him and
the place he got his degree from. IMO he learned a good deal more about
the subject than you did, and put it to use in a much more complicated
field.

Now you're lying. My comments about what an EE should know were not
directed to the OP, who has long since disappeared. The OP's question
was immediately answered by DerbyDad, I validated his answer and added that when
you re-purpose the white wire as a non-grounded conductor, it needs
to be marked at both ends to identify it. BTW, WTF did you add,
except a lot of incorrect BS, like arguing about not saying "current flow"?

On Wednesday, December 28, 2016 at 5:54:11 PM UTC-5, Ralph Mowery wrote:
In article ,
says...



Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

The capacitor will never fully charge.

It doesn't have to fully charge. The point is that a capacitor doesn't
dissipate energy, yet it has a voltage drop across it, just like a
resistor has a voltage drop across it. And the sum of all the voltage
drops in a circuit must be zero, per Kirchhoff's Law. THAT is what
Rafters doesn't get.

On Wednesday, December 28, 2016 at 6:55:36 PM UTC-5, James Wilkinson Sword wrote:
On Wed, 28 Dec 2016 23:48:58 -0000, Ralph Mowery wrote:

In article , says...

On Wed, 28 Dec 2016 23:21:52 -0000, Ralph Mowery wrote:

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

What if someone shorts the resistor for a bit?

The wire doing the shorting has some very low resistance, not counting
the wires from the source fo the voltage.
Just as there is no perfect insulator, there is no perfect conductor.

Then pre charge the capacitor with a bigger load. It is possible to make the capacitor and the battery have precisely the same voltage.

Try to keep up. Even if it was possible to count the number of
electrons in the capacitor and make them equal to the voltage source,
the dielectric in the capacitor will let a few electrons pass, creating
some current when it is hooked up.

Ever heard of the word negligible?

+1

This just keeps going farther and farther into the wilderness.
The real problem here is that Rafters does not understand Ohm's Law
and Kirchhoff's Laws. He denies there is voltage drop across a
capacitor that's in a circuit.

On Wednesday, December 28, 2016 at 9:18:30 PM UTC-5, FromTheRafters wrote:
trader_4 wrote on 12/28/2016 :
On Wednesday, December 28, 2016 at 4:33:37 PM UTC-5, James Wilkinson Sword
wrote:
On Wed, 28 Dec 2016 21:08:19 -0000, FromTheRafters
wrote:

After serious thinking James Wilkinson Sword wrote :
On Wed, 28 Dec 2016 20:19:47 -0000, Tony944 wrote:



"Uncle Monster" wrote in message
...

On Thursday, December 22, 2016 at 5:37:53 PM UTC-6, James Wilkinson Sword
wrote:
On Thu, 22 Dec 2016 19:57:23 -0000, Dave C wrote:

I have an AC wall socket, currently controlled by a wall switch. I
would like to change that outlet, so the one of the plugs is always
ON. The other plug would remaqin as is, ergo controlled by the wall
switch. I purchased a prior house with that configuartion for one
socket. Alas I never looked to see how that "split" outlet
configuration was implemented.

Can one inform me, how to make this change? Thanks
What century are you living in? All my sockets have switches on them.

You do know that North American wiring standards, voltages and frequency
of normal AC power for homes. business and industry are quite different
from those of Europe? Your inductive hum is lower and more masculine. I
suppose that's a point of pride for you. ヽ(ヅ)ノ

[8~{} Uncle Humming Monster



It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor. Person must have common sense in any
line of work and when you are ask question Simplicity is the right answer
not what degree you have or what the Ohms law is, these days you learn
that in third grade of school. Regarding school it is not always how much
of education have but do you know how to use it. There are many people
that have College Degrees but are sweeping floors because are; incapable
putting education in use!
Another subject there is many different Transformers, Most common is iron
core, that could be made of row iron stocked pieces that makes the core
and it is use up too 400 Hrz. Presently we have ferret iron cores
transformers that will work into very high Frequency depend on density of
iron/material that core is made of, it has became very popular. Then we
have Air core which is use in High frequency including Micro wave.
Voltage drops at 50or 60 Hrz. virtually dont exist in open circuit
unless you are running mile and miles of line. In normal use on open line
if there or no load/current there is no Voltage drop, you must have
current present to have Voltage drop.
In radio Frequency there are voltages drops example you radio Antenna can
be
an open circuit there are current and voltage drop present when
transmitter is on!
My question is why would I go into electrical theory to some one that
wants to know how or can he change duplex receptacle from single source
to double supply???
Unless he is trying too empress "us"; Stupid. Yes it is needless to say
€œStupid€�

The voltage always drops somewhere. In open circuit, it's across the gap.

No, it is not. The voltage across the gap (or cap) is the supply
voltage. If the voltage across the gap was less than the supply
voltage, as it would be when still charging, then you would have a drop
somewhere (the conductors). There is no current through the gap (or
cap) at all, and no dissipation, even when charging, and therefore no
voltage drop across it.

This appears to be a philosophical argument. As I see it, you have say 240
volts at the supply, and that voltage is dropped as you go round the
circuit. It could be 10 volts across some wires, and 230 across a heating
element. When it's switched off, all the drop is across the switch.

--
A group of white South Africans recently killed a black lawyer because he
was black. That was wrong. They should have killed him because he was a
lawyer.

Don't forget that Rafters came up with this gem:

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.

Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

Hell, you can't even spell it right.

Show me the capacitors.

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Let me get this right. You're again denying that Kirchhoff's laws
apply to circuits with capacitors? That there is no voltage drop
across a cap in a circuit? When your only source for
electrical engineering principles is wiki, I can see how you get
so ****ed up. Just because wiki only shows an example with resistors
for novices like you, doesn't mean the law doesn't apply to circuits
with capacitors and inductors too.

You really, really need to stop. You're like the guy that brings
a slingshot to a gun fight.

On Wednesday, December 28, 2016 at 9:29:11 PM UTC-5, FromTheRafters wrote:
Ralph Mowery pretended :
In article ,
says...



Voltage drop does not refer to energy lost or to dissipation.
Take a capacitor and a resistor in series with a voltage source.
urn on the voltage source.
Initially, all the voltage drop is across the resistor, it's equal
to the source voltage. As the cap charges, there will be voltage
drop across the cap, and less across the resistor. Add them
up and they will equal the source voltage. This is Kirchoff's voltage
law, the sum of all the voltage drops around the circuit must equal
zero. When the cap is fully charged, there will be a voltage drop
across it and it will equal the source. There is no current flowing,
no power dissipation, but Kirchoff's Law still works.

The capacitor will never fully charge. There is a time constant where
the capacitor charges to about 63.2 % . Usually 5 time constants are
concidered fully charged, but it never reaches a true full charge.

Don't forget that there is no perfect insulator on earth, so some
microscopic current is flowing all the time through the capacitor.


So say the nit pickers.

As long as we are back in the real world, my stipulation about ideal
voltage source and ideal capacitor is gone. I never argued about the
real world circuits.

Yes you did. You told us that there is no voltage drop across a
cap in a real world circuit, because there is no "dissipation",
no current flowing.

On Wednesday, December 28, 2016 at 10:51:10 PM UTC-5, FromTheRafters wrote:

You're the one who totally misunderstands concepts. You're even
here lecturing us about using the term "current flow", claiming
that it's actually electrons that are flowing.

It's a common misconception, and just by convention the word current is
used with flow to simplify things. I mentioned it in passing. It just
seemed to you like a lecture because you probably had to look it up.

First, let's note that you skipped all the sections where I smashed
all your technical nonsense. "I mentioned it in passing". No, you
brought it up like you, being the smart one, figured out something
that's incorrect. Actually, there is nothing incorrect about referring
to current flowing. Try taking a course in electrical engineering
and then you can correct the professor and impress the class. ROFL.




As if we don't
all know that. Good grief, please just stop already.

You seem to be having trouble with the word analogous now while trying
to move the goalposts. IMO, they should refer to that 'voltage drop' as
'reactive voltage drop' like they do with all of the other AC terms
that are not related to purely resistive devices.

We really don't care what your opinion is. You don't set the terms,
the definitions for EE. Voltage drop is referred to as just that.
The vast majority of the time, we don't need to specify whether it's
purely resistive or there is a reactance component. Even novices
know that if it's an ideal cap or inductor, then it's purely reactance.
Maybe YOU need it explained each time though.


Anyway, we weren't
discussing an analog to Ohm's Law and reactance.

We are now, because it's one way to smash your BS.


I only used the
capacitor as a way to slow down the act of opening the switch and
having the current start high and approach zero instead of suddenly
just being zero for demonstration purposes. It made it easier to see
the trend and the mathematical discontinuity.

I'll stop now as you seem to be getting overly emotional.

You should have stopped a long time ago, because you're just
embarrassing yourself. That BS about the term "current flow"
not being right to use, that to solve V = IR for I =0 requires
division by zero, those alone were way more than enough.

On Thursday, December 29, 2016 at 11:10:11 AM UTC-5, James Wilkinson Sword

Show me the capacitors.

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Where does it say you can't use them?


Exactly. It doesn't say it's limited to resistors, but apparently
Rafters thinks so, because they only show a simple example with resistors.

Better than that, Rafters own wiki reference says:

"KCL is applicable to any lumped network irrespective of the nature of the network; whether unilateral or bilateral, active or passive, linear or non-linear."

The above references only his current law, but it applies equally
to his voltage law. A quick check of any credible electrical
engineering course work will confirm that:

http://people.clarkson.edu/~jsvoboda...aws_foc_ac.pdf


Wow, caps and inductors too! What now, Rafters? How can that be?

I mean, this is so, so dumb, IDK why he keeps embarrassing himself.

James Wilkinson Sword brought next idea :

I think "FromTheRafters"'s head is up in the rafters in a daydream.

I know you are just trolling, it's what you do, but unlike trader_4
there might be some chance of you learning something. Anyone interested
might like to read this exchange while paying particular attention to
the opinions of Simon Bridge and Jim Hardy.

Avoid the urge to deflect by using the statement by Simon Bridge to

"Treat 'voltage drop' as an informal useage with no strict definition."

as it seems, considering his other contributions, that he is suggesting
only to 'treat' it that way in order to avoid confusing people like
trader_4 and the *everyone* trader_4 insists is silently agreeing with
him.

https://www.physicsforums.com/thread...e-drop.741405/

I won't be replying to you about this anymore either. I have posted
many references only to have them cherry-picked for things to deflect
about. Those who remain unconvinced will remain so despite anything
else I can add, and trader_4 only tries to deflect to my passing
mention of 'current' not actually flowing despite it being used in
sentences like "The current flows down the conductor" which is by
convention a normal everyday usage. My reason for even mentioning it at
all is because 'voltage drop' has fallen into the same category of
pervasive misuse of terminology.

Maybe 'voltage drop' is just another one of those things that people
with EE degrees don't need to know, but those studying physics or
electronics do need to know.

On Wed, 28 Dec 2016 21:13:44 -0500, FromTheRafters
wrote:

trader_4 used his keyboard to write :
On Wednesday, December 28, 2016 at 5:47:10 PM UTC-5, Ralph Mowery wrote:
In article ,
says...


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*


There is always some current flowing. It may only be a couple of
electrons and not detectable by any common means. In the simple circuit
where there is a switch and the insulation is several inches between
conductors when the switch is in the off position, there is still a very
, very small current flow across a very large resistance. Therefor as
they say 99 and 44/100 % of the voltage is dropped across the switch
when it is in the open position.

All of this is picking the nits off the nits.


I agree, but it's Rafter's specialty.



So ther is never a 0 in the equation.

Even in the ideal case, where there is zero current flow,
Ohm's Law, Kirchoff's Voltage Law, still work. For some bizarre
reason, Rafters claims that when we have V = IR, if I is zero,
some law of mathematics involving division by zero is violated,
so we can't solve for V. Everyone else here agrees we can and
the answer is zero.

You can't know that *everyone* agrees until *everyone* says so, and I
don't mean someone using "everyone" as a nym either.

What any of this has to do with wiring a switch, IDK, but here we are.

You should be able to stop at any time, but you can't can you? The
reason this switch thread devolved into this is because of your remark
about electrical engineers not being worth their salt if they don't
know about some minutia about cutting a plate on a duplex receptacle.

https://dengarden.com/home-improveme...alf-hot-outlet

Instead of helping the poster, you decided to attempt to make yourself
look better than him (as you almost always do) by denigrating him and
the place he got his degree from. IMO he learned a good deal more about
the subject than you did, and put it to use in a much more complicated
field.

Thank you for that terse rejoinde, oner that I did not feel that I
could personally post.

Let me assure Trader that I know how to design an EW ESM/ ECM Systems
During my job interview, we did not spend any time on Ohm's Law. I
recall I was asked to explain how I might impliment Cross Pole
jamming and other ECM techniques; along with ESM receiver
architectures. I was hired for that Senior position.

Equally irrelevant, during my qualifying interviewS, was how to do
House Wiring ! My original question here, re a spilt duplex box would
have been a Gotcha! I was unaware of the plug shorting shunt. Thanks
to the many Positive responders - my wiring change was in place Days
ago (w/o any Ohm's Law concerns)!

On 12/29/2016 1:07 PM, trader_4 wrote:
This just keeps going farther and farther into the wilderness.
The real problem here is that Rafters does not understand Ohm's Law
and Kirchhoff's Laws. He denies there is voltage drop across a
capacitor that's in a circuit.


Maybe were all just talking semantics here but I'm confused.

With respect to DC, is a capacitor like a rechargeable battery?

Is 'voltage' the same as 'voltage drop'?

Is there 'voltage drop' across a battery?

Is there 'voltage drop' across a rechargeable battery?

Is there 'voltage drop' across a capacitor?

On Thu, 29 Dec 2016 15:36:44 -0500, Confused
wrote:

On 12/29/2016 1:07 PM, trader_4 wrote:
This just keeps going farther and farther into the wilderness.
The real problem here is that Rafters does not understand Ohm's Law
and Kirchhoff's Laws. He denies there is voltage drop across a
capacitor that's in a circuit.


Maybe were all just talking semantics here but I'm confused.

With respect to DC, is a capacitor like a rechargeable battery?

Is 'voltage' the same as 'voltage drop'?

Is there 'voltage drop' across a battery?

Is there 'voltage drop' across a rechargeable battery?

Is there 'voltage drop' across a capacitor?
A capacitor acts "somewhat" like a battery.

Voltage is a measurement of electrical pressure - or difference in
electrical charge.
Voltage drop is loss of "pressure" or difference in charge
There is a voltage drop across a battery when under load. The amount
of drop is determined by the interior resistance of the battery and
explains the "droop" in voltage when a battery is placed under load.
Recahgeable or not, the story is the same
There is what es called the ESR of a capacitor - AKA the equivalent
series resistance of the capacitor, so the not so simple answer is
"yes"