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.