on 12/29/2016, Dave C supposed :
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.

You're welcome.

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)!

Thanks for the follow-up post. Some people ask questions and are never
heard from again. I suspected you had it all sorted and that you were
disinterested in the mess the thread had become.

On Wed, 28 Dec 2016 20:03:05 -0800, "Colonel Edmund J. Burke"
wrote:

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.

"Call an electrician" - DUH !! I successfully rewired my Simple Outlet
change Days ago - using advise offered one day, after my initial post!

I have enjoyed the on-going dialogue! Maybe Mr Olm (long deceased) is
likewise enjoying thiis discussion of his Law!!

Tommy Silva posted for all of us...



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

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

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

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

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

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

I called the township roadmaster and they fixed it in two days with a new
mailbox and posts. I could choose any mailbox I wanted. Since I knew him, I
was a taxpayer and it was begging for replacement I got a like replacement.

--
Tekkie

On Thu, 29 Dec 2016 19:23:13 -0000, FromTheRafters wrote:

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.

If I learn anything, it'll be from trader not you. He's the only one talking sense.

--
REALITY.EXE corrupt. Reboot universe (Y/N)?

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.

It seems to me that some of you guys trying to sale your knowledge here,
which in my opinion is very poor answer. 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 to 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 drop at 50 or 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!
Capacitor on the AC systems at 50 0r 60Hrz and up to micro wave is consider
dead short circuit, if there is voltage drop because it is something wrong
with capacitor, it has develop internal high resistance that makes it no
good, how ever checking capacitor for AC use it most have high resistance
because you are measuring with DC OHM meter.
Capacitor on DC is total difference but also should have high resistance to
€œInfinity€�,
it should never be voltage drop unless capacitor is leaking which it mean
that cap, is bad
and it should be replace. Note: if is leaking with no resistance in feed
line it will blow up or melt, in some all styles but normally it will blow.
To avoid that some lager capacitors; comes with build in fuse to avoid
blowing up.
My question is why would I go into electrical theory to some one that want
just 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 Thursday, December 29, 2016 at 2:23:19 PM UTC-5, FromTheRafters wrote:
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/

Show us where this person says that the voltage drops across a
resistor, capacitor and inductor in a circuit are not called
voltage drops. You can't, because he doesn't. Once again,
you're moving the goal posts. And he's wrong, because he says
a charge has to be moving to create a voltage drop. That is
wrong. There is voltage drop across a cap when it has
charge on it, even with no more current flowing. If there
were no voltage drop, we could not apply Kirchoff's Law.






I won't be replying to you about this anymore either.

Good thing, I told you to stop embarrassing yourself many posts ago.



I have posted
many references only to have them cherry-picked for things to deflect
about.

The problem is those referenced don't say what you claim they say.
The above is a prime example. The fellow, of unknown credentials
BTW, does not say that the voltage drops in a circuit across caps,
resistors, inductors, etc are not voltage drops.



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.

You seem to have a knack for trying to proclaim that you've found
something new, something that all of us in the EE world have not
known, and then making a big deal about it. The nonsense you
brought up about "current flow" not being right is a prime example
of the really, really stupid rat holes you venture down.




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.

It's obvious by now that you don't have a degree in any of the above.

We have a simple circuit consisting of a resistor, a capacitor and
a 60 hz voltage source. Do you deny that there is voltage drop on
the cap? According to you, there can be no voltage drop, unless
there is "energy dissipation". There is no energy dissipation in
the cap, so what's up with that?

The rest of us know that there is a voltage drop across the cap,
it follows Ohm's Law and the impedance of the cap. We can add
the sum of the voltage of the source, the voltage drop across the
resistor, and the voltage drop across the cap and we get ZERO.
Kirchoff's Law works.

On Thursday, December 29, 2016 at 3:00:02 PM UTC-5, Dave C wrote:
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.

If you bother to follow the thread at all, you would see that
I did not make that comment about an EE degree to you. I made it
to another poster who claimed that an EE can't figure out how to
wire a switched outlet because they don't teach that when you
get an EE degree. IDK where you got your degree. But where I got
my degree, we were taught electrical principles and from that
we could then solve all kinds of problems. That is the essence of
electrical engineering, applying electrical science to all kinds
of real world problems. If you can't figure
out how a simple switched AC circuit works and yet you have an
EE degree, then something is very wrong.





My original question here, re a spilt duplex box would
have been a Gotcha! I was unaware of the plug shorting shunt.

There is no plug shorting shunt. It's a shunt across the
two halves of the receptacle.


Thanks
to the many Positive responders - my wiring change was in place Days
ago (w/o any Ohm's Law concerns)!

Take that up with your buddy Rafters. He's the one that started in
with Ohms Law and chastising us for talking about "current flow",
which he says is incorrect, a term we can't use. Go figure.
I suppose you agree with him too, that you can't solve an equation
of the form Y = B*X when X is zero because it involves division.

BTW, DerbyDad correctly answered your question, it was the first
reply. Mine was second, where I agreed with what he had posted
and I told you about marking the white
wire if you re-purpose it as an ungrounded conductor. What did
Rafter contribute to solving your problem? NOTHING.

On Thursday, December 29, 2016 at 3:36:48 PM UTC-5, 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?


Yes, they have some similarities.



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

Voltage drop is the voltage across a component or from one
point in a circuit to another.



Is there 'voltage drop' across a battery?


There better be or Kirchoff's Law would not work. Let's say
a 12V battery drives two identical light bulbs. Now let's
take a meter and follow the circuit starting with the positive
terminal of the battery. Place the positive meter lead on it,
the neg meter lead on the other side of the first bulb. You
get +6V. That is the voltage drop across the first bulb.
Now keep going. Take the positive meter lead and put it between
the two bulbs and put the neg lead on far side of the second bulb.
Again you get +6V, the voltage drop across the second bulb.
Now complete the loop by taking the positive
lead and place it on the far side of the last bulb and the neg
meter lead back where you started, at the beginning of the loop,
which is the battery positive terminal. You get a reading of -12V .
Add up the voltage drops around the loop:

6 + 6 -12 = 0

Kirchhoffs Law

The drop across the battery is negative, meaning it's an increase,
not a drop.




Is there 'voltage drop' across a rechargeable battery?


Yes, if it's in a circuit.


Is there 'voltage drop' across a capacitor?

Yes, if it;s in a circuit.

Voltage is a difference in potential.
Voltage drop is a difference in potential across a particular component,
element, device in a circuit.


From Rafter's own source, wiki:

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

The directed sum of the electrical potential differences (voltage) around any closed network is zero, or:
More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop.

And note that it holds true whether they are resistors, caps, inductors
or even active components.


Rafter's specialty is perverting the straightforward and obvious.
Ask for his tutorial on how you can't say a 10A current is flowing
in a wire, because current doesn't flow.

On 12/30/2016 03:27 PM, trader_4 wrote:

[snip]

There better be or Kirchoff's Law would not work. Let's say
a 12V battery drives two identical light bulbs. Now let's
take a meter and follow the circuit starting with the positive
terminal of the battery. Place the positive meter lead on it,
the neg meter lead on the other side of the first bulb. You
get +6V. That is the voltage drop across the first bulb.
Now keep going. Take the positive meter lead and put it between
the two bulbs and put the neg lead on far side of the second bulb.
Again you get +6V, the voltage drop across the second bulb.
Now complete the loop by taking the positive
lead and place it on the far side of the last bulb and the neg
meter lead back where you started, at the beginning of the loop,
which is the battery positive terminal. You get a reading of -12V .
Add up the voltage drops around the loop:

6 + 6 -12 = 0

Kirchhoffs Law


Connect a third bulb in parallel with the second one above. The voltage
drops across the bulbs change, but the sum must remain 0.

In the same way, the sum of currents into a junction is 0. Still true
when the junction is in the middle of a 3-phase (wye) motor.

[snip]

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

Losing your faith is a lot like losing your virginity you don't realise
how irritating it was 'til it's gone.

Uncle Monster
Mon, 26
Dec 2016 23:44:34 GMT in alt.home.repair, wrote:

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

It's unmistakable if you've heard it before.



--
Sarcasm, because beating the living **** out of deserving people is
illegal.

trader_4
Mon, 26
Dec 2016 14:56:40 GMT in alt.home.repair, wrote:

On Friday, December 23, 2016 at 11:29:16 PM UTC-5,
wrote:
On Fri, 23 Dec 2016 20:12:44 -0800 (PST), Uncle Monster
wrote:


The old "Instant On" tube type TV sets used a lot of power on
standby to
keep the CRT and vacuum tube filaments half powered so the set
would come on more or less instantly. My LED TV might draw a few
milliamps when it's off to keep the IR remote control circuitry
active so it would be silly to unplug or kill the power to it.
?(€¢?€¢)?

[8~{} Uncle Instant Monster

"Smart" TVs will draw more because there is a processor running
all the time and even loafing along, it is still going to draw
something. Cranked up it can stream and process 1080 HD content
so that makes it a pretty powerful PC.

I would suspect that any digital TV today, even the most basic,
has multiple CPUs doing a variety of functions.

ROFL. You'd be surprised to find out that's typically not the case. A
single CPU can do the tasks required quite well. Especially when it's
a dedicated task. I've taken quite a few of the little *******s apart
because they tend to be unreliable for long term usage. They do
better if you turn it on and just leave it on than they do if you
turn them off (sleep mode) and wake them only to repeat the cycle
every single day.

If you're curious enough, try to locate a suitable mainboard for a
model tv set you own and you can usually find pics of the board
detailing every component on it. You'll find it has a single CPU that
may/may not have the transformers on it (primary transformer, baby
transformer for sleep mode and firing of a relay when you wake it;
which energizes the primary transformer). Oh, and some of the relays
are actually a tiny chip! instead of a small piece of plastic. If
that chip gets fuxored, it's difficult to find a suitable
replacement. I've had mixxed results soldering jumper leads from them
onto a real relay. Sometimes this works like a charm and the dead tv
is good to go, other times, it doesn't help. In those cases, I've had
to resort to replacing a board. That is, if the cost of the board is
still justified based on the age and size of the tv. Sometimes, it's
not worth doing. These new tvs are what I call 'super disposable'
grade.

If it doesn't, they'll be on a small board by themselves and some
filtering circuitry. You'll also find a seperate in most cases audio
amplifier board, and perhaps another board or two for switches and
ports. Most of the boards aren't very big. Very little physical space
inside the tv is spent on electronics. Most of it's the screen itself
and whatever backlight technology it's using.

You might also be shocked to find out that many times, that more
expensive tv you bought sitting beside the brand you've never heard
of that was a couple of c-notes cheaper have some interesting things
in common. Under the hood, they often share the same components;
right down to the ****ing model/part# on the board itself. You wind
up paying for the name on the case.

Like you sometimes would if you bought a Cobra radio over a Uniden
radio that happened to have the same features, but, looked different
outside. If you opened them, you'd find Uniden stamped on the
identical boards inside the chassis; only you paid more for the Cobra
'version'. Sometimes, a lot more.

Even in a PC, the display card or display processor on the MB was
one of the highest consumers of power.

The PC is responsible and capable of doing far more than the hardware
in your TV set could dream of. As a result, it's electrical power
demands will be higher. You aren't comparing this fairly. The CPU in
your smart tv has a small job to perform, compared to the CPU in your
desktop.


Digital TV would be similar.

Primarily the screen, backlight source AND audio amplifier board are
the power hungry components when the tv is running. But, in this
case, power hungry isn't really that many watts. One of my LED based
tvs consumes roughly 25watts when it's running. It doesn't consume a
measurable difference in power when watching bluray hdmi, dvd hdmi or
1080p HD vs crappy cable channels that aren't HD anything.

So, I wouldn't be surprised that there isn't much difference in
power with a smart PC.

Umm... There's a considerable amount of difference. You're comparing
a dedicated device with limited hardware specifically for that
purpose (the tv) to something else that can be configured to do
thousands of different things and has oodles more processor power as
well as storage space (which your tv itself isn't likely to be using
a mechanical hard disk, if it has what you'd call a hard disk at all)

Your smart tv isn't a big badass computer like your desktop. Not even
in the same ballpark. A smart phone isn't even comparable to todays
typical desktop in terms of configurability, processing power, and as
a result, power consumption. The smart tv and smart phone use very
little compared to your typical actual computer.

Some spec sheets would perhaps have the answers.

Yep. They do. You should consult with some before making such a
comparison again.




--
Sarcasm, because beating the living **** out of deserving people is
illegal.

trader_4
Mon, 26
Dec 2016 15:00:53 GMT in alt.home.repair, wrote:

Plus most people are not spec'ing out there own house and having
it built. So, you'd be stuck with whatever the builder decided to
do, whatever corners they cut to save a buck, etc. And there
would be even more variation in what you wind up with from home to
home. All in all, I think code is a good idea, including the new
reqt to have a neutral at all switch locations. How many people
have been in here over the years with problems trying to use a
smart switch like X10 at a switch where there is no neutral? How
many man hours have been wasted finding out that they can't work
with an LED, CFL, etc? All that is solved with a neutral there
and it's no big deal to run when you're installing a new circuit.

I find myself in complete agreement with you on this. I'm all for the
neutral requirement too.


--
Sarcasm, because beating the living **** out of deserving people is
illegal.

On Monday, February 6, 2017 at 7:31:25 AM UTC-5, Diesel wrote:


So, I wouldn't be surprised that there isn't much difference in
power with a smart PC.

Umm... There's a considerable amount of difference. You're comparing
a dedicated device with limited hardware specifically for that
purpose (the tv) to something else that can be configured to do
thousands of different things and has oodles more processor power as
well as storage space (which your tv itself isn't likely to be using
a mechanical hard disk, if it has what you'd call a hard disk at all)


Say what? AFAIK, typical smart TV doesn't have a hard drive. And
the vast majority of people aren't using the smart capability for
PC replacement, they are using it to stream video content to the TV.

But as always, the proof is in the spec sheet data. Here is an
example of two almost identical Samsung TVs, one smart one not:



40" non -smart:

http://www.bestbuy.com/site/samsung-...?skuId=8552023

Estimated Annual Operating Cost 11 United States dollars
Estimated Annual Electricity Use 99 kilowatt hours



40" smart:


http://www.bestbuy.com/site/samsung-...?skuId=6422016

Estimated Annual Operating Cost 11 United States dollars
Estimated Annual Electricity Use 93 kilowatt hours



Here are two from Insignia, 55"

Non-smart:

http://www.bestbuy.com/site/insignia...?skuId=4806800

Estimated Annual Operating Cost 28 United States dollars
Estimated Annual Electricity Use 233 kilowatt hours


Smart:

http://www.bestbuy.com/site/insignia...?skuId=4204506

Estimated Annual Operating Cost 30 United States dollars
Estimated Annual Electricity Use 250 kilowatt hours


Note that the smart model is also ultra HD too. So you have higher
resolution and smart capability at almost the same electric usage.


Your smart tv isn't a big badass computer like your desktop. Not even
in the same ballpark. A smart phone isn't even comparable to todays
typical desktop in terms of configurability, processing power, and as
a result, power consumption. The smart tv and smart phone use very
little compared to your typical actual computer.

Some spec sheets would perhaps have the answers.

Yep. They do. You should consult with some before making such a
comparison again.


I just did and have two comparisons, comparing 40" and 55" smart
and non-smart. One, pair, the
smart actually uses less power. The other, you get ultra resolution
and smart at almost power usage. So, apparently you're the one who
should do some consulting.

trader_4
Mon, 06
Feb 2017 16:02:10 GMT in alt.home.repair, wrote:

On Monday, February 6, 2017 at 7:31:25 AM UTC-5, Diesel wrote:


So, I wouldn't be surprised that there isn't much difference in
power with a smart PC.

Umm... There's a considerable amount of difference. You're
comparing a dedicated device with limited hardware specifically
for that purpose (the tv) to something else that can be
configured to do thousands of different things and has oodles
more processor power as well as storage space (which your tv
itself isn't likely to be using a mechanical hard disk, if it has
what you'd call a hard disk at all)


Say what? AFAIK, typical smart TV doesn't have a hard drive. And
the vast majority of people aren't using the smart capability for
PC replacement, they are using it to stream video content to the
TV.

You wrote smart pc. was this a typo on your part?



--
Sarcasm, because beating the living **** out of deserving people is
illegal.

On Tuesday, February 7, 2017 at 6:42:57 AM UTC-5, Diesel wrote:
trader_4
Mon, 06
Feb 2017 16:02:10 GMT in alt.home.repair, wrote:

On Monday, February 6, 2017 at 7:31:25 AM UTC-5, Diesel wrote:


So, I wouldn't be surprised that there isn't much difference in
power with a smart PC.

Umm... There's a considerable amount of difference. You're
comparing a dedicated device with limited hardware specifically
for that purpose (the tv) to something else that can be
configured to do thousands of different things and has oodles
more processor power as well as storage space (which your tv
itself isn't likely to be using a mechanical hard disk, if it has
what you'd call a hard disk at all)


Say what? AFAIK, typical smart TV doesn't have a hard drive. And
the vast majority of people aren't using the smart capability for
PC replacement, they are using it to stream video content to the
TV.

You wrote smart pc. was this a typo on your part?



That was my mistake, I meant smart TV. But if you look at the
thread at the point
of interest, it's clear we had been talking about smart TVs.
There was discussion about whether a smart TV used more power
than a similar non-smart TV. That was the context.
The term "smart PC" doesn't exist, but sorry for the confusion.